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galben [10]
3 years ago
6

Determine whether the sequence given below forms arithmetic progressions. If yes, identify the common difference d, and find the

26th term. -12, -6, 0, 6, 12
Mathematics
1 answer:
masha68 [24]3 years ago
6 0

Answer:

arithmetic progression with d = 6

Step-by-step explanation:

If the progression has a common difference d between consecutive terms then it is arithmetic

a₂ - a₁ = - 6 - (- 12) = - 6 + 12 = 6

a₃ - a₂ = 0 - (- 6) = 0 + 6 = 6

a₄ - a₃ = 6 - 0 = 6

a₅ - a₄ = 12 - 6 = 6

There is a common difference of 6 between consecutive terms.

Thus the sequence is arithmetic

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It took Amir 2 hours to hike 5 miles. On the first part of the hike, Amir averaged 3 miles per hour. For the second part of the
aivan3 [116]

Answer:

Times

first part   x  =  1,33 h       second part  y =  0,66 h

distances  d₁ (first part ) d₁ = 4 miles    d₂  (second part )  d₂ = 0,999 miles

Step-by-step explanation: IMPORTANT NOTE: EVEN PROBLEM STATEMENT DID NOT ASK ANY QUESTION WE WILL ASSUME QUESTION ARE: LENGTH ( IN MILES ) AND SPENT TIME  OF EACH PART OF THE HIKE

We formulate a two-equation system according to:

x = time for the first part

y = time for the second part

then      x  +  y  =  2           or      y = 2 - x

And 3 m/h * x(h)   +  1,5 m/h * y (h)  = 5

3*x  +  1,5*y  = 5

3*x + 1,5 * ( 2 - x )  = 5

3*x  + 3  - 1,5*x  =  5

1,5*x  = 2

x  = 2/1,5 (h)

x = 1,33 (h)

And  y  =  2 -  1,33       y  =  0,66 (h)

Distances are

first part  d₁ = 3*1,33          d₁ =  4 miles

second part  d₂ = 1,5*0,66        d₂  = 0,999 miles

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The percent change between the temperature at noon and the temperature at 5:00 p.m. is -20%

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Given that the At noon, the temperature was 80°F while At 5:00 p.m., the temperature was 64°F. Hence:

Percent change = [(64°F - 80°F) / 80°F] * 100%

Percent change = -20%

Therefore the percent change between the temperature at noon and the temperature at 5:00 p.m. is -20%

Find out more at: brainly.com/question/17968508

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