Answer:
stem: 5
leaf: 0, 6
stem: 6
leaf: 0, 2, 5, 6
stem: 7
leaf: 1, 5
stem: 8
leaf: 6
stem: 9
leaf: 9
i hope this helps <3
Step-by-step explanation:
Answer:
Option B R(2,4) is correct
Step-by-step explanation:
The equation of the circle is:

Where r = radius
a and b are coordinates of the center of circle.
To check which point lies on a circle, we need to verify the equation

We will check for each option.
Option A Q(1,11)
x=1 and y =11

So, Option A is incorrect
Option B R(2,4)
x =2 and y = 4

Option B is correct.
Option C S(4,-4)
x =4 and y =-4

Option C is incorrect
Option D T(9,-2)
x =9 and y =-2

Option D is incorrect.
Answer:
x = -5; y = 4, z = 1
Step-by-step explanation:
Given the row echelon form as:
![\left[\begin{array}{cccc}1&0&\ \ 4|&-1\\0&1&-1|&3\\0&0&\ \ 1|&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%5C%20%5C%204%7C%26-1%5C%5C0%261%26-1%7C%263%5C%5C0%260%26%5C%20%5C%201%7C%261%5Cend%7Barray%7D%5Cright%5D)
This matrix can be represented as:
![\left[\begin{array}{ccc}1&0&4\\0&1&-1\\0&0&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}-1\\3\\1\end{array}\right] \\\\Performing\ matrix\ multiplication\ gives:\\\\\left[\begin{array}{c}x+4z\\y-z\\z\end{array}\right] =\left[\begin{array}{c}-1\\3\\1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%264%5C%5C0%261%26-1%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CPerforming%5C%20matrix%5C%20multiplication%5C%20gives%3A%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%2B4z%5C%5Cy-z%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
Therefore:
z = 1
y - z = 3;
y = 3 + z = 3 + 1 = 4.
Hence, y = 4
x + 4z = - 1;
x = -1 - 4z = -1 - 4(1) = -5
x = -5