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Semmy [17]
1 year ago
14

On a bike ride, calvin starts at home and goes up a long hill for 30 minutes at just 6 mph. at the top, he turns around and ride

s home along the same path at a speed of 18 mph. what is his average speed for the round-trip?
Mathematics
2 answers:
Masja [62]1 year ago
5 0

9mph is the average speed  for the round-trip.

<h3>What is Average speed?</h3>

The average speed is the total distance traveled by the object in a particular time interval.

Given that Calvin's first leg of the trip took 0.5 hr (30 minutes) at 6 mph.  That means he travelled (0.5hr x 6 mph =) 3 miles.  

He travelled 3 miles on the return trip, by definition (same path home).  

At 18 mph, 3 miles would take (3 miles/18 mph =) 0.1667 hours.  

That makes a total of 6 miles

0.5 + 0.1667= 0.6667 hours.

(6 miles)/(0.6667 hours) = 9 mph

9mph is Calvin's average speed for the round trip.

To learn more on Average speed click:

brainly.com/question/12322912

#SPJ2

omeli [17]1 year ago
3 0

Answer:

<u>9 mph</u>

Step-by-step explanation:

Calvin's first leg of the trip took 0.5 hr (30 minutes) at 6 mph.  That means he travelled (0.5hr x 6 mph =) 3 miles.  He travelled 3 miles on the return trip, by definition (same path home).  At 18 mph, 3 miles would take (3 miles/18 mph =) 0.1667 hours.  That makes a total of 6 miles and (0.5 + 0.1667=) 0.6667 hours.

(6 miles)/(0.6667 hours) = 9 mph is Calvin's average speed for the round trip.

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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each verbal description of a sequen
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I think the question is wrong so, I will try and explain with some right questions

Step-by-step explanation:

We are give 6 sequences to analyse

1. an = 3 · (4)n - 1

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3. an = 2 · (3)n - 1

4. an = 4 + 2(n - 1)

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1. This is the correct sequence

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If this is an

Let know an+1, the next term

an+1=3•(4)^(n+1-1)

an+1=3•(4)^n

There fore

Common ratio an+1/an

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Then

First term is when n=1

an=3•(4)^(n-1)

a1=3•(4)^(1-1)

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2. This is the correct sequence

an=4•(2)^(n-1)

Therefore, let find an+1

an+1=4•(2)^(n+1-1)

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Common ratio=an+1/an

r=4•2ⁿ/4•(2)^(n-1)

r=2^(n-n+1)

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Then the common ratio is 2,

The first term is when n =1

an=4•(2)^(n-1)

a1=4•(2)^(1-1)

a1=4•(2)^0

a1=4

It is geometric progression with first term 4 and common ratio 2.

3. This is the correct sequence

an=2•(3)^(n-1)

Therefore, let find an+1

an+1=2•(3)^(n+1-1)

an+1= 2•3ⁿ

Common ratio=an+1/an

r=2•3ⁿ/2•(3)^(n-1)

r=3^(n-n+1)

r=3¹=3

Then the common ratio is 3,

The first term is when n =1

an=2•(3)^(n-1)

a1=2•(3)^(1-1)

a1=2•(3)^0

a1=2

It is geometric progression with first term 2 and common ratio 3.

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Let find an+1

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Let find common difference(d) which is an+1 - an

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an = 2 + 3(n - 1)

Let find an+1

an+1= 2+3(n+1-1)

an+1= 2+3n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=2+3n-(2+3(n-1))

d=2+3n-2-3(n-1)

d=2+3n-2-3n+3

d=3.

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Now, the first term is when n=1

an=2+3(n-1)

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6. I think this correct sequence so we will use it.

an = 3 + 4(n - 1)

Let find an+1

an+1= 3+4(n+1-1)

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This is not GP

Let find common difference(d) which is an+1 - an

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d=3+4n-(3+4(n-1))

d=3+4n-3-4(n-1)

d=3+4n-3-4n+4

d=4.

The common difference is 4

Now, the first term is when n=1

an=3+4(n-1)

a1=3+4(1-1)

a1=3+4(0)

a1=3

This is an arithmetic progression of common difference 4 and first term 3.

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