Question

10 valeria and zion also raced their spaceships. v(t) and z(t) represents the distances of valeria’s and zion’s spaceships after t seconds. here is some information about their race: - v(0) = z(0) - v(4) < z(4) - v(10) = 20 - v(13) = z(13) - v(15) > z(15) make a graph that could represent valeria’s distance traveled after t seconds. lesson practice a1.4.06 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 1/5 b. y to z 1/4 c. x to z 1/6 spiral review 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

Explanation:

Problem 5

Step1: 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 \( \frac{p(b)-p(a)}{b - a} \).

Step2: Identify the values for 1992 and 2000

From the table, for \( t = 1992 \), \( p(1992)=42700 \) and for \( t = 2000 \), \( p(2000)=33700 \). The time difference \( b - a=2000 - 1992 = 8 \).

Step3: Calculate the average rate of change

Substitute the values into the formula: \( \frac{p(2000)-p(1992)}{2000 - 1992}=\frac{33700 - 42700}{8}=\frac{- 9000}{8}=- 1125 \).

To have a negative rate of change, \( p(b)-p(a)<0 \) (i.e., \( p(b)<p(a) \) when \( b > a \)). Looking at the table:

• For \( t = 1992 \) (\( p = 42700 \)) and \( t = 1996 \) (\( p = 33100 \)), \( 33100<42700 \), so the interval \( 1992 \) to \( 1996 \) has a negative rate of change.
• Another example: \( t = 1992 \) and \( t = 2000 \) (we saw in problem 5 the rate is negative), or \( t = 2008 \) (\( p = 48400 \)) and \( t = 2012 \) (\( p = 40900 \)) since \( 40900<48400 \).

To have a positive rate of change, \( p(b)-p(a)>0 \) (i.e., \( p(b)>p(a) \) when \( b > a \)). Looking at the table:

• For \( t = 1996 \) (\( p = 33100 \)) and \( t = 2000 \) (\( p = 33700 \)), \( 33700>33100 \), so the interval \( 1996 \) to \( 2000 \) has a positive rate of change.
• Another example: \( t = 2000 \) (\( p = 33700 \)) and \( t = 2004 \) (\( p = 45000 \)) since \( 45000>33700 \), or \( t = 2012 \) (\( p = 40900 \)) and \( t = 2016 \) (\( p = 43000 \)) since \( 43000>40900 \).

Answer:

The average rate of change is \(-1125\) (people per year).

Problem 6