An asset is credited
an asset is debited
Answer:
$490
Explanation:
Let xR be the revenue function
xR = (4 + 0.2(x))(100 - 2x) = 400 + 12x - 2x²/5
Maximum revenue occurs when xR = 0:
xR = 12 - 4x/5 = 0
x = 15
Admission price = 4 + (0.2*15) = 4 + 3 = $7
Max revenue = $7 * (100 - (15*2) = 7 *70 = $490
Answer:
The property will be depreciated using the remaining 3 years of its life after the tax-free incorporation transfer year. This is because Dan had already depreciated the property for 2 years before the transfer.
Explanation:
Sec. 351 allows a tax-free incorporation transfer if certain requirements are met, including that the property must be transferred to Fleck Corporation by Dan in exchange for stock in Fleck Corporation, and, immediately after the exchange, the Fleck Corporation is in control.
Answer:
Value of x maximising profit : x = 5
Explanation:
Cost : C(x) = x^3 - 6x^2 + 13x + 15 ; Revenue: R(x) = 28x
Profit : Revenue - Cost = R(x) - C(x)
28x - [x^3 - 6x^2 + 13x + 15] = 28x - x^3 + 6x^2 - 13x - 15
= - x^3 + 6x^2 + 15x - 15
To find value of 'x' that maximises total profit , we differentiate total profit function with respect to x & find that x value.
dTP/dx = - 3x^2 + 12x + 15 = 0 ► 3x^2 - 12x - 15 = 0
3x^2 + 3x - 15x - 15 = 0 ► 3x (x +1) - 15 (x + 1) = 0 ► (x+1) (3x-15) = 0
x + 1 = 0 ∴ x = -1 [Rejected, production quantity cant be negative] ;
3x - 15 = 0 ∴ 3x = 15 ∴ x = 15/3 = 5
Double derivate : d^2TP/dx^2 = - 6x + 12
d^2TP/dx^2 i.e - 6x + 12 at x = 5 is -6(5) + 12 = - 30+ 12 = -8 which is negative. So profit function is maximum at x = 5
The trial period is called the probationary period. Typically last 90 days & the employee as well as the employer can end the work relationship without reason or cause. Sometimes the job fit is not right but at least this gives both parties to see how the situation works.
