Friday, August 21, 2009

5th Period Test 1

Everyone from 5rd period, this is where you need to comment with your solutions to the problems.

12 comments:

Laura Hawkins said...

pg 131 #36

6x^2+x-12
(3x-4)(2x+3)

nolanh said...

#125
Find an equation for the line that has x-intercept 4 and y-intercept 12


y=3x+12

carter williams said...

pg 131 #20
(a^2)^-3(a^3b)^2(b^3)^4
(a^0)(b^14)
(b^14)

Nicole Caston said...

Page 131 #16:

4^(1/4) x 324^(1/4) = 1296^(1/4)= 6

Since 4 and 324 have the same root, fourth root, they can be multiplied together. When they are, you get 1296^(1/4). Therefore, 6 x 6 x 6 x 6 is equivalent to 1296.

Chloe' said...

page 133 #98

x^3-4x^2-5x>2
-2>x^3-4x^2-5x>2
-2>x(x-5)(x+1)>2
-2>x=0,-1,5>2
x=(0,-1)

Megan9036 said...

Page 131 #60

(1/x+2)+(1/x^2-4)-(2/x^2-x-2)

(x-2)(x+1)+(x+1)-(2x+4)/(x+2)(x-2)(x+1)

-(x+3)/(x+2)

Malorie Swindle said...

x^2-2x-3 (x-3)(x+1)
-------- ---------
2x^2+5x+3 (2x+3)(x+1)

The (x+1)'s cancel out and you're left with

x-3
----
2x+3

Malorie Swindle said...

the problem was #54

Dylan Rigg said...

pg 131 #10

1- |1 -|-1||
1-|1-1|
1-0
1

Peyton Garett said...

#22

((r^2s^4/3)/(r^1/3s))^6

.......................

r^10s^2

Mr. James said...

Nolan, you need to check your slope again.

Mr. James said...

Chloe, I don't follow how the inequality turned into an equality. Is your answer the interval notation?