Mathematics, 21.06.2019 13:40 kemzzoo7206

Desmond wants to sell his car that he paid $8,000 for 2 years ago. the car depreciated, or decreased in value, at a constant rate each month over a 2-year period. if x represents the monthly depreciation amount, which expression shows how much desmond can sell his car for today? 8,000 + 24x 8,000 − 24x 8,000 + 2x 8,000 − 2x

Answers: 1

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Answer from: renelopez79318

8,000-24x

Step-by-step explanation:

there's 24 months in 2 years. If it's value decreased from 8,000 at a constant rate over a 2 year period, the equation would be 8,000-24x

Answer from: 12monkey6

B. 8,000-24x .

Step-by-step explanation:

We have been given that Desmond wants to sell his car that he paid $8,000 for 2 years ago. The car depreciated, or decreased in value, at a constant rate each month over a 2-year period.

Since the value of car depreciates at a constant rate each month, so the value of car depreciated in 2 years would be $2\times 12=24$ .

There are 24 months in two years, so value of car depreciated in 2 years would be 24x.

The initial value of car is $8,000, so value of car after 2 years would be 8,000-24x .

Therefore, option B is the correct choice.

Answer from: infoneetusinghoyg22o

8,000 - 24x
Step-by-step explanation:
Let
y > depreciated value of the car
x---> rate of depreciation
t > the time in months
we know that
The linear equation that represent this situation is
y=8,000-xt
For
t=2 years=2*12=24 months
substitute
y=8,,000-x(24)
y=8,000-24x

Answer from: IIIKiXIII

1. C. 8000 - 24x; 2 A. x + 2x + 2 = 40; 3. C. Step 3

Step-by-step explanation

1. Desmond

$\begin{array}{rcl}\text{Value of car two years ago} & = & 8000\\\\\text{Less depreciation = 24 mo} \times\dfrac{x}{\text{1 mo}} & = & -24 x\\\\\text{Current value} & = & \mathbf{8000 - 24x}\\\end{array}$

2. Avi and Sergi

$\begin{array}{rcl}\text{Sergi's money} & = & x\\\text{Double Sergi's money} & = & 2x\\\text{Avi's money} & = & 2x + 2\\\text{Avi's money + Sergi's money} & = & x + 2x + 2\\\text{Avi's money + Sergi's money} & = & 40\\x + 2x + 2 & = & 40\\\end{array}$

3. Isabella

$\begin{array}{crl} \textbf{Step} & \textbf{Work} & \textbf{Justification} \\ & 4x - 2x + 8 = 6(x + 4) & \\ 1 & 4x - 2x + 8 = 6x + 24 & \text{Distributive Property} \\ 2 & 2x + 8 = 6x + 24 & \text{Combine like terms} \\ 3 & -4x + 8 = 24 & \textbf{Subtraction Property of Equality}\\ 4 & -4x = 16 & \text{Subtraction Property of Equality} \\ 5 & x = -4 & \text{Division Property of Equality} \\ \end{array}$

Step 3 has the incorrect justification. Isabella subtracted 6x from each side, so she should have used the Subtraction Property of Equality

Answer from: anitadefrances

8,000 - 24x

Step-by-step explanation:

Let

y > depreciated value of the car

x---> rate of depreciation

t > the time in months

we know that

The linear equation that represent this situation is

y=8,000-xt

For t=2 years=2*12=24 months

substitute

y=8,,000-x(24)

y=8,000-24x

Answer from: Darkknighta26

8,000 - 24x

Step-by-step explanation:

Let

y > depreciated value of the car

x---> rate of depreciation

t > the time in months

we know that

The linear equation that represent this situation is

y=8,000-xt

For

t=2 years=2*12=24 months

substitute

y=8,,000-x(24)

y=8,000-24x

Answer from: Anshuman2002

8,000-24x

Step-by-step explanation:

Let

y > depreciated value of the car

x---> rate of depreciation

t > the time in months

we know that

The linear equation that represent this situation is

y=8,000-xt

For

t=2 years=2*12=24 months

substitute

y=8,,000-x(24)

y=8,000-24x

Answer from: allhailkingmilkdud

Step-by-step explanation:

The answer is B) 8,000-24x

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