Brainliest to the one who can asnwer my question and show me the work step by step so that i can understand. i want to understand.

approximate the arc length of one-quarter of the unit circle (which is π /2) by computing the length of the polygonal approximation with n=4 segments. (as seen in picture)

the length of the line segment on the interval 0≤x≤0.25
the length of the line segment on the interval 0.25≤x≤0.5
the length of the line segment on the interval 0.5≤x≤0.75
the length of the line segment on the interval 0.75≤x≤1

Edited 2018-03-09 07:49

Given unit circle, so radius=1.

We calculate lengths of vertical segments, with the help of Pythagoras Theorem, based on a right triangle radiating from circle centre O, and hypotenuse from O to a point on the circumference.
AO=1 (given unit circle)
BB'=sqrt(1^2-0.25^2)=0.968246
CC'=sqrt(1^2-0.5^2)=0.866025
DD'=sqrt(1^2-0.75^2)=0.661438
EE'=0

Now we proceed to calculate the segments approximating the arc.  Again, we use a right triangle in which the hypotenuse is the segment joining two points on the circumference.  The height is the difference between the two vertical segments, and the base is 0.25 for all four segments.
AB=sqrt((AO-BB)^2+0.25^2)=0.252009BC=sqrt((BB-CC)^2+0.25^2)=0.270091CD=sqrt((CC-DD)^2+0.25^2)=0.323042DE=sqrt((DD-0)^2+0.25^2)=0.7071068

giving a total estimation of the arc length
approximation of arc=AB+BC+CD+DE=1.55225

Period is the time it takes for one “cycle” to complete. in the parent sine function, period is 2*pi since sin(0) = 0, sin(pi/2) = 1, sin(pi) = 0, sin(3pi/2) = -1, and sin(2pi) = 0. at that point, the sequence repeats itself. i’m not sure if your graph is similar to the sine function since you didn’t provide a picture, but you can find the period by looking at a graph.