frac{(0.146+0.092)^{2}+(0.092-0.146)^{2}}{0.0 | vedclass

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

Question

(A) Let $a = 0.146$ and $b = 0.092$.Then $0.073 = \frac{a}{2}$ and $0.046 = \frac{b}{2}$.The expression becomes:$\frac{(a+b)^{2} + (b-a)^{2}}{(\frac{a

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let $a = 0.146$ and $b = 0.092$.Then $0.073 = \frac{a}{2}$ and $0.046 = \frac{b}{2}$.The expression becomes:$\frac{(a+b)^{2} + (b-a)^{2}}{(\frac{a}{2})^{2} + (\frac{b}{2})^{2}}$Since $(b-a)^{2} = (a-b)^{2}$,the numerator is $(a+b)^{2} + (a-b)^{2} = 2(a^{2} + b^{2})$.The denominator is $\frac{a^{2}}{4} + \frac{b^{2}}{4} = \frac{a^{2} + b^{2}}{4}$.Thus,the expression is $\frac{2(a^{2} + b^{2})}{\frac{a^{2} + b^{2}}{4}} = 2 	imes 4 = 8$.

$\frac{(0.146+0.092)^{2}+(0.092-0.146)^{2}}{0.073 \times 0.073+0.046 \times 0.046}$ is equal to

Similar Questions

The sum of three numbers is $2$. The $1^{\text{st}}$ number is $\frac{1}{2}$ times the $2^{\text{nd}}$ number and the $3^{\text{rd}}$ number is $\frac{1}{4}$ times the $2^{\text{nd}}$ number. The $2^{\text{nd}}$ number is

$5894 \div 15.01 + 590.01 - 111.98 = ?$

$3 \frac{6}{7} + 6 \frac{1}{4} - 5 \frac{1}{3} + 2 \frac{1}{2} = ?$

In a division sum,the quotient is $120$,the divisor is $456$,and the remainder is $333$. Find the dividend.

The least positive integer that should be subtracted from $3011 \times 3012$ so that the difference is a perfect square,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module