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Nadya [2.5K]
3 years ago
9

Canterbury Cycles sells Harleys and pays each salesperson a commission of $800 for each cycle sold. During the month of December

, a salesperson sold 3 cycles. The company pays commissions on the 5th day of the month following the sale. Which of the following statements is true?a. The salesperson will recognize commission revenue earned in the amount of 2400 in Decemberb. The company will recognize commission expense in the amount of $2,400 in December.c. The salesperson will recognize commission expense in the amount of $2,400 in January.d. The salesperson will recognize revenue in the same month that the cycle dealer recognizes expense.
Mathematics
2 answers:
Kitty [74]3 years ago
8 0

Answer:

b. The company will recognize commission expense in the amount of $2,400 in December.

Step-by-step explanation:

Based on the information given the statements that is true will be: THE COMPANY WILL RECOGNIZE COMMISSION EXPENSE IN THE AMOUNT OF $2,400 IN DECEMBER reason been that each salesperson were paid a commission in the amount of $800 for each cycle sold, which is 3 cycles during the month of December.

Calculation to determine the Commission Expense

Commission Expense=$800*3 cycles

Commission Expense=$2,400

Therefore The company will recognize commission expense in the amount of $2,400 in December.

belka [17]3 years ago
7 0

Answer: The company will recognize commission expense in the amount of $2,400 in December

Step-by-step explanation:

Based on the information given in the question, the company will recognize commission expense in the amount of $2,400 in December.

A commission is regarded as a fee which is paid to a salesperson by a company in exchange for completing a sale.

It should be noted that in the accrual basis of accounting, commission should be recorded in the same period as when the sale generated was generated.

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Please help me w the answer
evablogger [386]

Answer:

\frac{2(x-6)(x-10)}{(x-4)(x-5)}

Step-by-step explanation:

In essence, one needs to work their way backwards to solve this problem. Use the information to construct the function.

The function has verticle asymptotes at (x = 4) and (x = 5). This means that the denominator must have (x - 4) and (x - 5) in it. This is because a verticle asymptote indicates that the function cannot have a value at these points, the function jumps at these points. This is because the denominator of a fraction cannot be (0), the values (x - 4) and (x - 5) ensure this. Since if (x) equals (4) or (5) in this situation, the denominator would be (0) because of the zero product property (this states that any number times zero equals zero). So far we have assembled the function as the following:

\frac{()}{(x-4)(x-5)}

The function has x-intercepts at (6, 0), and (0, 10). This means that the numerator must equal (0) when (x) is (6) or (10). Using similar logic that was applied to find the denominator, one can conclude that the numerator must be ((x - 6)(x-10)). Now one has this much of the function assembled

\frac{(x-6)(x-10)}{(x-4)(x-5)}

Finally one has to include the y-intercept of (0, 120). Currently, the y-intercept is (60). This is found by multiplying the constants together. (6 * 10) equals (60). One has to multiply this by (2) to get (120). Therefore, one must multiply the numerator by (2) in order to make the y-intercept (120). Thus the final function is the following:

\frac{2(x-6)(x-10)}{(x-4)(x-5)}

6 0
3 years ago
Quadrilateral ABCD has vertices at: A(0,7),B(10,9),C(12,-1),D(2,-3)
Svetach [21]

Answer: the answer is c

Step-by-step explanation:

your welcome

7 0
3 years ago
If (ax^2 + 3x + 2b) - (5x^2+bx-3c)=3x^2-9, what is the value of A + B + C?
Rina8888 [55]

Answer:

We have the equation:

(ax^2 + 3x + 2b) - (5x^2+bx-3c)=  3x^2 - 9

First, move all to the left side.

(ax^2 + 3x + 2b) - (5x^2+bx-3c) - 3x^2 + 9 = 0

Now let's group togheter terms with the same power of x.

(a - 5 - 3)*x^2 + (3 - b)*x + (2b + 3c + 9) = 0.

This must be zero for all the values of x, then the things inside each parenthesis must be zero.

1)

a - 5 - 3 = 0

a = 3 + 5 = 8.

2)

3 - b = 0

b = 3.

3)

2b + 3c + 9 = 0

2*3 + 3c + 9 = 0

3c = -6 - 9 = -15

c = -15/3 = -5

Then we have:

a = 8, b = 3, c = -5

a + b + c = 8 + 3 - 5 = 6

8 0
3 years ago
Please help thank you
daser333 [38]

Answer:

21

Step-by-step explanation:

Add 34 to both sides.

14x = 11x + 63

Subtract 11x from both sides

3x = 63

Divide both sides by 3

21

Hope I helped :)

Please consider Brainliest :)

6 0
3 years ago
Which expression is equivalent to log w (x2-6)4
natka813 [3]
Log w (x^2-6)^4

Using log a b = log a + log b, with a=w and b=(x^-6)^4:
log w (x^2-6)^4 = log w + log (x^2-6)^4

Using in the second term log a^b = b log a, with a=x^2-6 and b=4
log w (x^2-6)^4 = log w + log (x^2-6)^4 =  log w + 4 log (x^2-6)

Then, the answer is:
log w (x^2-6)^4 = log w + 4 log (x^2-6)
7 0
3 years ago
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