Here's a problem that can easily be solved incorrectly if the order of operations is not used properly.
6 divided by 2(1+2)
The
answer is 9, some might think it's 1, but really it's 9. Do the work
in parenthesis first, so you have 6 divided by 2(3).. Now in the order
of operations multiplication comes before division.. IF you do it that
way you would get 1 because you'd multiply 2 and 3 to get 6, then 6
divided by 6 is 1. But the order of operations also states
multiplication and division is done from left to right, so the division
is first... IF the problem was 6/[2(1+2)] THEN it would be 6/[2(3)] =
6/6 = 1. But the problem doesn't state this.
Sunday, December 30, 2012
Thursday, December 27, 2012
Tuesday, December 25, 2012
We learned how to find a zero of a function using synthetic division. Now we will learn how to determine the number and nature of zeroes, (also known as roots) that a polynomial function has. To determine the number of positive and negative real roots, we use a technique founded by Rene Descartes, thus named Decartes' Rule of Signs.
To find the number of positive real roots, start with the sign of the coefficient of the term with the highest power of the variable. Count the sign changes as you proceed through the polynomial. The number of sign changes is the number of positive real roots or less than it by a multiple of two. For example, if there are 2 sign changes there are either 2 or (2 – 2) =
0 positive real roots.
Example: Find the number of positive real roots of f(x) = -4x^5 – 11x^4 + 2x^3 + 9x^2 - x + 3.
Starting with -4x^5 , there is a sign change at 2x^3, another at -x and a third at 3. Therefore there are 3 sign changes and either 3 or (3 – 2) = 1 positive real roots.
To find the number of negative real roots of f(x), substitute -x into f(x) to get f(-x). Then proceed in the same manner as you would when finding the number of positive real roots. In the above example,
f(-x) = -4(-x)^5 – 11(-x)^4 + 2(-x)^3 + 9(-x)^2 - (-x) + 3
f(-x) = 4x^5 - 11x^4 – 2x^3 + 9x^2 + x + 3
Starting with 4x^5 , there is a sign change at 11x^4 and another at 9x^2 . Therefore, there are 2 or (2 – 2) = 0 negative real roots.
• Note that the number of roots in a polynomial function equals the highest power. In the above example, there are 5 roots. The sum of the positive and negative roots will not always equal the highest power. In such cases, those polynomial functions have imaginary roots, which we will deal with in a later chapter.
Example: Find the number of positive and negative real roots of f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5.
For positive real roots f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -9x^3, another at 5x and a third at -5. Therefore there are 3 or (3 – 2) = 1 positive real roots.
For negative real roots f(-x) = 6(-x)^4 – 9(-x)^3 – 2(-x)^2 + 5(-x) – 5.
f(-x) = 6x^4 + 9x^3 – 2x^2 – 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -2x^2 and no other changes. Therefore there is 1 negative real root.
To find the number of positive real roots, start with the sign of the coefficient of the term with the highest power of the variable. Count the sign changes as you proceed through the polynomial. The number of sign changes is the number of positive real roots or less than it by a multiple of two. For example, if there are 2 sign changes there are either 2 or (2 – 2) =
0 positive real roots.
Example: Find the number of positive real roots of f(x) = -4x^5 – 11x^4 + 2x^3 + 9x^2 - x + 3.
Starting with -4x^5 , there is a sign change at 2x^3, another at -x and a third at 3. Therefore there are 3 sign changes and either 3 or (3 – 2) = 1 positive real roots.
To find the number of negative real roots of f(x), substitute -x into f(x) to get f(-x). Then proceed in the same manner as you would when finding the number of positive real roots. In the above example,
f(-x) = -4(-x)^5 – 11(-x)^4 + 2(-x)^3 + 9(-x)^2 - (-x) + 3
f(-x) = 4x^5 - 11x^4 – 2x^3 + 9x^2 + x + 3
Starting with 4x^5 , there is a sign change at 11x^4 and another at 9x^2 . Therefore, there are 2 or (2 – 2) = 0 negative real roots.
• Note that the number of roots in a polynomial function equals the highest power. In the above example, there are 5 roots. The sum of the positive and negative roots will not always equal the highest power. In such cases, those polynomial functions have imaginary roots, which we will deal with in a later chapter.
Example: Find the number of positive and negative real roots of f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5.
For positive real roots f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -9x^3, another at 5x and a third at -5. Therefore there are 3 or (3 – 2) = 1 positive real roots.
For negative real roots f(-x) = 6(-x)^4 – 9(-x)^3 – 2(-x)^2 + 5(-x) – 5.
f(-x) = 6x^4 + 9x^3 – 2x^2 – 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -2x^2 and no other changes. Therefore there is 1 negative real root.
Saturday, December 22, 2012
Many problems in algebra are in the form of words. In order to solve
them, we must know how to translate the words into an equation. The key
steps in solving a problem are analyzing the problem, forming the
equation, solving the equation and checking the result.
Words and phrases can represent different mathematical operations. It's important to be able to translate words and phrases into equations when solving word problems. Here are some common words and phrases and their associated mathematical operation.
John and Steve have been collecting baseball cards since 2003. They now have accumulated a total of 40,000 cards. If John has 7,500 more cards than Steve, how many cards do each of them have?
First, assign a variable for the number of cards that either John or Steve has.
Let x = number of cards that Steve has. Since John has 7,500 more cards than Steve, he has 7,500 + x cards. Therefore, the number of cards that Steve has + Number of cards that John has = total number of cards. The equation that must be solved then is x + 7,500 + x = 40,000.
Here's another example which can be applied to real life. Suppose you wish to decide between two long distance phone plans. The first plan is $0.20 for the first minute, $0.07 for each additional minute. The second plan is $0.15 for the first minute and $0.10 for each additional minute. How many minutes must you talk for the first plan to be cheaper than the second plan?
For the first plan, let m = total minutes. Therefore, the cost for first minute plus cost for each additional minute is 0.20 + 0.07(m - 1).
For the second plan, the cost for first minute plus cost for each additional minute is 0.15 + 0.10(m - 1). We set equations equal to each other and solve for m to see how many minutes will the cost be equal.
Suppose a school wants to have an enclosed rectangular playground and has enough money in the budget to afford 550 feet of fencing. If the length is to be 125 feet longer than the width, what are the dimensions of the fence that encloses the playground?
Let W = width of the fence. The length is W + 125, therefore 2 times the length + 2 times the width = 550. The equation we must solve is 2(W + 125) + 2W = 550. Once we get a value for W, substitute back into the equation to solve for the length.
The important thing to remember is to analyze the problem, assign variable, set up equations which make sense in the context of the problem and check solutions. Be sure to try and write the equation as you are reading the problem.
Words and phrases can represent different mathematical operations. It's important to be able to translate words and phrases into equations when solving word problems. Here are some common words and phrases and their associated mathematical operation.
- sum of, added to, increased, plus: addition
- minus, decreased, less than, reduced by: subtraction
- twice, product, multiplied by, times, of: multiplication
- quotient, divided by, ratio, into: division
John and Steve have been collecting baseball cards since 2003. They now have accumulated a total of 40,000 cards. If John has 7,500 more cards than Steve, how many cards do each of them have?
First, assign a variable for the number of cards that either John or Steve has.
Let x = number of cards that Steve has. Since John has 7,500 more cards than Steve, he has 7,500 + x cards. Therefore, the number of cards that Steve has + Number of cards that John has = total number of cards. The equation that must be solved then is x + 7,500 + x = 40,000.
Here's another example which can be applied to real life. Suppose you wish to decide between two long distance phone plans. The first plan is $0.20 for the first minute, $0.07 for each additional minute. The second plan is $0.15 for the first minute and $0.10 for each additional minute. How many minutes must you talk for the first plan to be cheaper than the second plan?
For the first plan, let m = total minutes. Therefore, the cost for first minute plus cost for each additional minute is 0.20 + 0.07(m - 1).
For the second plan, the cost for first minute plus cost for each additional minute is 0.15 + 0.10(m - 1). We set equations equal to each other and solve for m to see how many minutes will the cost be equal.
Suppose a school wants to have an enclosed rectangular playground and has enough money in the budget to afford 550 feet of fencing. If the length is to be 125 feet longer than the width, what are the dimensions of the fence that encloses the playground?
Let W = width of the fence. The length is W + 125, therefore 2 times the length + 2 times the width = 550. The equation we must solve is 2(W + 125) + 2W = 550. Once we get a value for W, substitute back into the equation to solve for the length.
The important thing to remember is to analyze the problem, assign variable, set up equations which make sense in the context of the problem and check solutions. Be sure to try and write the equation as you are reading the problem.
Wednesday, December 19, 2012
Here's a few examples of some word problems.
Michael has a budget of $1200 for the golf season. The course he plays costs $40 for a round of golf. He also wants to takes lessons which costs $70 each. He wants to play twice as many rounds as lessons taken.
How many rounds of golf can Michael play for the season? How many lessons can he have?
Solution:
Let x = Number of rounds played
x/2 = Number of lessons
Total cost of rounds played plus total number of lessons must be less than or equal to $1,200.
40x = Total cost of rounds played
70(x/2) = 35x = Total cost of lessons taken
40x + 35x ≤ 1,200
75x ≤ 1,200
x ≤ 16. Therefore, Michael can play 16 times and take 8 lessons for the season
The temperatures on a day in January satisfy the inequality |t - 30| ≤ 12. The temperatures on a day in August satisfy the equation |t - 70| ≤ 16. What are the temperature ranges for January and August and which month has the widest range of temperatures?
Solution:
Solve each inequality.
|t - 30| ≤ 12
Recall for absolute value there is a positive case and a negative case.
The positive case is t - 30 ≤ 12 and the negative case is -12 ≤ t – 30.
t - 30 ≤ 12, -12 ≤ t - 30
t ≤ 42 18 ≤ t
Therefore, the temperature ranges from 18 to 42 in January.
|t - 70| ≤ 16
t - 70 ≤ 16, -16 ≤ t - 70
t ≤ 86 54 ≤ t
Therefore, the temperature ranges from 54 to 86 in August.
The range for January temperatures is 24. (42 - 18 = 24)
The range for August temperatures is 32. (86 - 54 = 32)
Michael has a budget of $1200 for the golf season. The course he plays costs $40 for a round of golf. He also wants to takes lessons which costs $70 each. He wants to play twice as many rounds as lessons taken.
How many rounds of golf can Michael play for the season? How many lessons can he have?
Solution:
Let x = Number of rounds played
x/2 = Number of lessons
Total cost of rounds played plus total number of lessons must be less than or equal to $1,200.
40x = Total cost of rounds played
70(x/2) = 35x = Total cost of lessons taken
40x + 35x ≤ 1,200
75x ≤ 1,200
x ≤ 16. Therefore, Michael can play 16 times and take 8 lessons for the season
The temperatures on a day in January satisfy the inequality |t - 30| ≤ 12. The temperatures on a day in August satisfy the equation |t - 70| ≤ 16. What are the temperature ranges for January and August and which month has the widest range of temperatures?
Solution:
Solve each inequality.
|t - 30| ≤ 12
Recall for absolute value there is a positive case and a negative case.
The positive case is t - 30 ≤ 12 and the negative case is -12 ≤ t – 30.
t - 30 ≤ 12, -12 ≤ t - 30
t ≤ 42 18 ≤ t
Therefore, the temperature ranges from 18 to 42 in January.
|t - 70| ≤ 16
t - 70 ≤ 16, -16 ≤ t - 70
t ≤ 86 54 ≤ t
Therefore, the temperature ranges from 54 to 86 in August.
The range for January temperatures is 24. (42 - 18 = 24)
The range for August temperatures is 32. (86 - 54 = 32)
Sunday, December 16, 2012
Here's some tips for kids dealing with multiplication tables.
There is an important rule shown
in the multiplication tables. Any number multiplied by 0 is equal to
0, and is not included in the multiplication table. Any number
multiplied by 1 is itself.
Here
are the multiplication tables for 1 through 12
1
x 1 = 1 2 x 1 = 2 3 x 1 = 3 4 x 1 = 4
5 x 1 = 5 6 x 1 = 6 7 x 1 = 7
1
x 2 = 2 2 x 2 = 4 3 x 2 = 6 4 x 2 = 8
5 x 2 = 10 6 x 2 = 12 7 x 2 = 14
1
x 3 = 3 2 x 3 = 6 3 x 3 = 9 4 x 3 = 12
5 x 3 = 15 6 x 3 = 18 7 x 3 = 21
1
x 4 = 4 2 x 4 = 8 3 x 4 = 12 4 x 4 = 16
5 x 4 = 20 6 x 4 = 24 7 x 4 = 28
1
x 5 = 5 2 x 5 = 10 3 x 5 = 15 4 x 5 = 20
5 x 5 = 25 6 x 5 = 30 7 x 5 = 35
1
x 6 = 6 2 x 6 = 12 3 x 6 = 18 4 x 6 = 24
5 x 6 = 30 6 x 6 = 36 7 x 6 = 42
1
x 7 = 7 2 x 7 = 14 3 x 7 = 21 4 x 7 = 28
5 x 7 = 35 6 x 7 = 42 7 x 7 = 49
1
x 8 = 8 2 x 8 = 16 3 x 8 = 24 4 x 8 = 32
5 x 8 = 40 6 x 8 = 48 7 x 8 = 56
1
x 9 = 9 2 x 9 = 18 3 x 9 = 27 4 x 9 = 36
5 x 9 = 45 6 x 9 = 54 7 x 9 = 63
1
x 10 = 10 2 x 10 = 20 3 x 10 = 30 4 x 10 = 40 5 x
10 = 50 6 x 10 = 60 7 x 10 = 70
1
x 11 = 11 2 x 11 = 22 3 x 11 = 33 4 x 11 = 44 5
x 11 = 55 6 x 11 = 66 7 x 11 = 77
1
x 12 = 12 2 x 12 = 24 3 x 12 = 36 4 x 12 = 48 5 x
12 = 60 6 x 12 = 72 7 x 12 = 84
8
x 1 = 8 9 x 1 = 9 10 x 1 = 10 11 x 1 = 11
12 x 1 = 12
8
x 2 = 16 9 x 2 = 18 10 x 2 = 20 11 x 2 = 22
12 x 2 = 24
8
x 3 = 24 9 x 3 = 27 10 x 3 = 30 11 x 3 = 33
12 x 3 = 36
8
x 4 = 32 9 x 4 = 36 10 x 4 = 40 11 x 4 = 44
12 x 4 = 48
8
x 5 = 40 9 x 5 = 45 10 x 5 = 50 11 x 5 = 55
12 x 5 = 60
8
x 6 = 48 9 x 6 = 54 10 x 6 = 60 11 x 6 = 66
12 x 6 = 72
8
x 7 = 56 9 x 7 = 63 10 x 7 = 70 11 x 7 = 77
12 x 7 = 84
8
x 8 = 64 9 x 8 = 72 10 x 8 = 80 11 x 8 = 88
12 x 8 = 96
8
x 9 = 72 9 x 9 = 81 10 x 9 = 90 11 x 9 = 99
12 x 9 = 108
8
x 10 = 80 9 x 10 = 90 10 x 10 = 100 11 x 10 = 110
12 x 10 = 120
8
x 11 = 88 9 x 11 = 99 10 x 11 = 110 11 x 11 = 121
12 x 11 = 132
8
x 12 = 96 9 x 12 = 108 10 x 12 = 120 11 x 12 = 132
12 x 12 = 144
There are a few things to notice
when looking at the multiplication tables. Staring with the 1's,
each answer goes up by 1. We say that the multiples of 1 are
1,2,3,4,5,6, and so on.
For the 2's tables, each answer
goes up by 2. The multiples of 2 are 2,4,6,8,10,12, and so on.
For the 3's tables, each answer
goes up by 3. The multiples of 3 are 3,6,9,12,15,18, and so on.
