Question

Simplify (-2)^4*x^3*7^6 / (-2)^2*x^-2*y^-3...please! Algebra 2

Answer 1

I've been stuck on how to solve this. There are four possible answer choices:
a) -4xy^3
b) 64x^5*y^9
c)-64xy^3
d) 4x^5*y^9

Answer 2

Let's write it with the equation editor:\[\frac{ (-2)^4x^37^6 }{ (-2)^2x^-2y^-3}\]Now things are clearer. The (-2)^4=16 and (-2)^2=4 are not the problem, I think, it is the dividing using negative powers of x and y.
You can now use the rule:\[\frac{ a^b }{ a^c }=a^{b-c}\]for the powers of x.
BTW, are you sure there is a 7 and not a y in the numerator? It's in an unusual place...

Answer 3

Hint: because we've calculated the powers of -2 as 16 and 8, we're sure that the outcome doesn't have a minus sign, so it will be b or d.
And yes, the 7 in the numerator must be y to make sense of the answers.

Answer 4

@ZeHanz, an equation editor tip: if you have a negative exponent (or an expression as an exponent), enclose it in curly braces to protect it.

\[\frac{(-2)^4x^37^6}{(-2)^2x^{-2}y^{-3}}\]
\frac{(-2)^4x^37^6}{(-2)^2x^{-2}y^{-3}}

Answer 5

Ah, it's a y not a 7 in the numerator! Thank you! I'm kind of a newbie here.

Hmm. So then \[ \frac{ 16*x^3y^6 }{ 8*x^{-2}y^{-3}} \]

I understand why the ending to the answers would be \[x^5y^9\] but i don't know what to do with the 16 and 8...

Answer 6

Divide 16 by 8, what do you get?
Or think of 16 as 8*2...

Answer 7

Or think of 16 as 2^4 and 8 as 2^3...

Answer 8

But I'm not sure I see where you are getting an 8...

Answer 9

\[\frac{(-2)^4*x^3*y^6}{ (-2)^2*x^{-2}*y^{-3}}\] is the correct expression, right?

Answer 10

Yes, I'm a little frazzled now. >..<

I think I meant for my 8 to be a 4.

Answer 11

So I think the answer would be "d".

Answer 12

You could also think of it as \[(-2)^4/(-2)^2 = (-2)^{(4-2)} = (-2)^2 = 4\]

Answer 13

So you've got 16/4=4 and also x^5 and y^9. Now multiply everything...

Answer 14

\[4x^5y^9\]
Thanks everyone! I think I had the best first open study experience!

Answer 15

YW!