Resources

Resources

Compare linear and exponential growth

Write the equation

2(x+3)=2 x+6

as a system of equations.

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\begin{array}{l} y=\square x+\square \newline y=\square x+\square \end{array}

f(x)=2^{x}, f(x)=2 \cdot 2^{x}

f(x)=\frac{1}{2} \cdot 2^{x}

all have the same nuultiplier.

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a. What is the multiplier for these functions? Make an

x \rightarrow f(x)

table for each of the functions, using integer value of

x

x=-3

x=5

. How can you use the tables to determine the multipliers?

A line is represented by the equation

3x+2y=4

. What is another way to represent the same line?

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y=-\frac{3}{2}x+2

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y=\frac{3}{2}x+2

\newline

y=\frac{3}{2}x+4

\newline

y=-\frac{3}{2}x+4

David is writing an explicit function for the arithmetic sequence:

10, 13, 16, 19, \dots

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He comes up with

s(n) = 7 + 3n.

What domain should David use for

s

so it generates the sequence?

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1

\newline

n \geq 0

n

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n \geq 0

n

\newline

n \geq 1

n

\newline

n \geq 1

n

Select all of the true statements.

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Multi-select Choices:

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1 \times 10^{-2}

50

times as much as

2 \times 10^{-4}

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8 \times 10^{2}

0.02

times as much as

4 \times 10^{4}

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3 \times 10^{-3}

6

times as much as

5 \times 10^{-2}

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8 \times 10^{9}

50

0

times as much as

50

1

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9y^{3}+3y^{2}-2y+1

-5y^{2}+y-4

in a horizontal format.

h(x) = 5^x

change over the interval from

x = 8

x = 9

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\newline

h(x)

400\%

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h(x)

5

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h(x)

500\%

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h(x)

5

h(x) = 8^x

change over the interval from

x = 1

x = 2

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\newline

h(x)

800\%

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h(x)

increases by a factor of

8

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h(x)

8

\newline

h(x)

x = 1

0

h(t)

5

over every unit interval in

t

h(0) = 0

\newline

Which could be a function rule for

h(t)

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\newline

h(t) = \frac{t}{5}

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h(t) = t + 5

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h(t) = \frac{1}{5^t}

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h(t) = 5t

x^{2}+8x+6=0

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What are the solutions to the given equation?

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1

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x=-8-\sqrt{10} \text{ and } \ x=-8+\sqrt{10}

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x=-8-\sqrt{7} \text{ and } \ x=-8+\sqrt{7}

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x=-4-\sqrt{10} \text{ and } \ x=-4+\sqrt{10}

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x=-4-\sqrt{7} \text{ and } \ x=-4+\sqrt{7}

\begin{array}{l} P=2^{2} \times 3^{2} \times 5 \\ Q=2^{3} \times 3 \times 7 \end{array}

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Find the highest common factor (HCF) of

P

5 \boldsymbol{Q}

Consider the polynomial function

h(x)=x^{6}-3x^{5}-17x^{3}

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What is the end behavior of the graph of

h

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1

\newline

x \rightarrow \infty, h(x)\rightarrow \infty

x \rightarrow -\infty, h(x)\rightarrow \infty

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x \rightarrow \infty, h(x)\rightarrow -\infty

x \rightarrow -\infty, h(x)\rightarrow \infty

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x \rightarrow \infty, h(x)\rightarrow -\infty

x \rightarrow -\infty, h(x)\rightarrow -\infty

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x \rightarrow \infty, h(x)\rightarrow \infty

x \rightarrow -\infty, h(x)\rightarrow -\infty

Write and equation for a line passing through this point, and with this slope

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\begin{array}{l} (2,6) \\ m=3 \end{array}

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\begin{array}{l} y=3 x \\ y=3 x+2 \\ y=x+3 \\ y=\frac{1}{2} x+3 \end{array}

Solve the equation below for

x

a

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4(ax+3)-3ax = 25+3a

Based on the following calculator output, determine the range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=205.428571429 \\ \Sigma x=1438 \\ \Sigma x^{2}=334956 \\ S x=81.1887864648 \\ \sigma x=75.1662103852 \\ n=7 \\ \operatorname{minX}=78 \\ \mathrm{Q}_{1}=113 \\ \mathrm{Med}^{2}=214 \\ \mathrm{Q}_{3}=277 \\ \max \mathrm{X}=283 \end{array}

