thedudemanguyperson

No problem, keep the good work up! I hope to see you soon on the Alcumus hall of fame (or if it's a homework problem, on the forums).

While sam runs $3$ km, Ian runs $5+2=7$ km. So Ian is running $\frac{7}{3}$ as fast as Sam. Then $(s+4)=\frac{7}{3}s$. Now compute $s$ and finish.

Introduction to counting and probability is a great book ;)

Let's first calculate the probability for each club then average them.

5 student club - There are $\binom{5}{3}=10$ ways to choose the members. In $\binom{2}{2}\binom{3}{1}=3$ ways, the co-presidents are selected - the probability here is $\frac{3}{10}$.

7 student and 8 student clubs are analogous. Try it, I believe in you!

Noah rides at $n$ mph. Grady rides at $1.6n$ mph. Then $3.2n-2n=15\implies \boxed{n=12.5}$

$f(x)=y\implies 16x+3=y\implies x=\frac{y-3}{16}$ so $f^{-1}(x)=\frac{x-3}{16}$. The continuation is simple.

I like this one. Let the sides be $x,y,1-(x+y)$. By the triangle inequality, all of the following must be true:

$$x<1-x$$

$$y<1-y$$

$$1<2x+2y$$

Graph each of these on the coordinate plane and use the principle of geometric probability. It's $\frac14$. 

Yep, believe so. Nice job :)

Let's say there are $r$ reds and $y$ yellows. Then:
$\frac{r}{r+y}\cdot \frac{r-1}{r+y-1}=\frac{2}{15}$

Now solve the diophantine for $r$.

$\lim_{h\to 0}\frac{(x+h)^3-3(x+h)-x^3+3x}{h}=\lim_{h\to 0}\frac{-3 h + h^3 + 3 h^2 x + 3 h x^2}{h}=\lim_{h\to 0} -3+h^2+3hx+3x^2=3x^2-3$

At $x=2$, this is $3*4-3=9$.

Using this, you can easily find the tangent line equation so I will leave that to you.

For the first one, $\deg(denom)<\deg(num)$ so it goes to infinity

For the second one, $\deg(denom)>\deg(num)$ so it goes to zero.