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17Calculus - Calculus 2 - Practice Exam 1 (Semester C)

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This is the first of four exams for semester C of calculus 2. The material covered in this exam includes area between curves, moment and center of mass of a planar lamina, work moving water out of a container, volume of revolution, arc length and linear motion.

Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

2.5 hours

Questions

6

Total Points

100

Tools

Calculator

no

Formula Sheet(s)

2 pages, 8.5x11 or A4

Other Tools

ruler for drawing graphs

Instructions:
- Show all your work.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).
- Give exact, simplified answers.

[15 points] A stone is thrown vertically into the air off the edge of a bridge 160ft above a river. If the initial velocity of the rock is 48ft/sec (upward) and the acceleration is a constant -32ft/sec2 (downward), find the following. a. The velocity v(t) after t seconds. b. The position s(t) of the stone above the river. c. The velocity at which the stone hits the water.

Problem Statement

[15 points] A stone is thrown vertically into the air off the edge of a bridge 160ft above a river. If the initial velocity of the rock is 48ft/sec (upward) and the acceleration is a constant -32ft/sec2 (downward), find the following. a. The velocity v(t) after t seconds. b. The position s(t) of the stone above the river. c. The velocity at which the stone hits the water.

Final Answer

a. \(v(t)=-32t+48\) ft/sec
b. \(s(t)=-16t^2+48t+160\) ft
c. \(v(5)=-112\) ft/sec

Problem Statement

[15 points] A stone is thrown vertically into the air off the edge of a bridge 160ft above a river. If the initial velocity of the rock is 48ft/sec (upward) and the acceleration is a constant -32ft/sec2 (downward), find the following. a. The velocity v(t) after t seconds. b. The position s(t) of the stone above the river. c. The velocity at which the stone hits the water.

Solution

a. We know the acceleration is constant at \(a=-32\) ft/sec2. So we can integrate to get \(v(t)\).
\(v(t)=\int{a~dt} = \int{-32~dt} = -32t+C\)
To determine \(C\), we use the information about the initial velocity, \(48\) ft/sec as follows.
\(v(0)=-32(0)+C = 48 \to C=48 \to \) \(v(t)=-32t+48\) ft/sec

b. Now that we have the equation for velocity, we can integrate to get the equation for position.
\(s(t)=\int{v(t)~dt} =\) \(\int{-32t+38~dt} = -16t^2+48t+C\)
Like we did for velocity, we use the initial position given in the problem statement to determine the value of C as follows.
\(s(0)=-16(0)^2+48(0)+C=160 \to \) \(s(t)=-16t^2+48t+160\) ft

c. To determine the velocity when the stone hits the water, we use the velocity equation \(v(t)\). However, the parameter for velocity is time t. So we need to determine WHEN the stone hits the water.
We know the position of the stone \(s(t)\) and the position when the stone hits the water is when \(s(t)=0\). So we will set the position function equal to zero, solve for t and use that value in the velocity function.
\(\begin{array}{rcl} -16t^2+48t+160 & = & 0 \\ t^2-3t-10 & = & 0 \\ (t-5)(t+2) & = & 0 \end{array}\)
So \(t=5\) or \(t=-2\). The value of t that makes sense is \(t=5\).
\(v(5)=-32(5)+48=-160+48=-112 \to \) \(v(5)=-112\) ft/sec

Final Answer

a. \(v(t)=-32t+48\) ft/sec
b. \(s(t)=-16t^2+48t+160\) ft
c. \(v(5)=-112\) ft/sec

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[10 points] Calculate the length of the arc \(y=2x^{3/2}\) from \(x=1/3\) to \(x=8/3\).

Problem Statement

[10 points] Calculate the length of the arc \(y=2x^{3/2}\) from \(x=1/3\) to \(x=8/3\).

Final Answer

\(26/3\)

Problem Statement

[10 points] Calculate the length of the arc \(y=2x^{3/2}\) from \(x=1/3\) to \(x=8/3\).