The same pattern is true for all
the multiplication tables.
The answer to a multiplication
problem is also called the product.
Notice that any number multiplied
by 10 ends in 0.
It may seem hard to learn the
whole table, but notice that 1 x 2 is the same as 2 x 1, 3 x 2 is
the same as 2 x 3, and so on
Friday, December 14, 2012
When dealing with confidence intervals for proportions, the formula is
p^ +/- Zcritical(standard deviation)
Where p^ = r/n, r is the number of observations and n is the sample size
For a 95% confidence interval, Zcritical = 1.96
The standard deviation is square root[(p^)(1 - p^)/n]
The margin of error, denoted at E = Zcritical(standard deviation)
p^ +/- Zcritical(standard deviation)
Where p^ = r/n, r is the number of observations and n is the sample size
For a 95% confidence interval, Zcritical = 1.96
The standard deviation is square root[(p^)(1 - p^)/n]
The margin of error, denoted at E = Zcritical(standard deviation)
Wednesday, December 12, 2012
To find inflection points and concavity, take the second derivative and set equal to 0. Solve for x, then test a value on the left of the value for x and one on the right. If the second derivative of this value is less than zero, it's concave down on that interval, if the second derivative of this value is greater than zero, then it's concave up.
For example:
f(x) = 3x^3 + 2x^2 + 5x + 6
first derivative : f ' (x) = 9x^2 + 4x + 5
second derivative : f " (x) = 18x + 4
set the second derivative equal to 0 and solve for x
18x + 4 = 0
18x = -4
x = -4/18 = -2/9
Test a value to the left of -2/9, I choose -1
f " (-1) = -14
Test value to the right of -2/9, I choose 0
f " (0) = 4
Since f " (-1) is negative, the concavity is downward from negative infinity to -2/9
Since f " (0) is positive, the concavity is upward from -2/9 to infinity.
For example:
f(x) = 3x^3 + 2x^2 + 5x + 6
first derivative : f ' (x) = 9x^2 + 4x + 5
second derivative : f " (x) = 18x + 4
set the second derivative equal to 0 and solve for x
18x + 4 = 0
18x = -4
x = -4/18 = -2/9
Test a value to the left of -2/9, I choose -1
f " (-1) = -14
Test value to the right of -2/9, I choose 0
f " (0) = 4
Since f " (-1) is negative, the concavity is downward from negative infinity to -2/9
Since f " (0) is positive, the concavity is upward from -2/9 to infinity.
Monday, December 10, 2012
During a course of algebra, teachers discuss functions and composition of
functions. The topics can be confusing to many students, who also don't see any
practical uses beyond the classroom. The next few paragraphs will clear any
confusion you have on these topics.
Suppose we wish to represent the function g ( x ) by a token machine. The token machine yields one token for every quarter that is deposited into the machine. The tokens can be used to purchase prizes. We think of the quarters as the input x and the number of tokens as the output g ( x ). Suppose further that there is another machine that requires the use of tokens to obtain prizes. A certain number of tokens are needed to purchase each prize. We'll define the prize machine as f ( x ). The input is the number of tokens, which we defined as g ( x ). Notice that purchasing a prize out of the second machine is dependent on the number of tokens from the first machine. Such dependence can be interpreted in mathematical terms as composition of functions.
In the previous example, the domain x yields g ( x ), the number of tokens. Then g ( x ) becomes the input into f ( x ) to produce the output, which is the prize purchased. The end result is a composition function f º g , also noted as f ( g ( x )).
Next, notice the composition function f º g , also noted as f ( g ( x )), joining the two machines together as one machine which automatically deposits tokens into the prize generator, which ejects the appropriate prize corresponding to the number of tokens generated.
Here's a practical example using the composition of functions. A meteorologist predicts a low pressure area to move across the region over the next 36 hours. The current temperature of 60 degrees Fahrenheit is forecast to drop 1 degree every 3 hours. What is the composition function that expresses the Celsius temperature as a function of the number of hours from now? Note that expresses the Celsius temperature as a function of the number of hours from now? Note that C = (5/9)( F - 32).
Suppose we wish to represent the function g ( x ) by a token machine. The token machine yields one token for every quarter that is deposited into the machine. The tokens can be used to purchase prizes. We think of the quarters as the input x and the number of tokens as the output g ( x ). Suppose further that there is another machine that requires the use of tokens to obtain prizes. A certain number of tokens are needed to purchase each prize. We'll define the prize machine as f ( x ). The input is the number of tokens, which we defined as g ( x ). Notice that purchasing a prize out of the second machine is dependent on the number of tokens from the first machine. Such dependence can be interpreted in mathematical terms as composition of functions.
In the previous example, the domain x yields g ( x ), the number of tokens. Then g ( x ) becomes the input into f ( x ) to produce the output, which is the prize purchased. The end result is a composition function f º g , also noted as f ( g ( x )).
Next, notice the composition function f º g , also noted as f ( g ( x )), joining the two machines together as one machine which automatically deposits tokens into the prize generator, which ejects the appropriate prize corresponding to the number of tokens generated.
Here's a practical example using the composition of functions. A meteorologist predicts a low pressure area to move across the region over the next 36 hours. The current temperature of 60 degrees Fahrenheit is forecast to drop 1 degree every 3 hours. What is the composition function that expresses the Celsius temperature as a function of the number of hours from now? Note that expresses the Celsius temperature as a function of the number of hours from now? Note that C = (5/9)( F - 32).
To solve this we need to know the current
temperature and the rate of change of the temperature. We know the current
temperature is 60 and there is a 1/3 degree drop expected every hour. We will
represent time in hours since the temperature is 60 degrees at t. The
temperature in Fahrenheit at time t will be expressed by the function
F (t ). The temperature at time t expressed in Celsius will
be given by the composite function C (F ( t )).
The above example is just one application of composite functions in real life situations. The goal of the articles was to explain composition of functions as far as their structure is concerned and to show a real life application. I believe my explanation will clear questions one might have on these topics.
The above example is just one application of composite functions in real life situations. The goal of the articles was to explain composition of functions as far as their structure is concerned and to show a real life application. I believe my explanation will clear questions one might have on these topics.
Friday, December 7, 2012
On the long running hit game show, "Let's Make a Deal," Monty Hall would ask the
contestant which of three doors he or she wants. Behind each door is a prize,
only one of which is valuable. After the contestant chooses a door, Hall opened
one of the other doors and revealed a worthless prize. Then he would ask the
contestant if he or she wanted to switch doors. To maximize chances of winning,
one should always switch.
To analyze the problem, suppose I were to pick door number 1 and then was revealed door number 3 to be the worthless prize. Now I know the prize is either behind the door I chose or door number 2. It appears that the probability of winning is 50% and switching doesn't increase or decrease my chances.
Think of the problem this way. The probability that the prize is behind door number 1 given that the prize is not behind door number 3. The probability behind door number 1 equals 1/3 since there are 3 doors to choose from. The probability the prize is behind door number 1 or 2 knowing it is not behind door number 3 is 1/2.
The problem involves the use of conditional probability. Suppose you have events A and B, then the probability of A given B equals the probability of A and B divided by the probability of B. This is denoted as P(A/B) = P(A and B)/P(B).
Using the formula for conditional probability, you will find the probability of winning if switching is (1/3)/(1/2) = 2/3. Another way to think of this is that if you switch, the only way that you lose is if the prize was behind the door you initially picked, which has probability of winning 1/3. So the probability that you win is 1 - 1/3 = 2/3.
Although switching doesn't seem to give the best chance of winning, you can conduct an experiment. Play the game with someone assuming the role of Monty Hall and you'll find out as you increase the number of games you play, you'll win approximately 2 out 3 times.
To analyze the problem, suppose I were to pick door number 1 and then was revealed door number 3 to be the worthless prize. Now I know the prize is either behind the door I chose or door number 2. It appears that the probability of winning is 50% and switching doesn't increase or decrease my chances.
