Question

evaluate the following (-7/5) *(-2/7) in each power is -2


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Answer 1

Answer:

$\left(\frac{-7}{5}\right)^{-2}\times \left(\frac{-2}{7}\right)^{-2}=\frac{25}{4}$

Step-by-step explanation:

$\left(\frac{-7}{5}\right)^{-2}\times \left(\frac{-2}{7}\right)^{-2}\\=\left(\frac{-7\times (-2)}{5\times 7}\right)^{-2}=\left(\frac{2}{5}\right)^{-2}\\=\left(\frac{5}{2}\right)^{2}\\=\frac{5^{2}}{2^{2}}\\=\frac{25}{4}$

Therefore,

$\left(\frac{-7}{5}\right)^{-2}\times \left(\frac{-2}{7}\right)^{-2}=\frac{25}{4}$

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Answer 2

$\frac{25}{4}$ or $(\frac{5}{2}$$)^{2}$

Step-by-step explanation:

Given,

$(\frac{-7}{5}$$)^{-2}$ × $(\frac{-2}{7}$$)^{-2}$

To remove the negative exponent -2, the numerator and denominator are interchanged.

$(\frac{-5}{7}$$)^{2}$ × $(\frac{-7}{2}$$)^{2}$

$(\frac{-5}{7})$ ×  $(\frac{-5}{7})$ × $(\frac{-7}{2})$ × $(\frac{-7}{2})$

Removing the minus because there are two pairs of them:

$(\frac{5}{7})$ ×  $(\frac{5}{7})$ × $(\frac{7}{2})$ × $(\frac{7}{2})$

Cancelling 7 from both numerator and denominator:

$(\frac{5}{2})$ ×  $(\frac{5}{2})$

$\frac{25}{4}$ or $(\frac{5}{2}$$)^{2}$

Learn more:

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