6
 package for 
7^{\text{th}}
\qquad5.61-5.20=0.41
87\%
2\%
89\%
2\%
150
90\%
90\%
90\%
90\%
90\%
\red{89\pm2}
\qquad89-2=87
\qquad89+2=91
\red{\text{lectures}}
\red{\text{between }87\%\text{ and }91\%}
\blue{87\pm2}
\qquad87-2=85
\qquad87+2=89
\blue{\text{case studies}}
\blue{\text{between }85\%\text{ and }89\%}
87\%
89\%
(\text s)
(\text{cm})
5
50.5
10
42.0
40
12.0
5
h(t) = 60 - bt + ct^2
(\text{cm})
t
(\text s)
30 \, \text s
18
42
60
138
2
b
c
{-2}
b
c = \dfrac{1}{50}
b
c
\qquad h(t) \approx 60 - 2t + 0.02t^2
t=30
30
18
\qquad -5(x+3)>2x+7+5x
x < - \dfrac {11}{6}
x > - \dfrac 5 4
x < \dfrac 1 3
x>4
-5
\blue{7x}
\purple{15}
\pink{-12}
\qquad
x<-\dfrac{11}{6}
\sqrt3 \approx 1.732
\sin(65^\circ) \approx 0.91
\cos(65^\circ) \approx 0.42
\tan(65^\circ) \approx 2.14
\angle ADB
\angle CDB
\overline{AC}
AD+DC=AC
\overline{AD}
\overline{DC}
BCD
BC
DC
BD
DC
BD
ADB
30^\circ - 60^\circ - 90^\circ
AD
BD
\sqrt3
AC
AC
\overline{AC}
40
2007
3.8
3.8
2008
5.4
5.7
2009
7.6
8.4
2010
10.7
12.4
2011
15.1
18.3
2012
21.2
27
2013
29.9
40
2014
42.2
59.2
2015
59.4
87.5
2016
83.8
129.5
2017
118.1
191.7
2018
166.5
283.6
2019
234.7
419.7
2020
330.9
621.2
41\%
48\%
2007
(41\%)
(48\%)
2022
55
340
490
660
2020
330
41\%
41\%
41\%
2020
2021
1.41
2022
2022
660
\qquad\left(\dfrac23+\dfrac12i\right)\left(12-\dfrac13i\right)
a+bi
a
b
b
i=\sqrt{-1}
b=\dfrac{1}{6}
b=-\dfrac{1}{6}
b=\dfrac{49}{6}
b=\dfrac{52}{9}
\qquad\left(\dfrac23+\dfrac12i\right)\left(12-\dfrac13i\right)=\red{8}\blue{-\dfrac29i}\purple{+6i}\green{-\dfrac16i^2}
i^2=-1
a+bi
\qquad b=\dfrac{52}{9}
\qquad (3g-4)(2g-8)=ag^2+bg+c
a
b
c
a
-32
-16
5
6
\qquad 3g(2g)+3g(-8)-4(2g)-4(-8)
\qquad 6g^2-24g-8g+32
\qquad 6g^2\blue{-24g}\blue{-8g}+32
\qquad 6g^2-32g+32
a
g^2
6
 calling card, which was a gift from her grandmother. The rate to call Japan, where her boyfriend is living, is 
 per minute. Her family is in Turkey, where the calling rate is 
 per minute. Derin promised her grandmother she would spend at least 
 of her minutes on keeping in touch with family. Which of the following systems of inequalities represents the relationship between 
, the number of minutes Derin could call Japan, and 
 T \leq 208 - 1.6J
 T + J \geq 30
 T \leq 25 - 0.19J
 T + J \geq 30
 T \leq 208 - 1.6J
 T \geq 30
 T \leq 25 - 0.19J
 T \geq 30
30
\qquad T \geq 30
 T \leq 208 - 1.6J
 T \geq 30
\qquad 2(y-12)+y^2=0
y=a
y=b
ab-a-b
-20
-22
-24
-26
-24
2
-24
1
-24
-1
24
2
-12
-2
12
6
-4
-6
4
a=-6
b=4
ab-a-b
-22
4
l
w
l
w
4
w
l
\qquad w=\dfrac{1}{2}l+4
w=\dfrac{1}{2}l+4
y=mx+b
w=\dfrac{1}{2}l+4
\dfrac{1}{2}
y
4
\qquad \qquad \qquad \qquad \text{Big Island's Average Monthly Rainfall}
r
(\text{in})
m
m=1
m=2
4^{\text{th}}
0 \text{ in}
0.25 \text{ in}
0.60 \text{ in}
0.85 \text{ in}
0.85\text{ in} 
(4, 0.6)
0.6\text{ in.}
\qquad 0.85-0.60=0.25
0.25
2+2(8347-4783) = -2j
-j
-j
2
2
6
(\text{cm})
10\text{ cm}
90\%
90\%
0.90
\qquad\qquad V_{\text{soup}}\approx282.743\cdot0.90\approx254.469\text{ cm}^3
254\text{ cm}^3
792
20\%
z
m
792
\qquad z+m=792
20\%
1.2
20\%
\qquad z=1.2m
1.2m
z
2.2
\qquad m=360
360
360
792
360
1.2
20\%
432
2
2
432
\qquad v=74-32t
, of the t-shirt, in feet per second, 
-32
(\text{ft})
(\text{sec})
\qquad v\text{ ft per sec}=74 \text{ ft per sec}\pink{-\dfrac{32\text{ ft per sec}}{1\text{ sec}}}t\text{ sec}
-32
32
t
\red{\text{subtract }74}
\blue{\text{divide by }-32}
\gray{\text{multiplying by }\dfrac{-1}{-1}}
40
50
40
50
t
\red{\text{subtract }74}
\blue{\text{divide by }-32}
\green{\text{flip the inequality}}
0.75
1.06
40
50
v
(0,74)
t
v=74-32t
0
74
1
42
2
10
3
-22
4
-54
t
1
v
32
\qquad
\qquad
\bullet
-32
32
\bullet
\qquad t=\dfrac{74-v}{32}
\bullet
0.75
1.06
40
50
\bullet
\qquad
\qquad \dfrac{1}{f} = \dfrac{1}{o} + \dfrac{1}{i}
f
o
i
o=f-i
o=fi(i-f)
o = \dfrac{fi}{i-f}
o=\dfrac{1}{i-f}
o
o
\blue{\text{Subtract } \dfrac1i } \, 
\, \dfrac1f \,
\, \dfrac1i \,
 \, fi
-1
o = \dfrac{fi}{i-f}
\qquad\dfrac{2x}{x-3} - \dfrac{x+3}{x-3}
x\ne3
1 
\dfrac{x+3}{x-3} 
\dfrac{2x}{x-3}-1 
\dfrac{3x+3}{x-3} 
x+3
\qquad \dfrac{2x}{x-3} - \dfrac{x+3}{x-3}=\dfrac{2x-x-3}{x-3}
 \quad  \dfrac{2x-x-3}{x-3}=\dfrac{x-3}{x-3}=1 
\qquad\qquad 1
\overline{JK}
\overline{LM}
\overline{JL}
\overline{KM}
\overline{JL}
8
\overline{LM}
10
\overline{JM}
\overline{HK}
\dfrac45 \approx \tan(38.66^\circ)
\beta^\circ
30.93^\circ
38.66^\circ
48.33^\circ
51.34^\circ
\overline{JL}
\overline{LM}
\theta^\circ
\overline{JK}
\overline{LM}
\overline{HK}
\overline{JM}
\angle JKH
\angle LMJ
\beta^\circ
180
200
(\text{dB})
I
\text{dB}
60
120
t
tI
200 \text{ dB}
I
t = 0
200
(t, I) = (0, 200)
I
t = 60
t = 60
I
t = 60
. Which of the following functions best describes the additional amount of money Marta will have to spend if she increases each side of the fence by 
f(m) = 1.5m
f(m) = 3m
f(m) = 6m
f(m) = 12m
3m
3m
4
12m
\qquad f(m)=12m
f(m) = 12m
a
b
\qquad 2a
99\%
101\%
0.99
1.01
\qquad 0.99(2a) = 1.98a \space
\space 1.01(2a) = 2.02a
b
1.45
1
, what was the approximate cost, in dollars, of one gallon of gas?  (Note:  
 gallon is approximately 
3.785
\text  {liters}
 in 
15\text{ ft}
30\text{ ft}
800\text { ft}^2
3.4
5
6.7
17.5
x
A
2x+30
2x+15
A(x)=(2x+30)(2x+15)
800\text { ft}^2
3.4\text{ ft}
x = a(y + b)
a
b
x = a(y + b - 2)
y
2
y
y
3.5
y
5.5
h
T
(^\circ\text{F})
1.2
11^\circ\text{F}
1.2
5^\circ\text{F}
20\%
11^\circ\text{F}
20\%
5^\circ\text{F}
y=\red{a}\cdot \blue{b}^{\large\purple{x}}
\red a
\blue b
\purple x
1
\blue b >1
\purple{\dfrac{T}{5}}
1
\blue{1.2}
\blue{120\%}
20\%
T=0
\red{11}
\blue{1.2}
\purple{\dfrac{T}{5}}=1
20\%
5^\circ\text{F}
C
(\text{g/L})
s
10
8
\text{g/L}
sC
s
s
s = 10
s = 10
8
\text{g/L}
s \ge 10
C = 8
\qquad (3-i)^3
i=\sqrt{-1}
8-26i
18-26i
27-26i
30-26i
(3-i)(3-i)
i^2=-1
(8-6i)
(3-i)
i^2=-1
\qquad18-26i
Y
2
3
\dfrac{1}{5}
(a^m)^n=a^{mn}
\dfrac{1}{5}
a^m\cdot b^m=(a\cdot b)^m
Y=36
\qquad Y=36
11^\text{th}
2.70^\circ 
\qquad \dfrac{\left(\dfrac{ab}{xy}\right)}{\left(\dfrac{ij}{mn}\right)} = \left(\dfrac{ab}{xy}\right) \cdot \left(\dfrac{mn}{ij}\right)
\qquad = \dfrac{abmn}{xyij}
\qquad \dfrac{a(bm)n}{x(yi)j}
4
5
24
\dfrac14
\blue5
\green x
\purple4
\blue5
\green x
24
\dfrac14
\qquad\dfrac{1}{4}\green x = \blue5
\dfrac{4}{1}
4
\green x
\green x = \green {20}
\green{20}
\green{20}
\dfrac{3}{4}
\left(1 - \dfrac{1}{4}\right)
\red{15}
\red{15}
\purple4
\blue5
\green {20}
24
\red{15}
\purple4
19
\qquad4-4a-(5+a)(5-a)
(a-7)(-a+3)
(-a+7)(a+3)
(a-7)(a+3)
(a+7)(a-3)
25-a^2
\qquad4-4a-(25-a^2)
-1
\qquad4-4a\red{-1}(25-a^2)
\red{-1}
\qquad4-4a-25+a^2
\qquad\red4\blue{-4a}\red{-25}\purple{+a^2}
\qquad a^2-4a-21
-21
-4
-7
3
\qquad(a-7)(a+3)
H
E
0.3
E
B
0.5
0.84
\sin(0.84) \approx 0.745
\cos(0.84) \approx 0.667
\tan(0.84) \approx 1.116
P
T
0.72
0.89
1.07
1.20
0.84
0.8
\overline{PT}
PT
1.20
P
T
(\mu\text{V})
620
61
640
38
660
24
680
15
700
9
720
6
740
4
(\mu\text{V})
99.9\%
20 \mu\text{V}
20 \mu\text{V}
20 \mu\text{V}
20 \mu\text{V}
20\mu\text{V}
\red{\text{common difference}}
\blue{\text{common ratio}}
\red{\text{linear}}
\red{\text{common difference}}
20\mu\text{V}
\blue{\text{exponential}}
\blue{\text{common ratio}}
\red{\text{difference}}
\blue{\text{common ratio}}
0.63
20\,\mu\text{V}
\blue{\text{exponential}}
20 \mu\text{V}
30
2
1
9
10
20
21
j
h
j
h
30
\qquad j+h \le 30
1
2
h
2j
\qquad h \ge 2j
3
6
30-h
h
h\ge2j
h
2j
2j
\qquad 10
48 \% 
76 \%
18\%
7\%
29\%
 million on advertising and 
 million on advertising and 
 million on advertising and 
 million on advertising and 
18\%
7\%
29\%
3
\qquad R\geq 41.76
\qquad  \:\: A \geq 16.24
\qquad  \:\: R \leq 67.28 - A
\qquad 
\qquad
\qquad 
(20,45)
3
 million on advertising and 
\sin49^\circ
\cos \angle C
\sin \angle C
\cos \angle A
\sin \angle B
180^\circ
\angle A
\angle C
90^\circ
\sin\theta=\cos(90^\circ-\theta)
\sin49^\circ=\cos(90-49)^\circ=\cos41^\circ
\cos41^\circ
\cos\angle C
\sin49^\circ=\cos\angle C
\qquad\ \ \ \sin49^\circ=\dfrac{CB}{AC}\qquad
\ \ \cos\theta=\dfrac{\text{length of the side adjacent to } \theta}{\text{length of the triangle's hypotenuse}}
\qquad\ \ \ \cos\angle C=\dfrac{CB}{AC}\qquad
\sin49^\circ
\cos\angle C
xy
y=2(1.75)^x
y=2(0.75)^x
y=0.75(2)^x
y=1.75(2)^x
y
2
y=a(b)^x
a
y
b
y
2
\qquad  y=2(1.75)^x\ 
\qquad  y=2(0.75)^x
a
b
>1
a
b
0<b<1
0<0.75<1
y
2
\qquad y=2(0.75)^x
\qquad (t+1)^2+c=0
c
t=\dfrac32
t=-\dfrac{7}2
c
-\dfrac{729}4
-\dfrac{121}4
-\dfrac{25}4
-1
t =\dfrac32
c
-\dfrac{25}4
\qquad49x^2-64=0
x=\ \ \ \dfrac{64}{49}
x=\pm\dfrac{64}{49}
x=\ \ \ \dfrac{8}{7}
x=\pm\dfrac{8}{7}
0
x^2
\qquad (7x-8)(7x+8)=0
\qquad7x-8=0 \qquad \text{or} \qquad 7x+8=0
x
x=\dfrac{8}{7}
x=-\dfrac{8}{7}
x=\pm\dfrac{8}{7}
\qquad P(q)=-0.01(q-250)(q-80)
P(q)
q
80
80
80
80
P(q)=0
q
\qquad 0=-0.01(q-250)(q-80)
\qquad q-250=0
q-80=0
q=250
q=80
80
, so 
80
3
18
(\text{cm})
30^\circ
10\,\text{cm}
12\,\text{cm}
31\,\text{cm}
36\,\text{cm}
(18\,\text{cm})
\ell
36\,\text{cm}
\qquad (3y-2)(y+a)=3y^2+by-24
y
a
b
b
-38
12
34
36
y
(3a-2)
3ay-2y=(3a-2)y
\qquad 3y^2+(3a-2)y-2a=3y^2+by-24
3a-2=b
-2a=-24
a=12
3a-2=b
b
\qquad b=34
xy
\left(-\dfrac23,-\dfrac34\right)
5
 \left(x+\dfrac23\right)^2+\left(y+\dfrac34\right)^2=5
 \left(x-\dfrac23\right)^2+\left(y+\dfrac34\right)^2=25
 \left(x+\dfrac23\right)^2+\left(y-\dfrac34\right)^2=25
 \left(x+\dfrac23\right)^2+\left(y+\dfrac34\right)^2=25
(\green h,\purple k)
\red r
\qquad (x-\green h)^2+(y-\purple k)^2=\red r^2
\left(\green{-\dfrac23},\purple{-\dfrac34}\right)
\red 5
\qquad \left(x-\left(\green{-\dfrac23}\right)\right)^2+\left(y-\left(\purple{-\dfrac34}\right)\right)^2=\red 5^2
\qquad \left(x+\dfrac23\right)^2+\left(y+\dfrac34\right)^2=25
\qquad \left(x+\dfrac23\right)^2+\left(y+\dfrac34\right)^2=25
M
xy
(x+8)^2+(y+8)^2=100
(x+8)^2+y^2=100
(x+8)^2+(y-6)^2=100
(x+8)^2+(y+6)^2=100
(\green h,\purple k)
\red r
\qquad(x-\green h)^2+(y-\purple k)^2= \red r^2
M
(\green{-8},\purple{0})
10
\qquad(x-(\green {-8}))^2+(y-\purple {0})^2=\red {10}^2
\qquad(x+8)^2+y^2=100
 for a yearly membership. The first book is free with the membership, and any book after that costs 
 including  tax.  How much money, 
, does a student spend after buying 
m=7.60b
m=7.60(b - 1)
m=7.60b + 60
m=7.60(b - 1) + 60
 for the original membership.  If the student buys 
b-1
 per book.  Therefore, the cost of books is 
.  Let's add this to 
m=7.60(b - 1)+60
dw
d
(\text{dB})
w
6
7\,\text{dB}
0.8
1.25
\dfrac{1\times10^{-16}}{3}
\dfrac{2\times10^{-16}}{7}
k
w
1.25
6\,\text{dB}
7\,\text{dB}
x
 with which to purchase fencing that costs 
 per foot, what is the value of 
\qquad [\text{cost per foot}]\cdot [\text{number of feet}]=[\text{total cost of fencing}]
(2x+30)\text{ ft}
 per foot.  Finally, since Brett wants to make the fenced in area as wide as possible, he will be using his entire budget on fencing, and so the total cost of the fencing is 
\qquad 7(2x+30)=490
x
20\text { ft}
\left(2x^3y^4z^5\right)^3
6x^9y^{12}z^{15}
8x^9y^{12}z^{15}
6x^6y^7z^8
8x^6y^7z^8
(a^mb^n)^x=a^{m\cdot x}b^{n\cdot x}
\qquad\qquad\left(2x^3y^4z^5\right)^3=2^{1\cdot3}x^{3\cdot3}y^{4\cdot3}z^{5\cdot3}
2=2^1
2^3=8
\qquad\qquad2^{1\cdot3}x^{3\cdot3}y^{4\cdot3}z^{5\cdot3}=8x^9y^{12}z^{15}
\qquad 8x^9y^{12}z^{15}
\theta=\dfrac{\pi}{2} \text{radians}
\theta
45^{\circ}
90^{\circ}
135^{\circ}
180^{\circ}
2\pi
360^{\circ}
\qquad 2\pi \text{ radians}=360^{\circ}
4
\qquad \dfrac{\pi}{2} \text{ radians}=90^{\circ}
f(x)g(x)
\quad\qquad f(x) g(x)
\qquad =  \left( {\dfrac{1}{2}x} {-1}\right) (x^3+2x^2+4x+8)
\qquad =  \blue{\dfrac12x^4+x^3+2x^2+4x} \pink{-x^3-2x^2-4x-8}
\qquad =\dfrac12x^4 + 0 + 0 + 0 -8
\qquad =\dfrac12x^4  -8
80
(\text{in})
60\text{ in}
44\text{ in}
8\text{ in}
4\text{ in}
2\text{ in}
l
w
h
\qquad V= lwh
\dfrac{211{,}200}{64}=3{,}300
30
\text{5:30 a.m.}
\text{9:00 a.m.}
\text{7:30}
\text{7:30}
\text{7:30}
sC
C
s
s
100
0.25
C
s
20
s
C
20
70
40
80
60
90
80
100
C / s
s
C
C/s
20
70
70 / 20 = 3.50
40
80
80/40 = 2.00
60
90
90 / 60 = 1.50
80
100
100/80 = 1.25
s
C / s \,
\dfrac{9}{16}x^2-3x-4=0
\dfrac{9}{16}x^2-3x+4=0
\dfrac{16}{25}x^2-25=0
\dfrac{16}{25}x^2+25=0
b^2-4ac
0
4
1
\qquad b^2-4ac=(-3)^2-4\left(\dfrac{9}{16}\right)\left(-4\right)=18
2
\qquad b^2-4ac=(-3)^2-4\left(\dfrac{9}{16}\right)\left(4\right)=0
3
\qquad b^2-4ac=\left(0\right)^2-4\left(\dfrac{16}{25}\right)\left(-25\right)=64
4
\qquad b^2-4ac=\left(0\right)^2-4\left(\dfrac{16}{25}\right)\left(25\right)=-64
0
2
\dfrac{9}{16}x^2-3x+4=0
30\%
12
40\%
s
30\%
130\%
1.3s
12
s + 0
1.3s+12
40\%
140\%
s
120
\qquad 120+0=120
120
(x,y)
(-5,3)
(5,-3)
(16,5)
(0,-3)
(5,16)
(-3,0)
x
y
y
\qquad\qquad\qquad y^2-9=2y+6
 
y=5
y=-3
y
y
 
(16,5)
(0,-3)
(a,b)
a>0
a
x
y
(a,b)
y
y=3x
x
x
x^2
a>0
x
(a,b)
a
a=\dfrac{5}{4}
\overline{AB}
\overline{BC}
\overline{BT}
9
\overline{AC}
\overline{BC}
15
h
h
11
d
s
\qquad d(s) = 10.7 - 1.2s^2
d
d(s)
8
8 - 1.2(s-1.5)(s+1.5)
8 + 1.2(2.25 - s^2)
8(1 - 0.15s^2) + 2.7
11.9-1.2(s - 1)^2 - 2.4s
s
s
d
8
\qquad d(s) = k(s - r_1)(s - r_2)
k
r_1
r_2
d(r_1)
d(r_2)
0
\qquad d(s) = k(s-r_1')(s-r_2') + 8
d(r_1')
d(r_2')
8
\qquad 8 - 1.2(s-1.5)(s+1.5)
d(s)
\qquad 8-1.2(s-1.5)(s+1.5)
\qquad
xy
y=P(x)
P(x)=x^2+6x+8
P(x)=x^3+6x^2+8x
P(x)=x^2-6x+8
P(x)=x^3-6x^2+8x
x=0
x
2
x=\pink0
x=\gray{-4}
x=\purple{-2}
x=\red0
x=\green{4}
x=\blue{2}
x
2
\qquad P(x)=x^3+6x^2+8x
\qquad  I = 870\pi^2a^2f^2
I
\left( \dfrac{\text W}{\text m^2} \right)
a
(\text m)
f
(\text{Hz})
0.05 \, \dfrac{\text W}{\text m^2}
3 \times 10^{-5} \, \text{m}
8.77
80
6{,}500
4 \times 10^7
v
f\approx \pm 80.4 \, \text{Hz}
80 \, \text{Hz}
a
a
(x, y)
-\dfrac{3}{2}
\dfrac{5}{6}
3
y
y
a
a
a
(x, y)
\qquad-\dfrac{3}{2}
\quad S = 537.5 - 1.25p
S
p
125
537.5
537.5 - 1.25p
p
1.25p
537.5
537.5
537.5 - 1.25p
537.5 - 1.25p
11
\text{(ft)}
4.5 \text{ ft}
h
10.0
\sin(35^\circ) \approx 0.57
\tan(35^\circ) \approx 0.70
35^\circ
2
\ell
34.28
34
BC = 4
1.5
C
6
12
8\pi-6
16\pi-12
\qquad A = \dfrac{1}{2}\theta r^2
A
\theta
r
\theta
2\pi
\theta
2\pi
A
16\pi - 12
q(2q^4+12q^3+3q^2)
\qquad\quad(2q^5+7q)+(5q^5+3q^3)
\qquad =2q^5+7q+5q^5+3q^3
\qquad =2q^5+5q^5+3q^3+7q
\qquad =7q^5+3q^3+7q
q
q
\qquad =q(7q^4+3q^2+7)
\qquad
\overline{XY}
W
\overline{ZY}
5
V
\overline{XZ}
6
\overline{VW}
\overline{XY}
\overline{XY}
18
7
8
20
0.875
2.5
17.5
23
7:8
\qquad  \dfrac{7}{8}= \dfrac{x}{20}
x
17.5
20\ \text{rolls}
62
648
2\%
90\%
8\%
12\%
10\%
12\%
13\%
15\%
11\%
15\%
90\%
\qquad\dfrac{62}{648}=0.0956...\approx10\%
10\pm2
\qquad10-2=8
\qquad10+2=12
8\%
12\%
8\%
12\%
1\text{ ft}^2{:}\:900\text{ ft}^2
, which would be equivalent to 
 per 
 square foot 
 square foot, we can set up a proportion to determine the number of square feet, 
, that the property has if it sells for 
s
1\text{ ft}^2:900 \text{ ft}^2
d
1{,}800
\qquad \dfrac{1}{900 }=\dfrac{d}{1{,}800}
s
\qquad d=\dfrac{1}{900}\times 1{,}800=2
2
(f^2-2)
(f^7+2f^5+4f^3+8f)
\quad\qquad (\blue{f^2}\pink{-2}) \cdot (f^7+2f^5+4f^3+8f)
\qquad=f^9+2f^7-2f^7+4f^5-4f^5+8f^3-8f^3-16f 
\qquad = f^9 + 0 + 0 + 0 -16f
\qquad = f^9 -16f
f
\qquad = f(f^8 -16)
80
(\text{in})
12
14\,\text{in}
26\,\text{in}
34\,\text{in}
46\,\text{in}
12\,\text{in}
l\,\text{in}
\left(l-12\right)\,\text{in}
\qquad l+l+\left(l-12\right)+\left(l-12\right)=80
\qquad4l-24=80
26\,\text{in}
\qquad \dfrac{3}{i}+\dfrac{2}{i^2}
i=\sqrt{-1}
3i+2
3i-2
-3i+2
-3i-2
i^2=-1
2008
2009
2010
2011
2012
| 
|
|	
|	
 through 
\qquad11{,}864-11{,}223=641
\qquad i^{{101}}
i=\sqrt{-1}
1
-1
i
-i
i^2=-1
101
4
\qquad  i^{{101}}=i
2006
.  For each year before or after 
, the average price per square foot increased by approximately 
. In what years could  the average home price per square foot be 
2003
2009
2002
2010
2000
2012
1999
2013
y
\qquad3.50|y-2006|
98
\qquad  98+3.50|y-2006|
119
y
\qquad y=2012
y=2000
2000
2012
\qquad  x-1=(2-x)^2
0
\dfrac{5\pm\sqrt{5}}{2}
\dfrac{-1\pm\sqrt{21}}{2}
ax^2+bx+c=0
\qquad x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
\dfrac{5\pm\sqrt{5}}{2}
y=-\dfrac{1}{x}
x
y
y=\dfrac{1}{x}
0
0
x
-2
-1
-\dfrac{1}{2}
\dfrac{1}{2}
1
2
y
-\dfrac{1}{2}
-1
-2
2
1
\dfrac{1}{2}
y=-\dfrac{1}{x}
y=\dfrac{1}{x}
x
f(-x)=-f(x)
f(x)=\dfrac{1}{x}
y=-\dfrac{1}{x}
30^\circ
12
\text{(km)}
31 \text{ km}
60^\circ
d
\blue{x_1}
\blue{x_2}
\pink{y_1}
\pink{y_2}
\blue{x_1}
\blue{x_2}
\pink{y_1}
\pink{y_2}
42 \text{ km}
P
s
0 \le s \le 15
P
\qquad P = \dfrac{1}{3}s^2-5s+18
P
\qquad P = a(s-s_0)^2 + P_0
a
(s_0,P_0)
a
P
(7.5, -0.75)
0.75
7.5
7.5
\qquad\dfrac{7}{x-5} + \dfrac{4}{5-x}
x\neq5
\dfrac{11}{x-5} 
\dfrac{11}{5-x} 
\dfrac{3}{x-5} 
\dfrac{3}{5-x} 
5-x=\red{-1}(x-5)
\qquad\dfrac{7}{x-5} + \dfrac{4}{\red{-1}(x-5)}
\dfrac{4}{-1(x-5)}=\dfrac{-4}{x-5}
\qquad\dfrac{7}{x-5} + \dfrac{-4}{x-5}
\qquad \dfrac{7+-4}{x-5} =\dfrac{3}{x-5}
\qquad \dfrac{3}{x-5}
(x,y)
0
1
2
\qquad 4x+2y=1
\qquad 3y^2+5y-10=0
y=-\dfrac{5}{2}\pm\dfrac{\sqrt{145}}{2}
y=-\dfrac{5}{6}\pm\dfrac{\sqrt{145}}{6}
y=\dfrac{5}{6}\pm\dfrac{\sqrt{145}}{6}
y=\dfrac{5}{2}\pm\dfrac{\sqrt{145}}{2}
ax^2+bx+c=0
1
ax^2+bx+c=0
\qquad x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
y=-\dfrac{5}{6}\pm\dfrac{\sqrt{145}}{6}
 a month for a phone plan with a long distance rate of 
 per minute during off-peak hours.  Without this plan, her local telephone provider would charge her 
 per minute for long distance calls during off-peak hours. If Zoe makes calls only during off-peak hours and saved 
m
.  Therefore, 
 represents the cost, in dollars, for 
 for this plan, her total phone bill can be represented by the expression 
 per minute to call long distance during off-peak hours by her local provider.  Therefore, if she calls long distance for 
 minutes during non-peak hours, she will be charged 
 this month by purchasing the long distance plan, we can deduce that the cost of her long distance bill *without* the plan would have been 
m
128
 for each regular yard he mows, and he charges an extra 
 for each large yard that he mows. In one week he mowed 
 more large yards than regular yards and made 
.  If 
10(r+6)+15r=265
10(r+6) + 5r=265
10r+15(r+6)=265
10r + 5(r+6)=265
r
 for each yard that he mows, so he earns 
r
6
r+6
 for each large yard that he mows, he earns 
 total for mowing yards, an equation that could be used to solve for 
\qquad{10r}+{15\left(r+6\right)}={265}
\qquad 6{,}000 = bu + 4{,}000
b
600{,}000 + 100bu + 400{,}000
\qquad 600{,}000 + 100bu + 400{,}000
100
100bu + 400{,}000
600{,}000
\qquad c(ax-b)-d(b-ax)
(ax-b)(c+d)
(ax-b)^2(c+d)
(b-ax)^2(c-d)
(ax-b)(b-ax)(c-d)
(ax-b)
(b-ax)
-1
-1
\qquad c(ax-b)-d (-1)(-b+ax)
\qquad c(ax-b)+d(ax-b)
(ax-b)
\qquad c(ax-b)+d(ax-b)=(ax-b)(c+d)
(ax-b)
(c+d)
 c(ax-b)-d(b-ax)
\qquad(ax-b)(c+d)
7:4
4:7
3:4
7:3
11:4
7
7:4
7
4
\qquad 7\:\text{total} -4\:\text{in cups} = 3\:\text{remaining}
3
7
3
7:3
3a-5b\neq0
\qquad\dfrac{9a^2-30ab+25b^2}{3a-5b}
3a-5b
3a+5b
9a+25b^2
3a-30ab+5b
\qquad 9a^2-30ab+25b^2=(3a-5b)^2
\qquad\dfrac{(3a-5b)(3a-5b)}{(3a-5b)}
3a-5b
56
140
2{,}000
b
b\geq32
b\leq32
b\geq33
b\leq33
\qquad \text {Teddy's weight} + \text{weight of boxes} \leq 2{,}000
56b
b
\qquad 140 + 56b\leq2{,}000
140
\qquad 56b\leq1{,}860
56
\qquad b \leq 33.21
b\leq33
950
950
.   Which of the following could be the number of sandwiches sold in a month if  the owner's revenue decreased 
 s=905
s=995
 s=946
s=955
 s=500
s=1{,}400
 s=500
s=950
s
\qquad |s-950|
950
 for every sandwich above or below 
, for which the revenue decreases  
\qquad |s-950|=\dfrac{45}{0.10}=450
\qquad s-950=\pm450
\qquad s=500
s=1{,}400
\qquad s=500
s=1{,}400
A
70{,}045
23{,}684
B
74{,}577
17{,}046
C
64{,}019
19{,}998
D
60{,}490
x
E
69{,}154
15{,}569
5
5
\qquad\dfrac{23{,}684+17{,}046+19{,}998+x+15{,}569}{5}
18{,}990
(\ge)
18{,}990
\qquad\dfrac{23{,}684+17{,}046+19{,}998+x+15{,}569}{5}\ge18{,}990
\qquad60{,}490
\qquad64{,}019
\qquad70{,}045
\qquad74{,}577
\qquad15{,}569
\qquad23{,}684
\qquad69{,}154-18{,}522=50{,}632
50{,}632
A
16.7\%
70{,}045
\qquad0.167\cdot70{,}045=11{,}697.515
11{,}698
27.7\%
23{,}684
\qquad0.277\cdot23{,}684=6{,}560.468
6{,}560
19.5\%
A
b
69{,}154
b
\qquad\text{total active army bonuses}=69{,}154b
15{,}569
b
E
\bullet
\qquad\dfrac{23{,}684+17{,}046+19{,}998+x+15{,}569}{5}\ge18{,}990
\bullet
50{,}632
\bullet
19.5\%
A
\bullet
E
\qquad\sqrt{\dfrac{3380}{2000}}
\dfrac{13}{10}
\dfrac{13}{100}
\dfrac{169}{10}
\dfrac{169}{100}
\purple{\text{power of a quotient}}
\left(\dfrac xy\right)^a=\dfrac{x^a}{y^a}
\qquad\dfrac{13}{10}
4
w
h
\dfrac23
\dfrac23
\alpha
\beta
\dfrac\beta\alpha
\alpha
h
w
\beta
h \cdot \dfrac23
w \cdot \dfrac23
w : w \cdot \dfrac23
h : h\cdot\dfrac23
\alpha = \beta
1 = \dfrac\beta\alpha
\alpha
\beta
\dfrac\beta\alpha
a
a
(x, y)
y = 1.3
-2.1
-1.1
2
a
y
a
x
1.3
y
x
x
x
a = -2.1
y
 \qquad V = \pi r^2h
V
r
h
r =  \dfrac {\sqrt {Vh}}{ \pi }
r = \sqrt{ \dfrac {Vh}{ \pi}}
r =  \dfrac {\sqrt V}{ \pi h}
r = \sqrt{ \dfrac {V}{ \pi h}}
r^2
\pink{\text{dividing by } \pi}
\blue{\text{dividing by } h}
\pi h
r
\purple{\text{take the square root of both sides of the equation }}
\sqrt[2]{x}=\sqrt{x}
\qquad r = \sqrt{ \dfrac {V}{ \pi h}}
\qquad  B  =  0.55A  
B
A
B
A
0.55\%
0.45\%
55\%
45\%
A
0.55
B
55\%
45\%
\qquad 100\% - 55\% = 45\%
45\%
2000
68
16
14
75
2001
2006
2008
2014
16
14
\dfrac{8}{7}
t
2000
\dfrac{8}{7}t
t
2000
\dfrac{8}{7}t+68
t
2000
75
75
t
t\approx 6\large\frac{1}{8}
6
2006
75
2006
75
2
(\text{cm})
2
30\%
0.5\,\text{cm}
2
3
4
5
2\,\text{cm}
30\%
T(n)
\text{cm}
n
\qquad T(n)=2(1-0.30)^n=2(0.70)^n
n
T(n)\leq0.5\,\text{cm}
n=2
\qquad T(2)=2(0.70)^2\approx0.98
n=3
\qquad T(5)=2(0.70)^3\approx0.69
n=4
\qquad T(4)=2(0.70)^{4}\approx 0.48
n=4
4
c
c=12
c
-12
c
12
c
\left(\dfrac52\right)
y
y
\qquad\dfrac{c}{-4}=-3
c=12
c
12
c
12
\qquad w(x)=-\dfrac13x^2+3
x
x
\qquad
-\dfrac13(x^2-9)
 -\dfrac{1}{3}(x-3)(x+3)
 -\dfrac{1}{3}(x-1)(x+1)+\dfrac83
-\dfrac{1}{3}(x-2)(x+2)+\dfrac53
\qquad w(x) = -\dfrac13\left(x^2 - 9\right)
(x^2 -9 )
\qquad w(x) = -\dfrac{1}{3}(x-3)(x+3)
\qquad w(x) = -\dfrac{1}{3}(x-3)(x+3)
75\%
32
15\%
h
t
75\%
\qquad h=0.75t
32
h+32
15\%
1.15t
15\%
100\%
\qquad h+32=1.15t
h=0.75t
\qquad 0.75t + 32 = 1.15t
0.75t
\qquad 32 = 0.4t
0.4
t
t=80
t=80
h
\qquad h=0.75t
\qquad h=0.75(80) = 60
60
f
g
f(x) = 1+2x
g(f(x)) = x+3
g(5)
5
4
6
-3
g(f(x))
f(x)
1
g(1)
f(x)
1
x
g(f(x))
x = 2
f(2) = 1
g(1)
g(1) = 5
18
17\%
35
54
94
106
c
18
17\%
0.17c = 18
c
0.17
\qquad \dfrac{18}{0.17} \approx 105.9
106
\qquad -p+60 = -h+ 10{,}000
h
p=10
h
p=10
h
 5 <2x+3<11
-4x-6
-10
-22
-22
-10
1
4
4
1
 5 <2x+3<11 
 -4x-6 
 -4x-6=-2(2x+3) 
 -2(2x+3)=-4x-6 
-4x-6 
-2
a<x<b
b>a
x
a
b
x<2
x>5
5<x<2
-4x-6
-22
-10
(-32.7,-9.08)
\sqrt{10}
(x+32.7)^2+(y+9.08)^2=\sqrt{10}
(x+32.7)^2+(y+9.08)^2=\sqrt{20}
(x+32.7)^2+(y+9.08)^2=10
(x+32.7)^2+(y+9.08)^2=100
(\green h,\purple k)
\red r
(x-\green h)^2+(y-\purple k)^2=\red r^2
\qquad(x-(\green {-32.7}))^2+(y-(\purple {-9.08}))^2=\red {\sqrt{10 }}^ 2
(x+32.7)^2+(y+9.08)^2=10
x^3+b^3 = (x+b)(x^2-xb+9)
b
3
6
36
108
m^3+n^3
(m+n)(m^2-mn+n^2)
x^3+b^3
\qquad x^3+b^3=(x+b)(x^2-bx+b^2)
x^3+b^3 = (x+b)(x^2-xb+9)
b^2=9
\sqrt[2]{a}=\sqrt a
b
3
-3
35
n
s(n)
\qquad {s(n)=35^{\large n}}
\green{n+2}
\qquad {s(\green{n+2})=35^{\green{\large{n+2}}}}
{x^{\large y} \cdot x^{\large z}=x^{\large{y+z}}}
1225
y=\sqrt{x+3}+2
y=\sqrt{x}
\qquad\qquad\qquad
y=\sqrt{x-h}+k
h
k
y=\sqrt{x+3}+2
-3
2
y=\sqrt{x+3}+2
\qquad\qquad\qquad
\qquad T(r)=\dfrac{72}{r}
72
T(r)
r\%
72
72
72
1\%
72
72
2\%
r=1
\qquad T(1)=\dfrac{72}{1}=72
T(r)
r\%
72
1\%
\,r=72
1
72
1\%
\overline{AD}
\overline{BE}
\overline{AB}
\overline{DE}
\overline{CE}
4.2
6.3
6.8
10.2
\angle DCE
\angle BCA
\overline{AB}
\overline{DE}
\angle CAB
\angle CDE
\angle{ABC}
\angle{CED}
{ABC}
{DEC}
 ABC
{DEC}
 ABC
 DEC
 ABC
 DEC
DE
AB
7.5:2.5
\dfrac{7.5}{2.5}=3
DE
3
AB
 DEC
3
 ABC
\overline{CE}
3
\overline{AC}
\overline{CE}
6.3
m
0
1
\qquad {x^m \cdot x^n = x^{m + n}}
\qquad x^0 = 1 
x \ne 0
(x,y)
x
y
\dfrac{5}{6}
\dfrac{25}{36}
\dfrac{2}{3}
x + y
x
y
x
x
y
\qquad x = \dfrac{3}{4}y +\dfrac{5}{24}
x
y
(x, y) = \left( \dfrac{5}{12}, \dfrac{5}{18} \right)
x + y
\qquad \dfrac{5}{12} + \dfrac{5}{18} = \dfrac{15}{36} + \dfrac{10}{36} = \dfrac{25}{36} 
(x_1,y_1)
(x_2,y_2)
y_1+y_2
x
y
(x,y)
y=-3x
\qquad \dfrac{(-3x)-7}{6}=\dfrac{3x^2+x-5}{2}
6
3x
7
x
\qquad x = \dfrac{-b\pm\sqrt{b^2-4ac} }{2a}
ax^2+bx+c=0
x_1=-\dfrac{4}{3}
x_2=\dfrac{2}{3}
y
x
\qquad \left(-\dfrac{4}{3},4\right)
\left(\dfrac{2}{3},-2\right)
y
y_1+y_2
4+(-2)=2
r
1.3
h
5
4.5
r
\text{\red{at most}}
1.3
\text{\red{less than or equal}}
\qquad r \red{\leq} 1.3
h
\text{\blue{at least}}
5
\text{\blue{greater than or equal}}
\qquad h \blue{\geq} 5
\text{\green{at most}}
4.5
\text{\green{less than or equal}}
\qquad h \green{\leq} 4.5r
\qquad P(n)=n^2-5n-7
P(-3i)
i=\sqrt{-1}
-4+15i
-7+12i
-7+24i
-16+15i
-3i
n
\qquad P(-3i)=(-3i)^2-5(-3i)-7
i^2=-1
\qquad P(-3i)=-16+15i
848
178
1026
415
400
200
600
335
485
145
625
2483
1237
3570
2013
4.8\%
20
59
119
594
\text{\red{Spanish non-native speakers}}
\text{\blue{all non-native speakers}}
848
178
1026
\red{\text{Spanish}}
415
400
200
600
335
485
145
625
\blue{\text{Total}}
2483
3570
\text{\red{Spanish non-native speakers}}
\red x
\dfrac{\red x}{\blue{1237}}
4.8\%
0.048
\qquad\dfrac{\red x}{\blue{1237}}=0.048
0.048
\blue{1237}
\red{59}
E
t
1^{\text{st}}
2000
 on January 
, 
. Model 
 has tax expenditures which increase by 
 each year. Model 
 has tax expenditures which increase by a factor of 
 every 
 years. If the models have the same value on January 
, 
, what is the value of 
1^{\text{st}}
2010
t = 10
5 + 4(10) = 5 + 40 = 45
45
t = 10
a
a
1^{\text{st}}
2010
5a^2
45
a
a = 3
K
m
v
\qquad K=\dfrac{1}{2}mv^2
\dfrac{2}{3}
\dfrac{2}{3}
\dfrac{4}{9}
\dfrac{1}{3}
\dfrac{1}{9}
K
v
\dfrac{2}{3}
\dfrac{2}{3}
K
v^2
K
\left(\dfrac{2}{3}\right)^2
\dfrac{4}{9}
18
(\text{ft})
28\,\text{in}
\qquad\dfrac{12}{2}-\dfrac22=5
28.44\,\text{in}
s
18.3\,\text{ft}
n
C
125
 when 
 coat is purchased. Which of the following graphs in the 
-plane could represent the cost per coat when purchasing 
125
n  = 125
1
n = 1
n = 1
C
C
n = 125
C
\qquad 4+kp = 4p^2
k
0
k
0
1
2
4
ap^2+bp+c
p
0
b
-k
k
0
\quad P = 100h - 2{,}000
 a month for Mia to rent space to run her law firm. She charges 
 per hour for legal advice. The equation above gives Mia's profit, 
, per month, after working 
20
80
100
200
0
P
h
20
20
(
  | Profit 
in thousands of 
 | 
| 
, corresponding to 
 thousand dollars spent on advertising in a particular month. It is determined that for the region of interest, the profit can be modeled by a quadratic function of the form 
. If 
93\text{ m}
110\text{ m}
120\text{ m}
127\text{ m}
(60, 56)
(80, 88)
d=ax^2+bx
-4
3
b
 40=4800a
4800
a=\dfrac{1}{120}
a
b
a=\dfrac{1}{120}
b=\dfrac{13}{30}
\qquad d=\dfrac{1}{120}x^2+\dfrac{13}{30}x
100
x
d
d\approx126.7
100 \text{ km/h}
127 \text{ m}
\qquad \dfrac{1}{1-i}
i=\sqrt{-1}
2-2i
2+2i
\dfrac{1-i}{2}
\dfrac{1+i}{2}
i
1-i
1+i
\dfrac{1+i}{1+i}
i^2=-1
\qquad \dfrac{1+i}{2}
 for a haircut. His operating expenses are, on average, 
 per day. He calculates his profit by subtracting his operating costs from the money he earns from the haircuts he gives. In a given day, the barber expects to make a profit of at least 
. If the barber gives 
12h - 37 \geq 86
12(h - 37) \geq 86
12h + 37 \geq 86
12(h + 37) \geq 86
12h
\qquad 12h - 37 \geq 86
\qquad \dfrac{2.5dr^2}{6} + \dfrac{3dr^2}{4}
\dfrac{7dr^2}{6}
\dfrac{7dr^2}{12}
\dfrac{5.5dr^2}{10}
\dfrac{5.5dr^2}{12}
12
\blue{\dfrac 2 2}
\pink{\dfrac 3 3}
1
\qquad \dfrac{7dr^2}{6}
f(x)=\dfrac{2x-2}{x^2+1}
g(x)=x^2+1
f(1+g(1))
-4
-\dfrac{7}{4}
\dfrac{13}{2}
8
f(1+g(1))
g(1)
\qquad g(1)=1^2+1=2
\qquad f(1+g(1))=\dfrac25
\qquad \dfrac{1}{2}c^2+3c=1
c_1
c_2
c_1>c_2
c_1
-3+\sqrt{2}
-3+\sqrt{11}
3+\sqrt{2}
3+\sqrt{11}
2
c^2+6c=2
c^2
1
c
9
c=-3+\sqrt{11}
c=-3-\sqrt{11}
c_1=-3+\sqrt{11}
\qquad v=331.3\sqrt{1+\dfrac{T}{273.15}}
v
T
T=273.15\left(\dfrac{v^2}{331.3^2}-1\right)
T=273.15\left(\dfrac{v^2}{331.3^2}\right)-1
T=273.15(v-331.3)^2
T=273.15\left(\dfrac{v-331.3}{331.3}\right)^2
\red{\text{dividing by }331.3}
\green{\text{square}}
\blue{\text{subtract }1}
\purple{\text{multiply by }273.15}
\qquad T=273.15\left(\dfrac{v^2}{331.3^2}-1\right)
\qquad -2>\dfrac{3(b+4)}{-2}
b<-3
b< -\dfrac{16}{3}
b > -\dfrac{8}{3}
b>0
\gray{-2}
3
\pink{12}
\blue3
b>-\dfrac83
 per month in overhead cost. Each DVD costs 
 per night and each Hi-Def disc is 
 per night. The company wishes to make a profit of at least 
350
1200
900
900
1450
500
1875
450
D
H
\qquad 1.49D  + 1.89H
, so that means the expression for profit must be *greater than or equal to* 
. Turn this into an inequality and solve for 
1875
450
M
10
P
2.43
2.43
2.43
2.43
\red{a}\cdot \blue{b}^{\large\purple{x}}
\red a
\blue b
\purple x
1
\blue b<1,
P
M
\purple{\dfrac{M}{2.43}}
1
M
2.43
2.43
\qquad \sqrt{2x+4}=2+x
x
0
x=0
x=-2
x=\blue0
x=0
x = \blue{-2}
x=0
x=-2
0\cdot (-2)=0
0
1.4
(\text{m})
w
1.6\,\text{m}
1.7\,\text{m}
2.1\,\text{m}
2.4\,\text{m}
30^\circ-60^\circ-90^\circ
s
60^\circ
\sqrt3s.
