By the Pythagorean theorem,

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:
![\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Be%5E%7Bt%2F800%7DS%28t%29%5Cright%5D%3D%5Cdfrac8%7B100%7De%5E%7Bt%2F800%7D)
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.
Answer:
<h2>9 12/25.</h2>
Step-by-step explanation:
9.48 as a mixed number in simplest form:
<em><u>948/100</u></em>
<em><u>948/100= 948 ÷ 4/100 ÷ 4</u></em>
<em><u>948/100= 948 ÷ 4/100 ÷ 4= 237/25</u></em>
<em><u>948/100= 948 ÷ 4/100 ÷ 4= 237/25= 9 12/25</u></em>
<u>9 12/25 is the mixed number.</u>
We can say that ‘0’ is also a rational number since it can represent it in many forms of 0/1, 0/2, 0/3, etc. So yea it is hope this helped!
Domain values represent possible x values that are allowed to be plugged in and produce a y value. The values (in this case value) that are not allowed to be plugged in are what make the denominator zero (since you cannot divide by zero). Simply set the denominator equal to zero to figure out this value.
3x + 8 = 0
3x = -8
x= -8/3, this value is not in the domain.