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amm1812
4 years ago
8

Mr. Keller bought a car for $20,000. A study shows that a car will depreciate (go DOWN in value) by 15% each year. How much is M

r. Keller’s car worth after 5 years?
A) Formula used:
B)Substitute values:
C) Final answer
D) Does this final answer make sense compared to the original cost of the car? Why?
E) DESCRIBE (using a complete sentence or two) how you can solve this problem graphically.
Please answer all

Mathematics
1 answer:
Aneli [31]4 years ago
7 0
Average annual value lost: $7,390.65First-year depreciation: $3,000.00Total depreciation: $11,125.89Total depreciation percentage: 55.63%Value of vehicle at end of ownership period: $8,874.11
see attachment for graph


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The angle of depression from the top of a building to a point
AleksandrR [38]

The distance between point on the ground from the top of the building is 396 meter, if the building is 280 m high and The angle of depression from the top of a building to a point  on the ground is 45 degrees.

Step-by-step explanation:

The given is,

                 The angle of depression from the top of a building to a point  on the ground is 45 degrees.

                 Height of the building is 280 meter.

Step: 1

                 Given diagram is a right angled diagram,

                 For right angle triangle,

                              90° = 45° + 45°    

                                     = 90°  

                 Trignometric ratio,

                            sin ∅ = \frac{Opp}{Hyp}....................(1)

                For the above ratio take the bottom angle, that is angle of depression from the top of a building to a point  on the ground is 45 degrees.

                 Where, Opp side = 280 meters

                               Hyp side = x

                                           ∅ = 45°

                 Equation (1) becomes,

                                  sin 45° = \frac{280}{x}

                          0.70710678 =  \frac{280}{x}

                                            x =  \frac{280}{0.70710678}

                                            x = 395.979

               Distance between point on the ground from the top of the building,  x ≅ 396 meter                                                

                Trignometric ratio,

                                     cos ∅ = \frac{Adj}{Hyp}

                                   Cos 45 = \frac{Adj}{396}

                                         Adj = (0.70710678)(396)

              Bottom length, Adj = 280 meter

Result:

           The distance between point on the ground from the top of the building is 396 meter.

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