Plz help me with this question

Mathematics

1 answer:

solniwko [45]3 years ago

6 0

The answer is 13/25 13 is odd so it can't be evenly divided by anything so it is in its simplest form

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I do not know how to solve this problem or know what to do with the exponents

horsena [70]

K, remember
(ab)/(cd)=(a/c)(b/d) or whatever
also
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}= \frac{1}{x^m}$
and
$x^ \frac{m}{n}= \sqrt[n]{x^m}$
and
$( \frac{x}{y} )^m= \frac{x^m}{y^m}$
and
$(x^m)^n=x^{mn}$
and
(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)

so
$( \frac{-7x^ \frac{3}{2} }{5y^4} )^{-2}$ =
$( \frac{-7}{5} )^{-2}( \frac{x^ \frac{3}{2} }{y^4} )^{-2}$ =
$( \frac{(-7)^{-2}}{5^{-2}} )( \frac{(x^ \frac{3}{2})^{-2} }{(y^4)^{-2}} )$ =
$( \frac{ \frac{1}{(-7)^2} }{ \frac{1}{5^2} } )( \frac{x^ \frac{-6}{2} }{y^{-8}} )$ =
$( \frac{ \frac{1}{49} }{ \frac{1}{25} } )( \frac{(x^{-3}) }{\frac{1}{y^8}} )$
$( \frac{ \frac{1}{49} }{ \frac{1}{25} } )( \frac{\frac{1}{x^3} }{\frac{1}{y^8}} )$ =
$(\frac{25}{49} )( \frac{y^8}{x^3}$ =
$\frac{25y^8}{49x^3}$

4 0

3 years ago

How is 1/4 in porcent?

Archy [21]

25%, hope this helps

7 0

3 years ago

Read 2 more answers

The scale for the drawing below is 1 in = 2 in. Draw the larger figure and label the sides. By

arlik [135]

Answer:

i need the answer too :(