What is the simplified form of left parenthesis 27 x to the power of negative six end power y to the power of 12 end power right parenthesis to the power of start fraction two over three end fraction end power

- some.random.cool.kid

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- some.random.cool.kid

@zepdrix

- some.random.cool.kid

@triciaal

- some.random.cool.kid

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- some.random.cool.kid

@nincompoop @whpalmer4 @pooja195 @paki @ganeshie8

- some.random.cool.kid

mind helping?

- zepdrix

\[\large\rm \left[abc\right]^{n}=a^n b^n c^n\]When we have a bunch of stuff being raised to a power, we have to apply the exponent to each of them.

- zepdrix

\[\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=27^{2/3}(x^{-6})^{2/3}(y^{12})^{2/3}\]Ok with that step? :)

- some.random.cool.kid

mm got it am i supposed to add across? for the powers part?

- some.random.cool.kid

or exponents i should say?

- zepdrix

Have to apply one of your exponent rules from that point:\[\large\rm \color{orangered}{(x^a)^b=x^{a\cdot b}}\]When you have a power and another power, you `multiply`

- some.random.cool.kid

mmm ok ...

- zepdrix

So for the exponent on x, we'll multiply -6 and 2/3.

- some.random.cool.kid

-6 x 12? I belive.
what each of those two thirds?

- some.random.cool.kid

oh

- some.random.cool.kid

I belive that is -4? correct...

- zepdrix

\[\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=27^{2/3}x^{-4}(y^{12})^{2/3}\]k looks good :)

- some.random.cool.kid

mmm next step is ? -4 x 12 or -4 x the other 2/3rds?

- some.random.cool.kid

or is it -4 x 12 and 2/3 x 2/3?

- some.random.cool.kid

or is it just multiply across lol? i believe its what i said first though

- zepdrix

Notice that these two things have `different bases`.
One has a base of x,
the other a base of y,
that's letting us know that they won't interact with one another.
\(\large\rm x^{a}\cdot x^{b}=x^{a+b}\)
This rule that you're familiar with only works `when they have the same base`.

- zepdrix

So the -4 on the x isn't going to interact in any way with the powers on y.

- some.random.cool.kid

right so find the lcm?

- zepdrix

You have \(\large\rm (y^{12})^{2/3}\).
Our exponent rule tells us to `multiply` these powers, ya?
12 * 2/3?

- some.random.cool.kid

8

- zepdrix

\[\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=\color{orangered}{27^{2/3}}x^{-4}y^{8}\]k good :) almost done

- some.random.cool.kid

is this going to be 8 over 4?

- some.random.cool.kid

no

- zepdrix

Try to pay attention to what I've been saying.
`They have different bases`, so they won't interact with one another in any way.

- some.random.cool.kid

its not nvm i answered my own thing i belive you leave those alon now right so now i have 9 x^4y^8?

- zepdrix

yay good job \c:/

- some.random.cool.kid

:D mind helping me with a few more?

- zepdrix

stay cool kid (⌐▨_▨)
oh sure

- some.random.cool.kid

Ok great I have another one this one is a bit different then the last a little bit ...

- some.random.cool.kid

A student uses the quadratic formula to solve a quadratic equation and determines that one of the solutions is x equals negative five plus square root of negative 147 end square root. What are the values of a and b if this solution is written in the form x = a + bi, where a and b are real numbers?

- some.random.cool.kid

hold on i need to attach the numbers part this is too messy

- zepdrix

\[\large\rm x=-5+\sqrt{-147}\]

- some.random.cool.kid

##### 1 Attachment

- some.random.cool.kid

there we go :D

- zepdrix

Well, recall that we define the imaginary unit i, to be \(\large\rm \sqrt{-1}\)

- zepdrix

So we can do something clever with the square root in our problem.

- zepdrix

\[\large\rm \sqrt{-147}=\quad\sqrt{-1\cdot147}=\quad\sqrt{-1}\cdot\sqrt{147}\]I broke it up in a clever way to pull the sqrt(-1) outside like that.
Do you see the i?

- some.random.cool.kid

yep I have used this method before...

- zepdrix

\[\large\rm x=-5+\sqrt{-147}\]\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{147}\]

- some.random.cool.kid

first before you continue

- some.random.cool.kid

shouldnt I be trying to find the sqrt?

- some.random.cool.kid

and trying to figure out whetther its irrational or not?

- some.random.cool.kid

because I believe that is 12.124355653

- zepdrix

Yes, try to simplify the square root if you're able to.
In doesn't matter which order you do these steps in though :)
No no, we don't want an ugly decimal approximation.

- some.random.cool.kid

which is irrational so now you may preceed

- some.random.cool.kid

eh 12.12

- zepdrix

no...

- some.random.cool.kid

...

- zepdrix

You don't use your calculator for this step.

- some.random.cool.kid

no i rounded lol

- zepdrix

But you first used a calculator.. that's not what we're doing here.

- zepdrix

You try to simplify the root by pulling out a `perfect square`.
Example:
\(\large\rm \sqrt{72}=\sqrt{2\cdot36}=\sqrt{2}\cdot\sqrt{36}=\sqrt{2}\cdot 6\)

- some.random.cool.kid

oooo nvm its not 12.12

- zepdrix

So in my example here, I noticed that 72 was a multiple of 36,
36 can be taken out of the root since it's a perfect square.
it comes out as a 6

- some.random.cool.kid

its either 21 or 7sqrtof 3 I believe correct? not sure which one yet tough

- some.random.cool.kid

though*

- some.random.cool.kid

im not sure 6 is simplest of forms though?

- zepdrix

6sqrt2 is the simplified form of sqrt(72).
That was just an example though.

- zepdrix

Break down 147, what are some factors that we can use?
One little math trick is to add up the digits, 1 + 4 + 7 = 12.
12 is a multiple of 3,
so this number is divisible by 3.
So 147 = 3 * _

- some.random.cool.kid

/uh

- some.random.cool.kid

not sure

- zepdrix

Just use a calculator then :U
147 / 3

- some.random.cool.kid

49

- zepdrix

ok\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{147}\]\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{3\cdot49}\]\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{3}\cdot\sqrt{49}\]

- zepdrix

\[\large\rm x=-5+\color{orangered}{\mathcal i}\cdot\sqrt{3}\cdot\sqrt{49}\]What does sqrt(49) simplify to?

- some.random.cool.kid

ok

- some.random.cool.kid

7

- zepdrix

Good good good.
Let's write the i in the back, so it looks a lil bit cleaner.\[\large\rm x=-5+7\sqrt3~\mathcal i\]

- some.random.cool.kid

ok

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