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If we have 2 x 4 + pi, we'll have 8 + pi, or about 11.1416 for an answer.
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How do you multiply two times pi times two?
2 x pi x 2 = 4 x pi
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
How was it determined that the tangent of pi divided by 4 is 1?
As tan(x)=sin(x)/cos(x) and sin(pi/4) = cos(pi/4) (= sqrt(2)/2) then tan(pi/4) = 1
What is the area of circles with a radius of 4 inches and with a diameter of 4 inches and with a circumference of 4 inches?
The area (A) of a circle is pi (about 3.14) times the square of the radius (r). The circumference (c) of a circle is pi times its diameter (d), or pi times twice its radius. Knowing that, we can investigate the three circles. If rcircle = 4, Areacircle = pi r2 = 3.14 x 42 = 3.14 x 16 = 50.27 sq. in. If dcircle = 4, Areacircle = pi (d/2)2 = 3.14 x (4/2)2 = 3.14 x (2)2 = 3.14 x 22 = 3.14 x 4 = 12.57 sq. in. If c - 4, Areacircle = pi (c/2 pi)2 = pi (42/4 pi2) = pi x 16/ 4 x pi2 = 16/ 4 x pi = 4/3.14 = 1.27 sq. in.
How do you solve 2cosx plus 2sinx equals sqrt2 for solutions between 0 and 2pi?
Not easily and, given the limitations of the pathetic browser that we have at our disposal, even less easy. But here goes:You are given that 2cosx + 2sinx = sqrt(2)Consider "collapsing" the left hand side into a single trig function:r*sin(x+a) = sqrt(2) ............................................. (A)that is r*cosxsina + r*sinxcosa = sqrt(2)comparing coefficients,rsina = 2 and rcosa = 2thensina = 2/r = cosaand since sin^2 + cos^2 = 1, you have r = 2*sqrt(2)also, rsina/rcosa = 2/2 = 1 = tana which implies that a = pi/4.Therefore, equation (A) is 2*sqrt(2)*sin(x+pi/4) = sqrt(2)so that sin(x+pi/4) = 1/2.......................................(B)Since 0
The radius of a circle is x plus 4 what is the circumference?
Circumference = (2) x (pi) x (Radius)= (2 pi) times (X + 4)= same as (2 pi X) + (8 pi)= same as [ pi (2X + 8) ]
Sin x - cos x 0 0?
sin x - cos x = 0sin x = cos x(sin x)^2 = (cos x)^2(sin x)^2 = 1 - (sin x)^22(sin x)^2 = 1(sin x)^2 = 1/2sin x = ± √(1/2)sin x = ± (1/√2)sin x = ± (1/√2)(√2/√2)sin x = ± √2/2x = ± pi/4 (± 45 degrees)Any multiple of 2pi can be added to these values and sine (also cosine) is still ± √2/2. Thus all solutions of sin x - cos x = 0 or sin x = cos x are given byx = ± pi/4 ± 2npi, where n is any integer.By choosing any two integers , such as n = 0, n = 1, n = 2 we can find some solutions of sin x - cos x = 0.n = 0, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(0)(pi) = ± pi/4 ± 0 = ± pi/4n = 1, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(1)(pi) = ± pi/4 ± 2pi = ± 9pi/4n = 2, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(2)(pi) = ± pi/4 ± 4pi = ± 17pi/4
How do you multiply two times pi times two?
2 x pi x 2 = 4 x pi
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
How was it determined that the tangent of pi divided by 4 is 1?
As tan(x)=sin(x)/cos(x) and sin(pi/4) = cos(pi/4) (= sqrt(2)/2) then tan(pi/4) = 1
What is the area of circles with a radius of 4 inches and with a diameter of 4 inches and with a circumference of 4 inches?
The area (A) of a circle is pi (about 3.14) times the square of the radius (r). The circumference (c) of a circle is pi times its diameter (d), or pi times twice its radius. Knowing that, we can investigate the three circles. If rcircle = 4, Areacircle = pi r2 = 3.14 x 42 = 3.14 x 16 = 50.27 sq. in. If dcircle = 4, Areacircle = pi (d/2)2 = 3.14 x (4/2)2 = 3.14 x (2)2 = 3.14 x 22 = 3.14 x 4 = 12.57 sq. in. If c - 4, Areacircle = pi (c/2 pi)2 = pi (42/4 pi2) = pi x 16/ 4 x pi2 = 16/ 4 x pi = 4/3.14 = 1.27 sq. in.
How do you solve 2cosx plus 2sinx equals sqrt2 for solutions between 0 and 2pi?
Not easily and, given the limitations of the pathetic browser that we have at our disposal, even less easy. But here goes:You are given that 2cosx + 2sinx = sqrt(2)Consider "collapsing" the left hand side into a single trig function:r*sin(x+a) = sqrt(2) ............................................. (A)that is r*cosxsina + r*sinxcosa = sqrt(2)comparing coefficients,rsina = 2 and rcosa = 2thensina = 2/r = cosaand since sin^2 + cos^2 = 1, you have r = 2*sqrt(2)also, rsina/rcosa = 2/2 = 1 = tana which implies that a = pi/4.Therefore, equation (A) is 2*sqrt(2)*sin(x+pi/4) = sqrt(2)so that sin(x+pi/4) = 1/2.......................................(B)Since 0
What is radius of circle with area of 4?
1.2 Area = pi X radius^2 So set Area = 4, and figure it out: 4 = pi X radius ^2 4/pi = radius^2
What is the formula for the area of a sphere?
Area = 4 x pi x radius2.
What is the formula for an area of a circumference?
(i) Circumference = pi x Diameter (ii) Area = pi x Radius^2 From (i), D = C/pi R= (C/pi)/2 Substituting into (ii), Area = pi x ((C/pi)/2)^2 Simplifying: A = (C^2) / (4 x pi)
What is a polynomial for the area of a circle with radius(x plus 2) feet?
A = pi()*(x2 + 4x +4) square feet.
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).
