the equation y= 6 sin(3x-1)pi+4 has a fundamental period at
Assistance needed.
Please type your subject in the School Subject box. Any other words are likely to delay responses from a teacher who knows that subject well.
This is probably waaaaay to late,
but period can be determined using b=180/p
b=3
period is 60
To determine the fundamental period of the equation y = 6sin(3x-π) + 4, we need to consider the period of the standard sine function and how it affects the given equation.
The standard sine function, y = sin(x), has a fundamental period of 2π. This means that the graph of y = sin(x) repeats every 2π units along the x-axis.
In the given equation, y = 6sin(3x-π) + 4, the x-coordinate inside the sine function is (3x-π). To determine the fundamental period, we need to find the value of x that makes the argument of the sine function repeat.
To make the argument of the sine function repeat, we need to solve the equation 3x - π = 2π. Let's solve for x:
3x - π = 2π
3x = 3π
x = π
So, when x = π, the argument of the sine function repeats. Therefore, the fundamental period of the given equation is equal to the difference between two consecutive x-values where the argument of the sine function repeats.
The difference between two consecutive x-values where the argument of the sine function repeats is equal to the period of the standard sine function divided by the coefficient of x, which is 3 in this case. Thus, the fundamental period is:
Fundamental period = (2π) / 3
= (2/3)π
Therefore, the equation y = 6sin(3x-π) + 4 has a fundamental period of (2/3)π.