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# Sums and Differences of Single Variable Expressions
## Simplify expressions by combining like terms.
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Sums and Differences of Single Variable Expressions
Have you ever had your picture taken at an amusement park? They have those old time costumes that you can put on and everyone stands together for a photo?
Well, at the amusement park, Kelly has decided to gather a group of friends to do just that. When she first posed the question at lunch, she had five people say that they wanted to do it.
Then later in the day, four more people wanted to join it.
There is a fee per person if you want to be in the picture.
Kelly wants to figure out the total number in simple way and include the fee. How can she do this?
To do this, Kelly can write an expression using a single variable.
Do you know how to do this?
This Concept will show you how to write single variable expressions that include sums and differences. Then you will understand how to help Kelly.
### Guidance
In the last few Concepts, you learned how to write single-variable expressions and single variable equations. Now you are going to learn to work with single-variable expressions. The first thing that you are going to learn is how to simplify an expression.
What does it mean to simplify?
To simplify means to make smaller or to make simpler. When we simplify in mathematics, we aren’t solving anything, we are just making it smaller.
How do we simplify expressions?
Sometimes, you will be given an expression using variables where there is more than one term. A term is a number with a variable. Here is an example of a term.
4x\begin{align*}4x\end{align*}
This is a term. It is a number and a variable. We haven’t been given a value for x\begin{align*}x\end{align*}, so there isn’t anything else we can do with this term. It stays the same. If we have been given a value for x\begin{align*}x\end{align*}, then we could evaluate the expression. You have already worked on evaluating expressions.
When there is more than one LIKE TERM in an expression, we can simplify the expression.
What is a like term?
A like term means that the terms in question use the same variable.
4x\begin{align*}4x\end{align*} and 5x\begin{align*}5x\end{align*} are like terms. They both have x\begin{align*}x\end{align*} as the variable. They are alike.
6x\begin{align*}6x\end{align*} and 2y\begin{align*}2y\end{align*} are not like terms. One has an x\begin{align*}x\end{align*} and one has a y\begin{align*}y\end{align*}. They are not alike.
We can simplify expressions with like terms. We can simplify the sums and differences of expressions with like terms. Let’s start with sums.
5x+7x
First, we look to see if these terms are alike. Both of them have an x\begin{align*}x\end{align*}, so they are alike.
Next, we can simplify them by adding the numerical part of the terms together. The x\begin{align*}x\end{align*} stays the same.
5x+7x12x
You can think of the x\begin{align*}x\end{align*} as a label that lets you know that the terms are alike.
7x+2x+5y
First, we look to see if the terms are alike. Two of the terms have x\begin{align*}x\end{align*}’s and one has a y\begin{align*}y\end{align*}. The two with the x\begin{align*}x\end{align*}’s are alike. The one with the y\begin{align*}y\end{align*} is not alike. We can simplify the ones with the x\begin{align*}x\end{align*}’s.
Next, we simplify the like terms.
7x+2x=9x
We can’t simplify the 5y\begin{align*}5y\end{align*} so it stays the same.
9x+5y
This is our answer.
We can also simplify expressions with differences and like terms.
9y2y
First, you can see that these terms are alike because they both have y\begin{align*}y\end{align*}’s. We simplify the expression by subtracting the numerical part of the terms.
9 - 2 = 7
Our answer is 7y\begin{align*}7y\end{align*}.
Sometimes you can combine like terms that have both sums and differences in the same problem.
8x3x+2y+4y
We begin with the like terms.
8x3x2y+4y=5x=6y
Next, we put it all together.
5x+6y
This is our answer.
Remember that you can only combine terms that are alike!!!
Use your notebook and pencil to take some notes on how to identify like terms.
Try a few of these on your own. Simplify the expressions by combining like terms.
#### Example A
7z+2z+4z\begin{align*}7z+2z+4z\end{align*}
Solution: 13z\begin{align*}13z\end{align*}
#### Example B
25y13y\begin{align*}25y-13y\end{align*}
Solution: 12y\begin{align*}12y\end{align*}
#### Example C
7x+2x+4a\begin{align*}7x+2x+4a\end{align*}
Solution: 9x+4a\begin{align*}9x + 4a\end{align*}
Here is the original problem once again.
Well, at the amusement park, Kelly has decided to gather a group of friends to do just that. When she first posed the question at lunch, she had five people say that they wanted to do it.
Then later in the day, four more people wanted to join it.
There is a fee per person if you want to be in the picture.
Kelly wants to figure out the total number in simple way and include the fee. How can she do this?
To do this, Kelly can write an expression using a single variable.
Do you know how to to do this?
Now we can use the information that is provided in the dilemma to write a single variable expression.
First, 5 people wanted to be in the picture.
There is also a fee per person. Kelly doesn't know this amount. It is our variable x\begin{align*}x\end{align*}
5x\begin{align*}5x\end{align*}
Then 4 more people wanted to join the picture. The fee per person applies to them too.
5x+4x\begin{align*}5x + 4x\end{align*}
If we combine like terms, 9x\begin{align*}9x\end{align*} is the expression Kelly can use to figure out the total cost of the picture. Once she knows the cost per person, she will be able to substitute that into the given expression and solve for the total cost.
But wait, this is information we can use in another Concept.
### Guided Practice
Here is one for you to try on your own.
5x+2x1x+6y4y\begin{align*}5x + 2x - 1x + 6y - 4y\end{align*}
To simplify this expression, we simply combine the terms that are alike.
5x+2x1x=6x\begin{align*}5x + 2x - 1x = 6x\end{align*}
6y4y=2y\begin{align*}6y - 4y = 2y\end{align*}
Now we put those simplified terms together.
6x+2y\begin{align*}6x + 2y\end{align*}
This is our answer.
### Video Review
Here is a video for review.
### Explore More
Directions: Simplify the following expressions by combining like terms. If the expression is already in simplest form please write “already in simplest form.”
1. 4x+6x\begin{align*}4x+6x\end{align*}
2. 8y+5y\begin{align*}8y+5y\end{align*}
3. 9z+2z\begin{align*}9z+2z\end{align*}
4. 8x+2y\begin{align*}8x+2y\end{align*}
5. 7y+3y+2x\begin{align*}7y+3y+2x\end{align*}
6. 9xx\begin{align*}9x-x\end{align*}
7. \begin{align*}12y-3y\end{align*}
8. \begin{align*}22x-2y\end{align*}
9. \begin{align*}78x-10x\end{align*}
10. \begin{align*}22y-4y\end{align*}
11. \begin{align*}16x - 5x + 1x - 12y + 2y\end{align*}
12. \begin{align*}26x - 15x + 12x - 14y + 2y\end{align*}
13. \begin{align*}36x - 5x + 11x - 1x + 2y\end{align*}
14. \begin{align*}26x - 25x + 12x - 13y + 2y\end{align*}
15. \begin{align*}29x - 25x + 18x - 12x + 12y + 3y\end{align*}
### Vocabulary Language: English
Expression
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Simplify
Simplify
To simplify means to rewrite an expression to make it as "simple" as possible. You can simplify by removing parentheses, combining like terms, or reducing fractions. |
# How are fractions and division related?
One way to think of a fraction is as a division that hasn't been done yet. Why do we even use fractions? Why don't we just divide the two numbers and use the decimal instead? In this day of cheap calculators, that's a very good question.
Fractions were invented long before decimal numbers, as a way of showing portions less than 1, and they're still hanging around.
They're used in cooking, in building, in sewing, in the stock market – they're everywhere, and we need to understand them.
• Just to review, the number above the bar is called the numerator, and the number below the bar is called the denominator.
• We can read this fraction as three-fourths, three over four, or three divided by four.
Every fraction can be converted to a decimal by dividing. If you use the calculator to divide 3 by 4, you'll find that it is equal to 0.75.
Here are some other fractions and their decimal equivalents. Remember, you can find the decimal equivalent of any fraction by dividing.
Here are some terms that are very important when working with fractions.
Proper fraction When the numerator is less than the denominator, we call the expression a proper fraction. These are some examples of proper fractions.
1. Improper fraction An improper fraction occurs when the numerator is greater than or equal to the denominator. These are some examples of improper fractions:
2. Mixed number When an expression consists of a whole number and a proper fraction, we call it a mixed number. Here are some examples of mixed numbers:
We can convert a mixed number to an improper fraction. First, multiply the whole number by the denominator of the fraction. Then, add the numerator of the fraction to the product. Finally, write the sum over the original denominator. In this example, since three thirds is a whole, the whole number 1 is three thirds plus one more third, which equals four thirds.
Convert 1-1/3 to an improper fraction:
Equivalent fractions There are many ways to write a fraction of a whole. Fractions that represent the same number are called equivalent fractions.
This is basically the same thing as equal ratios. For example, ½, 2/4, and 4/8 are all equivalent fractions. To find out if two fractions are equivalent, use a calculator and divide.
If the answer is the same, then they are equivalent.
Reciprocal When the product of two fractions equals 1, the fractions are reciprocals. Every nonzero fraction has a reciprocal. It's easy to determine the reciprocal of a fraction since all you have to do is switch the numerator and denominator–just turn the fraction over. Here's how to find the reciprocal of three-fourths.
To find the reciprocal of a whole number, just put 1 over the whole number. For example, the reciprocal of 2 is ½.
## Dividing Fractions – Definition with Examples
A fraction is part of a whole number. It has two parts – a numerator and a denominator.
Dividing a fraction
Dividing a fraction by another fraction is the same as multiplying the fraction by the reciprocal (inverse) of the other. We get the reciprocal of a fraction by interchanging its numerator and denominator.
For example, the reciprocal of 25 is52
• Consider the following example:
• 1 2 ÷ 1 3
• Step 1:
• Find the reciprocal of the second fraction (divisor).
• Reciprocal of 1 3 is 3 1 or 3
• Step 2:
• Multiply the first fraction (dividend) by the reciprocal of the second fraction (divisor).
• 13 x 31
• Step 3:
• Multiply the numerators and denominators of the fractions.
• 12 x 31 = 1x32x1 = 32
• Dividing Fractions Song
• Take the fractions for the division to apply
• Flip the second one and then multiply
• In the end, you need to simplify.
• So next time you divide the fractions don’t forget to apply
• The simple rule is to flip and multiply.
1. Dividing a Fraction by a Whole Number
2. Follow a simple rule that a number divided by 1 is the number itself.
3. Steps to divide a fraction by a whole number:
• Convert the whole number into a fraction by using a denominator 1.
• Flip this number.
• Multiply by the fraction.
• Simplify the result, if needed.
Take an example, Divide 35 by 5
• Convert 5 into a fraction =51
• Flip 51by get 15
• Multiply the fractions: 35 x 15 = 325
Fun-facts Why do we turn the fraction upside down ? This is because the inverse operation of division is multiplication.
## Dividing Fractions
Turn the second fraction upside down, then multiply.
### There are 3 Simple Steps to Divide Fractions:
• Step 1. Turn the second fraction upside down (it becomes a reciprocal):
• 1 6 becomes 6 1
• Step 2. Multiply the first fraction by that reciprocal:
(multiply tops …)
1 2 × 6 1 = 1 × 6 2 × 1 = 6 2
(… multiply bottoms)
Step 3. Simplify the fraction:
6 2 = 3
### With Pen and Paper
1. And here is how to do it with a pen and paper (press the play button):
3. ♫ “Dividing fractions, as easy as pie,
Flip the second fraction, then multiply.
And don't forget to simplify,
4. Before it's time to say goodbye”
Another way to remember is: “leave me, change me, turn me over”
• 20 divided by 5 is asking “how many 5s in 20?” (=4) and so:
• 1 2 ÷ 1 6 is really asking:
• how many 1 6 s in 1 2 ?
Now look at the pizzas below … how many “1/6th slices” fit into a “1/2 slice”?
How many in ? Answer: 3
So now you can see why 1 2 ÷ 1 6 = 3
In other words “I have half a pizza, if I divide it into one-sixth slices, how many slices is that?”
1. Step 1. Turn the second fraction upside down (the reciprocal):
2. 1 4 becomes 4 1
3. Step 2. Multiply the first fraction by that reciprocal:
4. 1 8 × 4 1 = 1 × 4 8 × 1 = 4 8
5. Step 3. Simplify the fraction:
6. 4 8 = 1 2
### Fractions and Whole Numbers
• What about division with fractions and whole numbers?
• Make the whole number a fraction, by putting it over 1.
• Then continue as before.
1. Make 5 into 5 1 :
2. 2 3 ÷ 5 1
3. Then continue as before.
4. Step 1. Turn the second fraction upside down (the reciprocal):
5. 5 1 becomes 1 5
6. Step 2. Multiply the first fraction by that reciprocal:
7. 2 3 × 1 5 = 2 × 1 3 × 5 = 2 15
8. Step 3. Simplify the fraction:
9. The fraction is already as simple as it can be.
• Make 3 into 3 1 :
• 3 1 ÷ 1 4
• Then continue as before.
• Step 1. Turn the second fraction upside down (the reciprocal):
• 1 4 becomes 4 1
• Step 2. Multiply the first fraction by that reciprocal:
• 3 1 × 4 1 = 3 × 4 1 × 1 = 12 1
• Step 3. Simplify the fraction:
• 12 1 = 12
## Explainer: Dividing Fractions
In this explainer, we will learn how to divide proper fractions by proper fractions.
For this purpose, we are going to start with dividing fractions by whole numbers and vice versa before going into dividing fractions by fractions. We are going to look at the two different meanings such divisions can have, and, with the help of diagrams, we will find how to compute them. This will lead us to a general method to compute divisions by fractions.
Let us first recall that a fraction compares a part to a whole and describes what we call a proportion. The denominator of the fraction is the number of equal shares (or “portions”) the whole is split into, while the numerator is the number of these shares that make the part we are considering.
Now we are going to start with a very simple example of a division of a fraction by a whole number.
We are considering the fraction 23 here; that is, a whole has been divided into three equal shares, and the part we are considering is made of two of these shares.
We now want to divide the part (the shaded area in the diagram) into three parts. By dividing each third in three parts, we have our whole divided into 9 equal shares, and we see that 23÷3 is two of them, that is, two-ninths: 23÷3=29.
Notice here how dividing a fraction is further splitting the whole; the denominator, which is the number of shares, is indeed multiplied! So, dividing a half in three is creating sixths of the whole.
Let us look at another example where we can use this.
Which of the following division expressions has a quotient of 12?
1. 78÷19
2. 23÷16
3. 12÷57
4. 38÷34
5. 56÷47
We are looking for a division whose result is one-half; in other words, we are looking for a division where the dividend is half the divisor. In all the options given, the dividends and divisors are fractions. So, we are looking for a fraction divided by another fraction that is double its value.
Let us take two examples to see how a fraction can be one-half of another. First, 25 is half of 45 as both fractions have the same denominator but the numerator of the first is half that of the second.
And, second, 16 is half of 13 since both have the same numerator but the denominator has been divided by two in the second fraction, meaning that the number of shares has been halved, so their size has been indeed doubled.
Therefore, we are looking for one of these situations here: either the denominators are the same and the numerator of the dividend is half that of the divisor ????????÷2????????, or the numerators are the same and the denominator of the dividend is double that of the divisor ????2????÷????????.
Hence, the answer is 38÷34 (option D).
In the next example, we are looking at the division of a whole number by a fraction.
Evaluate 10÷12.
We are asked here to find the result of dividing 10 by 12. This can be understood as finding how many times there is 12 in 10.
There are two halves in 1, so there will be 10 times as many in 10; that is, 10×2=20.
Hence, 10÷12=10×2=20.
Let us look now at dividing a fraction by a fraction. For instance, we want to find 1213÷313, that is, how many 313 are in 1213. We can use a diagram to help us visualize this.
We see that, in this case, because both fractions have the same denominator, this division is simply 12 divided by 3: there are 4 times 313 in 1213.
That is, we have 1213÷313=12÷3=4.
From this, we see that a strategy to divide a fraction by another fraction with a different denominator is to rewrite the fractions so that they have the same denominator and then divide the numerators. It is a good strategy, for example, to compute 78÷34 since 34 can be easily rewritten as 68, and so 78÷34=78÷68=76.
See also: What is a möbius strip and how can you make one?
In the following, we are going to discover a general method equivalent to rewriting the fractions with the same denominator and dividing the numerators, but simpler.
So far, we have interpreted the division by a fraction as a measurement division: how many of this fraction are there in a given number?
There is another way to envision such a division. For instance, imagine that Oscar has written 12 pages for his assignment and he says that it is three-quarters of what is required. How can we find how many pages he is required to write for his assignment?
Let us use a diagram to help us.
We know that 12 is three-quarters 34 of what is required. So, if we start with a rectangle to represent what is required, we need to split it in 4 and the 12 pages already written are three of these shares.
Now, let us see how to find the number of pages required. First, we find the value of each quarter by dividing the 12 by 3. This gives 4. The number of pages required is 4 times this value (which is 4 quarters); that is, 4×4=16.
Now, we were told that 12 pages are three-quarters of what is required. We can write that as 12=34×.pagesrequired
Now, think of the complementarity between division and multiplication sentences. For instance, 2×5=10 (two groups of 5 make 10) means that 10÷2=5 (10 can be split in 2 groups of 5). Hence, we can rewrite the above equation as 12÷34=.pagesrequired
What we have found by reasoning on the diagram was nothing else as the result of 12÷34. And the stages were
• divide 12 by 3,
• multiply the result by 4.
We see that 12÷34=123×4. This equation can be rewritten as 12÷34=12×43.
We have found here a very important result: dividing by a fraction is mathematically the same as multiplying by the multiplicative inverse, or reciprocal, of the fraction (i.e., the numerator and denominator have been swapped over).
Let us look at a division of a fraction by a fraction. For instance, let us find a number knowing that four-fifths 45 of this unknown number equals 23. That is, 45×=23unknownnumber. For this, we are going to use three diagrams.
In the top diagram, 23 is represented in dark green, and the bigger rectangle represents our whole. We know that the dark green rectangle represents 45 of the unknown number. So, in as the first stage, we need to split our 23 in 4 shares of value ???? (second diagram): ????=23÷4=16.
Note that, without any diagram, we would have probably written ????=23÷4=212 and then simplified this fraction to 16. This is of course completely correct.
• The number we are looking for, called ????, is made of 5 of these shares (third diagram): ????=5×16=56.
• The unknown number is found to be five-sixths of our whole.
• Here, again, the two stages of the division by 45 were
• divide by 4,
• multiply the result by 5.
We have found that 23÷45=23×54=56.
Let us now interpret this division as a measurement division; that is, how many 45 are there in 23?
For this, we have already seen that it is helpful to rewrite the fractions with the same denominator. The lowest common multiple of 3 and 5 is 15. As 23=2×53×5=1015 and 45=4×35×3=1215, we have 23÷45=1015÷1215.
This step of rewriting the dividend and divisor with the same denominator can be visualized in the diagram.
Now that both fractions have the same denominator, finding how many 1215 there are in 1015 is simply finding how many 12s there are in 10: it is given by 10÷12=1012. This number is smaller than 1 because 12 is bigger than 10.
We have found that 23÷45=1012. Let us look back at how we got these numbers, 10 and 12, from 23÷45 when we renamed the fractions with the same denominator. The 10 was given by 2×5 and the 12 by 3×4. So, we did find that 23÷45=2×53×4;
that is, 23÷45=23×54.
We can write the general method we have found for dividing fractions by fractions.
Dividing by a fraction is equivalent to dividing by the numerator of the fraction and then multiplying by its denominator: ????÷????????=????÷????×????.
In other words, it is equivalent to multiplying by its multiplicative inverse or reciprocal. Hence, when a fraction is divided by another fraction, we can write ????????÷????????=????????×????????.
Let us look at an example where we are going to use our understanding of division by fractions.
Find 12÷23.
To find the result of dividing by a fraction, we use the fact that dividing by a fraction is equivalent to multiplying by its inverse (i.e., a fraction where the numerator and the denominator have been swapped). Hence, we have 12÷23=12⋅32.
Now, we just need to multiply the numerators together and the denominators together. We find 12÷23=34.
We have found out that one half is three-quarters of two-thirds and two-thirds of three-quarters!
Our last example is a word problem.
Natalie and Mason went out to get some ice cream. Natalie had 79 pt of chocolate chip ice cream, while Mason had 67 pt of strawberry-flavored ice cream. Determine how many times as much ice cream Mason had as Natalie.
Both Natalie and Mason have got a fraction of a pint of ice cream. We want to find how many times as much ice cream Mason had as Natalie. This is given mathematically by dividing the amount of ice cream Mason had by the amount of ice cream Natalie had, that is, 67÷79.
We know that when we divide by a fraction, this is equivalent to multiplying by its reciprocal, so 67÷79=67×97=5449.
This fraction is an improper fraction: the numerator is larger than the denominator. Hence, we will express it as a mixed number: 5449=49+549=1549.
1. A fraction compares a part to a whole. The denominator of the fraction is the number of equal shares (or “portions”) the whole is split into, while the numerator is the number of these shares that make the part we are considering.
2. If we know that a number is a fraction of another number, for instance, 12 is three quarters of an unknown number, we can write 12=34⋅unknownnumber. As we know, for instance, that 6=3⋅2 is equivalent to 6÷3=2, we can write that 12÷34=unknownnumber. Reasoning with the diagram, we find that, to find this unknown number, we need to divide 12 by 3 to find the value of one quarter and then multiply this by 4 to find the unknown number. So, we have 12÷34=12⋅43 since dividing by 3 and multiplying by 4 is the same as multiplying by 43.
## How to Convert a Division to a Fraction
••• Stockbyte/Stockbyte/Getty Images
Updated April 24, 2017
By Charlotte Johnson
Division is a mathematical process in which you calculate how many times a certain value will fit into another value. This process is the opposite of multiplication. The traditional way to write division problems is with a division bracket.
Another method of writing division calculations is to use fractions. In a fraction, the top number, or numerator, is divided by the bottom number, or denominator.
You may have to convert between traditional and fractional division forms in a high school or college math class.
Write down the dividend. This is the number that appears under the division bracket. This will be the numerator in the fraction.
Draw a dividing bar under the dividend or to the right of the dividend.
Write the divisor underneath the dividing bar or to the right of the bar. The divisor is the number to the left of the division bracket. This will be the denominator.
Write an equal sign, followed by the quotient if you are required to provide the solution to the problem. If you are looking at a completed division problem in traditional form, the quotient will appear on top of the division bracket. For example, if you have 50 divide by 5, you could write this as 50/5 = 10.
Charlotte Johnson is a musician, teacher and writer with a master's degree in education. She has contributed to a variety of websites, specializing in health, education, the arts, home and garden, animals and parenting.
## Dividing Fractions
Dividing fractions can be a little tricky. It’s the only operation that requires using the reciprocal. Using the reciprocal simply means you flip the fraction over, or invert it.
• For example, the reciprocal of 2/3 is 3/2.
• After we give you the division rule, we will show you WHY you have to use the reciprocal in the first place.
• But for now…
### Here’s the Rule for Division
To divide, convert the fraction division process to a multiplication process by using the following steps.
1. Change the “÷” (division sign) to “x” (multiplication sign) and invert the number to the right of the sign.
2. Multiply the numerators.
3. Multiply the denominators.
4. Re-write your answer in its simplified or reduced form, if needed
Once you complete Step #1 for dividing fractions, the problem actually changes from division to multiplication.
### Example 1: Dividing Fractions by Fractions
1. 1/2 ÷ 1/3 = 1/2 x 3/1
2. 1/2 x 3/1 = 3/2
3. Simplified Answer is 1 1/2
### Example 2: Dividing Fractions by Whole Numbers
• 1/2 ÷ 5 = 1/2 ÷ 5/1
• (Remember to convert
whole numbers to fractions, FIRST!)
• 1/2 ÷ 5/1 = 1/2 x 1/5
• 1/2 x 1/5 = 1/10
### Example 3: Dividing Whole Numbers by Fractions
1. 6 ÷ 1/3 = 6/1 ÷ 1/3
2. (Remember to convert
whole numbers to fractions, FIRST!)
3. 6/1 ÷ 1/3 = 6/1 x 3/1
4. 6/1 x 3/1 = 18/1 = 18
Now that’s all there is to it.
The main things you have to remember when you divide is to convert whole numbers to fractions first, then invert the fraction to the right of the division sign, and change the sign to multiplication.
The “divisor” has some other considerations you should keep in mind…
### Special Notes!
• Remember to only invert the divisor.
• The divisor’s numerator or denominator can not be “zero”.
• Convert the operation to multiplication BEFORE performing any cancellations.
I promised to try to explain why the rule requires inverting the divisor.
Here goes..
### Why Dividing Fractions Requires Inverting The Divisor
• Let’s use our simple example to actually validate this strange Rule for division.
• If you really think about it, we are dividing a fraction by a fraction, which forms what is called a “complex fraction” |
# AP Statistics Curriculum 2007 Hypothesis S Mean
(Difference between revisions)
Revision as of 21:03, 6 February 2008 (view source)IvoDinov (Talk | contribs) (→Example)← Older edit Current revision as of 04:36, 7 June 2013 (view source)IvoDinov (Talk | contribs) m (→Example) (14 intermediate revisions not shown) Line 3: Line 3: === Testing a Claim about a Mean: Small Samples=== === Testing a Claim about a Mean: Small Samples=== - The [[AP_Statistics_Curriculum_2007_Hypothesis_L_Mean | previous section discussed inference on the population mean for large smaples]]. Now, we show how to do hypothesis testing about the mean for small sample-sizes. + The [[AP_Statistics_Curriculum_2007_Hypothesis_L_Mean | previous section discussed inference on the population mean for large smaples]]. Now, we show how to do hypothesis testing of the mean for small sample-sizes. ===[[AP_Statistics_Curriculum_2007_Estim_L_Mean | Background]]=== ===[[AP_Statistics_Curriculum_2007_Estim_L_Mean | Background]]=== Line 9: Line 9: * For a given small $\alpha$ (e.g., 0.1, 0.05, 0.025, 0.01, 0.001, etc.), the $(1-\alpha)100%$ Confidence interval for the mean is constructed by * For a given small $\alpha$ (e.g., 0.1, 0.05, 0.025, 0.01, 0.001, etc.), the $(1-\alpha)100%$ Confidence interval for the mean is constructed by - : $CI(\alpha): \overline{x} \pm t_{df=n-1,\alpha\over 2} {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}$ + : $CI(\alpha): \overline{x} \pm t_{df=n-1,{\alpha\over 2}} {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}$ - : and $t_{df=n-1, \alpha\over 2}$ is the [[AP_Statistics_Curriculum_2007_Normal_Critical | critical value]] for a [[AP_Statistics_Curriculum_2007_StudentsT |T-distribution]] of df=(sample size - 1) at ${\alpha\over 2}$. + : and $t_{df=n-1, {\alpha\over 2}}$ is the [[AP_Statistics_Curriculum_2007_Normal_Critical | critical value]] for a [[AP_Statistics_Curriculum_2007_StudentsT |T-distribution]] of df=(sample size - 1) at ${\alpha\over 2}$. === Hypothesis Testing about a Mean: Small Samples=== === Hypothesis Testing about a Mean: Small Samples=== Line 16: Line 16: * Alternative Research Hypotheses: * Alternative Research Hypotheses: ** One sided (uni-directional): $H_1: \mu >\mu_o$, or $H_o: \mu<\mu_o$ ** One sided (uni-directional): $H_1: \mu >\mu_o$, or $H_o: \mu<\mu_o$ - ** Double sided: $H_1: \mu \not= \mu_o$ + ** Double sided: $$H_1: \mu \not= \mu_o$$ ====Normal Process with Known Variance==== ====Normal Process with Known Variance==== * If the population is Normally distributed and we know the population variance, then the [http://en.wikipedia.org/wiki/Hypothesis_testing#Common_test_statistics Test statistics] is: * If the population is Normally distributed and we know the population variance, then the [http://en.wikipedia.org/wiki/Hypothesis_testing#Common_test_statistics Test statistics] is: - : $Z_o = {\overline{x} - \mu_o \over \sigma} \sim N(0,1)$. + : $Z_o = {\overline{x} - \mu_o \over {\sigma \over \sqrt{n}}} \sim N(0,1)$. - ====(Approximately) Nornal Process with Unknown Variance==== + ====(Approximately) Normal Process with Unknown Variance==== - * If the population is approximately Normally distributed and we do not know the population variance, then the[http://en.wikipedia.org/wiki/Hypothesis_testing#Common_test_statistics Test statistics] is: + * If the population is approximately Normally distributed and we do not know the population variance, then the [http://en.wikipedia.org/wiki/Hypothesis_testing#Common_test_statistics Test statistics] is: - : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {\overline{x} - \mu_o \over {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}})} \sim N(0,1)$. + : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {\overline{x} - \mu_o \over {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}} \sim T_{(df=n-1)}$. ===Example=== ===Example=== - Let's use again the small-sample example of the [[AP_Statistics_Curriculum_2007_Estim_L_Mean | ''number of sentences per advertisement'']], where we measure of readability for magazine advertisements. A random sample of the number of sentences found in 10 magazine advertisements is listed below. Suppose we want to test at $\alpha=0.01$ a null hypothesis: $H_o: \mu=12$ against a one-sided research alternative hypothesis: $H_1: \mu > 12$. Recall that we had the following [[AP_Statistics_Curriculum_2007_Estim_S_Mean#Example |sample statistics: '''sample-mean=22.1, sample-variance=737.88 and sample-SD=27.16390579''']] for these data. + Let's use again the small-sample example of the [[AP_Statistics_Curriculum_2007_Estim_L_Mean | ''number of sentences per advertisement'']], where we measure the readability for magazine advertisements. A random sample of the number of sentences found in 10 magazine advertisements is listed below. Suppose we want to test at $\alpha=0.01$ a null hypothesis: $H_o: \mu=12$ against a one-sided research alternative hypothesis: $H_1: \mu > 12$. Recall that we had the following [[AP_Statistics_Curriculum_2007_Estim_S_Mean#Example |sample statistics: '''sample-mean=22.1, sample-variance=737.88 and sample-SD=27.16390579''']] for these data.
{| class="wikitable" style="text-align:center; width:75%" border="1" {| class="wikitable" style="text-align:center; width:75%" border="1" Line 36: Line 36: As the population variance is not given, we have to use the [[AP_Statistics_Curriculum_2007_StudentsT |T-statistics]]: $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} \sim T(df=9)$ As the population variance is not given, we have to use the [[AP_Statistics_Curriculum_2007_StudentsT |T-statistics]]: $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} \sim T(df=9)$ - : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {22.1 - 12 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(x_i-22.1)^2\over 9}}})}=1.176$. + : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {22.1 - 12 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(x_i-22.1)^2\over 9}}}}=1.176$. - : $p-value=P(T_{(df=9)}>T_o=1.176)=0.134882$ for this (one-sided) test. Therefore, we '''can not reject''' the null hypothesis at $\alpha=0.01$! The left white area at the tails of the T(df=9) distribution depict graphically the probability of interest, which represents the strenght of the evidence (in the data) against the Null hypothesis. In this case, this area is 0.134882, which is larger than the initially set [[AP_Statistics_Curriculum_2007_Hypothesis_Basics | Type I]] error $\alpha = 0.01$ and we can not reject the null hypothesis. + : $p-value=P(T_{(df=9)}>T_o=1.176)=0.134882$ for this (one-sided) test. Therefore, we '''cannot reject''' the null hypothesis at $\alpha=0.01$! The left white area at the tails of the T(df=9) distribution depicts graphically the probability of interest, which represents the strength of the evidence (in the data) against the Null hypothesis. In this case, this area is 0.134882, which is larger than the initially set [[AP_Statistics_Curriculum_2007_Hypothesis_Basics | Type I]] error $\alpha = 0.01$ and we cannot reject the null hypothesis.
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig4.jpg|600px]]
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig4.jpg|600px]]
- * You can see use the [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses (One-Sample T-Test)] to carry out these calculations as shown in the figure below. + * You can also use the [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses (One-Sample T-Test)] to carry out these calculations as shown in the figure below.
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig5.jpg|600px]]
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig5.jpg|600px]]
Line 48: Line 48: ====Cavendish Mean Density of the Earth==== ====Cavendish Mean Density of the Earth==== - A number of famous early experiments of measuring physical constants have later been shown to be biased. In the 1700's [http://en.wikipedia.org/wiki/Henry_Cavendish Henry Cavendish] measured the [http://www.jstor.org/view/02610523/ap000022/00a00200/0 Mean density of the Earth]. Formulate and test null and research hypotheses about these data regarding the now know exact mean-density value = 5.517. These sample statistics may be helpful + A number of famous early experiments of measuring physical constants have later been shown to be biased. In the 1700's [http://en.wikipedia.org/wiki/Henry_Cavendish Henry Cavendish] measured the [http://www.jstor.org/stable/pdfplus/107617.pdf mean density of the Earth]. Formulate and test null and research hypotheses about these data regarding the now known exact mean-density value = 5.517. These sample statistics may be helpful : n = 23, sample mean = 5.483, sample SD = 0.1904 : n = 23, sample mean = 5.483, sample SD = 0.1904
Line 73: Line 73: * Hypothesis (Significance) testing: Only one possible value for the parameter, called the hypothesized value, is tested. We determine the strength of the evidence (confidence) provided by the data against the proposition that the hypothesized value is the true value. * Hypothesis (Significance) testing: Only one possible value for the parameter, called the hypothesized value, is tested. We determine the strength of the evidence (confidence) provided by the data against the proposition that the hypothesized value is the true value. + + ===[[EBook_Problems_Hypothesis_S_Mean|Problems]]===
## General Advance-Placement (AP) Statistics Curriculum - Testing a Claim about a Mean: Small Samples
### Testing a Claim about a Mean: Small Samples
The previous section discussed inference on the population mean for large smaples. Now, we show how to do hypothesis testing of the mean for small sample-sizes.
### Background
• Recall that for a random sample {$X_1, X_2, X_3, \cdots , X_n$} of the process, the population mean may be estimated by the sample average, $\overline{X_n}={1\over n}\sum_{i=1}^n{X_i}$.
• For a given small α (e.g., 0.1, 0.05, 0.025, 0.01, 0.001, etc.), the (1 − α)100% Confidence interval for the mean is constructed by
$CI(\alpha): \overline{x} \pm t_{df=n-1,{\alpha\over 2}} {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}$
and $t_{df=n-1, {\alpha\over 2}}$ is the critical value for a T-distribution of df=(sample size - 1) at ${\alpha\over 2}$.
### Hypothesis Testing about a Mean: Small Samples
• Null Hypothesis: Ho:μ = μo (e.g., 0)
• Alternative Research Hypotheses:
• One sided (uni-directional): H1:μ > μo, or Ho:μ < μo
• Double sided: $$H_1: \mu \not= \mu_o$$
#### Normal Process with Known Variance
• If the population is Normally distributed and we know the population variance, then the Test statistics is:
$Z_o = {\overline{x} - \mu_o \over {\sigma \over \sqrt{n}}} \sim N(0,1)$.
#### (Approximately) Normal Process with Unknown Variance
• If the population is approximately Normally distributed and we do not know the population variance, then the Test statistics is:
$T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {\overline{x} - \mu_o \over {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}} \sim T_{(df=n-1)}$.
### Example
Let's use again the small-sample example of the number of sentences per advertisement, where we measure the readability for magazine advertisements. A random sample of the number of sentences found in 10 magazine advertisements is listed below. Suppose we want to test at α = 0.01 a null hypothesis: Ho:μ = 12 against a one-sided research alternative hypothesis: H1:μ > 12. Recall that we had the following sample statistics: sample-mean=22.1, sample-variance=737.88 and sample-SD=27.16390579 for these data.
16 9 14 11 17 12 99 18 13 12
As the population variance is not given, we have to use the T-statistics: $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} \sim T(df=9)$
$T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {22.1 - 12 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(x_i-22.1)^2\over 9}}}}=1.176$.
pvalue = P(T(df = 9) > To = 1.176) = 0.134882 for this (one-sided) test. Therefore, we cannot reject the null hypothesis at α = 0.01! The left white area at the tails of the T(df=9) distribution depicts graphically the probability of interest, which represents the strength of the evidence (in the data) against the Null hypothesis. In this case, this area is 0.134882, which is larger than the initially set Type I error α = 0.01 and we cannot reject the null hypothesis.
### Examples
#### Cavendish Mean Density of the Earth
A number of famous early experiments of measuring physical constants have later been shown to be biased. In the 1700's Henry Cavendish measured the mean density of the Earth. Formulate and test null and research hypotheses about these data regarding the now known exact mean-density value = 5.517. These sample statistics may be helpful
n = 23, sample mean = 5.483, sample SD = 0.1904
5.36 5.29 5.58 5.65 5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.1 5.27 5.39 5.42 5.47 5.63 5.34 5.46 5.3 5.75 5.68 5.85
### Hypothesis Testing Summary
Important parts of Hypothesis test conclusions:
• Decision (significance or no significance)
• Parameter of interest
• Variable of interest
• Population under study
• (optional but preferred) P-value
### Parallels between Hypothesis Testing and Confidence Intervals
These are different methods for coping with the uncertainty about the true value of a parameter caused by the sampling variation in estimates.
• Confidence intervals: A fixed level of confidence is chosen. We determine a range of possible values for the parameter that are consistent with the data (at the chosen confidence level).
• Hypothesis (Significance) testing: Only one possible value for the parameter, called the hypothesized value, is tested. We determine the strength of the evidence (confidence) provided by the data against the proposition that the hypothesized value is the true value. |
# How To Solve System Of Inequalities By Substitution References
How To Solve System Of Inequalities By Substitution. $$y=2x+4$$ $$3x+y=9$$ we can substitute y in the second equation with the. 1) y = 6x − 11 −2x − 3y = −7 (2, 1) 2) 2x − 3y = −1 y = x − 1 (4, 3) 3) y = −3x + 5 5x − 4y = −3 (1, 2) 4) −3x − 3y = 3 y = −5x − 17 (−4, 3) 5) y = −2 4x − 3y = 18 (3, −2) 6) y = 5x − 7 −3x − 2y = −12 (2, 3)
A system of inequalities is almost exactly the same, except you’re working with inequalities instead of equations! A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear.
### 22 Best Systems Of Equations Inequalities Images On
A way to solve a linear system algebraically is to use the substitution method. Add 2x + 1 to both sides.
### How To Solve System Of Inequalities By Substitution
Choose the easiest variable to eliminate and multiplyboth equations by different numbers so that the coefficients of that variableare the same.Click the blue arrow to submit.Divide both sides by 3.Eliminate the variable y y by adding equation (1) ( 1) and equation (2) ( 2) together:
Enter the system of equations you want to solve for by substitution.For example, if asked to.For example, to solve the system.Given a system of equations containing a line and a circle, find the solution.
Given a system of two equations in two variables, solve using the substitution method solve one of the two equations for one of the variables in terms of the other.How to solve systems of inequalities graphically.In light of this fact, it may be easiest to find a solution set for inequalities by solving the system graphically.In this tutorial, you’ll see how to solve a system of linear equations by substituting one equation into the other and solving for the variable.
Line up the equations so that the variables are linedup vertically.Make the coefficients of one of the variables the same in both equations.Noticing that parentheses are required.Now plug this into the second equation.
Plug that value into either equation to get the value for the other variable.Plug that value into either equation to get the value for the other variable.Plug the result of step 1 into the other equation and solve for one variable.Plug the result of step 2 into one of the original equations and solve for the other variable.
Plug this into either of the original given equations.Recall that a linear equation can take the form a x + b y + c = 0.Replace x with 1 in y = x + 3 to find that y = 1 + 3 = 4.So x = 1 + y.
Solve a system of linear equations by substitution.Solve for the remaining variable.Solve systems of linear equations by substitution.Solve the linear equation for one of the variables.
Solve the one variable system.Solve the second equation for x by adding y to both sides:Solve this linear system of equations by the substitution method.Solving systems of equations by substitution date_____ period____ solve each system by substitution.
Solving the first equation for x looks like a winner.Substitute the expression for this variable into the second equation, and then solve for the remaining variable.Substitute the expression obtained in step one into the equation for the circle.Substitute x x into equation ( 2) ( 2) and solve for y y.
Substitute y y back into equation ( 1) ( 1) and solve for x x.Take one of the equations and solve it for one of the variables.Take one of the equations and solve it for one of the variables.Take that value of x, and substitute it into the first equation given above (x + y = 3).
The coefficients of y y in the given equations are 1 1 and −1 − 1.The solution set for this system is {(1, 4)}.The solve by substitution calculator allows to find the solution to a system of two or three equations in both a point form and an equation form of the answer.Then plug that into the other equation and solve for the variable.
Then plug that into the other equation and solve for the variable.Then solve this equation for x.Then, see how to use that variable value to find the value of the other variable.There are many different ways to solve a system of linear equations.
This tutorial will introduce you to systems of inequalities.To solve such a system, you need to find the variable values that will make each inequality true at the same time.To solve using substitution, follow these four steps:Use either equation to express x x in terms of y y.
Use the simplest of the two given equations to express one of the.Use the substitution method to solve the system:We’re going to explain this by using an example.X + 4y = 16.
X^ {\msquare} \log_ {\msquare} \sqrt {\square} \nthroot [\msquare] {\square} \le.X^ {\msquare} \log_ {\msquare} \sqrt {\square} \nthroot [\msquare] {\square} \le. |
## Skills Review for Integration by Parts
### Learning Outcomes
• Apply the power rule
• Find the derivatives of the sine and cosine function.
• Find the derivatives of the standard trigonometric functions.
• Find the derivative of exponential functions
• Find the derivative of logarithmic functions
• Apply substitution integration shortcut formulas
In the Integration by Parts section, we will learn how to evaluate integrals where one part of the integral is easily differentiable while the other part is easily integrable. Here we will review some derivative-taking techniques along with substitution integration shortcuts.
## Apply the Power Rule
We know that
$\dfrac{d}{dx}\left(x^2\right)=2x$ and $\dfrac{d}{dx}\left(x^{\frac{1}{2}}\right)=\dfrac{1}{2}x^{−\frac{1}{2}}$
As we shall see, there is a procedure for finding the derivative of the general form $f(x)=x^n$. The following theorem states that this power rule holds for all non-variable powers of $x$.
### The Power Rule
Let $n$ be a number. If $f(x)=x^n$, then
$f^{\prime}(x)=nx^{n-1}$
Alternatively, we may express this rule as
$\dfrac{d}{dx}(x^n)=nx^{n-1}$
### Example: Applying Basic Derivative Rules
Find the derivative of the function $f(x)=x^{10}$ by applying the power rule.
### Example: Applying Basic Derivative Rules
Find the derivative of $f(x)=2x^5+7$.
### Example: Applying Basic Derivative Rules
Find the derivative of $f(x)=\sqrt{x}$.
### Try It
Find the derivative of $f(x)=2x^{-3}-6x^2+3$.
### Try It
Find the derivative of $f(x)=\sqrt{x^7}$.
## Find the Derivatives of the Standard Trigonometric Functions.
### The Derivatives of $\sin x$ and $\cos x$
The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.
$\frac{d}{dx}(\sin x)= \cos x$
$\frac{d}{dx}(\cos x)=−\sin x$
### Example: Differentiating a Function Containing $sinx$
Find the derivative of $f(x)=5x^3 \sin x$.
### The Derivatives of $\tan x, \, \cot x, \, \sec x$, and $\csc x$
The derivatives of the remaining trigonometric functions are as follows:
$\frac{d}{dx}(\tan x)=\sec^2 x$
$\frac{d}{dx}(\cot x)=−\csc^2 x$
$\frac{d}{dx}(\sec x)= \sec x \tan x$
$\frac{d}{dx}(\csc x)=−\csc x \cot x$
### Try It
Find the derivative of $f(x)= \cot x$.
## Find the Derivatives of Exponential and Logarithmic Functions with Base e
### Derivative of the Natural Exponential Function
Let $E(x)=e^x$ be the natural exponential function. Then
$E^{\prime}(x)=e^x$
In general,
$\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\prime}(x)$
If it helps, think of the formula as the chain rule being applied to natural exponential functions. The derivative of $e$ raised to the power of a function will simply be $e$ raised to the power of the function multiplied by the derivative of that function.
### Example: Differentiating An Exponential Function
Find the derivative of $f(x)=e^{\tan (2x)}$.
### Try It
Find the derivative of $f(x)=e^{5x^2}$.
Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.
### The Derivative of the Natural Logarithmic Function
If $x>0$ and $y=\ln x$, then
$\frac{dy}{dx}=\dfrac{1}{x}$
More generally, let $g(x)$ be a differentiable function. For all values of $x$ for which $g^{\prime}(x)>0$, the derivative of $h(x)=\ln(g(x))$ is given by
$h^{\prime}(x)=\dfrac{1}{g(x)} g^{\prime}(x)$
### Example: Differentiating A Natural Logarithmic Function
Find the derivative of $f(x)=\ln(x^3+3x-4)$
### Try It
Find the derivative of $g(x)=\ln(3x+7)$.
## Apply Substitution Integration Shortcut Formulas
When integrating certain functions using substitution, certain patterns can be noticed in the answers. Here, we will review some shortcut integration formulas that are a result of substitution.
### SUbstitution Integration Shortcut Formulas
Let $a$ be a constant, then
• $\displaystyle\int e^{ax} dx=\dfrac{e^{ax}}{a}+C$
• $\displaystyle\int \sin ax dx=-\dfrac{\cos ax }{a}+C$
• $\displaystyle\int \cos ax dx=\dfrac{\sin ax }{a}+C$
### Example: Using a Substitution Integration Shortcut Formula
Find $\displaystyle\int e^{10x} dx$.
### Try It
Find $\displaystyle\int \cos 2x dx$. |
## Algebraic Equations
Back
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Provides a quick overview of the topic selected!
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Practice and review the topic selected with illustrated flash cards!
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#### Study Guide Algebraic Equations Mathematics, Grade 6
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ALGEBRAIC EQUATIONS What are algebraic equations? Algebraic equations are mathematical equations that contain a letter or variable, which represents a number. To solve an algebraic equation, inverse operations are used. Inverse operations are performing the opposite operation to what is being performed in the equation. The inverse operation of addition is subtraction and the inverse operation of subtraction is addition. Multiplication and division are the inverse operations of each other. There are different types of algebraic equations. A one-step algebraic equation means that only one operation needs to be performed in order to solve the equation. A two-step algebraic equation means that two steps are needed in order to solve the equation. Another algebraic equation that is commonly used is the Interest equation, I = P · r · t. This equation is used to find simple interest, I, given the principle, P, the rate, r, and the time, t. When three of the four numbers are given, the fourth can be found using a two-step algebra. How to use algebraic equations: Algebraic equations are solved using the inverse operations. To solve the equation n - 52 = 38, look to see what operation is being performed in the problem, subtraction. The inverse operation of subtraction is addition. The problem can be solved using the inverse operation of addition as the example shows: n - 52 = 38 + 52 +52 n = 90 Since 52 was subtracted in the initial problem, it will be added with inverse operations. The rule when dealing with equations is whatever is done to one side of the equation must be done to the other side. So © Copyright NewPath Learning. All Rights Reserved. Permission is granted for the purchaser to print copies for non-commercial educational purposes only. Visit us at www.NewPathLearning.com.
52 is added to 38 on the other side of the equal sign. When both sides are added, the 52's on the left side of the equation cancel out, leaving only the n. On the right side of the equation, 38 and 52 are added, giving the answer of 90. Therefore n = 90. This is a one-step equation. Look at the two-step equation below and how to solve it. 3n + 12 = 27 - 12 -12 3n = 15 3 3 n = 5 In this equation, 12 is subtracted first from both sides. This because the variable needs to be alone before it can be solved. Now the equation is 3n = 15. Since the 3 is multiplied by n, division must be used in order to solve for n. When 3 is divided by both sides, the result is n = 5. T The interest equation, I = P·r·t, is also a two-step equation. When given the principle, rate and interest, multiplication is used to solve for the interest. If interest is given, along with two others, use inverse operations to solve. © Copyright NewPath Learning. All Rights Reserved. Permission is granted for the purchaser to print copies for non-commercial educational purposes only. Visit us at www.NewPathLearning.com.
Try this! 1. Solve the following: n - 4 = 16 56 + n = 134 n + 98 = 207 2n - 6 = 12 4n - 36 = 144 5n + 75 = 110 2. Solve for I, I = P·r·t: P = \$1000, r = 8%, t = 2 years P = \$5000, r = 4%, t = 1 year 3. Solve for the missing variable, I = P·r·t: I = \$500, r = 8%, t = 2 years I = \$50, P = \$2000, t = 1 year © Copyright NewPath Learning. All Rights Reserved. Permission is granted for the purchaser to print copies for non-commercial educational purposes only. Visit us at www.NewPathLearning.com. |
# CBSE Class 11 Mathematics Straight Lines Assignment Set B
Read and download free pdf of CBSE Class 11 Mathematics Straight Lines Assignment Set B. Get printable school Assignments for Class 11 Mathematics. Standard 11 students should practise questions and answers given here for Chapter 10 Straight Lines Mathematics in Grade 11 which will help them to strengthen their understanding of all important topics. Students should also download free pdf of Printable Worksheets for Class 11 Mathematics prepared as per the latest books and syllabus issued by NCERT, CBSE, KVS and do problems daily to score better marks in tests and examinations
## Assignment for Class 11 Mathematics Chapter 10 Straight Lines
Class 11 Mathematics students should refer to the following printable assignment in Pdf for Chapter 10 Straight Lines in standard 11. This test paper with questions and answers for Grade 11 Mathematics will be very useful for exams and help you to score good marks
### Chapter 10 Straight Lines Class 11 Mathematics Assignment
Chapter: - Straight Lines and Conic Sections
Q1.What point on x-axis is equidistant from the points A (1, 3) and B (2,-5)
Ans. (19/2, 0)
Q2. For what values of x, the area of the triangle formed the points (5, -1), (x, 4) and (6, 3) is 5.5 sq. units?
Ans. 7/2 or 9
Q3. A line cut off intercepts -3 and 4 on x and y-axis respectively. Find the slope and equation of the line.
Ans. 4/3, 4x -3y +12 =0
Q4. Find the equation of a line parallel to x-axis at a distance of 3 units above x-axis.
Ans. y=3
Q5. Find the value of c and m so that the line y = m x +c may pass through the points (-2, 3) and (4,-3)
Ans. c=1, m= -1
Q6. Mid-points of the sides of a triangle are (2, 2), (2, 3) and (4, 6). Find the equations of the sides of a triangle.
Ans. 3x- 2y =2, 2x-y =1, x=4
Q7. Find the equations of the medians of the triangle whose vertices are (2, 0), (0, 2) and (4, 6).
Ans. x=2, 5x-3y=2, x-3y +6 =0
Q8. Show that the points (a, 0) (0, b) and (3a, -2b) are collinear Also find the equation of the line containing them.
Ans. bx + ay =ab
Q9.Find the new coordinates of the point(3,-5) if origin is shifted to the point (2,3) by a transformation of axes.
Ans. (1,-8)
Q10. Find the equation of the line such that segment intercepted by the axes is divided by the points (-5,4) in the ratio 1:2.
Ans. 8 x -5y +60 =0
Q11. Find the equations of the medians of the triangle ABC whose vertices are A (2, 5), B (-4, 9) and C (-2,-1)
Ans. 8x-y+15=0, x-5y+23=0, 7x+4y-8=0
Q112. Find the image of the point (-8, 12) with respect to the line mirror 4x+7y+13=0.
Ans. (-16,-2)
Q13 Find the equation of the line passing through the intersection of the lines 3x-4y+1=0 and 5x+y-1=0 and cutting off equal intercepts on the coordinate axes.
Ans. 23x+23y=11
Q14. Find the coordinates of the foot of the perpendicular from a point (-1, 3) to the line 3x-4y=16
Ans. 68/25,-49/25)
Q15. Find the equation of the line parallel to y- axis and drawn through the point of intersection of x-7y+5=0 and 3x+7y=7
Ans. x=1/2
Q16. Find the equation of parabola with focus at (5, 0) and directrix x+5=0, Also find the length of latus rectum.
Ans. y2 =20x, 20
Q17. For the parabola y2=-12x, Find the coordinates of focus, the equation of directrix and length of latus rectum.
Ans. (-3, 0), x=3, 12
Q18. Find the equation of parabola with vertex at origin and having directrix y=2
Ans. x2=-8y
Please click the below link to access CBSE Class 11 Mathematics Straight Lines Assignment Set B
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SOLUTION 5: Draw a cube with edge lengths $x$, and assume that $x$ is a function of time $t$.
$\ \ \ \$ a.) The surface area (Add the areas of 6 square surfaces.) of a cube is $$S=x^2+x^2+x^2+x^2+x^2+x^2 \ \ \ \ \longrightarrow$$ $$S=6x^2$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = -2 \ cm/min.$
FIND: $\ \ \ \displaystyle{ dS \over dt }$ when $x =80 \ cm.$
Now differentiate the surface area equation with respect to time $t$, getting
$$D \{ S \} = D \{ 6x^2 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = 6 \cdot 2x \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$
$\Big($ Now let $\displaystyle{ dx \over dt } = -2$ and $x=80. \Big)$ $$\displaystyle{ dS \over dt } = 12(80)(-2) \ \ \ \longrightarrow$$ $$\displaystyle{ dS \over dt } = -1920 \ cm^2/min.$$
$\ \ \ \$ b.) The volume of a cube is $$V = (length)(width)(height) \ \ \ \ \longrightarrow$$ $$V = x^3$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = -2 \ cm/min.$
FIND: $\ \ \ \displaystyle{ dV \over dt }$ when $x =80 \ cm.$
Now differentiate the volume equation with respect to time $t$, getting
$$D \{ V \} = D \{ x^3 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = 3 x^2 \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$
$\Big($ Now let $\displaystyle{ dx \over dt } = -2$ and $x=80. \Big)$
$$\displaystyle{ dV \over dt } = 3(80)^2(-2) = -38,400 \ \ cm^3/min.$$ |
Any questions on the Section 2.1 homework that was due today?
Presentation on theme: "Any questions on the Section 2.1 homework that was due today?"— Presentation transcript:
Any questions on the Section 2.1 homework that was due today?
Comment on the 2.1 homework: Types of outcomes when solving linear equations in one variable: 1. One solution (nonzero). (most problems in 2.1) Example: 2x + 4 = 4(x + 3) Solution: x = -4 2. One solution (zero). (as in problems #6 & 20) Example: 2x + 4 = 4(x + 1) Solution: x = 0 3. Solution = “All real numbers ”. (as in problems #15,17,21) Example: 2x + 4 = 2(x + 2) Solution: All real numbers. (“R” on computer) 4. No solutions. (as in problems # 16,18,23) Example: 2x + 4 = 2(x + 3) Solution: No solution (“N” on computer)
Section 2.2 An Introduction to Problem Solving
Reminder: This homework assignment on section 2.2 is due at the start of next class period. Make sure you turn in the worksheet showing all your work for problems #5 -19 of this assignment. If you don’t turn this in, or if you don’t completely show your work on any problem/s, your online score will be reduced for those 15 problems out of the 24 total problems in the online assignment.)
Section 2.2 General strategy for problem solving 1)Understand the problem Read and reread the problem Choose a variable to represent the unknown Construct a drawing, whenever possible 2)Translate the problem into an equation 3)Solve the equation 4)Interpret the result Check solution State your conclusion
Example 1: The product of twice a number and three is the same as the difference of five times the number and ¾. Find the number. Understand Read and reread the problem. If we let x = the unknown number, then “twice a number” translates to 2x, “the product of twice a number and three” translates to 2x · 3, “five times the number” translates to 5x, and “the difference of five times the number and ¾” translates to 5x - ¾.
Example (cont.) Translate The product of · twice a number 2x2x and 3 3 is the same as = 5 times the number 5x5x and ¾ ¾ the difference of –
Example (cont.) Solve 2x · 3 = 5x – ¾ 6x = 5x – ¾ (simplify left side) 6x + (-5x) = 5x + (-5x) – ¾ (add –5x to both sides) x = - ¾ (simplify both sides) Now CHECK your answer: Left side: 2x·3= (2·-3/4)·3 = -6/4·3 = -3/2·3= -9/2 Right side: 5x-3/4 = 5·-3/4-3/4 = -15/4 – 3/4 = -18/4 = -9/2 (You can perform this check quickly by using your calculator.)
Sample problem from today’s homework:
Example 2: A car rental agency advertised renting a Buick Century for \$24.95 per day and \$0.29 per mile. If you rent this car for 2 days, how many whole miles can you drive on a \$100 budget? Understand Read and reread the problem. If we let x = the number of whole miles driven, then 0.29x = the cost for mileage driven
Example (cont.) Translate Daily costs 2(24.95) mileage costs 0.29x plus + are equal to = 100 maximum budget
Example (cont.) Solve 2(24.95) + 0.29x = 100 49.90 + 0.29x = 100 (simplify left side) 0.29x = 50.10 (simplify both sides) 49.90 – 49.90 + 0.29x = 100 – 49.90 (subtract 49.90 from both sides) (divide both sides by 0.29) x 172.75 (simplify both sides)
Example (cont.) Interpret Check: Recall that the original statement of the problem asked for a “whole number” of miles. If we replace “number of miles” in the problem with 173, then 49.90 + 0.29(173) = 100.07, which is over our budget. However, 49.90 + 0.29(172) = 99.78, which is within the budget. State: The maximum number of whole number miles is 172.
MATH 110 - Section 2.2 Homework Problem Tip: If you’re having trouble doing percent problems that give you a new value after a certain percent increase or decrease from an old value (such as sales tax problems), try thinking about it this way: Think about when you go shopping to buy, say, a TV. Usually you know how much the TV costs, for example \$400, and the percent tax rate, for example 5.5%. Normally what you do (or the salesclerk’s computer does) is calculate the TOTAL COST by taking 5.5% of \$400, then adding that amount back onto the \$400 price of the TV to get the total cost to you.
The working equation is PRICE + TAX = TOTAL COST. In words, here’s what you did (after writing the 5.5% as a decimal, 0.055): PRICE +.055 times PRICE = TOTAL COST Plugging in the numbers, we get 400 +.055 x 400 = 400 + 22 = 422. Notice that you’ve multiplied the OLD VALUE (the price before tax) by the.055.
The same basic format applies to anything with a percent increase or decrease from an original amount: Old amount +/- % of old amount = new amount (Remember to write the percent as a decimal.) This equation works for raises in pay, population increases or decreases, and many other percent change problems, especially where you’re given the new amount and the percent change and you need to work backwards to find out the old amount.
Example: After a 6% pay raise, Nora’s 2005 salary is \$39,703. What was her salary in 2004? (Round to the nearest dollar). Solution: Recall the equation: Old amount + % of old amount = new amount The “old amount” is her 2004 salary, which is unknown, so we’ll call it X. This gives us the equation X + 0.06X = 39703
Example (cont.) After a 6% pay raise, Nora’s 2005 salary is \$39,703. What was her salary in 2004? (Round to the nearest dollar). X + 0.06X = 39703 This simplifies to 1.06X = 39703 Divide both sides X = 39703 by 1.06 to get X. 1.06 Answer: Her 2004 salary was \$37,456
Now check your answer: 37456 +.06 x 37456 = 37456 + 2247 = 39703 NOTE that this DOES NOT give you the same answer as if you subtracted 6% of the new salary (39703) from the new salary. Try it and you’ll see that it doesn’t work. (It’s not real far off, but enough to give you the wrong answer, and the bigger the percentage, the farther off you’ll be.)
Sample problem from today’s homework:
Reminder: This homework assignment on section 2.2 is due at the start of next class period. Make sure you turn in the worksheet showing all your work. If you don’t turn this in, your online score will be reduced. If you don’t completely show your work on any problem/s, your online score will be reduced for those problems.
You may now OPEN your LAPTOPS and begin working on the homework assignment.
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# Basics of Coordinate Geometry
Coordinate Geometry is a very interesting chapter in mathematics. playing with coordinates and graphs, am I right?
This is an introduction or you can say basics of the coordinate geometry. I hope you will be enjoying learning coordinate geometry chapter.
Here we learn the concept of coordinate geometry. Basically the X-axis, Y-axis, Four Quadrants, concept of positive
and negative axis etc.
We learnt ;-
1. To locate the position of an object or a point in a plane, we require two perpendicular lines.
2. When Perpendicular lines intersect each other there is formation of four quadrants namely first quadrant (1st) ,
3. Intersecting point is called origin.
4. Horizontal line is called X-axis while vertical line is called y-axis.
5. X-axis is positive right side of the origin while negative in left side of the origin.
6. Y – axis is positive above the origin and negative below the origin.
7. Particular location is defined by the length traveled in X-axis called X-coordinate or abscissa followed by the length traveled in Y-axis c called y-coordinate or ordinate . Using comma( , ) we separate X & Y Coordinates.
8. IF the abscissa of a point is x and the ordinate is y then (x,y) are called coordinates of the point.
9. The coordinates of a point on the x-axis are of the form (x,o) & and that of the point on the y-axis are (0,y)
10. The coordinates of the origin is (0,0).
11.. The coordinates of the first quadrant are of the form ( + , + )
The coordinates of the second quadrant are of the form ( – , + )
The coordinates of the third quadrant are of the form ( – , – ).
The coordinates of the fourth quadrant are of the form ( – , + ).
12. If x is not equal to y then, (x,y) is not equal to (y , x ) and (x,y) equal to (y,x) if x=y.
I hope you understand how to graph the points also in a coordinate plane or Cartesian Plane.
next Video or lesson we will learn to graph the points with a lot of examples in different ways.
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# CBSE Class 10 Maths Formulas 2022 For Term 1 & Term 2 | Chapter-Wise Formulas PDF
CBSE Class 10 Maths Formulas 2022 For Term 1 & Term 2: You’ve probably heard from your parents, teachers, and seniors about the importance of CBSE Class 10. There is no denying that this class is important in establishing your career path as well as setting the foundation for it. Math is widely regarded as the most difficult subject in Class 10. This article will serve as a one-stop destination for all of the important CBSE Class 10 Math Formulas to assist you in solving problems more quickly. So sit back, chill, and read all the way to the end!
## Important CBSE Class 10 Maths Formulas 2022 For Term 1 & Term 2
Students who are appearing for upcoming Class 10 Board Exams can go through the CBSE Class 10 Maths Formulas as listed below.
### CBSE Class 10 Maths Formulas (Real Numbers)
Euclid’s Division Algorithm (lemma): According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r such that a = bq + r, where 0 ≤ r ≤ b. (Here, a = dividend, b = divisor, q = quotient and r = remainder.)
Here you can check the important formulas related to Class 10 Maths Real Numbers.
S. No Type of Numbers Description 1 Natural Numbers N = {1,2,3,4,5 >It is the counting numbers 2 Whole number W= {0,1,2,3,4,5>It is the counting numbers + zero 3 Integers All whole numbers including Negative number + Positive number ……-4,-3,-2,-1,0,1,2,3,4,5… so on.Like whole numbers, integers don’t include fractions or decimals. 4 Positive integers Z+ = 1,2,3,4,5, …… 5 Negative integers Z– = -1,-2,-3,-4,-5, …… 6 Rational Number A number is called rational if it can be expressed in the form p/q where p and q are integers (q> 0).Ex: P/q, 4/5 7 Irrational Number A number is called rational if it cannot be expressed in the form p/q where p and q are integers (q> 0).Ex: √2, Pi, … etc 8 Real Numbers A real number is a number that can be found on the number line. Real Numbers are the numbers that we normally use and apply in real-world applications.Real Numbers include Natural Numbers, Whole Numbers, Integers, Fractions, Rational Numbers and Irrational Numbers
HCF (Highest common factor)
HCF of two positive integers can be find using Euclid’s Division Lemma algorithm
We know that for any two integers a. b. we can write the following expression
• a=bq + r , 0≤r<b
If r=0 .then
HCF( a. b) =b
If r≠0 , then
HCF ( a. b) = HCF ( b.r)
Again expressing the integer b.r in Euclid’s Division Lemma, we get
b=pr + r1
HCF ( b,r)=HCF (r,r1)
Similarly, successive Euclid’s division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b
• HCF (a,b) =1 – Then a and b are co primes.
• Product of primes Theorem of Arithmetic – Composite Number = Product of Primes
• HCF and LCM by prime factorization method:
HCF = Product of the smallest power of each common factor in the numbers
LCM = Product of the greatest power of each prime factor involved in the number
• Important Formula HCF (a,b) X LCM (a,b) =a X b
• Important concept for rational Number – Terminating decimal expression can be written in the form p/2n5m
### CBSE Class 10 Maths Formulas (Polynomials)
• (a + b)2= a2 + 2ab + b2
• (a – b)2 = a2– 2ab + b2
• a2– b= (a + b) (a – b)
• (a + b)3 = a3+ b3 + 3ab(a + b)
• (a – b)3 = a3– b3 – 3ab(a – b)
• a3+ b3 = (a + b) (a– ab + b2)
• a3– b3 = (a – b) (a+ ab + b2)
• a4– b4 = (a2)2 – (b2)2 = (a2 + b2) (a2 – b2) = (a2 + b2) (a + b) (a – b)
• (a + b + c) = a2 + b2 + c2 + 2ab + 2bc + 2ac
• (a + b – c) = a2 + b2 + c2 + 2ab – 2bc – 2ca
• (a – b + c)2 = a2+ b2 + c2 – 2ab – 2bc + 2ca
• (a – b – c)2 = a2+ b2 + c2 – 2ab + 2bc – 2ca
• a3+ b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
### CBSE Class 10 Maths Linear Equations in Two Variables Formulas
For the pair of linear equations a1 + b1y + c1 = 0 and a2 + b2y + c2 = 0, the nature of roots (zeroes) or solutions is determined as follows:
• If a1/a2 ≠ b1/b2 then we get a unique solution and the pair of linear equations in two variables are consistent. Here, the graph consists of two intersecting lines.
• If a1/a2 ≠ b1/b2 ≠ c1/c2, then there exists no solution and the pair of linear equations in two variables are said to be inconsistent. Here, the graph consists of parallel lines.
• If a1/a2 = b1/b2 = c1/c2, then there exist infinitely many solutions and the pair of lines are coincident and therefore, dependent and consistent. Here, the graph consists of coincident lines.
### CBSE Class 10 Maths Quadratic Equation Formulas
For a quadratic equation, ax+ bx + c = 0
ax2+bx+c=0 where a ≠ 0 And x = [-b ± √(b2 – 4ac)]/2a
• Sum of roots = –b/a
• Product of roots = c/a
• If roots of a quadratic equation are given, then the quadratic equation can be represented as:
x2 – (sum of the roots)x + product of the roots = 0
• If Discriminant > 0, then the roots of the quadratic equation are real and unequal/unique.
• If Discriminant = 0, then the roots of the quadratic equation are real and equal.
• If Discriminant < 0, then the roots of the quadratic equation are imaginary (not real).
### CBSE Class 10 Maths Arithmetic Progression Formulas
• nth Term of an Arithmetic Progression: For a given AP, where a is the first term, d is a common difference, n is the number of terms, its nth term (an) is given as
a= a + (n−1)×d
• Sum of First n Terms of an Arithmetic Progression, Sn is given as:
Sn = n/2 [a + (n-1) d]
### CBSE Class 10 Maths The similarity of Triangles Formulas
• If two triangles are similar then the ratio of their sides is equal.
• Theorem on the area of similar triangles: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
### CBSE Class 10 Maths Coordinate Geometry Formulas
• Distance Formulae: Consider a line having two-point A(x1, y1) and B(x2, y2), then the distance of these points is given as:
AB= √[(x− x1)+ (y− y1)2]
• Section Formula: If a point p divides a line AB with coordinates A(x1, y1) and B(x2, y2), in ratio m:n, then the coordinates of the point p are given as:
P={[(mx2+nx1)/(m+n)],[(my2+ny1)/(m+n)]}
• Midpoint Formula: The coordinates of the mid-point of a line AB with coordinates A(x1, y1) and B(x2, y2), are given as:
P={(x1+x2)/2,(y1+y2)/2}
• Area of a Triangle: Consider the triangle formed by the points A(x1, y1) and B(x2, y2) and C(x3, y3) then the area of a triangle is given as-
∆ABC=½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|
### CBSE Class 10 Maths Trigonometry Formulas
In a right-angled triangle, the Pythagoras theorem states
(perpendicular )+ ( base )2 = ( hypotenuse )2
Important trigonometric properties: (with P = perpendicular, B = base and H = hypotenuse)
• SinA = P / H
• CosA = B / H
• TanA = P / B
• CotA = B / P
• CosecA = H / P
• SecA = H/B
Trigonometric Identities:
• sin2A + cos2A=1
• tan2A +1 = sec2A
• cot2A + 1= cosec2A
Relations between trigonometric identities are given below:
• Secθ = 1/cosθ
• Cotθ = 1/tanθ
• Cosecθ = 1/sinθ
• Tanθ = Sinθ/Cosθ
Trigonometric Ratios of Complementary Angles are given as follows:
• sin (90° – A) = cos A
• cos (90° – A) = sin A
• tan (90° – A) = cot A
• cot (90° – A) = tan A
• sec (90° – A) = cosec A
• cosec (90° – A) = sec A
Values of Trigonometric Ratios of 0° and 90° are tabulated below:
Angle 0° 30° 45° 60° 90° Sinθ 0 1/2 1/√2 √3/2 1 Cosθ 1 √3/2 1/√2 ½ 0 Tanθ 0 1/√3 1 √3 Undefined Cotθ Undefined √3 1 1/√3 0 Secθ 1 2/√3 √2 2 Undefined Cosecθ Undefined 2 √2 2/√3 1
### CBSE Class 10 Maths Circles Formulas
• Circumference of the circle = 2πr
• Area of the circle = πr2
• Area of the sector of angle θ = (θ/360) × πr2
• Length of an arc of a sector of angle θ = (θ/360) × 2πr (r = radius of the circle)
### CBSE Class 10 Maths Areas Related to Circles Formulas
• The equal chord of a circle is equidistant from the center.
• The perpendicular drawn from the center of a circle bisects the chord of the circle.
• The angle subtended at the center by an arc = Double the angle at any part of the circumference of the circle.
• Angles subtended by the same arc in the same segment are equal.
• To a circle, if a tangent is drawn and a chord is drawn from the point of contact, then the angle made between the chord and the tangent is equal to the angle made in the alternate segment.
The sum of the opposite angles of a cyclic quadrilateral is always 180o.
• Area of a Segment of a Circle: If AB is a chord that divides the circle into two parts, then the bigger part is known as the major segment and the smaller one is called the minor segment.
Here, Area of the segment APB = Area of the sector OAPB – Area of ∆ OAB
### CBSE Class 10 Maths Surface Areas and Volumes Formulas
The common formulas from the surface area and volumes chapter in Class 10 Maths include the following:
• Sphere Formulas
Diameter of sphere 2r Circumference of Sphere 2 π r The surface area of a sphere 4 π r2 Volume of Cylinder 4/3 πr2
• Cylinder Formulas
Circumference of Cylinder 2 πrh The curved surface area of Cylinder 2 πr2 The total surface area of Cylinder Circumference of Cylinder + Curved surface area of Cylinder = 2 πrh + 2 πr2 Volume of Cylinder π r2 h
• Cone Formulas
The slant height of a cone l = √(r2 + h2) The curved surface area of a cone πrl The total surface area of a cone πr (l + r) Volume of cone ⅓ π r2 h
• Cuboid Formulas
Perimeter of cuboid 4(l + b +h) Length of the longest diagonal of a cuboid √(l2 + b2 + h2) The total surface area of the cuboid 2(l×b + b×h + l×h) Volume of Cuboid l × b × h
Here, l = length, b = breadth and h = height In case of Cube, put l = b = h = a, as cube all its sides of equal length, to find the surface area and volumes.
### CBSE Class 10 Maths Statistics Formulas
Mean: The mean value of a variable is defined as the sum of all the values of the variable divided by the number of values.
Median: The median of a set of data values is the middle value of the data set when it has been arranged in ascending order. That is, from the smallest value to the highest value.
Median is calculated as
Where n is the number of values in the data. If the number of values in the data set is even, then the median is the average of the two middle values.
Mode: Mode of statistical data is the value of that variable that has the maximum frequency
For Grouped Data:
Mean: If x1, x2, x3,……xn are observations with respective frequencies f1, f2, f3,…..fn then mean is given as:
Median: For the given data, we need to have a class interval, frequency distribution, and cumulative frequency distribution. Then, the median is calculated as
Where
l = lower limit of median class,
n = number of observations,
cf = cumulative frequency of class preceding the median class,
f = frequency of median class,
h = class size (assuming class size to be equal)
Mode: Modal class: The class interval having the highest frequency is called the modal class and Mode is obtained using the modal class.
Where
l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f1 = frequency of the modal class,
f0 = frequency of the class preceding the modal class,
f2 = frequency of the class succeeding in the modal class.
## Other Important Links Related To CBSE Class 10 Maths 2022 For Term 1 & Term 2
We have provided the CBSE Class 10 Maths formulas according to the latest syllabus of the academic year. Go through the below-listed CBSE Class 10 Maths Formulas carefully and prepare well for the Class 10 Exams.
## FAQs on CBSE Class 10 Maths Formulas 2022 For Term 1 & Term 2
### From where can I download the chapter-wise CBSE Class 10 Maths Formulas PDF?
You can download CBSE Class 10 Maths Formulas PDF from the link mentioned in the above blog. |
# How to Find All The Factors of a Number (The Factor Pair Method)
## Here you shall be using the factor pair method to calculate all the factors of a particular number.
The factor pair method is probably the easiest way to work out all the factors of a particular number. Remember, that a factor divides exactly into a number without a remainder. With your factor pair method, begin with your 1 times tables and see if this goes into your number, then do the 2 times table, then the 3 times table and so on. If you come across a factor that is already found then there will be no more factors to find. Let’s take a look at a couple of question, of using the factor pair method to find factors.
Example 1
Find out all the factors of 16.
Begin with your 1 times table. 16 ÷ 1 is 16. So 1 and 16 are both factors of 16.
After this do your 2 times table. 16 ÷ 2 is 8. So 2 and 8 are both factors of 16.
After this do your 3 times table. 16 ÷ 3 = 5 remainder 1. So 3 is not a factor.
After this do your 4 times table. 16 ÷ 4 is 4. So 4 is a factor of 16.
After this do your 5 times table. 16 ÷ 5 is 3 remainder 1. So 5 is not a factor.
After this do your 6 times table. 16 ÷ 6 is 2 remainder 4. So 6 is not a factor.
After this do your 7 times table. 16 ÷ 7 is 2 remainder 2. So 7 is not a factor.
The next times table is 8, but we already have it as a factor, so there are no more factors to find.
Therefore, all the factors of 16 are:
1,2,4,8,16
Example 2
Find out all the factors of 42.
Begin with your 1 times table. 42 ÷ 1 is 42. So 1 and 42 are both factors of 42.
After this do your 2 times table. 42 ÷ 2 is 21. So 2 and 21 are both factors of 42.
After this do your 3 times table. 42 ÷ 3 is 14. So 3 and 14 are both factors of 42.
After this do your 4 times table. 42 ÷ 4 = 10 remainder 2. So 4 is not a factor.
After this do your 5 times table. 42 ÷ 5 = 8 remainder 2. So 5 is not a factor.
After this do your 6 times table. 42 ÷ 6 is 7. So 6 and 7 are both factors of 42.
The next times table is 7, but we already have it as a factor, so there are no more factors to find.
Therefore, all the factors of 42 are:
1,2,3,6,7,14,21,42.
For some easier questions on using the factor pair method to find factors then click here.
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# NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2
*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.
NCERT Solutions for Class 11 Maths Chapter 15 Statistics, contains solutions for all Exercise 15.2 questions. Chapter 15, Statistics of Class 11 Maths, is categorised under the CBSE Syllabus for the sessions 2023-24. By practising these questions, students can build their confidence level. NCERT Solutions also help the students to excel in their skills in statistics. Download these NCERT Solutions of Class 11 Maths now and start practising for the board exam.
## NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2
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### Access Other Exercise Solutions of Class 11 Maths Chapter 15 Statistics
Exercise 15.1 Solutions : 12 Questions
Exercise 15.3 Solutions : 5 Questions
Miscellaneous Exercise Solutions: 7 Questions
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NCERT Solutions for Class 11 Maths Chapter 15
NCERT Solutions for Class 11
#### Access NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2
Find the mean and variance for each of the data in Exercise 1 to 5.
1. 6, 7, 10, 12, 13, 4, 8, 12
Solution:-
So, xÌ… = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
Let us make the table of the given data and append other columns after calculations.
Xi Deviations from mean (xi – x̅) (xi – x̅)2 6 6 – 9 = -3 9 7 7 – 9 = -2 4 10 10 – 9 = 1 1 12 12 – 9 = 3 9 13 13 – 9 = 4 16 4 4 – 9 = – 5 25 8 8 – 9 = – 1 1 12 12 – 9 = 3 9 74
We know that Variance,
σ2 = (1/8) × 74
= 9.2
∴Mean = 9 and Variance = 9.25
2. First n natural numbers
Solution:-
We know that Mean = Sum of all observations/Number of observations
∴Mean, x̅ = ((n(n + 1))2)/n
= (n + 1)/2
and also WKT Variance,
By substitute that value of xÌ… we get,
WKT, (a + b)(a – b) = a2 – b2
σ2 = (n2 – 1)/12
∴Mean = (n + 1)/2 and Variance = (n2 – 1)/12
3. First 10 multiples of 3
Solution:-
First we have to write the first 10 multiples of 3,
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
So, xÌ… = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10
= 165/10
= 16.5
Let us make the table of the data and append other columns after calculations.
Xi Deviations from mean (xi – x̅) (xi – x̅)2 3 3 – 16.5 = -13.5 182.25 6 6 – 16.5 = -10.5 110.25 9 9 – 16.5 = -7.5 56.25 12 12 – 16.5 = -4.5 20.25 15 15 – 16.5 = -1.5 2.25 18 18 – 16.5 = 1.5 2.25 21 21 – 16.5 = – 4.5 20.25 24 24 – 16.5 = 7.5 56.25 27 27 – 16.5 = 10.5 110.25 30 30 – 16.5 = 13.5 182.25 742.5
Then, Variance
= (1/10) × 742.5
= 74.25
∴Mean = 16.5 and Variance = 74.25
4.
xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3
Solution:-
Let us make the table of the given data and append other columns after calculations.
Xi fi fixi Deviations from mean (xi – x̅) (xi – x̅)2 fi(xi – x̅)2 6 2 12 6 – 19 = 13 169 338 10 4 40 10 – 19 = -9 81 324 14 7 98 14 – 19 = -5 25 175 18 12 216 18 – 19 = -1 1 12 24 8 192 24 – 19 = 5 25 200 28 4 112 28 – 19 = 9 81 324 30 3 90 30 – 19 = 11 121 363 N = 40 760 1736
5.
xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3
Solution:-
Let us make the table of the given data and append other columns after calculations.
Xi fi fixi Deviations from mean (xi – x̅) (xi – x̅)2 fi(xi – x̅)2 92 3 276 92 – 100 = -8 64 192 93 2 186 93 – 100 = -7 49 98 97 3 291 97 – 100 = -3 9 27 98 2 196 98 – 100 = -2 4 8 102 6 612 102 – 100 = 2 4 24 104 3 312 104 – 100 = 4 16 48 109 3 327 109 – 100 = 9 81 243 N = 22 2200 640
6. Find the mean and standard deviation using short-cut method.
xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5
Solution:-
Let the assumed mean A = 64. Here h = 1
We obtain the following table from the given data.
Xi Frequency fi Yi = (xi – A)/h Yi2 fiyi fiyi2 60 2 -4 16 -8 32 61 1 -3 9 -3 9 62 12 -2 4 -24 48 63 29 -1 1 -29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 0 286
Mean,
Where A = 64, h = 1
So, x̅ = 64 + ((0/100) × 1)
= 64 + 0
= 64
Then, variance,
σ2 = (12/1002) [100(286) – 02]
= (1/10000) [28600 – 0]
= 28600/10000
= 2.86
Hence, standard deviation = σ = √2.886
= 1.691
∴ Mean = 64 and Standard Deviation = 1.691
Find the mean and variance for the following frequency distributions in Exercises 7 and 8.
7.
Classes 0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180 180 – 210 Frequencies 2 3 5 10 3 5 2
Solution:-
Let us make the table of the given data and append other columns after calculations.
Classes Frequency fi Mid – points xi fixi (xi – x̅) (xi – x̅)2 fi(xi – x̅)2 0 – 30 2 15 30 -92 8464 16928 30 – 60 3 45 135 -62 3844 11532 60 – 90 5 75 375 -32 1024 5120 90 – 120 10 105 1050 -2 4 40 120 – 150 3 135 405 28 784 2352 150 – 180 5 165 825 58 3364 16820 180 – 210 2 195 390 88 7744 15488 N = 30 3210 68280
8.
Classes 0 – 10 10 – 20 20 – 30 30 – 40 40 –50 Frequencies 5 8 15 16 6
Solution:-
Let us make the table of the given data and append other columns after calculations.
Classes Frequency fi Mid – points xi fixi (xi – x̅) (xi – x̅)2 fi(xi – x̅)2 0 – 10 5 5 25 -22 484 2420 10 – 20 8 15 120 -12 144 1152 20 – 30 15 25 375 -2 4 60 30 – 40 16 35 560 8 64 1024 40 –50 6 45 270 18 324 1944 N = 50 1350 6600
9. Find the mean, variance and standard deviation using short-cut method
Height in cms 70 – 75 75 – 80 80 – 85 85 – 90 90 – 95 95 – 100 100 – 105 105 – 110 110 – 115 Frequencies 3 4 7 7 15 9 6 6 3
Solution:-
Let the assumed mean, A = 92.5 and h = 5
Let us make the table of the given data and append other columns after calculations.
Height (class) Number of children Frequency fi Midpoint Xi Yi = (xi – A)/h Yi2 fiyi fiyi2 70 – 75 3 72.5 -4 16 -12 48 75 – 80 4 77.5 -3 9 -12 36 80 – 85 7 82.5 -2 4 -14 28 85 – 90 7 87.5 -1 1 -7 7 90 – 95 15 92.5 0 0 0 0 95 – 100 9 97.5 1 1 9 9 100 – 105 6 102.5 2 4 12 24 105 – 110 6 107.5 3 9 18 54 110 – 115 3 112.5 4 16 12 48 N = 60 6 254
Mean,
Where, A = 92.5, h = 5
So, x̅ = 92.5 + ((6/60) × 5)
= 92.5 + ½
= 92.5 + 0.5
= 93
Then, Variance,
σ2 = (52/602) [60(254) – 62]
= (1/144) [15240 – 36]
= 15204/144
= 1267/12
= 105.583
Hence, standard deviation = σ = √105.583
= 10.275
∴ Mean = 93, variance = 105.583 and Standard Deviation = 10.275
10. The diameters of circles (in mm) drawn in a design are given below:
Diameters 33 – 36 37 – 40 41 – 44 45 – 48 49 – 52 No. of circles 15 17 21 22 25
Calculate the standard deviation and mean diameter of the circles.
[Hint first make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]
Solution:-
Let the assumed mean, A = 42.5 and h = 4
Let us make the table of the given data and append other columns after calculations.
Height (class) Number of children (Frequency fi) Midpoint Xi Yi = (xi – A)/h Yi2 fiyi fiyi2 32.5 – 36.5 15 34.5 -2 4 -30 60 36.5 – 40.5 17 38.5 -1 1 -17 17 40.5 – 44.5 21 42.5 0 0 0 0 44.5 – 48.5 22 46.5 1 1 22 22 48.5 – 52.5 25 50.5 2 4 50 100 N = 100 25 199
Mean,
Where, A = 42.5, h = 4
So, x̅ = 42.5 + (25/100) × 4
= 42.5 + 1
= 43.5
Then, Variance,
σ2 = (42/1002)[100(199) – 252]
= (1/625) [19900 – 625]
= 19275/625
= 771/25
= 30.84
Hence, standard deviation = σ = √30.84
= 5.553
∴ Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553. |
Factors the 72 are the numbers, which offers the an outcome as 72 when multiplied together in a pair of two. For example, 2×3 = 6, claims that 2 and also 3 room the determinants of 6. Basically, multiples of 72 give an extended timetable the 72, such as 72, 144, 216, 288, 360, 432, 504, 576, 648 and so on. To discover the components of a number, 72, us will usage the administrate method. The components of 72 deserve to be stood for either in optimistic or an adverse forms. Yet the determinants of 72 can not be a decimal or fraction. For example, the factors of 72 deserve to be (1, 72) or (-1, -72). If us multiply a pair of a an adverse number, such together multiplying -1 and -72, that will an outcome in the initial number 72.
You are watching: List all the factors of 72
In this article, we room going to learn the factors of 72, and the pair factors and also the prime factors of 72 making use of the element factorization an approach with plenty of solved examples.
Table the Contents:
## What room the factors of 72?
The components of 72 space the numbers that division 72 precisely without leaving any kind of remainder. In various other words, the components of 72 are the number that room multiplied in pairs causing an original number 72. As the number 72 is a composite number, the has plenty of factors other than one and also the number itself. Hence the determinants of 72 space 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Factors that 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and also 72.Prime administrate of 72: 2 × 2 × 2 × 3 × 3 or 23 × 32
## Pair components of 72
The pair components of 72 room the pair the numbers, which are multiplied together leading to an initial number 72. The pair components of 72 have the right to be optimistic pair or an adverse pair. If us multiply the pair of an adverse numbers, the will an outcome in an original number 72. Thus, the positive and the an adverse pair components of 72 room as follows:
Positive Pair components of 72:
Positive determinants of 72 Positive Pair factors of 72 1 × 72 (1, 72) 2 × 36 (2, 36) 3 × 24 (3, 24) 4 × 18 (4, 18) 6 × 12 (6, 12) 8 × 9 (8, 9)
Hence, the hopeful pair factors of 72 room (1, 72), (2, 36), (3, 24), (4, 18), (6, 12) and (8, 9).
Negative Pair components of 72:
Negative components of 72 Negative Pair determinants of 72 -1 × -72 (-1, -72) -2 × -36 (-2, -36) -3 × -24 (-3, -24) -4 × -18 (-4, -18) -6 × -12 (-6, -12) -8 × -9 (-8, -9)
Hence, the an unfavorable pair factors of 72 space (-1, -72), (-2, -36), (-3, -24), (-4, -18), (-6, -12) and (-8, -9).
The number 72 is a composite number. Currently let us uncover the prime factorisation of this number.
The first step is to division the number 72 v the the smallest prime factor,i.e. 2.
72 ÷ 2 = 36
Again, division 36 through 2.
36 ÷ 2 = 18
18 ÷ 2 = 9
Now, if we divide 9 by 2 we acquire a portion number, which can not be a factor.
Now, continue to the following prime numbers, i.e. 3.
9 ÷ 3 = 3
3 ÷ 3 = 1
Links related to Factors Factors of 15 Factors of 36 Factors of 48 Factors that 18 Factors of 42 Factors that 60 Factors that 35 Factors of 24 Factors the 84 Factors the 50
### Examples
Example 1:
Find the usual factors that 72 and 71.
Solution:
The determinants of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and also 72.
The determinants of 71 space 1 and 71.
Thus, the usual factor the 72 and also 71 is 1.
Example 2:
Find the usual factors the 72 and also 73.
Solution:
Factors the 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Factors of 73 = 1 and also 73.
As 73 is a prime number, the common factor the 72 and also 73 is 1.
Example 3:
Find the usual factors that 72 and 70.
Solution:
The components of 72 space 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and also 72.
The components of 70 room 1, 2, 5, 7, 10, 14, 35, 70
Therefore, the typical factors that 70 and also 72 space 1 and 2.
### Practice Questions
Find the usual factors of 72 and also 36.What is the usual factor of 72 and also 144?What is the amount of all the factors of 72?What are the prime components of 72?Find the typical factors the 72 and 30.
See more: What Does Frfr Mean On Fb - What Does Frfr Mean On Facebook
Learn much more about factors and also prime components here v us in BYJU’S and additionally download BYJU’S – The Learning app for a much better experience. |
Rounding is making a number simple but keeping its value nearest to what it was. The result of round-off is less accurate and easier to use especially while doing arithmetical operations. Likewise, rounding a number to the nearest hundred means making the units and tens place zeros and either increasing or decreasing the remaining part of the number by 1. Students can get the solved examples questions, definitions, and round-off rules in the below-mentioned sections of this page.
## Round off to Nearest 100
Round off to the Nearest 100 is a process where you need to convert the given number into an easy form for various reasons. The obtained easy number is not the actual value but is an approximate value of the original number. Rounding to the nearest hundred means, you need to convert the given number to the nearest 100.
The purpose of rounding the numbers is it makes the numbers easier to understand and remember, the calculations become easier. The application of rounding is when you want to estimate an answer or try to find the most sensible guess, then rounding is used.
### Rules for Rounding Numbers to Nearest 100
• Rule 1: While round off to nearest 100, if the digit in the tens place is between 0 to 4 or <5, then the tens place is replaced by 0.
• Rule 2: If the digit in the unit’s place is equal to 5 or greater than 5, then the tens place is replaced by 0 and the hundreds place is increased by 1.
### How to Round Numbers to Nearest Hundred?
Get the detailed steps to rounding to the nearest 100. They are along the lines
• Get the number we want to round.
• Identify the digit in the tens place.
• If the digit in the tens place is less than 5, then place zero’s in the tens, units place of the number.
• If the digit in the tens place is more than or equal to 5, then place zero’s in the tens and units place, increase the hundredth place digit by 1.
• Now, write the obtained number.
Also, Read: Rounding Decimals to Nearest Whole Number
### Examples on Rounding Numbers to Nearest 100
Example 1:
Round the following numbers to the nearest 100.
(i) 148
(ii) 5520
(iii) 95
Solution:
(i) 148
Given number is 148
We see the digit in the tens place is 4 means that is less than 5. So, we round to the nearest multiple of a hundred which is less than the number. Keep zeros in the units and tens place.
Therefore, rounding of 148 to the nearest 100 is 100.
(ii) 5520
Given number is 5520
We can see that the digit in tens place is 2 which is less than 2. So, keep zero’s in the tens, units place, and write the remaining digits as it is.
Therefore, rounding of 5520 to the nearest 100 is 5500.
(iii) 95
Given number is 95
The tens digit of the given number is 9 that is more than 5. So, we need to place zeros in the tens, units place and increase the hundreds digit by 1. So, the obtained number is 100.
Example 2:
Round the following numbers to the nearest hundred.
(i) 696
(ii) 1,00,678
(iii) 12,05,896
Solution:
(i)696
Given number is 696
The digit in the tens place of the original number is 9. So, increase the hundreds digit by 1 and place zeros in the units, tens place.
Therefore, the obtained number is 700.
(ii) 1,00,678
Given number is 1,00,678
We can identify that the digit in the tens place of the number is 7 i.e > 5. Therefore, increase the digit in the hundreds place of the number by 1 and put zeros in the units, tens places.
Hence, the rounding off 1,00,678 to the nearest hundred is 1,00,700.
(iii) 12,05,896
Given number is 12,05,896
We see the digit in the tens place is 8, we round to the nearest multiple of hundred which is greater than the number. Hence, 12,05,896 is nearer to 12,05,900 than 12,05,800.
Example 3:
Round of To Nearest 100.
(i) 50
(ii) 255
(iii) 510
Solution:
(i) 50
Given number is 50
The digit in the tens position of the given number is 5. So, increase the hundreds position of the number by 1, place zeros in the tens, units place.
So, the round-off 50 to the nearest 100 is 100.
(ii) 255
Given number is 255
We choose the two multiple of 100 just greater than and just less than 255.
The nearest hundreds of 255 are 200, 300
255 – 200 = 55
300 – 255 = 45
As 300 has the lowest difference value.
So, 300 is the nearest 100 for 255.
(iii) 510
Given number is 510
We see the digit in the tens place is 1, we round to the nearest multiple of hundred which is less than the number. Hence, 510 is nearer to 500 than 600. |
How do you solve |10x + -8| <28?
Apr 28, 2017
See the entire solution process below:
Explanation:
First, rewrite the inequality as:
$\left\mid 10 x - 8 \right\mid < 28$
The absolute value function takes any positive or negative term and transforms it to its positive form. Therefore, you must solve the term within the absolute value function for both its positive and negative equivalent:
$- 28 < 10 x - 8 < 28$
$- 28 + \textcolor{red}{8} < 10 x - 8 + \textcolor{red}{8} < 28 + \textcolor{red}{8}$
$- 20 < 10 x - 0 < 36$
$- 20 < 10 x < 36$
$- \frac{20}{\textcolor{red}{10}} < \frac{10 x}{\textcolor{red}{10}} < \frac{36}{\textcolor{red}{10}}$
$- 2 < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}} x}{\cancel{\textcolor{red}{10}}} < 3.6$
$- 2 < x < 3.6$
Or
$x > - 2$ and $x < 3.6$
Or, in interval notation
$\left(- 2 , 3.6\right)$ |
# Vector spaces induced by matrices: column, row, and null spaces
Published:
Matrices are one of the fundamental objects studied in linear algebra. While on their surface they appear like simple tables of numbers, this simplicity hides deeper mathematical structures that they contain. In this post, we will dive into the deeper structures within matrices by discussing three vector spaces that are induced by every matrix: a column space, a row space, and a null space.
## Introduction
Matrices are one of the fundamental objects studied in linear algebra. While on their surface they appear like simple tables of numbers, as we have previously described, this simplicity hides deeper mathematical structures that they contain. In this post, we will dive into the deeper structures within matrices by showing three vector spaces that are implicitly defined by every matrix:
1. A column space
2. A row space
3. A null space
Not only will we discuss the definition for these spaces and how they relate to one another, we will also discuss how to best intuit these spaces and what their properties tell us about the matrix itself.
To understand these spaces, we will need to look at matrices from different perspectives. In a previous discussion on matrices, we discussed how there are three complementary perspectives for viewing matrices:
• Perspective 1: A matrix as a table of numbers
• Perspective 2: A matrix as a list of vectors (both row and column vectors)
• Perspective 3: A matrix as a function mapping vectors from one space to another
By viewing matrices through these perspectives we can gain a better intuition for the vector spaces induced by matrices. Let’s get started.
## Column spaces
The column space of a matrix is simply the vector space spanned by its column-vectors:
Definition 1 (column space): Given a matrix $\boldsymbol{A}$, the column space of $\boldsymbol{A}$, is the vector space that spans the column-vectors of $\boldsymbol{A}$
To understand the column space of a matrix $\boldsymbol{A}$, we will consider the matrix from Perspectives 2 and 3 – that is, $\boldsymbol{A}$ as a list of column vectors and as a function mapping vectors from one space to another.
Understanding the column space when viewing matrices as lists of column vectors
The least abstract way to view the column space of a matrix is when considering a matrix to be a simple list of column-vectors. For example:
The column space is then the vector space that is spanned by these three vectors. We see that in the example above, the column space is all of $\mathbb{R}^2$ since we can form any two-dimensional vector using a linear combination of these three vectors:
Understanding the column space when viewing matrices as functions
To gain a deeper understanding into the significance of the column space of a matrix, we will now consider matrices from the perspective of seeing them as functions between vector spaces. That is, recall for a given matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, we can view this matrix as a function that maps vectors from $\mathbb{R}^n$ to vectors in $\mathbb{R}^m$. This mapping is implemented by matrix-vector multiplication. A vector $\boldsymbol{x} \in \mathbb{R}^n$ is mapped to vector $\boldsymbol{b} \in \mathbb{R}^m$ via
$\boldsymbol{Ax} = \boldsymbol{b}$
Stated more explicitly, we can define a function $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ as:
$T(\boldsymbol{x}) := \boldsymbol{Ax}$
It turns out that the column space is simply the range of this function $T$! That is, it is the set of all vectors that $\boldsymbol{A}$ is capable of mapping to. To see why this is the case, recall that we can view matrix-vector multiplication between $\boldsymbol{A}$ and $\boldsymbol{x}$ as the act of taking a linear combination of the columns of $\boldsymbol{A}$ using the coefficients of $\boldsymbol{x}$ as coefficients:
Here we see that the output of this matrix-defined function will always be contained to the span of the column vectors of $\boldsymbol{A}$.
## Row spaces
The row space of a matrix is the vector space spanned by its row-vectors:
Definition 2 (row space): Given a matrix $\boldsymbol{A}$, the column space of $\boldsymbol{A}$, is the vector space that spans the row-vectors of $\boldsymbol{A}$
To understand the row space of a matrix $\boldsymbol{A}$, we will consider the matrix from Perspective 2 – that is, we will view $\boldsymbol{A}$ as a list of row vectors. For example:
The row space is then the vector space that is spanned by these vectors. We see that in the example above, the row space is a hyperplane:
Unlike the column space, the row space cannot be interpreted as either the domain or range of the function defined by the matrix. So what is the geometric significance of the row space in the context of Perspective 3 (viewing matrices as functions)? Unfortunately, this does not become evident until we discuss the null space, which we will discuss in the next section!
## Null spaces
The null space of a matrix is the third vector space that is induced by matrices. To understand the null space, we will need to view matrices from Perspective 3: matrices as functions between vector space.
Specifically, the null space of a matrix $\boldsymbol{A}$ is the set of all vectors that $\boldsymbol{A}$ maps to the zero vector. That is, the null space is all vectors, $\boldsymbol{x} \in \mathbb{R}^n$ for which $\boldsymbol{Ax} = \boldsymbol{0}$:
Definition 3 (null space): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, the null space of $\boldsymbol{A}$ is the set of vectors, $\{\boldsymbol{x} \in \mathbb{R}^n \mid \boldsymbol{Ax} = \boldsymbol{0}\}$
It turns out that there is a key relationship between the null space and the row space of a matrix: the null space is the orthogonal complement to the row space (Theorem 1 in the Appendix to this post). Before going further, let us define the orthogonal complement. Given a vector space $(\mathcal{V}, \mathcal{F})$, the orthogonal complement to this vector space is another vector space, $(\mathcal{V}’, \mathcal{F})$, where all vectors in $\mathcal{V}’$ are orthogonal to all vectors in $\mathcal{V}$:
Definition 4 (orthogonal complement): Given two vector spaces $(\mathcal{V}, \mathcal{F})$ and $(\mathcal{V}’, \mathcal{F})$ that share the same scalar field, each is an orthogonal complement to the other if $\forall \boldsymbol{v} \in \mathcal{V}, \ \forall \boldsymbol{v}’ \in \mathcal{V}’ \ \langle \boldsymbol{v}, \boldsymbol{v}’ \rangle = 0$
Stated more formally:
Theorem 1 (null space is orthogonal complement of row space): Given a matrix $\boldsymbol{A}$, the null space of $\boldsymbol{A}$ is the orthogonal complement to the row space of $\boldsymbol{A}$.
To see why the null space and row space are orthogonal complements, recall that we can view matrix-vector multiplication between a matrix $\boldsymbol{A}$ and a vector $\boldsymbol{x}$ as the process of taking a dot product of each row of $\boldsymbol{A}$ with $\boldsymbol{x}$:
$\boldsymbol{Ax} := \begin{bmatrix} \boldsymbol{a}_{1,*} \cdot \boldsymbol{x} \\ \boldsymbol{a}_{2,*} \cdot \boldsymbol{x} \\ \vdots \\ \boldsymbol{a}_{m,*} \cdot \boldsymbol{x} \end{bmatrix}$
If $\boldsymbol{x}$ is in the null space of $\boldsymbol{A}$ then this means that $\boldsymbol{Ax} = \boldsymbol{0}$, which means that every dot product shown above is zero. That is,
\begin{align*}\boldsymbol{Ax} &= \begin{bmatrix} \boldsymbol{a}_{1,*} \cdot \boldsymbol{x} \\ \boldsymbol{a}_{2,*} \cdot \boldsymbol{x} \\ \vdots \\ \boldsymbol{a}_{m,*} \cdot \boldsymbol{x} \end{bmatrix} \\ &= \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \\ &= \boldsymbol{0} \end{align*}
Recall, if the dot product between a pair of vectors is zero, then the two vectors are orthogonal. Thus we see that if $\boldsymbol{x}$ is in the null space of $\boldsymbol{A}$ it has to be orthogonal to every row-vector of $\boldsymbol{A}$. This means that the null space is the orthogonal complement to the row space!
We can visualize the relationship between the row space and null space using our example matrix:
$\begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & -1\end{bmatrix}$
The null space for this matrix is comprised of all of the vectors that point along the red vector shown below:
Notice that this red vector is orthogonal to the hyperplane that represents the row space of $\boldsymbol{A}$.
## Rank: the intrinsic dimensionality of the row and column space
The intrinsic dimensionality of the row space and column space are also related to one another and tell us alot about the matrix itself. Recall, the intrinsic dimensionality of a set of vectors is given by the maximal number of linearly independent vectors in the set. With this in mind, we can form the following definitions that describe the intrinsic dimensionalities of the row space and column space:
Definition 3 (column rank): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, the column rank of $\boldsymbol{A}$ is the maximum sized subset of the columns of $\boldsymbol{A}$ that are linearly independent.
Definition 4 (row rank): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, the row rank of $\boldsymbol{A}$ is the maximum sized subset of the rows of $\boldsymbol{A}$ that are linearly independent.
It turns out that intrinsic dimensionality of the row space and column space are always equal and thus the column rank will always equal the row rank:
Theorem 2 (row rank equals column rank): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, its row rank equals its column rank.
Because of the row rank and column rank are equal, one can simply talk about the rank of a matrix without the need to delineate whether we mean the row rank or the column rank.
Moreover, because the row rank equals the column rank of a matrix, a matrix of shape $m \times n$ can at most have a rank that is the minimum of $m$ and $n$. For example, a matrix with 3 rows and 5 columns can at most be of rank 3 (but it might be less!). In fact, we observed this phenomenon in our previous example matrix, which has a rank of 2:
As we can see, the column space spans all of $\mathbb{R}^2$ and thus, it’s intrinsic dimensionality is two. The row space spans a hyperplane in $\mathbb{R}^3$ and thus, it’s intrinsic dimensionality is also two.
## Nullity: the intrinsic dimensionality of the null space
Where the rank of a matrix describes the intrinsic dimensionality of the row and column spaces of a matrix, the nullity describes the intrinsic dimensionality of the null space:
Definition 5 (nullity): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, the nullity of $\boldsymbol{A}$ is the maximum number of linearly independent vectors that span the null space of $\boldsymbol{A}$.
There is a key relationship between nullity and rank: they sum to the number of columns of $\boldsymbol{A}$! This is proven in the rank-nullity theorem (proof provided in the Appendix to this post):
Theorem 3 (rank-nullity theorem): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, it holds that $\text{rank} + \text{nullity} = n$.
Below we illustrate this theorem with two examples:
On the left, we have a matrix whose rows span a hyperplane in $\mathbb{R}^3$, which is of dimension 2. The null space is thus a line, which has dimension 1. In contrast, on the right we have a matrix whose rows span a line in $\mathbb{R}^3$, which is of dimension 1. The null space here is a hyperplane that is orthogonal to this line. In both examples, the dimensionality of the row space and null space sum to 3, which is the number of columns of both matrices!
## Summarizing the relationships between matrix spaces
We can summarize the properties of the column space, row space, and null space with the following table organized around Perspective 2 (matrices as lists of vectors) and Perspective 3 (matrices as functions):
Moreover, we can summarize the relationships between these spaces with the following figure:
## The spaces induced by invertible matrices
We conclude by discussing the vector spaces induced by invertible matrices. Recall, that a square matrix $\boldsymbol{A} \in \mathbb{R}^{n \times n}$ is invertible if and only if its columns are linearly independent (see Theorem 4 in the Appendix to my previous blog post). This implies that for invertible matrices, it holds that:
1. The column space spans all of $n$ since they are linearly independent. This implies that the column rank is $n$
2. The row space spans all of $n$, since by Theorem 2 the row rank equals the column rank
3. The nullity is zero, since by Theorem 3 the nullity plus the rank must equal the number of columns
We call an invertible matrix full rank since the rank equals the number of rows and columns. The rank is “full” because it cannot be increased any further past the number of its columns/rows!
Moreover, we see that there is only one vector in the null space of an invertible matrix since its nullity is zero (a dimensionality of zero corresponds to a single point). If we think back on our discussion of invertible matrices as characterizing invertible functions, then this fact makes sense. For a function to be invertible, it must be one-to-one and onto. So if we use an invertible matrix $\boldsymbol{A}$ to define the function
$T(\boldsymbol{x}) := \boldsymbol{Ax}$
Then it holds that every vector, $\boldsymbol{b}$, in the range of the function $T$ has exactly one vector, $\boldsymbol{x}$, in the domain of $T$ for which $T(\boldsymbol{x}) = \boldsymbol{b}$. This must also hold for the zero vector. Thus, there must be only one vector, $\boldsymbol{x}$, for which $\boldsymbol{Ax} = \boldsymbol{0}$. Hence, the null space comprises just a single vector.
Now we may ask, what vector is this singular member of the null space. It turns out, it’s the zero vector! We see this by applying Theorem 1 from this previous blog post.
## Appendix
Theorem 1 (null space is orthogonal complement of row space): Given a matrix $\boldsymbol{A}$, the null space of $\boldsymbol{A}$ is the orthogonal complement to the row space of $\boldsymbol{A}$.
Proof:
To prove that the null space of $\boldsymbol{A}$ is the orthogonal complement of the row space, we must show that every vector in the null space is orthogonal to every vector in the row space. Consider vector $\boldsymbol{x}$ in the null space of $\boldsymbol{A}$. By the definition of the null space (Definition 5), this means that $\boldsymbol{Ax} = \boldsymbol{0}$. That is,
\begin{align*}\boldsymbol{Ax} &= \begin{bmatrix} \boldsymbol{a}_{1,*} \cdot \boldsymbol{x} \\ \boldsymbol{a}_{2,*} \cdot \boldsymbol{x} \\ \vdots \\ \boldsymbol{a}_{m,*} \cdot \boldsymbol{x} \end{bmatrix} \\ &= \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \\ &= \boldsymbol{0} \end{align*}
We note that for each row $i$, we see that $\boldsymbol{a}_{i,*} \cdot \boldsymbol{x} = 0$ implies that each row vector of $\boldsymbol{A}$ is orthogonal to $\boldsymbol{x}$.
$\square$
Theorem 2 (row rank equals column rank): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, the row rank equals the column rank
Proof:
This proof is described on Wikipedia, provided here in my own words.
Let $r$ be the row rank of $\boldsymbol{A}$ and let $\boldsymbol{b}_1, \dots, \boldsymbol{b}_r \in \mathbb{R}^n$ be a set of basis vectors for the row space of $\boldsymbol{A}$. Now, let $c_1, c_2, \dots, c_r$ be coefficients such that
$\sum_{i=1}^r c_i \boldsymbol{Ab}_i = \boldsymbol{0}$
Furthermore, let
$\boldsymbol{v} := \sum_{i=1}^r c_i\boldsymbol{b}_i$
We see that
\begin{align*} \sum_{i=1}^r c_i \boldsymbol{Ab}_i &= \boldsymbol{0} \\ \implies \sum_{i=1}^r \boldsymbol{A}c_i \boldsymbol{b}_i &= \boldsymbol{0} \\ \implies \boldsymbol{A} \sum_{i=1}^r c_i\boldsymbol{b}_i &= \boldsymbol{0} \\ \boldsymbol{Av} &= \boldsymbol{0} \end{align*}
With this in mind, we can prove that $\boldsymbol{v}$ must be the zero vector. To do so, we first note that $\boldsymbol{v}$ is in both the row space of $\boldsymbol{A}$ and the null space of $\boldsymbol{A}$. It is in the row space $\boldsymbol{A}$ because it is a linear combination of the basis vectors of the row space of $\boldsymbol{A}$. It is in the null space of $\boldsymbol{A}$, because $\boldsymbol{Av} = \boldsymbol{0}$. From Theorem 1, $\boldsymbol{v}$ must be orthogonal to all vectors in the row space of $\boldsymbol{A}$, which includes itself. The only vector that is orthogonal to itself is the zero vector and thus, $\boldsymbol{v}$ must be the zero vector.
This in turn implies that $c_1, \dots, c_r$ must be zero. We know this because $\boldsymbol{b}_1, \dots, \boldsymbol{b}_r \in \mathbb{R}^n$ are basis vectors, which by definition cannot include the zero vector. Thus we have proven that the only assignment of values for $c_1, \dots, c_r$ for which $\sum_{i=1}^r c_i \boldsymbol{Ab}_i = \boldsymbol{0}$ is the assignment for which they are all zero. By Theorem 1 in a previous post, this implies that $\boldsymbol{Ab}_1, \dots, \boldsymbol{Ab}_r$ must be linearly independent.
Moreover, by the definition of matrix-vector multiplication, we know that $\boldsymbol{Ab}_1, \dots, \boldsymbol{Ab}_r$ are in the column space of $\boldsymbol{A}$. Thus, we have proven that there exist at least $r$ independent vectors in the column space of $\boldsymbol{A}$. This means that the column rank of $\boldsymbol{A}$ is at least $r$. That is,
$\text{column rank of} \ \boldsymbol{A} \geq \text{row rank of} \ \boldsymbol{A}$
We can repeat this exercise on the transpose of $\boldsymbol{A}$, which tells us that
$\text{row rank of} \ \boldsymbol{A} \geq \text{column rank of} \ \boldsymbol{A}$
These statements together imply that the column rank and row rank of $\boldsymbol{A}$ are equal.
$\square$
Theorem 3 (rank-nullity theorem): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$, it holds that $\text{rank} + \text{nullity} = n$.
Proof:
This proof is described on Wikipedia, provided here in my own words along with supplemental schematics of the matrices used in the proof.
Let $r$ be the rank of the matrix. This means that there are $r$ linearly independent column vectors in $\boldsymbol{A}$. Without loss of generality, we can arrange $\boldsymbol{A}$ so that the first $r$ columns are linearly independent, and the remaining $n - r$ columns can be written as a linear combination of the first $r$ columns. That is, we can write:
$\boldsymbol{A} = \begin{pmatrix} \boldsymbol{A}_1 & \boldsymbol{A}_2 \end{pmatrix}$
where $\boldsymbol{A}_1$ and $\boldsymbol{A}_2$ are the two partitions of the matrix as shown below:
because the columns of $\boldsymbol{A}_2$ are linear combinations of the columns of $\boldsymbol{A}_1$, there exists a matrix $\boldsymbol{B} \in \mathbb{R}^{r \times n-r}$ for which
$\boldsymbol{A}_2 = \boldsymbol{A}_1 \boldsymbol{B}$
This is depicted below:
Now, consider a matrix
$\boldsymbol{X} := \begin{pmatrix} -\boldsymbol{B} \\ \boldsymbol{I}_{n-r} \end{pmatrix}$
That is, $\boldsymbol{X}$ is formed by concatenating the $n-r \times n-r$ identity matrix below the $-\boldsymbol{B}$ matrix. Now, we see that $\boldsymbol{AX} = \boldsymbol{0}$:
\begin{align*}\boldsymbol{AX} &= \begin{pmatrix} \boldsymbol{A}_1 & \boldsymbol{A}_1\boldsymbol{B} \end{pmatrix} \begin{pmatrix} -\boldsymbol{B} \\ \boldsymbol{I}_{n-r} \end{pmatrix} \\ &= -\boldsymbol{A}_1\boldsymbol{B} + \boldsymbol{A}_1\boldsymbol{B} \\ &= \boldsymbol{0} \end{align*}
Depicted schematically,
Thus, we see that every column of $\boldsymbol{X}$ is in the null space of $\boldsymbol{A}$.
We now show that these column vectors are linearly independent. To do so, we will consider a vector $\boldsymbol{u} \in \mathbb{R}^{n-r}$ such that
$\boldsymbol{Xu} = \boldsymbol{0}$
For this to hold, we see that $\boldsymbol{u}$ must be zero:
\begin{align*}\boldsymbol{Xu} &= \boldsymbol{0} \\ \implies \begin{pmatrix} -\boldsymbol{B} \\ \boldsymbol{I}_{n-r} \end{pmatrix}\boldsymbol{u} &= \begin{pmatrix} \boldsymbol{0}_r \\ \boldsymbol{0}_{n-r} \end{pmatrix} \\ \\ \implies \begin{pmatrix} -\boldsymbol{Bu} \\ \boldsymbol{u} \end{pmatrix} &= \begin{pmatrix} \boldsymbol{0}_r \\ \boldsymbol{0}_{n-r} \end{pmatrix} \end{align*}
By Theorem 1 in a previous post, this proves that the columns of $\boldsymbol{X}$ are linearly independent. So we have shown that there exists $n-r$ linearly independent vectors in the null space of $\boldsymbol{A}$, which means the nullity is at least $n-r$.
We now show that any other vector in the null space of $\boldsymbol{A}$ that is not a column of $\boldsymbol{X}$ can be written as a linear combination of the columns of $\boldsymbol{X}$. If we can prove this fact, we will have proven that the nullity is exactly equal to $n-r$ and is not greater.
We start by again considering a vector $\boldsymbol{u} \in \mathbb{R}^n$ that we assume is in the null space of $\boldsymbol{A}$. We partition this vector into two segments: one segment, $\boldsymbol{u}_1$, comprising the first $r$ elements and a second segment, $\boldsymbol{u}_2$, comprising the remaining $n-r$ elements:
$\boldsymbol{u} = \begin{pmatrix}\boldsymbol{u}_1 \\ \boldsymbol{u}_2 \end{pmatrix}$
Because we assume that $\boldsymbol{u}$ is in the null space, it must hold that $\boldsymbol{Au} = \boldsymbol{0}$. Depicted schematically:
Solving for $\boldsymbol{u}$, we see that
\begin{align*} \boldsymbol{Au} &= \boldsymbol{0} \\ \begin{pmatrix} \boldsymbol{A}_1 & \boldsymbol{A}_2 \end{pmatrix} \begin{pmatrix}\boldsymbol{u}_1 \\ \boldsymbol{u}_2 \end{pmatrix} &= \boldsymbol{0} \\\begin{pmatrix} \boldsymbol{A}_1 & \boldsymbol{A}_1\boldsymbol{B} \end{pmatrix} \begin{pmatrix}\boldsymbol{u}_1 \\ \boldsymbol{u}_2 \end{pmatrix} &= \boldsymbol{0} \\ \implies \boldsymbol{A}_1\boldsymbol{u}_1 + \boldsymbol{A}_1\boldsymbol{B}\boldsymbol{u}_2 &= \boldsymbol{0} \\ \implies \boldsymbol{A}_1 (\boldsymbol{u}_1 + \boldsymbol{Bu}_2) &= \boldsymbol{0} \\ \implies \boldsymbol{u}_1 + \boldsymbol{Bu}_2 &= \boldsymbol{0} \\ \implies \boldsymbol{u}_1 = -\boldsymbol{Bu}_2 \end{align*}
Thus,
\begin{align*}\boldsymbol{u} &= \begin{pmatrix}\boldsymbol{u}_1 \\ \boldsymbol{u}_2 \end{pmatrix} \\ &= \begin{pmatrix} -\boldsymbol{Bu}_2 \\ \boldsymbol{u}_2 \end{pmatrix} \\ &= \begin{pmatrix} -\boldsymbol{B} \\ \boldsymbol{I}_{n-r} \end{pmatrix}\boldsymbol{u}_2 \\ &= \boldsymbol{X}\boldsymbol{u}_2 \end{align*}
Thus, we see that $\boldsymbol{u}$ must be the linear combination of the columns of $\boldsymbol{X}$! Thus we have shown that:
1. There exists $n-r$ linearly independent vectors in the null space of $\boldsymbol{A}$
2. Any vector in the null space can be expressed as a linear combination of these linearly independent vectors
This proves that the nullity is $n-r$, and thus, the nullity $n-r$ plus the rank $r$, equals $n$.
$\square$
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How To Simplify Rational Exponents
This lesson covers how to simplify rational exponents in exponent form using the properties of exponents and is the second part to my blog post on Radical Form and Rational Exponents. Working with rational is an algebra 2, but the properties of exponents learned in algebra 1 with integer exponents apply. Here is a list of the properties of exponents for review:
• Product of Powers $a^{m} \cdot a^{n} = a^{m+n}$
• Power of Powers $(a^{m})^{n} = a^{mn}$
• Power of a Monomial $(ab)^{m} = a^{m}b^{m}$
• Negative Exponent $a^{-n} = \frac{1}{a^{n}}$
• Quotient of Powers $\frac{a^{m}}{a^{n}} = a^{m-n}$
• Power of a Quotient $(\frac{a}{b})^m = \frac{a^{m}}{b^{m}}$
The first section of the video reviews the above list of properties.
The second part of the video (example 4T part a) simplifying the expression $(-32)^{\frac{3}{5}}$. There are two methods to simplify this expression. The first method that is modeled is using the properties of exponents and the second method modeled is converting the rational exponent into a radical and a power. Either method that is used requires the problem solver to figure out “what number can be raised to the n power to get the original number?”
Example 4T part b is similar to part a, but the exponent is expressed as a mixed decimal number. To evaluate this expression, the mixed decimal number needs to be converted to an improper fraction. Then either method modeled in part a can be applied.
Finally, example 5T involve a more complicated expression to simplify and it involves negative exponents. The key concept to know is that all negative exponents need to be expressed as positive exponents.
Enjoy the video and feel free to post any questions in the comment section.
Regards,
Mr. Pi |
# Minors and Cofactors of Determinants
## Minors of Determinants
For a given square matrix $A$ of order $n \times n$, the minor $M_{ij}$ corresponding to the element $a_{ij}$ is the determinant of the submatrix that remains after removing the $i^{th}$ row and $j^{th}$ column from $A$.
For a 2x2 matrix
$A = \begin{vmatrix} a & b \\ c & d \end{vmatrix}$
The minors are simply the elements themselves, as removing a row and a column leaves a 1x1 matrix.
$M_{11} = a$
$M_{12} = b$
$M_{21} = c$
$M_{22} = d$
For a 3x3 Matrix
Consider a 3x3 matrix $A$ given by
$A = \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$
The minor $M_{11}$ corresponding to $a$ is the determinant of the 2x2 matrix obtained by removing the first row and first column:
$M_{11} = \begin{vmatrix} e & f \\ h & i \end{vmatrix} = ei - fh$
The minors $M_{ij}$ can be calculated as follows:
- $M_{11} = ei - fh$
- $M_{12} = di - fg$
- $M_{13} = dh - eg$
## Cofactors of Determinants
Cofactor of an element $a_{ij}$ , denoted by $A_{ij}$ is defined by
$A_{ij} = (–1)^{i + j} M_{ij}$ , where $M_{ij}$ is minor of $a_{ij}$
For a 2x2 matrix
$A = \begin{vmatrix} a & b \\ c & d \end{vmatrix}$
Minors are given as
$M_{11} = d$
$M_{12} = c$
$M_{21} = b$
$M_{22} = a$
Hence Cofactors are
$A_{11} = d$
$A_{12} = -c$
$A_{21} = -b$
$A_{22} = a$
For a 3x3 Matrix
Consider a 3x3 matrix $A$ given by
$A = \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$
The minors $M_{ij}$ can be calculated as follows:
- $M_{11} = ei - fh$
- $M_{12} = di - fg$
- $M_{13} = dh - eg$
Hence cofactors will be
- $A_{11} = ei - fh$
- $A_{12} = -(di - fg)$
- $A_{13} = dh - eg$
## Determinants Value in terms of Cofactors
The determinant $|A|$ is calculated as
$|A| = (-1)^{1 +1} a(ei - fh) + (-1)^{1 +2} b(di - fg) + (-1)^{1 +3}c(dh - eg)=a A_{11} + bA_{12} + cA_{13}$
Similary we can can be calculated by other five ways of expansion that is along R2, R3,C1, C2 and C3
Hence $\Delta$ = sum of the product of elements of any row (or column) with their corresponding cofactors
Important Point
If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero. For example
$d A_{11} + eA_{12} + fA_{13}=(-1)^{1 +1} d(ei - fh) + (-1)^{1 +2} e(di - fg) + (-1)^{1 +3}f(dh - eg)=dei -dfh -dei +efg + fdh - efg=0$
• Notes NCERT Solutions & Assignments
Go back to Class 12 Main Page using below links |
Question
# If $AB=7cm$, $BP=4cm$, $AP=5.4cm$, then compare the segments.A. $AB>BP>AB$B. $BPC.$AP>BPD. None of these
Hint:
We are given $AB=7cm$, $BP=4cm$, $AP=5.4cm$ and we are told to compare the segments. Comparing the segments means to arrange in order from bigger to smaller. So, see the numbers and arrange from bigger to smaller. Then taking the reference of numbers arrange the segments $AB,AP$ and $BP$. Try it, you will get the answer.
In geometry, a line segment is a part of a line that is bounded by two distinct endpoints, and contains every point on the line between its endpoints. A closed line segment includes both endpoints, while an open line segment excludes both endpoints; a half-open line segment includes exactly one of the endpoints.
Examples of line segments include the sides of a triangle or square. More generally, when both of the segment's end points are vertices of a polygon or polyhedron, the line segment is either an edge (of that polygon or polyhedron) if they are adjacent vertices, or otherwise a diagonal. When the end points both lie on a curve such as a circle, a line segment is called a chord (of that curve).
In geometry, a line can be defined as a straight one- dimensional figure that has no thickness and extends endlessly in both directions.
It is often described as the shortest distance between any two points.
A line is one-dimensional. It has zero width. If you draw a line with a pencil, examination with a microscope would show that the pencil mark has a measurable width. The pencil line is just a way to illustrate the idea on paper. In geometry however, a line has no width.
A straight line is the shortest distance between any two points on a plane.
Line, Basic element of Euclidean geometry. Euclid defined a line as an interval between two points and claimed it could be extended indefinitely in either direction. Such an extension in both directions is now thought of as a line, while Euclid’s original definition is considered a line segment. A ray is part of a line extending infinitely from a point on the line in only one direction. In a coordinate system on a plane, a line can be represented by the linear equation $ax+by+c=0$. This is often written in the slope-intercept form as $y=mx+b$, in which $m$is the slope and $b$ is the value where the line crosses the$y$ -axis. Because geometrical objects whose edges are line segments are completely understood, mathematicians frequently try to reduce more complex structures into simpler ones made up of connected line segments.
In the question we are given the length of segments $AB=7cm$, $BP=4cm$, $AP=5.4cm$.
Comparing the segments means to arrange in order from bigger to smaller.
Here, we can see that,
$AB=7cm$, $BP=4cm$, $AP=5.4cm$,
So, arranging the numbers from bigger to smaller we get,
$7>5.4>4$
So we get,
$AB>AP>BP$ i.e. \$BPSo we get the correct answer as option (B).
Note: Read the question in a careful manner. Also, do not miss any term while arranging. Do not confuse yourself with the greater and smaller signs. Your concept regarding segments should be cleared. You must know that, comparing the segments means to arrange in order from bigger to smaller. |
# Successive|Definition & Meaning
## Definition
Successive literally means to come after the other in some discernable order. For example, the days of a week are successive. Consecutive is not the same as successive. 1, 2, 3, and 4 are both successive and consecutive natural numbers because there are no gaps. However, 1, 3, 5, and 7 are only successive as there is a gap between them, but they still follow an order (+2).
In mathematics, successive means a sequence of values created in a particular order, typically via a mathematical procedure or algorithm. The sequence is generated by continually performing a mathematical function to a beginning value, yielding a new value with each iteration
The mathematical approach utilized to construct the sequence determines the ordering of the numbers. It is common practice to utilize successive integers to estimate solutions to issues that cannot be addressed exactly or to determine the maximum value or minimum value of a function.
Figure 1: Illustration of Successive Numbers
## Successive Numbers and Numerical Analysis
It is common practice in numerical analysis to employ successive numbers to approximately solve issues that could not be solved precisely. Finding the roots of problems and resolving differential equations are two frequent uses of successive numbers in numerical analysis.
### Finding Roots
Finding the roots of an integer is one of the most popular uses of successive values in numerical analysis. An equation’s root is a number that, when introduced into the equation, causes it to reach zero. For instance, the roots for the equation below will be x = 3 and x = -3.
f(x) = x2 – 9 = 0
Numerical techniques like the bisection method, secant method, and Newton-Raphson method are frequently employed to discover approximations of equations’ roots because it is usually impossible to calculate the equation’s roots analytically. These techniques work by repeatedly performing mathematical computations on an initial estimate to get repeated estimations of the root.
### Solving the Differential Equations
The solution of differential equations is yet another significant use of successive numbers in numerical analysis. Differential equations are mathematical expressions that explain how a phenomenon alters over time. They are employed in the modeling of a vast array of physical and biological processes. one popular approach for resolving differential equations is the Euler method. This method is focused on constructing successive estimations of the problems by repeatedly performing mathematical functions on a starting value.
## Successive Numbers in Calculus
Numerous calculus operations, such as successive derivatives and integrals, make use of successive numbers.
### Successive Derivatives
A function’s derivative in calculus is a measurement of how the function alters as its input alters. The second derivatives, also referred to as the second-order derivatives, are derivatives of the first derivatives. In the same manner, the derivative of (n-1)th order derivative is the nth order derivative. They are often referred to as successive derivatives and are employed to examine the local behavior of a function close to a point.
Figure 2: Illustration of Successive Derivatives
### Successive Integrals
The integral of functions in calculus represents the region beneath the curve of the functions. The function whose derivative is the original function is known as the definite integral and is commonly referred to as the antiderivative. in the same manner, the nth integral is the integral whose derivative is the (n-1)th integral. These can be used to determine the overall variation in a function during a specific period of time and are often referred to as successive integrals.
## Successive Numbers and Optimization
In the process of optimizing, we determine the maximum or minimum value of a function by using successive numbers. The process of determining which of a number of potential solutions is the most beneficial is known as optimization. In a variety of optimization methods, such as gradient descent, conjugate gradient, and Newton’s method, successive integers play an important role.
### Minimizing and Maximizing Functions
When determining the minimum or maximum value of a function, successive numbers are considered. It is possible for the function to be quite straightforward, such as a quadratic function, or rather complicated, such as a neural network. Various optimization strategies, such as gradient descent, conjugate gradient, and Newton’s method, make use of subsequent integers to continuously improve an initial approximation of the least or maximum value.
### Linear and Non-linear Optimisation
In both linear and non-linear optimization, successive numbers are essential. When it comes to optimization, linear optimization focuses on linear functions and limitations, while non-linear optimization focuses on non-linear functions and limitations. The non-linear optimization process is utilized in various domains, including control systems, image analysis, and computer vision.
## Generation of Successive Numbers Using Iterative Techniques
In mathematics, the generation of successive numbers frequently makes use of a method called the iterative approach. Iterative approaches entail continually performing a mathematical operation to the original value, which ultimately results in a different value for each iteration of the process. The mathematical algorithm that was used to construct the series of numbers is what decides the order of the numbers within it.
There are a variety of iterative approaches that can be utilized, such as the following, to produce successive numbers:
1. Fixed-Point Iteration
2. Newton-Raphson Method
3. Successive Over-Relaxation
4. Jacobi and Gauss-Seidel method
## Generation of Successive Numbers Using Recursive Methods
In mathematics, generating successive numbers can be accomplished through the application of recursive techniques. In recursive approaches, a sequence of numbers is defined in terms of the numbers that came before them in the sequence. This is often accomplished through the utilization of a recursive equation or algorithm.
There are a variety of recursive approaches that you can employ to produce successive numbers, such as the following:
### Fibonacci Numbers
The famous Fibonacci numbers are an iconic illustration of a recursive sequence. Each successive number in the series represents the addition of the two numbers that came before it. The digits 0 and 1 mark the beginning of the sequence, and from that point on, each succeeding number in the series is calculated by adding the two numbers that came before it.
### Recursive Algorithm
An algorithm is said to be recursive if it can invoke itself with data that is less complex. Computing the factorial of an integer, for instance, can be performed recursively.
### Recursive Formulas
A formula is said to be recursive if it describes a sequence of numbers by making reference to the numbers that came before them in the sequence.
## Generation of Successive Numbers Using Convergence Criteria
The application of convergence criteria allows one to ascertain whether or not a succession of successive numbers has, in fact, converged to a certain number or answer.
The criteria are employed to assess whether the sequence has acquired a level of stability, which is a state in which the values contained inside the sequence do not vary substantially from one iteration to the next.
For generating successive numbers, one can utilize any one of a number of distinct convergence criteria, such as the following:
### Absolute Error Criteria
This criterion is determined by the absolute deviation that exists between the sequence’s two most recent values. The sequence is deemed to have converged when the variation is smaller than a predetermined criterion for success.
### Relative Error Criteria
This criterion is determined by examining the degree of dissimilarity that exists between the sequence’s two most recent values. It is determined that the sequence converges when the disparity is smaller than a particular threshold.
### Ratio Test
This criterion evaluates how subsequent values in the sequence relate to one another in terms of their ratio. When the ratio gets close to a particular value, we say that the sequence has converged because it has reached that point.
## Example of Successive Values
Complete the series of following successive values
1. 1, 3, 5, ?, 9, 11
2. Monday, Tuesday, Wednesday, ?, Friday, Saturday, and Sunday
3. 2, 4, 6, ?, 10, 12
### Solution
In (a), we can clearly see that the series is of odd numbers. Thus, the completed series is as follows:
(a) 1, 3, 5, 7, 9, 11
In (b), we can clearly see that the series is on weekdays, so the completed series is as follows:
(b) Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday
Figure 3: Illustration of Successive Weekdays
In (c), we can clearly see that the series is of even numbers, so the completed series is as follows:
(c) 2, 4, 6, 8, 10, 12
All images/mathematical drawings were created with GeoGebra. |
# a differentiable function y =f(x) that has the given properties.
1. local minimum at (1,1), and local maximum at (3,3)
2. local minima at (1,1) and (3,3)
3. local maxima at (1,1) and (3,3)
I don't remember how to do this. Thank you.
http://en.wikipedia.org/wiki/Maxima_and_minima
I don't see how any function can have a local minima and maxima at the same point.
That is 3 different problems.
9 months ago
## No problem! I can help you with these questions. Let's go through each scenario one by one.
1. To find a function with a local minimum at (1,1) and a local maximum at (3,3), we can start by considering a quadratic function, as it can easily achieve these properties. Let's suppose our function is y = ax^2 + bx + c, where a, b, and c are constants.
To find the values of a, b, and c, we can set up a system of equations based on the conditions given. Since we want a local minimum at (1,1), we know that the derivative of the function at x = 1 should be zero, and the second derivative should be positive. Similarly, for the local maximum at (3,3), the derivative at x = 3 should be zero, and the second derivative should be negative.
Let's differentiate the quadratic function y = ax^2 + bx + c with respect to x to find the derivative:
dy/dx = 2ax + b
To find the second derivative, we differentiate the derivative with respect to x:
d^2y/dx^2 = 2a
Now, let's substitute x = 1 and x = 3 into the derivative and second derivative equations:
At x = 1:
dy/dx = 2a + b
d^2y/dx^2 = 2a
At x = 3:
dy/dx = 6a + b
d^2y/dx^2 = 2a
From the conditions given, we know that dy/dx = 0 at both x = 1 and x = 3. Hence, we have the following equations:
2a + b = 0 (equation 1)
6a + b = 0 (equation 2)
By solving this system of equations, we can find the values of a and b. Once we have a and b, we can substitute them into the original quadratic function to find c, using either of the points (1,1) or (3,3).
2. For a function with local minima at (1,1) and (3,3), we can use a similar approach. However, instead of having opposite signs for the second derivatives at the two points, we can consider the second derivatives to be positive at both points. The rest of the process would be the same as in scenario 1.
3. Similarly, for a function with local maxima at (1,1) and (3,3), we can consider the second derivatives to be negative at both points. The rest of the process remains the same.
Remember, this method is specific to quadratic functions and might not work for all types of functions. For more complex functions, you may need to use optimization techniques or consult mathematical software.
I hope this helps you understand how to approach these problems systematically. If you have any further questions, feel free to ask! |
Year 2
# Reasoning With 2D Shapes
Students explore properties of 2D shapes, then join together triangles or squares to create new shapes.
This is a classic reSolve sequence aligned with the Australian Curriculum V8.4. It is only available as a downloadable package.
This sequence starts with a task that focuses on developing students’ knowledge of the properties of shapes, particularly triangles, quadrilaterals, pentagons and hexagons. Students apply their knowledge of these properties to identify shapes that might not fit more common representations. Students then move on to work with transformations of two-dimensional shapes, using flips and turns. Students will learn to work systematically to find all the possible solutions to the task.
This sequence is for students who:
• have had some experience with common two-dimensional shapes.
• recognise polygons, or two-dimensional shapes, as plane, closed shapes with three or more straight sides and corners (or angles).
