I need help with 4 5 and 6

Currently the stock is $62 and has been decreasing 3% each month. So, if we call [tex]x=0[/tex] the first month then:
[tex]at \ x=0 \ y=62 \\ \\ at \ x=1 \ y=62-62\times 3\%=60.14 \\ \\ at \ x=2 \ y=60.14-60.14\times 3\%=58.3358 \\ \\ at \ x=3 \ y=58.3358-58.3358\times 3\%=56.5857 \\ \\ at \ x=4 \ y=56.5857-56.5857\times 3\%=54.8881 \\ \\ at \ x=5 \ y=54.8881-54.8881\times 3\%=53.2414[/tex]
So the Table is:
[tex]\left[\begin{array}{cc}x & y\\0 & 62\\1 & 60.14\\2 & 58.3358\\3 & 56.5857\\4 & 54.8881\\5 & 53.2414\end{array}\right][/tex]
This is an exponential function because it has a constant ratio.
From our values, we can conclude that the equation is:
[tex]y=62(100%-3%)^{x} \\ \\ \boxed{y=62(0.97)^{x}}[/tex]
In five months, the stock will be $53.2414
We are given two points:
[tex]P_{1}(0,5) \\ \\ P_{2}(3,135)[/tex]
We know that the exponential function is given by the form:
[tex]y=ab^{x}[/tex]
So:
[tex]\bullet \ If \ x=0 \ then \ y=5 \\ \\ 5=ab^{0} \therefore a=5 \\ \\ \\ \bullet \ If \ x=3 \ y=135 \\ \\ 135=ab^{3} \therefore ab^{3}=135 \\ \\ \\ Finding \ b: \\ \\ \frac{ab^{3}}{a}=\frac{135}{5} \\ \\ b^3=27 \therefore b=\sqrt[3]{27} \therefore \boxed{b=3}[/tex]
Then:
[tex]\boxed{y=5(3)^{x}}[/tex]
Here we are given two points:
[tex]A(4,9) \ and \ B(-2,3)[/tex]
So, the slope can be found as:
[tex]m=\frac{Change \ in \ y}{Change \ in \ x} \\ \\ m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ m=\frac{3-9}{-2-4} \\ \\ \boxed{m=1}[/tex]
Using the distance formula we can find the length:
[tex]d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} \\ \\ d=\sqrt{(-2-4)^2+(3-9)^2} \\ \\ d=\sqrt{36+36} \\ \\ d=\sqrt{72} \\ \\ \boxed{d=6\sqrt{2}}[/tex]
To find the midpoint, we have to use the midpoint formula:
[tex]Midpoint=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}) \\ \\ Midpoint=(\frac{4-2}{2},\frac{9+3}{2}) \\ \\ \boxed{Midpoint=(1,6)}[/tex]