the problem is a fraction
1/3x^-3
(one over three x with an expont of -3)
2006-10-18 23:33:40 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
just the x is being raised ....
so 1 in numerator .... 3 in denominator ...
when the exponent is negative, the "inverse" of that exponent is positive....(a neg 3 exp in den becomes pos 3 in numerator)
your answer is (x^3) / 3
2006-10-19 00:34:01 · answer #1 · answered by Brian D 5 · 0⤊ 0⤋
a million. any extensive style with an exponent of 0 is a million so #a million. = a million 2. (2x^5^)-2 simplifying, the expression could be a million ----------- (2x^5)^2 3. (x / 10)^-a million it may be: 10 ---- = answer x 4. 3^n * 3^2n = it may become: 9^3n , purely multiply the constants and then upload the exponents
2016-10-02 11:06:53 · answer #2 · answered by bungay 4 · 0⤊ 0⤋
Multiply the numerator and the denominator by x^3. This won't change the value of the fraction, but it will give you a different expression for it.
(x^3/x^3) * 1/(3*x^-3) = x^3/(3*x^-3*x^3) = x^3/(3 * x^0) = x^3/3
2006-10-19 00:19:10 · answer #3 · answered by Ted 4 · 0⤊ 0⤋
1/(1/3 x^3)
2006-10-18 23:41:50 · answer #4 · answered by QuriousGirl 2 · 0⤊ 0⤋
1/ 1/3x^3=3/x^3
2006-10-18 23:43:48 · answer #5 · answered by ??? 2 · 0⤊ 0⤋
Let's look at 1/x^-3
It can be rewritten as (x^-3)^-1 = x^3
Hence 1/3x^-3 = x^3/3
I hope this helps.
2006-10-18 23:41:42 · answer #6 · answered by scyxav 2 · 0⤊ 0⤋
1/3x^-3
=1/1/3x^3
=1*3x^3
=3x^3
2006-10-19 00:10:09 · answer #7 · answered by mzamo n 1 · 0⤊ 0⤋
1/(3x^-3) = (x^3)/3
2006-10-19 23:50:48 · answer #8 · answered by Anonymous · 0⤊ 0⤋
3/8.ab235.51.0
2006-10-19 00:28:06 · answer #9 · answered by chasen54 5 · 0⤊ 0⤋