On simplification of frac{(2.644)^{2}-(2.356) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The expression is in the form $\frac{a^2 - b^2}{a - b}$,where $a = 2.644$ and $b = 2.356$.Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The expression is in the form $\frac{a^2 - b^2}{a - b}$,where $a = 2.644$ and $b = 2.356$.Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we can rewrite the numerator as $(2.644 - 2.356)(2.644 + 2.356)$.Substituting this into the expression: $\frac{(2.644 - 2.356)(2.644 + 2.356)}{0.288}$.Since $2.644 - 2.356 = 0.288$,the expression becomes $\frac{0.288 	imes (2.644 + 2.356)}{0.288}$.Canceling $0.288$ from the numerator and denominator,we get $2.644 + 2.356 = 5$.

On simplification of $\frac{(2.644)^{2}-(2.356)^{2}}{0.288},$ we get

Similar Questions

$\frac{? \div 12}{0.2 \times 3.6} = 2$

What will be the remainder when $252^{126} + 244^{152}$ is divided by $10$?

The value of $3+\frac{1}{\sqrt{3}}+\frac{1}{3+\sqrt{3}}+\frac{1}{\sqrt{3}-3}$ is

$A$ boy was asked to multiply a given number by $\frac{8}{17}$. Instead,he divided it by $\frac{8}{17}$ and got the result $225$ more than what he should have got if he had multiplied the number by $\frac{8}{17}$. The given number was:

If $a+b+c=1$ and $ab+bc+ca=\frac{1}{3}$,then $a:b:c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module