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

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.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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 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
\[8y^5(\sqrt{40y^2})?\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
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
yeah that's all it takes is practice

Not the answer you are looking for?

Search for more explanations.

Ask your own question