Find the approximate value of sqrt{49.5}. | vedclass

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

Question

(B) Consider the function $y = \sqrt{x}$. Let $x = 49$ and $\Delta x = 0.5$. Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} = \sqrt{49.5} - \sqrt{49}

Vedclass · Accepted Answer

$7.036$

Vedclass · Answer

(B) Consider the function $y = \sqrt{x}$. Let $x = 49$ and $\Delta x = 0.5$. Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} = \sqrt{49.5} - \sqrt{49} = \sqrt{49.5} - 7$. This implies $\sqrt{49.5} = 7 + \Delta y$. We know that $dy$ is approximately equal to $\Delta y$,where $dy = \left(\frac{dy}{dx}ight) \Delta x$. Since $y = \sqrt{x}$,we have $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$. Substituting the values,$dy = \frac{1}{2\sqrt{49}} 	imes 0.5 = \frac{1}{2 	imes 7} 	imes 0.5 = \frac{0.5}{14} = \frac{1}{28} \approx 0.0357$. Therefore,the approximate value of $\sqrt{49.5} = 7 + 0.0357 = 7.0357 \approx 7.036$.

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

Similar Questions

The approximate value of $(1.0002)^{3000}$ is

An equilateral triangle has a side of $10$ units. If an error of $0.05$ units is made in measuring the side, then the percentage error in the area of the triangle is:

The approximate value of $x^3-2x^2+3x+2$ at $x=2.01$ is

Find the approximate value of $(32.15)^{\frac{1}{5}}$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module