CPhill

(a)  Assuming  that  we have an isosceles trapezoid

AB  = DC               sides of an isoceles trapezoid are equal

Angle ABC  = Angle DCB      base angles of an isoceles trapezoid are equal

Angle ABC + Angle DAB  = 180          opposite angles in an isoceles trapezoid are  supplementary

Angle  DCB  + Angle ADC  = 180

Angle DAB  = Ange  ADC                subtraction property of equality

AD  = AD                                       reflexive property       

Triangle  ABD  is congruent to Triangle DCA         By   SAS

(b)   

AD  is parallel to BC            bases of anisoscees trapezoid are  parallel

Angle ADB  = Angle DBC         a transversal  cutting parallel lines   makes alternate interior angles equal

Very nice, Gadfly   !!!!

x*y*z^3  =   24     (1) 

x*y^3*z   =  54     (2)

x^3*y*z  = 6         (3)

Divide   (1) by (3)   and we have that    z^2 / x^2  =  4   ⇒  z / x  =  ±2

So

z = 2x     or  z  = -2x

Divide (2)  by (3)    and we get that   y^2 / x^2  = 9  ⇒  y /x  = ±3

So

y = 3x    or  y  =  - 3x

Using (3) and substituting,we have that

x^3 *( 3x) ( 2x)  =  6           x^3 (-2x) (3x)  = 6        

x^5  = 1                              x^3  =  -1

x  = 1                                  x  = -1

And we have these real solutions

x       y         z

1       3         2

1      -3       -2

-1     -3        2

-1      3       -2 

Answered this just a few days ago.....

The guest's answer of 19  is correct    !!!!

Thanks, NewMember   !!!!

Note that if we had 3 people, A, B , C.....the number of handshakes  = 3

AB  AC

BC

Note that if we only had 4 people, A,B, C, D.....the number of handshakes  =  6

AB  AC  AD

BC BD

CD

This implies that for n peole, the  number of  handshakes    is     n ( n - 1)  / 2

So....for 100 people we have

100 (100 - 1)   / 2  =

50 * 99  =

4950 handshakes

(-x^3  + 4x)  =  x ( 4 -x^2)   =   x  ( 2 - x) ( 2 + x)

Since we have "hole"  at  x  = -2,   then   ( x + 2)  must be a factor of  q(x)

Also....since we have a vertical asymptote at  x =1, then  (x - 1)   must also be a factor of q(x)

Since we have no horizontal asymptote, then we must have  a slant asymptote, which means that we have a higher/ lower situation with respect to the degree of the polynomial in the numerator to the degree of the polynomial in the  denominator

And  since   q(3)   =  -30

Then we must have  that   

-30   = a (3 - 1) ( 3 + 2)

-30  = a (2)(5)

-30  = 10a

a=-30/10  =  -3

So

q(x)  must  be   -3 ( x -1 ) ( x + 2)  =   (-3) (x^2 + x - 2)  =  -3x^2 -3x  + 6

Nothing wrong with your method, EP......just shows another way to solve it

This is always good   !!!!

3.     (x^(4)+4x^(2)-6x-24)-(x+4)

Simplify  and we have     x^4 -4x^2  -7x - 28

WolframAlpha  shows that we have no rational roots for this  : 

https://www.wolframalpha.com/input/?i=%28x%5E4%2B4x%5E2-+6x+-+24%29+-+%28x%2B4%29

Are you sure that you have the correct function  ????

-12(x - 5)  =  (y + 2)^2

Vertex  =  ( 5, -2)

Opens to the left

4p = -12

p = -3

The   focus  is   l p l  =   l -3 l  =   3 units from the vertex

The directrix  is   2 l p l  =  2 l - 3 l  =  2(3)   =   6 units from the focus

Focus  =  (5 + 3 , - 2)  =  (5 - 3, 2)   =   (2, -2)

The equation of the directrix  is    x  = (5 + 3)   ⇒     x  =   8