Imnotamaster

Let's say smaller leg is X

The longer leg is x+8

Hypo is x+14 

a^2 + b^2 = c

so x^2 + (x+8)^2 = (x+14)^2. 
Expand
x^2 + x^2 + 16x + 64 = x^2 + 28x + 196.
Minus x^2 from both sides

x^2 + 16x + 64 = 28x+196
Minus 16x 

x^2 + 64 = 12x + 196
Minus 64

x^2 = 12x + 132.
or
x^2-12x-132 = 0.

From there, hopefully you can solve 

So this is a common cause of adding numbers together with large numbers (IDK what its called)
Basically, the equation is: Add the First and last term. (in your case it's 2009). Then divide by 2, and multiply by a number of numbers in that said sequence. 
So how do we how many numbers there are in the sequence? We can use the Explicit function: F(x) = 1+3 (x-1), where f(1)=1. This means that 2008-1 = 2007/3 = 669. This means there are 70 terms. So we have 2009/2) x 70. Or 2009 x 35 which is equal to 70315. 
To double-check my work, I used Wolfram Alpha! 

Welp, this is an easy case of the Chicken and cats problem (is it called that? Idk)
Let's start with calculating what would happen if only 123 adults came. 123 x 7 = 861 dollars. That's only 32 more dollars than they made.

Now, children cost 2 dollars less than adults (that sounds weird, but I'm too lazy to correct it). Therefore, if we divide 32 by 2, we would get 16. 

This means there are 

16 Children

107 adults. 

We double check this: 107 x 7, 16 x 5. 80 + 749 = 829. Thats the answer. 
(I could be wrong, Idk) 

IS it just me or can I not see the picture? I think theirs a picture, but I cannot see it. Can anybody else?

Alright, here we go. Lets count out how many there are for odd starting numbers, and then even starting numbers. 
For odd: The next two have to be even. So 5 possibilities per row, 5 rows, 25 possibilities for 1,3,5,7,9. So 125 for odd
For even its a bit more complicated. Zero doesn't count because that would be 2 digits. So we start with 2. 50 possibilities for that one. x 4. 200. Answers 325. I know I'm not clear but there's the answer. 






 

325. For all you people that don't want to read. 

 Dude. This is probably not what you wanted but here: x=5/9y - 1. 
y= 9/5 +9/5x. Hope that helped. If it didn't make your equation clearer. 

Alright here we go. We don't (really) care about the 3rd and 4th place, as the question asks how many times can A win? First we count all possibilities of which A won B and wins C. It would be ACBD or ACDB. Then we do the same with c. ADBC or ADCB. 4 times already. Thats it.






4 TIMES. I hope I'm right. If wrong, be sure to correct me. 

For number 2) We use algebra. Pretend the second piece is X. The first piece would be 3x, and the third would be x-20. Add these all up, you get 5x-20. 5x-20 is equal to 100 feet. Therefore, 5x=120 feet. So, x=24 feet. The second piece is 24 feet. We minus 20 from 24 and the shortest piece is 4 feet

1) Alright we have 3 possibilities. 3x+62=7x+30, 7x+30=5x+50, or 5x+50 = 3x+62. Lets see the first one. 
First one: If 3x+62=7x+30, 32=4x. x=8, and the perimeter would be: 24+62+56+30+40+50, or 272. 

Second one: 7x+30=5x+50. x=10. 100+100+92= 292.
Third one: 5x+50 = 3x+62. x=6. 80+80+72 = 232. So the least possible feet is 232. 

Stephan has been busy at work writing all the possible rearrangements of the letters in his name. He is one bored boy. If he can write twelve rearrangements of his name every minute, how many hours does it take to write all the possible rearrangements of his name?
 Well step number 1. Stephan has 7 letters, so 7!

2) 7!/20 minutes. 

3)7!/(20 x 60) = 4.2 I might be wrong.. but hey, I tried. 

IF YOU JUST WANT THE ANSWER BECAUSE YOU DONT WANT THE EXPLANATION it's 4.2
 

Thanks c-phil!

Thank you so much, c-phil!

Ironically, just as you answered one of the questions, I figured out one of them too.

That depends, where to you want to put the brackets?

Let's do this a different way! 
Group the terms in pairs. 
1000² - 999² = (1000+999)(1000-999) = 1999. 
998² - 997² = (998 + 997)(998-997) = 1995. 
. 
. 
. 
2²-1² = (2+1)(2-1) = 3. 

So we have to sum the arithmetic progression 
3 + 7 + ... + (4n-1) 
to 500 terms, since 1999 is the 500th term of 
this progression. 
So, how do we do this? 
Note that 
3 + 1999 = 2002 
7 + 1995 = 2002 
11 + 1991 = 2002 
. 
. 
. 
1999 + 3 = 2002 
Call the sum of each column S 
Then S + S = 500*2002 
S = 500*1001 = 500500.

Let the boxes be Box 1, Box 2, Box 3. 

consider the 3 white balls. They can be all of them in one box:

(3, 0, 0) (3 in Box 1, 0 in box 2 and 0 in box 0)
(0, 3, 0)
(0, 0, 3)

We can have 2 in one box, and 1 in one of the remaining boxes:

(2, 0, 1)
(2, 1, 0)
(0, 2, 1)
(1, 2, 0)
(0, 1, 2)
(1, 0, 2)

and there is only one way: (1, 1, 1) to place one white ball in each box

In total there are: 3+6+1=10 ways to place the white balls. Similarly there are 10 ways to place the black ones.

Since every placement of the white balls can be combined with any placement of the black balls, there are 10*10=100 ways to place the 3 white balls and the 3 black bals in the boxes.

Or if you just want the answer, its 100