Is u = 105 a solution to this equation

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

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

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

4 years ago

Ann will randomly select a bead from the package and then without replacing it randomly select a second bead based on the inform

san4es73 [151]

Answer:.......

Probability = $\frac{g}{g+o} + \frac{g-1}{g+o}$

Step-by-step explanation:

Let g represent the number of green beads , and o represent the number of orange beads.

due to the fact that Ann is selecting the bead without replacing i randomly

the probability (Pr) of Ann selecting a green bead and then an orange one = $\frac{g}{g+o} +\frac{g-1}{g+o}$

7 0

3 years ago

Does anyone know the answer?

motikmotik

Answer:

We conclude that the area of the right triangle is:

$A\:=p^2-2p-3$

Hence, option A is correct.

Step-by-step explanation:

From the given right-angled triangle,

Base = p+1

Perpendicular = 2p-6

Using the formula to determine the area of the right-angled triangle

Area of the right triangle A = 1/2 × Base × Perpendicular

$=\frac{1}{2}\left(p+1\right)\left(2p-6\right)$

Factor 2p-6: 2(p-3)

$=\frac{\left(p+1\right)\times \:2\left(p-3\right)}{2}$

Divide the number: 2/2 = 1

$=\left(p+1\right)\left(p-3\right)$

$\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$

$=pp+p\left(-3\right)+1\cdot \:p+1\cdot \left(-3\right)$

$=pp-3p+1\cdot \:p-1\cdot \:3$

$=p^2-2p-3$

Therefore, we conclude that the area of the right triangle is:

$A\:=p^2-2p-3$

Hence, option A is correct.

3 0

3 years ago

What single transformation was applied to triangle A to get triangle B? *

irakobra [83]

Answer: dilation

Step-by-step explanation:

the others are rigid transformations, the can’t change the size of the triangle.

6 0

3 years ago

6. Find OR.<br> 2<br> 3x + 22<br> 10x - 41

Sever21 [200]

3x+22=10x-41
-22 -22
----------------------
3x=10x-63
-10x -10x
-----------------
-7x=-63
divide by -7 on both sides
yep ur right its x=9

4 0

3 years ago