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1 year ago

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A business uses straight-line depreciation to determine the value y of an automobile over a 5-year period. Suppose the original

value (when t=0) is equal to $14,500 and the salvage value (when t=5) is equal to $4,000.
a) By how much has the automobile depreciated over the 5 years?
b) By how much is the value of the automobile reduced at the end of each of the 5 years?
c) Write the linear equation that models the value s of this automobile at the end of year t.

1 answer:

Elena-2011 [213]1 year ago

6 0

For this problem. it is found that:

a) The automobile has depreciated 75% in 5 years.

b) The value of the automobile was reduced by $10,500 in five years.

c) The linear function is: s(t) = 14,500 - 2,100t.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = ax + b

In which:

a is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.

b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

From the problem, we have that:

a) The automobile has depreciated 75% in 5 years, as 4000/14500 = 0.25, and 100% - 25% = 75%.

b) The value of the automobile was reduced by $10,500 in five years, as 14500 - 4000 = $10,500.

For the linear function, we have that:

The initial value is b = 14500.

The slope is of a = -10500/5 = -2100.

Hence the linear function is: s(t) = 14,500 - 2,100t.

More can be learned about linear functions at brainly.com/question/24808124

#SPJ1

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<h2>Hello!</h2>

The answer is:

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<h2>Why?</h2>

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We need to remember that we can calculate the area of any circle using the following formula:

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3 years ago

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3 years ago

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3 years ago

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3 years ago

Put the following in ascending order

Afina-wow [57]

Answer:

see explanation

Step-by-step explanation:

To be able to compare the fractions we require them to have a common denominator.

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-------------------------------------------------------

$\frac{7}{15}$ = $\frac{7(2)}{15(2)}$ = $\frac{14}{30}$

$\frac{2}{5}$ = $\frac{2(6)}{5(6)}$ = $\frac{12}{30}$

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$\frac{2}{3}$ = $\frac{2(10)}{3(10)}$ = $\frac{20}{30}$

----------------------------------------------------------------

Placing these in ascending order ( smallest to largest )

$\frac{10}{30}$ , $\frac{12}{30}$ , $\frac{14}{30}$ , $\frac{15}{30}$ , $\frac{20}{30}$ , $\frac{27}{30}$

----------------------------------------------------------------------

Listing the actual fractions corresponding to the equivalent fractions

$\frac{1}{3}$ , $\frac{2}{5}$ , $\frac{7}{15}$ , $\frac{1}{2}$ , $\frac{2}{3}$ , $\frac{9}{10}$

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3 years ago

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