CPhill

I know my answer is incorrect and geno's is correct.....but can someone spot my error??

I chose to see the problem this way.....

In each set of three people, we want to choose any 1 of 3 Americans and then choose any 2 of 16 of the other contestants

So the total possible sets - disregarding order, is  C(3,1)*C(16,2) = 3 * 120  = 360

And each of these sets can be ordered in 3!  = 6 ways

So......6 x 360 = 2160  which is 1/2 of geno's result.......????....where did i go wrong???

x+4y=8  2x-5y=29

Using the first equation, subtract 4y from both sides

So x  = 8 - 4y       and subbing this into the second equation foe x, we have

2(8 - 4y) - 5y = 29

16- 8y - 5y = 29   simplify

16 - 13y = 29       subtract 16 from both sides

-13y = 13             divide both sides by -13

y = -1 

And x = 8 - 4(-1)  = 8 + 4  = 12

Here's another way to see this.......

The cumulative sum of the total entries of n rows is given by.....n( n + 1)

And we can always figure out the row of the entry we're looking for by taking the ceiling function answer of the positive root of n^2 + n - k = 0   where k = the nth entry .... so

$${{\mathtt{n}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{n}}{\mathtt{\,-\,}}{\mathtt{40}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{161}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{n}} = {\frac{\left({\sqrt{{\mathtt{161}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = -{\mathtt{6.844\: \!288\: \!770\: \!224\: \!760\: \!2}}\\
{\mathtt{n}} = {\mathtt{5.844\: \!288\: \!770\: \!224\: \!760\: \!2}}\\
\end{array} \right\}$$

So ceiling(5.8)  = 6  ...so the 40th entry will be in row 6

And the value of each entry on row n = 2n......so....the 40th entry will be 2(6) = 12

In my fencing club there are twice as many boys who are in high school as there are boys who are in middle school. Of the students in the club who are in middle school, there are three times as many girls as there are boys. Half of the girls who are in the club are in high school. Everyone in my fencing club is in either middle school or high school. If there are 72 people in my fencing club, how many are middle school boys?

x = boys in middle school ....    2x =  boys in high school .... girls in middle school = 3x  = girls in high school

So we have

x + 2x + 3x + 3x = 72    combine like terms

9x = 72    divide by 9 on both sides

x = 8   this is the number of boys in middle school

I think you might mean....4x^2 + 3x - 2 =0   

This will not factor....using the onsite solver, we have

$${\mathtt{0.4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{305}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{305}}}}{\mathtt{\,-\,}}{\mathtt{15}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{8.116\: \!062\: \!299\: \!143\: \!245\: \!2}}\\
{\mathtt{x}} = {\mathtt{0.616\: \!062\: \!299\: \!143\: \!245\: \!2}}\\
\end{array} \right\}$$

The first person shakes 5 other hands........the second person shakes 4 other hands (he has already shaken the first person's hand)......the third person shakes three other hands  (since he has already shaken the hand of the first two), etc.

So we just have the sum of the first five numbers = 5 * 6 / 2  = 15 handshakes

The number is given by (126 + 129) - 200 =  55

Note that we can find the number taking either, but not both, by taking (126 - 55) =   71   and (129 - 55) = 74

So the ones taking just history plus the number taking both plus the number taking history is....

71 + 74 + 55 = 200

The expression is

x + y - z

To see this, we add the number in both classes......but we must subtract the number taking both to prevent "double-counting"

Happy Birthday, Alan.....!!!!!