Answer:
length= 3.7 ft.
Step-by-step explanation:
T= (square root of length) * k
1.8 seconds = (square root of 3) * k
k= 1.8 seconds / (square root of 3)
a.) T= (square root of 11) k
substituting k:
T=(square root of 11) * 1.8 seconds / (square of 3)
T= 3.4 seconds
b.) 2 seconds = (square root of length) k
substituting k:
2 seconds =(square root of length) 1.8 seconds / (square root of 3)
Divide the equation by 1.8 seconds / (square root of 3):
2 seconds / [1.8 seconds / (square root of 3)] = (square root of length)
10(square root of 3) / 9 = (square root of length)
square both sides:
100/7 ft= length
length= 3.7 ft.
Hope this helps.
Answer:
Characteristic equation:
Eigen values:
Step-by-step explanation:
We are given the matrix:
![\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-14%26-6%266%5C%5C28%2612%26-4%5C%5C0%260%264%5Cend%7Barray%7D%5Cright%5D)
The characteristic equation can be calculated as:
We follow the following steps to calculate characteristic equation:
![=det\Bigg(\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]-\lambda\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\Bigg)\\\\= det\Bigg(\displaystyle\left[\begin{array}{ccc}-14-\lambda&-6&6\\28&12-\lambda&-4\\0&0&4-\lambda\end{array}\right]\Bigg)\\\\=(-14-\lambda)[(12-\lambda)(4-\lambda)]+6[28(4-\lambda)]-6[(28)(0)-(12-\lambda)(0)]\\\\= -\lambda^3 + 2\lambda^2 + 8\lambda](https://tex.z-dn.net/?f=%3Ddet%5CBigg%28%5Cdisplaystyle%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-14%26-6%266%5C%5C28%2612%26-4%5C%5C0%260%264%5Cend%7Barray%7D%5Cright%5D-%5Clambda%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%5CBigg%29%5C%5C%5C%5C%3D%20det%5CBigg%28%5Cdisplaystyle%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-14-%5Clambda%26-6%266%5C%5C28%2612-%5Clambda%26-4%5C%5C0%260%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%5CBigg%29%5C%5C%5C%5C%3D%28-14-%5Clambda%29%5B%2812-%5Clambda%29%284-%5Clambda%29%5D%2B6%5B28%284-%5Clambda%29%5D-6%5B%2828%29%280%29-%2812-%5Clambda%29%280%29%5D%5C%5C%5C%5C%3D%20-%5Clambda%5E3%20%2B%202%5Clambda%5E2%20%2B%208%5Clambda)
To obtain the eigen values, we equate the characteristic equation to 0:
![p(\lambda) = -\lambda^3 + 2\lambda^2 + 8\lambda = 0\\-\lambda(\lambda^2-2\lambda-8) = 0\\-\lambda(\lambda^2-4\lambda+2\lambda-8) = 0\\-\lambda[(\lambda(\lambda-4)+2(\lambda-4)] = 0\\-\lambda(\lambda+2)(\lambda-4) = 0 \\\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4](https://tex.z-dn.net/?f=p%28%5Clambda%29%20%3D%20-%5Clambda%5E3%20%2B%202%5Clambda%5E2%20%2B%208%5Clambda%20%3D%200%5C%5C-%5Clambda%28%5Clambda%5E2-2%5Clambda-8%29%20%3D%200%5C%5C-%5Clambda%28%5Clambda%5E2-4%5Clambda%2B2%5Clambda-8%29%20%3D%200%5C%5C-%5Clambda%5B%28%5Clambda%28%5Clambda-4%29%2B2%28%5Clambda-4%29%5D%20%3D%200%5C%5C-%5Clambda%28%5Clambda%2B2%29%28%5Clambda-4%29%20%3D%200%20%5C%5C%5Clambda_1%20%3D%200%2C%20%5Clambda_2%20%3D%20-2%2C%20%5Clambda_3%3D%204)
We can arrange the eigen values as:
Answer:
c. (4,3)
Step-by-step explanation:
If you look at all the rest of the coordinates, compared to (4,3). None of them are in the red area. They are either in the blue, or the orange. To be a solution it has to be in both areas, (the red). Hope this helps!
Answer:
E < F < G
Step-by-step explanation:
hope this helps