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Based on the following calculator output, determine the inter-quartile range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=198.571428571 \\ \Sigma x=1390 \\ \Sigma x^{2}=280364 \\ S x=26.9249397576 \\ \sigma x=24.9276504128 \\ n=7 \\ \min \mathrm{X}=159 \\ \mathrm{Q}_{1}=175 \\ \mathrm{Med}^{2}=199 \\ \mathrm{Q}_{3}=209 \\ \max \mathrm{X}=244 \end{array}

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Based on the following calculator output, determine the mean of the dataset, rounding to the nearest

10

oth if necessary.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=169.285714286 \\ \Sigma x=1185 \\ \Sigma x^{2}=202903 \\ S x=19.5764678949 \\ \sigma x=18.1242874596 \\ n=7 \\ \min \mathrm{X}=140 \\ \mathrm{Q}_{1}=145 \\ \mathrm{Med}^{2}=177 \\ \mathrm{Q}_{3}=184 \\ \max \mathrm{X}=192 \end{array}

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Based on the following calculator output, determine the inter-quartile range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=64.4285714286 \\ \Sigma x=451 \\ \Sigma x^{2}=29415 \\ S x=7.72133716522 \\ \sigma x=7.14856914468 \\ n=7 \\ \operatorname{minX}=54 \\ \mathrm{Q}_{1}=57 \\ \mathrm{Med}^{2}=66 \\ \mathrm{Q}_{3}=73 \\ \operatorname{maxX}=74 \end{array}

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Based on the following calculator output, determine the inter-quartile range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=91.7142857143 \\ \Sigma x=642 \\ \Sigma x^{2}=69948 \\ S x=42.9484741123 \\ \sigma x=39.7625605877 \\ n=7 \\ \operatorname{minX}=33 \\ \mathrm{Q}_{1}=45 \\ \mathrm{Med}^{2}=92 \\ \mathrm{Q}_{3}=117 \\ \operatorname{maxX}=156 \end{array}

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Based on the following calculator output, determine the mean of the dataset, rounding to the nearest

10

oth if necessary.

\newline

\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=53.7142857143 \\ \Sigma x=376 \\ \Sigma x^{2}=20530 \\ S x=7.45462464323 \\ \sigma x=6.90164133096 \\ n=7 \\ \operatorname{minX}=44 \\ \mathrm{Q}_{1}=48 \\ \mathrm{Med}^{2}=53 \\ \mathrm{Q}_{3}=63 \\ \max \mathrm{X}=64 \end{array}

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Based on the following calculator output, determine the inter-quartile range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=216.857142857 \\ \Sigma x=1518 \\ \Sigma x^{2}=356184 \\ S x=67.0756502551 \\ \sigma x=62.0999852115 \\ n=7 \\ \operatorname{minX}=108 \\ \mathrm{Q}_{1}=135 \\ \mathrm{Med}^{2}=245 \\ \mathrm{Q}_{3}=267 \\ \operatorname{maxX}=278 \end{array}

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Based on the following calculator output, determine the range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=151.857142857 \\ \Sigma x=1063 \\ \Sigma x^{2}=172815 \\ S x=43.5715066354 \\ \sigma x=40.3393766204 \\ n=7 \\ \operatorname{minX}=109 \\ Q_{1}=122 \\ \mathrm{Med}^{2}=140 \\ \mathrm{Q}_{3}=162 \\ \max \mathrm{X}=241 \end{array}

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Based on the following calculator output, determine the range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=57 \\ \Sigma x=399 \\ \Sigma x^{2}=23479 \\ S x=11.0754984839 \\ \sigma x=10.2539191114 \\ n=7 \\ \operatorname{minX}=46 \\ Q_{1}=47 \\ \mathrm{Med}^{2}=56 \\ \mathrm{Q}_{3}=63 \\ \max \mathrm{X}=76 \end{array}

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Based on the following calculator output, determine the range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=60.2857142857 \\ \Sigma x=422 \\ \Sigma x^{2}=27106 \\ S x=16.6604750404 \\ \sigma x=15.4246026642 \\ n=7 \\ \operatorname{minX}=37 \\ \mathrm{Q}_{1}=46 \\ \mathrm{Med}^{2}=57 \\ \mathrm{Q}_{3}=76 \\ \max \mathrm{X}=84 \end{array}

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Based on the following calculator output, determine the range of the dataset.

