thedudemanguyperson

Pick two points. Find the change in the x value, and the change in the y value. The slope is $\frac{\text{change in y}}{\text{change in x}}$. Remember that you need signs, if $x$ goes up it's $+1$, if $x$ goes down it's $-1$.

Let's take this table:

The initial value is the first value, i.e. $x=0$ and $y=1$.

The rate of change (for linear functions) is the difference between the $y$ values when the $x$ value changes by $1$. Looking closely, as $x$ goes up by $1$, $y$ goes up by $2$. The rate of change is $\frac{\text{change in y}}{\text{change in x}}=\frac{2}{1}=\boxed{2}$.

Hope this helped.

Nope.

Take $y=x\{x\ge -1\}$, the initial value is $-1$ and the y-intercept is $0$.

The idea is to counter the misconception that "All graphs start at $x=0$".

The product of four digits is $336$. 

Obviously one must be seven so the product of three digits is $48$.

Listing factor pairs of $48$:

$(1,48)$ --> from here we can make the $48$ into $6,8$, so the pairs $(1,6,8)$ appear.

$(2,24)$ --> from here we can make $24$ into $8,3$ or $4,6$ to make $(2,4,6)$ and $(2,3,8)$

$(4,12)$ --> from here we can make $12$ into $2,6$ or $3,4$ so $(4,2,6)$, $(4,3,4)$

$(6,8)$ --> we can either change the $6$ into $1,6$ or $2,3$ or change $8$ into $1,8$ or $2,4$ to make $(6,1,6), (6,2,4), (1,6,8), (2,3,8)$.

So we can have $(1,6,8), (2,4,6), (2,3,8), (4,2,6), (4,3,4), (6,1,6), (6,2,4), (1,6,8), (2,3,8)$. We double counted a few, so lets take some out:

$(1,6,8), (2,4,6), (2,3,8), (4,3,4), (6,1,6)$.

So we can have $(7,1,6,8), (7,2,4,6), (7,2,3,8), (7,4,3,4), (7,6,1,6)$. Now, the first three pairs can be arranged in $4!=24$ ways. The last two can be arranged in $\frac{4!}{2!}=12$ ways. The answer is $24*3+12*2=24*4=\boxed{96}$.

Whew!

This is what I would do in contest to eliminate silly errors, but you can obviously just list triples of $48$ without my process.

We want to maximize $xy$ if $x\in[-5,3]$ and $y\in [-2,1]$. This either occurs when they are both negative or both positive. If both positive, then it's $3*1=3$. If both negative, it's $(-5)*(-2)=10$. Since $10>3$, the answer is $\boxed{10}$.

$f(2)=9\implies f(-(-2))=9$, the point is $(-2,9)$ and the sum is $9-2=\boxed{7}$.

The volume is proportional to $r^2$, so we multiply by $\left(\frac{\frac{9}{8}}{\frac{15}{16}}\right)^2=\left(\frac{6}{5}\right)^2=\frac{36}{25}$. It's proportional to the height itself, so we want to multiply by $\frac{4}{\frac{36}{25}}=\frac{25}{9}$. So $\frac{1}{16}*\frac{25}{9}=\boxed{\frac{25}{144}}$. 

Base radius $\frac{15}{2}$, multiplying by $\frac{3}{2}$, so height is $12\cdot \frac32=18$. 

The volume of a cone is $\pi r^2 \cdot \frac{h}{3}=\pi \cdot \frac{225}{4}\cdot 6=\boxed{\frac{675}{2}\pi}$

Rounded to the nearest tenth it is $\boxed{1060.3}$

$\triangle ADE$ has base $2$ and height $10$, so it has area $10$. The area of the rectangle is $6*10=60$, so the answer is $\frac{10}{60-10}=\frac{10}{50}=\boxed{\frac15}$.

From MATHCOUNTS? (only source that asks for common fraction, unless it's illegal AoPS homework)

Right triangle with base $10+8=18$, angle $36^{\circ}$ and we want the height, $x$.

So $\tan(36)=\frac{18}{x}\implies x=\frac{18}{\tan(36)}\approx 24.77$.

Note: You can derive the exact value of $\cos(36)$ with a pentagon, and then find that $\tan(36)=\sqrt{5+\sqrt{2}}$ exactly.