Mathematics: Afootball team ran four plays. the first play lost 5 yards. the second play lost 9 yards. the third play gained 12 yards. the last play lost 8 yards. express the total gain or loss of yards as an integer

Afootball team ran four plays. the first play lost

david1236544

27.05.2020

-10

Step-by-step explanation:

To find the the total gain or loss of yards, use an integer expression to write each play.

1st play = -5

2nd play = -9

3rd play = 12

4th play = -8

The expression is -5 + -9 + 12 + -8

-14 + 12 + -8

-2 + -8

-10

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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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