Yes. A is the right graph because the line must travel through two points: (0 ,1)and(-3, 0)
Y=1/x solve for x
x=1/y now reverse lables...
y=1/x
Yes that is the same equation you started with. What that means is the the inverse function in this case produces the same exact graph :)
Answer:
12...you already figured it out but alr
Step-by-step explanation:
9514 1404 393
Answer:
(a, b) = (-2, -1)
Step-by-step explanation:
The transpose of the given matrix is ...
![A^T=\left[\begin{array}{ccc}1&2&a\\2&1&2\\2&-2&b\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26a%5C%5C2%261%262%5C%5C2%26-2%26b%5Cend%7Barray%7D%5Cright%5D)
Then the [3,1] term of the product is ...
![(A\cdot A^T)_{31}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}1&2&2\end{array}\right]=a+2b+4](https://tex.z-dn.net/?f=%28A%5Ccdot%20A%5ET%29_%7B31%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%262%26b%5Cend%7Barray%7D%5Cright%5D%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%262%5Cend%7Barray%7D%5Cright%5D%3Da%2B2b%2B4)
and the [3,2] term is ...
![(A\cdot A^T)_{32}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}2&1&-2\end{array}\right]=2a-2b+2](https://tex.z-dn.net/?f=%28A%5Ccdot%20A%5ET%29_%7B32%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%262%26b%5Cend%7Barray%7D%5Cright%5D%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%26-2%5Cend%7Barray%7D%5Cright%5D%3D2a-2b%2B2)
Both of these terms in the product matrix are 0. We can solve the system of equations by adding these two terms:
(a +2b +4) +(2a -2b +2) = (0) +(0)
3a +6 = 0
a = -2
Substituting for 'a' in term [3,1] gives ...
-2 +2b +4 = 0
b = -1
The ordered pair (a, b) is (-2, -1).
Find a common denominator or the bottom number of a fraction. 5 and 5/6 - 11/6 - 4/6 - 3/4 = 140/24 - 44/24 - 16/24 - 18/24 = 96/24 - 16/24 - 18/24=62/24 or 2 and 7/12.