Answer: How to solve for FX and FY?
to find fx(x, y): keeping y constant, take x derivative; • to find fy(x, y): keeping x constant, take y derivative. f(x1,...,xi−1,xi + h, xi+1,...,xn) − f(x) h . ∂y2 (x, y) ≡ ∂ ∂y ( ∂f ∂y ) ≡ (fy)y ≡ f22. similar notation for functions with > 2 variables.
Explanation:
Answer:
The minimum frequency is 702.22 Hz
Explanation:
The two speakers are adjusted as attached in the figure. From the given data we know that
=3m
=4m
By Pythagoras theorem

Now
The intensity at O when both speakers are on is given by

Here
- I is the intensity at O when both speakers are on which is given as 6

- I1 is the intensity of one speaker on which is 6

- δ is the Path difference which is given as

- λ is wavelength which is given as

Here
v is the speed of sound which is 320 m/s.
f is the frequency of the sound which is to be calculated.

where k=0,1,2
for minimum frequency
, k=1

So the minimum frequency is 702.22 Hz
Explanation:
To find the volume of a box of Tissue (also known as a rectangular prism) just multiply the area of the base (length times width) by the height. The formula is V = l• w• h.
The frequency of the wave is 
Explanation:
The frequency, the wavelength and the speed of a wave are related by the following equation:

where
c is the speed of the wave
f is the frequency
is the wavelength
For the radio wave in this problem,


Therefore, the frequency is:

Learn more about waves here:
brainly.com/question/5354733
brainly.com/question/9077368
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