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labwork [276]
3 years ago
11

Every computer that a business owner purchases will lose most of its value after 5 years of use. the owner plans to purchase a c

omputer for $2.100 and replace it after 4 years.
The business owner bought a computer that cost $2,400 instead of $2,100. the owner plans to replace the computer after 3 years instead of 4 years. Comparing both methods, which statement is true about the remaining value of this computer after 3 years of use ?
Mathematics
1 answer:
pickupchik [31]3 years ago
5 0

Answer:

Step-by-step explanation:

a) Data and Calculations:

Estimated useful life of a computer = 5 years

Cost of a computer purchased = $2,400

Annual depreciation expense = $480 ($2,400/5)

This means that the business owner chooses to spread (expense) the cost of the computer over 5 years at a rate of $480.

After 3 years, the cost already expensed = $1,440 ($480 * 3)

The Remaining value of this computer after 3 years of use will be $960 ($2,400 - $1,440).

b) Depreciation is an accounting method for spreading (or expensing) the cost of a long-term asset over its useful life.

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Step-by-step explanation:

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4 years ago
Solve by substitution
Hunter-Best [27]

Answer:

A

Step-by-step explanation:

y = -9 ------(1)

9x -4y = -9 -------(2)

substitute (1) into (2)

9x - 4 (-9) = -9

9x = - 45

x = -5

there you go, x = -5 and y = -9

so. (-5 , -9 )

hope this helps

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7 0
3 years ago
Holly had 5000 in her bank account.she withdrew $800 to buy a new bike.what is the percent decrease in the balance of her accoun
vazorg [7]

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16%

Step-by-step explanation:

800 is 16% of 5000 so and it decreased by 800, so it decreased by 16%

4 0
3 years ago
Convert 32 feet per second to kilometers per hour. (Hint: There are 5280 feet in a mile, and one mile is 1.61 kilometers)
bogdanovich [222]

Answer:

35.113 km per hour is the correct answer

6 0
3 years ago
Solve the given initial-value problem. x' = 1 2 0 1 − 1 2 x, x(0) = 2 7
Ilia_Sergeevich [38]
I'll go out on a limb and guess the system is

\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x

with initial condition \mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top. The coefficient matrix has eigenvalues \lambda such that

\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12

The corresponding eigenvectors \eta are such that

\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}
\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}

\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}
\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}

So the characteristic solution to the ODE system is

\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}

When t=0, we have

\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}

from which it follows that C_1=2 and C_2=5, making the particular solution to the IVP

\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}

\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}
5 0
4 years ago
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