Question

3x^2(5XY)^2(XY)^-1 please help

Answer 1

\[3x^2*(5xy)^2*(xy)^{-1}\]I would first expand the items in parentheses, using the property that
\[(ab)^n = a^nb^n\]

Answer 2

Then I would simplify by using \[a^na^m = a^{n+m}\]

Answer 3

The answers I can use are 75x^4 y^2
75x^3y
15x4y^2
15x^3y

Answer 4

I am having a really hard time figuring this problem out.. I am in the middle of an online class that I am allowed to get help on.

Answer 5

Doesn't matter what the answer choices are, do what I suggested and you should get the correct answer :-)

Answer 6

Even when you don't have the luxury of seeing your choices, that is :-)

Answer 7

thank you.

Answer 8

so, let's start with the \[(5xy)^2\]

Answer 9

\[(5xy)^2 = 5^2*x^2*y^2 = \]

Answer 10

Thank you.

Answer 11

what do I do to the other side?

Answer 12

you simplify \[5^2*x^2*y^2\]what is \(5^2=\)

Answer 13

25

Answer 14

good, so \[5^2*x^2*y^2 = 25x^2y^2\]right?

Answer 15

Alright I understand that.

Answer 16

okay, next group is \((xy)^{-1}\)can you do the same thing there?

Answer 17

where did you get the (xy)-1 I am sorry I just don't see it..

Answer 18

You originally wrote 3x^2(5XY)^2(XY)^-1

Answer 19

Formatted, that's
\[3x^2(5xy)^2(xy)^{-1}\]

Answer 20

ohh okay I see that now

Answer 21

\[3x^2\] is already done, and we just did \[(5xy)^2 = 25x^2y^2\]so now we just need to expand \[(xy)^{-1}\]multiply all the pieces together and simplify

Answer 22

I think whats confusing me is the format.. I can't really understand it when its written like that. I'll split it up on my notes to make it easier

Answer 23

well, it could also be written as \[\frac{1}{(xy)^1}\]

Answer 24

because \[a^{-n} = \frac{1}{a^n}\]

Answer 25

but that's a less compact notation, and arguably makes it a little harder to simplify

Answer 26

in any case,
\[(xy)^{-1} = x^{-1}y^{-1}\]or\[\frac{1}{(xy)^1} = \frac{1}{x^1y^1}\]

Answer 27

If we line these all up, we get:

\[3x^2*25x^2y^2*x^{-1}y^{-1}\]now collect all the numbers together:
\[3*25*x^2*x^2y^2*x^{-1}y^{-1} = 75x^2*x^2y^2*x^{-1}y^{-1}\]now collect all of the \(x^n\) terms together by adding their exponents:\[75x^{2+2-1}*y^2*y^{-1}=75x^3y^2y^{-1}\]do the same for the \(y^n\) terms:\[75x^3y^{2-1}= 75x^3y\]

Answer 28

okay

Answer 29

If I had done the fraction way, it would look something like this:

\[3x^2*25x^2y^2*\frac{1}{x^1y^1} = \frac{75x^4y^2}{x^1y^1} = 75x^{4-1}y^{2-1} = 75x^3y\]Here we combined the exponents when dividing by subtracting them:
\[\frac{a^n}{a^m} = a^{n-m}\]

Answer 30

Now, if you're careful about the details, you could also see that you've got

\[3x^2*(5xy)^2*\frac{1}{(xy)^1} = 3x^2*5^2*(xy)^2*\frac{1}{(xy)^1}\]and apply the division exponent rule to the whole \((xy)\) group at once:
\[3x^2*5^2*(xy)^2*\frac{1}{(xy)^1} = 75x^2*(xy)^{2-1} =75x^2*(xy)^1 = 75x^3y\]

Answer 31

but until you're really comfortable with these sorts of manipulations, I would stick with carefully going through all the intermediate steps

Answer 32

okay thank you so much!!