The area of a circle with radius 5 is 25 pi. Concentric circles with radius 3 and 4 have areas of 9 pi and 16 pi. The concentric circle with radius four consumes the circle with radius 3. 25 pi minus 16 pi leaves 9 pi of the circle with radius 5 left over. 16 pi is slightly over three-fifths of the circle with radius 5.
Tan(Pi/5) = √(5-2*√(5)) ~= 0.7265
Pi / 5 would be in Quadrant I.
1.25
7*pi/5 = 1.4*pi pi < 1.4*pi < 1.5*pi and so the angle is in the third quadrant.
The area of a circle with radius 5 is 25 pi. Concentric circles with radius 3 and 4 have areas of 9 pi and 16 pi. The concentric circle with radius four consumes the circle with radius 3. 25 pi minus 16 pi leaves 9 pi of the circle with radius 5 left over. 16 pi is slightly over three-fifths of the circle with radius 5.
X = V0t + 1/2 a t250 pi = 5V0 + (3.1) (25)50 pi = 5V0 + 77.5-5 V0 = 77.5 - 50 piV0 = 10 pi - 15.5 = 10 (pi - 1.55)At start of 5 sec: V0 = 15.916 rad/secAt end of 5 sec: Vf = V0 + 5 a = 15.916 + 5(6.2) = 46.916rad/sec==========================================Check:Vavg = 1/2(Vi + Vf) = 1/2(15.916 + 46.916) = 1/2(62.832) = 31.416 rad/secTotal rads in 5 sec = 5 Vavg = 5 x 31.416 = 157.08 radsRevs = rads/(2 pi) = 157.08 / (2 pi) = 25.00006 revs. Close enough. Yay !
Tan(Pi/5) = √(5-2*√(5)) ~= 0.7265
Pi / 5 would be in Quadrant I.
1.25
Pi / 5 would be in Quadrant I.
pi = 4/1 - 4/3 + 4/5 - 4/7 ...
7*pi/5 = 1.4*pi pi < 1.4*pi < 1.5*pi and so the angle is in the third quadrant.
4-(4/3)+(4/5)-(4/7)+(4/9) etc. =pi
4-(4/3)+(4/5)-(4/7)+(4/9)-(4/11)....=pi There are no non-infinite serieses known for finding pi
The four roots are cos(theta)+i*sin(theta) where theta = pi/4, 3*pi/4, 5*pi/4 and 7*pi/4.
Yes. nonzero number: -4, -0.5, 5, pi, 30 absolute number: |-4| = 4 |-0.5| = 0.5 |5| = 5 |pi| = pi |30| = 30