(–6)² – (–7)² I need a step by step process.

The expressions in which the numbers, or variables, or both, are connected by operational signs (+, - etc.) are called algebraic expressions. For example 5, 4x, a+b, x−y.

4 0

3 years ago

Read 2 more answers

6. If the net investment function is given by

Pachacha [2.7K]

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 298.87.

(b) The number of years before the capital stock exceeds $100,000 is approximately 46.15 years.

Reasons:

(a) The given investment function is presented as follows;

$I(t) = 100 \cdot e^{0.1 \cdot t}$

(a) The capital formation is given as follows;

$\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot e^{0.1 \cdot t}} + C$

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;

$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

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;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

$\displaystyle \mathbf{1000 \cdot e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000$

$\displaystyle e^{0.1 \cdot t}} = 101$

$\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15$

The number of years before the capital stock exceeds $100,000 ≈ 46.15 years.

Learn more investment function here:

brainly.com/question/25300925

6 0

3 years ago