The 4th option.
The values with no x or y attached are the y intercept
6x - 7y = 21
Subtract 6x from both sides
6x - 7y - 6x = 21 - 6x
-7y = 21 - 6x
Divide both sides by - 7
-7y/-7 = 21/-7 - 6x/-7
Y = -3 +6x/7
The why intercept, -3 is the same as the one in the question
Consider the operation is 
.
Given:
The augmented matrix below represents a system of equations.
![\left[\left.\begin{matrix}1&0&1\\1&3&-1\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-9\\-2\end{matrix}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft.%5Cbegin%7Bmatrix%7D1%260%261%5C%5C1%263%26-1%5C%5C3%262%260%5Cend%7Bmatrix%7D%5Cright%7C%5Cbegin%7Bmatrix%7D-1%5C%5C-9%5C%5C-2%5Cend%7Bmatrix%7D%5Cright%5D)
To find:
Matrix results from the operation .
Step-by-step explanation:
We have,
After applying , we get
![\left[\left.\begin{matrix}1&0&1\\-3(1)&-3(3)&-3(-1)\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-3(-9)\\-2\end{matrix}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft.%5Cbegin%7Bmatrix%7D1%260%261%5C%5C-3%281%29%26-3%283%29%26-3%28-1%29%5C%5C3%262%260%5Cend%7Bmatrix%7D%5Cright%7C%5Cbegin%7Bmatrix%7D-1%5C%5C-3%28-9%29%5C%5C-2%5Cend%7Bmatrix%7D%5Cright%5D)
![\left[\left.\begin{matrix}1&0&1\\-3&-9&3\\3&2&0\end{matrix}\right|\begin{matrix}-1\\27\\-2\end{matrix}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft.%5Cbegin%7Bmatrix%7D1%260%261%5C%5C-3%26-9%263%5C%5C3%262%260%5Cend%7Bmatrix%7D%5Cright%7C%5Cbegin%7Bmatrix%7D-1%5C%5C27%5C%5C-2%5Cend%7Bmatrix%7D%5Cright%5D)
Therefore, the correct option is A.
Answer:
3
Step-by-step explanation:
Call that number “x”
The square of that number is x^2
Also as the cube is x^3
The ratio between these 2 values is 1:3, this means x^2 / x^3 = 1 / 3
<==> 1 / x = 1 / 3 (eliminate “x^2” for numerator and denominator of the fraction x^2 / x^3 )
<==> x = 1 * 3 / 1 = 3
So, that number is 3 ! This math problem seems simple ;)
If you were going to multiply those two numbers, the answer would be D, 192.
