
Michael McCabe
Assistant Professor
Mathematics
mccabem85@cod.edu(630)942-2152
BIC 3436B
Professor room number
Michael McCabe
Assistant Professor
Mathematics
mccabem85@cod.edu(630)942-2152
BIC 3436B
Professor room numberAlways available by appointment, don't hesitate to request a meeting.
(As of 5/23/2021)
Students develop the foundational mathematical skills necessary for general education mathematics courses (Math 1218 and Math 1220). Content features collaborative project-based and technology-enabled group work including modeling, problem solving, critical thinking, data analysis, algebra fundamentals, and both verbal and written communication of mathematical ideas.
(As of 5/23/2021)
At least two of the following:
For Mr. McCabe's session
(As of 5/23/2021)
Students develop the foundational mathematical skills necessary for general education mathematics courses (Math 1218 and Math 1220). Content features collaborative project-based and technology-enabled group work including modeling, problem solving, critical thinking, data analysis, algebra fundamentals, and both verbal and written communication of mathematical ideas.
(As of 5/23/2021)
At least two of the following:
For Mr. McCabe's session
The bearing from City A to City B is N$38^{\circ}$E. The bearing from City B to City C is S$52^{\circ}$E. An automobile driven at 65 miles per hour takes 1.4 hours to drive from City A to City B and takes 1.8 hours to drive from City B to City C. Find the distance from City A to City C.
Remember distance equals the product of speed and time. $$d=rt$$
Therefore, the distance from City A to City B is: $$d_1=m(A,B)=(65)(1.4)=91$$ and the distance from City B to City C is: $$d_2=m(B,C)=(65)(1.8)=117$$
In the drawing above the angle for B can be found by adding the two angles: $$38+52=90$$ This means triangle(A,B,C) is a right triangle. The legs of the right triangle is $d_1$ and $d_2$. The hypotenuse,
$r$, is the distance from City A to City C. Therefore, we want to solve for $r$:
We will use the positive distance.
Since the answer should be to the nearest mile the answer is: "The distance from City
A to City C is approximately 148 miles."
d14:65*1.4;
d24:65*1.8;
38+52;
float(solve((d14)^2+(d24)^2=r^2,r));
rat: replaced 21970.0 by 21970/1 = 21970.0
Two ships leave a port at the same time. The first ship sails on a bearing of $40^{\circ}$ at 12 knots (nautical miles per hour) and the second on a bearing of $130^{\circ}$ at 14 knots. How far apart are they after 1.5 hours?
Remember distance equals the product of speed and time. $$d=rt$$ Therefore, the distance for the first boat is $$d_1=(12)(1.5)=18$$ and the distance for the second boad is $$d_2=(14)(1.5)=21$$ These are both in units of nautical miles.
Two rays can be created: ray(Port,Ship 1) and ray(Port, Ship 2). The angles between the two rays is: $$130-40=90$$
This means triangle(Port, Ship 1, Ship 2) is a right triangle. The distance for each leg is $d_1$ and $d_2$. The hypotenuse,
$r$, is the distance away from each ship. Therefore, we want to solve for $r$:
We will use the positive distance.
Since the answer should be to the nearest nautical mile the answer is: "After 1.5
hours, the ships are 28 nautical miles apart."
d15:12*1.5;
d25:14*1.5;
130-40;
float(solve((d15)^2+(d25)^2=r^2,r));
rat: replaced 765.0 by 765/1 = 765.0
(As of 5/23/2021)
Students will be introduced to basic concepts of differential and integral calculus. This course is intended for students planning to major in business, or the behavioral, social, or biological sciences.
(As of 5/23/2021)
Functions
Differential calculus
Integral calculus
For Mr. McCabe's session