Question

prepare models using matrices to solve simple mixture problems​

Answer 1

$\huge\underline\orange{Answer}$

A matrix is sort of like a “box” of information where you are keeping track of things both right and left (columns), and up and down (rows). Usually a matrix contains numbers or algebraic expressions. You may have heard matrices called arrays, especially in computer science.

As an example, if you had three sisters, and you wanted an easy way to store their age and number of pairs of shoes, you could store this information in a matrix. The actual matrix is inside and includes the brackets:

Answer 2

Answer:

Step-by-step explanation:

From the above question,

They have given :

Mixture problems can be solved using matrices by setting up a system of equations based on the information given in the problem. Here is an example of how to use matrices to solve a simple mixture problem:

Example:

A tank contains 50 liters of a solution that is 25% acid. Another tank contains 30 liters of a solution that is 40% acid. How many liters of each solution should be mixed to obtain a new solution that is 30% acid?

We can set up a system of equations based on the information given in the problem. Let x be the number of liters of the first solution and y be the number of liters of the second solution that are mixed to obtain the new solution.

The equation for the acid percentage of the first solution is:

                          (0.25)(x) = (0.3)(x+y)

The equation for the acid percentage of the second solution is:

                           (0.40)(y) = (0.3)(x+y)

We can write this system of equations in matrix form as:

              |0.25x = 0.3x+0.3y|

              |0.40y = 0.3x+0.3y|

We can then use matrix algebra to solve for x and y.

First, we can subtract the first equation from the second equation to eliminate one of the variables:

              |0.25x = 0.3x+0.3y|

              |0.40y = 0.3x+0.3y|

              |0.15y = 0.15x|

Now we can solve for x or y by substituting the value of the other variable.

Let's solve for y:

              y = x / (1 - 0.15)

              y = 1.176 x

We can substitute this value of y back into the first equation and solve for x:

              0.25x = 0.3x + 0.3(1.176x)

              0.25x = 0.3x + 0.3528x

              x = 20

So, 20 liters of the first solution and 23.52 liters of the second solution should be mixed to obtain a new solution that is 30% acid.

In this way, you can use matrices to solve simple mixture problems. It's important to note that this method will only work for problems that can be set up as a system of linear equations.

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