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3 years ago

20 pt what do you see

Mathematics

1 answer:

GrogVix [38]3 years ago

8 0

A watermarked picture by Shutterstock, known as a stock photo with the photo displaying "LOL" and comic like explosion.

I have no idea what you are asking about so I will take your question literally.

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What is the difference between -3 and 7? *

Murrr4er [49]

Answer:

Step-by-step explanation:

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3 years ago

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What is The equivalent fraction of 2/5

Nat2105 [25]

Answer:

4/10 is equivalent to 2/5

Step-by-step explanation:

because 4/10 is written as 0.4 in decimal form and so is 2/5, which means they are equivalent

7 0

3 years ago

The angle 118 degrees is also equal to

LekaFEV [45]

I don't understand your question

8 0

3 years ago

EDIT// answer below thank for the help -.-

HACTEHA [7]

Step-by-step explanation:

Move it it up 4 and then find your x intercepts.

Find the x intercepts.

$-x^2 + 4 = 0$

Move the 4 over to the right

$-x^2 = -4$

Divide out the -

$\frac{-x^2}{-} = \frac{-4}{-}$

$x^2 = 4$

Take the square root of each side

$\sqrt{x^2} = \pm \sqrt{4}$

$x = \pm \sqrt 2$

Our x intercepts are at (-2,0) (2,0)

8 0

4 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 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