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Elanso [62]
3 years ago
5

X^2y^8/x^7y^6 if x= -2 and y= 4

Mathematics
1 answer:
damaskus [11]3 years ago
7 0

Answer:

- \frac{1}{2}

Step-by-step explanation:

Given

\frac{x^2y^8}{x^7y^6}

= \frac{y^{(8-6)} }{x^{(7-2)} }

= \frac{y^2}{x^5} , substitute values

= \frac{4^2}{(-2)^{5} }

= \frac{16}{-32}

= - \frac{1}{2}

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Simplify the following:
((x^2 - 11 x + 30) (x^2 + 6 x + 5))/((x^2 - 25) (x - 5 x - 6))

The factors of 5 that sum to 6 are 5 and 1. So, x^2 + 6 x + 5 = (x + 5) (x + 1):
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The factors of 30 that sum to -11 are -5 and -6. So, x^2 - 11 x + 30 = (x - 5) (x - 6):
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x - 5 x = -4 x:
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Factor -2 out of -4 x - 6:
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x^2 - 25 = x^2 - 5^2:
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Factor the difference of two squares. x^2 - 5^2 = (x - 5) (x + 5):
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((x - 6) (x + 1))/(-2 (2 x + 3))

Multiply numerator and denominator of ((x - 6) (x + 1))/(-2 (2 x + 3)) by -1:

Answer: (-(x - 6) (x + 1))/(2 (2 x + 3))
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