Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

i need the steps in simplifying: square root of 27x^8y^5

OpenStudy (anonymous):

You can break it up like this: sqrt(27)*sqrt(x^8)*sqrt(y^5)

OpenStudy (anonymous):

If you have something raised to an exponent, such as x^8, you can subtract 2 from the exponent and put one outside the square root. Sqrt(x^8) = x*sqrt(x^6)

OpenStudy (anonymous):

wait

OpenStudy (anonymous):

the square root of 27 is 5.19

OpenStudy (anonymous):

If you do that to your variables, and then break your 27 into its composite prime form, you'll be able to see everything you can take out.

OpenStudy (anonymous):

In simplifying, you don't usually calculate the square root, you just make it as small as possible. So if you have sqrt(8), you simplify it to 2 * sqrt(2)

OpenStudy (anonymous):

Because 8 is 2*2*2

OpenStudy (anonymous):

Do you understand how simplifying works?

OpenStudy (anonymous):

also 3 *3 ^2 right?

OpenStudy (anonymous):

i think i got it so the answer could be 3x^4y^2 square root 3y ??

OpenStudy (anonymous):

That's right

OpenStudy (anonymous):

can i give you one more?

OpenStudy (anonymous):

Yea

OpenStudy (anonymous):

\[63m ^{-2}v ^{3} OVER 15^{6}v ^{-5} \]63m^-2v

OpenStudy (anonymous):

forget about the bottom 63m

OpenStudy (anonymous):

I'm not quite sure what you mean. Do you mean this? (63m^(-2)*v^3)/(15^6*v^-5)?

OpenStudy (anonymous):

i am not sure either i just know i need to have my final answer be a binomial square.. hw haha

OpenStudy (anonymous):

But if you have something with a negative exponent like 1/n^(-3), it's equal to n^3

OpenStudy (anonymous):

oops wrong part soory!

OpenStudy (anonymous):

i need to simplify the following and only use positive exponents

OpenStudy (anonymous):

That's the first step. Negative exponents are a pain to deal with. Then if you have something like this: n^4/n^2 you subtract the smaller from the larger. So n^2. Remember that if you wrote it out it'd be (n*n*n*n)/(n*n). If you crossed them out one by 1, you end up with n*n = n^2

OpenStudy (anonymous):

can we do the problem i put down? its for a test today... i have no clue haha

OpenStudy (anonymous):

whoops, that wasn't for you

OpenStudy (radar):

Rewrite as:\[\sqrt{27x ^{8}y ^{5}}=\sqrt{9*3}\sqrt{x ^{8}}\sqrt{y ^{4}y}\] Now extract the square roots: \[3x ^{4}y ^{2}\sqrt{3y}\] I think that would be your answer. Use calculator to obtain square root of 3

OpenStudy (anonymous):

Omg, my computer is spazzing. That was for you, it was showing something from another window that confused me. And I can't read the problem you put down or I would.

OpenStudy (anonymous):

i got that thanks radar!

OpenStudy (anonymous):

can you help me with a diff prob? and no worries dwobwinkle ! i really appreciate it!

OpenStudy (radar):

Use to helpful, but have to review. You understand

OpenStudy (anonymous):

huh

OpenStudy (anonymous):

So I don't quite get it, what are you stuck on? I'm not able to interpret the question.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!