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ioda
3 years ago
15

On four exams, Wallace’s grades were 79,93,91, and 68. What grade must he obtain on his fifth exam to have an 80 average?

Mathematics
2 answers:
neonofarm [45]3 years ago
7 0
(79 + 93 + 91 + 68 + x) / 5 = 80
(331 + x) / 5 = 80.....multiply both sides by 5
331 + x = 80 * 5
331 + x = 400
x = 400 - 331
x = 69 <=== he would have to make at least a 69
Leona [35]3 years ago
7 0

(79 + 93 + 91 + 68 + x) / 5 = 80

Simplify

(331 + x)/5 = 80

Multiply 5 to both sides to isolate the x.

331 + x = 400

Isolate the x. Subtract 331 from both sides

x + 331 (-331) = 400 (-331)

x = 69

69 is your answer

hope this helps

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Step-by-step explanation:

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(x2 + 4)(x2 - 4) please help
VikaD [51]

(x^{2} + 4)(x^{2} - 4)

To solve this question you must FOIL (First, Outside, Inside, Last) like so

First:

(x^2 + 4)(x^2 - 4)

x^2 * x^2

x^4

Outside:

(x^2 + 4)(x^2 - 4)

x^2 * -4

-4x^2

Inside:

(x^2 + 4)(x^2 - 4)

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4x^2

Last:

(x^2 + 4)(x^2 - 4)

4 * -4

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Now combine all the products of the FOIL together like so...

x^4 - 4x^2 +4x^2 - 16

Combine like terms:

x^4 - 4x^2 +4x^2 - 16

- 4x^2 +4x^2 = 0

x^4 - 16 <<<This is your answer

Hope this helped!

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