Answer:
8 and 9
Step-by-step explanation:
Answer:
sorry you can send me picture of work
Step-by-step explanation:
ok send me and get answers
If you are graphing quadratic equations of the type
ax^2 + bx + c = 0
The equation will look like a "U" <span>if "a" is positive </span>or it will look like an upside-down "U" <span>if "a" is negative </span>
Answer:
LN=10
Step-by-step explanation:
if LM = 5 and MN = 5 then LN=10
Answer: D
<u>Step-by-step explanation:</u>
The first matrix contains the coefficients of the x- and y- values for both equations (top row is the top equation and the bottom row is the bottom equation. The second matrix contains what each equation is equal to.
![\begin{array}{c}2x-y\\x-6y\end{array}\qquad \rightarrow \qquad \left[\begin{array}{cc}2&-1\\1&-6\end{array}\right] \\\\\\\begin{array}{c}-6\\13\end{array}\qquad \rightarrow \qquad \left[\begin{array}{c}-6\\13\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7D2x-y%5C%5Cx-6y%5Cend%7Barray%7D%5Cqquad%20%5Crightarrow%20%5Cqquad%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-1%5C%5C1%26-6%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Bc%7D-6%5C%5C13%5Cend%7Barray%7D%5Cqquad%20%5Crightarrow%20%5Cqquad%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-6%5C%5C13%5Cend%7Barray%7D%5Cright%5D)
The product will result in the solution for the x- and y-values of the system.