\qquad \sqrt3\cdot 2\approx3.464
1.4\,\text{m}
1.4\,\text{m}
2.4\,\text{m}
2
67
6
2
61
73
2
61
73
2
61
67
2
67
73
2
61
73
 in change in her pocket. The 
21
24
25
29
, and an equal number of nickels and dimes, worth 
 and 
, respectively. Because the number of nickels and dimes is equal, we can use the expression 
, where 
16
\qquad 35A + 28P = 250
A
P
. The caricature artist charges 
 per hour and the face painter charges 
 per hour. What is the meaning of the 
35A
35A
35
A
35A
A
A
P
\blue{\text{units}}
35A
A
65
92
5{,}200
\pi \approx 3.14
4
5
8
25
r
h
\qquad V=\pi r^2h
V\leq 5{,}200
r\leq 5
 5{,}200
5
\qquad4+5m=4m+1+m
m=0
m=1
m
m
4=1
 for each regular driveway and he charges an extra 
 for each large driveway that he shovels. After a snowstorm he shovels 
 fewer large driveways than regular driveways and makes 
3
4
7
9
r
 for each one, so he earns 
r
3
r-3
 for each large driveway that he shovels, he earns 
7
9
4,1988
16
10,1987
17
15,1988
14
21,1987
5
22,1988
11
14,1988
3
1,1987
12
20,1989
10
2,1987
18
7,1987
15
20,1987
2007
2007
1
30
2007
1
30
1987
1989
2007
1987
6
11
2007
1987
2007
1987
1987
1986
1988
1989
1987
1989
9
11
1
30
2007
2007
1
30
2007
1
30
\ P(t)= 350\, (2)^{t} ,
P(t)
t
350
2
350
2
t=1
2
2
\qquad (8c+5)^3
512c^2+125
(8c+5)(64c^2-40c+25)
128c^3+1040c^2+220c+625
512c^3+960c^2+600c+125
\qquad (8c+5)^3=(8c+5)(8c+5)(8c+5)
8c+5
\qquad 512c^3+960c^2+600c+125
x
0
x
\qquad w(x)=-\dfrac14(x^2-12)
w(x)
x
\qquad
-\dfrac14x^2+3
  -\dfrac{1}{4}(x-\sqrt{12})(x+\sqrt{12})
 -\dfrac{1}{4}(x-2)(x+2)+2
 -\dfrac{1}{4}(x-4)(x+4)-1
y
(0,3)
3
\qquad w(x) = -\dfrac14x^2+3
6
(\text{in})
16\text{ in}
126 \pi
144 \pi
180 \pi
504 \pi
\qquad V = \pi r^2 h
V
r
h
\qquad V = \dfrac{4}{3} \pi r^3
V
r
6\text{ in}
\, \dfrac{6}{2} = 3\text{ in}
16 - 3 - 3 = 10\text{ in}
\qquad V_\text{tank} = \pi r^2 h + \dfrac{4}{3} \pi r^3
r
h
126 \pi \text{ in}^3
f
g
f(x)=g(x+5)+6
f(x)=g(x-5)+6
g(x)=f(x-5)+6
g(x)=f(x+5)+6
c>0
h(x)\rightarrow h(x+c)
h
c
h(x)\rightarrow h(x-c)
h
c
h(x)\rightarrow h(x)+c
h
c
h(x)\rightarrow h(x)-c
h
c
6
g
f
g
5
6
f
g
5
g(x)\rightarrow g(x-5)
6
g(x-5)\rightarrow g(x-5)+6
f(x)=g(x-5)+6
\qquad \dfrac16\sqrt{k}-3=1-\sqrt k
1.9
2.9
3.4
11.8
\sqrt k
k=\left(\dfrac{24}{7}\right)^2
k=\dfrac{576}{49}\approx 11.8
11.8
\qquad29=3(x+7)^2+41
i=\sqrt{-1}
x=-7+2i
x=-42-12i
x=-\dfrac76+\dfrac{\sqrt{443}}{6}i
x=-7-\dfrac{\sqrt{123}}{3}i
29
y
a=3
2
\qquad a=3,\qquad b=42,\qquad c=159
i=\sqrt{-1}
\qquad x=-7+2i
x=-7-2i
\qquad x=-7+2i
62
61.98
1
(^\circ\,\text{C})
0.00056
1.0003^\circ\,\text{C}
3.57^\circ\,\text{C}
10.003^\circ\,\text{C}
35.71^\circ\,\text{C}
d
\red{61.98}
0.00056
\green{0.00056d}
\qquad\red{61.98}+\green{0.00056d}
\purple{62}
\qquad\red{61.98}+\green{0.00056d}=\purple{62}
d
35.71^\circ\,\text{C}
O
5
X
Y
7
A
\qquad A = \dfrac{1}{2}\theta r^2
A
\theta
r
\qquad s = r \theta
s
r
\theta
s
r
\theta
\theta
r
A
A
17.5
\qquad\dfrac{3}{c+4} +\dfrac{1}{4}
c\neq-4
1
\dfrac{4}{c+8} 
\dfrac{c+16}{4c+16} 
\dfrac{c+16}{8c+32} 
4(c+4)
\dfrac{4}{4}
\dfrac{c+4}{c+4}
\qquad\qquad\dfrac{12}{4c+16}+\dfrac{c+4}{4c+16}=\dfrac{12+c+4}{4c+16}=\dfrac{c+16}{4c+16}
\qquad \dfrac{c+16}{4c+16}
3
8
(\text{cm})
16\,\text{cm}
12\,\text{cm}
9\,\text{cm}
3
(\text{cm}^3)
\pi\approx 3.14
804\,\text{cm}^3
1{,}055\,\text{cm}^3
1{,}859\,\text{cm}^3
7{,}436\,\text{cm}^3
r
h
\qquad V=\pi r^2h
\qquad V=\pi (4)^2(16)+\pi (4)^2(12)+\pi (4)^2(9)
\dfrac12(8)=4
1{,}859\,\text{cm}^3
\qquad  | -3x + 1 \: | \: >  -6 
 \dfrac53 < x < \dfrac73 
 x <  \dfrac53 \:\:\:\: \text{or} \:\:\:\: x > \dfrac73
\qquad  | -3x + 1 \: | \: >  -6 
x
x
\qquad k\left(k-\dfrac12\right)=-\dfrac1{16}
k=\dfrac18
k=\dfrac14
k=-\dfrac18
k=\dfrac18
k=-\dfrac14
k=\dfrac14
\qquad k=\dfrac14
4
4
40{-49}
4
5
9
50{-59}
24
60{-69}
4
23
50
59
45.8\%
45.8\%
100
0.458
24
11
\qquad 2 + \dfrac{4}{v} - \dfrac{3}{2v^2}
v<0
\dfrac{-12}{v^3}
\dfrac{2v^2 + 2v - 3}{2v^2}
\dfrac{4v^2 + 8v - 3}{2v^2}
\dfrac{4v^2 + 8v - 3}{6v^2}
2v^2
2
\qquad \dfrac 2 1 + \dfrac{4}{v} - \dfrac{3}{2v^2}
\blue{\dfrac {2v^2}{2v^2}}
\purple{\dfrac{2v}{2v}}
\qquad \dfrac{4v^2 + 8v - 3}{2v^2}
135^\circ
2
(\text{m})
2.4\,\text{m}
d
d>\dfrac{0.8\sqrt{3}}{3}\,\text{m}
d>0.4\sqrt2\,\text{m}
d>\dfrac{4.8\sqrt{3}}{3}\,\text{m}
d>2.4\sqrt2\,\text{m}
2
2.4-2=0.4
135^\circ
(135-90)^\circ=45^\circ
d
\qquad d>0.4\sqrt2\,\text{m}
\qquad4-\dfrac{1}{3}z=-7z+6
z=\dfrac3{10}
z=-\dfrac3{10}
z=\dfrac{20}3
z=\dfrac{20}{6}
z
\red{7z}
\blue{4}
\green{\dfrac3{20}}
z=\pink{\dfrac{3}{10}}
z=\pink{\dfrac{3}{10}}
(z)
\qquad  az+ b= cz + d
 a\ne c
z=\dfrac3{10}
z=\dfrac3{10}
20
(\text{cm})
10\ \text{cm}
1{,}250
r
h
\qquad V= \pi r^2h
1{,}250\text{ cm}^3
1{,}250\text{ cm}^3
\qquad\dfrac{6{,}283.19}{1{,}250}\approx5.03\approx5
5
\theta=\dfrac{4\pi}{9} \text{radians}
\theta
20^{\circ}
36^{\circ}
80^{\circ}
720^{\circ}
2\pi
360^{\circ}
\qquad 2\pi \text{ radians}=360^{\circ}
2\pi
1\text{ radian}=\dfrac{180}{\pi}^{\circ}
 2\pi \text{ radians}=360^{\circ}
9
 \dfrac{2\pi}{9} \text{ radians}=40^{\circ}
\dfrac{4\pi}{9}
2\cdot  \dfrac{2\pi}{9}
\dfrac{4\pi}{9}\text{ radians}
2\cdot 40^{\circ}=80^{\circ}
\qquad\dfrac{4\pi}{9}=80^{\circ}
1
48.5
4
t
p
48.5p < 4t
48.5p > 4t
4p < 48.5t
4p > 48.5t
48.5
1
t
4
1
144\text{ ft}^2
50 \text { ft}
x
x(72-x)=50
x(25-x)=144
x(144-2x)=50
x(50-2x)=144
x
y
A
P
\qquad  A=xy
\qquad  P=2x+2y
50 \text{ ft}
50\text{ ft}
144\text{ ft}^2
\qquad  xy=144
\qquad  2x+2y=50
y
y
y
x(25-x)=144
xy
y=2(x-4)^2+2
y=-(x-a)^2+12
a
a
\ \ \ 4
-4
\ \ \ 8
 -8
y=a(x-h)^2+k
a
h
k
x=h
y=2(x-4)^2+2
x=4
y=2(x-4)^2+2
y=-(x-a)^2+12
a
4
a=4
xy
(-2,4)
(1,1)
 x^2+y^2-4x-8y+2=0
 x^2+y^2+4x-8y+2=0
 x^2+y^2+4x+8y-2=0
 x^2+y^2+4x-8y+14=0
(\green h,\purple k)
\red r
\qquad (x-\green h)^2+(y-\purple k)^2=\red r^2
(\green{-2},\purple4)
\qquad (x-(\green{-2}))^2+(y-\purple4)^2=\red r^2
\qquad (x+2)^2+(y-4)^2=\red r^2
(1,1)
(x,y)
\red r^2
\qquad (x+2)^2+(y-4)^2=18
\qquad x^2+y^2+4x-8y+2=0
\left(z^{3+t}\right)^{4}=z^{20}
z
t
2
\sqrt{8}
4
p
\qquad (a^m)^n=a^{mn} 
z^{12+4t}=z^{20}
 \red{z}^{12+4t}
\red{z}^{20} 
\text{\red{equal bases}}
t
t=2
y
(\text{mpg})
x
(\text{mph})
\qquad\qquad \qquad \qquad \qquad\text{ Fuel Economy vs Speed}
46\text { mpg}
46\text{ mph}
46\text { mpg}
0 \text{ mph}
70\text { mph}
46\text{ mpg}
(46,30.7)
y
x
\text{mph}
y
\text{mpg}
(46, 30.7)
30.7\text { mpg}
46\text { mph}
46\text{ mph}
80
35\%
73
28.8\%
38.3\%
65.0\%
71.2\%
80
35\%
80
\qquad 0.35 \times 80 = 28
28
80
\qquad 80 - 28 = 52
(52)
73
\qquad73 - 52 = 21
\qquad\dfrac{21}{73} \times 100 \approx 28.8\%
28.8 \%
C
d
{\qquad C(d)=9300\cdot0.8^{\large d}}
{0.0000019\cdot(1-0.2)^{d-100}}
{9279\cdot(1-0.2)^{d+100}}
{9300\cdot(1-0.002)^{100d}}
100
\green{100d}
\dfrac{100}{100}
{\left(x^{\large y}\right)^{\large z}=x^{{\large y}\cdot {\large z}}}
\red{0.998}
1
\qquad1-\red{0.998}=\blue{0.002}
0.2\%
\red{0.998}
1-\blue{0.002}
\qquad{9300\cdot(1-\blue{0.002})^{\green{100d}}}
\qquad \dfrac{xy}a - \dfrac{x^2y^2}b
xy
\qquad\quad \dfrac{xy}a - \dfrac{x^2y^2}b
xy
xy
\angle{QRS}
\angle{QRS}
\theta
\theta
\qquad\qquad \cos\theta=\dfrac{\red{\text{Length of the side adjacent to }\theta}}{\blue{\text{Length of the triangle's hypotenuse}}}
QRS
\angle{QRS}
\red{\text{{adjacent}}}
\angle{QRS}
\red {5\sqrt5}
\blue{\text{{hypotenuse}}}
QRS
\blue{10\sqrt5}
\qquad\qquad \cos\theta=\dfrac{\red{5\sqrt5}}{\blue{10\sqrt5}}=\dfrac{1}{2}
0^\circ<\theta<90^\circ
\cos60^\circ
\dfrac{1}{2}
\theta=60^\circ
\angle{QRS}
60^\circ
\qquad \dfrac{3k}{4}\leq \dfrac{3+2k}{5}
\purple5
\blue{4}
\pink{8k}
\qquad (ax+5)(x+v)
\qquad ax^2 + 25x + 25
a
v
a
\qquad\quad (ax+5)(x+v)
\qquad =ax^2+5x+axv+5v
x
\qquad =ax^2+(5+av)x+5v
\qquad=ax^2 + 25x + 25
5+av=25
5v=25
5
v=5
a
\qquad \sqrt{x}=\sqrt{ 3x}
\qquad  \sqrt{x}=\sqrt {3x}=\sqrt 3\sqrt x
1-\sqrt 3
\sqrt x=0
x=0
x=\blue0
x=0
0
\qquad U=\dfrac12kx^2
U
x
k
x=\sqrt{\dfrac{kU}{2}}
x=\sqrt{\dfrac{2U}{k}}
x=\dfrac{\sqrt{2U}}{k}
x=\left({\dfrac{2U}{k}}\right)^2
\purple{\text{multiply both sides by } 2}
\dfrac 1 2
\red{\text{divide by }k}
x
\blue{\text{take the square root of both sides of the equation}}
\sqrt[2]{a}=\sqrt{a}
\qquad x=\sqrt{\dfrac{2U}{k}}
\qquad \dfrac{5}{z} - \dfrac{2z+4}{z+2} = - 3
z
\qquad z(z+2)
\qquad \dfrac{5\blue{(z+2)}}{z\blue{(z+2)}} + \dfrac{-\pink{z}(2z+4)}{\pink{z}(z+2)} = \dfrac{- 3 \pink{z}\blue{(z+2)}}{\pink{z}\blue{(z+2)}}
\qquad \dfrac{5z+10}{z\blue{(z+2)}} + \dfrac{-2z^2-4z}{\pink{z}(z+2)} = \dfrac{- 3z^2 - 6z}{\pink{z}\blue{(z+2)}}
\qquad 5z + 10 - 2z^2 - 4z = -3z^2 - 6z
\qquad z^2 + 7z + 10 = 0
\qquad (z + 5) ( z + 2) = 0
z
-5
-2
z = -5
z = -5
z = -2
0
z = -5
-5
200
A
w
\qquad A(w) = 100w-w^2
 -(w-50)^2 +2500
 -(w+20)^2 +140w+400
 -(w-10)^2 +80w+100
 -(w+10)^2 +120w+100
-1
w
-1
\qquad~~~~~~~~~~~~ A(w)=-(w-50)^2+2500
\qquad A(w) = -(w-50)^2 +2500
(
)
(
)
0.35
2.04
?
1.44
0.50
0
v
h_\circ
t
h(t) = h_\circ - vt - 16 t^2
t
1.44
t
h(t) = 1.44
h(t)
v
h_\circ
h(\purple{0.35}) = \purple{2.04}
 h(\pink{0.50}) = \pink 0
h_\circ
v
v
h_\circ
h(t) = 1.44
t
t = 0.4
0.4
1.44
y^M
y\neq 0
M
y^M
M
2
a^m\cdot a^n=a^{m+n}
\dfrac{a^m}{a^n}=a^{m-n}
a\neq 0
M=\dfrac{5}{12}
\qquad M=\dfrac{5}{12}
 per pound in 
.  The price per pound increased about 
 each year until 
, and is expected to do so for the next two years.    Which of the following graphs represents the relationship between years after 
, 
, and price per pound, 
 in 
, the graph goes through the point 
. In addition, the slope is 
\qquad P=0.18t+0.51
\quad g = 15 - \dfrac{m}{32} 
 g 
m
32
32
32
32
32
m=0
15
 g = 15 - \dfrac{m}{32} 
 15 - \dfrac{m}{32} 
15
 \dfrac{m}{32} 
m
m
m
\dfrac{1}{32}
\dfrac{1}{32}
32
20
40
(\text{mm})
0.08\,\text{mm}
20
 39.92\,\text{mm}
40.08\,\text{mm}
 399.2\,\text{mm}
400.8\,\text{mm}
 38.4\,\text{mm}
41.6\,\text{mm}
 384\,\text{mm}
416\,\text{mm}
t
20
t
\qquad |t-40|
\qquad \dfrac{|t-40|}{20}
0.08\,\text{mm}
t=38.4\,\text{mm}
t=41.6\,\text{mm}
\qquad 38.4\,\text{mm}
41.6\,\text{mm}
R = 60
\alpha^\circ \approx 26.5^\circ
\beta^\circ = 12^\circ
\sin(26.5^\circ) \approx 0.446
\cos(26.5^\circ) \approx 0.895
\tan(26.5^\circ) \approx 0.499
\sin(12^\circ) \approx 0.208
\cos(12^\circ) \approx 0.978
\tan(12^\circ) \approx 0.213
L
h
\beta^\circ
\alpha^\circ
h
L
\qquad\dfrac{5x^2-33x-14}{x^2-7x}
x<6
-26
\dfrac{5x-2}{-x}
\dfrac{5x-2}{x}
\dfrac{5x+2}{x}
5x^2-33x-14
(a+b)(c+d)
a\cdot c
5x^2
b\cdot d
-14
a\cdot d+b\cdot c
-33x
\qquad (5x+2)(x-7)
x
\qquad x^2-7x=x(x-7)
\qquad \dfrac{(5x+2)(x-7)}{x(x-7)}
x
0
x\neq\purple{7}
x<6
\dfrac{5x+2}{x}
x<6
h=61.41+2.32f
h
f
2003
160
(\text{cm})
f\le42.5
f\ge42.5
f<42.5
f>42.5
h=61.41+2.32f
160\,\text{cm}
h>160
\qquad61.41+2.32f>160
61.41
\qquad2.32f>98.59
2.32
\qquad f>42.5
\qquad f>42.5
s
 annual membership and pay an additional 
 per copy 
. The other is to pay 
 per copy 
 to represent the 
 cents per copy times however many copies we make. If the first scenario begins with a one-time 
 membership fee, the 
 is added to 
c
0.05c
xy
y=-\dfrac{1}{4}(x-3)^2+2
(x,y)
y = \dfrac{1}{2}(x+3)-7
(3,2)
(-3,-7)
(7,-2)
(-1,-6)
(3,2)
(-1,-6)
(7,-2)
(-3,-7)
y=\dfrac{1}{2}(x+3)-7
(-3,-7)
y=-\dfrac{1}{4}(x-3)^2+2
(-3,-7)
m=\dfrac{1}{2}
y=\dfrac{1}{2}(x+3)-7
(7,-2)
x
y
(-3,-7)
(7,-2)
y=-\sqrt{-x}
y=\sqrt{x}
0
x
0
1
4
9
y
0
1
2
3
x
-x
y
y=\sqrt{-x}
\sqrt{-x}
y
x
y=-\sqrt{-x}
y=-\sqrt{-x}
\qquad 8ix = -5 
x
i=\sqrt{-1}
-\dfrac{8i}{5}
\dfrac{8i}{5}
-\dfrac{5i}{8}
\dfrac{5i}{8}
x
i
i^2=-1
\qquad x=\dfrac{5i}{8}
60
(\text{ft})
w\text{ ft}
200
\text{ft}
60+w=200
60+2w=200
120+w=200
120+2w=200
60\text{ ft}
w\text{ ft}
2(60)+2(w)
200\text{ ft}
\qquad2(60)+2(w)=200
\qquad 120+2w=200
120+2w=200
y = h(x)
f(x)=\dfrac{1}{2}h\left(-\dfrac{x}{2}\right)
f
k \cdot f(x)
k
f(x)
y
k
\; \dfrac{1}{2}h(x)
h(x)
y
2
f(k \cdot x)
k
f(x)
x
 \dfrac{1}{k} 
 \dfrac{1}{2}h \left( \dfrac{x}{2} \right)
 \dfrac{1}{2} h(x)
x
2
f(- x)
f(x)
y
 \dfrac{1}{2}h \left(-\dfrac{x}{2} \right)
 \dfrac{1}{2} h \left( \dfrac{x}{2} \right)
y
 \dfrac{1}{2} h \left( -\dfrac{x}{2} \right)
10
(\text{dB})
1
2\,\text{dB}
26\,\text{dB}
t
2t+10\ge26
2t+10>26
2(t-1)+10\ge26
2(t-1)+10>26
10\,\text{dB}
2\,\text{dB}
t-1
2
t
2(t-1)+10
26\,\text{dB}
(>)
\qquad2(t-1)+10>26
\qquad2(t-1)+10>26
(x,y)
(2,28)
(1,14)
(-2,-28)
(-1,-14)
(-7,-98)
(7,98)
y
x
x
14x=y
y
x^2+54=y+5
x
\qquad\qquad\qquad x^2+49=14x
x=7
x
y
x
(7,98)
(x_1,y_1)
(x_2,y_2)
x
x_1\cdot x_2
x
y
(x,y)
x
x
y
y
y_1=6
y_2=-5
x
y
x
y_1=6
y_2=-5
(-4,6)
(-7,-5)
x
x_1\cdot x_2
\qquad -4\cdot-7=28
xy
(-2,2)
y - 3 = 0
y + 2 = 0
x + 2 = 0
y - 2 = 0
x - 2 = 0
y - 3 = 0
y = 3
y
xy
y = k
k
(-2, 2)
y
2
k
2
y = 2
y - 2 = 0
(\text{nA})
\left(\dfrac{\mu\text{mol}}{\text{L}}\right)
10
0
60
60
110
121
160
182
210
242
260
303
310
363
(\text{nA})
\left(\dfrac{\mu\text{mol}}{\text{L}}\right)
50\,\text{nA}
50\,\text{nA}
50\,\text{nA}
50\,\text{nA}
50\,\text{nA}
\red{\text{common difference}}
\blue{\text{common ratio}}
\red{\text{linear}}
\red{\text{common difference}}
50\,\text{nA}
\blue{\text{exponential}}
\blue{\text{common ratio}}
\blue{\text{ratio}}
0
60
\red{\text{common difference}}
60.5\,\dfrac{\mu\text{mol}}{\text{L}}
50\,\text{nA}
xy
(0,15)
(3, 2)
(x-15)^2+y^2=178
(x-15)^2+y^2=\sqrt{178}
x^2+(y-15)^2=178
x^2+(y-15)^2=\sqrt{178}
(\green h,\purple k)
\red r
(x-\green h)^2+(y-\purple k)^2=\red r^2
\red{\sqrt{178}}=\red r
(x-\green {0})^2+(y-\purple {15})^2=\red{\sqrt{178}}^2
x^2+(y- 15)^2=178
10
1\text{,}000
10
10
10
22
20
16.5
30
320
325
330
335
x
x
1
20
x
x \div 20
22.5
x
\qquad 22 - \dfrac{x}{20}
x
\qquad 16.5 - \dfrac{x}{30}
x
330
14h
18h
h^2
\gray{\text{rational exponents}}
y = p(x)
y = w(x)
4
3
x
y
w(x)
w(x) = p(x-3)+4
w(x) = p(x+3)+4
w(x) = p(x-3)-4
w(x) = p(x+3)-4
y = p(x)
y = w(x)
4
3
y = w(x)
y = p(x)
4
3
f(x) = g(x + a) + b
y = f(x)
y = g(x)
a
b
a
b
4
3
a = -3
b = -4
\qquad w(x) = p(x-3)-4
 \qquad ax^2+5x+2=0
a
x=-2
x=-\dfrac{1}{2}
a
-2
\ \ \ 2
\ \ \ 3
-3
 ax^2+5x+2=0
x
0
 ax^2+5x+2=0
x=-2
x=-\dfrac{1}{2}
x
0
-2
x
a
a
4a-8=0
\red{2}x^2+5x+2=0
x=-2
x=-\dfrac{1}{2}
a=2
\quad W = - \dfrac{1}{2}m+16
W
m
-\dfrac{1}{2}
\dfrac{1}{2}
2
\dfrac{1}{2}
\dfrac{1}{2}
m
W
W
m
m
W
0
16
1
15.5
2
15
3
14.5
4
14
\dfrac{1}{2}
\dfrac{1}{2}
\qquad\sqrt s+7=6+4\sqrt s
\sqrt s
s=\blue{\dfrac19}
s=\dfrac{1}{9}
\dfrac19
D
T
0.5
60
1.0
82
2.0
127
3.0
172
3.5
195
T
(^\circ \text{F})
D
(\text{km})
D
T
D
T
D
T
T
D
T
D
T
D
T
\Delta T
0.5
60
1
82
82-60 = 22
2
127
127-82=45
3
172
172-127 = 45
3.5
195
195-172 = 23
D
0.5
T
22.5
D
1
T
45
T
T
D
T
D
D
T
\Delta T
\Delta T / \Delta D
0.5
60
1
82
82-60 = 22
22/0.5 = 44
2
127
127-82=45
45/1 = 45
3
172
172-127 = 45
45/1 = 45
3.5
195
195-172 = 23
23/0.5 = 46
T
D
45
D
T
T
D
-2a - b
-5
-\dfrac13
\dfrac13
5
-2a-b
a
b
b
b
a
\qquad b = \dfrac{5}{2}a -1
b
a
(a, b) = \left( \dfrac{4}{3}, \dfrac{7}{3} \right)
-2a-b
\quad P = 21.25t + 525
1910
2010
P
t
1910
525
525{,}000
1910
2010
525{,}000
1910
2010
2010
525{,}000
1910
525{,}000
t=0
P=525
t=0
1910
P
1910
525{,}000
1
11
12
23
2
10
11
21
3
6
12
18
4
23
4
27
50
39
89
4
\dfrac{4}{39}
2
\dfrac{21}{89}
1
\dfrac{12}{23}
2
\dfrac{2}{3}
1
\gray{11}
12
\red{23}
2
\purple{11}
\green{21}
3
6
12
18
4
23
\pink4
27
50
\blue{39}
89
21
89
\blue{\text{analysis problems}}
\purple{\text{chapter }2}
\dfrac{\purple{11}}{\blue{39}}
\dfrac{21}{89}
12
23
12
1
\red{\text{chapter }1}
\gray{\text{skills problems}}
\dfrac{\gray{11}}{\red{23}}
\dfrac{12}{23}
2
3
\dfrac{12}{18}
\dfrac{2}{3}
\green{\text{chapter }2}
\dfrac{2}{3}
\blue{\text{analysis problems}}
\pink{\text{chapter }4}
{\dfrac{\pink4}{\blue{39}}}
BC
\overline{EB}
\overline{DC}
\angle BED
\angle EBA
\overline{EB}
\overline{DC}
\qquad\dfrac{7.5}{3}=\dfrac{10}{BC}
BC
\qquad BC=4
42
1000 \text { km}^2
8500 \text { km}^2
24
200
357
3035
42
1000 \text { km}^2
8500 \text { km}^2
357
357
8500 \text { km}^2
 \qquad (2x+5)(mx+9)=0
m
x=-\dfrac{5}{2}
x=\dfrac{3}{2}
m
-2
\ \ \ 2
\ \ \ 3
-3
(2x+5)(mx+9)=0
m
2x+5
x=-\dfrac{5}{2}
-\dfrac{9}{m}
x=\dfrac{3}{2}
\dfrac{3}{2}
x
x=-\dfrac{9}{m}
m
(2x+5)(\red{-6}x+9)=0
x=-\dfrac{5}{2}
x=\dfrac{3}{2}
m=-6
\qquad -2x^2+5x=17
b^2-4ac
\qquad -2x^2+5x-17=0
0
\qquad2x-1=-1+a\,x
a
a
2
3
\qquad \blue a \,x+\red b=\blue c \, x + \red d
\blue a\ne\blue c
a
2
a=2
a=3
\qquad 2x-1=3x-1
x=0
a=3
f
g
g(f(5))
3.5
5
7
9
f(5)
f(5)
9
g(f(5)) = g(9)
g(9)
7
g(f(5))
7
7
16
(\text{cm})
5\text{ cm}
4
21\text{ cm}
5\text{ cm}
m
1.25m=5
1.25m=21
16+1.25m-21=5
5+1.25m-16=21
5\text{ cm}
4
5\div 4=1.25\text{ cm}
1.25m
m
16\text{ cm}
16+1.25m
m
21\text { cm}
16+1.25m-21
5\text { cm} 
\qquad 16+1.25m-21=5
 16+1.25m-21=5
(\text{kW})
1{,}000
(\text{W})
(\text{kWh})
1\ \text{kW}
1
1\ \text{kWh}
 per kilowatt-hour, and a lightbulb rated at 
 operates in that city for one hour every day for 
 consecutive days. Which equation best models the cost in dollars, 
c=\dfrac{60}{1{,}000}\cdot 200\cdot 0.14
c=\dfrac{1{,}000}{60}\cdot 200\cdot 0.14
c=\dfrac{60}{1{,}000}\cdot \dfrac{200}{0.14}
c=60\cdot1{,}000\cdot 200\cdot 0.14
60\ \text{W}
1{,}000\ \text{W}
\dfrac{60}{1{,}000}\ \text{kW}
1
200
\dfrac{60}{1{,}000} \cdot 1 \cdot 200 =\dfrac{60}{1{,}000}\cdot 200
, the total cost 
c
c=\dfrac{60}{1000}\cdot 200\cdot 0.14
mf
(\text{m})
(\text{Hz})
0.5\,\text{m}
0.\overline6\,\text{m}
0.1\overline6
0.5
0.\overline6
0.75
k
f
0.75
0.5\,\text{m}
0.\overline6\,\text{m}
\qquad i^{11} + i^{13}
i=\sqrt{-1}
-2i
2i
0
2
i^2=-1
\qquad-i + i = 0
\qquad 0
\qquad (x^2+h^2)(x^2-h^2)
\qquad (1+m-p)x^4  -mp
h
m
p
b
m
\qquad\quad (x^2+h^2)(x^2-h^2)
x
1 = 1+m-p \qquad\qquad \ \ \ \  x^4\text{ term equation}
-h^4 = -mp\qquad\qquad\qquad\text{constant term equation}
x^4
m
m
h^2
-h^2
h^2
DS
S
(\text{km/h})
D
S
S
(30, 160)
D = 30
D
30
30 \text{ km}
190\%
5
762
1
365
1.1
2.2
320
397
190\%
100\%+190\%=290\%
2.9
5
\qquad \red A\cdot\blue{2.9}^{\large{t}}
\red A
t
\red{762}
1
2.2
y=\green mx+\red b
\red b
\green m
\qquad\blue1(365)+\red{762}=1{,}127
1.1
\qquad2.2-1.1=1.1
1.1
(x,y)
y>0
x
x
y
(x,y)
y
y
x
x
x
x=6
x=8
y
y
y>0
y
x
y
x=6
x=8
(6,-1)
\left(8,\dfrac{1}{3}\right)
x
y>0
x=8
(r,s)
r+s
(r,s)
r+s
6
r
s
\qquad r+s=\dfrac9{44}-\dfrac{15}{11}=-\dfrac{51}{44}<0
r+s
-\dfrac{j^2}{2}
-\dfrac{1}{2}j^2
-1
2\, \text{cm}
8 \, \text{cm}
S
x
S=x^2+14x-16
S=2x^2+28x-32
S=16+2x-x^2
S=32+4x-2x^2
S
\qquad S=2lw +2lh +2wh
\qquad S=2(x)(x-2) +2(x)(8) +2(x-2)(8)
\qquad S=2x^2+28x-32
3300
80
(\text{mph})
5
290\,\text{mph}
338\,\text{mph}
370\,\text{mph}
450\,\text{mph}
\qquad\text{distance}=\text{rate}\cdot\text{time}
x
x
5
\red{5x}
x-80
5
\green{5\left(x-80\right)}
3300
\purple{3300}
x
\qquad \red{5x}+\green{5\left(x-80\right)}=\purple{3300}
370\,\text{mph}
\qquad \left(v-\dfrac53\right)^2=49
v=-\dfrac{142}3
v=\dfrac{152}3
v=-44
v=54
v=-2
v=12
v=-\dfrac{16}3
v=\dfrac{26}3
v
\qquad v=-\dfrac{16}3
v=\dfrac{26}3
\qquad q=9r^2+16s^2r^2
r
q
s
r=\pm\dfrac{\sqrt{q}}{3+4s}
r=\pm\sqrt{\dfrac{q}{9+16s^2}}
r=\pm\sqrt{\dfrac{q}{25s^2}}
r=\pm\dfrac{\sqrt{q-16s^2}}{3}
g
\green{\text{factor } r^2 \text{ from the expression }9r^2+16s^2r^2}
r^2
\blue{\text{dividing by }9+16s^2}
r
\purple{\text{take the square root of both sides of the equation}}
\sqrt[2]{x}=\sqrt{x}
\qquad r=\pm\sqrt{\dfrac{q}{9+16s^2}} 
(\text{mi})
500
1950
125 \, \text{mi}
50{,}000
1950
20
1950
1990
3.68
10
107.36
103.68
1950
\qquad \dfrac{50{,}000}{500} = 100
1990
40
1950
10
\, \dfrac{40}{10} = 4
125
1990
20
1.2
1990
1990
\qquad \dfrac{103{,}680}{1{,}000} = 103.68
1950
1990
\qquad 103.68 - 100 = 3.68
3.68
\qquad \left(x+\dfrac{13}2\right)\left(x-\dfrac{13}2\right)=0
0
1
2
3
\qquad \left(x+\dfrac{13}2\right)\left(x-\dfrac{13}2\right)=0
x=-\dfrac{13}2
x=\dfrac{13}2
2
15^\text{th}
135^\text{th}
56
n
56n+135=365
56(n-1)+135=365
56n-1+135=365
56(n+135)=365
365
135
n-1
56(n-1)
n^\text{th}
135+56(n-1)
n^\text{th}
1
135+56(1-1)=135+0=135
2
135+56(2-1)=135+56=191
3
135+56(3-1)=135+112=247
\qquad56(n-1)+135=365
\qquad56(n-1)+135=365
11^{\text{th}}
\;\;\;\text{Class A Student Height Distribution in Inches}
\;\;\;\text{Class B Student Height Distribution in Inches}
\;\;\;\text{Class C Student Height Distribution in Inches}
\;\;\;\text{Class D Student Height Distribution in Inches}
\text{Class A, Class B,}
\text{Class C}
\text{Class D}
\left(70\right)
\text{Class D}
\;\;\;\;\qquad
\qquad\qquad\;\text{Class D Student Height Distribution in Inches}
\overline{FH}
12
\overline{JK}
7
\angle FGH
\theta
\dfrac57 \approx \tan(0.62) \approx \cot(0.95) \approx \sin(0.80) \approx \cos(0.78)
\theta
7
\overline{FH}
5
\overline{JK}
\theta
\theta
0.62
\dfrac{2}{3}
550
313
m
m\geq53.\overline{6}
m\ge158
m\ge303
m\ge366.\overline{6}
313
m
313+m
\dfrac{2}{3}
(\ge)
\qquad 313+m\ge\left(\dfrac{2}{3}\right)550
\qquad313+m\ge366.\overline{6}
313
\qquad m\ge53.\overline{6}
53.\overline{6}
(
54)
10\:\text{dB}
0.0000632 \:\text{Pa}
20\:\text{dB}
0.0002 \:\text{Pa}
30\:\text{dB}
0.000632 \:\text{Pa}
40\:\text{dB}
0.002 \:\text{Pa}
50\:\text{dB}
0.00632 \:\text{Pa}
60\:\text{dB}
0.02 \:\text{Pa}
70\:\text{dB}
0.0632 \:\text{Pa}
80\:\text{dB}
0.2 \:\text{Pa}
90\:\text{dB}
0.632 \:\text{Pa}
100\:\text{dB}
2 \:\text{Pa}
(\text{dB})
(\text{Pa})
(
\text{Pa})
110\,\text{dB}
20
10
90
0.632
90+20 = 110
0.632 \cdot 10 = 6.32
\qquad
6.32
1.3
12^\text{th}
90^\circ
3.3
\text{mi}
d
1.3 \text{ mi}
3.3 \text{ mi}
4.6 \text{ mi}
3.5468\ldots \text{ mi}
\quad 4.6 - 3.5486\ldots = 1.05317\ldots
1.1 \text{ mi}
\qquad \dfrac{x^2-16}{(x-3)(x+149)}\div \dfrac{(x+a)(a-x)}{x^2+cx-447}=-1
c-a
4
142
146
150
\qquad \dfrac{x^2-16}{(x-3)(x+149)}\cdot \dfrac{x^2+cx-447}{(x+a)(a-x)}
-1
\qquad \dfrac{(x+4)(x-4)}{(x-3)(x+149)}\cdot \dfrac{x^2+cx-447}{-1(x+a)(x-a)}
a=4
-1
c=146
c-a=146-4=142
\qquad (1+2i)+\dfrac{1}{9-i}
i=\sqrt{-1}
\dfrac{8+17i}{9-i}
\dfrac{12+18i}{9-i}
\dfrac{12+17i}{9-i}
\dfrac{2+2i}{9-i}
50\%
15
8
10
23
30
x
50\%
1.5x
\qquad 1.5x = 15
1.5
x = 10
10
108
\text{(kW)}
1.8 \text{ kW}
2.2 \text{ kW}
. The rental company is charging 
 for each speaker and 
40
30
12
54
26
13
38
22
S
L
1.8S+2.2L
S
L
75S+42L
108 \: \text{kW}
 and  
L
y
S
x
L
S
L
\qquad L\leq-\dfrac{1.8}{2.2}S+\dfrac{108}{2.2} \qquad
 \qquad L\leq-\dfrac{75}{42}S+\dfrac{3{,}300}{42}
\leq
\qquad 
(26,13)
\qquad 26
13
 regular haircut and a 
 deluxe haircut that includes a shave. On a certain day, the barber gave 
 fewer deluxe haircuts than he did regular haircuts, 
, and earned 
15.00h+20.00(h-3)=500.00
15.00h+20.00(h+3)=500.00
15.00(h-3)+20.00h=500.00
15.00(h+3)+20.00h=500.00
h
3
h-3
, and the price for a deluxe haircut is 
, we can use the expression 
15.00h+20.00(h-3)=500.00
\qquad f(x)=3x^2+4x-2
g(x)=7x^2-bx+3
f(x)-g(x)=-4x^2+6x-5
x
b
-10
-2
2
10
f(x)-g(x)
\qquad f(x)-g(x)=(3x^2+4x-2)-(7x^2-bx+3)
f
g
g
x
(4+b)
4x+bx=(4+b)x
f(x)-g(x)
\qquad -4x^2+(4+b)x-5=-4x^2+6x-5
4+b=6
b=2
\qquad b=2
t
6.43
s
5.39
\dfrac st \approx \tan(0.697) \approx \sin(0.994) \approx \cos(0.576)
\beta
s
t
3
s
s
\beta
\dfrac\pi2 - \beta
s
t
\tan\left(\dfrac\pi2 - \beta\right)
\left(\dfrac\pi2 - \beta\right)
t
s
\dfrac st \approx \tan(0.697)
\beta
0.7
8=2^3
\left(a^m\right)^n=a^{mn}
\dfrac{a^m}{a^n}=a^{m-n}
a\neq 0
\qquad \qquad \qquad \qquad \qquad \text{Gas Prices in 2014}
2014
g
1
d
1
2014
1
12
2014
255^{\text{th}}
t=255
g=3.50
 is the  best approximation for the price of 
 gallon of gas on September 
, 
8000
20
\qquad A = l \cdot w
20 \, \text{ft}
l = w + 20
\qquad A = (w + 20)w = w^2 + 20w
2
w = -100
w = 80
80
\qquad(-1-5n+an^2)-(7n-4n^2+9)=12n^2-12n-10
n
a
8
12
16
19
n^2
(4+a)
an^2+4n^2=(a+4)n^2
\qquad (a+4)n^2-12n-10=12n^2-12n-10
a+4=12
a=8
\qquad a=8
\qquad A = P \left(1 + r \right)^t
P
r
A
t
r = \dfrac{{\large\frac{A} {P}} - 1}{t}
r = \dfrac{A - P - 1}{t}
r = \left(\dfrac A P -1 \right)^{\Large \frac{1}{t}} 
r = \left(\dfrac A P \right)^{\Large \frac{1}{t}}-1 
r
\blue{\text{dividing both sides by } P}
\purple{\text{raise both sides to the } \dfrac1 t \text{ power}}
\pink{\text{subtract } 1}
\qquad r = \left(\dfrac A P \right)^{\Large \frac{1}{t}}-1 
\qquad f_D=\dfrac{343}{343-v}f
v
\left(\dfrac{\text{m}}{\text{s}}\right)
f
f_D
343\dfrac{\text{m}}{\text{s}} 
v
v
2v
f_D
\qquad f_D(2v)=\dfrac{343}{343-2v}f
f_D(v)
f_D(v)
343-2v
f_D(2v)
f_D(v)
70\,\text{ milliliters (mL)}
1000\,\text{ mL}
\text{mL}
h
f(h) = 930h
f(h) = 1000h
f(h) = 1000h - 70
f(h) = 1000 - 70h
1000 \,\text{mL} 
1000
70h
h
1000
\qquad f(h)=1000-70h
f(h) = 1000 - 70h
, and one pound of walnuts costs 
. Arturo can spend at most 
 on almonds and walnuts. The graph at left, in the 
-plane, relates the number of pounds, 
, of walnuts to the number of pounds, 
2
1
3
7
4
2
2
4
2
1
2
4
2
4
(2,3)
(3,2)
(2,4)
(4,2)
(2, 3)
(4.2,2.2)
(2,3)
(4,2)
(2,3)
(2,3)
(4,2)
(4,2)
(2,3)
\quad P = 0.7c
P
c
3\%
7\%
30\%
70\%
P=0.7c
0.7
0.7
70\%
70\%
70\%
30\%
100\%
30\%
r
\left(\dfrac{kH}{m^2}\right)
0^\circ
21^\text{st}
m
21^\text{st}
r
21^\text{st}
5
m=5
m=5
\qquad
\qquad
\sin(70^\circ) \approx 0.94
\cos(70^\circ) \approx 0.34
\tan(70^\circ) \approx 2.7
\angle ABC
\angle BCD
\overline{AB}
9.83
13.90
25.37
28.74
ABC
BCD
BC
CD
ABC
45^\circ-45^\circ-90^\circ
B
\overline{AB}
\overline{BC}
\qquad AB \approx 10
\overline{AB}
9.83
38^\circ
w
\sin(19^\circ) \approx 0.326
\tan(19^\circ) \approx 0.344
48\,\text{in}
38^\circ
38^\circ
w
19^\circ
31\,\text{in}
r
155^{\circ}
\red{\text{vertical angle}}
50^{\circ}
\blue{\text{interior angles of a triangle}}
180^{\circ}
x
\green{\text{supplementary angles}}
180^{\circ}
\qquad r=125
\qquad V=C(1-r)^t
V
t
C
r
t
t
\dfrac{V}{(1-r)^t}
\red{\text{dividing by }C}
t
\left(x^a\right)^b=x^{a\cdot b}
\blue{\text{subtract }1}
\purple{\text{multiply by }-1}
\qquad (3+i)(2-4i)
i=\sqrt{-1}
2-10i
2-14i