• are familiar with the fact that polygons can be classified according to properties, including the number of sides. Students should be able to use these properties to recognise and name two-dimensional shapes, including non-typical examples of shapes.
Students would benefit from some experience manipulating two-dimensional shapes.
### Lesson 1: Shape Makers
Students create 2D shapes by joining pins on a circular geoboard. As they create shapes, they develop an understanding of the properties of triangles and quadrilaterals. They also identify what makes a shape regular or irregular.
### Lesson 2: Joining Triangles
Students explore the different shapes that can be made by joining together a set number of identical equilateral triangles. Students first explore combinations of two, three and four triangles. They are then asked to find all possible shapes that can be made from five triangles. They are asked to justify that they have found all possible combinations.
### Lesson 3: Combining Squares
Students explore the different tetrominoes that can be made by joining together four squares. The students are asked to justify that they have found all five possible tetrominoes. The students then use two of each tetromino to create two different rectangles.
Last updated June 12 2020.
This is a classic reSolve sequence aligned with the Australian Curriculum V8.4. It is only available as a downloadable package. |
# 11.2 - Introduction to Bootstrapping
11.2 - Introduction to Bootstrapping
In this section, we will start by reviewing the concept of sampling distributions. Recall, we can find the sampling distribution of any summary statistic. Then, the method of bootstrapping samples to find the approximate sampling distribution of a statistic is introduced.
## Review of Sampling Distributions
Before looking at the bootstrapping method, we will need to recall the idea of sampling distributions. More specifically, let's look at the sampling distribution of the sample mean, $$\bar{x}$$.
Suppose we are interested in estimating the population mean, $$\mu$$. To do this, we find a random sample of size $$n$$ and calculate the sample mean, $$\bar{x}$$. But how do we know how good of an estimate $$\bar{x}$$ is? To answer this question, we need to find the standard deviation of the estimate.
Recall that $$\bar{x}$$ is calculated from a random sample and is, therefore, a random variable. Let's call the sample mean from above $$\bar{x}_1$$. Now suppose we gather another random sample of size $$n$$ and calculate $$\bar{x}$$ from that sample and denote it $$\bar{x}_2$$. Take another sample, and so on and so on. With many of these samples, we can construct a histogram of the sample means.
With theory and the central limit theorem, we have the following summary:
If the sample satisfied at least one of the following:
• The distribution of the random variable, $$X$$, is Normal
• The sample size is large; rule of thumb is $$n>30$$
...then the sampling distribution of $$\bar{X}$$ is approximately Normal with
• Mean: $$\mu$$
• Standard deviation: $$\frac{\sigma}{\sqrt{n}}$$
• Standard error: $$\frac{s}{\sqrt{n}}$$
Using the above, we can construct confidence intervals, and hypothesis test for the population mean, $$\mu$$.
What happens when we do not know the underlying distribution and cannot resample from the distribution? How could we estimate certain sample statistics? This is what we try to answer in the next section.
# 11.2.1 - Bootstrapping Methods
11.2.1 - Bootstrapping Methods
Point estimates are helpful to estimate unknown parameters but in order to make inference about an unknown parameter, we need interval estimates. Confidence intervals are based on information from the sampling distribution, including the standard error.
What if the underlying distribution is unknown? What if we are interested in a population parameter that is not the mean, such as the median? How can we construct a confidence interval for the population median?
If we have sample data, then we can use bootstrapping methods to construct a bootstrap sampling distribution to construct a confidence interval.
Bootstrapping is a topic that has been studied extensively for many different population parameters and many different situations. There are parametric bootstrap, nonparametric bootstraps, weighted bootstraps, etc. We merely introduce the very basics of the bootstrap method. To introduce all of the topics would be an entire class in itself.
Bootstrapping
Bootstrapping is a resampling procedure that uses data from one sample to generate a sampling distribution by repeatedly taking random samples from the known sample, with replacement.
Let’s show how to create a bootstrap sample for the median. Let the sample median be denoted as $$M$$.
Steps to create a bootstrap sample:
1. Replace the population with the sample
2. Sample with replacement $$B$$ times. $$B$$ should be large, say 1000.
3. Compute sample medians each time, $$M_i$$
4. Obtain the approximate distribution of the sample median.
If we have the approximate distribution, we can find an estimate of the standard error of the sample median by finding the standard deviation of $$M_1,...,M_B$$.
Sampling with replacement is important. If we did not sample with replacement, we would always get the same sample median as the observed value. The sample we get from sampling from the data with replacement is called the bootstrap sample.
Once we find the bootstrap sample, we can create a confidence interval. For a 90% confidence interval, for example, we would find the 5th percentile and the 95th percentile of the bootstrap sample.
You can create a bootstrap sample to find the approximate sampling distribution of any statistic, not just the median. The steps would be the same except you would calculate the appropriate statistic instead of the median.
#### Video: Bootstrapping
Sampling R Code from the Bootstrapping Video
sampling.distribution <- function(n = 100, B = 1000, mean = 5, sd = 1, confidence = 0.95) {
median <- rep(0, B)
for (i in 1:B) {
median[i] <- median(rnorm(n, mean = mean, sd = sd))
}
med.obs <- median(median)
c.l <- round((1 - confidence) / 2 * B, 0)
c.u <- round(B - (1 - confidence) / 2 * B, 0)
l <- sort(median)[c.l]
u <- sort(median)[c.u]
cat(c.l / 1000 * 100, "-percentile: ", l, "\n")
cat("Median: ", med.obs, "\n")
cat(c.u / 1000 * 100, "-percentile: ", u, "\n")
return(median)
}
bootstrap.median <- function(data, B = 1000, confidence = 0.95) {
n <- length(data)
median <- rep(0, B)
for (i in 1:B) {
median[i] <- median(sample(data, size = n, replace = T))
}
med.obs <- median(median)
c.l <- round((1 - confidence) / 2 * B, 0)
c.u <- round(B - (1 - confidence) / 2 * B, 0)
l <- sort(median)[c.l]
u <- sort(median)[c.u]
cat(c.l / 1000 * 100, "-percentile: ", l, "\n")
cat("Median: ", med.obs, "\n")
cat(c.u / 1000 * 100, "-percentile: ", u, "\n")
return(median)
}
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility |
# Solving Inequalities
Review how to solve an inequality.
To solve the inequality $$-\frac{3}{4}(x-16) > 18$$, follow these steps:
1. Distribute the coefficient $$-\frac{3}{4}$$ across the expression in parentheses:
$-\frac{3}{4}(x-16) > 18$
$-\frac{3}{4}x + \frac{3}{4}(16) > 18$
2. Simplify inside the parentheses:
$-\frac{3}{4}x + \frac{3}{4}(16) > 18$
$-\frac{3}{4}x + \frac{3}{4}(16) > 18$
$-\frac{3}{4}x + 12 > 18$
3. Subtract 12 from both sides to isolate the term with $$x$$:
$-\frac{3}{4}x + 12 - 12 > 18 - 12$
$-\frac{3}{4}x > 6$
4. Multiply both sides by $$-\frac{4}{3}$$ to solve for $$x$$. Remember to flip the inequality sign when multiplying or dividing by a negative number**:
$(-\frac{4}{3}) \times (-\frac{3}{4}x) < (-\frac{4}{3}) \times 6$
$x < -8$
5. Write the solution:
$x < -8$
So, the solution to the inequality $$-\frac{3}{4}(x-16) > 18$$ is $$x < -8$$. |
# 3.1 Derivatives. Derivative A derivative of a function is the instantaneous rate of change of the function at any point in its domain. We say this is.
## Presentation on theme: "3.1 Derivatives. Derivative A derivative of a function is the instantaneous rate of change of the function at any point in its domain. We say this is."— Presentation transcript:
3.1 Derivatives
Derivative A derivative of a function is the instantaneous rate of change of the function at any point in its domain. We say this is the derivative of f with respect to the variable x. If this limit exists, then the function is differentiable.
Symbols Used to Denote Derivatives
Note on Notations dx does not mean “d times x!!” dy does not mean “d times y!!”
Example Find the derivative of the function f(x) = x 3.
Note Your book talks about an “alternate definition.” Do not worry about using the “alternate definition.” You will never see it on an AP exam! If the directions on your HW say to use the alternate definition, use the regular definition of the derivative.
Example Find the derivative of (Multiply by the conjugate)
A Note from the Example From the previous example: What was the domain of f? [0, ∞) What was the domain of f’? (0, ∞) Significance??? Sometimes the domain of the derivative of a function may be smaller than the domain of the function.
Functions and Derivatives Graphically The function f(x) has the following graph: What does the graph of y’ look like? Remember: y’ is the slope of y.
Functions/Derivatives Graphically The derivative is defined at the end points of a function on a closed interval.
One-Sided Derivatives Since derivatives involve limits, in order for a derivative to exist at a certain point, its derivative from the left has to equal its derivative from the right.
Example Show that the following function has a left- and right-hand derivatives at x = 0, but no derivative there.
Download ppt "3.1 Derivatives. Derivative A derivative of a function is the instantaneous rate of change of the function at any point in its domain. We say this is."
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# MHF4U Unit Outline
```K. Stewart
MHF4U Unit Outline
Chapter 1: Polynomial Functions
Throughout this chapter, you will explore how the curves represented by polynomial functions are applied in various
design-related fields such as civil engineering, architectural design, computer graphics design, interior design, and
landscape architecture.
By the end of this chapter, you will be assessed on your ability to:
C1.0
D1.0
identify and describe some key features of polynomial functions, and make connections between the numeric,
graphical, and algebraic representations of polynomial functions.
demonstrate an understanding of average and instantaneous rate of change, and determine, numerically and
graphically, and interpret the average rate of change of a function over a given interval and the instantaneous rate
of change of a function at a given point.
Section
1.1 Power
Functions
1.2
Characteristics
of Polynomial
Functions
Learning goals
What are the key features of the graphs of
power function?
How can I recognize polynomial functions?
What is the connection between the graphs
and equations of power functions?
Vocab: power, exponent, power function,
polynomial function, trigonometric function,
rational function, exponential function,
square root function, coefficient, degree,
polynomial, key features, domain, range,
end behaviour, symmetry (line symmetry,
point symmetry), axis of symmetry,
intercepts
Notation: x , x –
interval notation – < x 0
set notation (– , 0]
graphical notation (on a number line)
domain notation e.g. {x| x }
range notation e.g. {y| y > 0, y }
Key equation: Power function is of the form
f(x) = axn where n is a whole number
Polynomial function is of the form
f(x)= an xn +an-1 xn-1 +an-2 xn-2 +…+a1 x1 +a0
where coefficients are real numbers and n is
a whole number
What are the key features of the graphs of
polynomial functions?
What is the relationship between finite
differences and the equation of a
polynomial function?
What is the connection between the graphs,
equations and tables of values of
polynomial functions?
Vocab: local minimum (or maximum), absolute
(or global) minimum (or maximum),
optimal value, finite difference, finite
difference table, factorial
Notation: n! is read n factorial and is the
product (n)(n – 1) (n – 2)…(2)(1)
Homework
Pg 11 #1 – 4 optional
Pg 12 #6, 7, 8, 13 mandatory
Skills: match graph with corresponding power function,
name function by degree, identify power functions from their
equations, use interval notation, set notation and graphical
notation interchangeably, identify key features of power
functions (domain and range, end behaviour, symmetry,
intercepts, degree, positive or negative leading coefficient),
solve problems involving power functions
Pg 26 #1 – 4, 12 optional
Pg 27 #5 – 8, 11 mandatory
Skills: match graph with corresponding polynomial function,
name function by degree, identify polynomial functions from
their equations, use interval notation, set notation and
graphical notation interchangeably, identify key features of
polynomial functions (domain and range, end behaviour,
symmetry, intercepts, degree, positive or negative leading
coefficient), solve problems involving polynomial functions,
determine value of finite difference using a difference table
and algebraically, use a regression analysis on a graphing
calculator to determine the equation of best fit.
(Continued next page)
K. Stewart
1.3 Equations
and Graphs of
Polynomial
Functions
1.4
Transformations
1.5 Slopes of
Secants and
Average Rate of
Change
1.6 Slopes of
Tangents and
Instantaneous
Rate of Change
What is the connection between the factored
form of a polynomial function and its
graph?
How can I sketch graphs of polynomial
functions from the equations?
How is symmetry represented in the
equation of a polynomial function?
Vocab: order of an intercept, even function,
odd function, constants are coefficients of
x0,
Key equations: even function f(x) = f(–x)
odd function –f(x) = f(–x)
(day 1)
Pg 39 #1, 2 optional
Pg 40 # 9, 11, 12 mandatory
What are the roles of a, k, d, and c in
polynomial functions of the form
y = a[k(x-d)]n+c where n N?
How can I describe transformations from an
equation and use them to sketch a graph?
How can I determine an equation give the
graph of a transformed function?
Vocab: transformation, reflection, translation,
stretch, compression, opening
Notation: y = a[k(x-d)] n+c
How can I connect average rate of change
and slope?
How can I calculate and interpret average
rates of change from any given
representation?
Vocab: average rate of change, slope, secant,
What is the connection between the slopes of
secants, the slope of a tangent and the
instantaneous rate of change?
How can I estimate an instantaneous rate of
change from any representation?
Vocab: instantaneous rate of change, tangent,
velocity
Pg 49 #1 – 7 optional
Pg 50 #8, 10, 11, 12 mandatory
Skills: write an equation of a polynomial function from a
description of its key features, identify zeros of an equation,
sketch a graph from an equation
(day 2)
Pg 39 #3 – 5 optional
Pg 40 #6, 7, 8, 10, 15 mandatory
Skills: write an equation from a graph of a polynomial
function, identify symmetry of graph from equation,
determine whether polynomial function is an even function,
odd function or neither.
Skills: applying transformations to sketch a graph, identify
and describe transformations from a graph or an equation,
determine equation from graph using transformations,
solving for parameters given graph or value of function at a
point.
Pg 62 #1 – 5, 9, 12 optional
Pg 62 #6, 7, 8, 10 mandatory
Skills: connect rate of change to slope, calculate and
interpret average rates of change from a graph, table of
values or an equation.
Pg 71 # 1, 2, 7, 8 optional
Pg 71 #3, 4, 5, 10, 11 mandatory
Skills: describe relationship between slope of secants and the
slope of a tangent, estimate instantaneous rate of change of a
point from a graph, table of values or an equation,
differentiate between the rate of change at a point and the
value of the function at that point.
The Chapter will wrap up with a review, the Chapter 1 Task and a written test. Class discussions, class work, homework
and quizzes will help you to determine how well you understand the course material in preparation for the summative
assessments.
Extra help is available with me upon request.
There is free tutoring and instructional videos available on the class moodle.
``` |
Lesson Explainer: Equation of a Straight Line in Space: Parametric Form Mathematics
In this explainer, we will learn how to find the parametric equations of straight lines in space.
Let us remind ourselves first about the different forms of equations of a straight line in the -plane (i.e., in two dimensions, 2D). The form gives us the slope and the -intercept . In other words, the direction vector of the line is and the line goes through the point .
From the form , we know that the direction vector of the line is and that the point of coordinates is on the line.
And, finally, when the equation of the line is in the form , we find that the direction vector of the line is (or or , etc.) and that the line passes through the point , where .
Whatever the form of the equation, the two key pieces of information that define a line are its direction vector and one of its points. Let us see how the reasoning works in 2D before moving on to three dimensions (3D).
If we have a line of direction vector that passes through two points and , then vector with components is collinear with vector . In other words, is a scalar multiple of . Therefore, we have where is a real number.
From the above equation, we find that
π₯βπ₯=πο‘ο (1)
and
π¦βπ¦=ππ.ο‘ο (2)
Substituting (1) into (2), we find that
Therefore, the slope of the line that goes through the points and is given by
Let us now consider a line in space whose direction vector is and that goes through the point . For any other point that lies on the line, and are collinear; therefore, , where is a real number. The figure below illustrates this vector equation, with being a point of the line such that .
We can find the same equation by rewriting as and, using the vector position for , and that for , , we find that ; that is, ; it is the equation of the line in vector form.
Considering the components of and , since , we find that is,
If we let vary from to , the above three equations describe the coordinates of all the points on the line. They describe the coordinates of point when .
This set of three equations is called the parametric equations of a straight line in space. Since there are infinitely many points that lie on the line and any vector is a direction vector of the line, there is not a unique set of parametric equations. However, they will all describe the coordinates of all the points on the line (as varies from to , there is no limitation!), and they all define unambiguously the same line.
Definition: Parametric Equations of a Straight Line in Space
The parametric equations of a line in space are a nonunique set of three equations of the form where are the coordinates of a point that lies on the line, is a direction vector of the line, and is a real number (the parameter) that varies from to .
Let us look at the first example.
Example 1: Finding the Parametric Equation of a Line Given a Point and Its Direction Vector
Give the parametric equation of the line on point , with direction vector .
The parametric equations of a line are of the form where are the coordinates of a point that lies on the line, is a direction vector of the line, and is a real number (the parameter) that varies from to .
Here, we know that lies on the line; therefore, we will substitute these coordinates into the equation for ; the components of the direction vector are , so we substitute these for . We find
This is a parametric equation of the line on point , with direction vector .
It is worth noting that this set of equations that defines the line on point , with direction vector is not unique. We could take, for instance, as the direction vector of the line and find the parametric equations
We could also find the coordinates of another point that lies on the line by choosing a value for . With, for example, , we find, using our first set of parametric equations, that the point of coordinates lies on the line. Using these coordinates gives another set of parametric equations; namely,
Let us now find the parametric equations of a line that passes through two given points.
Example 2: Finding the Parametric Equation of a Line Given Two Points
Write the equation of the straight line passing through the points and in parametric form.
1. , , , for
2. , , , for
3. , , , for
4. , , , for
5. , , , for
The parametric equations of a line are of the form where are the coordinates of a point that lies on the line, is a direction vector of the line, and is a real number (the parameter) that varies from to .
Here, we are given two points that lie on the line. To find the components of a direction vector, we simply need to find the components of, for example, . They are .
We could now substitute the coordinates of either or for and the components (or, e.g., ) for .
Since we have here a limited choice of answers, we can start by identifying the components of the direction vectors used in each set of equationsβthey are given by the coefficients of in each equation:
We see that only answer B uses a correct direction vector. Let us now check that the coordinates used in the equations in answer B are correctβthese are the constants in each equation, that is, the coordinates obtained when . We find , that is, the coordinates of . Hence, answer B is a correct set of parametric equations.
Note that the coordinates of used in the equations could be those of neither nor , but the equations could still be correct. In this case, having found a direction vector of the line (), we would have to check that the point of coordinates lies on the line. For this, we need to verify that the vector (or ) is collinear with , that is, verify that
We see that it is equivalent to checking that there exists a value such that the coordinates of verify the parametric equations:
How can we find parametric equations of a line from its Cartesian equations? Remember that the Cartesian equations of a line in space are in the form where are the coordinates of a point that lies on the line, and is a direction vector of the line where , , and are all nonzero real numbers. This form of equations is closely related to the set of parametric equations since they simply give the three expressions for that we get from each of the parametric equations: is equivalent to
When one component of the direction vector is zero, it means that the corresponding coordinate of all the points lying on the line is constant. For instance, if the direction vector of a line is and the point lies on the line, then the parametric equations of the line are
The Cartesian equations of the line are then
The line is perpendicular to the -axis and is in a plane parallel to the -plane.
If the direction vector is unidimensional, that is, two of its components are zero, then the line is parallel to one of the axes. For instance, if the line is parallel to the -axis and passes through the point , its parametric equations are and its Cartesian equations would be , . Compare this equation in 3D with the equation of a line in 2D that is parallel to the -axis: . The value of is the same for all points and nothing is said about the -coordinate because it can take up any value; the set of the -coordinates of all points lying on the line is . It is the meaning of as well when varies from to . Note that we could take as well βall values of would describe as well when varies from to .
Let us practice converting Cartesian equations of a line into parametric equations.
Example 3: Finding the Parametric Equation of a Line Given Its Cartesian Equations
Find the parametric equations of the straight line .
The Cartesian equations given here have been slightly rearranged from the standard form . However, it does not matter, as we simply need to write that to find a set of parametric equations by rearranging each equation. We find
Note that there is not a unique set of parametric equations as there is not a unique set of Cartesian equations of the same line either. Here, for instance, as we have found that the direction vector is , we could have chosen to take to write our parametric equations. This is equivalent to taking as parameter instead of , that is, having as Cartesian equations .
We will now look at a final example where we need to find the parametric equations of the diagonal of a cube.
Example 4: Finding the Parametric Equation of a Line in Two Steps
A cube of side length 3 sits with a vertex at the origin and three sides along the positive axes. Find the parametric equations of the main diagonal from the origin.
Let us start by drawing a diagram of the cube.
The main diagonal of the cube goes from the origin of coordinates to the furthest vertex from that at the origin, namely, the point , since the side of the cube is 3 length units.
The line that contains the diagonal has therefore as direction vector the vector that goes from the origin to the point , that is, the vector of components
The parametric equations of a line are of the form where are the coordinates of a point that lies on the line, is a direction vector of the line, and is a real number (the parameter) that varies from to .
Taking here the origin for the point that lies on the line, we find
Or we could have taken as direction vector, leading to the simplest equations
In short, any point on this line has equal -, -, and -coordinates.
In the previous example, we could have limited the possible values for the parameter t to describe only the diagonal of the cube, that is, the line segment that goes from to . With the equations , , , it means that . The range of depends on the equations used. If we take the parametric equations , , , then the range makes these equations describe the diagonal of the cube.
Key Points
• The parametric equations of a line are a nonunique set of three equations of the form where are the coordinates of a point that lies on the line, is a direction vector of the line, and is a real number (the parameter) that varies from to .
• When , , and are all nonzero real numbers, the parametric equations of a straight line can be derived from its Cartesian equations by writing and rearranging each of the three resulting equations.
• When one component of the direction vector is zero, it means that the corresponding coordinates of all the points lying on the line are constant. For instance, if the direction vector of a line is and the point lies on the line, then the parametric equations of the line are and the Cartesian equations of the line are then
• If two components of the direction vector are zero, the line is parallel to one axis, meaning that only one coordinate varies, while the other two are fixed. For instance, the parametric equations of a line parallel to the -axis and passing through point are and its Cartesian equations are , . |
# How to Calculate Multiplication and Division of Decimals by Powers of Ten
In this step-by-step guide, you will learn how to multiply and divide decimals by powers of ten.
## A step-by-step guide tomultiplication and division decimals by powers of ten
Whenever you want to multiply by the power of $$10$$, move the decimal over to the right-hand side with the identical number of spaces as the number of zeros or as the exponent.
To multiply any whole number via $$10$$, you must multiply the numbers and then put a $$0$$ after the last digit in this number.
Whenever you want to divide via the power of $$10$$, move its decimal toward the left-hand side of the identical number of spaces as the number of zeros or as the exponent.
To divide any multiple of $$10$$ by $$10$$, you must take out the last zero digit which is in one place from the number.
### Multiplication and Division of Decimals by Powers of Ten – Example 1
Divide: $$7.7÷10=$$__?
Solution:
Step 1: Count the zeros in $$10$$. There is $$1$$ zero in $$10$$.
Step 2: Move the decimal point $$1$$ place to the left in $$7.7$$. So, $$7.7→0.77$$
$$7.7÷10=0.77$$
### Multiplication and Division of Decimals by Powers of Ten – Example 2
Multiply: $$100×0.436=$$__?
Solution:
Step 1: Count the zeros in $$100$$. There are $$2$$ zeros in $$100$$.
Step 2: Move the decimal point $$2$$ places to the right in $$0.436$$. So, $$0.436→43.6$$
$$100×0.436=43.6$$
## Exercises forMultiplication and Division of Decimals by Powers of Ten
### Solve.
1. $$\color{blue}{0.065 ÷ 10}$$
2. $$\color{blue}{0.243 × 1000}$$
3. $$\color{blue}{0.8954 × 100}$$
4. $$\color{blue}{0.25 ÷ 100}$$
1. $$\color{blue}{0.0065}$$
2. $$\color{blue}{243}$$
3. $$\color{blue}{89.54}$$
4. $$\color{blue}{0.0025}$$
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# POWERS AND EXPONENTS
A power is the result of a repeated multiplication of the same factor.
For example, 125 is 5 to the 3rd power. Because
125 = 5 ⋅ 5 ⋅ 5
A power can be written in a form that has two parts.
A number called the base and a number called the exponent. The exponent shows the number of times the base is used a factor.
Number raised to the first power, such as 131, are usually written without the exponent.
We can read and write powers as given below.
Power
In words
Value
151
15 to the first power
151 = 15
(0.4)2
0.4 to the second power, or 0.4 squared
(0.4)(0.4) = 0.16
43
4 to the third power, or 4 cubed
⋅ 4 ⋅ 4 = 64
74
7 to the fourth power
⋅ 7 ⋅ 7 ⋅ 7 = 2401
## Using Exponents
Example 1 :
Write the following products using an exponent.
a. 15 ⋅ 15 ⋅ 15 ⋅ 15
b. (0.5)(0.5)(0.5)
c. x ⋅ x ⋅ x ⋅ x ⋅ x ⋅ x
d. y ⋅ y ⋅ y ⋅ y ⋅ y
Solution :
a. 154 The base 15 is used as a factor 4 times.
b. (0.5)3 The base 0.5 is used as a factor 3 times.
c. x6 The base x is used as a factor 6 times.
d. y5 The base y is used as a factor 5 times.
Example 2 :
Evaluate the expression m5 when m = 3.
Solution :
= m5
Substitute 3 for m.
= (3)5
Use 3 as a factor 5 times.
= 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3
Multiply.
= 243
Example 3 :
Evaluate the expression x4 when x = 0.5.
Solution :
= x4
Substitute 0.5 for x.
= (0.5)4
Use 0.5 as a factor 4 times.
= (0.5)(0.5)(0.5)(0.5)
Multiply.
= 0.0625
## Use Powers in Formulas
Example 4 :
An artist uses a cube-shaped block of ice to make an ice sculpture for a competition. Find the volume of the block of ice.
Solution :
Use the formula for volume of a cube :
V = s3
Substitute 15 for s.
= 153
Use 15 as a factor 3 times.
= 15 ⋅ 15 ⋅ 15
Multiply.
= 3375
The volume of the block of ice is 3375 cubic inches.
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WORD PROBLEMS
Word problems on simple equations
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Profit and loss shortcuts
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Remainder when 2 power 256 is divided by 17
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Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
2.3: Scale Drawings and Maps
Difficulty Level: At Grade Created by: CK-12
Introduction
Alex’s Garden Design
Now that Alex has figured out what he wants the garden to look like, he wants to make a drawing of the plot that is accurate.
What does this mean?
It means that Alex wants to use a scale to draw his design. When you use a scale, you choose a unit of measure to represent the real thing. For example, if you want to draw a picture of a ship that is 100 feet long, it doesn’t make sense to actually draw it 100 feet long. You have to choose a unit of measurement like an inch to help you.
Alex decides to use a 1” = 1 ft scale, but he is having a difficult time.
He has two pieces of paper to choose from that he wants to draw the design on. One is \begin{align*}8\frac{1}{2}” \times 11”\end{align*} and the other is \begin{align*}14\frac{1}{2}” \times 11”\end{align*}. He starts using a 1 inch scale and begins to measure the garden plot onto the \begin{align*}8\frac{1}{2}” \times 11”\end{align*} sheet of paper.
At that moment, Tania comes in from outside. She looks over Alex’s shoulder and says, “That will never fit on there. You are going to need a smaller scale or a larger sheet of paper.”
Alex is puzzled. He starts to rethink his work.
He wonders if he should use a \begin{align*}\frac{1}{2}”\end{align*} scale.
Keep in mind the measurements he figured out in the last lesson.
If he uses a 1” scale, what will the measurements be? Does he have a piece of paper that will work?
If he uses a \begin{align*}\frac{1}{2}”\end{align*} scale, what will the measurements be? Does he have a piece of paper that will work?
In this lesson you will learn all about scale and measurement, then you’ll be able to help Alex figure out his garden dilemma.
What You Will Learn
In this lesson, you will learn the following skills.
• Finding actual distances or dimensions given scale dimensions.
• Finding scale dimensions given actual dimensions.
• Solving real-world problems using scale drawings and maps.
Teaching Time
I. Finding Actual Distances or Dimensions Given Scale Dimensions
Maps represent real places. Every part of the place has been reduced to fit on a single piece of paper. A map is an accurate representation because it uses a scale. The scale is a ratio that relates the small size of a representation of a place to the real size of a place.
Maps aren’t the only places that we use a scale. Architects use a scale when designing a house. A blueprint shows a small size of what the house will look like compared to the real house. Any time a model is built, it probably uses a scale. The actual building or mountain or landmark can be built small using a scale.
We use units of measurement to create a ratio that is our scale. The ratio compares two things.
It compares the small size of the object or place to the actual size of the object or place.
A scale of 1 inch to 1 foot means that 1 inch on paper represents 1 foot in real space. If we were to write a ratio to show this we would write:
1” : 1 ft-this would be our scale.
If the distance between two points on a map is 2 inches, the scale tells us that the actual distance in real space is 2 feet.
We can make scales of any size. One inch can represent 1,000 miles if we want our map to show a very large area, such as a continent. One centimeter might represent 1 meter if the map shows a small space, such as a room.
How can we figure out actual distances or dimensions using a scale?
Let’s start by thinking about distances on a map. On a map, we have a scale that is usually found in the corner. For example, if we have a map of the state of Massachusetts, this could be a possible scale.
Here \begin{align*}\frac{3}{4}”\end{align*} is equal to 20 miles.
Example
What is the distance from Boston to Framingham?
To work on this problem, we need to use our scale to measure the distance from Boston to Framingham. We can do this by using a ruler. We know that every \begin{align*}\frac{3}{4}”\end{align*} on the ruler is equal to 20 miles.
From Boston to Framingham measures \begin{align*}\frac{3}{4}”\end{align*}, therefore the distance is 20 miles
If the scale and map were different, we could use the same calculation method. Let’s use another example that just gives us a scale.
Example
If the scale is 1”:500 miles, how far is a city that measures \begin{align*}5\frac{1}{2}”\end{align*} on a map?
We know that every inch is 500 miles. We have \begin{align*}5\frac{1}{2}”\end{align*}. Let’s start with the 5.
5 \begin{align*}\times\end{align*} 500 \begin{align*}=\end{align*} 2500 + \frac{1}{2} \times 500 \begin{align*}=\end{align*} 2750 miles
By using arithmetic, we were able to figure out the mileage.
Another way to do this is to write two ratios. We can compare the scale with the scale and the distance with the distance. Let’s look at an example that has an object in it instead of a map.
Example
If the scale is 2” : 1 ft, what is the actual measurement if a drawing shows the object as 6” long?
We can start by writing a ratio that compares the scale.
\begin{align*}\frac{1 \ ft}{2”}=\frac{x \ ft}{6”}\end{align*} Here we wrote a proportion. We don’t know how big the object really is, so we used a variable to represent the unknown quantity.
Notice that we compared the size to the scale in the first ratio and the size to the scale in the second ratio.
We can solve this logically using mental math, or we can cross multiply to solve it.
\begin{align*}1 \times 6 &= 6\\ 2(x) &= 2x\\ 2x &= 6 \qquad \text{“What times two will give us 6?”}\\ x &= 3 \ ft\end{align*}
The object is actually 3 feet long.
This may seem more confusing, but you can use it if you need to. If it is easier to solve the problem using mental math then that is alright too.
Here are a few problems for you to try on your own.
1. If the scale is 1” : 3 miles, how many miles does 5 inches represent?
2. If the scale is 2” : 500 meters, how many meters does 4 inches represent?
Take a few minutes to check your work with a peer.
II. Finding Scale Dimensions Using Actual Dimensions
In the last section, we worked on figuring out actual dimensions or distances when we had been given a scale.
Now we are going to look at figuring out the scale given the actual dimensions.
To do this, we work in reverse. To make a map, for instance, we need to “shrink” actual distances down to a smaller size that we can show on a piece of paper. Again, we use the scale. Instead of solving for the actual distance, we solve for the map distance. Let’s see how this works.
Example
Suppose we are making a map of some nearby towns. We know that Trawley City and Oakton are 350 kilometers apart. We are using a scale of 1 cm : 10 km. How far apart do we draw the dots representing Trawley City and Oakton on our map?
We use the scale to write ratios that make a proportion. Then we fill in the information we know. This time we know the actual distance between the two towns, so we put that in and solve for the map distance.
\begin{align*}\frac{1 \ cm}{10 \ km}=\frac{x \ cm}{350 \ km}\end{align*} Next we cross multiply to find the number of centimeters that we would need to draw on the map.
\begin{align*}1(350) &= 10x\\ 350 &= 10x\\ 35 &= x\end{align*}
Using our scale, to draw a distance of 350 km on our map, we need to put Trawley City 35 centimeters away from Oakton.
We can figure out the scale using a model and an actual object too.
Let’s look at an example
Example
Jesse wants to build a model of a building. The building is 100 feet tall. If Jesse wants to use a scale of 1” to 25 feet, how tall will his model be?
Let’s start by looking at our scale and writing a proportion to show the measurements that we know. \begin{align*}\frac{1”}{25 \ ft}=\frac{x}{100 \ ft}\end{align*}
To solve this proportion we cross multiply.
\begin{align*}1(100) &= 25(x)\\ 100 &= 25x\\ 4 &= x\end{align*} Jesse’s model will be 4 inches tall.
Our answer is \begin{align*}4”\end{align*}.
Real Life Example Completed
Alex’s Garden Design
Now that we have learned all about scales and scale drawing, we are ready to help Alex with his garden design.
Let’s begin by looking at the problem again.
Now that Alex has figured out what he wants the garden to look like, he wants to make a drawing of the plot that is accurate.
What does this mean?
It means that Alex wants to use a scale to draw his design. When you use a scale, you choose a unit of measurement to represent the real thing. For example, if you want to draw a picture of a ship that is 100 feet long, it doesn’t make sense to actually make a drawing 100 feet long. You have to choose a unit of measurement like an inch to help you.
Alex’s decides to use a scale of 1” = 1 ft., but he is having a difficult time.
He has two pieces of paper to choose from that he wants to draw the design on. One is \begin{align*}8\frac{1}{2}” \times 11”\end{align*} and the other is \begin{align*}14 \frac{1}{2}” \times 11”\end{align*}. He starts using a 1 inch scale and begins to measure the garden plot onto the \begin{align*}8\frac{1}{2}” \times 11”\end{align*} sheet of paper.
At that moment, Tania comes in from outside. She looks over Alex’s shoulder and says, “That will never fit on there. You are going to need a smaller scale or a larger sheet of paper.”
Alex is puzzled. He starts to rethink his work.
He wonders if he should a \begin{align*}\frac{1}{2}”\end{align*} scale.
Keep in mind the measurements he figured out in the last lesson.
If he uses a 1” scale, what will the measurements be? Does he have a piece of paper that will work?
If he uses a \begin{align*}\frac{1}{2}”\end{align*} scale, what will the measurements be? Does he have a piece of paper that will work?
First, let’s begin by underlining all of the important information in the problem.
Next, let’s look at the dimensions given each scale, a 1” scale and a \begin{align*}\frac{1}{2}”\end{align*} scale.
First, we start by figuring out the dimensions of the square. Here is our proportion.
\begin{align*}\frac{1”}{1 \ ft} &= \frac{x \ ft}{9 \ ft}\\ 9 &= x\end{align*} To draw the square on a piece of paper using this scale, the three matching sides would each be 9 inches.
Next, we have the short side. It is one foot, so it would be 1” long on the paper.
Now we can work with the rectangle.
If the rectangle is 12 ft \begin{align*}\times\end{align*} 8 ft and every foot is measured with 1”, then the dimensions of the rectangle are 12” \begin{align*}\times\end{align*} 8”.
You would think that this would fit on either piece of paper, but it won’t because remember that Alex decided to put the two garden plots next to each other.
If one side of the square is 9” and the length of the rectangle is 12” that equals 21”. 21 inches will not fit on a piece of \begin{align*}8\frac{1}{2}” \times 11”\end{align*} paper or \begin{align*}14\frac{1}{2}” \times 11”\end{align*} paper.
Let’s see what happens if we use a \begin{align*}\frac{1}{2}” = 1\end{align*} foot scale.
We already figured out a lot of the dimensions here.
We can use common sense and divide the measurements from the first example in half since \begin{align*}\frac{1}{2}”\end{align*} is half of 1”.
The square would be 4.5” on each of the three matching sides.
The short side of the square would be \begin{align*}\frac{1}{2}”\end{align*}.
The length of the rectangle would be 6”. The width of the rectangle would be 4”.
With the square and the rectangle side-by-side, the length of Alex's drawing would be 10.5". This will fit on either piece of paper.
Use your notebook to draw Alex’s garden design.
Use a ruler and draw it to scale.
The scale is \begin{align*}\frac{1}{2}” = 1\end{align*} foot.
When you have finished, check your work with a peer.
Vocabulary
Here are the vocabulary words from this lesson.
Scale
a ratio that compares a small size to a larger actual size. One measurement represents another measurement in a scale.
Ratio
the comparison of two things
Proportion
a pair of equal ratios, we cross multiply to solve a proportion
Technology Integration
Other Videos
http://www.teachertube.com/viewVideo.php?video_id=79418&title=PSSA_Grade_7_Math_19_Map_Scale – You will need to register with this website. This is a video about solving a ratio and proportion problem.
Time to Practice
Directions: Use the given scale to determine the actual distance.
Given: Scale 1” = 100 miles
1. How many miles is 2” on the map?
2. How many miles is \begin{align*}2\frac{1}{2}”\end{align*} on the map?
3. How many miles is \begin{align*}\frac{1}{4}”\end{align*} on the map?
4. How many miles is \begin{align*}\frac{1}{2}”\end{align*} on the map?
5. How many miles is \begin{align*}5 \frac{1}{4}”\end{align*} on the map?
Given: 1 cm = 20 mi
6. How many miles is 2 cm on the map?
7. How many miles is 4 cm on the map?
8. How many miles is \begin{align*}\frac{1}{2}”\end{align*} cm on the map?
9. How many miles is \begin{align*}1 \frac{1}{2}\end{align*} cm on the map?
10. How many miles is \begin{align*}4 \frac{1}{4}\end{align*} cm on the map?
Directions: Use the given scale to determine the scale measurement given the actual distance.
Given: Scale 2” = 150 miles
11. How many scale inches would 300 miles be?
12. How many scale inches would 450 miles be?
13. How many scale inches would 75 miles be?
14. How many scale inches would 600 miles be?
15. How many scale inches would 900 miles be?
Directions: Use the given scale to determine the scale measurement for the following dimensions.
Given: Scale 1” = 1 foot
16. What is the scale measurement for a room that is 8’ \begin{align*}\times\end{align*} 12’?
17. What is the scale measurement for a tree that is 1 yard high?
18. What is the scale measurement for a tower that is 360 feet high?
19. How many feet is that?
20. What is the scale measurement for a room that is \begin{align*}12’ \times 16 \frac{1}{2}’\end{align*}?
Directions: Use what you have learned about scale and measurement to answer each of the following questions.
21. Joaquin is building the model of a tower. He is going to use a scale of 1” = 1 foot. How big will his tower be in inchesif the actual tower if 480 feet tall?
22. How many feet high will the model be?
23. Is this a realistic scale for this model? Why or why not?
24. If Joaquin decided to use a scale of \begin{align*}\frac{1}{2}” = 1\end{align*} foot, what would the new height of the model be in inches?
25. How many feet tall will the model be?
26. If Joaquin decided to use a scale that was \begin{align*}\frac{1}{4}”\end{align*} for every 1 foot, how many feet high would his model be?
27. What scale would Joaquin need to use if he wanted his model to be 5 feet tall?
28. How tall would the model be if Joaquin decided to use \begin{align*}\frac{1}{16}” = 1\end{align*} foot?
29. If Joaquin’s model ends up being shorter than \begin{align*}2 \frac{1}{2}\end{align*} feet tall, did he use a scale that is smaller or larger than \begin{align*}\frac{1}{8}” = 1\end{align*} foot?
30. If Joaquin wants his model to be half the size of the real model, will it fit in his classroom or will he need to build it outside?
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# How do you solve the system x + y = 4 and –x + y = 2 by graphing?
Mar 26, 2018
$\left(1 , 3\right)$
(Graph in explanation)
#### Explanation:
First, we need to set the equation so that $y$ is by itself.
For the first equation $x + y = 4$, we can subtract $x$ from both sides and get $y = 4 - x$.
For the second equation $- x + y = 2$, we can add $x$ on both sides of the equation and get $y = x + 2$.
Now we can graph this. We graph both equations on the same graph because to solve by graphing, we have to find the point(s) where the two graphs intersect:
The red line represents the equation $y = 4 - x$.
The blue line represents the equation $y = x + 2$.
As you can see, the equations intersect at the point $\left(1 , 3\right)$.
Therefore, the answer is $\left(1 , 3\right)$.
Hope this helps! |
Resources to help teach and assess Common Core Eighth Grade Math
Engage NY (Eureka) Video Lessons and Homework Solutions
Khan Academy Engage NY (Eureka) Support
In Grade 8, instructional time should focus on three critical areas: (1) formulating and reasoning about expressions and equations, including modeling an association in bivariate data with a linear equation, and solving linear equations and systems of linear equations; (2) grasping the concept of a function and using functions to describe quantitative relationships; (3) analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem.
1. Students use linear equations and systems of linear equations to represent, analyze, and solve a variety of problems. Students recognize equations for proportions (y/x = m or y = mx) as special linear equations (y = mx + b), understanding that the constant of proportionality (m) is the slope, and the graphs are lines through the origin. They understand that the slope (m) of a line is a constant rate of change, so that if the input or x-coordinate changes by an amount A, the output or y-coordinate changes by the amount m·A. Students also use a linear equation to describe the association between two quantities in bivariate data (such as arm span vs. height for students in a classroom). At this grade, fitting the model, and assessing its fit to the data are done informally. Interpreting the model in the context of the data requires students to express a relationship between the two quantities in question and to interpret components of the relationship (such as slope and y-intercept) in terms of the situation.Students strategically choose and efficiently implement procedures to solve linear equations in one variable, understanding that when they use the properties of equality and the concept of logical equivalence, they maintain the solutions of the original equation. Students solve systems of two linear equations in two variables and relate the systems to pairs of lines in the plane; these intersect, are parallel, or are the same line. Students use linear equations, systems of linear equations, linear functions, and their understanding of slope of a line to analyze situations and solve problems.
2. Students grasp the concept of a function as a rule that assigns to each input exactly one output. They understand that functions describe situations where one quantity determines another. They can translate among representations and partial representations of functions (noting that tabular and graphical representations may be partial representations), and they describe how aspects of the function are reflected in the different representations.
3. Students use ideas about distance and angles, how they behave under translations, rotations, reflections, and dilations, and ideas about congruence and similarity to describe and analyze two-dimensional figures and to solve problems. Students show that the sum of the angles in a triangle is the angle formed by a straight line, and that various configurations of lines give rise to similar triangles because of the angles created when a transversal cuts parallel lines. Students understand the statement of the Pythagorean Theorem and its converse, and can explain why the Pythagorean Theorem holds, for example, by decomposing a square in two different ways. They apply the Pythagorean Theorem to find distances between points on the coordinate plane, to find lengths, and to analyze polygons. Students complete their work on volume by solving problems involving cones, cylinders, and spheres.
# The Number System
#### Know that there are numbers that are not rational, and approximate them by rational numbers.
CCSS.MATH.CONTENT.8.NS.A.1
Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
CCSS.MATH.CONTENT.8.NS.A.2
Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., π2). For example, by truncating the decimal expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.
# Expressions & Equations
#### Expressions and Equations Work with radicals and integer exponents.
CCSS.MATH.CONTENT.8.EE.A.1
Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 32 × 3-5 = 3-3 = 1/33 = 1/27.
CCSS.MATH.CONTENT.8.EE.A.2
Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational.
CCSS.MATH.CONTENT.8.EE.A.3
Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. For example, estimate the population of the United States as 3 times 108 and the population of the world as 7 times 109, and determine that the world population is more than 20 times larger.
CCSS.MATH.CONTENT.8.EE.A.4
Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology
#### Understand the connections between proportional relationships, lines, and linear equations.
CCSS.MATH.CONTENT.8.EE.B.5
Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed.
CCSS.MATH.CONTENT.8.EE.B.6
Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.
#### Analyze and solve linear equations and pairs of simultaneous linear equations.
CCSS.MATH.CONTENT.8.EE.C.7
Solve linear equations in one variable.
CCSS.MATH.CONTENT.8.EE.C.7.A
Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = aa = a, or a = b results (where a and b are different numbers).
CCSS.MATH.CONTENT.8.EE.C.7.B
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
CCSS.MATH.CONTENT.8.EE.C.8
Analyze and solve pairs of simultaneous linear equations.
CCSS.MATH.CONTENT.8.EE.C.8.A
Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.
CCSS.MATH.CONTENT.8.EE.C.8.B
Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6.
CCSS.MATH.CONTENT.8.EE.C.8.C
Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair.
# Functions
#### Define, evaluate, and compare functions.
CCSS.MATH.CONTENT.8.F.A.1
Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output (Function notation is not required for Grade 8).
CCSS.MATH.CONTENT.8.F.A.2
Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change.
CCSS.MATH.CONTENT.8.F.A.3
Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line.
#### Use functions to model relationships between quantities.
CCSS.MATH.CONTENT.8.F.B.4
Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
CCSS.MATH.CONTENT.8.F.B.5
Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally.
# Geometry
#### Understand congruence and similarity using physical models, transparencies, or geometry software.
CCSS.MATH.CONTENT.8.G.A.1
Verify experimentally the properties of rotations, reflections, and translations:
CCSS.MATH.CONTENT.8.G.A.1.A
Lines are taken to lines, and line segments to line segments of the same length.
CCSS.MATH.CONTENT.8.G.A.1.B
Angles are taken to angles of the same measure.
CCSS.MATH.CONTENT.8.G.A.1.C
Parallel lines are taken to parallel lines.
CCSS.MATH.CONTENT.8.G.A.2
Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them.
CCSS.MATH.CONTENT.8.G.A.3
Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates.
CCSS.MATH.CONTENT.8.G.A.4
Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them.
CCSS.MATH.CONTENT.8.G.A.5
Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the sum of the three angles appears to form a line, and give an argument in terms of transversals why this is so.
#### Understand and apply the Pythagorean Theorem.
CCSS.MATH.CONTENT.8.G.B.6
Explain a proof of the Pythagorean Theorem and its converse.
CCSS.MATH.CONTENT.8.G.B.7
Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
CCSS.MATH.CONTENT.8.G.B.8
Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.
#### Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres.
CCSS.MATH.CONTENT.8.G.C.9
Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.
# Statistics & Probability
#### Investigate patterns of association in bivariate data.
CCSS.MATH.CONTENT.8.SP.A.1
Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association.
CCSS.MATH.CONTENT.8.SP.A.2
Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line, and informally assess the model fit by judging the closeness of the data points to the line.
CCSS.MATH.CONTENT.8.SP.A.3
Use the equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. For example, in a linear model for a biology experiment, interpret a slope of 1.5 cm/hr as meaning that an additional hour of sunlight each day is associated with an additional 1.5 cm in mature plant height.
CCSS.MATH.CONTENT.8.SP.A.4
Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. Use relative frequencies calculated for rows or columns to describe possible association between the two variables. For example, collect data from students in your class on whether or not they have a curfew on school nights and whether or not they have assigned chores at home. Is there evidence that those who have a curfew also tend to have chores? |
Factoring and Multiples
# What are the common factors of 92?
001
###### 2016-10-16 07:09:36
The factors of 92 are 1, 2, 4, 23, 46, 92
For them to be common, they need to be compared to another set of factors.
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###### 2011-10-07 20:58:12
It is not possible to give a sensible answer to this question. The common factor refers to a factor that is COMMON to two or more numbers. You have only one number in the question!
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## Related Questions
The common factors of 42 and 92 are: 1 and 2
The factors of 114 are 1,2,3,6,19, 38, 57 and 114. The factors of 92 are 1,2,4,23,46 and 92. The common factors of 114 and 92 are therefor 1 and 2.
THe common factors of 48 and 92 are 1, 2 and 4
The factors of 38 are:1, 2, 19, 38The factors of 92 are:1, 2, 4, 23, 46, 92The common factors are:1, 2
Factors of 69: 1 3 23 69 Factors of 92: 1 2 4 23 46 92 Common factors of 69 and 92: 1 and 23
The factors of 84 are:1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84The factors of 92 are:1, 2, 4, 23, 46, 92The common factors are:1, 2, 4
The factors of 92 are: 1, 2, 4, 23, 46, 92The factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100The common factors are: 1, 2, 4
A factor is a divisor - a number that will evenly divide into another number. One way to determine the common factors (and greatest common factor) is to find all the factors of the numbers and compare them. The factors of 20 are 1, 2, 4, 5, 10, and 20. The factors of 92 are 1, 2, 4, 23, 46, and 92. The common factors are 1, 2, and 4.
The common factors are: 1, 2 The Greatest Common Factor (GCF) is: 2
Factors of 69: 1 3 23 69 Factors of 92: 1 2 4 2346 92 The GCF of 69 and 92 is 23
The factors of 92 are: 1, 2, 4, 23, 46, 92The factors of 180 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180The common factors are: 1, 2, 4
The factors of 92 are: 1, 2, 4, 23, 46, 92 The factors of 108 are: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108 The common factors are: 1, 2, 4 The Greatest Common Factor: GCF = 4
The factors of 46 are:1, 2, 23, 46The factors of 92 are:1, 2, 4, 23, 46, 92The factors of 138 are:1, 2, 3, 6, 23, 46, 69, 138The common factors are:1, 2, 23, 46The Greatest Common Factor (GCF) is:46
The GCF of 28 and 92 is 4.First, find the prime factors:28 = 2*2*792 = 2*2*23Then, compute the GCF by multiplying the common factors:GCF = 2*2 = 4
The factors of 92 are: 1, 2, 4, 23, 46 and 92.
The even factors of 92 are 2, 4, 46, and 92.
1,2,4,23,46,92 are all the factors of 92
###### Factoring and MultiplesMath and ArithmeticAlgebra
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# Unit 1
Theory of Numbers
Natural Numbers
The numbers 1, 2, 3, , which are used in counting are called natural numbers or positive integers.
## Basic Properties of Natural Numbers
In the system of natural numbers, we have two operations addition and multiplication with the
following properties.
Let x , y , z denote arbitrary natural numbers, then
1. x + y is a natural number i.e., the sum of two natural number is again a natural number.
2. Commutative law of addition x + y = y + x
3. Associative law of addition x + ( y + z ) = ( x + y ) + z
4. x y is a natural number i.e., product of two natural numbers is a natural number.
5. Commutative law of multiplication x y = y x
6. Associative law of multiplication x ( y z ) = ( x y ) z
7. Existence of multiplicative identity a 1 = a
8. Distributive law x ( y + z ) = xy + xz
Divisibility of Integers
An integer x 0 divides y, if there exists an integer a such that y = ax and thus we write as x | y (x divides y).
If x does not divides y, we write as x \| y (x does not divides y )
[This can also be stated as y is divisible by x or x is a divisor of y or y is a multiple of x].
Properties of Divisibility
1. x | y and y | z x | z
2. x | y and x | z x | (ky lz ) for all k , l z. z = set of all integers.
3. x | y and y | x x = y
4. x | y, where x > 0, y > 0 x y
5. x | y x | yz for any integer z.
6. x | y iff nx | ny, where n 0.
Test of Divisibility
Divisibility by certain special numbers can be determined without actually carrying out the process of
division. The following theorem summarizes the result :
A positive integer N is divisible by
2 if and only if the last digit (unit's digit) is even.
4 if and only if the number formed by last two digits is divisible by 4.
8 if and only if the number formed by the last three digits is divisible by 8.
3 if and only if the sum of all the digits is divisible by 3.
9 if and only if the sum of all the digits is divisible by 9.
5 if and only if the last digit is either 0 or 5.
25 if and only if the number formed by the last two digits is divisible by 25.
125 if and only if the number formed by the last three digits is divisible by 125.
11 if and only if the difference between the sum of digits in the odd places (starting from right) and
sum of the digits in the even places (starting from the right) is a multiple of 11.
Division Algorithm
For any two natural numbers a and b, there exists unique numbers q and r called respectively quotient
and remainder, a = bq + r , where 0 r < b.
Common Divisor
If a number c divides any two numbers a and b i.e., if c | a and c | b, then c is known as a common divisor
of a and b.
## Greatest Common Divisor
If a number d divides a and b and is divisible by all the common divisors of a and b, then d is known as
the greatest common divisor (GCD) of a and b or HCF of a and b.
The GCD of numbers a and b is the unique positive integer d with the following two properties.
(i) d | a and d | b
(ii) If c | a and c | b; then c | d
We write it as (a , b ) = d
For example, (12, 15) = 3 ; (7, 8) = 1
## Note 1. (a, b ) = (b, a ) 2. If a|b; then (a, b ) = a
Properties of GCD
1. If ( b , c ) = g and d is a common divisor of b and c, then d is a divisor of g.
2. For any m > 0, (mb , mc ) = m (b , c )
b c 1
3. If d |b and d |c and d > 0, then , = (b , c )
d d d
b c
4. If ( b , c ) = g, then , =1
g g
5. If ( b , c ) = g, then there exists two integers x and y such that g = xb + yc
6. If (a , b ) = 1 and (a , c ) = 1, then ( a , bc ) = 1
7. If a | bc = 1 and (a , b ) = 1, then a | c = 1
Theory of Numbers 3
## For example a = 6, b = 21, c = 10
6 | 21 10 but (6, 21) = 3
and (6, 10) = 2 and 6 divides neither 21 nor 10
LCM of two integers a, b is the smallest positive integer divisible by both a and b and it is denoted by
[a, b].
The Euclidean algorithm can be used to find the GCD of two integers as well as representing GCD as
in 5th property. Consider 2 numbers 18, 28
28 = 1.8 + 10 ; 18 = 110
. + 8
10 = 1 8 + 2 ; 8 = 4 2 + 0
(18, 28) = 2
(18, 28) = 2 = 10 1.8 = 10 (18 110
. )
= 210
. 118
. = 2 (28 118
. ) 118
.
= 228
. 318
. = 228
. + ( 3)18
Note The representation in 5th property is not unique. In fact we can represent (a, b ) as xa + yb in infinite
number of ways where x , y Z
Z = set of all integers.
In above example, 252 is LCM of 18 and 28.
252 = 9.28
252 = 1418
.
(18, 28) = 2 . 28 + ( 3) 18
= 2.28 + 252K + ( 3)18 252K
= ( 2 + 9K )28 + ( 3 14K )18
where K is any integer.
Unit
1 is called unit in the set of positive integers.
Prime
A positive integer P is said to be prime, if
(i) P > 1
(ii) P has no divisors except 1 and P i.e., A number which has exactly two different factors, itself and
one, is called a prime number.
Thus, 2, 3, 5, 7, 11, ... are primes. 2 is the only even number which is prime. All other primes
being odd.
But the converse is not true i.e., every odd number need not be prime.
Composite
Every number (greater than one) which is not prime is called composite number. i.e., a number which has
more than two different factors is called composite. For example 18 is a composite number because 2, 3,
6, 9 are divisors of 18 other than 1 and 18.
We can also define a composite number as : A natural number n is said to be composite, if there exists
integers l and m such that n = lm, where 1 < l < n and l < m < n.
Remark
l A prime number P can be written as a product only in one way namely P.1 .
l A composite number n can also be written as n.1. But composite number can be written in one more way
also as mentioned above.
l A composite number has at least three factors.
## Note 1 is neither prime nor composite.
Twin Primes
A pair of numbers is said to be twin primes, if they differ by 2.
e.g., 3, 5 are twin primes.
Perfect Number
A number n is said to be perfect if the sum of all divisors of n (including n) is equal to 2n.
For example 28 is a perfect number because divisors of 28 are 1, 2, 4, 7, 14, 28.
Sum of divisors of n (= 28) = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2n
Coprime Integers
Two numbers a and b are said to be coprime, if 1 is only common divisors of a and b.
i.e., if GCD of a and b = 1 i.e., if (a , b ) = 1
e.g., (4, 5) = 1, (8, 9) = 1.
Theorem 1 If a = qb + r , then (a , b ) = (b , r ).
## Proof Let (a , b ) = d and (b , r ) = e
Q (a , b ) = d d | a and d | b
d | a and d | qb d | (a qb)
i.e., d |r [Qa qb = r ]
d is common divisor of b and r.
d |e [Qe is the GCD of b and r] (i)
Again Q (b, r) = e
e e
e |b and e |r and
bq r
e|bq + r i.e., e|a [Qa = bq + r ]
e is a common divisor of a and b.
e|d [Qd is the GCD of a and b] (ii)
From Eqs. (i) and (ii), we have d = e
i.e., (a , b ) = (b , r )
Remark
The above result can also be stated as :
GCD of a and b is same as GCD of b and r, where r is remainder obtained on dividing a by b.
Corollary If (a , b ) = 1, then (b , r ) = (a , b ) = 1
i.e., if a is coprime to b, then r is coprime to b. r is remainder obtained on dividing a by b.
Theory of Numbers 5
Theorem 2 If d is the greatest common divisor of a and b, then there exists integers x and y such that
d = xa + yb and d is the least positive value of xa + yb.
Proof
Case I By successive application of division algorithm to numbers a and b.
Let r1 , r2 , K , rn be successive remainders.
Therefore, a = bq1 + r1 , 0 < r1 < b (Dividing a by b)
b = r1q 2 + r2 , 0 < r2 < r1 (Dividing b by remainder r1 )
r1 = r2q3 + r3 , 0 < r3 < r2 (Dividing r1 by remainder r2 )
M
M
M
rn 2 = rn 1qn + rn , 0 < rn < rn 1
rn 1 = rn qn +1 + rn + 1, 0 < rn +1 < rn
Since r1 > r2 > K is a set of decreasing integers, this process must terminate after a finite number of step.
i.e., remainder must be zero after some stage.
So let rn +1 = 0 rn 1 = rn qn +1 + 0 = rn qn +1
rn | rn 1 (rn 1 , rn ) = rn [If a | b, then (a, b ) = a ]
Now, (a, b ) = (b, r1 ) = (r1 , r2 ) = (r2 , r3 ) = K = (rn 1 , rn ) = rn (i)
i.e., GCD of a and b is rn .
From first of above equations r1 = a bq1 = ax 1 + by1, where x 1 = 1, y1 = q ,
Putting the value of r1 = a bq1 in r2 = b r1q 2
r2 = b r1q 2 = b (a bq1 ) q 2 = b aq 2 + bq1q 2
= aq 2 + b (1 + q1q 2 ) = ax 2 + by 2 , where x 2 = q 2
y 2 = 1 + q1q 2
Similarly, r3 = ax 3 + by3 and so on rn = ax n + byn
or rn = ax + by
where x n = x and yn = y
i.e., GCD of a and b = rn can be expressed as (a , b ) = d = ax + by [By Eq. (i)]
Case II (a , b ) = d (Q d | a and d | b)
d | (ax + yb ) for all values of x and y.
an integer k such that xa + yb = kd (ii)
But least value of k is 1.
Putting k = 1 in Eq. (ii), least value of xa + yb is d.
Corollary If a and b are coprime integers i.e., if (a , b ) = 1, then there exists integers x and y such that
ax + by = 1
## Example 1 If (a, b ) = d , then , = 1.
a b
d d
Solution Q (a, b ) = d
d | a and d | b [By definition of GCD].
There exists integers a1and b1 such that a = da1 (i)
b = db1 (ii)
Again Q (a, b ) = d
## There exists integers x and y such that
ax + by = d [By Theorem 2]
Putting values of a and b from Eqs. (i) and (ii)
da1x + db1y = d .
or a1x + b1y = 1
(a1, b1) = 1 [By corollary theorem 1]
a b a b
or , =1
QFrom Eq. (i), a1 = , from Eq. (ii), b1 =
d d d d
Remark
(a, b ) = d and a = a1d , b = b1d , then a1 and b1 are coprime i.e., (a1, b1) = 1
## Example 2 If a|bc and (a, b ) = 1, then a|c.
Solution Q a|bc
There exists an integer d such that bc = ad (i)
Q (a, b ) = 1
There exist integers m and n such that am + bn = 1 (ii)
Multiplying both sides of Eq.(ii) by c, acm + bcn = c (iii)
Putting bc = ad from Eq. (i) in Eq. (iii)
a(cm + dn ) = c
a|c
Note If a|bc and (a, b ) = 1, then a|c. This result is also known as Gauss Theorem.
## Example 3 Prove that every two consecutive integers are coprime.
Solution Let n and n + 1 be two consecutive integers.
Let (n, n + 1) = d
d |n and d |n + 1
d |(n + 1) n or d |1
d =1
(n, n + 1) = 1
i.e., n and (n + 1) are relatively prime.
## Example 4 Show that GCD of a + b and a b is either 1 or 2, if (a, b ) = 1.
Solution Let (a + b, a b ) = d
d | (a + b ) and d | (a b )
d | (a + b + a b ) and d | (a + b ) (a b )
or d | 2a and d | 2b.
i.e., d is a common divisor of 2a and 2b.
d | ( 2a, 2b ) [By definition of GCD]
Theory of Numbers 7
## i.e., d | 2(a, b ) [Q(ma, mb ) = m (a, b )]
But (a, b ) = 1 d|2
d = 1 or d = 2 .
Example 5 Find GCD of 858 and 325 and express it in the form m 858 + n 325.
Solution 858 = 325.2 + 208 (i)
Dividing 858 by 325
325 = 2081
. + 117 (ii)
Dividing 325 by 208
208 = 1171
. + 91 (iii)
Dividing 208 by 117
117 = 911
. + 26 (iv)
Dividing 117 by 91
91 = 26.3 + 13 (v)
Dividing 91 by 26
26 = 13.2
GCD of 858 and 325 is d = 13
From Eq. (v), d = 13 = 91 26.3
Substituting the value of 26 from Eq. (iv)
91 3(117 911
. ) = 91 3117
. + 3.91 = 4.91 3117
.
Substituting the value of 91 from Eq. (iii)
= 4( 208 117) 3117
. = 4.208 7117
.
Substituting the value of 117 from Eq. (ii)
= 4.208 7( 325 2081) = 4.208 7.325 + 7.208
= 11.208 7.325
= 11( 858 325.2) 7.325 [Putting the value of 208 from Eq. (i)]
= 11.858 22.325 7.325 = 11.858 29.325
= m 858 + n.325 where m = 11, n = 29
Example 6 If a and b are relatively prime, then any common divisor of ac and b is a divisor of c.
Solution a and b are relatively prime
integers x and y such that
ax + by = 1 (i)
Let d be any common divisor of ac and b.
Q d | ac, an integer m such that
ac = dm (ii)
Q d | b, an integer n such that
b = dn (iii)
Multiplying both sides of Eq. (i) by c
acx + bcy = c (iv)
Putting the values of ac and b from Eqs. (ii) and (iii) in Eq. (iv).
dmx + dncy = c
or d (mx + ncy ) = c
d|c
Example 7 If a and b are any two odd primes, show that (a 2 b 2 ) is composite.
Solution a 2 b 2 = (a b )(a + b )
Q a.0 and b are odd primes.
So, let a = 2k + 1
b = 2k + 1
a b = 2k + 1 2k 1 = 2k 2k = 2(k k ) is even
a + b = 2k + 1 + 2k + 1 = 2k + 2k + 2 = 2(k + k + 1) is even
Neither (a b ) nor (a + b ) is equal to 1.
Neither of the two divisors (a b ) and (a + b ) of (a 2 b 2 ) is equal to 1.
(a 2 b 2 ) is composite.
[Q Out of the two divisors of a prime number p, one must be equal to 1]
## Example 8 If a |c, b |c and (a, b ) = 1, then ab |c.
Solution Q a |c
There exists an integers d such that
Q b|c
There exists an integer e such that
c = be
Q (a, b ) = 1, therefore there exist integers m, n such that
am + bn = 1 (ii)
Multiplying both sides by c
acm + bcn = c (iii)
Putting c = be from Eq. (ii) in acm and c = ad from Eq. (i) in bcn, Eq. (iii) becomes
abem + band = c
ab (em + dn ) = c
ab | c
## Example 9 If a 2 b 2 is a prime number, show that a 2 b 2 = a + b, where a, b are natural numbers.
Solution a 2 b 2 = (a b )(a + b ) (i)
Q (a b ) is a prime number
2 2
## One of the two factors = 1
Q a b =1 [Qa b < a + b ]
Theory of Numbers 9
## Q The only divisor of a prime number are 1 and itself.
Eq. (i) becomes a 2 b 2 = 1 (a + b )
or a 2 b2 = a + b
e.g., 32 22 = 5 (which is prime)
32 22 = 3 + 2, 3, 2 N.
Example 10 Prove that an integer is divisible by 9 if and only if the sum of its digits is divisible by 9.
Solution Let a = an K a 3a 2a1 be an integer
[Note a is not the product of a1, a 2, a 3,, an but a1, a 2, a 3,,an are digits in the value of
a. For example 368 is not the product of 3, 6 and 8 rather 3, 6, 8 are digits in value
of 368
= 8 + 6 10 + 3 (10)2 ]
a = an a 3a 2a1
= a1 + (10)1a 2 + (10)2a 3 + (10)3a 4 + K + (10)n 1an
= a1 + 10a 2 + 100a 3 + 1000a 4 + K
= a1 + (a 2 + 9a 2 ) + (a 3 + 99a 3 ) + (a 4 + 999a 4 ) + ...
= (a1 + a 2 + a 3 + a 4 + K ) + ( 9a 2 + 99a 3 + 999a 4 + K ) (i)
or a = S + 9 (a 2 + 11a 3 + 111a 4 + ... )
where S = a1 + a 2 + a 3 + a 4 + K
is the sum of digits in the value of a
a S = 9 (a 2 + 11a 3 + 111a 4 + ... )
9 | (a S ) (ii)
Case I a is divisible by 9
i.e., 9 |a (iii)
9 | [a (a S )] [From Eqs. (ii) and (iii)]
i.e., 9|S i.e., sum of digits is divisible by 9.
Case II S (sum of digits) is divisible by 9
i.e., 9|S (iv)
From Eqs. (ii) and (iv), 9|[(a S ) + S ]
i.e., 9|a
i.e., the integer a is divisible by 9.
