Unit rate has to be per week or per day or per hr
so keeping in mind rate has to be 108/6 = 18 per week
        
             
        
        
        
Answer:
a) 72.25sec
b) 6.25secs
c) after 10.5secs and 2 secs 
Step-by-step explanation:
Given the height reached by the rocket expressed as;
s(t)= -4t^2 + 50t - 84
At maximum height, the velocity of the rocket is zero i.e ds/dt  = 0
ds/dt = -8t + 50
0 = -8t + 50
8t = 50
t = 50/8
t = 6.25secs
Hence it will reach the maximum height after 6.25secs
To get the maximum height, you will substitute t - 6.25s into the given expression
s(t)= -4t^2 + 50t - 84
s(6.25) = -4(6.25)^2 + 50(6.25) - 84
s(6.25) = -156.25 + 312.5 - 84
s(6.25)  = 72.25feet
Hence the maximum height reached by the rocket is 72.25feet
The rocket will reach the ground when s(t) = 0
Substitute into the expression
s(t)= -4t^2 + 50t - 84
0 = -4t^2 + 50t - 84
4t^2 - 50t + 84 = 0
2t^2 - 25t + 42 = 0
2t^2 - 4t - 21t + 42 = 0
2t(t-2)-21(t-2) = 0
(2t - 21) (t - 2) = 0
2t - 21 = 0 and t - 2 = 0
2t = 21 and t = 2
t = 10.5 and 2
Hence the time the rocket will reach the ground are after 10.5secs and 2 secs 
 
        
             
        
        
        
You have a 1 in 3 chance of drawing a queen. so about 33%
        
                    
             
        
        
        
 The answer is 253 square feet
        
                    
             
        
        
        
Answer:
C
Step-by-step explanation:
The secant- tangent angle MLK is half the difference of the intercepted arcs, that is
 (KN - KM) = 70° ( multiply both sides by 2 to clear the fraction )
 (KN - KM) = 70° ( multiply both sides by 2 to clear the fraction )
KN - KM = 140°
KN - 61° = 140° ( add 61° to both sides )
KN = 201° → C