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Stels [109]
3 years ago
6

I don’t understand this question. Can anyone help? I need answers ASAP. Thanks for all the help

Mathematics
1 answer:
zavuch27 [327]3 years ago
7 0

There's some unknown (but derivable) system of equations being modeled by the two lines in the given graph. (But we don't care what equations make up these lines.)

There's no solution to this particular system because the two lines are parallel.

How do we know they're parallel? Parallel lines have the same slope, and we can easily calculate the slope of these lines.

The line on the left passes through the points (-1, 0) and (0, -2), so it has slope

(-2 - 0)/(0 - (-1)) = -2/1 = -2

The line on the right passes through (0, 2) and (1, 0), so its slope is

(0 - 2)/(1 - 0) = -2/1 = -2

The slopes are equal, so the lines are parallel.

Why does this mean there is no solution? Graphically, a solution to the system is represented by an intersection of the lines. Parallel lines never intersect, so there is no solution.

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The number 40 cannot be decomposed into prime factors<br><br>True<br>either<br>False​
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For the polynomial function ƒ(x) = −x6 + 3x4 + 4x2, find the zeros. Then determine the multiplicity at each zero and state wheth
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There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.


You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.


-x^6 + 3x^4 + 4x^2

-x^2(x^4 - 3x^2 - 4)


Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.


-x^2(x^4 - 3x^2 - 4)

-x^2(x^2 - 4)(x^2 + 1)


Now we can use the factors of two perfect squares rule to factor the middle parenthesis.


-x^2(x^2 - 4)(x^2 + 1)

-x^2(x - 2)(x + 2)(x^2 + 1)


We would also want to split the term in the front.


-x^2(x - 2)(x + 2)(x^2 + 1)

(x)(-x)(x - 2)(x + 2)(x^2 + 1)


Now we would set each portion equal to 0 and solve.


First root

x = 0 ---> no work needed


Second root

-x = 0 ---> divide by -1

x = 0


Third root

x - 2 = 0

x = 2


Forth root

x + 2 = 0

x = -2


Fifth and Sixth roots

x^2 + 1 = 0

x^2 = -1

x = +/- \sqrt{-1}

x = +/- i

7 0
3 years ago
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