Answer:
y = 12
Step-by-step explanation:
y = 8 is the equation of a horizontal line, parallel to the x- axis.
Thus a parallel line will also be horizontal.
The equation of a horizontal line is
y = c
where c is the y- coordinates of the points it passes through.
The line passes through (- 6, 12) with y- coordinate of 12, thus
y = 12 ← equation of parallel line
Answer:
B
Step-by-step explanation:
If they have 168 cats, they have capacity for 32 more. If they take in 4 cats a day, then it will take 8 days for their facility to fill up. So less than 8.
Answer:
Remember, a basis for the row space of a matrix A is the set of rows different of zero of the echelon form of A.
We need to find the echelon form of the matrix augmented matrix of the system A2x=b2
![B=\left[\begin{array}{cccc}1&2&3&1\\4&5&6&1\\7&8&9&1\\3&2&4&1\\6&5&4&1\\9&8&7&1\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C4%265%266%261%5C%5C7%268%269%261%5C%5C3%262%264%261%5C%5C6%265%264%261%5C%5C9%268%267%261%5Cend%7Barray%7D%5Cright%5D)
We apply row operations:
1.
- To row 2 we subtract row 1, 4 times.
- To row 3 we subtract row 1, 7 times.
- To row 4 we subtract row 1, 3 times.
- To row 5 we subtract row 1, 6 times.
- To row 6 we subtract row 1, 9 times.
We obtain the matrix
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&-6&-12&-6\\0&-4&-5&-2\\0&-7&-14&-5\\0&-10&-20&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%26-6%26-12%26-6%5C%5C0%26-4%26-5%26-2%5C%5C0%26-7%26-14%26-5%5C%5C0%26-10%26-20%26-8%5Cend%7Barray%7D%5Cright%5D)
2.
- We subtract row two twice to row three of the previous matrix.
- we subtract 4/3 from row two to row 4.
- we subtract 7/3 from row two to row 5.
- we subtract 10/3 from row two to row 6.
We obtain the matrix
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&0&0\\0&0&3&2\\0&0&0&2\\0&0&0&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%260%260%260%5C%5C0%260%263%262%5C%5C0%260%260%262%5C%5C0%260%260%262%5Cend%7Barray%7D%5Cright%5D)
3.
we exchange rows three and four of the previous matrix and obtain the echelon form of the augmented matrix.
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&3&2\\0&0&0&0\\0&0&0&2\\0&0&0&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%260%263%262%5C%5C0%260%260%260%5C%5C0%260%260%262%5C%5C0%260%260%262%5Cend%7Barray%7D%5Cright%5D)
Since the only nonzero rows of the augmented matrix of the coefficient matrix are the first three, then the set
![\{\left[\begin{array}{c}1\\2\\3\end{array}\right],\left[\begin{array}{c}0\\-3\\-6\end{array}\right],\left[\begin{array}{c}0\\0\\3\end{array}\right] \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5C%5C3%5Cend%7Barray%7D%5Cright%5D%2C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C-3%5C%5C-6%5Cend%7Barray%7D%5Cright%5D%2C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5Cend%7Barray%7D%5Cright%5D%20%5C%7D)
is a basis for Row (A2)
Now, observe that the last two rows of the echelon form of the augmented matrix have the last coordinate different of zero. Then, the system is inconsistent. This means that the system has no solutions.
4. 54 + 63 + 54=171
5. 3+3+75=81
6. 20+68=89
7. 120+180=300 x 1.48=$444
13pi/12 lies between pi and 2pi, which means sin(13pi/12) < 0
Recall the double angle identity,
sin^2(x) = (1 - cos(2x))/2
If we let x = 13pi/12, then
sin(13pi/12) = - sqrt[(1 - cos(13pi/6))/2]
where we took the negative square root because we expect a negative value.
Now, because cosine has a period of 2pi, we have
cos(13pi/6) = cos(2pi + pi/6) = cos(pi/6) = sqrt[3]/2
Then
sin(13pi/12) = - sqrt[(1 - sqrt[3]/2)/2]
sin(13pi/12) = - sqrt[2 - sqrt[3]]/2