Mathematics: The exact times, in seconds, of four sprinters from a school track meet are as follows: sprinter sprinter a sprinter b sprinter c sprinter d time 132.38 132.31 132.29 132.25 if the stopwatch used was precise only to the tenths place, which sprinter s time is different from the others? your sprinter a sprinter b sprinter c sprinter d

The exact times, in seconds, of four sprinters

potatogirl9300

01.02.2022

Step-by-step explanation:

h(t) = (t + 3)² + 5

A) h(t) = (t + 3)² + 5 -2 ≤ t ≤ 0

average rate of change in h over the change in t is given as:

$\frac{h(0)-h(-2)}{0-(-2)}=\frac{(0+3)^2+5-[(-2+3)^2+5]}{0+2}=\frac{14-6}{2} =4$

It does not have a negative average rate of change

B) A) h(t) = (t + 3)² + 5 -4 ≤ t ≤ -3

average rate of change in h over the change in t is given as:

$\frac{h(-3)-h(-4)}{-3-(-4)}=\frac{(-3+3)^2+5-[(-4+3)^2+5]}{-3+4}=\frac{5-6}{1} =-1$

It have a negative average rate of change

c) A) h(t) = (t + 3)² + 5 -3 ≤ t ≤ 4

average rate of change in h over the change in t is given as:

$\frac{h(4)-h(-3)}{4-(-3)}=\frac{(4+3)^2+5-[(-3+3)^2+5]}{4+3}=\frac{54-5}{7} =7$

It does not have a negative average rate of change

D) h(t) = (t + 3)² + 5 1 ≤ t ≤ 4

average rate of change in h over the change in t is given as:

$\frac{h(4)-h(1)}{4-(1)}=\frac{(4+3)^2+5-[(1+3)^2+5]}{4-1}=\frac{54-21}{3} =11$

It does not have a negative average rate of change

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