BuilderBoi

The girls can be chosen in ${10 \choose 2}$ ways. The boys can be chosen in $8 \choose 2$ ways. of the 4 members, there is 4 ways to choose a leader.

So... the answer is ${10 \choose 2} \times {8 \choose 2} \times 4 = 5040$ ways.

What's the problem?

What kind of a questions is this? Whoever wrote this problem should feel ashamed of themselves.

That will be useful in 3 years for me.

Everything after p(5) is $x^2$. This means that p(7) is 49.

In order to get the smallest value, you want as big of a negative number x as big of a positive number, so the answer is a super small negative. The way we can do this is by having a = -2, and b = 5. 

Plug these values in:

${({-1 \over 2} + {1\over 5})}* {({1 \over 5} - {-1 \over 2})}$. This yields $\color{red} -21/100$.

12 + 6 = 18

12 - 6 = 6

The x-coordinate of the other x-intercept is $\color{red} 6$ by symmetry

Think about it... Try and make 2 equations that would intersect at an infinite amount of points. Find similarities between the two equations and apply them to this problem. 

Subtract $3$ from both sides:

$-3x^2+24x \leq 0$

Factor the equation by $3x$:

$3x(-x+8) \leq 0$

Both $3x$ and $-x+8$ have to be less than or equal to $0$. This means your answer is $\color{red}x \leq 0$ or $\color {red}x \geq 8$

${{5 \over 8} * {1 \over 5}} = {1 \over 8}$

So, $1 \over 8$ of the tractors are in working conditions.