Think of the problem this way. The probability that the prize is behind door number 1 given that the prize is not behind door number 3. The probability behind door number 1 equals 1/3 since there are 3 doors to choose from. The probability the prize is behind door number 1 or 2 knowing it is not behind door number 3 is 1/2.
The problem involves the use of conditional probability. Suppose you have events A and B, then the probability of A given B equals the probability of A and B divided by the probability of B. This is denoted as P(A/B) = P(A and B)/P(B).
Using the formula for conditional probability, you will find the probability of winning if switching is (1/3)/(1/2) = 2/3. Another way to think of this is that if you switch, the only way that you lose is if the prize was behind the door you initially picked, which has probability of winning 1/3. So the probability that you win is 1 - 1/3 = 2/3.
Although switching doesn't seem to give the best chance of winning, you can conduct an experiment. Play the game with someone assuming the role of Monty Hall and you'll find out as you increase the number of games you play, you'll win approximately 2 out 3 times.
Tuesday, December 4, 2012
Sunday, December 2, 2012
Over the course of mathematics, we learn how to raise a quantity to an exponent.
We know, for example, that xn means we multiply x by itself n times.
But suppose n is zero. How do we multiply a number by itself zero times? We are
taught that any number raised to the zero power equals one, but rarely does
anyone explain why this is the case.
One way to explain the zero exponent phenomenon is to use rules for exponents. When multiplying like bases, exponents are added. For example, 35 x 30 = 3(5 + 0). Therefore 35 x 30 = 35, so 30 must equal 1.
When dividing like bases, exponents are subtracted. For example, (46)/(40) = 4(6 - 0). Therefore (46)/(40) = 46, so 40 must equal 1.
Another way to show that any base raised to the zero power is one is to examine some patterns. Notice the pattern in the following:
Finally, I can use the concept of limits to show that any number raised to the zero power is one. Suppose we take 5n, and we start with n = 1. We know that 51 = 1. Take the square root of 5, which is equivalent of 5(1/2), which is approximately 2.24. Now take a smaller value for n, such as 1/3, which gives us 5(1/3) = 1.71. Continue to take values for n smaller and smaller but not less than or equal to zero. You'll start to notice what is happening, 5(1/10) = 1.17 , 5(1/1000) = 1.002, 5(1/100000) = 1.00002. Notice how the result is getting closer and closer to 1. We say that the limit as n approaches 0 of 5n = 1.
I just explained three methods that clarify why any number raised to the zero power equals one. Next time someone is puzzled by this fact, you can explain the reason behind this somewhat vague and often explained topic in mathematics.
One way to explain the zero exponent phenomenon is to use rules for exponents. When multiplying like bases, exponents are added. For example, 35 x 30 = 3(5 + 0). Therefore 35 x 30 = 35, so 30 must equal 1.
When dividing like bases, exponents are subtracted. For example, (46)/(40) = 4(6 - 0). Therefore (46)/(40) = 46, so 40 must equal 1.
Another way to show that any base raised to the zero power is one is to examine some patterns. Notice the pattern in the following:
- 24 = 16
- 23 = 8
- 22 = 4
- 21 = 2
- 20 = ?
Finally, I can use the concept of limits to show that any number raised to the zero power is one. Suppose we take 5n, and we start with n = 1. We know that 51 = 1. Take the square root of 5, which is equivalent of 5(1/2), which is approximately 2.24. Now take a smaller value for n, such as 1/3, which gives us 5(1/3) = 1.71. Continue to take values for n smaller and smaller but not less than or equal to zero. You'll start to notice what is happening, 5(1/10) = 1.17 , 5(1/1000) = 1.002, 5(1/100000) = 1.00002. Notice how the result is getting closer and closer to 1. We say that the limit as n approaches 0 of 5n = 1.
I just explained three methods that clarify why any number raised to the zero power equals one. Next time someone is puzzled by this fact, you can explain the reason behind this somewhat vague and often explained topic in mathematics.
Friday, November 30, 2012
The use of Pi in geometry is very common. It is seen in formulas for area and circumference of circles, volume of cones and cylinders and more. Most of us know the value of Pi to be around 3.14, with many people memorizing Pi up to many digits. But where does the value of Pi come from?
The distance around the outside a circle is its circumference. Consider the formula for the circumference of a circle, Circumference = Pi x Diameter. Solving for Pi we get, Pi = Circumference divided by Diameter. Therefore, for a circle of any size, the circumference divided by the diameter is approximately 3.14. But how do we know this?
We can start approximating Pi but considering the perimeter of an regular n-sided figure and dividing that by the length of a diagonal. Take a square with side equal to one. By using the Pythagorean Theorem, or the knowledge of 45-45-90 triangles, the diagonal is approximately 1.41. Therefore the perimeter divided by the diagonal is approximately 2.82.
Now take an octagon, with sides of length equal to one. The length of the diagonal found by taking 2 times the apothem is approximately 2.61. Taking the perimeter of 8 divided by 2.61, we get an approximation for Pi to be 3.06. Notice that this is fairly close to the value of Pi or 3.14.
As the number of sides of a regular polygon increases, the shape more closely approximates that of a circle. When considering a regular polygon with 500 sides, each side of length one, we see calculate the diagonal to be approximately 159.155 and the approximation for Pi to be approximately 3.14157. This is extremely close to the value of Pi at 5 digits, which is 3.14159.
One can continue this process by increasing the number of sides of a regular polygon to get as close of an approximation as possible to the actual value of Pi. I have a book which has the value of Pi calculated out to 1 million digits. Why someone needs to know that, I don't know. For all practical purposes, 3.14 works fine. Also, the fraction 22/7 is often used in calculations involving Pi.
Tuesday, November 27, 2012
I was working with a student on completing the square and it tends to be a more tricky method of factoring than using the quadratic formula. There are several steps to remember.
Recall how we factored and solved quadratic equations using the reverse FOIL method. An example of this type of factoring is x^2 + 5x + 4 = 0 factored is (x +4)(x + 1) = 0. Sometimes it's quite difficult to solve quadratic equations using this method, so we can solve by a method known as completing the square.
The idea behind completing the square is to turn a binomial into a perfect square trinomial. For example, consider the binomial x2 + 6x. The perfect square trinomial with the first two terms x2 + 6x is x2 + 6x + 9 because (x + 3)(x + 3) = x^2 + 6x + 9. Notice how we added a 9 to x^2 + 6x. The question we ask ourselves is, “What number squared equals 9?” We know that 3^2 = 9. Also notice that 3 is half of the coefficient of the middle term 6x. We take half of the middle term, square it and
add it to form the perfect square trinomial. Then we factor the trinomial.
Examples: Complete the square and factor the perfect square trinomial.
1. x^2 + 8x
Step one: Take half the coefficient of the middle term. (1/2)(8) = 4.
Step two: Square the result in step one. 4^2 = 16.
Step three: Add the result in step two to the binomial to form the trinomial x^2 + 8x + 16.
Step four: Factor the trinomial. (x + 4)(x + 4) or (x + 4)^2.
x^2 – 7x
Step one: Take half of the coefficient of the middle term. (1/2)(-7) = -7/2.
Step two: Square the result in step one. (-7/2)^2 = 49/4.
Step three: Add the result in step two to the binomial to form the trinomial x^2 – 7x + 49/4.
Step four: Factor the trinomial. (x – 7/2)(x – 7/2) or (x -7/2)^2
To solve quadratic equations by completing the square we must remember the following:
1. Make sure the coefficient of the squared term is one. If it is not one, we must divide both sides of the equation by the coefficient of that term. For example, if the term is 2x2, we must divide both sides of the equation by 2.
2. Get all variables on one side of the equation and the constants on the other side. This makes sure we have a binomial in the form x2 + bx.
3. Take half of the coefficient of the middle term, square it and add it to both sides of the equation.
4. Factor the perfect square trinomial.
5. Solve the equation using the square root property and check answers by substituting into the original equation.
Note if the coefficient in front of the x^2 term is not 1, must divide the equation by the coefficient before completing the score.
