Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of
(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min
and flows out at a rate of
(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min
Then the net flow rate is governed by the differential equation
Solve for S(t):
The left side is the derivative of a product:
Integrate both sides:
There's no sugar in the water at the start, so (a) S(0) = 0, which gives
and so (b) the amount of sugar in the tank at time t is
As , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.
f(x) = (3x+5)/7
Step 1:
we write f(x) as y
y=(3x+5)/7
Step 2:
switch y by x and x by y
x=(3y+5)/7
Step 3:
solve for y,
x=(3y+5)/7
multiply both sides by 7
7x=3y+5
subtract 5 from left side
3y=7x-5
divide both sides by 3
y= (7x-5)/3
f⁻¹ (x) =(7x-5)/3
Answer:
Yes
Step-by-step explanation:
Let's take any random multiple of 4 as an example:
8 = 2 • 4
64 = 2 • 32
20 = 2 • 10
Try any multiple of 4—this pattern continues.