3 years ago

Twice a number decreased by four is greater than ten, what numbers satisfy this condition

Mathematics

1 answer:

bezimeni [28]3 years ago

6 0

The values are represented by the inequality x>7

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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

25(3+x)+10x5 what is an equivalent expression

gogolik [260]

10x^5+25x+75 that will be your answer

4 0

3 years ago

What is the surface area of the right prism given below?

Sloan [31]

Answer:

10*15= 150

150 *8= 1,200