4/5 cube root of 125y^5..
2006-12-12 10:46:11 · 4 answers · asked by itsallforyou 2 in Science & Mathematics ➔ Mathematics
4/5 cube root of 125x^5
=4/5*5x^(5/3)
=4x^(5/3)
2006-12-12 10:52:23 · answer #1 · answered by raj 7 · 0⤊ 0⤋
A root is just a fractional power. So the square root of something is that thing raised to the 1/2 power.
In your problem, the cube root of 125y^5 is all that to the 1/3 power. Since 125 is a cube (5^3) you get 5y^(5/3) for that part. When you multiply by 4/5 you get 4y^(5/3)
2006-12-12 18:53:12 · answer #2 · answered by xaviar_onasis 5 · 0⤊ 0⤋
Since cube root is the same as bringing something to the 1/3 power, I'll rewrite your question.
(4/5) (125y^5)^(1/3)
Note that (ab)^n = (a^n)*(b^n). That's exactly what we do here; split up the cube root.
(4/5) (125)^(1/3) * (y^5)^(1/3)
The cube root of 125 is 5.
(4/5) (5) * (y^5)^(1/3)
The 4/5 and 5 can be muliplied to make 4.
(4) * (y^5)^(1/3)
Split up y^5 into y^3 * y^2
(4) * (y^3 * y^2)^(1/3)
(4) * (y^3)^(1/3) * (y^2)^(1/3)
Note that the cube root of y^3 is y.
4y * (y^2)(1/3)
And that whenever you take a power to a power, you multiply the exponents.
4y * y^(2/3)
And that's the final simplification.
2006-12-12 18:53:12 · answer #3 · answered by Puggy 7 · 0⤊ 0⤋
4cbrt(125y^5)/5 = 4cbrt(5^3y^3y^2)/5 = 20ycbrt(y^2)/5 = 4cbrt(y^2)
2006-12-12 18:49:19 · answer #4 · answered by kindricko 7 · 0⤊ 0⤋