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Which of the following is equal to
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07 May 2022, 00:38
07 May 2022, 00:38
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Which of the following is equal to \(\left( \begin{array}{cc}\frac{\sqrt{12}}{5}\end{array} \right)\)\(\left( \begin{array}{cc}\frac{\sqrt{60}}{2^4}\end{array} \right)\)\(\left( \begin{array}{cc}\frac{\sqrt{45}}{3^2}\end{array} \right)\) ?
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
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Re: Which of the following is equal to
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07 May 2022, 05:14
07 May 2022, 05:14
1
Carcass wrote:
Which of the following is equal to \(\left( \begin{array}{cc}\frac{\sqrt{12}}{5}\end{array} \right)\)\(\left( \begin{array}{cc}\frac{\sqrt{60}}{2^4}\end{array} \right)\)\(\left( \begin{array}{cc}\frac{\sqrt{45}}{3^2}\end{array} \right)\) ?
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
Given: \(\left( \begin{array}{cc}\frac{\sqrt{12}}{5}\end{array} \right)\)\(\left( \begin{array}{cc}\frac{\sqrt{60}}{2^4}\end{array} \right)\)\(\left( \begin{array}{cc}\frac{\sqrt{45}}{3^2}\end{array} \right)\)
Simplify each root to get: \(\left( \begin{array}{cc}\frac{2\sqrt{3}}{5}\end{array} \right)\)\(\left( \begin{array}{cc}\frac{2\sqrt{15}}{2^4}\end{array} \right)\)\(\left( \begin{array}{cc}\frac{3\sqrt{5}}{3^2}\end{array} \right)\)
Multiply the numerators and the denominators: \(\frac{12\sqrt{225}}{(5)(2^4)(3^2)}\)
Simplify the numerator: \(\frac{(12)(15)}{(5)(2^4)(3^2)}\)
Rewrite the numerator as follows: \(\frac{(2^2)(3^2)(5)}{(2^4)(3^2)(5)}\)
Simplify: \(\frac{1}{2^2} = \frac{1}{4}\)
Answer: C
Re: Which of the following is equal to
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28 Oct 2023, 10:02
28 Oct 2023, 10:02
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
