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lidiya [134]
3 years ago
7

A car was purchased for $12,000 with a salvage value of $3000. It is

Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer:

<h2>B.$9,000</h2>

Step-by-step explanation:

The sum of years method is used to calculate the depreciation of an asset. Depreciation is the decrease value of an asset, because it's assumed that its value decreases over time.

We need to calculate a Depreciation Base or Depreciable Cost, and Depreciation Fraction, as the formula attached indicates.

First, the Depreciation Base is calculated by subtracting the total amount $12,000 with the salvage value $3,000 which results in $9,000.

<em>(the salvage value is an estimation of the book value of the asset after the depreciation).</em>

Second, we know that the Remaining Useful Life of the Asset is 5 years, and the sum of the years digits is 1 + 2 + 3 + 4 + 5 = 15, which give us 5/15 as Depreciation Fraction. Then, we are able to calculate the Depreciation Expense.

Depreciation Expense = (5/15)x9000 = $3,000.

So, the book value at the end of the first year will show: $12,000 - $3000 = $9,000.

This means that the fist year the asset will lose $3,000 of its value. Companies calculate the depreciation of all the assets to know the productivity.

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Which of the following statements must be true about this diagram?<br> Check all that apply.
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2 years ago
Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3√5 , KE = 8, LM:KN = 3:5 Find: KM, LM, KN, Area of KLMN
lora16 [44]
Q1)Find KM
As ME is perpendicular to KN, ∠KEM is a right angle
Therefore ΔKEM is a right angled triangle 
KE is given and and ME is also given, we need to find KM
for this we can use Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides.
KM² = KE² + ME²
KM² = 8² + (3√5)²
       = 64 + 9x5
KM = √109
KM = 10.44

Q2)Find LM
It is said that ratio of LM:KN is 3:5
Therefore if we take the length of one unit as x
length of LM is 3x
and the length of KN is 5x
KN is greater than LM by 2 units 
If we take the figure ∠K and ∠N are equal. 
Since the angles on opposite sides of the bases are equal then this is called an isosceles trapezoid. So if we draw a line from L that cuts KN perpendicularly at D, ΔKEM and ΔLDN are congruent therefore KE = DN
since KN is greater than LM due to KE and DN , the extra 2 units of KN correspond to 16 units 
KN = LM + 2x 
2x = KE + DN
2x = 8+8
x = 8
LM = 3x = 3*8 = 24

Q3)Find KN 
Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have;
same height ME = LD perpendicular distance between the 2 parallel sides 
same right angle when the perpendicular lines cut KN
∠K = ∠N 
when 2 angles and one side of one triangle is equal to two angles and one side on another triangle then the 2 triangles are congruent according to AAS theorem (AAS). Therefore KE = DN 
the distance ED = LM
Therefore KN = KE + ED + DN
 since ED = LM = 24
and KE + DN = 16
KN = 16 + 24 = 40
another way is since KN = 5x and x = 8
KN = 5 * 8 = 40

Q4)Find area KLMN
Area of trapezium can be calculated using the following general equation 
Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)
where perpendicular height - ME
2 parallel sides are KN and LM
substituting values into the general equation
Area = 1/2 * ME * (KN+ LM) 
         = 1/2 * 3√5 * (40 + 24)
         = 1/2 * 3√5 * 64
         = 3 x 2.23 * 32
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8 0
3 years ago
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I don’t understand need help
RideAnS [48]

Answer: Choice D

(a-e)/f

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

Explanation:

Points D and B are at locations (e,f) and (a,0) respectively.

Find the slope of line DB to get

m = (y2-y1)/(x2-x1)

m = (0-f)/(a-e)

m = -f/(a-e)

This is the slope of line DB. We want the perpendicular slope to this line. So we'll flip the fraction to get -(a-e)/f and then flip the sign from negative to positive. That leads to the final answer (a-e)/f.

Another example would be an original slope of -2/5 has a perpendicular slope of 5/2. Notice how the two slopes -2/5 and 5/2 multiply to -1. This is true of any pair of perpendicular lines where neither line is vertical.

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