Algebra

(9)   ( 2 + 3i) (2 − 3i)  =   4  − 9i^2   = 4 − 9(−1)  =  4 − (−9)   =  2^2 + 3^2 =  13

A real number

The product of a complex number and its conjugate always produces this real number →

(a + bi) (a −bi)  = a^2 − (b^2 )(i^2)   = a^2 − ( b^2) ( −1)  = a^2  − ( −b^2)  = a^2 + b^2    !!!!!

2)

\(f(x)=3x-1 \qquad g(x)=x-1\qquad find (f \circ g)(x)\\~\\ (f \circ g)(x)=f(g(x))=f(x-1)\\ \text{So replace the x in the f function with (x-1) }\\ f(x-1)=3(x-1)-1=3x-3-1=3x-4\\ (f \circ g)(x)=3x-4 \)

If I  read it correctly ....here's   6

$[5x]      for    0 ≤ x ≤ 10

$[50]      for    x > 10

10

I think there is a problems with this question.

3 points is enough to define a parabola

I am assuming this is supposed to be a standard concave up parabola.

Now (-5,5) and (4,5) both have the same y value so the x half way between them should be the axis of symmetry.

\(x=\frac{-5+4}{2}=\frac{-1}{2}=-0.5\)

The axis of symmetry is x= -0.5

Since one root is at x=-1, the other root should be at x= 0

So the parabola is  

\(y=a(x-0)(x--1)\\ y=a(x)(x+1)\\ \text{Another point is (4,5) so sub that in to find }a\\ 5=a(4)(4+1)\\ 5=a(4)(5)\\ 5=20a\\ a=0.25 so\\ y=0.25(x^2+x)\\ \)

Here is the graph - you can see that this parabola cannot include the point (2,2)

Vertex.  

\(x=-0.5\\ y=0.25*(0.25-0.5)\\ y=0.25*-0.25\\ y=\frac{-1}{16}\\ \text{Vertex is } (\frac{-1}{2},\frac{-1}{16})\)

\(\)