Monday, July 25, 2011
Practice problems with complex numbers a + bi
a] (4 + i) + (7 - i)
b] (4 + i) - (7 - i)
c] (7 - i) - (4 + i)
d] (4 + i) * (7 - i)
e] (4 + i) / (7 - i)
f] (7 - i) / (4 + i)
Answers in the comments.
Wednesday, July 20, 2011
Two 4x4 sudoku problems
Problem 1.
__________
|b| ||c| |
| |a|| | |
| |c||_|b|
| | ||a| |
Problem 2.
__________
| |
||c| |
| |a|| | |
|b| ||_| |
| | ||d| |
Answers in the comments.
Link to degrees of freedom problems with contingency tables.
The post says "tomorrow's exam, but it's a post from an earlier session. Contingency problems will be on the homework, not the test on July 21, 2011.
Here is the link.
Tuesday, July 19, 2011
Monday, July 18, 2011
Tautology practice for Summer 2011.
When a logical statement is a tautology, the bitstring created when performing all the necessary operations is all 1s. In the following problems we will have two logical variables p and q, so p = 1100 and q = 1010. The AND operator is ^.
The OR operator is v.
The NOT operator is ~.
The IMPLIES operator should be an arrow pointing right, but since that single symbol is not an option in .html, I will use => instead. Remember that p => q can be changed to ~p v q.
Determine if each of these logical statements is a tautlogy or not.
1) p v (p => q)
2) p ^ (p => q)
3) (p v ~p) => q
4) (p ^ ~p) => q
Answers in the comments.
Thursday, July 14, 2011
Wednesday, July 13, 2011
Practice for the quiz on Thursday 7/14:
polygons and time
The sum of the interior angles of a polygon with n sides is (n – 2)180 °. Use this formula in the following problems.
a) What is the sum of the interior angles of a 10-sided polygon (decagon)?
b)If the interior angles of a polygon add up to 1260 °, how many sides does it have?
The angle measure of a regular n-sided polygon is (180 - 360/n) °.
c) What is the angle measure of a regular 10-sided polygon (decagon)?
d) If the angle measure of a regular polygon is 170°, how many sides does it have?
time problems.
e) How many minutes in 5 days?
f) 510 hours is __________ days and ________ hours.
g) 6,502 seconds is ________ hours, ________ minutes and __________ seconds.
h) 6,502 hours is ________ weeks, ________days and __________ hours.
i) After a flu outbreak, a hospital has decided quarantine will last 66 hours. If it starts at 5:00 pm on Tuesday, give the day and time when it will be over.
Day: ________ time: ___________ a.m or p.m? __________
Answers in the comments.
a) What is the sum of the interior angles of a 10-sided polygon (decagon)?
b)If the interior angles of a polygon add up to 1260 °, how many sides does it have?
The angle measure of a regular n-sided polygon is (180 - 360/n) °.
c) What is the angle measure of a regular 10-sided polygon (decagon)?
d) If the angle measure of a regular polygon is 170°, how many sides does it have?
time problems.
e) How many minutes in 5 days?
f) 510 hours is __________ days and ________ hours.
g) 6,502 seconds is ________ hours, ________ minutes and __________ seconds.
h) 6,502 hours is ________ weeks, ________days and __________ hours.
i) After a flu outbreak, a hospital has decided quarantine will last 66 hours. If it starts at 5:00 pm on Tuesday, give the day and time when it will be over.
Day: ________ time: ___________ a.m or p.m? __________
Answers in the comments.
Monday, July 11, 2011
Practice with angles of triangles and classification
Here are two earlier posts that have some practice problems with answers.
Look at Part B) in this post.
Look at problems 3) and 4) in this post.
Here are some more problems.
1) You are told you have a right triangle. If the second angle is given, either find the third angle or say it is not possible.
a) 84°
b) 31°
c) 102°
2) You are told you have an isosceles triangle. If one angle less than 180° is given, there is always an isosceles triangle that has that measure as the size of one angle. Sometimes, there is also a second possible triangle with that angle measure. Find both triangles if possible, or find the only possible isosceles triangle with that angle measure and write NO SECOND TRIANGLE POSSIBLE as the second answer.
a) 84°
b) 31°
c) 102°
Answers to these new problems in the comments.
Look at Part B) in this post.
Look at problems 3) and 4) in this post.
Here are some more problems.
1) You are told you have a right triangle. If the second angle is given, either find the third angle or say it is not possible.
a) 84°
b) 31°
c) 102°
2) You are told you have an isosceles triangle. If one angle less than 180° is given, there is always an isosceles triangle that has that measure as the size of one angle. Sometimes, there is also a second possible triangle with that angle measure. Find both triangles if possible, or find the only possible isosceles triangle with that angle measure and write NO SECOND TRIANGLE POSSIBLE as the second answer.
a) 84°
b) 31°
c) 102°
Answers to these new problems in the comments.
