Miran Anwar, mth 235 ss21 1 Hw17-5.2-SDE-2x2PP. Due 042620

Miran Anwar, mth 235 ss21 1: Hw17-5.2-SDE-2x2PP. Due:
04/26/2021 at 11:00pm EDT.
See in LN, § 5.2, See Example 5.2.4.
1. (10 points)
Note: You have only 5 attempts to solve this problem.
Consider the system of differential equations x′ = A x, where x
=
[
x1
x2
]
and the 2×2 matrix A has eigenpairs
λ± = α±β i, α = 0.25, β = 2,
v± = a±b i, a =
[
2
1
]
, b =
[
−1/2

1
]
,
Draw on paper the solution x(t) = eαt
(
a cos(βt)−b sin(βt)
)
on the x1x2-plane and then answer the ques-
tions below.
(a) Based on your graph, select the correct solution curve from
the interactive graph below.
• Select One
• Curve 1
• Curve 2
• Curve 3
• Curve 4
• None
(b) Use the graph below to find lim
t→+∞
‖x(t)‖, where ‖x(t)‖ is the length of the solution vector x( t).
• Select One
• Zero
• Infinity
• None
(c) Introduce the unit vectors u1 = a/‖a‖ and u2 = b/‖b‖. Use the
graph below to find the lim

t→+∞
u(t),
where u(t) =
x(t)
‖x(t)‖
, is a unit vector in the direction of the solution vector x(t).
[Select One/U1/-U1/U2/-U2/None]
(d) Use the graph below to find lim
t→−∞
‖x(t)‖.
1
• Select One
• Zero
• Infinity
• None
(e) Use the graph below to find the lim
t→−∞
u(t), where u(t) =
x(t)
‖x(t)‖
.

(f) Characterize the zero solution, x0 = 0.
• Select One
• Source Node
• Source Spiral
• Sink Node
• Sink Spiral
• Saddle
• Center
• None
Comments on the graph below:
• The graph is interactive.
• You can click on the boxes to turn on and off possible solution
curves.
• For each possible solution we display: the possible solution
curve, the possible solution vector
x(t), and the associated unit vector u(t) = x(t)/‖x(t)‖.
• You can move the time slider to see how each possible
solution vector x(t) and unit vector u(t)
change in time.
• You can turn on or off the ellipse formed by vectors a and b.
1
2. (10 points)
=
[

x1
x2
]
λ± = α±β i, α = 0.45, β = 2,
v± = a±b i, a = −
[
2
1
]
, b = −
[
−1/2
1
]
,
(
a sin(βt)+ b cos(βt)
)
tions below.
2

• Select One
• Curve 1
• Curve 2
• Curve 3
• Curve 4
• None
t→+∞
‖x(t)‖, where ‖x(t)‖ is the length of the solution vector x(t).
• Select One
• Zero
• Infinity
• None
t→+∞
u(t),
where u(t) =
x(t)
‖x(t)‖

t→−∞
‖x(t)‖.
• Select One
• Zero
• Infinity
• None
t→−∞
u(t), where u(t) =
x(t)
‖x(t)‖
.
• Select One
• Source Node
• Source Spiral
• Sink Node
• Sink Spiral
• Saddle
• Center
• None
3

curves.
change in time.
1
3. (10 points)
=
[
x1
x2
]
λ± = α±β i, α = −0.4, β = 2,
v± = a±b i, a =
[
2

1
]
, b = −
[
−1/2
1
]
,
(
a cos(βt)−b sin(βt)
)
tions below.
• Select One
• Curve 1
• Curve 2
• Curve 3
• Curve 4
• None
t→+∞

• Select One
• Zero
• Infinity
• None
t→+∞
u(t),
where u(t) =
x(t)
‖x(t)‖
4
t→−∞
‖x(t)‖.
• Select One
• Zero
• Infinity
• None
t→−∞

u(t), where u(t) =
x(t)
‖x(t)‖
.
• Select One
• Source Node
• Source Spiral
• Sink Node
• Sink Spiral
• Saddle
• Center
• None
curves.
change in time.
1

4. (10 points)
=
[
x1
x2
]
λ± = α±β i, α = −0.3, β = 2,
5
v± = a±b i, a = −
[
2
1
]
, b =
[
−1/2
1
]
,

