Simplifying
7 + -8n = 1
Solving
7 + -8n = 1
Solving for variable 'n'.
Move all terms containing n to the left, all other terms to the right.
Add '-7' to each side of the equation.
7 + -7 + -8n = 1 + -7
Combine like terms: 7 + -7 = 0
0 + -8n = 1 + -7
-8n = 1 + -7
Combine like terms: 1 + -7 = -6
-8n = -6
Divide each side by '-8'.
n = 0.75
Simplifying
n = 0.75
Answer:
Step-by-step explanation:
For Compound Interest with continuous deposit,
We have that:
![\frac{dS}{dt}=rS+k](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdt%7D%3DrS%2Bk)
![\frac{dS}{dt}-rS=k](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdt%7D-rS%3Dk)
Using Integrating factor:
=![e^{-rt}](https://tex.z-dn.net/?f=e%5E%7B-rt%7D)
![\frac{dS}{dt}e^{-rt}=ke^{-rt}](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdt%7De%5E%7B-rt%7D%3Dke%5E%7B-rt%7D)
Taking Integrals from 0 to t
![Se^{-rt}=\frac{k}{-r}e^{-rt}+C\\S=\frac{k}{-r}+Ce^{rt}\\When t=0\\C=S_{0}+\frac{k}{r}](https://tex.z-dn.net/?f=Se%5E%7B-rt%7D%3D%5Cfrac%7Bk%7D%7B-r%7De%5E%7B-rt%7D%2BC%5C%5CS%3D%5Cfrac%7Bk%7D%7B-r%7D%2BCe%5E%7Brt%7D%5C%5CWhen%20t%3D0%5C%5CC%3DS_%7B0%7D%2B%5Cfrac%7Bk%7D%7Br%7D)
Now substituting back C
We have:
![Se^{-rt}=\frac{k}{-r}e^{-rt}+S_{0}+\frac{k}{r}\\S(t)=S_{0}e^{rt}+\frac{k}{r}-\frac{k}{r}e^{rt}\\S(t)=S_{0}e^{rt}+\frac{k}{r}(1-e^{rt})](https://tex.z-dn.net/?f=Se%5E%7B-rt%7D%3D%5Cfrac%7Bk%7D%7B-r%7De%5E%7B-rt%7D%2BS_%7B0%7D%2B%5Cfrac%7Bk%7D%7Br%7D%5C%5CS%28t%29%3DS_%7B0%7De%5E%7Brt%7D%2B%5Cfrac%7Bk%7D%7Br%7D-%5Cfrac%7Bk%7D%7Br%7De%5E%7Brt%7D%5C%5CS%28t%29%3DS_%7B0%7De%5E%7Brt%7D%2B%5Cfrac%7Bk%7D%7Br%7D%281-e%5E%7Brt%7D%29)
![Se^{-rt}=\frac{k}{-r}e^{-rt}+S_{0}+\frac{k}{r}\\S(t)=S_{0}e^{rt}+\frac{k}{r}-\frac{k}{r}e^{rt}\\S(t)=S_{0}e^{rt}+\frac{k}{r}(1-e^{rt})](https://tex.z-dn.net/?f=Se%5E%7B-rt%7D%3D%5Cfrac%7Bk%7D%7B-r%7De%5E%7B-rt%7D%2BS_%7B0%7D%2B%5Cfrac%7Bk%7D%7Br%7D%5C%5CS%28t%29%3DS_%7B0%7De%5E%7Brt%7D%2B%5Cfrac%7Bk%7D%7Br%7D-%5Cfrac%7Bk%7D%7Br%7De%5E%7Brt%7D%5C%5CS%28t%29%3DS_%7B0%7De%5E%7Brt%7D%2B%5Cfrac%7Bk%7D%7Br%7D%281-e%5E%7Brt%7D%29)
If initial deposit =$750, Continuous Deposit = 750, t= 4X 25 years=100
![S(t)=S_{0}e^{rt}+\frac{k}{r}(1-e^{rt})\\120000=750e^{100r}+\frac{750}{r}(1-e^{100r})\\160=e^{100r}+\frac{750}{r}-\frac{750}{r}e^{100r}\\](https://tex.z-dn.net/?f=S%28t%29%3DS_%7B0%7De%5E%7Brt%7D%2B%5Cfrac%7Bk%7D%7Br%7D%281-e%5E%7Brt%7D%29%5C%5C120000%3D750e%5E%7B100r%7D%2B%5Cfrac%7B750%7D%7Br%7D%281-e%5E%7B100r%7D%29%5C%5C160%3De%5E%7B100r%7D%2B%5Cfrac%7B750%7D%7Br%7D-%5Cfrac%7B750%7D%7Br%7De%5E%7B100r%7D%5C%5C)
Answer:
If a graphical plot is made with sufficient resolution linear equations may be solved by observing the solution point(s) on the graph coordinates.
In this case, the lines
y
=
x
+
2 and y
=
x
2
−
2
should be plotted.
They will intersect at the point (-8, -6) when the resolution of the graph is properly adjusted.
Step-by-step explanation: