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kykrilka [37]
3 years ago
10

Mark is solving the following system.

Mathematics
2 answers:
fredd [130]3 years ago
6 0
I think its he didnt eliminate the same variables
miskamm [114]3 years ago
6 0

Solving the system of linear equations Mark tries to apply elementary transformations in order to eliminate one variable.

He makes such steps:

1. He multiplies equation (1) x+y+z=2 by 7 and adds it to equation (3) 4x-y-7z=16. This gives him:

7x+7y+7x+4x-y-7z=14+16,

11x+6y=30.

2. He multiplies equation (3)  4x-y-7z=16 by 2 and adds it to equation (2) 3x+2y+z=8. This step gives him:

8x-2y-14z+3x+2y+z=32+8,

11x-13z=40.

Thus, he did not eliminate the same variables in steps 1 and 2.

Answer: correct choice is C

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Answer:

Zeros : 1 , -1, 3

Degree : 4

End Behaviour : At x-> ∞ f(x) -> ∞ and x->-∞ f(x) -> ∞

Y - intercept : -3

Extra Points: (0,-3), (2,-3)

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f(x) = 0 to find the zeros

Therefore (x+1)(x-1)^{2} (x-3) = 0

Clearly x = -1,1,3

Here 1 is a repeating root as it is (x-1)²

Degree is highest power of x in f(x)

Clearly it is x*x²*x = x⁴ is the maximum power of x

Thus degree is 4

Looking at end behavior we substitute x->∞ and x-> -∞

Clearly f(x)>0 as all terms are positive and f(x)->∞

Similarly when x->-∞

f(x)>0 as 2 terms are -ve and their product is positive thus f(x)-> ∞

Y-Intercept is f(0)

f(0) = (0+1)(0-1)²(0-3) = 1*1*-3 = -3

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Substitute x = 0 , 2 for extra points

Thus f(0) = -3

and f(2) = -3

Thus points on the graph (0,-3), (0,2)

We can use all this information to draw a graph remember that 1 is a repeating root so that will be a point of minima. The graph is a parabola that passes through x-axis at x = -1, 3.

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