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elixir [45]
3 years ago
15

Equipment was purchased for $85,000 on january 1, 2015. freight charges amounted to $3,500 and there was a cost of $10,000 for b

uilding a foundation and installing the equipment. it is estimated that the equipment will have a $15,000 salvage value at the end of its 5-year useful life. what is the amount of accumulated depreciation at december 31, 2016, if the straight-line method of depreciation is used?
Mathematics
1 answer:
Elena-2011 [213]3 years ago
3 0
85000+3500+10000=98500 is a historic price of equipment. Annual depreciation is (98500-15000)/5=83500/5=16700. At december 31 2016 will be gone 2 years-2015 and 2016. So the accumulated depreciation will be 2*16700=33400
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What the heck does this mean lol
dangina [55]

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Step-by-step explanation:

The significant figures of a number that carry meaningful contribution to its measurement resolution

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A teacher has 3 hours to grade all the papers submitted by the 35 students in her class. She gets through the first 5 papers in
Andreyy89
So, she has 3hrs to grade all papers, for 35 students.. alrite.

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now, she has still 2 hours and a half, or 150 minutes, to do the remaining 30 papers... she has to work at a rate of 30/150 then... which is 1/5 simplified.

now if we take 1/6 as the 100%, what is 1/5 in percentage then?

\bf \begin{array}{ccllll}
rate&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
\frac{1}{6}&100\\\\
\frac{1}{5}&x
\end{array}\implies \cfrac{\frac{1}{6}}{\frac{1}{5}}=\cfrac{100}{x}\implies \cfrac{1}{6}\cdot \cfrac{5}{1}=\cfrac{100}{x}\implies \cfrac{5}{6}=\cfrac{100}{x}
\\\\\\
x=\cfrac{6\cdot 100}{5}\implies x=120

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8 0
3 years ago
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Find the area and the circumference of the circle with the radius 6yd.
fgiga [73]

Answer:

<h2>A = 36π yd² ≈ 113.04 yd²</h2><h2>C = 12π yd ≈ 37.68 yd</h2>

Step-by-step explanation:

The formula of an area of a circle:

A=\pi r^2

The formula of a circumference of a circle:

C=2\pi r

<em>r</em><em> - radius</em>

<em />

We have <em>r = 6yd</em>.

Substitute:

A=\pi(6^2)=36\pi\ yd^2

C=2\pi(6)=12\pi\ yd

If you want round the answers, then use <em>π ≈ 3.14</em>

A\approx(36)(3.14)=113.04\ yd^2

C\approx(12)(3.14)=37.68\ yd

3 0
3 years ago
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