(1.06+0.04)^{2}-?=4 times 1.06 times 0.04 | vedclass

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Question

(A) Let the missing value be $x$.The given equation is $(1.06+0.04)^{2}-x=4 	imes 1.06 	imes 0.04$.Let $a = 1.06$ and $b = 0.04$.The equation become

Vedclass · Accepted Answer

$1.0404$

Vedclass · Answer

(A) Let the missing value be $x$.The given equation is $(1.06+0.04)^{2}-x=4 	imes 1.06 	imes 0.04$.Let $a = 1.06$ and $b = 0.04$.The equation becomes $(a+b)^{2}-x=4ab$.Rearranging for $x$,we get $x = (a+b)^{2}-4ab$.Using the algebraic identity $(a+b)^{2}-4ab = (a-b)^{2}$,we have:$x = (a-b)^{2} = (1.06-0.04)^{2}$.$x = (1.02)^{2} = 1.0404$.

$(1.06+0.04)^{2}-?=4 \times 1.06 \times 0.04$

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