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Paul [167]
3 years ago
6

A polynomial has one root that equals 2+i.Name one other root of this polynomial.

Mathematics
2 answers:
weeeeeb [17]3 years ago
8 0
2-i. Complex roots always come in pairs (complex conjugates)
andrew11 [14]3 years ago
4 0
ANSWER

One other root of this polynomial is
2 - i


EXPLANATION


One property of the complex root of a polynomial is the conjugate root property.


If one root of a polynomial is
a + bi
then the conjugate is also a root of this polynomial.

The conjugate of
a + bi
is
a - bi

Also the conjugate of
a - bi

is
a + bi



Therefore if
2 - i
is a root of the polynomial, then its complex conjugate
2 + i
is also a root of that polynomial.

The same applies to purely imaginary complex roots too.

Thus, if
bi
is a root then its conjugate
- bi
is also a root.
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Using its concept, it is found that the probability that the lawyer will flip to a page number that is not a multiple of 4 is given by:

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Probability is a field of mathematics that deals with the numerical description of the likelihood that an event will occur or that a statement is true. The probability of an event is a number between 0 and 1, with approximately 0 indicating the impossibility of the event and 1 indicating certainty.

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7 0
2 years ago
determine the equation of the circle if its center is (8,-6) and which passes through the points (5,-2).
ser-zykov [4K]

Answer:

(x - 8)² + (y + 6)² = 25

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

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Here (h, k ) = (8, - 6 ) , then

(x - 8)² + (y - (- 6))² = r² , that is

(x - 8)² + (y + 6)² = r²

The radius is the distance from the centre to a point on the circle

Calculate r using the distance formula

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = (8, - 6 ) and (x₂, y₂ ) = (5, - 2 )

r = \sqrt{(5-8)^2+(-2-(-6))^2}

  = \sqrt{(-3)^2+(-2+6)^2}

  = \sqrt{9+4^2}

  = \sqrt{9+16}

   = \sqrt{25}

   = 5

Then equation of circle is

(x - 8)² + (y + 6)² = 5² , that is

(x - 8)² + (y + 6)² = 25

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