The approximate value of (7.995)^{1/3} correc | vedclass

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

Question

(B) We can write $(7.995)^{1/3}$ as $(8 - 0.005)^{1/3}$.Using the binomial expansion $(1 - x)^n \approx 1 - nx + \frac{n(n-1)}{2!}x^2$ for small $x$:$

Vedclass · Accepted Answer

$1.9996$

Vedclass · Answer

(B) We can write $(7.995)^{1/3}$ as $(8 - 0.005)^{1/3}$.Using the binomial expansion $(1 - x)^n \approx 1 - nx + \frac{n(n-1)}{2!}x^2$ for small $x$:$(8 - 0.005)^{1/3} = 8^{1/3} \left(1 - \frac{0.005}{8}ight)^{1/3} = 2 \left(1 - \frac{0.005}{8}ight)^{1/3}$.Applying the expansion $(1 - x)^n \approx 1 - nx$:$2 \left(1 - \frac{1}{3} 	imes \frac{0.005}{8}ight) = 2 \left(1 - \frac{0.005}{24}ight)$.$2 \left(1 - 0.00020833...ight) = 2(0.99979166...)$.$= 1.9995833...$.Rounding to four decimal places,we get $1.9996$.

The approximate value of $(7.995)^{1/3}$ correct to four decimal places is

Similar Questions

The approximate value of $(66)^{\frac{1}{3}}$ is

The time $T$ of oscillation of a simple pendulum of length $L$ is governed by $T=2 \pi \sqrt{\frac{L}{g}}$,where $g$ is constant. The percentage by which the length must be changed in order to correct an error of loss equal to $2$ minutes of time per day is

Find the approximate value of $\sqrt{0.6}$ using differentials.

If $y=5x^2+6x+6$,$x=2$,and $\Delta x=0.001$,then the values of $\Delta y$ and $dy$ respectively are:

For a given function $y=f(x)$,$\delta y$ denotes the actual error in $y$ corresponding to an actual error $\delta x$ in $x$,and $dy$ denotes the approximate value of $\delta y$. If $y=f(x)=2x^2-3x+4$ and $\delta x=0.02$,then the value of $\delta y - dy$ when $x=5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module