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some.random.cool.kid
 one year ago
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
 one year ago
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

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some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0@nincompoop @whpalmer4 @pooja195 @paki @ganeshie8

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0mind helping?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \left[abc\right]^{n}=a^n b^n c^n\]When we have a bunch of stuff being raised to a power, we have to apply the exponent to each of them.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \left[27x^{6}y^{12}\right]^{2/3}=27^{2/3}(x^{6})^{2/3}(y^{12})^{2/3}\]Ok with that step? :)

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0mm got it am i supposed to add across? for the powers part?

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0or exponents i should say?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Have 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
 one year ago
Best ResponseYou've already chosen the best response.0mmm ok ...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So for the exponent on x, we'll multiply 6 and 2/3.

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.06 x 12? I belive. what each of those two thirds?

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0I belive that is 4? correct...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \left[27x^{6}y^{12}\right]^{2/3}=27^{2/3}x^{4}(y^{12})^{2/3}\]k looks good :)

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0mmm next step is ? 4 x 12 or 4 x the other 2/3rds?

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0or is it 4 x 12 and 2/3 x 2/3?

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0or is it just multiply across lol? i believe its what i said first though

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Notice 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
 one year ago
Best ResponseYou've already chosen the best response.1So the 4 on the x isn't going to interact in any way with the powers on y.

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0right so find the lcm?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You have \(\large\rm (y^{12})^{2/3}\). Our exponent rule tells us to `multiply` these powers, ya? 12 * 2/3?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \left[27x^{6}y^{12}\right]^{2/3}=\color{orangered}{27^{2/3}}x^{4}y^{8}\]k good :) almost done

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0is this going to be 8 over 4?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Try 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
 one year ago
Best ResponseYou've already chosen the best response.0its not nvm i answered my own thing i belive you leave those alon now right so now i have 9 x^4y^8?

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0:D mind helping me with a few more?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1stay cool kid (⌐▨_▨) oh sure

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0Ok great I have another one this one is a bit different then the last a little bit ...

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0A 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
 one year ago
Best ResponseYou've already chosen the best response.0hold on i need to attach the numbers part this is too messy

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm x=5+\sqrt{147}\]

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0there we go :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Well, recall that we define the imaginary unit i, to be \(\large\rm \sqrt{1}\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So we can do something clever with the square root in our problem.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \sqrt{147}=\quad\sqrt{1\cdot147}=\quad\sqrt{1}\cdot\sqrt{147}\]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
 one year ago
Best ResponseYou've already chosen the best response.0yep I have used this method before...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm x=5+\sqrt{147}\]\[\large\rm x=5+\color{orangered}{\sqrt{1}}\cdot\sqrt{147}\]

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0first before you continue

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0shouldnt I be trying to find the sqrt?

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0and trying to figure out whetther its irrational or not?

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0because I believe that is 12.124355653

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0which is irrational so now you may preceed

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You don't use your calculator for this step.

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0no i rounded lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1But you first used a calculator.. that's not what we're doing here.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You 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
 one year ago
Best ResponseYou've already chosen the best response.0oooo nvm its not 12.12

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.0its either 21 or 7sqrtof 3 I believe correct? not sure which one yet tough

some.random.cool.kid
 one year ago
Best ResponseYou've already chosen the best response.0im not sure 6 is simplest of forms though?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.16sqrt2 is the simplified form of sqrt(72). That was just an example though.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Break 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 * _

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Just use a calculator then :U 147 / 3

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1ok\[\large\rm x=5+\color{orangered}{\sqrt{1}}\cdot\sqrt{147}\]\[\large\rm x=5+\color{orangered}{\sqrt{1}}\cdot\sqrt{3\cdot49}\]\[\large\rm x=5+\color{orangered}{\sqrt{1}}\cdot\sqrt{3}\cdot\sqrt{49}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm x=5+\color{orangered}{\mathcal i}\cdot\sqrt{3}\cdot\sqrt{49}\]What does sqrt(49) simplify to?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Good good good. Let's write the i in the back, so it looks a lil bit cleaner.\[\large\rm x=5+7\sqrt3~\mathcal i\]
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