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Serggg [28]
3 years ago
14

Kenya plans to make a down payment plus monthly payments in order to buy a motorcycle. At one dealer she would pay $2,250 down a

nd $175 each month. At another dealer, she would pay $2,850 down and $150 each month. After how many months would the total amount paid be the same for both dealers? What would that amount be?
Mathematics
1 answer:
gavmur [86]3 years ago
5 0
<span>We use X to represent the number of months involved. In the first scenario, the expression for the amount paid is 2250 + 175X. In the second, the expression is 2850 + 150X. We set them equal to each other, because we want to know when the amount paid is the same. 2250 + 175X = 2850 + 150X. We move like to like and get 2850 - 2250 = 175X - 150X. This leaves 25X = 600, and so X = 600/25 = 24. It would take 24 months for this to happen, and subbing back in, the amount is 2250 + 175(24) = 6450.</span>
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We know the area of a small square is 24 in².

Assuming the small rectangles are all equal in area, the total area of the window is equal to 6 x 24 = 144 in².

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144 = 18 x width
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8 0
3 years ago
Please hlelp meeeeeeeeeeeeee i need help
monitta

Answer:

3/4

Step-by-step explanation:

Total number of cases = 8  [1,2,4,2,5,3,1,2]

No. of number less than 4 = 6 [1,2,2,3,1,2]

Probability = 6/8 = 3/4

4 0
3 years ago
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N
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Answer:

x = 4 and x = 6  

Step-by-step explanation:

There are two solutions to an absolute value equation.

If |x - 5| = 1, then  

x - 5 = -1 and  

x - 5 = +1.

1. Solve x - 5 = -1

Add 5 to each side.

x = 4

2. Solve x - 5 = 1

Add 5 to each side.

x = 6

The solutions are x = 4 and x = 6.

3. Number line

The number line would have markers at 4 and 6.

It should look like the image below.

6 0
3 years ago
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Mrac [35]
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3 years ago
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How to find the limit
Ludmilka [50]
\displaystyle\lim_{n\to\infty}\left(k!+\frac{(k+1)!}{1!}+\cdots+\frac{(k+n)!}{n!}\right)=\lim_{n\to\infty}\dfrac{\displaystyle\sum_{i=0}^n\frac{(k+i)!}{i!}}{n^{k+1}}=\lim_{n\to\infty}\frac{a_n}{b_n}

By the Stolz-Cesaro theorem, this limit exists if

\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}

also exists, and the limits would be equal. The theorem requires that b_n be strictly monotone and divergent, which is the case since k\in\mathbb N.

You have

a_{n+1}-a_n=\displaystyle\sum_{i=0}^{n+1}\frac{(k+i)!}{i!}-\sum_{i=0}^n\frac{(k+i)!}{i!}=\frac{(k+n+1)!}{(n+1)!}

so we're left with computing

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)!}{(n+1)!\left((n+1)^{k+1}-n^{k+1}\right)}

This can be done with the help of Stirling's approximation, which says that for large n, n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n. By this reasoning our limit is

\displaystyle\lim_{n\to\infty}\frac{\sqrt{2\pi(k+n+1)}\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\sqrt{2\pi(n+1)}\left(\dfrac{n+1}e\right)^{n+1}\left((n+1)^{k+1}-n^{k+1}\right)}

Let's examine this limit in parts. First,

\dfrac{\sqrt{2\pi(k+n+1)}}{\sqrt{2\pi(n+1)}}=\sqrt{\dfrac{k+n+1}{n+1}}=\sqrt{1+\dfrac k{n+1}}

As n\to\infty, this term approaches 1.

Next,

\dfrac{\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\left(\dfrac{n+1}e\right)^{n+1}}=(k+n+1)^k\left(\dfrac{k+n+1}{n+1}\right)^{n+1}=e^{-k}(k+n+1)^k\left(1+\dfrac k{n+1}\right)^{n+1}

The term on the right approaches e^k, cancelling the e^{-k}. So we're left with

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}

Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.

\displaystyle\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{n^{k+1}+(k+1)n^k+\cdots+1+n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}

Divide through the numerator and denominator by n^k:

\dfrac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}=\dfrac{1+\cdots+\left(\frac{k+1}n\right)^k}{(k+1)+\cdots+\frac1{n^k}}

So you can see that, by comparison, we have

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\lim_{n\to\infty}\frac1{k+1}=\frac1{k+1}

so this is the value of the limit.
8 0
3 years ago
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