Answer:
41.4%
Step-by-step explanation:
The formula for determining the frequency: [tex]f(h)=440(2)^{\frac{h}{12}}[/tex] --A
where h is the number of half-steps from the A above middle C on the keyboard.
A note is six half-steps away from the A above middle C.
Now we are supposed to find How much greater is the frequency of the new note compared with the frequency of the A above middle C?
Now initially there is no half steps .
So, substitute h =0
[tex]f(0)=440(2)^{\frac{0}{12}}[/tex]
[tex]f(0)=440[/tex]
Now we are given that A note is six half-steps away from the A above middle C
So, substitute h =6
[tex]f(6)=440(2)^{\frac{6}{12}}[/tex]
[tex]f(6)=622.25[/tex]
Now To find change percentage
Formula: [tex]=\frac{\text{final} - \text{initial}}{\text{Initial}} \times 100[/tex]
[tex]=\frac{622.25- 440}{440} \times 100[/tex]
[tex]=0.414 \times 100[/tex]
[tex]=41.4\%[/tex]
Hence the frequency of the new note is 41.4% greater with the frequency of the A above middle C.
