left[frac{8(3.75)^{3}+1}{(7.5)^{2}-6.5}right] | vedclass

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Question

(C) The given expression is $\frac{8(3.75)^{3}+1}{(7.5)^{2}-6.5}$.We can rewrite $8(3.75)^{3}$ as $(2 	imes 3.75)^{3} = (7.5)^{3}$.So,the expression

Vedclass · Accepted Answer

$8.5$

Vedclass · Answer

(C) The given expression is $\frac{8(3.75)^{3}+1}{(7.5)^{2}-6.5}$.We can rewrite $8(3.75)^{3}$ as $(2 	imes 3.75)^{3} = (7.5)^{3}$.So,the expression becomes $\frac{(7.5)^{3} + 1^{3}}{(7.5)^{2} - 6.5}$.Note that $6.5 = 7.5 - 1$,so the denominator is $(7.5)^{2} - (7.5 - 1) = (7.5)^{2} - 7.5 + 1$.Actually,using the identity $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$,where $a = 7.5$ and $b = 1$:Numerator $= (7.5 + 1)((7.5)^{2} - (7.5)(1) + 1^{2}) = (8.5)((7.5)^{2} - 7.5 + 1)$.Denominator $= (7.5)^{2} - 6.5 = 56.25 - 6.5 = 49.75$.Wait,let us re-evaluate the denominator: $(7.5)^{2} - 6.5 = 56.25 - 6.5 = 49.75$.Using the identity: $\frac{(7.5)^{3} + 1^{3}}{(7.5)^{2} - 7.5 + 1} = \frac{(7.5+1)(7.5^{2} - 7.5 + 1)}{7.5^{2} - 7.5 + 1} = 7.5 + 1 = 8.5$.

$\left[\frac{8(3.75)^{3}+1}{(7.5)^{2}-6.5}\right]$ is equal to

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