Question

standard 15 homework
score: 6/21 answered: 7/20
question 9
consider the equation ( y = 3 sin left( \frac { 5 pi } { 3 } x + \frac { 10 pi } { 3 }
ight) + 7 ).
the amplitude is
the period is
the phase shift is units to the
the midline is ( y = )

Explanation:

Step1: Recall the formula for the period of a sine function

The general form of a sine function is \(y = A\sin(Bx - C)+D\). The period \(T\) is given by \(T=\frac{2\pi}{|B|}\).
In the given function \(y = 3\sin(\frac{5\pi}{3}x+\frac{10\pi}{3})+7\), \(B = \frac{5\pi}{3}\).
So, \(T=\frac{2\pi}{\frac{5\pi}{3}}\).

Step2: Calculate the period

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Step3: Recall the formula for the mid - line of a sine function

For the function \(y = A\sin(Bx - C)+D\), the mid - line is \(y = D\).
In the given function \(y = 3\sin(\frac{5\pi}{3}x+\frac{10\pi}{3})+7\), \(D = 7\).

Answer:

The period is \(\frac{6}{5}\) (or \(1.2\)), and the mid - line is \(y = 7\).