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Sveta_85 [38]
3 years ago
6

The first tern of an arithmetic sequence is -3 and the fifteenth term id 53. What is the common difference of the sequence?

Mathematics
1 answer:
arsen [322]3 years ago
5 0
The common difference is 4. U know that the formula is the first term + d(n -1). N-1 is 14 so u know that 14 times a number needs to equal 56 because the first term is -3 and 56-3 is 53. 14 * 4 is 56. So the common difference is 4
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Answer:

In 2026 car will have a value of $5,000.

Step-by-step explanation:

We have been given that Jack bought a new car in 2014 for 28,000. If the value of the car decreases by 14% each year.

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

a = Initial value,

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Upon substituting a =28,000 and r=0.14 in exponential decay function we will get,

y=28,000(1-0.14)^x, where x represents number of years after 2014.

Therefore, the function y=28,000(0.86)^x represents the value of car x years after 2014.

To find the number of years it will take to car have the value of $5,000, we will substitute y=5,000 in our function.

5,000=28,000(0.86)^x

Let us divide both sides of our equation by 28,000.

\frac{5,000}{28,000}=\frac{28,000(0.86)^x}{28,000}

0.1785714285714286=(0.86)^x

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

ln(0.1785714285714286)=ln((0.86)^x)

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

ln(0.1785714285714286)=x*ln(0.86)

\frac{ln(0.1785714285714286)}{ln(0.86)}=\frac{x*ln(0.86)}{ln(0.86)}

\frac{-1.7227665977411033893}{-0.1508228897345836}=x

x=11.422447\approx 12  

As in the 12th year after 2014 car will have a value of $5,000, so we will add 12 to 2014 to find the year.

2014+12=2026

Therefore, in 2026 car will have a value of $5,000.

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