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1 year ago

The football team gained 8 yards on the first play. Lost 6 yards on the next. Gained 3 yards on the

Mathematics

2 answers:

Marrrta [24]1 year ago

8 0

Answer:

-7 yards

Step-by-step explanation:

Might help to line everything up

First play: gained 8 yards (+8)

Second play: lost 6 yards (-6)

Third play: gained 3 yards (+3)

Fourth play: lost 12 yards (-12)

Then put it all together and solve (I'll solve left to right):

(8-6)+3-12

(2+3)-12

5-12

-7

The final change in yardage was -7 yards

Aliun [14]1 year ago

5 0

The difference is negative seven I believe

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Question is below (wasn't sure about the answer)

Paha777 [63]

Answer: Choice A)

$H = \frac{SA}{2(L+W)} - \frac{LW}{L+W}\\\\$

=================================================

Work Shown:

$SA = 2(LW + LH + HW)\\\\2(LW + LH + HW) = SA\\\\LW + LH + HW = \frac{SA}{2}\\\\LH + HW = \frac{SA}{2} - LW\\\\H(L + W) = \frac{SA}{2} - LW\\\\H = \frac{\frac{SA}{2} - LW}{(L + W)}\\\\H = \frac{SA}{2(L+W)} - \frac{LW}{L+W}\\\\$

This is not the only way to solve for H.

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2 years ago

A factory makes candy bars and ships them in cases

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Answer:

b, 24 bars per box

Step-by-step explanation:

you just tale 144÷6

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2 years ago

Maya plots the yards lost during a football game on a number line and labels it A. She plots Point B to the right of zero, a dis

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

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2 years ago

Fernell and Dabney shared the driving

Agata [3.3K]

Answer:

15 hours

Step-by-step explanation:

Given:

Fernell's speed $v_F= 50$ mph

Dabney's speed $v_D= 64$ mph

Denote:

$D_F$ - distance covered by Fernell

$D_D$ - distance covered by Dabney

$t_F$ - Fernell's time

$t_D$ - Dabney's time

1. If Fernell drove for 3 hours longer than Dabney, then his time is 3 hours more than Dabney's time and

$t_F=t_D+3$

2. If Fernell covered 18 miles less than Dabney, then

$D_D-D_F=18$

Use formula $D=t\cdot v$

$D_D=v_D\cdot t_D\Rightarrow D_D=64\cdot t_D\\ \\D_F=v_F\cdot t_F\Rightarrow D_F=50\cdot (t_D+3)$

Subtract from the first equation the second equation and equate it to 18:

$64t_D-50(t_D+3)=18\\ \\64t_D-50t_D-150=18\\ \\14t_D=168\\ \\t_D=12\ \text{hours}$

$t_F=t_D+3=12+3=15\ \text{hours}$

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2 years ago