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klasskru [66]
3 years ago
15

Which expression is equivalent to sqrt128x^5y^6/2x^7y^5 ? Assume x 0 and y > 0.

Mathematics
2 answers:
sergiy2304 [10]3 years ago
5 0

Answer:

\frac{8\sqrt{y}}{x}

Step-by-step explanation:

\sqrt{\frac{128x^5y^6}{2x^7y^5}}

we have division inside the square root .

LEts simplify it

128 divide by 2 is 64

use property of exponents to simplify the exponents

\frac{a^m}{a^n} =a^{m-n}

\frac{x^5}{x^7} =x^{5-7}=x^{-2}

\frac{y^6}{y^5} =y^{6-5}=y

\sqrt{\frac{128x^5y^6}{2x^7y^5}}

\sqrt{\frac{64x^{-2}y}{1}}

To make the exponent positive move x to the denominator

\sqrt{\frac{64y}{x^2}}

Now we take square root

square root (64) is 8

square root of x^2 is x

\frac{8\sqrt{y}}{x}

kolbaska11 [484]3 years ago
4 0

<u>Answer:</u>

\frac{8\sqrt{y} }{x}

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

We are given the following expression and we are to simplify it:

\sqrt {\frac {128x^5y^6} {2x^7y^5} }

To make it easier to solve, we can also write this expression as:

\sqrt {\frac{128}{2} * \frac {x^5} {x^7} * \frac {y^6} {y^5}}

Now we will cancel out the like terms to get:

\sqrt {64*\frac{1} {x^2}*y }

Taking the square root of the terms to get:

8*\frac {1}{x} .\sqrt{y}

\frac{8\sqrt{y} }{x}

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