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
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