2sin s = cos s; whence, 4sin2 s = cos2 s = 1 - sin2s, and 5sin2 s = 1. Therefore sin s = ±√0.2 = 0.4472, approx. Check: cos2 s = 1 - 0.2 = 0.8, whence, cos s = ±√0.8 = ±2√0.2 = 2 sin s.
Given the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + c
Area of the cone part is pi*r*s (s is the slant height which is the hypotenuse of the triangle formed by the radius and the altitude). cosine(theta) = r/s, so s = r/cos(theta), and A = pi*r*s = pi*r^2 / cos(theta). The area of the circle part is given by pi*r^2. Then just add them together.
You must mean, either,(1) sec2 x = sec x + 2, or(2) sec(2x) = sec x + 2.Let's first assume that you mean:sec2 x = sec x + 2;whence,if we let s = sec x, we have,s2 = s + 2,s2 - s - 2 = (s - 2)(s + 1) = 0, ands = 2 or -1; that is,sec x = 2 or -1.As, by definition,cos x = 1/sec x ,this means thatcos x = ½ or -1.Therefore, providing that the first assumption is correct,x = 60°, 180°, or 300°; or,if you prefer,x = ⅓ π, π, or 1⅔ π.Now, let's assume, instead, that you mean:sec(2x) = sec x + 2;whence,1/(cos(2x) = (1/cos x) + 2.If we let c = cos x,then we have the standard identity,2c2 - 1 = cos (2x); and,thus, it follows that1/(cos(2x) = 1/(2c2 - 1)= (1/c) + 2 = (1 + 2c)/c.This gives,1/(2c2 - 1) = (1 + 2c)/c;(2c2 - 1)(2c + 1) = 4c3 + 2c2 - 2c - 1 = c; and4c3 + 2c2 - 3c - 1 = (c + 1)(4c2 - 2c - 1) = 0.As our concern is only with real roots,c = cos x = -1; and,therefore, providing that the second assumption is correct,x = 180°; or,if you like,x = π.
9.5 inches
Cos 295 fall s in the 4th quadrant where cosine is positive cos 295 = cos (360-295) = cos 65 = 0.4226
At time t = 0.485 s, the object is 0.610 m away from the equilibrium position, because the displacement at any time is given by x(t) = A * sin(2πt/T), where A is the amplitude and T is the period. Substituting t = 0.485 s, A = 0.610 m, and T = 1.210 s into the equation will give you the position of the object at that time.
2sin s = cos s; whence, 4sin2 s = cos2 s = 1 - sin2s, and 5sin2 s = 1. Therefore sin s = ±√0.2 = 0.4472, approx. Check: cos2 s = 1 - 0.2 = 0.8, whence, cos s = ±√0.8 = ±2√0.2 = 2 sin s.
S-waves actually have a higher amplitude, despite being slower than P-waves. It is this amplitude that indicates stress, which is why S-waves can't travel through liquids, as liquids cannot support the stresses of S-waves
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Max data rate of 2133MB/s
When the amount of energy in an S wave increases, the amplitude of the wave also increases. This means that the S wave will have a greater maximum displacement from its resting position as it carries more energy.
r=a(1+cos x) r^2=a^2(1+cos x)^2 = a^2 + 2[a^2][cos(x)] + [a^2][cos^2(x)] dr/dx = r' = -a sin(x) (r')^2 = [a^2][sin^2 (x)] Therefore perimeter (s) of curve r=a(1+cos x) in polar coordinate with x vary from 0 to Pi (due to curve is symmetry on axis x=0) is s = 2 Int { sqrt[r^2 + (r')^2] } dx where x vary from 0 to Pi. Thus sqrt[r^2 + (r')^2] = sqrt { a^2 + 2[a^2][cos(x)] + [a^2][cos^2 (x)] + [a^2][sin^2 (x)] } = sqrt { (2a^2)[1+cos(x)] } = [sqrt(2)]a {sqrt [1+cos(x)]} Then s = 2[sqrt(2)]a . Int {sqrt [1+cos(x)]} dx Let 1+cos(x) = 1+2cos^2 (x/2) - 1 = 2cos^2 (x/2) s = 2[sqrt(2)]a . Int {sqrt [2cos^2 (x/2)]} dx s = 4a . Int [cos(x/2)] dx where x vary from 0 to Pi s = 4a [sin(x/2)]/(1/2) s = 8a [sin(Pi/2) - sin(0)] s = 8a
Cos he`s into the music genre!!! I dont realy know, probably cos some game designer liked blue?
Given the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + c
cos its 's a town full of inbred twats.
cos that s life