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lesantik [10]
3 years ago
11

Use technology to approximate the solution(s) to the system of equations to the nearest tenth of a unit.

Mathematics
2 answers:
Vikentia [17]3 years ago
7 0

Answer:

(2.5, -1)

(-2.5, 1)

(-1, 2.5)

Step-by-step explanation:

alukav5142 [94]3 years ago
4 0

Answer:

(2.5, -1)

(-2.5, 1)

(-1, 2.5

Step-by-step explanation:

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X is a natural number and a factor of 18 whats the value of X​
miskamm [114]

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Step-by-step explanation:

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Find the difference of 11.3 and 1.75.
Hitman42 [59]

Answer:

9.55

Step-by-step explanation:

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5 0
3 years ago
If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?
Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor  (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

5 0
3 years ago
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