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Nice answer, TJM.........!!!

You started at the bottom – and it’s been downhill ever since.

oh snap epic anon burn

use ^ (shift+6 on most keyboards) or x^y

Isolate the variable. If it was 3x+4=5x-7 then you would subtract 3x from both sides leaving you with 4=2x-7. Then you would Add 7 to both sides to get 11=2x. Then you would divide by 2 and get X=5.5.

well its actually 4, but if youre going by the 1+1=window then yeah its fish. 3+3 is snowman i think

Can you give an example? 

im not sure if you mean $${\frac{{\mathtt{27}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}\left({\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)$$ or$${\frac{{\mathtt{27}}}{\left({\mathtt{3}}{\mathtt{\,-\,}}\left({\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)\right)}}$$but im assuming its the first one. ill do both anyway

for the first one:

$${\frac{{\mathtt{27}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}\left({\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)$$

order of operations, 4 + 3/4 = 19/4

27/3 - 19/4 = 108/12 - 57/12

$${\mathtt{108}}{\mathtt{\,-\,}}{\mathtt{57}} = {\mathtt{51}}$$

$${\frac{{\mathtt{51}}}{{\mathtt{12}}}} = {\frac{{\mathtt{17}}}{{\mathtt{4}}}} = {\mathtt{4.25}}$$

for the other one

$${\frac{{\mathtt{27}}}{\left({\mathtt{3}}{\mathtt{\,-\,}}\left({\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)\right)}}$$

first figure out the denominator deal of course

4+(3/4) =19/4

$${\frac{{\mathtt{19}}}{{\mathtt{4}}}} = {\mathtt{4.75}}$$

3-4.75=-1.75

$${\frac{{\mathtt{27}}}{-{\mathtt{1.75}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{108}}}{{\mathtt{7}}}} = -{\mathtt{15.428\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

Anonymous please do not double thread. This was already posted.

@ anonymous. Use a ^. so put it like a^largenumberhere

9-4.75 or 4.25. Is that right?

Better question. Why are you asking that in a math forum?

The square root of 15 is approximately 3.8729. 

There is something missing in the question. If you mean, what is the probability oh drawing aa white or a gray marble then it is:

For the white marbles 2 out of a total of 9 or 2÷9.

For gray it is 1÷9.

Let x be the price of the children's ticket and y be the price of the adult...so we have

12x + 3y = 70

12x + 12y = 216    subtract the first equation from the second...this gives

9y = 156  divide both sides by 9

y = $16.22    this is the approx cost of the adult ticket

And using the first equation to find the cost of the child's ticket, we have

12x + 3(16.22) = 70

12x + 48.66 = 70     subtract 48.66 from each side

12x = 21.34     divide both sides by 12

x = $1.78  and this is the cost of the child's ticket

$$\\3.2=\frac{3.2}{1}=\frac{32}{10}=\frac{16}{5}\\
or\\
3.2=3\frac{2}{10}=3\frac{1}{5}$$

oh cool thanks

It is not a problem - I over reacted anyway.  I'll put your post back.  :)

thanks sasflower, i have seen other people do it like you and it has always made me wonder.

This is how i would do it

$$\\\frac{3}{8}\;of\;56\\\\
=\frac{3}{8}\times \frac{56}{1}\\\\
=\frac{3}{\not{8}}\times \frac{\not{56}^7}{1}\\\\
=21$$

That is easier don't you think?

eh id say about like 6.2/10 but i dont know you well enough. first impressions yknow. not enough info yo.

A new amusement park is going to be built near 2 major highways. On a cordinate grid of the area, with the scale 1 unit representing 1 km, the park is located at P(3,4). Highway 2 is represented by the equation y=2x+5, and highway 0 is represented by the equation y=-0.5x+2. Determine the coordiantes of the exits that must be built on each highway to result in shortest road to the park.

We first need to find equations of the lines that are perpendicular to these two lines and pass through (3,4)

The equation for the  perpendicular line to Highway 2 is  y - 4 = -.5(x - 3) →  y = .5x + 5.5   (1)

The equation of the perpendicular line to Highway 0 is y - 4 = 2(x -3) →  y = 2x - 2       (2)

We could set the equation for Highway 2 equal to (1) and the equation for Highway 0 equal to (2)  and find the intersection points, but here's a graphical solution...

:) hello kitty also nice to meet you. and sorry for the misunderstanding melody