The square root of frac{(0.75)^{3}}{1-0.75}+l | vedclass

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(B) Let the given expression be $E = \frac{(0.75)^{3}}{1-0.75} + (0.75 + (0.75)^{2} + 1)$.We know that $1 - 0.75 = 0.25$.So,$E = \frac{(0.75)^{3}}{0.2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the given expression be $E = \frac{(0.75)^{3}}{1-0.75} + (0.75 + (0.75)^{2} + 1)$.We know that $1 - 0.75 = 0.25$.So,$E = \frac{(0.75)^{3}}{0.25} + (0.75^{2} + 0.75 + 1)$.We can rewrite the expression as:$E = \frac{(0.75)^{3} + 0.25(0.75^{2} + 0.75 + 1)}{0.25}$.Since $0.25 = 1 - 0.75$,we have:$E = \frac{(0.75)^{3} + (1 - 0.75)(1^{2} + 1 	imes 0.75 + 0.75^{2})}{0.25}$.Using the algebraic identity $a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$,where $a = 1$ and $b = 0.75$:$(1 - 0.75)(1^{2} + 1 	imes 0.75 + 0.75^{2}) = 1^{3} - 0.75^{3} = 1 - 0.75^{3}$.Substituting this back into the expression for $E$:$E = \frac{0.75^{3} + (1 - 0.75^{3})}{0.25} = \frac{1}{0.25} = 4$.The square root of the expression is $\sqrt{4} = 2$.

The square root of $\frac{(0.75)^{3}}{1-0.75}+\left(0.75+(0.75)^{2}+1\right)$ is

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