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vekshin1
3 years ago
15

Rich is comparing the cost of maintaing his car with the depreciating value of the car. When will the cost and value be the same

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
8 0

Complete question:

Rich is comparing the cost of maintaining his car with the depreciation value of the car.

The value starts at $20,000 and decreases by 15% each year. The maintanance cost is $500 the first year and increases by 28% per year.

When will the maintenance cost and the value be the same.

Answer:

9 years

Step-by-step explanation:

Depreciation is modeled by an exponential function :

A = p(1 - r)^t [decrease, '-']

A = 20,000(1 - 0.15)^t

A = 20000(0.85)^t - - - (1)

Maintainace cost :

A = p(1 + r)^t ; [increase, '+']

A = 500(1 + 0.28)^t

A = 500(1.28)^t - - - (2)

Equating (1) and 2

20000(0.85)^t = 500(1.28)^t

1.28^t / 0.85^t = 20000/500

(1.28 / 0.85)^t = 40

Take log of both sides

Log (1.28 / 0.85)^t = log 40

t * 0.1777910 = 1.6020599

t = 1.6020599 / 0.1777910

t = 9.01

t = 9 years

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Answer:

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Step-by-step explanation:

<u>Step 1: Define Systems</u>

y = 3

y = -3x + 6

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Substitute in <em>y</em>:                                                                                                 3 = -3x + 6
  2. [Subtraction Property of Equality] Subtract 6 on both sides:                      -3 = -3x
  3. [Division Property of Equality] Divide -3 on both sides:                                1 = x
  4. Rewrite/Rearrange:                                                                                        x = 1

<u>Step 3: Solve for </u><em><u>y</u></em>

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Natalija [7]
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Anon25 [30]

An equation is formed of two equal expressions. The correct options are C, D, and E.

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Substitute the value of x in other option to check if the equations are equal are not. Therefore, the options can be written  as,

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