Do the operations and simplify
a ^-2 b ^5 . a ^6 b ^3
_____________________
x ^-4 y . x ^4 y ^2
some of the answers I have to choose from are
a ^4 b ^8
_______
xy ^3
a ^4 b ^8
_________
y ^2
a ^0 b ^6
________
y ^3
a ^4 b ^8
________
y ^3
2007-05-06 03:20:22 · 6 answers · asked by Simone J 1 in Science & Mathematics ➔ Mathematics
a ^4 b ^8
________
y ^3
The exponents are added to arrive at this.
2007-05-06 03:25:01 · answer #1 · answered by ignoramus 7 · 0⤊ 0⤋
When multiplying powers you add items with common variables. For example x^2*x^9 = x^11. In the case above you have some negative numbers that play into it. Your answer to the problem above is the last one. Just add common powers. "a" has the powers of -2 and 6, so it would result in a^4, and so on and so on. Since the denominator has x^-4 and x^4 they cancel out and equal x^0 which equals 1.
2007-05-06 10:28:19 · answer #2 · answered by bklyn_tim 1 · 0⤊ 0⤋
The rule for multiplying is to add the exponents of each variable.
a ^-2 b ^5 . a ^6 b ^3
_____________________
x ^-4 y . x ^4 y ^2
would be a^(-2+6)b^(5+3) over x(-4 + 4) y^(1+2)
which is a^4b^8 over y^3 since x^0 is just 1
2007-05-06 10:26:41 · answer #3 · answered by hayharbr 7 · 0⤊ 0⤋
Do the operations and simplify
a ^-2 b ^5 . a ^6 b ^3=a^4b^8
_____________________
x ^-4 y . x ^4 y ^2=y^3
some of the answers I have to choose from are
a ^4 b ^8
_______
xy ^3
a ^4 b ^8
_________
y ^2
a ^0 b ^6
________
y ^3
2007-05-06 10:29:26 · answer #4 · answered by Anonymous · 0⤊ 0⤋
well assuming that "." is a multiplication sign, then you would get a^(-2+6)*b^(5+3)/[x^(-4+4)*y^(1+2)] which equals a^4*b^8/y^3
2007-05-06 10:26:14 · answer #5 · answered by 7Mathbaby 2 · 0⤊ 0⤋
a) a^-2 . b^5 = b^5/a^2
b^5/a^2 . a^6b^3 = a^4b^8 - exponentials are additive
b)x^-4 . y = y/x^4 . x^4y^2 = y^3
2007-05-06 10:27:21 · answer #6 · answered by Rich W 2 · 0⤊ 0⤋