The nearest approximate value of sqrt{2023} i | vedclass

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(B) We know that $45^2 = 2025$.Since $2023 < 2025$, we have $\sqrt{2023} < \sqrt{2025}$, which means $\sqrt{2023} < 45$.Using the linear approximation

Vedclass · Accepted Answer

$44.9778$

Vedclass · Answer

(B) We know that $45^2 = 2025$.Since $2023 Using the linear approximation formula $\sqrt{x + \Delta x} \approx \sqrt{x} + \frac{\Delta x}{2\sqrt{x}}$, let $x = 2025$ and $\Delta x = -2$.Then, $\sqrt{2023} = \sqrt{2025 - 2} \approx \sqrt{2025} + \frac{-2}{2\sqrt{2025}}$.$\sqrt{2023} \approx 45 - \frac{2}{2(45)} = 45 - \frac{2}{90} = 45 - \frac{1}{45}$.$\sqrt{2023} \approx 45 - 0.0222 = 44.9778$.Thus, the nearest approximate value is $44.9778$.

The nearest approximate value of $\sqrt{2023}$ is

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