Advances in Applied Clifford Algebras was created in 1991.
Assuming the shape concerned is a circle, Radius = Circumference/(2*pi)and thenArea = pi*(radius)2= pi*(circumference)2/(2*pi)2= (circumference)2/(4*pi)Assuming the shape concerned is a circle, Radius = Circumference/(2*pi)and thenArea = pi*(radius)2= pi*(circumference)2/(2*pi)2= (circumference)2/(4*pi)Assuming the shape concerned is a circle, Radius = Circumference/(2*pi)and thenArea = pi*(radius)2= pi*(circumference)2/(2*pi)2= (circumference)2/(4*pi)Assuming the shape concerned is a circle, Radius = Circumference/(2*pi)and thenArea = pi*(radius)2= pi*(circumference)2/(2*pi)2= (circumference)2/(4*pi)
C = 2 pi R = 9.42 R = 9.42 / (2 pi) A = pi R2 = pi [ 9.42 / (2 pi) ]2 = (9.42)2 pi / 4 pi2 = (9.42)2 / (4 pi) = 7.0614 (rounded) ======================================== I just thought of something: C = 2 pi R A = pi R2 = 1/2 (2 pi R) x (R) = 1/2 (2 pi R) x (1/2pi) (2 pi R) = C/2 x C/(2 pi) = C2 / (4 pi)Let's see if this gives the same answer as above: C2 / (4 pi) = (9.42)2 / (4 pi) = 7.0614 Yay ! Next time, I'll remember that the area is (circumference2) divided by (4 pi).
11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4
Circumference = 2 pi R = 33R = 33 / (2 pi)Area = pi R2 = pi (33)2 / (2 pi)2= (33)2 / (4 pi) = 86.66 square units
Maria Fragoulopoulou has written: 'Topological algebras with involution' -- subject(s): Topological algebras 'An introduction of the representation theory of topological *-algebras' -- subject(s): Topological algebras, Representations of algebras
Helmut Strade has written: 'Modular lie algebras and their representations' -- subject(s): Modules (Algebra), Lie algebras, Representations of algebras
Advances in Applied Clifford Algebras was created in 1991.
Assuming the shape concerned is a circle, Radius = Circumference/(2*pi)and thenArea = pi*(radius)2= pi*(circumference)2/(2*pi)2= (circumference)2/(4*pi)Assuming the shape concerned is a circle, Radius = Circumference/(2*pi)and thenArea = pi*(radius)2= pi*(circumference)2/(2*pi)2= (circumference)2/(4*pi)Assuming the shape concerned is a circle, Radius = Circumference/(2*pi)and thenArea = pi*(radius)2= pi*(circumference)2/(2*pi)2= (circumference)2/(4*pi)Assuming the shape concerned is a circle, Radius = Circumference/(2*pi)and thenArea = pi*(radius)2= pi*(circumference)2/(2*pi)2= (circumference)2/(4*pi)
V. N. Gerasimov has written: 'Three papers on algebras and their representations' -- subject(s): Associative algebras, Radical theory, Representations of algebras
2 pi / 2 = pi.
Ion Suciu has written: 'Function algebras' -- subject(s): Function algebras
Bertram Yood has written: 'Banach algebras' -- subject(s): Banach algebras
G. R. Krause has written: 'Growth of algebras and Gelfand-Kirillov dimension' -- subject(s): Associative algebras, Dimension theory (Algebra), Lie algebras
No. It is circumference/(2*pi)No. It is circumference/(2*pi)No. It is circumference/(2*pi)No. It is circumference/(2*pi)
Marcelo Aguiar has written: 'Coxeter groups and Hopf algebras' -- subject(s): Coxeter groups, Hopf algebras 'Monoidal functors, species, and Hopf algebras' -- subject(s): Categories (Mathematics), Quantum groups, Combinatorial analysis, Hopf algebras, Symmetry groups
Yurij A. Drozd has written: 'Finite dimensional algebras' -- subject(s): Associative algebras