sqrt{11449} times sqrt{6241} - (54)^{2} = sqr | vedclass

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

Question

(B) First,calculate the square roots:$\sqrt{11449} = 107$$\sqrt{6241} = 79$Next,calculate the squares:$(54)^{2} = 2916$$(74)^{2} = 5476$Substitute the

Vedclass · Accepted Answer

$3721$

Vedclass · Answer

(B) First,calculate the square roots:$\sqrt{11449} = 107$$\sqrt{6241} = 79$Next,calculate the squares:$(54)^{2} = 2916$$(74)^{2} = 5476$Substitute these values into the equation:$107 	imes 79 - 2916 = \sqrt{?} + 5476$$8453 - 2916 = \sqrt{?} + 5476$$5537 = \sqrt{?} + 5476$Isolate $\sqrt{?}$:$\sqrt{?} = 5537 - 5476$$\sqrt{?} = 61$Finally,square both sides to find $?$:$? = (61)^{2} = 3721$

$\sqrt{11449} \times \sqrt{6241} - (54)^{2} = \sqrt{?} + (74)^{2}$

Similar Questions

The smallest number which is a perfect square and contains $7936$ as a factor is

The square root of $0 . \overline{4}$ is

The square root of $(7+3 \sqrt{5})(7-3 \sqrt{5})$ is

The value of $\sqrt{0.01} + \sqrt{0.81} + \sqrt{1.21} + \sqrt{0.0009}$ is

If $a^{2}=2,$ then $(a+1)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module