Answer:
How many drinks should be sold to get a maximal profit? 468
Sales of the first one = 345 cups
Sales of the second one = 123 cups
Step-by-step explanation:
maximize 1.2F + 0.7S
where:
F = first type of drink
S = second type of drink
constraints:
sugar ⇒ 3F + 10S ≤ 3000
juice ⇒ 9F + 4S ≤ 3600
coffee ⇒ 4F + 5S ≤ 2000
using solver the maximum profit is $500.10
and the optimal solution is 345F + 123S
Answer:
0.24 its opposite value is - 0.24
1/4 its opposite value is -1/4
0
-5 its opposite value is 5
240 which is positive and its opposite value is -240
Step-by-step explanation:
hey
from the given statement we can see that question demand is about the opposite values. so
as our first value is 0.24 its opposite value is - 0.24
then move on next value is 1/4
its opposite value is -1/4
0 is a neutral value it neither positive or negative
move on to the next value which is -5 who is already negative
so we take a positive value that is 5
again move to 240 which is positive and its opposite value is -240
Hope it will be brainliest to you
hello
Step-by-step explanation:
I don’t know, I forgot how to solve problems like this, look it up
Answer:
The value of AB is ![\left[\begin{array}{ccc}21\\11\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
and it's not possible to multiply BA.
Step-by-step explanation:
Consider the provided matrices.
![A=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C2%261%5Cend%7Barray%7D%5Cright%5D)
, ![B=\left[\begin{array}{ccc}3\\5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
Two matrices can be multiplied if and only if first matrix has an order m × n and second matrix has an order n × v.
Multiply AB
Matrix A has order 2 × 2 and matrix B has order 2 × 1. So according to rule we can multiply both the matrix as shown:
![AB=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right] \left[\begin{array}{ccc}3\\5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C2%261%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}2\times 3+3\times 5\\2\times 3+1\times 5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5Ctimes%203%2B3%5Ctimes%205%5C%5C2%5Ctimes%203%2B1%5Ctimes%205%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}6+15\\6+5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%2B15%5C%5C6%2B5%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}21\\11\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
Hence, the value of AB is
Now calculate the value of BA as shown:
Multiply BA
Matrix B has order 2 × 1 and matrix A has order 2 × 2. So according to rule we cannot multiply both the matrix.
We can multiply two matrix if first matrix has an order m × n and second matrix has an order n × v.
That means number of column of first matrix should be equal to the number of rows of second matrix.
Hence, it's not possible to multiply BA.
