some.random.cool.kid
  • some.random.cool.kid
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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some.random.cool.kid
  • some.random.cool.kid
@zepdrix
some.random.cool.kid
  • some.random.cool.kid
@triciaal
some.random.cool.kid
  • some.random.cool.kid

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some.random.cool.kid
  • some.random.cool.kid
@nincompoop @whpalmer4 @pooja195 @paki @ganeshie8
some.random.cool.kid
  • some.random.cool.kid
mind helping?
zepdrix
  • 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
  • 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
  • some.random.cool.kid
mm got it am i supposed to add across? for the powers part?
some.random.cool.kid
  • some.random.cool.kid
or exponents i should say?
zepdrix
  • 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
  • some.random.cool.kid
mmm ok ...
zepdrix
  • zepdrix
So for the exponent on x, we'll multiply -6 and 2/3.
some.random.cool.kid
  • some.random.cool.kid
-6 x 12? I belive. what each of those two thirds?
some.random.cool.kid
  • some.random.cool.kid
oh
some.random.cool.kid
  • some.random.cool.kid
I belive that is -4? correct...
zepdrix
  • 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
  • some.random.cool.kid
mmm next step is ? -4 x 12 or -4 x the other 2/3rds?
some.random.cool.kid
  • some.random.cool.kid
or is it -4 x 12 and 2/3 x 2/3?
some.random.cool.kid
  • some.random.cool.kid
or is it just multiply across lol? i believe its what i said first though
zepdrix
  • 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
  • zepdrix
So the -4 on the x isn't going to interact in any way with the powers on y.
some.random.cool.kid
  • some.random.cool.kid
right so find the lcm?
zepdrix
  • 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
  • some.random.cool.kid
8
zepdrix
  • 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
  • some.random.cool.kid
is this going to be 8 over 4?
some.random.cool.kid
  • some.random.cool.kid
no
zepdrix
  • 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
  • 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
  • zepdrix
yay good job \c:/
some.random.cool.kid
  • some.random.cool.kid
:D mind helping me with a few more?
zepdrix
  • zepdrix
stay cool kid (⌐▨_▨) oh sure
some.random.cool.kid
  • 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
  • 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
  • some.random.cool.kid
hold on i need to attach the numbers part this is too messy
zepdrix
  • zepdrix
\[\large\rm x=-5+\sqrt{-147}\]
some.random.cool.kid
  • some.random.cool.kid
some.random.cool.kid
  • some.random.cool.kid
there we go :D
zepdrix
  • zepdrix
Well, recall that we define the imaginary unit i, to be \(\large\rm \sqrt{-1}\)
zepdrix
  • zepdrix
So we can do something clever with the square root in our problem.
zepdrix
  • 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
  • some.random.cool.kid
yep I have used this method before...
zepdrix
  • zepdrix
\[\large\rm x=-5+\sqrt{-147}\]\[\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{147}\]
some.random.cool.kid
  • some.random.cool.kid
first before you continue
some.random.cool.kid
  • some.random.cool.kid
shouldnt I be trying to find the sqrt?
some.random.cool.kid
  • some.random.cool.kid
and trying to figure out whetther its irrational or not?
some.random.cool.kid
  • some.random.cool.kid
because I believe that is 12.124355653
zepdrix
  • 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
  • some.random.cool.kid
which is irrational so now you may preceed
some.random.cool.kid
  • some.random.cool.kid
eh 12.12
zepdrix
  • zepdrix
no...
some.random.cool.kid
  • some.random.cool.kid
...
zepdrix
  • zepdrix
You don't use your calculator for this step.
some.random.cool.kid
  • some.random.cool.kid
no i rounded lol
zepdrix
  • zepdrix
But you first used a calculator.. that's not what we're doing here.
zepdrix
  • 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
  • some.random.cool.kid
oooo nvm its not 12.12
zepdrix
  • 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
  • some.random.cool.kid
its either 21 or 7sqrtof 3 I believe correct? not sure which one yet tough
some.random.cool.kid
  • some.random.cool.kid
though*
some.random.cool.kid
  • some.random.cool.kid
im not sure 6 is simplest of forms though?
zepdrix
  • zepdrix
6sqrt2 is the simplified form of sqrt(72). That was just an example though.
zepdrix
  • 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
  • some.random.cool.kid
/uh
some.random.cool.kid
  • some.random.cool.kid
not sure
zepdrix
  • zepdrix
Just use a calculator then :U 147 / 3
some.random.cool.kid
  • some.random.cool.kid
49
zepdrix
  • 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
  • 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
  • some.random.cool.kid
ok
some.random.cool.kid
  • some.random.cool.kid
7
zepdrix
  • 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
  • some.random.cool.kid
ok

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