(
a sin(βt)+ b cos(βt)
)
tions below.
• Select One
• Curve 1
• Curve 2
• Curve 3
• Curve 4
• None
t→+∞
• Select One
• Zero
• Infinity
• None
t→+∞
u(t),

where u(t) =
x(t)
‖x(t)‖
t→−∞
‖x(t)‖.
• Select One
• Zero
• Infinity
• None
t→−∞
u(t), where u(t) =
x(t)
‖x(t)‖
.
6
• Select One

• Source Node
• Source Spiral
• Sink Node
• Sink Spiral
• Saddle
• Center
• None
curves.
change in time.
1
See in LN, § 5.2, See Examples 5.2.1, 5.2.2, 5.2.3.
5. (10 points)
=
[
x1
x2

]
λ1 = 1.5, v1 =
[
1
1
]
, and λ2 = 0.75, v2 =
[
−1
1
]
.
Draw on paper the solution x(t) =−v1 eλ1t +v2 eλ2t on the
x1x2-plane and then answer the questions below.
• Select One
• Curve 1
• Curve 2
• Curve 3
• Curve 4
• None
t→+∞

7
• Select One
• Zero
• Infinity
• None
(c) Introduce the unit vectors u1 = v1/‖v1‖ and u2 = v2/‖v2‖. Use
the graph below to find the
lim
t→+∞
u(t), where u(t) =
x(t)
‖x(t)‖
t→−∞
‖x(t)‖.
• Select One
• Zero
• Infinity
• None
t→−∞

u(t), where u(t) =
x(t)
‖x(t)‖
.
• Select One
• Source Node
• Source Spiral
• Sink Node
• Sink Spiral
• Saddle
• Center
• None
curves.
change in time.
• You can move the eigenvectors v1 and v2 by dragging them
from the endpoint, and then see how
the curves would change.

1
6. (10 points)
8
Consider the system of differential equations x′ = A x, w here x
=
[
x1
x2
]
λ1 = 0.5, v1 =
[
1
1
]
, and λ2 =−1.75, v2 =
[
−1
1
]
.

Draw on paper the solution x(t) = v1 eλ1t + v2 eλ2t on the
• Select One
• Curve 1
• Curve 2
• Curve 3
• Curve 4
• None
t→+∞
• Select One
• Zero
• Infinity
• None
lim
t→+∞
u(t), where u(t) =
x(t)
‖x(t)‖

t→−∞
‖x(t)‖.
• Select One
• Zero
• Infinity
• None
t→−∞
u(t), where u(t) =
x(t)
‖x(t)‖
.
9
• Select One
• Source Node
• Source Spiral
• Sink Node
• Sink Spiral
• Saddle
• Center

• None
curves.
change in time.
1
7. (10 points)
=
[
x1
x2
]

λ1 =−1, v1 =
[
1
1
]
, and λ2 =−1.5, v2 =
[
−1
1
]
.
Draw on paper the solution x(t) =−v1 eλ1t +v2 eλ2t on the
• Select One
• Curve 1
• Curve 2
• Curve 3
• Curve 4
• None
t→+∞
10

• Select One
• Zero
• Infinity
• None
lim
t→+∞
u(t), where u(t) =
x(t)
‖x(t)‖
t→−∞
‖x(t)‖.
• Select One
• Zero
• Infinity
• None
t→−∞
u(t), where u(t) =
x(t)
‖x(t)‖

.
• Select One
• Source Node
• Source Spiral
• Sink Node
• Sink Spiral
• Saddle
• Center
• None
curves.
change in time.
1
See in LN, § 5.2, See Examples 5.2.1, 5.2.2, 5.2.3, 5.2.4.

