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.
Answer:
300 g of sugar
Step-by-step explanation:
15 divided by 5 is 3 so 3 x 100 is 300. Hope this helps! :)
The function is L = 10m + 50
Here, we want to find out which of the functions is required to determine the number of lunches L prepared after m minutes
In the question, we already had 50 lunches prepared
We also know that he prepares 10 lunches in one minute
So after A-lunch begins, the number of lunches prepared will be 10 * m = 10m
Adding this to the 50 on ground, then we have the total L lunches
Mathematically, that would be;
L = 10m + 50
Answer:
since the rate at which concentration inside the cell is proportional to the difference in the concentration of the solute in the blood stream and the concentration within the cell, then the rate of change of concentration within the cell is equals to K(L-C).
Thus, the required differential equation is Δc/Δt = K( L - C ).
Step-by-step explanation: