sqrt{10} times sqrt{15} = ? | vedclass

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

Question

(A) To solve $\sqrt{10} 	imes \sqrt{15}$,we use the property $\sqrt{a} 	imes \sqrt{b} = \sqrt{a 	imes b}$.$\sqrt{10} 	imes \sqrt{15} = \sqrt{10

Vedclass · Accepted Answer

$5 \sqrt{6}$

Vedclass · Answer

(A) To solve $\sqrt{10} 	imes \sqrt{15}$,we use the property $\sqrt{a} 	imes \sqrt{b} = \sqrt{a 	imes b}$.$\sqrt{10} 	imes \sqrt{15} = \sqrt{10 	imes 15} = \sqrt{150}$.Now,factorize $150$ to simplify the square root:$150 = 25 	imes 6$.Therefore,$\sqrt{150} = \sqrt{25 	imes 6} = \sqrt{25} 	imes \sqrt{6}$.Since $\sqrt{25} = 5$,the expression simplifies to $5 \sqrt{6}$.

$\sqrt{10} \times \sqrt{15} = ?$

Similar Questions

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

$\sqrt{\frac{96.8}{0.578}} = ?$

$\frac{\sqrt{4375}}{\sqrt{7}} = ?$

$\frac{\sqrt{24}+\sqrt{216}}{\sqrt{96}}=?$

$\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$ simplifies to

Vedclass Test Series

Exam Paper Generator

Online Exam Module