14a^2bc^4/21abc^6 = 2a/3b^2
1)
C ≈ 3.14 • 9
C ≈ 28.26
The circumference is 28.3 cm to the nearest tenth of a centimeter
2)
C ≈ 3.14 (2 • 13)
C ≈ 3.14 • 26
C ≈ 81.64
The circumference is 81.6 in to the nearest tenth of a inch
3)
C = 3.14 (13)
C = 40.82
C = 40.8
4)
C = 3.14 (2 • 5)
C = 3.14 (10)
C = 31.4
5)
C = 3.14 (2 • 1.5)
C = 3.14(3)
C = 9.42
C = 9.4
17. The letter is J
The letter appears 3 times
18. Lefthanded students = 4 students
<h3>How to determine the factors</h3>
17. We have that 1/4 of the months of the year start with the same letter
It is important to note that:
January, June and July all start with the same letter
The letter is J
The letter appears 3 times
18. Total number of students = 24
5/ 6 are right handed
Right handed students = 24 × 5/ 6 = 4 × 5 = 20 students
Lefthanded students = 24 - righthanded students
Lefthanded students = 24 - 20
Lefthanded students = 4 students
Thus, we can see that the letter that appear in 3 months of the year is J
Learn more about word problems here:
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It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is

It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.
Multiply the sum by 1/3 and you have

Now subtracting this from

gives

That is, all the matching terms will cancel. Now solving for

, you
have

