Slope is

if we plug in your coordinates we get:

Therefore, the slope of (0,1) and (2,7) is 3.
Answer:
t=18 years
Step-by-step explanation:
A year has two semesters, then
n = 2v(t)=p(1+r/2)2t
3875.79 = 1900∗(1+(0.04/2))^2t
2.0398895 = (1+0.042)^2t
Apply natural logarithm on both sides
ln(2.0398895) = ln[(1+0.042)^2t]
Then simplify,
0.712896 = 2t∗ln(1.02)
t = 0.712896 / (2∗ln(1.02))
t=18 years
Remember:
-When multiplying numbers with decimals just forget about the decimals for a second and pretend you are multiplying a 3 digit number times a 2 digit number.
-one digit at a time. Multiply going from right to left.
First:
-Multiply the 3 with the top numbers.
-Pretend you are seeing this:
5.72 (3)
- 2*3=6. 7*3=21. (carry the 2). 5*3 +2=17. (leave the 1).
Second:
-write below and move a digit to the left.
-pretend you are seeing this: 5.72(6).
- 2*6 =12. (carry the 1). 7*6 +1=43. (carry the 4) 5*6+4=34. (leave the 3).
Third: since in the problem the decimal is placed in the hundredths only once, place the decimal in the hundredths as your solution.
Your solution should be 360.36
Here is your answer; ITS CALLED GOOGLE.
a = amount invested at 7%
b = amount invested at 9%
we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000. We can also say that

since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.
![\begin{cases} a + b = 36000\\\\ 0.07a+0.09b=2920 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{a + b = 36000\implies \underline{b = 36000-a}}~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.07a~~ + ~~0.09(\underline{36000-a})~~ = ~~2920} \\\\\\ 0.07a+3240-0.09a=2920\implies 3240-0.02a=2920\implies -0.02a=-320 \\\\\\ a=\cfrac{-320}{-0.02}\implies \boxed{a=16000}~\hfill \boxed{\stackrel{36000~~ - ~~16000}{20000=b}}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20a%20%2B%20b%20%3D%2036000%5C%5C%5C%5C%200.07a%2B0.09b%3D2920%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20the%201st%20equation%7D%7D%7Ba%20%2B%20b%20%3D%2036000%5Cimplies%20%5Cunderline%7Bb%20%3D%2036000-a%7D%7D~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20on%20the%202nd%20equation%7D%7D%7B0.07a~~%20%2B%20~~0.09%28%5Cunderline%7B36000-a%7D%29~~%20%3D%20~~2920%7D%20%5C%5C%5C%5C%5C%5C%200.07a%2B3240-0.09a%3D2920%5Cimplies%203240-0.02a%3D2920%5Cimplies%20-0.02a%3D-320%20%5C%5C%5C%5C%5C%5C%20a%3D%5Ccfrac%7B-320%7D%7B-0.02%7D%5Cimplies%20%5Cboxed%7Ba%3D16000%7D~%5Chfill%20%5Cboxed%7B%5Cstackrel%7B36000~~%20-%20~~16000%7D%7B20000%3Db%7D%7D)