CPhill

Let C be the cost of a child's ticket.....and 2C be the cost of an adult ticket....then

2 child's tickets cost 2(C)  = 2C

And 2 adult tickets cost 2(2C)  = 4C

So the total cost is given by

2C + 4C = 24   simplify

6C = 24     divide both sides by 6

C = 4    and the cost of a child's ticket in dollars

And 2(4) = 8  = the cost of an adult ticket in dollars

Let N be the number......so we have

N^2 - N  = 4N - 1      add 1 to both sides and subtract 4N from both sides

N^2 - 5N + 1  = 0  using the onsite calculator and substituting x for N, we have

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{21}}}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{21}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.208\: \!712\: \!152\: \!522\: \!08}}\\
{\mathtt{x}} = {\mathtt{4.791\: \!287\: \!847\: \!477\: \!92}}\\
\end{array} \right\}$$

Those are the two solutions......!!!!

"PM" = Private Messages.....it's basically our "email" for members.....!!!

P(first is a heart) = 13/52  = 1/4

P(second is a spade)  = 13/51

P(third is a heart)= 12/50

So

(1/4)(13/51)(12/50)  = 156/10200 = 13/850

Thanks for this, Melody.....I want to look at this one a little more....

The number is.......

s = rθ     where s is the arc length, r is the radius and θ  is in radians

Using the point-slope form rearranged, we have

y = 3(x - 5) + 1

y = 3x - 15 + 1

y = 3x - 14