10-10i
10-14i
(3+i)(2-4i)
i^2=-1
\qquad10-10i
(p, q)
0
1
p
0
\qquad P(x)=(x+7)(x-10)
x=-70
x=-3
x=1
x=-1
x=3
x=70
x=-7
x=10
x=-10
x=7
0
\qquad\green{(x+7)}\pink{(x-10)}=0
0
x
x=\green{-7}
x=\pink{10}
y
p
0
6
1
10
2
20
3
36
4
64
5
102
6
144
7
156
8
164
p
y
6
c
c
p
(y,p) = (0,11)
c
c
10
CI
I
C
(C,I) = (400, 0)
(C,I) = (2800, 1800)
400
2800
75
400
2800
25
400
2800
75
400
2800
25
0.75
400 < C < 2800
0
C < 400
C
400
0.75
C
2800
400
2800
75
400
2800
75
\qquad \qquad \qquad \text{Millions of international migrants,}\ (y)
\qquad\qquad\qquad\qquad\qquad \text{Years since} \ 1990,\ (x)
1990
2013
y=144.99x+3.73
y=-144.99x+3.73
y=3.73x+144.99
y=-3.73x+144.99
\qquad y=mx+b
y
\qquad m=\dfrac{y_2-y_1}{x_2-x_1}
(15,200)
(23,230)
\qquad\qquad\qquad \text{Millions of international migrants,}\ (y)
\qquad\qquad\qquad\qquad\qquad \text{Years since} \ 1990,\ (x)
(15,200)
(23,230)
\qquad \dfrac{230-200}{23-15}=\dfrac{30}{8}=3.75
y
145
\qquad y=3.73x+144.99
\qquad\qquad\qquad \text{Millions of international migrants,}\ (y)
\qquad\qquad\qquad\qquad\qquad \text{Years since} \ 1990\ (x)
\qquad y=3.73x+144.99
\qquad 2x^2-\dfrac{11}2x-\dfrac32=0
x=-\dfrac14
x=2
x=-\dfrac14
x=3
x=2
x=3
x=-\dfrac14
x=2
x=3
(2xa)(2x+b)
2xb+2xa
11
b=-3
a=1
x=-\dfrac14
x=3
\qquad x=-\dfrac14
x=3
\qquad
\overline{BC}
\overline{AD}
(\text{cm})
\overline{BF}
\text{cm}
\overline{BC}
\overline{AD}
\red{\text{alternate interior angles}}
\overline{FC}
70^\circ
\qquad
\overline{AD}
180
\blue r
\qquad
\Delta BCF
\overline{BC}
\overline{BF}
\qquad BF=BC=10\text{ cm}
4
2
6\large\frac{1}{2}
2
4
5\large\frac{1}{2}
1
1
1
1\large\frac{1}{2}
2
2\large\frac{1}{2}
x
y
4
2
6\large\frac{1}{2}
\qquad 2 x + 4 y = 6\large\frac{1}{2}
2
4
5\large\frac{1}{2}
\qquad 4 x + 2 y = 5\large\frac{1}{2}
x
y
x
y
x
x
y
1
1
2
\qquad 
300
100
20
t
c
240
100
360
180
240
260
120
180
t
(t-100)
c
(c-20)
300
(0,420)
(420,0)
(80,0)
(500,420)
\qquad
t=360
c=180
360
180
70\%
100
(\text{ml})
90\%
70\%
\text{ml}
\qquad\purple{\text{volume of alcohol}}=\red{\%\text{ of alcohol}}\cdot\green{\text{volume of solution}}
90\%
\qquad0.9(100)=90
\purple{90\,\text{ml}}
w
100\,\text{ml}
w
\green{100+w}
\red{70\%}
\qquad\purple{90}=\red{0.7}\green{(100+w)}
w
28.6\,\text{ml}
39\ \text{inches (in)}
78\ \text{inches (in)}
H
D
R
x
B
R
\angle HRB
\angle DRC
B
\overline{AC}
15\ \text{in}
B
x
7\ \text{in}
9\ \text{in}
11\ \text{in}
13\ \text{in}
\triangle HRB
\triangle DRC
180^\circ
\triangle HRB
\triangle DRC
\triangle HRB
\triangle DRC
\triangle HRB
\triangle DRC
\qquad \dfrac{ \overline{DC} }{ \overline{RC} }=\dfrac{\overline{HB}}{\overline{BR}} 
B
\overline{AC}
\overline{BC}=39\ \text{in} 
\overline{RC}=39-x
\overline{BR}
\qquad \dfrac{39}{39-x} =\dfrac{ 15 }{ x }
x
\qquad 54\cdot x=585
x\approx11\ \text{in}
2:1
4:5
3:4
2:3
3:5
480\ \text{Hz}
800\ \text{Hz}
480:800
480:800=6:10=3:5
\qquad \dfrac{1}{4}(a-2)^2-10=50
a=2\pm4\sqrt{10}
a=2\pm4\sqrt{15}
a=-2\pm4\sqrt{10}
a=-2\pm4\sqrt{15}
Y=(a-2)
\dfrac{1}{4}Y^2-10=50
Y
Y^2
Y
a
(a-2)
Y
a-2=\pm4\sqrt{15}
2
a=2\pm4\sqrt{15}
a=2\pm4\sqrt{15}
\qquad|x-4|-x>3
 x<\dfrac12
 \dfrac12<x <\dfrac72
 x<\dfrac12 \:\: \text{ or } \:\:  x >7  
2
 -1
(1)
x
4
3
x
\pink{x}
\blue{4}
(2)
2
(1)
\qquad\text{No solution} \quad \text{or} \quad x<\dfrac12
x
 \dfrac12
x
 \dfrac12
x<\dfrac12
a
a
a
0.8
a
-0.8
a=0.8
a
a
\qquad g=255(100-m)
g
m
m=25{,}500-g
m=g-25{,}500
m=\dfrac{g}{255}-100
m=100-\dfrac{g}{255}
\red{\text{dividing by }255}
\blue{\text{subtract }100}
\green{\text{multiply by }-1}
\qquad m=100-\dfrac{g}{255}
N(t)
t
\qquad  N(t)=N_0\cdot(1-r)^{\large\text{ t}/\tau}
N_0
t=0
r
(0<r<1)
\tau
\tau
4
4
4
4
\tau=4
t=0
t=4
0<r<1
0<(1-r)<1
\qquad N_0\cdot(1-r)<N_0\ 
4
N(t)
4
24
30\%
25
M
h
M(h)=25\cdot (0.3)^{\large{h}}
24
25(0.3)
h=24
S(h)=25(0.3). 
\qquad S(24)=25(0.3)
h=48
\qquad S(48)=25(0.3)(0.3)=25(0.3)^2
\qquad S(72)=25(0.3)^2(0.3)=25(0.3)^3
h
\qquad x^2 + 5x - 3x - 2x - 100
6x^2-2x-100
6x^2-5x-100
x^2-10x-100
x^2-100
\qquad x^2+0x-100
0x
\qquad x^2-100
\qquad x^2-100
w
C
1
30
2
60
4
120
8
240
16
480
32
960
32
C
w^{\text{th}}
C
w
30
100
30
100
w
C
wC
(1, 30)
\qquad (1,30) \quad (2,60) \quad (4,120) \quad (8,240) \quad (16,480) \quad (32, 960)
(w_0, C_0)
(2w_0, 2C_0)
(4w_0, 4C_0)
\qquad \dfrac{2C_0 - C_0}{2w_0-w_0} = \dfrac{C_0}{w_0}
\qquad \dfrac{4C_0 - 2C_0}{4w_0-2w_0} = \dfrac{2C_0}{2w_0} = \dfrac{C_0}{w_0}
(w_0, C_0)
\, \dfrac{C_0}{w_0} \,
\,\dfrac{30}{1} = 30
w + 1
30
w
30
1
1980
2008
(17.24,1.33)
\qquad\qquad\qquad\qquad\qquad \text{Years since 1980}, (x)
1997
1
1997
1
1997
1
1997
1
y
x=17.24,\  y=1.33
\qquad\qquad\qquad\qquad\qquad \text{Years since 1980}, (x)
x
y
1
x
1980
y
1
(17.24,1.33)
1980+17.24=1997.24
1997
1
xP
x
P
x
300
 and a maximum profit of 
300
900
300
900
300
900
x
300
900
\ell < -2
n>4
0
1
{\ell^4n^5} 
\dfrac{1}{\ell^4n^5} 
1
\,1
\dfrac x x
1
1x
1
\qquad\dfrac{1}{\purple{\ell \cdot \ell\cdot \ell\cdot \ell} \cdot \blue{ n\cdot n \cdot n \cdot n \cdot n }}
\qquad  \dfrac{1}{l^4n^5} 
35
7
1.5
d
7d>35
7d\geq35
7d+3>35
7d+3d\geq35
\qquad 7\times\text{days worked} + 1.5\times2\times\text{days worked} \geq35
1.5\times2 = 3
3d
\qquad 7d+3d\geq35
m
\overline{AC}
\overline{DB}
6\text{ cm}
A
D
B
\Delta ADB
\overline{DB}
\angle D
\angle B
\qquad m=71
|x|>8
\qquad\dfrac{\left(\dfrac{2x^2+4x}{x^3-5x^2+4x}\right)}{\left(\dfrac{4x^2-16}{x^2-6x+8}\right)}
\dfrac{2}{x-1}
\dfrac{1}{2(x-1)}
\dfrac{2(x+2)}{(x-1)(x-2)}
\dfrac{x+2}{2(x-1)(x-2)}
\qquad\dfrac{2x^2+4x}{x^3-5x^2+4x}\div\dfrac{4x^2-16}{x^2-6x+8}
\qquad \dfrac{2x(x+2)}{x(x-4)(x-1)} \div \dfrac{\red{4(x-2)(x+2)}}{\blue{(x-2)(x-4)}}
\qquad \dfrac{2x(x+2)}{x(x-4)(x-1)}  \cdot \dfrac{\blue{(x-2)(x-4)}}{\red{4(x-2)(x+2)}}
x
0
x\neq{\green{-2},\pink{0},\purple2,\gray4}
|x|>8
2
4
2
\qquad \dfrac{1}{2(x-1)}
1950
0.320
1960
0.612
1970
1.169
1980
?
1990
4.393
2000
8.390
2010
16.03
7
1980
1.726
2.262
2.781
3.218
1980
1.169 \cdot 2 = 2.338
2.338
1980
2.262
\qquad (4y-9)^2-(3y+c)=16y^2-75y+73
y
c
-73
-8
8
73
(4y-9)
3y+c
\qquad 16y^2-75y+(81-c)
\qquad 16y^2-75y+(81-c)=16y^2-75y+73
81-c=73
c
c=8
\qquad c=8
f
f(c) = (c-k)(c^2-4c+4)
k
2
f
f(c)
(c-2)
2
f
(c-2)
f(c)
f(c)
(c-2)
0
xy
x=3
(4,5)
\sqrt2
(3,3)
(3,4)
(3,5)
(3,7)
(\green h,\purple k)
\red r
\qquad (x-\green h)^2+(y-\purple k)^2=\red r^2
x=3
(\green 3,\purple k)
\qquad (x-\green{3})^2+(y-\purple k)^2=(\red {\sqrt 2})^2
\qquad (x-3)^2+(y-\purple k)^2=2
(4,5)
(x,y)
\purple k
\purple k
\qquad (\purple k-4)(\purple k-6)=0
0
\qquad k=4
k=6
(3,4)
(3,6)
(3,4)
(3,4)
4.5
(\text{in})
1.25
(\text{in})
6.0
(\text{m})
1.07\ \text{m}
1.25\ \text{m}
1.67\ \text{m}
1.83\ \text{m}
\left( 6\ \text{m}\right)
\left(x \right)
\left( 4.5\ \text{in}\right)
\left( 1.25\ \text{in}\right)
\quad \Large\frac{6}{x}=\Large\frac{4.5}{1.25}
1.67\ \text{m}
x=\Large \frac{6\times1.25}{4.5}\approx1.67\ \text{m}
1.67 \text{ m}
406
452
2
436
390
400
410
420
x
\qquad \dfrac{x}{100}=\dfrac{406}{452}
x
89.8\%
2
91.8\%
91.8\%
436
y
\qquad \dfrac{91.8}{100}=\dfrac{y}{436}
y
400
400
(x,y)
x
\qquad\quad
y
x
x
\qquad\qquad\qquad 4x^2+x-29=x-4
x
x
 
x
\dfrac{5}{2}\cdot-\dfrac{5}{2}=-\dfrac{25}{4}
(x+6)
(x+1)
x^2+7x+6
x^2+6x+7
8x+6
7x+7
\qquad(x+6)(x+1)
\qquad x\cdot x + x\cdot 1 + 6\cdot x + 6\cdot 1
\qquad x^2 + 1x + 6x + 6
\qquad x^2+7x+6
\qquad x^2+7x+6
24
\qquad\qquad \pi r^2h_\text{cylinder}=\dfrac13 \pi r^2h_\text{cone}
24\text{ cm}
24
h_\text{cone}
\qquad\qquad \pi r^2h_\text{cylinder}=\dfrac13 \pi r^2\cdot24
\pi
r^2
8
\quad C = 1.5b + 67.5
C
b
1.5
1.5
 when 
, we know that 
, we know that our cost increases by 
 is the amount it costs to produce each book after the base cost of 
xy
(0,0)
(3,1)
(-3,1)
 x^2+y^2+10y=0
 x^2+y^2-10y=0
 x^2-y^2-10y=20
 x^2+y^2-5 y=0
(\green h,\purple k)
\red r
\qquad (x-\green h)^2+(y-\purple k)^2=\red r^2
(x,y)
\green h
\purple k
\red r
(0,0)
\qquad (0-\green{h})^2+(0-\purple k)^2=\red r^2
(3,1)
\qquad (3-\green h)^2+(1-\purple k)^2=\red r^2
(-3,1)
\qquad  (-3-\green h)^2+(1-\purple k)^2=\red r^2
\qquad (3-\green h)^2-(-3-\green h)^2=0
\qquad \purple k^2=\red r^2
\qquad 9 +(1-\purple k)^2=\red r^2
(0,5)
(0,0)
5
\qquad x^2+(y-5)^2=25
\qquad x^2+y^2-10y+25=25
\qquad x^2+y^2-10y=0
\qquad x^2+y^2-10y=0
x = -\dfrac{1}{3}y
y
0
y
0
(0,0)
x
x
1
y
(-3)1 = -3
(1, -3)
 per hour, during which customers are allowed to pick as many apples as possible.  If Yves can pick approximately 
 apples in 
 hour, and he picks 
75
50
\dfrac{75}{50}=1.5
 per hour, Yves can expect to pay 
k
(0,3)
(5,0)
k
3
-\dfrac{3}{5}
\dfrac{3}{5}
-3
(5,0)
x
(0,3)
x=0
y
y=3
x=0
y=3
k
k
-\dfrac{3}{5}
5
\left(x^a\right)^b = x^{ab}
n^{2\cdot3\cdot5} k^{2\cdot5\cdot7}
\left(n^2k^{5}\right)^{3}=n^{2\cdot3}k^{5\cdot3}\quad 
n
k
k
\left(n^{15}k^{35}\right)^2=n^{15\cdot2} k^{35\cdot2}=n^{3\cdot5\cdot2} k^{5\cdot7\cdot2}\quad 
\left(n^6k^{14}\right)^5
\qquad \left(n^{15}k^{35}\right)^2
\qquad \dfrac{5}{k-2} - \dfrac{2k-2}{k-1} = - 1
k
\qquad (k-2)(k-1)
\qquad 5k -5 +-2k^2 + 6k -4 = - k^2 + 3k - 2
\qquad k^2 - 8k +7 = 0
\qquad (k-7)(k-1) = 0
k
7
1
k = 7
k = 7
k = 1
0
k = 7
7
g(x)=x^2-5
f(g(x))=\sqrt{x^2+4}
f(x)
f(x)=\sqrt{x+1}
f(x)=\sqrt{x+9}
f(x)=\sqrt{x^2+1}
f(x)=\sqrt{x^2+9}
f
f(x^2-5)=\sqrt{x^2+4}
f(g(x))=\sqrt{x^2+4}
f
x^2-5
9
x^2-5
x^2+4
f(x)=\sqrt{x+9}
f(x)=\sqrt{x+9}
g(x)=x^2-5
f(g(x))
g(x)
x
f
f(x)=\sqrt{x+9}
275
10
15
(\text{mins})
15
10\text{ mins}
5
(\text{hrs})
50\text { mins}
50
75
200
300
x
5\text{ hrs}
50\text { mins}, 
350\text{ mins}
350-x\text{ mins}
10\text{ pages}
15\text{ mins}
350-x
\dfrac{10}{15}(350-x)
15\text{ pages}
10\text{ mins}
x\text { mins}
\dfrac{15}{10}x
275
50\text{ mins}
\left(\dfrac{15\text{ pages}}{10\text{ mins}}\right)
\dfrac{15}{10}\cdot 50
75
75
\qquad\dfrac{ \left( \dfrac{3x^2-2x-5}{x^2+x} \right) }{ \left( \dfrac{15-9x}{3x} \right) }
x<-2
1
-1
\dfrac{(3x-5)(5-3x)}{x^2}
-\dfrac{x-1}{x+1}
\qquad \dfrac{3x^2-2x-5}{x^2+x} \cdot \dfrac{3x}{15-9x}
5-3x=\red{-1}(3x-5)
\qquad \dfrac{(3x-5)(x+1)}{x(x+1)} \cdot \dfrac{3x}{(\red{-1})(3)(3x-5)}
5-3x=\red{-1}(3x-5)
x
0
x\neq{\green{-1},\pink{ \dfrac53},\blue0}
x<-2
\qquad -1
\qquad \left(5-7i+i^2\right)+\left(8i^3+12\right)
a+bi
a
b
a
i=\sqrt{-1}
16
17
18
19
i
i
\sqrt{-1}
i^2
-1
i^3
-\sqrt{-1}=-i
i^4
1
\qquad (5-7i+\red{i^2})+(8\blue{i^3}+12)=(5-7i+\red{(-1)})+(8\blue{(-i)}+12)
a+bi
\qquad a=16
\qquad 1.75I - 0.75P = B
P
I
B
1.75
1.75
B
2
1.75I
\qquad\qquad a^2 = b^2+c^2
a
b
c
c
a
b
c = a-b
c = \sqrt{a^2}-b^2
c = \sqrt{a^2-b^2}
c = \sqrt{a^2+b^2}
c^2
\blue{\text{subtracting } b^2 \text{ from both sides of the equation}}
c
\purple{\text{take the square root of both sides of the equation}}
\sqrt[2]{x}=\sqrt{x}
\qquad  c = \sqrt{a^2-b^2}
5\%
5\%
4\%
4\%
 is invested in the bank and 
 is invested with the stockbroker, after 
5\%
5\%
4
4\%
4\%
4
t
0.41
2.5
0.8
0.4
20\%
2.5
20\%
0.4
y=\red{a}\cdot \blue{b}^{\large\purple{x}}
\red a
\blue b
\purple x
1
\blue b <1
\purple{\dfrac{2t}{5}}
1
\blue{0.8}
\blue{80\%}
20\%
t=0
\red{0.41}
\blue{0.8}
\purple{\dfrac{2t}{5}}=1
20\%
2.5
f
\dfrac{2}{3}
t
f
t = 3
t = 3 - \dfrac{1}{3}f
t = 3 + \dfrac{2}{3}f
t = 3 + f
f
f
9
\dfrac{2}{3}
\qquad\dfrac{2}{3}(9-f)
\qquad 9 -f- \dfrac{2}{3}(9-f)
\qquad9 - f-6 + \dfrac{2}{3}f = 3 - \dfrac{1}{3}f
\qquad t=3-\dfrac13 f
\qquad t = 3- \dfrac{1}{3}f
xy
(-41,69)
(31,-85)
(x+41)^2+(y-69)^2=28900
(x+5)^2+(y+8)^2=7225
(x-31)^2+(y+85)^2=28900
(x+10)^2+(y+16)^2=7225
(\green h,\purple k)
\red r
\qquad(x-\green h)^2+(y-\purple k)^2=\red r^2
\green h
x
\qquad \green h=\dfrac{-41+31}{2}=\green{-5}
\purple k
y
\qquad \purple k=\dfrac{69+(-85)}{2}=\purple{-8}
\qquad\sqrt{28900}=170
170
\red{85}
\qquad(x-(\green {-5}))^2+(y-(\purple {-8}))^2= \red{ 85}^2
\qquad(x+5)^2+(y+8)^2=7225
 per pound and peanuts cost 
 per pound.  The owner wants to create 
 pounds of the product that cost 
 per pound. Which of the following systems of equations can be used to determine the number of pounds of peanuts, 
, and the number of pounds of raisins, 
\qquad p + r = 20
\qquad 2.50p + 3.50r
20
\qquad\dfrac{2.50p+3.50r}{20}
19
20
59
60
\qquad\dfrac{x^2+6x-27}{9-x^2}
x<0
-\dfrac{x+9}{3+x}
-\dfrac{x-9}{x-3}
\dfrac{x+9}{3+x}
6x-3
(x+9)(x-3)
9-x^2
\qquad (3-x)(3+x)
\qquad \dfrac{(x+9)(x-3)}{(3+x)(3-x)}
3-x=\red{-1}(x-3)
x
0
x\neq\purple3
x<0
-\dfrac{x+9}{3+x}
x<0
23
60
13
60
h
10
13-\dfrac{23h}{60}=10
13-\dfrac{60h}{23}=10
\dfrac{13-60h}{23}=10
\dfrac{13-23h}{60}=10
\qquad \dfrac{\small{\text{[Miles per hour]}\cdot  \text{[Num. of hours driven]}}}{[ \text{ Miles per gallon}]}=\text{[Gallons used]}
60
h
23
\dfrac{60h}{23}
1
60\cdot 1=60
\dfrac{60}{23}\approx2.6
13
10
h
\qquad\qquad 13-\dfrac{60h}{23}=10
13-\dfrac{60h}{23}=10
8
60
5280
1
90\ \text{ft}
700\ \text{ft}
750\ \text{ft}
800\ \text{ft}
60
8
88\times8=704\ \text{ft}
\approx 700\ \text{ft}
\qquad (6x+5)-(12-4x^2)
4x^2+6x-7
-4x^2+6x-7
24x^3+20x^2-72x-60
-24x^3-20x^2+72x+60
\qquad 4x^2+6x-7
\qquad 4x^2+6x-7
18^\circ
36
\text{(ft)}
y
11 \text{ ft}
5
5
5
5
5
5
5
5
5
5
2004
28\%
2006
34\%
2008
43\%
2010
56\%
2012
78\%
13\%
13\%
2
40\%
40\%
2
2
2
2
2004
28\%
2006
34\%
6\%
2008
43\%
11\%
2010
56\%
13\%
2012
78\%
22\%
2
2004
28\%
2006
34\%
\approx 1.21
2008
43\%
\approx 1.26
2010
56\%
\approx 1.30
2012
78\%
\approx 1.69
2
1.4
1.4
40\%
2
(j,k)
0
1
2
y
\left(\dfrac{3}{22}\text{ and} \,\dfrac{16}{22}\right)
 \qquad A = \dfrac 1 2 (b_1 + b_2) h
A
h
b_1
b_2
2
  h = \dfrac A 2 (b_1 + b_2)
h =  \dfrac{2}{A(b_1+b_2)} 
h =  \dfrac{A}{2(b_1+b_2)} 
h =  \dfrac{2A}{(b_1+b_2)} 
\purple{\text{multiply both sides by } 2}
\dfrac 1 2
\pink{\text{divide both sides by } (b_1 + b_2)}
\qquad h =  \dfrac{2A}{(b_1+b_2)} 
\qquad T = -110 + 0.13x(21 - x)
x
T
x = 0
\qquad -110 + 0.13(0)(21-0) = -110 + 0(21) = -110
x
-110
x = 0
x = 21
21 - 0=
21
21
y=-2(x-1)^2+5
y=a(x-h)^2+k
(h,k)
y=-2(x-1)^2+5
(1,5)
a<0
\left((0,3), (2, 3), (3,-3)\right)
\qquad 
2x+y=3
y=-2x+3
-2
y
(0,-3)
\qquad
8v
6t
4v
3t
a=(yt)^{3t}
b=(lv)^{4v}
\qquad a^2-b^2 = (a+b)(a-b)
\qquad m(t_2) - m(t_1) = 15(t_2 - t_1)
4^\circ
m(t)
(\text{g})
t
t_1
t_2
t
t_1 < t_2
15 \, \text{g}
1 \, \text{g}
15
15
15
\qquad \dfrac{m(t_2) - m(t_1)}{t_2-t_1} = 15
t_1
t_2
t_1
t_2
m
t
15
m
t
15
m
t
15 \, \text{g}
15 \, \text{g}
t
\qquad h(t)=-4.9t^2+19.6t
-4.9(t-2)^2+19.6
-4.9(t-3)^2-9.8t+44.1
-4.9(t-1)^2+9.8t+4.9
\qquad h(t)=-4.9t^2+19.6t=-4.9t(t-4)
h(t)=0
t=0
t=4
0\leq t\leq 4
4
\overline{TR}
12
\sin(r^\circ) \approx 0.927
\cos(r^\circ) \approx 0.375
\tan(r^\circ) \approx 2.472
\overline{RS}
4.9
12.9
29.7
32.0
QTR
12
QR
SQR
QR
\dfrac{12}{\sin(r^\circ)}
\tan
RS
\overline{RS}
f(x)=\sqrt{x+3}-4
g(x)=\sqrt{x+3}+b
xy
b
f
g
h(x)=\sqrt{x}
0
x
0
1
4
9
y
0
1
2
3
g
h
3
5
h(x) \rightarrow h(x+3)
h
3
h(x+3)\rightarrow h(x+3)+5
5
g
g(x)=\sqrt{x+3}+5
g(x)=\sqrt{x+3}+b
b=5
\qquad b=5
\qquad \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}
P
V
T
(1)
(2)
V_1 = V_2
 V_2 = \dfrac{P_1 T_2 V_1}{ ({T_1}/{P_2})} 
 V_2 = \dfrac{P_1 T_2 V_1}{P_2 T_1} 
 V_2 = \dfrac{P_1 V_1}{ {T_1}} - \dfrac {T_2}{P_2} 
\blue{\text{multiply both sides by } T_2} \, 
\pink{\text{divide both sides by } P_2}
\qquad V_2 = \dfrac{P_1 T_2 V_1}{P_2 T_1} 
3(r+300)=6
r+300-2
(r+300)
3
0
k
k
(p, q)
-2
0
2
q
k
k = 0
0
k
\dfrac{2}{3}
p
k
0
k
q
-\dfrac{2}{5} = -\dfrac{2}{5}
k = -2
k
\qquad \sqrt{\ell(\ell-5)}=6
\ell
0
\ell=9
\ell=-4
\ell=\blue9
\ell=9
\ell= \blue{-4}
\ell=9
\ell=-4
9+(-4)=5
5
4
79\%
h
b
h = 4 \cdot 0.79^{(b-1)}
h = 4 \cdot 0.79^b
h = 4 - 0.79 \cdot b^2
h = 4 \cdot ( 1- 0.21 \cdot b^2 )
4
b=0
h=4
79\%
4
b=1
h = 4 \cdot 0.79
79\%
4\cdot0.79
b=2
h=4\cdot0.79\cdot0.79
h=4\cdot0.79^2
\qquad h = 4 \cdot 0.79^b
xy
-3
5
-6\le x\le6
\text{I}
\text{I}
\text{II}
\text{II}
\text{III}
\text{I}
\text{II}
\text{III}
0
0
x
x
x
-3
5
\text{I}
x
-3
5
\text{II}
x
-3
5
\text{III}
x
5
-3
\text{I}
\text{II}
\qquad w = 3 +2.3t 
3
w
t
2.3
3
1
2.3
2000
9
5
21
\%
6\%
x
15
\qquad\dfrac{9}{21}=\dfrac{15}{x}
x=35
35
15
1
5
1
\qquad \dfrac{2.25 \text{ euros}}{1\text{ box}}\cdot \dfrac{1 \text{ box}}{5\text{ servings}}
0.45
1
\qquad 0.45x=0.55
x
x\approx 1.22
35
15
1
y=3x+33
(x,y)
y =-3(x+10)^2+9
(-11,0)
(-8,9)
(-12,-3)
(-13,-6)
y =-3(x+10)^2+9
(-10,9)
U
3
(-12,-3)
(-12,-3)
(-12,-3)
\qquad \qquad \text{World Population Aged 80 and Over:  1950 - 2050}
80
1950
2050
t
1950
n
80
80
2030
18
38
180
380
t
1950
2030
t=80
t=80
n=180
2030
180
80
180
80
2030
s=a\cdot a\cdot a\cdot a
s=(a-1)(a-1)
s=a\cdot a\cdot a\cdot a\cdot a
s=(a-1)(a-1)(a-1)
a=0
a
\qquad
s=a\cdot a\cdot a\cdot a
\qquad
\qquad
s=a\cdot a\cdot a\cdot a\cdot a
a
s=a\cdot a\cdot a\cdot a=a^4
a
s
0
s=a\cdot a\cdot a\cdot a\cdot a=a^5
a
s
a
s=a\cdot a\cdot a\cdot a\cdot a
, and the median price for non-musical plays is 
 on e-books and paperback books combined.  Each e-book he buys costs 
, and each paperback book he buys costs 
.  If Hiroto buys 
 e-books and 
5x+12y=68
12x+5y=68
17(x+y)=68
17xy=68
[\text{cost per book}]\cdot [\text{number of books}]=[\text{total cost for books}]
12x
5y
5x+12y=68
10
x
\qquad x = y^2 - 4y + 5
y
\qquad x = a(y-y_0)^2+x_0
a
(x_0,y_0)
x
a
1
x
x_0 = 1
1
 per page, plus a 
 setup fee. The cost of printing the manuscript at a photocopy store is 
140
175
230
350
\qquad 350
(x, y)
y
(x, y)
y
(x, y)
(x, y)
y
y
\qquad 2x - 1  = (3x - 1)
x
x
x
x
y
y
(x, y)
y
6
5
30
36
60
72
60
(\text{min})
1
30
6
(\text{mi})
5
\qquad \dfrac{6\text{ mi}}{5\text{ min}}=\dfrac{m\text{ mi}}{30\text{ min}}
m
36
\qquad \dfrac{1.6xy^2 + 0.8y}{0.4xy}
y>6
x>8
\dfrac{3y}{5}
\dfrac{2xy+1}{4x}
\dfrac{4xy+2}{x}
{4y+2}
\qquad \dfrac{1.6xy^2 + 0.8y}{0.4xy} \quad = \quad \dfrac{0.4y(4xy + 2)}{0.4y(x)}
\qquad \dfrac{4xy+2}{x}
3
6
m
6m+6\left(2m\right)=3
6m+6\left(\dfrac{m}{2}\right)=3
6m+6\left(2m\right)=180
6m+6\left(\dfrac{m}{2}\right)=180
6
m
\red{6m}
6
\dfrac{m}{2}
\green{6\left(\dfrac{m}{2}\right)}
6
\red{6m}+\green{6\left(\dfrac{m}{2}\right)}
3
60
 =1
\purple{180}
\qquad\red{6m}+\green{6\left(\dfrac{m}{2}\right)}=\purple{180}
\qquad{6m}+{6\left(\dfrac{m}{2}\right)}={180}
\qquad N = 2 \cdot 3^t
N
t
2
3
2
3
t
N
2
3
3
2
1
2
2
(\text{GR}
)
(M)
(N)
\,N = 10^{a - b M}
 N
\geq M
a
b
10^a
2f
\dfrac f2
111{,}000 \pi
\left(\text{m}^3\right)
\sqrt[3]{83{,}250}
\sqrt[3]{166{,}500}
2 \sqrt[3]{83{,}250}
2 \sqrt[3]{166{,}500}
\qquad V = \dfrac{4}{3} \pi r^3
V
r
\qquad V_\text{hemisphere} = \dfrac{2}{3}\pi r^3
r
r
d = 2 \sqrt[3]{166{,}500}
2 \sqrt[3]{166{,}500} \text{ m}
1280\,\text{m}
\qquad h(x)=0.000371(x^2-1280x)+152
x
\qquad
640\,\text{m}
152\,\text{m}
0.000371\,\text{m}
1280\,\text{m}
y
x=0
\qquad h(0)=0.000371(0)^2-1280(0)+152
152\,\text{m}
5
6
11
2
10
12
6
11
17
13
27
40
40
65\%
68\%
65\%
83\%
65\%
68\%
65\%
83\%
\red{\text{proportion of trips from Route C that were late}}
\blue{\text{proportion of all trips that were late}}
5
6
11
2
10
12
6
\red{11}
\red{17}
13
\blue{27}
\blue{40}
\red{\text{proportion of trips from Route C that were late}}
11
17
\blue{\text{proportion of all trips that were late}}
27
40
\red{0.65}
\red{65\%}
=\blue{0.68}
\blue{68\%}
(\red{65\%} < \blue{68\%})
\purple{\text{proportion of late trips that were from Route C}}
\green{\text{proportion of all trips that were from Route C}}
5
6
11
2
10
12
6
\purple{11}
\green{17}
13
\purple{27}
\green{40}
\purple{\text{proportion of late trips that were from Route C}}
11
27
\green{\text{proportion of all trips from Route C}}
\purple{0.41}
\purple{41\%}
\green{0.43}
\green{43\%}
\purple{\text{proportion of late trips from Route C}}
\green{\text{proportion of all trips from Route C}}
65\%
68\%
\qquad\qquad\text{U.S. annual per capita hfcs consumption in pounds,  } (y)
\qquad\qquad\qquad\qquad \text{Years since 1970, } (x)
(\text{hfcs})
1970
1985
y=0.201x^2-0.264x+0.969
y=-0.201x^2-0.264x+0.969
y=201.00x^2-264.00x+969.00
y=-201.00x^2-264.00x+969.00
\qquad y=-0.201x^2-0.264x+0.969
\qquad y=-201.00x^2-264.00x+969.00
\qquad y=201.00x^2-264.00x+969.00
\qquad y=0.201x^2-0.264x+0.969
(0,0)
y
0
y
(969.00)
y=201.00x^2-264.00x+969.00
y=0.201x^2-0.264x+0.969
y
0.969
\qquad y=0.201x^2-0.264x+0.969
\qquad\qquad\text{U.S. annual per capita hfcs consumption in pounds,  } (y)
\qquad\qquad\qquad\qquad \text{Years since 1970, } (x)
\qquad y=0.201x^2-0.264x+0.969
y=-(x+2)^3
y=x^3
x
-2
-1
0
1
2
y
-8
-1
0
1
8
f(x)=x^3
 f(x)\rightarrow f(x+2)
f(x+2)\rightarrow -f(x+2)
x
y=x^3
y=-(x+2)^3
y=-(x+2)^3
450
6
11
s
11s + 6>450
11s + 6<450
6s + 11>450
6s + 11<450
\qquad \text{students traveling by bus} + \text{students traveling by car} >450
6
11
s
11s
\qquad 11s + 6>450
\qquad r = 9.4t 
r
t
9.4
9.4
9.4
9.4
9.4
\blue{\text{units}}
9.4
9.4
x
(-3,0)
(7,0)
x=h
h
x
x
7
-3
x=2
\qquad h=2
9
(\text{ft})
3\,\text{ft}
\ell
d
\ell=d+3
15\,\text{ft}
 -20 <8x+10<-8
4x+5
-4
-10
-10
-4
-\dfrac{9}{4}
-\dfrac{15}{4}
-\dfrac{15}{4}
-\dfrac{9}{4}
 -20 <8x+10<-8 
 4x+5 
 4x+5=\dfrac{8x+10}{2} 
 4x+5=\dfrac{8x+10}{2} 
4x+5 
2
a<x<b
b>a
x
a
b
x<2
x>5
5<x<2
4x+5
-10
-4
2x^2+4x+3
2(x+1)^2+1
2(x+1)^2+3
2(x+1)^2+5
2(x+1)^2+4x+1
x
1
2 \cdot 1
2
\qquad 2x^2+4x+3\:=\:2(x+1)^2+1
300
30
45
t
t\leq6\large\frac{1}{6}
t\geq6\large\frac{1}{6}
t\leq6\large\frac{2}{3}
t\geq6\large\frac{2}{3}
300
45
300-45t
\geq
30
\dfrac{1}{2}\cdot45
22.5
300-45t\geq22.5
t
300
\qquad-45t\geq-277.5
-45
\qquad t\leq6\large\frac{1}{6}
\qquad t\leq6\large\frac{1}{6}
P
S
P
 per dollar of 
. Which of the following graphs in the 
-plane could be the graph of a call option with a strike price of 
P
S
15
S
15
P
 per dollar of 
. This creates a line segment with a slope of 
 and 
(15, -4)
(25, 6)
\qquad \dfrac{6 - (-4)}{25 - 15} = \dfrac{10}{10} = 1
7.9
(\text{cm})
3.0\,\text{cm}
1.5\,\text{cm}
80
\left(\text{cm}^3\right)
80
4.7\,\text{cm}^3
9.5\,\text{cm}^3
11.9\,\text{cm}^3
28.4\,\text{cm}^3
l
w
h
\qquad V=\dfrac{lwh}{3}
80
11.9
\qquad (0.8)(11.9)=9.52\,\text{cm}^3
9.5\,\text{cm}^3
th
h
t
6
7
9
10
10
110
10
6
\qquad x = 78(7.7 - t)
x
t
7.7
10.1
78.0
600.6
x = 0
 ( 7.7 - t ) = 0 \, 
\qquad 7.7-t = 0 \qquad \Rightarrow \qquad t=7.7
7.7
5
5
10
5
100
10
25
p
5
10{,}000{,}000{,}000{,}000{,}000{,}000{,}000_5
22
5^{22}
p
22
\qquad(x+55)^2+(y-11.5)^2=121
xy
(\green h,\purple k)
\red r
\qquad(x-\green h)^2+(y-\purple k)^2=\red r^2
\red r^2=121
\qquad\red r=\pm\red{11}
22
w+2, 2w-5,
3w+4
6w^3-40
6w^3+40
6w^3+5w^2-34w-40
6w^3+11w^2-26w-40
\qquad (w+2)(2w-5)(3w+4)
3w+4
w+2, 2w-5
3w+4
\qquad 6w^3+5w^2-34w-40
2007
50\%
2005
2013
0.5
t
2005
r
2005
2013
r-50=-0.5(t-2)
r-50=-0.5(t-7)
r-2=-0.5(t-50)
r-7=-0.5(t-50)
r
0.5
y-y_1=m(x-x_1)
m
(x_1,y_1)
m=-0.5
t
r
(2, 50)
\qquad r-50=-0.5(t-2)
2005
2013
\qquad  r-50=-0.5(t-2)
2
10\%
15\%
15\%
25\%
35\%
 is 
 of what is 
x
\qquad 0.10x=400
x=4000
12
\quad
17
27
61
111
86-25 = 61
897
\text{(ft)}
17^\circ
d
10
d
260 \text{ ft}
ABCD
t^\circ
30^\circ
40^\circ
50^\circ
60^\circ
180
s^\circ
\angle ADC
90
k^\circ
j^\circ
180
r^\circ
360
t^\circ
t^\circ = 40^\circ
y \ge 0
\; y \ge 0
\qquad -\left(\dfrac{12}8-b\right)=-b^2
b=s
b=t
st
-\dfrac32
-\dfrac23
\dfrac23
\dfrac32
s
t
-s \cdot -t
st
st
-\dfrac32
(20y+5)(4y-30)=80y^2+by-150
y
b
-580
-40
0
620
b
y
(20y+5)(4y-30)
80y^2-580y-150
(20y+5)(4y-30)
\qquad 80y^2-580y-150=80y^2+by-150
b=-580
\qquad b=-580
8\text{ in}\times10\text{ in}
x
A
(
\text{ in}^2)
A=(8-x)(10-x)
A=(8-2x)(10-2x)
A=(8+x)(10+x)
A=(8+2x)(10+2x)
10+2x
8+2x
A=lw
A=(8+2x)(10+2x)
(a,b)
a>0
a
x
y
(a,b)
y
x
\qquad x = \dfrac{-b\pm\sqrt{b^2-4ac} }{2a}
ax^2+bx+c=0
(a,b)
a>0
a
x
a=\dfrac{3}{7}
\qquad y - \sqrt{7y-31} = 3
-13
-3
3
13
\qquad (y-5)(y-8) = 0
\qquad y = 5
\qquad y = 8
y = 5
y=5
y = 8
y=8
5
8
\qquad 5 + 8 = 13
1080
720
10.2
5.66\,\text{cm}
6.8\,\text{cm}
8.49\,\text{cm}
15.3\,\text{cm}
1080
720
(\text{px})
d
d
1298\,\text{px}
\ell
5.66\,\text{cm}
xy
(4,13)
 (x+6)^2+(y-10)^2=397
 (x-10)^2+(y-6)^2=1{,}588
 (x-10)^2+(y+6)^2=1{,}588
 (x-10)^2+(y+6)^2=397
(\green h,\purple k)
\red r
\qquad(x-\green h)^2+(y-\purple k)^2=\red r^2
\green h
x
\qquad \green h=\dfrac{16+4}{2}=\green{10}
\purple k
y
\qquad \purple k=\dfrac{-25+13}{2}=\purple{-6}
(\green{10},\purple{-6})
d
\qquad (x-10)^2+(y+6)^2=397
xy
(-23,15)
(1,-55)
(x+23)^2+(y-15)^2=1369
(x+23)^2+(y-15)^2=5476
(x+11)^2+(y+20)^2=1369
(x+11)^2+(y+20)^2=5476
(\green h,\purple k)
\red r
\qquad(x-\green h)^2+(y-\purple k)^2=\red r^2
\green h
x
\qquad \green h=\dfrac{-23+1}{2}=\green{-11}
\purple k
y
\qquad \purple k=\dfrac{15+(-55)}{2}=\purple{-20}
\qquad\sqrt{5476}=74
74
\red{37}
\qquad(x-\green {(-11)})^2+(y-\purple {(-20)})^2=\red {37}^2
\qquad(x+11)^2+(y+20)^2=1369
60
(\text{dB})
0
(\text{m})
30 \, \text{dB}
0 \, \text{m}
8
60
3
3
30
3
3
\qquad 7.5 - 6 = 1.5
\qquad at^2+\dfrac72t-4=0
t=-8
t=1
a
-1
-\dfrac{1}{4}
\dfrac{1}{2}
1
t=1
a
\dfrac12
\qquad-81x^2=-11
x= \dfrac{\sqrt{11}}{81}
x=-\dfrac{\sqrt{11}}{81}
x=\dfrac{\sqrt{11}}{81}
x=-\dfrac{\sqrt{11}}{9}
x=\dfrac{\sqrt{11}}{9}
x=\dfrac{\sqrt{11}}{9}
-81x^2=-11
ax^2=c
a
c
x
x^2
81x^2-11
0
x^2
x=-\dfrac{\sqrt{11}}{9}
x=\dfrac{\sqrt{11}}{9}
\qquad 7.99C + 9.50M = S
S
C
M
S
7.99C
9.50M
M
M
\blue{\text{units}}
9.50M
\qquad \dfrac12x^2-\dfrac16x-\dfrac13=0
x=1
x=-\dfrac23
x=\dfrac16-\sqrt{\dfrac{23}{36}}
x=\dfrac16+\sqrt{\dfrac{23}{36}}
x=2
x=-\dfrac43
x=\dfrac16-\sqrt{-\dfrac{23}{36}}
x=\dfrac16+\sqrt{-\dfrac{23}{36}}
a
b
ab=-2
3xb+1xa=-1x
3b+a=-1
a=2
b=-1
1
-\dfrac23
0
0
x=1
x=-\dfrac23
15\%
m
f(m) = 0.15m
f(m) = 0.85m
f(m) = 1 - 0.15m
f(m) = 1 - 0.85m
15\%
85\%
m
\qquad f(m) = 0.85m
f(m)=0.85m
(a,b)
a\cdot b
-\dfrac{1{,}276}{243}
\dfrac{29}{9}
\dfrac{44}{27}
\dfrac{1{,}276}{243}
a
b
a
b
a=\dfrac13\left(\dfrac{29}{9}\right)+\dfrac59=\dfrac{44}{27}
a\cdot b=\left(\dfrac{44}{27}\right)\left(\dfrac{29}{9}\right)=\dfrac{1{,}276}{243}.
