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yanalaym [24]
4 years ago
11

Courtney would like to buy a new ca in 4 years. It is estimated that a new car will cost her 25,000 and that she will get 13,000

for her used car provided she takes good care of it. Based on these estimates, what is the least amount she should save each month in order to pay cash for her new car?
Mathematics
1 answer:
KiRa [710]4 years ago
3 0

4 years=48 months

A new car costs 25,000, but she will get 13,000 for her used car as long as she takes good car of it. If she takes good care of her used car and gets 13,000 for it, she will only need to save 12,000 total. We can divide 12,000 by 48 to find out how much Courtney should save minimum each month to afford her new car. 12,000/48=250

You can check to see if this is correct by multiplying 48*250=12,000. Paying 250 for each month of the 48 months will equal the $12,000 that Courtney needs.

Courtney needs to save at least $250 per month to afford her new car in 4 years.

I hope this helps :)

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Considering her change, it is found that she will donate $0.28 to the animal shelter.

<h3>How to find the amount of change in a purchase?</h3>

The amount of change is given by the subtraction of the total paid and the price of the products.

In this problem, the price of the products bought is given by:

$15.74 + $3.98 = $19.72

She paid $20, hence the change is given by:

$20 - $19.72 = $0.28.

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Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }

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Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }

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Following the pattern, we can see that for f^{n}(x),

f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} }  \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}

This applies for n ≥ 1, Expressing f(x) in summation, we have

\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}

Combining ln2 with the rest of series, we have

f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

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