5x to the -2/5 power times 7x to the 3/4 power
2006-12-20 12:35:07 · 7 answers · asked by Kiwi 1 in Science & Mathematics ➔ Mathematics
[5x^(-2/5)][7x^(3/4)]
We can combine all constants by muliplying them out, so let's start by doing that.
35 [x^(-2/5)] [x^(3/4)]
And, whenever we have the same base but different exponents and are multiplying them together, we can add the exponents.
35 x^(-2/5 + 3/4)
Which becomes
35 x^(-8/20 + 15/20), and simplifies to
35 x^(7/20)
2006-12-20 12:42:10 · answer #1 · answered by Puggy 7 · 3⤊ 0⤋
(5x^ -2/5)(7x^3/4)
First: multiply the whole numbers:
5(7) = 35
Second: both coefficients have the same base, which is "x" > next, you add exponents:
x^ -2/5 + 3/4 = 7/20
Third: rewrite the expression by combining your results:
= 35x^7/20
2006-12-20 13:35:52 · answer #2 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
35 x ^ (7/20)
2006-12-20 13:26:38 · answer #3 · answered by syikin_310 2 · 0⤊ 0⤋
5x^(-2/5) x 7x^(3/4) = 35x^(3/4 - 2/5) = 35x^(15/20 - 8/20)
= 35x^(7/20)
2006-12-20 12:39:08 · answer #4 · answered by Mr.Math 1 · 2⤊ 2⤋
35 x ^ (7/20)
2006-12-20 12:38:19 · answer #5 · answered by begasaka1 1 · 1⤊ 1⤋
your problem statement is a little vague. Is it
A) (5x)^-0.4*(7x)^.75 or
B) 5x^-0.4*7x^.75
A) (5x)^-0.4*(7x)^.75
5^-.4*7^.75*x^(.75-.4)=2.261x^.35
B) 5x^-0.4*7x^.75
35x^.35
2006-12-20 13:08:31 · answer #6 · answered by yupchagee 7 · 1⤊ 0⤋
384.32
2006-12-20 12:37:42 · answer #7 · answered by Dustin K 2 · 0⤊ 2⤋