Question

simplify: 5 over square root of k^7

Answer 1

\[\sqrt[5]{k ^{7}}\]

Answer 2

how do i simplify this?

Answer 3

in expo notation/.

Answer 4

k to the power of (7/5) is the best you can do with it.

k^(7/5)

Answer 5

so for any problem that looks like this i take the outter number and make it the denominator of the inner #?

Answer 6

ie 7 square root k^4 woudl be k^ 4/7

Answer 7

exactly

Answer 8

so here is one\[63m^{-2}v ^{3} \over 15m ^{6}v ^{-5}\] more if you don't mind....

Answer 9

i can only use positive expo

Answer 10

move 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^(-2-6)) (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 :)

Answer 11

\[2x+3 \over x \] MINUS \[x-2 \over x+1\]

YOU ARE A lifesaver. thanks so much if you have time here is one more that i am stuck on,

Answer 12

glorified 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 (x-2)(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

Answer 13

i'm lost...

Answer 14

back 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.

Answer 15

so you had the (2x+3)(x+1) AND THEN you had that equal something, where did you get that?

Answer 16

i'm terribly confused.

Answer 17

where did the x and x-2 go?

Answer 18

when 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.

Answer 19

(2x + 3) (x+1) (2x^2 +5x +3)
------- ----- = ------------
(x) (x+1) (x)(x+1)

Answer 20

(x-2) (x) (x^2 -2x)
----- -- = -------
(x+1) (x) (x)(x+1)

Answer 21

(2x^2 +5x +3) - (x^2 -2x)
------------------------
(x)(x+1)

Answer 22

(2x^2 +5x +3) - (x^2 -2x) is the same as:

2x^2 +5x +3 -x^2 +2x so combine like terms now

(2-1)x^2 +(5+2)x +3 = x^2 +7x +3

x^2 +7x +3
----------
(x)(x+1)

x^2 +7x +3
----------
x^2 + x