Fill in the blanks to make the following stat | vedclass

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

Question

(A) To solve the expression $\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{6}$,we use the property of radicals $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$.Firs

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) To solve the expression $\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{6}$,we use the property of radicals $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$.First,multiply the terms inside the square roots: $\sqrt{2} \cdot \sqrt{3} = \sqrt{2 \cdot 3} = \sqrt{6}$.Now,substitute this back into the original expression: $\sqrt{6} \cdot \sqrt{6}$.Using the property $\sqrt{a} \cdot \sqrt{a} = a$,we get $\sqrt{6} \cdot \sqrt{6} = 6$.Therefore,the correct answer is $6$.

Fill in the blanks to make the following statement true: $\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{6} = \dots$

Similar Questions

Simplify each of the following expressions:
$(i)$ $(3+\sqrt{3})(2+\sqrt{2})$
$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})$
$(iii)$ $(\sqrt{5}+\sqrt{2})^{2}$
$(iv)$ $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

Divide $8 \sqrt{15}$ by $2 \sqrt{3}$.

Add $4 \sqrt{3}+2 \sqrt{5}$ and $6 \sqrt{3}-4 \sqrt{5}$.

Rationalise the denominator of $\frac{1}{7+3 \sqrt{2}}$

If $x = 5 + 2\sqrt{6}$,then find the value of $x^{2} + \frac{1}{x^{2}}$ and $x^{3} + \frac{1}{x^{3}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module