92 times 576 div 2 sqrt{1296} = (?)^{3} + sqr | vedclass

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Question

(C) Given equation: $(?)^{3} + \sqrt{49} = 92 	imes 576 \div 2 \sqrt{1296}$Step $1$: Simplify the square roots.$\sqrt{1296} = 36$ and $\sqrt{49} = 7$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Given equation: $(?)^{3} + \sqrt{49} = 92 	imes 576 \div 2 \sqrt{1296}$Step $1$: Simplify the square roots.$\sqrt{1296} = 36$ and $\sqrt{49} = 7$.Step $2$: Substitute these values into the equation.$(?)^{3} + 7 = 92 	imes 576 \div (2 	imes 36)$Step $3$: Perform the division.$(?)^{3} + 7 = 92 	imes 576 \div 72$$(?)^{3} + 7 = 92 	imes 8$Step $4$: Multiply the values.$(?)^{3} + 7 = 736$Step $5$: Solve for $(?)^{3}$.$(?)^{3} = 736 - 7 = 729$Step $6$: Find the cube root.$? = \sqrt[3]{729} = 9$Therefore,the correct option is $C$.

$92 \times 576 \div 2 \sqrt{1296} = (?)^{3} + \sqrt{49}$

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