## Theorem 3 Prove that the product of any r consecutive numbers is divisible by r ! .
Proof Let
Pn = n (n + 1)(n + 2) K (n + r 1) (i)
be the product of r consecutives integers beginning with n.
We shall prove the theorem by Induction method.
For r = 1, Pn = n is divisible by 1! for all n.
The theorem is true for r = 1 i.e., the product of 1 (consecutive) integer is divisible by 1!.
Let us assume the theorem to be true for the product of (r 1) consecutive integers.
## i.e., every product of (r 1) consecutive integers is divisible by (r 1)! .
Changing n to n + 1 in Eq. (i)
Pn +1 = (n + 1)(n + 2)K (n + r )
Multiplying both sides by n
nPn +1 = n (n + 1)(n + 2) K (n + r )
= n (n + 1)(n + 2) K (n + r 1)(n + r )
nPn +1 = (n + r )Pn
= nPn + rPn
or n (Pn + 1 Pn ) = r Pn
P
or Pn + 1 Pn = r n
n
n (n + 1)(n + 2) K (n + r 1)
=r [Using value of Pn ]
n
or Pn +1 Pn = r (n + 1)(n + 2) K (n + r 1)
or Pn +1 Pn = r
Product of (r 1) consecutive integers.
= rP (i)
Where P denotes the product of (r 1) consecutive integers.
But the product P of (r 1) consecutive integers is divisible by (r 1)! . [By assumption]
P = k (r 1)!
Eq. (i) becomes Pn +1 Pn = rk (r 1)! = kr (r 1) ! = k (r )!
i.e., r ! | (Pn +1 Pn ), n
Put n =1
r !| (P2 P1 )
But P1 = 1 2 3 K r = r ! is divisible by r !
i.e., r1 !| P1
r !| (P2 P1 ) + P1 i.e., r !| P2
Put n = 2,
r !| (P3 P2 )
But r !| P2
r !| (P3 P2 ) + P2
i.e., r !| P3 and so on.
Generalising we can say that r !| Pn for all n.
n
Corollary Cr is an integer
n! n (n 1)(n 2) K (n r + 1)(n r )!
n
Cr = =
r !(n r )! r! (n r )!
n (n 1)(n 2) K (n r + 1)
=
r!
the product of r consecutive integers
= = An integer
r!
(QThe product of r consecutive integer is divisible by r !) .
Note Therefore the product of two consecutive integers is divisible by 2! = 2 ; the product of any three
consecutive integers is divisible by 3 = 6! and so on.
Theory of Numbers 11
Example 1 Prove that product of two odd numbers of the form 4n + 1 is of the form ( 4n + 1) .
Solution Let a = 4k + 1, b = 4k + 1
be two numbers of the form ( 4n + 1)
ab = ( 4k + 1)( 4k + 1)
= 16kk + 4k + 4k + 1
= 4( 4kk + k + k ) + 1 = 4l + 1
(where l = 4kk + k + k )
Which is in form ( 4n + 1) .
## Solution Let n = 2m + 1 be an odd number .
n 2 = ( 2m + 1)2 = 4m 2 + 4m + 1
= 4m(m + 1) + 1
Now, m(m + 1) being product of two consecutive integers, is divisible by 2! = 2
m(m + 1) = 2j
n 2 = 4( 2j ) + 1 = 8j + 1
Example 3(a) Show that sum of an integer and its square is even.
Solution Let n be any integer.
So, we have to prove that n 2 + n is even.
n 2 + n = n(n + 1) which is product of two consecutive numbers n and n + 1 and
hence divisible by 2! = 2
Hence, n 2 + n is an even number.
## Example 3(b) If n is an integer. Prove that product n(n 2 1) is multiple of 6.
Solution n(n 2 1) = n(n 1)(n + 1) = (n 1)n(n + 1)
Which being the product of three consecutive integers is divisible by 3! = 6
n(n 2 1) is divisible by 6.
i . e., n(n 2 1) is a multiple of 6.
Note If n is a multiple of p, we shall write
n = M ( p) .
## Example 4 If r is an integer, show that r (r 2 1)( 3r + 2) is divisible by 24.
Solution r (r 2 1)( 3r + 2) = r (r 1)(r + 1)( 3r + 2)
= (r 1)r (r + 1){3(r + 2) 4}
= 3(r 1)r (r + 1)(r + 2) 4(r 1)r (r + 1)
(r 1)r (r + 1)(r + 2) being the product of four consecutive integers is divisible by
4! = 24
3(r 1)r (r + 1)(r + 2) is also divisible by 24.
## Again (r 1)r (r + 1) is divisible by 3! = 6
4(r 1)r (r + 1) is also divisible by 4 6 = 24
r (r 2 1)( 3r + 2) = 3(r 1)r (r + 1)(r + 2) 4(r 1) r (r + 1)
is also divisible by 24.
## Example 5 If m, n are positive integers, show that (m + n )! is divisible by m ! n ! .
(m + n )! 1 2 3 K m(m + 1)(m + 2) K (m + n )
Solution =
m!n ! m!n !
m !(m + 1)(m + 2) K (m + n )
=
m!n !
(m + 1)(m + 2) K (m + n )
=
n!
The product of n consecutive integers
=
n!
= An integer
(m + n )! is divisible by m ! n !.
## Example 6 If ( 4x y ) is a multiple of 3, show that 4x 2 + 7xy 2y 2 is divisible by 9.
Solution Q 4x y is a multiple of 3.
4x y = 3m
y = 4x 3m
On putting value of y in 4x 2 + 7xy 2y 2
= 4x 2 + 7x ( 4x 3m ) 2 ( 4x 3m )2
= 4x 2 + 28x 2 21xm 2 (16x 2 + 9m 2 24xm )
= 4x 2 + 28x 2 21xm 32x 2 18m 2 + 48xm
= 27mx 18m 2 = 9m ( 3x 2m )
4x 7x 2y 2 is divisible by 9.
2
## Example 7 If n is an integer, prove that n (n + 1) ( 2n + 1) is divisible by 6.
Solution n (n + 1) ( 2n + 1) = n (n + 1) [(n + 2) + (n 1)]
= n (n + 1) (n + 2) + n (n + 1) (n 1)
= n (n + 1) (n + 2) + (n 1) n (n + 1)
Each of the two (n 1) n (n + 1) and (n + 2) (n + 1) n being the product of three
consecutive integers is divisible by 3! = 6
n (n + 1) ( 2n + 1) is also divisible by 6.
## Example 8 Prove that 4 does not divide (m 2 + 2) for any integer m.
Solution Q m is an integer.
Either m is even or m is odd.
Case I m is even
So let m = 2k
Theory of Numbers 13
m 2 + 2 = ( 2k )2 + 2 = 4k 2 + 2 = 2 ( 2k 2 + 1)
= (2 an odd integer)
Which is not divisible by 4.
Case II m is odd
Let m = 2k + 1
m 2 + 2 = ( 2k + 1)2 + 2 = 4k 2 + 1 + 4k + 2
= ( 2 + 4k + 4k 2 ) + 1.
Which being an odd integer is not divisible by 4.
## Note Two important formulae.
1. If n is either even or odd,
x n y n = ( x y )( x n 1 + x n 2y + x n 3y 2 + K + y n 1)
2. If n is odd,
x n + y n = ( x + y )( x n 1 x n 2y + x n 3y 2 K + y n 1)
## Example 9 Prove that 8n 3n is divisible by 5.
Solution 8n 3n = ( 8 3)( 8n 1 + 8n 2 . 3 + K + 3n 1)
or 8n 3n = 5 ( 8n 1 + 8n 2.3 K + 3n 1)
8n 3n is divisible by 5.
## Example 10 If p > 1 and 2p 1 is prime, then prove that p is prime.
Solution If possible, let p be not prime
p is composite. (Q p > 1)
p = mn, where m > 1 and n > 1
2p 1 = 2mn 1 = ( 2m )n 1n
Putting 2m = a = an 1n , where a = 2m > 2
= (a 1)(an 1 + an 2 + K + 1n 1)
Now, each of the two factors on RHS is greater than 1.
2p 1 is composite.
But it is contrary to given
p must be prime.
Remark
l
But converse is not true i.e., when p is prime, 2p 1 need not be prime.
l
For example, p = 11 is prime but 211 1 is divisible by 23 and hence is not prime.
## Theorem 4 The number of primes is infinite.
Proof If possible suppose that the numbers of prime is finite.
the greatest prime say q.
Let b denote the product of these primes 2, 3, 5, , q.
i.e., let b = 2 35Kq (i)
let a =b + 1 (ii)
Surely, a 1 (Qa = b + 1 > 1)
## The number a must have a prime say factor p i.e., p | a.
Now, p is one of the primes 2, 3, 5, 7 q (Because according to our assumption 2, 3, 5, 7, q are the only
primes).
p |b (Qb = 235
. . K q)
Now p | a and p | b
p |a b or p |1 [Qfrom Eq. (ii), a b = 1]
p =1 (which is impossible) (Q1 is not prime)
So our supposition is false.
The number of primes is infinite.
Theorem 5 Fundamental Theorem of Arithmetic each natural number greater than 1 can be
expressed as a product of primes in one and only one way (except for the order of the factors).
## Example 1 Every natural number other than 1 admits of a prime factor.
Solution Suppose that n 1 is a natural number.
If n itself is a prime number, the example is proved in as much as the prime number
n is a factor of itself.
If n is composite, then n must have factors other than 1 and n.
Let l be the least of these factors of n other than 1 and n.
i.e., 1 < l < n and l | n
Now, we have to prove, l is prime.
If possible let l be not prime.
But l >1
l is composite
integers l1 and l2 such that
l = l1l2 where 1 < l1 < l and 1 < l2 < l
l1| l but l | n
l1| n where 1 < l1 < l < n
i.e., l1( < l ) is a divisor of n other than 1 and n.
But this is a contradiction because l1 < l and l is the least divisor of n other than 1
and n.
Our supposition is wrong.
l is a prime factor of n.
Example 2 Show that every odd prime can be put either in the form 4k + 1 or 4k + 3 (i.e., 4k 1),
where k is a positive integer.
Solution Let n be any odd prime. If we divide any n by 4, we get
n = 4k + r
where 0 r < 4 i.e., r = 0, 1, 2, 3
Either n = 4k or n = 4k + 1
or n = 4k + 2 or n = 4k + 3
Clearly, 4n is never prime and 4n + 2 = 2( 2n + 1)
cannot be prime unless n = 0 (Q 4 and 2 can not be factors of an odd prime)
Theory of Numbers 15
## An odd prime n is either of the form
4k + 1 or 4k + 3.
But 4k + 3 = 4(k + 1) 4 + 3 = 4k 1 (where k = k + 1)
An odd prime n is either of the form 4k + 1 or ( 4k + 3) i.e., 4k 1.
Note 1. Every number of the form 4k + 3 is of the form 4k 1 and conversely.
2. Every number of the form 4k + 1 or 4k 1 is not necessarily prime.
3. The above result should be committed to memory.
Example 3 Show that there are infinitely many primes of the form 4n + 3.
Solution If possible, let number of primes of form ( 4n + 3) be finite.
These primes are 3, 7, 11,, q (put n = 0, 1, 2, )
Let q be the greatest of these primes of the form ( 4n + 3) .
Let a = 3, 7,11, K , q be the product of all primes of the form ( 4n + 3) .
Let b = 4a 1 (i)
b >1 [Qa 3, b = 4a 1, b 11]
By fundamental theorem b can be expressed as a product of primes say
p1. p2. p3 K pr .
i.e., b = p1. p2 K pr (ii)
Now, b = 4a 1 is odd and hence 2 can't be a factor of b.
None of the prime factors in RHS of Eq. (ii) is 2.
i.e., Every prime factor in RHS of Eq. (ii) is odd.
Each of p1, p2, K , pr is of the form ( 4n + 1) or ( 4n + 3) .
Again, all p1, p2, K , pr can't be of the form ( 4n + 1).
[Q If it were so, then b (their product) will also be of the form ( 4n + 1)] .
But this is contrary to Eq. (i) as b = 4a 1 is of the form ( 4n + 3) .
At least one of p1, p2 K pr (say p) is (a prime factor of b) of the form ( 4n + 3) i.e.,
p|b.
Also p|a [Q p is one prime of the form ( 4n + 3) and a is product of all such primes] .
p|4a
p | ( 4a b )
p |1 [Q from Eq. (i), 4a b = 1]
Which is impossible [Q p being prime > 1 ]
Our supposition is wrong.
Number of primes of the form ( 4n + 3) is infinite.
## Theorem 6 The number of divisors of a composite number n : If n is a composite number of the
order n = p1 1 . p 2 2 p k k , then the number of divisors denoted by d (n ) is
(1 + 1)( 2 + 1) K ( k + 1)
Proof Let n be any composite number, let d (n ) denote the number of divisors of composite number n
by Fundamental Theorem of Arithmetic. n can be expressed as the product of the powers of primes.
n = p1 1 . p 2 2 Kp k k , where p1 , p 2 , K , p k are distinct primes and 1 , 2, k are non negative integers.
Qp1 is a prime number, therefore, the only divisors of p1 1 are 1, p1, p12 , p13 , K , p1 1
The number of these divisors of p1 1 = 1 + 1
Similarly, the number of divisors of p 2 2 = 2 + 1
The number of divisors of p k k = k + 1
Therefore, the total number of divisors of n = p1 1 p 2 2 K p k k = (1 + 1) ( 2 + 1) K ( k + 1)
i
[Q Every divisor of pi (1 i k ) is a divisor of n]
i.e., d (n ) = (1 + 1)( 2 + 1)( k + 1).
Note Let n = p11 p2 2 p3 3 [ p1, p2, p3 are distinct prime numbers]
Let d (n ) denotes number of divisor .
1. If n is a perfect square then d (n ) is odd
(Q all the di are even)
2. If n is not a perfect square then, d (n ) is even.
[The number of ways of writing n are the product of two factors.]
d (n ) + 1
If n is a perfect square, then number of ways are equal to .
2
d (n )
If n is not a perfect square, then number of ways are equal to .
2
Theorem 7 The sum of the divisors of any composite number n is denoted by (n ) which is equal to
p 1 + 1 1 p2 + 1 1 k + 1 1
1 2 K pk .
p1 1 p2 1 pk 1
Proof Let n be any composite number and let (n ) is sum of positive divisors of n. By Fundamental
Theorem of Arithmetic n can be expressed as the product of the powers of primes.
n = p1 1 . p 2 2 K p k k
[where p1 , p 2 , K , p k are distinct primes and 1 , 2 , K , k are non-negative integers]
Q p1 is prime number
divisors of p1 1 are only 1, p1 , p12 , K , p1 1
1 [ p1 + 1 1]
Sum of these divisors of p1 1 = 1 + p1 + p12 + K + p1 1 =
p 1
a (r n 1)
Q RHS is a Geometric Progression with a = 1, r = p1 , n = 1 + 1 and Sn = r 1
+1
(p2 2 1)
Similarly sum of divisors of p 2 2 =
p2 1
+1
(pk k 1)
Sum of divisors of p k k =
pk 1
(n ) = Sum of divisors of n = p1 1 p 2 2 ... p k k
+1 +1 +1
( p1 1 1) ( p 2 2 1) (p k 1)
= . K k .
( p1 1) ( p 2 1) ( p k 1)
i
[QEvery divisor of pi (1 i k ) is divisor of n]
Theory of Numbers 17
## Note If dk (n ) denotes the sum of k th power of divisor of n, then
( p1k (1 + 1) 1) ( pk2 ( 2 + 1) 1) ( pm
k ( m + 1)
1)
dk (n ) = K
( p1 1)
k
( p2 1)
k
( pm 1)
k
## Example 1 Find the sum of the cubes of divisor of 12.
Solution 12 = 22 3
23(2 + 1) 1 33(1 + 1) 1
dk (12) = = 2044
23 1 33 1
## Example 2 Find the number of divisor of 600.
Solution 600 = 23 31 52
1 = 3, 2 = 1, 3 = 2
Number of divisors = ( 1 + 1)( 2 + 1)( 3 + 1)
= ( 3 + 1)(1 + 1)( 2 + 1) = 4 2 3 = 24
## Greatest Integer Function
Greatest integer function is also known as Bracket Function.
If x is any real number, then the largest integer which does not exceed x is called the integral part of x
and will be denoted by [x ] .
The function which associates with each real number x, the integer [x ] is often called the bracket
function.
For example, [3] = 3, [4] = 4, [37
. ]=3
. ] = 5, = 1, [ ] = 4
5
[42
3
## Note 1. [ x ] is the largest integer x.
2. If a and b are positive integer, such that
a = qb + r , 0 r < b
a r r
Then, = q + , where 0 < < 1
b b b
a
=q
b
a
i.e., is the quotient in the division of a by b.
b
## Properties of Greatest Integer Function
(1) [x ] x < [x ] + 1 and x 1 < [x ] x , 0 x , 0 x [x ] < 1
(2) If x 0, [x ] = 1
1 i x
(3) [x + m ] = [x ] + m , If m is an integer.
(4) [x ] + [y ] [x + y ] [x ] + [y ] + 1
(5) [x ] + [x ] = 0, If x is an integer = 1 otherwise.
[x ]
(6) = , if m is a positive integer.
x
m m
(7) [x ] is the least integer greater than or equal to x.
This is denoted as (x ) (read as ceiling x)
. ) = 3, (25
For example, ( 25 . )= 2
(8) [x + 05
. ] is the nearest integer to x.
If x is midway between two integers, [x + 05
. ] represents the larger of the two integers.
(9) The number of positive integers less than or equal to n and divisible by m is given by
n
.
m
## (10) If p is a prime number and e is the largest exponent of p such that
n
pe | n ! , then e =
i =1 p
i
[a ] a
Theorem 1 If a is real number, c is natural number, then
c = c
Proof Let [a ] = n i.e., n is largest integer a
a = n + r, 0 r < 1 (i)
[a ] n
c = c = m
Let
n
= m + s, where 0 s < 1
c
n = mc + cs, where 0 cs < c (ii)
[a ]
LHS = = = m
n
Now,
c c
n + r
RHS = =
a
[Putting value of a from Eq. (i)]
c c
## Putting the value of n from Eq. (ii).
mc + cs + r
=
c
cs + r
= m +
c
From Eq. (ii), cs (c 1) and r < 1
Adding cs + r < c 1 + 1
cs + r < c
cs + r
<1
c
cs + r
RHS = m + =m
c
LHS = RHS.
Theory of Numbers 19
x + x + 1 = [x ]
2 2
## Proof First suppose that x = 2m + y , where m is an integer and 0 y < 1.
[x ] = 2m , = m
x
2
x + 1 = 2m + 1 + y = m
2 2
1 1+y
Since <1
2 2
x + x + 1 = [x ]
2 2
## Next, let x = (2m + 1) + y , where m is an integer and 0 y < 1.
Then, x = m , x + 1 = (2m + 2) + y = m + 1
2 2 2
[x ] = 2m + 1
x + x + 1 = [x ]
2 2
## Example 1 Find the highest power of 3 contained in 1000!.
Solution p = 3, n = 1000
n 1000 1
p = 3 = 333 3 = 333
n 333
p 2 = 3 = [111] = 111
n 111
p 3 = 3 = [ 37] = 37
n 37 1
p 4 = 3 = 12 3 = 12
n 12
p 5 = 3 = [ 4] = 4
n 4 1
p 6 = 3 = 1 3 = 1
n 1
p 7 = 3 = 0
Highest power of 3 contained in 1000!
n n n n n n n
= + 2 + p3 + p4 + p5 + p6 + p7
p p
= 333 + 111 + 37 + 12 + 4 + 1 + 0 = 498
n + 1 2n
If n and k are positive integers and k is greater than 1, then +
n
Theorem 3
k k k
Proof Let n = qk + r , q and r are integers and
0 r k 1
n r n+1 r +1
Then, =q + , =q + ,
k k k k
2n 2r
= 2q +
k k
n + 1
(i) r < k 1, then = q , = q , 2q
n 2n
k k k
The desired result is immediate.
(ii) r = k 1, then
n = q , n + 1 = q + 1,
k k
2n = 2q + 2(k 1) = 2q + 1
k k
n + n + 1 = 2n
k k k
## Theorem 4 If n be any positive integer, then show that
n + 1 + n + 2 + n + 4 + n + 8 + K = n
2 4 8 16
## Proof We know that,
x + 1
[x ] = +
x
2 2
n n n n
Applying above formula to n , , , , ,K
2 4 8 16
n + 1
[n ] = +
n
2 2
n = n + (n / 2) + 1
2 4 2
n = n + (n / 4) + 1
4 8 2
n = n + (n / 8) + 1
8 16 2
Adding corresponding sides and cancelling out the terms , , , from both sides, we have
n n n
2 4 8
n + 1 n + 2 n + 4
n= + + +K
2 4 8
[n ] = n
Theory of Numbers 21
## Theorem 5 For every real number x
[x ] + x + + x + + K +
1 2 x + n 1 = [nx ]
n n n
Proof Let x = [x ] + y , where 0 y < 1
Let p be an integer such that
p 1 ny < p
(This is always possible because given a real number, we can always find two consecutive integers
between which the number lies).
k k
Now, x + = [x ] + y +
n n
k p 1+ k p+k
Also, y + lies between and
n n n
p 1+ k
So long as < 1,
n
i.e., k < n (p 1)
k
y + is less than 1 and consequently
n
k
x + n = [x ]
k
x + n = [x ]
i.e., for k = 0, 1, , n p
k
But x + = [x ] + 1, for k = n p + 1,, n 1
n
n 1
[x ] + x + + K + x +
1
n n
= [x ] + K + [x ](n p + 1 times) + ([x ] + 1) + ([x ] + 1) + K ( p 1) times
= n[x ] + ( p 1) (i)
Also, [nx ] = [n[x ] + ny ] = n[x ] + ( p 1)
Since p 1 ny < p (ii)
From Eqs. (i) and (ii)
n 1
[x ] + x + + x + + K + x +
1 2
= [nx ]
n n n
## Theorem 6 The highest power of a prime number p contained in n ! is given by
n n n
k (n !) = + 2 + 3 + K
p p p
Proof Let k (n !)denote the highest power of p contained in n !n ! is the product of the factors 1, 2, 3, , n.
The factors in n ! which will be divisible by p are
n
p , 2p , 3p , K p
p
n n
k (n !) = + k ! (i)
p p
n
Changing n to in Eq. (i)
p
n n n
k ! = 2 + k 2 ! (ii)
p p p
Putting the value from Eq. (ii) in Eq. (i)
n n n
k (n !) = + 2 + k 2 !
p p p
n n n
Continuing this process, we get k (n !) = + 2 + 3 + K
p p p
This process must end after a finite number of steps.
Congruences
If a and b are two integers and m is a positive integer, then a is said to be congruent to b modulo m, if m
divides a b denoted by m | (a b ).
In notation form we express it as a b mod m or a b 0 mod m.
## Note 1. a b mod m, then m | (a b ) or (a b ) is a multiple of m.
2. If m | (a b ) [m does not divides (a b )], then a is said to be incongruent to b mod m and this
fact is expressed as a is not congruent to b mod m.
3. If m | a, then a 0 mod m
For example :
(i) 13 1 mod 4 (Q 4 | (13 1) = 12)
(ii) 4 1 mod 5 (Q 5 | ( 4 ( 1)) = 5)
(iii) 12 0 mod 4 (Q 4|12)
(iv) 17 is not congruent to 3 mod 5 (Q 5 | (17 3))
## Theorem 7 If a b mod m, then
(i) a + c = b + c mod m
(ii) ac bc mod m, where c is any integer.
Proof (i) Qa b mod m
m | (a b)
m | {(a + c ) (b + c )} [Qa + c ( b + c ) = a b ]
a + c b + c mod m
(ii)Qa b mod m
m | (a b )
m | c (a b )
m | (ac bc )
ac bc mod m
## Note The converse of theorem 15 (ii) is not true.
Theorem states that if a b mod m, then ac bc mod m.
i.e., a congruence can always be multiplied by an integer
But the converse is not true i.e.,. It is not always possible to cancel a common factor from a
congruence.
For example : 16 8 mod 4 [Q 4|16 8]
But if we cancel the common factor 8 from numbers 16 and 8, we get 2 1 mod 4 which is a false
result because 4 | ( 2 1)
Theory of Numbers 23
## Theorem 8 If a b mod m and c d mod m, then
(i) a + c b + d mod m (ii) a c b d mod m (iii) ac bd mod m
Proof (i) Qa b mod m and c d mod m
m | (a b ) and m| (c d )
m | ((a b ) + (c d ))
or m | ((a + c ) (b + d ))
a + c b + d mod m
(ii) a b mod m and c d mod m
m | (a b ) and m | (c d )
m | (a b ) (c d ) or m | (a c ) (b d )
a c b d mod m
(iii)Qa b mod m and c d mod m
m | (a b ) and m | (c d )
There exists integer h and k such that
a b = mh and c d = mk
a = b + mh and c = d + mk
Multiplying the two equations, we get
ac = (b + mh )(d + mk ) = bd + mbk + mhd + m 2hk
ac bd = m (bk + hd + mhk )
By definition of divisibility m | (ac bd )
or ac bd mod m
## Corollary If a b mod m, then a 2 b 2 mod m
Q a b mod m and again
a b mod m
Multiplying the two congruence a 2 b 2 mod m
Theorem 9 (i) Prove that a a mod m i.e., every integer is congruent to itself.
(ii) If a b mod m, then prove that b a mod m
(iii) If a b mod m, b c mod m prove that a c mod m
Proof (i) We know that m|0 (m 0)
m | (a a )
a a mod m [by definition of congruence]
(ii) Let a b mod m
m | (a b )
m | (a b ) or m | (b a )
b a mod m
(iii) Let a b mod m and b c mod m
m | ( a b ) and m | (b c )
m | ( a b ) + (b c ) or m | ( a c )
a c mod m
## Proof We know that,
a k b k = (a b )(a k 1 + a k 2b + a k 3b 2 + K + b k 1 ) (i)
But a b mod m m | (a b )
There exists an integer t such that
a b = mt (ii)
Putting this value of (a b ) from Eq. (ii) in Eq. (i)
a k b k = mt (a k 1 + a k 2b + K b k 1 )
m| (a k b k )
a k b k mod m
1 2
Theorem 11 If a b mod m and f (x ) = p 0x n + p1x n + p 2x n + K +pn 1x + pn is an integral
rational function of an indeterminate x with integral coefficients, then f ( a ) f ( b ) mod m
1 2
Proof Q f ( x ) = p 0x n + p1x n + p 2x n + K + pn 1 x + pn
Putting x = a
1 2
f ( a ) = p 0an + p1an + p 2an + K + pn 1 a + pn (i)
Putting x = b
1 2
f (b ) = p 0bn + p1bn + p 2bn + K + pn 1b + pn (ii)
## Subtract Eq. (i) from Eq. (ii), we get
1 1
f (a ) f (b ) = p 0 (an bn ) + p1 (an bn ) + K + pn 1 (a b)
n 1 n 2 n 1
= p 0 (a b )(a +a b + K+ b )+
n 2 n 3 n 2
p1 (a b )(a +a b + K+ b ) + K + pn 1 (a b)
1 2 2 2 3 2
or f (a ) f (b ) = (a b )[p 0 (an + an b + K + bn ) + p1 (an + an b + K + bn )
+ K + pn 1] (iii)
= (a b ) t (say)
But a b mod m (given)
m| (a b )
an integer k such that a b = mk
Putting this value of a b = mk in Eq. (iii)
f (a ) f (b ) = mkt
m |f (a ) f (b )
f (a ) f (b ) mod m
## Theorem 12 Fermat Theorem
If p is prime, then
(a + b ) p = (a p + b p ) mod p.
## Proof Expanding by binomial theorem
(a + b ) p = a p + pC1a p 1b + pC2ap 2b 2 + K + pCp 1abp 1 + bp
p 1
or (a + b ) p = (ap + bp ) + p
Cr ap r br (i)
r =1
Theory of Numbers 25
p!
Now, p
Cr = ; 1 r (p 1)
r !(p r )!
But p ! = 123
. . p is divisible by p
p is coprime to r !
Q p is coprime to 1, 2, 3, r (Qr < p , p is prime)
p is co prime to their product = r !
Also for the same reason p is coprime to (p r )!
p!
p
Cr = is divisible by p.
r !(p r )!
an integer k r such that
p
Cr = pk r
p
Putting this value of Cr in Eq. (i)
p 1
(a + b )p (ap + bp ) = p kr ap r br
r =1
which is divisible by p.
(a + b )p (ap + bp ) mod p
Generalization
If p is a prime number, prove that
(a1 + a 2 + a3 + K + an ) p
(a1p + ap2 + a3p + K + anp ) mod p
(a1 + a 2 + a3 + K + an ) p = (a1 + b1 ) p
where b1 = a 2 + a3 + K + an
(a1p + b1p ) mod p.
[a1p + (a 2 + a3 + K + an ) p ] mod p
[a1p + (a 2 + c 2 ) p ] mod p
where c 2 = a3 + a 4 + K + an (a1p + ap2 + c p2 ) mod p
continuing like this, we get
(a1 + a 2 + a3 + K + an ) p (a1p + ap2 + K + anp ) mod p
ap = a mod p
## Proof We know that,
(a1 + a 2 + a3 + K + an ) p (a1p + ap2 + a3p K + anp ) mod p (i)
Putting a1 = a 2 = a3 = K = an = 1 in Eq. (i)
(1 + 1 + 1 + K + 1) p
(1p + 1p + 1p + K + 1p ) mod p
or n (1 + 1 + 1 + K + 1) mod p
p
## or n n mod p for every natural number n.
p
Replacing n by a.
ap = a mod p
## Theorem 14 Fermat Little Theorem
If p is a prime number and (a , p ) 1, prove that ap 1 = 1 mod p.
Proof As p is prime.
ap = a mod p
Cancelling a from both sides. [a is coprime to p]
p 1
We have a 1 mod p.
2 n 1 n
Theorem 15 n ! = nn n C1 (n 1)n + n C2 (n 2)n ... + (1)n Cn 22
n
+ (1)n Cn 1
## Proof Expanding by binomial theorem
1) x 2) x n 2 2x n 1 x
(e x 1)n = enx n C1e (n + n C2e (n K + n Cn 2 (1) e + n Cn 1 (1) e + (1)n (i)
2 3
We know that e = 1 + + + +K
1! 2! 3!
Using this expansion of e , Eq. (i) becomes
n
x x 2 x3 xn
1 + + + + K+ + K 1
1 ! 2 ! 3 ! n !
(nx )2 (nx )n (n 1)x [(n 1)x ]2 [(n 1)x ]n
+ K n C1 1 +
nx
= 1 + + + K+ + + K+ + K
1! 2! n!
1! 2! n!
(n 2) x [(n 2) x ]2 [(n 2) x ]n
+ n C2 1 + + +K + + K
1! 2! n!
n 2 2x (2x )2 (2x )n
+ K + n Cn 2 (1) 1 + + +K+ + K
1! 2! n
1n x x2 xn
+ (1)n Cn 1
1 + + + K+ + K + (1)n
1! 2! n!
n
Comparing coefficient of x on both sides
nn n (n 1)n (n 2)n
1= C1 + n C2 K+
n! n! n!
2n (1)n 1 n Cn 1
(1)n 2 n Cn 2 +
n! n!
Multiplying both sides by n !
2 n 1 n
n ! = nn n C1 (n 1)n + n C2 (n 2)n K + (1)n Cn 22
n
+ (1)n Cn 1
## Theorem 16 Wilson Theorem
If p is prime, then (p 1)! + 1 0 mod p
Proof
Case I when p = 2
(2 1)! + 1 0 mod 2 [Putting p = 2 in (p 1)! + 1 0 mod p ]
1 ! + 1 0 mod 2
2 0 mod 2
which is true
Wilson theorem is true for p = 2 .
Theory of Numbers 27
## Case II If p is an odd prime.
2 n 1 n
n ! = nn n C1 (n 1)n + n C2 (n 2)n ... + (1)n Cn 22
n
+ (1) n Cn 1
## Put n = p 1 on both sides
( p 1)! = ( p 1) p 1 p 1
C1 ( p 2) p 1
p 1
+ C2 ( p 3) p 1 + .. + (1) p 3 p 1
Cp 32p 1 + (1) p 2 p 1
Cp 2 (i)
Q p is prime.
p is coprime to all numbers < p.
i.e., p is coprime to p 1, p 2, p 3, 2, 1.
Putting a = p 1, p 2, p 3, 2 in Fermat Theorem
ap 1 mod p
( p 1) p 1 1 mod p
or ( p 1) p 1 1 = M(p)
( p 1) p 1 = M ( p ) + 1
Similarly, ( p 2) p 1 = M ( p ) + 1
( p 3) p 1 = M ( p ) + 1
2p 1 = M ( p ) + 1
Putting these values of ( p 1) p 1 , ( p 2) p 1, , 2p 1 in Eq. (i).
p 1 p 1
( p 1)! = [M ( p ) + 1] C1[M ( p ) + 1] + C2[M ( p ) + 1] + .. + (1) p 3 p 1Cp 3
[M ( p ) + 1] + (1) p 2 p 1Cp 2
p 1
p 1 p 1
or ( p 1)! = M ( p ) + 1 C1 + C2 K + (1) p 3 Cp 3 + (1) p 2 p 1Cp 2
## Adding and Subtracting (1) p 1 in RHS.
( p 1)! = M (p ) + [(1) p 1 p 1
C1 + p 1
C2 p 1
C3 + K + (1) p 3 p 3Cp 3
## + (1) p 2 p 1Cp 2 + (1) p 1 ] (1) p 1
( p 1)! = M ( p ) + (1 1) p 1 (1) p 1
( p 1)! = M ( p ) + 0 (1) p 1 = M ( p ) 1
Qp is odd. p 1 is even.
(1) p 1 = 1
or ( p 1)! + 1 = M ( p )
( p 1)! + 1 is divisible by p.
( p 1)! + 1 0 mod p.
## Theorem 17 Converse of Wilson Theorem
If p > 1 and ( p 1)! + 1 0 mod p, then p is a prime number.
Proof If possible let p be not prime.
p is composite (Qp > 1).
So let p = p1p 2, where (1 < p1 < p , 1 < p 2 < p ) or 1 < p1 p 1, 1 < p 2 p 1
Now, 1 < p1 p 1
p1 is one of the factors in the value of ( p 1)! and therefore p1 divides ( p 1)! .
Also p = p1p 2 (i)
p1 | p (ii)
But ( p 1)! + 1 0 mod p
p | ( p 1)! + 1 (iii)
From Eqs. (ii) and (iii)
p1 | ( p 1)! + 1 (iv)
From Eqs. (iv) and (i)
p1 | ( p 1)! + 1 (p 1)!
i.e., p1 | 1
But this is impossible. (Qp1 > 1)
p is a prime number.
Eulers Function
Definition : The number of integers n and coprime to n is called Euler's function for n and is denoted by
(n ).
Examples
(1) = 1
[Q1 is the only integer 1 and coprime to 1].
( 2) = 1
[Q1 is the only integer < 2 and coprime to 2].
(8) = 4
[Q1, 3, 5, 7 are the only four integers < 8 and coprime to 8].
Remark
l
If p is a prime number, then 1, 2, 3, ( p 1) are all less than p and coprime to p and are ( p 1) in total.
( p) = p 1
1 1 1
Theorem 18 Prove that (n ) = n 1 1 K 1 where p1 , p 2 , ... , pr are distinct prime
p1 p2 pr
factors of n.
Proof Q p1 , p 2 , K , pr are distinct prime factors of n.
k k
n = p1 1 . p 2 2 K prkr
k k k k
(n ) = ( p1 1 . p 2 2 ... prkr ) = ( p1 1 ) ( p 2 2 ) K ( prkr )
[Qp1 , p 2 , K , pr are distinct primes and hence are coprime to each other and (ab ) = (a ) (b ), if a and b are
coprime to each other.]
k 1 k 1 1
= p1 1 1 p 2 2 1 K prkr 1
p1 p2 pr
k k 1 1 1
= p1 1 . p 2 2 K prkr 1 1 K 1
p1 p 2 pr
1 1 1 k k
= n 1 1 K 1 [Qn = p1 1 . p 2 2 K prkr ]
p1 p2 pr
Theory of Numbers 29
1
Theorem 19 Prove that ( p k ) = p k 1 , where p is prime.
p
Proof Number of integers from 1 to p k which are not coprime to p k are p.1, p.2, p.3, p. p k 1.
Total number of such integers, which are not coprime to p k = p k 1.
(p k ) = Number of integers coprime to p k and < p k .
= p k p k 1 = p k (1 1 / p )
Remark
If a and b are coprime to each other, then (ab ) = (a ) (b ).
Example 1 Find the number of positive integers 3600 that coprime to 3600.
Solution n = 3600 = 24 32 52
(n ) = ( 3600) = ( 24 32 52 )
1 1 1 1 1 1
= n 1 1 1 = 3600 1 1 1
p1 p2 p3 2 3 5
[Here p1 = 2, p2 = 3, p3 = 5]
1 2 4
= 3600
2 3 5
( 3600) = 960
## Solution If (a, m ) = 1, then (m a, m ) = 1
Integers coprime to m occur in pairs of type a and m a .
(m ) is even.
## Example 3 For what values of m is (m ) odd.
Solution If m > 2, (m ) is even.
(1) = 1
( 2) = 1
Only for m = 1, m = 2
(m ) is odd.
10n 1 10n 1
Concept Let a = =
10 1 9
10n 1
We can express any a of the form in terms of perfect square.
9
10n 1
a= 9a = 10n 1
9
9a + 1 = 10n
Let b = 9a + 1
c = 8a + 1
Now, consider 4ab + c = 4a (9a + 1) + 8a + 1 = 36a 2 + 12a + 1 = (6a + 1)2
## Verification (6 1 + 1)2 = 72 = 49 = (6 11 + 1)2 = 672
(6 111 + 1)2 = 6672
Now, consider
(a 1) b + c = (a 1)(9a + 1) + 8a + 1
= 9a 2 + a 9a 1 + 8a + 1 = 9a 2 = (3a )2
Verification a = 1 32
a = 11 (33)2
Now, consider (16ab + c )
16a (9a + 1) + 8a + 1
(12a + 1)2
This is also a perfect square.
Concept Prove that every number of the sequence 49, 4489, 44489, 4448889 is a perfect square.
If there are n fours and (n 1) eight and one 9.
Let us denote 444889 as 43829.
Consider 667 written as 627
We know 444889 = (667)2.
We develop (6nm 17)
2
= 4n 8n 19
## If this is true then
2 2 2
6(10n 1) 6 10n + 3 2 . 10n + 1
(6n 17)
2
= + 1 = =
9 9 9
. 2n
410 . n + 1 40n 140n 1 + 1
+ 410
= = = 4n 8n 19
9 9
Example 1 Let n be the natural number. If 2n + 1 and 3n + 1 are perfect square. Then prove that n
is divided by 40.
Solution 40 = 23 5. It is sufficient to prove that n is divisible by 8 and 5.
Let 2n + 1 = x 2 (i)
and 3n + 1 = y 2 (ii)
x 2 is odd.
x is odd.
Let x = 2a + 1
( 2n + 1) = ( 2a + 1)2
2n + 1 = 4a 2 + 4a + 1
n = 2a 2 + a
n is even.
If n is even 3n + 1 is odd
y 2 is odd y is odd
Let y = 2b + 1
Subtract Eq. (i) from Eq. (ii), we get
n = y2 x2 (iii)
Theory of Numbers 31
n = ( 2b + 1)2 ( 2a + 1)2
We know square difference of odd number is always divisible by 8.
n is divisible by 8. (iv)
If we eliminate n between 1 and 2
3x 2 2y 2 = 1
Since square of odd number ends with 1, 5 or 9
3x 2 ends with 3, 5 or 4, 7
2y 2 ends with 2, 0, 8
x 2 ends with 1 and y 2 ends with 1
n = y2 x2 [from Eq. (iii)]
=0
It is divisible by 5.
Example 2 Prove that there are infinitely many squares in the sequence 1, 3, 6, 10, 15, 21, 28,
Solution Suppose Tn is a square
n(n + 1)
Let Tn of the above sequence be
2
n(n + 1)
Tn =
2
If it is a square then Tn = (m )2
n(n + 1)
= (m )2
2
n(n + 1) = 2(m )2
Also, T4n (n + 1) is also a square.
4n(n + 1)[ 4(n )(n + 1) + 1]
T4n (n + 1) =
2
4( 2m 2 )[ 4n 2 + 4n + 1]
= = 4m 2( 2n + 1)2
2
T4n (n + 1)is also a perfect square.
perfect squares are T1 = 1
T8 = 36 is a perfect square.
T288 is also a perfect square.
## Example 3 If N = 123 34 52, find the total number of even factor of N.
Solution If N = 123 34 52
Then, N = 26 37 52
Total Number of factors are = ( 6 + 1)(7 + 1)( 2 + 1) = 7 8 3 = 168
In above factors, some of these are odd multiple and some are even.
The odd multiples are formed only with combination of 35 and 55.
So total number of odd multiples is
(7 + 1)( 2 + 1) = 24
Even multiples = 168 24 = 144
## Example 4 Show that n 2 3n 19 is not a multiple of 289 for any integer n.
Solution Suppose 172 | n 2 3n 19
Since n 2 3n 19 = (n + 7)(n 10) + 51
17| (n + 7)(n 10);
172 | (n + 7)(n 10) (Q n + 7 n 10 (mod 17))
172 | (n 2 3n 19) (n + 7)(n + 10)
i.e., 172 |51 which is a contradiction
Consequently n 2 3n 19 is not a multiple of 289.
## Example 5 Determine all integers n such that n 4 n 2 + 64 is the square of an integer.
Solution Since n 4 n 2 + 64 > n 4 2n 2 + 1 = (n 2 1)2 for some non negative integer k,
(n 4 n 2 + 64) = (n 2 + k )2 = n 4 + 2n 2k + k 2
64 k 2
i.e., n 2 = from which we find that the possible values 64, 1, 0 for n 2 are
2k + 1
obtained when k = 0, 7, 8 respectively.
Hence, n ( 0, 1, 8)
## Example 6 Let a, b, c, d , e be consecutive positive integers such that b + c + d is a perfect square
and a + b + c + d + e is a perfect cube. Find smallest possible value of c.
Solution a, b, c, d, e are consecutive positive integer b + c + d = 3c and a + b + c + d + e = 5c
Now, 3|3c 32 | 3c (Q 3c is a square)
3|c 3|5c 33 |5c (Q 5c is a cube)
Also 5 | 5c 5 |5c 5 | c
3 2
## 3352|c i.e., 675|c
675 being a possible value of c is the smallest of such numbers.
## Example 7 If 11 + 11 11a 2 + 1 is an odd integer where a is a rational number. Prove that a is
perfect square.
Solution Let = 11 + 11 11a 2 + 1
Then, ( 11)2 = 112(11a 2 + 1)
Simplifying, we get ( 22) = 113a 2
r
Putting | a | = , r , s N such that (r , s ) = 1 gives ( 22)s 2 = 113r 2.
s
Since 112| s because otherwise 11 would divide r,11|. Writing 11 = , we get
( 2)s 2 = 11r 2
Since 11| s for otherwise we would have 11|r. It follows that s = 1. Thus we have
( 2) = 11r 2. Since 2 and are consecutive odd integers they are relatively
prime.
If 11| 2, then is a square of form 11n + 2 which is not possible.
11| and hence = 11n 2 for some n N.
Thus, we have = 11 = 112n 2
Theory of Numbers 33
Example 8 Determine all pairs of positive integers (m, n ) for which 2m + 3n is a perfect square.
Solution Let 2m + 3n = k 2
Since ( 1)m 2m k 2 1 (mod 3) (Q 3 | k )
m is even, say 2p.
Now, (k 2p ) (k + 2p ) = 3n
k 2 = 1 and k + 2 = 3n 2p + 1 + 1 = 3n
p p
## Since ( 1)n 3n (mod 4) = 2p + 1 + 1 1, n is even, say 2q.
Now ( 3q 1)( 3q + 1) = 2p + 1 3q 1 = 2
3q = 3 q = 1 and hence p = 2
So, we have only one solution (4, 2).
## Example 9 Determine the set of integers n for which n 2 + 19n + 92 is a square.
Solution Let n 2 + 19n + 92 = m 2, m is a non negative integer. Then, n 2 + 19n + 92 m 2 = 0
1
Solving for n, we get n = ( 19 4m 2 7 )
2
4m 2 7 is a square i.e., 4m 2 7 = p 2
Where p N
( 2m p )( 2m + p ) = 7
2m + p being positive therefore (2m+p) is 7 and 2m p = 1
Hence, 4m = 8 m = 2
Thus, we have n 2 + 19n + 92 = 4
n 2 + 19n + 88 = 0
(n + 8)(n + 11) = 0
n = 8 or 11
## Example 10 Find n, if 2200 2192 31 + 2n is a perfect square.
Solution 2200 2192 31 + 2n = 2192( 28 31) + 2n = 2192( 256 31) + 2n = 2192 225 + 2n
For some m N
2n = m 2 2192 225 = m 2 ( 296 15)2 = (m 296 15)(m + 296 15)
So, m = 296 15 = 2 and m + 296 15 = 2 + for some non negative integers , .
Hence, 297 15 = 2 + 2 = 2 ( 2 1)
2 = 297 and 2 1 = 15.
i.e., = 97 and = 4
n = 2 + = 198
Example 11 Find the number of values of n for which 211 + 28 + 2n is a perfect square.
Solution We can write 211 + 28 + 2n as
28 ( 23 + 1) + 2n
28 9 + 2n
2n ( 28n 9 + 1)
Note that for any k < 8, 2k ( 28 k 9 + 1) is not a square, when k is odd, 2k is not a square
and in the other case, the second factor is not a square. Hence n 8. Now write
211 + 28 + 2n as 28( 9 + 2n 8 ). Then the problem is to find the number of non negative
integers k such that 9 + 2k is a square.
9 + 2k = t 2 2k = (t 3)(t + 3)
t 3 = 2p and t + 3 = 2p + q for some non negative integers p and q.
2p ( 2q 1) = 6 implying p = 1 from which it follows that t = 5.
Hence, there is a unique solution.
## Example 12 Find all positive integers n for which n 2 + 96 is a perfect square.
Solution Suppose m is a positive integer, such that n 2 + 96 = m 2
Then, m 2 n 2 = 96 (m n )(m + n ) = 96
since m n < m + n and m n,m + n must be both even [as m + n = (m n ) + 2n.
Therefore m n, m + n must be both odd or both even; also if both of them are odd,
then the product cannot be even.
Only possibilities are
m n = 2, m + n = 48 m = 25, n = 23
m n = 4,m + n = 24 m = 14, n = 10
m n = 6, m + n = 16 m = 11, n = 5
m n = 8,m + n = 12 m = 10, n = 2
Example 13 Give with justification, a natural number n for which 39 + 312 + 315 + 3n is a perfect
cube .
Solution 39 + 312 + 315 + 3n = 39(1 + 34 + 36 + 3n 9 )
= ( 33 )3{1 + 3 32 + 3( 32 )2 + ( 32 )3 + 3n 9 3( 32 )2}
= ( 33 )3(1 + 32 )3 provided 3n 9 35 = 0
= ( 270)3 provided 3n 9 = 35
i.e., provided n = 14
So, given number is a perfect cube when n = 14
## Example 14 Prove that 2p + 3p is not a perfect power if p is a prime number.
Solution If p = 2, 2p + 3p = 22 + 32 = 13 (not a perfect power)
Let now p be a prime > 2..
x + a divides x p + a p , whenever p is odd [factor theorem]
2p + 3p is divisible by 2 + 3 = 5. We shall show that 2p + 3p is not divisible by 52.
x p + 3p = ( x + 3)( x p 1 3x p 2 + 32 x p 3 + K + ( 3)p 1)
When x = 3, then
x p 1 3 x p 2 + K + ( 3 )p 1
= ( 3 )p 1 3 ( 3 )p 2 + K + ( 3 )p 1
= p 3p 1
Theory of Numbers 35
## Showing that x + 3 does not divide
x p 1 3 x p 2 + 3 2 x p 3 + K + ( 3 )p 1
Consequently ( x + 3)2 does not divide x p + 3p. So, ( 2 + 3)2 does not divide
2p + 3p . Since 2p + 3p is a multiple of 5 but is not a multiple of 52.
it cannot be a perfect power.
Example 15 A 4 digit number has the following properties (I) It is a perfect square (II) its first 2 digit
are equal to each other (III) its last two digit are equal to each other.Find all such four
digit number.
Solution We want to find positive integers x and y such that1 x 9,0 y 9 and xxyy is a
perfect square. Since,102 = 100, 1002 = 10000. It follows that xxyy must be the square
of a 2 digit number. Suppose that (ab )2 = xxyy .
The number xxyy is clearly a multiple of 11.
Since, it is a perfect square it must be a multiple of 112 i.e., 121.
It must be of the form
121 1, 121 4, 121 9, 121 16, 121 25,
121 36,121 49,121 64,121 81
Out of these 121 64 i.e., 7744 is of the form xxyy, we conclude that 7744 is the
desired number.
Example 16 Show that for any integer n, the number n 4 20n 2 + 4 is not a prime number.
Solution n 4 20n 2 + 4 = (n 4 4n 2 + 4) 16n 2
= (n 2 2)2 16n 2
= (n 2 4n 2)(n 2 + 4n 2) (i)
Note It can be easily seen that none of the factors n 2 4n 2 ,n 2 + 4n 2 can have the value 1,
whatever integral value n may have. Here four cases arises.
4 28
(i) n 2 4n 2 = 1 n =
2
4 20
(ii) n 4n 2 = 1 n =
2
2
4 28
(iii) n + 4n 2 = 1 n =
2
2
4 20
(iv) n + 4n 2 = 1 n =
2
2
From the above four cases, we find that whatever integral value n may have, n 4 20n 2 + 4 is the
product of the integers n 2 4n 2 and n 2 + 4n 2 neither of which equals 1..
Example 17 Prove that the product of four consecutive natural numbers cannot be a perfect cube.
Solution Consider the product
P = n (n + 1) (n + 2)(n + 3), where n is a natural number.
If possible, that P is a perfect cube = k 3 Two cases arises.
Case I If n is odd. n, (n + 1), (n + 3) are all prime to n + 2
Now, we know that every common divisor of n + p and n + q must divide q p.
## n + 2 and n(n + 1)(n + 3) are relatively prime.
Since, their product is a perfect cube, each of them must be a perfect cube.
Since, n 3 < n(n + 1)(n + 3) < (n + 3)3
n(n + 1)(n + 3) = (n + 1)3 or (n + 2)3
As n(n + 1)(n + 3) and (n + 2)3 are relatively prime, so second possibility ruled out.
Also n(n + 1)(n + 3) = (n + 1)3 n = 1. Since P = 24, when n = 1 which is not a perfect
cube. So the possibility n = 1 is also ruled out. So n cannot odd.
Case II If n is even.
Then n + 1 is prime to n, n + 2 and n + 3. Consequently n + 1 is relatively prime to
n(n + 2)(n + 3). Since the product of relatively prime numbers n + 1 and
n(n + 2)(n + 3) is a perfect cube, each of them must be a perfect cube.
n 3 < n(n + 2)(n + 3) < (n + 3)3
n(n + 2)(n + 3) = either (n + 1)3
or (n + 2) since n(n + 2)(n + 3) and n + 1 are relatively prime
3
## First possibility ruled out.
Also n(n + 2)(n + 3) = (n + 2)3 n + 4 = 0 which is out of question. Consequently n
cannot be even.
Thus, we find that the product of 4 consecutive integers cannot be a perfect cube.
Example 18 Find all primes p for which the quotient ( 2p 1 1)| p is a square.
Solution Suppose m (m + 1) = 7n 2, m and n are integers since m and m + 1are relatively prime.
m and m + 1 must be the numbers 7p 2, q 2
(in some order) p and q are relatively prime and pq = n; Since the product of 2
consecutive integer is even.
m(m + 1) is even, which means that one of the numbers m, m + 1 must be even.
Suppose m = q 2 (so that m + 1 = 7p 2 ). Since every square number is of one of the
forms 4k , 4k + 1. Consequently m + 1 must be of one of the forms 4k + 1, 4k + 2.
However this is not possible for if p is even, then 7p 2 is of the form 4k. If p is odd,
then 7p 2 is of the form 4k + 3.
m + 1 7p 2. So m = 7p 2 and m + 1 = q 2
## Concept of Finding Number of Positive Integral Solutions for
the Equation of the Form x2 + y2 = k
We know that,
(2n )2 0 mod 4
and ( 2n + 1)2 1 mod 4
Now, if
(a) x and y are both even then,
x 2 + y 2 0 mod 4
(b) x and y are both odd then,
x 2 + y 2 2 mod 4
Theory of Numbers 37
## (c) one is even and other is odd then,
x 2 + y 2 1 mod 4
{x 2 + y 2 0, 1, 2 mod 4 and x 2 + y 2
/ 3 and 4}
## the above discussion implies, if
x 2 + y2 = k
and if k is of the form of (4m + 3), then x 2 + y 2 = k does not have any integral solution.
e.g., suppose, we are asked to find integral solution for equation
x 2 + y 2 = 19, then it will not have any integral solution because 19 is of the form (4m + 3).
Now, if we have x 4 + y 4 = k , then we know that
(2n )4 0 mod 16 and (2n + 1)4 1 mod 16
Again, if (a) x and y are both even then,
x 4 + y 4 0 mod 16
(b) x and y are both odd then,
x 4 + y 4 2 mod 16
(c) one is even and other is odd, then
x 4 + y 4 1 mod 16
x 4 + y 4 0, 1, 2 mod 16
and x 4 + y4
/ i mod 16,
where i = (3, 4, 5, .. , 15)
So, the above discussion implies, if
x 4 + y4 = k
and k is of the form (16m + i ), where i = 3, 4, 5, K , 15, then x 4 + y 4 = k will not have any integral solution.
e.g., suppose we are asked to find integral solutions for equation.
x 4 + y 4 = 16003, then it will not give any integral solution because 16003 is of the form (16m + 3).
The concept can be extended to more than two variables expression, suppose the equation is
x 14 + x 24 + x 34 + x 44 + ... + x 14
4
= 1599
now, we know that
(2n )4 0 mod 16
and (2n + 1)4 1 mod 16
14
x i4 0, 1, 2 K 14 mod 16
i =1
## but our RHS is 1599 15 mod 16.
No integral solution can be obtained for the above equation.
Reason
[If all the variables are considered to be odd, then maximum remainder which can come out is 14 and if
any of the variable is an even number then remainder will be less than 14.]
Now, let us consider another discussion.
If we have a 2 + b 2 + c 2 = a 2b 2
we know, ( 2n )2 0 mod 4 and ( 2n + 1)2 1 mod 4
## Case I If a, b and c all are odd then,
a 2 + b 2 + c 2 3 mod 4
whereas a 2b 2 1 mod 4.
It will never give an equality, so the given equation has no integral solution.
Case II If two numbers are odd and one is even, then
a 2 + b 2 + c 2 2 mod 4
whereas a 2b 2 0, 1 mod 4
Again we get no integral solution.
Case III If two even and one odd.
a 2 + b 2 + c 2 1 mod 4
whereas a 2b 2 0 mod 4
Again, no integral solution
Case IV If all are even then
a 2 + b 2 + c 2 0 mod 4
whereas a 2b 2 0 mod 4
Now, there is a possibility to have a solution and the only possible solution is (0, 0, 0) which is the only
trivial solution.
Now, let us come again to the discussion of
x 2 + y2 = k
we have seen, if k = 4m + 3 there is no integral solution now, if k is even then it will be either of the form
k = 4m
or k = 4m + 2
considering k = 4m
If m can be expressed as i 2 + j 2 , where i and j are non-negative integers such that
(i) i j , then there will be four integral solutions and 8 ordered pairs.
(ii) if i = j or m can be written as i 2 + 02, then there will be two integral solutions and 4 ordered pairs.
Let us consider some examples.
e.g., x 2 + y 2 = 20
here, 20 is of the form 4(5) and 5 can be expressed as 5 = 12 + 22 which gives i = 1 and j = 2 so, it implies
there are 4 integral solutions, which will be of the form 2i and 2j also we have 8 ordered pairs.
Therefore in this case we have (2, 4) as one of the solutions and other solutions are
(2, 4), (2, 4), and (2, 4) also (4, 2), (4, 2), (4, 2) and (4, 2) keep this thing in mind there are only 4
integers used.
e.g., x 2 + y2 = 8
Here, 8 = 4(2)
and 2 = 12 + 12
which gives i = j .
So, it implies there are 2 integral solutions which will be of the form 2i and 2j also we have 4 ordered
pairs therefore in this case we have our solutions are
(2, 2), (2, 2), (2, 2) and (2, 2)
keep this thing in mind there are only 2 integers used.
Theory of Numbers 39
e.g., x 2 + y2 = 4
Here, 4 = 4(1)
and 1 = 12 + 02
which gives i = 1 and j = 0
So, it implies there are 2 integral solutions, which are of the form 2i and 2j also we have 4 ordered pairs.
Therefore in this case the solutions are
( 2, 0), (2, 0), (0, 2) and (0, 2).
e.g., x 2 + y 2 = 24
Here, 24 = 4 6
and 6 can't be represented as i 2 + j 2 so it will not have any integral solution.
e.g., x 2 + y 2 = 12
Here, 12 = 4 3
and 3 again can't be expressed as i 2 + j 2 so it will also not have any integral solution.
Now, considering k = 4m + 2 and if m can be written as (i 2 + j 2 + i + j ) and if i j , then these will be
4 integers and 8 ordered pairs of solutions.
And if i = j , then there will be 2 integers and 4 ordered pairs of solutions.
e.g., x 2 + y 2 = 10
Here, 10 = 4 ( 2) + 2
and 2 = (1)2 + (0)2 + (1) + (0)
Therefore there will be four integral solutions which will be given as (2i + 1) and (2 j + 1) and in this
case the solutions are 3 and 1 which will give eight ordered pairs as (3, 1), (3, 1), (3, 1) and (3, 1)
also (1, 3), (1, 3), (1, 3) and (1, 3).
e.g., x 2 + y2 = 2
Here, 2 = 4 (0) + 2
and 0 = (0)2 + (0)2 + (0) + (0)
Therefore there are only two integral solutions which will again be given as ( 2i + 1) and ( 2 j + 1)
and in this case the solutions are 1 and 1, which will give four ordered pairs as (1, 1), (1, 1), (1, 1) and
(1, 1).
e.g., x 2 + y 2 = 18
Here, 18 = 4 (4) + 2
and 4 = (1)2 + (1)2 + (1) + (1)
So i = 1 and j = 1
i = j
Therefore it will have 2 integral solutions which will be given as ( 2i + 1) and ( 2 j + 1) and in this case
the solutions are 3 and 3 which gives four ordered pairs as
( 3, 3), ( 3, 3), (3, 3) and (3, 3)
e.g., x 2 + y 2 = 14
Here, 14 = 4 (3) + 2
and 3 can't be expressed as (i 2 + j 2 + i + j ) as it is always an even number and an even number can't be
equal to an odd number. So it implies if right hand side is (4m + 2) and m is an odd number. So the
equation will never produce any integral solution.
Now, we will extend this concept for an odd number in right hand side of the equation.
x 2 + y2 = k
i.e., k = 4m + 1
or k = 4m + 3
As it has been already discussed (k = 4m + 3) will not produce any integral solutions.
So, considering k = 4m + 1, only.
If m = i 2 + j 2 + j,
then there will be an odd integer and an even integer, if i 0 and j 0 or i 0 and j = 0 , then there are
four integers and 8 ordered pairs which will satisfy the equation.
So, one of the integral solution is 2i and other is ( 2 j + 1).
Now, if i = 0, j 0, then there are two integers and four ordered pairs which will satisfy the equations.
So, one of the integral solution is 0 and other is ( 2 j + 1) .
e.g., x 2 + y 2 = 21
Here, 21 = 4 5 + 1
But 5 cannot be written as i 2 + j 2 + j , so it will not give any integral solution.
Exceptional case :
If x 2 + y2 = k
and k is an odd and a perfect square, then perform the following test always.
Take square root of k, which will come out to be as k, now subtract 1 from it we get ( k 1) always
double it, so it becomes 2( k 1), now add 1 to it which becomes 2( k 1). If this value is a perfect
square say, it is a 2 , then the equation will always have 6 integers and 12 ordered pairs as its solutions
and the integers will be a , ( k 1), k and 0. Always keep this thing in mind k is an integer.
And if the test fails, then equation will be solved by the method discussed earlier.
e.g., x 2 + y 2 = 169
here 169 = 4(42) + 1, which is of the form (4m + 1) and also it is an odd perfect square so we will have to
perform the mentioned test.
e.g., 169 = 13
13 1 = 12
12 2 = 24
24 + 1 = 25
and 25 = 52
we will have 4 integers in which 5, 12, will form 8 ordered pairs (5, 12), (5, 12), (5, 12), (5, 12), (12, 5),
(12, 5), (12, 5), (12, 5) also there will be three 13, 0 which will form four pairs (13, 0),(13, 0), (0, 13),
(0, 13)
e.g., x 2 + y 2 = 49
here 49 = 4 12 + 1, which is of the form (4m + 1) and also an odd perfect, so we will again perform the
mentioned test.
49 = 7
71= 6
6 2 = 12
12 + 1 = 13
but 13 is not a perfect square therefore the solution will be checked by the earlier method.
12 = (0)2 + (3)2 + (3)
Here, i = 0 and j = 3
so the solutions will be 0 and 7. Also the ordered pairs will be (0, 7), (0, 7), (7, 0) and (7, 0)
Theory of Numbers 41
## Concept Solving of the equation of the form xy = n.
If we are asked to find the number of positive integral solution for xy = n, we first write n is the form
p1 1 . p 2 2 . p3 3 . The number of positive integral solution is same as the number of divisors of n which is
equal to (1 + 1)( 2 + 1)(3 + 1)
Let us consider example xy = 8 = 23
Number of divisors of 8 are 3 + 1 = 4. So there are 4 integral solution and 4 ordered pair namely (1, 8)
(8, 1) (2, 4) (4, 2)
Now let us consider another example xy = 72(x + y )
we can write it as (x 72)( y 72) = 722.
Let x 72 = X , y 72 = 722
XY = 722 = 26.34
Number of solutions are 35.
A
Concept The area of a formed by pythogorean triplet with
integer sides is always divisible by 3.
1
Area of ABC = BC AB
2 k2 + 1
1
= 2x (x 2 1)
2 k2 1
= (x x )
3
Let p (x ) = x 3 x
B C
2k
For x > 1 we use induction
p(2) = 8 2 = 6 , p(2) is true
Let p (x ) be true for n = m
p (m ) = m3 m = 3c
p (m + 1) = (m + 1)3 (m + 1)
= (m + 1)3 (m + 1) = m3 + 3m 2 + 2m
= 3c + m + 3m 2 + 2m = 3(c + m + m 2 )
P (m + 1) is true.
Concept The radius of the circumcircle of a formed by pythogorean A
triplet cannot be integer.
The hypotenuse of the ABC is the diameter of the circle.
Let us consider the pythogorean triplet x 2 1, 2x , x 2 + 1
Here x 2 + 1 is hypotenuse. Since x is an even number its square is also even,
therefore an even number plus one is an odd number.
x 2 + 1 is an odd number B C
x2 + 1
2
Concept For any natural number x for
x = 0, 1, 2, ... , n
2n + 1, 2x (22x 2x 1), 2x (22n 2x
+ 1)
2n 2x
AC = [2 (2
2 x
+ 1)]
2
4x 2x + 1
= 22x (24n + 22n + 1) C
4 x + 2x
= (24n + 22n 2x + 1 + 2x + 22x )
2x +1
AC 2 = (24n + 22n + 2 2x )
2x +1 2
2x(22n2x + 1) 2x + 1
AB 2 + BC 2 = [22x (22n 1)2 ] + (2n )
4x 2x + 2
= 22x (24n 2 22n + 1) + 22n
2x
= 2 4n + 22x + 22n 22 22n 2
2x A 2x (22n2x 1) B
= 2 4n + 22x + 4 22n 22
. 2n
2x
= 2 4n + 22x + 22
. 2n
2x +1
= 2 4n + 22x + 22n
Example 1 Prove that there are no natural numbers, which are solutions of15x 2 7y 2 = 9.
Solution 15x 2 9 = 7y 2
3( 5x 2 3) = 7y 2
7y 2 is a multiple of 3.
y is a multiple of 3.
Let y = 3z
3( 5x 2 3) = 7 9z 2
5x 2 3 = 21z 2
5x 2 = 21z 2 + 3
5x 2 is a multiple of 3.
x is a multiple of 3
Let x = 3u
15u 2 = 1 + 7z 2
15u 6z 2 = 1 + z 2
2
1 + z 2 is a multiple of 3.
But for any z between 0 to 9, 1 + z 2 is not a multiple of 3.
For any z, the given equation has no integral solution.
Aliter
Since RHS is odd, x and y must be opposite
i.e., one even and one odd.
As 3|15 and 3|9
3 must divide 7y 2.
## Again, since 3 divide 21 so 3 must divide 5x 2.
Let x = 3x 1 , we get
15x 12 7y12 = 1
15x 12 = 7y12 + 1
Theory of Numbers 43
## Last digit of perfect square y12 may be one of these values 0, 1, 4, 9, 6, 5.
Hence, last digit of 7y12 + 1 will be 1, 8, 9, 4, 3, 6 respectively.
But 15x 12 ends in 0 or 5.
15x 12 = 7y12 + 1 has no solutions.
## Example 2 Show that x 2 + 1 = 3y has no solutions in integers.
Solution Since LHS cannot be a multiple of 3 for any x between 0 to 9.
RHS is always a multiple of 3.
x 2 + 1 = 3y has no integral solutions.
## Example 3 Show that 21x 2 10y 2 = 9 has no solution.
Solution 21x 2 9 = 10y 2
3 (7x 2 3) = 10y 2
10y 2 is a multiple of 3.
y is a multiple of 3.
Let y = 3y1
3 (7x 3) = 10 9y12
2
7x 2 3 = 30y12
7x 2 = 3 + 30y12 7x 2 = 3 (1 + 10y12 )
7x 2 is a multiple of 3
So, x is multiple of 3.
