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Contact [7]
3 years ago
6

Solving systems of equations has been used consistently throughout this course in order to shed light on all kinds of different

situations, i.e. finding the intersection point of a system of linear equations, finding eigenvectors, etc. List at least two different types of problems that require solving a system of linear equations, explain how solving a system of linear equations is used, and what information can be learned from the solution.
Mathematics
1 answer:
Margarita [4]3 years ago
3 0

Answer:

Types of problems that require solving a system of linear equations are:

Calculation of mileage rate

Calculations of ages of children born to a family

Calculation of income over a range of period.

Step-by-step explanation:

Solving a system of linear equations have many real life applications. Many real life problems can be used to set up a system of linear equations.

The following are the steps:

Understand the problem

Break the problem down into parts and represent each part with unknown variables such as: X1, X2, X3,...

Combine the unknown variables mathematically, from the statements of the problem(That is, 2X1 + X2 - 6X3, X1 + 2X2 + 4X3,....)

The results of the solved linear equation system (after solving) will give the solution to the problem. The solutions to the real life problems can be learned by the nature, analysis, and relationship of the solutions of the system of the formed linear equation

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Bhavik bought 3 liters of milk and 5 loaves of bread for a total of $11. A month later, he bought 4 liters of milk and 4 loaves
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Step-by-step explanation:

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The radius of a circle is changing at the rate of 1/π inches per second. At what rate, in square inches per second, is the circl
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Answer:

10 square inches per second.

Step-by-step explanation:

The radius of the circle is given by the equation:

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Remember that the area of a circle of radius R is written as:

A = π*R^2

Then the area of our circle will be:

A(t) = π*( (1/π  in/s)*t)^2 = π*(1/π  in/s)^2*(t)^2

Now we want to find the rate of change (the first derivation of the area) when the radius is equal to 5 inches.

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Now let's find the first derivate of A(t)

dA(t)/dt = A'(t) = 2*(π*(1/π  in/s)^2*t = (2*π*t)*(1/π  in/s)^2

Now we need to evaluate this in the time such that the radius is equal to 5 inches, we will get:

A'(5*π s) = (2*π*5*π s)*((1/π  in/s)^2

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The rate of change is 10 square inches per second.

4 0
3 years ago
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