Recall how we factored and solved quadratic equations using the reverse FOIL method. An example of this type of factoring is x^2 + 5x + 4 = 0 factored is (x +4)(x + 1) = 0. Sometimes it's quite difficult to solve quadratic equations using this method, so we can solve by a method known as completing the square.
The idea behind completing the square is to turn a binomial into a perfect square trinomial. For example, consider the binomial x2 + 6x. The perfect square trinomial with the first two terms x2 + 6x is x2 + 6x + 9 because (x + 3)(x + 3) = x^2 + 6x + 9. Notice how we added a 9 to x^2 + 6x. The question we ask ourselves is, “What number squared equals 9?” We know that 3^2 = 9. Also notice that 3 is half of the coefficient of the middle term 6x. We take half of the middle term, square it and
add it to form the perfect square trinomial. Then we factor the trinomial.
Examples: Complete the square and factor the perfect square trinomial.
1. x^2 + 8x
Step one: Take half the coefficient of the middle term. (1/2)(8) = 4.
Step two: Square the result in step one. 4^2 = 16.
Step three: Add the result in step two to the binomial to form the trinomial x^2 + 8x + 16.
Step four: Factor the trinomial. (x + 4)(x + 4) or (x + 4)^2.
x^2 – 7x
Step one: Take half of the coefficient of the middle term. (1/2)(-7) = -7/2.
Step two: Square the result in step one. (-7/2)^2 = 49/4.
Step three: Add the result in step two to the binomial to form the trinomial x^2 – 7x + 49/4.
Step four: Factor the trinomial. (x – 7/2)(x – 7/2) or (x -7/2)^2
To solve quadratic equations by completing the square we must remember the following:
1. Make sure the coefficient of the squared term is one. If it is not one, we must divide both sides of the equation by the coefficient of that term. For example, if the term is 2x2, we must divide both sides of the equation by 2.
2. Get all variables on one side of the equation and the constants on the other side. This makes sure we have a binomial in the form x2 + bx.
3. Take half of the coefficient of the middle term, square it and add it to both sides of the equation.
4. Factor the perfect square trinomial.
5. Solve the equation using the square root property and check answers by substituting into the original equation.
Note if the coefficient in front of the x^2 term is not 1, must divide the equation by the coefficient before completing the score.
Saturday, November 24, 2012
For B to be the multiplicative inverse of A, AB = I and BA = I. So we begin by multiplying the first row of A with the
first column of B to get
(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9
first column of B to get
(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9
For those having difficulty finding the inverse of a matrix, here's part of a chapter from my book on this topic.
Multiplicative Inverses of Matrices
Recall that the multiplicative inverse of any real number x is the number when multiplied by x equals 1. In this case the multiplicative inverse of x is 1/x. Suppose we have matrix A of the form
If we multiply A by what is known as the identity matrix, we still get A. We will call the identity matrix I, note the illustration below of AI = A.
If A is an n by n matrix, then there exists another matrix A-1, called A inverse, so that AA-1 = I and A-1A = I. So we basically are looking for a matrix that when multiplied by the original matrix, equals the identity matrix and vice versa. The identity matrix is always a matrix with 1's along the diagonal from upper left to lower right and 0's everywhere else.
Example: Show that B is the multiplicative inverse of A where
For B to be the multiplicative inverse of A, AB = I and BA = I. So we begin by multiplying the first row of A with the first column of B to get
(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9
= 15/27 + 12/27
= 1
Next we multiply the first row of A with the second column of B to get
(-3)(4/27) + (4)(1/9) = -12/27 + 4/9
= -12/27 + 12/27
= 0
Now proceed to multiply the second row of A with the first row of B to get
(3)(-5/27) + (5)(1/9) = -15/27 + 5/9
= -15/27 + 15/27
= 0
Finally, multiply the second row of A by the second column of B to get
(3)(4/27) + (5)(1/9) = 12/27 + 5/9
= 12/27 + 15/27
= 1
Therefore B is the multiplicative inverse of A. Notice the illustration of this below.
Friday, November 23, 2012
Here's a little review on graphing and finding equations of perpendicular lines.
Example: Find the equation of the line that passes through the point (2, 3) and is perpendicular to the line
y = 3x - 5.
Solution:
The slope of the line perpendicular to y = 3x – 5 must multiply by 3 equal -1. We think of the slope of the first line as m1 and the slope of the second line as m2, therefore 3(m1) = -1, so m1= -1/3.
Now we know that m = – 1/3, x = 2 and y =3.
Use the formula y = mx + b
3 = (-1/3)(2) + b
3 = -2/3 + b
3 2/3 = b
11/3 =b
The equation of the line perpendicular to y = 3x – 5 that passes through the point (2, 3) is y = (-1/3)x + 11/3.
Notice the graph of both lines below.
Example: Find the equation of the line that passes through the point (2, 3) and is perpendicular to the line
y = 3x - 5.
Solution:
The slope of the line perpendicular to y = 3x – 5 must multiply by 3 equal -1. We think of the slope of the first line as m1 and the slope of the second line as m2, therefore 3(m1) = -1, so m1= -1/3.
Now we know that m = – 1/3, x = 2 and y =3.
Use the formula y = mx + b
3 = (-1/3)(2) + b
3 = -2/3 + b
3 2/3 = b
11/3 =b
The equation of the line perpendicular to y = 3x – 5 that passes through the point (2, 3) is y = (-1/3)x + 11/3.
Notice the graph of both lines below.
Tuesday, November 20, 2012
Here's a math puzzle
Suppose you go to a hotel and ask for a room and the clerk says that there are no room available. The clerk then states that if I ask everyone to move to the room next them, then a room will become available. For example, the people in room 1 will move to room 2, from room 2 to room 3 and so on. The people in the last room will move to room 1. How does a room open up???
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This is a hotel with an infinite number of rooms, therefore noone ever moves into room 1. It's kind of an absurd concept, which leads to the concept of infinity. It's a very uncertain concept in and of itself.
Suppose you go to a hotel and ask for a room and the clerk says that there are no room available. The clerk then states that if I ask everyone to move to the room next them, then a room will become available. For example, the people in room 1 will move to room 2, from room 2 to room 3 and so on. The people in the last room will move to room 1. How does a room open up???
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This is a hotel with an infinite number of rooms, therefore noone ever moves into room 1. It's kind of an absurd concept, which leads to the concept of infinity. It's a very uncertain concept in and of itself.
Saturday, November 17, 2012
Suppose we wish to graph the equation y = -3(x - 3)^2 + 5. We recognize this is a parabola with the vertex at (3, 5) because we can rewrite this as y - 5 = -3(x - 3)^2, which is in the form y - k = a(x - h)^2 (vertex at (h, k)).
The parabola opens up or down, and since a is negative, the parabola opens down.
Interesting, we can use calculus to determine whether (3, 5) is a maximum or a minimum. If it's a maximum, the parabola opens down, if it's a minimum, the parabola opens up.
y' = -6(x - 3) = -6x + 18
We now find the derivative for a value of x < 3 and one for a value for x > 3
for x = 2, y' = -6(2-3) = -6(-1) = 6
for x = 4, y' = -6(4-3) = -6(1) = -6
Since the slope of the curve at x = 2 is positive and the slope of the curve at x = 4 is negative we have a maximum at x = 3. Therefore (3, 5) is a maximum and the parabola opens downward.
Just another way to determine how a parabola opens.
The parabola opens up or down, and since a is negative, the parabola opens down.
Interesting, we can use calculus to determine whether (3, 5) is a maximum or a minimum. If it's a maximum, the parabola opens down, if it's a minimum, the parabola opens up.
y' = -6(x - 3) = -6x + 18
We now find the derivative for a value of x < 3 and one for a value for x > 3
for x = 2, y' = -6(2-3) = -6(-1) = 6
for x = 4, y' = -6(4-3) = -6(1) = -6
Since the slope of the curve at x = 2 is positive and the slope of the curve at x = 4 is negative we have a maximum at x = 3. Therefore (3, 5) is a maximum and the parabola opens downward.
Just another way to determine how a parabola opens.
