frac{(0.188+0.077)^{2}+(0.188-0.077)^{2}}{0.1 | vedclass

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Question

(D) Let $a = 0.188$ and $b = 0.077$.The given expression is in the form $\frac{(a+b)^{2} + (a-b)^{2}}{a^{2} + b^{2}}$.Expanding the numerator using al

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Let $a = 0.188$ and $b = 0.077$.The given expression is in the form $\frac{(a+b)^{2} + (a-b)^{2}}{a^{2} + b^{2}}$.Expanding the numerator using algebraic identities:$(a+b)^{2} = a^{2} + 2ab + b^{2}$$(a-b)^{2} = a^{2} - 2ab + b^{2}$Adding these two expressions:$(a+b)^{2} + (a-b)^{2} = (a^{2} + 2ab + b^{2}) + (a^{2} - 2ab + b^{2}) = 2a^{2} + 2b^{2} = 2(a^{2} + b^{2})$.Substituting this back into the original fraction:$\frac{2(a^{2} + b^{2})}{a^{2} + b^{2}} = 2$.Therefore,the value of the expression is $2$.

$\frac{(0.188+0.077)^{2}+(0.188-0.077)^{2}}{0.188 \times 0.188+0.077 \times 0.077} = ?$

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