The value of sqrt{24.99} is | vedclass

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

Question

(C) To find the approximate value of $\sqrt{24.99}$,we use the concept of differentials.Let $f(x) = \sqrt{x}$.We know that $24.99 = 25 - 0.01$. Here,$

Vedclass · Accepted Answer

$4.999$

Vedclass · Answer

(C) To find the approximate value of $\sqrt{24.99}$,we use the concept of differentials.Let $f(x) = \sqrt{x}$.We know that $24.99 = 25 - 0.01$. Here,$x = 25$ and $\Delta x = -0.01$.The formula for the approximation is $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.First,calculate $f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$.At $x = 25$,$f(25) = \sqrt{25} = 5$ and $f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{2 	imes 5} = \frac{1}{10} = 0.1$.Now,substitute these values into the formula:$f(24.99) \approx f(25) + f'(25) 	imes \Delta x$$f(24.99) \approx 5 + (0.1) 	imes (-0.01)$$f(24.99) \approx 5 - 0.001$$f(24.99) \approx 4.999$.

The value of $\sqrt{24.99}$ is

Similar Questions

Using differentiation,the approximate value of $f(x) = x^2 - 2x + 1$ at $x = 2.99$ is ....

Find the approximate value of $f(5.001),$ where $f(x)=x^{3}-7x^{2}+15$.

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

Electric current $(I)$ is measured by a galvanometer, where the current is proportional to the tangent of the angle $( \theta)$ of deflection. If the deflection is read as $45^{\circ}$ and an error of $1\%$ is made in reading the angle, the percentage error in the current is:

If the increase in the side of a square is $6 \%$,then the approximate percentage increase in its area is: (in $\%$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module