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Bogdan [553]
3 years ago
8

You purchase a stereo system for $830. The value of the stereo system decreases 13% each year. a. Write an exponential decay mod

el for the value of the stereo system in terms of the number of years since the purchase. b. What is the value of the system after 2 years? c. When will the stereo be worth half the original value?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

a. y=830*(0.87)^x

b. The value of stereo system after 2 years will be $628.23.

c. After approximately 4.98 years the stereo will be worth half the original value.

Step-by-step explanation:

Let x be the number of years.

We have been given that you purchased a stereo system for $830. The value of the stereo system decreases 13% each year.

a. Since we know that an exponential function is in form: y=a*b^x, where,

a = Initial value,

b = For decay b is in form (1-r), where r is rate in decimal form.

Let us convert our given rate in decimal form.

13\%=\frac{13}{100}=0.13

Upon substituting our given values in exponential decay function we will get

y=830*(1-0.13)^x

y=830*(0.87)^x

Therefore, the exponential model y=830*(0.87)^x represents the value of the stereo system in terms of the number of years since the purchase.

b. To find the value of stereo system after 2 years we will substitute x=2 in our model.

y=830*(0.87)^2

y=830*0.7569

y=628.227\approx 628.23

Therefore, the value of stereo system after 2 years will be $628.23.

c. The half of the original price will be \frac{830}{2}=415.

Let us substitute y=415 in our model to find the time it will take the stereo to be worth half the original value.

415=830*(0.87)^x

Upon dividing both sides of our equation by 830 we will get,

\frac{415}{830}=\frac{830*(0.87)^x}{830}

0.5=0.87^x

Let us take natural log of both sides of our equation.

ln(0.5)=ln(0.87^x)

Using natural log property ln(a^b)=b*ln(a) we will get,

ln(0.5)=x*ln(0.87)

\frac{ln(0.5)}{ln(0.87)}=\frac{x*ln(0.87)}{ln(0.87)}

\frac{ln(0.5)}{ln(0.87)}=x

\frac{-0.6931471805599}{-0.139262067}=x

x=4.977286\approx 4.98

Therefore, after approximately 4.98 years the stereo will be worth half the original value.

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Answers:

  • Satellite is approximately <u>2446.43 km</u> from station A.
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=========================================================

Explanation:

I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.

The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths

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Focusing on triangle ACD, we can apply the tangent rule to isolate h.

tan(angle) = opposite/adjacent

tan(A) = CD/AD

tan(86.4) = h/x

x*tan(86.4) = h

h = x*tan(86.4)

We'll use this later in the substitution below.

--------------------

Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say

tan(angle) = opposite/adjacent

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tan(B) = CD/(DA+AB)

tan(85) = h/(x+60)

tan(85)*(x+60) = h

tan(85)*(x+60) = x*tan(86.4) .............  apply substitution; isolate x

x*tan(85)+60*tan(85) = x*tan(86.4)

60*tan(85) = x*tan(86.4)-x*tan(85)

60*tan(85) = x*(tan(86.4)-tan(85))

x*(tan(86.4)-tan(85)) = 60*tan(85)

x = 60*tan(85)/(tan(86.4)-tan(85))

x = 153.612786190499

--------------------

We'll use this approximate x value to find h

h = x*tan(86.4)

h = 153.612786190499*tan(86.4)

h = 2441.60531869599

h = 2441.61 km  is how high the satellite is above the ground.

Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC

cos(angle) = adjacent/hypotenuse

cos(A) = AD/AC

cos(86.4) = x/AC

cos(86.4) = 153.612786190499/AC

AC*cos(86.4) = 153.612786190499

AC = 153.612786190499/cos(86.4)

AC = 2446.43279498247

AC = 2446.43 km is the distance from the satellite to station A.

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