\qquad a\cdot b=\dfrac{1{,}276}{243}
\alpha
\alpha
{\large\ell}:s
\sin(\alpha) : 1
1 : \cos(\alpha)
\sin(\alpha)^2 : 1
1: \cos(\alpha)^2
\alpha
d
{\large\ell}
\alpha
d
s
\alpha
{\large\ell}:s
{\large\ell}
{\large\ell}
\qquad C = 0.4m + 68 
m
C
0.4m
0.4m
0.4m
0.4m
0.4m
C
2
0.4m
68
 term gets larger or smaller as the mileage increases or decreases, and it is a portion of the total cost 
0.4m
120^\circ
2\dfrac12\,\text{in}
2\dfrac78\,\text{in}
4\dfrac13\,\text{in}
8\dfrac23\,\text{in}
120^\circ
h
\qquad 2\cdot 4.33= 8.66
8\dfrac23\,\text{in}
\qquad\qquad
\angle {MLN}
\angle {MLN}
\theta
\theta
\qquad\qquad \sin\theta=\dfrac{\red{\text{Length of the side opposite from }\theta}}{\blue{\text{Length of the triangle's hypotenuse}}}
MLN
\angle{MLN}
\red{\text{{opposite}}}
\angle{MLN}
\red {10\sqrt3}
\blue{\text{{hypotenuse}}}
MLN
\blue{10\sqrt6}
\qquad\qquad \sin\theta=\dfrac{\red{10\sqrt3}}{\blue{10\sqrt6}}
0^\circ<\theta<90^\circ
\sin45^\circ
\dfrac{\sqrt2}{2}
\theta=45^\circ
\angle{MLN}
45^\circ
f(x)=8\left(\dfrac{2}{5}\right)^x
g(x)=8(b)^x
xy
b
f
g
y
f(x)=8\left(\dfrac{2}{5}\right)^x
x
-x
y
b=\dfrac{5}{2}
g(x)=8(b)^x
x
\qquad b=\dfrac{5}{2}
(
=0)
8
50\%
2
6
6^{\text{th}}
17
45
71
135
6
8
8+6y
y
\qquad 8+6(6)=44
6
44
50\%
2
2
150\%
1.5
2
1.5
0
8
2
8(1.5)
4
8(1.5)(1.5)=8\cdot1.5^2
6
8(1.5)(1.5)(1.5)=8\cdot1.5^3=27
6
27
\qquad44+27=71
71
6^{\text{th}}
1.2^\circ
5
\sin(1.2^\circ) \approx 0.02094
\cos(1.2^\circ) \approx 0.99978
1.26
60.01
165.58
238.75
\qquad \cos(\theta^\circ)=\dfrac{\text{length of adjacent side}}{\text{length of hypotenuse}}
d
12
\qquad5.001(12)=60.012
60.01
(
1.00)
\text{pH}
4
\text{pH}
0.5
\text{pH}
4
\text{pH}
0.10
\text{pH}
0.40
2
6
3.75
4.25
0
8
3.20
4.80
\text{pH}
4\,\text{pH}
\qquad |\text{pH}-4|
\qquad \dfrac{0.10}{0.5}|\text{pH}-4|=0.2|\text{pH}-4|
0.40
\text{pH}
\qquad  \text{pH}=2
 \text{pH}=6
\qquad  \text{pH}=2
 \text{pH}=6
.  Out-of-pocket expenses include any deductibles and coinsurance that the client pays beyond the normal monthly premium.  The client has a deductible of 
.  This means that the client has to pay all of the first 
 of expenses insured by the policy.  After that, the client has a 
 co-insurance, meaning the client pays 
 of the insured expenses and the insurance company pays the remainder.  Which inequality represents how much the total insured expenses, 
x<26{,}000
x<25{,}000
x<24{,}000
x<23{,}750
20\%
.  We can represent this with 
.  The deductible is a flat rate of 
.  This means that the total the client pays will be equal to or less than 
250
\qquad0.2(x-250)<4750
0.2
\qquad x-250<23{,}750
250
\qquad x<24{,}000
y=-36x^2-50
(x,y)
y =120x+50
\left(\dfrac{2}{3},-70\right)
\left(-\dfrac{4}{3},-110\right)
\left(-\dfrac{5}{3},-150\right)
\left(\dfrac{4}{3},-114\right)
y=120x+50
y
(0,50)
m=120
(-1,-70)
(-2,-190)
x=-1
x=-2.5
\left(-\dfrac{4}{3},-110\right)
\left(-\dfrac{5}{3},-150\right)
\left(-\dfrac{4}{3},-110\right)
x=-\dfrac{4}{3}
y=-110
\left(-\dfrac{4}{3},-110\right)
\left(-\dfrac{5}{3},-150\right)
x=-\dfrac{5}{3}
y=-150
x
y
\left(-\dfrac{5}{3},-150\right)
x
8\sqrt3
16
16\sqrt2
32
ABC
\overline{BC}
\overline{BC}
y
\theta
\qquad\qquad \tan\theta=\dfrac{\red{\text{Length of the side opposite from }\theta}}{\blue{\text{Length of the side adjacent to }\theta}}
ABC
\angle B=30^\circ
\red{\overline{AC}}
\red{\text{{opposite}}}
\angle B
\red{4}
\blue{\overline{BC}}
\blue{\text{{adjacent}}}
\angle B
\blue{y}
\qquad\qquad \tan30^\circ=\dfrac{\red{4}}{\blue{y}}
\tan30^\circ
\dfrac{\sqrt3}{3}
x
\overline{BD}
\angle B
BCD
\theta
\qquad\qquad \cos\theta=\dfrac{\blue{\text{Length of the side adjacent to }\theta}}{\green{\text{Length of the triangle's hypotenuse }}}
BCD
\angle B=30^\circ
\blue{\overline{BD}}
\blue{\text{{adjacent}}}
\angle B
\blue{x}
\green{\overline{BC}}
\green{\text{{hypotenuse}}}
BCD
\green{\dfrac{12}{\sqrt3}}
\qquad\qquad \cos30^\circ=\dfrac{\blue{x}}{\left(\green{\dfrac{12}{\sqrt3}}\right)}
\cos30^\circ
\dfrac{\sqrt3}{2}
x
6
xy
(x-10)^2+(y-60)^2=841
x
-10
y
x
y
y-60=21
60
y=81
y-60=-21
60
y=39
y
39
81
16
1
4
4
2^\text{nd}
3^\text{rd}
4^\text{th}
5^\text{th}
m
p
1m
m
4m
p
\qquad 16+m=4+4m
m
\qquad 16 = 4 + 3m
4
\qquad 12=3m
3
\qquad m=4
Y
v
1995
88
1996
132
1997
198
1998
297
1999
446
v
A
B
Y
44 \%
50 \%
44
50
v
Y
v
\Delta v
1995
88
1996
132
132-88=44
1997
198
198-132=66
1998
297
297-198=99
1999
446
446-297=149
Y
v
\Delta v
\Delta v / v_i
1995
88
1996
132
44
44/88=0.5
1997
198
66
66/132=0.5
1998
297
99
99/198=0.5
1999
446
149
149/297 \approx 0.5
50 \%
50 \%
\qquad \dfrac{ 4\ell^2 + 4\ell - 3}{ 2\ell^2 - 7\ell - 15}
\ell > 6
\dfrac{3}{20}
-\dfrac{3}{20}
\dfrac{2\ell + 1}{\ell + 5}
\dfrac{2\ell - 1}{\ell - 5}
\dfrac{ \pink{ 4\ell^2 + 4\ell - 3}}{ 2\ell^2 - 7\ell - 15} 
\dfrac{ { 4\ell^2 + 4\ell - 3}}{ \purple{2\ell^2 - 7\ell - 15}}
1
\,\ell
\,0
\ell>6
\qquad \dfrac{2\ell - 1}{\ell - 5}
l
h
\, \dfrac{\sqrt{6}}{3} l
l
\dfrac{\sqrt{2}}{12} l^3
\dfrac{\sqrt{3}}{12} l^3
\dfrac{\sqrt{2}}{4} l^3
\dfrac{\sqrt{3}}{4} l^3
\qquad V = \dfrac{1}{3} Bh
V
B
h
l
\qquad B = \dfrac{\sqrt{3}}{4} l^2
B
\, \dfrac{\sqrt{6}}{3}l
h
\, \dfrac{\sqrt{2}}{12} l^3
\qquad\dfrac{3h}{h-2} + \dfrac{6}{h+2} = -1
\qquad4(h + 4)(h - 1) = 0
\qquad h = -4
h = 1
h = -4
h = -4
h = 1
h = 1
-4 + 1 = -3
m>0
\qquad\dfrac{\left(\dfrac{7m^2+59m+24}{m^2+4m-5}\right)}{\left(\dfrac{m^2+4m-32}{3m+15}\right)}
\dfrac{7m^2+59m-8}{3m+10}
\dfrac{-32(7m^2+59m+24)}{-15(m+5)}
\dfrac{21m+9}{(m-1)(m-4)}
\dfrac{7m+3}{(m-1)(m-4)}
\qquad\dfrac{7m^2+59m+24}{m^2+4m-5}\div\dfrac{m^2+4m-32}{3m+15}
\qquad \dfrac{(7m+3)(m+8)}{(m+5)(m-1)} \div \dfrac{\red{(m-4)(m+8)}}{\blue{3(m+5)}}
\qquad \dfrac{(7m+3)(m+8)}{(m+5)(m-1)} \cdot \dfrac{\blue{3(m+5)}}{\red{(m-4)(m+8)}}
m
0
m\neq{\purple{-8},\green{-5}}
m>0
\qquad\dfrac{21m+9}{(m-1)(m-4)}
f(x)=-2.5(x+2)^2+8
g(x)=-2.5(x+2)^2+b
xy
a
y
f
b
y
g
a
5
b
b
y=a(x-h)^2+k
(h,k)
f(x)=-2.5(x+2)^2+8
(-2,8)
a=8
g(x)=-2.5(x+2)^2+b
(-2,b)
a
5
b
a=b-5
a=8
8=b-5
b=13
\qquad b=13
y=-2x^3+24x
xy
2
(-2,-32)
(2,32)
x<-2
-2<x<2
x>2
x<-2
-2<x<2
x>2
x<-32
-32<x<32
x>32
x<-32
-32<x<32
x>32
y=-2x^3+24x
(-2,-32)
x<-2
(2, 32)
-2<x<2
(2,32)
x>2
x<-2
-2<x<2
x>2
(x_1,y_1)
(x_2,y_2)
y
y_1
y_2
x
y
(x,y)
y
9
x
x^2
x
x_1=\dfrac{8}{3}
x_2=-\dfrac{8}{3}
y
x
y
x_1=\dfrac{8}{3}
x_2=-\dfrac{8}{3}
\left(\dfrac{8}{3},\dfrac{13}{6} \right)
\left(-\dfrac{8}{3},-\dfrac{1}{2} \right)
y
y_1+y_2
\qquad\dfrac{13}{6}+\left(-\dfrac{1}{2}\right)=\dfrac{5}{3}
y=x^3-3x
xy
2
(-1,2)
(1,-2)
x<-1
-1<x<1
x>1
x<-1
-1<x<1
x>1
x<-2
-2<x<2
x>2
x<-2
-2<x<2
x>2
y=x^3-3x
(-1,2)
x<-1
(1,-2)
-1<x<1
(1,-2)
x>1
x<-1
-1<x<1
x>1
b
m
n
(x,y)
-2
\qquad -14m+8n=20
b=-14
b
-14
92.5\%
x
y
88\%
91\%
x
y
0.075x+.12y=0
0.015x-0.03y=0
0.925x+0.88y=91
0.925x+0.88y=0.91xy
\quad [\text{grams of silver alloy}]\cdot [\text{percent silver}]=[\text{grams of pure silver}]
x
92.5\%
0.925x
y
88\%
0.88y
88\%
0.925x+0.88y
x+y
91\%
0.91(x+y)
\qquad 0.925x+0.88y=0.91(x+y)
x
y
\qquad 0.015x-0.03y=0
\qquad-my = 1{,}500-4{,}500
m
1{,}500 + my
-1{,}500
1{,}500
3{,}000
4{,}500
1{,}500 + my
4{,}500
my
da
a
(\text{m/s}^2)
d
a
d
d
d
\text{ m/s}^2
d
\text{ m/s}^2
d
d
a
0
d
d
d
20\%
500
25
3
20\%
y=\green m\purple x+\red b
\red b
\green m
\green{25}
\red {500}
\qquad\text{number of visitors}=\green{25}(\purple{\text{months since SEO ended}})+\red{500}
3
\qquad\green{25}(\purple3)+\red{500}=575
3
575
y=\red{a}\cdot \blue{b}^{\large\purple{x}}
\red a
\blue b
\purple x
1
20\%
\blue{120\%}
\blue{1.2}
\qquad\text{hypothetical number of visitors}=\red{500}\cdot\blue{1.2}^{\purple{\text{months since SEO ended}}}
\qquad\red{500}\cdot\blue{1.2}^{\purple3}=864
864
3
\qquad864-575=289
289
40
(\text{mm})
(\text{mm}^3)
2{,}667\,\text{mm}^3
5{,}333\,\text{mm}^3
10{,}667\,\text{mm}^3
16{,}746\,\text{mm}^3
x
\qquad 2x^2=d^2
\,x=\dfrac{d}{\sqrt2}
\,x^2=\dfrac{d^2}{2}
\dfrac{(40)^2}2 =800
h
A
V=\dfrac13 Ah
20
\qquad V=\dfrac13\cdot800\cdot 20=5{,}333\,\text{mm}^3
5{,}333\,\text{mm}^3
td
d
(\text{cm})
h
d
28 \text{ cm}
28 \text{ cm}
28 \text{ cm}
28 \text{ cm}
16
d
d
(t, d) = (0, 28)
28 \text{ cm}
0
28 \text{ cm}
\qquad (1+i)(1-i)
i=\sqrt{-1}
2-2i
2i
0
2
i^2=-1
\qquad 2
y=3(x+6)^2+5
y
(0,b)
b
y
y
y
0
x
y
y
y
(0,113)
y
(0,b)
b=113
\qquad b=113
8
1
8
52
13
8
(\text{mi})
1
m
13
1
=60
(\text{min})
\qquad \dfrac{8\text{ mi}}{60\text{ min}}=\dfrac{13\text{ mi}}{m\text{ min}}
m
8
52
m
13
\qquad \dfrac{8\text{ mi}}{52\text{ min}}=\dfrac{13\text{ mi}}{m\text{ min}}
m
97.5-84.5=13
y=-2\left(\dfrac{4}{3}\right)^x
y=ab^x
a
y
(0,a)
y
(0,-2)
\qquad
\quad
y=2\left(\dfrac{4}{3}\right)^x
y=-2\left(\dfrac{4}{3}\right)^x
a>0
b>1
y=2\left(\dfrac{4}{3}\right)^x
-1
x
y=-2\left(\dfrac{4}{3}\right)^x
x
y
1
-\dfrac{8}{3}
-1
-\dfrac{3}{2}
y=-2\left(\dfrac{4}{3}\right)^x
\qquad
36
35\%
75\%
5.85
12.6
17.55
23.4
35\%
35\%
36
\qquad 0.35\times36=12.6
12.6
36
\qquad 36-12.6=23.4
23.4
\qquad 0.75 \times23.4=17.55
17.55
23.4
\qquad 23.4-17.55=5.85
5.85
350
150
3:4
3:7
4:5
4:7
350-150=200
150:200
3:4
\qquad 3x^2+2x=11
b^2-4ac
\qquad 3x^2+2x-11=0
0
5tm^2
7tm^2
91
5
7tm^2
P(t)
t
2f
\dfrac f2
P(t)
\dfrac{t}{1.06}=1
t=1.06
1.06
1.1
1.1
v
2
5
3
2v^2 - 13v + 17 = 0
3v^2 - 15v - 4 = 0
3v^2 - 11v + 10 = 0
3v^2 - 19v + 10 = 0
v = \dfrac{d}{t}
v
t_1
t_2
t = t_1 + t_2
3
2
v
2
5
v - 5
\qquad t_2 = \dfrac{2}{v - 5}
\qquad3 = \dfrac{2}{v} + \dfrac{2}{v -5 }
v
v \ne 5
v \ne 0
\qquad 3v^2 - 19v + 10 = 0
100\%
30\%
50\%
41\%
30\%
100\%
32\%
32\%
51\%
32\%
100\%
48\%
41\%
32\%
48\%
100\%
1.475
h
t
12.5
1.5
1.6
3.0
3.1
4.9
y
y=a(x-b)^2+c
(b,c)
12.5
1.5
(1.5,12.5)
\qquad h(t)=a(t-1.5)^2+12.5
a
1.475
(0,1.475)
a
a
h
t
\qquad h(t)=-4.9(t-1.5)^2+12.5
0
t
h=0
t\approx3.10
t\approx-0.10
3.1
\left(4x^5-8x^3+\dfrac{1}{3}x^2\right)
\left(\dfrac{1}{2}x^3 - 4x^2\right)
4x^5 - 12x^3 + \dfrac{5}{6}x^2
4x^5 - \dfrac{15}{2}x^3 - \dfrac{11}{3}x^2
4x^5 - \dfrac{7}{2}x^3 - x^2
4x^5 - \dfrac{17}{2}x^3 + \dfrac{11}{3}x^2
0
\qquad 4x^5 - \dfrac{15}{2}x^3 - \dfrac{11}{3}x^2
\qquad\left(v+\dfrac{1}5\right)^2-9=0
-\dfrac{3}5
-\dfrac{2}5
-\dfrac{1}5
0
v
-\dfrac{2}5
y=(x-3)^3
y=x^3
y=(x-h)^3+k
h
k
y=(x-3)^3
3
0
y=(x-3)^3
\qquad\sqrt{4x+20} = x + 2
\qquad x = 4
\qquad x = -4
x=4
x= 4
x=-4
x = -4
x=4
4
196
(\text{cm})
250\,\text{cm}
98\,\text{cm}
170\,\text{cm}
217\,\text{cm}
339\,\text{cm}
196\,\text{cm}
h
60^\circ
170\,\text{cm}
4
3
85
3
5
105
a
r
4
a
\pink{4a}
3
r
\blue{3r}
85
\qquad \pink{4a} + \blue{3r} = 85
3
\purple{3a}
5
\green{5r}
105
\qquad \purple{3a} + \green{5r} = 105
\qquad \left(t+\dfrac83\right)\left(t+b\right)=0
b
-\dfrac83
\dfrac{13}3
b
-\dfrac{13}3
-\dfrac{8}3
\dfrac{8}3
\dfrac{13}3
t=-\dfrac83
t=\dfrac{13}3
0
\left(t+\dfrac83\right)
t=-\dfrac83
\left(t+b\right)
t=-b
t=\dfrac{13}3
-b=\dfrac{13}3
b
-\dfrac{13}3
1{,}000
 \qquad 2(x-3)^2-b=0
b
x=3\pm\sqrt5
b
-2
\ \ \ 2
\ \ \ 3
-3
2(x-3)^2-b=0
b
(x-3)^2
x=3\pm\sqrt5
10
b
3\pm\sqrt{\dfrac{b}{2}}
3\pm\sqrt5
b=10
 for every 
 minutes of play time and the friends spent 
 in total. On average, the friends played 
 ping pong game every 
(\text{min})
 for every 
 minutes of playtime. We can set up a proportion to determine how many total minutes 
 they spent playing if they paid 
1
20
g
120
\qquad \dfrac{1\text{ game}}{20 \text{ min}}=\dfrac{g\text{ games}}{120\text{ min}}
g
6
(x,y)
(4,-4)
(-2,2)
(-4,4)
(2,-2)
(2,-4)
(4,-2)
y
x
x
y=-x
y
y+2x^2=3x+16
x
\qquad\qquad\qquad -2x^2+3x+16=-x
 
x=4
x=-2
x
x
 
(4,-4)
(-2,2)
100\ \text{dollars}
960
\text{pesos}
\text{pesos}
\text{dollars}
24{:}1
\text{dollars}
\text{yen}
1{:}105
330\ \text{yen}
4200\ \text{yen}
6000\ \text{yen}
6300\ \text{yen}
24\ \text{pesos}
1
40
\qquad  \dfrac{960}{24}= \dfrac{x}{1},\ \text{ so }\ x=40
100 - 40 = 60
\qquad \dfrac{60}{y}= \dfrac{1}{105}
y=6300\ \text{yen}
6300\ \text{yen}
\qquad 7x = 13\sqrt{x}
x
0
\dfrac{49}{169}
\dfrac{169}{49}
\dfrac{13}{7}
x
\sqrt{x}
7x=13\sqrt x
2
0
0
\dfrac{169}{49}
x
\dfrac{169}{49}
\qquad\qquad\qquad\qquad\qquad\qquad \text{Day } 1
\qquad\qquad\qquad\qquad\qquad\qquad \text{Day } 2
1
2
25^\text{th}
75^\text{th}
50^\text{th}
1
1.6
2
2.0
2.0 - 1.6 = 0.4
RI
I
R
I = 4
0
4
4
4
4
I
4
R
I
4
R
R
10
I
4
4
4
75^\circ \text C
22^\circ \text C
t
\qquad T(t) =22 + 53 \,(0.74)^{t}
2
22^\circ \text C
51^\circ \text C
74^\circ \text C
75^\circ \text  C
75^\circ \text C
22^\circ \text C
22^\circ \text C
T(t)
t=2
\qquad T(2) =22 + 53 \,(0.74)^{2}\approx 51^\circ \text C
51^\circ\text C.
2
51^\circ \text C.
 per pound as well as 
 from selling pork. On Sunday, she makes the same amount of money from selling an equivalent 
 pounds of beef at 
 per pound as well as 
 from selling pork. Which system of equations can be used to find out how many 
 pounds of beef she made for a total of 
 per pound, we can represent that as 
. The 
 in pork can be represented by adding 
 to 
. Therefore, we can write 
 per pound is 
. When we add the pork, the equation becomes 
ABCD
HIJK
l^\circ
60^\circ
80^\circ
90^\circ
100^\circ
80
ABCD
\overline{AD}
\overline{BC}
\overline{AD}
\overline{BC}
80
l^\circ
\qquad l^\circ = 80^\circ
24v^3w^8
O
XYZ
10.8
XOZ
1.8
r
\sqrt{12}
6
\dfrac{50}{3}
19.44
\qquad s = r \theta
s
r
\theta
s
\theta
r
r
6
, between layers of molecules to the angle, 
, of incoming light rays with wavelength 
d= \dfrac{\lambda}{2\sin\theta}
d= \dfrac{2\lambda}{\sin\theta}
d= \dfrac{2\sin\theta}{\lambda}
d= \dfrac{\sin\theta}{2\lambda}
\pink{ \text{dividing both sides by } 2 }
\blue{\text{divide both sides by } \sin\theta}
\qquad d = \dfrac{\lambda}{2\sin\theta}
\qquad FV(t)=1{,}500(1.045)^{\large{t}}
FV
t
\qquad FV(t)=1{,}500(1.045)^{\large{t}}
t=0
\qquad FV(0)=1{,}500(1.045)^0=1{,}500
  -18\geq6-12ax 
1-2ax
1-2ax\geq-23
1-2ax\leq-23
1-2ax\leq-3
1-2ax\geq-3
 -18\geq6-12ax 
 1-2ax 
 1-2ax=\dfrac{6-12ax}{6} 
 \dfrac{6-12ax}{6}=1-2ax 
1-2ax 
6
1-2ax\leq-3
1-2ax
\qquad\qquad 1-2ax\leq-3
\theta=315^{\circ}
\theta
\dfrac{5\pi}{4}
\dfrac{3\pi}{2}
\dfrac{5\pi}{3}
\dfrac{7\pi}{4}
2\pi
360^{\circ}
\qquad 2\pi \text{ radians}=360^{\circ}
360
1^{\circ}=\dfrac{\pi}{180}\text{ radians}
 2\pi \text{ radians}=360^{\circ}
8
 \dfrac{\pi}{4} \text{ radians}=45^{\circ}
315^{\circ}
7\cdot 45^{\circ}
 7\cdot \dfrac{\pi}{4}
\qquad315^{\circ}=\dfrac{7\pi}{4}
7
3
2
5
2
3
5
7
d
7
3
7+3(d-1)
3
4
\green{3d+4}
2
d
d-2
5
\red{5\left(d-2\right)}
\qquad\green{3d+4}=\red{5\left(d-2\right)}
7
p
1760
2000
y
p = 740+
^{(y-1760)}
a
800
1890
1800
1980
a
0.032
1.032
2.032
3.032
0.032^{(y-1760)}
0.032 < 1
1.032^{(y-1760)}
2.032^{(y-1760)}
2.032 > 2
y=1900
2^{(1900-1760)} = 2^{140}
p=820
3.032^{(y-1760)}
\qquad p = 740+1.032^{(y-1760)}
\qquad \left(2b^{-5}\right)^3
b\neq0
\dfrac{2}{b^{15}}
\dfrac{8}{b^{15}}
\dfrac{1}{2b^{15}}
\dfrac{1}{8b^{15}}
\qquad (x\cdot y)^m=x^my^m
\qquad  \left(2b^{-5}\right)^3=2^3\cdot \left(b^{-5}\right)^3
(x^m)^n=x^{mn}
\qquad 2^3\cdot \left(b^{-5}\right)^3=2^3\cdot b^{-15}
x^{-n}=\dfrac{1}{x^n}
x\neq0
b\neq0
\qquad \dfrac{8}{b^{15}}
4
16
(\text{c})
1
(\text{gal})
25\%
33\%
67\%
75\%
\qquad \text{total volume} = 1\:\text{gal}
4
\qquad \text{volume taken away} = 4\:\text{c}
\qquad \text{percent of total that is remaining}
\leftarrow
\qquad \text{remaining volume} = \text{total volume} - \text{volume taken away}
\qquad \text{remaining volume} = 1\:\text{gal} - 4\:\text{c}
1
16
\qquad \text{remaining volume} = 16\:\text{c} - 4\:\text{c}
\qquad \text{remaining volume} = 12\:\text{c}
75\%
l
60
w
30
200
\text{\red{at most}}
60
\text{\red{less than or equal}}
60
\qquad l \red{\leq} 60
\text{\blue{at least}}
30
\text{\blue{greater than or equal}}
30
\qquad w \blue{\geq} 30
l+l+w+w
2l+2w
\text{\green{no more than}}
200
2l+2w
\text{\green{less than or equal to}}
200
\qquad 2l+2w \green{\leq} 200
2
(r,s)
r+s
-5.2
-5.1
0.1
5.1
r
s
r
s
s
r=-52s
-k+78 = 98 - 20
k
k
1859
24
1920
10
R
t
1859
R(t)=24+\left(1.3845\right)t
R(t) =24\left( 1.3845\right)^t
R(t)=24\left(0.6155\right)^t
R(t)=10^{10}\left(1.3845\right)^t
\qquad R(t)=24+\left(1.3845\right)t
10\,\text{billion}
1920
\qquad R(t)=24\left(0.6155\right)^t
\qquad R(t)=10^{10}\left(1.3845\right)^t
t=0
\qquad R(t)=24\left( 1.3845\right)^t
1920
61
1859
R(61)\approx 10\,\text{billion}
t
1859
\qquad R(t)=24\left( 1.3845\right)^t
27{,}000
1
301
1
1{,}300
27{,}000
r
301
\left(\text{mi}^2\right)
\qquad \dfrac{27{,}000\text{ residents}}{1\text{ mi}^2}=\dfrac{r\text{ residents}}{301\text{ mi}^2}
r
1
1{,}309
s
8{,}224{,}825
\qquad \dfrac{1\text{ subway car}}{1{,}300\text{ residents}}=\dfrac{s\text{ subway cars}}{8{,}127{,}000\text{ residents}}
s
6{,}300
\qquad100-121k^2=0
k=\dfrac{100}{121}
k=-\dfrac{100}{121}
k=\dfrac{100}{121}
k=\dfrac{10}{11}
k=-\dfrac{10}{11}
k=\dfrac{10}{11}
0
x^2
\qquad (10-11k)(10+11k)=0
\qquad10-11k=0 \qquad \text{or} \qquad 10+11k=0
k
k=\dfrac{10}{11}
k=-\dfrac{10}{11}
k=-\dfrac{10}{11}
k=\dfrac{10}{11}
\left(\text{in} \, \dfrac{\text m}{\text s}\right)
(\text{in N}
 | 
| 
 meters per second 
, it experiences a drag force of 
 newtons 
. The table above represents the design engineer's measurements. If the drag force can be modeled by a quadratic equation of the form 
, what value of speed, in meters per second, corresponds a drag force of 
b
c
b
c
v
2
F = bv + cv^2
2
2
b
b
, we can set the 
 expressions for 
 equal to each other and solve for 
c
b
\qquad F = 10v + 10v^2
F = 120
v
\qquad 120 = 10v + 10v^2
av^2 + bv + c = 0
v
(v + 4)
(v - 3)
0
v
v = 3.0 \,\dfrac {\text m} {\text s}
\qquad 15y- 5\sqrt{y} = 0
y
0
\dfrac{1}{3}
\dfrac{1}{9}
3
5\sqrt{y}
y=\sqrt y\cdot \sqrt y
\qquad 5\sqrt{y}\left(3\sqrt y - 1\right) = 0
2
0
0
0
\dfrac{1}{9}
0
p>0
\left(p^m\right)^{2n}\cdot \left(p^{2m}\right)^n=16^{mn}
p
2
\sqrt{8}
4
p
\qquad \left(a^m\right)^n=a^{mn}\qquad 
\qquad a^m\cdot a^n=a^{m+n}
p^{4mn}=16^{mn}
(p^4)^{mn}=16^{mn}
\qquad p^4=16
p
p=\pm2
p>0
p=2
L =21
y=11
20^\circ
\overline{AB}
\overline{CD}
\overline{AB}
5.5
\sin(20^\circ) \approx 0.342
\cos(20^\circ) \approx 0.940
\tan(20^\circ) \approx 0.363
x
\overline{CD}
\tan(20^\circ) \approx 0.363
\overline{AB}
\overline{AB}
20^\circ
20^\circ
x
x
6
1^{\text{st}}
2^\text{nd}
17
1^{\text{st}}
2^{\text{nd}}
1^{\text{st}}
17
2^{\text{nd}}
2^{\text{nd}}
\sin{\dfrac{\pi}{2}}
-1
0
\dfrac{\sqrt{2}}{2}
1
\theta
P
P= (\cos{\theta}, \sin{\theta})
\dfrac{\pi}{2}
(0,1)
y
\sin{\dfrac{\pi}{2}}=1
15\%
2
y=\green m\purple x+\red b
\red b
\green m
 billion from 
\qquad\green{179{,}500{,}000}(2)+\red{2{,}100{,}000{,}000}=2{,}459{,}000{,}000
y=\red{a}\cdot \blue{b}^{\large\purple{x}}
\red a
\blue b
\purple x
1
15\%
\blue{115\%}
\blue{1.15}
\quad \text{hypothetical revenue}=\red{2{,}100{,}000{,}000}\cdot\blue{1.15}^{\purple{\text{years since 2 years ago}}}
\qquad\red{2{,}100{,}000{,}000}\cdot\blue{1.15}^{\purple2}=2{,}777{,}250{,}000
15\%
\dfrac{1}{3}
 to spend this week, the florist purchases 
 bushels of flowers at 
 per bushel and 
 gallons of nutrient-rich water at 
120
20
100
80
60
40
80
10
\dfrac{1}{3}
w
f
(0,0)
(120,40)
 for a bushel of flowers and 
 for a gallon of water.  The florist has 
 to spend, so the cost must be *less than or equal to* 
 to 
f=60, w=40
60
40
\qquad \sqrt{2y^3z}\cdot \sqrt{8y^{13}z^8}
y, z \geq 0
4y^4z^3
8y^4z^3
4y^8z^4\sqrt{z}
8y^8z^4\sqrt{z}
y, z \geq 0
\sqrt{2y^3z}
\sqrt{8y^{13}z^8}
\sqrt{a}
\sqrt{b}
 \sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}
a^m\cdot a^n=a^{m+n}
 \sqrt{16y^{16}z^9}
 \sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}
\sqrt{y^{16}}=y^8
y^8\cdot y^8=y^{16}
\sqrt{z^{8}}=z^4
z^4\cdot z^4=z^{8}
4y^8z^4\sqrt{z}
y = \dfrac{1}{12} - x
xy
x + y = 1
y
y
x
y
x + y = 1
y = -x + 1 = (-1)x + 1
m = -1
y
b = \dfrac{1}{12}
y = \dfrac{1}{12} - x
y
\dfrac{1}{3}(7s^3+7s^2-6s-2)
\dfrac{1}{3}(7s^3+7s^2-6s+2)
s
s
-7s
s
-2
\dfrac43
s
-2s^0
\dfrac43s^0
\qquad\quad (s-2)+\left(\dfrac{7}{3}s^3+\dfrac{7}{3}s^2-7s+\dfrac{4}{3}\right)
\qquad =\left(\dfrac{7}{3}{s}^3+\dfrac{7}{3}{s}^2-6s\blue{ - \dfrac23}\right)
\dfrac13
y=-(x+4)^2+5
\qquad y=-3x+b
b
(-4,5)
(-7,-4)
(-6,1)
(-1,-4)
(-2,1)
(-4,5)
y=-3x+b
m=-3
(-4,5)
(-1,-4)
(-1,-4)
x
y
(-1,-4)
x
y
(-1,-4)
(-1,-4)
\qquad\sqrt{3m^2 + 24} = 2m + 2
-10
-8
2
8
\qquad m = -10
\qquad m = 2
m= 2
m= 2
m= -10
m= -10
m=2
2
\qquad(4x^3-5x)+(8x-3x^3)
x^3+3x
x^3-3x
7x^3-3x
7x^3+3x
\qquad x^3 + 3x
x^3+3x
a(x)
x
\qquad a(x) = 0.14x-0.0014x^2
\qquad a(x) = k(x - x_0)^2+h_0
k
(x_0, h_0)
k
(50, 3.5)
3.5
50
3.5
C
4\text{ meters (m)}
x^{\circ}
20 \text{ m}
x
5
80
286
304
s
\theta
r
s=r\theta
\theta
\qquad 20=4\cdot \theta
\theta
\theta=5
\theta
2\pi
360^{\circ}
\qquad 2\pi \text{ radians}=360^{\circ}
2\pi
1\text{ radian}=\dfrac{180}{\pi}^{\circ}
5 \text{ radians}
\dfrac{180}{\pi}
x
286
\text{sin}(r\pi) = \text{cos}(4r\pi)
0< r<1
r
r
\qquad 0.1
\dfrac {1}{10}
\quad
\left(\dfrac{\text{bbl}}{\text{day}}\right)
30\,\dfrac{\text{bbl}}{\text{day}}
30
\dfrac{\text{bbl}}{\text{day}}
30
\dfrac{\text{bbl}}{\text{day}}
30
\dfrac{\text{bbl}}{\text{day}}
30
\dfrac{\text{bbl}}{\text{day}}
30
\dfrac{\text{bbl}}{\text{day}}
11
30
(
11
0)
8
30
11
8
30
\dfrac{\text{bbl}}{\text{day}}
\qquad (2b-1)^3
2b^3-1
8b^3-1
4b^3-10b^2+4b-1
8b^3-12b^2+6b-1
\qquad (2b-1)^3=(2b-1)(2b-1)(2b-1)
2b-1
8b^3-12b^2+6b-1
3^{\text{rd}}
t
d
\qquad t^2=3.98\cdot10^{-20}\cdot d^3
4
\qquad t^2=\red{3.98\cdot10^{-20}\cdot d^3}
d
\green{4d}
\qquad\sqrt{64t^2}=8t
8
2004
16
223{,}357{,}000
66\%
16
46.4\%
16
2004
68{,}400{,}000
104{,}000{,}000
147{,}400{,}000
157{,}000{,}000
16
2004
66\%
223{,}357{,}000
66\%
223{,}357{,}000
\qquad  0.66\times223{,}357{,}000=147{,}415{,}620
147{,}415{,}620
16
2004
46.4\%
46.4\%
147{,}415{,}620 
\qquad  0.464\times147{,}415{,}620=68{,}400{,}847
68{,}400{,}000
2004
\qquad(x-h)^2+(y-k)^2=r^2
x
y
(h,k)
r
y
x
h
k
r
y=r-x+h+k
y=r-(x-h)^2+k
y=\pm\sqrt{r^2-(x-h)^2}+k
y=\pm\sqrt{r^2-(x-h)^2+k}
y
\red{\text{subtracting }(x-h)^2}
\green{\text{add }k}
\qquad y=\pm\sqrt{r^2-(x-h)^2}+k
\qquad\dfrac{x^2+6x-30}{30-6x-x^2}
x^2+6x-30\neq0
-1
1
0
\dfrac{(x+6)(x-5)}{(6-x)(5+x)}
30-6x-x^2
-1(x^2+6x-30)
\qquad \dfrac{(x^2+6x-30)}{-1(x^2+6x-30)}
x
0
x^2+6x-30\neq\purple0
-1
x^2+6x-30\neq 0
xy
x^2+y^2-10x+32y+272=0
(10,-32)
4\sqrt{17}
(-10,32)
4\sqrt{17}
(-5,16)
3
(5,-16)
3
\qquad(x^2-10x)+(y^2+32y)=-272
x
y
\qquad(x^2-10x\pink{+25})+(y^2+32y\blue{+256})=-272\pink{+25}\blue{+256}
\qquad(x-5)^2+(y+16)^2=9
(\green h,\purple k)
\red r
\qquad(x-\green h)^2+(y-\purple k)^2=\red r^2
\qquad(x-\green5)^2+(y-(\purple{-16}))^2=\red3^2
(5,-16)
3
A_1
V_1
A_2
V_2
\qquad A_1V_1=A_2V_2
\dfrac{1}{4}
\dfrac{1}{4}
\dfrac{3}{4}
\dfrac{4}{3}
4
A_1
A_2
\dfrac{1}{4}
\dfrac{3}{4}
\purple{\dfrac{3}{4}A_1}
A_2
\qquad A_1V_1=\purple{\dfrac{3}{4}A_1}V_2
V_2
\dfrac{4}{3}
 | 
 | 
 | 
 | 
 \dfrac{\text{rent}}{\text{square feet}} 
(x,y)
x
y
x
x
\qquad\qquad\qquad x^2+x+5=2x+7
x
x
 
x
2\cdot-1=-2
\qquad \sqrt{-2p-1}=p+2
-6
-5
-1
1
p
0
p=-5
p=-1
p=\green{-5}
p=-5
p=\green{-1}
p=-1
-1
-1
y-3=\left(\dfrac{1}{3}\right)^x
y
y=\left(\dfrac{1}{3}\right)^x+3
y=\left(\dfrac{1}{3}\right)^x
3
y=\left(\dfrac{1}{3}\right)^x
y=ab^x
a>0
0<b<1
\qquad
(x-3)^2+(y-2)^2=16
(x-h)^2+(y-k)^2=r^2
(h,k)
r
(x-3)^2+(y-2)^2=16
(3,2)
4
\qquad 
\qquad
W
T
Q
W
T
48.5
T
Q
20
Q
W
36
91
W
T
W
Q
19
38
68
123
91
W
T
91
W
T
48.5
W
Q
W
Q
36
K
68
W
Q
ABCD
PQRS
a
PQRS
ABCD
PQRS
ABCD
\overline{BC}
\overline{AD}
\angle ADB
\angle CBD
\qquad \angle CBD = \dfrac{5\pi}8
\pi
a
\qquad a = 0.39
\qquad \quad \text{Natural Gas Usage Versus Average Daily Temperature}
G
T
(^{\circ}\text{F})
\qquad G= 14-0.2T
y=mx+b
m
y
(0,b)
G
T^{\circ}\text{F}
 G= 14-0.2T
-0.2
\qquad m=\dfrac{-0.2}{1}=\dfrac{\Delta G}{\Delta T}=\dfrac{\Delta \text{ natural gas used (in therms)}}{\Delta \text{ temperature} (^{\circ}\text{F}) }
1^{\circ}\text{F}
0.2
(\text{temperature in } ^{\circ}\text{F}, \text{ gas usage in therms})
G
(0,14)
0^{\circ}\text{F}
14
G
4.2
49^{\circ}\text{F}
29^{\circ}\text{F}
8.2
29^{\circ}\text{F}
31
 or 
1^{\circ}F
0.2
G
0^{\circ}F
14
4.2
49^{\circ}F
 \qquad (x+4)(2x-a)=0
a
x=4
x=-4
a
-2
\ \ \ 2
\ \ \ 3
-3
(x+4)(2x-a)=0
a
x+4
x=-4
\dfrac{a}{2}
x=4
4
x
x=\dfrac{a}{2}
a
(x+4)(2x-\red{8})=0
x=-4
x=4
a=8
\qquad \dfrac{          3x^2 + x - 2      }{     3x^2 - 5x + 2     }
x>2
-1
\dfrac 2 3
\dfrac{x+1}{x-1}
\dfrac{x-2}{2-5x}
\dfrac{ \pink{ 3x^2 + x - 2 } }{ 3x^2 - 5x + 2 } 
1
x
0
x>2
\qquad \dfrac{x+1}{x-1}
y=\dfrac{x+4}{2x-5}
y=\dfrac{ax+b}{cx+d}
x
y=\dfrac{a}{c}\;
y=\dfrac{x+4}{2x-5}
2x-5=0
x=\dfrac{5}{2}
x
x+4=0
(-4,0)
y=\dfrac{1}{2}
y=\dfrac{x+4}{2x-5}
\qquad
800
1
4
30{,}000
(\text{c})
1
(\text{hr})
800
1
4
800\cdot4=3{,}200
3{,}200
1
h
30{,}000
\qquad \dfrac{3{,}200\text{ c}}{1\text{ hr}}=\dfrac{30{,}000\text{ c}}{h\text{ hrs}}
h
9.4
30{,}000
(3c-4c^2+c^3)-(5c^2+8c^3-6c)
7c^3 + 9c^2 - 3c
7c^3 + c^2 - 3c
-7c^3 + c^2 + 9c
-7c^3 - 9c^2 + 9c
(3c-4c^2+c^3)
300
1
6
50
n
180
180\ge\dfrac{n}{6}+\dfrac{5n}{6}
180>\dfrac{n}{6}+\dfrac{5n}{6}
300\le6n+\dfrac{6}{5}n
300<6n+\dfrac{6}{5}n
50
\dfrac{50}{60}
\dfrac{5}{6}
n
6
1
\dfrac{1}{6}
n
\dfrac{n}{6}
\dfrac{5}{6}
\qquad\text{time to address }n\text{ invitations}\ge\dfrac{5}{6}n
180
\qquad180\ge\dfrac{n}{6}+\dfrac{5n}{6}
\qquad180\ge\dfrac{n}{6}+\dfrac{5n}{6}
xy
y=2(x-3)^2-7
y=-(x+a)^2+2
a
a
\ \ \ 3
-3
\ \ \ 6
-6
y=a(x-h)^2+k
a
h
k
x=h
y=2(x-3)^2-7
x=3
y=-(x+a)^2+2
x=3
a
-3
a=-3
\qquad 9 \left| -\dfrac z 3 \right| + 6 <  30 
-6<z<6
-8<z<8
-12<z<12
\qquad\text{[lower limit]} < \text{[variable]} < \text{[upper limit]}
2
-1
 \qquad z > -8  \;\; \text{ and } \;\; z < 8 
 \qquad -8 < z < 8 
yD
D
y
D
175 \text{ m}
225
D
(y, D) = (0, 225)
D
D
0
\qquad s(s-1)=2
s=-2
s=1
s=0
s=-1
s=0
s=1
s=2
s=-1
-2
-1
-2
1
2
-1
s=-2
s=1
R(p) = (24627 - p) (287 + 48p/1000)
287
. For each drop of  
 in the average electric vehicle's purchase price, it is expected that the monthly purchase rate will increase by 
.  Which of the following best models the monthly electric vehicle revenue in this country as a function of 
 price reductions of 
R(p) = 24{,}627 (287 - 952p)
R(p) = 1{,}000(24{,}627 - p) (287 + 48p)
R(p) = (24{,}627 - p) (287 + p)
R(p) = (24{,}627 - 1{,}000p) (287 + 48p)
p
 reductions in vehicle price and 
 vehicles to be sold each month at 
 more electric vehicles for each 
 reduction in their average price, we see that a vehicle price of 
 corresponds to 
\qquad R(p)=(24{,}627 - 1{,}000p) (287 + 48p)
g(x)=3\left(\dfrac{1}{2}\right)^{-x}+2
xy
g
y
g
(0,2)
x
g
(0,3)
g
y
\qquad g(x)=3\left(\dfrac{1}{2}\right)^{-x}+2
f(x)=\left(\dfrac{1}{2}\right)^x
x
-2
-1
0
1
2
f(x)
4
2
1
\dfrac{1}{2}
\dfrac{1}{4}
f(x)=\left(\dfrac{1}{2}\right)^x
g(x)=3\left(\dfrac{1}{2}\right)^{-x}+2
f
g
f
y
{\blue{f_2}})
3
(
\red{f_3})
2
(
\green{g})
g
k\ge33
\qquad\dfrac{4k^2-12k+9}{2k^2+19k-33}\cdot\dfrac{k^2+8k-33}{k^2-3k}
\dfrac{2k-3}{k}
\dfrac{8(4k^2+13)}{(2k+19)}
\dfrac{(2k^2+9)(2k^2-33)}{4k^4-57k^2-33}
\dfrac{2k^4+20k^3-73k^2+53k-297}{k(k^3+13k^2-30k+11)}
k
0
k\ge33
\qquad\dfrac{2k-3}{k}
1.5
1.75
18
p
p>41
p>54
p>72
p>90
1.5
18
1.5p+18
1.75
1.75p
(<)
\qquad1.5p+18<1.75p
1.5p
\qquad18<0.25p
0.25
\qquad72<p
p
p
72
\qquad p>72
 and the monthly membership fee for an adult is 
 in membership fees for the month of January, which of the following represents the relationship between the number of student memberships, 
, and the number of adult memberships, 
25x+40y=12{,}250
40x+25y=12{,}250
\dfrac{x}{25}+\dfrac{y}{40}=12{,}250
\dfrac{x}{40}+\dfrac{y}{25}=12{,}250
\qquad [\text{money from student memberships}]+[\text{money from adult memberships}]=[\text{total collected from membership fees}]
 per month and 
 students were members that month, we have that 
 per month, and 
 adults were members that month, we have that 
\qquad  25x+40y=12{,}250
x
y
\qquad 25x+40y=12{,}250
\qquad \dfrac{2}{1+i}
i=\sqrt{-1}
1-i
-1-i
1+i
1-i
a+bi
a
b
1-i
\qquad \dfrac{2}{1+i}=1-i
 per day in addition to tips of 
 of all her customer receipts, 
f(t) = 50 + 20t
f(t) = 300 + 20t
f(t) = 50 + 0.2t
f(t) = 300 + 0.2t
50
6
6
0.2
t
20\%
300
0.2t
f(t)
\qquad f(t)=300+0.2t
f(t) = 300 + 0.2t
xy
\left(\dfrac{5}{8},-\dfrac{6}{5}\right)
\dfrac{7}{10}
\left(x+\dfrac{5}{8}\right)^2+\left(y-\dfrac{6}{5}\right)^2=\dfrac{49}{100}
\left(x+\dfrac{5}{8}\right)^2+\left(y-\dfrac{6}{5}\right)^2=\dfrac{49}{400}
\left(x-\dfrac{5}{8}\right)^2+\left(y+\dfrac{6}{5}\right)^2=\dfrac{49}{100}
\left(x-\dfrac{5}{8}\right)^2+\left(y+\dfrac{6}{5}\right)^2=\dfrac{49}{400}
(\green h,\purple k)
\red r
\qquad(x-\green h)^2+(y-\purple k)^2=\red r^2
\qquad\dfrac{1}{2}\cdot\dfrac{7}{10}=\dfrac{7}{20}
\qquad\left(x-\green {\dfrac{5}{8}}\right)^2+\left(y-\left(\purple {-\dfrac{6}{5}}\right)\right)^2=\red {\left(\dfrac{7}{20}\right)}^2
\qquad\left(x-\dfrac{5}{8}\right)^2+\left(y+\dfrac{6}{5}\right)^2=\dfrac{49}{400}
(x_1,y_1)
(x_2,y_2)
y_1
y_2
y
x
y
y_1=0 \quad \text{or}\quad y_2=-\dfrac{100}{7}
y_1
y_2
\quad 0\cdot-\dfrac{100}{7}=0
\qquad\dfrac{2d^2-3d}{16d^4-81}
d>2
\dfrac{d(2d-3)}{(4d^2-9)^2}
\dfrac{d}{(2d-3)(4d^2+9)}
\dfrac{d}{(2d+3)(4d^2-9)}
\dfrac{d}{(2d+3)(4d^2+9)}
2d^2-3d
d
d(2d-3)
16d^4-81
(4d^2-9)(4d^2+9)
(4d^2-9)(4d^2+9)
(4d^2-9)
\qquad (2d-3)(2d+3)(4d^2+9)
\qquad \dfrac{d(2d-3)}{(2d-3)(2d+3)(4d^2+9)}
d
0
d\neq\purple{\dfrac32}
d>2
\dfrac{d}{(2d+3)(4d^2+9)}
d>2
\qquad f=12gh+15g
g
f
h
g=\dfrac{f}{12h+15}
g=\dfrac{f}{12h-15}
g=\dfrac{f-15}{12h}
g=\dfrac{f}{27h}
g
\green{\text{factor } g \text{ from the expression }12gh+15g}
g
\blue{\text{dividing by }12h+15}
\qquad g=\dfrac{f}{12h+15}
s
150
80
t
t = s + 230
t = 2s + 230
t = \dfrac{5}{2}s + 230
t = 4s + 230
s
s
 s + 150
2s
80
s + s + 150 + 2s + 80
4s + 230
\qquad t=4s+230
t=4s+230
\quad
5
10
5
\qquad(x-1)^2 + (x+1)^2
2x
-2x
2x^2+2
2x^2
-2x
2x
0
\qquad 2x^2+2
t
\qquad {18\cdot16^{0.0125t}}
\blue{A}\cdot \red{B}^{\purple {\large m}\green {\large t}}
\red B
\purple {m}\cdot \green t=1
\red B
2
16
2^4=16
\qquad {18\cdot\left( 2^4\right)^{0.0125t}} 
{\left(x^{\large y}\right)^{\large z}=x^{{\large y}\cdot {\large z}}}
\qquad{18\cdot2^{4\cdot0.0125t}=\blue{18}\cdot\red 2^{\purple{0.05}\green{t}}}
\purple m\cdot \green t=1
\purple m
\purple{0.05}
\purple{0.05}
20
2y=-4x+8
2y=-4x+16
xy
y
y
x
y=mx+b
m
b
y
-2
y
4
-2
y
-8
\qquad\dfrac{2}{x^2-9} + \dfrac{1}{x^2+9x+18}
x>-2
\dfrac{3}{2x^2+9x+9} 
\dfrac{2x+13}{x^2+9x+18} 
\dfrac{3}{x^2+3x-18} 
\dfrac{3}{x^2+9x+18} 
(x-3)(x+3)(x+6)
\qquad \dfrac{2}{(x-3)(x+3)}+\dfrac{1}{(x+3)(x+6)}
(x-3)(x+3)(x+6)
\green{\dfrac{x+6}{x+6}}
\red{\dfrac{x-3}{x-3}}
\qquad\green{\dfrac{x+6}{x+6}}\cdot \dfrac{2}{(x-3)(x+3)}+\red{\dfrac{x-3}{x-3}}\cdot \dfrac{1}{(x+3)(x+6)}
\qquad  =\dfrac{2x+12}{(x-3)(x+3)(x+6)} + \dfrac{x-3}{(x-3)(x+3)(x+6)}
 \qquad  =\dfrac{3x+9}{(x-3)(x+3)(x+6)} 
 \quad  =\dfrac{3(x+3)}{(x-3)(x+3)(x+6)} 
x
0
x\neq{\purple{-3}}
x>-2
\qquad \dfrac{3}{x^2+3x-18}
x
X
y
Y
X
Y
y
x
X
Y
x
y
5
3
xy
5
(5,0)
3
(0,3)
\qquad
y
x
200
15
2
r
2r-2\left(r+15\right)=200
2r-2\left(r-15\right)=200
2r+2\left(r+15\right)=200
2r+2\left(r-15\right)=200
\text{distance}=\text{rate}\cdot\text{time}
r
2
\red{2r}
r+15
2
\green{2\left(r+15\right)}
200
\purple{200}
\qquad\red{2r}+\green{2\left(r+15\right)}=\purple{200}
\qquad{2r}+{2\left(r+15\right)}={200}
80
(^\circ\,\text{F})
40^\circ\,\text{F}
mT
T
m
80
m=0
T=80
m
\red{(0,80)}
40
\qquad
40^\circ\,\text{F}
\qquad
40^\circ\,\text{F}
\qquad
\qquad
^2
^2
180
90\%
360
0.031
M
k
t
t=360
M=0.031
k
t=0
M
3.1
35
, which system can be used to find the number of 
 dimes and 
d
n
35
d+n=35
0.05n
0.1n
0.05n+0.1d=3.3
100
5n+10d=330
\qquad\dfrac{12x-x^2}{x^2-10x-24}
x<10
-\dfrac{x}{x-2}
\dfrac{x}{x-2}
\dfrac{x}{x+2}
-\dfrac{x}{x+2}
x
\qquad 12x-x^2=x(12-x)
\qquad x^2-10x-24=(x-12)(x+2)
\qquad \dfrac{x(12-x)}{(x-12)(x+2)}
12-x=\red{-1}(x-12)
x
0
x\neq\purple12
x<10
-\dfrac{x}{x+2}
x<0
\sqrt{32}+\sqrt{8}=a\sqrt{2}
a
\sqrt{x}
\sqrt{y}
\qquad \sqrt{x\cdot y}=\sqrt{x}\cdot \sqrt{y}
\sqrt{32}
\sqrt{8}
\sqrt{32}+\sqrt{8}=a\sqrt{2}
a=6
\qquad a=6
260
(\text{ft})
3
1
(\text{mi})
1\ \text{mi}=5{,}280\ \text{ft}
21
24
25
29
s
260\text{ ft}
3
\dfrac{260}{3}s
s
1\ \text{mi}
5{,}280\ \text{ft}
\qquad \dfrac{260}{3}s=5{,}280
1\ \text{mi}
61
\qquad\left(x-\sqrt{3}\right)^2(x-\sqrt{7})
x=-\sqrt{3}
x=-\sqrt{7}
x=\sqrt{3}
x=\sqrt{7}
x=3
x=\sqrt{7}
x=-3
x=-\sqrt{7}
0
0
\qquad\blue{(x-\sqrt{3})}\red{(x-\sqrt{3})}\green{(x-\sqrt{7})}=0
\blue{x-\sqrt{3}}=0
\sqrt{3}
\qquad x=\blue{\sqrt{3}}
\red{x-\sqrt{3}}=0
\sqrt{3}
\qquad x=\red{\sqrt{3}}
\green{x-\sqrt{7}}=0
\sqrt{7}
\qquad x=\green{\sqrt{7}}
x=\blue{\sqrt{3}}
x=\green{\sqrt{7}}
\qquad A=\dfrac{TP}{TP+FP+FN+IP}
A
TP
FP
FN
IP
IP=-FP-FN
IP=\dfrac{1}{A}-1-FP-FN
IP=\dfrac{TP}{A}-TP+FP+FN
IP=\dfrac{TP}{A}-TP-FP-FN
IP
\red{\text{multiplying by}}
\blue{\text{divide by }A}
IP
\green{\text{subtracting }(TP+FP+FN)}
\qquad IP=\dfrac{TP}{A}-TP-FP-FN
150
120
24
16
67\%
73\%
78\%
80\%
\qquad \dfrac {\text{total free-throws made}}{\text{total free-throws attempted}}\times 100
136
174
\qquad \dfrac {136}{174}\times 100
78
78\%
\qquad S = 27+6\cdot t\ 
S
t
2010
2010
6
27
6\%
6
y=mx+b
m
x
S=27+6\cdot t
6
S
6
6
3
2
100
50
1
1
4
50
2{,}500
100
2
100
\qquad \dfrac{ 2\:\text{weeks} }{ 100\:\text{blocks} }
50
1
50
b
\qquad \dfrac{ 50\:\text{weeks} }{ b\:\text{blocks} }
b
2{,}500
1
\qquad d = 1.06 ( 212 - t )
d
t
212
212
212
212
212
1.06
(212 - t)
t = 212
d
d
212
212
\qquad\qquad\qquad\qquad\text{U.S. prime lending rates}, (y) 
\qquad\qquad\qquad\qquad\qquad \text{Years since 2004}, (x)
2004
2009
y=-0.65x^2-3.74x+3.17
y=0.65x^2-3.74x+3.17
y=-0.65x^2+3.74x+3.17
y=0.65x^2+3.74x+3.17
2004
2007
2008
2009
y=ax^2+bx+c
a,b
c
a<0
\qquad y=-0.65x^2-3.74x+3.17\qquad
\qquad y=-0.65x^2+3.74x+3.17
a
\qquad y=-0.65x^2-3.74x+3.17\qquad
\qquad y=-0.65x^2+3.74x+3.17
a
c
a
y
c
b
x=\dfrac{-b\ \ }{2a}
x=2.9
\qquad y=-0.65x^2+3.74x+3.17
\qquad\qquad\qquad\qquad\text{U.S. prime lending rates}, (y) 
\qquad\qquad\qquad\qquad\qquad \text{Years since 2004}, (x)
\qquad \dfrac{3|3x+4|}{2}\geq3
-\dfrac43 \leq x \leq \dfrac43
x\le-2  \;\; \text{ or } \;\;  x\ge-\dfrac23
2
2
3
2
-1
\qquad x\leq-2  \;\; \text{ or } \;\;  x\geq-\dfrac23
\qquad x^2-2x+k=0
k
k
1-\sqrt{2}
1+\sqrt{2}
-1
-\dfrac{1}{2}
\dfrac{1}{2}
1
x_1
x_2
x^2-2x+k=0
x^2-(x_1+x_2)x+x_1x_2=0
k
k=\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)=-1
x=1\pm\sqrt{2}
k=-1
110
20
k
j
k
j
110
k+j = 110
20
k=j+20
\qquad\dfrac{d-5}{d-3} + \dfrac{5d-2}{4d-12} = \dfrac{3}{2}
d = \dfrac{3}{4}
d = \dfrac{4}{3}
d = \dfrac{3}{2}
d = 3
4d - 12
\qquad 9d - 22 = 6d - 18
d
d = \dfrac{4}{3}
\qquad\quad \left(w-\dfrac32\right)
\qquad =\left(\dfrac{2w}2-\dfrac32\right)
\qquad =\left(\dfrac{2w-3}2\right)
\qquad\quad \left(\dfrac{2w+7}2\right)
 \qquad ax^2+5x+2=0
a
x=-2
x=-\dfrac{1}{2}
a
-2
\ \ \ 2
\ \ \ 3
-3
 ax^2+5x+2=0
x=-2
x=-\dfrac{1}{2}
 ax^2+5x+2
a
 ax^2+5x+2=0
\qquad(x+2)(2x+1)=0
\red{2}x^2+5x+2=0
a
2
\qquad a=2
2010
2010
645{,}149
48.43
2{,}695{,}598
227.13
362{,}470
35.67
1{,}517{,}550
135.09
2010
13{,}321
11{,}868
10{,}161
11{,}233
2010
12
15
s
f(s) = 12s + 15
f(s) = 15s + 12
f(s) = 12s + 12
f(s) = 15s + 15
s
12s
15
\qquad f(s)=12s+15
f(s) = 12s + 15
\qquad \dfrac{f(t-1)}{f(t)} = 1.0044
f(t)
t
f(t)
\qquad\dfrac23-5x= b x+\dfrac13
b
b
5
0
-5
\dfrac23
\qquad \blue a x+\red b=\blue c  x + \red d
\blue a=\blue c
\red b\neq \red d
\red b=\red d
x
\red {\dfrac23}\neq \red {\dfrac13}
\blue b=\blue {-5}
\blue {5}{x}
x
b,
b=-5
\qquad\sqrt{3}t^2+5t+\sqrt{27}=0
i=\sqrt{-1}
t=\dfrac{-4\sqrt{11}i}{6}
t=\dfrac{-3\sqrt{33}i}{6}
t=\dfrac{-5+\sqrt{11}i}{6}
t=\dfrac{-5\sqrt{3}+\sqrt{33}i}{6}
\qquad a=\sqrt{3},\qquad b=5,\qquad c=\sqrt{27}
\dfrac{\sqrt{3}}{\sqrt{3}}
i=\sqrt{-1}
t=\dfrac{-5\sqrt{3}\green{+}\sqrt{33}i}{6}
t=\dfrac{-5\sqrt{3}\red{-}\sqrt{33}i}{6}
\qquad t=\dfrac{-5\sqrt{3}+\sqrt{33}i}{6}
\text{(cm)}
d
d
4.8
1{,}000
\left(\text{m}^3\right)
10\sqrt[3]{3\pi^2}
20\sqrt[3]{3\pi^2}
30\sqrt[3]{3\pi^2}
40\sqrt[3]{3\pi^2}
\qquad V = \dfrac{1}{3}\pi r^2h
V
r
h
1{,}000 \text{ m}^3
20 \sqrt[3]{3\pi^2}
y<0
z<0
\qquad\dfrac{y-10}{14z^2}\cdot\dfrac{3z}{10-y}
\dfrac{3}{14z}
-\dfrac{3}{14z}
\dfrac{42z^3}{y^2-100}
\dfrac{42z^3}{y^2-20y+100}
\qquad\dfrac{(y-10)}{14\cdot z\cdot z}\cdot\dfrac{3\cdot z}{-1(y-10)}
y
z
0
\qquad -\dfrac{3}{14z}
\qquad  -7(x-8)^2=-63
x=k
x=l
k \cdot l
7
12
13
19
Y=(x-8)
-7Y^2=-63
Y
-7
Y
\qquad Y=\pm 3
x
x
x-8
Y
\qquad x-8=\pm 3
8
\qquad  x=8\pm 3
k=8+3=11
l=8-3=5
\qquad 11\cdot5=55
k\cdot l=55
y=a(x-h)^2+k
(h,k)
a
y=-(x-2)^2+6
(2,6)
a
y+\dfrac{1}{2}x=-5
y=mx+b
m
b
y
-\dfrac{1}{2}
y
-5
(2,6)
-\dfrac{1}{2}
y
-5
25
(\text{g})
40
(\text{ml})
\left(\dfrac{\text{g}}{\text{ml}}\right)
200\ \text{ml}
m
200m=(25)(40)
200m=\dfrac{25}{40}
\dfrac{8}{5}=\dfrac{m}{200}
\dfrac{5}{8}=\dfrac{m}{200}
\dfrac{\text{mass}}{\text{volume}}
\qquad\qquad\dfrac{\text{grams}}{\text{milliliters}}
25\ \text{g}
40\ \text{ml}
\qquad \dfrac{25}{40}=\dfrac58
m\ \text{grams}
200\ \text{ml}
\dfrac{m}{200}
\qquad\qquad\dfrac{5}{8}=\dfrac{m}{200}
\dfrac{5}{8}=\dfrac{m}{200}
5{,}000
500
5{,}000
   Which of the following could be the number of units sold in a month if  the company's profit decreased 
100
900
1{,}000
9{,}000
460
540
4{,}600
5{,}400
u
\qquad |u - 5|
u
5
 for every 
 units more or less than 
u
 if 
 or 
1{,}000
9{,}000
\qquad\qquad a^2 = b^2+c^2-2bc\cos A
A
B
C
a
b
c
\cos A
a
b
c
\cos A = \dfrac{a^2}{b^2+c^2-2bc}
\cos A = a^2-b^2-c^2+2bc
\cos A = \dfrac{a^2-b^2-c^2}{2bc}
\cos A = \dfrac{a^2-b^2-c^2}{-2bc}
-2bc\cos A
\blue{\text{subtracting } a^2 \text{ and}\;\ b^2}
\green{\text{divide both sides by the expression } -2bc}
\cos A
\qquad  \cos A=\dfrac{a^2-b^2-c^2}{-2bc}
xy
4x^2+4y^2-24x=28
(51,14)
(7,10)
(0,0)
(0,7)
(\green h,\purple k)
\red r
\qquad (x-\green h)^2+(y-\purple k)^2=\red r^2
4
4
\qquad x^2+y^2-6x =7
x
x
\qquad(x-3)^2+y^2=16
\sqrt {16}=4
8
8
\angle ADB
\angle CDB
{AD}={CD}=2x-3
ABCD
13\ \text{cm}
\overline{AB}
2\dfrac{1}{2}\ \text{cm}
3\ \text{cm}
3\dfrac{1}{2}\ \text{cm}
4\ \text{cm}
\overline{AD}
\overline{CD}
\angle CDB
\angle ADB
\triangle ABD
\triangle BCD
\overline{AB}
\overline{BC}
ABCD=13\ \text{cm}
(x-1)+(x-1)+(2x-3)+(2x-3)=13
x
\qquad 6x-8=13 \quad\Rightarrow\quad x=\dfrac{7}{2}=3\dfrac{1}{2}
\overline{AB}=x-1
\overline{AB}=2\dfrac{1}{2}\ \text{cm}
 per bottle. The grocery store is currently having a sale on lemonade which advertises 
 bottles for only 
 for 
, so the sale price is 
\sin(x^\circ) = 0.9
\cos(90^\circ-x^\circ)
0.45
0.90
1.12
1.96
90^\circ-x^\circ
x^\circ
(
90^\circ)
\sin(x^\circ) = \cos(90^\circ-x^\circ)
\cos(90^\circ-x^\circ) = 0.9
\qquad\qquad
2006
2007
2008
2009
2010
2011
2012
2013
14.6
16.9
18.6
18.5
18.9
16.6
17.0
17.7
35{,}000
2006
2006
2007
2010
2006
2010
2011
2010
2006
2013
2006
2007
2006
2007
0.16
36
0.03
w
36
 \dfrac{|w-0.16|}{36}=0.03
 \dfrac{|w-5.76|}{36}=0.03
36|w-0.16|=0.03
 36|w-5.76|=0.03
36
\qquad 36(0.16)=5.76
w
w
36
\qquad |w-5.76|
\qquad \dfrac{|w-5.76|}{36}
0.03
0.03
\qquad \dfrac{|w-5.76|}{36}=0.03
\qquad \dfrac{|w-5.76|}{36}=0.03
f(x)=-2\left(\dfrac{1}{3}\right)^x
g(x)=2\left(\dfrac{1}{3}\right)^x
xy
(a,b)
f
g
(a,b)
(-a,b)
(a,-b)
(-a,-b)
f(x)=-g(x)
(x,y)
f
(x, -y)
g
(a,b)
f
(a,-b)
g
1950
2013
\qquad f(x)=0.07x+2.34
x
1950
2040
\qquad\qquad\text{World population in billions of people}, f(x) 
\qquad\qquad\qquad\quad\qquad  \text{Years since 1950}, (x)
5.2
2.34
1{,}252.29
8.64
\qquad f(x)=0.07x+2.34
x
1950
f(x)
f(x)
2040
2040-1950=90
90
x
f(90)
2040
8.64
\qquad\qquad\text{World population in billions of people}, f(x) 
\qquad\qquad\qquad\quad\qquad  \text{Years since 1950}, (x)
a^\circ+b^\circ
73^\circ
a^\circ
180^\circ
180-107=73^\circ
180^\circ
180^\circ
\qquad a^\circ+b^\circ+73^\circ=180^\circ
a^\circ+b^\circ=107^\circ
1836
\qquad h(x) = -0.57x^2+7.98x
h
x
0.57\,\text{m}
7.98\,\text{m}
14\,\text{m}
28\,\text{m}
x
x
(0,0)
(14,0)
14\,\text{m}
14\,\text{m}
x
\angle QSR
\angle RST
90^\circ
m\angle QSR=\theta
m\angle QST=90^\circ
\green{m\angle RST=90^\circ-\theta}
\sin\theta=\cos(90^\circ-\theta)
x
\overline{QR}
\angle QSR
QRS
\theta
\qquad\qquad \sin\theta=\dfrac{\text{Length of the side opposite from }\theta}{\text{Length of the triangle's hypotenuse}}
\theta
\qquad\qquad \cos\theta=\dfrac{\text{Length of the side adjacent to }\theta}{\text{Length of the triangle's hypotenuse}}
QRS
RST
QRS
QR=x
\theta
RS=6\sqrt2
\qquad\qquad \sin\theta=\dfrac{x}{6\sqrt2}
RST
RS=6\sqrt2
90^\circ-\theta
ST=12
\qquad\qquad \cos(90^\circ-\theta)=\dfrac{6\sqrt2}{12}
\sin\theta=\cos(90^\circ-\theta)
\dfrac{x}{6\sqrt2}=\dfrac{6\sqrt2}{12}
x
6
\qquad\dfrac{8c^2+2c^3}{2c^2-32}
c>1
-\dfrac{c}{4}
\dfrac{c^2}{c+4}
\dfrac{c^2}{c-4}
-\dfrac{c^2}{c-4}
8c^2+2c^3
2c^2
2c^2(4+c)
2c^2-32
2
2(c^2-16)
2(c^2-16)
(c^2-16)
\qquad 2(c+4)(c-4)
\qquad \dfrac{2c^2(4+c)}{2(c+4)(c-4)}
\dfrac{4+c}{c+4}=1
c
0
c\neq\purple{-4}
c>0
\dfrac{c^2}{c-4}
c>0
1050
490
70
7
8
14
15
70p
p
\qquad 450 + 70x < 1050
450
\qquad 70x < 600
70
\qquad x < 8.57
8.57
8
8
7
\qquad\dfrac{x^2}{x-2}+\dfrac{4}{2-x}
x\neq2
x+2
x-2
\dfrac{x^2-4}{2-x}
\dfrac{x^2+4}{x-2}
-1
-1
\qquad \dfrac{x^2}{x-2}-\dfrac{4}{x-2}=\dfrac{x^2-4}{x-2}
x
0
x\neq{\green{2}}
\qquad x+2
xy
x=3
y=0
y=\dfrac{1}{x+3}
y=\dfrac{1}{x-3}
y=\dfrac{1}{x}+3
y=\dfrac{1}{x}-3
y=\dfrac{a}{x-k}+h
a
k
h
k\neq x
x=k
y=h
x=3
y=0
y=\dfrac{1}{x-3}
y=\dfrac{1}{x-3}
h
0
y=0
x=3
y=0
\qquad\qquad\qquad y=\dfrac{1}{x-3}\   
4\%
1984
2014
2014
15
1994
13
16
20
29
A(n)
\text{million km}^2
n
\qquad A(n)=A_0(0.96)^n
A_0
1984
3
A(n)=15
\qquad A(3)=15 = A_0(0.96)^3\approx A_0(0.88)
A_0
A_0\approx17.05
1984
17.05\,\text{million km}^2
1994
16\,\text{million km}^2.
16\,\text{million km}^2
\qquad\qquad 
8
5
160
h
5
8
12
40
5
\qquad V=(\text{area of base})(5)=160
32
b
h
\dfrac 12 bh
8
\qquad 32=\dfrac12(8)h
h=8
8
100
250
(\text{K})
360 \, \text{K}
T_I
T_F
T_I - T_F
5
0
-5
5
\qquad \dfrac{T_F}{250} = \dfrac{360}{T_F}
T_F
5
p
15\,\text{ft}^2.  
