The capital formation of the investment function over a given period is the
accumulated capital for the period.
- (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.
- (b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.
Reasons:
(a) The given investment function is presented as follows;

(a) The capital formation is given as follows;

From the end of the second year to the end of the fifth year, we have;
The end of the second year can be taken as the beginning of the third year.
Therefore, for the three years; Year 3, year 4, and year 5, we have;

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87
(b) When the capital stock exceeds $100,000, we have;
![\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Cmathbf%7B%5Cleft%5B1000%20%5Ccdot%20%20e%5E%7B0.1%20%5Ccdot%20t%7D%7D%20%2B%20C%20%5Cright%5D%5Et_0%7D%20%3D%20100%2C000)
Which gives;




The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.
Learn more investment function here:
brainly.com/question/25300925
Slope is the picture.
Point-slope form is y-2=-5/6*(x+1)
The y=mx+b is y=-5/6x+7/6
i hope its not confusing lol
First 3 terms are a^2 + n a^(n-)1 b + n(n-1)/2 * a^(n-2) b^2
So q^2 / pr = (n^2 * a^(2n-2) * b^2 ) / (1/2 * a^n * (n(n-1) * a^(n-2) * b^2 )
= n^2 * a^2n-2 * b^2
-----------------------------------
1/2 n(n-1) * a^(2n-2) * b^2
= 2n / n - 1 as required
given p = 4, q=32 and r = 96:-
32^2 / 4*96 = 2n / n-1
2n / n-1 = 8/3
6n = 8n - 8
2n = 8
n = 4 answer
Answer:
the answer is 45
Step-by-step explanation:
all you have to do is multiply 30*2*3/4
to get 45
Answer:
I think the answer is B not 100% sure
Step-by-step explanation:
Sorry if its wrong which it probably is