First recall that

\(\displaystyle ln(e^{-2})=-2\)

To find the tangent line of \(\displaystyle y=3e^{-x}\) at \(\displaystyle x=-2\), we first determine the slope of \(\displaystyle y=3e^{-x}\). To do so, we must find its derivative. 

Recall that derivatives of exponential functions involving \(\displaystyle e\) are given as:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}Ae^{g(x)}= A*g'(x)e^{g(x)}\), where \(\displaystyle A\) is a constant and \(\displaystyle g(x)\) is any function of \(\displaystyle x\)

In our case, \(\displaystyle A=3, g(x)=-x, g'(x)=-1\),. 

\(\displaystyle \frac{d}{dx}3e^{-x}= -3e^{-x}\)

At \(\displaystyle x=-2\),

 \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=-3e^{-(-2)}}= -3e^2=m\), where \(\displaystyle m\) is the slope of the tangent line.

To use point-slope form, we need to know the value of the original function at \(\displaystyle x=-2\), 

\(\displaystyle y=3e^{-x}\)

\(\displaystyle y(-2)=3e^{-(-2)}=3e^2\)

Therefore,

\(\displaystyle (y_{appx}-3e^2)=-3e^2(x+2)\)

At \(\displaystyle x=ln(e^3)=3\), 

\(\displaystyle (y_{appx}-3e^2)=-3e^2(3+2)\)

\(\displaystyle y_{appx}=-15e^2+3e^2=-12e^2\)

The population size \(\displaystyle P(t)\) after some time \(\displaystyle t\) is given by:

\(\displaystyle P(t)= P_ie^k^t\)

where \(\displaystyle P_i\) is the initial population.

At the start, there were 30 blood cells.

\(\displaystyle P_i=P(0)=30\)

Substitute this value into the given formula.

\(\displaystyle P(t)= 30e^k^t\)

After 2 hours, 45 blood cells were present.  Write this in mathematical form.

\(\displaystyle P(2)=45\)

Substitute this into \(\displaystyle P(t)= 30e^k^t\), and solve for \(\displaystyle k\).

\(\displaystyle 45= 30e^k^(^2^)\)

\(\displaystyle \frac{3}{2}= e^2^k\)

\(\displaystyle ln\left(\frac{3}{2}\right)=ln(e^2^k)\)

\(\displaystyle 2k=ln\left(\frac{3}{2}\right)\)

\(\displaystyle k=\frac{1}{2}ln\left(\frac{3}{2}\right)\)

Direct constant of proportionality for any given function y, between any x values, is given by

\(\displaystyle r=\frac{\Delta y}{\Delta x}\), where \(\displaystyle r\) is the direction constant of proportionality

In the case of a linear function 

\(\displaystyle \frac{\Delta y}{\Delta x}\) is the same thing as the slope. 

Therefore, the constant of proportionality is \(\displaystyle m\)

To determine the direct constant of proportionality, we determine the rate of change from \(\displaystyle x=1\) and \(\displaystyle x=2.5\) for \(\displaystyle x^2\).

Rate of change is determined by

\(\displaystyle r=\frac{\Delta y}{\Delta x}\).

In our case, \(\displaystyle y=x^2\) between \(\displaystyle x=1\) and \(\displaystyle x=2.5\), the rate of change is

\(\displaystyle r=\frac{2.5^2-1}{2.5-1}=\frac{5.25}{1.5}=3.5\).

Direct constant of proportionality \(\displaystyle r\) is given by

\(\displaystyle r=\frac{\Delta y}{\Delta x}\).

Since \(\displaystyle y=ln(3x)\) and \(\displaystyle \Delta x= 4-2=2\)

\(\displaystyle r= \frac{\Delta y}{\Delta x}=\frac{ln(3*4)-ln(3*2)}{2}= \frac{ln(2)}{2}\)

The equation for population growth is given by \(\displaystyle P=Ae^{kt}\). \(\displaystyle P\) is the population, \(\displaystyle A\) is the intial value, \(\displaystyle t\) is time, and \(\displaystyle k\) is the growth constant. We can plug in the values we know at time \(\displaystyle t=0\) and solve for \(\displaystyle A\) .

\(\displaystyle 100=Ae^{0k}=A\)

Now that we solved for \(\displaystyle A\), we can plug in what we know for time \(\displaystyle t=30\) and solve for \(\displaystyle k\).

\(\displaystyle 1000=100e^{30k}\)

\(\displaystyle e^{30k}=10\)

\(\displaystyle 30k=\ln(10)\)

\(\displaystyle k=\frac{\ln(10)}{30}\)

The equation for population growth is given by \(\displaystyle P=Ae^{kt}\). P is the population, \(\displaystyle A\) is the intial value, \(\displaystyle t\) is time, and \(\displaystyle k\) is the growth constant. We can plug in the values we know at time \(\displaystyle t=0\) and solve for \(\displaystyle A\) .

\(\displaystyle 50=Ae^{0k}=A\)

Now that we have solved for \(\displaystyle A\) we can solve for \(\displaystyle k\) at \(\displaystyle t=7\)

\(\displaystyle 200=50e^{7k}\)

\(\displaystyle e^{7k}=4\)

\(\displaystyle 7k=\ln(4)\)

\(\displaystyle k=\frac{\ln(4)}{7}\)

The equation for population growth is given by \(\displaystyle P=Ae^{kt}\). \(\displaystyle P\) is the population, \(\displaystyle A\) is the intial value, \(\displaystyle t\) is time, and \(\displaystyle k\) is the growth constant. We can plug in the values we know at time \(\displaystyle t=0\) and solve for \(\displaystyle A\).

\(\displaystyle 200=Ae^{0k}=A\)

Now that we have \(\displaystyle A\) we can solve for \(\displaystyle k\) at \(\displaystyle t=6\).

\(\displaystyle 1600=200e^{6k}\)

\(\displaystyle e^{6k}=8\)

\(\displaystyle 6k=\ln(8)\)

\(\displaystyle k=\frac{\ln(8)}{6}\)

Direct constant of proportionality \(\displaystyle k\) is given by 

\(\displaystyle k=\frac{\Delta y}{\Delta x}\), where \(\displaystyle \Delta y\) is the change in the \(\displaystyle y\) position and \(\displaystyle \Delta x\) is the change in the \(\displaystyle x\) position. 

Since \(\displaystyle y=3sec(x)\), and we're going from \(\displaystyle x=0\) to \(\displaystyle x=\frac{\pi}{3}\)

\(\displaystyle \Delta y=3sec(\frac{\pi}{3})-3sec(0)=6-3=3\)

\(\displaystyle \Delta x=\frac{\pi}{3}-0=\frac{\pi}{3}\)

\(\displaystyle k=\frac{3}{\frac{\pi}{3}}=\frac{9}{\pi}\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population decreased by 7 percent between 2009 and 2011, we can solve for this constant of proportionality:

\(\displaystyle (1-0.07)y_0=y_0e^{k(2011-2009)}\)

\(\displaystyle 0.93=e^{2k}\)

\(\displaystyle 2k=ln(0.93)\)

\(\displaystyle k=\frac{ln(0.93)}{2}=-0.036\)