60% of what number is 13.5?

Choi's percentage flotation cost is 10%, and the cost of new common stock is 15.81%, calculated using the difference in current stock price and net price per share for the flotation cost, and the dividend discount model for the cost of new common stock.

Calculating Choi's Flotation Cost and Cost of New Common Stock

The percentage flotation cost of Choi Company's new stock is 90.28% and the cost of new common stock is 15.94%.

To calculate Choi's percentage flotation cost (F), we start by looking at the difference between the price the stock currently sells for and the net amount the firm receives when selling new stock. The current stock price is $36, and the firm can net $32.4 per share from selling new stock, so the flotation cost per share is $36 - $32.4 = $3.6 per share. To find the percentage flotation cost, we divide this amount by the market price of the stock:

F = Flotation cost per share / Current stock price = $3.6 / $36 = 0.10 or 10%

Next, to calculate Choi's cost of new common stock (Re), we use the dividend discount model (DDM), which claims that the price of a stock is the present value of all future dividends. The expected dividend, D1, is $3.18 and the growth rate, g, is 6%. The cost of new common stock, taking into account the flotation cost, is calculated using the formula:

Re = (D1 / Net price per share) + g = ($3.18 / $32.4) + 0.06 = 0.0981 + 0.06 = 0.1581 or 15.81%

The equations of the tangent and normal to f(x) at x = 2 are -

Tangent → y = 5x - 7

Normal → y = -1/2 x + 3.4

A tangent is a line touching the curve at one distinct point, and this distinct point is called the point of contact.

The normal line to a curve at a particular point is the line through that point and perpendicular to the tangent.

Given is a function f(x) and f(2) = 3 and f'(2) = 5.

We can find the tangent to the curve at x = 2 by finding the value of the derivative of f(x) at x = 2. The slope of the tangent at x = 2 is -

m[T] = (dy/dx) [at x = 2] = f'(2) = 5

Now. the equation of tangent can be written as -

(y - y₁) = m [T] (x - x₁)

and

f(2) = 3

At x = 2, y =3 →

y - 3 = 5 (x - 2)

y - 3 = 5x - 10

y = 5x - 10 + 3

y = 5x - 7

This is the equation of the tangent.

In order to find the slope of Normal at x = 2, use the following formula -

m [T] x m [N] = -1

m [T] = 5

Then -

m [N] = - 1/5

The equation of the normal is given by -

(y -  y₁) = m [N] (x - x₁)

(y - 3) = -1/5 (x - 2)

y - 3 = -x/2 + 2/5

y = -x/2 + 2/5 + 3

y = -1/2 x + 3.4

This is the equation of Normal.

Therefore, the equations of the tangent and normal to f(x) at x = 2 are -

Tangent → y = 5x - 7

Normal → y = -1/2 x + 3.4

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Correct Answer: maturity date

Answer: The answer is ∠XWZ.

Step-by-step explanation:  Given that WXYZ is a parallelogram. We are to name an angle that is congruent to ∠XYZ.

Attached herewith, an image of parallelogram WXYZ.

We know that the opposite angles in a parallelogram are congruent to each other. So, in the parallelogram WXYZ, we must have

∠WXY ≅ ∠WZY and ∠XYZ ≅ ∠XWZ.

Therefore, angle congruent to ∠XYZ is ∠XWZ.

Thus, the answer is ∠XWZ.

Final answer:

Angle congruent to angle XYZ in parallelogram WXYZ would be angle W since opposite angles in a parallelogram are congruent.

Explanation:

If WXYZ is a parallelogram and you are asked to name an angle congruent to angle XYZ, you should look at the properties of a parallelogram. One key property is that opposite angles are congruent. Therefore, the angle congruent to angle XYZ would be angle W. If we consider the vertices in order around the parallelogram W, X, Y, Z, angle XYZ corresponds to angle WZY, so its opposite angle is angle WXZ (or simply angle W).

It's important to remember that in a parallelogram, consecutive angles are supplementary (their sum is 180 degrees). This supplementary property could help you in other problems related to angles in a parallelogram, but for this particular case—identifying a congruent angle—the opposite angles property directly gives us the answer.

When p equals 1/3, 9p^2 calculates to 1 after squaring p and multiplying by 9.

The question involves calculating the expression 9p^2 given that p = 1/3. First, we square the value of 'p' and then multiply it by 9. Following the process, we have:

[tex](\frac{1}{3} )^2 = \frac{1}{9} \\9 \times (\frac{1}{9} ) = 1[/tex]

Therefore, 9p^2 is equal to 1 when p is equal to 1/3.

start off by finding 10% of $14. in order to find 10%, divide it by 70 which gives 2. you now need to make the 30% to add to the 70%, by timesing the 2 by 3to give 6. now all you need to do it add the 6 to 14 to make 20!

so your answer is $20

Answer:

The original price was $20.00

Step-by-step explanation:

Joe paid $14.00 for a board game.

This is the 70% of the original price.

Let the original price of a board game be x

70% of x = $14.00

0.70x = 14.00

x = [tex]\frac{14}{0.7}[/tex] = $20.00

The original price was $20.00

Let

x--------> the length side of the rectangular plot (assume side along the river)

y------> the width side of the rectangular plot

we know that

the perimeter of the rectangular plot is equal to

[tex]P=x+2y[/tex]

[tex]P=7,000\ m[/tex]

so

[tex]7,000=x+2y[/tex]

Clear variable y

[tex]y=(7,000-x)/2[/tex] -------> equation [tex]1[/tex]

The area of the rectangular plot is

[tex]A=x*y[/tex] ------> equation [tex]2[/tex]

substitute equation [tex]1[/tex] in equation [tex]2[/tex]

[tex]A=x*[(7,000-x)/2][/tex]

[tex]A=x*[(7,000-x)/2]\\\\ A=3,500x-0.5x^{2}[/tex]

we know that

To find the larger area that can be enclosed------> Find the vertex of the quadratic equation

The quadratic equation is a vertical parabola open down

so

the vertex is a maximum

using a graphing tool

see the attached figure

the vertex is the point [tex](3,500,6,125,000)[/tex]

that means

For [tex]x=3,500\ m[/tex]

the largest area is [tex]6,125,000\ m^{2}[/tex]

Find the value of y

[tex]y=(7,000-3,500)/2=1,750\ m[/tex]

the dimensions of the rectangular plot are

[tex]Length=3,500\ m\\Width=1,750\ m[/tex]

therefore

the answer is

The largest area that can be enclosed is [tex]6,125,000\ m^{2}[/tex]

The pair of sets, {x | x < 0} and {x | x > 0}, has no solution because no number can be simultaneously less than zero and greater than zero. The other pairs do have solutions as they include the number zero.

The set of mathematical expressions provided in the question represent intervals along the number line. Understanding which intervals overlap and which intervals don't is the key to comprehending which of these options potentially has a solution. To consider when these sets will have no solution, you must find where no numbers exist that can satisfy both conditions simultaneously.

The first pair, {x | x < 0} and {x | x > 0}, means 'the set of all x such that x is less than 0' and 'the set of all x such that x is greater than 0'. Because no number can be both less than zero and greater than zero at the same time, this pair has no solution.

The other pairs, {x | x ≤ 0} and {x | x ≥ 0} and {x | x ≤ 0} or {x | x ≥ 0} do have solutions because they both include 0 itself, so there are numbers that can satisfy both conditions.

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Answer:

and only if they have the same slope. Any two vertical lines are parallel. Therefore, the slope of AB is -8 if AB is parallel to CD.

Step-by-step explanation: