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\(\dfrac{\sqrt{180}+\sqrt{45}}{\sqrt{135}}\)

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by M7MBA » Wed Jul 01, 2020 1:48 am

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\(\dfrac{\sqrt{180}+\sqrt{45}}{\sqrt{135}}\)

A. \(\sqrt3\)

B. \(3\sqrt3\)

C. \(5\sqrt3\)

D. \(3\sqrt5\)

E. \(5\sqrt5\)

[spoiler]OA=A[/spoiler]

Source: Veritas Prep
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Source: — Problem Solving |

M7MBA wrote:
Wed Jul 01, 2020 1:48 am
\(\dfrac{\sqrt{180}+\sqrt{45}}{\sqrt{135}}\)

A. \(\sqrt3\)

B. \(3\sqrt3\)

C. \(5\sqrt3\)

D. \(3\sqrt5\)

E. \(5\sqrt5\)

[spoiler]OA=A[/spoiler]

Source: Veritas Prep
So, we have \(\dfrac{\sqrt{180}+\sqrt{45}}{\sqrt{135}}\)

\(=> \dfrac{\sqrt{36*5}+\sqrt{9*5}}{\sqrt{15*9}}\)

\(=> \dfrac{6\sqrt{5}+3\sqrt{5}}{3\sqrt{15}}\)

\(=> \dfrac{9\sqrt{5}}{3\sqrt{15}}\)

\(=> \dfrac{3\sqrt{5}}{\sqrt{3*5}}\)

\(=> \dfrac{3\sqrt{5}}{\sqrt{3}\times \sqrt{5}}\)

\(=> \sqrt{3}\)

Correct answer: A

Hope this helps!

-Jay
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M7MBA wrote:
Wed Jul 01, 2020 1:48 am
\(\dfrac{\sqrt{180}+\sqrt{45}}{\sqrt{135}}\)

A. \(\sqrt3\)

B. \(3\sqrt3\)

C. \(5\sqrt3\)

D. \(3\sqrt5\)

E. \(5\sqrt5\)

[spoiler]OA=A[/spoiler]

Solution:

Simplifying the expression, we have:

(√36√5 + √9√5) / (√9√3√5)

(6√5 + 3√5) / (3√3√5)

9/(3√3)

3/√3

√3

Answer: A

Scott Woodbury-Stewart
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