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evablogger [386]
4 years ago
8

Which expression is equivalent to Assume x 0 and y > 0.

Mathematics
2 answers:
vlada-n [284]4 years ago
7 0
The answer is the last option. 
motikmotik4 years ago
6 0

Given radical expression: \sqrt{\frac{25x^9y^3}{64x^6y^{11}}}

\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}\:=\:x^{a-b}

\frac{x^9}{x^6}=x^{9-6}=x^3

\frac{y^3}{y^{11}}=\frac{1}{y^{11-3}}=\frac{1}{y^8}

\sqrt{\frac{25x^9y^3}{64x^6y^{11}}} =\sqrt{\frac{25x^3}{64y^8}}

=\frac{\sqrt{25}\sqrt{x^3}}{\sqrt{64}\sqrt{y^8}}

\sqrt{25}=5

\sqrt{64}=8

=\frac{5\sqrt{x^3}}{8\sqrt{y^8}}

\sqrt{x^3}=x\sqrt{x}

\sqrt{y^8}=y^4

Therefore,

\frac{5\sqrt{x^3}}{8\sqrt{y^8}} =\frac{5x\sqrt{x}}{8y^4}.

Therefore, \sqrt{\frac{25x^9y^3}{64x^6y^{11}}}=\frac{5x\sqrt{x}}{8y^4}.



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