200\,\text{ft}^2
a
a = 15(200 - p)
a = 200 - (15 + p)
a = 200p - 15
a = 200 - 15p
15p
p
200
\qquad a=200-15p
a = 200 - 15p
(100^\circ \text C)
20^\circ \text C
60^\circ \text C
4
T
t
T(t)=20+80\cdot(0.84)^t
T(t)=80+20\cdot(0.84)^t
T(t)=80\cdot(0.84)^t
T(t)=100\cdot(0.84)^t
t=0
T(t)
100
\qquad T(t)=80\cdot(0.84)^t
\qquad T(t)=100\cdot(0.84)^t
t=0
t=0
t=4
\qquad T(t)=20+80\cdot(0.84)^t
\qquad T(t)=20+80\cdot(0.84)^t
 and expects to sell 
 increase 
 in the price of the sneakers will lose the company 
, based on the number, 
, of 
\green A(x-\red h)^2+\blue k
x=\red h
\blue k
R=\green{-750}(x-\red4)^2+\blue{192{,}000}
x=\red 4
\green A
\red4
 increase over the original price of 
, or 
 \qquad r=\sqrt{x^2+y^2}
r
(x,y)
x
y
x =  \sqrt{r^2}-y^2
x =  \sqrt{r^2-y^2}
x =  \sqrt{r^2+y^2}
x = r-y
x^2+y^2
\purple{\text{square both sides of the equation}}
\sqrt[2]{a}=\sqrt{a}
x^2
\blue{\text{subtracting } y^2 \text{ from both sides of the equation}}
x
\red{\text{take the square root of both sides of the equation}}
x
\qquad x =  \sqrt{r^2-y^2}
59
(^{\circ}F)
48.3^{\circ}F
3000 \text{ feet} (\text{ft})
T
^{\circ}F
h
59^{\circ}F
T(h)=59-3.6h
T(h)=59-10.7h
T(h)=59-0.0107h
T(h)=59-0.0036h
f(x)=mx+b
m
b
59^{\circ}F
10.7^{\circ}F
3000 \text{ ft}
-\dfrac{10.7^{\circ}F}{3000\text{ ft}}
-0.0036^{\circ}F \text{ per foot}
\qquad T(h)=-0.0036h+59
T(h)=59-0.0036h
h
59^{\circ}F
\qquad(x^a)^b = x^{ab}
\qquad\dfrac1{x^t}=x^{-t}
\qquad 7^{3b}
4^{-3a}
7^{3b}
20\%
x
y
40\%
10\%
x
y
0.4x+0.1y=20
0.4x+0.1y=x+y
0.20(x+y)=x+y
0.2(x+y)=0.4x+0.1y
\qquad[\text{amount of liquid}]\cdot [\text{percent juice}]=[\text{total amount of juice}]
x
40\%
0.4x
y
10\%
0.1y
0.4x+0.1y
20\%
x+y
20\%
0.2(x+y)
\qquad 0.4x+0.1y=0.2(x+y)
x
y
\qquad 0.4x+0.1y=0.2(x+y)
48
(\text{ft})
65\,\text{ft}
x\,\text{ft}
A(x) = (48 - x)(65 - x)
A(x) = (48 - 2x)(65 - 2x)
A(x) = (48 + x)(65 + x)
A(x) = (48 + 2x)(65 + 2x)
\qquad A(x) = (\text{lengh})(\text{width})
x
\red{L}
\red{W}
\red{LW}
\red{L} = 65 - 2x
\red{W} = 48 - 2x
\red{LW} = (65 - 2x) (48 - 2x)
\qquad A(x) = (65 - 2x) (48 - 2x)
\cos(210^{\circ})
-\dfrac{\sqrt{3}}{2}
-\dfrac{1}{2}
\dfrac{1}{2}
\dfrac{\sqrt{3}}{2}
\theta
P
P= (\cos{\theta}, \sin{\theta})
210^{\circ}
\left(-\dfrac{\sqrt{3}}{2},-\dfrac{1}{2}\right)
x
\cos{210^{\circ}}=-\dfrac{\sqrt{3}}{2}
\qquad9x+3 =cx + 3+4x
c
c
5
9
6
7
\qquad \blue a x+\red b=\blue c  x + \red d
\blue a=\blue c
\red b=\red d
x
\red 3
x
\qquad \blue{c+4}=\blue{9}
c=5
\qquad w=\sqrt{108w}
w
0
w=0
w=108
w=\blue0
w=0
x = \blue{108}
w=0
w=108
0+108=108
108
\qquad \qquad \qquad\qquad \text{World Production of Copper}
c
1900
2010
t
1900
1995
0.5
9
11
13
t=95
c=11.2
1995
11
 \qquad A = \dfrac{ \pi r^2 \theta }{ 360 } 
A
r
\theta
r = \sqrt{\dfrac{360A}{\pi \theta}}
r = \dfrac{360A}{ \pi r \theta}
r = \sqrt{\dfrac{\pi \theta}{360A}}
r = \dfrac{ \pi \theta}{360Ar}
\blue{\text{multiply by } 360}
\pink{\text{divide both sides by } \pi \theta}
\qquad r = \sqrt{\dfrac{360A}{{\pi \theta}}} 
\qquad(a-4)(6a^2+10a-5)
6a^3-14a^2-45a+20
6a^3-14a^2-35a+20
16a^3-24a^2-45a+20
16a^3-24a^2-35a+20
(a-4)
(6a^2+10a-5)
\qquad a(6a^2) + a(10a)+a(-5)-4(6a^2)-4(10a)-4(-5)
\qquad 6a^3 + 10a^2-5a-24a^2-40a+20
\qquad 6a^3-14a^2-45a+20
A
\overline{AC}
8
\overline{AD}
\overline{BC}
E
\overline{DE}
1
\overline{BC}
2\sqrt{15}
8
2\sqrt{17}
9
\overline{AB}
\overline{AC}
\overline{AD}
8
\overline{AE}
AD - ED = 8 - 1 = 7
\overline{AD}
\overline{BC}
AEC
E
\qquad AE^2+CE^2 = AC^2
AC
AE
CE
\overline{AC}
\overline{AB}
AEC
AEB
AEC
AEB
\qquad BE = CE = \sqrt{15}
\overline{BC}
BE
CE
\overline{BC}
2\sqrt{15}
(x,y)
y=x-7 
(4,1)
(9,6)
(5,-2)
(8,1)
(5,-2)
(9,6)
(3,6)
(8,1)
y
x
x
\qquad\qquad\qquad (x-6)^2-3=x-7
 (x-6)^2-3=x-7
 
x=5
x=8
x
x
 
(5,-2)
(8,1)
(x,y)
y=(x-6)^2-3
y=x-7
hm
m
h
h
12
20
20
20
12
h
(h,m) = (20, 0).
0
20
h = 20
20
 \qquad Q=\sqrt{\dfrac{2KD}{F}}
Q
K
D
F
D = \sqrt{ \dfrac{2FQ}{K}}
D =  \sqrt{\dfrac{FQ}{2K}}
D =  \dfrac{2FQ^2}{K}
D =  \dfrac{FQ^2}{2K}
\purple{\text{squaring both sides of the equation}}
\sqrt[2]{x}=\sqrt{x}
F
\blue{\text{multiplying both sides of the equation by } F}
\green{\text{divide both sides of the equation by } 2\cdot K}
\qquad D =  \dfrac{FQ^2}{2K}
\qquad
175
120
50
\pi\approx 3.14
\qquad V_{\text{cylinder}}+ V_{\text{cone}}\approx2{,}166{,}600
2{,}166{,}600
\qquad \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}
V
T
(1)
(2)
T_1 = T_2
T_1 = V_1 - \dfrac{V_2}{T_2}
T_1 = \dfrac{V_2}{T_2V_1}
T_1 = \dfrac{T_2V_1}{V_2}
T_1
\blue{\text{multiplying both sides by } T_1}
\pink{\text{multiply both sides by } T_2}
\green{\text{divide both sides by } V_2}
\qquad T_1 = \dfrac{T_2V_1}{V_2}
\qquad A(x)=8-\dfrac{1}{4}x
B(x)=\dfrac{2}{5}-\dfrac{x}{16}
A(x)-4B(x)
-\dfrac{32}{5}
\dfrac{32}{5}
-\dfrac{32}{5}+\dfrac{1}{2}x
\dfrac{32}{5}+\dfrac{1}{2}x
\qquad A(x)-4B(x)=\left(8-\dfrac{1}{4}x\right)-4\left(\dfrac{2}{5}-\dfrac{x}{16}\right)
-4
\qquad A(x)-4B(x)=\dfrac{32}{5}
x
-\dfrac{1}{2} + \dfrac{1}{4}x=10
\qquad 42
\dfrac{1}{2} <7ax+\dfrac{1}{4}<8
14ax+\dfrac{1}{2} 
4
\dfrac{1}{4}
\dfrac{1}{4}
4
16
1
1
16
 \dfrac{1}{2} <7ax+\dfrac{1}{4}<8 
 14ax+\dfrac{1}{2} 
14ax+\dfrac{1}{2}=2\left(7ax+\dfrac{1}{4}\right) 
14ax+\dfrac{1}{2}=2\left(7ax+\dfrac{1}{4}\right) 
14ax+\dfrac{1}{2} 
2
a<x<b
b>a
x
a
b
x<2
x>5
5<x<2
14ax+\dfrac{1}{2} 
1
16
\qquad12s\leq4(s-2)
s\leq-1
s\leq-\dfrac23
s\geq-\dfrac14
 s \geq 8 
\purple{\dfrac14}
\pink s
\blue 2
1
1
1
 x - 6 + 6 
 6 
s\leq-1
 for the first pound of a package and 
 for each additional pound of the package. If Robin wants to spend under 
 on shipping, which inequality best represents how many pounds, 
3.46 + 1.53p<100
3.46 + 1.53(p-1)<100
3.46 + 1.53p - 1<100
3.46(p-1) + 1.53<100
\text{lb}
\text{cost of first lb} + \text{additional lbs } (\text {cost per additional lb}) <\text{total}
. The number of additional pounds, however, is not 
, because we would double count the first pound in this case. We can actually represent the number of additional pounds using 
\qquad3.46+1.53(p-1)<100
\qquad \sqrt{16x^2y^{16}+25x^{16}y^2}
x
y
4xy^8+5x^8y
4xy^4+5x^4y
xy\sqrt{16y^{14}+25x^{14}}
x^2y^2\sqrt{16y^{14}+25x^{14}}
x^2y^2 
\qquad \sqrt{16x^2y^{16}+25x^{16}y^2}=\sqrt{x^2y^2\left(16y^{14}+25x^{14}\right)}
a
b
\qquad \sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{b}
\qquad xy\sqrt{16y^{14}+25x^{14}}
200
200
19^\text{th}
20^\text{th}
14
4
20
20^\text{th}
200
4
20
24
200
200
19^\text{th}
20^\text{th}
\red x
\blue{14}
4
\green{20}
\purple{24}
\red{200+\text{ page books from the }20^\text{th}\text{ century}}
\green{\text{all books at least }200\text{ pages}}
\dfrac{\red x}{\green{20}}
\red x
200
20^\text{th}
\blue{20^\text{th}\text{ century books}}
\purple{\text{all books}}
\dfrac{\blue{14}}{\purple{24}}
58.\overline3\%
\dfrac{\red x}{\green{20}}
58.\overline3\%
\qquad \dfrac{\red x}{\green{20}} > 0.58\overline3
\green{20}
11.\overline6
12
\red{200+\text{ page books from the }20^\text{th}\text{ century}}
\blue{\text{all books from the }20^\text{th}\text{ century}}
\dfrac{\red x}{\blue{14}}
\green{\text{books more than }200\text{ pages}}
\purple{\text{all books}}
\dfrac{\green{20}}{\purple{24}}
8\overline3\%
\dfrac{\red x}{\blue{14}}
8\overline3\%
\dfrac{\red x}{\blue{14}} > 0.8\overline3
\blue{14}
11.\overline6
12
200
20^\text{th}
12
13
14
\qquad V=\dfrac{s^2h}{3}
h
s
30\%
30\%
\,69\%
\,169\%
\,300\%
30\%
s
1.3s
\qquad V=\dfrac{(1.3s)^2h}{3}=\dfrac{(1.69)s^2h}{3}=1.69\left(\dfrac{s^2h}{3}\right)
69\%
69\%
O
1.4
AC
2.1
OB
AC
AOC
l
AC
2.8 \sin(0.75)
2.8 \cos(0.75)
1.4 \sin(1.5)
1.4 \cos(1.5)
\qquad s = r \theta
s
r
\theta
s
r
\theta
AOC
1.5
OB
AOC
BOC
AOC
\dfrac{1.5}{2} = 0.75
OBC
BOC
BC
AO
CO
OBA
OBC
OB
OBA
OBC
AB
BC
1.4 \sin(0.75)
AC
2(1.4 \sin(0.75)) = 2.8 \sin(0.75)
\qquad F = P + Prt
F
t
P
r
P = \dfrac{F}{1 + rt}
P = \dfrac{F}{1 - rt}
P = \dfrac{F + P}{rt}
P = \dfrac{F - P}{rt}
P
\blue{\text{divide both sides by } (1 + rt)}
\qquad P = \dfrac{F}{1 + rt}
\qquad A(t) = t^2 - t - 30
B(t)=t-6
A(t)\cdot B(t)
t^2+6t+180
t^2-30t+180
t^3-7t^2-24t+180
t^3-5t^2-36t+180
A(t)
B(t)
\qquad(t^2-t-30)(t-6)
\qquad t^2(t)+t^2(-6)-t(t)-t(-6)-30(t)-30(-6)
\qquad t^3-6t^2-t^2+6t-30t+180
\qquad t^3\blue{-6t^2}\blue{-t^2}\red{+6t}\red{-30t}+180
\qquad t^3\blue{-7t^2}\red{-24t}+180
\qquad t^3-7t^2-24t+180
\overline{DB}
6
(\text{ft})
\overline{AE}
20\,\text{ft}
\sin(\theta) \approx 0.616
\cos(\theta) \approx 0.788
\overline{EC}
12.32\,\text{ft}
15.76\,\text{ft}
25.38\,\text{ft}
32.47\,\text{ft}
\pi
AEC
AE
20
EC
\overline{EC}
3b(b-4)
6(b-8)
3b(1-2b-16)
3b(b-2-16)
3(b^2-2b-16)
3(b^2-2-16b)
\qquad3b(b-4) = 3b^2-12b
\qquad6(b-8) = 6b-48
0
\qquad3b^2-6b-48
3
\qquad3(b^2-2b-16)
JKL
RST
t^\circ
50^\circ
55^\circ
60^\circ
65^\circ
q^\circ
\overline{RT}
180
q^\circ
180
t^\circ
t^\circ
60^\circ
\qquad 1-9b^2=0
b=\pm\dfrac{1}{3}
b=\pm\dfrac{1}{9}
b=\dfrac{1}{3}
b=\dfrac{1}{9}
0
\qquad (1-3b)(1+3b)=0
\qquad 1-3b=0 \qquad \text{or} \qquad 1+3b=0
b
b=\dfrac{1}{3}
b=-\dfrac{1}{3}
b=\pm\dfrac{1}{3}
\qquad\quad \left(2+\dfrac{1}{3}d\right)\cdot\left(5d+\dfrac{1}{3}d^2\right)
\qquad =5d\left(2+\dfrac{1}{3}d\right)+\dfrac{1}{3}d^2\left(2+\dfrac{1}{3}d\right)
\qquad =5d\cdot2+5d\cdot\dfrac{1}{3}d+\dfrac{1}{3}d^2\cdot 2+\dfrac{1}{3}d^2 \cdot \dfrac{1}{3}d
\qquad =10d+\dfrac{5}{3}d^2+\dfrac{2}{3}d^2+\dfrac{1}{9}d^3 
\qquad =10d+\left(\dfrac{5}{3}+\dfrac{2}{3}\right)d^2+\dfrac{1}{9}d^3 
\qquad =10d+\dfrac{7}{3}d^2+\dfrac{1}{9}d^3 
p\cdot p+p
\qquad\quad\;\: p\cdot p+p
\qquad=\left(\dfrac{1}{3}h\right) \cdot \left(\dfrac{1}{3}h\right) + \left(\dfrac{1}{3}h\right)
\qquad =\dfrac{1}{3}h\cdot \dfrac{1}{3}h+ \dfrac{1}{3}h \cdot 1
\dfrac13h
\qquad =\dfrac{1}{3}h\left(\dfrac{1}{3}h + 1\right)
g(x)=8x-5
g(g(x))
64x-10
64x-45
64x^2+25
64x^2-80x+25
g(g(x))
g(x)
g
\qquad g(g(x))=8(g(x))-5
g(x)=8x-5
g(g(x))=64x-45
2^{20}
2^{625}
32^2
32^{12}
(a^m)^n=a^{mn}
32=2^5
32^4=\left(2^5\right)^4
(a^m)^n=a^{mn}
2^{20}
2^{20}
(x,y)
y
x
y
(x,y)
x
3
y
\qquad y = \dfrac{-b\pm\sqrt{b^2-4ac} }{2a}
ay^2+by+c=0
y
\qquad y=\dfrac{3}{2}
\qquad =\blue{u^2}(u+t)+\blue{2ut}(u+t)+\blue{t^2}(u+t)
(u+t)
u^3
t^3
3u^2t+3ut^2
\qquad 3u^2t+3ut^2=3ut(u+t)
\qquad m^2+6m+10=0
i=\sqrt{-1}
\text{I}
\text{I}
\text{II}
\text{I}
\text{III}
\text{I}
\text{II}
\text{III}
\qquad a=1,\qquad b=6,\qquad c=10
i=\sqrt{-1}
\qquad m=-3+i, m=-3-i
\text{I}
\text{II}
\qquad \text{I}
\text{II}
(x,y)
x
\ \ \ 50
-50
-5
\ \ \ 5
x
y
y
\qquad\qquad\qquad y^2+25=10y
y
y=5
y
x
y
x
50
70\%
 per share in 
.  Knowing that such growth is not sustainable, the adviser makes a prediction for 
 based on the average rate of change, which was 
 per year.  To the nearest dollar, how much higher would the financial adviser's prediction be for 
 if the adviser assumed the earnings would continue to grow 
y
2011
 and increase by 
, 
 in 
a\cdot b^{\large{t}}
a
b
t
a
 and 
 each year, then each year they are worth 
 of their worth from the previous year.    That is the same as multiplying the earnings by 
 each year.  So, 
 in 
\qquad133-36=97
 higher if the adviser assumed the earnings would continue to grow 
 9<15mx-8<27
m
 \dfrac{8}{3}-5mx 
-3
-9
-9
-3
-\dfrac{7}{3m}
-\dfrac{17}{15m}
-\dfrac{17}{15m}
 -\dfrac{7}{3m}
 9<15mx-8<27
 \dfrac{8}{3}-5mx 
 \dfrac{8}{3}-5mx=\dfrac{15mx-8}{-3} 
 \dfrac{8}{3}-5mx=\dfrac{15mx-8}{-3} 
 \dfrac{8}{3}-5mx  
-3
a<x<b
b>a
x
a
b
x<2
x>5
5<x<2
 \dfrac{8}{3}-5mx 
-9
-3
a
(\text{m/s}^2)
0
9.8
\text{m/s}^2
4
(\text{mi})
d
da
0
0
0 \, \text{m/s}^2
(0, 0)
d
(4, \;9.8)
(4, \;9.8)
a
-\dfrac{9}{14}
-\dfrac{28}{9}
\dfrac{9}{14}
\dfrac{28}{9}
(
a\neq0)
y
y
\qquad a=\dfrac{28}{9}
A
A = \pi(R^2-r^2)
R
r
\pi
R
A
r
R = \sqrt{\dfrac{A+r^2}{\pi}}
R = \sqrt{\dfrac{A}{\pi}}+r
R = \sqrt{\dfrac{A}{\pi}-r^2}
R = \sqrt{\dfrac{A}{\pi}+r^2}
R^2-r^2
\blue{\text{dividing both sides of the equation by }\pi}
R^2
\green{\text{adding } r^2 \text{ to both sides of the equation}}
R
\purple{\text{take the square root of both sides of the equation}}
\sqrt[2]{x}=\sqrt{x}
\qquad R = \sqrt{\dfrac{A}{\pi}+r^2}
\qquad 11-x=\sqrt{11-x}
x=0
x=1
x=10
x=11
x=11
x=12
x=1
x=10
x
0
x=10
x=11
x=\blue{10}
x=0
x = \blue{11}\,
x=10
x=11
x=10
x=11
5
(\text{ft})
h
n^{\text{th}}
0.3\text{ ft}
0.5\text{ ft}
1\%
12\%
A(t)=A_0\cdot (1-r)^t
A_0
r
A
t
5\text{ ft}
(4,3)
r
0.12
12\%
(1-r)^{4}=0.6
r
4^{\text{th}}
\dfrac{1}{4}
12\%
h
m
\qquad h(m) = -\dfrac{1}{50}m^2 +200
m
 -\dfrac{1}{50}(m+100)(m-100)
 -\dfrac{1}{50}(m^2 - 10{,}000)
 -\dfrac{1}{50}(m+40)(m-40) +168
-\dfrac{1}{50}(m+50)(m-50) +150
\qquad h(m) = -\dfrac{1}{50}(m^2 - 10{,}000)
(m^2 -10{,}000)
\qquad h(m) = -\dfrac{1}{50}(m+100)(m-100)
\qquad h(m)=-\dfrac1{50}(m+100)(m-100)
\qquad4-3u\geq|9u+2|
 -1 \leq u \leq \dfrac16 
 -\dfrac12\leq u\leq\dfrac16
-2\leq u \leq2
\qquad\text{[lower limit]} \leq \text{[variable]} \leq \text{[upper limit]}
2
 -1
(1)
\blue{3u}
\purple2
\pink{12}
(2)
\blue{3u}
\purple2
\gray6
 \qquad u \leq \dfrac16  \;\; \text{ and } \;\; u\geq -1 
 \qquad-1 \leq u \leq \dfrac16 
\qquad2(x+55)(x-17)
x=-55
x=17
x=-55
x=-2
x=17
x=-17
x=55
x=-17
x=2
x=55
0
0
\qquad\blue2\red{(x+55)}\green{(x-17)}=0
\red{x+55}=0
55
\qquad x=\red{-55}
\green{x-17}=0
17
\qquad x=\green{17}
\blue{2}=0
x
x=\red{-55}
x=\green{17}
\qquad {\large\ell}(x) = x^4+36x^2-10{,}000
(x+10)
{\large\ell}(x)
(x+10)
{\large\ell}(-10)
{\large\ell}(x)
(x+10)
3{,}600
v
1.3
2
0.2
0.2v^2 - 2.2v - 2.6 = 0
1.3v^2 - 2v - 0.2 = 0
0.2(v^2 - 3) + 2.6 = 0
3(0.2 - v) (2.6 - v) = 0
v = \dfrac{d}{t}
v
t_1
t_2
t = t_1 + t_2
0.2
1.3
v
1.3
2
v + 2
\qquad t_2 = \dfrac{1.3}{v +2}
\qquad 0.2 = \dfrac{1.3}{v} + \dfrac{1.3}{v + 2}
v
v \ne 0
v \ne -2
\qquad 0.2v^2 - 2.2v - 2.6 = 0
c
y
z
(y,z)
4
\qquad 4y-12z=12
c=12
c
12
308
31.3\%
96
212
676
984
m
\qquad 0.313m=308
0.313
\qquad m = 984
984
tP
P
t
1^{\text{ st}}
1^{\text{ st}}
1^{\text{ st}}
0
1^{\text{ st}}
t
0
P(0)
P
P
P
17
17
\text{Mark}
\text{Daniel}
\text{Gallons of fuel used}
13
11
\text{Miles traveled}
286
m
m
(\text{mi})
(\text{gal})
286:13
m
11
\qquad \dfrac{286\text{ mi}}{13 \text{ gal}}=\dfrac{m\text{ mi}}{11\text{ gal}}
m
242\text{ miles}
111
3
136
26
53
258
17
\red{\text{least value}}
\blue{\text{greatest value}}
\red3
\blue{258}
\qquad\blue{258}-\red{3}=255
255
\qquad \dfrac{x}{x-2} + \dfrac{x}{(x-2)(x-3)} = \dfrac{4}{x-3}
2\ 
4\ 
2
3
2
4
\qquad\blue{(x-2)}\pink{(x-3)}
\qquad \dfrac{\pink{(x-3)}x}{\pink{(x-3)}(x-2)} + \dfrac{x}{(x-2)(x-3)} = \dfrac{\blue{(x-2)}4}{\blue{(x-2)}(x-3)}
\qquad \dfrac{x^2 -3x}{\pink{(x-3)}(x-2)} + \dfrac{x}{(x-2)(x-3)} = \dfrac{4x - 8}{\blue{(x-2)}(x-3)}
\qquad x^2 - 3x + x = 4x - 8
x = 4
x = 2
x = 4
x=4
x = 2
0
x=2
x=4
\qquad x^2-10x+21=0
x_1
x_2
x_1>x_2
x_1-x_2
-1
-\dfrac{1}{4}
\dfrac{1}{2}
1
b^2-4ac
x^2
21
-10x
\qquad x^2-10x+21=(x-7)(x-3)
\qquad x-7=0\qquad \text{or}\qquad x-3=0
x=7
x=3
x_1-x_2
x_1>x_2
x_1-x_2=4
2
40
7\text{ ft}
8\text{ ft}
16\text{ ft}
18\text{ ft}
2
a
b
A
\qquad  A=a \cdot b
40 \, \text{ft}
b
a
\qquad b = 40 - a
b
{-1}
a^2
y = a(x - h)^2 + k
(h, k)
400
2^3=8
n+p
n-p
\left(\dfrac{2p-2n}4\right)
2
\left(\dfrac{p-n}2\right)
\left(\dfrac{n-p}2\right)
\qquad r=0.5(1.2)^{\theta}
\theta
x
r
0.5
\qquad
0.5
2\pi
0.5
2\pi
0.5
\theta=0
0.5
\theta=2\pi
\qquad r=0.5(1.2)^{\theta}
\theta=0
\qquad r=0.5(1.2)^{0}=0.5
0.5
\theta=0
0.5
\theta=0
\dfrac{1}{3}
\dfrac{1}{3}
\dfrac{1}{2}
580
570
b
g
\dfrac{1}{2}
\rightarrow \dfrac{1}{2}b
 \dfrac{1}{3}
\rightarrow \dfrac{1}{3}  g
\qquad \dfrac{1}{2} b + \dfrac{1}{3} g = 580
\qquad \dfrac{1}{3}b + \dfrac{1}{2}g = 570
2f
\dfrac f2
D
P
D
2D
20
260
300
13.2
9.5
1.5
10
6
6
10
14
11
11
14
20
13.2
1.5
13.2 \cdot 1.5
260
10
6
(x_1,y_1)
(x_2,y_2)
y_1
y_2
x
y
(x,y)
y
y
x
x_1=5
x_2=8
x
y
x_1=5
x_2=8
x_1=5
x_2=8
(5,65)
(8,74)
y
y_1+y_2
\qquad74+65=139
 hats.  If Aiden purchases only t-shirts, he will get 
 t-shirts.  Which of the following best models the relationship between the number of hats, 
, and the number of t-shirts, 
40h+80t=400
80h+40t=200
5h+2.5t=200
2.5h+5t=200
[\text{cost per hat}]\cdot [\text{number of hats}]+[\text{cost per t-shirt}]\cdot [\text{number of t-shirts}]=[\text{total cost}]
 on hats, he will get 
 hats.  Therefore, we can divide 
 by 
 to see that each hat costs 
 on t-shirts, he will get 
 t-shirts.  Dividing 
 by 
 shows us that each t-shirt costs 
\qquad 5h+2.5t=200
5h+2.5t=200
h
t
60^\circ
67
(\text{in})
33.5\sqrt{3}\,\text{in}
33.5\sqrt{2}\,\text{in}
33.5\,\text{in}
\dfrac{134\sqrt{3}}{3}\,\text{in}
30^\circ-60^\circ-90^\circ
67
60^\circ
h
33.5\sqrt{3}\,\text{in}
6
50\%
6
11\,\text{p.m.}
20\,\text{mg}
2\,\text{oz}
80\,\text{mg}
5\,\text{a.m.}
11\,\text{a.m.}
11{:}30\,\text{a.m.}
12\,\text{p.m.}
C(h)
h
\qquad C(h)=80\cdot(0.50)^{\large\frac h6}
\,5\,\text{a.m.}
h=18
\qquad C(18)=80\cdot(0.50)^{\large\frac {18}6}=10\,\text{mg}
2
(h=12)
C(h)=20
12
11{:}30\,\text{a.m.}
12\,\text{p.m.}
h=12
11\,\text{a.m.}
11\,\text{a.m.}
\qquad \sqrt {y+k}=y+1
k
y=1
y=\red1
k
3
\qquad \sqrt {y+3}=y+1
0
y=1
y=-2
y=\green1
y=1
y=\green{-2}
y=-2
y=1
k=3
y=1
\qquad 1\,\text{T} = 1 \dfrac{\text{Wb}}{\text{m}^2}
P
6 \cdot 10^6
1
10^6
6 \cdot 10^{-6}
6 \cdot 10^0
6 \cdot 10^{12}
6 \cdot 10^{18}
6 \cdot 10^6
S
\dfrac{\text{Wb}}{\text{m}^2}
\text{M}
1{,}000{,}000
10^6
S
6 \cdot 10^{-6}
\qquad\dfrac{-7}{9-20p} = \dfrac{3}{4}
p
-\dfrac{12}{11}
-\dfrac{11}{12}
\quad \dfrac{11}{12}
\quad\dfrac{12}{11}
\qquad(-7)(4) = (3)(9-20p)
\qquad -28 = 27 - 60p
p
\qquad\dfrac{11}{12} = p
\qquad p = \dfrac{11}{12}
\qquad \qquad \qquad \qquad \quad \text{ Online Radio Trends}
2007
2015
P
t
2007
t
P
P=t+20
P=2.3t+20
P=0.23t+18
P=4.3t+18
y=mx+b
m
(0,b)
y
y
P
P
(0, 18)
(0, 18)
(3,30)
P
P=mt+b
P=4t+18
P=4.3t+18
t
P
P=4.3t+18
12
1
(\text{in})
4
(\text{hrs})
4\ \text{in}
480
500
560
600
4
12
240
12
480
\qquad240+240=480
4
\qquad\dfrac{240\ \text{min}}{12\ \text{in}} = \dfrac{x}{4\ \text{in}}
80
4
\qquad x=\dfrac{4\times240}{12}=80
480+80=560
\dfrac13
x
(\text{cm})
s
\dfrac13
a=\dfrac s3
b=\dfrac{2s}{3}
30\,\text{cm}
12\,\text{cm}
f(x)=2x+3
g(x)=x^2-4x
g(f(x))
4x^2+4x-3
4x^2-8x-3
4x^2+4x+21
2x^2-8x+3
g(f(x))
f(x)
g
\qquad g(f(x))=(f(x))^2-4(f(x))
f(x)=2x+3
g(f(x))=4x^2+4x-3
0
0.00
0.00
0.00
5
0.25
0.27
0.28
10
0.50
0.54
0.56
15
0.75
0.81
0.83
20
1.00
1.08
1.11
25
1.25
1.35
1.39
110\text{ in}:3\text{ in}
9
(\text{in})
28
37
330
3{,}960
110\text{ in }{:}\:3 \text{ in }
9
l
\qquad \dfrac{110\text{ in} }{3\text{ in} }=\dfrac{l}{9\text{ in}}
l
12
1
12
\qquad\dfrac{330}{12}=27.5\approx28
28
-7
\qquad\quad 4^2(20k)
\qquad=4^2\cdot20k
\qquad=20k \cdot 4^2
\qquad 20k\cdot4^2 - 7
\qquad \dfrac{5+7i}{6-3i}
i=\sqrt{-1}
\dfrac{9+57i}{45}
\dfrac{9+57i}{3}
\dfrac{51+57i}{45}
\dfrac{51+57i}{3}
\green{\text{complex conjugate}}
a+bi
a-bi
6-3i
\qquad 6-(-3i)=\green{6+3i}
\blue{\text{ identity}}
i=\sqrt{-1}
i^2=-1
\qquad\dfrac{9+57i}{45}
400
?
