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amplitude of the function y =-3 sin 3x
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How do you find the amplitude maximum minimum and period for y equals -1 plus 3sin4x?
y = -1 + 3 sin 4xLet's look at the equation of y = 3 sin 4x, which is of the form y = A sin Bx, wherethe amplitude = |A|, and the period = (2pi)/B.So that the amplitude of the graph of y = 3 sin 4x is |3| = 3, which tell us that the maximum value of y is 3 and the minimum value is -3, and the period is (2pi)/4 = pi/2, which tell us that each cycle is completed in pi/2 radians.The graph of y = -1 + 3 sin 4x has the same amplitude and period as y = 3 sin 4x, and translates the graph of y = 3 sin 4x one unit down, so that the maximum value of y becomes 2 and the minimum value becomes -4.
Find the inverse of f x 3x plus 9?
f(x) = 3x + 9 y = 3x + 9 so 3x = y - 9 x = y/3 - 3 Therefore, the inverse function is g(x) = x/3 - 3
How do you solve sin parenthesis arcsin parenthesis 2 over 3 closed parenthesis and another closed parenthesis?
sin(arcsin(2/3)) = 2/3, since sin is the inverse function of arcsin.
Put x-3y equals -9 in function form?
y=1/3x+3
How do you graph y equals 4 sin 3x?
The simplest way is to use a graphing calculator such as a TI-83. To enter in the graph do the following... 1) Hit "Y=" (It should be located in the upper left hand corner) 2) Enter the function = 4 sin (3x) Use the X,T,Theta,N button for "x" 3) Hit "Graph" Please note, make sure your calculator is in Degree Mode, and the graph is set to a "Functional" graph. To check this hit the mode button. Degree and Func should be highlighted. -------------------------------------------------------------------------- You can also draw this by hand here's how... First you need to understand the important values of sin x sin(0) = 0 sin(30) = ½ sin(60) = √3 / 2 sin(90) = 1 sin(120) = √3 / 2 sin(150) = ½ sin(180) = 0 These are important because they are part of the unit circle. Notice the repeating pattern. The important points are 0, 30, 90, 150, 180 We can plot those on a graph then we see an oscillating wave that repeats. But this would be for ƒ(x) = sin (x) Instead the 3x on the inside means we are looking for values which make our sin the same We find these by dividing the special points by 3. 0,10,30,50,60 So on those x values we will put a coordinates. Now we have to determine the y values of the coordinates. To find these we just multiply by the coefficient 4. 4 sin (3*00) = 0 4 sin (3*10) = 4/2 = 2 4 sin (3*30) = 4 4 sin (3*50) = 4/2 = 2 4 sin (3*60) = 0 Now we have our points (00,00) (10,02) (30,04) (50,02) (60,00) We plot these and then connect them on a graph to create an oscillating wave...
What is the amplitude of the function y 3 sin 3x?
3
Parentheses before -1 and after 3x and before 3 and after 3x What is the answer to -2 cos2 3x-3 sin 3x equals 0?
I think you mean to solve:(-2 cos23x) - (3 sin 3x) = 0cos2 x + sin2 x = 1⇒ cos2x = 1 - sin2 x⇒ -2 cos2 3x - 3 sin 3x = -2(1 - sin2 3x) - 3 sin 3x = 0⇒ 2 sin2 3x - 3 sin 3x - 2 = 0⇒ (2 sin 3x + 1)(sin 3x - 2) = 0⇒ sin 3x = -1/2 or 2sin 3x = 2 is impossible as the range of sine is -1 ≤ sine ≤ 1Thus:sin 3x = -1/2⇒ 3x = 2nπ - π/6 or 2nπ - 5π/6⇒ x = 2/3nπ -π/18 or 2/3nπ -5π/18
How do you use the Euler's formula to obtain the sin3X in terms of cosX and sinX?
Using Euler's Formula, you use (cos(x) + i sin(x))^n = cos (nx) + i sin(nx) Now you let n=3 (cos(x) + i sin (x))3 = cos(3x) + i sin (3x) (cos(x))3 + 3(cos(x))2 * i sin(x) + 3cos(x) * i2 (sin(x))3 = cos(3x)+ i sin(3x) (cos(x))3 + i(3sin(x)(cos (x))2) - 3cos(x)(sin(x)2) - i(sin(x))3 = cos (3x) + i sin(3x) Now only use the terms with i in them to figure out what sin(3x) is... 3sin(x)(cos(x))2 - (sin(x))3 = sin(3x) Hope this helps! :D
Find the derivatives of y equals 2 sin 3x and show the solution?
y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x
Find the derivatives of y equals 2 sin 3x?
y=2 sin(3x) dy/dx = 2 cos(3x) (3) dy/dx = 6 cos(3x)
How would you evaluate the indefinite integral -2xcos3xdx?
Integrate by parts: ∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx Let u = -2x Let v = cos 3x → u' = d/dx -2x = -2 → ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx = -2x/3 sin 3x - ∫ -2/3 sin 3x dx = -2x/3 sin 3x - 2/9 cos 3x + c
What is the inverse function of y equals 3x-3?
6
What are function rules?
y=3x=3
How do you find the amplitude maximum minimum and period for y equals -1 plus 3sin4x?
y = -1 + 3 sin 4xLet's look at the equation of y = 3 sin 4x, which is of the form y = A sin Bx, wherethe amplitude = |A|, and the period = (2pi)/B.So that the amplitude of the graph of y = 3 sin 4x is |3| = 3, which tell us that the maximum value of y is 3 and the minimum value is -3, and the period is (2pi)/4 = pi/2, which tell us that each cycle is completed in pi/2 radians.The graph of y = -1 + 3 sin 4x has the same amplitude and period as y = 3 sin 4x, and translates the graph of y = 3 sin 4x one unit down, so that the maximum value of y becomes 2 and the minimum value becomes -4.
How do you write this equation 3x equals 2y plus 3 in function form?
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
How do you write 3X equals 2Y plus 3 in function form?
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
Integration cos3 x?
Take f(x) = cos(3x) ∫ f(x) dx = ∫ cos(3x) dx Take u=3x → du = 3dx = ∫ 1/3*cos(u) du = 1/3*∫ cos(u) du = 1/3*sin(u) + C, C ∈ ℝ = 1/3*sin(3x) + C