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\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=46.2857142857 \\ \Sigma x=324 \\ \Sigma x^{2}=15426 \\ S x=8.45998986828 \\ \sigma x=7.83242866393 \\ n=7 \\ \operatorname{minX}=34 \\ \mathrm{Q}_{1}=41 \\ \mathrm{Med}^{2}=45 \\ \mathrm{Q}_{3}=55 \\ \max \mathrm{X}=59 \end{array}

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Follow the steps to compute the volume of the solid obtained by rotating the region bounded by

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y=x^{2} \quad \text { and } \quad y=6 x

\newline

x=0

using the method of disks or washers.

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a. Using the method of disks or washers, set up the integral.

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\begin{array}{l} V=\int_{a}^{b}\square \\ \text { with } a=\square \text { and } b= \end{array}

Three points on the graph of the function

f(x)

\{(0,3),(1,9),(2,27)\}

. Which equation represents

f(x)

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f(x)=6 x+3

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f(x)=3(3)^{x}

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f(x)=18 x-9

\newline

f(x)=x^{2}+3

Three points on the graph of the function

f(x)

\{(0,2),(1,4),(2,6)\}

. Which equation represents

f(x)

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f(x)=x^{2}+2

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f(x)=2 x+4

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f(x)=2 \cdot 2^{x}

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f(x)=2 x+2

h(x)=\begin{cases} 5x, & \text{for } x < -2 \ x^{3}-2, & \text{for } x \geq -2 \end{cases}

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\lim_{x \to -2}h(x)

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1

\newline

-10

\newline

-2

\newline

10

ax^{3}+bx^{2}+cx+d=0

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In the equation above,

\newline

a,b,c,

\newline

d

\newline

If the equation has roots

\newline

-1,-3,

5

, which of the following is a factor of

\newline

ax^{3}+bx^{2}+cx+d

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x-1

\newline

x+1

\newline

x-3

\newline

x+5

An IV administers medication to a patient's bloodstream at a rate of

3

cubic centimeters per hour.

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At the same time, the patient's organs remove the medication from the patient's bloodstream at a rate proportional to the current volume

V

of medication in the bloodstream.

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Which equation describes this relationship?

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1

\newline

\frac{d V}{d t}=3-k V

\newline

\frac{d V}{d t}=-3 k V

\newline

\frac{d V}{d t}=3 k-V

\newline

\frac{d V}{d t}=k-3 V

Baylee tried to solve the differential equation

\frac{d y}{d x}=\frac{e^{x}}{y^{2}}

. This is her work:

\newline

\frac{d y}{d x}=\frac{e^{x}}{y^{2}}

\newline

1

\quad \int y^{2} d y=\int e^{x} d x

\newline

2

\quad \frac{y^{3}}{3}=e^{x}

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3

\quad y^{3}=3 e^{x}

\newline

4

\quad y=\sqrt[3]{3 e^{x}}+C

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Is Baylee's work correct? If not, what is her mistake?

\newline

1

\newline

(A) Baylee's work is correct.

\newline

1

is incorrect. The separation of variables wasn't done correctly.

\newline

2

is incorrect. The right-hand side of the equation should be

e^{x}+C

\newline

4

is incorrect. The right-hand side of the equation should be

\sqrt[3]{3 e^{x}}+\sqrt[3]{C}

Anastasia tried to solve the differential equation

\frac{d y}{d x}=y^{2} \sin (x)

. This is her work:

\newline

\frac{d y}{d x}=y^{2} \sin (x)

\newline

1

\quad \int y^{2} d y=\int \sin (x) d x

\newline

2

\quad \frac{y^{3}}{3}=-\cos (x)+C_{1}

\newline

3

\quad y^{3}=-3 \cos (x)+C

\newline

4

\quad y=\sqrt[3]{-3 \cos (x)+C}

\newline

Is Anastasia's work correct? If not, what is her mistake?

\newline

1

\newline

(A) Anastasia's work is correct.

\newline

1

is incorrect. The separation of variables wasn't done correctly.

