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Marysya12 [62]
3 years ago
15

A new car worth 24000 is depreciating in value by 3000 per year after how many years will the car be 12000

Mathematics
2 answers:
AveGali [126]3 years ago
7 0

Answer:

4 years

Step-by-step explanation:

need to see how many times 3000 can go in 24000 and count how many years tell u get 12000

hopes I helped

Oliga [24]3 years ago
5 0

Answer:

4 years

Step-by-step explanation:

subtract 3000 from 24000 then 3000 from the sum of each answer till you get to 12,000

after <u>year 1</u>: $21,000

24,000 - 3,000 = 21,000

after <u>year 2</u>: $18,000

21,000 - 3,000 = 18,000

after <u>year 3</u>: $15,000

18,000 - 3,000 = 15,000

after<u> year 4</u>: $12,000

15,000 - 3,000 = 12,000

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The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

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u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

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W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

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