Thursday, November 15, 2012
I am noticing a disturbing trend in schools today regarding mathematics and teaching of mathematics. The majority of students I work with tell me how poor their teachers are. Many basically just read things from the book, don't take time to really explain concepts in class or take the time to work with student after school. I even heard of one teacher that simply posts youtube videos on the topic and hands out worksheets with the answer key. Making matters worse, the school has a tutoring center, but that has been deemed pretty useless as well. When I was in school, teachers seemed more apt and wanting to help students. Are teachers, in general, getting more lazy? I know there are still plenty of excellent teachers and certainly want to give credit to those fine educators that take their jobs seriously. But I can't overlook what seems to be a higher percentage of poor teachers today than 25 to 30 years ago.
Wednesday, November 14, 2012
Testing a hypothesis with proportions can be easy.
For example: Suppose the manufacturers of a certain brand of candy says that 40% of the pieces in the bag are red and the rest are green. Suppose a bag of 50 pieces of candy is opened and 15 of them are green. We can run a test to see if our claim is true based on the sample size and sample proportion. We will chose to test at 5% significance.
Step 1: state your null and alternate hypothesis. The null hypothesis is what is claimed.
The null hypothesis is that 40% are green. The alternate hypothesis is proportion of green is not 40%
Step 2: calculate the test statistic
Z = (p^ - p)/standard deviation
p^ is the sample proportion of candy that is green = 15/50 = .30
p is the claimed proportion of candy that is green = .40
square root[(p)(1-p)/n] = standard deviation = 0.0693
test statistic = (p^ - p)/standard deviation = -1.44
Step 3: compare test statistic with critical value for the test. For 95% and 2 tailed test, this value is 1.96 and -1.96
If the test statistic is greater than 1.96 or less than -1.96 we reject the null hypothesis, otherwise we accept the null hypothesis.
In this case we accept the null hypothesis.
For example: Suppose the manufacturers of a certain brand of candy says that 40% of the pieces in the bag are red and the rest are green. Suppose a bag of 50 pieces of candy is opened and 15 of them are green. We can run a test to see if our claim is true based on the sample size and sample proportion. We will chose to test at 5% significance.
Step 1: state your null and alternate hypothesis. The null hypothesis is what is claimed.
The null hypothesis is that 40% are green. The alternate hypothesis is proportion of green is not 40%
Step 2: calculate the test statistic
Z = (p^ - p)/standard deviation
p^ is the sample proportion of candy that is green = 15/50 = .30
p is the claimed proportion of candy that is green = .40
square root[(p)(1-p)/n] = standard deviation = 0.0693
test statistic = (p^ - p)/standard deviation = -1.44
Step 3: compare test statistic with critical value for the test. For 95% and 2 tailed test, this value is 1.96 and -1.96
If the test statistic is greater than 1.96 or less than -1.96 we reject the null hypothesis, otherwise we accept the null hypothesis.
In this case we accept the null hypothesis.
Monday, November 12, 2012
Sometimes there is confusion among students when graphing lines such as x = 5 and y = 4.
The x-axis is the horizontal axis and the y-axis is the vertical axis, yet the graph of x = 5 is a vertical line through 5 on the x-axis. Likewise the graph of y = 4 is a horizontal line through 4 on the y-axis.
This seems to be opposite of what one might expect, but there is a simple reason for this.
For the line x =5, think first of the point (5,0). Now draw the vertical line through this point and pick another point on the line, for instance (5,2), another point would be (5, -1) and so on. Notice that for every point, the only value that changes is the y coordinate, the x coordinate is always 5. So the vertical line is correct because the x coordinate for every point on that line is 5. The slope of a vertical line is undefined.
The same argument holds true for the equation y = 4. It's a horizontal line because for every point on that line, the y coordinate is 4. The point (0,4) is on the line, as are points (1, 4), (-2, 4) and so on.
The x-axis is the horizontal axis and the y-axis is the vertical axis, yet the graph of x = 5 is a vertical line through 5 on the x-axis. Likewise the graph of y = 4 is a horizontal line through 4 on the y-axis.
This seems to be opposite of what one might expect, but there is a simple reason for this.
For the line x =5, think first of the point (5,0). Now draw the vertical line through this point and pick another point on the line, for instance (5,2), another point would be (5, -1) and so on. Notice that for every point, the only value that changes is the y coordinate, the x coordinate is always 5. So the vertical line is correct because the x coordinate for every point on that line is 5. The slope of a vertical line is undefined.
The same argument holds true for the equation y = 4. It's a horizontal line because for every point on that line, the y coordinate is 4. The point (0,4) is on the line, as are points (1, 4), (-2, 4) and so on.
Saturday, November 10, 2012
Consider the concept of "theoretical probability".
You roll a fair 6 sided die. It is equally likely that any one of the 6 sides, denoted {1, 2, 3, 4, 5, 6} will appear face up.But the actual side that lands face up is not certain. If we roll the die 5 times, the possible outcomes are called the sample space and any occurrence in which the outcome is not known is called an experiment. Rolling a die is an example of an experiment. Therefore, for a single roll of the die, the sample space, denoted by S would be S ={1, 2, 3, 4, 5, 6}.
Suppose the result on a single roll of a die is 2. The next roll of the die is a 4. Each of these results are a subset of the sample space. A subset is called an event and denoted by E. Therefore, the subsets noted above are E ={2} and E = {4}.
The way in which we calculate theoretical probabilities is we divide the number of outcomes in the event divided by the total number of outcomes in the sample space.
What is the probability that a single roll of a fair 6 sided die lands 1 face up?
P(E) = (number of outcomes that result in 1 lands face up)/ (total number of possible outcomes) = 1/6
Example: A coin is flipped 3 times.
a. What is the probability that 2 of the 3 coins land heads up?
First we will figure out the sample space and the number of outcomes in the event.
We'll define the event E = { (H, H, T), (H, T, H), (T, H, H) }
The sample space is S = { (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, T, H), (T, H, T), (T, T, T) }
There are 3 events and 8 outcomes in the sample space, therefore the probability that 2 of the 3 coins land heads up is 3/8.
• Note that an easy way to determine how large the sample space is in this problem is we take the number of
possibilities per coin toss and raise it to the power of the number of tosses (23 = 8).
b. What is the probability that at least 2 coins land tails up?
In this case, we define the event E = { (H, T, T), (T, T, H), (T, H, T), (T, T, T) }
The sample space is the same as in part a, therefore the probability that at least 2 coins land heads up is 4/8 = ½.
Notice the events that 0, 1, 2 and 3 heads appear are listed below with their associated probabilities.
0 heads: E = { (T, T, T) } P(E) = 1/8
1 head: E = { (H, T, T), (T, H, T), (T, T, H) } P(E) = 3/8
2 heads: E = { (H, H, T), (H, T, H), (T, H, H) } P(E) = 3/8
3 heads: E = { (H, H, H) } P(E) = 1/8
Notice if we add the probabilities of all the events, we get 1. The sum of theoretical probabilities of all possible outcomes in a sample space equals 1.
Example: Two fair 6 sided dice as rolled. What is the probability of getting a sum of 6, 7 or 8 ?
Since each die has equally likely outcomes, there are 6 ∙6 = 36 possible outcomes in the sample space.
Notice all of the possible outcomes below.
S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
The outcomes in S that give us a sum of 6, 7 or 8, which we denote as event E are
E = { (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 4), (4, 3), (1, 6), (6, 1), (2, 5), (5, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4) }
There are 16 outcomes in E and 36 outcomes in S. Therefore, the probability of getting a sum of 6, 7 or 8 is
16/36 =4/9
You roll a fair 6 sided die. It is equally likely that any one of the 6 sides, denoted {1, 2, 3, 4, 5, 6} will appear face up.But the actual side that lands face up is not certain. If we roll the die 5 times, the possible outcomes are called the sample space and any occurrence in which the outcome is not known is called an experiment. Rolling a die is an example of an experiment. Therefore, for a single roll of the die, the sample space, denoted by S would be S ={1, 2, 3, 4, 5, 6}.
