Detailed derivation:
dA/dt = 6 - 0.02A
dA/dt = -0.02 (A - 300)
Rearranging terms.
dA / (A - 300) = -0.02 dt
Integrate both sides.
ln(A - 300) = -0.02t + C
Isolate A.
A - 300 = Ce^(-0.02t)
A = 300 + Ce^(-0.02t)
Apply initial condition to determine C.
50 = 300 + Ce^(-0.02 × 10)
50 = 300 + Ce^(-0.2)
-250 = Ce^(-0.2)
C = -250e^(0.2)
A = 300 - 250e^(0.2)e^(-0.02t)
A = 300 - 250e^(0.2 - 0.02t)
Response:
Count of bottles = 6
Count of cans = c
Detailed reasoning:
Based on the following information:
Containers for selling water:
17 oz bottle
11 oz can
Let the number of bottles be denoted as b
Count of cans represented as c
b + c = 11 - - - (1)
17b + 11c = 157 - - - (2)
b = 11 - c
Insert into (2)
17(11 - c) + 11c = 157
187 - 17c + 11c = 157
187 - 6c = 157
-6c = 157 - 187
-6c = -30
c = 30/6
c = 5
From; b = 11 - c
b = 11 - 5
b = 6
Thus,
Count of bottles = 6
Count of cans = c
X = -m/b.
By subtracting z from both sides, you find
-m= bx
Next, divide every term by b to obtain
-m/b= x
Solution:
Given that
G being the midpoint of FH implies
Additionally, we have
Hence, we can express
Therefore, 
