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vovangra [49]
3 years ago
12

How do I solve question 13 and 14?

Mathematics
1 answer:
xxMikexx [17]3 years ago
6 0

Answer:

13. 13 or 17 (not sure about this, sorry)

14. 28

Step-by-step explanation:

13.

a_{n} = \{a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, ... \} \\

if counting the sequence from a_{0} then a_{4} is 17, but if

a_{n} = \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, ... \} \\

counting from a_{1}, then a_{4} is 13.

14.

a_{10} = 1 + 3(10 - 1) = 1 + 3*9 = 28

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Jon recently drove to visit his parents who live 280 miles away. On his way there his average speed was 9 miles per hour faster
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Answer:

11 mph and 20 mph

Step-by-step explanation:

Represent his average speed going by r1 and his average speed returning by r2.  We know that r1 = r2 + 9.

Recall that distance = rate times time, so time = distance / rate.

Time spent going was (280 mi) / r1, or (280 mi) / (r2 + 9 mph).

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The total time was 14 hrs, so (280 mi) / (r2 + 9 mph) + (280 mi) / r2 = 14 hrs

Note that there is only one variable here:  r2.  Find r2, and then from r2, find r1:

Dividing all 3 terms by 14 hrs yields:

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The LCD here is r2(r2 + 9).  Thus, we have:

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Then 20(r2) = (r2)^2 + 9(r2).  This is reducible by dividing all terms by r2:

20 = r2 + 9, or 11 = r2.  Then r1 = 11 + 9, or 20.

The two rates were 11 mph and 20 mph.

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