Solution

\(s=\int_{a}^{b}{ \sqrt{1+[f'(x)]^2} ~dx }\)
\(f(x)=y=2x^{3/2}\)
\(f'(x)=2(3/2)x^{1/2} = 3x^{1/2}\)
\(\displaystyle{ s = \int_{1/3}^{8/3}{\sqrt{1+9x} ~dx} }\)
Use integration by substitution.
\(u=1+9x \to du=9dx\)
Change the limits of integration.
\(x=1/3 \to u=4\)
\(x=8/3 \to u=25\)

\(\displaystyle{s=\int_{4}^{25}{u^{1/2}~\frac{du}{9}} }\)

\(\displaystyle{ \left. \frac{1}{9} \frac{u^{3/2}}{3/2} \right|_{4}^{25} } \)

\(\displaystyle{ \frac{2}{27}(25)^{3/2}-\frac{2}{27}(4)^{3/2} }\)

\(\displaystyle{ \frac{2}{27}[125-8] }\)

\(\displaystyle{ \frac{2}{27}(117) = \frac{26}{3}}\)

Final Answer

\(26/3\)

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[10 points] Consider the region bounded by \(y=\ln(x)\) and the line with slope \(m=(3+\ln(5))/5\) and y-intercept at \(x=-3\) above the x-axis as shown in the plot. Set up and but do not evaluate one integral that calculates the shaded area.

Problem Statement

[10 points] Consider the region bounded by \(y=\ln(x)\) and the line with slope \(m=(3+\ln(5))/5\) and y-intercept at \(x=-3\) above the x-axis as shown in the plot. Set up and but do not evaluate one integral that calculates the shaded area.

Final Answer

Area = \(\displaystyle{\int_{0}^{\ln(5)}{\frac{5(y+3)}{3+\ln(5)} - e^y ~ dy}}\)

Problem Statement

[10 points] Consider the region bounded by \(y=\ln(x)\) and the line with slope \(m=(3+\ln(5))/5\) and y-intercept at \(x=-3\) above the x-axis as shown in the plot. Set up and but do not evaluate one integral that calculates the shaded area.

Solution

The key here is to notice that the question asks for ONE integral. The only way to do that is to evaluate in the y-direction.
The equation of the straight line is \(y=[(3+\ln(5))/5]x-3\). All of this information is given in the problem statement. In order to find the point of intersection of the two curves, you can pull right off the graph that the point of intersection occurs at \(x=5\). So, \(y=[(2+\ln(5))/5](5)-3 = \ln(5) \to (5,\ln(5))\).
The limits of integration are from \(y=0\) to \(y=\ln(5)\).
Our general equation for area between curves is \(\int_{c}^{d}{f(y)-g(y)~dy}\). In our case, \(f(y)\) is the straight line and \(g(y)\) is \(y=\ln(x)\).
The integral is in terms of y, so we need to solve both equations for y.
\(y=\ln(x) \to e^y = e^{\ln(x)} \to e^y = x\)
\(y=[(3+\ln(5))/5]x-3 \to y+3=[(3+\ln(5))/5]x \to x=(y+3)[5/(3+\ln(5))]\)
So the integral is
\(\displaystyle{ \int_{0}^{\ln(5)}{\frac{5(y+3)}{3+\ln(5)} - e^y ~ dy}}\)

Final Answer

Area = \(\displaystyle{\int_{0}^{\ln(5)}{\frac{5(y+3)}{3+\ln(5)} - e^y ~ dy}}\)

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[20 points] Consider the region R bounded by \(y=x^3\) and \(y=\sqrt{x}\).
a. Calculate the area of the region R. Sketch a plot and shade the region. b. Calculate the moments Mx and My. c. Determine the center of mass.

Problem Statement

[20 points] Consider the region R bounded by \(y=x^3\) and \(y=\sqrt{x}\).
a. Calculate the area of the region R. Sketch a plot and shade the region. b. Calculate the moments Mx and My. c. Determine the center of mass.

