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egoroff_w [7]
3 years ago
6

Two cars are on the same straight road. Car A moves east at 55 mph and sounds its horn at 625 Hz. Rank from highest to lowest th

e frequency observed by Car B’s driver in each of these cases: Car B is moving east at 70 mph and is (a) behind and (b) in front of Car A. Car B is moving west at 70 mph and is (c) behind and (d) in front of Car A.
Mathematics
1 answer:
Fantom [35]3 years ago
8 0

Answer:

1. Car B is moving east at 70MPh and is in from of car A

2.car B is moving west at 70 mph and is behind car A

Step-by-step explanation:

Use the cardinal point system to evaluate the problem first

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A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 1
Jet001 [13]

Answer:

The answer is "$238".

Step-by-step explanation:

Current worth= \$ 17,000

depreciates by \frac{1}{2} in 3 years.

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Using formula:

\to \text{Worth=  Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}

\to A_t=A_0(1-\frac{r}{100})^t

calculates depreciate value in 3 year = \frac{1}{2} \times 17,000

                                                              = 8,500

so,

A_t=8,500\\\\A_0=17,000\\\\t=3\ years

\to A_t=A_0(1-\frac{r}{100})^t\\\\\to 8,500= 17,000(1-\frac{r}{100})^3\\\\\to \frac{8,500}{17,000}= (1-\frac{r}{100})^3\\\\\to \frac{1}{2}= (1-\frac{r}{100})^3\\\\\to (\frac{1}{2})^{\frac{1}{3}}= (1-\frac{r}{100})\\\\\to 0.793700526 = (1-\frac{r}{100})\\\\\to \frac{r}{100} = (1-0.793700526)\\\\\to \frac{r}{100} = (1-0.8)\\\\\to r= 0.2 \times 100 \\\\\to r= 20 \%

depreciates rate= 20%

\to \text{Worth=  Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}

= \$ 17,000 (1- \frac{20}{100})^{19}\\\\= \$ 17,000 (1-0.2)^{19}\\\\= \$ 17,000 (0.8)^{19}\\\\= \$ 17,000 \times 0.014\\\\= \$ 238

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3 years ago
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