our class is working on monomials and exponent stuff. please help me with this problem:
(8y^3)(-3x^2y^2)(3/8xy^4)
this needs to be simplified.
-no fractions
-each base appears only once
-parenthesis are gone
NOTE: ^(#) MEANS IT IS RAISED TO THAT POWER
2007-03-10 02:57:00 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Just so you know guys, you have all been wrong so far!
2007-03-10 07:48:29 · update #1
(8y^3)(-3x^2y^2)(3/8xy^4)
Multiply first two paren and the 3 out of the third paren to get the numerator.
(-72 x^2 y^5) / (8 x y^4)
Divide the -72 by 8.
(-9 x^2 y^5) / (x y^4)
Divide numerator and denominator by x y^4
-9 xy
2007-03-10 03:04:08 · answer #1 · answered by ecolink 7 · 0⤊ 0⤋
Ok i'm pretty sure this is the answer and I suggest you learn this and not just copy
1)8y^3 times -3x^2y^5 = -24x^2Y^5
it's negative because a positive times a negative number is negative
2)-24x^2Y^5 times 3/8xy^4 = -9x^3y^9
you multiply -24 times 3 then divded by 9
2007-03-10 12:15:10 · answer #2 · answered by sexy sexy 2 · 0⤊ 0⤋
Ok i'm pretty sure this is the answer and I suggest you learn this and not just copy
1)8y^3 times -3x^2y^5 = -24x^2Y^5
it's negative because a positive times a negative number is negative
2)-24x^2Y^5 times 3/8xy^4 = -9x^3y^9
you multiply -24 times 3 then divded by 9
2007-03-10 11:16:15 · answer #3 · answered by briana11us2004 2 · 0⤊ 1⤋
(8y^3)(-3x^2y^2)(3/8xy^4) =
(-24x^2y^5)(3/8xy^4) =
-72x^2y^5 / 8xy^4 =
-9xy
2007-03-10 11:51:26 · answer #4 · answered by Steve A 7 · 0⤊ 0⤋