The approximate value of sqrt[3]{0.026} is | vedclass

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

Question

(B) Let $f(x) = x^{\frac{1}{3}}$.Then,$f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$.We choose $a = 0.027$ because $\sqrt[3]{0.02

Vedclass · Accepted Answer

$0.2963$

Vedclass · Answer

(B) Let $f(x) = x^{\frac{1}{3}}$.Then,$f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$.We choose $a = 0.027$ because $\sqrt[3]{0.027} = 0.3$,and $h = -0.001$ such that $a + h = 0.026$.Using the linear approximation formula $f(a + h) \approx f(a) + hf'(a)$:$f(a) = (0.027)^{\frac{1}{3}} = 0.3$.$f'(a) = \frac{1}{3(0.027)^{\frac{2}{3}}} = \frac{1}{3(0.09)} = \frac{1}{0.27}$.Substituting these values:$f(0.026) \approx 0.3 + (-0.001) 	imes \frac{1}{0.27}$.$f(0.026) \approx 0.3 - \frac{0.001}{0.27} \approx 0.3 - 0.0037 = 0.2963$.

The approximate value of $\sqrt[3]{0.026}$ is

Similar Questions

If $1^{\circ} = 0.0175 \text{ radians}$,then the approximate value of $\sec 58^{\circ}$ is

The approximate value of $\log _{10} 1002$ is (Given $\log _{10} e = 0.4343$)

If $y=2x^3-2x^2+3x-5$, then for $x=2$ and $\Delta x=0.1$, the value of $\Delta y$ is:

If the percentage error in the radius of a circle is $3\%$, then the percentage error in its area is (in $\%$)

If the percentage error in the radius of a circle is $3\%$,then the percentage error in its area is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module