react-components/test/katex-a11y-math.txt

React components used by Khan Academy

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