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3 years ago

The height of the tide of the ocean is measured along a cliff. The water

Mathematics

1 answer:

Whitepunk [10]3 years ago

3 0

Answer:

its 500 cuz u have to dived it in to a fraction

Step-by-step explanation:

thats how its done

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gregori [183]

A. Zero points of intersection

the lines are parallel

7 0

3 years ago

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Given the point (3,4) the slope of 6, find y, when x=27

vitfil [10]

Answer:

y = 148

Step-by-step explanation:

First, we use the point-slope form of the equation of a line to find the equation of this line.

y - y1 = m(x - x1)

y - 4 = 6(x - 3)

y - 4 = 6x - 18

y = 6x - 14

Now we let x = 27 and find y.

y = 6(27) - 14

y = 162 - 14

y = 148

6 0

3 years ago

An equation for a line parallel to the line with equation y= 10x + 17

azamat

To be parallel with that line the equation has to have the same slope as the equation given, like y = 10x + 3.

3 0

3 years ago

Help me on this :( pls

Doss [256]

X = 12 dhucdjvdsiksdjfbhfud gimme brainliest

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

2 years ago