286
350
290
5
325
299
\qquad i^3 + i^2
i=\sqrt{-1}
-1
-2
-1 + i
-1 - i
i^2 = -1
\qquad -1-i
\qquad T = \dfrac{(d - 15)^2}{300} + 20
T
d
1
\qquad T = a(d - d_0)^2+T_0
a
(d_0, T_0)
(d, T) = (15, 20)
20
20
24
c
a
24
24
\qquad a + c \leq 24
c
a
a
c
\qquad a \leq c
\qquad
xy
\text{I}
\text{II}
\text{I}
\text{III}
\text{I}
\text{II}
\text{III}
2
x
2
\text{II}
y=(x-4)^2
2
x
(
1
3)
y=(x-3)(x+3)
3
-3
\text{I}
y=(x+2)(x+7)
-2
-7
\text{III}
\text{I}
(
\text s)
(
\text {ft})
1
90
3
730
(\text{ft})
t
h(t) = at^2 + c
2
a
c
(t, h) = (1 \, \text{s}, 90 \, \text{ft})
(t,h) = (3, 730)
{-1}
c
, and plugging that into 
 and 
 equal 
 and 
2
t
2
330
2
\qquad A = 4 \pi r^2
A
r
8{,}000
(\text{mi})
10 \text{ mi}
64{,}000{,}000 \pi
160{,}000{,}000 \pi
640{,}000{,}000 \pi
2{,}560{,}000{,}000 \pi
V
A
t
\qquad V = At
\qquad A = 4 \pi r^2
A
r
A
V
\qquad V = 4 \pi r^2t
t
10 \text{ mi}
2
\, \dfrac{8{,}000}{2} = 4{,}000 \text{ mi}
t
r
640{,}000{,}000 \pi \text{ mi}^3
\qquad {\sqrt{4p^7q^2 + p^4q^2}}
p
q
{2\sqrt{p^{11}q^4}}
{2q\sqrt{p^7+p^4}}
{ p^2 q \sqrt{4p^3 + 1} }
\qquad \sqrt{xy} = \sqrt{x} \sqrt{y}
\qquad { \sqrt{x^{{2}}} = x \, , \, x > 0}
\qquad {\sqrt{4p^7q^2 + p^4q^2} = p^2q\sqrt{4p^3+1}}
\qquad R = \dfrac{C - P}{P}
R
C
P
R = C - 1
P = C - R - 1 
P = \dfrac C R +1
P = \dfrac{C}{R+1}
P
\blue{\text{multiplying both sides by } P}
P
\green{\text{denominator}}
P
\pink{ \text{adding } P \text{ to both sides}}
P
\gray{ \text{divide both sides by } (R+1) }
\qquad P = \dfrac{C}{R+1}
1{,}830
\text{(mi)}
605 \text{ mi}
x
1727 \text{ mi}
(x,y)
x
y
-2
-1
0
x + y
x
y
x
2y+1
x
y
y
y = -1
x
(x, y) = (-1, -1)
x+y
\qquad -1 + (-1) = -2
x
y
-2
\overline{WY}
6
4\sqrt3
16
16\sqrt2
\angle WXY=90^\circ
\angle WXZ=45^\circ
\angle ZXY
90^\circ-45^\circ = 45^\circ
\overline{WX}
\overline{WY}
x
y
180^\circ
x
\overline{WX}
WXZ
\theta
\qquad\qquad \cos\theta=\dfrac{\red{\text{Length of the side adjacent to }\theta}}{\blue{\text{Length of the triangle's hypotenuse}}}
WXZ
\angle WXZ=45^\circ
\red{\overline{XZ}}
\red{\text{{adjacent}}}
\angle{WXZ}
\red{8}
\blue{\overline{WX}}
\blue{\text{{hypotenuse}}}
WXZ
\blue{x}
\qquad\qquad \cos45^\circ=\dfrac{\red{8}}{\blue{x}}
\cos45^\circ
\dfrac{\sqrt2}2
\overline{WX}
y
\overline{WY}
WXY
WXY
m\angle YWX=45^\circ
{\overline{WX}}
\angle{YWX}
8\sqrt2
\overline{WY}
WXY
y
\qquad\qquad \cos45^\circ=\dfrac{8\sqrt2}{y}
\overline{WY}
16
12{,}000
97\%
90\%
2.4\%
88\%
92\%
96\%
100\%
90\%
97\%\pm2.4\%
\qquad97-2.4=96.6
\qquad97+2.4=99.4
96.6\%
99.4\%
96\%
\qquad\dfrac{5m^2+7m}{2m-9}-\dfrac{2m}{2m-9}
\dfrac{3m^2+7m}{2m-9}
\dfrac{5m^2+5m}{2m-9}
\dfrac{5m+9m}{2m-9}
\dfrac{5m^2+7m-1}{2m-9}
\qquad \dfrac{5m^2+7m-2m}{2m-9}
\qquad \dfrac{5m^2+5m}{2m-9}
7260
1200
 acre 
 square feet
(\text{ft}^2)
\text{length}\times \text{width}
\qquad 7260\text{ ft}\times1200 \text{ ft}=8{,}712{,}000 \text{ ft}^2
43{,}560:1
a
8{,}712{,}000
\qquad \dfrac{43{,}560\text{ ft}^2}{1 \text{ acre}}=\dfrac{8{,}712{,}000\text{ ft}^2}{a\text{ acres}}
a
\qquad\left(x-4\right)\left(x-5\right)=0
x=s
x=t
|s-t|
-9
-1
1
9
\left(x-4\right)\left(x-5\right)=0
\left(x-4\right)=0
\left(x-5\right)=0
x=4
x=5
s=4
t=5
|s-t|
|s-t|
1
\qquad \left( \dfrac{3}{4} \right) ^{1.5} \cdot \left( \dfrac{2}{3} \right)^{2.5}
3^{-3.75} \cdot 2^{-7.5}
\dfrac{1}{16}
\dfrac{1}{3\sqrt{2}}
\qquad \left( \dfrac{x}{y} \right)^m = \dfrac{x^m}{y^m}
\qquad \left( \dfrac{3}{4} \right) ^{1.5} \cdot \left( \dfrac{2}{3} \right)^{2.5} = \dfrac{3^{1.5}}{4^{1.5}} \cdot \dfrac{2^{2.5}}{3^{2.5}}
\qquad { \dfrac{1}{x^m} = x^{-m}}
\qquad { \dfrac{3^{1.5}}{4^{1.5}} \cdot \dfrac{2^{2.5}}{3^{2.5}} = 3^{1.5} \cdot 4^{-1.5}\cdot 2^{2.5} \cdot 3^{-2.5}}
\qquad { x^m \cdot x^n = x^{m+n}} \space
\space { \left(x^m\right)^n = x^{mn}}
\qquad { \left( \dfrac{3}{4} \right) ^{1.5} \cdot \left( \dfrac{2}{3} \right)^{2.5} = \dfrac{1}{3\sqrt{2}}}
1{,}594\,\dfrac{\text{kJ}}{\text{h}}
2{,}726\,\dfrac{\text{kJ}}{\text{h}}
77\,\dfrac{\text{L}}{\text{h}}
130\,\dfrac{\text{L}}{\text{h}}
49\,\dfrac{\text{g}}{\text{h}}
r
55
r
27 \,\dfrac{\text{g}}{\text{h}}
29 \,\dfrac{\text{g}}{\text{h}}
83 \,\dfrac{\text{g}}{\text{h}}
85 \,\dfrac{\text{g}}{\text{h}}
1{,}594:2{,}726
77:130
49:r
r
r
r
r
r
49
130
77
83 \,\dfrac{\text{g}}{\text{h}}
ta
8
a
t
8
156
184
0.93
1.08
26
28
8
\dfrac{a_2-a_1}{t_2-t_1}
\qquad\dfrac{96-70}{184-156}=\dfrac{26}{28}\approx0.93
8
{0.93}
156
184
9{,}000
3{,}600
147{,}000
12{,}600
58{,}800
159{,}600
367{,}500
\green d
58{,}800
\qquad3(t+1)=-\dfrac12-5t
t=-\dfrac7{16}
t=-\dfrac74
t
\red{5t}
\green 3
t=-\dfrac7{16}
t=-\dfrac7{16}
(t)
\qquad  a t+ b= c  t+ d
 a\ne c
t=-\dfrac7{16}
t=-\dfrac7{16}
\qquad 
3
\qquad 
1280\,\text{m}
\qquad h(x)=0.000371(x-640)^2
x
\qquad
0.000371x^2-0.47488x+151.9616
0.000371(x^2-1280x+409{,}600)
0.000371((x-600)^2-80x+49{,}600)
0.000371((x-650)^2+20x-12{,}900)
\qquad ax^2+bx+c
c
x=0
\qquad 0.000371x^2-0.47488x+151.9616
1.67
2:1
 per gallon for the Glyphosate in the weed killer and charges 
 per gallon for the other ingredients in the weed killer, then, rounded to the nearest cent, how much does the manufacturer charge for the 
2{:}1
(\text{gal})
g
1.67
\qquad\dfrac{2\text{ total gal of weed killer}}{1 \text{ gal of Glyphosate}}=\dfrac{1.67\text{ total gal of weed killer}}{g\text{ gal of Glyphosate}}
g
\qquad g=0.835\text{ gal of Glyphosate}
0.835
1.67
1.67-0.835=0.835
0.835
 per gallon of Glyphosate, then it charges 
 gallons of other ingredients and the manufacturer charges 
 per gallon of the other ingredients, then it charges 
 for the 
\qquad x^2+\dfrac{13}2x+\dfrac{15}2=0
x=a
x=b
ab
-15
-\dfrac{15}2
\dfrac{15}2
15
a
b
x^2+\dfrac{13}2x+\dfrac{15}2
(x-a)(x-b)
ab
\dfrac{15}2
ab
ab
\dfrac{15}2
P
30
2\pi
3\pi
5\pi
10\pi
\qquad 5\pi
a(x+b)^2 = 3x^2+36x+c
c
3
6
36
108
\qquad a(x+b)^2 = 3x^2+36x+c
3
a
3
3
\qquad (x+b)^2=x^2+12x+\dfrac{c}{3}
\left(\dfrac{1}{2}(12)\right)^2
\dfrac{c}{3}
36
\dfrac {c}{3}=36
c=108
\qquad \sqrt{x^2-4x+4}=0
x
x=2
2
\qquad 243p^{36m}-75q^{16n}
3(9p^{6m}+5q^{4n})(9p^{6m}-5q^{4n})
3(9p^{18m}+5q^{8n})(9p^{18m}-5q^{8n})
3(9p^{6m}+5q^{4n})\cdot 3(9p^{6m}-5q^{4n})
3(9p^{18m}+5q^{8n})\cdot3(9p^{18m}-5q^{8n})
3
\qquad 243p^{36m}-75q^{16n}=3\left(81p^{36m}-25q^{16n}\right)
\sqrt{p^{36m}}=p^{18m}
p^{18m}\cdot p^{18m}=p^{36m}
\sqrt{q^{16n}}=q^{8n}
q^{8n}\cdot q^{8n}=q^{16n}
243p^{36m}-75q^{16n}
\qquad 3(9p^{18m}+5q^{8n})(9p^{18m}-5q^{8n})
3
(\text{in})
2.5\text{ in}
2\text{ in}
\qquad V_\text{cube} = s^3
V_\text{cube}
s
\qquad V_\text{prism} = lwh
V_\text{prism}
l
w
h
s
l
w
h
s
w
h
l
l
s
w
h
5.4\text{ in}
(x,y)
(-4,10)
(3,3)
(4,2)
(-3,9)
(4,-3)
(-4,3)
y
x
x
y=x^2-2x-6
y
y+x=6
x
\qquad\qquad\qquad x^2-2x-6=-x+6
 
x=4
x=-3
x
x
 
(4,2)
(-3,9)
 that he will put toward the trip. To save more money for the trip, Felipe gets a job where each month he can add 
 to his savings for the trip. Let 
 be the number of months that Felipe has worked at his new job.  If Felipe needs to save 
250m-350=2700
250m+350=2700
350m-250=2700
350m+250=2700
 for *each* month that he works. We can represent this with the expression 
, and we do not need to multiply 
 by 
\qquad350m+250=2700
U
1
h
t
U = 9.8h
U
t
1
2
2
h
\qquad U = 9.8(2) = 19.6\;
U
U
19.6
\qquad\dfrac{x}{x-1} + \dfrac{4}{x-2} = \dfrac{4}{(x-1)(x-2)}
x
-4
-2
1
2
\qquad\dfrac{x^2-2x}{x^2 -3x +2} + \dfrac{4x-4}{x^2 -3x +2} = \dfrac{4}{x^2 -3x +2}
\qquad\dfrac{x^2+2x-4}{x^2 -3x +2} = \dfrac{4}{x^2 -3x +2}
\qquad x^2+2x-4 = 4 
4
\qquad x^2+2x-8 = 0 
\qquad (x + 4) (x - 2)=0
x
-4
2
x=2
x
x
-4
k
x
y
y
\qquad  -\dfrac{3.2}{k} = -1.6
k=2
y
k
2
(
  | Profit 
in hundreds of 
 | 
| 
, corresponding to 
 hundred dollars spent on advertising last month. It is determined that the profit can be modeled by a quadratic function of the form 
. If 
P(20)
 **spent on advertising**. Before we can do this, we have to build our quadratic function by finding 
 and 
(10, 450)
(40, 240)
P(x) = c + bx - 0.5x^2
c
{-1}
b
c
c
b
P(20)
20
(
 is spent on advertising, a profit of 
 hundreds 
or 
(\text{cm})
1{,}019\,\text{cm}
1{,}089\,\text{cm}
1{,}105\,\text{cm}
1{,}120\,\text{cm}
\purple{\text{slant height}}
\red{\text{pyramid height}}
\purple{\text{slant height}}
80\,\text{cm}
40\,\text{cm}
\purple{\text{slant height}}
\red{\text{pyramid height}}
40\,\text{cm}
\red{70\,\text{cm}}
\green{\text{slant height}}
\pink {\text{height}}
\qquad \dfrac{290\,\text{cm}-80\,\text{cm}}{2}=105\,\text{cm}
\green{\text{slant height}}
\pink{\text{height}}
105\,\text{cm}
\qquad\red{70}+\pink {1{,}019}=1{,}089
1{,}089\,\text{cm}
-
-
l
s
40
l=2s+1
l=1+2s
40
l+s=40
l
l
\qquad (1+2s)+s = 40
\qquad 1+3s = 40
1
\qquad 3s=39
3
\qquad s=13
13
13
l
\qquad l+ \: s \: =40
\qquad l+13=40
\qquad l=27
27
30
\qquad
\qquad\qquad\qquad \text{Online Course Enrollment}
n
2002
2011
t
2002
2002
2011
130{,}000
580{,}000
1{,}725{,}000
58{,}000{,}000
(2, 2.5)
(4.5, 4)
m=\dfrac{0.6}{1}=\dfrac{\Delta n}{\Delta t}
0.6
0.6
600{,}000
580{,}000
2002
2011
580{,}000
\qquad6=-s+77
1+5(77-s)
-739
-29
31
741
31
10\text{ ft}
5 \text{ ft}
x
y
3 \text{ ft}
y=3(x-5)^2+10
y=-3(x-5)^2+10
y=\dfrac{7}{25}(x-5)^2+10
y=-\dfrac{7}{25}(x-5)^2+10
y
y=a(x-h)^2+k
(h,k)
10 \text{ ft}
5 \text{ ft}
(5,10)
\qquad y=a(x-5)^2+10
a
3 \text { ft}
(0,3)
a
a
\qquad y=-\dfrac{7}{25}(x-5)^2+10
6
\text{(mi)}
14{,}000
\text{(ft)}
5{,}280
1
14{,}000 \text{ ft}
4
d
7 \text{ mi}
7
(\text{ft})
5\text{ ft}
V
h
A
V=Ah
A
\dfrac12 sh_{\small{\triangle}}
s
h_{\small{\triangle}}
\dfrac{\sqrt3}{2}s
5\text{ ft}
7\text{ ft}
76\text{ ft}^3
3
4
h
w
h = \dfrac{4}{3}w
h = 3w
h = 4w
h = 12w
4
3
12
12
w
w
h = 12w
5
-3
\dfrac{9}{11}
(x-5)(x+3)(11x-9)
(x+5)(x-3)(11x+9)
(x-5)(x+3)(9x-11)
(x+5)(x-3)(9x+11)
0
\blue5
\red{-3}
\blue{x-5}
\red{x+3}
\green{11x-9}
\qquad\blue{(x-5)}\red{(x+3)}\green{(11x-9)}
\qquad 2x^2+5x-k=0
k
k
-\dfrac{25}{8}
-\dfrac{5}{4}
\:\:\:\dfrac{5}{4}
\:\:\dfrac{25}{8}
b^2-4ac
k
0
\qquad 25+8k=0
k
k=-\dfrac{25}{8}
xy
14
99
\qquad -21
\qquad x^2-ax+24
8
0
8
a
a
\qquad x^2-11x+24=\blue{(x-3)}\red{(x-8)}
x
\blue3
\red8
\blue3
4
0.12
30
m
d
d=4-0.12m
d=4+0.12m
d=4-0.004m
d=4+0.004m
30
0.12\text{ GB}
\;\dfrac{0.12\text{ GB}}{30\text{ minutes}}\;
0.004 \text{ GB per minute}
[\text{GB remaining}]=[\text{initial GB}]-[\text{GB used per min}]\cdot [\text{no. of mins}]
4\text{ GB}
m
\qquad d(m)=4-0.004m
d(m)=4-0.004m
\qquad\qquad H = \dfrac{kA\ \Delta T}{ L}
H
kA
\Delta T
L
kA = \dfrac{\Delta T}{HL}
kA = \dfrac{HL}{\Delta T}
kA = HL\ \Delta T
kA = \dfrac{H\  \Delta T}{L}
L
\blue{\text{multiplying both sides of the equation by } L}
\green{\text{divide both sides of the equation by } \Delta T}
\qquad kA = \dfrac{HL}{\Delta T}
l =30
(\text{cm})
t = 1 \text{ cm}
d = 4 \text{ cm}
120 \pi
150 \pi
270 \pi
600 \pi
\qquad V = \pi r^2 h
V
r
h
4\text{ cm}
1 + 4 + 1 = 6\text{ cm}
2\text{ cm}
3\text{ cm}
l
30\text{ cm}
\qquad V_\text{inner} = \pi (2)^2(30)= 120 \pi
\qquad V_\text{outer} = \pi (3)^2(30)= 270 \pi
\qquad V_\text{pipe} = 270 \pi - 120 \pi = 150 \pi
150 \pi \text{ cm}^3
320
\dfrac{1}{4}
p
c
p = 80c
p = 160c
p = 240c
p = 320c
\dfrac{1}{4}
\dfrac{3}{4}
\dfrac{3}{4}
320
240
240c
\qquad p=240c
p = 240c
\qquad (6+\sqrt{3})^2
15
39
39+12\sqrt{3}
45+12\sqrt{3}
\qquad (6+\sqrt{3})^2=(6+\sqrt{3})\cdot (6+\sqrt{3})
\qquad 39+12\sqrt{3}
50\%
6
64\ \text{mg}
3
12
16\ \text{mg}
32\ \text{mg}
48\ \text{mg}
96\ \text{mg}
\qquad 3 \times 64 \text{ mg}=192 \text{ mg}
6
50\%
\qquad 0.50 \times 192 \text{ mg} = 96 \text{ mg}
12
6
12
50\%
6
\qquad 0.50 \times 96 \text{ mg} = 48 \text{ mg}
2
 per hour and the second job pays 
 hours, but she needs to make at least 
 hours at the first job and 
40
a
b
40
(0,40)
(40,0)
b
, and she must make *at least* 
b
\qquad i(7-3i)
i=\sqrt{-1}
4i
10i
7i-3
7i+3
i^2=-1
\qquad 7i+3
12
\qquad S= I\cdot t - (r+p)\cdot t\  
S
t
I
r
p
I>r+p
I
r
p
S
t
I\cdot t
(r+p)\cdot t
I
r
p
m
\qquad m=I-(r+p)
I>(r+p)
S
\qquad\dfrac{3w}{w+4} - \dfrac{8}{w} = 1
\qquad 2(w-8)(w+2) = 0
\qquad w = 8
w = -2
w = 8
w = 8
w = -2
w = -2
8 + (-2) = 6
\qquad \left(x^{\left({4^6}\right)}\right) ^ {\left(4^1\right)}
s
t
s+t
\qquad\quad \left(x^{\left({\large{4^6}}\right)}\right) ^ {\left(\large{4^1}\right)}
\left(x^a\right)^b = x^{ab}
4^6\cdot4^1
k^a\cdot k^b = k^{a+b}
s+t
7
24
(\text{in})
15\text{ in}
12\text{ in}
900
\qquad V = lwh
2.5\text{ in}
2.5\text{ in}
35
(\text{ft})
195\text{ ft}
21
24
25
29
35\text{ ft}
35+4s
s
195\text{ ft}
35+4s=195
40\text{ ft}
20\%
30\%
 to buy a ball.  This weekend, Suraj is planning to use 
 balls at Roycefield and 
 balls at Hunterdon.  If he only has 
5
10
15
20
14
b
14 \leq b
, and therefore the cost of 
b
b=15
70
(\text{cm})
20\,\text{cm}
40\,\text{cm}
(\text{ms})
x
0\leq x\leq70
0.2
12\,\text{cm}
28\,\text{cm}
19.2\,\text{cm}
20.8\,\text{cm}
32\,\text{cm}
48\,\text{cm}
39.2\,\text{cm}
40.8\,\text{cm}
\qquad | x - 20|
0.2
\qquad 40\dfrac{\text{cm}}{\text{ms}}(0.2\,\text{ms})=8\,\text{cm}
\qquad x=12
x=28
12\,\text{cm}
28\,\text{cm}
 when the operator guessed 
105
115
105
125
115
125
115
135
5
120
\quad |x-120| = 5
c
m
c(m)=\dfrac{3}{2}\left(0.5^{60m}-1\right)
c(m)=\dfrac{3}{2}\left(1.2^{60m}-1\right)
\blue{A}\cdot \red{B}^{\green {\large x}}
\red B
\green x
1
\green x
1
\red B
60
\dfrac{6}{6}
{\left(x^ {\large y}\right)^{\large z}=x^{{\large y}\cdot {\large z}}}
\red{729}
(2,5)
(5,2)
(-2,4)
(-4,2)
(-2, 4)
(2.1,4.9)
(-2,4)
(2,5)
(-2,4)
(-2,4)
(2,5)
(2,5)
(-2,4)
(x_1,y_1)
(x_2,y_2)
y_1+y_2
x
y
(x,y)
y
y=4x+2
x
x
\qquad x = \dfrac{-b\pm\sqrt{b^2-4ac} }{2a}
ax^2+bx+c=0
x_1=\dfrac{5}{3}\,
x_2=-\dfrac{1}{2}
x
y
x
x_1=\dfrac{5}{3}
x_2=-\dfrac{1}{2}
y
y_1+y_2
\dfrac{26}{3}
a^m\cdot a^n=a^{m+n}
x
y
x
y
90
60
480
1
2
S
c
90
90
c
90
S=480
c=60
m
c
1
S
2
S
2
S
2
x
\dfrac{9}{25} 
\dfrac{3}{5} 
0
\dfrac{9}{25}
0
\dfrac35
2
0
0
x
0
\dfrac{9}{25}
\qquad 16x^4-81
(4x^3-9)(4x+9)
(2x-3)(2x+3)(4x^2-9)
(2x-3)(2x+3)(4x^2+9)
(2x-3)^4
a^2-b^2
(a-b)(a+b)
16x^4
81
16x^4-81
\qquad 16x^4-81=(4x^2-9)(4x^2+9)
(4x^2-9)
\qquad (4x^2-9)(4x^2+9)=(2x-3)(2x+3)(4x^2+9)
16x^4-81
\qquad (2x-3)(2x+3)(4x^2+9)
\qquad\dfrac{\sqrt{75x^4}}{\sqrt{12x^7}}
\dfrac{5x\sqrt{3x}}{6}
\dfrac{5\sqrt{3x}}{6x^2}
\dfrac{5}{2\sqrt{x}}
\dfrac{5\sqrt x}{2x^2}
\purple{\text{power of a quotient}}
\qquad \left(\dfrac xy\right)^a=\dfrac{x^a}{y^a}
\sqrt{25}=5
\qquad\dfrac{5}{2x\sqrt x}
\dfrac{\sqrt x}{\sqrt x}
\qquad\dfrac{5\sqrt x}{2x^2}
\qquad x^2 + y^2 + z^2 = R^2
3
3
(x,y,z)
R
y
x
z
R  
y = R-x-z
y = \sqrt{R - x - z}
y = \sqrt{R^2 - x^2 - z^2}
y = \dfrac{R^2 - x^2 - z^2}{y}
y^2
\pink{\text{subtracting } x^2 \text{ and } y^2}
\qquad y =  \sqrt{R^2 -x^2 - z^2}
 -8-8y=6-2y
y
y=-\dfrac37
y=\dfrac37
y=\dfrac73
y=-\dfrac73
\red{2y}
\blue{8}
\green{-\dfrac16}
y=\pink{-\dfrac73}
y=\pink{-\dfrac73}
(y)
\qquad  ay+ b= c y + d
 a\ne c
y=-\dfrac73
y=-\dfrac73
2009
2013
r
2010
 billion, and in 
, it was approximately 
 represents years since 
, which of the following best models the print advertising revenue obtained by U.S. newspapers from 
 to 
f(t)=22.8-0.55t
f(t)=23.4-0.55t
f(t)=22.8-1.83t
f(t)=24.6-1.83t
t=0
2009
(t, f(t))
\qquad P_1=(1, 22.8) \text{ and  } P_2=(4, 17.3)
2009
2013
1.83
y=mx+b
f(t)=-1.83t+b
b
2009
t
f(t)
b
f(t)=-1.83t+24.6
f(t)=24.6-1.83t
2009
2013
x>5
\qquad\dfrac{x^2-x-6}{x^2+6x-27} \cdot \dfrac{4x-32}{x^2+11x+18}
4x-32
\dfrac{4x-32}{{x^2+81}} 
\dfrac{(x+2)^2}{4x-32}
\dfrac{4x-32}{{(x+9)}^2} 
\qquad \dfrac{(x-3)(x+2)}{(x-3)(x+9)} \cdot \dfrac{4(x-8)}{(x+2)(x+9)}
x
0
x\neq{\green{-2},\pink{ 3}}
x>5
\qquad \dfrac{4x-32}{(x+9)^2}
\qquad \dfrac{1+2i}{1-2i}\div\dfrac{1-2i}{1+2i}
i=\sqrt{-1}
1
-1
-\dfrac{7}{25}+\dfrac{24}{25}i
-\dfrac{7}{25}-\dfrac{24}{25}i
a+bi
a
b
-3+4i
\qquad \dfrac{1+2i}{1-2i}\div\dfrac{1-2i}{1+2i}=-\dfrac{7}{25}-\dfrac{24}{25}i
\qquad  (x+3)(x-5)=5
x=-3
x=5
x=2
x=10
x=1- 2\sqrt{5}
x=1+ 2\sqrt{5}
x=1- \sqrt{21}
x=1+ \sqrt{21}
x^2
1
x
x
\qquad x^2-2x=20
1
\qquad x^2-2x+1=20+1
x=1+\sqrt{21} 
x=1-\sqrt{21} 
(-5, b)
b
(-5, 1)
(-2,4)
(-5,1)
(-5,1)
x
-5
b=1
10
3.5
 and an 
 hour job would cost 
.  If the construction company completes a job in one day, which of the following functions best models the cost, 
, in dollars, for an 
c(h)=50h
c(h)=60h
c(h)=50h+75
c(h)=60h+10
c
h
\left(h, c(h)\right)
\qquad   P_1=(3.5, 250) \qquad \text{ and}  \qquad P_2=(8, 475)
y=mx+b
c(h)=50h+b
b
c
h
b
c(h)=50h+75
h
t
(1
=\,52
)
1-0.25^{26}
1.25^{26}-1
26\cdot0.25
26\cdot1.25
26
52
\dfrac{26}{26}
{\left(x^{\large y}\right)^{\large z}=x^{{\large y}\cdot {\large z}}}
1
1.25^{26}
100\%
1
\qquad\green{1.25^{26}-1}\approx329.87
\green{1.25^{26}-1}
1^{\text{st}}
2014
. Each year the value of the bond increases linearly by 
. Which graph below represents 
, the dollar value of the bond, as a function of 
, the number of years after January 
, 
0
, so the 
-intercept is 
 each year at a linear rate. Therefore, the slope of 
 with respect to 
 is 
\qquad v = 1000 + 75t
t = 6
v
(0, 1000)
(6, 1450)
x
10
15
1.15\left(\dfrac{x}{10}\right)
0.85\left(\dfrac{x}{10}\right)
0.15\left(\dfrac{x}{10}\right)
1.15x
\qquad 1.15\left(\dfrac{x}{10}\right)
3.60 \times 10^7
3{,}030
439.82
9.29 \times 10^7
7{,}921
288.15
	| 
	| 
  
	| 
	| 
 
	| 
	| 
 
[\text{Mar's distance}] : [\text{Saturn's distance}]
4:25
[\text{Mar's distance}] \approx 1.42 \times 10^8
[\text{Earth's distance}] : [\text{Mar's distance}]
\dfrac{[\text{Earth's distance}]}{[\text{Mar's distance}]} \approx \dfrac{[9.29\times10^7]}{[1.42\times10^8]} \approx 0.65
\left( \dfrac23 , \dfrac32, \dfrac49, \dfrac94 \right)
0.65
2:3
0.0084\%
0.0086\%
2.79 \cdot 10^9
0.887 \cdot 10^9
\qquad
1.9 \cdot 10^9
2:3
1.9 \cdot 10^9
P \approx R(6.7 \times 10^{21})
\qquad \qquad \qquad \qquad \text{North Dakota Population}
p
2000
2014
t
2000
p=0.78(t-2.8)^2+638.4
p=0.78(t+2.8)^2+638.4
p=0.78(t-638.4)^2+2.8
p=0.78(t+638.4)^2+2.8
y=a(x-h)^2+k
(h,k)
(3, 638)
p=0.78(t-2.8)^2+638.4
(3,638)
a
(8, 660)
p=0.88(t-3)^2+638
p=0.78(t-2.8)^2+638.4
\qquad \dfrac{2}{3}x^2-\dfrac{1}{2}x-\dfrac{3}{4}=0
x_1
x_2
x_1+x_2
-\dfrac{3}{2}
-\dfrac{3}{4}
0
\dfrac{3}{4}
12
8x^2-6x-9=0
b^2-4ac
8x^2
-9
-6x
\qquad 8x^2-6x-9=(4x+3)(2x-3)
\qquad 4x+3=0\qquad \text{or}\qquad2x-3=0
x=-\dfrac{3}{4}
x=\dfrac{3}{2}
x_1+x_2= \dfrac{3}{4}
x
f(x)
g(x)
1
2
2
2
3
2
3
5
1
5
6
4
f(g(2))?
-3
-6
2
1
f(g(2))
g(2)
f
g(2)
2
f(g(2)) = f(2)
3
3
p(x) = (x-a)(x-15)(x-20)+15
a
p(7)=15
a
7
a
\qquad P = a \cdot (1+b)^{kt}
P
t
a
b
k
P
P
k > 0
P
a  > 1
P
k > 0
P
a > 1
\qquad f(x) = A \cdot C^x
A > 0
C > 0
C < 1
C > 1
\qquad P = a \cdot ((1+b)^k)^t
A = a
C = (1+b)^k
C > 1
(1+b)^k > 1
b
b > 0
1
\qquad 1 + b > 1
(1+b)^k
1
k
(1+b)^k
1
k
k
C > 1
P
k > 0
\qquad\qquad
x
\angle YXZ
\angle WXY
90^\circ
\angle YXZ=\theta
\angle WXZ=90^\circ
\green{\angle WXY=90^\circ-\theta}
\cos\theta=\sin(90^\circ-\theta)
x
\overline{XY}
WXY
\theta
XYZ
\theta
\qquad\qquad \cos\theta=\dfrac{\text{Length of the side adjacent to }\theta}{\text{Length of the triangle's hypotenuse}}
\theta
\qquad\qquad \sin\theta=\dfrac{\text{Length of the side opposite from }\theta}{\text{Length of the triangle's hypotenuse}}
WXY
XYZ
XYZ
\overline{XY}
\theta
XY=x
\overline{XZ}
XZ=49
\qquad\qquad \cos\theta^\circ=\dfrac{x}{49}
WXY
\overline{WY}
90^\circ-\theta
WY=36
\overline{XY}
XY=x
\qquad\qquad \sin(90^\circ-\theta^\circ)=\dfrac{36}{x}
\cos\theta=\sin(90^\circ-\theta)
\dfrac{x}{49}=\dfrac{36}{x}
x
42
\qquad R(q)=-0.31(q-260)^2+18{,}500
R(q)
q
\qquad R(q)=-0.31(q-260)^2+18{,}500
(260,18{,}500)
\qquad 4(2+3p)-2(1+3p)\geq32
p \leq -\dfrac{23}3
p \geq \dfrac{11}3
p \geq \dfrac{13}3
p\geq-\dfrac32 \:\:\:\: \text{or} \:\:\:\: p\geq-\dfrac13
\purple4
\pink{-2}
\blue6
\gray6
p \geq \dfrac{13}3
246{,}400
25
61{,}600
p
25p+61{,}600\ge246{,}400
25p+61{,}600\le246{,}400
25p-61{,}600\ge246{,}400
25p-61{,}600\le246{,}400
25
61{,}600
25p-61{,}600
246{,}400
\qquad25p-61{,}600\ge246{,}400
\qquad25p-61{,}600\ge246{,}40
xy
\left(44,-34\right)
\sqrt3
 \left(x+34\right)^2+\left(y-44\right)^2=3
 \left(x+34\right)^2+\left(y-44\right)^2=\sqrt3
 \left(x-44\right)^2+\left(y+34\right)^2=3
\left(x-44\right)^2+\left(y+34\right)^2=\sqrt 3
(\green h,\purple k)
\red r
\qquad (x-\green h)^2+(y-\purple k)^2=\red r^2
\left(\green{44},\purple{-34}\right)
\red {\sqrt3}
\qquad \left(x-\green{44}\right)^2+\left(y-\left(\purple{-34}\right)\right)^2=(\red {\sqrt 3})^2
\qquad \left(x-44\right)^2+\left(y+34\right)^2=3
\qquad \left(x-44\right)^2+\left(y+34\right)^2=3
50
\text{(yd)}
12\text{ yd}
50-12
d
45.5
60
90
72
21
24
25
29
60
90
1.5
60+1.5t
t
72
\qquad 60+1.5t=72
8
\qquad \qquad \qquad \qquad \text{Dial-up Internet Trends}
P
t
2005
2005
2012
P=30(0.7)^t
P=30(1.5)^t
P=-30(0.7)^t
P=-30(1.5)^t
y=a\cdot b^x
a
y
b>0
a>0
0<b<1
a>0
b>1
P=-30(0.7)^t
P=-30(1.5)^t
-30
P=30(1.5)^t
P=30(0.7)^t
(4,7)
(5,5)
P=a\cdot b^t
a
a=\dfrac{7}{b^4}
\dfrac{7}{b^4}
a
b
b=0.71
7=a(0.71)^4
a
P=28(0.71)^t
P=30(0.7)^t
2005
2012
td
d
t
2
0
0
0
\qquad
\quad\qquad \left[(9f+9)+(9f+9+1)\right]\cdot f
\qquad =\left[9f+9+9f+9+1\right]\cdot f
\qquad =\left[9f+9f+9+9+1\right]\cdot f
\qquad =\left[18f+19\right]\cdot f
f
\qquad =18f^2+19f
\qquad A = t^2+5.8t+9.41
A
t
1
2011
t
1
2014
1
2014
1
(t - (-2.9))^2 + 1
(t - (-1))^2  + 3.8t + 8.41
(t - (-3.9))(t - (-1.9)) + 2
t(t - (-5.8)) + 9.41
t
-2.9
1
1
\qquad A = (t + 2.9)^2 + 1
10\large\frac{1}{2}
3
1.5
x
3+1.5x<10.5
3+1.5x\leq10.5
3x+1.5<10.5
3x+1.5\leq10.5
3
1.5
3+1.5x
10.5
10.5
10.5
(\leq)
\qquad3+1.5x\leq10.5
\qquad3+1.5x\leq10.5
\qquad a=\dfrac{5bc}{20}+10
b
a
c
b=\dfrac{4a}{c}-200
b=\dfrac{4a-40}{c}
b=\dfrac{4a}{c}-10
b=\dfrac{4a+40}{c}
\green{\text{subtracting } 10 \text{ from both sides of the equation}}
\dfrac{5bc}{20}
\red{\text{multiply}}
\red{\text{by }20}
\red{20}
b
\blue{\text{dividing by }5c}
\qquad b=\dfrac{4a-40}{c}
\qquad\sqrt{2x^2 +7}  +5 = 0
0
1
2
2
5
x
2
x = 3
x=3
x = -3
x = -3
0
1{,}500
25
399
635{,}000
25
9{,}500
39{,}000
127{,}000
169{,}000
\green d
169{,}000
25
w
w(x) = ax^2+bx+c
a
b
c
a=3
w(3)=w(15)=0
b
w(3)=0
w(15)=0
w
h(x)
w(x) = 3x^2+bx+c
2
h(x)=k
k=3
-k \cdot 18 = b
3
b
54
\qquad 256a^4-1
(4a-1)^4
(16a-1)(16a^3+1)
(4a-1)(4a+1)(16a^2-1)
(4a-1)(4a+1)(16a^2+1)
x^2-y^2
(x-y)(x+y)
256a^4
1
256a^4-1
\qquad 256a^4-1=(16a^2-1)(16a^2+1)
(16a^2-1)
\qquad (16a^2-1)(16a^2+1)=(4a-1)(4a+1)(16a^2+1)
256a^4-1
\qquad (4a-1)(4a+1)(16a^2+1)
2005
3.8\ \text{billion}
1\ \text{billion}
118\ \text{billion}
152\ \text{billion}
380\ \text{billion}
3.8\ \text{billion}
1\%
380
31\%
118
\quad 31\%
380 = 0.31\times380\approx117.8
118
hv
v
h
8.5
9
v = 9
9
9
\qquad f(x)=-g(x)-7
g
f
g
x
7
g
7
x
g
y
7
g
7
y
f
g(x)\rightarrow -g(x)
g
x
-g(x)\rightarrow -g(x)-7
7
g
f
g
x
7
\qquad \dfrac{2n^3m}{5\ell^2} - \dfrac{5n^3m}{7\ell^2}
\ell < -2
\dfrac{7n^3m }{35\ell^2}
\dfrac{-3n^3m }{35\ell^2}
\dfrac{-11n^3m }{35\ell^2}
\dfrac{-10n^6m^2 }{35\ell^4}
35\ell^2
\purple{\dfrac{7}{7}}
\pink{\dfrac{5}{5}}
\qquad \dfrac{14n^3m - 25n^3m}{35\ell^2} \quad = \quad \dfrac{-11n^3m }{35\ell^2}
\qquad \dfrac{-11n^3m }{35\ell^2}
xy
x
y
7
10
40
48
y=0
x
x
(40,0)
x
40
40
5
6
11
2
10
12
6
11
17
13
27
40
40
\dfrac{3}{20}
\dfrac{2}{9}
\dfrac{3}{5}
\dfrac{6}{11}
\purple6
5
\purple6
\green{11}
2
10
12
6
11
17
13
{27}
40
\green{11}
\dfrac{\purple6}{\green{11}}
\dfrac{6}{11}
\qquad \qquad \qquad \qquad \qquad \text{Online Sales Revenue}
R
2004
2014
t
2004
2004
2014
2\%
3\%
10\%
30\%
\qquad (0, 7), (1, 9), (2, 11), (3, 15), (4, 19)
(t, R)
t
2004
2004
7
2005
9
2006
11
2007
15
2008
19
\qquad \dfrac{9}{7}\approx 1.3
\qquad \dfrac{11}{9}\approx 1.2
\qquad \dfrac{15}{11}\approx 1.4
\qquad \dfrac{19}{15}\approx 1.3
1.3
130\%
30\%
30\%
\dfrac{12}{13} \approx \tan(42.7^\circ) \approx \sin(67.4^\circ) \approx \cos(22.6^\circ)
\alpha
180^\circ
\dfrac{12}{13} \approx \cos(22.6^\circ)
\alpha
23
\qquad5(v-3)=6v+2
v=17
v=-17
v=5
v=-5
v
\red {5v}
\blue{2}
v=\pink{-17}
v=\pink{-17}
(v)
\qquad  av+ b= cv + d
 a\ne c
v=-17
v=-17
\qquad \qquad \quad \text{Average New Home Sale Prices in the US}
p
2003
2009
t
2003
 each year between 
 and 
3
 in 
2003
(3.5,310)
(t, p)
t
2003
p
2006
 in 
14{:}00
12{:}00
8{:}00
7{:}00
12.5\%
14.3\%
14.3\%
16.7\%
14.3\%
12.5\%
16.7\%
14.3\%
14.3\%
\qquad \dfrac{12-14}{14}=- \dfrac{2}{14}\approx- 0.1429
\qquad\Rightarrow\quad 14.3\%
12.5\%
\quad  \dfrac{7-8}{8}=-\dfrac{1}{8}=-0.125
  \qquad\Rightarrow\quad 12.5\%
14.3\%
12.5\%
y=\dfrac{1}{6}(x+12)
xy
y=\dfrac{1}{6}x+3
12y=2x+3
x+6y=18
y
y=\dfrac{1}{6}x+12
y=\dfrac{1}{6}(x+12)
\dfrac{1}{6}
\dfrac{1}{6}
y
2
y=\dfrac{1}{6}x+3
y=\dfrac{1}{6}x+3
y=\dfrac{1}{6}x+2
12y=2x+3
12y=2x+3
y=mx+b
y=\dfrac{1}{6}x+\dfrac{1}{4}
y=\dfrac{1}{6}x+3
12y=2x+3
1970
1990
101
41
25
60
53
55
57
59
x
1970
x
1970
y
1970
101
1990
1970
20
y
1970
25
1990
1970
20
x
1970
16
x
53
14
x
2
y
3
30
2
14
\qquad x+y=14
2x
3y
\qquad 2x+3y=30
2
x
\qquad -y=-2
\qquad y=2
\qquad x+2=14
\qquad x=12
12
\qquad C=1.2272M+3.0556F
C
M
F
M=\dfrac{C-3.0556F}{1.2272}
M=\dfrac{C}{1.2272}-3.0556F
M=\dfrac{3.0556F-C}{1.2272}
M=\dfrac{3.0556F}{1.2272}-C
M
\red{\text{subtract }3.0556F}
\green{\text{divide by}}
M
\green{1.2272}
\qquad M=\dfrac{C-3.0556F}{1.2272}
\qquad A(x)=\dfrac{1}{2}x^2-\dfrac{1}{3}x-7
B(x)=\dfrac{3}{2}-\dfrac{2}{3}x-7x^2
A(x)
B(x)
-\dfrac{7}{2}x^2+\dfrac{2}{9}x-\dfrac{21}{2}
-6.5x^2-x-5.5
2x^2-x-14
7.5x^2-x-8.5
\qquad \left(\dfrac{1}{2}x^2-\dfrac{1}{3}x-7\right)+\left(\dfrac{3}{2}-\dfrac{2}{3}x-7x^2\right)
A(x)
B(x)
A(x)
B(x)
\qquad -6.5x^2-x-5.5
\qquad E=\dfrac{A-V}{A}
E
A
V
V=A-EA
A=-V(E-1)
A=V(E-1)
A=\dfrac{V}{1-E}
A=\dfrac{V}{E-1}
A
\red{\text{multiplying by }A}
A
\blue{\text{subtracting }A}
A
\green{\text{divide by }E-1}
\purple{\text{multiplying by the identity }\dfrac{-1}{-1}}
\qquad A=\dfrac{V}{1-E}
. Fake mustaches cost 
 each, and party hats cost 
 each. If he must purchase a combination of at least 
 mustaches and party hats, which of the following systems of inequalities best models the relationship between the number of mustaches, 
, and party hats, 
\text{\red{at most}}
. Since fake mustaches cost 
 apiece and party hats cost 
 apiece, we can set up the following 
\text{\blue{at least}}
15
m
p
m+p
\text{\blue{greater than or equal}}
15
\qquad m+p \blue{\geq} 15
  she earned from her summer job into an account and will use it to pay for school expenses. She withdraws 
 of the remaining balance each month to pay for part of her living expenses.  Which of the following functions models the balance, 
, of Wyatt's money (in dollars) 
B(t)=450\cdot(1.1)^t
B(t)=450\cdot(0.9)^t
B(t)=450+1.1t
B(t)=450-0.9t
10\%
90\%
\qquad B(1)=450\cdot(0.9)
2
t
\qquad B(t)=450\cdot(0.9)^t
\qquad B(t)=450\cdot(0.9)^t
\dfrac{1}{2}+\dfrac25s=s-\dfrac34
s
s=\dfrac3{4}
s=\dfrac{25}{12}
s=-\dfrac{25}{12}
s=-\dfrac34
\red s
\blue{\dfrac12}
\green{-\dfrac53}
s=\pink{\dfrac{25}{12}}
s=\pink{\dfrac{25}{12}}
(s)
\qquad  as+ b= c s + d
 a\ne c
s=\dfrac{25}{12}
s=\dfrac{25}{12}
\qquad \dfrac{3}{2+i}
i=\sqrt{-1}
2-i
2+i
\dfrac{6+3i}{5}
\dfrac{6-3i}{5}
i
2+i
2-i
\dfrac{2-i}{2-i}
i^2=-1
\qquad \dfrac{6-3i}{5}
hd
d
h
35
35
35
35
2
0
3
\green{\text{home}}
\red{\text{work}}
\purple{\text{supermarket}}
\green{\text{home}}
8.5
16
35
35
\qquad V(m) - V(m-1) = 622
V(m)
m
m
V(m)
\qquad V(m) = V(m-1)+622
m
622
(a,b)
b<0
a
x
y
(a,b)
y
y
x
x
x
x=3
x=-2
y
x
y
x=3
x=-2
(3,-9)
(-2,36)
(a,b)
b<0
(3,-9)
a
a=3
\qquad 3+5x=\sqrt{k+6x}
k
x=0
x=\red0
k
9
\qquad 3+5x=\sqrt{9+6x}
0
x=0
x=-\dfrac{24}{25}
x=\green0
x=0
x=\green{-\dfrac{24}{25}}\,
x=-\dfrac{24}{25}
x=0
k=9
x=0
2008
211
150
2009
163
125
2008
2009
2008
2009
17\%
22\%
26\%
38\%
2008
150-211=-61\ \text{billion dollars}
2009
125-163=-38\ \text{billion dollars}
2008
2009
2008
2009
2008
\Large \frac{-61-(-38)}{-61}=\Large \frac{-23}{-61}\approx0.38=38\%
38\%
2008
2009
\qquad 4(80+n) = (3k)n
k
k
\qquad\;\, 4(80+n) = (3k)n
n
n
(
1=0)
\qquad 4\cdot80+4n = (3k)n
4n=(3k)n
4=3k
k
k=\dfrac43
n
\qquad\qquad\: 4\cdot80 = 0
k=\dfrac43
k=\dfrac43
2008
