Answer
16π cm ≈ 50.2655 cm
Step-by-step explanation
To find the circumference of a circle, we can use the equation C = 2πr.
C stands for the circumference while r stands for the radius. We can see that there is a proportional positive linear relationship between radius and circumference for all circles, and that to find circumference when we have a radius value, we multiply the radius value by 2π.
The value of π, also called pi, is a constant and is the ratio of a circle's circumference to its diameter (the diameter is twice the radius, hence the 2 in the equation). Note that π is a constant and applies to all circles because all circles are similar.
Since we know the value of r, or the radius, given as 8 cm in the question, we can plug this value into the equation C = 2πr from earlier.
C = 2πr (plug in 8 cm for the radius)
C = 2π * 8
C = 16π cm
Since the radius is in units of cm (centimeters), the circumference is also in units of cm (centimeters).
16π cm is the exact value of the circumference. However, if we want this circumference in decimal form, we would multiply 16 by the decimal form of π which is approximately 3.1416. Note that π actually has an infinite amount of decimals and that this 3.1416 is actually a rounded π value
C = 16π
C ≈ 16 * 3.1416
C ≈ 50.2655 cm rounded to four decimal places
Answer:
Step-by-step explanation:
Area of a Circle: 
The r is the length of the radius of the circle
The radius of the shown circle is 9 m
Answer:
y = - 4
Step-by-step explanation:
Given
y = 2x + ( - 2) , that is
y = 2x - 2 ← substitute x = - 1
y = 2(- 1) - 2 = - 2 - 2 = - 4
This is confirmed by the point (- 1, - 4 ) on the graph
Answer:
a.8y³-6y
taking common
<u>2y(4y²-3)</u>
is a required answer
b.
3x²-20x+12
doing middle term factorization
3x²-18x-2x+12
3x(x-6)-2(x-6)
<u>(x-6)(3x-2)</u>
<u>is</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u><u> </u><u>.</u>
20:5
The total number of jelly beans is 20, and there are 5 orange jelly beans. This is a total to part ratio because it’s comparing the entirety to a part.