Final Answer

a. Area = \(5/12\)
b. \(M_x = 5\rho/28\), \(M_y = \rho/5\)
c. \((\bar{x}, \bar{y} ) = (12/25, 3/7)\)

Problem Statement

[20 points] Consider the region R bounded by \(y=x^3\) and \(y=\sqrt{x}\).
a. Calculate the area of the region R. Sketch a plot and shade the region. b. Calculate the moments Mx and My. c. Determine the center of mass.

Solution

a. The plot is shown here. The green curve is \(y=x^3\) and the red curve is \(y=\sqrt{x}\). We choose to evaluate in the x-direction. The point of intersection is \((1,1)\) so our integral is
Area = \(\displaystyle{ \int_{0}^{1}{\sqrt{x}-x^3~dx} = }\) \(\displaystyle{ \left[ \frac{x^{3/2}}{3/2} - \frac{x^4}{4} \right]_{0}^{1} = }\) \(\displaystyle{ \frac{2}{3} - \frac{1}{4} = \frac{5}{12} }\)

Area = \(5/12\)

b. The general equations for moments of a planar lamina are
\(\displaystyle{ M_x = \rho \int_{a}^{b}{(1/2)[f(x)+g(x)][f(x)-g(x)]~dx} }\)
\(\displaystyle{ M_y = \rho \int_{a}^{b}{x[f(x)-g(x)]~dx} }\)
where \(f(x) \geq g(x)\)
Since the density is not given, we can assume a constant density of \(\rho\).

Calculate \(M_x\)

\(\displaystyle{ \int_{a}^{b}{(\rho/2)[f(x)+g(x)][f(x)-g(x)]~dx} }\)

\( \rho/2 \int_{a}^{b}{(\sqrt{x}+x^3)(\sqrt{x}-x^3)~dx} \)

\( \rho/2 \int_{0}^{1}{x-x^6~dx} \)

\( \displaystyle{ \frac{\rho}{2}\left[ \frac{x^2}{2}-\frac{x^7}{7} \right]_{0}^{1} } \)

\( \displaystyle{ \frac{\rho}{2} \left[ \frac{1}{2} - \frac{1}{7} \right] } \)

\( \displaystyle{ \frac{\rho}{2} \left[ \frac{7}{14} - \frac{2}{14} \right] } \)

\( \displaystyle{ \frac{5\rho}{28} } \)

\(M_x = 5\rho/28\)

\(\begin{array}{rcl} M_y & = & \rho\int_{0}^{1}{x(\sqrt{x}-x^3)~dx} \\ & = & \rho \int_{0}^{1}{x^{3/2}-x^4~dx} \\ & = & \displaystyle{\rho \left[\frac{x^{5/2}}{5/2} - \frac{x^5}{5} \right]_{0}^{1} } \\ & = & \displaystyle{\rho \left[ \frac{2}{5}-\frac{1}{5} \right] } \\ & = & \displaystyle{ \frac{\rho}{5} } \end{array}\)

\(M_y = \rho/5\)

c. To calculate the center of mass we use the moments \(M_x\) and \(M_y\) and we need the total mass. Since the density is constant, the mass is just the density times area, i.e. \(m=5\rho/12\).
\(\displaystyle{ \bar{x} = \frac{M_y}{m} = \frac{\rho}{5} \cdot \frac{12}{5\rho} = \frac{12}{25}}\)

\(\displaystyle{ \bar{y} = \frac{M_x}{m} = \frac{5\rho}{28} \cdot \frac{12}{5\rho} = \frac{3}{7}}\)

\((\bar{x},\bar{y}) = (12/25, 3/7)\)

Final Answer

a. Area = \(5/12\)
b. \(M_x = 5\rho/28\), \(M_y = \rho/5\)
c. \((\bar{x}, \bar{y} ) = (12/25, 3/7)\)

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[20 points] For the region R bounded by the graphs of \(y=2x\), the x-axis and \(x=1\), consider the solid generated by revolving the region about the line \(y=3\). Sketch a different plot of region R for each method, clearly showing the region by shading, the axes of revolution, the representative rectangle, R, r, p and h.
a. Set up the integral to calculate the volume using the disc-washer method. b. Set up the integral to calculate the volume using the shell-cylinder method. c. Calculate the volume of the solid by evaluating one of the integrals you set up in parts a and b.

