Question

The product of (3/4)^-4 and (-2/3)^2 is divided by a number so that the quotient is( 3/5) ^ -2 . find the number​

Answer 1

i got upto this......

Answer 2

$\frac{1024}{2025}$ is the divisor for the product  $(\frac{3}{4})^{-4} \ and \ (\frac{-2}{3})^{2}$.

Given:

When the product $(\frac{3}{4})^{-4} \ and \ (\frac{-2}{3})^{2}$ is divided by a number it gives a quotient $(\frac{3}{5})^{-2}$.

To Find:

The divisor number.

Solution:

Let D be the divisor

$\frac{(\frac{3}{4} )^{-4}\times (\frac{-2}{3} )^{2} }{D} = (\frac{3}{5} )^{-2}$

$D=\frac{(\frac{3}{4} )^{-4}\times (\frac{-2}{3} )^{2} }{(\frac{3}{5} )^{-2}} \\\\D = (\frac{3}{4} )^{-4}\times (\frac{-2}{3} )^{2} \times (\frac{3}{5} )^{2}}$

Applying the Formula $a^{-1} = \frac{1}{a}$

$D = (\frac{4}{3} )^{4}\times (\frac{-2}{3} )^{2} \times (\frac{3}{5} )^{2}}\\\\D = \frac{256}{81} \times\frac{4}{9} \times \frac{9}{25} \\\\D= \frac{1024}{2025}$

Therefore, $\frac{1024}{2025}$ is the divisor of the product  $(\frac{3}{4})^{-4} \ and \ (\frac{-2}{3})^{2}$.

#SPJ2

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