Let x = 3x1
7 9x12 = 3 (1 + 10y12 )
21 x12 = 1 + 10y12
21 x12 9y12 = 1 + y12
3 (7x12 3y12 ) = 1 + y12
1 + y12 is a multiple of 3.
But 1 + y12 is not a multiple of 3.
The given equation has no integral solution.
Note Every integer m can be written in the form x 2 + y 2 5z 2.
If m = 2n, then
= 2n = (n 2)2 + ( 2n 1)2 5(n 1)2
If m = 2n + 1 = (n + 1)2 + ( 2n )2 5n 2
Verification,
7 = 2( 3) + 1 = ( 3 + 1)2 + ( 2 3)2 5( 3)2
= 16 + 36 45
= 52 45 = 7
Similarly, every integer can be written in the form of
x 2 + y 2 + z 2 5u 2
1 2n
Example 4 Prove Cn is an integer.
n +1
## Solution If a and b are integers and a b = c, then c is also an integer.
Let a= 2n
Cn ; b = 2n
Cn 1
2n ! 2n ! 2n ! n
2n
Cn 2n
Cn 1 = = 1
n ! n ! (n + 1)!(n 1)! n ! n ! n + 1
1
2n !
= n + 1 it is an integer.
n !n !
(kn )! kn !
kn
Cn knCn 1 =
(kn n )! n ! (kn n + 1)!(n 1)!
(kn )! (kn n )! n !
= 1
(kn n )! n ! (kn n + 1)!(n 1)!
(kn )! n
= 1 kn n + 1
(kn n )! n !
kn n + 1 n kn kn 2n + 1
= knCn = Cn kn n + 1 .
kn n + 1
It is an integer.
## Example 5 If xy = 22 34 57( x + y ) , find the number of integral solution.
Solution Let N = 22 34 57
xy = N( x + y )
xy = Nx + Ny
xy Nx Ny = 0
( x N )( y N ) = N 2
( x N )( y N ) = 24.38.514
Number of integral solution
= ( 4 + 1)( 8 + 1)(14 + 1) = 5 9 15 = 675
## Example 6 Find all positive integers x, y satisfying
1 1 1
+ =
x y 20
Solution Suppose x, y are two positive integers such that
1 1 1
+ = (i)
x y 20
1 1 1 x 20
then = =
y 20 x 20 x
1 x + 20 4 5x
=
y 20x
Implying that 5x is rational.
Theory of Numbers 45
## Now x N 5x N. Hence 5x is the square of an integer which is divisible by 5.
5x = ( 5a )2 for some a N i.e., x = 5a 2 similarly y = 5b 2 for some b N.
Now, Eq. (i) becomes
1 1 1
+ = 2(a + b ) = ab
a b 2
(a 2)(b 2) = 4
(a, b ) {( 3, 6), ( 4, 4), ( 6, 3)}.
Solution set is {(45, 180), (80, 80), (180, 45)}.
Example 7 Find the number of solutions in positive integers of the equation 3x + 5y = 1008.
Solution Let x , y N such that 3x + 5y = 1008 then 3 | 5y 3|y y = 3k for some k N
Now, 3x + 15k = 1008
x + 5k = 336
5k 335
k 67
Thus, any solution pair is given by ( x , y ) = ( 336 5k , 3k ) where 1 k 67.
Number of solutions is 67.
## Example 8 Prove that there do not exist positive integrs x , y , z satisfying
2xz = y 2 and x + z = 997.
Solution 2|y 2 4|2xz 2|x or z 2|x and z [Q 2 | ( x + z )]
Let x = 2x1, y = 2y1 and z = 2z1
Then 2x1z1 = y12 and x1 + z1 = 997
Again y1 and one of x1, z1, say x1, are even writing x1 = 2x 2 and y1 = 2y 2
We have x 2z1 = y 22 and 2x 2 + z1 = 997
Since, 997 is a prime, x 2 and z1 are relatively prime.
Each is a square since their product is square.
Since any square is of the form 8n, 8n + 1 or 8n + 4, 2x 2 0 or 2(mod 8) and z1 1
(mod 8) (Q z1 is odd).
Hence 2x 2 + z1 1 or 3 (mod 8).
A contradiction as 997 5 (mod 8).
## Example 9 Show that the equation 3x 10 y 10 = 1991 has no integral solution.
Solution Suppose the existence of x , y Z such that 3x 10 y 10 = 1991. Note that 11|1991.
Neither x nor y is divisible by 11 for otherwise 11 would divide both .
1110|3x 10 y 10 = 1110|1991
an impossibility.
Hence x and y are prime to 11.
x 10 y 10 1 (mod 11)
1991 = 3x 10 y 10 3 1 = 2 a contradiction.
## Example 10 Find all integral solutions of
x 4 + y 4 + z 4 t 4 = 1991
Solution Let n be any integer ; when it is odd n 4 1 = (n 1)(n + 1)(n 2 + 1) is divisible by 16 as
n 1, n + 1, n 2 + 1 are all even and n 1,n + 1being consecutive even integers one of
them is divisible by 4 . When n is even n 4 0 (mod 16).
Thus, n 4 0 or 1, Now for any x , y , z , t , Z
x4 + y4 + z4 t4
where {1, 0, 1, 2, 3}, since 1991 7
x 4 + y 4 + z 4 t 4 1991
Example 11 For n N, let s(n) denote the number of ordered pairs ( x , y ) of positive integers for
1 1 1
+ = . Determine the set of positive integers n for which s(n ) = 5 .
which
x y n
1 1 1
Solution + = x, y > n
x y n
x = n + a and n + b, a, b N.
1 1 1
Now, + =
n+a n+b n
(n + b + n + a )n = (n + a )(n + b )
n 2 = ab
s(n ) is the number of divisors of n 2
Let n = p1 1 K pmm be prime factorization of n where 1 K m .
Then s(n ) = (1 + 2 1) K (1 + 2 m )
s(n ) = 5
1 + 2 1 = 5
and m =1
n = p12
Required set is {p 2 : p is prime}.
1 1 1
Example 12 If + = ;a, b, c are positive integers with no common factors. Prove that (a + b ) is a
a b c
square.
a+b 1
Solution By the hypothesis =
ab c
i.e., (a + b )c = ab
Let p be any prime which divides (a + b ); then p divides one of a, b and therefore
both.
Since gcd (a, b, c ) = 1; p does not divide c.
for any k N, pk | a pk | b
Hence the maximum power of p which divides a + b is the square of the maximum
power of p which divides a.
a + b is a square.
Theory of Numbers 47
3n 5
Example 13 Find all integers n such that is also an integer.
n +1
3n 5 8
Solution Since, = 3 is an integer.
n +1 n +1
8
{ 1, 2, 4, 8}
n +1
8
and 3 is a square.
n +1
8
Consequently = 1 or 2 ;
n +1
i.e., n = 9 or 3
## Example 14 Find the number of pairs of integers (a, b ) such that
a 3 + a 2b + ab 2 + b 3 + 1 = 2002.
Solution a 3 + a 2b + ab 2 + b 3
= (a + b ) (a 2 + b 2 ) = 2001
= 3 23 29
a + b is therefore one of the three numbers 3, 23, or 29.
If a + b = 3, then a = 1, b = 2
(or a = 2, b = 1) so that a 2 + b 2 = 5.
But in this case a 2 + b 2 = 23 29
a + b is not 3.
If a + b = 23, then a 2 + b 2 = 87
so that both a and b will be less than 10 and a + b < 20, a contradiction.
If a + b = 29, then a 2 + b 2 = 69 so that both a and b will be less than 10 and
a + b < 20, a contradication.
Thus, the number of pairs (a, b ) satisfying the given condition is zero. |
# Fraction word problems
Here you will learn about fraction word problems, including solving math word problems within a real-world context involving adding fractions, subtracting fractions, multiplying fractions, and dividing fractions.
Students will first learn about fraction word problems as part of number and operations—fractions in 4 th grade.
## What are fraction word problems?
Fraction word problems are math word problems involving fractions that require students to use problem-solving skills within the context of a real-world situation.
To solve a fraction word problem, you must understand the context of the word problem, what the unknown information is, and what operation is needed to solve it. Fraction word problems may require addition, subtraction, multiplication, or division of fractions.
After determining what operation is needed to solve the problem, you can apply the rules of adding, subtracting, multiplying, or dividing fractions to find the solution.
For example,
Natalie is baking 2 different batches of cookies. One batch needs \cfrac{3}{4} cup of sugar and the other batch needs \cfrac{2}{4} cup of sugar. How much sugar is needed to bake both batches of cookies?
You can follow these steps to solve the problem:
Step-by-step guide: Adding and subtracting fractions
Step-by-step guide: Subtracting fractions
Step-by-step guide: Multiplying and dividing fractions
Step-by-step guide: Multiplying fractions
Step-by-step guide: Dividing fractions
## Common Core State Standards
How does this relate to 4 th grade math to 6 th grade math?
• Grade 4: Number and Operations—Fractions (4.NF.B.3d)
Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem.
• Grade 4: Number and Operations—Fractions (4.NF.B.4c)
Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem.
For example, if each person at a party will eat \cfrac{3}{8} of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie?
• Grade 5: Number and Operations—Fractions (5.NF.A.2)
Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem.
Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result \cfrac{2}{5}+\cfrac{1}{2}=\cfrac{3}{7} by observing that \cfrac{3}{7}<\cfrac{1}{2} .
• Grade 5: Number and Operations—Fractions (5.NF.B.6)
Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
• Grade 5: Number and Operations—Fractions (5.NF.B.7c)
Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem.
For example, how much chocolate will each person get if 3 people share \cfrac{1}{2} \: lb of chocolate equally? How many \cfrac{1}{3} cup servings are in 2 cups of raisins?
• Grade 6: The Number System (6.NS.A.1)
Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem.
For example, create a story context for \cfrac{2}{3} \div \cfrac{4}{5} and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that \cfrac{2}{3} \div \cfrac{4}{5}=\cfrac{8}{9} because \cfrac{3}{4} of \cfrac{8}{9} is \cfrac{2}{3}. (In general, \cfrac{a}{b} \div \cfrac{c}{d}=\cfrac{a d}{b c} \, )
How much chocolate will each person get if 3 people share \cfrac{1}{2} \: lb of chocolate equally? How many \cfrac{3}{4} cup servings are in \cfrac{2}{3} of a cup of yogurt? How wide is a rectangular strip of land with length \cfrac{3}{4} \: m and area \cfrac{1}{2} \: m^2?
## How to solve fraction word problems
In order to solve fraction word problems:
1. Determine what operation is needed to solve.
2. Write an equation.
3. Solve the equation.
## Fraction word problem examples
### Example 1: adding fractions (like denominators)
Julia ate \cfrac{3}{8} of a pizza and her brother ate \cfrac{2}{8} of the same pizza. How much of the pizza did they eat altogether?
1. Determine what operation is needed to solve.
The problem states how much pizza Julia ate and how much her brother ate. You need to find how much pizza Julia and her brother ate altogether, which means you need to add.
2Write an equation.
\cfrac{3}{8}+\cfrac{2}{8}= \, ?
3Solve the equation.
To add fractions with like denominators, add the numerators and keep the denominators the same.
\cfrac{3}{8}+\cfrac{2}{8}=\cfrac{5}{8}
The last step is to go back to the word problem and write a sentence to clearly say what the solution represents in the context of the problem.
Julia and her brother ate \cfrac{5}{8} of the pizza altogether.
### Example 2: adding fractions (unlike denominators)
Tim ran \cfrac{5}{6} of a mile in the morning and \cfrac{1}{3} of a mile in the afternoon. How far did Tim run in total?
Determine what operation is needed to solve.
Write an equation.
Solve the equation.
### Example 3: subtracting fractions (like denominators)
Pia walked \cfrac{4}{7} of a mile to the park and \cfrac{3}{7} of a mile back home. How much farther did she walk to the park than back home?
Determine what operation is needed to solve.
Write an equation.
Solve the equation.
### Example 4: subtracting fractions (unlike denominators)
Henry bought \cfrac{7}{8} pound of beef from the grocery store. He used \cfrac{1}{3} of a pound of beef to make a hamburger. How much of the beef does he have left?
Determine what operation is needed to solve.
Write an equation.
Solve the equation.
### Example 5: multiplying fractions
Andre has \cfrac{3}{4} of a candy bar left. He gives \cfrac{1}{2} of the remaining bit of the candy bar to his sister. What fraction of the whole candy bar does Andre have left now?
Determine what operation is needed to solve.
Write an equation.
Solve the equation.
### Example 6: dividing fractions
Nia has \cfrac{7}{8} cup of trail mix. How many \cfrac{1}{4} cup servings can she make?
Determine what operation is needed to solve.
Write an equation.
Solve the equation.
### Teaching tips for fraction word problems
• Encourage students to look for key words to help determine the operation needed to solve the problem. For example, subtracting fractions word problems might ask students to find “how much is left” or “how much more” one fraction is than another.
• Provide students with an answer key to word problem worksheets to allow them to obtain immediate feedback on their solutions. Encourage students to attempt the problems independently first, then check their answers against the key to identify any mistakes and learn from them. This helps reinforce problem-solving skills and confidence.
• Be sure to incorporate real-world situations into your math lessons. Doing so allows students to better understand the relevance of fractions in everyday life.
• As students progress and build a strong foundational understanding of one-step fraction word problems, provide them with multi-step word problems that involve more than one operation to solve.
• Take note that students will not divide a fraction by a fraction as shown above until 6 th grade (middle school), but they will divide a unit fraction by a whole number and a whole number by a fraction in 5 th grade (elementary school), where the same mathematical rules apply to solving.
• There are many alternatives you can use in place of printable math worksheets to make practicing fraction word problems more engaging. Some examples are online math games and digital workbooks.
### Easy mistakes to make
• Misinterpreting the problem
Misreading or misunderstanding the word problem can lead to solving for the wrong quantity or using the wrong operation.
• Not finding common denominators
When adding or subtracting fractions with unlike denominators, students may forget to find a common denominator, leading to an incorrect answer.
• Forgetting to simplify
Unless a problem specifically says not to simplify, fractional answers should always be written in simplest form.
### Practice fraction word problem questions
1. Malia spent \cfrac{5}{6} of an hour studying for a math test. Then she spent \cfrac{1}{3} of an hour reading. How much longer did she spend studying for her math test than reading?
Malia spent \cfrac{1}{2} of an hour longer studying for her math test than reading.
Malia spent \cfrac{5}{18} of an hour longer studying for her math test than reading.
Malia spent \cfrac{1}{2} of an hour longer reading than studying for her math test.
Malia spent 1 \cfrac{1}{6} of an hour longer studying for her math test than reading.
To find the difference between the amount of time Malia spent studying for her math test than reading, you need to subtract. Since the fractions have unlike denominators, you need to find a common denominator first.
You can use 6 as the common denominator, so \cfrac{1}{3} becomes \cfrac{3}{6}. Then you can subtract.
\cfrac{5}{6}-\cfrac{2}{6}=\cfrac{3}{6}.
\cfrac{3}{6} can then be simplified to \cfrac{1}{2}.
Finally, you need to choose the answer that correctly answers the question within the context of the situation. Therefore, the correct answer is “Malia spent \cfrac{1}{2} of an hour longer studying for her math test than reading.”
2. A square garden is \cfrac{3}{4} of a meter wide and \cfrac{8}{9} of a meter long. What is its area?
The area of the garden is 1\cfrac{23}{36} square meters.
The area of the garden is \cfrac{27}{32} square meters.
The area of the garden is \cfrac{2}{3} square meters.
The perimeter of the garden is \cfrac{2}{3} meters.
To find the area of a square, you multiply the length and width. So to solve, you multiply the fractional lengths by mulitplying the numerators and multiplying the denominators.
\cfrac{3}{4} \times \cfrac{8}{9}=\cfrac{24}{36}
\cfrac{24}{36} can be simplified to \cfrac{2}{3}.
Therefore, the correct answer is “The area of the garden is \cfrac{2}{3} square meters.”
3. Zoe ate \cfrac{3}{8} of a small cake. Liam ate \cfrac{1}{8} of the same cake. How much more of the cake did Zoe eat than Liam?
Zoe ate \cfrac{3}{64} more of the cake than Liam.
Zoe ate \cfrac{1}{4} more of the cake than Liam.
Zoe ate \cfrac{1}{8} more of the cake than Liam.
Liam ate \cfrac{1}{4} more of the cake than Zoe.
To find how much more cake Zoe ate than Liam, you subtract. Since the fractions have the same denominator, you subtract the numerators and keep the denominator the same.
\cfrac{3}{8}-\cfrac{1}{8}=\cfrac{2}{8}
\cfrac{2}{8} can be simplified to \cfrac{1}{4}.
Therefore, the correct answer is “Zoe ate \cfrac{1}{4} more of the cake than Liam.”
4. Lila poured \cfrac{11}{12} cup of pineapple and \cfrac{2}{3} cup of mango juice in a bottle. How many cups of juice did she pour into the bottle altogether?
Lila poured 1 \cfrac{7}{12} cups of juice in the bottle altogether.
Lila poured \cfrac{1}{4} cups of juice in the bottle altogether.
Lila poured \cfrac{11}{18} cups of juice in the bottle altogether.
Lila poured 1 \cfrac{3}{8} cups of juice in the bottle altogether.
To find the total amount of juice that Lila poured into the bottle, you need to add. Since the fractions have unlike denominators, you need to find a common denominator first.
You can use 12 as the common denominator, so \cfrac{2}{3} becomes \cfrac{8}{12}. Then you can add.
\cfrac{11}{12}+\cfrac{8}{12}=\cfrac{19}{12}
\cfrac{19}{12} can be simplified to 1 \cfrac{7}{12}.
Therefore, the correct answer is “Lila poured 1 \cfrac{7}{12} cups of juice in the bottle altogether.”
5. Killian used \cfrac{9}{10} of a gallon of paint to paint his living room and \cfrac{7}{10} of a gallon to paint his bedroom. How much paint did Killian use in all?
Killian used \cfrac{2}{10} gallons of paint in all.
Killian used \cfrac{1}{5} gallons of paint in all.
Killian used \cfrac{63}{100} gallons of paint in all.
Killian used 1 \cfrac{3}{5} gallons of paint in all.
To find the total amount of paint Killian used, you add the amount he used for the living room and the amount he used for the kitchen. Since the fractions have the same denominator, you add the numerators and keep the denominators the same.
\cfrac{9}{10}+\cfrac{7}{10}=\cfrac{16}{10}
\cfrac{16}{10} can be simplified to 1 \cfrac{6}{10} and then further simplified to 1 \cfrac{3}{5}.
Therefore, the correct answer is “Killian used 1 \cfrac{3}{5} gallons of paint in all.”
6. Evan pours \cfrac{4}{5} of a liter of orange juice evenly among some cups.
He put \cfrac{1}{10} of a liter into each cup. How many cups did Evan fill?
Evan filled \cfrac{2}{25} cups.
Evan filled 8 cups.
Evan filled \cfrac{9}{10} cups.
Evan filled 7 cups.
To find the number of cups Evan filled, you need to divide the total amount of orange juice by the amount being poured into each cup. To divide fractions, you mulitply the first fraction (the dividend) by the reciprocal of the second fraction (the divisor).
\begin{aligned}& \cfrac{4}{5} \div \cfrac{1}{10}= \, ? \\\\ & \downarrow \downarrow \downarrow \\\\ & \cfrac{4}{5} \times \cfrac{10}{1}=\cfrac{40}{5} \end{aligned}
\cfrac{40}{5} can be simplifed to 8.
Therefore, the correct answer is “Evan filled 8 cups.”
## Fraction word problems FAQs
What are fraction word problems?
Fraction word problems are math word problems involving fractions that require students to use problem-solving skills within the context of a real-world situation. Fraction word problems may involve addition, subtraction, multiplication, or division of fractions.
How do you solve fraction word problems?
To solve fraction word problems, first you need to determine the operation. Then you can write an equation and solve the equation based on the arithmetic rules for that operation.
How are fraction word problems similar to decimal word problems?
Fraction word problems and decimal word problems are similar because they both involve solving math problems within real-world contexts. Both types of problems require understanding the problem, determining the operation needed to solve it (addition, subtraction, multiplication, division), and solving it based on the arithmetic rules for that operation.
## Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. |
## Polite Subtraction
Here is a nifty trick for speeding up subtraction of large numbers, generally three digit numbers or a combination of three-digit and two-digit numbers.
For example, to subtract 64 from 260, many of us would scribble on paper or grab a calculator. If the number was 60, many of us would do the equation in our heads, but that 4 takes a little extra effort.
Try this helpful strategy. Round 64 up to100, then subtract 100 from 260.
260 – 100 = 160.
Now add the difference between 100 and 64. Let’s see, 100-64 = 36, so 36 is the difference. Then add 160 + 36, which equals 196.
“But that was more difficult than just writing it our or using a computer!” you might say.
Hold on. This is the fun part. What seems difficult becomes quick and easy if we use complements. (Yes, be nice to your math and it will be nice to you.)
Look below at the numbers subtracted from 100 and the answers. See a pattern?
100 100 100 100
-81 -68 -33 -24
19 32 67 76
Each of the numbers in the in the ones columns add up to 10. Each of the numbers in the tens columns add up to 9, always.
Oh, that is not exactly true, but the only exceptions are equations that end in zeros, such as 90 – 10 =80, which most of us can do anyway.
Now that you understand complements, try them out with these problems. Check them against a calculator or on paper. Does the method work?
245 416 759
-78 -29 -82
(Let’s do the first one together. Round 78 up to 100. 245 – 100 = 145. The complement of 78 is 22. 145 + 22 = 167)
You can make this even simpler by adding left to right = 145 + 20 = 165, 165 + 2 = 167.
Try these out, then write your own and try to do them in your head. Before long you will be as fast as your calculator!
## Quicker Addition: Left to Right Math
Dr. Benjamin Arthur ,of the book The Secrets of Mental Math, encourages kids and adults alike to try adding left to right instead of right to left. The reason is simple. We are taught to read left to right, we write left to right, and are therefore more adept at thinking left to right. Only in math do we reverse the routine.
Here is how left to right addition works. Take an addition problem, such as 34 + 45. Add the tens columns first (34 + 40 = 74). Now add the ones column (74 + 5 = 79).
Try these examples in your head. I will pitch in some three digit numbers as well.
______23 62 134
_____+42 +26 + 63
This is all very good, but what if I throw in much bigger numbers, such as…
___256
__+289
Not only are the numbers bigger, but they include carrying-over. Here’s a trick.
Round-up the bottom number to 300. Now add 300 to 256. Then subtract the difference between 300 and 289, which is 11.
Hence, 256 + 289 = 256 + 300 – 11.
Then 256 + 300 = 556
556 -11 = 556 – 10 -1
556-10 = 546 and finally 546 -1 =545 Your answer is 545.
Try these: 645 + 276 and 327 + 499
What we learned:
* Adding left to right can be faster.
* With big numbers it is often quicker to round up the bottom number and subtract the difference from the sum.
On Friday I will share how you can build on left to right math with subtraction! |
# HSC Extension 1 and 2 Mathematics/2-Unit/Preliminary/Basic arithmetic and algebra
The start of the 2-unit mathematics course is really just revision of year 10 work.
## Converting between fractions, decimals and percentages
### Converting recurring decimals to fractions
So, you're in an exam and the test paper gives you a recurring decimal, say 2.35555..., and asks you to convert it into a fraction. Well, that's easy, it's just 2.3 plus 5 9ths divided by 10. simple enough. But then you get another question asking you to convert 5.676767... into a fraction. What fraction makes the repeating 0.676767...? Ahh, now you're stuck.
That's what would have happened if you hadn't learnt how to convert recurring decimals into fractions algebraically before sitting for your exam. What use does converting recurring decimals into fractions have? Can't calculators convert between decimal and fraction easily? Yes they can. This topic is probably most useful in getting you to think like a mathematician, applying maths (in this case algebra) to problems. There isn't much more use for this, other than providing an endless source of questions to test your basic algebra.
So let's start off with our recurring decimal number 5.676767... We are going to assign that value to x, so 'x = 5.676767.... Now we have something to work with. To convert it into a fraction, we need to get rid of all those annoying decimal places. To get rid of them we are going to need to multiply x by 100 to give us 100x = 567.676767..., and then we can subtract x, or 5.676767..., from that.
$x = 5. \dot 6 \dot 7$
$100x = 567. \dot 6 \dot 7$
$100x - x = 567. \dot 6 \dot 7 - 5. \dot 6 \dot 7$
$99x = 562 \;$
$x = \frac{562}{99}$
$x = 5 \frac{67}{99}$
And there you go. This is the method of removing the first period
When looking at problems with recurring decimals, you need to see how many numbers are in the repeating part. In this example, the start of the repeating part was 0.67, and there were 2 numbers in each repeat. Because there were two numbers in the repeating part being repeated, we had to move the number 2 places to the left, by multiplying the number by 100, or 102. If you have a number with a non-repeating part a and the start of the repeating part r, and a repeating length, or period, P, then we can write a general list of equations for converting the recurring decimal number into a fraction.
$x = a + \dot r$
where the dotted r is the repeating part of the equation that we are going to get rid of.
$10^Px = 10^Pa + 10^Pr + \dot r$
where 10Pr is the start of the repeating part, shifted P places to the left.
$10^Px - x = 10^Pa + 10^Pr + \dot r - a - \dot r$
Now we subtract one equation from the other to get rid of the repeating part by making it subtract from itself, leaving just the first period of the repeating part which we shifted left P places.
$(10^P-1)x = (10^P-1)a + 10^Pr \;$
Factorize both sides
$x = \frac{(10^P-1)a + 10^Pr}{(10^P-1)x}$
make x the subject
If the non-repeating number a had a decimal part, like in the number 2.35555..., where a was 2.3, then the numerator would be a decimal (but not a recurring decimal) and you would have to multiply the numerator and denominator by a power of ten to make both numbers integers, and then you would have to simplify it. Calculators make this easy with their fraction buttons which can convert divisions into their simplest fractions. You could make an even more general formula if you let D, for decimal, equal the amount of decimal places that the non-repeating part has. Then you can just multiply the numerator and denominator in your final answer by 10D.
$x = \frac{[(10^P-1)a + 10^Pr] \times 10^D}{(10^P-1)x \times 10^D}$
It is really more important that you remember the basic method and idea behind it, cancelling out the repeating part by subtracting it from itself. This formula is just for fun.
An interesting thing here is that the numerator will be equal to the non-repeating number with its decimal place removed with the repeating part of the number on the end, minus the non-repeating number with its decimal place removed without the repeating part on the end. So for 2.3555..., a = 2.3, r = 0.05, P = 1, D = 1. This makes the numerator (10×2.3 - 2.3 + 0.5)×10, so 235 - 23. 235 is a with no decimals and with r on the end, and then you are just subtracting a without decimals from that. Another thing to notice is that for every recurring number in the recurring part of the decimal, there will be a 9 in the denominator. 10P - 1 is a 1 followed by zeros when P is greater than 1. So when you take one away, all you are left with is nines. 100 - 1 = 99, 1000 - 1 = 999, etc. Also, for every decimal place in the non-repeating part, there will be a zero on the end of the denominator, caused by the multiplication of 10D.
This is where the general algorithm of subtracting the non-repeating plus start of repeating as an integer from the non-repeating as integer to get the numerator, and then adding nines for every repeating decimal number and zeros for every non-repeating decimal number comes from.
## Scientific notation and approximation
significant figures
Rules:
1. All non-zero digits are significant. 2. In a number without a decimal point, only zeros BETWEEN non-zero digits are significant. 3. In a number with a decimal point, all zeros to the right of the first non-zero digit are significant.
For example: 1. 1500 has two significant figures because the two zeros after 15 do not count. If the question asks to round to 1 significant figure then the answer is 2000.
2. 1050 has three significant figures since the 0 between 1 and 5 counts. If the question asks to round to 2 significant figures then the answer is 1100.
3. 0.001500 has four significant figures because the 3 zeros on the left hand side of the 1 do not count. However, the 2 zeros on the right hand side of 5 DO count, thus there are four significant figures. If the question asks to round to 1 significant figure then the answer is 0.002.
## Evaluation of expressions involving all this stuff
If The N+1 Number Of S.f More Than 5 We Add 1+ To N Number , If It Is Less Than 5 nothing is change
## Surds
A surd is any irrational expression, i.e. it cannot be expressed rationally (as a fraction of two integers). Common examples include √2 and $\pi$. These numbers have no known patterns in their digits, and so cannot be expressed as the fraction of two integers, they can only be expressed as a decimal with the decimal places continuing on forever. Because of this, surds can only ever be evaluated as approximations to a number of decimal places. The word 'surd' actually comes from the Arabic surdus which means root. The Arab mathematicians saw numbers like plants, growing out of their roots.
### Like and Unlike surds
Basically, two surds are like if they have the same number under the radical, which is called the radicand. For example, $3\sqrt{3}$ and $8\sqrt{3}$ are like surds. Unlike surds do not have the same radicand, and hence, for example, $3\sqrt{37}$ and $3\sqrt{43}$ are unlike.
### Simplifying surds
Some surds can be simplified as a product of a rational number and a surd. For example,
$\sqrt{45}$ can be simplified to $3\sqrt{5}$
This happens when the number within the surd is expressed as a product of two numbers, one of them being a perfect square:
$\sqrt{45}$ $= \sqrt{9\times5}$
Using the law of surds, this surd can be expressed as a product of two different, separate surds:
$\sqrt{9}\times\sqrt{5}$
We can then evaluate $\sqrt{9}$ as $3$:
$3\times\sqrt{5}$ $= \sqrt{9\times5}$
This method of simplification can also be applied to roots of any power.
### Addition and subtraction of surds
$a\sqrt b \pm c\sqrt b = \left( {a \pm c} \right)\sqrt b$
### Multiplication of surds
$\sqrt a \times \sqrt b = \sqrt {ab}$
### Division of surds
One surd (root) divided by another surd (root) is equal to the root of the fraction of the two radicands.
### Rationalizing the denominator
The correct form of writing a fraction requires that the denominator be rational, and hence, not a surd. To rationalise the denominator, we multiply both numerator and denominator by the denominator, resulting in a rational denominator.
## Inequalities and absolute values
### Inequations
Inequations are mathematical statements that don't use the equals sign to express relationships between the two sides of the statement, but instead use signs such as greater than, greater than or equal to, less than, and less than or equal to.
greater than $> \$ greater than or equal to $\ge$ less than $< \$ less than or equal to $\le$
While equations give exact values for pronumerals, such as x = 2, where x is exactly 2, inequations give a range of values which the pronumeral can have, such as x > 2, where x can posses any value greater than 2.
Just like when dealing with equations, adding or subtracting an amount to both sides doesn't change the inequation. If y + 2 > 1 then by subtracting 2 from both sides y > -1.
When multiplying or dividing both sides by positive number it also works the same as an equation, for example if $8x \ge 4$ then by dividing both sides by 8 you get $x \ge \begin{matrix} \frac{1}{2} \end{matrix}$.
When multiplying or dividing both sides of an inequation by a negative number the sign reverses. for example if you have $-2x \le 3$ and you divide both sides by -2, the less than or equal sign changes to a greater than or equal sign as $x \ge -1 \begin{matrix} \frac{1}{2} \end{matrix}$. This is because if one number $a\$ is greater than another number $b\$, i.e. $a > b \$, then on a number line $a\$ is further away from 0 in the positive direction (right) than $b$. But if you multiply the two numbers by -1, they both move to the left of the number line. Both numbers are still the same distance from 0, but because they are on the negative side of 0 $a\$ is now more negative than $b \$, and so $a. This applies to any two numbers, positive or negative.
### Absolute Values
What are absolute values? Lets say you have a number x, and you put it on a number line. The value of x is the value that it has on the number line, lets say -3. The absolute value of x, expressed in mathematics as |x|, is the distance of x from 0. Because distance cannot be negative, |x| is always greater than or equal to 0. This means that if x = -3 then |x| would be 3, because the distance of -3 from 0 is 3. So if x is greater than or equal to 0, |x| will just be its normal value x. But if x is less than 0, then |x| will be its negative value (the negative of a negative number), and so will be positive. This can be expressed as
$| x \vert = \begin{cases} x, & x \ge 0 \\ -x, & x<0 \end{cases}$
where |x| is the absolute value of x.
$|ab| = |a| . |b|$
$|a + b| \le |a| + |b|$
## Algebraic manipulation
### Simplification
Removing grouping symbols and collecting like terms...
### Substitution
Evaluating expressions using giving values and formulas. Usually requires you to make something the subject first.
### Factorization
distributive law?
#### Difference of two squares
x2 - y2. This can be factorized. Can you guess how? It's probably a good idea to add some maths in here instead of just giving you the formula, so lets look at the geometrical interpretation of this. Both pronumerals are squared. The terms squared and cubed come from the formula for area of a square and volume of a cube. The area of a square is equal to the length of one side multiplied by the length of another, and because the sides of a square are all equal, the area is the length of one side multiplied by itself, so it is raised to the power of 2, or squared =).
So if you have one squared number subtracted from another squared number, you could interpret this as one area of a square being subtracted from another area of another square. One square has side length x, while the other square has side length y. Does it matter which square is larger? Not really, if the area being subtracted is larger, you will end up with a negative area, and even though area, like distance, cannot be negative, we are not talking about real area, and x2 - y2 is allowed to be negative because it is an algebraic expression. If you don't like that, then think of it this way x2 - y2 = -(y2 - x2), so if y is larger than x, then just subtract the area of the x square from the area of the y square to get a positive area, then make that negative.
So, think about this in terms of a larger square with sides x having a smaller square with sides y subtracted from it. Subtracting a square from a square is going to leave a square shaped hole in the larger square, lets say in the corner. Now what used to be the larger square is going to have 6 edges. The two edges that were not cut will still have a length x. The other two sides of the original square are going to have a length of x-y, and the two inner edges created by the hole are going to have a length y. From this we can calculate the remaining area. There are two rectangles, each with area (x-y)×y, and one square with area (x-y)(x-y). If we expand and simplify these we get 2xy - 2y2 + x2 - 2xy + y2, which can be written as y2 + 2xy + x2 - 2xy, which factorizes to y(y+2x) + x(x-2y), then y(x+y) - xy - x(x-y) + xy, which can be further factorizes to (x-y)(x+y) simplifies to x2-y2. So
$x^2 - y^2 = (x+y)(x-y) \;$
You could try the same thing with x2+y2, but it doesn't get you anything really useful, just a(a-b) + b(a+b).
Another way to looks at this kind of factorisation problem, especially if there aren't any geometrical methods to factorize it, is to look for an expression that looks similar, but gives slightly different results, and then figure out what needs to be done in order to get from that expression to the first.e.g.
$(x+y)^2 = x^2 + 2xy + y^2$
$x^2 - y^2 = x^2 +2xy + y^2 + a$
$a = x^2 - y^2 - (x^2 +2xy + y^2)$
$a = - 2xy - 2y^2$
$x^2 - y^2 = x^2 +2xy + y^2 - 2xy - 2y^2$
$x^2 - y^2 = x^2 +xy + y^2 - xy - 2y^2 + xy - xy$
$x^2 - y^2 = x(x + y) - y(y + x)$
$x^2 - y^2 = (x - y)(x + y)$
A third way to look at it is to add something to both sides.
$xy - xy = 0$
$x^2 - y^2 = x^2 - y^2$
$x^2 - y^2 + 0 = x^2 + xy - xy - y^2$
$x^2 - y^2 = x(x + y) - y(x + y)$
$x^2 - y^2 = (x - y)(x + y)$
probably went into too much detail on this. Awell.
#### The sum and difference of two cubes
a3–b3 = (a-b) (a2+ab+b2)
a3+b3 = (a+b) (a2-ab+b2)
### Algebraic fractions
#### Reduction
factorizing numerator and denominator and canceling any common factors
LCM
## Linear equations
$ax + b = 0,\,a \ne 0$
## Linear inequalities
if both sides of an inequality are multiplied by a negative number, the direction of the inequality is reversed.
$ax^2 + bx + c = 0, a \ne 0$
solution by factorisation, completing the square, and formula.
equation 1
4t+1=19 4t +1=19 take 1 to the other side 4t=19-1 4t=18 t=18/4 simplyfy it divide the both numbers by 2 the answer is 9/2 this is a linear equation!
linear |
## The Mode, Mean and Median, are types of averages.
The mean is the average calculated by adding the numbers and dividing by the number of items in the data set. The median is the middle value in a data set. To calculate the median, put the numbers in order, and the median will be the middle number. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers.
See our example below. The mode is the most frequently occurring number. If no number is repeated, then there is no mode.
Examples
Find the median, mode and mean of the following list:
6, 7, 8, 12, 14, 6, 7, 10
Find the mean
First add the numbers
6 + 7 + 8 + 12 + 14 + 6 + 7 + 10 = 70
There are 8 numbers in the list, so divide by 9
70/8 = 8.75 = mean
Find the median
First put the numbers in order
6, 6, 7, 7, 8, 10, 12, 14,
The data set has an even number of numbers, so the median is the average of 7 and 8. (7 + 8)/2 = 7.5
Find the mode
The mode is the most frequently occurring number. Here
6 and 7 both occur twice, so they are both considered the
mode.
1. Find the mean of these set of numbers – 200,000, 10,020, 30,000, 15,000 1080
a. 1080
b. 15,000
c. 256,100
d. 51,220
2. Find the mean of these set of numbers – 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
a. 55
b. 5.5
c. 11
d. 10
3. Find the mean of these set of numbers – 2.5, 10.2, 4.5, 1.25, 7.05, 20.8
a. 7.6
b. 45.6
c. 7
d. 1.25
## Median Questions
4. Find the median of the set of numbers – 3, 1, 9, 7, 13, 11, 15, 21, 15, 9, 7
a. 11
b. 1
c. 9
d. 90
5. Find the median of these set of test scores taken from a class of students – 90, 80, 77, 86, 50, 91, 73, 66, 69, 45, 43, 65, 75
a. 13
b. 73
c. 9
d. 706
6. Find the median of 32, 64, 109, 67 and 3
a. 64
b. 3
c. 275
d. 5
## Mode Questions
7. Find the mode from these numbers – 7,2,3,9,6,5,1,4,8
a. 1
b. 5
c. 9
d. None of the above
8. Find the mode from these numbers – 190, 280, 177, 186, 180, 291, 177, 166, 169, 165, 243, 165, 177, 243, 190
a. 190
b. 177
c. 165
d. 243
9. Find the mode from these test results – 2, 4, 2, 6, 4, 9, 6, 7, 2, 9, 7, 6, 4, 10, 10, 2, 6, 7, 9
a. 2 and 9
b. 2
c. 2 and 6
d. 2 and 7
1. D
First add all the numbers 200,000 + 10,020 + 30,000 + 15,000 + 1080 = 256,100. Then divide by 5 (the number of data provided) = 256,100/5 = 51,220
2. C
First add all the numbers 1 + 2 + 3 + 4 + 5 +6 + 7 +8 + 9 + 10 = 55. Then divide by 10 (the number of data provided) = 55/5 = 11
3. A
First add all the numbers 2.5 + 9.5 + 4.5 + 1.25 + 7.05 + 20.8 = 45.6. Then divide by 6 (the number of data provided) = 45.6/6 = 7.6
4. C
First arrange the numbers in a numerical sequence – 1, 3, 7, 7, 9, 9, 11, 13, 15, 15, 21. Next find the middle number. The median = 9
5. B
First arrange the numbers in a numerical sequence – 43, 45, 50, 65, 66, 69, 73, 75, 77, 80, 86, 90, 91. Next find the middle number. The median = 73
6. A
First arrange the numbers in a numerical sequence – 3, 32, 64, 67, 109. Next find the middle number. The median = 64
7. D
Simply find the most recurring number. All the numbers in the series appeared only once. The answer is No Mode
8. D
Simply find the most recurring number. The most occurring number in the series is 177
9. C
Simply find the most recurring number. The most occurring numbers in the series is 2 and 6
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## Basic Geometry Practice
Note: Figure not drawn to scale
1. Calculate the length of side x.
a. 6.46
b. 8.48
c. 3.6
d. 6.4
Note: figure not drawn to scale
2. What is the length of each side of the indicated square
above? Assume the 3 shapes around the center triangle
are square.
a. 10
b. 15
c. 20
d. 5
3. Reflect the parallelogram ABCD with the given mirror
line m.
4. In a class, there are 8 students who take only dancing lessons. 5 students take both dancing and singing lessons. The number of students taking singing lessons is 1 less than 2 times the number of students taking dancing lessons only. If 4 students take neither of these courses,
how many students are in the class?
a. 19
b. 25
c. 27
d. 42
5. The interior angles of a triangle are given as 2x + 5, 6x
and 3x – 23. Find the supplementary of the largest angle.
a. 640
b. 720
c. 1000
d. 1080
6. In the right triangle above, |AB| = 2|BC| and |AC| =
15 cm. Find the length of |AB|.
a. 5 cm
b. 10 cm
c. 5√5 cm
d. 6√5 cm
Note: figure not drawn to scale
7. What is the perimeter of the equilateral ΔABC above?
a. 18 cm
b. 12 cm
c. 27 cm
d. 15 cm
Note: figure not drawn to scale
8. What is the perimeter of the above shape, assuming
the bottom portion is square?
a. 22.85 cm
b. 20 π cm
c. 15 π cm
d. 25 π cm
9. What year was Euclidian geometry disproven, and by
whom?
a. Thales – BC 500s
b. Pythagor BC 500s
c. Pierre De Fermat 1600s
d. Nikolai Lobachevsky 1830s
10. Which postulate below disproves Euclidean
geometry?
a. Through any two points, there is exactly one line.
b. If equals are added to equals, the wholes are equal.
c. Parallel postulate
d. Things which coincide with one another are equal to
one another (Reflexive property).
1. B
In the question, we have a right triangle formed inside the circle. We are asked to find the length of the hypotenuse of this triangle. We can find the other two sides of the triangle by using circle properties:
The diameter of the circle is equal to 12 cm. The legs of the
right triangle are the radii of the circle; so they are 6 cm
long.
Using the Pythagorean Theorem:
(Hypotenuse)2 = (Adjacent Side)2 + (Opposite Side)2
x2 = r2 + r2
x2 = 62 + 62
x2 = 72
x = √72
x = 8.48
2. B
We see that there are three squares forming a right triangle in the middle. Two of the squares have the areas 81 m2 and 144 m2. If we denote their sides a and b respectively:
a2 = 81 and b2 = 144. The length, which is asked, is the hypotenuse; a and b are the opposite and adjacent sides of the right angle. By using the Pythagorean Theorem, we can find the value of the asked side:
Pythagorean Theorem
:
(Hypotenuse)2 = (Opposite Side)2 + (Adjacent Side)2
h2 = a2 + b2
a2 = 81 and b2 = 144 are given. So,
h2 = 81 + 144
h2 = 225
h = 15 m
3.
We reflect points A, B, C and D against the mirror line m at
right angle and we connect the new points A’, B’,C’ and D’.
4. C
The class can be shown by set E. Let us say, set D represents dancing lessons, and set S represents singing lessons. 8 students take only dancing lessons: s(DS) = 8
5 students take both dancing and singing lessons: s(D∩S) = 5
The number of students taking singing lessons:
s(S) = 2 * 8 – 1 = 15
s(S) = s(SD) + s(D∩S) and also s(D) = s(DS) + s(D∩S)
4 students take neither of these courses: s(E) – s(D ∪ S) = 4
s(D ∪ S) can be found by s(DS) + s(S) or s(SD) + s(D)
We are asked to find s(E):
s(E) = 4 + s(DS) + s(S) = 4 + 8 + 15 = 27
5. B
The interior angles of a triangle sum up to 1800:
(2x + 5) + (6x) + (3x – 23) = 180
82 NYSTCE® Mathematics Skill Practice!
11x – 18 = 180
11x = 198
x = 180
The largest angle is 6x = 6 * 18 = 1080
The supplementary of an angle is the angle which plus the
angle gives 1800. Then, the supplementary of 1080 is:
180 – 108 = 720
6. D
In the right triangle above, AB and BC are the legs and AC is the hypotenuse. For side lengths, Pythagorean Theorem is applied:
|AB|2 + |BC|2 = |AC|2
Let us say that |BC| = x. Then, |AB| = 2x:
(2x)2 + x2 = 152
5x2 = 225
x2 = 45
x = √45 = 3√5 cm
|AB| = 2x → |AB| = 6√5 cm
7. C
The perimeter of an equilateral triangle with 9 cm. sides will be
= 9 + 9 + 9 = 27 cm.
8. A
The question is to find the perimeter of a shape made by merging a square and a semi circle. Perimeter = 3 sides of the square + ½ circumference of the circle.
= (3 x 5) + ½(5 π)
= 15 + 2.5 π
Perimeter = 22.85 cm
9. D
Euclidian geometry was disproven by Nikolai Lobachevsky in 1830s.
10. C
Euclidian geometry supports parallel geometry. On the contrary, Non-Euclidian geometry is the study of geometry with curved spaces that is elliptic and hyperbolic geometry. In elliptic geometry; the inner angles of a triangle do not sum up to 1800; the sum is equal to 1800 plus the area of
the triangle. In hyperbolic geometry; the sum is equal to 1800 minus the area of the triangle. non-Euclidian geometry inspires from the shape of the world: If two meridians are selected; both intersect with the equator by 900. There is also a vertex angle in the pole. So, the inner angles sum up to 90 + 90 + vertex pole which is higher than 1800. Another example; a person walks 10 m south, 10 m west and then 10 m north. He sees that he is where he started moving. Normally, we would say that he would be 10 m east from the starting point. Think that he is on the North Pole.
If he goes 10 m down, 10 m west and then 10 m north, he is
again on the pole.
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## Operations with Polynomials
1. (x3 + 5x + 3x2 +2) + (4x3 + 3x2+ 14)
a. 5x+ 9x+ 16
b. 4x3+6x2+x+14
c. 5x3+6x2+5x+16
2. (2x3+ 5x4 + 3x2 +12) + (7x3 + 4x2+ 3)
a. 7x3+5x2+15
b. 5x3+5x4+12+x2
c. 5x4+9x3+7x2+15
Subtracting Polynomials
3. (-4x2 – 5y2 – 8) – (2x2 + 3y2 + 8)
a. -4x2-5y2+8
b. -4x2-4y2-16
c. -6x2-8y2-16
4. (x2 + 4x) + (x2 + 8x) – (3x – 7)
a. 2x2+9x+7
b. X2-3x+7
c. 4x2-12x-7
5. Simplify (2x 2 + 3x)(5x – 7)
a. 10x3-14x2-21x
b. 10x3+x2-21x
c. 10x2+15x-21
6. Simplify (- 3x 3 +3)(- 4x 2 – 6)
a. -12x5+12x2-18
b. 12x5+18x3-12x2-18
c. 12x5+18x4-x2-18 |
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# Find the market price of the refrigerator, of its selling price along with goods and services tax is given. $S.P=Rs16,275$ and rate of goods and services tax is $5\%$.
Last updated date: 15th Jun 2024
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Hint: We have a relation between the Selling Price$\left( S.P \right)$ , cost price$\left( C.P \right)$ and tax percentage as $Tax\%=\dfrac{S.P-C.P}{C.P}\times 100$. In the above relation we will substitute the given values $S.P=Rs16,275$ and tax percentage$=5\%$ and simplify the above equation to get the value of Cost Price$\left( C.P \right)$.
Given that,
Selling Price of the refrigerator is $S.P=Rs16,275$.
Goods and service tax percentage is $Tax\%=5\%$.
Let the Cost Price is $C.P=x$.
Now we have relation between Selling Price$\left( S.P \right)$ , cost price$\left( C.P \right)$ and tax percentage as $Tax\%=\dfrac{S.P-C.P}{C.P}\times 100$. Substituting the values of $S.P$ and $Tax\%$ in the above equation, then we will have
\begin{align} & Tax\%=\dfrac{S.P-C.P}{C.P}\times 100 \\ & \Rightarrow 5=\dfrac{16,275-x}{x}\times 100 \\ \end{align}
Dividing the above equation with $100$ on both sides, then we will have
\begin{align} & \dfrac{5}{100}=\dfrac{16,275-x}{x}\times \dfrac{100}{100} \\ & \Rightarrow \dfrac{1}{20}=\dfrac{16,275-x}{x} \\ \end{align}
Multiplying $x$ on both sides, then we will have
$x\times \dfrac{1}{20}=\dfrac{16275-x}{x}\times x$
we know that $\dfrac{1}{x}\times x=1$, then we will get
$\Rightarrow \dfrac{x}{20}=16275-x$
Adding $x$ on both sides, then we will have
$\Rightarrow x+\dfrac{x}{20}=16275-x+x$
We know that $x-x=0$, then
$\Rightarrow x+\dfrac{x}{20}=16275$
Taking $x$ common on the left side of the equation, then we will have
\begin{align} & \Rightarrow x\left( 1+\dfrac{1}{20} \right)=16275 \\ & \Rightarrow x\left( 1+0.05 \right)=16275 \\ & \Rightarrow 1.05x=16275 \\ \end{align}
Dividing the above equation with $1.05$ on both sides, then we will have
\begin{align} & x=\dfrac{16275}{1.05} \\ & \Rightarrow x=15,500 \\ \end{align}
$\therefore$ Cost Price of the Refrigerator is Rs.$15,500$.
Note: In this problem we have given the values of Selling Price and the Tax percentage. In some cases, they give any two variables from the Selling Price, Cost Price and Tax percentage and ask to calculate the third one. In these cases, also we will use the above relation to solve the problem. |
# Complex logarithms
So, another thought popped into my head today. This thought also involving complex numbers.
So we know what putting a complex number to another complex number is, but what about the opposite? What about complex logarithms?
$\large\log_{z_1}{z_2}$
So how do we go about solving this one?
Well first let's change the base to $e$, we'll keep the complex numbers as they are for now.
$\large \log_{z_1}{z_2} = \frac{\ln{z_2}}{\ln{z_1}}$
Now we'll convert the complex numbers into a different form.
$\large \frac{\ln{r_2e^{\theta_2 i}}}{\ln{r_1e^{\theta_1 i}}}$
Then using the rules of logarithms we'll simplify.
$\large \frac{\ln{r_2} + \ln{e^{\theta_2 i}}}{\ln{r_1} + \ln{e^{\theta_1 i}}}$
$\large \frac{\ln{r_2} + \theta_2 i}{\ln{r_1} + \theta_1 i}$
Now we need to multiply the top and bottom halves by $\boxed{\ln{r_1} - \theta_1 i}$ in order to make the bottom of the fraction a real number rather than a complex one.
$\large \frac{(\ln{r_2} + \theta_2 i)(\ln{r_1} - \theta_1 i)}{(\ln{r_1} + \theta_1 i)(\ln{r_1} - \theta_1 i)}$
$\large \frac{\ln{(r_2)}\ln{(r_1)} - \ln{(r_2)}\theta_1 i + \ln{(r_1) \theta_2 i + \theta_2\theta_1}}{(\ln{(r_1)})^2 + \theta_1^2}$
$\large \frac{(\ln{(r_2)\ln{(r_1)} + \theta_2\theta_1) + i(\ln{(r_1)}\theta_2 - \ln{(r_2)}\theta_1)}}{(\ln{(r_1)})^2 + \theta_1^2}$
And now the messy bit, substitution.
$\large \log_{z_1}{z_2} = \frac{\left(\ln{\sqrt{a_2^2 + b_2^2}}\ln{\sqrt{a_1^2 + b_1^2}} + \arctan{\frac{b_2}{a_2}}\arctan{\frac{b_1}{a_1}}\right) + i\left(\ln{\sqrt{a_1^2 + b_1^2}}\arctan{\frac{b_2}{a_2}} - \ln{\sqrt{a_2^2 + b_2^2}}\arctan{\frac{b_1}{a_1}}\right)}{\left(\ln{\sqrt{a_1^2 + b_1^2}}\right)^2 + \arctan{\left(\frac{b_1}{a_1}\right)}^2}$
So that's that, not as difficult as calculating this but still pretty tedious.
Hope you enjoyed the note.
Note by Jack Rawlin
5 years, 2 months ago
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## Comments
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Wow! That's a gigantic formula. I'm not sure if it's applicable or not because I seldom (never) see such forms of $\log_{z_1} z_2$ before.
Why don't you put your working into one of these wikis here?
- 5 years, 2 months ago
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# What are the steps in solving multi-step problem?
May 23, 2020 Off By idswater
## What are the steps in solving multi-step problem?
Here are steps to solving a multi-step problem: Step 1: Circle and underline. Circle only the necessary information and underline what ultimately needs to be figured out. Step 2: Figure out the first step/problem in the paragraph and solve it. Last step: Find the answer by using the information from Steps 1 and 2.
## What is a multi-step word problem?
A multi-step-word problem is like a puzzle with lots of pieces. Multi-step word problems are math problems that have more than one operation. An operation is addition, subtraction, multiplication, or division.
## What is an example of a multi-step problem?
Multi-Step Word Problems Examples. Robert had 16 marbles. His brother gave him 3 more bags of marbles. If each bag contained 5 marbles, how many marbles does Robert have now?
## What is the multi-step routine?
A multi-step addition and subtraction problem is one in which there are several steps that must be performed in order to get a final answer. You may need to add some quantities and subtract others, and you need to carefully read the problem in order to determine what you need to do to solve it.
## What are 2 step equations?
A two-step equation is an algebraic equation that takes you two steps to solve. You’ve solved the equation when you get the variable by itself, with no numbers in front of it, on one side of the equal sign.
## What are two-step problems?
A two-step problem is a word problem that requires two operations to solve it, for example: I buy a magazine costing 83p and a pencil costing 45p. I pay with a voucher that gives me 20p off the things I am buying.
## How do you solve two-step equations step by step?
Solving Two-Step Equations
1. 1) First, add or subtract both sides of the linear equation by the same number.
2. 2) Secondly, multiply or divide both sides of the linear equation by the same number.
3. 3)* Instead of step #2, always multiply both sides of the equation by the reciprocal of the coefficient of the variable.
## How to solve multi-step problems using the four operations?
Multi-Step Problems Using the Four Operations 4.OA.A.3 Application Mini -Assessment by Student Achievement Partners OVERVIEW This mini-assessment is designed to illustrate the standard 4.OA.A.3, which sets an expectation for students to solve multi-step word problems using the four operations.
## What are the words for multi step word problems?
Focus Area: Multi-Step Word Problems Nouns: word problems, operations, problems, equations, letters, unknown quantity, answers, mental computation, estimation strategies, rounding Verbs: solve, use, represent, assess Vocabulary: multi-step, operations, represent, equations, unknown quantity, mental computation, estimation, rounding Activity/Lesson:
## What to do in multi step Unit 4?
Multi-Step Word Problem Unit Grade 4 In this unit you will find: ๏a complete, comprehensive unit plan ๏12, two-step word problems and solutions ๏6 word problems and solutions for students to analyze ๏student Planning Sheet ๏assessment and solutions Suggestions:
## Do you need to model a multi step problem?
Mathematics is not only about answer-getting. Students need to model situations and attend to meanings of quantities. Traditionally, multi -step word problems have focused almost entirely on the solution; however, standard 4.OA.A.3 has other key components to highlight.
Multi-Step Problems Using the Four Operations 4.OA.A.3 Application Mini -Assessment by Student Achievement Partners OVERVIEW This mini-assessment is designed to illustrate the standard 4.OA.A.3, which sets an expectation for students to solve multi-step word problems using the four operations.
## How to solve multistep word problems in math?
Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity.
Mathematics is not only about answer-getting. Students need to model situations and attend to meanings of quantities. Traditionally, multi -step word problems have focused almost entirely on the solution; however, standard 4.OA.A.3 has other key components to highlight.
## Which is the best standard for multi-step problem solving?
A CLOSER LOOK Standard 4.OA.A.3 is a capstone standard in the development of problem solving skills using the four operations, with multiplicative compare situations being the most recently introduced situation type. Questions 4 and 5 highlight clearly worded multiplicative comparison, using the phrases times as many as and times as long as. |
# How do you solve x^(2/3) - 3x^(1/3) - 4 = 0?
Jan 14, 2016
Set $z = {x}^{\frac{1}{3}}$ When you find the $z$ roots, find $x = {z}^{3}$
Roots are $\frac{729}{8}$ and $- \frac{1}{8}$
#### Explanation:
Set ${x}^{\frac{1}{3}} = z$
${x}^{\frac{2}{3}} = {x}^{\frac{1}{3} \cdot 2} = {\left({x}^{\frac{1}{3}}\right)}^{2} = {z}^{2}$
So the equation becomes:
${z}^{2} - 3 z - 4 = 0$
Δ=b^2-4ac
Δ=(-3)^2-4*1*(-4)
Δ=25
z_(1,2)=(-b+-sqrt(Δ))/(2a)
${z}_{1 , 2} = \frac{- \left(- 4\right) \pm \sqrt{25}}{2 \cdot 1}$
${z}_{1 , 2} = \frac{4 \pm 5}{2}$
${z}_{1} = \frac{9}{2}$
${z}_{2} = - \frac{1}{2}$
To solve for $x$:
${x}^{\frac{1}{3}} = z$
${\left({x}^{\frac{1}{3}}\right)}^{3} = {z}^{3}$
$x = {z}^{3}$
${x}_{1} = {\left(\frac{9}{2}\right)}^{3}$
${x}_{1} = \frac{729}{8}$
${x}_{2} = {\left(- \frac{1}{2}\right)}^{3}$
${x}_{2} = - \frac{1}{8}$
Jan 14, 2016
x = 64 or x = -1
#### Explanation:
note that ${\left({x}^{\frac{1}{3}}\right)}^{2} = {x}^{\frac{2}{3}}$
Factorising ${x}^{\frac{2}{3}} - 3 {x}^{\frac{1}{3}} - 4 = 0$ gives ;
$\left({x}^{\frac{1}{3}} - 4\right) \left({x}^{1 / 3} + 1\right) = 0$
$\Rightarrow \left({x}^{\frac{1}{3}} - 4\right) = 0 \mathmr{and} \left({x}^{\frac{1}{3}} + 1\right) = 0$
$\Rightarrow {x}^{\frac{1}{3}} = 4 \mathmr{and} {x}^{\frac{1}{3}} = - 1$
'cubing' both sides of the pair of equations :
${\left({x}^{\frac{1}{3}}\right)}^{3} = {4}^{3} \mathmr{and} {\left({x}^{\frac{1}{3}}\right)}^{3} = {\left(- 1\right)}^{3}$
$\Rightarrow x = 64 \mathmr{and} x = - 1$ |
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# Lesson 4.
1 Polynomial Function
Objectives:
1. Identify a polynomial function from a given set of functions.
2. Determine the degree of a polynomial function.
Study Guide:
In mathematics, a polynomial is an expression of finite length constructed from
variables (also known as in determinates) and constants, using only the operations of
addition, subtraction, multiplication, and non-negative integer exponents. It is a function
that can be written in a form
P(x) = a0xn + a1xn-1 + a2xn-2 + + an
P(x) has the following properties:
a0 the first non-zero coefficient
n the highest exponent is the degree
is
called
the
coefficient
## an the constant term
Exercise 4.1
A. Identify which of the following are polynomial functions. Explain your answer.
1.
P(x) = 4x-2
2.
P(x) = 2x
3.
P(x) = 7x3+4x-3
4.
f(x) =
5.
g(x) = (x-3)3
h(x) = 10x+2
f(x) = 5+7x-3x2-4x3
P(x) = 25-4x2
6.
7.
8.
9.
f(x) =
10.
h(x) =
B. For each function, determine the leading coefficient (LC), the leading term (LD),
constant term (CT) and the degree (D) of each of the given polynomials.
LC
4
## f(x) = 3x +5x +17x -25x+10
f(x) = 2x3+3x2-2x-5
f(x) = x5-5x3+6x-2
LD
CT
## 4.2 Synthetic Division
Objectives:
1. Use synthetic division to find the quotient and the remainder wh a polynomial is
divided by a linear expression of the form (x-c)
Study Guide:
Synthetic Division is also called Horners Method. Following are the steps in
finding the quotient of P(x) divided by x-c using synthetic division:
a. Arrange the coefficients in descending powers of x in the first row, placing zeros
for the missing terms.
b. Bring down the leading coefficient in the third row.
c. Multiply the entries in the third row by c, put the result in the second row under
the next column, and add. Put the sum in the third row under the present column.
d. Repeat the third step until the column of the constant term is reached.
e. Write the quotient, Q(x), using the entries in the third row as the coefficient of the
terms of x. The quotient is 1 degree lower than the degree of the dividend. The
entry in the last column is the value of R(x).
Exercises 4.2
Divide P(x) by D(x) using long division then express P(x) in the form P(x) = Q(x)D(x)
+R(x)
1. P(x) = x4+5x3-2x+8
D(x) = x-2
2. P(x) = 2x5-7x4+5x3-4x2-x+5
D(x) = x+1
3. P(x) = 3x5+5x4-3x3+x2+5x-2
D(x) = x+2
4. P(x) = 3x6+2x5-4x4+7x3-5x2+8x+1
D(x) = 3x-1
5. P(x) = 4x5-5x3+9
D(x) = 2x+3
Critical Thinking
What must be multiplied by x-1 to get x4-x3-3x2+10x-7? Explain your answer.
## 4.3 The Remainder Theorem and the Factor Theorem
Objectives:
1. State and illustrate the Factor Theorem
2. Find P(r) by synthetic Division and Remainder Theorem
Study Guide:
In Algebra, the polynomial remainder theorem is an application of polynomial
long division. It states that the remainder of a polynomial f(x) divided by a linear divisor
x-a is equal to f(a).
Exercises 4.3
A. Answer the following questions below and put your answer to the corresponding by:
*Note: To check your answer, make sure that the sum of vertical, horizontal and
diagonal is equal to 15.
A.
7
9
D.
A.
B.
C.
D.
E.
B.
C.
1
3
E.
(5x3+3x2-10x+2)/(x-2)
(x4-5x3+7x2+9x-8)/(x-5)
(x3-12x2+8x+60)/(x-10)
(2x3+3x2-4x+5)/(x+2)
(3x4-4x2+3x-2)/(x+1)
6 = 212
2 = 34
8 = -6
4=9
5 = -60
B. Determine if the given binomial is a factor of the given polynomial, then choose the
letter of the correct answer.
1. f(x) = 5x4+16x3-15x2+8x+16
A. x-2
2. f(x) = x3+2x2-5x-6
B. x-3
3. f(x) = x5-2x4+3x3-6x2-4x+8
C. x+4
4. f(x) = 2x3+3x2-8x-12
D. x+1
5. x3-3x2+4x-12
E. 2x+3
## 1. Is x-1 a factor of f(x) = 2x4+3x2-5x+7?
2. Is (2x-3)(x-1) are factors of f(x) = 2x2-5x+3?
3. Is (x+2) a factor of f(x) = x4-3x-5?
4. Is (x+3) a factor of 2x4+5x3-2x2+5x+3
Critical Thinking
1. Determine the value of k so that x-2 is a factor of x 4+3x3+kx2+x-14?
2. If x-1 and x-2 are both factors of x4+Ax2-5x2+Bx+4, what are the values of A and
B? Why?
## 4.4 Roots and their Multiplicities
Objectives:
1. Find the zeros of polynomial functions of degree greater than 2 using the factor
theorem, synthetic division, Depressed Equations and Factoring.
Study Guide:
Example: Find the zeros of P(x) = (x-2)2(x+1)3(x+3)
Let (x-2)2(x+1)3(x+3)=0
Since the polynomial is already in factored form, then just equate each factor to zero
and solve for x.
(x-2)2 = 0
(x+1)3 = 0
x+3 = 0
x-2 = 0
x+1 = 0
x = -3
x=2
x = -1
## It can be observed that x=2 is a zero of multiplicity of 2 of P(x) since (x-2)2 is a
factor of P(x), while x=-1 is a zero of multiplicity 3 since (x+1)3 is also a factor of P(x).
Therefore, the zeros of P(x) = (x-2)2(x+1)3(x+3) are 2 of multiplicity 2, -1 of multiplicity 3
and -3.
Exercises 4.4
A. Find all zeros of each polynomial function with their multiplicities.
1.
P(x) = (x-3)(x+2)2
2.
P(x) = (x+1)2(x-3)2(x+5)2
3.
P(x) = x2(x+3)2(x-5)3
4.
P(x) = (2x+3)2(3x-4)3(x+4)
5.
P(x) = (x+4)2(x2-9)
6.
P(x) = (2x-5)4(x2-2x-15)
7.
P(x) = (4x-1)2(x-4)3(x2+5x-14)
8.
P(x) = x3(5x+2)(2x2+5x+3)
9.
P(x) = x(x-4)2(3x+2)3
10.
P(x) = x2(3x-4)2(5x2-9x+4)
B. Use the factor theorem, synthetic division or factoring techniques to determine all the
zeros of the following polynomial functions.
1.
P(x) = x3+4x2-11x-30
2.
P(x) = x4-x3-7x2+x+c
3.
P(x) = x5-6x4+x3-40x2+16x
4.
P(x) = x3-2x2-29x+42
5.
P(x) = x4-5x3-14x2
Critical Thinking:
Determine the polynomial function of lowest integral coefficient if one of the zeros of
P(x) is
## 4.5 The Rational Roots Theorem and Descartes Rule of Signs
Objectives:
1. State and illustrate the Rational Zero Theorem
2. Use Descartes Rule of Signs in finding the possible number of positive and
negative real zeros of a given polynomial function
Study Guide:
The Rational Zero Theorem states that the possible rational zeros of a
polynomial must be equal to a factor p of the constant term divided by a factor of the
leading coefficient. This means that the possible rational zeros are equal to p/q.
Descartes Rule of Signs
1. The number of positive real zeros of f(x) is either equal to the number of sign
changes or variations in signs of f(x) or is less than that number by an even
integer. Note that if there is only one sign change in f(x), then f(x) has exactly one
positive real zero.
2. The number of negative real zeros of f(x) is either equal to the number of sign
changes or variations in signs of f(-x) or is less than that number by an even
integer. Note that if f(-x) has only one sign change, then f(x) has exactly one
negative real zero.
Exercises 4.5
A. Determine all the possible zeros of the following polynomials.
1. P(x)=2x4-4x3-x2+4x-1
2. P(x)=4x3+11x2-2x+6
3. P(x)=4x4-3x3-10x2+3x-6
4. P(x)=2x5+2x4+3x3-3x2-6x-1
5. P(x)=5x6-45x4+14x3-90x+27
B. Make a chart summarizing the possible combinations of positive, negative and
imaginary roots of each polynomial equations using Descartes Rule of Signs.
Number of Positive Real
Roots
1.
2.
3.
4.
x2-x-5=0
x3-x2+3x+4=0
x3+4x2-x-12=0
3x4+4x3-x2+x+4=0
Roots
## Number of Imaginary Roots
5. 4x4-2x3+x2+2x-4=0
## 4.6 The Graphs of Polynomial Functions
Objective:
1. Draw the graphs of polynomial functions of degree greater than 2.
Study Guide:
The graph of polynomial function is a curve with turning points depending upon
the degree of the polynomial.
Exercises 4.6
A. Prepare a table of values then sketch the graph of each polynomial function.
1.
P(x)=x3+x2-5x+3
2.
P(x)=-10-3x+6x2-x3
3.
P(x)=x4-4x3-2x2+12x+9
4.
P(x)=x4-8x3+22x2-24x+9
5.
P(x)=8x2+4x3-2x4-x5
B. Communicating Mathematics
1. Based on the graphs in A, describe the graph of the polynomial function when:
a. The degree of the polynomial is even and a n0.
___________________________________________________________________
b. The degree of the polynomial is even and a n0.
___________________________________________________________________
c. The degree of the polynomial is odd and a n0.
___________________________________________________________________
d. The degree of the polynomial is odd and a n0.
___________________________________________________________________
2. At most, how many zeros does a polynomial functions have? Is it possible for a
polynomial to have no zero? Explain.
## CHAPTER 5: EXPONENTIAL AND LOGARITHMIC FUNCTIONS
5.1 The Nature of Exponential Functions
Objective:
1. Identify real-life relationships which are exponential in nature.
Study Guide:
A certain situation or occurrence shows exponential change of the original
amount is multiplied by a fixed factor.
Exercises 5.1
A. Tell whether the situations show exponential change or not by completing the tables
below.
1. A man buys a pair of signature pants worth P3000 with his credit card on condition
that he pays his bill within a month or he will be charged an interest of 1% per month
accumulated over the period he does not pay his bill.
Time (months)
Amount of Bill
2. A faucet leaks such that water drips from it at a fixed rate of 12 droplets per minute.
Time (minutes)
No. of Droplets
12
B. Analyze each situation and tell whether it is related to exponential change or not.
1. A new convenience store has initially 20 costumers and each week 2 new
customers are coming.
2. A population of months increases by half of its population size every week.
3. If a student forgets to return a library book on the date it is due, he is fined P5 on
the first day and P2 more each day thereafter.
4. The population of a certain type of microorganism doubles every hour.
5. A man accepts a position at P12000 a month with the understanding that he will
receive P500 increase every year.
## 5.2 Other Exponential Functions
Objectives:
1. Describe some properties of the exponential function.
2. Sketch the graph of an exponential function.
Study Guide:
Domain is the set of all first elements of the ordered pairs in a relation. Range is
the set of all second elements of the ordered pairs in a relation. Intercept is the
intersection of the graph with either the x-or-y-axis.
Exercises 5.2
A. 1. Consider a piece of string. Fold it into two then cut it. . Observe that there are now
two pieces of string. Put the two pieces of string together, fold them again into two
then cut the pieces. How many pieces of string are there now.
No. of folds (x)
## No. of string y=f(x)
2.Draw the graph of the table above.
3.What mathematical sentence describes the relationship above?
4.What is the domain of the function?
5.Describe the graph obtained.
B. Sketch the graph of the exponential function of the form f(x)=a x , where a1 and x is a
real number for the domain -3x3.
1. Draw the graphs of f(x)=2x, f(x)=3x and f(x)=4x on one set of axes.
x
-3
-2
-1
f(x)=2x
f(x)=3x
f(x)=4x
## 2.What do the three graphs have in common?
C. Draw the graphs of f(x)=(1/2)x, f(x)=(1/3)x and f(x)=(1/4)x on one set of axes.
x
f(x)=(1/2)x
f(x)=(1/3)x
f(x)=(1/4)x
-3
-2
-1
## 5.3 Solving Exponential Equations
Objectives:
1. Use the law of exponents to transfer exponential equations into algebraic
equations.
2. Solve exponential equations.
Study Guide:
Laws of Exponents
## An exponential equation is an equation involving exponential functions. To solve
an exponential equation, express both sides of the equation in the same base and solve
for the value of the missing term.
A. Determine the solution/s to each exponential equation.
1. 3x=812x+5
6. 22x=1/128
2. 2x+1=32-2x+5
7. 3x^2+4x=1/27
3. 162x-1=645x+3
8. 53x=125-x
4. 253x+1=125x+3
9. 42x64x=1/512
5. 7x+5=1/49
10. 125=(1/5)x+5
B. Solve for x.
1.
2.
3.
4.
6.
5.
C. Communicating Mathematics
1. Can you find a value of x such that f(x)=2 x will be equal to zero? Justify your answer.
______________________________________________________________________
2. For any exponential function, f(x)=a x, is there any value of x so that ax=0? Explain
______________________________________________________________________
## 5.4 Inverse Functions
Objectives:
1. Define inverse functions
2. Determine the inverse of a given function.
Study Guide:
Two functions f and g are inverse functions if and only if f(g(x))=x and g(f(x))=x.
The inverse function of f is often denoted as f -1.
Exercises 5.4
A. Show the inverse of each of the following functions. Is the inverse still a function?
1. {(-2,4),(-1,1),(0,0),(1,1),(2,4)}
__________
2. {(0,1),(1,2),(2,4),(3,8),(4,16)}
__________
3. {(-2,-8),(-1,-1),(0,0),(1,1),2,8)}
__________
## B. Find the inverse of each relation.
1. f(x)= 5x-2
2. f(x)=
7. f(x)=
8. f(x)=
3. f(x)= 1/5(x-2)
4. f(x)=x2+2x-2-4
5. f(x)=
6. f(x)= 5(3x)
9. f(x)=
10. f(x)=
## 5.5 The Logarithmic Function
Objectives:
1. Define the logarithmic function as the inverse of the exponential function is the
same base.
2. Describe the properties of logarithmic functions.
Study Guide:
The mathematics of logarithms and exponentials occurs naturally in many
branches of science. It is very important in solving problems related to growth and
decay. Therefore, we need to have some understanding of the way in which logs and
exponentials work.
The formula y=logbbx is said to be written in logarithmic form and x=by is said to
be written in exponential form. In working with these problems, it is most important to
remember that y=log bx and x=by are equivalent statements.
Exercises 5.5
Table Completion: Complete the table by converting the exponential to logarithmic form
in Table A while convert the following logarithm to exponential form in Table B. Write the
correct answer on the table.
A.
Exponential Form
Logarithmic Form
49=7
1.
(1/9)-=3
2.
23=8
3.
(1/5)-2=25
4.
42=16
5.
33=1/27
6.
(1/4)-1=4
7.
43/2=8
8.
zy=x
9.
64=9
10.
B.
Logarithmic Form
Exponential Form
log366=
1.
log82=1/3
2.
log 327=3
3.
log 1/749=-2
4.
log3/41=0
5.
log100.01=-2
6.
log2=-1
7.
log10100=2
8.
log264=6
9.
log1/8=3
10.
## 5.6 The Laws of Logarithms
Objective:
1. State and apply the laws of logarithms.
Study Guide:
Four Basic Properties of Logarithms
1. logbxy = logbx + logby
2. logb = logbx-logby
3. logbxn = nlogbx
4. logbx = logax/logab
Exercises 5.6
A. logarithm of product
D. logarithm of quotient
G. log7
B. Logarithm of root
E. logarithm of power
H. log72
C. logarithm
F. log20
I. log6
## 1. It is logarithm that is equal to p times the logarithm of number.
2. It is logarithm that is equal to the logarithm of the dividend minus the logarithm of
the divisor.
3. log5+log4 = x
4. log14-log2 = x
5. It is logarithm that the two numbers equals the sum of the logarithms of the
numbers.
6. log2+log3 = x
7. What do call to another name for an exponent?
8. 2log3+3log2 = x
9. It is logarithm that is equal to the logarithm of the number divided by r.
B. Direction: Express the following as single logarithms. Reach the star to get a bonus
of 5 points. Write your answer on the box.
1. log 5 + log 4=
2. log 14 log 2=
3. 2log 3 + 3log 2=
4. 2log 6 + log 2=
5.
6. log 7 + log 3=
7. log 10 log 2=
8. 2log 10 log 5=
9. 4log 6 2log 2=
10. 3log 3 + 2log 5=
## 5.7 Common Logarithms
Objective:
1. Compute common logarithms using a:
a. Calculator
b. Table of logarithms
Study Guide:
A logarithm to the base 10 is called the common logarithm. TO simplify the
rotation needed to write them, we shall agree that when the base of a logarithm is not
written, it is understood to be 10. That is:
log y = log10y
If log x = y, then x is called the antilogarithm of y. in symbols, x = antilog y.
Exercises 5.7
A. Between what two consecutive integers do each of the following logarithms lie? Do
not use calculator.
1. log 236
6. log 0.03
2. log 178
7. log 2.3
3. log 67
8. log 0.005
4. log 3002
9. log 1.1
5. log 12000
## 10. log 7.5
B. Determine the value of each of the following common logarithms to six decimal
places.
1. log 5
6. log 17.7
2. log 11
7. log 245
3. log 3.48
8. log 387
4. log 5.12
9. log 14320
5. log 23.42
10.
Communicating Mathematics
1. Your calculator displays an error when you try to find log 0. Why?
2. Why does your calculator display an error when you try to find log(-10)?
5.8 Solving Logarithmic Equations
Objectives:
1. Solve logarithmic equations.
2. Use the laws of logarithms to solve logarithmic equations.
Study Guide:
Logarithmic equations are equations involving logarithmic functions. To solve
logarithmic equations, apply the laws of logarithms/exponents.
Exercises 5.8
A. Solve for the unknown
1. logx27 = 3
2. log2/3x = 2
3. loge20 = x
4. log1080 = x
5. log4(x+3) = 2
6. log (2x-1) = log (4x-3) log x
7. log y = log 5x
8. log2x = 4.5
9. log23x = 4.5
10. log9x = 1
B. Loop the words that have a connection in logarithm.
Q
## 5.9 Problem Solving with Logarithmic and Exponential Equations
Objective:
1. Solve problems involving logarithms and exponential equations.
Study Guide:
Exponential and logarithmic functions have many applications not only in science
but also in business.
The use of calculator is a great help in solving many of the problems involving
exponential and logarithmic functions.
Exercises 5.9
Solve the following problems.
1.
Principal
Rate (%)
Time in Years
Compounded
1. P60000
Every 6 mos.
2. P120000
10
1.5
Yearly
3. P240000
Every 2 mos.
4. P15000
2.5
Yearly
Amount
2. If P50000 is invested at 5% today, how much will it be worth at the end of 3 years if it
is compounded
a. annually?
b. semi-annually?
c. quarterly?
d. monthly?
3. On her 7th birthday, Princess parents placed P20000 in time deposit at 5% interest
compounded monthly. In ten years, how much money would be available for her
educational expenses?
4. A certain city has a population of 2000 and a growth rate of 2.5%. What will be the
expected population after 5 years?
5. If the half-life of a certain radioactive substance is 100 years, what fraction of the
original amount of substance will remain after 400 years? after 600 years?
CHAPTER 6: Circular Functions
6.1 Measuring Angles in Radians
Objectives:
1. Convert angle measures from degrees to radians, and vice versa.
2. Illustrate angles in standard position.
3. Determine the coterminal angle or angles and the reference angle of an angle.
Study Guide:
A central angle whose arc is equal in length to the radius of the circle is called a
radian. The radian measure of is defined to be the ratio of the arc length S to radius r:
where r is the radius, S is the arc length and is the measure of the angle in
If is a complete revolution, S=2r and
## If is a complete revolution, the degree measure of is 360.
Exercises 6.1
A. Radian. The Snowman.
The snowman picture has a unit circle for its base. You are to label 16 points in
the unit circle with the radian measure inside the circle and the coordinates of the
points outside the circle. Then color and decorate the snowman.
B. Wrute True in the space provided if the underlined statement is correct, if not write
False.
1. Radian is the ration between the length of an arc and its radius.
2. Radian is the standard unit of angular measure.
3. It is widely used in English.
4. The unit was formerly an SI complementary unit.
5. The radian is represented by the symbol dian.
6. Radian describes the plane angle subtended by a circular arc as the length of the
arc divided by the radius of the arc.
7. Radian is credited to Roger Cotes.
8. The term radian first appeared in print on June 5, 1873.
9. Radian is a false number.
10. The magnitude in radians of such a subtended angle is equal to the ratio of the
arc length to the length of the radius of the circle.
C. Convert the following Radians to Degrees.
1. 5
2. 321
7.
3. 12
4. 412
5. 2
6.
8.
9.
10.
1. 0
6. 120
2. 30
7. 135
3. 45
8. 150
4. 60
9. 180
5. 90
10. 360
## 6.2 The Sine and Cosine Functions
Objectives:
1. Define the sine and cosine functions of an angle , given a point P() on the unit
circle.
2. Evaluate the sine and cosine functions of special angles, and use identities to
quickly derive the others.
Study Guide:
Given a right triangle, ABC with its parts labeled, the ratio of the side opposite to the
hypotenuse is called the sine of the measure of .
The ratio of the side adjacent to the hypotenuse is called the cosine of the measure of
.
Exercise 6.2
A. Fill the table with the correct values of special angles.
sin
cos
tan
sec
csc
cot
30
45
60
90
2
1
## 6.4 The Other Circular Functions
Objectives:
1. Define the four other circular functions.
2. Evaluate the circular functions of special angles
Study Guide:
We define the other four circular functions-tangent function, cosecant function,
secant function and cotangent function in terms of the sine and cosine functions.
Exercise 6.4
A. Direction: Match the given trigonometric functions on column A with the given
values on column B.
6. sin 30
COLUMN A
7. sec 45
1. cos 90
8. sec 60
2. csc 60
9. tan 90
3. sin 45
10.
4. cot 30
5. sec 0
COLUMN B
csc 0
a. Undefined
f. 0
b.
g.
c. 1
d. 2
e.
h.
i.
B. Illustration
Direction: Draw, label and put description on the given angle of triangles.
1-3. 60-30-90
4-6 45-45-90
7-10. 30-60-90
## 6.5 Graphs of Other Circular Functions
Objective:
1. Sketch the graphs of the circular functions.
Study Guide:
The secant and cosine functions are reciprocals. Hence the secant function can
be graphed by making use of the cosine since
(cos0).
Exercise 6.5
1. Complete the table.
Function
Domain
Range
Period
Amplitude
Sine
Cosine
Tangent
Cotangent
Secant
Cosecant
2. Complete the table.
Function
y = 2 sin 3
y = sin /2
y = 4 sin 5
y = 3 cos /3
y = cos 4
3. Sketch the graph of each of the following functions on the interval 0 2. Give
the period and amplitude of each function.
a. y = sin 3
d. y = csc 3
b. y = cos /2
e. y = -cos
c. y = sec 2
4. Sketch the graph of each of the following functions on the interval .
a. y = tan 2
d. y = cot 2
b. y = tan
e. y = -2 tan
c. y = 3 tan
## 6.6 The Fundamental Trigonometric Identities
Objective:
1. State the fundamental trigonometric identities and use them to prove other
identities.
Study Guide:
A trigonometric identity is an equation involving trigonometric functions that can
be solved by any angle. Trigonometric identities have less to do with evaluating
functions at specific angles than they have to do with relationships between functions.
Reciprocal Identities:
Quotient Identities:
Pythagorean Identities:
sin2 + cos2 = 1
tan2 + 1 = sec2
1 + cot2 = csc2
cos(-) = -cos
tan(-) = -tan
Negative-Angle Identities:
sin(-) = -sin
Exercises 6.6:
A. For each trigonometric expression in Column A, choose the expression from Column
B that completes a fundamental identity.
Column A
Column B
a. sin2 x + cos2 x
1.
b. cot x
2. tan x
c. sec2 x
3. cos(-x)
4. tan2 x + 1
d.
5. 1
e. cos x
## B. Direction: Use the fundamental identities to get an equivalent expression then
simplify it.
1. cot sin
2. cos csc
3. sin2 (csc2 1)
4. tan + cot
5. sin (csc sin )
6. sin2 + tan2 + cos2
7. sec cot sin
8. cot2 (1 + tan2 )
9. (sec 1)(sec + 1)
10. (sec + csc )(cos sin )
## Lesson 6.7 Other Common Identities
Objective:
1. State other trigonometric identities.
Study Guide:
The following identities are derived from the fundamental trigonometric identities:
sin csc = 1
cos sec = 1
tan cot = 1
tan cos = sin
cot sin = cos
Exercises 6.7
Direction: Verify the following:
1.
2.
## 3. 1 + cot = csc (cos + sin )
4. sec4 sec2 = tan4 + tan2
## Lesson 6.8 Double-Angle, Half-Angle, Sum-to-Product, and Product-to-Sum
Formulas
Objectives:
1. Use the appropriate identities to evaluate expressions containing circular
functions.
Study Guide:
Double-Angle Identities:
cos 2A = cos2 A sin2 A
## sin 2A = 2 sin A cos A
cos 2A = 1 2 sin2 A
cos 2A = 2 cos2 A 1
Half-Angle Identities
Exercises 6.8
Directions: Use the identities to complete the following and simplify.
1.
5.
2.
3. cos 14 = 1 2 sin2 ____
6.
4.
## Lesson 6.9 Trigonometric Equations
Objective:
1. Solve simple trigonometric equations.
Study Guide:
Conditional equations with trigonometric (or circular) functions can usually be
solved by using algebraic methods and trigonometric identities. For example:
2 sin + 1 = 0
2 sin = -1
sin = -
Exercises 6.9
Directions: Solve each equation for solutions in the interval (0, 2) by first solving for the
trigonometric function.
1. 2 cot x + 1 = -1
2. 2 sec x + 1 = sec x + 3
3. 2 cos4 x = cos2 x
4. sin x + 2 = 3
5. tan2 x + 3 = 0
6. -2 sin2 x = 3 sin x + 1
7. 2 sin x + 3 = 4
8. sec2 x + 2 = -1
9. cos2 x + 2 cos x + 1 = 0
10. tan3 x = 3 tan x
Lesson 6.10 Functions Derived from the Sine and the Cosine Functions
Objective:
1. Describe the properties of functions derived from the sine and cosine functions.
Study Guide:
Sum and Difference Identities:
sin(A + B) = sin A cos B + cos A sin B
sin(A B) = sin A cos B cos A sin B
cos(A + B) = cos A cos B sin A sin B
cos(A B) = cos A cos B + sin A sin B
Exercises 6.10
Directions: Find the value of the following.
1. cos 15
2. cos 75
3. sin 105
4. sin 165
5. sin (-345) |
## What are the three formulas of Ohm’s law?
3-4: A circle diagram to help in memorizing the Ohm’s Law formulas V = IR, I = V/R, and R= V/I. The V is always at the top.
## What is Ohm’s law in simple terms?
Ohm’s law says that in an electrical circuit, the current passing through a resistor between two points, is related to the voltage difference between the two points, and are related to the electrical resistance between the two points.
## What is Ohm’s law mathematically?
Ohm’s law may be expressed mathematically as V/I = R. That the resistance, or the ratio of voltage to current, for all or part of an electric circuit at a fixed temperature is generally constant had been established by 1827 as a result of the investigations of the German physicist Georg Simon Ohm.
## What is the formula for current?
Current is usually denoted by the symbol I. Ohm’s law relates the current flowing through a conductor to the voltage V and resistance R; that is, V = IR. An alternative statement of Ohm’s law is I = V/R.
## What is Ohm’s law with diagram?
Ohm’s Law tells us that if a conductor is at a constant temperature, the current flowing through the conductor is directly proportional to the voltage across it. This means that if we plot voltage on the x-axis of a graph and current on the y-axis of the graph, we will get a straight-line.
## What is the basic principle of Ohm’s law?
Ohm’s Law definition Ohm’s Law states that the current flowing in a circuit is directly proportional to the applied potential difference and inversely proportional to the resistance in the circuit. In other words by doubling the voltage across a circuit the current will also double.
## What is meant by 1 ohm?
ohm. [ ōm ] The SI derived unit used to measure the electrical resistance of a material or an electrical device. One ohm is equal to the resistance of a conductor through which a current of one ampere flows when a potential difference of one volt is applied to it.
## How do I calculate resistance?
If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm’s Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps. The total resistance RT = 9 volts / 3 amps = 3 Ω.
## What is Watt’s law?
Watt’s Law states that: Power (in Watts) = Voltage (in Volts) x Current (in Amps) P = V I Combining with Ohm’s law we get two other useful forms: P = V*V / R and P = I*I*R Power is a measurement of the amount of work that can be done with the circuit, such as turning a motor or lighiting a light bulb.
## What is the SI unit of resistivity?
ohm
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## What is Ohm’s law class 10th?
Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions remain unchanged. Current is directly proportional to voltage difference through a resistor.
## What is work formula?
Work is done when a force that is applied to an object moves that object. The work is calculated by multiplying the force by the amount of movement of an object (W = F * d).
## What is the symbol for current?
The conventional symbol for current is I, which originates from the French phrase intensité du courant, (current intensity). Current intensity is often referred to simply as current. The I symbol was used by André-Marie Ampère, after whom the unit of electric current is named, in formulating Ampère’s force law (1820).
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Assignment #1:
I will begin by graphing some of these equations. I think that the most effective way to start this project is to graph each equation individually, and then to go from there.
It is pretty obvious to us all that the first equation is that of the unit circle. This is something we have been learning about for most of our math education.
Our next equation on the other hand, that's not one that I have ever seen graphed before.
This is pretty interesting. What happens here is that the graph is in Quadrant I when x & y range from 0 to 1. After that, one value (x or y) is positive while the other one is negative. Well, let's see. Yes, that makes sense because both values must add to one, so when a negative is cubed, you will still get a negative, so I can see how this graph works.
Well let's see what we got. Our next graph has both x & y raised to an even power, so what should happen here? Well, a negative raised to an even power will be positive, so then this graph should look similar to our last one, except that x & y should both be between 0 and 1. Let's see what happens when we graph it.
You know, this looks like our first graph. Well, this one is a little more square at the corners, but it centers around the origin. That must be because of the even powers.
So let's think about this, if this is a pattern developing here, then our next equation should look similar to equation 2. Is that right? Hmmm! Well, we are dealing with an odd power again, so when we raise a negative to an odd power we get a negative. That was the same situation, so we should, and do wind up with the same result.
Now, what should happen with our last two. I've never attempted to graph anything to a power that high. I think that there is a pattern developing here. I suspect that when we graph these next two equations we are going to see a pretty similar situation. Well there is only one way to find out.
That's pretty much what we expected to happen. The only difference is that as the exponents get that high, the edges start to get pretty close to being square.
So, what is the point of doing this? That is really a good question.
It could be that this is a very good method to explore the graphs of odd exponents as opposed to even exponents. Well, we learned that the difference is that an even exponent graph will be closed and centered around the origin, whereas an odd exponent graph will stay between o and 1 until one value becomes negative. Is that it?
It could also be used to show that just because our values started to get really high, the underlying shape of each upheld. The only difference was that the edges were squared out as the numbers increased.
I think another way we could go from here is to predict what will happen when our exponents get mixed:
If we think about this, we can tell what should happen. Let's graph the first two equations and see what the results are.
The graphs here make sense. It would also seem to follow that the graphs of the last two equations should be similar except that the edges would be more square.
Well I think that this has been a pretty interesting lesson. I got to see that there are similarities between graphs dealing with exponents. I want to leave this project by showing all of our original graphs displayed at the same time, just in case any one is curious to see what it all looked like put together. |
## Creating Differential Equations Solution2
In this page creating differential equations solution2 we are going to see solutions of some practice questions.
Form a differential equations by eliminating arbitrary constants given in brackets against each.
(iii) x y = c² {c}
Solution:
here we have only one arbitrary constant,so we can differentiate the given equation with respect to x only once.
x y = c²
x (dy/dx) + y (1) = 0
x y' + y = 0
Therefore the required equation is x y' + y = 0.
(iv) (x²/a²) + (y²/b²) = 1 {a , b}
Solution:
(x²/a²) + (y²/b²) = 1
(x²b² + y²a²)/a²b² = 1
x²b² + y²a² = a²b²
In this question we have two arbitrary constants. So we can differentiate the given equation two times.
differentiate the given equation with respect to x
2 x b² + 2 y y' a² = 0
divide the whole equation by 2
x b² + y y' a² = 0 ------ (1)
again differentiate the given equation with respect to x
we are going to differentiate y y' using product rule
u = y v = y'
u' = y' v' = y''
formula for product rule:
d (u v) = u v' + v u'
= y y'' + y' (y')
= y y'' + (y')²
(1) b² + [y y'' + (y')²] a² = 0
b² + [y y'' + (y')²] a² = 0 ----- (2)
=
x y y' 1 y''+yy'
x (y'² + y y'') - y y' = 0
Therefore the required equation is x (y'² + y y'') - y y' = 0.
(v) y = A e^(2x) + Be(-5x) {A , B}
Solution:
Here we have two arbitrary constants. So we can differentiate the given equation two times.
y = A e^(2x) + Be(-5x)
y' = 2 A e^(2 x) - 5 Be^(-5 x)
y'' = 4 A e^(2 x) + 25 B e^(-5 x)
=
y e2x e-5x y' 2e2x -5e-5x y'' 4e2x 25e-5x
taking e^(2x) and e^(-5x) commonly from second and third column,we get
=e2xe-5x
y 1 1 y' 2 -5 y'' 4 25
e^(2x-5x) y(50 + 20) - y'(25 - 4) + y''(-5-2) = 0
e^(-3x) y(70) - y'(21) + y''(-7) = 0
70 y - 21 y' - 7 y'' = 0
now we are going to divide the whole equation by 7
10 y - 3 y' - y'' = 0
y'' + 3 y' - 10 y = 0
Therefore the required equation is y'' + 3 y' - 10 y = 0. |
# SubsectionThe Assignment
• Read chapter 3 section 4 of Strang.
# SubsectionLearning Goals
Before class, a student should be able to:
• Identify a particular solution to a matrix-vector equation $Ax=b$. (Provided there is one.)
• Find the complete solution to a matrix-vector equation $Ax=b$ as a parametrized object. (Provided there is one.)
After class, a student should be able to:
• Describe the complete solution to a matrix-vector equation $Ax=b$ as an implicit object, cut out by equations.
• Describe the possibilities for the number of solutions to a matrix-vector equation $Ax=b$ in terms of the shape of the matrix.
# Subsection Discussion: The Complete Solution to a System of Equations
This is the big day! We finally learn how to write out the general solution to a system of linear equations. We have spent so much time understanding things related to this, that it should go pretty quickly.
The tiny little facts underneath the analysis for this section are these: For a matrix $A$, vectors $v$ and $w$ and a scalar $\lambda$, all chosen so that the equations make any sense, \begin{equation*} \begin{array}{rcl} A(v+w) &= &Av + Aw \\ A(\lambda v) &= &\lambda ( Av ) \end{array} \end{equation*}
The first is a kind of distributive property, and the second is a kind of commutative property. When taken together, these things say that the operation of “left-multiply by the matrix $A$” is a special kind of function. The kind of function here is important enough that we have a special word for this combined property: it is called linearity. That is, left-multiplication by $A$ is a linear operation or a linear transformation.
The linearity property makes it possible to check the following two results.
And if we put these two theorems together, we find this result which sounds fancier, but has exactly the same content.
• Form the augmented matrix $\left( A \mid b \right)$ and use Gauss-Jordan elimination to put it in reduced row echelon form $\left( R \mid d \right)$.
• Use the information from the RREF to find a particular solution $x_p$ by solving for the pivot variables from the vector $d$ and setting the free variables to zero.
• Use the special solutions $s_1, s_2, \dots, s_k$ (if any exist!) to describe the nullspace $\mathrm{null}(A)$.
• Write down the resulting general solution: \begin{equation*} x = x_p + a_1 s_1 + a_2 s_2 + \dots + a_k s_k, \quad \text{for any scalars } a_i \in \mathbb{R}. \end{equation*}
# SubsectionSageMath and Solving General Systems
SageMath has many built-in methods for solving systems of linear equations. We will investigate three common ones with a single example considered several times.
# SubsubsectionMethod One: RREF and the Nullspace
First we find a particular solution.
This clearly has three pivots, and all belong in the original matrix. So there will be a solution. We pull out the particular solution.
Since we typed that in by hand, we should check our work.
Now we need to find the nullspace and the special solutions.
The basis has only one row, so there is only one special solution. This matches our expectation. Our system is $3\times 4$ and has rank $3$. So there is only one free column, and hence only one special solution.
Now we can check the “general solution”.
# SubsubsectionMethod Two: A SageMath built-in
SageMath has a built-in method that looks like “Matrix division”. Here we “left divide” by the matrix. This is odd notation, and is just something SageMath allows.
It is weird, but this works even if A is not invertible, like now.
The downside to this particular method is that in only gives you one particular solution. It does not produce the complete solution. You have to do that bit for yourself, maybe like the above.
# SubsubsectionMethod Three: Another SageMath Built-in
Finally, SageMath will also try to solve the system if you apply the .solve_right() method to A. You have to supply the vector b as an argument to the command.
Again, this only pulls out a single particular solution. It is up to you to figure out the rest.
# SubsectionExercises
(Strang ex 3.4.4) Find the complete solution (also called the general solution) to \begin{equation*} \begin{pmatrix} 1 & 3 & 1 & 2 \\ 2 & 6 & 4 & 8 \\ 0 & 0 & 2 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}. \end{equation*}
(Strang ex 3.4.6) What conditions on $b_1$, $b_2$, $b_3$ and $b_4$ make each of these systems solvable? Find a solution in those cases.
1. \begin{equation*} \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ 2 & 5 \\ 3 & 9 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{pmatrix}. \end{equation*}
2. \begin{equation*} \begin{pmatrix} 1 & 2 & 3\\ 2 & 4 & 6\\ 2 & 5 & 7\\ 3 & 9 & 12\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{pmatrix}. \end{equation*}
(Strang ex 3.4.11) It is impossible for a $1 \times 3$ system of equations to have $x_p = (2,4,0)$ and $x_n = \text{ any multiple of } (1,1,1)$. Explain why.
(Strang ex 3.4.13) Each of the statments below is false. Find a $2\times 2$ counterexample to each one.
1. The complete solution is any linear combination of $x_p$ and $X_n$.
2. A system $Ax=b$ has at most one particular solution.
3. The solution $x_p$ with all free variables zero is the shortest solution, in that it has the minimum norm $||x_p||$.
4. If $A$ is an invertible matrix, there is no solution $x_n$ in the nullspace.
(Strang ex 3.4.21) Find the complete solution in the form $x_p + x_n$ to these full rank systems.
1. \begin{equation*} x+y+z = 4\end{equation*}
2. \begin{equation*} \begin{array}{ccccccc} x & + & y & + & z & = & 4 \\ x & - & y & + & z & = & 4 \end{array} \end{equation*}
(Strang ex 3.4.24) Give examples of matrices $A$ for which the number of solutions to $Ax = b$ is
1. $0$ or $1$, depending on $b$;
2. $\infty$, regardless of $b$;
3. $0$ or $\infty$, depending on $b$;
4. $1$, regardless of $b$.
1. The only solution of $Ax = \left(\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right)$ is $x = \left(\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right)$.
2. The only solution of $Bx = \left(\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right)$ is $x = \left(\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right)$.
(Strang ex 3.4.33) The complete solution to the equation $Ax = \left(\begin{smallmatrix} 1 \\ 3\end{smallmatrix}\right)$ is $x = \left(\begin{smallmatrix} 1 \\ 0\end{smallmatrix}\right) + c\left(\begin{smallmatrix}0\\ 1 \end{smallmatrix}\right)$. Find the matrix $A$. Write the set of equations that corresponds to $Ax = b$. (This is the implicit description of this set!) |
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• ## Related Books
### Identity Tan(squared)x+1=Sec(squared)x
Main formulas:
• For any angle x for which the tangent and secant are defined, we have tan2 x + 1 = sec2 x.
• For any angle x for which the tangent and secant are defined, we have cot2 x + 1 = csc2 x.
Example 1:
Prove the identity tan2 x + 1 = sec2 x.
Example 2:
Given that tanθ = − 4.21 and (π/2) < θ< π , find secθ .
Example 3:
Prove the following trigonometric identity:
cscθ− cotθsecθ− 1 = cotθ
Example 4:
Prove the identity cot2 x + 1 = csc2 x.
Example 5:
Given that secθ = (13/12) and 270° < θ < 360° , find tanθ .
### Identity Tan(squared)x+1=Sec(squared)x
Given that tanθ = - 5.3 and [(3π)/2] <θ < 2π, find secθ.
• Use the Pythagorean Identity to solve for secθ: 1 + tan2θ = sec2θ
• 1 + ( − 5.31)2 = sec2θ ⇒ 1 + 28.1961 = sec2θ ⇒ sec2θ = 29.1961 ⇒ secθ = ±√{29.1961}
• secθ = ± 5.4 Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV which means the same holds true for secant values.
secθ = 5.4 because θ is in quadrant IV
Given that secθ = [13/5] and 180°<θ < 270° , find tanθ.
• Use the Pythagorean Identity to solve for tanθ: 1 + tan2θ = sec2θ
• 1 + tan2θ = ( [13/5] )2 ⇒ 1 + tan2θ = [169/25] ⇒ tan2θ = [169/25] - [25/25] ⇒ tanθ = ±√{[144/25]}
• tanθ = ±[12/5] Remember the mnemonic ASTC. This tells which quadrant tangent values will be positive. Tangent is only positive in quadrants I and III.
tanθ = [12/5] because θ is in quadrant III
Given that tanθ = - 3.23 and [(π)/2] <θ <π, find secθ.
• Use the Pythagorean Identity to solve for secθ: 1 + tan2θ = sec2θ
• 1 + ( − 3.23)2 = sec2θ ⇒ 1 + 10.4329 = sec2θ⇒ sec2θ = 11.4329 ⇒ secθ = ±√{11.4329}
• secθ = ± 3.38126 Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV which means the same holds true for secant values.
secθ = - 3.4 because θ is in quadrant II
Given that secθ = − [5/4] and 90°<θ < 180°, find tanθ.
• Use the Pythagorean Identity to solve for tanθ: 1 + tan2θ = sec2θ
• 1 + tan2θ = ( − [5/4] )2 ⇒ 1 + tan2θ = [25/16] ⇒ tan2θ = [25/16] - [16/16] ⇒ tanθ = ±√{[9/16]}
• tanθ = ±[3/4] Remember the mnemonic ASTC. This tells which quadrant tangent values will be positive. Tangent is only positive in quadrants I and III.
tanθ = − [3/4] because θ is in quadrant II
Verify the following identity: (tan2θ + 1)(cos2x - 1) = - tan2x
• Try to get the left hand side to look like the right hand side because it is the more complicated side
• (sec2θ)( - sin2x) = - tan2x by the Pythagorean Identities of sin2θ + cos2θ = 1 and 1 + tan2θ = sec2θ
• [1/(cos2x)] · - sin2x = - tan2x by the Reciprocal Identity
• [( − sin2x)/(cos2x)] = - tan2x by multiplying
• ( [( − sinx)/cosx] )2 = - tan2x by the Rule of Exponents
- tan2x = - tan2x by the Quotient Identity
Verify the following identity: secθ + tanθ = [(cosθ)/(1 − sinθ)]
• Try to get the right hand side to look like the left hand side bcause it is the more complicated side
• secθ + tanθ = [(cosθ)/(1 − sinθ)] · [(1 + sinθ)/(1 + sinθ)], multiple by the conjugate
• secθ + tanθ = [(cosθ+ cosθsinθ)/(1 − sin2θ)], multiplication of fractions
• secθ + tanθ = [(cosθ+ cosθsinθ)/(cos2θ)], Pythagorean Identity sin2θ + cos2θ = 1
• secθ + tanθ = [(cosθ)/(cos2θ)] + [(cosθsinθ)/(cos2θ)], Separate Fractions
• secθ + tanθ = [1/(cosθ)] + [(sinθ)/(cosθ)], Simplifying
secθ + tanθ = secθ + tanθ, Reciprocal Identities
Verify the following identity: tanθ + cotθ = secθcscθ
• Try to get the left hand side to look like the right hand side bcause it is the more complicated side
• [(sinθ)/(cosθ)] + [(cosθ)/(sinθ)] = secθcscθ, Quotient Identity
• [(sin2θ+ cos2θ)/(cosθsinθ)] = secθcscθ, Adding Fractions
• [1/(cosθsinθ)] = secθcscθ, Pythagorean Identity
• [1/(cosθ)] · [1/(sinθ)] = secθcscθ, Product of Fractions
secθcscθ = secθcscθ, Reciprocal Identities
Verify the following identity: [(cot2θ)/(cscθ)] = cscθ - sinθ
• Try to get the left hand side to look like the right hand side bcause it is the more complicated side
• [(csc2θ− 1)/(cscθ)] = cscθ - sinθ, Pythagorean Identity: csc2θ - 1 = cot2θ
• [(csc2theta)/(cscθ)] - [1/(cscθ)] = cscθ - sinθ, Separate Fractions
• cscθ - [1/(cscθ)] = cscθ - sinθ, Simplifying
cscθ - sinθ = cscθ - sinθ, Reciprocal Identity
Verify the following identity: cot2θ(sec2θ - 1) = 1
• Try to get the left hand side to look like the right hand side bcause it is the more complicated side
• cot2θtan2θ = 1, Pythagorean Identityq: tan2θ = sec2θ - 1
• ( [(cos2θ)/(sin2θ)] ) ( [(sin2θ)/(cos2θ)] ) = 1, Reciprocal Identities
1 = 1, Simplifying
Verify the following identity: [(cot2θ)/(1 + cscθ)] = [(1 − sinθ)/(sinθ)]
• [(csc2θ− 1)/(1 + cscθ)] = [(1 − sinθ)/(sinθ)], Pythagorean Identity on the left hand side
• [((cscθ− 1)(cscθ+ 1))/(1 + cscθ)] = [(1 − sinθ)/(sinθ)], Factor the left hand side
• cscθ - 1 = [(1 − sinθ)/(sinθ)], Simplify the left hand side
• cscθ - 1 = [1/(sinθ)] - [(sinθ)/(sinθ)], Separate Fractions on the right hand side
cscθ - 1 = cscθ - 1, Reciprocal Identity on the right hand side
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Identity Tan(squared)x+1=Sec(squared)x
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Main Formulas 0:19
• Companion to Pythagorean Identity
• For Cotangents and Cosecants
• How to Remember
• Example 1: Prove the Identity 1:40
• Example 2: Given Tan Find Sec 3:42
• Example 3: Prove the Identity 7:45
• Extra Example 1: Prove the Identity
• Extra Example 2: Given Sec Find Tan
### Transcription: Identity Tan(squared)x+1=Sec(squared)x
We are talking about the trigonometric identities in particular to the Pythagorean identities for (tan) and (cot) of (theta), and (sec) and (cosec).0000
Here we are being asked to prove the identity (cot)2 +1 = (cosec)2.0012
The trick there is to remember the original Pythagorean identity.0020
Let me write that down to start with, sin2x + cos2x = 1, that is the original Pythagorean identity.0024
That should be very familiar to you because you use that all the time in trigonometry.0035
We will start with that and I know I’m trying to find (cot) and (cosec) in this somehow.0043
What I’m going to do is divide both sides by sin2x.0049
Divide everything here by sin2x and that gives me 1.0063
Now (cos)2/(sin)2 that is (cos)/(sin)2, that is (cot)2x.0072
1/(sin)2 that is 1/(sin), that is (cosec) by definition.0081
There you have it, that is a pretty quick one.0090
We start with the original Pythagorean identity and we just divide both sides by (sin)2x to make it look exactly like what we are asked for.0093
I can rearrange terms here and I can get (cot)2x + 1 is equal to (cosec)2x.0104
This just comes back to knowing the original Pythagorean identity, if you remember that original Pythagorean identity.0119
And if you remember the definitions of sec, tan, cosec, and cot, then you can pretty quickly derive the new one cot2 + 1 = cosec2.0125
You might not even really need to memorize that one as long as you know the others vey well because you can work it out quickly.0136
Hi we are given (sec) of an angle here, 13/12 and we are given the angle in terms of degrees is between 270 and 360 and we are asked to find the tan(theta).0000
This is pretty clearly asking us to use the Pythagorean identity 10 2(theta) +1= sec2(theta).0012
Sec2(theta) is we plug in 13/12, that is 13/122 and that is 169/144.0023
102(theta) if we subtract 1 from both sides is 169/144 – 1 which is 169-144/144 which is 25/144.0038
Tan(theta) if we take the square root of both sides is + or – the square root of 25/144, which is + or -, those are both perfect squares so it comes out neatly 5/12.0067
Now the question is, whether it is the positive one or the negative one and we need to figure that out for the problem.0088
There is extra information given in the problem that we have not use yet.0094
We have not use the fact that (theta) is between 270 and 360 degrees .0098
Let me draw our (theta) would be approximately, here is my unit circle, in degrees that is 0, 90, 180, 270, and 360.0104
(theta) is between 270 and 360, (theta) is down there somewhere, so (theta) is an angle around there.0123
If you remember all students take calculus, that tells you which of these basic functions are positive and negative in each quadrants.0133
In the first quadrant they are all positive, in the second quadrant only (sin) is, third quadrant only (tan) is, fourth quadrant only (cos) is.0145
Our (theta) is in the fourth quadrant, let me write that as (theta) is in quadrant 4.0156
Tan(theta) only cos is positive there, tan is not, so tan(theta) is negative.0177
When we are choosing between these positive and negative square root here, we know we have to pick the negative one, -5/12.0188
There are really two steps to figuring out how to do this problem, one is to remember the Pythagorean identity tan2(theta) + 1=sec2(theta).0210
Then we take the value of sec(theta) that we are given, we plug it in, we work it down through and we try to solve for tan(theta) and we end up taking the square root.0221
We get plus or minus in there and we do not know what we want if positive or negative.0232
The second step there was to use the information about what quadrant (theta) is in and once we know that (theta) is in quadrant 4, we know that its tan it has to be negative.0237
That is from the all student take calculus rule and tan(theta) is negative, we know we need to take the negative square root.0253
That is how we get tan(theta) is equal to -5/12.0260
That is our lesson on the Pythagorean identity tan2(theta) + 1= sec2(theta).0265
These are the trigonometry lectures for www.educator.com.0271
OK, welcome back to the trigonometry lectures on educator.com, and today we're talking about the identity, tan2(x)+1=sec2(x).0000
You really want to think about this as a kind of companion identity to the main Pythagorean identity, which is that sin2(x)+cos2(x)=1.0010
Hopefully, you've really memorized this identity, sin2(x)+cos2(x)=1, that's one that comes up everywhere in trigonometry.0022
Then sort of the companion Pythagorean identity to that for secants and tangents is tan2(x)+1=sec2(x).0034
There's another related identity, essentially just the same for cotangents and cosecants, is that cot2(x)+1=csc2(x).0045
You may wonder how you'll remember this identities.0058
For example, how do you remember whether it's tan2+1=sec2, or maybe it's the other way around, sec2+1=tan2.0062
Well, really the answer to that lies in the exercises that we're about to do.0072
We'll show you how to derive those identities from the original one sin2+cos2=1.0077
As long as you can remember sin2+cos2=1, you'll learn how you can use that to figure out the others and you really won't have to work hard to remember this new Pythagorean identities.0085
Let's jump right into that.0099
The first example here is to prove the identity tan2(x)+1=sec2(x).0102
The trick there is to remember the original Pythagorean identity, which is cos2(x)+sin2(x)=1, that's the original Pythagorean identity.0109
That's one that you really should memorize and remember throughout all your work with trigonometry.0127
What you do to manipulate this into the new identity, is you just divide both sides by cos2(x).0136
On the left, we get cos2(x)/cos2(x) plus, let me write it as (sin(x)/cos(x)2=1/cos2(x).0152
Then of course, cos2(x)/cos2(x) is just 1, sin/cos, remember that's the definition of tangent, so this is tan2(x), 1/cos, that's the definition of sec(x), so sec2(x).0173
You can just rearrange this into tan2(x)+1=sec2(x).0191
That shows that this new identity is just a straight consequence of the original Pythagorean identity, so really if you can remember that Pythagorean identity, you can pretty quickly work out this new Pythagorean identity for tangents and cotangents.0203
Let's try using the Pythagorean identity for tangents and cotangents.0224
In this problem, we're given the tan(θ)=-4.21, and θ is between π/2 and π, we want to find secθ.0230
Remembering the Pythagorean identity, tan2(θ)+1=sec2(θ).0238
If we plug in what we're given here, tan2(θ), that's (-4.212)+1=sec2(θ).0246
Well, 4.212, that's something I'm going to workout on my calculator, is 17.7241.0259
So, 17.7241+1=sec2(θ), that's 18.7241=sec2(θ).0274
If I take the square root of both sides, I get sec(θ) is equal to plus or minus the square root of 18.7241.0290
Again, I'm going to do that square root on the calculator, and I get approximately 4.33.0303
The question here is whether we want the positive or negative square root, and that's where we use the other piece of information given in the problem.0323
Θ is in the second quadrant here, θ is between π/2 and π, so θ is somewhere over here.0332
Now, sec(θ) remember, is 1/cos(θ).0348
In the second quadrant, if you remember All Students Take Calculus.0353
In the second quadrant, sine is positive, but cosine is not, cosine is negative.0358
That means, sec(θ) is negative.0365
θ is in quadrant 2, cos(θ) is negative, sec(θ), because that's 1/(θ), is negative.0370
So, sec(θ), we want to take the negative square root there, and we get an approximate value of -4.33.0394
That negative is very important.0411
The key to that problem is first of all invoking this Pythagorean identity, tan2+1=sec2.0415
That's very important to remember.0423
We plug in the value that we're given and then we work it through and we'll try to solve for sec(θ).0426
We get this plus or minus at the end because we don't know if we want the positive square root or the negative square root.0431
That's where we use the fact that θ is in the second quadrant.0437
Since θ is in the second quadrant, cos(θ) must be negative, sec(θ), remember, is just 1/cos(θ), must also be negative.0448
That's how we know we want the negative square root there, so we get -4.33.0458
In our next example, we're given trigonometric identity and we're asked to prove it, (csc(θ)-cot(θ))/(sec(θ)-1) = cot(θ).0467
I'm going to start with the left-hand side of this trigonometric identity and I'm going to manipulate it.0480
I'm going to try and manipulate it into the right-hand side.0490
The left-hand side which is (csc(θ)-cot(θ))/(sec(θ)-1).0496
Now, I'm going to do something clever here and it's based on this old principle of whenever you have an (a-b) in the denominator, try multiplying by the conjugate, (a+b)/(a+b).0511
If it's the other way around, if you have (a+b), you multiply by the conjugate (a-b)/(a-b).0524
The reason you do that is that it makes the denominator (a2-b2).0530
We're taking advantage of that old formula from algebra, the difference of squares formula, and a lot of times the (a2-b2) is something that it'll simplify into something nice.0538
That's what's going to happen in here.0549
We're going to multiply, since we have (sec(θ)-1 in the denominator, by sec(θ)+1.0552
What that turns into in the denominator is this (a2-b2) form, so sec2(θ)-1.0565
The numerator really doesn't have anything very good happening yet, (csc(θ)-cot(θ))×sec(θ)+1, nothing very good happening in the numerator yet.0577
In the denominator, we're going to take advantage of the fact that tan2(θ)+1=sec2(θ), that's the Pythagorean identity that we're studying today.0593
If you move that around, you get sec2(θ)-1=tan2(θ).0607
That converts the denominator into tan2(θ), so that's very useful.0614
In the numerator, we have csc(θ)-cot(θ) and sec(θ)+1.0623
I think now I'm going to translate everything into sines and cosines because those will be easier to understand than cosecants and cotangents.0628
So, cosecant, remember is 1/sin(θ), minus cotangent is cos(θ)/sin(θ), sec(θ)+1, well, sec(θ) is 1/cos(θ), and tan(θ), if we translate into sines over cosines, that's sin2(θ)/cos2(θ).0637
Now, I've got fractions divided by fractions, I think the easiest thing to do here is to bring the denominator, flip it up and multiply it by the numerator.0670
We get cos2(θ)/sin2(θ), that's flipping up the denominator, and then the old numerator is, if I combined the first two terms of the first one, it's (1-cos(θ))/sin(θ).0681
In the second one, we have 1/cos(θ)+1.0706
Now, I think what I'm going to do is I'm going to look at this cosine squared, write it as cos(θ)×cos(θ).0715
Then pull one of those cosines over and multiply it by this term, and try to simplify things a little bit that way.0726
That will leave me with just one cosine left, still have sin2 in the denominator, (1-cos(θ)/sin(θ)) times, now, cos(θ)×(1/cos(θ) is 1, plus 1×cos(θ).0736
Now, look, we have (1-cos(θ))×(1+cos(θ)).0767
We're going to use this pattern again, the (a-b)×(a+b)=a2-b2.0772
If cos(θ)/sin2(θ), now multiply (1-cos)×(1+cos) gives us (1-cos2(θ)) and sin(θ) in the denominator.0780
Now, we're going to use the other Pythagorean identity, the original one which are right over here, sin2(θ)+cos2(θ)=1, if you switched that around 1-cos2(θ)=sin2(θ).0798
I'll plug that in, cos(θ)/sin(θ), 1-cos2(θ), now translates into sin2(θ) all divided by sin(θ).0819
I forgot my squared there, that's from the line above.0836
Now, we have some cancellations, sin2(θ) cancel, we get cos(θ)/sin(θ), but that's cot(θ).0840
By definition of cotangent, cotangent is defined to be cos/sin.0852
That's exactly equal to the right-hand side of the identity.0858
When you're proving this trigonometric identities, you pick one side and you multiply it as much as you can.0864
You try to manipulate it into the other side.0870
It does takes some trial and error.0874
I worked this problem out ahead of time, I tried a couple of different things and I finally found a way that gets us through it relatively quickly.0876
It's not like you'll always know exactly which path to follow, it takes a little bit of experimentation.0882
There are some patterns that you see over and over again, and the two patterns that we really invoke strongly in this one are these algebra pattern where (a-b)×(a+b)=a2-b2.0888
Essentially, whenever you see an (a-b) or an (a+b), think about multiplying it by the conjugate, and then taking advantage of that pattern.0904
That's what really got us started on the first step here.0913
We had a sec(θ)-1, and I thought, okay, let's multiply that by sec(θ)+1, that gives us sec2-1.0917
The second pattern that we use there was to invoke these Pythagorean identities, tan2+1=sec2, and sin2+cos2=1.0927
We invoked that one here, converting sec2-1 into tan2.0942
Then later on, when he had another (a-b)×(a+b), it converted into 1-cos2 and that in turn, by the Pythagorean identity converted into sin2.0947
It's just a lot of manipulation but you can let your manipulation be kind of guided by these common algebraic identities and these common Pythagorean identities.0962
Then you just try to keep manipulating until you get to the other side of the equation.0973
We'll try some more examples later on.0978 |
# How do you use partial fraction decomposition to decompose the fraction to integrate (x^3+x^2+x+2)/(x^4+x^2)?
Sep 25, 2015
See the explanation.
#### Explanation:
${x}^{4} + {x}^{2} = {x}^{2} \left({x}^{2} + 1\right)$
$\frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C x + D}{{x}^{2} + 1} =$
$= \frac{A x \left({x}^{2} + 1\right) + B \left({x}^{2} + 1\right) + \left(C x + D\right) {x}^{2}}{{x}^{2} \left({x}^{2} + 1\right)} =$
$= \frac{A {x}^{3} + A x + B {x}^{2} + B + C {x}^{3} + D {x}^{2}}{{x}^{2} \left({x}^{2} + 1\right)} =$
$= \frac{\left(A + C\right) {x}^{3} + \left(B + D\right) {x}^{2} + A x + B}{{x}^{2} \left({x}^{2} + 1\right)}$
$A + C = 1$
$B + D = 1$
$A = 1$
$B = 2$
$C = 1 - A = 0$
$D = 1 - B = - 1$
$\int \frac{{x}^{3} + {x}^{2} + x + 2}{{x}^{4} + {x}^{2}} \mathrm{dx} = \int \frac{\mathrm{dx}}{x} + 2 \int \frac{\mathrm{dx}}{x} ^ 2 - \int \frac{\mathrm{dx}}{{x}^{2} + 1} =$
$= \ln | x | - \frac{2}{x} - \arctan x + C$ |
Month: January 2020
# How to Find Crossing Angle in Railway – Mathematical Problem & Solution
## Find the Crossing Angle – Problem & Solution
Mathematical Problem:
Find the crossing angle of 1 in $8\frac{1}{2}$ crossing by using all the three methods(Right angle, centre line and isosceles triangle method):
Solution:
1. By using right angle method
We know, $\cot \alpha = N = 8.5$ [N = $8\frac{1}{2}$ = 8.5]
So, Angle of Crossing = $\alpha = 6^{\circ} {42}’ {35}”$
2. By using centre line method
We know,
$\cot \frac{\alpha }{2} = 2N = 2 \times 8.5 = 17$
So, Angle of Crossing = $\alpha = 6^{\circ} {43}’ {59}”$
3. By using Isosceles triangle method
We know,
Or, $\alpha = 6^{\circ} {44}’ {41}”$
Angle of Railway Crossing Methods
Loop Line in Railway
# Loop Line in Railway and its Types
## Loop Line in Railway?
When a branch line leaves from a mainline and again this branch line terminates at the same mainline later, it is called a loop line. The main purpose of providing a loop line is to provide a bypass route. It helps to pass the fast-moving trains on time. The arrangement of the loop line can be either of the following types:
1. Split turnout
2. Trailing turnout
3. Straight and loop
➤ A split turnout is not suitable for fast-moving trains as they have to reduce their speed while nearing the facing points. Therefore, the use of split turnout is unacceptable when considering fast-moving trains.
➤ A trailing turnout is less objectionable as compared to a split turnout because there is a slight reduction in the speed of fast trains while leaving over a reverse curve.
➤ When considering fast through trains, straight and loop arrangement is always acceptable because there is no reduction in speed. Additional loops and other facilities are provided whenever needed.
Triangle – Railway Track Junction
Railway Water Columns
Switch Angle in Railway
# Triangle – Railway Track Junction
## Triangle – Railway Track Junction
The triangle is a junction of a railway track, which is constructed for changing the direction of engines. They require a large area and therefore they are constructed where there is enough land available for their construction.
Turntables are also used to change the direction of engines, but they are very costly. So, if enough area is available, then the construction of triangles is preferable to turntable installation. The triangle’s maintenance cost is less than that of the turntable.
Generally, a triangle junction consists of three tracks, they are PQ, QS, and SP, which are shown in the above fig. Usually, PQ and SQ tracks are curved and the PS track is straight. But sometimes all three tracks are laid on curves as shown in the figure.
A dead-end siding QR is provided at point Q to accommodate the length of the engine. If space permits, the QR length should be kept slightly longer than the lengths of two locomotives.
### Working Principle of Triangle
The working of triangle is very simple and it is very easy to understand. If an engine is standing at S facing towards P and the engine moves in the direction of arrows(i.e along SQ track) as shown[Now, in a simple way we can say a train moves from S to Q, then Q to R, then back R to Q, and then Q to P], it will be facing S when it reaches point P.
### Importance Features of Triangle
i. It consists of two simple turnouts (i.e., at S and P) and one symmetrical split at Q.
ii. This type of track junction has three acute angle crossings as shown in the figure.
iii. This is mainly used for turning the faces of engines where the provision of the turntable is costly.
Permanent Way in Railway
Types of Railway Platform
Types of Rails – Double Headed, Bull Headed, and Flat Footed Rails
Switch Angle in Railway
# Types of Railway Platform || Railway Engineering
## Types of Railway Platform
The platform is a raised level surface of a railway station, from where either passengers can get into the train, or the loading and unloading of goods are done. Generally, the following two types of railway platform are provided:
1. Passenger Platform
2. Goods Platform
### 1. Passenger Platform
As the name suggests, this type of platform is constructed for the movement of passengers who are using the railway. The following are the essentials of a passengers platform:
i. The passengers’ platform should be covered for a minimum distance of 60 to 61 m of their length.
ii. These types of railway platforms should be paved with a minimum width of 3.65 m.
iii. A ramp should be provided at the end of the platform of maximum slope 1: 6.
iv. Sufficient arrangement of light should be made for the efficient and safe running of trains at night.
v. The drinking water facility should be provided on the passenger platform, and also necessary sanitary arrangements should be made.
vi. The names of stations should be written on a R.C.C. or stone board in bold letters in Hindi, English, and the regional language. The station’s name is written on both sides of the board.
vii. The station’s name should be written on a yellow background in black letter with a size of 300 mm. The spacing between the letters, and also at the top and the bottom letter is 150 mm.
viii. The name board should be kept at least a height of 1.8 m from the platform level.
### Dimensions of the passenger platform
a. Length and Width of the passenger platform
• The minimum length recommended for passenger platform in case of all gauges is 180 m. But the desirable length for B.G is 305 m.
• The minimum permissible width in front of a station building for all gauge is 3.60 m.
b. Clearance between the center line of the adjacent track and the edge of the platform
• In India, the clearance is kept 1676 mm for Broad Gauge.
• 1346 mm for Meter Gauge.
• And 1219 mm for narrow gauge.
c. Height or elevation of the passenger platform
Generally, in India, there are three types of platform elevation, namely, the rail level, the low level and high-level platform. The height above rail level for different gauges and platform are as follows:
d. Slope of Platform
The platforms are provided with a slope of 1 in 30 across its width.
### 2. Goods Platform
As the name suggests, these types of railway platforms are used for the loading and unloading of goods. The following are the essentials of a goods platform.
i. Generally goods platforms are higher and the minimum width is 3 m.
ii. The goods sheds should be provided on the goods platforms and at the same time an arrangements for weighing the goods should be provided on these platforms.
iii. Proper drainage facility is very essential for a goods platform, hence proper arrangements for drainage should be made on these platforms.
The Permissible height of the goods platforms above the rail level is as follows:
Facilities Required at Railway Stations
Railway Gauge
Purpose of Providing Railway Station
# Calculate the Correct Length of the Tape – Mathematics Problem & Solution
Mathematics Problem – Example:
A line of true length 500 m, when measured by a 20 m tape, is reported to be 502 m long. The correct length of tape is?
Solution:
Given,
• True Length = 500 m
• Measured Length = 502 m
The line is measured long (502-500) = 2 m. Hence, the tape is short by some length and the actual length is not 20 m.
Hence,
True Length of Line = $\frac{{l}’}{l}\times$ measured length
Or, $500 = \frac{{l}’}{20}\times 502$
Or, $\frac{500\times 20}{502} = {l}’$
Or, ${l}’ = 19.92 m$
Different Types of Map Symbols
# 5 Purposes of Providing Superelevation in Roads
## 5 Purposes of Providing Superelevation
Following are the 5 purposes or advantages of providing superelevation on the curve portion of a road.
1. When a vehicle is moving on a curved path it is subject to an outward force known as centrifugal force. In order to counteract the effect of the centrifugal force, superelevation is provided.
Know more –Maximum And Minimum SuperElevation
2. It helps to keep your speed constant with comfort on a curve path.
3. It reduces the number of accidents.
4. It helps to drain off rainwater towards the inner side. so no need to provide camber at this portion of roads…(Analysis of Superelevation Formula)
5. It also helps to keep the vehicles to their proper side on the roads and thus prevents head collision of vehicles moving in opposite directions.
# Facilities Required at Railway Stations
## Facilities Required at Railway Stations
The features or requirements of a railway station can be grouped as under.
1. Public Requirements.
2. Traffic Requirements.
3. Requirements of the locomotive department.
4. General Requirement.
### 1. Public Requirements
i. A booking office for tickets.
ii. A platform for goods and passengers.
iii. A minimum platform covering.
iv. Proper lighting for night times.
v. Proper arrangement of drinking water.
vi. A refrigerator to supply cold water in hot weather if possible.
vii. Proper sanitary arrangements.
viii. Bathrooms
ix. Refreshment Rooms.
x. Waiting rooms and retiring rooms.
xi. Public telephone.
xii. Microphones to announce the arrival and departure of trains.
xiii. Guide map of the town.
xiv. A big board to show the timetable of the trains.
xv. Police office to help the passengers, and to avoid any violation.
xvi. Inquiry office.
xvii. Boards showing reservation charts.
### 2. Traffic Requirements
i. Proper arrangements for booking tickets, issuing of luggage labels and goods receipts, and also, the arrangement to collect all these things at the end of the journey should be made.
ii. The way for controlling and reporting the movement of trains by means of signals should be made.
iii. A sufficient number of sidings and platforms should be constructed to handle goods and traffic.
v. In the case of a big station, for moving luggage and heavy goods through lifts or underground passages should be constructed.
vi. Arrangements should be made to accommodate the staff of the traffic department.
### 3. Requirements of the Locomotive Department
Following are the requirements of the locomotive department.
i. Proper arrangements for supplying fuel and water to locomotives should be made.
ii. In a proper way, cleaning and examining the locomotives should be provided.
iii. The facility should be provided for maintaining and inspecting the locomotive.
### 4. General Requirement
General requirements or requirements for the development of railways are as follows:
i. Easy and suitable approach roads should be available towards the stations from the surrounding areas.
ii. Clocks for the correct time should be provided.
iii. Availability of coolies
iv. Foot bridges for connecting various platforms should be constructed.
Purpose of Providing Railway Station
Angle of Crossing in Railway
Top 10 Most Beautiful Railway Stations in the World
# Angle of Crossing in Railway & Methods Used to Determine it
## Angle of Crossing in Railway
The angle which is formed between the gauge faces of the Vee is known as the crossing angle or angle of the railway crossing. Following are the three methods used to determine the angle of crossing in railway.
### 1. Right Angle or Cole’s Method
In this method, a right angle triangle is used to determine the angle of crossing.
Notation:
• $\alpha$= Angle of Crossing
• N = Crossing Number
From this Figure, we can write
$\tan \alpha = \frac{1}{N}$
Or, $\cot \alpha = N$ …………(1.a)
Or, $N = \cot \alpha$ …………(1.b)
Equation (1.a) shows the angle of crossing.
Equation (1.b) shows the number of crossings.
Note: This is the standard method mostly adopted by the Indian Railways.
### 2. Centre Line Method
In this method, the measurement is taken along a line bisecting the crossing angle as shown in Fig.
Now,
$\tan \frac{\alpha}{2} = \frac{\frac{1}{2}}{N} = \frac{1}{2N}$
Or, $\cot \frac{\alpha}{2} = 2N$ …………..(2.a)
Or, $N = \frac{1}{2}.\cot \frac{\alpha}{2}$ ……….(2.b)
Equation (2.a) shows the angle of crossing.
Equation (2.b) shows the number of crossings.
Note: This method is adopted in some countries like U.K. and U.S.A.
### 3. Isosceles Triangle Method
In this case, the measurement is taken along one side of the isosceles triangle as shown in fig.
Then,
$\sin \frac{\alpha}{2} = \frac{\frac{1}{2}}{N} = \frac{1}{2N}$
Or, $cosec \frac{\alpha}{2} = 2N$ …………..(3.a)
Or, $N = \frac{1}{2}. cosec \frac{\alpha}{2}$ ……….(3.b)
Equation (3.a) shows the angle of crossing.
Equation (3.b) shows the number of crossings.
Note: This method is implemented in the case of tramway layout.
More the angle of crossing lesser will be permissible speed and vice versa |
# Geometric Interpretation of Complex Algebraic Proof of Sum of Squares Statement
When I see answers regarding proofs such as the one mentioned here, it seems that there is a considerable diversity of ways to attempt to look at this proof. Similarly, although this sum of squares question is very different, I was wondering if there was a geometric interpretation to the following problem:
Prove that given $a,b,c,d \in \mathbb{Z} \ \ \Rightarrow \ \ \exists \ u,v, \in \mathbb{Z} \ \ s.t.$ $$(a^{2} + b^{2})(c^{2} + d^{2}) = u^{2} + v^{2}$$
I have mentioned my solution to the problem using basic tools from my complex analysis class.
We write the statement as the product of complex conjugates to prove this theorem:
$a^{2} + b^{2} = a^{2} - b^{2} (-1) = a^{2} - b^{2} i^{2}$
$= a^{2} - (bi)^{2} = a^{2} - (ab) i + (ab) i - (bi)^{2}$
$= a(a-bi) + bi (a-bi)$
$= (a+bi)(a-bi)$
Similarly, we repeat the procedure for $c^{2} + d^{2}$ to get the expression:
$(c+di)(c-di)$
Taking the product:
$= [(a+ bi)(a- bi)] [(c+di)(c-di)]$
$= [(a+bi)(c+di)][(a-bi)(c-di)]$
The numbers in both terms are complex conjugates So, we can express them in terms of $u$ and $v$. This gives us: $(u+vi)(u-vi)$
$\therefore u + vi = (a +bi)(c+di)$
$\therefore a(c+di) + bi(c+di)$
$\therefore (ac) + (ad)i + (bc)i + (bd)(i^{2})$
$\therefore (ac) + (ad)i + (bc)i + (bd)(-1)$
$\therefore (ac - bd) + (ad + bc)i \equiv u + vi$
We we are left with expressions to determine u and v for all integers a,b,c,d.
$u = ac - bd$
$v = ad + bc$
These satisfy our original statement. For example:
$(2^{2} + 3^{2})(4^{2} + 5^{2}) = (4+9)(16+25) = 533$
$u = (2)(4) - (3)(5) = 8 - 15 = -7$
$v = (2)(5) + (3)(4) = 10 + 12 = 22$
$u^{2} + v^{2} = (-7)^{2} + (22)^{2} = 533$
Note that during the process of this proof, we were taking products of two complex numbers. A geometric interpretation of that is shown below (reproduced from Complex Analysis. Bak, Newman). However, given that this is a number theoretic problem and the complex numbers cancel out to give integers, do complex numbers still help us give a geometric explanation?
• This result actually holds over general commutative rings, not just the integers, and the proof just uses the fact that we can pick $u=ac-bd$ and $v=ad+bc$ and the equation will be satisfied. Complex numbers were not used in an essential way. This doesn't answer your question though. – Matt Samuel Jan 28 '16 at 0:12
I. Algebraic preliminaries. Laplace's Identity for 3D vectors $\vec V, \vec W$ states that cross products and dot products are related by $$(1) \qquad ||\vec V||^2\ ||\vec W||^2 = ||\vec V \times \vec W||^2 + ||\vec V \cdot \vec W||^2$$
It can be proved by brute force algebraically.
Specializing to the case that both vectors lie in the plane, by setting
$\vec V= <a,b,0>, \vec W=<c,d,0>$ proves what you wanted. In more detail, if $\vec V= <a,b,0>$ and $\vec W= <c,d>$ we get $(a^2+ b^2)(c^2+d^2) = ||\vec V||^2 ||\vec W||^2 =$ sum of two squared terms, namely $||\vec V\cdot \vec W||^2 + ||\vec V \times \vec W||^2$. That's the algebraic method.
The algebraic structure of this identity (its scaling properties) makes it clear that if is true for unit-length vectors it must be true in general.
II. Geometric method. Now we can prove Laplace's Identity geometrically, at least in the special case where $V$ and $W$ are unit vectors. By the parallelogram law of vector addition, the points $0, V, W, V+W$ form the vertices of a rhombus, and the diagonals $A=V-W$ and $B=V+W$ are therefore orthogonal vectors. (This orthogonality can be checked easily by taking the dot product of $A$ and $B$, checking that you get zero). Next solve for $V$ and $W$ in terms of two orthogonal components $A$ and $B$. For example, $V=\frac{1}{2} (A+B)$ After a rotation of space we can assume everything lies in the plane, and $A$ and $B$ are parallel to the $x$ and $y$ axes: $A=<a,0,0>, B= <0, b,0>$. Then $2V= <a,b,0>$ and $2W= <-a,b,0>$. Then you can verify fairly easily that the dot and cross products behave as desired.
III. Generalization to higher dimensions. The cross product of two vectors is not defined for vectors in higher dimensional Euclidean space (the cross product is no longer a vector). However there is a natural extension of the cross product to higher dimensions, called the wedge product $\vec V\wedge \vec W$. (Basically it is collection of all ${{n}\choose {2}}$ possible $2\times 2$ sub determinants extracted from the $2\times n$ matrix whose two rows are $\vec V$ and $\vec W$.) Then Laplace's Identity states that $||\vec V||^2 ||\vec W||^2 = (\vec V\cdot \vec W)^2 + ||\vec V \wedge \vec W||^2$.
The complex explanation of this doesn't depend on how the angles behave in complex multiplication. It is merely the observation that for complex numbers, $|wz| = |w||z|$, or by squaring, $|wz|^2$ = $|w|^2|z|^2$. Combine this with the fact that if two complex numbers have integers for both real and imaginary parts (such complex numbers are called Gaussian Integers), then so does their product, and you have the result.
To make this a geometric picture, then you need to explain both multiplying moduli and preservation of Gaussian Integers under multiplication geometrically. The picture you provided gives a geometric interpretation of how the arguments add when you multiply, but that isn't important here. It doesn't provide a good geometric understanding that the moduli multiply, which is what is needed.
I am not personally aware of a good way of showing that moduli multiply geometrically. Perhaps there is one, but personally I think you would be better off looking elsewhere for a strictly geometric interpretation of the proof. |
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Is 101 a Prime Number?
## 101 is a prime number or not?
For a number to be classified as a prime number, it has to have exactly two factors. To understand whether 101 is a prime or a composite number, it is important to find its factors. The answer to whether 101 is a prime number is: “yes, 101 is a prime number”.
Lesson Highlights:
Is 101 a prime number? Yes, 101 is a prime number. What are the factors of 101 Factors of 101 are 1 and 101. Why 101 is a prime number? 101 is prime since it has only 2 factors i.e. 1 and 101 itself.
## Why 101 is a Prime Number?
101 is a prime number since it has only two factors i.e. 1 and 101 itself. Since it has exactly two factors, it satisfies the definition of prime numbers. To recall, a prime number is a positive whole number which has exactly two factors or divisors.
Check for Yourself: Prime Numbers Calculator
### How to Confirm 101 is Prime or Not?
To check whether 101 is prime or not, the steps mentioned in how to find prime numbers have to be followed. For 101, follow the following steps to determine why is it a prime number.
• Step 1: Check whether the units digit is either 0, 2, 4, 6 and 8. In this case, it is not.
• Step 2: Check is the sum of digits in 101 are divisible by 3. Here, 1+0+1=2 which is not divisible by 3
• Step 3: Take the square root of 101 which is √101 = 10.04
• Step 4: Check the divisibility of 101 with numbers below 10.
There are no numbers below 10 which divide 101 perfectly and hence, it can be classified as a prime number.
Another way of finding whether a number is prime or not is by factorization. By knowing the factors of a number, it can be easily determined whether a number is prime or not. For 101, there are only two factors which are 1 and itself.
### How Can 101 be Classified?
Among the various classification of numbers, some of the categories in which 101 can be classified are:
• A positive integer
• A natural number
• A whole number
• A prime number
• An odd number
• A rational number
Also Check: How to Find Prime Numbers
### Similar Questions:
Stay tune with BYJU’S and get more such maths articles on various topics. Also, download BYJU’S- The Learning App to access numerous maths video lessons and learn in a more personalized, engaging and efficient way. |
# How do you prove Sec(x) - cos(x) = sin(x) * tan(x)?
Jul 10, 2016
Knowing that $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ and $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$, rewriting the equation yields
$\frac{1}{\cos} \left(x\right) - \cos \frac{x}{1} = \sin \left(x\right) \cdot \left(\sin \frac{x}{\cos} \left(x\right)\right)$
Rewriting the left fraction by using the property
$\frac{a}{b} - \frac{c}{d} = \frac{a d - b c}{b d}$
and simplifying the right side of the equation yields
$\frac{1 - {\cos}^{2} \left(x\right)}{\cos} \left(x\right) = \frac{{\sin}^{2} \left(x\right)}{\cos} \left(x\right)$
Note that in this case, we can make use of the identity
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$, since $1 - {\cos}^{2} \left(x\right) = {\sin}^{2} \left(x\right)$, giving us
${\sin}^{2} \frac{x}{\cos} \left(x\right) = {\sin}^{2} \frac{x}{\cos} \left(x\right)$, which is true, therefore we have proven that
$\sec \left(x\right) - \cos \left(x\right) = \sin \left(x\right) \tan \left(x\right)$ |
L-10(M-9) torque and rotational inertia We consider the rotation of rigid bodies. A rigid body is an extended object in which the mass is distributed.
Presentation on theme: "L-10(M-9) torque and rotational inertia We consider the rotation of rigid bodies. A rigid body is an extended object in which the mass is distributed."— Presentation transcript:
L-10(M-9) torque and rotational inertia We consider the rotation of rigid bodies. A rigid body is an extended object in which the mass is distributed spatially. Where should a force be applied to make it rotate (spin)? The same force applied at different locations produces different results. 1 AXLE
TORQUE – Greek letter tau To make an object rotate, a force must be applied in the right place. the combination of force and point of application is called TORQUE The lever arm L is the distance from the axis of rotation to the point where the force is applied If the line of action of F passes through the axis of rotation, it produces no torque. Force, F lever arm: L Axis 2
force F in Newtons, N lever arm L in meters, m Torque in units of N m Torque: (Greek tau) Torque = force (F) x lever arm (L) F L 3
Torque example F L What is the torque on a bolt applied with a wrench that has a lever arm: L= 20 cm with a force: F = 30 N? Solution: F L = 30 N (1/5) m = 6 N m For the same force, you get more torque with a bigger wrench the job is easier! 4
Torque wrench 5 A torque wrench is a wrench that applies a calibrated torque to a bolt. It prevents a bolt from being over-tightened and possibly breaking.
Homer attempts to straighten out the leaning tower of Pisa fulcrum lever 6
Net Force = 0, Net Torque ≠ 0 10 N > The net force = 0, since the forces are applied in opposite directions so it will not accelerate. > However, together these forces will make the rod rotate in the clockwise direction. 7
Net torque = 0, net force ≠ 0 The rod will accelerate upward under these two forces, but will not rotate. 8
Balancing torques 10 N 20 N 1 m 0.5 m Left torque = 10 N x 1 m = 10 n m Right torque = 20 N x 0.5 m = 10 N m 9
Equilibrium To ensure that an object does not accelerate or rotate two conditions must be met: – net force = 0 – net torque = 0 this results in the practical 4-1 “ladder rule” 10
When is an object stable? If you can tip it over a bit and it doesn’t fall The object may wobble a bit but it eventually stops and settles down to its upright position. A thinner object is easier to topple An object that is thicker at its base is more stable 11
Why do tall objects tend to fall over Every object has a special point called the center of gravity (CG). The CG is usually in the center of the object. if the center of gravity is supported, the object will not fall over. The lower the CG the more stable an object is. stable not easy to knock over! 12
Condition for stability If the CG is above the edge of the table, the object will not fall off. CG 13
Why makes an object tip over? For the wide object, the dashed line extending from the CG down is to the left of the point of contact; the torque due to the weight tends to rotate the object counterclockwise For the narrow object, the dashed line extending from the CG down is to the right of the point of contact, the torque due to the weight tends to rotate the object clockwise. 14 STABLE UNSTABLE D D CG
Stable structures Structures are wider at their base to lower their center of gravity 15
If the center of gravity is supported, the blocks do not fall over Playing with blocks CG 16
Coin Stack 17
As more stuff is loaded into a semi, its center of gravity moves upward, making it susceptible to tipping over in high winds. High Profile Vehicles wind 18
Rotational Inertia (moment of inertia) symbol I A rigid body is characterized by a parameter called its rotational inertia The rotational inertia of a RB depends on how its mass is distributed relative to the axis of rotation The rotational inertia of a RB is the parameter that is analogous to inertia (mass) for a non- extended object For a RB, the rotational inertia determines how much torque is needed to produce a certain amount of rotational acceleration (spin). 19
rotational inertia examples Rods of equal mass m and length L axis through center axis through end 20 The rod with the axis through the end requires more torque to get it rotating.
How fast does it spin? For spinning or rotational motion, the rotational inertia of an object plays the same role as ordinary mass for simple motion For a given amount of torque applied to an object, its rotational inertia determines its rotational acceleration the smaller the rotational inertia, the bigger the rotational acceleration 21
Big rotational inertia Small rotational inertia Same torque, different rotational inertia spins slow spins fast 22 L = R F = mg F L = mgR
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## Number Series Questions With Solution For SBI Clerk
Number Series Questions With Solution For SBI Clerk.If you are preparing for Bank and other competitive exams, you will come across Number series section on Quantitative Aptitude. If you are preparing for Bank and other competitive exams, you will come across Number series section on Quantitative Aptitude.
Today, we are providing you with Quant Quiz on Number Series based on the latest pattern for your daily practice. This Quant subject quiz based number series is important for other banking exams such as SBI Clerk, SBI PO, IBPS PO, IBPS Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, and other competitive exams.
## Number Series Questions With Solution For SBI PO | SBI Clerk : Set- 27
1.The following numbers form a series. Find the odd one out.
20, 22, 34, 70, 150, 320
A 70
B 34
C 150
D 320
E None of these
D. 320
The series follows the pattern:
20 + (13 + 12) = 22,
22 + (23 + 22) = 34,
34 + (33 + 32) = 70,
70 + (43 + 42) = 150,
150 + (53 + 52) = 300
2.The following numbers form a series. Find the odd one out.
8, 16, 48, 240, 1800, 18480
A 18480
B 1800
C 48
D 240
E None of these
B. 1800
The series follows the pattern:
8 × 2 = 16,
16 × 3 = 48,
48 × 5 = 240,
240 × 7 = 1680,
1680 × 11 = 18480
3.The following numbers form a series. Find the odd one out.
40320, 20160, 13440, 10080, 8064, 6520
A 10080
B 8064
C 13440
D 6520
E None of these
D. 6520
4.The following numbers form a series. Find the odd one out.
-1, 9, 27, 53, 87, 164
A 164
B 53
C 27
D 87
E None of these
A. 164
The series follows the pattern:
(1 × 2) – 3 = -1,
(3 × 4) – 3 = 9,
(5 × 6) – 3 = 27,
(7 × 8) – 3 = 53,
(9 × 10) – 3 = 87,
(11 × 12) – 3 = 129
5.The following numbers form a series. Find the odd one out.
20, 23, 31, 46, 94, 105
A 94
B 46
C 105
D 31
E None of these
A. 94
The series follows the pattern:
20 + 12 + 2 = 23,
23 + 22 + 4 = 31,
31 + 32 + 6 = 46,
46 + 42 + 8 = 70,
70 + 52 + 10 = 105
6.The following numbers form a series. Find the odd one out.
800, 400, 600, 1500, 4750, 23625
A 600
B 23625
C 4750
D 1500
E None of these
C. 4750
The series follows the pattern:
800 × 0.5 = 400,
400 × 1.5 = 600,
600 × 2.5 = 1500,
1500 × 3.5 = 5250,
5250 × 4.5 = 23625
7.The following numbers form a series. Find the odd one out.
3, 63, 195, 399, 675, 1050
A 399
B 1050
C 675
D 195
E None of these
B. 1050
Consecutive odd numbers have been multiplied while skipping an odd number in between.
1 × 3 = 3,
7 × 9 = 63,
13 × 15 = 195,
19 × 21 = 399,
25 × 27 = 675,
31 × 33 = 1023
8.The following numbers form a series. Find the odd one out.
30000, 33000, 36300, 38080, 43923, 48315.3
A 38080
B 48315.3
C 43923
D 36300
E None of these
A. 38080
The series follows the pattern:
30000 × 1.1 = 33000,
33000 × 1.1 = 36300,
36300 × 1.1 = 39930,
39930 × 1.1 = 43923,
43923 × 1.1 = 48315.3
9.The following numbers form a series. Find the odd one out.
121, 101, 85, 73, 69, 61
A 69
B 85
C 73
D 61
E None of these
A. 69
The series follows the pattern:
02 + 112 = 121,
12 + 102 = 101,
22 + 92 = 85,
32 + 82 = 73,
42 + 72 = 65,
52 + 62 = 61
10.The following numbers form a series. Find the odd one out.
-7, -23, -55, -109, -191, -310
A -109
B -310
C -55
D -191
E None of these
B. -310
The series follows the pattern:
12 – 23 = -7,
22 – 33 = -23,
32 – 43 = -55,
42 – 53 = -109,
52 – 63 = -191,
62 – 73 = -307 |
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# Is ${\left( {10} \right)^{1000}} - 1$ divisible by both $9$ and $11$ ?
Last updated date: 22nd Feb 2024
Total views: 338.4k
Views today: 6.38k
Answer
Verified
338.4k+ views
Hint: In this type of question we will try to observe its pattern which it follows. Now if we talk about ${\left( {10} \right)^n}$then we know that ${10^1} = 10,{10^2} = 100,{10^3} = 1000$. We observe that the number which is in power that no. of zeroes are written after digit 1. So, in that way we can write ${\left( {10} \right)^n} = 1000...................n{\text{ }}terms$. Now, in the same way we can write ${\left( {10} \right)^{1000}}$ that is after digit 1, $1000$ zero must be written. Then we have to subtract 1 from it. So
$\begin{array}{c} {\left( {10} \right)^{1000}} = {10^1} \times {10^2} \times {10^3}...............................{\text{ }}upto{\left( {10} \right)^{1000}}\\ = 10000.......................................000 = 1001{\text{ }}digits \end{array}$
Now, subtract it by 1 we will get $999999.......upto{\text{ }}1000{\text{ }}digits$
Then we can check the divisibility of $9{\text{ }}and{\text{ }}11$ and can reach upto our answer.
Step by step Solution:
So, applying above concept we will get
$\begin{array}{c} {\left( {10} \right)^{1000}} = {10^1} \times {10^2} \times {10^3}...............................{\text{ }}upto{\left( {10} \right)^{1000}}\\ = 10000.......................................000 = 1001{\text{ }}digits \end{array}$
Now, subtract it by 1 we will get $1000$digits that is
(i)For divisibility of 9
We can see that each digit is fully divisible by 9 without leaving remainder. So, if we write it as
$\dfrac{{999.........99}}{9} = 111......111 = 1000{\text{ }}digit$, here each digit will get divided.
So we can say that ${\left( {10} \right)^{1000}} - 1$ is divisible by $9$
(ii)Divisibility by 11
As we have seen above ${\left( {10} \right)^{1000}} - 1 = 9999..............99 = 1000{\text{ }}digits$. We can see that 11 divides the first two digits that is 99 fully without leaving any remainder.
So, we can also write $99999.........99 = as{\text{ }}500{\text{ }}pairs{\text{ }}of{\text{ }}99{\text{ }}to{\text{ }}get$
Now, dividing it by 11, we get
$\dfrac{{999........999}}{{11}} = 909090........09$
0 comes in between because, after dividing the first two digits we need to add zero to get the next two digits for division at the same time.
So, from above calculations we can write ${\left( {10} \right)^{1000}} - 1{\text{ }}is{\text{ }}divisible{\text{ }}by{\text{ }}both{\text{ }}9{\text{ }}and{\text{ }}11$.
Note:
One more method to check divisibility by 9 is that the total sum of the digits of the number must be a multiple of 9 or we can say that it must be fully divisible by 9 without leaving remainder.while expanding any power we must write all digits very carefully. |
### Online Programming Contest, 5-6 February 2005
Solution to Problem 2: Critical Intersections
```This problem is just a cloaked version of a well known problem
about graphs. Let us first see, how to model this problem via
graphs.
Let us recall, from the solution to the "The Great Escape
Problem" of October 2004, the basic definitions regarding graphs:
A graph, say G, consists of two sets, a set of vertices or nodes
(V) and a set of edges (E). The set of vertices could be given
finite set. Each edge is a set consisting of two vertices. We
may interpret the set of edges as describing which pairs of
vertices are "neighbours".
Example: Here is a graph G = (V,E) with 7 vertices and 7 edges:
V = { 1,2,3,4,5,6,7 }
E = { {1,2},{1,3},{2,3},{3,4},{5,6},{5,7},{6,7} }
It is often convenient to draw graphs with boxes or circles to
represent the nodes and lines/curves connecting boxes/circles to
denote edges:
--- --- ---
| 1 |---------| 2 | | 7 |
--- /--- /---\
| / / \
| /-----/ / \
| / / \
---/ --- ---/ \ ---
| 3 |--------| 4 | | 5 |------------| 6 |
--- --- --- ---
Note that the placement of the boxes or the style of drawing the
edges are irrelevant. The "real" graph is given by the two sets V
and E and the picture is just to help us think.
A path in a graph is a sequence of vertices v1,v2,...,vk such
that {v1,v2}, {v2,v3} ... {v{k-1}vk} are edges. For example,
1,2,3,4 is a path in the above graph. The length of a path is the
number of edges in the path. For example, the length of the path
1,2,3,4 is 3.
We say that two vertices u and v are connected if there is a path
u=v_1, v_2, ..., v_k = v. In this case we say that v is
reachable from u (and u is reachable from v, since v_k v_{k-1}
... v_1 is a path from v to u).
We can think of the intersections and roads of Siruseri as a
graph. The vertices of this graph are the intersections of
Siruseri and the edges are the streets. Thus, an intersection is
reachable from another if and only if it is reachable in the
graph theoretic sense (described above).
Notice that we have been assured that initially every
intersection is reachable from every other intersection. A graph
in which every vertex is reachable from every other vertex is
called a "connected graph".
(The example graph defined above is not connected, since there is
no path from vertex 1 to vertex 7, for example, and thus it
cannot represent the intersections and streets of Siruseri.)
What is a critical intersection? Recall, that an intersection Y
is critical if there are two other intersections X and Z such
that all the routes from X to Z go through Y. (Since streets are
bidirectional, it follows that all the routes from Z to X also go
through Y). In the language of graphs, a vertex Y is a critical
intersection if there are vertices X and Z such that all the
paths from X to Z go through Y (i.e. Y appears in any path
beginning at X and ending in Y). Such vertices are called
"Articulation Points" in the language of graphs. Our aim is to
count the number of articulation points in the graph representing
the intersections and roads of Siruseri.
So how do we find whether a vertex Y in a graph G is an
articulation point? Suppose we drop Y (and all the edges incident
on it) from G and get the graph
G-Y = ( V - {Y}, E - {{X,Y} | X in V} ).
If Y is an articulation point then, G-Y is no longer a connected
graph. Conversely, if Y is any vertex in a connected graph G
such that G-Y is not connected then Y is an articulation point of
G. Thus, to determine whether Y is an articulation point, it
suffices to check if the graph G-Y is connected.
This can be done, for example, by the breadth-first search
algorithm described in the solution to the "The Great Escape"
problem from October 2004 problem set in time proportional to N*N
where N is the number of vertices. We can repeat this for each Y
to determine if it is an articulation point. The overall time
taken would thus be N*N*N.
We now describe a different algorithm called depth-first search
algorithm to explore the vertices of a graph and determine
whether it is connected. Like breadth-first search it takes time
proportional to N*N to determine if the graph is connected
(actually time proportional to the number of edges, just like
breadth-first search, but we leave that discussion for later.)
However, it is easier to implement than the breadth-first search
and has other interesting properites.
Recall that the breadth-first search algorithm visits the
vertices of the graph as follows: visit every neighbour of the
current vertex, then visit all the neighbours of the neighbours
and so on. It unravels the graph level by level --- first the
vertex itself, then the vertices at distance 1, then the vertices
at distance 2 and so on. On the other hand, the depth-first
search explores the graph by going as far as possible via a
picked neighbour (cutting through many levels) before
backtracking and exploring other neighbours.
The depth-first search algorithm is rather easy to understand and
here is its code:
============================================================================
/* Mark all vertices as unvisited */
for each vertex v in G {
Visited[v] = 0
}
dfs(u) /* start the dfs algorithm at some vertex u */
/* if the graph is connected then the dfs would have visited
all the vertices. Otherwise there would be unvisited vertices */
if there is vertex v with Visited[v]=0 then {
print "Graph Not Connected"
}
else {
print "Graph is Connected"
}
/* The dfs algorithm */
dfs(v) {
if (Visited[v]=1) then return /* v has been visited in the past*/
Visited[v] = 1 /* visit v */
DFSNo[v] = curr /* v is the curr'th vertex visited */
curr = curr + 1
for each neighbour w of v do {
dfs(w)
}
}
--------------------------------------------------------------------------
To illustrate how this algorithm works, here is a graph and a
description of a run of the dfs algorithm on this graph:
--- --- ---
| A |---------| B |------| F |
--- /--- ---
| /
| /-----/
| /
---/ --- ---
| C |--------| D |--------| E |
--- --- ---
Suppose we call dfs at the vertex (C) then, the sequence of calls
placed inside square parantheses)
dfs(C)
|
|-------> dfs(A)
| |
| |--------> dfs(B)
| | |
| | |-------> [dfs(C)]
| | |-------> [dfs(B)]
| | |-------> dfs(F)
| | |
| | |-------> [dfs(B)]
| |
| |--------> [dfs(C)]
|
|-------> [dfs(B)]
|
|
|-------> dfs(D)
|
|--------> dfs(E)
| |
| |-------> [dfs(D)]
|
|---------> [dfs(C)]
The order in which the vertices are visited is C, A, B, F, D,
E. Starting at the vertex C and having picked the neighbour A,
the dfs visited all the vertices reachable via A (namely A, B,
and F) before returning to explore the graph via the other
neighbour \$D\$.
Notice, that each dfs(v) is called as many times as there are
edges incident on v. Thus each vertex is examined as many times
as its degree and thus the overall running time is proportional
to the number of edges in G. However, that is true only if we can
examine each neighbour of a vertex in time proportional to its
degree. However, if we maintain the graph as an adjacency matrix
then this algorithm will take time proportional to N*N. Thus the
overall time taken to find the articulation points is N*N*N.
(This will get 100% marks for the range of inputs mentioned in
the problem statement.)
Exercise: Try and write a iterative version of the dfs algorithm
(i.e. try to program it without recursion.) Hint: Recall that
the vertices were maintained in a queue in the BFS
algorithm. Here you would need a "stack". i.e. an array into with
vertices are inserted and removed where the element removed is
the last one to be inserted.
It turns out that a single DFS of the graph G can determine all
the articulation points in the graph! However, we postpone a
description of that idea to a different day.
```
Copyright (c) IARCS 2003-2018; Last Updated: 17 Jul 2005 |
# a and b multiplied
Visit the d/dx art shop
When we first begin to learn about multiplication it can often be a help to get our the box of wooden bricks to explore the idea that multiplying 2 by 3 is the same essentially as building a rectangle of three rows of two bricks like this:
Then we probably move on and relegate the insight to the past, which is a pity, since it has a lot to offer as we move more deeply into multiplication involving algebra. Anyone who has been fortunate enough to brush up against Montessori education will know that very young children can quickly become comfortable with the idea of calculations like ${{(a+b)}^{2}}$ using flat tiles or even ${{(a+b)}^{3}}$ using colourful blocks to build a cube.
But why bother? After a while, most students become familiar with the idea that multiplying out ${{(a+b)}^{2}}$ results in ${{a}^{2}}+2ab+{{b}^{2}}$, so what is the point of messing about with tiles or drawings. Well for me at least, the answer is twofold: firstly I love to see the parallels between the realm in which mathematics operates and the physical world, especially because so often the mathematics casts new light on why the world is as it is. The second reason is almost the opposite, that finding ways to express something on paper often casts light back on the way the mathematics works.
That’s the case here, where the three shapes in the design cast light on three simple equations which are useful to remember as building blocks for more complicated work:
• $\displaystyle {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
• $\displaystyle (a+b)\times (a-b)={{a}^{2}}-{{b}^{2}}$
• $\displaystyle {{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
The first shape in the design illustrates equation 1), with the value of a represented by the brown bar and b by the yellow bar – the bars are only there to show the length of a and b. There is no particular significance to the lengths chosen other than to create a pleasing design.
It shouldn’t be difficult to see that what we have here is a square with each side equal to the lengths of a and b added together. And by connecting the places where a and b join we end up with two squares which represent ${{a}^{2}}$ and ${{b}^{2}}$ plus two rectangles which each have one side as a and one side as b, so their size is $a\times b$ or, in other words, $ab$. So there in one simple shape you have the physical reality behind the equation $\displaystyle {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
The second shape is little more messy, but there’s a good reason.
When you’re multiplying together terms which involve minus signs you often end up with a result in which the same value is present in different quantities as positive and negative items. In this case, going the long way around multiplying $\displaystyle (a+b)\times (a-b)$, where we actually multiply all the terms one by one gives the result $\displaystyle {{a}^{2}}-ab+ab-{{b}^{2}}$. Expressing that as $\displaystyle {{a}^{2}}-{{b}^{2}}$is quite correct but we lose some information about the way the thing works. So the second design is an attempt to show the correct final result (which is the area surrounded by the dotted line) without losing those pluses and minuses. What the design says to us is that if we were to take the whole rectangle $\displaystyle a\times (a+b)$ and divide it up as shown (stay calm) then we end up with the following rectangles: $\displaystyle {{a}^{2}}$and $\displaystyle ab$. to get to $\displaystyle (a+b)\times (a-b)$ we have to subtract $\displaystyle ab$and $\displaystyle {{b}^{2}}$, giving us $\displaystyle {{a}^{2}}-ab+ab-{{b}^{2}}$which is the same as $\displaystyle {{a}^{2}}-{{b}^{2}}$. Once again, expressing the calculation as a picture gives us a better idea of what is going on than simply writing down $\displaystyle (a+b)\times (a-b)={{a}^{2}}-{{b}^{2}}$.
(Just as an aside, even if we had drawn $\displaystyle {{a}^{2}}-{{b}^{2}}$ in the shorthand form
all the information we need is actually there, since the rectangles involved are $\displaystyle {{(a-b)}^{2}}$ and two copies of $\displaystyle b\times (a-b)$. Multiplying all those out we get $\displaystyle {{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and twice $\displaystyle b\times (a-b)$= $2ab-2{{b}^{2}}$. Unsurprisingly, adding it all together gives ${{a}^{2}}-2ab+2ab+{{b}^{2}}-2{{b}^{2}}={{a}^{2}}-{{b}^{2}}$.)
If you’ve coped with everything so far, the final shape in the design should be obvious. It shows that ${{(a-b)}^{2}}$ is actually made up of ${{a}^{2}}$ with two copies of $b\times (a-b)$and one ${{b}^{2}}$ subtracted. Multiplying all that out and adding them all together gives ${{a}^{2}}-2\times ab-2\times (-{{b}^{2}})-{{b}^{2}}={{a}^{2}}-2ab+2{{b}^{2}}-{{b}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$which is correct.
With a little imagination we can get straight to the final form if we recognize that there are actually two ab rectangles meeting in the bottom right corner. If we take them both at full value to give us $-2ab$ then we have taken off too much and we have to add back the square in the corner, which is ${{b}^{2}}$.
This design is reminder that expressing equations on paper or in physical shapes can often demystify them. Perhaps you’d like to try ${{(a+b+c)}^{3}}$ on paper or get out the woodworking tools (or the 3D printer) and produce one of those cubes beloved of children in Montessori nurseries. |
Word Lesson: Ratios
In order to solve problems involving ratios, you should be able to:
• work with fractions
• simplify fractions
A ratio is a way of expressing a relationship between two numbers that can be expressed in the following three common ways:
as a fraction
with a colon (:)
using the word “to
Suppose we are told that there are three nurses for every ten patients at a hospital. The three ways this ratio of nurses to patients can be written are:
• as a fraction:
• with a colon: 3:10
• with “to”: 3 to 10
Notice that in each case, the number of nurses is written first (or in the numerator of the fraction). This is because the problem stated that we were to express the ratio of nurses to patients. Typically the first thing mentioned in a ratio problem is what is listed first.
If the problem had asked us to express the number of patients to nurses, each of our ratios would be written in the opposite order:
• as a fraction:
• with a colon: 10:3
• with “to”: 10 to 3
When working with ratios, it is extremely important to pay attention to order. Read each problem carefully to find out what is being asked.
Examples
At the Technology Convention there were 17 males for every 12 females. Express each of the following relationships as a ratio in fraction form: a. the number of males to females attending the Technology Convention.b. the number of males to the total number of people present at the Technology Convention. What is your answer?
Out of every 100 students at a two-year college, 52 are taking a math class. Express each of the following relationships as a ratio in fraction form: a. the number of students taking a math class to the total number of students.b. the number of students not taking a math class to the number of students taking a math class. What is your answer?
Examples
Cameron went to a restaurant where there were 80 seats for non-smokers and 10 seats for smokers. Express the ratio of the number of seats for smokers to the total number of seats. What is your answer?
A waiter at a local restaurant notes that during the day his customers ordered 3 diet drinks for every 2 non-diet drinks. Express the ratio of diet drinks to non-diet drinks as a fraction. What is your answer?
Being able to write correct ratios is important when comparing two quantities. It is also an important and essential skill in solving proportions where more than one ratio is involved. Make sure that you completely understand ratios before going on to work with proportions.
S Taylor
Show Related AlgebraLab Documents |
# Video: Using the Sine Rule to Calculate Unknown Lengths in a Triangle
The scale of a map is 1 cm : 1.35 km. The position of three towns on a map form a triangle. Towns B and C are 17 cm apart, and the angles of towns A and B are 83ยฐ and 65ยฐ respectively. Find the actual distance between towns A and B and between towns A and C giving the answer to the nearest kilometre.
04:23
### Video Transcript
The scale of a map is one centimetre to 1.35 kilometres. The position of three towns on a map form a triangle. Towns B and C are 17 centimetres apart, and the angles of towns A and B are 83 degrees and 65 degrees, respectively. Find the actual distance between towns A and B and between towns A and C, giving the answer to the nearest kilometre.
There are two things we should do before we perform any tricky calculations. The first is to calculate the measure of the angle at C. Since we know that angles in a triangle add to 180 degrees, we can find the measure of the angle at C by subtracting 83 and 65 from 180. 180 minus 83 plus 65 is 32 degrees.
Next, since weโre being asked to find the actual distances between the towns, we should convert the scale measurement of the distance between towns B and C into the actual measurement. The scale is one centimetre to 1.35 kilometres. So we can calculate the actual distance between towns B and C by multiplying 17 by 1.35. That gives us a distance of 22.95 kilometres.
Next, letโs fully label our triangle. We know that the side opposite the angle A can be denoted as lowercase ๐, the side opposite angle B is lowercase ๐, and the side opposite angle C is lowercase ๐. So we have a non-right-angled triangle, for which we know all three angles and the length of one of its sides.
We can use the law of sines to calculate the lengths of the two missing sides. We know that weโre not going to use the law of cosines since that requires at least two known side lengths. The law of sine says ๐ over sin A equals ๐ over sin B which equals ๐ over sin C. Alternatively, that can be written as sin A over ๐ equals sin B over ๐, which equals sin C over ๐.
We can use either of these equations. However, since weโre trying to find the length of the missing sides, weโll use the first equation. By choosing this one, weโll be minimizing the amount of rearranging we need to do to solve our equations.
Letโs start by calculating the distance between towns A and B. Weโve called that lowercase ๐. At this stage, weโre not interested in the measure of the angle at B nor the side ๐. So weโre going to use these two parts of the equation: ๐ over sin A equals ๐ over sin C.
Substituting what we know into this formula gives us 22.95 over sin of 83 equals ๐ over sin of 32. Notice that weโve used the actual distance between the towns rather than the scale measurement. We can solve this equation by multiplying both sides by sin of 32. ๐ is, therefore, equal to 22.95 over sin of 83 degrees multiplied by sin of 32, which is 12.252. Correct to the nearest kilometre, the distance between towns A and B then is 12 kilometres.
Next, letโs calculate the distance between towns A and C. Weโve called that lowercase ๐. This time, weโll use ๐ over sin A equals ๐ over sin B. We could have chosen to use ๐ over sin C instead of ๐ over sin A. However, that would have involved using some rounded answers, which we want to avoid wherever possible.
This time, when we substitute the values into our equation, we get 22.95 over sin of 83 degrees equals ๐ over sin of 65 degrees. And weโre gonna solve in the exact same way. Weโre gonna multiply by sin of 65 degrees. That gives us 22.95 over sin of 83 multiplied by sin of 65, which is 20.955.
Correct to the nearest kilometre, the distance between towns A and C is 21 kilometres.
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### Theory:
Theorem:
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Explanation:
Let us take the composite number $$N$$.
Decompose the number $$N$$ into the product of primes.
Here, the number $$N = x_1 \times x_2$$. But, both $$x_1$$ and $$x_2$$ are again a composite number. So, factorise it further to obtain a prime number.
The prime factors of $$x_1 = p_1 \times p_2$$.
The prime factors of $$x_2 = p_3 \times p_4$$.
We get, $$N = p_1 \times p_2 \times p_3 \times p_4$$ where $$p_1$$, $$p_2$$, $$p_3$$ and $$p_4$$ are all prime numbers.
If we have repeated primes in a product, then we can write it as powers.
In general, given a composite number $$N$$, we factorise it uniquely in the form $N={{p}_{1}}^{{q}_{1}}×{{p}_{2}}^{{q}_{2}}×{{p}_{3}}^{{q}_{3}}×...×{{p}_{n}}^{{q}_{n}}$ where ${p}_{1},\phantom{\rule{0.147em}{0ex}}{p}_{2},\phantom{\rule{0.147em}{0ex}}{p}_{3},\phantom{\rule{0.147em}{0ex}}...\phantom{\rule{0.147em}{0ex}}{p}_{n}$ are prime numbers, and ${q}_{1},\phantom{\rule{0.147em}{0ex}}{q}_{2},\phantom{\rule{0.147em}{0ex}}{q}_{3},\phantom{\rule{0.147em}{0ex}}...\phantom{\rule{0.147em}{0ex}}{q}_{n}$ are natural numbers.
Thus, every composite number can be expressed as a product of primes apart from the order.
Example:
Consider a composite number $$26950$$.
Let us factor this number using the factor tree method.
The prime factor of $$26950$$ $$=$$ $$2 \times 5 \times 5 \times 7 \times 7 \times 11$$.
That is, $$26950 = 2 \times 5^2 \times 7^2 \times 11$$.
Here, a composite number $$26950$$ is written as a product of prime numbers.
If we change the order of the prime numbers, then also the answer will be the same composite number.
We can write $$26950 = 2 \times 7^2 \times 5^2 \times 11$$ or $$26950 = 11 \times 7^2 \times 5^2 \times 2$$.
Thus, the prime factorisation of a natural number is unique, except for the order of its factors.
Significance of the fundamental theorem of arithmetic
1. If a prime number $$p$$ divides $$ab$$, then $$p$$ divides either $$a$$ or $$b$$. That is, $$p$$ divides at least one of them.
Example:
Let us take a prime number $$3$$ divides $$5 \times 6$$.
$\frac{5×6}{3}$
Here, $$3$$ cannot divide $$5$$, but it divides $$6$$.
That is, a prime number $$p$$ divides at least one of them.
2. If a composite number $$n$$ divides $$ab$$, then $$n$$ neither divide $$a$$ nor $$b$$.
Example:
Let us take a composite number $$4$$ divides $$2 \times 6$$.
$\frac{2×6}{4}$
Here, $$4$$ neither divide $$2$$ nor divide $$6$$. But, it divides the product of $$2 \times 6 = 12$$.
Thus, if a composite number $$n$$ divides $$ab$$, then $$n$$ neither divide $$a$$ nor $$b$$.
Important!
Recall:
HCF $$=$$ Product of the smallest power of each common prime factor in the numbers.
LCM $$=$$ Product of the greatest power of each prime factor involved in the numbers. |
# Newton-Raphson Method
A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022
## Newton-Raphson Method
The Newton-Raphson Method is a different method to find approximate roots. The method requires you to differentiate the equation you’re trying to find a root of, so before revising this topic you may want to look back at differentiation to refresh your mind.
A Level
## Using the Newton-Raphson Method
Finding roots of an equation in the form $f(x)=0$, requires you to find $f'(x)$ and then use the following formula:
$\Large{x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}}$
This method iteratively finds the $x$-intercept of the tangent to the graph of $f(x)$ at $x_n$ and then uses this value as $x_{n+1}$.
A Level
## Why the Newton-Raphson Method Can Fail
• Similar to other iteration formulas, if your starting point of $x_0$ is too far away from the actual root, the Newton-Raphson method may diverge away from the root.
• The Newton-Raphson method can also fail if the gradient of the tangent at $x_n$ is close or equal to $\textcolor{red}{0}$. This is shown in the diagram below, where the tangent has a gradient very close to $0$, so the point where it meets the $x$-axis will be very far away from the root, so the sequence of iterations may diverge.
• Furthermore, if the tangent at a point on $f(x)$ is horizontal, i.e. $x_n$ is a stationary point, then the Newton-Raphson method will fail. This is because the tangent will never meet the $x$-axis, so there will be no further iterations. In addition to this, the tangent is horizontal when $f'(x)=0$, so the formula would not work as you cannot divide by $0$.
A Level
A Level
## Example: The Newton Raphson Method
Find a root of the equation $x^2-8x+11=0$ to $5$ decimal places using $x_0=6$
First we need to differentiate $f(x)=x^2-8x+11$:
$f'(x)=2x-8$
Substituting this into the Newton-Raphson formula:
$x_{n+1}=x_n-\dfrac{x^2-8x+11}{2x-8}$
Starting with $x_0=6$:
$x_1=6-\dfrac{6^2-8(6)+11}{2(6)-8}=6.25$
Using the formula again to find the following iterations:
$x_2=6.25-\dfrac{6.25^2-8(6.25)+11}{2(6.25)-8}=6.236111111$
$x_3=6.236111111-\dfrac{6.236111111^2-8(6.236111111)+11}{2(6.236111111)-8}=6.236067978$
$x_4=6.236067978-\dfrac{6.236067978^2-8(6.236067978)+11}{2(6.236067978)-8}=6.236067977$
Thus a root of $x^2-8x+11=0$ is $6.23607$ to $5$ decimal places.
A Level
## Example Questions
Firstly we need to differentiate $f(x)=x^3-2x^2-5x+8$
$f'(x)=3x^2-4x-5$
Now we need to apply the Newton-Raphson formula, starting with $x_0=1$:
$x_1=1-\dfrac{1^3-2(1)^2-5(1)+8}{3(1)^2-4(1)-5}=\dfrac{4}{3}$
$x_2=\dfrac{4}{3}-\dfrac{(\dfrac{4}{3})^3-2(\dfrac{4}{3})^2-5(\dfrac{4}{3})+8}{3(\dfrac{4}{3})^2-4(\dfrac{4}{3})-5}=1.362962963$
$x_3=1.362962963-\dfrac{(1.362962963)^3-2(1.362962963)^2-5(1.362962963)+8}{3(1.362962963)^2-4(1.362962963)-5}=1.36332811$
$x_4=1.36332811-\dfrac{(1.36332811)^3-2(1.36332811)^2-5(1.36332811)+8}{3(1.36332811)^2-4(1.36332811)-5}=1.363328238$
So a root of $x^3-2x^2-5x+8=0$ is $1.36333$ to $5$ decimal places.
Firstly, we need to rearrange the equation so it is in the form $f(x)=0$:
$3x\ln{x}-7=0$
Then we need to differentiate $f(x)=3x\ln{x}-7$, to do this we will need to use the product rule:
\begin{aligned} f'(x) &=3\ln{x}+3x\times \dfrac{1}{x} \\ &=3\ln{x}+3 \\ &=3(\ln{x}+1) \end{aligned}
Now we need to apply the Newton-Raphson formula starting with $x_0=2$:
$x_1=2-\dfrac{3(2)\ln{2}-7}{3(\ln{2}+1)}=2.559336473$
$x_2=2.559336473-\dfrac{3(2.559336473)\ln{2.559336473}-7}{3(\ln{2.559336473}+1)}=2.522322342$
$x_3=2.522322342-\dfrac{3(2.522322342)\ln{2.522322342}-7}{3(\ln{2.522322342}+1)}=2.522182638$
$x_4=2.522182638-\dfrac{3(2.522182638)\ln{2.522182638}-7}{3(\ln{2.522182638}+1)}=2.522182636$
So the root of $3x\ln{x}=7$ is $2.522$ to $4$ significant figures.
Differentiating $f(x)=-x^2+x+12$:
$f'(x)=-2x+1$
Substituting in $x=0.5$:
$f'(0)=-2(0.5)+1=0$
Thus, the Newton-Raphson method will fail because you cannot divide by $0$
A Level
A Level
A Level
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Find the remainder when is divided by(i) (ii) (iii) (iv) (v)
Remainder Theorem: If f(x) is a polynomial and it is divided by another polynomial g(x), the remainder of this division equals to the value f(a), where a is the solution of polynomial g(x) = 0
for example: if x2 + 2 is divided by x - 1, then to find remainder.
put x - 1 = 0
x = 1 and now putting this value in x2 + 1, we get 1 + 1 =2 as the remainder.
(i) f(x) = x3 + 3 x2 + 3 x + 1
Now let g(x) = x + 1
So for finding remainder, put g(x) = 0
x + 1 = 0
x = -1
So f(- 1) will be the remainder when f(x) is divided by g(x)
f (- 1) = (- 1)3 + 3 (- 1)2 + 3 (- 1) + 1
f (- 1) = -1 + 3 -3 + 1
f (- 1) = 0
Hence, Remainder = 0
(ii) f(x) = x3 + 3 x2 + 3 x + 1
Now let g(x) = x - 1/2
So for finding remainder, put g(x) = 0
x - 1/2 = 0
x = 1/2
so, f(1/2) will be the remainder when f(x) is divided by g(x)
f(1/2) = (1/2)3 + 3 (1/2)2 + 3 (1/2) + 1
Hence, Remainder = 27/8
(iii) f(x) = x3 + 3 x2 + 3 x + 1
Now let g(x) = x
So for finding remainder, put g(x) = 0
x = 0
So, f(0) will be the remainder when f(x) is divided by g(x)
f(0) = (0)3 + 3 (0)2 + 3 (0) + 1
f(0) = 1
Hence, Remainder = 1
(iv) f(x) = x3 + 3 x2 + 3 x + 1
Now let g(x) = x + Π
So for finding remainder, put g(x) = 0
x + Π = 0
x = - Π
So, f(- Π) will be the remainder when f(x) is divided by g(x)
f(- Π) = (- Π)3 + 3 (- Π)2 + 3 (- Π) + 1
f(- Π) = - Π3 + 3 Π2 - 3Π + 1
- Π3 + 3 Π2 - 3Π + 1 will be the remainder
(v) f(x) = x3 + 3 x2 + 3 x + 1
Now let g(x) = 5 + 2x
So for finding the remainder, g(x) = 0
5 + 2x = 0
x = -5/2
So, f(-5/2) will be the remainder when f(x) is divided by g(x)
f(-5/2) = -27/8 will be the remainder .
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# How to Calculate the Gross Area of a Ceiling
eHow may earn compensation through affiliate links in this story.
Whether ornate or plain, determining the gross area of the ceiling is straightforward.
Gross means "with no deductions," so the gross area of a ceiling is the total ceiling area including lights, extractor fans and all other fittings. It usually matches the gross floor area and is calculated in the same way. When the ceiling area matches the floor it is possible to determine the area by measuring the more accessible floor, however, the method is identical whether the measurements are taken at ceiling or floor level.
Video of the Day
## Step 1
Determine the length and width of the ceiling in a rectangular room by taking measurements from corner to corner. Multiply the length by the width to obtain the gross ceiling area. For example, a room 15 feet long and 10 feet wide has a gross ceiling area of 150 square feet -- 15 times 10 = 150.
## Step 2
Divide irregular shaped ceilings into more than one rectangle. If necessary divide the ceiling into other regular shapes too, such as triangles and circles. Calculating the area of a regular shape is much easier than attempting to calculate that of an irregular shape. Determine the area of each section of the ceiling and then combine them to obtain the gross ceiling area. For example, a ceiling in a rectangular room containing a chimney breast, positioned centrally on one wall, may be divided into three rectangles; the main ceiling and two smaller rectangular ceiling areas on either side of the chimney breast.
## Step 3
Find the area of triangle with the formula: area = (width by length) / 2 and of a circle with the formula: area = pi by radius squared. For example, a semi-circular ceiling area above a window bay with a radius of 5 feet has an area of 39.27 square feet -- 3.1415 ( 5^2) = 78.54, and half of 78.54 is 39.27.
#### Tip
If the floor shape and size match those of the ceiling then it is both easier and safer to measure the floor.
#### Warning
If you use a ladder to reach the ceiling, follow appropriate safety guidelines; don't lean too far and be aware of your center of balance at all times. |
## Projectile Motion
### Learning Outcomes
• Define projectile motion
• Solve a polynomial function that represents projectile motion
• Interpret the solution to a polynomial function that represents projectile motion
Projectile motion happens when you throw a ball into the air and it comes back down because of gravity. A projectile will follow a curved path that behaves in a predictable way. This predictable motion has been studied for centuries, and in simple cases, an object’s height from the ground at a given time, $t$, can be modeled with a polynomial function of the form $h(t)=at^2+bt+c$, where $h(t)$ represents the height of an object at a given time, $t$. Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile’s motion, as in the water in the image of the fountain below.
Parabolic water trajectory in a fountain.
Parabolic motion and its related functions allow us to launch satellites for telecommunications and rockets for space exploration. Recently, police departments have even begun using projectiles with GPS to track fleeing suspects in vehicles rather than pursuing them by high-speed chase [1].
In this section, we will use polynomial functions to answer questions about the parabolic motion of a projectile. The real mathematical model for the path of a rocket or a police GPS projectile may have different coefficients or more variables, but the concept remains the same. We will also learn to interpret the meaning of the variables in a polynomial function that models projectile motion.
### Example
A small toy rocket is launched from a $4$-foot pedestal. The height (h, in feet) of the rocket t seconds after taking off is given by the function $h(t)=−2t^{2}+7t+4$. How long will it take the rocket to hit the ground?
In the next example, we will solve for the time that the rocket reaches a given height other than zero.
### Example
Use the formula for the height of the rocket in the previous example to find the time when the rocket is $4$ feet from hitting the ground on its way back down. Refer to the image.
$h(t)=−2t^{2}+7t+4$
In the following video, we show another example of how to find the time when a object following a parabolic trajectory hits the ground.
1. "Cops' Latest Tool in High-speed Chases: GPS Projectiles." CBSNews. CBS Interactive, n.d. Web. $14$ June $2016$. |
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# Median and Mean
## Median
The median is the second of the three 'm' words which children associate with averages and easily confuse... As with the mode, it's an easy figure to find although it can take a little working out. The median is the middle figure. The one which stands in the middle when you put all the data in order from smallest to largest (or largest to smallest, it really doesn't make any difference!)
For example, if you want to find the median figure from the following set of data, you need firstly to arrange the figures in order.
2 7 6 3 8 9 3
Once arranged from smallest to largest, we get:
2 3 3 6 7 8 9
As there are seven figures, the one in the middle will be the fourth. In this case, it is '6' and this is the median.
The rule for finding the median is easy as long as the number of items in the list is an odd number. In the example, there was a clear 'middle' figure so we could simply state it. However, it is conceivable that a paper may ask for a median figure from an even-numbered set of data. In this case, you need to find the halfway point between the two central figures and give this, rather than list both figures as you would with the mode.
What is the median figure in the following data?
4 5 9 3 9 2
Rearrange the numbers in order:
2 3 4 5 9 9
There are six numbers so the central one doesn't exist; the midway point is between the third and fourth number. Halfway between 4 and 5 is 4.5 ( 4 + 5 divided by 2 ).
The median figure is 4.5
## Mean (Average)
The mean is the posh way of saying the average. It means the total divided by the number of items. Unlike the mode, range and median, the mean will require some proper maths to work out.
Find the mean of the following figures:
6 4 8 10 9 5
There are six terms in the data set. They can be added up together (6 + 4 + 8 + 10 + 9 + 5) to give 42.
The mean is the total divided by the number of items, so 42 ÷ 6 = 7. The mean is 7.
The answer that a child gets should always be looked at in retrospect and compared to the data set. If the average is outside of the range of the figures given, it is wrong. It is impossible to have an answer which is lower than the lowest figure in the series or is higher than the highest figure. Some common sense can save lots of marks and only takes a second to apply! Similarly, if the number is very close to either extreme of the figures, check them carefully.
What is the mean height of five children, whose heights are given below:
1.04m 1.16m 1.30m 1.14m 1.26m
Firstly, as all of the heights are 1.xx we can discount this first digit and just put it back on at the end. We can only do this because all of the figures are 1.xx and if even one of them is different, it is not possible.
So, we can add them all together and divide by the number of heights (5)
Using our short-cut, all we really need to add are 4, 16, 30, 14 and 26. These add up to 90. 90 ÷ 5 = 18 so the average of these figures is 18; if we now replace the 1.xx that we took off to start with, the average height is 1.18m. If this short-cut complicates things for you, simply add all the heights together without amending them, and divide by 5. The answer will be the same. |
# How to Evaluate Expressions with Substitution
Instructor: Laura Pennington
Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.
In this lesson, we will review the definition of an algebraic expression and order of operations. We will then look at how to evaluate algebraic expressions using substitution and practice the process on multiple examples.
## Finding Area
Suppose you are redecorating your house, and you need to know the area of a rectangular rug to know if it will be a good fit for your dining room. You know that to calculate the area of a rectangle, you need to multiply the length of the rectangle by the width. That is, area = l*w, where l = length and w = width. You use a tape measure to measure the rug's dimensions and find that the length is 11 feet and the width is 8 feet. To calculate the area of the rug, all we need to do is plug 11 for l and 8 for w into l*w, and simplify.
We see that the area of the rug is 88ft 2. This will fit perfectly in your dining room!
The process of plugging the length and width into the expression l*w is an example of evaluating an algebraic expression using substitution.
## Algebraic Expressions and PEMDAS
An algebraic expression is an expression containing more than one number, variable, and arithmetic operation. Some examples of algebraic expressions are shown.
We can evaluate an algebraic expression using substitution by simply plugging in values for the variables and simplifying in the appropriate order. That statement probably left you wondering what was meant by 'the appropriate order'. When we evaluate or simplify an algebraic expression, we have an order of operations that we follow. This order is illustrated in the following image:
A great way to remember this order is by remembering the word PEMDAS. As shown in the image, this word's letters each represent an operation, and they are in the correct order.
For example, consider the expression -3(1 + 4) - 82/4. To evaluate this expression, we don't just work left to right like we would read a book, we have to follow our order of operations. That is, we have to go in the order PEMDAS.
First, we address the P in PEMDAS, or the parentheses. Thus, the first thing we want to do is simplify what is in the parentheses to get -3(4 + 1) - 8 2/4 = -3(5) - 8 2/4. We also have a division bar, which is an inclusion, or grouping, symbol. This falls under P in PEMDAS as well, so we want to simplify the numerator and denominator to get -3(5) - 8 2/4 = -3(5) - 64/4.
At this point, there are no exponents, so we can move onto the M in PEMDAS. That is multiplication. We perform any multiplication we see to get -3(5) - 64/4 = -15 - 64/4.
The next letter in PEMDAS is D, which stands for division, so we perform any division in the expression to get -15 - 64/4 = -15 - 16.
There is no addition at this point, so we move onto the S in PEMDAS, which is subtraction. Thus, we have -15 - 16 = -31. This is all illustrated in a compact form in the image.
We see that if we evaluate the expression in the correct order, we get -31. Now that we've looked at the order in which we should simplify or evaluate expressions, let's look at evaluating algebraic expressions using substitution.
## Evaluating Algebraic Expressions Using Substitution
As we said, when we found the area of the rug for your dining room, we evaluated the algebraic expression l*w using substitution. We plugged values in for the variables and then evaluated. In general, to evaluate algebraic expressions using substitution, we follow these steps.
1. Plug in given values for variables in the expression.
2. Use PEMDAS to evaluate the expression.
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# Convergence and Divergence
Convergence and Divergence – Introduction to Series
Convergence and Divergence – Introduction to Series
## Example Questions
### Example Question #61 : Ratio Test
Use the ratio test to find out if the following series is convergent:
Determine the convergence of the series based on the limits.
Solution:
1. Ignore constants and simplify the equation (canceling out what you can).
2. Once the equation is simplified, take .
### Example Question #62 : Ratio Test
Use the ratio test to find out if the following series is convergent:
Determine the convergence of the series based on the limits.
Solution:
1. Ignore constants and simplify the equation (canceling out what you can).
2. Once the equation is simplified, take .
### Example Question #63 : Ratio Test
Use the ratio test to find out if the following series is convergent:
Determine the convergence of the series based on the limits.
Solution:
1. Ignore constants and simplify the equation (canceling out what you can).
2. Once the equation is simplified, take .
### Example Question #64 : Ratio Test
Use the ratio test to find out if the following series is convergent:
Determine the convergence of the series based on the limits.
Solution:
1. Ignore constants and simplify the equation (canceling out what you can).
2. Once the equation is simplified, take .
### Example Question #65 : Ratio Test
Use the ratio test to find out if the following series is convergent:
Determine the convergence of the series based on the limits.
Solution:
1. Ignore constants and simplify the equation (canceling out what you can).
2. Once the equation is simplified, take .
### Example Question #1 : Ratio Test And Comparing Series
Determine if the following series is divergent, convergent or neither.
Divergent
Inconclusive
Convergent
Neither
Both
Convergent
In order to figure out if
is divergent, convergent or neither, we need to use the ratio test.
Remember that the ratio test is as follows.
Suppose we have a series . We define,
Then if
, the series is absolutely convergent.
, the series is divergent.
, the series may be divergent, conditionally convergent, or absolutely convergent.
Now lets apply the ratio test to our problem.
Let
and
Now
Now lets simplify this expression to
.
Since
.
We have sufficient evidence to conclude that the series is convergent.
### Example Question #2 : Ratio Test And Comparing Series
Determine if the following series is divergent, convergent or neither.
Divergent
Convergent
Inconclusive
Both
Neither
Divergent
In order to figure if
is convergent, divergent or neither, we need to use the ratio test.
Remember that the ratio test is as follows.
Suppose we have a series . We define,
Then if
, the series is absolutely convergent.
, the series is divergent.
, the series may be divergent, conditionally convergent, or absolutely convergent.
Now lets apply the ratio test to our problem.
Let
and
Now
.
Now lets simplify this expression to
.
Since ,
we have sufficient evidence to conclude that the series is divergent.
### Example Question #3 : Ratio Test And Comparing Series
Determine if the following series is divergent, convergent or neither.
Both
Neither
Inconclusive
Convergent
Divergent
Divergent
In order to figure if
is convergent, divergent or neither, we need to use the ratio test.
Remember that the ratio test is as follows.
Suppose we have a series . We define,
Then if
, the series is absolutely convergent.
, the series is divergent.
, the series may be divergent, conditionally convergent, or absolutely convergent.
Now lets apply the ratio test to our problem.
Let
and
.
Now
.
Now lets simplify this expression to
.
Since ,
we have sufficient evidence to conclude that the series is divergent.
### Example Question #66 : Ratio Test
Determine if the series converges or diverges:
Neither converges nor diverges.
Converges
Conditionally converges.
Diverges
There is not enough information to determine convergency.
Converges
The ratio test states that if you take the n+1 term of the series and divide it by the n term, and then take the limit as n approaches infinity and if you take the absolute value of your answer and if it less than 1, it converges.
If it is greater than 1, it diverges.
If it is 1, the test is inconclusive.
The n+1 term is .
Note that you are substituting n+1 for n and so you will distribute the term by 2 and 3 respectively. Dividing the n+1 term by the n term gives you the following: , which when multiplied out gives us the following:
.
To use the ratio test, we must take the limit of this term as n approaches infinity. From inspection, we can see that the denominator is increasing much faster than the numerator (there are more n terms) and so the limit as n appraoches infinity is 0. Since the absolute value of 0 is less than 1, the series converges.
### Example Question #67 : Ratio Test
Determine what the limit is using the Ratio Test.
To determine what this series converges to
we need to remember the ratio test.
The ratio test is as follows.
Suppose we a series . Then we define,
.
Now lets apply this to our situtation.
Let
and
Now
We can rearrange the expression to be
Now lets simplify this.
When we evaluate the limit, we get.
.
Thus the limit of this series is using the ratio test.
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# Factoring Trinomials: The ac Method
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1 6.7 Factoring Trinomials: The ac Method 6.7 OBJECTIVES 1. Use the ac test to determine whether a trinomial is factorable over the integers 2. Use the results of the ac test to factor a trinomial 3. For a given value of x, evaluate f(x) before and after factoring The product of two binomials of the form ( x )( x ) will be a trinomial. In your earlier mathematics classes, you used the FOIL method to find the product of two binomials. In this section, we will use the factoring by grouping method to find the binomial factors for a trinomial. First, let s look at some factored trinomials. Example 1 Matching Trinomials and Their Factors Determine which of the following are true statements. (a) x 2 2x 8 (x 4)(x 2) This is a true statement. Using the FOIL method, we see that (x 4)(x 2) x 2 2x 4x 8 x 2 2x 8 x 2 6x 5 (x 2)(x 3) Not a true statement, because (x 2)(x 3) x 2 3x 2x 6 x 2 5x 6 (c) x 2 5x 14 (x 2)(x 7) True, because (x 2)(x 7) x 2 7x 2x 14 x 2 5x 14 (d) x 2 8x 15 (x 5)(x 3) False, because (x 5)(x 3) x 2 3x 5x 15 x 2 8x 15 CHECK YOURSELF 1 Determine which of the following are true statements. (a) 2x 2 2x 3 (2x 3)(x 1) 3x 2 11x 4 (3x 1)(x 4) (c) 2x 2 7x 3 (x 3)(2x 1) 431
2 432 CHAPTER 6 POLYNOMIALS AND POLYNOMIAL FUNCTIONS The first step in learning to factor a trinomial is to identify its coefficients. To be consistent, we first write the trinomial in standard ax 2 bx c form, then label the three coefficients as a, b, and c. Example 2 Identifying the Coefficients of ax 2 bx c When necessary, rewrite the trinomial in ax 2 bx c form. Then label a, b, and c. (a) x 2 3x 18 a 1 b 3 c 18 x 2 24x 23 a 1 b 24 c 23 (c) x x First rewrite the trinomial in descending order. x 2 11x 8 Then, a 1 b 11 c 8 CHECK YOURSELF 2 When necessary, rewrite the trinomial in ax 2 bx c form. Then label a, b, and c. (a) x 2 5x 14 x 2 18x 17 (c) x 6 2x 2 Not all trinomials can be factored.to discover if a trinomial is factorable, we try the ac test. Rules and Properties: The ac Test A trinomial of the form ax 2 bx c is factorable if (and only if) there are two integers, m and n, such that ac mn and b m n In Example 3, we will determine whether each trinomial is factorable by finding the values of m and n. Example 3 Using the ac Test Use the ac test to determine which of the following trinomials can be factored. Find the values of m and n for each trinomial that can be factored. (a) x 2 3x 18 First, we note that a 1, b 3, and c 18, so ac 1( 18) 18. Then, we look for two numbers, m and n, such that mn ac and m n b. In this case, that means mn 18 m n 3
3 FACTORING TRINOMIALS: THE ac METHOD SECTION We will look at every pair of integers with a product of 18. We then look at the sum of each pair. mn 1( 18) 18 2( 9) 18 3( 6) 18 m n 1 ( 18) 17 2 ( 9) 7 3 ( 6) 3 We need look no further because we have found two integers whose mn product is 18 and m n sum is 3. m 3 n 6 x 2 24x 23 We see that a 1, b 24, and c 23. So, ac 23 and b 24. Therefore, we want mn 23 m n 24 We now work with integer pairs, looking for two integers with a product of 23 and a sum of 24. mn m n 1(23) ( 23) 23 1 ( 23) 24 We find that m 1 and n 23. (c) x 2 11x 8 We see that a 1, b 11, and c 8. So, ac 8 and b 11. Therefore, we want mn 8 m n 11 mn m n 1(8) (4) ( 8) 8 1 ( 8) 9 2( 4) 8 2 ( 4) 6 There is no other pair of integers with a product of 8, and none has a sum of 11. The trinomial x 2 11x 8 is not factorable. (d) 2x 2 7x 15 We see that a 2, b 7, and c 15. So, ac 30 and b 7. Therefore, we want mn 30 m n 7 mn m n 1( 30) 30 1 ( 30) 29 2( 15) 30 2 ( 15) 13 3( 10) 30 3 ( 10) 7 5( 6) 30 5 ( 6) 1 6( 5) 30 6 ( 5) 1 10( 3) ( 3) 7
4 434 CHAPTER 6 POLYNOMIALS AND POLYNOMIAL FUNCTIONS There is no need to go further. We have found two integers with a product of 30 and a sum of 7. So m 10 and n 3. In this example, you may have noticed patterns and shortcuts that make it easier to find m and n. By all means, use those patterns. This is essential in mathematical thinking. You are taught a step-by-step process that will always work for solving a problem; this process is called an algorithm. It is very easy to teach a computer an algorithm. It is very difficult (some would say impossible) for a computer to have insight. Shortcuts that you discover are insights. They may be the most important part of your mathematical education. CHECK YOURSELF 3 Use the ac test to determine which of the following trinomials can be factored. Find the values of m and n for each trinomial that can be factored. (a) x 2 7x 12 x 2 5x 14 (c) 2x 2 x 6 (d) 3x 2 6x 7 So far we have used the results of the ac test only to determine whether a trinomial is factorable. The results can also be used to help factor the trinomial. Example 4 Using the Results of the ac Test to Factor a Trinomial Rewrite the middle term as the sum of two terms, then factor by grouping. (a) x 2 3x 18 We see that a 1, b 3, and c 18, so ac 18 b 3 We are looking for two numbers, m and n, so that mn 18 m n 3 In Example 3, we found that the two integers were 3 and 6 because 3( 6) 18 and 3 ( 6) 3. That result is used to rewrite the middle term (here 3x) as the sum of two terms. We now rewrite the middle term as the sum of 3x and 6x. x 2 3x 6x 18 Then, we factor by grouping: x 2 3x 6x 18 x(x 3) 6(x 3) (x 3)(x 6) x 2 24x 23 We use the results of Example 3, in which we found m 1 and n 23, to rewrite the middle term of the expression. x 2 24x 23 x 2 x 23x 23 Then, we factor by grouping: x 2 x 23x 23 x(x 1) 23(x 1) (x 1)(x 23)
5 FACTORING TRINOMIALS: THE ac METHOD SECTION (c) 2x 2 7x 15 From example 3(d ), we know that this trinomial is factorable and that m 10 and n 3. We use that result to rewrite the middle term of the trinomial. 2x 2 7x 15 2x 2 10x 3x 15 2x(x 5) 3(x 5) (x 5)(2x 3) CHECK YOURSELF 4 Rewrite the middle term as the sum of two terms, then factor by grouping. (a) x 2 7x 12 x 2 5x 14 (c) 2x 2 x 6 (d) 3x 2 7x 6 Not all product pairs need to be tried to find m and n. A look at the sign pattern will eliminate many of the possibilities. Assuming the lead coefficient to be positive, there are four possible sign patterns. Pattern Example Conclusion 1. b and c are both positive. 2x 2 13x 15 m and n must be positive. 2. b is negative and c is x 2 3x 2 m and n must both be negative. positive. 3. b is positive and c is x 2 5x 14 m and n are of opposite signs. negative. (The value with the larger absolute value is positive.) 4. b and c are both negative. x 2 4x 4 m and n are of opposite signs. (The value with the larger absolute value is negative.) Sometimes the factors of a trinomial seem obvious. At other times you might be certain that there are only a couple of possible sets of factors for a trinomial. It is perfectly acceptable to check these proposed factors to see if they work. If you find the factors in this manner, we say that you have used the trial and error method. This method is discussed in Section 6.7*, which follows this section. To this point we have been factoring polynomial expressions. When a function is defined by a polynomial expression, we can factor that expression without affecting any of the ordered pairs associated with the function. Factoring the expression makes it easier to find some of the ordered pairs. In particular, we will be looking for values of x that cause f(x) to be 0. We do this by using the zero product rule. Rules and Properties: Zero Product Rule If 0 ab, then either a 0, b 0, or both are zero. Another way to say this is, if the product of two numbers is zero, then at least one of those numbers must be zero.
