CPhill

Thanks, h7......

b / 15  = $120    multiply both sides by 15

b = $1800

Just multiply all three numbers together......the answer will be in cubic inches

7x^(1/3)-x(-2/3)=0    factor

x^(-2/3) (7x- 1) = 0

The first factor is never 0

For the second one, we have

7x - 1  = 0

Add 1 to both sides

7x = 1    divide both sides by 7

x = 1/ 7 = about .1428

x^(-4/3)-x^(-2/3)-8=0   let u = x^(-2/3)   then we have

u^2 - u -  8  = 0          add 8 to both sides

u^2 - u    =  8             complete the square on both sides

u^2 -  u  + 1/4  = 8.25        factor

(u - 1/2)^2  = 8.25              take both roots

u - 1/2 = ±√8.25                   add 1/2 to both sides

u = ±√8.25 + 1/2

So u = √8.25 + 1/2   or  u = -√8.25 + 1/2

Then, back-substituting, we have

x^(-2/3) = √8.25 + 1/2    raise both sides to the (-3) power

x*2 = (√8.25 + 1/2)^(-3) 

Take both roots

x = ±√((√8.25 + 1/2)^(-3))  = ±(√8.25 + 1/2)^(-3/2) = about ± .161

The other solution, u = -√8.25 + 1/2   leads to a non-real answer

I assume we have this

1-csc (x) sin^3 (x)=cos^2x     and we'd like to show that it's an identity.....working on the ledt, we have

1 - (i1/ sin x)sin^3x =

1 - sin^2 x  =

cos^2x      which is equal to the right side

-6=2x / 7     multiply both sides by 7

-42 = 2x     divide both sides by 2

-21 = x

6 + 2y = 5     subtract 6 from both sides

2y = -1         divide both sides by 2

y = -1/2

and there you go.....!!!

If you think about it, a regular hexagon is just comprised of six equilateral triangles.

The area of one of these triangles is given by

(1/2)s^2 *(√3/2) = (√3/4)*(s^2)   where s is the side length of the hexagon

So.....6 times this is just

(3/2)(√3) s^2