frac{(0.146+0.092)^{2}+(0.092-0.146)^{2}}{0.0 | vedclass

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Question

(A) Let $a = 0.146$ and $b = 0.092$.Then $0.073 = \frac{a}{2}$ and $0.046 = \frac{b}{2}$.The expression becomes:$\frac{(a+b)^{2} + (b-a)^{2}}{(\frac{a

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let $a = 0.146$ and $b = 0.092$.Then $0.073 = \frac{a}{2}$ and $0.046 = \frac{b}{2}$.The expression becomes:$\frac{(a+b)^{2} + (b-a)^{2}}{(\frac{a}{2})^{2} + (\frac{b}{2})^{2}}$Since $(b-a)^{2} = (a-b)^{2}$,the numerator is $(a+b)^{2} + (a-b)^{2} = 2(a^{2} + b^{2})$.The denominator is $\frac{a^{2}}{4} + \frac{b^{2}}{4} = \frac{a^{2} + b^{2}}{4}$.Thus,the expression is $\frac{2(a^{2} + b^{2})}{\frac{a^{2} + b^{2}}{4}} = 2 	imes 4 = 8$.

$\frac{(0.146+0.092)^{2}+(0.092-0.146)^{2}}{0.073 \times 0.073+0.046 \times 0.046}$ is equal to

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