Circumference of a circle - derivation
This page describes how to derive the formula for the circumference of a circle.
Recall that the definition of pi (π) is the circumference c of any circle divided by its diameter d. Put as an equation, pi is defined as
π
=
c
d
Rearranging this to solve for c we get
c
=
π
d
The diameter of a circle is twice its radius, so substituting 2r for d
c
=
2
π
r
If you know the area
Recall that the area of a circle is given by
area
=
π
r
2
Solving this for r
r
2
=
a
π
So
r
=
√
a
π
The circumference c of a circle is
c
=
2
π
r
Answer:
1. 6x + 25
2. 60 + 28k
3. 5 + 19y
4. 3d + 1
5. 5v
6. w - 14.04
Step-by-step explanation:
These are what I got. I'm very sorry if I'm wrong.
Answer:
Thank you, person.
I hope this is good enough:
you want to talk about anything
Answer:
1. D. 20, 30, and 50
2. A. 86
3. B. 94
Step-by-step explanation:
1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.
The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.
Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.
Thus:
Q1 = (85 + 87)/2 = 86
Q3 = (93 + 95)/2 = 94
IQR = Q3 - Q1 = 94 - 86
IQR = 8
Outliers in the data set are data values below the lower limit or above the upper limit.
Let's find the lower and upper limit.
Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74
The data values below the lower limit (74) are 20, 30, and 50
Let's see if we have any data value above the upper limit.
Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106
No data value is above 106.
Therefore, the only outliers of the data set are:
D. 20, 30, and 50
2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.
Thus:
Q1 = (85 + 87)/2 = 86
3. Q3 = (93 + 95)/2 = 94
