The statement is false, as the system can have no solutions or infinite solutions.
<h3>
Is the statement true or false?</h3>
The statement says that a system of linear equations with 3 variables and 3 equations has one solution.
If the variables are x, y, and z, then the system can be written as:
Now, the statement is clearly false. Suppose that we have:
Then we have 3 parallel equations. Parallel equations never do intercept, then this system has no solutions.
Then there are systems of 3 variables with 3 equations where there are no solutions, so the statement is false.
If you want to learn more about systems of equations:
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*See diagram in the attachment
Answer:
27 inches
Step-by-step explanation:
Combined perimeter = perimeter of rectangle + perimeter of triangle = 63 inches
Perimeter of rectangle = 2(L + W)
L = 11 in.
W = 7 in.
Perimeter of rectangle = 2(11 + 7) = 36 in.
Let perimeter of the triangle be represented by x
Therefore:
36 + x = 63
Subtract 36 from each side
x = 63 - 36
x = 27
Perimeter of triangle = 27 inches
Answer:
4
Step-by-step explanation:
Since 78² then 8² = 64 which indicates 4 is the unit digit
Okay okay baby boo I don’t remember how to
Suppose we wish to determine whether or not two given polynomials with complex coefficients have a common root. Given two first-degree polynomials a0 + a1x and b0 + b1x, we seek a single value of x such that
Solving each of these equations for x we get x = -a0/a1 and x = -b0/b1 respectively, so in order for both equations to be satisfied simultaneously we must have a0/a1 = b0/b1, which can also be written as a0b1 - a1b0 = 0. Formally we can regard this system as two linear equations in the two quantities x0 and x1, and write them in matrix form as
Hence a non-trivial solution requires the vanishing of the determinant of the coefficient matrix, which again gives a0b1 - a1b0 = 0.
Now consider two polynomials of degree 2. In this case we seek a single value of x such that
Hope this helped, Hope I did not make it to complated
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