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3 years ago

5

What is -1 as a decimal

2 answers:

vovikov84 [41]3 years ago

8 0

It is -1.0 cause im a smart chicky nuggy

hammer [34]3 years ago

5 0

Answer:

-1.0

Step-by-step explanation:

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A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time

user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

$a(t) = 24t + 2$

Velocity:

$v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K$

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

$v(t) = 12t^{2} + 2t + 13$

Position:

$s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K$

Since s(0) = 3

$s(t) = 4t^{3} + t^{2} + 13t + 3$

What is its position at time t=5?

This is s(5).

$s(t) = 4t^{3} + t^{2} + 13t + 3$

$s(5) = 4*5^{3} + 5^{2} + 13*5 + 3$

$s(5) = 593$

It's position at time t = 5 is 593.

3 0

3 years ago

A model of a building is 18 inches tall. If the building is really 678 feet tall, how tall is a window that is 1/9 in on the mod

ipn [44]

Answer:

50.2in or 4.2ft

Step-by-step explanation:

This is a problem about proportions so I generally set up my problems like below. I just set some variable here to make it easier to refer back to the measurements for each. You don't have to do this, I am just showing it for clarity. I will convert everything to inches too. We need to find the real window height.

Mb = model building = 18in

Mw= model window = $\frac{1}{9}$ in

Rb = real building = 678ft = 8136in

Rw = real window = ?

$\frac{Mw}{Mb}$ = $\frac{Rw}{Rb}$

$\frac{1/9}{18}$ = $\frac{Rw}{8136}$

$\frac{1}{9*18}$ = $\frac{Rw}{8136}$

$\frac{1}{162}$ = $\frac{Rw}{8136}$

$\frac{1*8136}{162}$ = $\frac{8136*Rw}{8136}$

$\frac{8136}{162}$ = Rw

$\frac{452}{9}$ = Rw

Rw ≈ 50.2in ≈ 4.2ft

5 0

2 years ago

How do you simplify 12 + 15 ÷ 3 × 6 – 4 step by step?

tatuchka [14]

Seperate the questions 12+15=25 ÷ by 3×6= 18-4= the answer

6 0

3 years ago

PLEASE HELP. EIGHTH GRADE MATH.

vampirchik [111]

Answer:

Cant see the picture can you zoom in?

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Step-by-step explanation:

5 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

3 years ago