8. (10 points)
11
Match each vector field with its differential equation.
? 1. x′ =
[
1 1
−1 1
]
x
? 2. x′ =
[
0 −1
1 0
]
x
? 3. x′ =
[
0 3
3 0
]
x

? 4. x′ =
[
−4 1
−1 −4
]
x
A B
C D
12
9. (10 points)
Match each solution (in red) with its initial value problem.
? 1. x′ =
[
0 3
3 0
]
x, x(0) =
[
0
−2

]
? 2. x′ =
[
0 3
3 0
]
x, x(0) =
[
−2
0
]
? 3. x′ =
[
0 3
3 0
]
x, x(0) =
[
0
2
]
? 4. x′ =
[
0 3
3 0
]

x, x(0) =
[
2
0
]
13
A B
C D
10. (10 points)
14
? 1. x′ =
[
5 1
−1 5
]
x, x(0) =
[

2
4
]
? 2. x′ =
[
−5 1
−1 −5
]
x, x(0) =
[
2
4
]
? 3. x′ =
[
5 1
−1 5
]
x, x(0) =
[
3
0
]
? 4. x′ =
[
−5 1

−1 −5
]
x, x(0) =
[
3
0
]
A B
C D
11. (10 points)
15
? 1. x′ =
[
2 1
1 2
]
x, x(0) =
[

2
0
]
? 2. x′ =
[
−2 1
1 −2
]
x, x(0) =
[
0
2
]
? 3. x′ =
[
−2 1
1 −2
]
x, x(0) =
[
2
0
]
? 4. x′ =
[
2 1

1 2
]
x, x(0) =
[
0
2
]
A B
C D
16
12. (10 points)
? 1. x′ =
[
0 −3
−3 0
]
x, x(0) =
[
2

0
]
? 2. x′ =
[
0 −3
−3 0
]
x, x(0) =
[
0
2
]
? 3. x′ =
[
0 3
3 0
]
x, x(0) =
[
0
2
]
? 4. x′ =
[
0 3
3 0

]
x, x(0) =
[
2
0
]
17
A B
C D
Generated by c©WeBWorK, http://webwork.maa.org,
Mathematical Association of America
18
PK-PD Primer Homework
Max 4pages only answers
PK-Pharmacology Primer
Instructions for the PK problems:
Please answer the questions below as best as you can by
examining the outcomes of the pharmacoki netic models
provided in the Excel file “CLRE-238 SP21 PK-PD
Homework.xlxs.” In many cases, you will need to find the
requested value of parameters by trial and error, changing some
of the values in bold red in the worksheets and looking at the
resulting plots. You do not need to perform any calculations.
All the answers are very short. Do not be intimidated by the

Excel worksheets; this is a lot easier than it appears at first
glance.
General comment: Because you are expected to obtain your
answers by visually examining the pharmacokinetic plots in the
associated Excel file, your quantitative answers will be graded
as correct if they were reasonably close to the exact answer.
There is no requirement to be precise.
Problem A:
The antibiotic genericomycin (not a real one) is administered by
iv bolus to hospitalized patients. The antibiotic has an
elimination half-life of 4.0 h in people. Previous experience has
shown that to be effective at reducing the bacterial burden its
plasma concentration must equal or higher than 5 µg/ml (Ceff).
Question 1: Using the “Intravenous” tab of the Excel file,
determine the duration of action of the following iv bolus doses
and enter the values in the table below.
Hint: You can eyeball the time values from the chart or read
them from the list of Cp vs time for the Plot (blue numbers)
Answer 1:
Dose
(mg)
Duration of Action
(h)
140
4
280
8
560
12
1120
16
Question 2: What is the effect of doubling the dose on the
duration of action?

Answer 2: it adds the duration of action by 4 hours
Question 3: What would be the necessary iv bolus dose to
establish plasma concentration 5 µg/ml at 24 h? Is this dose
realistic? Assume genericomycin costs $10/mg.
Answer 3:
Problem B:
You have created a new company, NovoAntibiot, Inc. with the
intention of creating a new antibiotic with much better
properties than genericomycin. Your clinical advisors suggest
that the new antibiotic should be administered by iv bolus once
a day with a dose of at most 560 mg per person (your new
therapeutic product profile).
Question 4: What elimination half-life should you design in
your new antibiotic to achieve a plasma concentration of at least
5 µg/ml 24 h after a 560 mg iv bolus dose?
Answer 4: 40
Welcome to drug discovery! After working for 2 years and
spending $15 M from your Angel and series A investors you
were unable to increase the half-life of your new antibiotics. All
of them still have halflives around 4 h. But you got lucky, some
of your new compounds have higher potency. Drug candidate
NABT 813, is much more potent than genericomycin. In vitro
experiments and animal models suggest that the minimum
effective concentration (Ceff) in humans will be 1 µg/ml.
Question 5: What iv bolus dose of NABT 813 will treat the
patients for 24h if its elimination half-life is 4.0 h and its Ceff
is 1 µg/ml? Is it realistic? (compare with your Answer 3
above).)
Answer 5: 140mg

