(3x^2 y)^3/(6x^-2 y^5) please help medal + fan and please explain it.

- baby456

(3x^2 y)^3/(6x^-2 y^5) please help medal + fan and please explain it.

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- hlambach

Your missing a part there after 6x

- hlambach

oops *You're haha

- baby456

-2

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## More answers

- baby456

is right there

- hlambach

What power is 6x being raised to? 2y^5?

- baby456

-2

- baby456

let me draw it

- hlambach

Okay.

- baby456

|dw:1435273567323:dw|

- hlambach

Got it. :) So you're simplifying it, right?

- baby456

yes

- anonymous

use your basic properties of exponents. \(\dfrac{a^M}{a^N}=a^{(M-N)}\\\) etc.

- baby456

ok

- baby456

|dw:1435273964905:dw|

- hlambach

Tip: First simplify \[(3x ^{3}y)^{3}\]

- baby456

9x^9y^3

- hlambach

Good, now you can use the formulas Jenny gave you.

- hlambach

Hold on, 3^3 is what?

- baby456

|dw:1435274627143:dw|

- anonymous

3^3=3*3*3
3^3=/=3*3

- hlambach

Again, what Jenny said... I was wrong to say that you were right in your simplification.

- baby456

oh i simpfieed it 27X^6y^3

- baby456

|dw:1435274935189:dw|

- hlambach

There you go.

- baby456

ok so how about the bottom

- hlambach

Well, now all you have to use is the Exponent Rule:
\[\frac{ x ^{a} }{ x ^{b} } = x ^{a-b}\]

- baby456

how about the 6

- baby456

do you subtract 27-6

- hlambach

Well, this is kind of hard to explain in text, but I'll try. :)
You have (ignoring the y)
27x^9
------
6x^-2
To divide it into 27, make it 2(3x^-2) since 3 can be divided into 27. Then use the exponent rule.

- baby456

how? again

- baby456

|dw:1435275710920:dw|

- hlambach

|dw:1435275705062:dw|
SOrry for my terrible drawing skills haha. Now just divide out the 3x^-2

- baby456

|dw:1435275773927:dw|

- baby456

how

- hlambach

Just divide
27x^9
------
3x^-2
You will leave the 2 at the bottom.

- baby456

9x^7

- hlambach

remember its
x^a
--- = x^a - b
x^b

- hlambach

9 - (-2) is what?

- baby456

x to 6 not x to the 9

- anonymous

split it up like this \(\dfrac{(3x^2 y)^3}{(6x^{-2} y^5) }=\dfrac{27x^6y^3}{6x^{-2}y^5}=\dfrac{27}{6}*\dfrac{x^6y^3}{x^{-2}y^5}\)

- baby456

i am so confused

- anonymous

read your textbook. she's explained the property several times over

- baby456

i dont have a textbook

- baby456

still confused @jenny1994 is confusing me

- hlambach

Let's go with Jenny's way
For now, focus on simplifying
x^6 y^3
-------
x^-2y^5

- hlambach

Using the rule that I am too lazy to write out. (Just look above)

- anonymous

you can't follow a simple pattern?
\(\large \dfrac{x^a}{x^b}=a^{(a-b)}\implies \dfrac{x^6}{x^{-2}}=?\)

- anonymous

x^(a-b)*

- hlambach

We can do this Jenny... haha Patience is a virtue. I wish I could put "OpenStudy Math Tutor on my transcript.

- anonymous

Don't mention openstudy, but say you tutor. Resumes are all about embellishment

- hlambach

Ha, I was joking, but that's a good idea!

- hlambach

Sorry, back on subject.

- baby456

who way should i follow

- hlambach

Go with Jenny's
Simplify
x^6 y^3
-------
x^-2 y^5

- baby456

you lost me at what to do about the bottom part
please explain more

- hlambach

Using the rule that has been mentioned several times, thus I won't write it out again, simplify.
EXAMPLE
x^4
--- = x^(4-2) = x^2
x^2
Get it? But to that do your formula

- hlambach

*but do that to your formula.

