Square root of 8y^5 X the square root of 40y^2

- anonymous

Square root of 8y^5 X the square root of 40y^2

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- anonymous

to multiply variables with exponents, you add the exponents together

- anonymous

oh there's a square root

- anonymous

\[8y^5(\sqrt{40y^2})?\]

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## More answers

- anonymous

you can pull a y out of the square root because \[\sqrt{y^2}=y\]

- anonymous

you can also pull a 2 out of the square root because of \[\sqrt{2^2*10}\]

- anonymous

\[\sqrt{8y^{5}} * \sqrt{40y ^{2}}\]no its

- anonymous

ok

- anonymous

you can put it all under one square root

- anonymous

such as this \[\sqrt{(8y^5)(40y^2)}\]

- anonymous

then i got \[\sqrt{320y ^{7}}\]

- anonymous

yep

- anonymous

now you have to pull numbers and variables out

- anonymous

as a hint, 64*5=320

- anonymous

then change 320 to \[8\sqrt{5} \] then \[8\sqrt{5y ^{7}}\] but i dont know what to do next

- anonymous

yes \[8\sqrt{y^7}\]

- anonymous

im sorry \[8\sqrt{5y^7}\]

- anonymous

now you can factor y

- anonymous

if there was a y^2 in there, you would just take out the y, correct?

- anonymous

into 4 and 3

- anonymous

yeah that too

- anonymous

so what if you factored \[y^7\] as \[y^6y\]

- anonymous

i thought i have to factor it into a perfect square?

- anonymous

\[y^6=(y^3)^2\], if you remember the rules of exponents

- anonymous

so can you pull y^6 out now?

- anonymous

\[8\sqrt{5y ^{6}}y\]

- anonymous

yes, but now \[y^6=(y^3)^2\], which is a perfect square

- anonymous

\[8\sqrt{5y^6y}\]

- anonymous

\[=8\sqrt{5(y^3)^2y}\]

- anonymous

if i said y^3=x, then you would see an x^2 inside of the square root

- anonymous

and you'd pull the x out of \[\sqrt{x^2}\]

- anonymous

see what i'm talking about?

- anonymous

but then i have 8y^3 square root of 5y^3y

- anonymous

nope, you'd pull the y^3 out of the root, leaving you with only the other y

- anonymous

\[8\sqrt{5(y^3)^2y} = 8y^3\sqrt{5y} = \]

- anonymous

forget that last = sign, that is it. it's done now

- anonymous

there's no way to make perfect squares from factors of 5 or y

- anonymous

understand that when you raise a power to a power, you multiply the exponents

- anonymous

i kind of get it i think..i just have to have more practice because it's still kind of confusing but ill get it

- anonymous

yeah that's all it takes is practice

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