(21.98)^{2} - (25.02)^{2} + (13.03)^{2} = ? | vedclass

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

Question

(A) To solve this,we approximate the given values to the nearest integers:$21.98 \approx 22$$25.02 \approx 25$$13.03 \approx 13$Now,substitute these v

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) To solve this,we approximate the given values to the nearest integers:$21.98 \approx 22$$25.02 \approx 25$$13.03 \approx 13$Now,substitute these values into the expression:$(22)^{2} - (25)^{2} + (13)^{2}$$= 484 - 625 + 169$$= (484 + 169) - 625$$= 653 - 625$$= 28$The value $28$ is closest to $25$ among the given options.

$(21.98)^{2} - (25.02)^{2} + (13.03)^{2} = ?$

Similar Questions

$958 \times 21 \div 4 = ?$

$72 \%$ of $390 + 28 \%$ of $165 = (x - 3)$

The nearest whole number to one million which is divisible by $537$ is

$73 \times 18 + 486 = ? + (13)^{2}$

What is the least number that must be added to $4042$ to make it a perfect square?

Vedclass Test Series

Exam Paper Generator

Online Exam Module