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lisov135 [29]
3 years ago
8

I need help, i cant do this easly i suck at math so was wandering if yall can give me an answer for these or tell me how to do i

t.
I NEED HELP ASAP

Mathematics
2 answers:
AVprozaik [17]3 years ago
8 0
This is what I got from Photomath
Scorpion4ik [409]3 years ago
4 0
On all of them sorry but if i dont know wich one to do then....
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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}=
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( \frac{ \frac{1}{(-7)^2} }{ \frac{1}{5^2} } )( \frac{x^ \frac{-6}{2} }{y^{-8}} )=
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( \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}
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Six friends each het 3/8 of a pound of candy. how much candy will the friends get in all
tamaranim1 [39]
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4 years ago
The pre–image, A, was dilated about the origin. It was then transformed in another way to get A'.
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Step-by-step explanation:

i hope this helps

6 0
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X-y=3<br> X+2y=-6<br> Solve using substitution
Tresset [83]
X-y=3=>x=3+y
X+2y=-6
=========
3+y+2y=-6
3+3y=-6
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3y=-9
y=-3

X+2y=-6
X+2(-3)=-6
X-6=-6
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290÷10=29 ........................

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