how would i go about solving
24k^6 over 6k^2
algebraic fractions with powers?
2009-06-14 9:33 am
回答 (6)
2009-06-14 11:27 am
✔ 最佳答案
24k^6/(6k^2)= (24/6)(k^6/k^2)
= 4[k^(6 - 2)]
= 4k^4
2009-06-14 4:57 pm
= 24k⁶/6k²
= 4k⁴
Answer: 4k⁴
= 4k⁴
Answer: 4k⁴
2009-06-17 6:57 am
simplify the k's to get k^4 on the top and 1 on the bottom
then simplify 24 over 6 to get 4 over 1
then it would look like this: 4k^4/1
which is the same as 4k^4
........simple algebra................
then simplify 24 over 6 to get 4 over 1
then it would look like this: 4k^4/1
which is the same as 4k^4
........simple algebra................
2009-06-14 4:51 pm
24k^6
--------
6k^2
=4k^4 answer//
--------
6k^2
=4k^4 answer//
2009-06-14 4:41 pm
You can solve this using logarithms.
2009-06-14 4:39 pm
It's not an equation, so you can't solve it. But you can simplify it using the rule of exponents that x^m / x^n = x^(m - n).
24k^6 / 6k^2 = (24/6) * (k^6 / k^2) = 4 * k^(6 - 2) = 4k^4
24k^6 / 6k^2 = (24/6) * (k^6 / k^2) = 4 * k^(6 - 2) = 4k^4
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