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2 years ago

3 * n + 2 = ? 2 * n + 7 = ?

Mathematics

1 answer:

natali 33 [55]2 years ago

8 0

Answer:

1.3n+2

2.2n+7

Step-by-step explanation:

for number 1.you are multiplying 3 and n which gives you 3n but 2 you cant because it dosent have a variable to times with it says plus instead of times.

and for number 2 it the same thing 2 times n equals 2n + 7 is 2n + 7 .

If this doesnt help then im sorry but this is how i know and yeah have a wonderful day.

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6. If the net investment function is given by

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The capital formation of the investment function over a given period is the

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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.

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