HELP PLEASE!!!

The bear population in Canada was 380,000 in the year 2015, and environmentalists think that the population is increasing at a rate of 2.5% per year.
Consider the function that represents the exponential growth of the bear population in Canada.

Part A: Defines a variable for the function and state what the variable represents.

Part B: What is a reasonable domain for the situation?

Part C: Write the function that represents the exponential growth of the bear population.

Part D: What will the bear population estimated to be in 2050?

The inverse Laplace transform of 1 / (3s² + 5s + 1) is f(t) = 1/2 × [tex]e^{(-t)[/tex]- 1/2 × [tex]e^{(-t/3)[/tex].

To find the inverse Laplace transform of the function F(s) = 1 / (3s² + 5s + 1), we can use partial fraction decomposition and reference tables for Laplace transforms.

Step 1: Factorize the denominator

Factorize the denominator of the function 3s² + 5s + 1 to find its roots:

3s² + 5s + 1 = (s + 1)(3s + 1)

Step 2: Write the partial fraction decomposition

Write the function F(s) as a sum of partial fractions:

F(s) = A / (s + 1) + B / (3s + 1)

Step 3: Determine the values of A and B

To find the values of A and B, we can multiply both sides of the equation by the common denominator and equate the numerators:

1 = A(3s + 1) + B(s + 1)

Expand the right side and collect like terms:

1 = (3A + B)s + (A + B)

By equating the coefficients of s and the constant terms on both sides, we get a system of equations:

3A + B = 0

A + B = 1

Solving this system of equations, we find A = 1/2 and B = -1/2.

Step 4: Write the inverse Laplace transform

Using the partial fraction decomposition, we can now write the inverse Laplace transform:

F(s) = 1/2 × (1 / (s + 1)) - 1/2 × (1 / (3s + 1))

Referring to Laplace transform tables, we find that the inverse Laplace transform of 1 / (s + a) is [tex]e^{(at)[/tex], and the inverse Laplace transform of 1 / (s - a) is [tex]e^{(at)[/tex]. Therefore, applying these results, we have:

f(t) = 1/2 × [tex]e^{(-t)[/tex] - 1/2 × [tex]e^{(-t/3)[/tex]

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Answer:

2. there is no E it would be AB=CD

There are 168 ways for 6 adults and 3 children to stand together in a line with no adjacent children.

we can treat the 6 adults as distinct entities and place them in the line first. There are 6! (factorial) ways to arrange the adults. Now, we have 7 spaces between the adults (including the ends) where the children can be placed. We need to choose 3 out of these 7 spaces for the children, which can be done in 7C3 ways (7 choose 3). Therefore, the total number of arrangements is 6! × 7C3 = 720 × 35 = 16800. However, we need to exclude the cases where two or more children are adjacent. By considering the possible positions of two adjacent children, there are 48 such cases. Subtracting these cases from the total gives us 16800 - 48 = 16752. Therefore, there are 16752 ways for 6 adults and 3 children to stand together in a line with no adjacent children.

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Step-by-step explanation:

V≈75.4

A≈100.53

The function f(z) = u(x, y) + iv(x, y) simplifies to f(z) = -8x³y + 8xy³.

The function f(z) corresponding to the given function u(x, y) is f(z) = -8x³y + 8xy³.

To prove that the function u(x, y) = -8x³y + 8xy³ is harmonic, we need to show that it satisfies the Laplace's equation:

∇²u = ∂²u/∂x² + ∂²u/∂y² = 0

Let's calculate the second partial derivatives of u with respect to x and y:

∂u/∂x = -24x²y + 8y³

∂²u/∂x² = -48xy

∂u/∂y = -8x^3 + 24xy²

∂²u/∂y²= 48xy

Now, substitute these values into the Laplace's equation:

∂²u/∂x² + ∂²u/∂y² = -48xy + 48xy = 0

Since the Laplace's equation is satisfied, the function u(x, y) = -8x³y + 8xy³ is harmonic.

b) To find the conjugate harmonic function v, we need to find a function v(x, y) such that f(z) = u(x, y) + iv(x, y) is analytic, where z = x + iy.

Since the given function u(x, y) = -8x³y + 8xy³ is a real-valued function, the imaginary part of the conjugate harmonic function v(x, y) must be zero. Therefore, v(x, y) = 0, and the conjugate harmonic function is purely real.

Hence, the function f(z) = u(x, y) + iv(x, y) simplifies to f(z) = -8x³y + 8xy³.

Therefore, the function f(z) corresponding to the given function u(x, y) is f(z) = -8x³y + 8xy³.

