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Marrrta [24]
4 years ago
5

Emery Mining Inc. recently reported $147,500 of sales, $75,500 of operating costs other than depreciation, and $10,200 of deprec

iation. The company had $16,500 of outstanding bonds that carry a 7.25% interest rate, and its federal-plus-state income tax rate was 25%. How much was the firm's net income
Mathematics
1 answer:
Molodets [167]4 years ago
6 0

Answer:

net income is $48452.81

Step-by-step explanation:

Sales =$147500

subtract operating expenses

-$75500 =$72000

subtract non- operating costs

depreciation -$10200 =$61800

-interest expense payable (16500*7,23%)$1196.25=63603.75

from profit before tax deduct income taxes =63603.75*25%=15150.9375

Net Income is therefore $63603.75-$15150.9375 = $48452.81

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150 miles in 6 gallons in unit rates
vazorg [7]

Answer:

1g/25m

Step-by-step explanation:

divide both of them by 6 so one of them can be one

Please give me Brainliest

8 0
3 years ago
What is another way to group the factors (3x2)x5
Elza [17]
3*(2*5)
You're Wlecome, 
6 0
3 years ago
David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below. GFC of 80, 32, and 48: 16 GCF o
BlackZzzverrR [31]

Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :(b^4,b^2,b^4)

b^4= b × b × b × b

b^2 = b × b

b^4 =b × b × b × b

GCF of variables :(b^4,b^2,b^4) is b^2

3) GCF of c^3 and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.

4) 80b^4-32b^2c^3+48b^4c

=16b^2(5b^2-2c^3+3b^2c)

David applied the distributive property.

7 0
3 years ago
Read 3 more answers
A certain circle can be represented by the following equation. X^2+y^2-6x-6y+2=0.
pav-90 [236]

Answer

       

Step-by-step explanation:

7 0
3 years ago
This is Matrix for pre calc
gavmur [86]

The given system of equations in augmented matrix form is

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]

If you need to solve this, first get the matrix in RREF:

  • Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]

  • Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]

  • Add 164(row 3) to -91(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]

  • Multiply row 4 by 1/13080:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]

  • Add -153(row 4) to row 3:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]

  • Multiply row 3 by -1/91:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add 6(row 3) and -8(row 4) to row 2:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 2 by 1/5:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 1 by 1/3:

\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

So the solution to this system is (w,x,y,z)=(1,-2,4,-3).

6 0
4 years ago
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