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Ad libitum [116K]
2 years ago
13

A corporation purchased 1,000 shares of its own $5 par common stock at $10 and subsequently sold 500 of the shares at $17. What

is the amount credited to paid in capital - treasury stock?
Mathematics
1 answer:
zaharov [31]2 years ago
4 0

The amount credited to paid in capital - treasury stock is $3,500.

<h3><u>Capital-treasury stocks</u></h3>

To determine what is the amount credited to paid in capital - treasury stock, the following calculation must be performed:

  • 1000 x 5 - 500 x 17 = X
  • 5000 - 8500 = X
  • -3500 = X

Therefore, the amount credited to paid in capital - treasury stock is $3,500.

Learn more about capital - treasury stocks in brainly.com/question/25715888

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The measure of an interor angle of a regular polygon is given. Find the number of sides in the polygon.
EleoNora [17]

Let side be n

#1

\\ \sf\longmapsto 120n=(n-2)180

\\ \sf\longmapsto 120n=180n-360

\\ \sf\longmapsto -60n=-360

\\ \sf\longmapsto n=6

#2

\\ \sf\longmapsto 160n=(n-2)180

\\ \sf\longmapsto 160n=180n-240

\\ \sf\longmapsto -20n=-240

\\ \sf\longmapsto n=12

#3

\\ \sf\longmapsto 144n=(n-2)180

\\ \sf\longmapsto 144n=180n-360

\\ \sf\longmapsto -36n=-360

\\ \sf\longmapsto n=10

6 0
3 years ago
I'm so confused. If you answer, please explain.
Ahat [919]

Answer:

its SAS

Step-by-step explanation:

HF = GF

HGJ angle = FGJ angle

JG = JG

3 0
3 years ago
A boy purchased​ (bought) a​ party-length sandwich 51 in. long. He wants to cut it into three pieces so that the middle piece is
shepuryov [24]

Answer:

Step-by-step explanation:

we have a 51 in sandwich

let a piece be x

we have x= shorter piece

another piece is x+6

another piece (x+6)-9=x-3

x+x+6+x-3=51

3x=51-3

3x=48

x=48/3=16

x=16, longer piece is 16+6=22, shorter piece =10

6 0
2 years ago
A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a
Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 =  0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = 2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

W_2 =  \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0
2 years ago
The sides of a square field are 28 meters. A sprinkler in the center of the field sprays a circular area with a diameter that co
Stella [2.4K]
If a field is 900 meters squared and one side is twice as big as the other how big is each side
6 0
2 years ago
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