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Step-by-step explanation:

no this is a test I cant hlep i will tell ms.jules you are searching for the answers

Answer:

V≈7853.98

Step-by-step explanation:

V=πr2h

r=10

h=25

Solution

V=πr2h=π·102·25≈7853.98163

Answer:

3

Step-by-step explanation:

12/4=3

The net financing need considering the three accounts is approximately $185,753.42.

To determine the average investment in accounts receivable, inventories, and accounts payable, we need to calculate the Inventory Period, Receivables Period, and Payment Period.

Inventory Period:

The Inventory Period is the average number of days it takes for finished goods to be sold. In this case, the average production process time is given as 40 days, and finished goods are kept on hand for an average of 15 days before they are sold. Therefore, the Inventory Period is the sum of these two periods:

Inventory Period = Production Process Time + Days Kept on Hand

Inventory Period = 40 days + 15 days

Inventory Period = 55 days

Receivables Period:

The Receivables Period is the average number of days it takes for accounts receivable to be collected. It is given that accounts receivable are outstanding for an average of 40 days. Therefore, the Receivables Period is 40 days.

Payment Period:

The Payment Period is the number of days the firm receives credit on its purchases from suppliers. It is given that the firm receives 40 days of credit. Therefore, the Payment Period is 40 days.

Now, we can calculate the average investment in accounts receivable, inventories, and accounts payable.

Average Investment in Accounts Receivable:

Average Investment in Accounts Receivable = (Net Sales / 365) * Receivables Period

Average Investment in Accounts Receivable = ($1,200,000 / 365) * 40

Average Investment in Accounts Receivable ≈ $131,506.85

Average Investment in Inventories:

Average Investment in Inventories = (Cost of Goods Sold / 365) * Inventory Period

Average Investment in Inventories = ($900,000 / 365) * 55

Average Investment in Inventories ≈ $142,191.78

Average Investment in Accounts Payable:

Average Investment in Accounts Payable = (Cost of Goods Sold / 365) * Payment Period

Average Investment in Accounts Payable = ($900,000 / 365) * 40

Average Investment in Accounts Payable ≈ $87,945.21

Finally, to calculate the net financing need, we subtract the average investment in accounts payable from the sum of the average investment in accounts receivable and inventories:

Net Financing Need = Average Investment in Accounts Receivable + Average Investment in Inventories - Average Investment in Accounts Payable

Net Financing Need = $131,506.85 + $142,191.78 - $87,945.21

Net Financing Need ≈ $185,753.42

Therefore, the net financing need considering the three accounts is approximately $185,753.42.

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Answer:

The width of the actual fence is 90ft

Step-by-step explanation:

If 2in = 10ft, you want to find the scale factor. To do this, you need to divide 10/2, so for every 5ft, it is 1in. Now, you would take how many inches you have and multiply it by 5. (in our case, 18in)

18 x 5 = 90.

So the actual fence length is 90ft.

Step 1: To tell the null and alternative hypotheses for the test:

Step 2: The test statistic for this scenario is the z-score, calculated as:

Step 3 involves comparing the chosen significance level (1% in this instance) with the test statistic to decide whether to reject or accept the null hypothesis. In case the test statistic falls within the critical region, the null hypothesis is rejected, but if it doesn't, the null hypothesis is not rejected.

We can also determine the p-value related to the test statistic and assess its compatibility with the selected level of significance. If the significance level exceeds the p-value, we do not reject the null hypothesis, but if the p-value is lower, then we reject the null hypothesis.

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Answer:

This example is not a random example due to the fact that it is only questioning the eighth-graders. Random sample means that the sample is chosen simply randomly and everyone has an equal opportunity to be apart of the sampling.

Step-by-step explanation:

.....um...i think u can't do this

Answer:

u should probably contact support or the creator of this site

Step-by-step explanation:

Answer:

Here is the image you need to replicated

(っ◔◡◔)っ ♥ Hope It Helps ♥

Answer:

42 units²

Step-by-step explanation:

Answer:

It has been answered for you

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Answer:

[tex]t=382[/tex]

Step-by-step explanation:

[tex]\mathrm{Subtract\:}38\mathrm{\:from\:both\:sides}[/tex]

[tex]t+38-38=420-38[/tex]

[tex]=382[/tex]

Answer:

[tex] \displaystyle t = 382[/tex]

Step-by-step explanation:

we are given a equation

we want to figure out t

to do so

recall that

therefore

if we substract 38 from both sides it won't affect the equality

so

[tex] \displaystyle t + 38 - 38 = 420 - 38 \\ \therefore \: t = 382[/tex]

let's verify it

[tex] \displaystyle382 + 38\stackrel{?}{=}420[/tex]

simplify addition:

[tex] \displaystyle420\stackrel{ \checkmark}{=}420[/tex]

hence, verified!

Answer:

NLE Choppa noooooooooooooooooo

Step-by-step explanation:

THE BEST RAPPPERRR!!!!!!!!!

Answer:

He wasnt in jail? I dont think cuz he was makin new music

Step-by-step explanation:

The cοrrect answer is (c) Never true, i.e., false. "The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns is never less than the rank οf the augmented matrix" is nοt always true.

A linear equatiοn is an algebraic equatiοn that represents a straight line when graphed οn a Cartesian cοοrdinate plane. It is an equatiοn in which the highest pοwer οf the variable(s) is 1.

The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns represents the maximum number οf linearly independent equatiοns in the system. It gives us infοrmatiοn abοut the dimensiοn οf the sοlutiοn space.

On the οther hand, the rank οf the augmented matrix οf the system includes bοth the matrix οf cοefficients and the cοlumn vectοr οf cοnstants. It represents the maximum number οf linearly independent rοws in the augmented matrix.

In general, the rank οf the matrix οf cοefficients can be less than, equal tο, οr greater than the rank οf the augmented matrix. It depends οn the specific system οf equatiοns.

Therefοre, the statement "The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns is never less than the rank οf the augmented matrix" is nοt always true.

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The exponential function passing through the points (-3, 1/2) and (3, 32) can be represented by the equation [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].

To find the exponential function passing through the given points, we can start by assuming the general form of an exponential function, [tex]f(x) = a * (b^x)[/tex], where a and b are constants to be determined. Plugging in the coordinates (-3, 1/2) into this equation gives us [tex]1/2 = a * (b^{(-3)})[/tex], and plugging in (3, 32) gives us [tex]32 = a * (b^3)[/tex].

Now, we can solve this system of equations to find the values of a and b. Taking the ratio of the two equations, we get [tex](1/2) / 32 = (a * (b^{(-3)})) / (a * (b^3))[/tex], which simplifies to [tex]1/64 = 1/b^6[/tex]. Solving for b, we find [tex]b = 2^{(1/6)[/tex].

Substituting this value back into either of the original equations, we can solve for a. Using the equation [tex]1/2 = a * (2^{(-3/6)})[/tex], we find [tex]a = 2^{(5/6)[/tex].

Therefore, the exponential function passing through the given points is [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].

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Step-by-step explanation:

division fractions that's how u will rewrite them

Answer:

16 cans

Step-by-step explanation:

Hey,

I have to admit, this problem is pretty complicated. But I got you :).

To begin, we have to represent the small and large boxes with two separate variables. So...

y = large

x = small

(It really doesn't matter what variable you use, I just used these)

Now, we know that Estelle fills 3 large boxes and 5 small boxes and had a total of 170 cans in total. We can represent this by writing...

3y + 5x = 170

Then, we know that the boy worker filled 4 large boxes and 4 small boxes and had a total of 184 cans. We can represent this by writing...

4y + 4x = 184

As you can see, this is a system of equations.

3y + 5x = 170

4y + 4x = 184

We want to know how many cans each small box can hold, so we have to find a common number for x since x represents the small box.

To do this, we have to multiply the first equation by 4 and the second by 3. Here's what I mean...

4 (3y + 5x = 170)

3 (4y + 4x = 184)

When we do this you get...

12y + 20x = 680

12y + 12x = 552

Notice how the y's are now both 12. We had to do that in order to get rid of y, they had to equal the same number. Now, subtract...

8x = 128

Divide by 8...

x = 16

That means that...

YOUR ANSWER: Each small box can hold up to 16 cans.

I hope this helps :)

Answer:

The angle is 58 degrees

Step-by-step explanation:

Given

See attachment for kite

Required

The angle at the tail end

Represent this angle with x.

From the attached kite, we have:

1 angle = 122

2 angles = right-angled

So, we have:

[tex]x + 122 +90+90 = 360[/tex] --- sum of angles in a kite

[tex]x + 302 = 360[/tex]

Solve for x

[tex]x =- 302 + 360[/tex]

[tex]x =58^\circ[/tex]

Answer:

The amount invested in ADB is ¢1363.[tex]\overline 3[/tex] =

The amount invested in Barclays is ¢3,427.[tex]\overline 3[/tex]

The amount invested in GCB is ¢1,713.[tex]\overline 3[/tex]

Step-by-step explanation:

The parameters of the investment Stonewall made are;

