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
The missing side is 17 and rounded it is 20
Answer:
y = (-1/3)(x + 10)
Step-by-step explanation:
The slope of the new (perpendicular) line is the negative reciprocal of the slope of the given line, which appears to be 3. Thus, the perpendicular line has the slope -1/3.
Using the slope-intercept form y = mx + b, and substituting the givens, we obtain:
y = mx + b => -6 = (-1/3)(8) + b, or
-6 = -8/3 + b. We must solve for the y-intercept, b:
Multiplying all three terms by 3 removes the fraction:
-18 = -8 + 3b. Thus, -10 = 3b, and so b must be -10/3.
The desired equation is
y = (-1/3)x - 10/3, or
y = (-1/3)(x + 10)