= what?
You have to provide us with an answer before we can solve for the unknown. In fact, you have to provide us with another equation as well, as there are two unknowns.
Sadly, these will both remain unknown.
Pssst! Samantha! You're not supposed to give them a plausible answer! Where's the fun in that?
2007-08-14 15:28:01
·
answer #1
·
answered by BotanyDave 5
·
1⤊
0⤋
X= 1/27y is the solution
X^6= 27y^3
X= 27y^3/6
X=27y^1/5
X=1/27y
2007-08-14 22:27:00
·
answer #2
·
answered by cowpony 3
·
1⤊
0⤋
x-to-the-sixth minus 27-y-cubed
I don't know, what are x and y supposed to be?
what do you want me to do with it?
If x=2 and y=4, then x^6-27y^3= -1664
2007-08-14 22:28:48
·
answer #3
·
answered by Anonymous
·
1⤊
0⤋
As you have X and Y, you need two equations to solve your problem.
Nevertheless, you can find the value of X according to the value of Y or vice versa.
2007-08-16 05:46:57
·
answer #4
·
answered by fredericus 3
·
0⤊
0⤋
It's a difference of cubes problem.
(x²)³ - (3y)³
Factors to
(x²-3y)(x^4 + 3y + 9y²)
2007-08-14 22:29:00
·
answer #5
·
answered by hogan.enterprises 5
·
1⤊
0⤋
looks like math
2007-08-14 22:22:08
·
answer #6
·
answered by LoveUSA 2
·
3⤊
0⤋
a math equasion.
2007-08-14 22:22:21
·
answer #7
·
answered by melissa 3
·
3⤊
0⤋
=(x^3-9y)(x^3+9y)
=(x-3y^1/3)(x^2+3x(y^1/3)+y^2/3)(x^3+9y)
2007-08-14 22:29:39
·
answer #8
·
answered by bashah1939 4
·
1⤊
0⤋
no math! ah!
2007-08-14 22:22:30
·
answer #9
·
answered by erwafredsfdsfsdf 5
·
1⤊
0⤋
6.339, or not.
2007-08-14 22:23:03
·
answer #10
·
answered by Anonymous
·
2⤊
0⤋