The value of sqrt{24.99} is | vedclass

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(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,$

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$4.999$

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(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

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