It’s the first one because the parabola is open
Answer:
y=4x+9
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(1-(-11))/(-2-(-5))
m=(1+11)/(-2+5)
m=12/3
m=4
y-y1=m(x-x1)
y-(-11)=4(x-(-5))
y+11=4(x+5)
y=4x+20-11
y=4x+9
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.
Answer:
A.
Step-by-step explanation:
100 is equal to $1.
100/5=20
Answer:
6 and 9
Step-by-step explanation:
set up an equation and solve for x
(x+3) * x = 54
easiest way to to try some numbers.
if you know your multiplication tables...
what two number multiplied together are 54
be sure the numbers you are thinking of meet the rules
is one 3 more than the other
do they multiply together to get 54