What is the greatest common factor of 14 and 6?
2 answers:
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Answer:
Each term in the pattern will be odd.
Step-by-step explanation:
The first differences are ...
7 -3 = 4
15 -7 = 8
These differ by 4, and the second is double the first.
These relationships between the first differences give rise to two possible sequences: a) an exponential sequence; b) a quadratic sequence. We can use a graphing calculator to find the coefficients of each of the patterns.
<u>Exponential Sequence</u>
a[n] = 2·2^n -1
The sequence is 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047.
The 10th term is 2047.
<u>Quadratic Sequence</u>
a[n] = 2n^2 -2n +3
The sequence is 3, 7, 15, 27, 43, 63, 87, 115, 147, 183.
The 10th term is 183.
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Based on the above, the only thing we can say about these sequences is that they are all odd numbers.
Clearly, the 10th term is not 30. 7 is not a multiple of 3, so that observation fails immediately. 15 is not prime, so that observation also fails immediately.
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<em>Additional comment</em>
The differences of first differences are called "second differences". The differences of those are "third differences". and so on. If you keep taking differences of a sequence described by a polynomial, the n-th differences being constant means the sequence is described by an n-th degree polynomial.
Here, if we assume the 2nd differences are constant at 4, then the sequence is described by a 2nd-degree polynomial. (Similarly, constant first-differences are described by a 1st-degree polynomial--a linear function.)
If sequential levels of differences are all exponential sequences, then the sequence itself is an exponential sequence. As here, it may have a vertical offset. The base of the exponential will be the common ratio of the differences. (Here, the first differences have a ratio of 8/4 = 2, so the sequence could be exponential with a base of 2.)
Answer:
The depreciation each year is option B. $1500.00
Step-by-step explanation:
Initial value of car = $9600
Salvage value = $600
Salvage value is the final value of product after all the depreciation have been applied on the product.
So, total depreciation on the car =
Also, it is given that the depreciation occurs at a straight line method i.e. same depreciation occurs every year.
Depreciation in 6 years = $9000
Depreciation in 1 year =
Hence, option B. $1500 is the correct answer for the depreciation every year.