At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

to multiply variables with exponents, you add the exponents together

oh there's a square root

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

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

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

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

ok

you can put it all under one square root

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

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

yep

now you have to pull numbers and variables out

as a hint, 64*5=320

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

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

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

now you can factor y

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

into 4 and 3

yeah that too

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

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

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

so can you pull y^6 out now?

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

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

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

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

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

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

see what i'm talking about?

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

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

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

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

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

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

yeah that's all it takes is practice