Answer: a) 1/64
Step-by-step explanation:
Answer:
![P(x) = 190 x -x^2](https://tex.z-dn.net/?f=%20P%28x%29%20%3D%20190%20x%20-x%5E2%20)
In order to maximize the last equation we can derivate the function in term of x and we got:
![\frac{dP}{dx} = 190 -2x](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7Bdx%7D%20%3D%20190%20-2x)
And setting this derivate equal to 0 we got:
![\frac{dP}{dx} = 190 -2x=0](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7Bdx%7D%20%3D%20190%20-2x%3D0)
And solving for x we got:
![x = 95](https://tex.z-dn.net/?f=%20x%20%3D%2095)
And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:
![P(x=95)= 95(190-95)= 95*95 = 9025](https://tex.z-dn.net/?f=P%28x%3D95%29%3D%2095%28190-95%29%3D%2095%2A95%20%3D%209025)
Step-by-step explanation:
For this case we have the following function for the profit:
![P(x) = x(190-x)](https://tex.z-dn.net/?f=%20P%28x%29%20%3D%20x%28190-x%29)
And we can rewrite this expression like this:
![P(x) = 190 x -x^2](https://tex.z-dn.net/?f=%20P%28x%29%20%3D%20190%20x%20-x%5E2%20)
In order to maximize the last equation we can derivate the function in term of x and we got:
![\frac{dP}{dx} = 190 -2x](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7Bdx%7D%20%3D%20190%20-2x)
And setting this derivate equal to 0 we got:
![\frac{dP}{dx} = 190 -2x=0](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7Bdx%7D%20%3D%20190%20-2x%3D0)
And solving for x we got:
![x = 95](https://tex.z-dn.net/?f=%20x%20%3D%2095)
And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:
![P(x=95)= 95(190-95)= 95*95 = 9025](https://tex.z-dn.net/?f=P%28x%3D95%29%3D%2095%28190-95%29%3D%2095%2A95%20%3D%209025)
Answer:
D.
Step-by-step explanation:
According to the Exterior Angle Theorem, the exterior angle is equal to the sum of the interior angles opposite from the exterior angle. In other words:
![68+x=100](https://tex.z-dn.net/?f=68%2Bx%3D100)
Solve for <em>x</em>. Therefore:
![m\angle x=32^\circ](https://tex.z-dn.net/?f=m%5Cangle%20x%3D32%5E%5Ccirc)
Our answer is D.
![9\cos(2t)=6\implies\cos(2t)=\dfrac23](https://tex.z-dn.net/?f=9%5Ccos%282t%29%3D6%5Cimplies%5Ccos%282t%29%3D%5Cdfrac23)
Using the fact that cos is 2π-periodic, we have
![\cos(2t)=\dfrac23\implies2t=\cos^{-1}\left(\dfrac23\right)+2n\pi](https://tex.z-dn.net/?f=%5Ccos%282t%29%3D%5Cdfrac23%5Cimplies2t%3D%5Ccos%5E%7B-1%7D%5Cleft%28%5Cdfrac23%5Cright%29%2B2n%5Cpi)
That is,
for any
and integer
.
![\implies t=\dfrac12\cos^{-1}\left(\dfrac23\right)+n\pi](https://tex.z-dn.net/?f=%5Cimplies%20t%3D%5Cdfrac12%5Ccos%5E%7B-1%7D%5Cleft%28%5Cdfrac23%5Cright%29%2Bn%5Cpi)
We get 2 solutions in the interval [0, 2π] for
and
,
![t=\dfrac12\cos^{-1}\left(\dfrac23\right)\text{ and }t=\dfrac12\cos^{-1}\left(\dfrac23\right)+\pi](https://tex.z-dn.net/?f=t%3D%5Cdfrac12%5Ccos%5E%7B-1%7D%5Cleft%28%5Cdfrac23%5Cright%29%5Ctext%7B%20and%20%7Dt%3D%5Cdfrac12%5Ccos%5E%7B-1%7D%5Cleft%28%5Cdfrac23%5Cright%29%2B%5Cpi)