PLease help me on this one? <3

Mathematics

1 answer:

MariettaO [177]3 years ago

6 0

Answer:

A. (1, 0)

Reason:

When the first equation is graphed: y < x + 1, the left side of the line is shaded and when the second equation is graphed: y < x^2 -3x, the bottom is shaded.

The green section does not contain solutions. So D is automatically out.

The red section does not contain solutions as well. So C is automatically out.

For choice B, the point goes out of the shaded blue section, so that's out also.

As for choice A its in the blue shaded section, which makes that answer correct.

(when the two equations are within each other and combine a color, then whatever points is within the shaded part is the right answer)

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Find the complex zeros of x^3+27. write f in factored form

aliina [53]

$x^3+27=0
\\x^3+3^3=0
\\(x+3)(x^2-3x+9)=0
\\x^2-3x+9=0
\\a=1,b=-3,c=9
\\
\\x_{1,2}= \frac{-b\pm \sqrt{b^2-4ac} }{2a} = \frac{-(-3)\pm \sqrt{(-3)^2-4\times1\times9} }{2\times1} =\frac{3\pm \sqrt{-27} }{2} =\frac{3\pm 3\sqrt{3}i }{2} 
\\
\\(x+3)(x-\frac{3+ 3\sqrt{3}i }{2})(x-\frac{3- 3\sqrt{3}i }{2})=0

$