2 years ago

Https://www.analyzemath.com/high_school_math

1 answer:

3 0

Answer:

Where is the question?

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denis23 [38]

<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.

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

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

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

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

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

7 0

3 years ago

posledela

Refer to the attached file

For your understanding,I am sending this too so that u can understand the concept

8 0

3 years ago

Andre45 [30]

It would be A) The angles are angle bisectors
I am 90% sure

5 0

3 years ago

Ivenika [448]

Answer:

▶◀

Step-by-step explanation:

n/a bc you didn't explain wut to do

7 0

3 years ago

Aloiza [94]

Answer:

I might be able to help you if you add a picture

8 0

3 years ago