Maria runs 7 miles in 80 minutes. at the same rate, how many miles would she run in 64 minutes
1 answer:
Answer:
5.6 miles
Step-by-step explanation:
Firstly, we need to find the unit rate (the number of miles in 1 minutes).
To get this, we need to divide total miles ran (7 miles) by the total minutes taken (80 minutes). Thus, we will have MILES PER MINUTE.
7 miles ÷ 80 minutes = 7/80 miles per minute
Now,
We want number of miles in 64 minutes. Since we know "PER MINUTE", we simply multiply that with 64 to get number of miles in that amount of time.
THus,
So, in 64 minutes, Maria can run 5.6 miles
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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,
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