Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

For problems 1–5, simplify the expression. Answers written in decimal form will not be accepted. Each of these problems is worth 1 point. "v/" is a radical by the way ___ 1.v/96 ______ 2. 8 v/63x^5 ___________ 3. v/128x^5y^2 ___ 4. ^3v/32 ________ 5. ^3v/56x^14

See more answers at
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


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

So, I need them in radical form.
Factor everything first and then apply your radical rules... for example \[\sqrt[3]{x^3} = x\]
Problem is that I have no idea what you're saying. ._.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Ok for example \[\sqrt[8]{256x^4}\] becomes, if you factor \[\sqrt[8]{2^{8}x^4}\] which then becomes \[2\sqrt[8]{x^4}=2 x^{\frac{4}{8}}=2x^{1/2}=2\sqrt{2}\]
This uses the idea that \[\sqrt[a]{x} = x^{1/a}\]
just wondering, would you mind using bigger font? I can't see the exponents. To do that, put what you are saying in the curly braces in \(\huge\text{}\.) but take out the . at the end.
the ^ means exponents. like, 2^3 is 2 cubed.
\(\Large\text{But again, I have no idea how to do this.}\)
is ^3v/56x^14 supposed to be \[\bigg(\sqrt{56x^{14}}\bigg)^3\] then?
So, \(\huge\sqrt[8]{256x^4}\) turns into \(\huge\sqrt[8]{2^{8}x^4}\) which turns into \(\huge2\sqrt[8]{x^4}=2 x^{\frac{4}{8}}=2x^{1/2}=2\sqrt{2}\) But how? \(\huge\text{:|}\)
Sorry that should be \[2\sqrt{x}\]
and no, it's supposed to be \[^{3}\sqrt{56x^{14}}\]
ok the basic theorems that you need for these type of problems are that \[\sqrt[a]{x} = x^{\frac{1}{a}}\] So for example \[\large \sqrt[3]{x^{10}}\to x^{\frac{10}{3}}\to x^3x^{\frac{1}{3}}\to x^3\sqrt[3]{x}\]
I understand the first half of your example equation, but not the second half.
So any complicated radical you are given you can convert to exponents, use the rules of exponents shamelessly and then convert back to radical. So \[\large \sqrt[3]{56x^{14}}\to \sqrt[3]{(7)(8)x^{14}}\to 7^{1/3}8^{1/3}x^{14/3}\] This then becomes \[ 7^{1/3}8^{1/3}x^{14/3}\to 7^{1/3}(2^3)^{1/3}x^{12/3}x^{2/3}\] becomes \[7^{1/3}2x^{4}x^{2/3}\to 2x^4 7^{1/3}x^{1/3}\to 2x^4\sqrt[3]{7x}\]
my brain is about to explode. e_o
So you don't understand how \[x^{10/3} \to x^3x^{1/3}\]? This results from the fact that \[X^aX^b = X^{ab}\] and the reverse. So if you have \[x^{10/3}\] you have \[x^{9/3 + 1/3}\to x^{9/3}x^{1/3}\to x^3x^{1/3}\]
again, might I suggest the larger font size? Maybe Large would be a good replacement for huge though.

Not the answer you are looking for?

Search for more explanations.

Ask your own question