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s344n2d4d5 [400]
3 years ago
11

An investment property now worth $180,000 was purchased seven years ago for $142,000. At the time of the purchase, the land was

valued at $18,000. Using a 39-year life for straight-line depreciation purposes, the present book value of the property isa. $95,071.35.b. $113,071.00.c. $126,000.50.d. $119,743.59.
Mathematics
1 answer:
Andru [333]3 years ago
8 0

Answer:

d. $119,743.59

Step-by-step explanation:

actual value (AV)=$180,000

purchase price (PP) =$142,000

intial value (IV) =$18,000

useful live (UL)= 39 years

First, we subtract the value of the property from the purchase value or IV to know the value to be depreciated:

PP-IV= $142,000-$18,000 = $124,000

Then we find out the yearly depreciation by dividing $124,000 into useful live (UL) years:

$124,000/39 = $3,179.49 This is the amount that the property depreciates every year.

But after 7 years the depreciation is: $3,179.49*7= $22,256.41  

We subtract the depreciation in the 7 years from the purchase price (PP) and that's the present book value of the property:  

$142,000-$22,256.41=$119,743.6

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When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2. D
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<h3>Answer:   7x - 5</h3>

============================================

Explanation:

We're dividing some unknown polynomial P(x) over (x+2) to get a quotient Q1(x) and remainder -19. That must mean

P(x)/(x+2) = Q1(x) - 19/(x+2)

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Note that plugging in x = -2 leads to P(-2) = -19. This is an example of the remainder theorem.  

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since dividing by (x-1) leads to a remainder of 2. We have P(1) = 2.

We'll use P(-2) = -19 and P(1) = 2 in the later sections below.

----------------

We want to find the value of r such that

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Let's say that r = ax+b

Meaning we now have

P(x) = Q3*(x+2)(x-1) + r

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The goal from here is to find 'a' and b so we can get the remainder ax+b.

----------------

Plug in x = -2 and solve for b.

P(x) = Q3*(x+2)(x-1) + ax+b

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----------------

Plug in x = 1 to get...

P(x) = Q3*(x+2)(x-1) + ax+b

P(1) = Q3*(1+2)(1-1) + a(1)+b

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----------------

We found that a = 7 and b = -5.

The remainder ax+b updates to the final answer 7x-5

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