Answer:
16022 square foot
Step-by-step explanation:
The world's largest cookie was baked in North Carolina in 2003. Its diameter was 100 feet and it contained about 6,000 pounds of chocolate chips! If the cookie was a cylinder 1 foot tall, and you wanted to cover it with icing, how many square inches would you have to ice? Use 3.14 for .
We find the surface area of the cylinder
= 2πr(r + h)
r = Diameter /2 = 100 feet/2
r = 50 feet
h = 1 feet
Hence
2 × 3.14 × 50(50 + 1)
= 16022.122533 square foot
= 16022 square foot
Answer:
A: ⅚
B: ⅚
Step-by-step explanation:
City A: 4, 3.5, 5, 5.5, 4, 2
Mean: sum/n
= 24/6 = 4
|x - mean|: 0, 0.5, 1, 1.5, 0 , 2
MAD = sum|x - mean|/n
5/6 = 0.833
City B: 5, 6, 3.5, 5.5, 4, 6
Mean: 30/6 = 5
|x - mean|: 0, 1, 1.5, 0.5, 1, 1
MAD = sum|x - mean|/n
5/6 = 0.833
8.4
triangle ABC is 1.2 times bigger than triangle XYZ
10 x 1.2 = 12
5 x 1.2 = 6
so 7 x 1.2 = 8.4
Answer:
![y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5B3cos%28%5Csqrt%7B6%7Dx%29%2B%5Cfrac%7B2%5Csqrt%7B6%7D%7D%7B3%7Dsin%28%5Csqrt%7B6%7Dx%29%5D)
(See attached graph)
Step-by-step explanation:
To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation 
where the values of 
are the roots:
Since the values of are complex conjugate roots, then the general solution is ![y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B%5Calpha%20x%7D%5BC_1cos%28%5Cbeta%20x%29%2BC_2sin%28%5Cbeta%20x%29%5D)
where 
.
Thus, the general solution for our given differential equation is ![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D)
.
To account for both initial conditions, take the derivative of 
, thus, ![y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%27%28x%29%3D-2e%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%2Be%5E%7B-2x%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7Dx%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7Dx%29%5D)
Now, we can create our system of equations given our initial conditions:
![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%280%29%3De%5E%7B-2%280%29%7D%5BC_1cos%28%5Csqrt%7B6%7D%280%29%29%2BC_2sin%28%5Csqrt%7B6%7D%280%29%29%5D%3D3%5C%5C%5C%5CC_1%3D3)
![y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2](https://tex.z-dn.net/?f=y%27%28x%29%3D-2e%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%2Be%5E%7B-2x%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7Dx%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%27%280%29%3D-2e%5E%7B-2%280%29%7D%5BC_1cos%28%5Csqrt%7B6%7D%280%29%29%2BC_2sin%28%5Csqrt%7B6%7D%280%29%29%5D%2Be%5E%7B-2%280%29%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7D%280%29%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7D%280%29%29%5D%3D-2%5C%5C%5C%5C-2C_1%2B%5Csqrt%7B6%7DC_2%3D-2)
We then solve the system of equations, which becomes easy since we already know that 
:
Thus, our final solution is:
![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%28x%29%3De%5E%7B-2x%7D%5B3cos%28%5Csqrt%7B6%7Dx%29%2B%5Cfrac%7B2%5Csqrt%7B6%7D%7D%7B3%7Dsin%28%5Csqrt%7B6%7Dx%29%5D)