.  For each year before or after 
, the average price per square foot increased by 
. Which of the following equations below could be used to determine in which year, 
, the price per square foot was  
2.75|y-2008|=126
115+2.75|y-2008|=126
2.75|y-126|=2008
115-2.75|y-126|=2008
y
2008
y
\qquad 2.75|y-2008|
115
\qquad  115+2.75|y-2008|
126
\qquad  115+2.75|y-2008|=126
\qquad 115+2.75|y-2008|=126
16
2
10
13
14
16
m
t
m
t
t
m
\, m \,
\, \dfrac{m}{2}
\qquad t \ge \dfrac{m}{2}
\qquad t \le 16 - m
t
\space 16 - m \space
m
m
2
m
2
14
t
\space \dfrac{m}{2} \space
\space \dfrac{m}{2} \space
m
m
2
1
\qquad 14 - 1 = 13
18
3{,}300
30{,}000
26
1{,}131
1{,}280
1{,}633
1{,}665
\red{\text{unloaded truck}}
\blue{\text{cargo}}
\red{30{,}000}
w
\blue{26w}
\qquad\red{30{,}000}+\blue{26w}
18
\qquad18(3{,}300)=\purple{59{,}400}
\purple{59{,}400}
w
1{,}131
\qquad 16B + 3.5W = 22
22
B
W
16B
16B
16B
16B
16B
\blue{22 \text{ miles long}}
\qquad 16B + 3.5W = \blue{22}
B
16B
16B
10 \text{ cm}
500 \text { cm}^3
4\text{ cm}
8\text{ cm}
12.5\text{ cm}
15.9\text{ cm}
\qquad V=\pi r^2h
\qquad 500=10\pi r^2
r
10\pi
r\approx\pm4
4\text{ cm}
\qquad P(x)=3x^2+4x-5
Q(x)=7x^3-2x-5
P(x)-Q(x)
-7x^3+3x^2+6x
-7x^3+3x^2+6x - 10
-4x^2+6x
-4x^2+6x-10
0\text{s}
0
0
y^2+2y=15+x
(x,y)
y =\dfrac{1}{2}x+3
(-12,1)
(0,3)
(-12,-3)
(0,-5)
(-12,-3)
(0,3)
(-12,1)
(0,-5)
y=\dfrac{1}{2}x+3
(0,3)
y
m=\dfrac{1}{2}
1
2
(0,3)
y=\dfrac{1}{2} x+3\,
(0,3)
(-12,-3)
x
y
\left(0,3\right)
\left(-12,-3\right)
(-12,-3)
(0,3)
\qquad (8-2i)^2(8+2i)
i=\sqrt{-1}
60
68
480 - 120i
544-136i
\qquad (8-2i)(8-2i)(8+2i)=(8-2i)(64-4i^2)
i=\sqrt{-1}
i^2=-1
\qquad 544-136i
x
8\sqrt3
16
16\sqrt2
32
ABC
\overline{AC}
ABC
ACD
\theta
\qquad\qquad \tan\theta=\dfrac{\red{\text{Length of the side opposite from }\theta}}{\blue{\text{Length of the side adjacent to }\theta}}
ABC
\angle B=30^\circ
\red{\overline{AC}}
\red{\text{{opposite}}}
\angle B
\blue{\overline{AB}}
\blue{\text{{adjacent}}}
\angle B
\blue{16\sqrt3}
\red y
\red{\overline{AC}}
\qquad\qquad \tan30^\circ=\dfrac{\red{y}}{\blue{16\sqrt3}}
\tan30^\circ
\dfrac{\sqrt3}{3}
180^\circ
x
\overline{DC}
ACD
\theta
\qquad\qquad \sin\theta=\dfrac{\red{\text{Length of the side opposite from }\theta}}{\green{\text{Length of the triangle's hypotenuse}}}
ACD
\angle D=45^\circ
\red{\overline{AC}}
\red{\text{{opposite}}}
D
\red{16}
\green{\overline{DC}}
\green{\text{{hypotenuse}}}
ADC
\green{x}
\qquad\qquad \sin45^\circ=\dfrac{\red{16}}{\green{x}}
\sin45^\circ
\dfrac{\sqrt2}{2}
x
16\sqrt2
ds
s
d
s
25
25
25
25
0
s
(d, s)=(0, 25)
25
0
0
s
25
2700
(\text{ft}^3)
25\text{ ft}^3
1
10\%
44\%
56\%
93\%
25\text{ ft}^3
60
1
1500 \text { ft}^3
1
x
1
\qquad \dfrac {x}{100} = \dfrac{1500}{2700}
\qquad x
56
56\%
1
x
\angle
 is 
 and 
CBD
27^\circ
\angle
 must be 
\sin(\angle
 
 , so since 
ABD = 63^\circ
\sin63^\circ
= \dfrac{x}{10}
\sin63^\circ = 0.89
0.89 = \dfrac{x}{10}
x
8.9
0.89
10
\qquad Q(x)=\dfrac{P(2x)}{2}
Q
P
y = P(x)
y = Q(x)
y = P(x)
y = Q(x)
y = P(x)
x
2
y
 \dfrac{1}{2} 
y = P(x)
x
 \dfrac{1}{2}
y
 \dfrac{1}{2}
y = P(x)
x
 \dfrac{1}{2}
y
2
f(x) = a \cdot g(bx)
y = f(x)
y = g(x)
x
 \dfrac{1}{b} 
y
a
a = \dfrac{1}{2} 
b = 2
Q
P
 \dfrac{1}{2} 
Q
P
 \dfrac{1}{2} 
x
 \dfrac{1}{2} 
y
\qquad f(x)=-0.12x+24.91
-20
\qquad\qquad \qquad \text{Water temperature in degrees Celsius}, f(x) 
\qquad\qquad\qquad\quad \text{Water depth in meters}, (x)
27.31
24.91
374.25
210
\qquad f(x)=-0.12x+24.91
x
f(x)
x
-20
-20
f(x)
x
374.25
-20
\qquad\qquad \qquad \text{Water temperature in degrees Celsius}, f(x) 
\qquad\qquad\qquad\quad \text{Water depth in meters}, (x)
(x,y)
x
y
-1
1
\dfrac{13}{6}
xy
x
y
y
y
x
\qquad y = -\dfrac{1}{12} +\dfrac{1}{2}x
y
x
(x, y) = \left( \dfrac{3}{2}, \dfrac{2}{3} \right)
xy
\qquad \left( \dfrac{3}{2} \right) \left( \dfrac{2}{3} \right) = 1
2001
2012
40\%
40\%
11:18
3:1
27.2\%
27.8\%
37.1\%
37.9\%
\qquad \text{percentage} = \dfrac{\text{part}}{\text{total}} \cdot 100\%
\qquad P_w = \dfrac{\text{densely vegetated land in west}}{\text{total land in west}} \cdot 100\%
\qquad P_w = \dfrac{\text{11 parts}}{\text{11 parts + 18 parts}} \cdot 100\%
\qquad P_w = \dfrac{\text{11 parts}}{\text{(11 + 18) parts}} \cdot 100\%
\qquad P_w = \dfrac{11}{29} \cdot 100\%
\qquad P_w = 0.379310345... \cdot 100\%
\qquad P_w = 37.9310345...\cdot\%
\qquad P_w \approx 37.9\%
\qquad P_c = 75\%
\qquad P_c - P_w
\qquad \approx 75\% - 37.9\%
\qquad = 37.1\%
37.1\%
\qquad\left(4s^2t\right)^3 + \left(p-3\right)
\left(4^3s^{23}t^3\right) + \left(p-3\right)
\left(p-3\right) + \left(4s^2t^3\right)
\left(12s^2t\right) + \left(p-3\right)
\left(p-3\right) + \left(64s^6t^3\right)
(p-3)
3
\left(4s^2t\right)^3
\left(4s^2t\right)^3
3
\qquad\left(p-3\right) + \left(64s^6t^3\right)
\qquad\dfrac{2x}{x+3} - \dfrac{4}{x-2} = -3
x
-2
1
3
5
\qquad\dfrac{2x\blue{(x-2)}}{(x+3)\blue{(x-2)}} - \dfrac{4\pink{(x+3)}}{\pink{(x+3)}(x-2)} = -3
\qquad 2x^2 -8x -12 = -3(x^2 + x - 6)
\qquad 2x^2 -8x -12 = -3x^2 -3x + 18
\qquad 5x^2  - 5x - 30 = 0
5
\qquad 5(x^2  - x - 6) = 0
\qquad 5(x-3)(x+2)
\qquad x = -2
\qquad x = 3
x
-3
2
x
\qquad  -2 + 3 = 1
1
3
4
2
6
8
5
7
12
8
16
24
\dfrac{3}{16}
\dfrac{1}{8}
\dfrac{1}{2}
\dfrac{3}{4}
\green4
1
\blue3
\green4
2
6
8
5
7
12
8
16
24
\blue3
\dfrac{\blue3}{\green4}
3
4
46
5
15:184
10:69
69:10
184:15
3
4
\qquad r:b = 3:4
\qquad \dfrac{r}b = \dfrac34
46
5
\qquad \dfrac{l}s = \dfrac{46}5
l
s
1
\qquad 1:\dfrac{69}{10}
10
\qquad 10:69
10:69
\qquad\qquad\text{ Average monthly temperature in Chicago, IL}, \ f(x)
\qquad\qquad\qquad\qquad\text{Months of the year during 2013, }  (x)
2013
f(x)=-1.67x^2-23.04x-2.86
f(x)=-1.67x^2+23.04x-2.86
f(x)=1.67x^2-23.04x-2.86
f(x)=1.67x^2+23.04x-2.86
(
1)
(
7)
(
8)
(
12)
f(x)=ax^2+bx+c
a,b
c
a<0
\qquad f(x)=-1.67x^2-23.04x-2.86\qquad
\qquad f(x)=-1.67x^2+23.04x-2.86
a
\qquad f(x)=-1.67x^2-23.04x-2.86\qquad
\qquad f(x)=-1.67x^2+23.04x-2.86
a
c
a
y
c
b
x=\dfrac{-b\ \ }{2a}
x=6.9
\qquad f(x)=-1.67x^2+23.04x-2.86
\qquad\qquad\text{ Average monthly temperature in Chicago, IL}, \ f(x)
\qquad\qquad\qquad\qquad\text{Months of the year during 2013, }  (x)
2014
(22)
22
22
\dfrac{0\cdot5+1\cdot2+2\cdot2+3\cdot4+4\cdot1+5\cdot1+8\cdot1+10\cdot1+12\cdot1+16\cdot2+20\cdot1+22\cdot1}{22}
22
\qquad\dfrac{109}{21}=5.\overline{190476}\approx5
(a,b)
b-a
0
1
2
3
a
b
a
b
a=-5b=0
(0,0)
b-a=0-0=0
p
t
(t-10)
(22-t)
(t+10)
(20+t)
p(t)
y=p(t)
ty
p(t)
y=p(t)
t
t
t
t
14 \text{ mm}
10 \text{ mm}
132 \text{ mm}
d
10d+14\leq132
14d+20\leq132
20d+14\leq132
14d+10\leq132
{14\,\text{mm}}
{20\,\text{mm}}
d
{14}d+{20}
\qquad{14}d+{20}\leq{132}
\qquad{14}d+{20}\leq{132}
0
1
2
(0, 0)
1
x<0
\qquad\dfrac{x^2+2x-63}{x^2-49}
\dfrac{x-9}{x-7}
\dfrac{x+9}{x+7}
\dfrac{2x-63}{49}
\dfrac{2x-63}{-49}
(x+9)(x-7)
x^2-49
\qquad (x+7)(x-7)
\qquad \dfrac{(x+9)(x-7)}{(x+7)(x-7)}
x
0
x\neq\purple7
x<0
\dfrac{x+9}{x+7}
x<0
ABCD
EFGH
 d^\circ =  f^\circ
d^\circ = g^\circ
d^\circ = 90^\circ - g^\circ
d^\circ = 180^\circ - f^\circ
4
\quad \:\:\:ABCD \qquad\qquad\qquad \:\:\:\:\: EFGH
 d^\circ
 d^\circ =  b^\circ
 b^\circ =  f^\circ
 d^\circ =  f^\circ
xy
(3,4)
(5,0)
(-5,0)
(4,3)
y={2^{x-1}}
x^2+y^2=25
(3,4)
3
x
4
y
(-5,0)
y={2^{x-1}}
y=0
y
0
(-5,0)
-5
x
0
y
(3,4)
\quad
30
13-16
17-20
16
16
16
16
16
16
AFG
ABC
\angle AGF
102^\circ
\angle AFG=65^\circ
180^\circ
\angle AGF=x
\qquad x+37^\circ+65^\circ=180^\circ
180^\circ
x
x=78^\circ
\angle AGF=78^\circ
3
(\text{in})
216\text{ in}^3
\qquad V = lwh
V
l
w
h
h
l
w
\qquad A = lw
A
lw
\qquad V = Ah
72 \text{ in}^2
f
g
g(f(2))
-6
-2
0
2
g(f(2))
f(2)
f(2)
f
x=2
f(2)=0
0
f(2)
\qquad g(f(2)) = g(0)
g(0)
g(0)
g
x=0
g(0)=-6
g(f(2))=g(0) = -6
1
\qquad { \dfrac{1}{x} = x^{-1}}
\qquad { \left(x^m\right)^n = x^{mn}}
\qquad x^mx^n = x^{m+n}
\:\: x
\quad \sqrt{x}
\:\: 1
\quad 1
\:\: 2
\quad 1.414 \ldots
\:\: 3
\quad 1.732 \ldots
\:\: 4
\quad 2
\:\: 5
\quad 2.236 \ldots
\qquad x^2 - x - 1 = 0
\qquad (x + \text{[ ? ]} )(x - \text{[ ? ]} ) = x^2 - x - 1
ax^2 + bx + c = 0
2
x
\qquad x = \dfrac{-b \pm \sqrt{b^2 - 4ac} }{2a}
x^2 - x - 1 = 0
a = 1
b = {-1}
c = {-1}
x^2 - x - 1 = 0
x = 1.618 \ldots
1.62
xy
(11,12)
(13,14)
 x^2+y^2-22x+24y=-257
 x^2+y^2+22x-24y=-257
 x^2+y^2+22x+24y=-257
 x^2+y^2-22x-24y=-257
(\green h,\purple k)
\red r
\qquad (x-\green h)^2+(y-\purple k)^2=\red r^2
(\green{11},\purple{12})
\qquad (x-\green{11})^2+(y-\purple{12})^2=\red r^2
(13,14)
(x,y)
\red r^2
\qquad (x-11)^2+(y-12)^2=8
\qquad x^2+y^2-22x-24y=-257
m>1
n>1
\qquad \dfrac{49m^4n-21m^6n^2}{7m^2n^4}
7m^2n^4-3m^4n^2
7m^2n^3-3m^4n^2
\dfrac{7m^2-3m^3}{n^2}
\dfrac{7m^2-3m^4n}{n^3}
7m^4n(7-3m^2n)
\qquad \dfrac{7\cdot m\cdot m\cdot m\cdot m\cdot n(7-3m^2n)}{7\cdot m\cdot m \cdot n \cdot n \cdot n \cdot n}
\dfrac{7m^2-3m^4}{n^3}
m>1
n>1
\dfrac{3}{\sqrt{5}}
\dfrac{3\sqrt{5}}{5}
\dfrac{3\sqrt{5}}{25}
\dfrac{\sqrt{15}}{5}
\dfrac{\sqrt{15}}{25}
\sqrt{5}
\dfrac{3}{\sqrt{5}}\;
\dfrac{3\sqrt{5}}{5}
2
30
1
3
3
C
x
30
C=ax^2+bx
6
6
9
15
18
x
30
C
(2,1)
(3,3)
C=ax^2+bx
-3
2
b
\qquad3=6a
\qquad
\qquad a=\dfrac{1}{2}
a
b
a=\dfrac{1}{2}
b=-\dfrac{1}{2}
\qquad C=\dfrac{1}{2}x^2-\dfrac{1}{2}x
6
x
C
C=15
15
6
\qquad(p+1)^2 - 2^2
\qquad a^2 - b^2 = (a+b)(a-b)
a=p+1
b=2
\qquad(p+1)^2 - 2^2 = (p+1+2)(p+1-2)
tv
v
t
\quad
\text{\red{first half}}
\text{\green{second half}}
\qquad
\text{\red{first half}}
\text{\green{second half}}
\qquad
v
t
\qquad
 a(x) = \left(\dfrac89 x^2+\dfrac{15}9 x\right)
 b(x) = \left(\dfrac 23 x^2+4x\right)
a(x)-b(x)
\dfrac x9 (6x-21)
\qquad\quad a(x)-b(x)
a(x)
b(x)
\qquad =\left(\dfrac89 x^2+\dfrac{15}9 x\right)- \left(\dfrac 23 x^2+4x\right)
\qquad =\left(\dfrac89 x^2+\dfrac{15}9 x\right)+\left(\left(-\dfrac 23\right) x^2+(-4)x\right)
\qquad =\dfrac89 x^2+\dfrac{15}9 x+\left(-\dfrac 23\right) x^2+(-4)x
\qquad =\dfrac89 x^2+\left(-\dfrac 23\right) x^2+\dfrac{15}9 x+(-4)x
\qquad =\left(\dfrac89-\dfrac 23\right) x^2+\left(\dfrac{15}9 -4\right)x
\dfrac x9
\qquad =\dfrac x9\left(2 x-21\right)
c
s
a(s)
a(s)
s
a(s) = c \cdot \left[1 - \left(\dfrac12\right)^{\large s} \right]
a(s) = c \cdot \left[ 1 - \left(\dfrac1s\right)^2 \right]
a(s) = c \cdot  \left(\dfrac12\right)^{\large s}
a(s) = c \cdot  \left(\dfrac1s\right)^2
c(s)
\qquad c(s) = c  \left(\dfrac12\right)^{\large s}
s
\quad
s
c
a(s)
c(s)
a(s)
s
c
-\dfrac{x}{3.5} + \dfrac{y}{12.5} = 1
xy
12.5
(3.5, 12.5)
x
y
x
y
12.5
(3.5, 12.5)
(3.5, 12.5)
x
0
y
x
y
0
x
y
0
y
0
x
x
y
 for the first mile, 
 for each of the next 
 miles, and 
 miles 
f(m) = 2.05 + 1.56(m-4)
f(m) = 2.05 + 2.78(m-4)
f(m) = 3.61 + 1.22(m-4)
f(m) = 6.73 + 1.22(m-4)
4
2.05
1.56
3
6.73
1.22\,(m-4)
\qquad f(m)=6.73+1.22(m-4)
f(m) = 6.73 + 1.22(m-4)
t(x)
u(x)
y = t(x)
y = u(x)
u(x) = -t(2x)
u(x) = -t(-2x)
u(x) = -t \left(\dfrac{x}{2} \right)
u(x) = -t \left(-\dfrac{x}{2}\right)
u(x) = -t(kx)
k
y = u(x)
y = t(kx)
x
k
y = t(x)
x
y = u(x)
y = -t(x)
2
x
y = -t(x)
x
k > 0
x
\dfrac{1}{k}
k = 2
y = -t \left( \dfrac{x}{2} \right)
y = u(x)
u(x) = -t \left(\dfrac{x}{2} \right)
u(x) = -t \left(\dfrac{x}{2} \right)
73.5^\circ
1.8
\sin(73.5^\circ) \approx 0.96
\cos(73.5^\circ) \approx 0.28
\tan(73.5^\circ) \approx 3.38
0.51\,\text{m}
0.53\,\text{m}
1.27\,\text{m}
1.73\,\text{m}
73.5^\circ
x
1.8\,\text{m}
x
0.53\text{ m}
1990
2010
 less 
 times the number of years either before or after 
1993
2005
1993
2007
1995
2005
1995
2003
\qquad 750 - 575 = 175
2000
\qquad 175 \div 35 = 5
5
2000
\qquad |x-2000| = 5
~~~~
ABC
AB
BC
DEF
ABC
p
ABC
\angle BAC
\angle BCA
t^\circ
q^\circ
ABC
180
t^\circ
t^\circ
q^\circ
180
p^\circ
q^\circ
\qquad p^\circ = 80^\circ
\qquad (5-i)^2
i=\sqrt{-1}
24-10i
24+10i
26-10i
26+10i
5-i
i^2=-1
\qquad24-10i
\qquad y=mx+b
m
y
b
m
m=\dfrac{y}{x+b}
m = \dfrac{y + b}{x}
m = \dfrac{y - b}{x}
m = \dfrac{y}{x}-b
mx
\blue{\text{subtracting } b}
m
\pink{\text{dividing both sides of the equation by } x}
\qquad m=\dfrac{y-b}{x}
c
c
(a, b)
0
\dfrac{5}{17}
\dfrac{5}{3}
b
c
c = 0
0
c
\dfrac{2}{5}
a
c
0
c
b
-\dfrac{14}{25} \ne \dfrac{14}{25}
c = \dfrac53
\qquad w=6+\dfrac{3xy}{75}
y
w
x
y=\dfrac{25w}{x}-6
y=\dfrac{25w+150}{x}
y=\dfrac{25w-150}{x}
y=\dfrac{25w-2}{x}
\green{\text{subtracting } 6 \text{ from both sides of the equation}}
\dfrac{3xy}{75}
\red{\text{multiply}}
\red{\text{by }75}
\red{75}
y
\blue{\text{dividing by }3x}
\qquad y=\dfrac{25w-150}{x}
\qquad x(x-a)-b(a+b)=0
a
b
x=-b
x=(a-b)
x=-b
x=(a+b)
x=a
x=(a-b)
x=a
x=(a+b)
x
x
S
T
S+T=-a
ST=-(a+b)b
S=-(a+b)
T=b
\qquad x=-b
x=(a+b)
{\qquad p=14.63\cdot0.8^a}
(\text{psi})
a
{0.8^a}
0.8\,\text{psi}
0.8\,\text{psi}
20\%
20\%
a
|0.8|<1
a
{0.8^a}
0
0\,\text{psi}
0.8\,\text{psi}
0
0
a
0
14.63\,\text{psi}
0\,\text{psi}
1
a=0
a=1
0.8
80\%
20\%
20\%
\qquad D = 10 \cdot 2.5^{-t}
10
D
t
D
D
0
D
0
D
0
D
0
D
\qquad D = \dfrac{10}{2.5^{t}}
D
2.5
D
D
0
10
2.5
D
0
2
4
.  It costs 
 for 
 hot dogs and 
h
d
\qquad2h+4d=8
\qquad3h+2d = 7
h
\qquad h=4-2d
-1
0
1
2
0
\qquad {4^0 = 1}
0
0
1
1
\qquad {1^{-5} = 1}
2
150^{\circ}
15 \pi
(\text{cm})
0.3 \text{ cm}
2.6 \text{ cm}
5 \text{ cm}
18 \text{ cm}
s
\theta
r
s=r\theta
\theta
150^{\circ}
2\pi
360^{\circ}
\qquad 2\pi \text{ radians}=360^{\circ}
12
\dfrac {\pi}{6}=30^{\circ}
30^{\circ}=\dfrac {\pi}{6}\text{ radians}
150^{\circ}=\dfrac {5\pi}{6}\text{ radians}
18 \text{ cm}
2
3
4
446
x
b
x
b
\dfrac{x}3 + \dfrac{b}4 \leq 446
\dfrac{x}3 + \dfrac{b}4 \geq 446
3x + 4b \leq 446
3x + 4b \geq 446
a
3
\dfrac a3
\dfrac b4
446
x
x
a
a
(x \leq a)
446
a
a
(p, q) = (1, 1)
0
1
2
(1, 1)
(1, 1)
(1, 1)
a
(1, 1)
(1, 1)
a
(1, 1)
1
2010
12{,}000
2011
5\%
2010
2012
7\%
2011
2
240
882
1440
1482
2011
5\%
2010
2011
2011
5\%
12{,}000
12{,}000
\qquad  0.05\times 12{,}000+12{,}000=12{,}600
12{,}000
105\%
12{,}000
\qquad 12{,}000\times (0.05+1)
\qquad \text{or}
\qquad12{,}000\times 1.05=12{,}600
2011
12{,}600
2012
7\%
2011
2012
2012
7\%
12{,}600
12{,}600
107\%
12{,}000
\qquad  0.07\times 12{,}600+12{,}600=13{,}482
\qquad \text{or}
\qquad 12{,}6000\times 1.07=13{,}482
2012
13{,}482
2012
2010
1482
\qquad0=32-50x^2
x=\ \ \ \dfrac{4}{5}
x=\pm\dfrac{4}{5}
x=\ \ \ \dfrac{16}{25}
x=\pm\dfrac{16}{25}
0=32-50x^2
2
2
0
\qquad 0=2(4-5x)(4+5x)
\qquad4-5x=0 \qquad \text{or} \qquad 4+5x=0
x
x=\dfrac{4}{5}
x=-\dfrac{4}{5}
x=\pm\dfrac{4}{5}
\qquad W(x)=0.25x^2+0.2x
V(x)=100x+40
W(x)\cdot V(x)
0.25x^2+99.8x+40
25x^3+30x^2+8x
0.25x^2+20x+40
25x^3+10x^2+20x+8
\qquad (0.25x^2+0.2x)(100x+40)
25x^3+30x^2+8x
\qquad W(x)\cdot V(x)=25x^3+30x^2+8x
(PV)
, which of the following functions models the present value, 
, to be invested in a savings account earning 
 interest compounded annually for 
PV(t)=10{,}000(1.05)^{\large{t}}
PV(t)=10{,}000(1.05)^{\large{-t}}
PV(t)=10{,}000(1+0.05t)
PV(t)=10{,}000(1-0.05t)
A(t)
t
\qquad A(t)=PV (1+0.05)^t
PV
(
A(t)
PV
  after  
 years, let's  solve for 
 is the present value of 
 after 
 years earning 
\qquad PV=\dfrac{10{,}000}{(1.05)^{\large{t}}}=10{,}000(1.05)^{\large{-t}}
\qquad PV(t)=10{,}000(1.05)^{\large{-t}}
\qquad Q(t)=2-0.3t+0.8t^2
R(t)=1.5t^2+0.7t-6.3
R(t)-Q(t)
-0.7t^2-t+8.3
-0.5t^2+t-7.1
0.7t^2+0.4t-4.3
0.7t^2+t-8.3
\qquad(1.5t^2+0.7t-6.3)-(2-0.3t+0.8t^2)
\qquad R(t)-Q(t)=0.7t^2+t-8.3
A(x)=4x-5
B(x)=3x-7
A(x)\cdot B(x)
-31x+35
12x^2+35
7x^2-13x+35
12x^2-43x+35
A(x)\cdot B(x)
4x-5
A(x)
3x-7
B(x)
12x^2-43x+35
A(x)\cdot B(x)=12x^2-43x+35
2014
1986
1986
27.4
22.0
18.1
27.6
27.2
77.35
77{,}350
27.2
(\text{1986 dollars})
1
1986
2.13
2014
1
2014
0.76
2014
S
2014\,\text{euros}
27.2
1980
45
\qquad\qquad
70\,\text{cm}
\pi\approx 3.14
5{,}792\,\text{cm}^3
46{,}338\,\text{cm}^3
179{,}594\,\text{cm}^3
1{,}436{,}755\,\text{cm}^3
r
2\pi r
70
\qquad r=\dfrac{70}{2\pi}=\dfrac{35}{\pi}
5{,}792\,\text{cm}^3
14{,}697
 per day in November of last year, as compared with 
 per day in November two years ago.  The estimates had a margin of error of 
 at the 
 and 
95\%
 and 
91\%
99\%
95\%
 and 
\quad
19
250
9{,}597
250
250
\qquad\qquad \qquad\quad  \text{Cost for a }
 second 
 advertisement during a major sporting event from 
 to 
, where 
 is years since 
 and 
30\text { sec}
1970
30\text { sec}
2010
1970
2010
30\text { sec}
1970
2010
30\text { sec}
10\%
30\text { sec}
1970
100{,}000
30\text { sec}
2010
 and 
, the cost to run a 
 Super Bowl advertisement increased by about 
A(t)=A_0\cdot (1+r)^t
A_0
r
A
t
90
(10, 210)
r
r
9\%
30\text { sec}
9\%
1970
2010
30\text { sec}
10\%
\qquad \dfrac{1}{3} - \dfrac{1}{2u^2} + \dfrac{2}{u^3}
u<-5
\dfrac{2}{3u^3}
\dfrac{-2}{6u^5}
\dfrac{2u^3 + 9}{6u^3}
\dfrac{2u^3 - 3u + 12}{6u^3}
6u^3
\blue{\dfrac {2u^3}{ 2u^3}}
\pink{\dfrac {3u} {3u}}
\purple{ \dfrac 6 6 }
1
\qquad \dfrac{2u^3 - 3u + 12}{6u^3}
1{,}500{,}000
(\text{kg})
363\,\text{kg}
6\,\text{kg}
2
b
c
1{,}500{,}000
b
363\,\text{kg}
c
6\,\text{kg}
2
2
2
\qquad
4
4.5
5
5.5
30
15^\text{th}
16^\text{th}
15^\text{th}
4
16^\text{th}
5.
\qquad \dfrac{4+5}{2} = 4.5
p
t
\qquad p = 50{,}000(2t-1)(t-5)
t
t
0
p = 0
t
p
0
\qquad 2t - 1 = 0 \qquad t - 5 = 0
\qquad t = \dfrac{1}{2} \qquad \qquad t = 5
t = \dfrac{1}{2} \;
t = 5
t = \dfrac{1}{2}
EH
3
E
H
(\text{J})
5 \text{ J}
3
H
H
(20, 155)
H
155
155 \text{ J}
(-a, -1)
a
(-3, -1)
(1,3)
(-3,-1)
(
y
-1
)
(-3,-1)
y
-1
-a=-3
a=3
\qquad a=3
\qquad (5+i^2)(2-2i)
i=\sqrt{-1}
6-8i
8-8i
10-8i
12-8i
i^2=-1
i^3=-i
\qquad 8-8i
9.6
0.8
\pi = 3.14
9.6 \, \text{cm}^3
0.8 \, \text{cm}
\qquad V = \pi r^2 h
V
h
\qquad 9.6 \, \text{cm}^3= \pi r^2 (0.8 \, \text{cm})
\pi
0.8 \, \text{cm}
r^2
r
\qquad r = 2 
y = f(x)
y = g(x)
2
x
g(x) = 9^x
f(x)
f(x) = 81^x
f(x) = 2 \cdot 9^x
f(x) = 9^{x+2}
f(x) = 3^x
y = g(x)
x
k > 0
x
 \dfrac{1}{k} 
k = 2
a^{bc} = (a^b)^c
f(x) = 3^x
11:6
2.75
1.5
4.5
5
15
1
n
\qquad \dfrac{11}{6}=\dfrac{2.75}{n}
n
\qquad n=1.5
1.5
3
4.5
3
\gray{\text{rational exponents}}
x
\blue{\text{quotient of a power}}
\dfrac{x^a}{x^b}=x^{a-b}
\pink{\text{negative exponent}}
x^{-a}=\dfrac{1}{x^a}
\gray{\text{radical}}
4^{3}
64^3
8^{18}
64^{24}
\qquad \left(x^m\right)^n = x^{mn}
\qquad \left(10-8i^3\right)-(6+i)
i=\sqrt{-1}
4-7i
4+7i
4+9i
4-9i
i
i
\sqrt{-1}
i^2
-1
i^3
-\sqrt{-1}=-i
i^4
i^2 \cdot i^2 = 1
-1
\qquad 4+7i
(8r^4+4r^2+2r)-(8r^4+4r^2+1)
\qquad\quad (8r^4+4r^2+2r)-(8r^4+4r^2+1)
\qquad =(8r^4+4r^2+2r)+((-8)r^4+(-4)r^2+(-1))
\qquad =8r^4+4r^2+2r+(-8)r^4+(-4)r^2+(-1)
\qquad =8r^4+(-8)r^4+4r^2+(-4)r^2+2r+(-1)
\qquad =(8-8)r^4+(4-4)r^2+2r+(-1)
\qquad =(0)r^4+(0)r^2+2r-1
\qquad =2r-1
x
\angle
 and the measure of 
BAD=30^\circ
\angle
 must be 
180^\circ
x
\overline{CA}
CAD
\theta
\qquad\qquad \cos\theta=\dfrac{\red{\text{Length of the side adjacent to }\theta}}{\blue{\text{Length of the triangle's hypotenuse}}}
CAD
\angle CAD=60^\circ
\red{\overline{AD}}
\red{\text{{adjacent}}}
\angle{CAD}
\red6
\blue{\overline{AC}}
\blue{\text{{hypotenuse}}}
CAD
\blue{x}
\qquad\qquad \cos60^\circ=\dfrac{\red{6}}{\blue{x}}
\cos60^\circ
\dfrac12
x
12
\qquad \sqrt{6r }=\sqrt{r^2-16}
r
0
r=8
r=-2
r=\green8
r=8
r=\green{-2}
\sqrt{-12}
r=-2
r=8
8
8
5x+10y=30
xy
x
3
y
6
\dfrac{1}{2}
-\dfrac{2}{1}
x
6
y
3
x
y
5x+10y=30
x
0
y
x
x
5x+10y=30
(6,0)
y
0
x
y
y
5x+10y=30
(0,3)
x
6
y
3
\qquad
\angle{EFG}
\angle EFG
\theta
\theta
\qquad\qquad \tan\theta=\dfrac{\red{\text{Length of the side opposite from }\theta}}{\blue{\text{Length of the side adjacent to }\theta}}
EFG
\angle EFG
\red{\text{{opposite}}}
\angle EFG
\red {7\sqrt6}
\blue{\text{{adjacent}}}
\angle EFG
\blue{7\sqrt6}
\qquad\qquad \tan\theta=\dfrac{\red{7\sqrt6}}{\blue{7\sqrt6}}=1
0^\circ<\theta<90^\circ
\tan45^\circ
1
\theta=45^\circ
EFG
45^\circ
55
65
(\text{mph})
 penalty is assessed for each 
49
66 \,\text{mph}
49
71 \,\text{mph}
54
66 \,\text{mph}
54
71 \,\text{mph}
6\,\text{mph}
\qquad \dfrac{1}{1-6i}-\dfrac{1}{1+6i}
i=\sqrt{-1}
\dfrac{12}{37}i
-\dfrac{12}{37}i
\dfrac{12}{37}
-\dfrac{12}{37}
\qquad(1-6i)(1+6i)=1-(6i)^2=1-36(-1)=37
\qquad \dfrac{1}{1-6i}-\dfrac{1}{1+6i}=\dfrac{12}{37}i
\qquad \dfrac{1}{1-6i}-\dfrac{1}{1+6i}=\dfrac{12}{37}i
\qquad F = G \dfrac{Mm}{r^2}
F
M
m
r
G
m
M
 m = G \dfrac{FM}{r^2}
m = \dfrac{FGr^2 } {M}
m = \dfrac{Fr^2 } {GM}
m = \dfrac{F} {GMr^2 }
m
\pink{\text{multiplying both sides by } r^2}
\blue{\text{divide both sides by } GM}
\dfrac{r^2}{GM}
\qquad m = \dfrac{Fr^2 } {GM}
(3,0)
(-2,0)
(2,0)
(0,-6)
(4,6)
(0,-6)
(-6,0)
(6,4)
(4,6)
(0,-6)
(4,6)
(0,-6)
58
(\text{mg})
C
\text{ mg}
t
t
C
C=58-5t
C=58-9t
C=58(0.84)^t
C=58(1.18)^t
y=a\cdot b^x
a
y
(C
)
b>0
a>0
0<b<1
a>0
b>1
C=58-5t
C=58-9t
C=58(1.18)^t
a>0
b>0
C=58(0.84)^t
58
(13, 6)
b^{13}=0.84
b
13^{\text{th}}
\dfrac{1}{13}
C=58(0.84)^t
C=58(0.84)^t
C
t
10
500
40\%
5
15
10
500
40\%
5
60\%
0.6
5
15
3
\qquad 500\,\text{words}\cdot(0.6)(0.6)(0.6)=500\cdot0.6^3=108\,\text{words}
\qquad 650-108=542
542
15
1
1
0
150
1{,}000
1
\qquad \dfrac {2}{2-i} -\dfrac{2}{2+i}
i=\sqrt{-1}
\dfrac {4i}{5}
\dfrac {-4i}{5}
\dfrac {2i}{3}
\dfrac {-2i}{3}
\qquad \dfrac{4i}{5}
500
DE
E
(\text{m})
D
(\text{km})
\text{m/km}
D = 0
E = 0
(0, 0)
(6.5, 130)
20 \text{ m/km}
s = (t+3)^2(t+2)(t+1)(t)(t-1)
st
t
t
t
t
-3
-2
-1
0
1
t
0
\qquad \dfrac{\left(     \dfrac{10x^2}{7y^3}         \right) }{\left(      \dfrac{5x}{2y^6}      \right)}
x>1
y>2
\dfrac{50x^4}{14y^9} 
\dfrac{7}{4xy^3} 
\dfrac{4xy^3}{7} 
\dfrac{4x^2y^2}{7} 
1
1
\dfrac x x
1
1x
1
\qquad  \dfrac{4xy^3}{7} 
x^2+y^2=4
x^2+y^2=r^2
(0,0)
r
x^2+y^2=4
(0,0)
2
\left(x^2+y^2=2^2\right)
\qquad 
y=x^2-1
y=a(x-h)^2+k
(h,k)
y=x^2-1
(0,-1)
a>0
\left((1,0), (2, 3), (-1,0)\right)
45
5
2
x
2
45x + 5\leq120
45x + 5\geq120
45 + 5(x-1)\leq120
45 + 5x\leq120
2
120
\qquad \text{original time} + \text{additional time per week} \leq120 \space \text{minutes}
x
5x
x-1
\qquad 45 + 5(x-1)\leq120
\qquad 45 + 5(x-1)\leq20
\qquad(x+20)^2+(y-30)^2=225
xy
y
(20,-30)
(-20,30)
(400,-900)
(-400,900)
(\green h,\purple k)
\red r
\qquad(x-\green h)^2+(y-\purple k)^2=\red r^2
\qquad(x-\green{20})^2+(y-\purple{30})^2=\red{15}^2
y
\purple{30}
\qquad \left(\dfrac{1}{2}\right)^{-2}+3^0
-\dfrac{1}{4}
\dfrac{3}{4}
4
5
x\neq0
x^{-n}=\dfrac{1}{x^n}
-n^{\text{th}}
n^{\text{th}}
\left(\dfrac{1}{2}\right)^{-2}
\left(\dfrac{1}{2}\right)^{2}
\left(\dfrac{1}{2}\right)^{2}=\dfrac{1}{4}
\left(\dfrac{1}{2}\right)^{-2}=4 
x^0=1
x\neq0
x\neq0
x^0=1
5
\qquad\dfrac{5m}{m^2-24mn+144n^2}+\dfrac{2n}{m^2-144n^2}
\dfrac{5m^2+60mn+2n^2}{(m-12n)(m+12n)}
\dfrac{5m^2+62mn-24n^2}{(m-12n)^2(m+12n)}
\dfrac{5m-24mn+2n}{(m-12n)(m-12n)}
\dfrac{5m^2-58mn+24n^2}{(m-12n)(m+12n)^2}
\qquad\dfrac{5m}{(m-12n)^2}+\dfrac{2n}{(m+12n)(m-12n)}
\qquad (m-12n)^2(m+12n)
\dfrac{(m+12n)}{(m+12n)}
\dfrac{(m-12n)}{(m-12n)}
\qquad\dfrac{5m(m+12n)}{(m-12n)^2(m+12n)}+\dfrac{2n(m-12n)}{(m-12n)^2(m+12n)}
\qquad\dfrac{5m^2+62mn-24n^2}{(m-12n)^2(m+12n)}
\text{{Time in minutes}}
\text{{Number of runners}}
18
 1
19
2
20
3
21
6
22
 r
23
1
24
2
3
21
r
1
2
3
6
21
\qquad 21=\dfrac{18\cdot1+19\cdot2+20\cdot3+21\cdot6+22\cdot r +23\cdot1+24\cdot2}{15+r}
2
22
180^{\circ}
195^{\circ}
210^{\circ}
225^{\circ}
2\pi
360^{\circ}
\qquad 2\pi \text{ radians}=360^{\circ}
2\pi
1\text{ radian}=\dfrac{180}{\pi}^{\circ}
 2\pi \text{ radians}=360^{\circ}
24
 \dfrac{\pi}{12} \text{ radians}=15^{\circ}
\dfrac{13\pi}{12}
13\cdot  \dfrac{\pi}{12}
\dfrac{13\pi}{12}\text{ radians}
13\cdot 15^{\circ}
195^{\circ}
\qquad\dfrac{13\pi}{12}=195^{\circ}
\left(3v+2\right)-\left(\dfrac{2}{3}v-\dfrac{1}{4}\right)
\dfrac{7}{3}v+\dfrac{7}{4}
\dfrac{7}{3}v+\dfrac{9}{4}
2v^2+\dfrac{1}{2}
2v^2+\dfrac{7}{12}x-\dfrac{1}{2}
\qquad \left(3v+2\right)-\left(\dfrac{2}{3}v-\dfrac{1}{4}\right)=3v+2-\dfrac{2}{3}v+\dfrac{1}{4}
\qquad \dfrac{7}{3}v+\dfrac{9}{4}
y=2x^2-5x-3
x
y=a(x-h)^2+k
(h,k)
y=2x^2-5x-3
y=2x^2-5x-3
x^2
1
2 \cdot \dfrac{25}{16}\;
\dfrac{25}{8}\;
\dfrac{25}{8}\;
y=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{49}{8}
\left(\dfrac{5}{4},-\dfrac{49}{8}\right)
x
y=ax^2+bx+c
x=-\dfrac{b}{2a}
y=2x^2-5x-3
x
\dfrac{5}{4}
2002
 per pound and the price of tuna fish was 
 per pound.  For the following 
 months, the price of ground beef increased at the rate of 
 per month and the price of tuna fish decreased at 
 per month. In approximately how many months after the beginning of January 
7.5
7.5
10
10
x
2002
 per pound and increased 
 each month.  The price 
 per pound and decreased 
 each month.  The price 
10
2002
10
\qquad 1.70 + 0.03x
\qquad 1.70 + 0.03\cdot 10
\qquad 1.70 + 0.30
\qquad 2.00
10
3
. If the cost of the hat costs 
3x<94
3x>94
4x<94
4x>94
\qquad \text{cost of coat} + \text{cost of hat} <94
3
, then the coat costs 
\qquad 4x<94
2014
1\text{,}000
560
56\%
56\%
56\%
56\%
1\text{,}000
56\%
56\%
56\%
56\%
56\%
\qquad\qquad\qquad\qquad \text{Original data} 
\qquad \text{Data without the lowest and highest heights } 
y=8\left(\dfrac{3}{4}\right)^x
y=\dfrac{1}{8}\cdot\left(\dfrac{7}{4}\right)^x
xy
y=8\left(\dfrac{3}{4}\right)^x
y=\dfrac{1}{8}\cdot\left(\dfrac{7}{4}\right)^x
y=8\left(\dfrac{3}{4}\right)^x
y=\dfrac{1}{8}\cdot\left(\dfrac{7}{4}\right)^x
y=ab^x
a
b
0< b<1
b>1
a
y
0< \dfrac{3}{4}<1
y=8\left(\dfrac{3}{4}\right)^x
\dfrac{7}{4}>1
y=\dfrac{1}{8}\cdot\left(\dfrac{7}{4}\right)^x
y=8\left(\dfrac{3}{4}\right)^x
y=\dfrac{1}{8}\cdot\left(\dfrac{7}{4}\right)^x
c<0
\qquad\dfrac{3c^2-300}{5c^2-52c+20}
\dfrac{c+10}{5c-2}
\dfrac{3c+30}{5c-2}
\dfrac{c-20}{5c+4}
\dfrac{3c-60}{5c+4}
3
\qquad3c^2-300=3(c^2-100)=3(c+10)(c-10)
\qquad5c^2-52c+20=(5c-2)(c-10)
c
0
c<0
\qquad \dfrac{3c+30}{5c-2}
6
56
.  Brand B has 
 rolls, each with 
 sheets, for 
6
56
6\cdot 46=336
8
48
8\cdot 48=384
439
25\%
110
351
549
1756
25\%
0.25
439
x
\qquad 0.25x = 439
0.25
x = 1756
1756
\qquad45, 45, 56, 56, 73, 73, 73
56
\qquad \dfrac{45+45+56+56+73+73+73}{7} = 60.143\approx 60.14
\qquad 60.14 - 56 = 4.14
tP
P
(\text{nW})
t
t
t
P
0
P
0 \text{ nW}
\left(1-p\right)
\left(\dfrac{1}{2}-p\right)
p
\left(1-p\right)
\left(\dfrac{1}{2}-p\right)
p
\left(1-p\right) \cdot \left(\dfrac{1}{2}-p\right)
p
\quad\qquad \left(1-p\right) \cdot \left(\dfrac{1}{2}-p\right)-p
\dfrac12
\dfrac12
\qquad =\dfrac12\left[ \left(1-p\right) \cdot \left(1-2p\right)-2p\right]
\dfrac12
\qquad =\dfrac12\left[ (1-p-2p+2p^2)-2p\right]
\qquad =\dfrac12\left[ 1-p-2p+2p^2-2p\right]
\qquad =\dfrac12\left[ 1\blue{-p}\blue{-2p}+2p^2\blue{-2p}\right]
\qquad =\dfrac12\left( 1\blue{-5p}+2p^2\right)
\qquad n=\left(\dfrac{1.96\sigma}{E}\right)^2
n
\sigma
E
\sigma=\sqrt{\dfrac{En}{1.96}}
\sigma=\dfrac{E\sqrt{n}}{1.96}
\sigma=\sqrt{\dfrac{1.96n}{E}}
\sigma=\dfrac{1.96\sqrt{n}}{E}
\red{\text{multiply by }E}
\blue{\text{divide by }1.96}
\qquad\sigma=\dfrac{E\sqrt{n}}{1.96}
\qquad w = 500 - 6.2t
w
t
6.2t
6.2t
6.2t
6.2t
t
6.2t
t
500
6.2t
w
6.2t
6.2t
500
6.2t
t
\qquad\dfrac{3k}{k+4}-\dfrac{4k}{k-2}=-5
k
-2
5
-2
4
-2
5
\qquad(k+4)(k-2)
\qquad\dfrac{3k\pink{(k-2)}}{(k+4)\pink{(k-2)}}-\dfrac{4k\blue{(k+4)}}{\blue{(k+4)}(k-2)}=\dfrac{-5\pink{(k-2)}\blue{(k+4)}}{\pink{(k-2)}\blue{(k+4)}}
\qquad\dfrac{3k^2 - 6k}{(k+4)(k-2)}-\dfrac{4k^2+16k}{(k+4)(k-2)}=\dfrac{-5(k^2 + 2k - 8)}{(k-2)(k+4)}
k
\qquad 4(k - 5)(k + 2) = 0
k = 5
k = -2
k = 5
k =5
k = -2
k =-2
k = 5
k= -2
f(x)=\dfrac{2}{3x-1}
g(x)=\dfrac{x+4}{x}
(g\circ f)(x)
x\neq\dfrac{1}{3}
6x-1
6x-2
\dfrac{x}{x+6}
\dfrac{2x+8}{3x^2-x}
(g\circ f)(x)=g(f(x))
g(f(x))
f(x)
g
\qquad g(f(x))=\dfrac{(f(x))+4}{(f(x))}
f(x)=\dfrac{2}{3x-1}
\qquad g(f(x))=\dfrac{\dfrac{2}{3x-1}+4}{\dfrac{2}{3x-1}}
3x-1
\qquad g(f(x))=\dfrac{\dfrac{2}{3x-1}+4}{\dfrac{2}{3x-1}}\cdot \dfrac{\dfrac{3x-1}{1}}{\dfrac{3x-1}{1}}
1
(g\circ f)(x)
x
f
f(x)
g
f
x\neq\dfrac{1}{3}
f(x)
g
f(x)
g
f(x)\neq0
\dfrac{2}{3x-1}\neq0
(g\circ f)(x)=6x-1
x\neq\dfrac{1}{3}
y=\dfrac{2}{5}x+2
y=\dfrac{2}{5}x-2
y=-\dfrac{2}{5}x+2
y=-\dfrac{2}{5}x-2
y
y=mx+b
y
(0,-2)
y
-2
\dfrac{y_2-y_1}{x_2-x_1}
(0,-2)
(5,-4)
\qquad \dfrac{-4-(-2)}{5-0}=-\dfrac{2}{5}
y=mx+b
m
b
y
\qquad y=-\dfrac{2}{5}x-2
y=-\dfrac{2}{5}x-2
\qquad\qquad\qquad
20
4
\pi\approx 3.14
452
904
1{,}072
8{,}574
20
10
4
\qquad r=10-4=6
\qquad\qquad
6
452
\qquad\dfrac{2.5+3.0}{2}=2.75
2.75
2.9\overline{6}
0.2