Answer: option C
Explanation: THIS CAN BE REPRESENTED AS FOLLOWS :-
If we eliminate the product there would be no sales, no variable expenses and therefore, no contribution.
sales = nil
-variable expenses= <u>nil</u>
contribution = nil
- fixed expenses = <u>56,000</u>
NET LOSS = <u> (56000)</u>
.
NOTE :-
Fixed expense = (140,000)*(40%)= 56,000
.
.
Thus increase in loss would be 56000- 50,000=6000
Answer:
p = 59.11 dollars
Explanation:
Given
Price: p(x) = 8eˣ (0 ≤ x ≤ 2)
Revenue; R = x*p = 8xeˣ
p = ? when R be at maximum
We can apply
dR/dx = d(x*p)/dx = 0
⇒ d(8xeˣ)/dx = 8*(1*eˣ + x*eˣ) = 0
⇒ eˣ*(1 + x) = 0 ⇒ x = - 1
as x = - 1 ∉ [0, 2]
then, we have
p(0) = 8e⁰ = 8
R = 0*8 = 0
If x = 1
p(1) = 8e¹ ≈ 21.74
R = 1*21.74 = 21.74
If x = 2
p(2) = 8e² ≈ 59.11
R = 2*59.11 = 118.22
Implies that, R(x) is maximum at x = 2.
Thus, the price that maximize the revenue of the company is 59.11 dollars.
168,000 is amount of the gain is Ethan allowed to exclude from his gross income
Solution:
Ethan's post 2009 non-qualified use is 2 years.
He owned the property for 10 years so he is not allowed to exclude 20% of the gain
= $210,000 × 20% = $42,000
He is allowed to exclude = ($210,000 - $42,000)
= $168,000
Answer:
Total sum at te end of 6 years=$ 73,138.97
Explanation:
<em>The total sum in Lee's account at the end of year would be determined as follows:</em>
FV= A × (1+r)^n
FV- Future sum?,
A- deposit amount ,
r- interest rate - 9%/2 = 4.5% per 6 months
n- number of years is 6
First deposit for 6 years
FV = 15,000× 1.045^(2×6)
= $25,438.22
Last 40,000 for 2 years
FV = 40,000 1.045^(2×2)
=47,700.74403
Total sum at the end of 6 years
= $25,438.22 + $47,700.74
=$ 73,138.97
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