Answer:
6 Chickens
Step-by-step explanation:
Let H represent the number of horses and C represent the number of chickens. Since there are 32 legs altogether this means that on equation to use would be given by:
2C + 4 H = 32 (Eq. 1)
Another equation can be made from the fact that there are 11 heads or 11 animals altogether which can be written as:
C + H = 11 (Eq. 2)
From Eq. 2, solve for the number of chickens:
C + H = 11
C = 11 - H (Eq. 3)
Substituting Eq. 3 in Eq. 1, the number of horses can be determined:
2 C + 4 H = 32
2 ( 11 − H ) + 4 H = 32
22 − 2 H + 4 H = 32
2H = 32 − 22
2 H = 10
H = 5 Eq.4
Putting Eq.4 in Eq.1
2C + 4*5 = 32
2C = 32 - 20
2C = 12
C = 12/2
C = 6
I think it would be 28100% i might be wrong
Answer:
Intercepts:
x = 0, y = 0
x = 1.77, y = 0
x = 2.51, y = 0
Critical points:
x = 1.25, y = 4
x = 2.17
, y = -4
x = 2.8, y = 4
Inflection points:
x = 0.81, y = 2.44
x = 1.81, y = -0.54
x = 2.52, y = 0.27
Step-by-step explanation:
We can find the intercept by setting f(x) = 0
![4sin(x^2) = 0](https://tex.z-dn.net/?f=4sin%28x%5E2%29%20%3D%200)
![sin(x^2) = 0](https://tex.z-dn.net/?f=sin%28x%5E2%29%20%3D%200%20)
where n = 0, 1, 2,3, 4, 5,...
![x = \sqrt(n\pi)](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%28n%5Cpi%29)
Since we are restricting x between 0 and 3 we can stop at n = 2
So the function f(x) intercepts at y = 0 and x:
x = 0
x = 1.77
x = 2.51
The critical points occur at the first derivative = 0
![f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0](https://tex.z-dn.net/?f=f%5E%7B%27%7D%28x%29%20%3D%204cos%28x%5E2%292x%20%3D%208xcos%28x%5E2%29%20%3D%200)
![xcos(x^2) = 0](https://tex.z-dn.net/?f=xcos%28x%5E2%29%20%3D%200)
or
![cos(x^2) = 0](https://tex.z-dn.net/?f=cos%28x%5E2%29%20%3D%200)
where n = 0, 1, 2, 3
![x = \sqrt{\pi(n+1/2)}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%7B%5Cpi%28n%2B1%2F2%29%7D)
Since we are restricting x between 0 and 3 we can stop at n = 2
So our critical points are at
x = 1.25, ![y = f(1.25) = 4sin(1.25^2) = 4](https://tex.z-dn.net/?f=y%20%3D%20f%281.25%29%20%3D%204sin%281.25%5E2%29%20%3D%204)
x = 2.17
, ![y = f(2.17) = 4sin(2.17^2) = -4](https://tex.z-dn.net/?f=y%20%3D%20f%282.17%29%20%3D%204sin%282.17%5E2%29%20%3D%20-4)
x = 2.8, ![y = f(2.8) = 4sin(2.8^2) = 4](https://tex.z-dn.net/?f=y%20%3D%20f%282.8%29%20%3D%204sin%282.8%5E2%29%20%3D%204)
For the inflection point, we can take the 2nd derivative and set it to 0
![f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0](https://tex.z-dn.net/?f=f%5E%5B%27%27%7D%28x%29%20%3D%208%28cos%28x%5E2%29%20-%20xsin%28x%5E2%292x%29%20%3D%208cos%28x%5E2%29%20-%2016x%5E2sin%28x%5E2%29%20%3D%200)
![cos(x^2) = 2x^2sin(x^2)](https://tex.z-dn.net/?f=cos%28x%5E2%29%20%3D%202x%5E2sin%28x%5E2%29)
![tan(x^2) = \frac{1}{2x^2}](https://tex.z-dn.net/?f=tan%28x%5E2%29%20%3D%20%5Cfrac%7B1%7D%7B2x%5E2%7D)
We can solve this numerically to get the inflection points are at
x = 0.81, ![y = f(0.81) = 4sin(0.81^2) = 2.44](https://tex.z-dn.net/?f=y%20%3D%20f%280.81%29%20%3D%204sin%280.81%5E2%29%20%3D%202.44)
x = 1.81, ![y = f(1.81) = 4sin(1.81^2) = -0.54](https://tex.z-dn.net/?f=y%20%3D%20f%281.81%29%20%3D%204sin%281.81%5E2%29%20%3D%20-0.54)
x = 2.52, ![y = f(2.52) = 4sin(2.52^2) = 0.27](https://tex.z-dn.net/?f=y%20%3D%20f%282.52%29%20%3D%204sin%282.52%5E2%29%20%3D%200.27)
It should be 95,040. On the first day, you have 12 choices. On the second, eleven, and so on, until day 5. 12x11x10x9x8.
Answer:
A i think its a A try. it it that looks correct