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Pie is 3.14 etc etc etc.
Its used in all the formulas to calculate the sums of a circle.
Including Diameters, Radiuses and arcs & etc.
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∙ 14y ago
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When was Advances in Applied Clifford Algebras created?
Advances in Applied Clifford Algebras was created in 1991.
How do you work out the area if given the circumference?
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)
If the circumference of a circle is 9.42 what is its area?
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).
What is the exact value using a sum or difference formula of the expression cos 11pi over 12?
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
What is the area of a circle with a circumference of 33 units?
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
What has the author Maria Fragoulopoulou written?
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
What has the author Helmut Strade written?
Helmut Strade has written: 'Modular lie algebras and their representations' -- subject(s): Modules (Algebra), Lie algebras, Representations of algebras
When was Advances in Applied Clifford Algebras created?
Advances in Applied Clifford Algebras was created in 1991.
How do you work out the area if given the circumference?
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)
What has the author V N Gerasimov written?
V. N. Gerasimov has written: 'Three papers on algebras and their representations' -- subject(s): Associative algebras, Radical theory, Representations of algebras
What is 2 pi over 2?
2 pi / 2 = pi.
What has the author Ion Suciu written?
Ion Suciu has written: 'Function algebras' -- subject(s): Function algebras
What has the author Bertram Yood written?
Bertram Yood has written: 'Banach algebras' -- subject(s): Banach algebras
What has the author G R Krause written?
G. R. Krause has written: 'Growth of algebras and Gelfand-Kirillov dimension' -- subject(s): Associative algebras, Dimension theory (Algebra), Lie algebras
Is the radius one fourth of the circumference?
No. It is circumference/(2*pi)No. It is circumference/(2*pi)No. It is circumference/(2*pi)No. It is circumference/(2*pi)
What has the author Marcelo Aguiar written?
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
What has the author Yurij A Drozd written?
Yurij A. Drozd has written: 'Finite dimensional algebras' -- subject(s): Associative algebras