- baby456

so is it x4 since its 6-2

- hlambach

I'm sorry, I typed it wrong, it's
x^9 y^3
------
x^-2 y^5
So it would be x^11 (since 9 - (-2) = 11)
But you grasped the concept. Now what about the y?

- baby456

where are you getting x^9 from

- hlambach

From when you simplified (3x^3 y)^3 = 27x^9 y^3

- baby456

(3x^2 y)^3= 27X^6y^3

- baby456

2^3=6

- zepdrix

Baby the subtraction comes from the rule,
the bottom number also had a negative
so don't let that extra negative get lost in the process! :)\[\Large\rm \frac{x^{\color{orangered}{6}}}{x^{\color{royalblue}{-2}}}=x^{\color{orangered}{6}-\color{royalblue}{-2}}\]You don't get x^4, see that?

- zepdrix

2^3 does not equal 6.
I think you meant 2*3 = 6 :)

- hlambach

Oh, I'm sorry.
But remember cubed mean 2 * 2 *2

- baby456

OPPS SORRY

- baby456

SO THEN X=8

- hlambach

Hey, I gotta run.
@zepdrix could you take over?

- zepdrix

you mean for the final power on x?
for the 6 - - 2?
ya that looks better baby :)

- zepdrix

ya you get outta here missy :P hehe

- hlambach

haha, thanks, I;m already running late. :)

- zepdrix

So you understood this first step, yes?\[\Large\rm \frac{(3x^2 y)^3}{6x^{-2} y^5}=\frac{27x^6 y^3}{6x^{-2} y^5}\]And then did you understand the part with the x's?\[\Large\rm \frac{27\color{orangered}{x^6} y^3}{6\color{orangered}{x^{-2}} y^5}=\frac{27\color{orangered}{x^8}y^3}{6y^5}\]
I'm just trying to figure out where we're at :O
This thread is so long and messy

- baby456

YEP

- zepdrix

\[\Large\rm \frac{27x^8\color{royalblue}{y^3}}{6\color{royalblue}{y^5}}=\frac{27x^8\color{royalblue}{y^?}}{6}\]What about the y's? :)
What do you think will happen there?

- baby456

3-5=-2

- zepdrix

\[\Large\rm \frac{27x^8\color{royalblue}{y^{-2}}}{6}\]Mmmm good.
As a final step, you want to notice that the 27 and 6 share some factor.

- baby456

3

- zepdrix

\[\Large\rm \frac{27x^8y^{-2}}{6}=\frac{9\cdot3x^8y^{-2}}{2\cdot3}\]Yah that sounds right

- zepdrix

\[\Large\rm \frac{9\cdot\cancel3x^8y^{-2}}{2\cdot\cancel3}\]

- zepdrix

Sometimes your teacher will ask for the final answer with NO NEGATIVE POWERS.
Do you understand how to deal with the y if you have to get rid of the negative power? :o

- baby456

NOPE

- zepdrix

mm k, well our exponent rule tells us to flip it.
if its in the numerator, it gets tossed down the denominator and the sign changes on the power.
if it's in the denominator it goes up to the numerator and the sign changes on the power.
examples:\[\Large\rm \frac{1}{2x^{-7}}=\frac{x^7}{2}\]
\[\Large\rm \frac{x^{-4}}{3}=\frac{1}{3x^4}\]

- baby456

OH I REMEMBER THIS YOU JUST SWITCH IT LIKE A RECIPROCAL

- zepdrix

good good :) reciprocal, and change from negative to positive

- baby456

OK THANKS I HAVE TO GO NOW

- baby456

THANKS FOR ALL YOUR HELP

- baby456

sorry about caplocks

- zepdrix

\[\Large\rm \frac{9x^8y^{-2}}{2}=\frac{9x^8}{2y^{2}}\]so ya that will fix the y :)

- zepdrix

np

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