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Scilla [17]
3 years ago
15

The value of a car will “depreciate” over time. For example, a car that was worth $24 000 when it was new, is being sold for $13

500 three years later. Determine the annual depreciation rate on this car. Express your final answer as a percent, rounded to one decimal place.
Mathematics
1 answer:
LekaFEV [45]3 years ago
7 0

Answer:

The car will depreciate at a rate of 21.14% per year.

Step-by-step explanation:

Given that the value of a car will “depreciate” over time, and, for example, a car that was worth $ 24,000 when it was new, is being sold for $ 13,500 three years later, to determine the annual depreciation rate on this car the following calculation must be performed:

13,500 x (1 + X) ^ 1x3 = 24,000

13,500 x (1 + 0.2114) ^ 3 = 24,000

X = 21.14%

Therefore, the car will depreciate at a rate of 21.14% per year.

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<span>Compatible </span>numbers for 256...
1. Rounded to the nearest hundred = 200
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Compatible numbers for 321...
1. Rounded to the nearest hundred = 300
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4 years ago
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Determine the equation of the line that passes through A(2,-5) and B(6,-3).
yaroslaw [1]

Answer:

y = 1/2x - 6

Step-by-step explanation:

y2 - y1 / x2 - x1

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2/4

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3 years ago
1. Writing an equation for an exponential function by
Naya [18.7K]

Answer: 1) t(n)=0.6(2)^n

2) f(x)=10(5)^x

Step-by-step explanation:

1) Let the function that shows the thickness of the paper after n folds,

t(n) = ab^n         ---------(1)

Since, According to the question,

Initially the thickness of the paper = 0.6

That is, at n = 0, t(0) = 0.6

By equation (1),

0.6 = a(b)^0\implies 0.6 = a

Hence the function that shows the given situation,

t(n) = 0.6 b^n       -----------(2)

Again when we fold the paper the thickness of the paper will be doubled.

Thus, at n = 1, t(1) = 1.2

By equation (2),

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Thus, the complete function is,

t(n) = 0.6 (2)^n    

2) Let the function that is passing through the points (-2, 2/5) and (-1,2),

f(x) = ab^x         ---------(1)

For f(x) = 2, x = -1

By equation (1),

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Also, For f(x) = 2/5, x = -2

Again, By equation (1),

\frac{2}{5}= a(b)^{-2}

\implies \frac{2}{5}=ab^{-1}b^{-1}=2b^{-1}

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\implies 2b=10

\implies b = 5

By substituting this value in equation (2),

We get, a = 10

Hence, from equation (1), the function that is passing through the points (-2, 2/5) and (-1,2),

f(x) = 10(5)^x

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