Avocado and cottage cheese are both proteins.
Answer:
.
Explanation:
The frequency 
of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)
The wave in this question travels at a speed of 
. In other words, the wave would have traveled 
in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be
How many wave cycles can fit into that ? The wavelength of this wave
gives the length of one wave cycle. Therefore:
.
That is: there are 
wave cycles in of this wave.
On the other hand, Because that of this wave goes through that point in each second, that wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be
Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:
.
The calculations above can be expressed with the formula:
,
where
represents the speed of this wave, and 
represents the wavelength of this wave.
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
