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anonymous
 5 years ago
simplify: 5 over square root of k^7
anonymous
 5 years ago
simplify: 5 over square root of k^7

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do i simplify this?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0k to the power of (7/5) is the best you can do with it. k^(7/5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so for any problem that looks like this i take the outter number and make it the denominator of the inner #?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ie 7 square root k^4 woudl be k^ 4/7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so here is one\[63m^{2}v ^{3} \over 15m ^{6}v ^{5}\] more if you don't mind....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can only use positive expo

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0move everything to the top except for the (15) and make their exponents the opposite sign: (63/15) (m^2)(m^6) (v^3)(v^5) When you multiply like bases the exponents add together: (63/15) (m^(26)) (v^(3+5)) (63/15) m^8 v^8 ; now everything with a negative exponent gets dropped to the bottom and reverse the sign. (63 v^8) / (15 m^8) reduce your fraction to: (12 v^8)/ (5 m^8) as long as I havent any typos, that should be it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2x+3 \over x \] MINUS \[x2 \over x+1\] YOU ARE A lifesaver. thanks so much if you have time here is one more that i am stuck on,

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0glorified fractions is all these are. which means we have to get like denominators. The like denominator with simply be (x)(x+1) (2x+3)(x+1) = 2x^2 +5x +3 (over (x)(x+1) of course) and (x2)(x) = x^2 2x (over (x)(x+1)) (2x^2 +5x +3)  (x^2 2x) = 2x^2 +5x +3 x^2 +2x lets clean that up to read: (x^2 +7x +3) over (x^2 +x) That should work

radar
 5 years ago
Best ResponseYou've already chosen the best response.0back to your original problem there is a conflict between what you state and the way you expressed by notation. You say 5 over the square root of k^7l, but you indicate the fifth root (not square root) of k^7. I didn't follow the rest as I want to know what you are trying to simplify.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you had the (2x+3)(x+1) AND THEN you had that equal something, where did you get that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm terribly confused.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did the x and x2 go?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when dealing with fraction; the only way to add or subtract them together is to get a common denominator. the common denominator of (x) and (x+1) is simply (x)(x+1) so we have to modify these fractions by muliplying them by either (x+1) over (x+1) ... OR (x)/(x) to get them to have common denominators.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(2x + 3) (x+1) (2x^2 +5x +3)   =  (x) (x+1) (x)(x+1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(x2) (x) (x^2 2x)   =  (x+1) (x) (x)(x+1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(2x^2 +5x +3)  (x^2 2x)  (x)(x+1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(2x^2 +5x +3)  (x^2 2x) is the same as: 2x^2 +5x +3 x^2 +2x so combine like terms now (21)x^2 +(5+2)x +3 = x^2 +7x +3 x^2 +7x +3  (x)(x+1) x^2 +7x +3  x^2 + x
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