Problem C
Unfortunately, your new antibiotic NABT 813 decreased the
number and body weight of rat pups in formal GLP repro-tox
studies in rats. The compound cannot be used in women who
might not know whether they are pregnant at the time they are
admitted to the hospital with a systemic infection. This is a
crushing blow and your investors abandon you. The new drug
discovery program is terminated.
However, your PK consultant suggests that you might be able to
formulate genericomycin for subcutaneous administration by
creating an insoluble salt and adding some excipients that will
retard its absorption, allowing (perhaps) once a day dosing.
Your intellectual property attorney states that you might be able
to obtain patent protection for the new formulation if you
demonstrate substantial advantages over the generic iv
formulation. You apply for and get an SBIR grant to
demonstrate proof of (PK) concept.
Question 6: In the “Subcutaneous” tab of the Excel file, set the
elimination half-life of genericomycin to 4.0 h, its Ceff at 5
µg/ml, and its dose to 280 mg. Examine the effect of changing
the absorption half-life of genericomycin from your new
formulations by replacing the value of t1/2 (abs) and complete
the table below.
Answer 6:
t1/2 abs (h)
Tmax (h)
Cmax (µg/ml)

0.05
0.25
18
1
2.5
13
3
5
8.5
6
7
5.9
10
9
4.2
Hint: You can eyeball the above values from the chart or read
them from the list of Cp vs time for the Plots (blue numbers).
No need to be very precise.
Question 7: Why does the PK profile for absorption with a very
short half-life (0.05 h) look like the ivbolus profile? (This never
happens in real life….)
Answer 7: Clearance is more rapid and half-lives are shorter
Question 8: What is the effect of increasing the absorption half-
life on Tmax and Cmax?
Answer 8: increasing tmax increases absorption while
increasing cmax decreases absorption
Question 9: Is there any value of absorption half-life that will
allow you to have plasma concentration higher or equal to Ceff
at 24 h when dosing sc 280 mg?
Answer 9: yes

Question 10: What is the duration of action of genericomycin at
a dose of 280 mg if it is administered in the formulation with
absorption half-life of 10 hours?
Answer 10: 4.3
Problem X (bonus):
Your microbiology consultant tells you that an analogue of
genericomycin with longer eliminationhalflife was developed in
the Soviet Union in the 1960’s, but it never reached the West
and was forgotten. Its name was bolshoimitsina and had an
elimination half-life in humans of 7.8 h, and Ceff also 5 µg/ml.
You obtain a sample of bolshoimitsina via a website vendor in
Bulgaria and find that it also has a sc absorption half-life of
10.0 h in your preferred formulation.
Question 11: What sc dose of formulated bolshoimitsina will
provide a plasma concentration equal to Ceff 5 µg/ml at 24 h?
Answer 11: 9.43mg
Question 12: At the dose of bolshoimitsina you found in Answer
11, when does the drug start having its effect in the patient?
Answer 12: 0.5hrs
PK-Pharmacology Primer
Homework with Answers
Student Name:
Instructions for the Pharmacology problems:
Please extract the requested information from the papers
provided. You do not need to read the papers in detail. This is
an exercise in identifying the required information quickly
while checking its validity. Your answers will be very short, a
value (with units!) or a few words.
Problem P:

Look at the provided paper by Busnelli et al., “Selective and
Potent Agonists and Antagonists for Investigating the Role of
Mouse Oxytocin Receptors”, and answer the following
questions:
Question 31: What was the Ki value for compound “LVA” at
the mOTR (mouse oxytocin receptor)?
Answer 31:
Question 32: Does LVA have higher or lower affinity than “OT”
(oxytocin) for the mOTR?
Answer 32:
Look in Materials and Methods, section on Ligand Binding
Assays.
Question 33: How were the Ki values obtained: Schild analysis
or from IC50 values?
Answer 33:
Question 34: What compound was the radioligand for measuring
binding to the mOTR, and at what concentration was used?
Answer 34:
Question 35: What was the Kd value for the radioligand at the
mOTR?
Answer 35:
Look in the main body of the paper.
Question 36: Which figure shows the competition binding
experiments for compound “OTA1”?
Answer 36:
Question 37: Looking at said figure, would you say that OTA1
has higher affinity for the mV1aR (mouse vasopressin type 1a
receptor) than for the mOTR?
Answer 37:

