E. 1/2
With trig functions, multiple x values correspond with the same y value.
Using an initial x value (the principle value), we can find other x values for the same y value, this is what we are are being asked to find in this question.
There are slightly different ways to find these x values (also known as solutions) for each of the basic trig functions.
The x values are in degrees for the basic trig functions.
For cosine, the rule is as follows:
360 - principle value;
this will give what I, personally, like to think of as a secondary principle value (this value is not actually recognised as a secondary principle value, I just like to think of it as such). Anyway, all other solutions can the be found by adding or substrating any integer multiple of 360 to/from the PV and 'secondary PV'.
For your question:
cos60 = 1/2
60 is the x value (PV)
so...
360 - 60 = 300 is the 'secondary PV'
Just to be clear, this means that if I were to find the cos300, I would get 1/2.
That is sufficient for explaining the answer to this particular question but if you wanted to find any other solution, you would just have to do either:
60 + or - n(360)
or...
300 + or - n(360),
where n = any integer
Answer:
5v+4
Step-by-step explanation:
X / 6 + 2 = 9
x/6 = 9 - 2
x/6 = 7
x = 7 * 6
x = 42 <== ur number
Answer:
Carrie has 0.125 more than Larry and Larry has 3/4 more than Harry. Between them Harry, Larry and Carrie have 151. How many does Harry have? Carrie=63, Larry=56, Harry=32.
Step-by-step explanation:
Answer:
The rate of change of the height is 0.021 meters per minute
Step-by-step explanation:
From the formula
Differentiate the equation with respect to time t, such that
To differentiate the product,
Let r² = u, so that
Then, using product rule
![\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%5Bu%5Cfrac%7Bdh%7D%7Bdt%7D%20%2B%20h%5Cfrac%7Bdu%7D%7Bdt%7D%5D)
Since 
Then, 
Using the Chain's rule
∴ ![\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%5Bu%5Cfrac%7Bdh%7D%7Bdt%7D%20%2B%20h%28%5Cfrac%7Bdu%7D%7Bdr%7D%20%5Ctimes%20%5Cfrac%7Bdr%7D%7Bdt%7D%29%5D)
Then,
![\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%5Br%5E%7B2%7D%20%5Cfrac%7Bdh%7D%7Bdt%7D%20%2B%20h%282r%29%20%5Cfrac%7Bdr%7D%7Bdt%7D%5D)
Now,
From the question
At the instant when 
and 
We will determine the value of h, using
Now, Putting the parameters into the equation
![236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]](https://tex.z-dn.net/?f=236%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%5B%2899%29%5E%7B2%7D%20%5Cfrac%7Bdh%7D%7Bdt%7D%20%2B%20%28%5Cfrac%7B20%7D%7B363%5Cpi%20%7D%29%20%282%2899%29%29%20%287%29%5D)
![236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]](https://tex.z-dn.net/?f=236%20%5Ctimes%203%20%3D%20%5Cpi%20%5B9801%20%5Cfrac%7Bdh%7D%7Bdt%7D%20%2B%20%28%5Cfrac%7B20%7D%7B363%5Cpi%20%7D%29%201386%5D)
Hence, the rate of change of the height is 0.021 meters per minute.