\newline

2

is incorrect. Anastasia didn't integrate

\sin (x)

\newline

4

is incorrect. The right-hand side of the equation should be

\sqrt[3]{-\cos (x)}+C

Daniela tried to solve the differential equation

\frac{d y}{d x}=2 x \cdot e^{-y}

. This is her work:

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\frac{d y}{d x}=2 x \cdot e^{-y}

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1

\quad \int e^{-y} d y=\int 2 x d x

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2

\quad-e^{-y}=x^{2}+C_{1}

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3

\quad e^{-y}=-x^{2}+C

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4

\quad \ln \left(e^{-y}\right)=\ln \left(-x^{2}+C\right)

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5

\quad-y=\ln \left(-x^{2}+C\right)

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6

\quad y=-\ln \left(-x^{2}+C\right)

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Is Daniela's work correct? If not, what is her mistake?

\newline

1

\newline

(A) Daniela's work is correct.

\newline

1

is incorrect. The separation of variables wasn't done correctly.

\newline

2

is incorrect. Daniela didn't integrate

e^{-y}

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4

is incorrect. The right-hand side of the equation should be

\ln \left(-x^{2}\right)+C

The base of a solid is the region enclosed by the graphs of

y=\ln (x)

y=0.1 x^{3}-0.5 x^{2}

x=2

x=5

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Cross sections of the solid perpendicular to the

x

-axis are rectangles whose height is

6-x

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Which one of the definite integrals gives the volume of the solid?

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1

\newline

\int_{2}^{5}\left[0.1 x^{3}-0.5 x^{2}-\ln (x)\right](6-x) d x

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\int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right](6-x) d x

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\int_{2}^{5}\left[\ln ^{2}(x)+\left(0.1 x^{3}+0.5 x^{2}\right)^{2}\right] d x

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\int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right]^{2} d x

Harold tried to find all the points on the curve given by

x y^{2}-x^{2} y=-54

where the line tangent to the curve is horizontal. This is his solution:

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1

: Finding an expression for

\frac{d y}{d x}

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\frac{d y}{d x}=\frac{y(2 x-y)}{x(2 y-x)}

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2

: Forming a system of equations.

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\left\{\begin{array}{l} x y^{2}-x^{2} y=-54 \\ x(2 y-x)=0 \\ y(2 x-y) \neq 0 \end{array}\right.

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3

: Solving the system.

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(6,3)

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Is Harold's solution correct? If not, at which step did he make a mistake?

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1

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(A) The solution is correct.

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1

\newline

2

\newline

3

Kendall tried to find all the equations of vertical lines tangent to the curve given by

x^{2}+2 x y^{2}=25

. This is her solution:

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1

: Finding an expression for

\frac{d y}{d x}

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\frac{d y}{d x}=\frac{-x-y^{2}}{2 x y}

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2

: Forming a system of equations.

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\left\{\begin{array}{l} x^{2}+2 x y^{2}=25 \\ 2 x y=0 \\ -x-y^{2} \neq 0 \end{array}\right.

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3

: Solving the system.

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x=-5, x=0

x=5

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Is Kendall's solution correct? If not, at which step did she make a mistake?

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1

\newline

(A) The solution is correct.

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1

\newline

2

\newline

3

The base of a solid is the region enclosed by the graphs of

y=\ln (x)

y=0.1 x^{3}-0.5 x^{2}

x=2

x=5

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Cross sections of the solid perpendicular to the

x

-axis are rectangles whose height is

6-x

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Which one of the definite integrals gives the volume of the solid?

\newline

1

\newline

\int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right]^{2} d x

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\int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right](6-x) d x

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\int_{2}^{5}\left[0.1 x^{3}-0.5 x^{2}-\ln (x)\right](6-x) d x

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\int_{2}^{5}\left[\ln ^{2}(x)+\left(0.1 x^{3}+0.5 x^{2}\right)^{2}\right] d x

The base of a solid is the region enclosed by the graphs of

y=2+\frac{x}{2}

y=2^{\frac{x}{2}}

y

x=4

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Cross sections of the solid perpendicular to the

x

-axis are rectangles whose height is

4-x

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Which one of the definite integrals gives the volume of the solid?

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1

\newline

\int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}-2\right](4-x) d x

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\int_{0}^{4}\left[2^{\frac{x}{2}}+\frac{x}{2}+2\right](4-x) d x

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\int_{0}^{4}\left[2+\frac{x}{2}-2^{\frac{x}{2}}\right](4-x) d x

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\int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}+2\right](4-x) d x

William was given this problem:

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r(t)

of a sphere is increasing at a rate of

7

5

meters per minute. At a certain instant

t_{0}

, the radius is

5

meters. What is the rate of change of the surface area

S(t)

of the sphere at that instant?

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Which equation should William use to solve the problem?

