Friday, August 21, 2009

7th Period Test 1

Everyone from 7rd period, this is where you need to comment with your solutions to the problems.

6 comments:

Courtney Meadows said...

pg. 131 #44

a^4B^2 ab^5
= ab(a^3b + ab^4)

Summer Emmons said...
This comment has been removed by the author.
Summer Emmons said...

Page 132 #27

8r^(1/2)s^(-3) /
2r^(-2)s^(4) =
4r^(1/2)r^(2) /
s^(3)s^(5) =

4r^(5/2) / s^(7)

*(1/2)+(2/1)= (1/2)+(4/2)=
(5/2) is how you get the ending
answer.

Morgan Roper said...

pg. 131 #20

(a^2)^(-3)(a^3b)^2(b^3)^4
a^(-6)a^6b^2b^12
a^6b^2b^12/a^6
=b^14

Mr. James said...

Courtney, if we pull another b out, we'll get ab^2*(a^3 + b^3) and (a^3 + b^3) can factor too.

Mr. James said...

Summer, remember we don't want to leave any fractional exponents in our answer.