That's not correct. The terms 2a and 3b are not like terms, so we cannot combine them to get 5ab. We simply leave it as 2a+3b.
If you had 2a+3a, then it would simplify to 5a
Similarly, 2b+3b = 5b
Or you could have 2ab+3ab = 5ab
The key is that the variable portions must match up to be able to add them.
Answer:
the lower right matrix is the third correct choice
Step-by-step explanation:
Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).
Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.
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We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.
The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...
![\left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%261%2629000%5C%5C1%2B2%261-3%261-1%261000%2829%2B1%29%5C%5C0%260.15%260.15%262100%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%261%2629000%5C%5C3%26-2%260%2630000%5C%5C0%260.15%260.15%262100%5Cend%7Barray%7D%5Cright%5D)
Already, we see that the second row matches that in the lower right matrix.
The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.
The correct choices here are the two you have selected, and <em>the lower right matrix</em>.
You can not see the problem
Answer
the second one, and 5th one
Step-by-step explanation:
because it is liquid