What is the fractional equivalent of the repeating decimal n = 0.5151...?
2 answers:
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Part a.
The function f(x) = sqrt(x-1) has the domain [1, infinity). We would solve x-1 >= 0 for x to get x >= 1 to ensure that the (x-1) expression is never negative. So the smallest x value we can plug in is x = 1. Recall that applying the square root to a negative number is not defined (assuming you are ignoring complex or imaginary numbers).
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Part b.
Pick any number you want. Then add on some other number. Let's say we pick 7 as our first number. Then let's say we add on 4. That gets us to 11. Add on 4 again and we jump up to 15. Do it again twice more and you have this sequence
7, 11, 15, 19, 23
which is arithmetic since we increase by the same amount (4) each time. The first term is 7 and the common difference is 4.
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Part c.
There are lot of options here. All we need to do is ensure that the slopes of each line are different. This will guarantee that the lines are not parallel. Non-parallel lines will always cross each other one time and one time only.
So one system we could have is
the slopes 2 and 6 are different so the system will have exactly one solution.