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miss Akunina [59]
3 years ago
5

6. The first six terms in a quadratic pattern are

Mathematics
1 answer:
dexar [7]3 years ago
5 0
Since 83-59=24, 83 is added by 28 so your answer would be 111.
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Multiply. Express your answer in simplest form.
kodGreya [7K]

Step-by-step explanation:

1.) First, you need to take the whole numbers (9 and 1) and set them aside.

2.) Multiply the numerators from each fraction by one another. The numerators are the numbers on the top of the fraction. (Whatever you get on that will be the numerator of the answer!)

3.) Multiply the denominators by each other as well. The denominators are the numbers on the bottom. This will be the denominator of your answer.

4.) Now, you need to multiply your whole numbers (9 and 1) together.

5.) Then, simplify or reduce your answer. (this can't be simplified, as it is already in its simplest form)

Your answer will be 9 1/66.

Your work should look like this in the end:

4 0
3 years ago
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 <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;

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 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

6 0
2 years ago
Which statement about sound is true because it is a mechanical wave?
maw [93]

Answer:

It must travel through a medium

Step-by-step explanation: Because I said so >:(

4 0
3 years ago
Read 2 more answers
Find the diffrence 3/4 - 1/3
gtnhenbr [62]

Answer: 0.4

Step-by-step explanation: 3/4 - 1/3 = 5/12

converted to decimal: 5 ÷ 12 = 0.4

3 0
3 years ago
Read 2 more answers
State how many imaginary and real zeros the function has.
sammy [17]

Answer:

Step-by-step explanation:

If you'd graph this function on a graphing calculator or graphing utility, you'd see quickly that the graph never touches or crosses the x-axis.  This tells us immediately that there are no real roots; all four roots are complex or imaginary.

6 0
3 years ago
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