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Anettt [7]
3 years ago
7

For which rational expression is 8 an excluded value ? check all that apply

Mathematics
1 answer:
denis23 [38]3 years ago
7 0

<u>Answer:</u>

The correct answer options are C. \frac{x^2+5}{x-8} and D. \frac{x^2-x-56}{x^2-64}.

<u>Step-by-step explanation:</u>

The values which make the denominator equal to zero are called the excluded values.

Here, we can substitute 8 for x and check if it makes the denominator 0.

\frac{x-8}{x+8} = \frac{8-8}{8+8} =\frac{0}{16} =0

\frac{x-2}{x^2-4} = \frac{8-2}{8^2-4} =\frac{6}{60} =\frac{1}{10}

\frac{x^2+5}{x-8} = \frac{8^2+5}{8-8} =\frac{69}{0}

\frac{x^2-x-56}{x^2-64} = \frac{8^2-8-56}{8^2-64} = \frac{0}{0} =0

\frac{8x^2-2}{x^2-16} = \frac{8(8)^2-2}{8^2-16} =\frac{510}{48}

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D(4,-3)

Step-by-step explanation:

Given three of the vertices of the square: A(4, -7), B(8, -7),C(8, -3)

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\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\$Therefore$:\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\$Similarly$\\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3

The coordinates of the fourth vertex is D(4,-3)

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