What is her speed when she gets to the bottom of the slope ?

Her potential energy at the top of the slope = (mass) x (gravity) x (height)

Her kinetic energy at the bottom of the slope = (1/2) x (mass) x (speed)² .

The only possible way to handle this problem without making it
grossly complicated is to assume that the slope is frictionless. 
Then ALL of the potential energy she had at the top shows up
as kinetic energy at the bottom, so we can write:

              (1/2) x (mass) x (speed)²  =  (mass) x (gravity) x (height)

Divide each side
by  (mass) :              (1/2) x (speed)²  =  (gravity) x (height)

Isn't that interesting ! 
The speed at the bottom only depends on the height of the hill
and what planet she's skiing on.  Her mass/weight doesn't matter. 
A beanpole and a tons-o'-fun will both have the same speed when
they reach the bottom !

                         (1/2) x (speed)²  =  (gravity) x (height)

Multiply each side
by  2 :                             speed²  =  (2) x (gravity) x (height)

Square root
each side:                     Speed  =  √ (2 g h)

                                                     =  √ (19.6 h)

                                                     = about  4.43 x √(height of the hill, meters) 

Ithink the answer would be a

Make them both to where they have the denominator of 18that makes 3/2 equal 27/18 and 1/9 equal 2/18so now you have  27/18x+5=2/18subtract 5 from the other sideso now you have 27/18x+2/18-5then subtract 2/18 from 27/18 to get 25/18x=-5then divide 25/18 by -5 and your answer will be  -5/18