$\displaystyle 3r+5t=\frac{-4Q+5z}{2}$

Multiply by $\displaystyle 2$ on each side

$\displaystyle 2(3r+5t)=-4Q+5z$

Expand right side

$\displaystyle 6r+10t=-4Q+5z$

Subtract $\displaystyle 5z$ on each side

$\displaystyle -4Q=6r+10t-5z$

Divide by $\displaystyle -4$ on each side

$\displaystyle Q=\frac{6r+10t-5z}{-4}$

To figure out what the equation is, we need to use the point-slope form.

$\displaystyle y-y_0=m(x-x_0)$, where $\displaystyle m$ is the slope, and $\displaystyle (x_0, y_0)$ is a point.

$\displaystyle \text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

In this example, $\displaystyle y_2=500, y_1=0, x_2=5, x_1=0$

$\displaystyle \text{Slope}=\frac{500-0}{5-0}=100$

$\displaystyle y-0=100(x-0)$

$\displaystyle y=100x$

The x-intercepts are the points on the graph where the $\displaystyle y=0$ line touches or intersects the x axis. By looking at the graph, there are two points. One at $\displaystyle x=2$, and then another one at $\displaystyle x=3$. So there are two x intercepts. 

Use the formula for exponential growth $\displaystyle y = A*(1+r)^t$ where y is the current value, A is the initial value, r is the rate of growth, and t is time. Between 1997 and 2015, 18 years passed, so use $\displaystyle t=18$. The stuffed animal was originally worth $6, so $\displaystyle A=6$. It is now worth $1,015, so $\displaystyle y=1015$.

Our equation is now:

$\displaystyle 1015 = 6*(1+r)^{18}$ divide by 6:

$\displaystyle 169.1\overline{6}=(1+r)^{18}$ take both sides to the power of $\displaystyle \frac{1}{18}=0.0\overline{5}$:

$\displaystyle 1.329 \approx 1+r$ subtract 1

$\displaystyle 0.329 = r$

As a percent, r is about 33%.

To construct a function that describes this situation first identify what is known.

Since this particular situation is talking about population decrease, the function will be an exponential decay.

Recall that an exponential decay function is in the form,

$\displaystyle P(t)=I(1-r)^{t}$

where,

$\displaystyle \\P(t)=\text{ Finial population} \\I=\text{ Initial population} \\r=\text{ Rate} \\t=\text{Time}$

Since the statements says that the population decreases every 50 years we can rewrite the general form to,

$\displaystyle P(t)=I(1-r)^{\frac{t}{50}}$

Now substituting in the known values, the function can be written. 

$\displaystyle P(t)=120,000(0.85)^{\frac{t}{50}}$

We can determine the demand equation by using point slope form.

Point slope form is $\displaystyle y-y_0=m(x-x_0)$, where $\displaystyle (x_0, y_0)$, and $\displaystyle m$ is the slope, where $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$.

Let $\displaystyle y_2=50$, $\displaystyle y_1=10$, $\displaystyle x_2=0$, and $\displaystyle x_1=10$.

$\displaystyle m=\frac{50-10}{0-10}=-4$

Now we have

$\displaystyle y-y_0=-4(x-x_0)$

Choose a point $\displaystyle (0,50)$,

$\displaystyle y-50=-4(x-0)$

$\displaystyle y=-4x+50$

To solve this, all we need to do is set the equations equal to each other.

$\displaystyle -4x+50=5x$

Now solve for $\displaystyle x$

$\displaystyle 50=9x$

$\displaystyle x=\frac{50}{9}$

This is an exponential growth problem, so let's recall the equation for exponential growth.

$\displaystyle P(t)=P(1+r)^t$, where $\displaystyle P$ is the starting amount of ants, $\displaystyle r$ is the growth rate, and $\displaystyle t$ is the time in years.

First step is to convert $\displaystyle 20\%$ into a decimal. $\displaystyle 20\%=0.20$

$\displaystyle P(t)=5000(1+0.20)^t$

$\displaystyle P(t)=5000(1.20)^t$

$\displaystyle P(6)=5000(1.20)^6=14930$

So in 6 years, Amanda will have $\displaystyle 14930$ ants.

The first step is to plug in all the values into the equation.

$\displaystyle 1.34\times10^{-12} N=\frac{6.67408\times10^{-11}m^3kg^{-1}s^{-2}\cdot 10kg\cdot 20kg}{r^2}$

Now we will solve for $\displaystyle r$.

$\displaystyle r^2=-\frac{6.67408\times10^{-11}m^3kg^{-1}s^{-2}\cdot 10kg\cdot 20kg}{1.34\times10^{-12} N}$

$\displaystyle r^2=\frac{1.334816\times10^{-8}m^3kg^{-1}s^{-2} kg^2}{1.34\times10^{-12} N}$

$\displaystyle r^2=9961.313\:m^2$

Take the square root on each side

$\displaystyle r=\sqrt{9961.313\:m^2}=99.80\ m$

$\displaystyle g(x) = 3x+2$

$\displaystyle g(-2) = 3(-2)+2=-6+2=-4$