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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
 one year ago
 one year ago
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
 one year ago
 one year ago

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poopsiedoodleBest ResponseYou've already chosen the best response.1
So, I need them in radical form.
 one year ago

freewilly922Best ResponseYou've already chosen the best response.1
Factor everything first and then apply your radical rules... for example \[\sqrt[3]{x^3} = x\]
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
Problem is that I have no idea what you're saying. ._.
 one year ago

freewilly922Best ResponseYou've already chosen the best response.1
Ok for example \[\sqrt[8]{256x^4}\] becomes, if you factor \[\sqrt[8]{2^{8}x^4}\] which then becomes \[2\sqrt[8]{x^4}=2 x^{\frac{4}{8}}=2x^{1/2}=2\sqrt{2}\]
 one year ago

freewilly922Best ResponseYou've already chosen the best response.1
This uses the idea that \[\sqrt[a]{x} = x^{1/a}\]
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
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.
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
the ^ means exponents. like, 2^3 is 2 cubed.
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
\(\Large\text{But again, I have no idea how to do this.}\)
 one year ago

freewilly922Best ResponseYou've already chosen the best response.1
is ^3v/56x^14 supposed to be \[\bigg(\sqrt{56x^{14}}\bigg)^3\] then?
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
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? \(\huge\text{:}\)
 one year ago

freewilly922Best ResponseYou've already chosen the best response.1
Sorry that should be \[2\sqrt{x}\]
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
and no, it's supposed to be \[^{3}\sqrt{56x^{14}}\]
 one year ago

freewilly922Best ResponseYou've already chosen the best response.1
ok the basic theorems that you need for these type of problems are that \[\sqrt[a]{x} = x^{\frac{1}{a}}\] So for example \[\large \sqrt[3]{x^{10}}\to x^{\frac{10}{3}}\to x^3x^{\frac{1}{3}}\to x^3\sqrt[3]{x}\]
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
I understand the first half of your example equation, but not the second half.
 one year ago

freewilly922Best ResponseYou've already chosen the best response.1
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 \[7^{1/3}2x^{4}x^{2/3}\to 2x^4 7^{1/3}x^{1/3}\to 2x^4\sqrt[3]{7x}\]
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
my brain is about to explode. e_o
 one year ago

freewilly922Best ResponseYou've already chosen the best response.1
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 \[x^{9/3 + 1/3}\to x^{9/3}x^{1/3}\to x^3x^{1/3}\]
 one year ago

poopsiedoodleBest ResponseYou've already chosen the best response.1
again, might I suggest the larger font size? Maybe Large would be a good replacement for huge though.
 one year ago
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