The frequency of a4 is 440.00 hz. what is the frequency of a6, which is two octaves above a4?

To solve this problem you must apply the proccedure shown below:

 1. You have the following information given in the problem above: The frequency of A4 is 440.00 Hz.

 2. So, you must apply the following formula to calculate the frequency of A6, which is two octaves above:

 (2^n)f

 Where n is the octaves above and f is the known frequency.

 3. Therefore, you have:

 A6=(2^n)f
 A6=(2^2)(440.00 Hz)
 A6=1760 Hz

 Therefore, the answer is: A6=1760 Hz.

The formula for getting the unknown frequecy is given by, (2∧n)×f.
Where n ⇒the number of octavos above A4 and 
            f ⇒the given frequency

So, the frequency of  A6 is calculated as follows. 

Frequency = (2∧2)×440.00
                  = 1760.00 Hz

I’m going to infer you said parallel.solve for yy= -1/2x+7/4lines that are parallel have the same slope, different y intercept.so we know m is -1/2then from there solve for y intercept.

The answer to your question is d 68%