The correct option is C 417√(?+483)÷ 6 =125^1/3⇒√(x+483)/6=5 ⇒√(x+483)=30 ⇒ x+483=(30)^2⇒ 900 483=417 ...

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Algebra of Limits

√?+483÷ 6=125...

Question

$\sqrt{? + 483} \div 6 = 125^{1 / 3}$

429

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411

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423

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413

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417

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Solution

The correct option is E 417
$\sqrt{? + 483} \div 6 = 125^{1 / 3} \Rightarrow \frac{\sqrt{x + 483}}{6} = 5 \Rightarrow \sqrt{x + 483} = 30 \Rightarrow x + 483 = (30)^{2} \Rightarrow 900 - 483 = 417$

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{(\frac{64}{125})}^{1 / 3}

Find

$125^{\frac{1}{3}}$

\frac{\sqrt[6]{2} {[{(625)}^{\frac{3}{5}} \times {(1024)}^{- \frac{6}{5}} \div {(25)}^{\frac{3}{5}}]}^{\frac{1}{2}}}{[{(\sqrt[3]{128})}^{- \frac{5}{2}}] \times {(125)}^{\frac{1}{5}}} \div \frac{{(10^{3})}^{2} \div {(10^{2})}^{3}}{{(10^{2})}^{3} \div {(10^{3})}^{2}}

(27 \times 125)^{\frac{1}{3}}

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