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3 years ago

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Freddy wants an average grade of 93 from his four algebra exams. The scores for his first three exams are 94, 89, and 96. What g

rade would Freddy need to earn on his fourth exam to have an average grade of 93?

1 answer:

White raven [17]3 years ago

8 0

He would need a 93% on his fourth exam.

94+89+96+93= 372, 372 divided by 4= 93

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Answer:

Length of third side is: $\mathbf{8m^3-4m^2+3m}$

Step-by-step explanation:

Perimeter of triangle = $18m^3+2m^2-m$

Length of side 1= $5m^3+3m^2-2m$

Length of side 2= $5m^3+3m^2-2m$

We need to find length of third side.

The formula used is: $Perimeter \:of\:triangle=Sum\:of\:all\:sides$

Putting values and finding length of third side.

$Perimeter \:of\:triangle=Sum\:of\:all\:sides\\18m^3+2m^2-m=5m^3+3m^2-2m+5m^3+3m^2-2m+Length\:of\:3rd\:side\\18m^3+2m^2-m=5m^3+5m^3+3m^2+3m^2-2m-2m+Length\:of\:3rd\:side\\18m^3+2m^2-m=10m^3+6m^2-4m+Length\:of\:3rd\:side\\Length\:of\:3rd\:side=18m^3+2m^2-m-(10m^3+6m^2-4m)\\Length\:of\:3rd\:side=18m^3+2m^2-m-10m^3-6m^2+4m\\Length\:of\:3rd\:side=18m^3-10m^3+2m^2-6m^2-m+4m\\Length\:of\:3rd\:side=8m^3-4m^2+3m$

So, Length of third side is: $\mathbf{8m^3-4m^2+3m}$

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