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Tcecarenko [31]
3 years ago
8

A four-door sedan costs$35000 with a residual value of $2000. Its service life is five years. Using the declining balance method

at twice the straight-line rate, the book value at the end of year 2 is?
Mathematics
1 answer:
larisa [96]3 years ago
3 0

Answer:

At the end of the 2 year, the book value of the truck is $12,600

Step-by-step explanation:

we know that

<u><em>Double declining balance method</em></u> is a form of an accelerated depreciation method in which the asset value is depreciated at twice the rate it is done in the straight-line method.

Step 1

Determine the straight-line depreciation rate

Divide the total cost by the number of years in the asset's useful life.

\frac{\$35,000}{5}=\$7,000

Step 2

Then, multiply that number by 2 and that is your Double-Declining Depreciation Rate

7,000(2)=\$14,000 -----> is the depreciation for Year 1

Step 3

At the end of the first year, the book value of the truck is

\$35,000-\$14,000=\$21,000

Step 4

For Year 2, we will apply the same formula to the book value of the truck by the end of Year 1

\frac{\$21,000}{5}=\$4,200

4,200(2)=\$8,400 -----> is the depreciation for Year 2

therefore

At the end of the 2 year, the book value of the truck is

\$21,000-\$8,400=\$12,600

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V125BC [204]

The radian measure of 205 degees is \frac{ 41 \pi }{36} radians

<em><u>Solution:</u></em>

Given that we have to express 205 degrees as radian

<em><u>To convert from degrees to radians use the following:</u></em>

x^{\circ} \times \frac{ \pi}{180}

where x is the value in degrees to be converted

Here, x = 205 degrees

Substituting we get,

\rightarrow 205 \times \frac{ \pi }{180}\\\\\rightarrow 41 \times \frac{ \pi }{36}\\\\\rightarrow \frac{41 \pi}{36}

Thus the radian measure of 205 degees is \frac{ 41 \pi }{36} radians

8 0
3 years ago
Perform long division on the​ integrand, write the proper fraction as a sum of partial​ fractions, and then evaluate the integra
worty [1.4K]

Answer:

x^2+4x -\frac{1}{2}  lnx  + \frac{2}{13}  ln(x-2) + C

Step-by-step explanation:

Given the integrand \int\limits{\dfrac{2x^3 - 2x + 1}{x^2-2x} } \, dx, before evaluating the integral function, we will need to simplify the function first by applying long division as shown in the attachment.

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Integrating its partial sum

\int\limits \dfrac{2x^3 - 2x + 1}{x^2-2x} }dx  = \int\limits  (2x+4 + \frac{6x+1}{x^2-2x})\ dx\\\\= \int\limits {2x} \, dx + \int\limits {4} \, dx + \int\limits {\frac{6x+1}{x^2-2x} \, dx\\ = \frac{2x^2}{2}+4x -\frac{1}{2}  \int\limits{\frac{1}{x} } \, dx  + \frac{2}{13}  \int\limits{\frac{1}{x-2} } \, dx

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3 years ago
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alex41 [277]
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2 5/8 = 16/8 + 5/8 = 21/8
so now we can operate easily, to calculate we need to multiply 21/8 by 2/3 that is because, if 1/3 is used then 2/3 are left over (1/3 + 2/3 = 1, the total), so we have:
(2/3)(21/8) = (2*21)/(3*8)
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