The amount in interest he receives from ADB, Barclays and GCB = ¢250

The amount of interest ADP pays = 2% per annum

The amount of interest Barclays pays = 4% per annum

The amount of interest GCB pays = 5% per annum

The amount invested in Barclays = The amount invested in ADB and GCB + ¢350

The amount invested in Barclays = 2 × The amount invested in GCB

a) Let 'x', represent the amount invested in ADB, 'y' represent the amount invested in Barclays, and 'z', represent the amount invested in GCB

We have;

y = x + z + 350

y = 2·z

0.02·x + 0.04·y + 0.05·z = 250

Therefore, we get the three linear equations as follows;

-x + y - z = 350...(1)

y - 2·z = 0...(2)

0.02·x + 0.04·y + 0.05·z = 250...(3)

Using Matrix inversion, we have;

[tex]\left[\begin{array}{ccc}-1&1&-1\\0&1&-2\\0.02&0.04&0.05\end{array}\right] \times \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}350&0&250\end{array}\right][/tex]

The transpose of the 3 by  3 matrix [tex]M^T[/tex] is given as follows;

[tex]M^T = \left[\begin{array}{ccc}-1&0&0.02\\1&1&0.04\\-1&-2&0.05\end{array}\right][/tex]

The Adjugate Matrix is given as follows;

[tex]Adj = \left[\begin{array}{ccc}0.13&-0.09&-1\\-0.04&-0.03&-2\\-0.02&0.06&-1\end{array}\right][/tex]

The inverse of the matrix = Adj/Det where, Det = -0.15, is therefore;

[tex]M^{-1} = \left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right][/tex]

We therefore, get the solution as follows;

[tex]\left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right]\times \left[\begin{array}{c}350&0&250\end{array}\right] = \left[\begin{array}{c}\dfrac{4,090}{3} \\&\dfrac{10,280}{3} \\ & \dfrac{5,140}{3} \end{array}\right][/tex]

[tex]\begin{array}{c}x = \dfrac{4,090}{3} \\&y = \dfrac{10,280}{3} \\ & z = \dfrac{5,140}{3} \end{array}[/tex]

The amount invested in ADB, x = ¢4,090/3 = ¢1363.[tex]\overline 3[/tex]

The amount invested in Barclays, y = ¢10,282/3 = ¢3,427.[tex]\overline 3[/tex]

The amount invested in GCB, z = ¢5,140/3 = ¢1,713.[tex]\overline 3[/tex]

a The net income is $3505000

b The number of common shares outstanding is 399,000

c Earnings per share is $8.79

d Book value of common stock is $46.62

e. Price-earnings ratio (P/E ratio) is 6.14

a. Net income available to common stockholders:

Net income = $3,595,000

Dividends to preferred stockholders = (Number of preferred shares * Par value * Dividend rate) = (10,000 * $100 * 0.09) = $90,000

Net income available to common stockholders = Net income - Dividends to preferred stockholders

= $3,595,000 - $90,000

= $3,505,000

b. Weighted average number of common shares outstanding:

Number of common shares outstanding = 399,000

c. Earnings per share: EPS = Net income available to common stockholders / Weighted average number of common shares outstanding

= $3,505,000 / 399,000

≈ $8.79

d. Book value of common stock:

Total stockholders' equity = $19,600,000

Preferred stock equity = Number of preferred shares * Par value = 10,000 * $100 = $1,000,000

Common stock equity = Total stockholders' equity - Preferred stock equity

= $19,600,000 - $1,000,000

= $18,600,000

Book value of common stock = Common stock equity / Number of common shares outstanding

= $18,600,000 / 399,000

≈ $46.62

e. Price-earnings ratio (P/E ratio):

Price-earnings ratio = Price per share / Earnings per share

= $54 / $8.79

≈ 6.14

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$7 per hour plus $9 off

$7x5=$35-$9 discount=$26

$26+$7=$33

6hours=$42-$9=$33

Joe could rent the boat for 6 hours

Equation system that corresponds to the graph:

1. y ≤ 3x 2. y > –2x – 1

To find the system of equations that corresponds to the provided graph, we must first analyze it and locate the regions that fulfil the specified requirements.

1. Begin by locating the darkened region underneath the line y 3x. This line has a slope of 3 and intersects the origin (0,0). Shade the area beneath the line.

2. After that, locate the darkened region above the line y > -2x - 1. The slope of this line is -2, while the y-intercept is -1. The area above the line should be shaded.

3. The solution space that meets both requirements is represented by the overlapping shaded region between the two lines. The common area is located below y 3x and above y > -2x - 1.

4. The equation system that corresponds to this common area is: - y 3x - y > -2x - 1

The space for the addition in the open area next to the present building is defined by these two formulae.

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