sqrt[3]{4663} + 349 = ? div 21.003 | vedclass

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

Question

(D) Step $1$: Approximate the cube root. Since $16^3 = 4096$ and $17^3 = 4913$,$\sqrt[3]{4663} \approx 16.7 \approx 17$.Step $2$: Approximate the divi

Vedclass · Accepted Answer

$7680$

Vedclass · Answer

(D) Step $1$: Approximate the cube root. Since $16^3 = 4096$ and $17^3 = 4913$,$\sqrt[3]{4663} \approx 16.7 \approx 17$.Step $2$: Approximate the divisor. $21.003 \approx 21$.Step $3$: Set up the equation: $17 + 349 = ? \div 21$.Step $4$: Simplify: $366 = ? \div 21$.Step $5$: Solve for $?$: $? = 366 	imes 21 = 7686$.Step $6$: Comparing with the given options,the closest value is $7680$.

$\sqrt[3]{4663} + 349 = ? \div 21.003$

Similar Questions

$348 \div 29 \times 15 + 156 = (?)^{3} + 120$

The sum of the square of the first number and the cube of the second number is $568$. Also,the square of the second number is $15$ less than the square of $8$. What is the value of $\frac{3}{5}$ of the first number? (Assuming both numbers are positive)

Find the cube root of $15.625$. (in $.5$)

$\sqrt[3]{54872} \times (304 \div 8) = (?)^{2}$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module