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3 years ago

Two buses leave a station at the same time and travel in opposite directions. One bus travels

Mathematics

1 answer:

natima [27]3 years ago

4 0

Answer:

63 miles per hour

54 miles per hour

Step-by-step explanation:

Let the two buses be represented as bus A and bus B.

bus A and bus B leave a station at the same time and travel in opposite directions.

Let the speed of bus A be A miles per hour.

Let the speed of bus B be B miles per hour.

One bus travels 9 miles/hour slower than the other.

Let bus A be the faster bus.

This means that the speed of bus B is B = A - 9

If the two buses are 585 miles apart after 5hours, this means that after 5hours, the total distance travelled by bus A and bus B is 585 miles.

Distance travelled by bus A in 5 hours = speed × time = 5A

Distance travelled by bus B in 5 hours = speed × time = 5B

5A + 5B = 585 - - - - - -1

Substituting B = A - 9 in equation 1

5A + 5(A - 9) = 585

5A + 5A - 45 = 585

10A = 585 + 45 = 630

A = 630/10 = 63 miles per hour

B = A - 9 = 63 - 9 = 54 miles per hour

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3 years ago

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4 years ago

Find the first five terms of the sequence in which a1 = 6 and an = –3an – 1 – 12, if n ≥ 2.

KatRina [158]

Answer:

$\huge\boxed{6,\ -30;\ 78;\ -246;\ 690}$

Step-by-step explanation:

$a_1=6\\a_n=-3a_{n-1}-12\qquad n\geq2\\\\\text{Substitute}\ n=2,\ n=3,\ n=4,\ n=5\ \text{to}\ a_n=-3a_{n-1}-12:\\\\\text{for}\ n=2\\\\a_2=-3a_{2-1}-12=-3a_1-12\\\\\text{substitute}\ a_1=6\\\\a_2=-3(6)-12=-18-12=-30\\\\\text{for}\ n=3\\\\a_3=-3a_{3-1}-12=-3a_2-12\\\\\text{substitute}\ a_2=-30\\\\a_3=-3(-30)-12=90-12=78$

$\text{for}\ n=4\\\\a_4=-3a_{4-1}-12=-3a_3-12\\\\\text{substitute}\ a_3=78\\\\a_4=-3(78)-12=-234-12=-246\\\\\text{for}\ n=5\\\\a_5=-3a_{5-1}-12=-3a_4-12\\\\\text{substitute}\ a_4=-246\\\\a_5=-3(-246)-12=702-12=690$

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3 years ago

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