What is a^11 times b^2 times b^8 times a^-3? Please explain how you did this.?

Answer 1

a^11 x b^2 x b^8 x a^-3 = a^8 x b^10

first you have to group the numbers like this:

a^11 x a^-3 x b^2 x b^8

then you just add or subtract the exponents:

11-3=8 so its a^8, 2+8=10 so its b^10

.. pretty easy stuff.

Answer 2

General rule: when you multiply like bases, you add the exponents.
    m^x + m^y = m^(x + y)

So a^11*b^2*b^8*a^-3 = a^11* a^-3 * b^2 * b^8    [move the a's next to each other]

   = a^8 * b^10     [for the a's 11 + -3 = 8, for the b's 2 + 8 = 10]

Answer 3

whenever you multioly with exponents you add the exponents that come with the figure.

a^11 x b^2 x b^8 x a^-3= a^8 b^10

Answer 4

(a^11)(b^2)(b^8)(a^-3)
Because of commutativity and associativity of multiplication, ab=ba and (ab)c=a(bc) respectively,
(a^11)(b^2)(b^8)(a^-3)=[(a^11) (a^-3)] [(b^2) (b^8)]

Because of one of the rules about exponents, to wit: (a^p)(a^q)=a^(p+q)
[(a^11)(a^-3)][(b^2)(b^8)] = [a^(11+ -3)] [b^(2+8)]

Because of arithmetic,
[a^(11+ -3)][b^(2+8)]=(a^8) (b^10)

Learn these rules.. and a few others, there really aren't many, and you'll be ok. If you don't learn the rules and understand them, and where they come from, you'll find yourself stuck in this predicament every time you do your homework or take a test.

Answer 5

a^11*b^2*b^8*a^-3 when you multiply, you add exponents
a^(11-3)*b^(2+8)=a^8*b^10

Answer 6

Firstly, I will write the terms out neatly in letter order.

(a^11)(a^-3)(b^2)(b^8)

Using this property:

(a^x)(a^y) = a^(x+y)

I can solve:
(a^11)(a^-3)(b^2)(b^8)
(a^(11+-3))(b^(2+8))
(a^8)(b^10)

Hence:

(a^8)(b^10)