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Elina [12.6K]
3 years ago
12

The graph represents the direct variation function between earings in dollars and hours worked. Which equation can be used to de

scribe the direct variation function between E, the total earings in dollars, and h, the number of hours worked?
Mathematics
1 answer:
atroni [7]3 years ago
5 0

Answer:The formula y=kxn y = k x n is used for direct variation. The value k is a nonzero constant greater than zero and is called the constant of variation.

Step-by-step explanation:

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Find the nth term of the sequence 7,25,51,85,127​
olya-2409 [2.1K]

Let <em>a </em>(<em>n</em>) denote the <em>n</em>-th term of the given sequence.

Check the forward differences, and denote the <em>n</em>-th difference by <em>b </em>(<em>n</em>). That is,

<em>b </em>(<em>n</em>) = <em>a </em>(<em>n</em> + 1) - <em>a </em>(<em>n</em>)

These so-called first differences are

<em>b</em> (1) = <em>a</em> (2) - <em>a</em> (1) = 25 - 7 = 18

<em>b</em> (2) = <em>a</em> (3) - <em>a</em> (2) = 51 - 25 = 26

<em>b </em>(3) = <em>a</em> (4) - <em>a</em> (3) = 85 - 51 = 34

<em>b</em> (4) = <em>a </em>(5) - <em>a</em> (4) = 127 - 85 = 42

Now consider this sequence of differences,

18, 26, 34, 42, …

and notice that the difference between consecutive terms in this sequence <em>b</em> is 8:

26 - 18 = 8

34 - 26 = 8

42 - 34 = 8

and so on. This means <em>b</em> is an arithmetic sequence, and in particular follows the rule

<em>b</em> (<em>n</em>) = 18 + 8 (<em>n</em> - 1) = 8<em>n</em> + 10

for <em>n</em> ≥ 1.

So we have

<em>a </em>(<em>n</em> + 1) - <em>a </em>(<em>n</em>) = 8<em>n</em> + 10

or, replacing <em>n</em> + 1 with <em>n</em>,

<em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 8 (<em>n</em> - 1) + 10

<em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 8<em>n</em> + 2

We can solve for <em>a</em> (<em>n</em>) by iteratively substituting:

<em>a</em> (<em>n</em>) = [<em>a</em> (<em>n</em> - 2) + 8 (<em>n</em> - 1) + 2] + 8<em>n</em> + 2

<em>a</em> (<em>n</em>) = <em>a </em>(<em>n</em> - 2) + 8 (<em>n</em> + (<em>n</em> - 1)) + 2×2

<em>a</em> (<em>n</em>) = [<em>a</em> (<em>n</em> - 3) + 8 (<em>n</em> - 2) + 2] + 8 (<em>n</em> + (<em>n</em> - 1)) + 2×2

<em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 3) + 8 (<em>n</em> + (<em>n</em> - 1) + (<em>n</em> - 2)) + 3×2

and so on. The pattern should be clear; we end up with

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 8 (<em>n</em> + (<em>n</em> - 1) + … + 3 + 2) + (<em>n</em> - 1)×2

The middle group is the sum,

\displaystyle 8\sum_{k=2}^nk=8\sum_{k=1}^nk-8=\frac{8n(n+1)}2-8=4n^2+4n-8

so that

<em>a</em> (<em>n</em>) = <em>a</em> (1) + (4<em>n</em> ² + 4<em>n</em> - 8) + 2 (<em>n</em> - 1)

<em>a</em> (<em>n</em>) = 4<em>n</em> ² + 6<em>n</em> - 3

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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