Respuesta :

frika

Answer:

D

Step-by-step explanation:

Use exponents property:

[tex]\dfrac{x^a}{x^b}=x^{a-b}[/tex]

1. Note that

[tex]\dfrac{x^9}{x^6}=x^{9-6}=x^3[/tex]

and

[tex]\dfrac{y^3}{y^{11}}=y^{3-11}=y^{-8}=\dfrac{1}{y^8}[/tex]

2. Now

[tex]\sqrt{25}=5\\ \\\sqrt{64}=8\\ \\\sqrt{x^3}=x\sqrt{x}\\ \\\sqrt{\dfrac{1}{y^8}}=\dfrac{1}{y^4}[/tex]

So

[tex]\sqrt{\dfrac{25x^9y^3}{64x^6y^{11}}}=\dfrac{\sqrt{25}\sqrt{x^3}}{\sqrt{64}\sqrt{y^8}}=\dfrac{5x\sqrt{x}}{8y^4}[/tex]

because [tex]x>0,\ y>0[/tex]

luisejr77

Answer: Last option.

Step-by-step explanation:

You need to remember the Quotient of powers property:

[tex]\frac{a^m}{a^n}=a^{(m-n)}[/tex]

 Applying this property, we know that:

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

Descompose 25 and 64 into their prime factors:

[tex]25=5*5=5^2\\64=8*8=8^2[/tex]

Since:

[tex]\sqrt[n]{a^n}=a[/tex]

And according to the Product of powers property:

[tex](a^m)(a^n)=a^{(m+n)}[/tex]

You can simplify. So, the equivalent expression is:

[tex]\sqrt{\frac{5^2x^2*x}{8^2y^8}}=\frac{5x\sqrt{x} }{8y^4}[/tex]

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