A, B and D are already rational numbers, 0.4 itself is a rational number, two rational numbers make a product of rational number. 
So A, B, D are the correct choices. 
Are you sure it asks for rational numbers, not irrational number?
        
                    
             
        
        
        
Answer: BC = 5.83
Step-by-step explanation:
Luckily, the triangle is placed on the graph nicely so we can count the legs of the triangle:
AB = 5
AC = 3
BC = ?
To find BC, we can simply use the Pythagorean Theorem:

5^2 + 3^2 = c^2
25 + 9 = c^2
34 = c^2
Now square root to find c, or BC.

c = 5.83 (rounded by nearest hundredth)
 
        
             
        
        
        
Answer:
y = - 2x + (1/2 38) + 2   = y =  -2x + 19x + 2  at any second  and y = -2x^2 / 2  + 19/2 + 2/2 = <u>- x^2 + 19/2 + 2 </u>=  - x (4) + 19/2(4) + 1 = -4+ 38 + 1 = 35 seconds  
Step-by-step explanation:    We see that 38-2ft = 36ft and y intercept  = 2 and then -2x allows us to represent the starting point 2 as -2 (1) to allow a descend to our back to 0 for y  one we find y intercept we know - x^2  is our simplified equation and an input into this to find the static and slowed descend back to 0  if you keep inputting at 5 and 6 you see the equation speed up   Anyway at (4) substitute = 4 seconds we divide our simplified equation a = -x2 into  a b c and divide each by 2  before working out the<u> </u><u>decline</u> of the equation<u> (as equation still represents all ascending and descending) </u>for the 4th second as the height was <u>already said to be at its max at 38 feet see equation in answer to find 4 as substitute for x </u>
 
        
             
        
        
        
Remark
I would have had the answer a whole lot sooner if I would have read the question properly. The figure in the circle is called a cyclic quadrilateral. It has the odd property that the angles that are opposite each other add up to 180o. 
So DEB + DCB = 180o
DEB = 180 - 87
DEB = 93o
Note: The arcs marked 60 and 76 have nothing whatever to do with this problem.