Answer:
Dh/dt  = 0.082 ft/min
Step-by-step explanation:
As a perpendicular cross section of the trough is in the shape of an isosceles triangle the trough has a circular cone shape wit base of  1 feet and height     h = 2 feet.
The volume of a circular cone is:
V(c)  = 1/3 * π*r²*h
Then differentiating on both sides of the equation we get:
DV(c)/dt   = 1/3* π*r² * Dh/dt   (1)
We know that DV(c) / dt   is  1 ft³ / 5 min      or     1/5  ft³/min
and  we are were asked how fast is the water rising when the water is 1/2 foot deep. We need to know what is the value of r at that moment
By proportion we know
r/h  ( at the top of the cone  0,5/ 2)   is equal to  r/0.5  when water is 1/2 foot deep
Then      r/h   =   0,5/2   =  r/0.5
r  =  (0,5)*( 0.5) / 2        ⇒   r  =  0,125 ft
Then in equation (1) we got
(1/5) / 1/3* π*r² =  Dh/dt
Dh/dt  = 1/ 5*0.01635
Dh/dt  = 0.082 ft/min
 
        
             
        
        
        
Answer:
True
Step-by-step explanation:
0     .      0                0                     0
whole     Tenths      Hundredths    Thousandths
 
        
             
        
        
        
Line b is the answer, hope this helped
        
                    
             
        
        
        
Answer:
243cm squared
Step-by-step explanation:
 
        
             
        
        
        
<span>1/2, 2/4, 3/6, 4/8, 5/10, 6/12, 7/14, 8/16, 9/18, 10/20, and so on ...
Hoped this Helped! 
-Hayden </span>