(35)^{2} div sqrt[3]{125} + (25)^{2} div 125 | vedclass

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

Question

(B) Given expression: $(35)^{2} \div \sqrt[3]{125} + (25)^{2} \div 125 = ?$Step $1$: Calculate the squares and the cube root.$(35)^{2} = 1225$$\sqrt[3

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(B) Given expression: $(35)^{2} \div \sqrt[3]{125} + (25)^{2} \div 125 = ?$Step $1$: Calculate the squares and the cube root.$(35)^{2} = 1225$$\sqrt[3]{125} = 5$$(25)^{2} = 625$Step $2$: Substitute these values into the expression.$1225 \div 5 + 625 \div 125 = ?$Step $3$: Perform the divisions.$1225 / 5 = 245$$625 / 125 = 5$Step $4$: Add the results.$245 + 5 = 250$Therefore,the correct answer is $250$.

$(35)^{2} \div \sqrt[3]{125} + (25)^{2} \div 125 = ?$

Similar Questions

The least perfect cube which is completely divisible by $21, 24$ and $27$ is

$(4 \times 4)^{3} \div (512 \div 8)^{4} \times (32 \times 8)^{4} = (2 \times 2)^{?} + ?^{4}$

Find the cube root of $0.000729$.

Find the cube root of $0.000216$.

The smallest five-digit number which is a perfect cube is

Vedclass Test Series

Exam Paper Generator

Online Exam Module