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MaRussiya [10]
2 years ago
13

If the perimeter of a rectangle is 6x^2-2x+14 and the length is x^2-x+8 what is the measure of the width?

Mathematics
1 answer:
s2008m [1.1K]2 years ago
7 0

Answer:

w = 2x^2-2x+15

Step-by-step explanation:

Since the formula to find the perimeter is P=2l+2w, use this to set up your equation:

6x^2-2x+14 = 2(x^2-x+8) + 2w

Then, distribute the 2 to everything inside the parentheses.

6x^2-2x+14 = 2x^2-2x+16 + 2w

Subtract 2x^2-2x+16 from both sides, leaving you:

4x^2-4x+30 = 2w

Divide 2 from both sides.

2x^2-2x+15

The answer is 2x^2-2x+15

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Answer:

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Step-by-step explanation:

The equation of a circle in standard form is

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where (h, k) are the coordinates of the centre and r is the radius

To obtain this form use the method of completing the square

Given

x² - 8x + y² - 2y - 8 = 0 ( add 8 to both sides )

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add ( half the coefficient of the x/y terms )² to both sides

x² + 2(- 4)x + 16 + y² + 2(- 1)y + 1 = 8 + 16 + 1

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Answer:

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<u>Right side</u>:

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