Wednesday, July 6, 2011
Links to the stories about the computer scientists.
Stories about John Von Neumann, Alan Turing, Grace Hopper and Donald Knuth. There is also a section about Edsger Dijkstra, but there won't be any questions about his life story.
Tuesday, July 5, 2011
The triangle inequality, Heron's formula and the test for acute, right or obtuse using side lengths.
You probably already learned in high school that the interior angles of a triangle always add up to 180°. Just as importantly, the Triangle Inequality tells us that any two side lengths of a triangle must add up to more than the third side length. Another way to say this is that every side length must be less than half the total perimeter.
The idea of half the perimeter or semi perimeter shows up in Heron's Formula, a way to find the area of a triangle if you are given the three side lengths.
We also have a test using side lengths to see if a triangle is acute, right or obtuse. If we have three side u, v and w and we declare that w is the long side, then we can use the sum of the squares of the short side to see how a triangle is classified.
u² + v² < w² : The triangle is obtuse
u² + v² = w² : The triangle is right
u² + v² > w² : The triangle is acute
Here are some practice problems for area and classification.
For each of these triples of numbers:
1) Determine if the triangle is equilateral, isosceles or scalene.
2) Determine if the triangle is acute, right or obtuse.
3) Find the area as a simplified square root
4) Find the area rounded to the nearest thousandth
a) 1, 1, 1
b) 1, 2, 2
c) 1, 3, 3
d) 1, 4, 4
e) 2, 2, 2
f) 2, 2, 3
g) 2, 3, 3
h) 2, 3, 4
i) 2, 4, 4
j) 3, 3, 3
k) 3, 3, 4
L) 3, 4, 4
m) 4, 4, 4
Answers in the comments.
The idea of half the perimeter or semi perimeter shows up in Heron's Formula, a way to find the area of a triangle if you are given the three side lengths.
We also have a test using side lengths to see if a triangle is acute, right or obtuse. If we have three side u, v and w and we declare that w is the long side, then we can use the sum of the squares of the short side to see how a triangle is classified.
u² + v² < w² : The triangle is obtuse
u² + v² = w² : The triangle is right
u² + v² > w² : The triangle is acute
Here are some practice problems for area and classification.
For each of these triples of numbers:
1) Determine if the triangle is equilateral, isosceles or scalene.
2) Determine if the triangle is acute, right or obtuse.
3) Find the area as a simplified square root
4) Find the area rounded to the nearest thousandth
a) 1, 1, 1
b) 1, 2, 2
c) 1, 3, 3
d) 1, 4, 4
e) 2, 2, 2
f) 2, 2, 3
g) 2, 3, 3
h) 2, 3, 4
i) 2, 4, 4
j) 3, 3, 3
k) 3, 3, 4
L) 3, 4, 4
m) 4, 4, 4
Answers in the comments.
Sunday, July 3, 2011
Practice with square roots for homework due July 6.
The editor for Blogger doesn't have a square root sign available, so I will use sqrt(2) to signify the square root of 2, for example.
Here are two sides of a right triangle. In these problems, c is always the hypotenuse and a and b are the short sides, also called the legs. Find the missing side using the Pythagorean Theorem, write it as a square root in simplified form and give the approximation to the nearest thousandth.
1) a = 7, b = 6, c = _________
2) b = 6, c = 7, a = __________
Here are some fractions with the square root in the denominator. Write them in standard form and simplify.
3) 20/sqrt(10)
4) 15/sqrt(6)
Find the distance between the two given points using the formula Distance = sqrt((x1 - x2)² + (y1 - y2)²). Write the number as a square root in simplified form and give the approximation to the nearest thousandth.
5) (3, 7) and (-2, 6)
6) (3, 1) and (-5, -9)
Answers in the comments.
Here are two sides of a right triangle. In these problems, c is always the hypotenuse and a and b are the short sides, also called the legs. Find the missing side using the Pythagorean Theorem, write it as a square root in simplified form and give the approximation to the nearest thousandth.
1) a = 7, b = 6, c = _________
2) b = 6, c = 7, a = __________
Here are some fractions with the square root in the denominator. Write them in standard form and simplify.
3) 20/sqrt(10)
4) 15/sqrt(6)
Find the distance between the two given points using the formula Distance = sqrt((x1 - x2)² + (y1 - y2)²). Write the number as a square root in simplified form and give the approximation to the nearest thousandth.
5) (3, 7) and (-2, 6)
6) (3, 1) and (-5, -9)
Answers in the comments.
Labels:
distance,
practice problems,
Pythagorean theorem
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