CPhill

4725 = 3² x 5² x 7

A = s² = (1/4)in²

So, the side is the square root of (1/4)in² = 1/2 in

And the perimeter is 4 times this = 2 in

-11/4+2/3|3y-6|=-2 solve the absolute value equation

Add 11/4 to both sides

(2/3)l3y-6l = 3/4       multiply both sides by 3/2

l3y-6l = 9/8        So we have two equations to solve

3y-6 = 9/8     and  -(3y-6) = 9/8

For the first one, add 6 to both sides

3y = 57/8       divide by 3 on both sides

y = 19/8

For the second equation, we have (multiplying both sides by -1)

3y - 6 = -9/8     add 6 to both sides

3y = 39/8         divide both sides by 3

y = 13/8

9/23  = about .391 (or about 39.1%)

.3 x .1   = .03

4 inches = 1/3ft

So in one load, there are  400 x 1/3 = 400/3 ft³

And the number of cubic feet required  = 80 x 70 x 1/3  = 5600/3 ft³

So the number of loads required =

(5600/3) /  (400/3) = 5600/400 = 14

25b/4c divided by 5/2b =

(25b/4c) x (2b/5) =

(25b/5) x (2b/4c) =

(5b) x (b/2c) =

5b² / 2c

(2t - 13) (t +1)

As rosala would say......."You just made my head hurt, Melody."

(LOL).......

Thanks, Anonymous......I see what you mean!!!

However, I might point out that we don't have to actually use logs here.....

x^2/3 = 8       just exponeniate  both sides to the (3/2)

x = 8^3/2

And you can get the answer by using the onsite calculator.......