I know that x^-3 simplifies to 1/x^3...but when it gets to 6n^-1, I get confused. So I need someone to explain to me how that works and how I would figure out 3/2t^-2. And I know a lot of people on here hate when the kid just asks for the answers so if you dont want to give me the answer and would rather just explain it to me instead, that works :D Thanks!
2007-07-23 07:00:32 · 6 answers · asked by ellyceqt 1 in Science & Mathematics ➔ Mathematics
The exponent in 6n^-1 applies only to n. So
6n^-1 = 6 * (n^-1) = 6 * (1/n) = 6/n
Likewise, 2/(2t^-2) =
(2/2) * (1 / t^-2) = 1 * t^2 = t^2
It would be a different story if you had (6n)^-1. In that case, this is (6^-1)(n^-1) = 1 / 6n
2007-07-23 07:04:18 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Looking at your example 6n^-1 consider each part seperately. 6 * n^-1. There is not an exponent on the 6, so you should leave it alone. n^-1 is just like what you did with the x^-3, so n^-1= 1/n^1 or just 1/n. Now put the two pieces back together: 6*1/n or 6/n.
The second example will be the same. Just break it apart first. (3*1/2*t^-2 )
Hope this helps.
2007-07-23 14:12:42 · answer #2 · answered by Jenn 1 · 0⤊ 0⤋
It seems you already understand the basics. Remember that the exponent only applies to the variable. So 6n^-1 would just be 6/n^1. The 6 is left alone since it is not part of the variable. BUT if it was (6n)^-1 then it would be equivalent to 1/6n.
3/2t^-2. Basically, t^-2 is just 1/t^2. So just multiply that 3/2 and you get (3t^2)/2.
Hope that helps!
2007-07-23 14:07:25 · answer #3 · answered by Aaron 3 · 0⤊ 0⤋
I am not sure this helps or not but you just have to take the reciprocal of the number and put the power of whatever number is there. And be sure whether there is a bracket or not. For example: As you have given If it is 6(n^-1) = 6(1/n) and if it is (6n)^-1 = 1/(6n) and the power to n is 1 in both the cases.Just try practicing it.
2007-07-23 14:12:41 · answer #4 · answered by sumanth 2 · 0⤊ 0⤋
6n^-1 can be written as
6n^-1 = (6) * (n)^(-1)
In this form, you can see that the exponent (-1) is applied to (n).
2t^-2 = (2) * (t)^(-2)
exponent (-2) is applied to (t) only.
3/2t^-2 = 3 / ((2) * (t)^(-2))
= (3/2) * (1/t^-2)
2007-07-23 14:09:11 · answer #5 · answered by buoisang 4 · 0⤊ 0⤋
You are correct when you say x^(-3) = 1 / x³
Similarly:-
6 n^(-1) = 6 x (1/n) = 6 / n
(3/2) t^(-2) = (3/2) x 1/ t ² = 3 / (2t ²)
2007-07-23 14:23:53 · answer #6 · answered by Como 7 · 0⤊ 0⤋