A)
![\bf 1990-1910=80\leftarrow t \\\\\\ P(t)=8000(2)^{-\frac{t}{29}}\implies P(80)=8000(2)^{-\frac{80}{29}} \\\\\\ P(80)=8000\cdot \cfrac{1}{2^{\frac{80}{29}}}\implies P(80)=\cfrac{8000}{\sqrt[29]{2^{80}}}](https://tex.z-dn.net/?f=%5Cbf%201990-1910%3D80%5Cleftarrow%20t%0A%5C%5C%5C%5C%5C%5C%0AP%28t%29%3D8000%282%29%5E%7B-%5Cfrac%7Bt%7D%7B29%7D%7D%5Cimplies%20P%2880%29%3D8000%282%29%5E%7B-%5Cfrac%7B80%7D%7B29%7D%7D%0A%5C%5C%5C%5C%5C%5C%0AP%2880%29%3D8000%5Ccdot%20%5Ccfrac%7B1%7D%7B2%5E%7B%5Cfrac%7B80%7D%7B29%7D%7D%7D%5Cimplies%20P%2880%29%3D%5Ccfrac%7B8000%7D%7B%5Csqrt%5B29%5D%7B2%5E%7B80%7D%7D%7D)
and surely you know how much that is.
b)
since in 1910 t = 0, 174 years later from 1910, is 2084, so in 2084 they'll be 125 exactly, so the next year, 2085, will then be the first year they'd fall under that.
Answer:
We get the value of Principal amount i.e initial investment = $61640
Step-by-step explanation:
Interest rate r = 3.59% or 0.0359
Compounded quarterly n = 4
Future Amount A = 117,300
Time t = 18 years
We need to find initial investment i,e Principal Amount P
The formula used is: 
Putting values and finding P
So, We get the value of Principal amount i.e initial investment = $61640
The answer to this question is 375
Answer:
0.75 or 3/4
Step-by-step explanation:
The sum of 1/2 and 1/2 is 1. (1/2 + 1/2 = 1)
The product of 1/2 and 1/2 is (1*1)/(2*2) = 1/4.
If you subtract 1/4 from 1, you get 3/4.
Answer:
24
Step-by-step explanation:
The question is asking for the net area from x=3 to x=10.
It gives you the net area from x=3 to x=5 being -18.
It gives you the net area from x=5 to x=10, being 42.
Together those intervals make up the interval we want to find the net area for.
-18+42=42-18=24
