Solve the given two equations and select the | vedclass

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

Question

(B) For equation $I$: $676 x^{2} - 1 = 0$$676 x^{2} = 1$$x^{2} = \frac{1}{676}$$x = \pm \frac{1}{26}$So,$x = \frac{1}{26} \approx 0.0384$ or $x = -\fr

Vedclass · Accepted Answer

if $x < y$

Vedclass · Answer

(B) For equation $I$: $676 x^{2} - 1 = 0$$676 x^{2} = 1$$x^{2} = \frac{1}{676}$$x = \pm \frac{1}{26}$So,$x = \frac{1}{26} \approx 0.0384$ or $x = -\frac{1}{26} \approx -0.0384$.For equation $II$: $y = \frac{1}{\sqrt[3]{13824}}$Since $24^{3} = 13824$,we have $\sqrt[3]{13824} = 24$.Thus,$y = \frac{1}{24} \approx 0.0416$.Comparing the values:$x = 0.0384$ or $-0.0384$ and $y = 0.0416$.In both cases,$x < y$.

Solve the given two equations and select the correct option.
$I.$ $676 x^{2}-1=0$
$II.$ $y=\frac{1}{\sqrt[3]{13824}}$

Similar Questions

Solve the given two equations and select the correct option.

Solve the given two equations and select the correct answer from the given options.
$I.$ $2x + 5y = 6$
$II.$ $5x + 11y = 9$

Solve the given two equations and choose the correct option.
$I.$ $99 x^{2} + 149 x + 56 = 0$
$II.$ $156 y^{2} + 287 y + 132 = 0$

The condition that ${x^3} - 3px + 2q$ may be divisible by a factor of the form ${x^2} + 2ax + {a^2}$ is

Find the sum of all integral values of $a$ such that the equation $||x - 2| - |3 - x|| = 2 - a$ has a solution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct option.$I.$ $676 x^{2}-1=0$$II.$ $y=\frac{1}{\sqrt[3]{13824}}$