Here's the problem:
(-64/343)^2/3
That is the fraction 64 OVER 343 RAISED to the power of the fraction 2/3.
2007-01-30 05:16:19 · 8 answers · asked by jaguarj23 1 in Science & Mathematics ➔ Mathematics
"the fraction 64 OVER 343 RAISED to the power of the fraction 2/3":
(64/343) ^ (2/3) = .186 ^ .667 = .327
2007-01-30 05:26:58 · answer #1 · answered by rich h 3 · 0⤊ 0⤋
Think of this as
[(-64/343)^1/3]^2
(-4/7)^2
16/49
Do the cube root first then square it.
(-4)^3 = -64 (7)^3 = 343
(-4)^2 = 16 (7)^2 = 49
2007-01-30 05:26:00 · answer #2 · answered by Tony O 2 · 0⤊ 0⤋
The fractional exponent tells you to take the 3rd route (that's the denominator part of the exponent) and then square (the numerator part of the exponent) the fraction:
(64/343)^(1/3) = 4/7
(4/7)^2 = 16/49
2007-01-30 05:30:14 · answer #3 · answered by Little_Platypus 1 · 0⤊ 0⤋
How's that an Algebra question? Where's the variable?
I did it on my calculator, the answer was 16/49.
(-64/343)^2/3 = sqcube((-64/343)^2) = (4/7)^2 = 16/49
No idea what happened to the negative sign.
2007-01-30 05:26:11 · answer #4 · answered by Anonymous · 0⤊ 2⤋
You have -64 in 1st line but have 64 in second line?
Will go with second line of (64/343)^(2/3):-
Take cube root first to give (4/7)² = 16/49
2007-01-30 05:27:46 · answer #5 · answered by Como 7 · 0⤊ 0⤋
(-64/343)^2/3 = (-4/7)^2 = 16/49
2007-01-30 05:28:00 · answer #6 · answered by tenniscolleggrl24 2 · 0⤊ 0⤋
4/2x + 2/3x = a million/6 12/6x + 4/6x = a million/6 ^the following, you calculate their lowest hardship-loose denominator. Calculate the least hardship-loose different of their denominators to discover it. 16/6x = a million/6 x = a million/6 . 16/6 x = a million/3 . 4/3 (simplify) x = 4/9 3(x+a million) + 2 = 6(x+a million) 3x + 3 + 2 = 6x + 6 (the following, you multiply what's contained in the brackets via the large style outside. continuously attempt to sparkling up what's in brackets first, before multiplying) 3 + 2 + 6 = 6x - 3x 11 = 3x 11/3 = x
2016-12-03 06:03:44 · answer #7 · answered by ? 4 · 0⤊ 0⤋
Take the cube root of each number, and then square it. The result should be positive.
16/49
2007-01-30 05:23:53 · answer #8 · answered by bequalming 5 · 1⤊ 0⤋