(3y^3)(2y^2)^2 / (y^4)^3 X (y^3)^o
2007-01-09 10:39:04 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I assume you are trying to simplify
[(3y^3)(2y^2)^2 / (y^4)^3] . (y^3)^0
That is, I assume the o should be an 0 and the X means multiplication and (y^3)^0 is a separate term rather than being on the denominator.
Now, remember the power rules:
(a^b)^c = a^(bc)
(ab)^c = a^c b^c
a^b a^c = a^(b+c)
a^b / a^c = a^(b-c)
a^0 = 1.
Let's start with (y^3)^0. By the last rule this is 1, so we can remove it. So we have
(3y^3)(2y^2)^2 / (y^4)^3
Now let's do the bracketed terms that are raised to powers:
(2y^2)^2 and (y^4)^3. We use the rules (ab)^c = a^c b^c and (a^b)^c = a^(bc) to handle these:
= (3y^3) (2^2 (y^2)^2) / (y^(4.3)
= (3y^3) (2^2) (y^(2.2)) / y^12
= 3 y^3 . 4 . y^4 / y^12
Now use a^b . a^c = a^(b+c) and a^b / a^c = a^(b-c) to get
= 12 y^(3+4) / y^12
= 12 y^7 / y^12
= 12 (y^7 / y^7) / (y^12 / y^7)
= 12 (1) / (y^(12 -7))
= 12 / y^5.
2007-01-09 10:52:33 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
=(3y^3)(4y^4) / (y^12)(1) << anything to the power 0 is 1.
=12y^7 / y^12
= 12 / y^5
2007-01-09 18:43:24 · answer #2 · answered by teekshi33 4 · 0⤊ 0⤋
12 / y^10
2007-01-09 18:44:32 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(3y^3)(4y^4) / (y^12)(1)
12y^7 / y^12
12 / y^5 OR 12y^-5
2007-01-09 18:54:41 · answer #4 · answered by Anonymous · 0⤊ 0⤋