Find the approximate value of sqrt{25.3}. | vedclass

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

Question

(B) Let $y = \sqrt{x}$.We choose $x = 25$ and $\Delta x = 0.3$,so that $x + \Delta x = 25.3$.Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} = \sqrt{2

Vedclass · Accepted Answer

$5.030$

Vedclass · Answer

(B) Let $y = \sqrt{x}$.We choose $x = 25$ and $\Delta x = 0.3$,so that $x + \Delta x = 25.3$.Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} = \sqrt{25.3} - \sqrt{25} = \sqrt{25.3} - 5$.This implies $\sqrt{25.3} = 5 + \Delta y$.Using the approximation $dy \approx \Delta y$,we have $dy = \left(\frac{dy}{dx}ight) \Delta x$.Since $y = x^{1/2}$,$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.Substituting the values,$dy = \frac{1}{2\sqrt{25}} 	imes 0.3 = \frac{1}{2 	imes 5} 	imes 0.3 = \frac{0.3}{10} = 0.03$.Therefore,$\sqrt{25.3} \approx 5 + 0.03 = 5.03$.

Find the approximate value of $\sqrt{25.3}$.

Similar Questions

The approximate value of $(3 \sqrt[3]{126} + \sin 61^{\circ})$ correct to three decimal places,obtained by taking $1^{\circ} = 0.0174$ radians,is

Find the approximate value of $\sqrt{49.5}$.

Using differentials,find the approximate value of $(33)^{-\frac{1}{5}}$.

The approximate value of $\tan ^{-1}(0.999)$ (up to $4$ decimal places) is

Using differentials,find the approximate value of $\sqrt{0.082}$. Let $f(x) = \sqrt{x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module