frac{0.1 times 0.1 times 0.1 + 0.02 times 0.0 | vedclass

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Question

(A) Let $x = 0.1$ and $y = 0.02$.The given expression is $\frac{x^3 + y^3}{(2x)^3 + (2y)^3}$.We know that $(2x)^3 = 8x^3$ and $(2y)^3 = 8y^3$.Substitu

Vedclass · Accepted Answer

$0.125$

Vedclass · Answer

(A) Let $x = 0.1$ and $y = 0.02$.The given expression is $\frac{x^3 + y^3}{(2x)^3 + (2y)^3}$.We know that $(2x)^3 = 8x^3$ and $(2y)^3 = 8y^3$.Substituting these into the expression,we get $\frac{x^3 + y^3}{8x^3 + 8y^3}$.Factoring out $8$ from the denominator,we get $\frac{x^3 + y^3}{8(x^3 + y^3)}$.Canceling the common term $(x^3 + y^3)$,we are left with $\frac{1}{8}$.Calculating the decimal value,$\frac{1}{8} = 0.125$.

$\frac{0.1 \times 0.1 \times 0.1 + 0.02 \times 0.02 \times 0.02}{0.2 \times 0.2 \times 0.2 + 0.04 \times 0.04 \times 0.04}$ is equal to

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