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1

\newline

S(t)=2 \pi \cdot r(t)

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S(t)=\pi[r(t)]^{2}

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S(t)=4 \pi[r(t)]^{2}

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S(t)=\frac{4}{3} \pi[r(t)]^{3}

Consider the following problem:

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The air gap (the distance from the bottom of the bridge to the surface of the water) under a bridge is changing at a rate of

r(t)=0.6 \sin \left(\frac{\pi t}{6}\right)

meters per hour (where

t

is the time in hours). At time

t=13

, the air gap is

62

meters. What is the air gap when

t=15

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Which expression can we use to solve the problem?

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1

\newline

r^{\prime}(13)+62

\newline

r^{\prime}(15)

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62+\int_{13}^{15} r(t) d t

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62+\int_{13}^{15} r^{\prime}(t) d t

Consider the following problem:

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The number of people in a cafeteria is changing at a rate of

r(t)=1920-160 t

people per hour (where

t

is the time in hours). At time

t=11.5

60

people in the cafeteria. How many people were in the cafeteria at hour

12

5

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Which expression can we use to solve the problem?

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1

\newline

\int_{11.5}^{12.5} r^{\prime}(t) d t

\newline

60+\int_{11.5}^{12.5} r^{\prime}(t) d t

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\int_{11.5}^{12.5} r(t) d t

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60+\int_{11.5}^{12.5} r(t) d t

Consider the following problem:

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The velocity of a pendulum changes at a rate of

a(t)=0.9^{t} \cos (0.2 t)

meters per second squared (where

t

is the time in seconds). At time

t=6

, the velocity of the pendulum is

0

8

meters per second. By how much does the velocity change between

t=6

t=12

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Which expression can we use to solve the problem?

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1

\newline

\int_{6}^{12} a(t) d t

\newline

a^{\prime}(6)+0.8

\newline

a^{\prime}(12)

\newline

a^{\prime}(12)-a^{\prime}(6)

Consider the following problem:

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The total number of subscribers Zhang Wei has for his video page is changing at a rate of

r(t)=21-2 t

subscribers per week (where

t

is the time in weeks). At time

t=8

weeks, Zhang Wei has

120

subscribers. How many subscribers does Zhang Wei have by week

20

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Which expression can we use to solve the problem?

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1

\newline

r(20)-r(8)+120

\newline

\int_{8}^{20} r(t) d t+120

\newline

r(20)

\newline

\int_{20}^{20} r(t) d t

Consider the following problem:

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The velocity of a pendulum changes at a rate of

a(t)=0.9^{t} \cos (0.2 t)

meters per second squared (where

t

is the time in seconds). At time

t=6

, the velocity of the pendulum is

0

8

meters per second. By how much does the velocity change between

t=6

t=12

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Which expression can we use to solve the problem?

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1

\newline

a^{\prime}(6)+0.8

\newline

a^{\prime}(12)

\newline

a^{\prime}(12)-a^{\prime}(6)

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\int_{6}^{12} a(t) d t

\lim _{x \rightarrow 2} \frac{x^{4}-4 x^{3}+4 x^{2}}{x-2}

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1

\newline

-4

\newline

0

\newline

4

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(D) The limit doesn't exist

Kingsley knows that

1

2

54

centimeters. He wants to write an equation he can use to convert any given length in inches

(i)

(c)

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How should Kingsley write his equation?

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1

\newline

c=2.54 i

\newline

i=\frac{c}{2.54}

\newline

\frac{c}{i}=2.54

Maya is running a

10 \mathrm{~km}

race. She wants to write an equation for her average rate

(r)

given the time it takes her to finish

(t)

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How should Maya write her equation?

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1

\newline

10=r t

\newline

\frac{10}{t}=r

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\frac{10}{r}=t

Abby knows that

1

1

61

kilometers. She wants to write an equation she can use to convert any given distance in miles

(m)

(k)

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How should Abby write her equation?

\newline

1

\newline

1.61=\frac{k}{m}

\newline

1.61 \mathrm{~m}=k

\newline

\frac{k}{1.61}=m

Muneera is planning a

300 \mathrm{~km}

trip. She wants to write an equation that shows how many hours the trip will take

(t)

in terms of the rate at which she travels

(r)

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How should Muneera write her equation?

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1

\newline

\frac{300}{t}=r

\newline

\frac{300}{r}=t

\newline

300=r t

...