Question

spiral review

1. determine the average rate of change for each interval on the graph of $g(r)$.

interval | average rate of change
$r=-2$ to $r=1$ |
$r=-2$ to $r=5$ |
$r=-2$ to $r=8$ |

1. functions $c(t)$ and $k(t)$ represent the distance of two cats from home after $t$ seconds. select all the true statements.

$square$ a. $k(5) > c(5)$
$square$ b. $k(t)$ and $c(t)$ have the same domain and range
$square$ c. $k(t)$ keeps increasing from 0 to 13 seconds
$square$ d. $k(11) = c(11)$
$square$ e. both cats return home

1. tickets to the state fair cost $10 each. the function $c(x)=10x$ gives the total cost in dollars, $c(x)$, for the number of tickets purchased, $x$. select all the values that are possible outputs for $c(x)$.

$square$ a. 0
$square$ b. 70
$square$ c. 105
$square$ d. 680
$square$ e. 905

reflection

1. put a star next to a problem where you revised your thinking.
2. use the space to ask a question or share something youre proud of.

Explanation:

Problem 8

Step1: Recall average rate formula

Average rate of change = $\frac{g(r_2)-g(r_1)}{r_2-r_1}$

Step2: Get values for $r=-2$ to $r=1$

From graph: $g(-2)=6$, $g(1)=6$
$\text{Rate} = \frac{6-6}{1-(-2)} = \frac{0}{3}=0$

Step3: Get values for $r=-2$ to $r=5$

From graph: $g(-2)=6$, $g(5)=2$
$\text{Rate} = \frac{2-6}{5-(-2)} = \frac{-4}{7}=-\frac{4}{7}$

Step4: Get values for $r=-2$ to $r=8$

From graph: $g(-2)=6$, $g(8)=8$
$\text{Rate} = \frac{8-6}{8-(-2)} = \frac{2}{10}=0.2$

• A. At $t=5$, $k(5)=40$, $c(5)=30$, so $k(5)>c(5)$ is true.
• B. Domain of both is $[0,25]$, but range of $k(t)$ is $[0,40]$, range of $c(t)$ is $[0,60]$: false.
• C. $k(t)$ decreases after $t=10$: false.
• D. At $t=11$, $k(11)=40$, $c(11)=40$, so $k(11)=c(11)$ is true.
• E. Both functions reach 0 at $t=25$: true.

The function $c(x)=10x$ has outputs that are multiples of 10 (since $x$ is a non-negative integer, number of tickets). 0,70,680 are multiples of 10; 105, 603 are not.

Answer:

• $r=-2$ to $r=1$: $0$
• $r=-2$ to $r=5$: $-\frac{4}{7}$
• $r=-2$ to $r=8$: $0.2$

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Problem 9