CPhill

f(x)  =  sqrt (-x)

Because the quantity under the root must be greater or equal to 0....then...

The domain of A is (-inf, 0 ]

The domain of B =  g(x)  is exactly the same

So

C.) The domain of both functions is the set of real numbers less than or equal to 0.

Function A has two x intercepts  since the points (3, 0) and (1,0)  are on the graph

To find out the number  of x intercepts that B has

l x l - 2 = 0

l x l = 2

Two values of x satify this   x = 2  and x = 2

So....B has two x intercepts

So

Both functions have the same number of x-intercepts.

[ But not the SAME x intercepts ]  !!!!

Draw a square of some side

Draw a rhombus with the same side length(s)  as the square

The figures will be in proportion but they wll not be similar  !!!

3(4x+6)=9x+12

12x + 18   =  9x + 12       subtract 9x, 18 from both sides

3x = - 6       divide both sides by 3

x = -6 / 3   =   -2

Excellent, EW!!!!

A 2 : 3 ratio means that this point is (2/5) of the way from A to B

We can find the point thusly :

 [ 2 + (6 - 2)(2/5) , 7 + (2 - 7)(2/5) ] =

[ 2 + (4)(2/5)  , 7 + (-5)(2/5) ] =

[ 2 + 8/5  , 7 - 2   ]    =

( 10/5 + 8/5 , 5 ) =

(18/5, 5)

The locations of the friends will form vertexes of a triangle

If  we construct perpendicular bisectors of each side of this triangle.....they willl neet at the circumcenter.....

This will be the center of a circle that will pass  through each vertex....so.....the radial distance from this point to each of his friends will be equal

[BTW...this point could lie outside the triangle   !!! ]

Here's a good resource to show this : http://www.mathopenref.com/trianglecircumcenter.html

You need to draw a line through the apex point of the triangle that is parallel to the triangle's base

This shows how to do this :

http://www.mathopenref.com/printparallel.html

We can then use the fact that alternate interior angles are equal to complete the proof

If angle OAL =  70°   GO would be parallel to LA

To see this.....extend LA in the direction of A to   E

Then angle OAL + angle  OAE  sum to 180°

But angle OAE =  180 - angle OAL   =  180 - 70 =  110

Then angles GOA and OAE are equal

And if this is so....then transversal OA makes alternate interior angles GOA and OAE equal....so....

A transversal cutting lines (segments) so that alternate interior angles are equal means that the lines (segments) are parallel

So...GO is parallel to LA.....and a trapezoid must have one pair of sides parallel

Segment PT || segment QS,                                  Given

segment PT ≅ segment QS,

∠T ≅ ∠S

Angle TPQ = Angle SQR                      PR is a transversal  cutting parallel segments SQ and TP

                                                              So....it makes corresponding angles TPQ and SQR equal   

ΔPQT   ≅    ΔQRS                                ASA congruency