Problem Q:
Look at the provided paper by Ichinose et al., “Development of
a Highly Potent Analogue and a Long-Acting Analogue of
Oxytocin for the Treatment of Social Impairment-Like
Behaviors”, and answer the following questions:
Question 38: What was the EC50 at the hOTR (human oxytocin
receptor) of compound 4?
Answer 38:
Question 39: What was the efficacy of compound 4?
Answer 39:
Question 40: Of the new compounds prepared in this paper
(Compounds 1 to 6), which one was more potent at the hOTR?
Which one was the most potent at the hV1aR?
Answer 40:
Look in Results and Discussion, section on Agonist-Induced
Increase in Intracellular Ca2+ Concentrations.
Question 41: Which second messenger was measured to
determine receptor activation?
Answer 41:
Look at Table 2, data on Ki at the hOTR and EC50 at the hOTR
Question 42: Would you suspect that there might be
considerable “receptor reserve” for activation of the hOTR by
OT? What about carbetocin?
Answer 42:
Page 1 of 5

Instructions:
A company called Nastech was involved in developing
intranasal formulations of drugs to treat a variety of different
diseases. Dr Steven Quay was a founder of Nastech; he (and his
colleagues, including Dr Alexis Leonard) filed several patent
applications covering his inventions; some of these patents were
subsequently licensed or acquired by other companies. One of
these patent applications was for intranasal formulations of
carbetocin (an oxytocin analog), for possible treatment of
autism and other conditions.
Question 1:
What is the patent number of the US granted patent described
above?
Answer 1:
Question 2:
What is the number of the corresponding US published
application?
Answer 2:
Question 3:
Compare the claims in the granted patent vs. the published
application. What is/are the major difference(s) between the
claims in these two documents? (one or two sentences; hint:
focus on Claim 1)
Answer 3:
Question 4:

What is one key piece of translational research data presented in
the patent documents that supports the claims? (one or two
sentences; hint: do any of the “Examples” contain translational
research data?)
Answer 4:
Miran Anwar, mth 235 ss21 1: Hw18-5.3-5.4-SDE-2x2NS-CS.
Due: 04/26/2021 at 11:00pm
EDT.
Similar problem in LN, § 5.3, Example 5.3.6, 5.3.7.
1. (10 points)
Find the critical points (also called equilibrium solutions) of the
predator-prey system
x′ =−3 x + 2 x y
y′ = 7 y−8 x y
Critical Points:
Note: A point is an ordered pair (x,y), and your answer must be
a comma separated list of points.
Similar problem in LN, § 5.3, Example 5.3.6, 5.3.7.
2. (10 points)
Find the critical points (also called equilibrium solutions) of the
competing species system

x′ = 3 x−x2 −2 x y
y′ = 2 y−y2 −x y
Equilibrium Points:
Note: A point is an ordered pair (x,y), and your answer must be
a comma separated list of points.
See in LN, § 5.3, See Examples 5.3.8-5.3.10.
3. (10 points)
Part 1: Critical Points
Consider the two-dimensional autonomous system
x′ =−9 x + x3
y′ =−2 y
(a) The critical points of the system above have the form
x0 =
[
0
0
]
, x1 =
[
x1
0
]
, x2 =
[

x2
0
]
.
where x1 > 0 > x2. Find these components.
x1 = x2 =
Part 2: The Jacobian Matrix
1
Part 3: The Jacobian Matrix at X0
4. (10 points)
x′ =−16 x + x3
y′ = 3 y
x0 =
[
0
0
]

, x1 =
[
x1
0
]
, x2 =
[
x2
0
]
.
where x1 > 0 > x2. Find these components.
x1 = x2 =
5. (10 points)
x′ = 4 y−y3
y′ =−9 x−y2
x0 =

[
0
0
]
, x1 =
[
x1
y1
]
, x2 =
[
x2
y2
]
.
where y1 > 0 > y2. Find these components.
x1 = y1 =
x2 = y2 =
2

