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scoundrel [369]
3 years ago
7

Can someone solve this for me???

Mathematics
1 answer:
kolezko [41]3 years ago
4 0

Answer:

Part A: x^2 + 5x  - 8

Part B: 5x^3 + 3x^2 - 6x + 5

Step-by-step explanation:

Part A:

To find the total of side 1 and 2, add the two expressions which represent each side. Combine like terms to simplify.

3x^2 − 2x − 5 + (7x - 2x^2 − 3)

3x^2 - 2x^2 -2x + 7x -5 + -3

x^2 + 5x  - 8

Part B:

To solve for the third side of a triangle using its perimeter, subtract the lengths of two of its sides from its perimeter. The perimeter is 5x^3 + 4x^2 − x − 3. Subtract 3x^2 − 2x − 5  and 7x - 2x^2 − 3. The remaining expression is the third side. Combine like terms to simplify.

5x^3 + 4x^2 − x − 3 - (3x^2 − 2x − 5) - (7x - 2x^2 − 3)

5x^3 + 4x^2 - x - 3 - 3x^2 + 2x + 5 - 7x + 2x^2 + 3

5x^3 + (4x^2 -3x^2 + 2x^2) + (-x + 2x - 7x) + (-3 + 5 + 3)

5x^3 + 3x^2 - 6x + 5

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3 years ago
What is the solution of -8 / 2 y - 8 equals 5 / y + 4 - 7 y + 8 / Y 2 - 16​
AysviL [449]

Answer:

The value of y is 6.

Step-by-step explanation:

The given equation is

\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}

It can be rewritten as

\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}+\dfrac{7y+8}{y^2-4^2}

\dfrac{-4}{(y-4)}=\dfrac{5}{y+4}+\dfrac{7y+8}{(y+4)(y-4)}

Multiply both sides by (y+4)(y-4).

(y+4)(y-4)(\dfrac{-4}{(y-4)})=(y+4)(y-4)(\dfrac{5}{y+4})+(y+4)(y-4)(\dfrac{7y+8}{(y+4)(y-4)})

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-4y-16=5y-20-(7y+8)

-4y-16=5y-20-7y-8

-4y-16=-2y-28

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Divide both sides by -2.

y=6

Therefore, the value of y is 6.

5 0
3 years ago
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