problems 5–7: this table shows the population of a city from 1988 to 2016. 5. determine the average rate of change for p(t) between 1992 and 2000 6. state two values of t that create an interval with a negative rate of change. 7. state two values of t that create an interval with a positive rate of change. 8. match each interval to its average rate of change. interval average rate of change a. x to y $\frac{1}{5}$ b. y to z $\frac{1}{4}$ c. x to z $\frac{1}{6}$ 9. jada is walking to school. the function d(t) gives the distance from school, in meters, t minutes since jada left home. which equation represents this statement? jada is 600 meters from school after 5 minutes. a. $d(5) = 600$ b. $d(600) = 5$ c. $t(5) = 600$ d. $t(600) = 5$ 10. complete the arithmetic sequence with the missing terms ______ 6 ______ 22, 30

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Accepted Answer

### Problem 5:
# Explanation:
## Step 1: Recall the formula for average rate of change
The average rate of change of a function $ p(t) $ between $ t = a $ and $ t = b $ is given by $ \frac{p(b)-p(a)}{b - a} $.

## Step 2: Identify the values for 1992 and 2000
From the table, in 1992 ($ t = 1992 $), $ p(1992)=42700 $, and in 2000 ($ t = 2000 $), $ p(2000)=33700 $. The time difference $ b - a=2000 - 1992 = 8 $.

## Step 3: Calculate the average rate of change
Substitute into the formula: $ \frac{p(2000)-p(1992)}{2000 - 1992}=\frac{33700 - 42700}{8}=\frac{-9000}{8}=- 1125 $

# Answer: $-1125$

### Problem 6:
# Brief Explanations:
A negative rate of change occurs when the population decreases over time, i.e., $ p(t_2)<p(t_1) $ for $ t_2 > t_1 $. Looking at the table, from 1992 ($ p = 42700 $) to 1996 ($ p = 33100 $), the population decreases. So $ t_1 = 1992 $ and $ t_2 = 1996 $ (other valid pairs: 1992 - 2000, 2008 - 2012 etc.).

# Answer: $ t = 1992 $ and $ t = 1996 $ (or other valid pairs like 1992 and 2000, 2008 and 2012)

### Problem 7:
# Brief Explanations:
A positive rate of change occurs when the population increases over time, i.e., $ p(t_2)>p(t_1) $ for $ t_2 > t_1 $. From 2000 ($ p = 33700 $) to 2004 ($ p = 45000 $), the population increases. So $ t_1 = 2000 $ and $ t_2 = 2004 $ (other valid pairs: 2004 - 2008, 2012 - 2016 etc.).

# Answer: $ t = 2000 $ and $ t = 2004 $ (or other valid pairs like 2004 and 2008, 2012 and 2016)

### Problem 8:
First, find the coordinates of $ X $, $ Y $, $ Z $ from the graph: $ X(2,2) $, $ Y(8,3) $, $ Z(12,4) $

## For interval $ X $ to $ Y $:
# Explanation:
## Step 1: Use average rate of change formula
$ \frac{y_Y - y_X}{x_Y - x_X}=\frac{3 - 2}{8 - 2}=\frac{1}{6} $? Wait, no, wait the graph: Wait, maybe I misread. Wait the graph has $ X(2,2) $, $ Y(8,3) $, $ Z(12,4) $? Wait no, the points: $ X(2,2) $, $ Y(8,3) $, $ Z(12,4) $? Wait the average rate of change for $ X $ to $ Y $: $ \frac{3 - 2}{8 - 2}=\frac{1}{6} $? But the options are $ \frac{1}{5},\frac{1}{4},\frac{1}{6} $. Wait maybe the points are $ X(2,2) $, $ Y(8,3) $, $ Z(12,4) $? Wait no, let's recalculate:

- $ X $ to $ Y $: $ x $ from 2 to 8, $ y $ from 2 to 3. Rate: $ \frac{3 - 2}{8 - 2}=\frac{1}{6} $
- $ Y $ to $ Z $: $ x $ from 8 to 12, $ y $ from 3 to 4. Rate: $ \frac{4 - 3}{12 - 8}=\frac{1}{4} $
- $ X $ to $ Z $: $ x $ from 2 to 12, $ y $ from 2 to 4. Rate: $ \frac{4 - 2}{12 - 2}=\frac{2}{10}=\frac{1}{5} $

So:

- $ X $ to $ Y $: $ \frac{1}{6} $
- $ Y $ to $ Z $: $ \frac{1}{4} $
- $ X $ to $ Z $: $ \frac{1}{5} $

# Answer:
a. $ X $ to $ Y $: $ \frac{1}{6} $

b. $ Y $ to $ Z $: $ \frac{1}{4} $

c. $ X $ to $ Z $: $ \frac{1}{5} $

### Problem 9:
# Brief Explanations:
The function $ d(t) $ gives distance from school at time $ t $ minutes. So when $ t = 5 $ minutes, $ d(5) $ is the distance, which is 600 meters. So $ d(5)=600 $, which is option A.

# Answer: A. $ d(5) = 600 $

### Problem 10:
# Explanation:
## Step 1: Recall arithmetic sequence formula
An arithmetic sequence has the form $ a_n=a_1+(n - 1)d $, where $ d $ is the common difference.

## Step 2: Find the common difference $ d $
Given terms: $ a_2 = 6 $, $ a_4 = 22 $, $ a_5 = 30 $. The difference between $ a_5 $ and $ a_4 $ is $ 30 - 22 = 8 $, so $ d = 8 $.

## Step 3: Find $ a_1 $ and $ a_3 $
For $ a_2 = 6 $, $ a_2=a_1 + d\Rightarrow 6=a_1+8\Rightarrow a_1=6 - 8=- 2 $

For $ a_3 $, $ a_3=a_2 + d=6 + 8 = 14 $

So the sequence is $ -2,6,14,22,30 $

# Answer: The missing terms are $-2$ and $14$

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QUESTION IMAGE

Question

Explanation:

Problem 5:

Step 1: Recall the formula for average rate of change

Step 2: Identify the values for 1992 and 2000

Step 3: Calculate the average rate of change

Answer:

Problem 6: