Solve the second order differential equation if f ''(x) = 3x^3 + x^2 + 4x + 5, f'(2) = 2/3 and f(3) = 10
To find f '(x), integrate f ''(x)
Therefore f '(x) = (3/4)x^4 + (1/3)x^3 + 2x^2 + 5x + C
Now substitute 2 for x and 2/3 for f ' (x)
2/3 = (3/4)(2^4) + (1/3)(2^3) + 2(2^2) + 5(2) + C
2/3 = 12 + 8/3 + 8 + 10 + C
2/3 = 32 2/3 + C
C = -32
f ' (x) = (3/4)x^4 + (1/3)x^3 + 2x^2 + 5x - 32
To find f(x), integrate f ' (x)
Therefore f(x) = (3/20)x^5 + (1/12)x^4 + (2/3)x^3 + (5/2)x^2 - 32x + C
Now substitute 10 for f(x) and 3 for x
10 = (3/20)(3^5) + (1/12)(3^4) + (2/3)(3^3) + (5/2)(3^2) - 32(3) + C
10 = 729/20 + 81/12 + 18 + 45/2 - 96 + C
10 = -12.3 + C
22 3/10 = C
f(x) = (3/20)x^5 + (1/12)x^4 + (2/3)x^3 + (5/2)x^2 - 32x + 22 3/10.
Wednesday, February 27, 2013
Monday, February 25, 2013
Remember when dealing with conversions, if you are going from a smaller unit to a larger unit, divide. When going from a larger unit to a smaller unit, multiply.
For example:
2 tons = ? feet
A ton is larger than a foot, so multiply 2,000 by 2 since there are 2,000 pounds in a ton. The answer is 2 tons = 4,000 feet.
1,500 meters = ? kilometers
A meter is smaller than a kilometer, so divide 1,5000 by 1,000 since there are 1,000 meters to 1 kilometer. The answer is 1,500 meters = 1.5 kilometers.
For example:
2 tons = ? feet
A ton is larger than a foot, so multiply 2,000 by 2 since there are 2,000 pounds in a ton. The answer is 2 tons = 4,000 feet.
1,500 meters = ? kilometers
A meter is smaller than a kilometer, so divide 1,5000 by 1,000 since there are 1,000 meters to 1 kilometer. The answer is 1,500 meters = 1.5 kilometers.
Friday, February 22, 2013
When dealing with trig identities, you don't have to memorize the double angle formulas.
If you know sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
you can get
sin(2a) = sin(a + a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a)
That's how the double angle identities are derived.
cos(2a) = cos(a + a) = cos(a)cos(a) - sin(a)sin(a) = cos^2(a) - sin^2(a)
The same holds true for tan(a + b), you can derive tan(2a) = tan(a + a).
If you know sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
you can get
sin(2a) = sin(a + a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a)
That's how the double angle identities are derived.
cos(2a) = cos(a + a) = cos(a)cos(a) - sin(a)sin(a) = cos^2(a) - sin^2(a)
The same holds true for tan(a + b), you can derive tan(2a) = tan(a + a).
Tuesday, February 19, 2013
n elementary school, children will learn the con cept of money and
the coins used to represents different amounts of money. It is important
to recognize and know how to count and add coins. This article will
help assist children in this learning process as well as give parents a
few tips to get involved in helping their child master this concepts.
The basic coins used in United States currency are the penny, nickel, dime and quarter. The value of these coins are in units called "cents". A total of 100 cents equals one "dollar". One cent is shown as 1 c or $ .01.
5 pennies = 1 nickel
10 pennies = 1 dime
25 pennies = 1 quarter
2 nickels = 1 dime
5 nickels = 1 quarter
2 dimes and 1 nickel = 1 quarter
Here's a few examples of adding coins:
4 pennies and a 1 n ickel = 1 cent + 1 cent + 1 cent + 1 cent + 5 cents = $.01 + $.01 + $.01 + $.01 + $.05 = $. 09 .5 pennies and 1 nickel = 1 cent + 1 cent + 1 cent + 1 cent + 1 cent + 5 ce nts = $.01 + $.01 + $.01 + $.01 + $.01 + $.05 = $.10. Th is is 10 cents , which is t he same as a dime.2 dimes and 1 nickel = 10 cents + 10 cents + 5 cents = $.10 + $.10 + $.05 = $.25. This is 25 ce nts, which is the same as a quarter.
It's easy to learn how to count money, bu t s ometimes when we purchase items , we don't have the exact money and have to give more than what is needed. For example, suppose a pen and pencil costs $0.76. You look and notice you have 3 quarters and a nickel, which adds to $0.25 + $0.25 + $0.25 + $0.05 = $0.80. Since you gave more than is needed, you get money back. You subtract the cost from what you paid to get the amount of change. So you will get back $0.80 - $0.76 = $0.04.
See another example, this time with buying 3 items. The items are a pencil, pen and small ball. The pencil costs $ 0.24, the pen costs $0.45 and the small ball costs $ 0.23. The total cost of the items is $0.92. You don't have the exact amount of money, so you pay with 3 quarters and 2 dimes for a total of $ 0.75 + $ 0.20 = $ 0.95. The amount of change you get back is $ 0.95 - $ 0.92 = $ 0.03.
When item's add up to more than $1, often they are paid with paper money instead of coins. Paper money comes in $1, $5, $10, $20 and higher bills, which have the numbers showing how much they are worth. If an item is $1, many times it is paid with the $1 bill instead of 4 quarters of 10 dimes, which also equals $1.
Suppose two items cost $1.25 and you pay with a $1 bill and 3 dimes. How much change do you get?
3 dimes = $ 0.30 plus $1 equals $ 1.30. The change is $ 1.30 - $ 1.25 = $ 0.05, which will be either 1 nickel or 5 pennies.
Suppose three items cost $1.56 and you pay with a $1 bill, 1 quarter, 3 dimes and 1 nickel. How much change do you get?
$1 bill = $ 1.00, 1 quarter = $ 0.25, 3 dimes = $ 0.30, 1 nickel = $ 0.05. Add the money to get $1.60. The amount of change you get is $ 1.60 - $ 1.56 = $ 0.04 or 4 pennies.
*Tips for parents
When in the store, ask your child how much a couple items cost and what the change will be if you pay with a certain amount. Play games involving the use of money such as Monopoly and Life and h ave your child be th e banker. This will give him or her pra ctice counting change in more re al life situations. Keep practicing these exercises until your child can complete them correctly on a consistent basis.
This guide should help any child who has difficulties understanding how to count and add coins.
The basic coins used in United States currency are the penny, nickel, dime and quarter. The value of these coins are in units called "cents". A total of 100 cents equals one "dollar". One cent is shown as 1 c or $ .01.
5 pennies = 1 nickel
10 pennies = 1 dime
25 pennies = 1 quarter
2 nickels = 1 dime
5 nickels = 1 quarter
2 dimes and 1 nickel = 1 quarter
Here's a few examples of adding coins:
4 pennies and a 1 n ickel = 1 cent + 1 cent + 1 cent + 1 cent + 5 cents = $.01 + $.01 + $.01 + $.01 + $.05 = $. 09 .5 pennies and 1 nickel = 1 cent + 1 cent + 1 cent + 1 cent + 1 cent + 5 ce nts = $.01 + $.01 + $.01 + $.01 + $.01 + $.05 = $.10. Th is is 10 cents , which is t he same as a dime.2 dimes and 1 nickel = 10 cents + 10 cents + 5 cents = $.10 + $.10 + $.05 = $.25. This is 25 ce nts, which is the same as a quarter.
It's easy to learn how to count money, bu t s ometimes when we purchase items , we don't have the exact money and have to give more than what is needed. For example, suppose a pen and pencil costs $0.76. You look and notice you have 3 quarters and a nickel, which adds to $0.25 + $0.25 + $0.25 + $0.05 = $0.80. Since you gave more than is needed, you get money back. You subtract the cost from what you paid to get the amount of change. So you will get back $0.80 - $0.76 = $0.04.
See another example, this time with buying 3 items. The items are a pencil, pen and small ball. The pencil costs $ 0.24, the pen costs $0.45 and the small ball costs $ 0.23. The total cost of the items is $0.92. You don't have the exact amount of money, so you pay with 3 quarters and 2 dimes for a total of $ 0.75 + $ 0.20 = $ 0.95. The amount of change you get back is $ 0.95 - $ 0.92 = $ 0.03.
When item's add up to more than $1, often they are paid with paper money instead of coins. Paper money comes in $1, $5, $10, $20 and higher bills, which have the numbers showing how much they are worth. If an item is $1, many times it is paid with the $1 bill instead of 4 quarters of 10 dimes, which also equals $1.
Suppose two items cost $1.25 and you pay with a $1 bill and 3 dimes. How much change do you get?
3 dimes = $ 0.30 plus $1 equals $ 1.30. The change is $ 1.30 - $ 1.25 = $ 0.05, which will be either 1 nickel or 5 pennies.
Suppose three items cost $1.56 and you pay with a $1 bill, 1 quarter, 3 dimes and 1 nickel. How much change do you get?
$1 bill = $ 1.00, 1 quarter = $ 0.25, 3 dimes = $ 0.30, 1 nickel = $ 0.05. Add the money to get $1.60. The amount of change you get is $ 1.60 - $ 1.56 = $ 0.04 or 4 pennies.
*Tips for parents
When in the store, ask your child how much a couple items cost and what the change will be if you pay with a certain amount. Play games involving the use of money such as Monopoly and Life and h ave your child be th e banker. This will give him or her pra ctice counting change in more re al life situations. Keep practicing these exercises until your child can complete them correctly on a consistent basis.
This guide should help any child who has difficulties understanding how to count and add coins.
Sunday, February 17, 2013
Tips to teach children two digit subtraction.
After learning basic subtraction using
numbers from 1 to 10, children will learn how to subtract two digit numbers with
and without borrowing. This article will explain the procedure to subtract two
digit numbers and how to borrow. Parents will be given tips on how to help their
children on this topic
.
Two digit subtraction is done much the same way as two digit addition. Subtract the ones digit numbers first and the tens digit next. For example, if the problem is 65 - 24, do 5 - 4 first, which is 1. Then do 6 - 2, which is 4. The answer is 41. When subtracting money, keep the decimal place in line and subtract the same as in the last problems. For example, $0.75 - 0.41 = $0.34 and $0.66 - 0.04 = $0.62.
But how do we subtract a problem such as 66 - 19? Notice in the previous examples, the second number had a ones digit smaller than the ones digit in the first number. But that is not the case when solving 66 - 19. When subtracting 9 from 6 in the ones place, we get a number less than 0. There are two ways to solve this problem. The first is the method of borrowing.
When borrowing in the previous example, take 1 away from the tens place. That turns the 6 into 5. In essence, we are borrowing 10 and adding it to the number in the ones place. Place the 1 in front of the 6 in the ones place to make it 16. Subtract 9 from 16 to get 7. Then subtract 1 from 5 on the tens place to get 4. The answer is 47.
We can also solve the problem by subtracting 9 from 6. We can use a number line to show how to subtract 9 from 6. Place a dot at 6 and move 9 places to the left. After moving 6 places, we are at 0. You have to move 3 more places so there are negative numbers. The first negative number counting by 1 is -1, then -2, -3 and so on. Doing this, we get 6 - 9 = -3.
Subtract 1 from 6 in the tens place to get 5 and add the 0 to get 50. Next write -3 under the 0 in 50 to get 50 - 3. Subtract to get 47. Notice the answer is the same.
We can subtract 3 two digit numbers by subtracting 2 of them, then subtracting the 3 rd. To solve 99 - 43 - 18, take 99 - 43 = 56. Then take 56 - 18 = 38. You can also subtract 18 from 99 first and then subtract 43 and get the same answer. Take 99 - 18 = 81, and then 81 - 43 = 38.
Two digit subtraction is done much the same way as two digit addition. Subtract the ones digit numbers first and the tens digit next. For example, if the problem is 65 - 24, do 5 - 4 first, which is 1. Then do 6 - 2, which is 4. The answer is 41. When subtracting money, keep the decimal place in line and subtract the same as in the last problems. For example, $0.75 - 0.41 = $0.34 and $0.66 - 0.04 = $0.62.
But how do we subtract a problem such as 66 - 19? Notice in the previous examples, the second number had a ones digit smaller than the ones digit in the first number. But that is not the case when solving 66 - 19. When subtracting 9 from 6 in the ones place, we get a number less than 0. There are two ways to solve this problem. The first is the method of borrowing.
When borrowing in the previous example, take 1 away from the tens place. That turns the 6 into 5. In essence, we are borrowing 10 and adding it to the number in the ones place. Place the 1 in front of the 6 in the ones place to make it 16. Subtract 9 from 16 to get 7. Then subtract 1 from 5 on the tens place to get 4. The answer is 47.
We can also solve the problem by subtracting 9 from 6. We can use a number line to show how to subtract 9 from 6. Place a dot at 6 and move 9 places to the left. After moving 6 places, we are at 0. You have to move 3 more places so there are negative numbers. The first negative number counting by 1 is -1, then -2, -3 and so on. Doing this, we get 6 - 9 = -3.
Subtract 1 from 6 in the tens place to get 5 and add the 0 to get 50. Next write -3 under the 0 in 50 to get 50 - 3. Subtract to get 47. Notice the answer is the same.
We can subtract 3 two digit numbers by subtracting 2 of them, then subtracting the 3 rd. To solve 99 - 43 - 18, take 99 - 43 = 56. Then take 56 - 18 = 38. You can also subtract 18 from 99 first and then subtract 43 and get the same answer. Take 99 - 18 = 81, and then 81 - 43 = 38.
Tips for parents:
Make up some subtraction problems with borrowing. Allow your child to look at the example problems in this article for the first few problems so he or she can understand the procedures. Then have your child attempt problems without the use of the examples. Also, make up some fun activity type problem such as subtraction problems where you fill in the blank to solve. Another type of activity would be to match answers to subtraction problems with numbers drawn on cars or balloons or blocks. Once your child can consistently solve these types of problems correctly, he or she is ready for basic multiplication.
This article should help children master the concept of subtraction with borrowing and enable parents to help their children accomplish this goal.
Make up some subtraction problems with borrowing. Allow your child to look at the example problems in this article for the first few problems so he or she can understand the procedures. Then have your child attempt problems without the use of the examples. Also, make up some fun activity type problem such as subtraction problems where you fill in the blank to solve. Another type of activity would be to match answers to subtraction problems with numbers drawn on cars or balloons or blocks. Once your child can consistently solve these types of problems correctly, he or she is ready for basic multiplication.
This article should help children master the concept of subtraction with borrowing and enable parents to help their children accomplish this goal.
Wednesday, February 13, 2013
Making Sense of Radical Equations
Just as students learn to solve linear equations and
quadratic equations, they will also need to learn
how to solve radical equations. To solve equations involving radicals, we must use what is known as the power rule, which states that if x, y and n are real numbers and x = y, then xn = yn. This is important because to solve radical equations we have to remove the radical. In order to remove the radical, we must raise the radical to the index of the radical. For example, if we have 3√(x+3), we raise 3√(x+3) to the 3rd power to remove the radical. To remove a radical involving a square root, we raise it to the 2nd power and so on.
Sometimes when we raise both sides of an equation to the same power, we will get equations that are not equivalent. The reason for this is that their solution sets are different.
For example, if x = 4, the solution set is {4}. Now if we raise both sides to the 4th power, we get x4 = 256, with a solution set of {4, -4}. Notice the solution sets are different. The first equation is not satisfied by x = -4, as it is in the second equation. Therefore, we must always check each solution in the original equation to make sure it satisfies the equation. Any solution that doesn't satisfy the equation is discarded. Those discarded solutions are also known as extraneous solutions.
A radical equation is an equation with a radical expression. Some examples of radical equations are
√(x - 3) = 8, √(2x + 5) = √(x + 2), 3√(8x) = 9, 3√(3x + 4) = 3√(x2).
The first step in solving a radical equation is the same as for solving a linear equation. We want to isolate the radical expression to one side of the equation. Next, we raise both sides of the equation to the power that equals the index of the radical. Then we solve the equation for the variable and check all solutions to make sure they satisfy the original equation. Eliminate those solutions which don't satisfy the equation.
Example: Solve √(2x + 7) = 7.
First step is to isolate the radical to one side. The problem is set up with it already isolated. Next step is to square both sides to remove the radical.
[√(2x + 7)]2 = (7)2
2x + 7 = 49
Now solve for x.
2x = 42
x = 21.
Check to make sure 21 satisfies the original equation.
√(2x + 7) = 7
√(2∙21 + 7) = 7
√(42 + 7) = 7
√49 = 7
7 = 7. Therefore 21 is the solution to the equation.
Example: Solve √(4x - 11) + 4 = x.
First isolate the radical.
√(4x - 11) + 4 = x
√(4x - 11) + 4 - 4 = x - 4
√(4x - 11) = x - 4
Next, square both sides of the equation to remove the radical.
[√(4x - 11)]2 = (x - 4)2
4x - 11 = (x - 4)2
Now we solve the equation for x.
4x - 11 = x2 - 8x + 16 (by the FOIL method (x - 4)(x - 4) )
-11 = x2 - 12x + 16
0 = x2 - 12x + 27
Factor to get
0 = (x - 3)(x - 9) (-9 ∙ (-3) = 27 and -9 + (-3) = -12)
0 = (x - 3) or 0 = (x - 9). Therefore x = 3, 9
Check both answers to make sure they satisfy the original equation.
For x = 3, √(4 ∙ 3 - 11) + 4 = 3 For x = 9, √(4 ∙ 9 - 11) + 4 = 9
√1 + 4 = 3 √25 + 4 = 9
5 ≠ 3, 9 = 9
Therefore 9 is a solution to the equation, but 3 is not a solution.
Example: Solve 3√(x3 + 8) = x + 2.
The radical is already isolated on the left side of the equation, so now we have to raise both sides to the 3rd power to remove the radical.
[3√(x3 + 8)]3 = (x + 2)3
x3 + 8 = (x + 2)(x + 2)(x + 2)
x3 + 8 = (x2 + 4x + 4)(x + 2)
x3 + 8 = x3 + 6x2 + 12x + 8
8 = 6x2 + 12x + 8
0 = 6x2 + 12x
0 = 6x(x + 2)
0 = 6x or 0 = x + 2
x = 0, -2
Check solutions in original equation.
For x = 0, 3√(03 + 8) = 0 + 2 For x = -2 3√( ( -2)3 + 8) = -2 + 2
2 = 2, 0 = 0. Therefore, 0 and -2 are both solutions to the equation.
Example: Solve √(2x + 1) + √x = 5.
First we have to isolate a radical on one side. Subtract √x from both sides of the equation.
√(2x + 1) = 5 - √x
Now square both sides to remove the radical from the left side of the equation.
[√(2x + 1)]2 = (5 - √x)2
2x + 1 = 25 - 10√x + x (Note that (5 - √x)2 = (5 - √x)(5 - √x) = 25 - 5√x - 5√x + √(x)2 )
Since we still have a radical in the equation, we need to isolate the radical again.
2x + 1 - 25 = - 10√x + x
2x - 24 = -10√x + x
2x - x - 24 = -10√x
x - 24 = -10√x
Now square both sides of the equation to remove the radical.
(x - 24)2 = (-10√x)2
x2 - 48x + 576 = 100x (Notice that (x - 24)2 = (x - 24)(x - 24) and (-10√x)2 = (-10)(-10)(√x)(√x) )
x2 - 148x + 576 = 0
(x - 144)(x - 4) = 0 (Note that you can factor 576 into 2 ∙ 288 = 2 ∙ 2 ∙ 144 = 4 ∙ 144 and -4 + (-144) = -148).
x - 144 = 0 or x - 4 = 0
x = 4, 144
When substituting the values back into the original equation, 144 does not work, but 4 does. Therefore, 4 is a solution to the equation.
This guide to radical equations with examples solved in step by step fashion should ease any confusion on this topic.
how to solve radical equations. To solve equations involving radicals, we must use what is known as the power rule, which states that if x, y and n are real numbers and x = y, then xn = yn. This is important because to solve radical equations we have to remove the radical. In order to remove the radical, we must raise the radical to the index of the radical. For example, if we have 3√(x+3), we raise 3√(x+3) to the 3rd power to remove the radical. To remove a radical involving a square root, we raise it to the 2nd power and so on.
Sometimes when we raise both sides of an equation to the same power, we will get equations that are not equivalent. The reason for this is that their solution sets are different.
For example, if x = 4, the solution set is {4}. Now if we raise both sides to the 4th power, we get x4 = 256, with a solution set of {4, -4}. Notice the solution sets are different. The first equation is not satisfied by x = -4, as it is in the second equation. Therefore, we must always check each solution in the original equation to make sure it satisfies the equation. Any solution that doesn't satisfy the equation is discarded. Those discarded solutions are also known as extraneous solutions.
A radical equation is an equation with a radical expression. Some examples of radical equations are
√(x - 3) = 8, √(2x + 5) = √(x + 2), 3√(8x) = 9, 3√(3x + 4) = 3√(x2).
The first step in solving a radical equation is the same as for solving a linear equation. We want to isolate the radical expression to one side of the equation. Next, we raise both sides of the equation to the power that equals the index of the radical. Then we solve the equation for the variable and check all solutions to make sure they satisfy the original equation. Eliminate those solutions which don't satisfy the equation.
Example: Solve √(2x + 7) = 7.
First step is to isolate the radical to one side. The problem is set up with it already isolated. Next step is to square both sides to remove the radical.
[√(2x + 7)]2 = (7)2
2x + 7 = 49
Now solve for x.
2x = 42
x = 21.
Check to make sure 21 satisfies the original equation.
√(2x + 7) = 7
√(2∙21 + 7) = 7
√(42 + 7) = 7
√49 = 7
7 = 7. Therefore 21 is the solution to the equation.
Example: Solve √(4x - 11) + 4 = x.
First isolate the radical.
√(4x - 11) + 4 = x
√(4x - 11) + 4 - 4 = x - 4
√(4x - 11) = x - 4
Next, square both sides of the equation to remove the radical.
[√(4x - 11)]2 = (x - 4)2
4x - 11 = (x - 4)2
Now we solve the equation for x.
4x - 11 = x2 - 8x + 16 (by the FOIL method (x - 4)(x - 4) )
-11 = x2 - 12x + 16
0 = x2 - 12x + 27
Factor to get
0 = (x - 3)(x - 9) (-9 ∙ (-3) = 27 and -9 + (-3) = -12)
0 = (x - 3) or 0 = (x - 9). Therefore x = 3, 9
Check both answers to make sure they satisfy the original equation.
For x = 3, √(4 ∙ 3 - 11) + 4 = 3 For x = 9, √(4 ∙ 9 - 11) + 4 = 9
√1 + 4 = 3 √25 + 4 = 9
5 ≠ 3, 9 = 9
Therefore 9 is a solution to the equation, but 3 is not a solution.
Example: Solve 3√(x3 + 8) = x + 2.
The radical is already isolated on the left side of the equation, so now we have to raise both sides to the 3rd power to remove the radical.
[3√(x3 + 8)]3 = (x + 2)3
x3 + 8 = (x + 2)(x + 2)(x + 2)
x3 + 8 = (x2 + 4x + 4)(x + 2)
x3 + 8 = x3 + 6x2 + 12x + 8
8 = 6x2 + 12x + 8
0 = 6x2 + 12x
0 = 6x(x + 2)
0 = 6x or 0 = x + 2
x = 0, -2
Check solutions in original equation.
For x = 0, 3√(03 + 8) = 0 + 2 For x = -2 3√( ( -2)3 + 8) = -2 + 2
2 = 2, 0 = 0. Therefore, 0 and -2 are both solutions to the equation.
Example: Solve √(2x + 1) + √x = 5.
First we have to isolate a radical on one side. Subtract √x from both sides of the equation.
√(2x + 1) = 5 - √x
Now square both sides to remove the radical from the left side of the equation.
[√(2x + 1)]2 = (5 - √x)2
2x + 1 = 25 - 10√x + x (Note that (5 - √x)2 = (5 - √x)(5 - √x) = 25 - 5√x - 5√x + √(x)2 )
Since we still have a radical in the equation, we need to isolate the radical again.
2x + 1 - 25 = - 10√x + x
2x - 24 = -10√x + x
2x - x - 24 = -10√x
x - 24 = -10√x
Now square both sides of the equation to remove the radical.
(x - 24)2 = (-10√x)2
x2 - 48x + 576 = 100x (Notice that (x - 24)2 = (x - 24)(x - 24) and (-10√x)2 = (-10)(-10)(√x)(√x) )
x2 - 148x + 576 = 0
(x - 144)(x - 4) = 0 (Note that you can factor 576 into 2 ∙ 288 = 2 ∙ 2 ∙ 144 = 4 ∙ 144 and -4 + (-144) = -148).
x - 144 = 0 or x - 4 = 0
x = 4, 144
When substituting the values back into the original equation, 144 does not work, but 4 does. Therefore, 4 is a solution to the equation.
This guide to radical equations with examples solved in step by step fashion should ease any confusion on this topic.
Monday, February 11, 2013
In statistics, a measure of central tendency is a value used measure position in
the center of the data. This measure then is used to describe the set of data by
the central value. The measure of central tendency most familiar to people is
the mean, which is the average value of the set of data. Other measures of
central tendency are the median and the mode.
All three measures may be used to describe a set of data, but under certain circumstances, one measure will be better suited to use than the others. What are the definitions of mean, median, and mode and what circumstances dictate which is most appropriate to use?
Mean
The mean is the most commonly used measure of central tendency. One can think of the mean as the "average", which is found by adding the values of the data and dividing by the number of values. For example, if the set of data is {2, 5, 6, 8, 8, 10, 11, 11, 12} then the mean is (2 + 5 + 6 + 8 + 8 + 10 + 11 + 11 + 12)/9 = 8.1.
The mean calculated above is the sample mean, which differs from the population mean. Many times the mean will not include one of the values in the data, but in its calculation, uses every value of the data set. A disadvantage of the mean is it is affected by outliers, which are values that are much larger or much smaller than the rest of the data values.
Median
The median is the center value in the set of data when the data is arranged from largest to smallest or smallest to largest. For example, in the above data, the middle value is 8. Suppose we added the value 9 to the data, the set would look like {2, 5, 6, 8, 8, 9, 10, 11, 11, 12}. Then the middle values are 8 and 9. The values 2, 5, 6, and 8 would fall below and 10, 11, 11, and 12 would fall above. Then the median is the average of the middle values, which would be (8 + 9)/2 = 8.5.
An advantage of the median is it is not affected by outliers. For example, if the data set above had a value of 99 instead of 12, the median would still be 8.5.
Mode
The mode is the value in the data set that occurs most frequently. In the data set used above, the mode is 8 and 11, since 8 and 11 both occur twice in the set.
The mode is generally not used with continuous data, such as time, weight. For instance, when comparing data for time of competitors in a marathon, it's extremely unlikely that two runners will have the exact same time.
Another problem with the mode as a measure of central tendency is when the mode is a value that is not close to the rest of the data. For example, if the data set is {1, 5, 7, 8, 10, 11, 44, 44, 44}, the mode is 44. But that is obviously not representative of the center of the data set.
Which measure of central tendency is best to use?
The only situation where it's best to use the mode is when the dealing with nominal data. This would be data in which there are categories and the number of data values are the frequency under each category. For example, if we are going to classify where people live in a certain state by city. Another example would be a car dealer classifying their types of cars into categories such as sports cars, vans, mid size sedans, etc.
The mean is best used with interval or ratio data that is not skewed. Data is skewed when a large number of values tend to be on the upper end or lower end of the data. When the data is skewed or dealing with ordinal data, then the median is best used.
If the distribution is normal, which is symmetric about the mean, then then mean, median and mode will be identical. The normal distribution is represented by the classic bell-shaped curve.
Having read this article, understanding the measures of central tendency on which are best to use should be much more clear to those who had confusion on this topic.
All three measures may be used to describe a set of data, but under certain circumstances, one measure will be better suited to use than the others. What are the definitions of mean, median, and mode and what circumstances dictate which is most appropriate to use?
Mean
The mean is the most commonly used measure of central tendency. One can think of the mean as the "average", which is found by adding the values of the data and dividing by the number of values. For example, if the set of data is {2, 5, 6, 8, 8, 10, 11, 11, 12} then the mean is (2 + 5 + 6 + 8 + 8 + 10 + 11 + 11 + 12)/9 = 8.1.
The mean calculated above is the sample mean, which differs from the population mean. Many times the mean will not include one of the values in the data, but in its calculation, uses every value of the data set. A disadvantage of the mean is it is affected by outliers, which are values that are much larger or much smaller than the rest of the data values.
Median
The median is the center value in the set of data when the data is arranged from largest to smallest or smallest to largest. For example, in the above data, the middle value is 8. Suppose we added the value 9 to the data, the set would look like {2, 5, 6, 8, 8, 9, 10, 11, 11, 12}. Then the middle values are 8 and 9. The values 2, 5, 6, and 8 would fall below and 10, 11, 11, and 12 would fall above. Then the median is the average of the middle values, which would be (8 + 9)/2 = 8.5.
An advantage of the median is it is not affected by outliers. For example, if the data set above had a value of 99 instead of 12, the median would still be 8.5.
Mode
The mode is the value in the data set that occurs most frequently. In the data set used above, the mode is 8 and 11, since 8 and 11 both occur twice in the set.
The mode is generally not used with continuous data, such as time, weight. For instance, when comparing data for time of competitors in a marathon, it's extremely unlikely that two runners will have the exact same time.
Another problem with the mode as a measure of central tendency is when the mode is a value that is not close to the rest of the data. For example, if the data set is {1, 5, 7, 8, 10, 11, 44, 44, 44}, the mode is 44. But that is obviously not representative of the center of the data set.
Which measure of central tendency is best to use?
The only situation where it's best to use the mode is when the dealing with nominal data. This would be data in which there are categories and the number of data values are the frequency under each category. For example, if we are going to classify where people live in a certain state by city. Another example would be a car dealer classifying their types of cars into categories such as sports cars, vans, mid size sedans, etc.
The mean is best used with interval or ratio data that is not skewed. Data is skewed when a large number of values tend to be on the upper end or lower end of the data. When the data is skewed or dealing with ordinal data, then the median is best used.
If the distribution is normal, which is symmetric about the mean, then then mean, median and mode will be identical. The normal distribution is represented by the classic bell-shaped curve.
Having read this article, understanding the measures of central tendency on which are best to use should be much more clear to those who had confusion on this topic.
Friday, February 8, 2013
There are four types of conic sections, one of which is an ellipse. An
ellipse can sometimes resemble a circle or will be more long and narrow,
almost oval shaped. The ellipse is a set of all points in a single
plane where the total distance from two fixed points is constant.
The equation of the ellipse in standard form that is symmetric with both axes with a center of (0, 0) is given by (x 2)/a 2+ (y 2)/ b 2 = 1 where a > 0 and b > 0. The intercepts of the graph are (a , 0), (-a, 0), (0, b) and (0, -b). In an ellipse that is elongated horizontally, a > b. This is easy to remember because the larger number will be under the x2 and the elongation is along the x- axis. If a < b, the ellipse is elongated vertically. This is also easy to see since the larger number will be under the y2 and the elongation is along the y- axis. The foci (plural of focus) are at (c, 0) and (-c, 0), where c2 = a2 - b2 . The vertices are the endpoints of each axis. The line segment joining the vertices of the elongated side is called the major axis and the line segment joining the vertices of the shorter side is called the minor axis. The following examples will illustrate all of the points above about the ellipse centered at (0, 0).
Example:
Graph: (x 2)/25 + (y 2)/9 = 1.
We will first plot the intercepts by solving for a and b. From the standard equation of the ellipse, we know that a2 = 25 and b2 = 9. Therefore a = 5 and b = 3. The intercepts are then (5, 0), (-5, 0), (0, 3) and (0, -3). Since a > b we know that major axis is the x- axis and the minor axis is the y- axis. We calculate the value of c from c2 = b2 - a2 to plot the foci. Therefore c2 = 25 - 9 = 16, so c = 4. The foci are at (4, 0) and (-4, 0). We can draw the ellipse through the 4 points or we can add more points by substituting values in for x or y. For practical purposes, there is no need to add extra points
Example: Graph 36x2 + y2 = 36.
In this example, we must first notice that the equation is not in standard form. To get the equation in standard form, we must divide the entire equation by 36. This will set the equation equal to 1.
(36 x 2 + y 2 = 36)/36 = x 2 + (y 2)/36 = 1.
Now we can plot the intercepts by solving for a and b. We know that a2 = 1 and b2 = 36, therefore a = 1 and b = 6. The intercepts are (1, 0), (-1, 0), (0, 6) and (0, -6). Since a < b, we know that the major axis is the y- axis and the minor axis is the x- axis. We calculate the value of c from c2 = b2 - a2 to plot the foci. Therefore c2 = 36 -1 = 35, therefore c ≈ 5.9 and the foci are (0, 5.9) and (0, -5.9).
If the equation is in the form (x 2)/b 2 + (y 2)/a 2 = 1, then the major axis is the y- axis. Oftentimes an ellipse will not be centered at the origin. The equation of an ellipse in standard form with the center at (h, k) is given by
( x - h ) 2/a 2 + ( y - k ) 2 /b 2 = 1, where a > 0 and b > 0.
Example: Graph (x - 2) 2/ 9 + ( y - 4) 2/ 16 = 1.
We will first plot the vertices by solving for a and b. Since a2 = 9 and b2 = 16, we know that a = 3 and b = 4. The vertices are 3 units in both directions from the x- coordinate of the center and 4 units in both directions from the y- coordinate of the center. The center is at (2, 4), therefore the vertices are (5, 4), (-1, 4), (2, 8) and (2, 0). We know that the y- axis is the major axis and the x- axis is the minor axis. We calculate the value of c from c2 = a2 - b2 to plot the foci. Therefore c2 = 16 - 9 = 7, so c ≈ 2.65. Remember that the foci are along the major axis, so we move c units in either direction from the center along the y- axis. The foci are (2, 4 + 2.65) and (2, 4 - 2.65) or (2, 6.65) and (2, 1.35).
Sometimes information about an ellipse is given and you must write the equation.
Example: Find the equation of an ellipse in standard form with foci (6, -2) and (-2, -2) and major axis of length 10.
Since the y- coordinates of the foci are the same, we know the major axis is the x- axis. Recall that the foci are the same distance from the center. Since the foci are 8 units apart (6 - (-2) = 8), we move four units from either focus point to get the center (6 -4, -2) = (2, -2) or (-2 + 4, -2) = (2, -2). Therefore c = 4 and the center of the ellipse is at (2, -2). Since x is the major axis and the length of the axis is 10, we know that a = 5. Knowing a and c enables us to find b. We know that c2 = a2 - b2, therefore 42 = 52 - b2 and b = 3.
From the center (2, -2) we can find the vertices using a = 5 and b = 3. Find the vertices by moving 5 units in both directions along the x- axis from the center and 3 units in both directions along the y- axis from the center. Therefore the vertices are (7, -2), (-3, -2), (2, 1) and (2, -5). Substitute (2, -2) for (h, k), 5 for a and 3 for b to get the equation as ( x - 2) 2/ 25 + ( y + 2) 2/ 9 = 1.
This guide on the ellipse should assist any student having difficulty understanding how to graph and write the equation of an ellipse.
The equation of the ellipse in standard form that is symmetric with both axes with a center of (0, 0) is given by (x 2)/a 2+ (y 2)/ b 2 = 1 where a > 0 and b > 0. The intercepts of the graph are (a , 0), (-a, 0), (0, b) and (0, -b). In an ellipse that is elongated horizontally, a > b. This is easy to remember because the larger number will be under the x2 and the elongation is along the x- axis. If a < b, the ellipse is elongated vertically. This is also easy to see since the larger number will be under the y2 and the elongation is along the y- axis. The foci (plural of focus) are at (c, 0) and (-c, 0), where c2 = a2 - b2 . The vertices are the endpoints of each axis. The line segment joining the vertices of the elongated side is called the major axis and the line segment joining the vertices of the shorter side is called the minor axis. The following examples will illustrate all of the points above about the ellipse centered at (0, 0).
Example:
Graph: (x 2)/25 + (y 2)/9 = 1.
We will first plot the intercepts by solving for a and b. From the standard equation of the ellipse, we know that a2 = 25 and b2 = 9. Therefore a = 5 and b = 3. The intercepts are then (5, 0), (-5, 0), (0, 3) and (0, -3). Since a > b we know that major axis is the x- axis and the minor axis is the y- axis. We calculate the value of c from c2 = b2 - a2 to plot the foci. Therefore c2 = 25 - 9 = 16, so c = 4. The foci are at (4, 0) and (-4, 0). We can draw the ellipse through the 4 points or we can add more points by substituting values in for x or y. For practical purposes, there is no need to add extra points
Example: Graph 36x2 + y2 = 36.
In this example, we must first notice that the equation is not in standard form. To get the equation in standard form, we must divide the entire equation by 36. This will set the equation equal to 1.
(36 x 2 + y 2 = 36)/36 = x 2 + (y 2)/36 = 1.
Now we can plot the intercepts by solving for a and b. We know that a2 = 1 and b2 = 36, therefore a = 1 and b = 6. The intercepts are (1, 0), (-1, 0), (0, 6) and (0, -6). Since a < b, we know that the major axis is the y- axis and the minor axis is the x- axis. We calculate the value of c from c2 = b2 - a2 to plot the foci. Therefore c2 = 36 -1 = 35, therefore c ≈ 5.9 and the foci are (0, 5.9) and (0, -5.9).
If the equation is in the form (x 2)/b 2 + (y 2)/a 2 = 1, then the major axis is the y- axis. Oftentimes an ellipse will not be centered at the origin. The equation of an ellipse in standard form with the center at (h, k) is given by
( x - h ) 2/a 2 + ( y - k ) 2 /b 2 = 1, where a > 0 and b > 0.
Example: Graph (x - 2) 2/ 9 + ( y - 4) 2/ 16 = 1.
We will first plot the vertices by solving for a and b. Since a2 = 9 and b2 = 16, we know that a = 3 and b = 4. The vertices are 3 units in both directions from the x- coordinate of the center and 4 units in both directions from the y- coordinate of the center. The center is at (2, 4), therefore the vertices are (5, 4), (-1, 4), (2, 8) and (2, 0). We know that the y- axis is the major axis and the x- axis is the minor axis. We calculate the value of c from c2 = a2 - b2 to plot the foci. Therefore c2 = 16 - 9 = 7, so c ≈ 2.65. Remember that the foci are along the major axis, so we move c units in either direction from the center along the y- axis. The foci are (2, 4 + 2.65) and (2, 4 - 2.65) or (2, 6.65) and (2, 1.35).
Sometimes information about an ellipse is given and you must write the equation.
Example: Find the equation of an ellipse in standard form with foci (6, -2) and (-2, -2) and major axis of length 10.
Since the y- coordinates of the foci are the same, we know the major axis is the x- axis. Recall that the foci are the same distance from the center. Since the foci are 8 units apart (6 - (-2) = 8), we move four units from either focus point to get the center (6 -4, -2) = (2, -2) or (-2 + 4, -2) = (2, -2). Therefore c = 4 and the center of the ellipse is at (2, -2). Since x is the major axis and the length of the axis is 10, we know that a = 5. Knowing a and c enables us to find b. We know that c2 = a2 - b2, therefore 42 = 52 - b2 and b = 3.
From the center (2, -2) we can find the vertices using a = 5 and b = 3. Find the vertices by moving 5 units in both directions along the x- axis from the center and 3 units in both directions along the y- axis from the center. Therefore the vertices are (7, -2), (-3, -2), (2, 1) and (2, -5). Substitute (2, -2) for (h, k), 5 for a and 3 for b to get the equation as ( x - 2) 2/ 25 + ( y + 2) 2/ 9 = 1.
This guide on the ellipse should assist any student having difficulty understanding how to graph and write the equation of an ellipse.
Wednesday, February 6, 2013
When solving equations for angles between [0, 360] (measured in degrees), remember which quadrants sin and cos are positive and which quadrants they are negative.
First quadrant : sin +
cos +
Second quadrant: sin +
cos -
Third quadrant: sin -
cos -
Fourth quadrant: sin-
cos +
Remember when solving equations involving sec and csc, that sec = 1/cos and csc = 1/sin. Easiest to change back to sin and cos and then finding the angle.
First quadrant : sin +
cos +
Second quadrant: sin +
cos -
Third quadrant: sin -
cos -
Fourth quadrant: sin-
cos +
Remember when solving equations involving sec and csc, that sec = 1/cos and csc = 1/sin. Easiest to change back to sin and cos and then finding the angle.
Monday, February 4, 2013
When information is given about sides and angles of a triangle, sometimes there are no triangles possible with the given information, sometimes there is 1 triangle possible and other times 2 triangles possible.
Information given
C = 55 degrees, a = 10, c = 14
Use the law of sine to find angle a
c/Sin C = a/Sin A
14/sin 55 = 10/sin A
0.585 = Sin A
A = 36 degrees
So we know there is at least 1 triangle with angles, 55, 36, and 89 degrees.
But angle A could also be 180 - 36 = 144 since sine is positive in both the first and second quadrants. But that would be impossible because 144 + 55 (measure of angle C) = 189 which is greater than 180 degrees that all 3 triangles must add up to.
So therefore there is only 1 triangle possible.
If in the first step when solving for angle A, you get a value outside of -1 to 1, then there is no solution and no triangle exists.
If 180 - A plus measure of angle C is less than 180, then there are two triangles possible.
Information given
C = 55 degrees, a = 10, c = 14
Use the law of sine to find angle a
c/Sin C = a/Sin A
14/sin 55 = 10/sin A
0.585 = Sin A
A = 36 degrees
So we know there is at least 1 triangle with angles, 55, 36, and 89 degrees.
But angle A could also be 180 - 36 = 144 since sine is positive in both the first and second quadrants. But that would be impossible because 144 + 55 (measure of angle C) = 189 which is greater than 180 degrees that all 3 triangles must add up to.
So therefore there is only 1 triangle possible.
If in the first step when solving for angle A, you get a value outside of -1 to 1, then there is no solution and no triangle exists.
If 180 - A plus measure of angle C is less than 180, then there are two triangles possible.
Saturday, February 2, 2013
You can approximate the area under the curve by taking the upper sum and the lower sum and then getting the limit of each as n approaches infinity.
to get the upper sum..
f(Mi)* delta X
If the interval over which we are integrating is [a,b], then delta X = (b - a)/n, where n is the number of partitions
Mi = 2i/n for upper limit
and 2(i -1)/n for lower limit
f(x) = x^2, n = 4, interval is [0, 2]
Mi= 2i/4 = i/2
f(Mi) = (i/2)^2
= i^2/4
delta X = 1/2
Use special summation formulas for sum of f(Mi)* delta (x)
The area will be 8/3
The easier way is just to take the integral of x^2 from 0 to 2
(1/3)x^3 evaluated at 2 = 8/3
(1/3)x^3 evaluated at 0 = 0
8/3 - 0 = 8/3
to get the upper sum..
f(Mi)* delta X
If the interval over which we are integrating is [a,b], then delta X = (b - a)/n, where n is the number of partitions
Mi = 2i/n for upper limit
and 2(i -1)/n for lower limit
f(x) = x^2, n = 4, interval is [0, 2]
Mi= 2i/4 = i/2
f(Mi) = (i/2)^2
= i^2/4
delta X = 1/2
Use special summation formulas for sum of f(Mi)* delta (x)
The area will be 8/3
The easier way is just to take the integral of x^2 from 0 to 2
(1/3)x^3 evaluated at 2 = 8/3
(1/3)x^3 evaluated at 0 = 0
8/3 - 0 = 8/3
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