Suppose the result on a single roll of a die is 2. The next roll of the die is a 4. Each of these results are a subset of the sample space. A subset is called an event and denoted by E. Therefore, the subsets noted above are E ={2} and E = {4}.
The way in which we calculate theoretical probabilities is we divide the number of outcomes in the event divided by the total number of outcomes in the sample space.
What is the probability that a single roll of a fair 6 sided die lands 1 face up?
P(E) = (number of outcomes that result in 1 lands face up)/ (total number of possible outcomes) = 1/6
Example: A coin is flipped 3 times.
a. What is the probability that 2 of the 3 coins land heads up?
First we will figure out the sample space and the number of outcomes in the event.
We'll define the event E = { (H, H, T), (H, T, H), (T, H, H) }
The sample space is S = { (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, T, H), (T, H, T), (T, T, T) }
There are 3 events and 8 outcomes in the sample space, therefore the probability that 2 of the 3 coins land heads up is 3/8.
• Note that an easy way to determine how large the sample space is in this problem is we take the number of
possibilities per coin toss and raise it to the power of the number of tosses (23 = 8).
b. What is the probability that at least 2 coins land tails up?
In this case, we define the event E = { (H, T, T), (T, T, H), (T, H, T), (T, T, T) }
The sample space is the same as in part a, therefore the probability that at least 2 coins land heads up is 4/8 = ½.
Notice the events that 0, 1, 2 and 3 heads appear are listed below with their associated probabilities.
0 heads: E = { (T, T, T) } P(E) = 1/8
1 head: E = { (H, T, T), (T, H, T), (T, T, H) } P(E) = 3/8
2 heads: E = { (H, H, T), (H, T, H), (T, H, H) } P(E) = 3/8
3 heads: E = { (H, H, H) } P(E) = 1/8
Notice if we add the probabilities of all the events, we get 1. The sum of theoretical probabilities of all possible outcomes in a sample space equals 1.
Example: Two fair 6 sided dice as rolled. What is the probability of getting a sum of 6, 7 or 8 ?
Since each die has equally likely outcomes, there are 6 ∙6 = 36 possible outcomes in the sample space.
Notice all of the possible outcomes below.
S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
The outcomes in S that give us a sum of 6, 7 or 8, which we denote as event E are
E = { (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 4), (4, 3), (1, 6), (6, 1), (2, 5), (5, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4) }
There are 16 outcomes in E and 36 outcomes in S. Therefore, the probability of getting a sum of 6, 7 or 8 is
16/36 =4/9
Friday, November 9, 2012
When graphing a parabola, sometimes it's confusing which way the parabola opens. Does it open up, down, left or right?
A parabola in the form y = x^2 will open up. For every value of y, there will be 2 values for x.
When x = 3, y = 9.
x = -3, y = 9
x = 2, y = 4
x = -2, y = 4
No matter what value you substitute for x, you will get a positive value for y.
If you plot these points in a graph you will notice the graph is symmetric about the y axis. This means the y axis cuts the parabola in half, mirror images on both sides of the axis.
The graph opens down if in the form -y = x^2. This is because all the values which were positive before become negative.
You can also think of the parabola opening towards the linear term. In this case, the linear term is y and the parabola opens up or down, and the y axis runs vertically.
The exact opposite argument is true for parabolas in the form x = y^2 or -x = y^2. In this case the x axis is the axis of symmetry and the parabolas open either right or left (horizontally) and the x axis is the horizontal axis.
A parabola in the form y = x^2 will open up. For every value of y, there will be 2 values for x.
When x = 3, y = 9.
x = -3, y = 9
x = 2, y = 4
x = -2, y = 4
No matter what value you substitute for x, you will get a positive value for y.
If you plot these points in a graph you will notice the graph is symmetric about the y axis. This means the y axis cuts the parabola in half, mirror images on both sides of the axis.
The graph opens down if in the form -y = x^2. This is because all the values which were positive before become negative.
You can also think of the parabola opening towards the linear term. In this case, the linear term is y and the parabola opens up or down, and the y axis runs vertically.
The exact opposite argument is true for parabolas in the form x = y^2 or -x = y^2. In this case the x axis is the axis of symmetry and the parabolas open either right or left (horizontally) and the x axis is the horizontal axis.
Tuesday, November 6, 2012
Use implicit differentiation to find dy/dx. We differentiate each x term as usual and when differentiating each y term, you add dy/dx since we are differentiating with respect to x. Then isolate dy/dx to solve.
x^3 + y^3 - 3xy = 0
3x^2 + 3y^2(dy/dx) - [3x(dy/dx) + 3y] = 0
3y^2(dy/dx) - 3x(dy/dx) = -3x^2 + 3y
(dy/dx)(3y^2 - 3x) = -3x^2 + 3y
dy/dx = (-3x^2 + 3y)/(3y^2 - 3x)
dy/dx = (x^2 - y)/(y^2 - x)
x^3 + y^3 - 3xy = 0
3x^2 + 3y^2(dy/dx) - [3x(dy/dx) + 3y] = 0
3y^2(dy/dx) - 3x(dy/dx) = -3x^2 + 3y
(dy/dx)(3y^2 - 3x) = -3x^2 + 3y
dy/dx = (-3x^2 + 3y)/(3y^2 - 3x)
dy/dx = (x^2 - y)/(y^2 - x)
Friday, November 2, 2012
When multiplying numbers raised to fractional exponents, remember that if the numbers are the same, simply add the exponents. If the exponents are the same and the numbers are different, multiply the numbers and keep the exponent the same.
For example:
6^(2/3) * 6^(3/4) The bases are the same (6), therefore keep the base and add the exponents
6^(2/3 + 3/4)
2/3 + 3/4 = 8/12 + 9/12 = 17/12
The answer is 6^(17/12)
Another example:
4^(1/3) * 9^(1/3) The exponents are the same, so keep the exponent as is and multiply the bases
4 * 9 = 36
The answer is 36^(1/3)
For example:
6^(2/3) * 6^(3/4) The bases are the same (6), therefore keep the base and add the exponents
6^(2/3 + 3/4)
2/3 + 3/4 = 8/12 + 9/12 = 17/12
The answer is 6^(17/12)
Another example:
4^(1/3) * 9^(1/3) The exponents are the same, so keep the exponent as is and multiply the bases
4 * 9 = 36
The answer is 36^(1/3)
Thursday, November 1, 2012
The sum of three positive integers is 16. If the first integer is doubled, the sum is 22. If the third integer is tripled, the sum is 20. What are the 3 integers?
Solution:
Let x = First integer
y = Second integer
z = Third integer
x + y + z = 16 (Sum of the integers is 16)
2x + y + z = 22 (Double the first integer and the sum is 22)
x + y + 3z = 20 (Triple the third integer and the sum is 20)
We can eliminate both the y and z variable by subtracting the second equation from the first, therefore
x + y + z = 16
-(2x + y + z = 22)
-x = -6, x = 6
Subtract third equation from the first equation to eliminate the x and y variable, therefore
x + y + z = 16
-(x + y + 3z = 20)
-2z = -4, z = 2
Now substitute 6 for x and 2 for z in the first equation and solve for y.
6 + y + 2 = 16
8 + y = 16
y = 8
The three integers are 6, 8 and 2.
Solution:
Let x = First integer
y = Second integer
z = Third integer
x + y + z = 16 (Sum of the integers is 16)
2x + y + z = 22 (Double the first integer and the sum is 22)
x + y + 3z = 20 (Triple the third integer and the sum is 20)
We can eliminate both the y and z variable by subtracting the second equation from the first, therefore
x + y + z = 16
-(2x + y + z = 22)
-x = -6, x = 6
Subtract third equation from the first equation to eliminate the x and y variable, therefore
x + y + z = 16
-(x + y + 3z = 20)
-2z = -4, z = 2
Now substitute 6 for x and 2 for z in the first equation and solve for y.
6 + y + 2 = 16
8 + y = 16
y = 8
The three integers are 6, 8 and 2.
