0
y = A sin Bx y = A Cos Bx
Amplitude = A
Period = 2pi/B
so....
Amplitude = 1
Period = 2pi/8
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∙ 14y ago
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Given 2sin s equals cos s find all possible values of sin s?
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.
What is the integral of 1 divided by cos x multiplied by sin cubed x?
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
How do you find the area of a cone with only the angle degree and base radius given?
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.
Solve sec2x equals secx plus 2?
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.
What is the length in inches of a simple pendulum whose period s 1 s?
9.5 inches
What is cos 295?
Cos 295 fall s in the 4th quadrant where cosine is positive cos 295 = cos (360-295) = cos 65 = 0.4226
Do P-waves have a higher amplitude than S-waves?
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
A bob of mass 50 g oscillates as a simple pendulum with an amplitude 5 cm and period 2 s Calculate the instantaneous velocity of the bob and the tension in the supporting thread?
10000000000000000000000000000000000000to33333333333333333333333333333333333333330000000000000000000000000000000
Given 2sin s equals cos s find all possible values of sin s?
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.
An object is undergoing simple harmonic motion with period 1.210 s and amplitude 0.610 m. At t equals 0 the object is at x equals 0. How far is the object from the equilibrium position at time 0.485 s?
I got -0.495 m. I can't promise you this is correct, but here's my method:the position as a function of time is x(t)=A*cos(sqrt(k/m)*t)you already have A and t values, and you can solve for sqrt(k/m) by using the period they gave you.....T=2pi/(sqrt(k/m))sqrt(k/m)=2pi/TPlug and chug. Bada bing.
What is the oxidation number of CoS?
+2 for Co, -2 for S
Find the perimeter of the cord ra1 cos?
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
What is the data connection speed of AGP 8x?
Max data rate of 2133MB/s
Why is megaman blue?
Cos he`s into the music genre!!! I dont realy know, probably cos some game designer liked blue?
What is the integral of 1 divided by cos x multiplied by sin cubed x?
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
Is COS polar or non polar?
The Lewis structure for COS is as follows: .. .. O=C=S .. .. Therefore, the molecular geometry is linear. However, even though linear geometries normally represent nonpolar Lewis structures, O is much more electronegative than S, so COS is polar.
Why are Macclesfield Town rubbish?
cos its 's a town full of inbred twats.