See in LN, § 5.4, See Examples 5.4.1, 5.4.2.
6. (10 points)
x′ = x (1− x− y)
y′ = y
(1
2
−
1
4
y−
3
4
x
)
x0 =
[
0
0
]
, x1 =
[

0
y1
]
, x2 =
[
x2
0
]
, x3 =
[
x3
y3
]
.
y1 = x2 =
x3 = y3 =
7. (10 points)

x′ = x (1− x− y)
y′ = y
(3
4
− y−
1
2
x
)
x0 =
[
0
0
]
, x1 =
[
0
y1
]
, x2 =
[
x2
0

]
, x3 =
[
x3
y3
]
.
y1 = x2 =
x3 = y3 =
3
8. (10 points)
x′ = x (
3
2
− x−

1
2
y)
y′ = y
(
2− y−
3
4
x
)
x0 =
[
0
0
]
, x1 =
[
0
y1
]
, x2 =
[
x2

0
]
, x3 =
[
x3
y3
]
.
y1 = x2 =
x3 = y3 =
9. (10 points)
x′ = x (
3
2
− x−
1
2

y)
y′ = y
(
2−
1
2
y−
3
2
x
)
x0 =
[
0
0
]
, x1 =
[
0
y1
]
, x2 =
[

x2
0
]
, x3 =
[
x3
y3
]
.
y1 = x2 =
x3 = y3 =
4
5

Miran Anwar, mth 235 ss21 1: Hw19-5.4-SDE-2x2NS-PP-NP.
Due: 04/26/2021 at 11:00pm
EDT.
See in LN, § 5.4, See Subsection 5.4.3.
1. (10 points)
x′ =−
1
2
x + xy
y′ =
3
2
y−
1
2
xy
x0 =
[
0
0
]
, x1 =
[

x1
y1
]
.
Find these components.
x1 =
y1 =
2. (10 points)
x′ =−
1
4
x +
1
2
xy
y′ = y−
1
2
xy

x0 =
[
0
0
]
, x1 =
[
x1
y1
]
.
x1 =
y1 =
1
3. (10 points)

x′ =−
1
4
x +
1
2
xy
y′ = y−
1
2
y2 −
1
2
xy
x0 =
[
0
0
]
, x1 =
[
0

y1
]
, x2 =
[
x2
y2
]
.
y1 =
x2 =
y2 =
4. (10 points)
x′ =−x + xy
y′ =
9
8
y− y2 −

1
2
xy
x0 =
[
0
0
]
, x1 =
[
0
y1
]
, x2 =
[
x2
y2
]
.
y1 =
x2 =

y2 =
2
5. (10 points)
Nonlinear Pendulum: No Friction Case
The equation of a pendulum having an attached mass m > 0, a
massless rod of length l > 0 is
mlθ′′ =−mg sin(θ),
where θ(t) is the angular position of the pendulum as function
of time, measured from the vertical down-
wards position, positive counter-clockwise, and g is the
acceleration of gravity. We consider the particular
case
g
l
= 1 and m = 1.
Part 1: First Order Reduction
(a) Find the first order reduction of the equation above where u
= θ and v = θ′.
u′ =
v′ =

Part 3: The Derivative Matrix
Part 4: The Derivative Matrix at Even Critical Points
Part 5: The Derivative Matrix at Odd Critical Points
6. (10 points)
Nonlinear Pendulum: Small Friction Case
massless rod of length l > 0, and swinging
in a medium with damping constant d > 0 is
mlθ′′ =−mg sin(θ)−dlθ′,
acceleration of gravity. We consider the following
particular case:
•
g
l
= 1 and m = 1.
• Small friction, 0 < d < 2.
3
= θ and v = θ′.

u′ =
v′ =
7. (10 points)
Nonlinear Pendulum: Large Friction Case
massless rod of length l > 0, and swinging
in a medium with damping constant d > 0 is
mlθ′′ =−mg sin(θ)−dlθ′,
acceleration of gravity. We consider the following
particular case:
•
g
l
= 1 and m = 1.
• Large friction, d > 2.
= θ and v = θ′.
u′ =

v′ =
4

Miran Anwar, mth 235 ss21 1 Hw17-5.2-SDE-2x2PP. Due 042620

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