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maxonik [38]
3 years ago
7

A car costs £5000 when new. Each year the value decreases by 20%. Person A said that the car would be worth nothing in 5 years.

The answer is not 0. What is the answer? Why is Person A wrong?
Mathematics
1 answer:
den301095 [7]3 years ago
4 0

Answer:

£1638.40

Step-by-step explanation:

Step one:

Given data

A car costs £5000 when new

Each year the value decreases by 20%.

Step two:

let us compute the worth in the next 5 years

year 1

worth= 5000- 0.2*5000

worth= 5000-1000

worth=  £4000

year 2

worth= 4000- 0.2*4000

worth= 4000-800

worth=  £3200

year 3

worth= 3200- 0.2*3200

worth= 3200-640

worth=  £2560

year 4

worth= 2560- 0.2*2560

worth= 2560-512

worth=  £2048

year 5

worth= 2048- 0.2*2048

worth= 2048-409.6

worth=  £1638.40

The person is wrong because after 5 years the car is worth  £1638.40

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Which set of equations is enough information to prove that lines a and b are parallel lines cut by transversal f? m∠4 = 110° and
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4 0
3 years ago
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Write 23 hundredths in scientific notation
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A company manufactures running shoes and basketball shoes. The total revenue (in thousands of dollars) from x1 units of running
Alborosie

Answer:

x_1 =2 , x_2=7

Step-by-step explanation:

Consider the revenue function given by R(x_1,x_2) = -5x_1^2-8x_2^2 -2x_1x_2+34x_1+116x_2. We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).

\frac{dR}{dx_1} = -10x_1-2x_2+34 =0

\frac{dR}{dx_2} = -16x_2-2x_1+116 =0

From the first equation, we get, x_2 = \frac{-10x_1+34}{2}.If we replace that in the second equation, we get

-16\frac{-10x_1+34}{2} -2x_1+116=0= 80x_1-2x_1+116-272= 78x_1-156

From where we get that x_1 = \frac{156}{78}=2. If we replace that in the first equation, we get

x_2 = \frac{-10\cdot 2 +34}{2}=\frac{14}{2} = 7

So, the critical point is (x_1,x_2) = (2,7). We must check that it is a maximum. To do so, we will use the Hessian criteria. To do so, we must calculate the second derivatives and the crossed derivatives  and check if the criteria is fulfilled in order for it to be a maximum. We get that

\frac{d^2R}{dx_1dx_2}= -2 = \frac{d^2R}{dx_2dx_1}

\frac{d^2R}{dx_{1}^2}=-10, \frac{d^2R}{dx_{2}^2}=-16

We have the following matrix,  

\left[\begin{matrix} -10 & -2 \\ -2 & -16\end{matrix}\right].

Recall that the Hessian criteria says that, for the point to be a maximum, the determinant of the whole matrix should be positive and the element of the matrix that is in the upper left corner should be negative. Note that the determinant of the matrix is (-10)\cdot (-16) - (-2)(-2) = 156>0 and that -10<0. Hence, the criteria is fulfilled and the critical point is a maximum

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