Problem Statement

[20 points] For the region R bounded by the graphs of \(y=2x\), the x-axis and \(x=1\), consider the solid generated by revolving the region about the line \(y=3\). Sketch a different plot of region R for each method, clearly showing the region by shading, the axes of revolution, the representative rectangle, R, r, p and h.
a. Set up the integral to calculate the volume using the disc-washer method. b. Set up the integral to calculate the volume using the shell-cylinder method. c. Calculate the volume of the solid by evaluating one of the integrals you set up in parts a and b.

Final Answer

a. \(V=\pi\int_{0}^{1}{3^2-(3-2x)^2~dx}\)
b. \(V=2\pi\int_{0}^{2}{(3-y)(1-y/2)~dy}\)
c. \(V = 14\pi/3\)

Problem Statement

[20 points] For the region R bounded by the graphs of \(y=2x\), the x-axis and \(x=1\), consider the solid generated by revolving the region about the line \(y=3\). Sketch a different plot of region R for each method, clearly showing the region by shading, the axes of revolution, the representative rectangle, R, r, p and h.
a. Set up the integral to calculate the volume using the disc-washer method. b. Set up the integral to calculate the volume using the shell-cylinder method. c. Calculate the volume of the solid by evaluating one of the integrals you set up in parts a and b.

Solution

a. The general equation is \(V=\pi\int_{a}^{b}{R^2-r^2~dx}\).
\(y+r=3 \to r=3-y \to r=3-2x \) and \(R=3\)

\(V=\pi\int_{0}^{1}{3^2-(3-2x)^2~dx}\)

b. The general equation is \(V=2\pi\int_{c}^{d}{ph~dy}\)
\( y+p=3 \to p=3-y \)
\(x+h=1 \to h=1-x = 1-y/2 \)

\(V=2\pi\int_{0}^{2}{(3-y)(1-y/2)~dy}\)

c. The integral from part a is
\(\begin{array}{rcl} V & = & \pi\int_{0}^{1}{3^2-(3-2x)^2~dx} \\ & = & \pi \int_{0}^{1}{9 - (9-12x+4x^2)~dx} \\ & = & \pi \int_{0}^{1}{12x-4x^2 ~dx} \\ & = & \displaystyle{ \pi \left[ \frac{12x^2}{2} - \frac{4x^3}{3} \right]_{0}^{1} } \\ & = & \displaystyle{ \pi \left[ 6 - \frac{4}{3} \right] } \\ & = & \displaystyle{ \frac{14\pi}{3} } \end{array}\)
The integral from part b is
\(\begin{array}{rcl} V & = & 2\pi\int_{0}^{2}{(3-y)(1-y/2)~dy} \\ & = & \displaystyle{ 2\pi \int_{0}^{2}{3-\frac{3y}{2}-y+\frac{y^2}{2}~dy} } \\ & = & \displaystyle{ 2\pi \left[ 3y - \frac{5}{2}\frac{y^2}{2} + \frac{y^3}{6} \right]_{0}^{2}} \\ & = & \displaystyle{ 2\pi \left[ 6 - \frac{5}{4}4 +\frac{8}{6} \right] } \\ & = & \displaystyle{ 2\pi\left[ 1 +\frac{4}{3}\right] } \\ & = & \displaystyle{ 2\pi\frac{7}{3} = \frac{14\pi}{3} } \end{array}\)

Final Answer

a. \(V=\pi\int_{0}^{1}{3^2-(3-2x)^2~dx}\)
b. \(V=2\pi\int_{0}^{2}{(3-y)(1-y/2)~dy}\)
c. \(V = 14\pi/3\)

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[25 points] A swimming pool is 20 m long and 10 m wide, with a bottom that slopes uniformly from a depth of 1 m at one end to a depth of 2 m at the other end. Assuming the pool is full, how much work is required to pump the water to a level 1 m above the top of the pool?

