Let q be the number of 25 cent coins.
Let d be the number of 10 cent coins.
0.25q+0.10d= 3.95...(1)
q-d=6...(2)
(2)->    q-d= 6
       q-d+d= 6+d
              q= 6+d...(2a)
(2a)-> (1)      0.25q+0.10d=3.95
              0.25(6+d)=0.10d= 3.95
           1.95+0.25d+0.10d= 3.95
                               0.35d= 3.95-1.5
                       0.35d/0.35= 2.45/0.35
                                      d= 7...(3)
(3)->(2)     q-d= 6
                 q-7= 6
                    q=6+7
                    q= 13
There are 13 quarters and 7 dimes.
        
             
        
        
        
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:
r = 7t
or not mathematically, 
r(t) = 7t
r(t) means, r, which is a function of t.
Step-by-step explanation:
Initial size of the radius = 0 cm, at t = 0 s
Rate of increase of the radius of the circle = 7 cm/s
dr/dt = 7
dr = 7 dt 
∫ dr = 7 ∫ dt
r = 7t + C (C is the constant of integration)
At t = 0, r = 0,
0 = 0 + C
C = 0
r = 7t.
r(t) = 7t