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Take a look at this you tube clip.

$${\mathtt{28}}{\mathtt{\,\times\,}}{\mathtt{12}} = {\mathtt{336}}$$ $$\mbox{inches}$$ 

28*12 inches

Thanks Rosala,

I've never seen beard art before.  That is brilliant!

That cloud it is not real surely?    I want to see a real cloud like that one!

Yes that is exactly how you enter a mixed number

For more info look in the 

"How do I type in fractions thread" which is in the sticky notes on the right hand side of your page. 

$$\left({\mathtt{6}}{!}{\mathtt{\,-\,}}{\mathtt{3}}{!}\right) = {\mathtt{714}}$$

((6! - 3!) - ((6 × 100) - ((6 ÷ 3) × 100))) / 100

(714 - (600 - (2 × 100))) / 100

(714 - (600 - 200)) / 100

(714 - 400) / 100

Except for the missing bracket, you are correct!    

The least common denomonater of 18 and 9 is 9.

Thanks for the pics, rosala....I'm going to use that one of the guy "holding" the moon as my Facebook avatar....also loved the "wedding dress" (cake) one......where do you get these ???

You could multiply it by  68600 squared. That would work!

4725 = 3² x 5² x 7

An excellent answer Dragon 

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

what is 18 out of 39

what about it?  (2,6) is not a solution to y=x+3

-11/4+2/3|3y-6|=-2

do you mean

$$\begin{array}{rlll}
\frac{-11}{4}+\frac{2}{3}\times |3y-6|&=&-2\qquad &\mbox{}\\\\
\frac{2}{3}\times |3y-6|&=&-2+\frac{11}{4} \qquad &\mbox{I've added 11/4 to both sides}\\\\
\frac{2}{3}\times |3y-6|&=&-2+2\frac{3}{4} \qquad &\mbox{I've simplified}\\\\
\frac{2}{3}\times |3y-6|&=&\frac{3}{4} \qquad &\mbox{I've simplified}\\\\
|3y-6|&=&\frac{3}{4}\times \frac{3}{2} \qquad &\mbox{I've mult both sides by 3/2}\\\\
|3y-6|&=&\frac{9}{8} \qquad &\mbox{I've simplified}\\\\
\end{array}$$

Now there are 2 equations to be solved.

$$3y-6=\frac{9}{8}\qquad and \qquad 3y-6=\frac{-9}{8}$$

You should be able to finish it from here.  

21 = 4x -9 -x                 ---Combine 4x and -x to get 3x

21 = 3x - 9                  ---Add 9 to both sides to get rid of the -9 on the right side

+9        + 9

--------------------

30 = 3x                  ---Divide both sdies by 3 to find x!

30 ÷ 3 = 3 ÷ 3x

10 = 1x

10 = x

x = 10

0.00044 fakhdflashgoihasdg

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

Rose, you can get the formula for the area of a circle on the from the formulary page on this site.  It is

$$A=\pi r^2 \quad cm^2$$

Now if the radius is doubled it becomes 2r  the new area will be

$$\\A=\pi\times (2r)^2\\
A=\pi \times 2^2 \times r^2\\
A=4\pi r^2\quad cm^2\\$$

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)