Problem Statement

[25 points] A swimming pool is 20 m long and 10 m wide, with a bottom that slopes uniformly from a depth of 1 m at one end to a depth of 2 m at the other end. Assuming the pool is full, how much work is required to pump the water to a level 1 m above the top of the pool?

Final Answer

Total Work \(W = 1600\rho g/3 ~\text{J} \)

Problem Statement

[25 points] A swimming pool is 20 m long and 10 m wide, with a bottom that slopes uniformly from a depth of 1 m at one end to a depth of 2 m at the other end. Assuming the pool is full, how much work is required to pump the water to a level 1 m above the top of the pool?

Solution

For this problem, we need to set up a plot of the side of the pool and establish our coordinate system. We have chosen to flip the side, so that the origin is at bottom of the deep end of the pool, as shown here. The sides of the pool are outlined in blue and the red line is at \(y=3\), the level to where the water is to be lifted.
The general equation for work that we need is \(W=\int_{a}^{b}{\rho g A(y)D(y)~dy}\).
From this plot, you should see that we need two integrals to calculate work, one from 0 to 1, the other from 1 to 2. This is true because the profile, and thus the area, is different in those two sections. But first let's discuss the equation for the distance moved, \(D(y)\).
If we think about a slice of water, \(dy\), at some point within the pool, the slice will be at position y on our axes. We would then need to move that slice of water up to 3. This would mean the slice is moved \(3-y\). This is our equation for \(D(y)=3-y\).
Now we need to describe the area of one slice of water. The area is different depending on the position y. Let's start with a slice of water when \(1 \leq y \leq 2\). Since the depth of the pool width of the pool is 10m and the length is always 20m in this range of y, the area is just a rectangle and \(A(y)=10(20) = 200\).
So we have one of our integrals, i.e.
\(\displaystyle{\int_{1}^{2}{\rho g (200)(3-y)~dy}}\)
Let's go ahead and evaluate this integral.
\(\begin{array}{rcl} W_{12} & = & \displaystyle{\int_{1}^{2}{\rho g (200)(3-y)~dy}} \\ & = & \displaystyle{200\rho g\int_{1}^{2}{(3-y)~dy}} \\ & = & \displaystyle{ 200\rho g \left[ 3y - \frac{y^2}{2} \right]_{1}^{2}} \\ & = & 200\rho g \left[ (6-2) - (3-1/2) \right] \\ & = & 300\rho g \end{array}\)
Okay, so that is the work required to move the top 1m of water out of the pool, up to one meter above the pool. Now let's determine the integral for moving the bottom 1m of water.
We already know the equation for \(D(y)=3-y\). We just need the equation for \(A(y)\). Notice that at each slice, the area is rectangular with width of 10m. But the length is different depending on the height y.
The equation of the line from \((0,0)\) to \((20,1)\) is \(y=x/20\) so \(x=20y\). This x is the length. So our area is \(A(y)=(10)(20y)=200y\). Now we are ready to set up our integral and evaluate.
\(\begin{array}{rcl} W_{01} & = & \displaystyle{\int_{0}^{1}{\rho g (200y)(3-y)~dy}} \\ & = & \displaystyle{ 200\rho g\int_{0}^{1}{3y-y^2~dy}} \\ & = & \displaystyle{ 200\rho g \left[ \frac{3y^2}{2} - \frac{y^3}{3} \right]_{0}^{1}} \\ & = & \displaystyle{ 200\rho g \left[ \frac{3}{2} - \frac{1}{3} \right] } \\ & = & \displaystyle{ 200 \rho g \left[ \frac{9}{6} - \frac{2}{6} \right] } \\ & = & \displaystyle{ \frac{700\rho g}{3} } \end{array}\)
Now we combine the results of the two integrals to get the total work.
\(\displaystyle{ W = W_{01} + W_{12} = \frac{700\rho g}{3} + 300\rho g = \frac{1600\rho g}{3} }\) with units of Joules.

Final Answer

Total Work \(W = 1600\rho g/3 ~\text{J} \)

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You CAN Ace Calculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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