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AngelaB97
 one year ago
how do you solve this radical problem
AngelaB97
 one year ago
how do you solve this radical problem

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AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441681695839:dw

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0in thorough steps cuz i don't seem to understand this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So you will need to simplify each radical and then multiply them together. Let's start with the first one. We know that 2^3 is 8 and 3^3 is 27, so it only follows that the cube root of 15 must be in between 2 and 3 as 15 is between 8 and 27 and the opposite of an exponent is a root. But where is it between 2 and 3? Well, to figure that out we must approximate. But that is a very long process. So instead, let's go through the firstdecimaldigit possibilities. Let's start with ones closer to 2 as 8 is closer to 15. \[2.2^{3}=10.648\] Not quite. \[2.4^{3}=13.824\] Closer... let's try one more. \[2.5^{3}=15.625\] That's a bit too far. So we know it must be a bit smaller than 2.5 but quite a bit bigger than 2.4 You could keep repeating this process until you get it perfect, but I quite honestly do not feel like it. So I plugged it into my calculator and get approximately 2.466 Now for the next one. Cube root of 75. Well, 8^3=64 and 9^3=81, so it must be between 8 and 9. You could repeat the process above, but the calculator informs me it is about 4.217 Now for the easy part! \[2.466\times4.217=10.399\]

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0but its still supposed to be in radical form not decimal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah... that's different.

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0can you explain in that way?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find the prime factorization of both numbers dw:1441682872453:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You're taking a cube root, so look for group of 3 numbersdw:1441683029075:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Going off of what peachpi said, then you can multiply those under the cube root\[\sqrt[3]{5\times3\times5\times5\times3}\] and then see that there are 3 5's. Because it is a cube root, only one 5 can stay, and we pull it out of the cube root, so we are left with \[5\sqrt[3]{3\times3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Write the grouped number on the outside of the radical, multiply the numbers left in the radicaldw:1441683084511:dw

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0doesn't the rad 9 come out of the square root

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No. Because it is a cube root and not a square root. 2^3 is only 8, while 3^3 is 27. This is the simplest you will be able to get in radical form =)

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0okay can i ask for your help for one more problem

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441683413491:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You'll need to do something similar here, but I have not learned how to find roots of fractions yet. I could do it in decimal form, but I know that's not what you need. I'm very sorry!!

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0it's fine thanks for your help!
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