Tuesday, October 30, 2012
When numbers get very large or very small, they become difficult to work with and read. Examples of such
numbers are 0.0000000000043425 and 546,245,000,000,000,000,000. Oftentimes these types of numbers appear in science, such as the distance planets are from Earth or the mass of atoms. Scientific notation is a method to make using such numbers easier by writing them in a simpler form. A positive number in the form N times 10^x , where 1 ≤ N < 10 and x is an integer, is said to be in scientific notation.
When a number is in scientific notation, the decimal point is always after the first non zero number. Some
examples of numbers in scientific notation are
3.5 X 10^3, 5.89 X 10^(-4) and 2.9123 X 10^6
To change a number into scientific notation, we move the decimal point between the first two non zero
numbers. Then we count the number and direction that the decimal point must move to get the original number.
The number of decimal places moved will be the exponent. If we move the decimal point to the right, the
exponent is positive. If we move the decimal point to the left, the exponent is negative.
Examples: Write the following numbers in scientific notation.
1. 567,325,000,000,000,000
First place the decimal point between the 5 and 6. Count how many decimal places we have to move to
the right to get to the end of the number. Notice we have to move 17 decimal places. Therefore,
567,325,000,000,000,000 written in scientific notation is 5.67325 X 10^17.
•Note that the zeros after the 5 are not written when changing to scientific notation. When all the rest of
the numbers are zero they are not written.
2. 0.0000000007982
First place the decimal point between the 7 and 9. Count how many decimal places we have to move to
the left to get back to the beginning of the number. Notice we have to move 10 decimal places.
Therefore 0.0000000007982 written in scientific notation is 7.982 X 10^(-10) .
Be sure to be careful with numbers such as 65.8 X 10^3 and 254.69 X 10^(-4). At first glance, these appear to be in scientific notation, but the decimal point is not after the first non zero number.
To write in scientific notation, multiply the problem out and then convert to scientific notation.
65.8 X 10^3 = 65.8 X 1000 = 65,800
Another way to simplify the above is to move the decimal point 3 places to the right since the exponent is 3.
Now we can change 65,800 into scientific notation, which is 6.58 X 10^4.
In the example 254.69 X 10^(-4), multiply to get 254.69 X 0.0001 = 0.025469.
Another way to simplify the above is to move the decimal point 4 places to the left since the exponent is -4.
Now we can change 0.025469 into scientific notation, which is 2.5469 X 10^(-2) .
In some cases, using scientific notation makes multiplying and dividing very large or very small numbers easier.
Saturday, October 27, 2012
I will show you 3 methods for finding the vertex and axis of symmetry of a parabola.
Suppose an equation is as follows:
y = (x - 1)(x + 5)
First we can find the x intercepts by setting x - 1 and x + 5 equal to 0 and solve for x. When doing so, we get the intercepts to be (1, 0) and (-5, 0). The x coordinate of the vertex is halfway between 1 and -5. Therefore the x coordinate of the vertex is -2. The y coordinate of the vertex is found by substituting -2 for x in the equation.
y = (-2 - 1)(-2 + 5) = -3(3) = -9
Therefore the vertex of the parabola is (-2, -9). The axis of symmetry is the line through the x coordinate of the vertex, or x = -2 in this case.
Suppose the equation is
y = x^2 + 4x - 5.
If the equation is in the form y = ax^2 + bx + c, the x coordinate of the vertex is -b/2a.
a = 1, b = 4. Therefore the x coordinate of the vertex is -4/2(1) = -2.
The y coordinate of the vertex is y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9.
The vertex is at (-2, -9) and the axis of symmetry is x = -2.
Suppose the equation is
y + 9 = (x + 2)^2
If the equation is in the form y - k = (x - h)^2, the vertex is (h, k)
Therefore the vertex is (-2, -9) and the axis of symmetry is x = -2.
The equations y = (x -1)(x + 5), y = x^2 + 4x - 5 and y + 9 = (x + 2)^2 are the same, just written in different forms.
Suppose an equation is as follows:
y = (x - 1)(x + 5)
First we can find the x intercepts by setting x - 1 and x + 5 equal to 0 and solve for x. When doing so, we get the intercepts to be (1, 0) and (-5, 0). The x coordinate of the vertex is halfway between 1 and -5. Therefore the x coordinate of the vertex is -2. The y coordinate of the vertex is found by substituting -2 for x in the equation.
y = (-2 - 1)(-2 + 5) = -3(3) = -9
Therefore the vertex of the parabola is (-2, -9). The axis of symmetry is the line through the x coordinate of the vertex, or x = -2 in this case.
Suppose the equation is
y = x^2 + 4x - 5.
If the equation is in the form y = ax^2 + bx + c, the x coordinate of the vertex is -b/2a.
a = 1, b = 4. Therefore the x coordinate of the vertex is -4/2(1) = -2.
The y coordinate of the vertex is y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9.
The vertex is at (-2, -9) and the axis of symmetry is x = -2.
Suppose the equation is
y + 9 = (x + 2)^2
If the equation is in the form y - k = (x - h)^2, the vertex is (h, k)
Therefore the vertex is (-2, -9) and the axis of symmetry is x = -2.
The equations y = (x -1)(x + 5), y = x^2 + 4x - 5 and y + 9 = (x + 2)^2 are the same, just written in different forms.
Thursday, October 25, 2012
I have an easy way to remember the reflex, symmetric and transitive properties of equality.
When thinking of reflexive, think of the word "reflection". When you see your reflection, you see yourself. So reflexive would be something along the lines of "a = a" or "b = b".
When thinking of symmetric, think of the word "symmetry". If you cut something in half and have two equal parts, the object is symmetrical about the cut.
In geometry this is like saying "a = b, then b = a".
The transitive property involves 3 things, notice "trans" meaning three, is part of transitive.
In geometry this is like saying "if a = b and b = c, then a = c"
When thinking of reflexive, think of the word "reflection". When you see your reflection, you see yourself. So reflexive would be something along the lines of "a = a" or "b = b".
When thinking of symmetric, think of the word "symmetry". If you cut something in half and have two equal parts, the object is symmetrical about the cut.
In geometry this is like saying "a = b, then b = a".
The transitive property involves 3 things, notice "trans" meaning three, is part of transitive.
In geometry this is like saying "if a = b and b = c, then a = c"
Wednesday, October 24, 2012
Tuesday, October 23, 2012
Monday, October 22, 2012
Finding the inverse of a function and its graph.
Suppose we have a function f(x) and know that point (c, d) lies on its graph. If f(x) has an inverse, f -1(x), the point (d, c) lies on its graph. The points (c, d) and (d, c) are equidistant from the line y = x. Such points are called mirror images of each other. Notice the graph and the points with respect to the line y = x.
With respect to the line y = x, the points on the graphs of f(x) and f -1(x) are mirror images of each other, so it makes sense to draw the conclusion that the graphs of f(x) and f -1(x) are also mirror images of each other with respect to y = x
.
Examples: a. Find the inverse of f(x) = 2x + 4 and graph f(x) and f -1(x) on one rectangular coordinate system.
Before we find the inverse of f(x), we must determine if an inverse exists. If the function is one-to-one, then it has an inverse. Since f(x) is linear, then it does have an inverse. Recall, to find the inverse we interchange y and x and then solve the equation for y.
f(x) = 2x + 4
y = 2x + 4
x = 2y + 4
x – 4 = 2y
(x – 4)/2 = y
f -1(x) = (x – 4)/2
To graph f(x) = 2x + 4, we know the y- intercept is (0, 4). To find the x- intercept, rewrite the function as the equation y = 2x + 4 and substitute 0 for y. Then solve for x.
y = 2x + 4
0 = 2x + 4
-4 = 2x
-2 = x. Therefore the x- intercept is (-2, 0).
To graph f -1(x) = (1/2)x – 2, we know the y- intercept is (0, -2). To find the x- intercept, rewrite the inverse function as the equation y = (1/2)x – 2 and substitute 0 for y. Then solve for x.
y = (1/2)x – 2
0 = (1/2)x – 2
2 = (1/2)x, 4 = x. Therefore the x- intercept is (4, 0)
Notice the graph of f(x) and its inverse.
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