6 436 CHAPTER 6 POLYNOMIALS AND POLYNOMIAL FUNCTIONS Example 5 Factoring Polynomial Functions Given the function f(x) 2x 2 7x 15, complete the following. (a) Rewrite the function in factored form. From Example 4(c) we have f(x) (x 5)(2x 3) Find the ordered pair associated with f(0). f(0) (0 5)(0 3) 15 The ordered pair is (0, 15). (c) Find all ordered pairs (x, 0). We are looking for the x value for which f(x) 0, so 0 (x 5)(2x 3) By the zero product rule, we know that either (x 5) 0 or (2x 3) 0 which means that x 5 or x 3 2 The ordered pairs are ( 5, 0) and 3 2, 0. ordered pairs are associated with that function. Check the original function to see that these CHECK YOURSELF 5 Given the function f(x) 2x 2 x 6, complete the following. (a) Rewrite the function in factored form. Find the ordered pair associated with f(0). (c) Find all ordered pairs (x, 0). CHECK YOURSELF ANSWERS 1. (a) False; true; (c) true 2. (a) a 1, b 5, c 14; a 1, b 18, c 17; (c) a 2, b 1, c 6 3. (a) Factorable, m 4, n 3; factorable, m 7, n 2; (c) factorable, m 4, n 3; (d) not factorable 4. (a) x 2 3x 4x 12 (x 3)(x 4); x 2 7x 2x 14 (x 7)(x 2); (c) 2x 2 x 6 2x 2 4x 3x 6 (2x 3)(x 2); (d) 3x 2 7x 6 3x 2 9x 2x 6 (3x 2)(x 3) 3 2, 0 5. (a) f(x) (2x 3)(x 2); (0, 6); (c) and (2, 0)
7 Name 6.7 Exercises Section Date In exercises 1 to 8, determine which are true statements. 1. x 2 2x 3 (x 1)(x 3) ANSWERS x 2 2x 8 (x 2)(x 4) 3. 2x 2 5x 4 (2x 1)(x 4) 4. 3x 2 13x 10 (3x 2)(x 5) 5. x 2 x 6 (x 5)(x 1) 6. 6x 2 7x 3 (3x 1)(2x 3) 7. 2x 2 11x 5 ( x 5)(2x 1) 8. 6x 2 13x 6 (2x 3)( 3x 2) In exercises 9 to 16, when necessary, rewrite the trinomial in ax 2 bx c form, then label a, b, and c. 9. x 2 3x x 2 2x x 2 5x x 2 x x 1 2x x 3x x 3x x x In exercises 17 to 24, use the ac test to determine which trinomials can be factored. Find the values of m and n for each trinomial that can be factored. 17. x 2 3x x 2 x x 2 2x x 2 7x
8 ANSWERS x 2 3x x 2 10x x 2 5x x 2 x In exercises 25 to 70, completely factor each polynomial expression. 25. x 2 7x x 2 9x x 2 9x x 2 11x x 2 15x x 2 13x x 2 7x x 2 7x x 2 10x x 2 13x x 2 7x x 2 15x x 2 8xy 15y x 2 9xy 20y x 2 16xy 55y x 2 9xy 22y x 2 11x x 2 9x x 2 18x x 2 20x x 2 23x x 2 30x 7 438
9 ANSWERS 47. 4x 2 20x x 2 24x x 2 19x x 2 17x x 2 24x x 2 14x x 2 7x x 2 5x x 2 40x x 2 45x x 2 17xy 6y x 2 17xy 12y x 2 30xy 7y x 2 14xy 15y x 2 24x x 2 10x x 2 26x x 2 39x x 3 31x 2 5x 66. 8x 3 25x 2 3x 67. 5x 3 14x 2 24x 68. 3x 4 17x 3 28x x 3 15x 2 y 18xy x 3 10x 2 y 72xy
10 ANSWERS 71. (a) (c) 72. (a) (c) 73. (a) (c) In exercises 71 to 76, for each function, (a) rewrite the function in factored form, find the ordered pair associated with f(0), and (c) find all ordered pairs (x, 0). 71. f(x) x 2 2x f(x) x 2 3x f(x) 2x 2 3x f(x) 3x 2 11x (a) (c) (a) (c) (a) (c) 75. f(x) 3x 2 5x f(x) 10x 2 13x 3 Certain trinomials in quadratic form can be factored with similar techniques. For instance, we can factor x 4 5x 2 6 as (x 2 6)(x 2 1). In exercises 77 to 88, apply a similar method to completely factor each polynomial. 77. x 4 3x x 4 7x x 4 8x x 4 5x y 6 2y x 6 10x x 5 6x 3 16x 84. x 6 8x 4 15x x 4 5x x 4 5x x 6 6x x 6 2x 3 3 In exercises 89 to 96, determine a value of the number k so that the polynomial can be factored. 89. x 2 5x k 90. x 2 3x k 440
11 ANSWERS 91. 6x 2 x k 92. 4x 2 x k 93. x 2 kx x 2 kx x 2 kx x 2 kx The product of three numbers is x 3 6x 2 8x. Show that the numbers are consecutive even integers. (Hint: Factor the expression.) 98. The product of three numbers is x 3 3x 2 2x. Show that the numbers are consecutive integers (a) 100. (a) 101. (a) In each of the following, (a) factor the given function, identify the values of x for which f(x) 0, (c) graph f(x) using the graphing calculator and determine where the graph crosses the x axis, and (d) compare the results of and (c). 99. f(x) x 2 2x f(x) x 2 3x (a) 101. f(x) 2x 2 x f(x) 3x 2 x 2 In exercises 103 and 104, determine the binomials that represent the dimensions of the given figure Area 2x 2 7x 15? Area 3x 2 11x 10??? 441
12 Answers 1. True 3. False 5. False 7. False 9. a 1, b 3, c a 2, b 5, c a 2, b 1, c a 3, b 2, c Factorable, m 5, n Not factorable 21. Not factorable 23. Factorable, m 4, n (x 3)(x 4) 27. (x 8)(x 1) 29. (x 10)(x 5) 31. (x 10)(x 3) 33. (x 6)(x 4) 35. (x 11)(x 4) 37. (x 3y)(x 5y) 39. (x 11y)(x 5y) 41. (3x 4)(x 5) 43. (5x 2)(x 4) 45. (3x 5)(4x 1) 47. (2x 5) (5x 6)(x 5) 51. (5x 6)(x 6) 53. (2x 3)(5x 4) 55. (4x 5) (7x 3y)(x 2y) 59. (4x y)(2x 7y) 61. 3(x 5)(x 3) 63. 2(x 4)(x 9) 65. x(6x 1)(x 5) 67. x(x 4)(5x 6) 69. 3x(x 6y)(x y) 71. (a) (x 3)(x 1); (0, 3); (c) (3, 0) and ( 1, 0) 1 2, (a) (2x 1)(x 2); (0, 2); (c) and ( 2, 0) 75. (a) (x 4)(3x 7); (0, 28); (c) ( 4, 0) and 7 3, (x 2 1)(x 2 2) 79. (x 2 11)(x 2 3) 81. (y 3 5)(y 3 3) 83. x(x 2 2)(x 2 8) 85. (x 3)(x 3)(x 2 4) 87. (x 2)(x 2 2x 4)(x 3 2) , 1, or , 5, 1, or , 7, 17, 17, 3, or x(x 2)(x 4) (a) (x 4)(x 2); 4, (a) (2x 3)(x 1); 1 2, 103. (2x 3) and (x 5) 442
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# The continued product of three numbers in G.P is 216, and the sum of the product of them in pairs is 156. Find the numbers.
Last updated date: 20th Jun 2024
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Hint: First of all take three numbers in G.P as $\dfrac{a}{r},a,ar$. Then use the given information that the product of these numbers is equal to 216 to find the value of a. Further, use the value of the sum of the product of them in pairs to get the values of r. Hence, find the numbers.
Complete step by step solution:
We are given that the continued product of three numbers in G.P is 216 and the sum of the product of that in pairs is 156. We have to find the numbers.
First of all, we know that the general term of G.P is $a{{r}^{n-1}}$. So, let us consider three terms in G.P as,
$\dfrac{a}{r},a,ar$
We are given that the product of these three numbers in G.P is 216. So, we get
$\dfrac{a}{r}.a.ar=216$
By canceling the like terms, we get,
${{a}^{3}}=216$
By taking the cube root on both sides of the above equation, we get,
$\Rightarrow a=\sqrt[3]{216}$
Or, a = 6
Now by substituting the value of a = 6 in terms $\dfrac{a}{r},a,ar$. We get three terms as,
$\dfrac{6}{r},6,6r$
Also, we are given that the sum of the products of each pair of the given terms is 156. So, we get,
$\left( \dfrac{6}{r} \right).\left( 6 \right)+\left( 6 \right).\left( 6r \right)+\left( 6r \right).\left( \dfrac{6}{r} \right)=156$
By simplifying the above equation and taking 6.6 = 36 common, we get,
$36\left( \dfrac{1}{r}+r+\dfrac{r}{r} \right)=156$
Or, $36\left( \dfrac{1+{{r}^{2}}+r}{r} \right)=156$
By dividing both sides by 12, we get,
$3\left( \dfrac{1+{{r}^{2}}+r}{r} \right)=13$
By cross multiplying the above equation, we get,
$3+3{{r}^{2}}+3r=13r$
Or, $3{{r}^{2}}+3r-13r+3=0$
$\Rightarrow 3{{r}^{2}}-10r+3=0$
We can write 10r = 9r + r. So, we get,
$3{{r}^{2}}-9r-r+3=0$
We can also write the above equation as,
$3r\left( r-3 \right)-1\left( r-3 \right)=0$
By taking (r – 3) common, we get,
$\left( r-3 \right)\left( 3r-1 \right)=0$
Therefore, we get r = 3 and $r=\dfrac{1}{3}$.
By substituting the value of r = 3 in terms $\dfrac{6}{r},6,6r$. We get three terms as
$\dfrac{6}{3},6,6\times 3$
$=2,6,18$
Also, by substituting the value of $r=\dfrac{1}{3}$ in terms $\dfrac{6}{r},6,6r$. We get three terms as,
$=\dfrac{6}{\dfrac{1}{3}},6,6.\left( \dfrac{1}{3} \right)$
$=18,6,2$
Hence, we get three numbers in G.P as 2, 6, 18, or 18, 6, 2.
Note: Students should always take 3 numbers in G.P as $\dfrac{a}{r},a,ar$ to easily solve the given problem. Also, students often make this mistake of writing three terms as a, ar and $a{{r}^{2}}$ after getting the values of a and r. But they must note that as we have taken the terms as $\dfrac{a}{r},a,ar$. So they must substitute a and r in these terms to get 3 numbers in G.P. So, this mistake must be avoided. |
## Real Mathematics – Life vs. Maths #5
Drunkard’s Way Back Home
Finally we are back with Steve the accountant. So far I’ve talked about one dimensional random walk in order to understand drunkard’s walk. Although we are all aware of the fact that right and left are not the only options when we take a walk.
That is why Steve’s walk should be thought in two dimensions.
There are four choices of movement in a two-dimensional plane: Right, left, up and down. They all have same probability (just like in one dimension). It is ¼. These four directions are familiar to us as we already learned about Cartesian coordinates. In the end Steve’s walk turned out to be a walk in Cartesian coordinates. Let me choose origin of the coordinate system as Steve’s starting point. For the first random step there will be these four options:
To facilitate Steve the accountant’s walk, one could use a 12-sided dice. When it is thrown:
Take your step upwards for 1-2-3,
take your step downwards for 4-5-6,
take your step to right for 7-8-9,
and take your step to left for 10-11-12.
Q: How far from the starting point one would be after taking N random steps in a two-dimensional plane?
Answer to this question is same with one-dimensional random walk: After N random steps in two dimensions, one would be √N steps away from the starting point. That is same as imagining a circle that has its center at the origin (starting point) and has the radius √N. A two-dimensional random walk would likely end in this circle.
Recall that for one-dimensional random walks we found three conclusions. Third one was saying that we are likely to be back to the starting point if N is large enough of a number. Same conclusion can be made for two-dimensional random walks too; the more steps we take, the most likely we would be close to the starting point.
“A drunkard will find his way home, but a drunken bird may get lost forever.”
Shizou Kakutani
This result shows that Steve the accountant will likely make circles and return to his starting point. But he will eventually reach his home.
Let me take this one step further: If Steve the accountant’s walk is long enough he would have visited all the streets in his neighborhood. This is why we say two-dimensional random walks are recurrent just like one-dimensional ones. But if we go into three dimensions, things change. A three-dimensional random walk is not recurrent which is why it is possible to get lost in 3-D.
Biased Random Walk
We already know that in one-dimensional random walk there are two choices with equal probabilities: Right (1/2) and left (1/2).
Q: How can we find a one-dimensional random walk that is biased?
For such a random walk I can keep the probabilities of right or left unchanged. But I’ll arrange the number of steps taken. What I mean is that whenever we get a right; let’s take two steps instead of one but for a left we continue to take one step. Outcome used to be +1 or -1 for a step. Now it is either +2 or -1.
This is how a biased one-dimensional random walk can be created.
The reason this random walk is biased is that after taking N random steps, one will most likely be on the right side of the starting point. We can test that with a coin toss like we did in previous articles: Tails are +2, heads are -1.
After 10 coin tosses my path turned out to be as follows:
One wonders…
You make your own experiment with a coin and compare your results.
Ps. It is not over yet. To be continued…
M. Serkan Kalaycıoğlu
## Real Mathematics – Life vs. Maths #3
Drunkard’s Walk Back Home
Steve the accountant finished another working week. He usually spends his weekends in peace. But this specific weekend was different: He was supposed to meet his old college mates whom he hasn’t seen for ages. That night they talked about old days, laughed and drank until morning. Steve the accountant has never been a heavy drinker. At the end of the meeting even though he was barely standing he insisted that he can walk back to his home by himself.
Steve the accountant started walking in random directions: “This street looks familiar… Oh that building looks just like mine…”
Q: Can a drunkard make his way home using a random walk?
Random Walk in One Dimension
I have to talk about random walk in one dimension before I answer the fate of Steve the accountant. In one dimension walking path is something we are familiar since we are kids: The number line.
There are two directions on the number line: Right and left.
Rules:
• In one dimension one step to right means +1, one step to left means -1.
• Let’s assume that the probability of choosing right and left is same.
• Because of the previous assumption taking a step towards right or left has the probability of ½.
Now let’s draw a number line and choose zero as the starting point. First step can be taken towards +1 or -1. Their probabilities are equal: ½.
Taking two steps at once will be a little bit more complicated than taking one step. After taking only one step we concluded that there can be only two possibilities: +1 and -1. But when we try to take two steps at once, there will be four possibilities:
0 -> +1 -> +2
0 -> +1 -> 0
0 -> -1 -> -2
0 -> -1 -> 0.
In every possibility, probability will be equal: ¼.
After the second step, we may be standing on either one of +2, 0 or -2 with probabilities ¼, 2/4 and ¼ in order.
We already know what the probabilities are after two steps. According to our findings third step can start either at +2, 0 or -2.
• If we take our step from +2, we can go to +3 or +1. Their probabilities will be half of the probability of +2. Hence it will be 1/8 each.
• If we take our step from -2, we can go to -3 or -1. Their probabilities will be half of the probability of -2. Hence it will be 1/8 each.
• If we take our step from 0, we can go to +1 or -1. Their probabilities will be half of the probability of 0. Hence it will be 2/8 each.
Our calculations show that after the third step we could stand on:
+3 with the probability of 1/8.
-3 with the probability of 1/8.
+1 with the probability of 1/8 + 2/8 = 3/8.
-1 with the probability of 1/8 + 2/8 = 3/8.
In case we continue using the same logic, fourth and fifth steps would look like the following:
After 100 steps, final position and its probability is shown as follows:
So far we can make these conclusions about a random walk in one dimension:
1. When we take even number of steps, we stop on an even number. When we take odd number of steps, we stop on an odd number.
2. As we increase the number of steps probability of stopping around the starting point gets higher.
3. Previous argument indicates that if we take more steps, probability of returning to the starting point will increase as well.
Game of Random Walk in One Dimension
So far we understood that a random walk in one dimension has two possible outcomes. In order to simulate a one dimensional random walk we can use coin toss since there are only two possible outcomes for a coin toss: Heads or tails.
If we have a fair coin, probability of getting heads or tails will be equal to each other: 1/2. Then let’s assign tails to -1 and heads to +1.
1. Toss a coin 10 times. Where did you finish your random walk?
2. Do the same thing for 30 times and compare your result with the previous one.
I could hear you saying: “What about Steve the accountant?”
A little bit of a patience. We’ll get there in the upcoming articles.
M. Serkan Kalaycıoğlu |
Percentage Change
In this topic, we are going to discuss the concepts of percentage change. We see percentage change in our everyday life. From the increase in price to decrease in price, from more discounts, there is a change in percentage everywhere.
Percentage Change
The change in the value of product refers to as the amount of percentage change in the product. In percentage whenever the value of measured changes, the change can be captured in two ways. They are:
1. Absolute value change
2. Percentage change
1. Absolute Value Change
Absolute value change refers to the actual change in the measured quantity. It means the direct change in the product. For example, if the sales of the product in 1st year is 200 crore and in the second year it 220 crores, then the absolute value of the change will be 100 crores.
2. Percentage Change
This change in percentage can be obtained through the following formula.
Percentage change = Absolute value change/ original quantity x 100
= 20/200 x 100 = 10%
It is important to note here that the percentage change calculations are usually calculated on the original quantity unless it is mentioned in the question.
We will understand the concept of percentage change through some examples.
Example Based on Population
Q. The population of a city grew from 40 lakh to 44 lakh. Find the
a) percentage change
b) percentage change based on the final value of the population.
Here, we are required to find two things. First, we need to find the change in percentage from 40 to 44, then based on the final value we need to find the percentage change.
Change in percentage = absolute value change/ original quantity x 100
The increase in population is 4 lakh and the original population was 40 lakh. So, the change in percentage = 4/40 x 100 = 20%.
Now, we are required to find the change in percentage but the base should be final value.
Change in percentage = 4/44 x 100 = 9.09%.
Example Based on Price Hike
Q. Due to a 25% hike in the price of sugar, a person is able to purchase 20 kg less of sugar for Rs. 500. What was the initial price of sugar?
A. Rs. 4
B. Rs. 5
C. Rs. 6
D. Rs. 7
In this question, one thing you need to remember is that with the increase in price there will be the decrease in consumption. This is how the percentage works. Since the prices are rising by 25%, in this case, the consumption of the sugar will have to be reduced by 20% so that budget remains the same. But here it is given that there is an actual reduction of 20 kg. So, 20% of the decrease in consumption will be equal to 20 kg drop in the consumption.
So, the original consumption was 100 kg of sugar. While the total money spent on sugar was Rs. 500. So, the initial price of the sugar was Rs. 5/kg. So, the correct answer is B.
Through product constancy table, you can easily learn hike and reduction of consumption. We have discussed it in our other article.
Example Based on Area
Q. If the breadth of the rectangle is increased by 20 % and the length of the rectangle is decreased by 10 %, then what is the percent change increase in the area of the rectangle?
A. 6 % B. 8 % C. 10 % D. 12 %
Ans: The area of a rectangle is l x b.
So, due to increase and decrease in breadth and length of the rectangle respectively the new area will, 0.9l x 1.2b = 1.08 lb. Thus the increase in new area is 8 %. So, the correct answer is B.
Practice Questions
Q. The price of the ticket from Delhi to Mumbai increased by 8% and thereby one ends up paying 5940 for the journey. Find the rate of tickets before the increase in prices.
A. Rs. 5500 B. Rs. 5600 C. Rs. 5700 D. Rs. 5800
Ans: The correct answer is A.
Q. The price of the petrol goes down by 10 % and the consumption increases by 20 %. What was the original change in the expenditure on petrol?
A. -8 % B. +8% C. -10% D. +10%
Ans: The correct answer is B.
Q. The number of applicants for the government jobs increased by 75% from 1990 to 1995, From 1995 to 2000 there was an increase of 100 % in the number of applicants. Find the percentage increase in the number of applicants from 1990 to 2000.
A. 200 % B. 220 % C. 240 % D. 250 %
Ans: The correct answer is D.
Q. Raman’s salary was reduced by 30 % and then increased by 30 %. What was the net change in the salary of Raman?
A. -6 % B. -4 % C. +6 % D. No change
Ans: The correct answer is A.
Q. The candidate got 56 % of the votes cast in an election. He won by 144 votes. What was the total number of voters on the voting list if 80 % people cast their vote and there was not a single invalid vote?
A. 360 B. 1500 C. 1800 D. 720
Ans: The correct answer is B.
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One response to “Percentage Change”
1. kiran says:
Total wrong teaching and examples are wrong change the mistakes of every example, I found this
change in percentage = 4/40 x 100 = 20%. |
# Ex.11.2 Q4 Constructions Solution - NCERT Maths Class 10
Go back to 'Ex.11.2'
## Question
Draw a pair of tangents to a circle of radius $$5 \,\rm{cm}$$ which are inclined to each other at an angle of $$60^\circ$$.
## Text Solution
#### Steps:
Steps of construction:
(i) With $$O$$ as centre and $$5 \,\rm{cm}$$ as radius draw a circle.
(ii) Take a point $$A$$ on the circumference of the circle and join $$OA$$.
(iii) Draw $$AX$$ perpendicular to $$OA$$.
(iv) Construct \begin{align}\angle \rm{AOB}=120^{\circ} \end{align} where $$B$$ lies on the circumference.
(v) Draw $$BY$$ perpendicular to $$OB$$.
(vi) Both $$AX$$ and $$BY$$ intersect at $$P$$.
(vii) $$PA$$ and $$PB$$ are the required tangents inclined at \begin{align}60^{\circ}\end{align} .
Proof:
$$\angle {{OAP}} = \angle {{OBP}} = 90^\circ$$ (By construction)
$$\angle {{AOB}} = 120^\circ$$ (By construction)
In quadrilateral $$OAPB$$,
\begin{align} \angle {{APB}}& = 360^\circ - [\angle {{OAP}} + \angle {{OBP}} + \angle {{AOB}}]\\ &= 360^\circ - [90^\circ + 90^\circ + 120^\circ ]\\ &= 360^\circ - 300^\circ \\ &= 60^\circ \end{align}
Hence $$PA$$ and $$PB$$ are the required tangents inclined at \begin{align}60^{\circ}\end{align}.
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# Did you solve it? The birthday birthday problem
The solution to today’s puzzle
In my puzzle blog earlier today I set you the following problem:
Ariel, Balthazar and Chastity are great mates, genius logicians and they always tell the truth. Neither Ariel nor Balthazar know the day or the month of Chastity’s birthday, so she decides to tell them in the following way:
First, she says out loud, so both Ariel and Balthazar can hear her: “The day (of the month) of my birthday is at most the number of the month of my birthday.”
Then she whispers the day to Ariel and the month to Balthazar
Ariel says “Balthazar cannot know Chastity’s birthday.”
Balthazar thinks a bit, then says: “Ariel also cannot know Chastity’s birthday.”
They carry on like this, each saying these exact same sentences in turn until Balthazar announces: “Both of us can now know what Chastity’s birthday is. Interestingly, that was the longest conversation like that we could have possibly had before both figuring it out.”
When is Chastity’s birthday?
Solution
There are 12 months in the year. So, the maximum number for the month of Chastity’s birthday is 12. Since the day of the birthday is at most the number of the month of the birthday, the maximum number for the day must also be 12. In other words, both Ariel and Balthazar are whispered a number between 1 and 12.
We are told that the day of the birthday is at most the number of the month. This is the same as saying that the day can be any number less than or equal to the month.
Ariel knows the day of the birthday, Balthazar the month. The only way for Balthazar to also know the day would be if the month is January, since then he could deduce that her birthday was January 1. But for Ariel to be able to say that Balthazar cannot not know the birthday, he must know that Balthazar does not have January. How does he know that? Well, he must have a number bigger than 1. Since if he had a 1, he would be able to deduce nothing about what Balthazar has. But if he had anything bigger than 1, Balthazar cannot have January since the day is always less than the month. So, we can eliminate 1 for Ariel, and 1 (January) for Balthazar.
Now to the next statement. The only way for Ariel to know the month of the birthday would be if he had been told 12, since he would be able to deduce that the month would also be 12, or December. But following the above reasoning, for Balthazar to know that Ariel does not have 12, he must have a month with a number 11 or under. So we can eliminate both 12 and 12 (December).
If this exchange were to happen again, we can eliminate February and 2, and November and 11.
For this to continue for as long as possible it will continue until Ariel eliminates 6 and 6. Once that happens the only solution left is July 7, (7/7)
I hope you enjoyed the puzzle. Thanks again to the Art of Problem Solving for suggesting it. I’ll be back in two weeks.
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
I’m the author of several books of popular maths, including the puzzle books Can You Solve My Problems? and Puzzle Ninja. I also co-write the children’s book series Football School. |
# How do you solve the inequality (x^2-16)/(x^2-4x-5)>=0?
Apr 16, 2018
The solution is $x \in \left(- \infty , - 4\right] \cup \left(- 1 , 4\right] \cup \left(5 , + \infty\right)$
#### Explanation:
The numerator is
${x}^{2} - 16 = \left(x - 4\right) \left(x + 4\right)$
The denominator is
${x}^{2} - 4 x - 5 = \left(x - 5\right) \left(x + 1\right)$
Let
$f \left(x\right) = \frac{\left(x - 4\right) \left(x + 4\right)}{\left(x - 5\right) \left(x + 1\right)}$
Let's build a sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$$4$$\textcolor{w h i t e}{a a a a a}$$5$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$color(white)(aa)-$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 5$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$color(white)(aa)-$\textcolor{w h i t e}{a}$color(white)(aaa)-$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$+$
graph{((x+4)(x-4))/((x-5)(x+1)) [-14.24, 14.23, -7.12, 7.12]}
Therefore,
$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - 4\right] \cup \left(- 1 , 4\right] \cup \left(5 , + \infty\right)$ |
Sarah Taylor
HS-PS2-1
# Position vs. Time Graph Study Guide
A position vs time graph conveys where the object can be discovered after a period of time.
# INTRODUCTION
Many people are afraid of graphs and become uneasy when they have to use them. Position graphs are not terrifying creatures. They can be attractive, and they are an effective way to visually express a large quantity of information regarding an object's motion in a little space. Let's take a closer look at the position graphs and their characteristics.
## DEFINITION OF POSITION
• Definition of position can be stated as the location of an object at any given time.
• We should be able to define the position of any item in order to explain its motion.
• The position vs. time graph, which lets us describe the motion of an object, is among the most basic types of graphs in mechanics.
• In such graphs, the vertical axis represents the object's location, and the horizontal axis represents the time spent.
• We can examine the trajectory and position of the object exactly by interpreting a position-time graph.
Position-time graph represents the position x of a particle on the y-axis and the time t on the x-axis. Graphs in the x-y plane can be drawn by keeping the following in mind.
• On the x-axis, always include an independent variable.
• On the y-axis, dependent variables must be represented.
• The dependent variable is reliant on the independent variable due to a mathematical function.
## SLOPE OF A POSITION TIME GRAPH
• The gradient of a position vs. time graph represents the velocity of the object.
• As a result, the value of the slope at a given moment shows the object's velocity at that moment.
• The speed is determined by the numeric values of the tangent's slope.
• Another thing to remember is that the instantaneous velocity is determined by the slope of a position vs. time graph at a given point in time.
• You may get the average velocity between two distinctive points in time by computing the mean slope between them.
• The average velocity does not have to match the instantaneous velocity.
• If, on the other hand, the slope remains constant over time, the instantaneous velocity equals the average velocity between two points.
• The graph above is an example of a position vs. time graph.
• Here, the velocity of the object in motion from 0 to 5 seconds will be 5 √2 m/s.
• In between the time frame of 5 to 10 seconds, the instantaneous velocity will be equal to the average velocity as the gradient here remains constant.
# SUMMARY
• Position can be defined as the location of an object at any given time
• A position vs time graph depicts motion by plotting position relative to the starting point on the y-axis and time on the x-axis.
• The velocity is represented by the gradient of a position vs time graph. The faster the motion changes, the steeper the slope.
## FAQs
1. How do you graph a position vs. time graph?
A position vs. time graph can be plotted by determining the time of the motion on the x-axis and the position on the y-axis.
2. Why is position vs. time graphs important?
Position vs. time graphs is important as we can examine the trajectory and position of the object precisely using this.
3. What does the slope of position vs. time graph mean?
The gradient/slope of a position vs. time graph represents the velocity of the object.
We hope you enjoyed studying this lesson and learned something cool about Position vs. Time Graph! Join our Discord community to get any questions you may have answered and to engage with other students just like you! We promise, it makes studying much more fun!😎
## REFERENCE
1. Position Time Graphs: https://flexbooks.ck12.org/cbook/ck-12-physics-flexbook-2.0/section/2.8/primary/lesson/position-time-graphs-ms-ps/. Accessed 8th April 2022.
2. Position vs Time Graphs Examples: https://physexams.com/lesson/position-vs-time-graph-examples-for-high-schools_48Accessed 8th April 2022 |
## Pythagorean Triples
A Pythagorean triple is a triplet $(a,b,c)$, where $a, b, c$ are integers such that $a^2+b^2=c^2$.
It is called a Pythagorean triple because it can represent 3 sides of a right triangle. You probably already know two Pythagorean triples: $(3,4,5)$ and $(5,12,13)$. Are there other triples? How do we find them?
Let’s consider the equation
$a^2+b^2=c^2$
where $a, b, c$ are integers. If $a, b, c$ have a common factor $d$, say $a=da_1$, $b=db_1$, and $c=dc_1$. Then we have
$d^2a_1^2+d^2b_1^2=d^2c_1^2$
which is true if and only if
$a_1^2+b_1^2=c_1^2.$
So let’s assume that $a, b, c$ have no common factors. Such triples will be called primitive. Since odd perfect squares are congruent to 1 mod 4 and even squares are congruent to 0 mod 4, then $c$ must be odd and exactly one of $a$ or $b$ is even. Let’s suppose that $b$ is even: $b=2k$ for some integer $k$. Then
$4k^2=b^2=c^2-a^2=(c+a)(c-a).$
So both $c + a$ and $c - a$ must be even. Say $c + a = 2r$ and $c - a = 2s$. Then $rs=k^2$ and we have $c = r + s$ and $a = r - s$. Since $a$ and $c$ have no common factors, we must have that $r$ and $s$ have no common factors. Since $rs$ is a square, then both $r$ and $s$ are squares. Say $r=m^2$ and $s=n^2$. So $a=m^2-n^2, b=2mn, c=m^2+n^2$ if (a,b,c) is a primitive triple.
Hence, any Pythagorean triple, with $b$ even, is either of the form $(m^2-n^2,2mn,m^2+n^2)$ or it is a multiple of that form. You can plug in different values for $m$ and $n$ to get different right triangles with integer sides. Fun! |
# Geometric Sequences and Series A
Document Sample
``` George & Anh Tu Tuesday 12-1 QUAD1048 (Week 4)
Geometric Sequences and Series
A geometric sequence is a list of numbers arranged so that each term is formed by multiplying the
preceding term by a constant. This constant is called the common ratio.
a, ar, ar2, ar3……. arn-1 where: a = first term (a ≠ 0)
r = common ratio
A geometric series is the sum of a geometric sequence.
S = a + ar + ar2 + ar3 + ……. + arn-1
a(r n − 1) a(1 − r n )
S= for r > 1 OR S= for r < 1 where: n = number of terms
r−1 1−r
Example 1
What are the first 5 terms of a geometric sequence where the first term is 4 and the common ratio is
1
? (Ans 4, 2, 1, 0.5, 0.25) What is the sum of the geometric series? (Ans 7.75)
2
Annuity
An annuity can be defined in the following ways:
It consists of payments (or repayments) which are of equal value,
The payments are made at regular time intervals,
For an ordinary annuity, payments are made at the end of each period,
For an annuity due, payments are made at the beginning of each period.
Present value usually applies to the value of loans or mortgages as these are based on the money
borrowed at the start.
Future value usually applies to savings, superannuation and sinking funds.
Ordinary Annuity
Ordinary annuity refers to a stream of equal payments which are made regularly at the end of
each period. We can express the regular stream of payments on the time line below; where the
regular payments, R, are made at the end of each period.
PV FV
R R R R
0 1 2 3 n
Using the FV formula, S = R(1+r)n-1 + R(1+r)n-2 + …..+ R(1+r) + R
George & Anh Tu Tuesday 12-1 QUAD1048 (Week 4)
Rearranging this, S = R + R(1+r) + …..+ R(1+r)n-2 + R(1+r)n-1 . This is the sum of a geometric
series with the first term R and common ratio of (1+r). Applying the formula, we get
(1 + r ) n − 1
FV of ordinary annuity = S = R
r
Where R = regular payment per period, r = interest rate per period, n = number of periods.
Example 2
Suppose you deposit \$200 in a savings account at the end of every year for the next 4 years at 8%
p.a. compounded annually. What is the accumulated amount after 4 years? (Ans \$901.22)
Example 3
When Jonathan entered University, his parents decided that they would reward him upon graduation
with a trip to Europe. They estimated that the trip would cost \$9,000. They wish to make equal
deposits on the last day of each month, starting March 31, 2004, in a bank that pays 12% p.a.
interest compounded monthly. The last deposit will be made on December 31, 2008. How much
should they deposit each month? (Ans \$115.25)
The formula for the present value of an ordinary annuity can also be found from the sum of a
geometric series. It is
1 − (1 + r ) − n
PV of ordinary annuity = A = R
r
Example 4
Suppose you can afford to pay \$200 at the end of each year for the next 4 years at 8% p.a.
compounded annually. What is the largest loan you could obtain? (Ans \$662.43)
Example 5
Annual gym membership fees can be paid by two methods. A member can either pay the whole fee
immediately or pay \$12 per week starting at the end of the fourth week of the year. If the club
charges interest for the whole year on unpaid fees and interest is compounded weekly at 8% p.a.,
what immediate payment would make the two options equivalent? (Ans \$563.36)
Annuity Due
An annuity due is a stream of equal payments which are made regularly at the beginning of each
period.
PV FV
R R R R R
0 1 2 3 n-1 n
George & Anh Tu Tuesday 12-1 QUAD1048 (Week 4)
The formula for the future value of an annuity due is
(1 + r ) n − 1
FV of annuity due = S = R (1+r)
r
The formula for the present value of an annuity due is
1 − (1 + r ) − n
PV of annuity due = A = R (1+r)
r
Example 6
A manufacturer wishes to establish a sinking fund that will provide \$500,000 to replace some of its
machinery. It immediately pays into the fund a payment of \$42,000 followed by the same amount at
the beginning of every 6 months. How long will it take to reach its goal if interest is 7% p.a.
compounded semi-annually? (Ans 5 years)
Example 7
Olivia wants to have a big party for her 30th birthday that will be in 15 months. She estimates that
the cost of hiring a cruise boat on Sydney Harbour and catering expenses will be \$3,200. Olivia
begins to save for the party by making monthly deposits, the first made immediately, into a savings
account which has an interest rate of 6% p.a. compounded monthly. How much should each deposit
be? (Ans \$147.30)
Sinking Fund
A sinking fund is a fund into which periodic payments are made in order to satisfy a future
obligation. The future value of an ordinary or annuity due formula is used.
Example 8
A fish-monger wishes to establish a sinking fund to provide \$350,000 to replace his fishing trawler
in 4 years time. He opens a saving account and decides to make equal deposits at the end of every
quarter. How much is each deposit if the rate of interest is 9% p.a. compounded quarterly? (Ans
\$18,415.82)
Example 9 (Continued from example 8)
After making deposits for 3 years, there is a change in the interest rate. If the new rate is 6% p.a.
compounded quarterly, what will the new quarterly deposit be if the fish-monger is to achieve its
\$350,000 target as planned? (Ans \$20,565.78)
Loan Amortisation
A loan is amortised when both the principal and the interest are paid back by a series of equal
periodic payments. That is, each periodic payment can be split into an interest and principal
component.
Example 10
Joseph borrowed \$8,000 at an interest rate of 12% p.a. compounded annually which is to be repaid
over 4 years with equal payments due at the end of each year. Complete the table.
George & Anh Tu Tuesday 12-1 QUAD1048 (Week 4)
Year end (a) (b) (c) Principal repaid
Principal outstanding at Repayment Interest on principal (b) – (c)
beginning of period = 0.12 x (a)
1
2
3 4,451.37 2,633.88 534.26 2,099.72
4 2,351.65 2,633.88 282.20 2,351.68
8,000.03
Principal Outstanding
Principal outstanding is the amount of money still owed at a certain point during the loan period. It
is equivalent to the present value of the payments yet to be made.
1 − (1 + r ) − p
PO = R where p = number of payments yet to be made
r
Important Rules
Interest paid in any period = interest rate x principal outstanding at beginning of that period
Principal repaid in any period = payment – interest paid
The total interest paid during a loan is called the finance charge:
Finance charge = total amount paid – loan amount = nR - A
Example 11 (modified from QMA final examination, Nov 2002)
William borrows \$100,000 from a bank to buy a new speed boat and agrees to repay the debt by
quarterly instalments at the end of each quarter over a period of 10 years. If the interest rate is 9%
p.a. compounded quarterly,
a) Compute his quarterly repayment. (Ans \$3,817.74)
b) Compute his principal outstanding at the beginning of the 17th quarter (this means he has just
paid the 16th instalment. (Ans \$70,204.56)
c) Compute the interest component in the 17th instalment. (Ans \$1,579.60)
d) Compute the principal repaid in the 17th instalment. (Ans \$2,238.14)
e) After 5 years, the interest rate increases to 12% p.a. compounded quarterly. If the debt must be
paid off by the original date agreed upon, find the new quarterly instalment. (Ans \$4,096.48)
f) Compute the total amount paid over 10 years. (Ans \$158, 284.40)
g) Compute the finance charge on the loan. (\$58, 284.40)
Example 12 (modified from QMA final examination, Nov 2002)
A pie maker is estimated to yield a net annual return of \$75,000 for the next 10 years at which time
it can be sold for \$10,000. The company wants to earn 9% on its investment and also set up a
sinking fund to replace the purchase price. If money is placed in the fund at the end of each year
and earns 5% p.a. compounded annually, find the price the company should pay for the pie maker.
(Ans \$447,156.34)
```
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This page provides information to support educators and families in teaching K-3 students about rounding. It is designed to complement the Rounding topic page on BrainPOP Jr.
This movie will introduce estimation and rounding and explain how to round numbers to the nearest tens and hundreds place. We recommend watching the movie and exploring the features over the course of a week or more. You may want to watch the movie again after children are more familiar with the skill and then stop the movie when there is a number on the screen. Encourage children to round the numbers that are on the screen before Annie and Moby do!
Review with children that when you estimate, you use what you know or see to make a reasonable guess about an amount. Estimate different amounts together, such as the number of books on a shelf, number of marbles in a bowl, or the length of a desk. Go through reasonable and unreasonable estimates together. Explain that when people estimate, they round numbers to make them easier to work with. This means changing numbers so that they end in a zero (or several zeroes).
Show the number 38 on a number line from 30 through 40. Guide children to see that 38 is closer to 40 than it is to 30. So, we round 38 up to 40. Then show the number 31 on the number line. Help children see that 31 is closer to 30 than to 40. So, we round 31 down to 30. Then show the number 35 on the number line. Some children may point out that the number is right in the middle between 30 and 40. Explain that when a number ends in 5, it is right in the middle of two tens, and we round up. So, 35 rounds up to 40. If a two-digit number ends in 0, then it is already rounded to the nearest ten!
To show children how to round without a number line, present a number to children and point out the ones place. Explain that when you round to the nearest ten, you look at the digit in the ones place to help you decide whether to round up or round down. If the digit is less than 5, you round down to the nearest ten. If the digit is 5 or greater, you round up to the nearest ten. It may be helpful for children to imagine a hill with the numbers 1 through 10 going up and over the hill, with the 5 at the top of the hill. When you reach the top of the hill (5), it’s easier to go over to the higher number (10). Practice different examples together using number lines, hundred charts, and/or place value charts.
After children feel comfortable with rounding to the nearest ten, challenge them to round to the nearest hundred. Show the number 413 on a number line between 400 and 500. Which hundred is 413 closer to, 400 or 500? Help children see that 413 is closer to 400, so we round 413 down to 400. Then present 489 on the number line. Ask children if they should round the number down to 400 or up to 500? Why? Repeat the activity with 450. Remind children that when a number is right in the middle between two numbers, they should round up. Explain that when they round to the nearest hundred, they look at the digit in the tens place to help them decide whether to round up or round down. If the digit in the tens place is less than 5, round down. If the digit in the tens place is 5 or greater than 5, round up. If there are zeroes in the tens and ones places, then the number is already rounded to the nearest hundred! Round different three-digit numbers together and have children explain how and why they rounded up or down.
Discuss with children why rounding numbers might be useful. Challenge children to come up with numbers and have partners round them to the nearest ten or hundred. Give plenty of opportunities where children can practice the skill and become experts! |
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# If (1, 0), (0, 1), (1, 1) are the coordinates of vertices of a triangle. The triangle is _______ triangle.(a) Isosceles(b) Obtuse angled(c) Acute angled(d) Equilateral
Last updated date: 24th May 2024
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Hint: Take the points (1, 0), (0, 1), (1, 1) as A, B, C respectively, obtains the values of AB, BC and CA using formula $\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$, if $\left( {{x}_{1}},{{x}_{2}} \right)$ and $\left( {{y}_{1}},{{y}_{2}} \right)$ are points. Then try to find similarities and relations to get desired results.
Let us take A as (1,0), B as (0,1) and C as (1,1).
Then we will find length of AB, BC and CA using following formula:
If points are $\left( {{x}_{1}},{{y}_{2}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then its distance is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .
So, if A is (1,0) and B is (0,1), then
$AB=\sqrt{{{\left( 0-1 \right)}^{2}}+{{\left( 1-0 \right)}^{2}}}=\sqrt{1+1}$
Hence, $AB=\sqrt{2}$
Now if B is (0,1) and C is (1,1), then
$BC=\sqrt{{{\left( 1-0 \right)}^{2}}+{{\left( 1-1 \right)}^{2}}}=\sqrt{{{1}^{2}}+{{0}^{2}}}$
Hence, BC = 1.
Now, if C is (1,1) and A is (1,0), then
$CA=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( 0-1 \right)}^{2}}}=\sqrt{{{0}^{2}}+{{1}^{2}}}$
Hence, CA = 1.
So, we can say that $BC=CA\ne AB$.
Therefore, we can say that it is an isosceles triangle and not an equilateral triangle because in isosceles triangle two sides are equal.
Now we will apply Pythagoras theorem,
\begin{align} & B{{C}^{2}}+C{{A}^{2}}=A{{B}^{2}} \\ & {{1}^{2}}+{{1}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\ \end{align}
2 = 2.
Here the left hand side is equal to the right hand side. So, the triangle ABC is a right angled triangle.
Therefore, triangle ABC is an acute angled isosceles triangle and not an obtuse angled triangle.
Hence, the correct answer is option (a) and (c).
Note: Another approach is by plotting the points (1, 0), (0, 1) and (1,1) on the graph and joining the points one will just compare the lengths of triangles to understand what will be the answer. Students often stop their solution once they get to know that the triangle is an isosceles triangle. We should check for acute and obtuse angled conditions too. |
If half of the smaller of two consecutive even integers is 2 more than the larger, what are the integers?
Oct 3, 2015
The integers are $\left(- 8\right)$ and $\left(- 6\right)$
Explanation:
Let the two consecutive even integers be
$\textcolor{w h i t e}{\text{XXX}} 2 n$ and $2 n + 2$
We are told:
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} \cdot \left(2 n\right) = \left(2 n + 2\right) + 2$
Simplifying
$\textcolor{w h i t e}{\text{XXX}} n = 2 n + 4$
$\textcolor{w h i t e}{\text{XXX}} - n = 4$
$\textcolor{w h i t e}{\text{XXX}} n = - 4$
The numbers are
$\textcolor{w h i t e}{\text{XXX}} 2 n = - 8$
and
$\textcolor{w h i t e}{\text{XXX}} 2 n + 2 = - 6$ |
When writing out measurements, it is best to use numerals and show the units of measurement. Some units have standard abbreviation which can be used although these may need to be written out fully for their first use in the paper if it is not obvious for the field.
Common error: Pluralizing abbreviated units
Example of error: The measured weight was 3 gs.
Fixed – option 1: The measured weight was 3 grams.
Fixed – option 2: The measured weight was 3 g.
Common error: Pluralizing measured (or uncounted) quantities
Example of error: Then, 3.3 mL of the solution were separated.
Fixed: Then, 3.3 mL of the solution was separated.
Example of error: 12% of the time were spent undercover.
Fixed: 12% of the time was spent undercover.
This rule refers to measured quantities (as opposed to counted) where there may be decimal places or things are not counted one by one. For individual items that can be counted, use the plural.
Example of good use: Four solutions were considered.
Example of good use: 30% of participants were over 40 years old.
When writing measurements, it’s generally better to use whole numbers or decimals as these are likely what the measurement tool uses and is easier for readers to compare. For example, it takes a moment to figure out whether 3/7 or 4/9 is higher, but it’s clear that 0.43 is lower than 0.44.
### Exercise 6 – plurals in measurements:
1. Which is correct?
(a) We found that 14% of hats were black.
(b) We found that 14% of hats was black.
2. Which is correct?
(a) There were 8 g of silicone.
(b) There was 8 g of silicone.
3. Which is correct?
(a) The results showed there were a 20% decrease.
(b) The results showed there was a 20% decrease.
1. (a) is correct because we can count the number of hats that are black.
2. (b) is correct because the quantity is measured rather than counted.
3. (b) is correct because we cannot count the decrease; we can only measure or calculate it
Common error: Not including leading zeros
Example of error: The dosage was .34 mL.
Fixed: The dosage was 0.34 mL.
This is clearer because the reader can easily notice the decimal point.
There are some exceptions to this rule such as when a range is between 0 and 1, such as with some statistical tests. In these cases, check the relevant style guide. |
Atwood Machine
An Atwood machine is a laboratory set-up of two objects connected by a massless string that is passing over an ideal pulley.
Ideal pulley is massless and frictionless. It affects nothing on the object’s motion. The tension is also constant in the whole string.
To solve problems of objects in pulley, the following guides must be followed:
1. Draw a free-body diagram of each mass on the string.
2. Indicate the direction of the motion of the object.
3. Set-up the formula or equation to be used to solve the problem using Newton’s second law of motion and the equation for the different types of forces.
Consider two objects $$m_1$$ and $$m_2$$ hanging on an Atwood machine.
Since the two objects are connected by the same string over, the tension is the same. Object $$m_1$$ is heavier than $$m_2$$, thus $$m_1$$ accelerates downward pulling $$m_2$$ upward, indicating a negative and positive direction, respectively. Thus, the equation for the net force acting on each mass will be expressed as:
For $$m_1$$:
$$F_{net}=T-m_1 g=-m_1 a \implies T=m_1 g – m_1 a$$
For $$m_2$$:
$$F_{net}=T-m_2 g=m_2 a \implies T=m_2 g + m_2 a$$
Since the tension is the same in all the objects, the above equations can be combined as
$$m_1 g – m_1 a= m_2 g + m_2 a$$
Expressing in terms of $$g$$ and $$a$$
$$m_1 g – m_2 g = m_1 a + m_2 a \implies g(m_1 – m_2)=a(m_1 + m_2)$$
It follows that the acceleration of the entire system is the same. Therefore, the acceleration can be solved as
$$a=g \begin{pmatrix} \frac {m_1 – m_2}{m_1 + m_2} \end{pmatrix}$$
Example 1.
What is the acceleration of the objects in the pulley as illustrated below.
Given:
$$m_1=10\;kg\\ m_2=5\;kg$$
Solution 1.
Draw a separate free-body diagram for each mass in the pulley. Indicate the direction of the acceleration of the masses.
Mass 1 accelerates downward. This is considered so based on the mass of the objects. Since $$m_1 > m_2$$, it indicates that $$m_1$$ will pull the string downward causing $$m_2$$ to accelerate upward, giving a positive direction to $$m_2$$.
Solution 2.
Calculate for the acceleration of the masses in the entire system using $$a=g \begin{pmatrix} \frac {m_1 – m_2}{m_1 + m_2} \end{pmatrix}$$.
$$a=9.8\;m/s^2 \begin{pmatrix} \frac {10\;kg – 5\;kg}{10\;kg + 5\;kg} \end{pmatrix}\\$$
$$=9.8\;m/s^2 \begin{pmatrix} \frac {5\;kg}{15\;kg} \end{pmatrix}\\$$
$$=9.8\;m/s^2 \begin{pmatrix} \frac {1}{3} \end{pmatrix}\\$$
$$=3.27\;m/s^2$$
$$m_1$$ accelerates at a rate of $$3.27\;m/s^2\;\text{downward}$$ and $$m_2$$ at $$3.27\;m/s^2 \;\text{upward}$$.
Example 2.
Consider the system of masses below. Let $$a$$ be the mass of the first object and $$b$$ the mass of the second object.
Given:
$$a= 58\;kg\\ b= 36\;kg$$
Solution 1.
Create a free-body diagram for each object in the system.
Solution 2.
The acceleration of the objects is calculated by the equation $$a=g\begin{pmatrix}\frac{a-b}{a+b} \end{pmatrix}$$.
$$a=9.8\;m/s^2 \begin{pmatrix} \frac {58\;kg – 36\;kg}{58\;kg + 36\;kg } \end{pmatrix}\\$$
$$=9.8\;m/s^2 \begin{pmatrix} \frac {22\;kg}{94\;kg } \end{pmatrix}\\$$
$$=9.8\;m/s^2 (0.23)\\$$
$$=2.25\;m/s^2$$
$$a$$ accelerates $$2.25\;m/s^2 \;\text{downward}$$ and $$b$$ at $$2.25\;m/s^2 \;\text{upward}$$. |
Point Estimate Calculator
# Point Estimate Calculator
Maximum Likelihood Estimation (MLE):
Laplace Method:
Jeffrey's Method:
Best Estimate:
Best Method:
# Precision Unlocked : Using the Point Estimate Calculator
In the world of statistics and data analysis, accuracy is key. Making informed decisions often hinges on precise estimates. When dealing with proportions and percentages, a powerful tool at your disposal is the Point Estimate Calculator. In this article, we’ll dive into what a Point Estimate Calculator is, how it works, and why it’s a vital tool for statisticians, data scientists, and decision-makers alike.
### What is a Point Estimate Calculator?
A Point Estimate Calculator is a statistical tool that provides a single, specific value as an estimate of a population parameter. It’s used when we want to make predictions or infer characteristics about a larger group (population) based on a smaller sample of that group. Point estimates help us draw conclusions without having to examine every member of the population.
### The Four Point Estimate Methods
The Point Estimate Calculator employs four distinct methods to provide these precise estimates:
1. Maximum Likelihood Estimation (MLE):
• Formula: x / n, where x is the number of successes in the sample, and n is the sample size.
• When to Use: MLE is applicable when the ratio of successes to the sample size is less than or equal to 0.5.
2. Wilson:
• Formula: (x + z^2/2) / (n + z^2), where x is the number of successes, n is the sample size, and z is the z-score associated with a level of confidence.
• When to Use: The Wilson method is employed when the ratio of successes to the sample size falls between 0.5 and 0.9.
3. Laplace:
• Formula: (x + 1) / (n + 2), where x is the number of successes, and n is the sample size.
• When to Use: Laplace or Jeffrey’s method is applied when the ratio of successes to the sample size is between 0.9 and 1.0. The smallest of these estimates is used.
4. Jeffrey’s:
• Formula: (x + 0.5) / (n + 1), where x is the number of successes, and n is the sample size.
• When to Use: Jeffrey’s method, like Laplace, is used when the ratio of successes to the sample size is between 0.9 and 1.0. The smallest of these estimates is chosen.
### Why is the Point Estimate Calculator Important?
1. Efficiency: Calculating estimates for large populations is time-consuming and often impractical. Point estimates allow you to make informed decisions with a manageable amount of data.
2. Decision-Making: Whether you’re in marketing, finance, or any field reliant on data, point estimates empower you to assess trends, make predictions, and allocate resources wisely.
3. Accuracy: These methods are designed to provide reasonably accurate estimates even with limited data. This is especially valuable when working with small sample sizes.
### Who Should Use the Point Estimate Calculator?
• Data Analysts: Professionals who work with data to extract valuable insights can employ the Point Estimate Calculator to make informed decisions.
• Researchers: Researchers in various fields, from social sciences to medicine, can use point estimates to draw conclusions from sample data.
• Business Analysts: Business analysts can leverage point estimates to assess market trends, forecast demand, and allocate budgets effectively.
• Statisticians: Statisticians rely on point estimates to conduct hypothesis testing and determine the reliability of data.
### Using the Point Estimate Calculator
1. Gather Data: First, you’ll need data from a representative sample. This sample should be randomly selected to ensure unbiased results.
2. Select the Appropriate Method: Choose the point estimate method based on the ratio of successes to the sample size, as explained earlier.
3. Plug in the Values: Enter the number of successes (x), sample size (n), and the confidence level’s z-score into the calculator.
4. Get Your Estimate: Click the “Calculate” button, and the calculator will provide you with the point estimates according to the chosen method(s).
### Best Practices for Point Estimates
Understand Your Data: Ensure your sample data is representative of the population you’re estimating.
Use Confidence Intervals: While point estimates are valuable, combining them with confidence intervals provides a more comprehensive picture of the data’s reliability.
Choose the Right Method: Select the point estimate method that best fits your data’s characteristics.
Communicate Clearly: When presenting point estimates, make it clear which method you’ve used and what the estimate represents.
### In Conclusion
The Point Estimate Calculator is a powerful tool for anyone dealing with data analysis and statistics. Whether you’re making business decisions, conducting research, or simply seeking more accurate insights. Understanding and using point estimates can significantly enhance your decision-making capabilities.
By employing methods like Maximum Likelihood, Wilson, Laplace, and Jeffrey’s, you can unlock precision in your data analysis. Helping you make more informed and confident choices in an increasingly data-driven world.
In summary, the Point Estimate Calculator isn’t just about numbers; it’s about making better decisions based on data-backed insights. |
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# How do you factor ${{x}^{2}}-8x+16$ ?
Last updated date: 20th Jun 2024
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Hint: In the question we need to factorise the given equation. We can observe that the given equation is a quadratic equation. Generally, we will follow different methods for different types of equations. For the quadratic equation which is in the form of $a{{x}^{2}}+bx+c$, we will break the middle term into two parts like $b={{x}_{1}}+{{x}_{2}}$ such that ${{x}_{1}}.{{x}_{2}}=ac$. So, we will first calculate the value of $ac$ and we will write the factors of the calculated $ac$ and choose two factors such that they satisfy the above-mentioned conditions. After getting the values of ${{x}_{1}}$, ${{x}_{2}}$ we will break the middle term and simplify the equation by taking appropriate terms as common. After taking the terms as common we will get our required result.
Given that, ${{x}^{2}}-8x+16$.
Comparing the above equation with $a{{x}^{2}}+bx+c$, then we can write
Coefficient of ${{x}^{2}}$ is $a=1$.
Coefficient of $x$ is $b=-8$.
Constant is $c=16$.
Now the value of $ac$ is given by
\begin{align} & ac=1\left( 16 \right) \\ & \Rightarrow ac=16 \\ \end{align}
Here we have the value of $ac$ as $16$. We know that the factors for $16$ are $1$, $2$, $4$, $8$, $16$, $45$.
From the above factors we can observe that
\begin{align} & -4-4=8 \\ & -4\times -4=16 \\ \end{align}
So, we can break the middle term $-8x$ as $-8x=-4x-4x$. From this value, we are going to write the given equation as
$\Rightarrow {{x}^{2}}-8x+16={{x}^{2}}-4x-4x+16$
Taking $x$ as common from ${{x}^{2}}-4x$ and $-4$ as common from $-4x+16$, then the above equation is modified as
$\Rightarrow {{x}^{2}}-8x+16=x\left( x-4 \right)-4\left( x-4 \right)$
Again, taking $x-4$ as common on RHS of the above equation, then we will get
$\Rightarrow {{x}^{2}}-8x+16=\left( x-4 \right)\left( x-4 \right)$
Hence the factors of the given equation are $x-4$ and $x-4$.
We can also write the above equation as
$\Rightarrow {{x}^{2}}-8x+16={{\left( x-4 \right)}^{2}}$
Note:
In this problem, they have only asked to factorize the given equation, so we have ended our procedure after calculating the factors. If they have asked to solve the equation or asked to find the roots, then we need to equate the given equation to zero and we will find the values of $x$. |
# Khan Academy Slope Intercept Form
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Khan Academy Slope Intercept Form – There are many forms that are used to represent a linear equation, the one most frequently used is the slope intercept form. You may use the formula for the slope-intercept to identify a line equation when that you have the straight line’s slope as well as the y-intercept. This is the coordinate of the point’s y-axis where the y-axis crosses the line. Learn more about this particular line equation form below.
## What Is The Slope Intercept Form?
There are three basic forms of linear equations, namely the standard one, the slope-intercept one, and the point-slope. Though they provide the same results when utilized, you can extract the information line generated faster by using the slope intercept form. The name suggests that this form employs an inclined line, in which its “steepness” of the line reflects its value.
The formula can be used to determine the slope of a straight line, the y-intercept, also known as x-intercept which can be calculated using a variety of available formulas. The line equation in this formula is y = mx + b. The straight line’s slope is indicated in the form of “m”, while its y-intercept is signified via “b”. Each point of the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” must remain as variables.
## An Example of Applied Slope Intercept Form in Problems
The real-world in the real world, the slope-intercept form is commonly used to depict how an object or problem changes in its course. The value provided by the vertical axis is a representation of how the equation addresses the degree of change over the value given with the horizontal line (typically the time).
An easy example of the application of this formula is to determine how many people live in a particular area as the years go by. If the population of the area increases each year by a certain amount, the value of the horizontal axis will rise one point at a moment for every passing year, and the point worth of the vertical scale will increase in proportion to the population growth by the fixed amount.
You may also notice the beginning value of a particular problem. The starting point is the y value in the yintercept. The Y-intercept is the point at which x equals zero. In the case of the problem mentioned above the beginning point could be when the population reading starts or when the time tracking starts, as well as the associated changes.
The y-intercept, then, is the point in the population when the population is beginning to be documented in the research. Let’s assume that the researcher began with the calculation or measurement in the year 1995. In this case, 1995 will be”the “base” year, and the x=0 points will occur in 1995. This means that the population in 1995 is the y-intercept.
Linear equation problems that utilize straight-line formulas are almost always solved this way. The starting value is depicted by the y-intercept and the change rate is expressed as the slope. The main issue with this form is usually in the horizontal variable interpretation, particularly if the variable is attributed to the specific year (or any kind or unit). The key to solving them is to ensure that you comprehend the variables’ meanings in detail. |
# How do you integrate int 1/sqrt(4x^2-12x+8) using trigonometric substitution?
Mar 25, 2018
$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 8}} \mathrm{dx} = \frac{1}{2} \text{arcosh"(2x-3)+"c}$
#### Explanation:
We want to find $\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 8}} \mathrm{dx}$
We start by transforming the integral
$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 8}} \mathrm{dx} = \int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 9 - 1}} \mathrm{dx} = \int \frac{1}{\sqrt{{\left(2 x - 3\right)}^{2} - 1}} \mathrm{dx}$.
Now let $u = 2 x - 3$ and $\mathrm{du} = 2 \mathrm{dx}$. The integral becomes
$\int \frac{1}{\sqrt{{\left(2 x - 3\right)}^{2} - 1}} \mathrm{dx} = \frac{1}{2} \int \frac{1}{\sqrt{{u}^{2} - 1}} \mathrm{du}$
The integral is a standard integral and evaluates to
$\frac{1}{2} \int \frac{1}{\sqrt{{u}^{2} - 1}} \mathrm{du} = \frac{1}{2} \text{arcosh"u + "c"=1/2"arcosh"(2x-3)+"c}$ |
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## Number and algebra
• The Number System and Place Value
• Calculations and Numerical Methods
• Fractions, Decimals, Percentages, Ratio and Proportion
• Properties of Numbers
• Patterns, Sequences and Structure
• Algebraic expressions, equations and formulae
• Coordinates, Functions and Graphs
## Geometry and measure
• Angles, Polygons, and Geometrical Proof
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## Probability and statistics
• Handling, Processing and Representing Data
• Probability
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## For younger learners
• Early Years Foundation Stage
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## Area and Perimeter KS2
This collection is one of our Primary Curriculum collections - tasks that are grouped by topic.
## Making Boxes
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
## Numerically Equal
Can you draw a square in which the perimeter is numerically equal to the area?
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
## Torn Shapes
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
## Twice as Big?
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
## Area and Perimeter
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
## Ribbon Squares
What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether?
## Through the Window
My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
Area worksheets 5th grade introduce students to various types of shapes and how to calculate their areas. As the scope of the topic is very vast, hence, it is necessary for the students to get the right type of worksheet that will clear their concepts rather than confuse them.
## Benefits of 5th Grade Area Worksheets
Area worksheets 5th grade tend to start with using simple shapes so that students can understand the application of the formulas. Gradually the level starts increasing wherein students have to solve more complex problems that might involve formula manipulations. By solving problems with a gradual increase in the level of difficulty, students can get a better understanding of the concepts associated with the topic. The benefit of these 5th grade math worksheets is that it provides a visual representation of the problems for students to understand the solutions at each level.
## Printable PDFs for Grade 5 Area Worksheets
• Math 5th Grade Area Worksheet
• 5th Grade Area Math Worksheet
• Grade 5 Math Area Worksheet
Explore more topics at Cuemath's Math Worksheets .
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## Calculating Areas: Year 5 – Measurement – Maths Challenge
Resource Collection Challenge Maths
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These calculating area worksheets provide extra challenge for Y5 children. A variety of problems are spread across three sections, enabling you to use the whole sheet during a single lesson or to select specific problems for different teaching sessions, or home learning.
Y5 maths objective: calculate and compare the area of rectangles (including squares), and including using standard units, square centimetres (cm2) and square metres (m2) and estimate the area of irregular shapes. NOTE: THESE SHEETS MUST BE PRINTED AT 100% IN ORDER FOR THE MEASUREMENTS TO BE ACCURATE.
This resource is part of the Challenge Maths collection. View more from this collection
• NB THESE SHEETS MUST BE PRINTED AT 100% IN ORDER FOR THE MEASUREMENTS TO BE ACCURATE.
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## Area and Perimeter Word Problems
Area and perimeter worksheets with word problems for grade 5 where you calculate with the given dimensions of the shape. this is an excellent resource for grade 5 to understand real life problems. subscribe to www.grade1to6.com for \$25 / rs 2000 a year or buy this grade 5 math workbook for \$6 / rs 450 here.
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## Area of Rectangles Lesson
This Year 5 Area of Rectangles lesson covers the prior learning of counting squares, before moving onto the main skill of finding the area of rectangles.
The lesson starts with a prior learning worksheet to check pupils’ understanding. The interactive lesson slides recap the prior learning before moving on to the main skill. Children can then practise further by completing the activities and can extend their learning by completing an engaging extension task.
National Curriculum Objective
Mathematics Year 5: (5M7b) Calculate and compare the area of rectangles (including squares), and including using standard units, square centimetres (cm2) and square metres (m2) and estimate the area of irregular shapes
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## Lesson Slides
These lesson slides guides pupils through the prior learning of counting squares, before moving onto the main skill of area of rectangles. There are a number of questions to check pupils' understanding throughout.
## Modelling PowerPoint
This powerpoint can be used to model the questions that the children will complete on the Varied Fluency and Reasoning & Problem Solving worksheets as part of this lesson.
These are the same as the lesson slides on Classroom Secrets. You can assign this as an activity for pupils to access individually in school or remotely from home.
## Video Tutorial
Jade introduces a simple formula for calculating area in this Area of Rectangles Video Tutorial.
## 1 Prior Learning
This worksheet recaps prior learning of counting squares, before moving onto the main skill of finding the area of rectangles.
## Flash Cards Activity
These Year 4 Find the Area Flashcards are designed to check pupils’ understanding of finding area by counting squares.
## Interactive Animation
This Counting Squares Interactive Animation supports pupils’ understanding of area by counting squares.
## Interactive Activity
This Year 4 Counting Squares Game includes three questions designed to check pupils’ understanding of finding area by counting squares.
## 2 Varied Fluency
This worksheet includes varied fluency questions for pupils to practise the main skill for this lesson.
This Year 5 Area of Rectangles Game checks pupils’ understanding of calculating the area of rectangles using different methods.
## 2 Reasoning & Problem Solving
This area of rectangles extension task includes a challenge activity which can be used to further pupils' understanding of the concepts taught in the area of rectangles lesson.
## Mixed Practice
This worksheet includes varied fluency, reasoning and problem solving questions for pupils to practise the main skill of area of rectangles.
This differentiated worksheet includes reasoning and problem solving questions to support the teaching of this step.
## Discussion Problem
This Discussion Problems worksheet includes two discussion problems which can be used in pairs or small groups to further pupils' understanding of the concepts taught in this lesson.
This Year 5 Area of Rectangles Maths Challenge will help your pupils develop their problem-solving skills using their knowledge of calculating the area of rectangles.
This differentiated worksheet includes varied fluency and reasoning and problem solving questions to support the teaching of this step.
## Learning Video Clip
Alfie is helping Uncle Paul work out how much paint to order for the living room by calculating the area of the rectangular walls. They then decide to order some rectangular tiles for the bathroom floor and need help working out which sizes to purchase.
## Consolidation
This resource is aimed at Year 5 Expected and has been designed to give children the opportunity to consolidate the skills they have learned in Autumn Block 5 – Geometry Perimeter and Area.
## Home Learning Pack
This Autumn week 12 Maths pack contains varied fluency, reasoning and problem solving worksheets.
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## Word Problems Worksheets | Area of Rectangles
Promote mathematical interest in children of grade 3 through grade 6 by relating the concept to the real-life situation with this batch of printable word problems worksheets on area of rectangles. The word problems offer two levels of difficulty; level 1 comprises word problems to find the area of rectangles and the missing parameters, while level 2 has scenarios to find the area involving unit conversions or finding the cost. Also, included here are word problems to find the area of rectangular shapes. Delve into practice with the free worksheets here!
Area Word Problems | Level 1
Equip children to bring together reality and math with these pdf word problems on area of rectangles. Instruct 3rd grade and 4th grade kids to plug in the dimensions in the formula A = length * width to compute the area.
Area Word Problems | Level 2
Visualize the scenario and highlight the length and width. Children are expected to convert to the specified units and then solve for the area of rectangles. Some word problems require calculating the cost.
Find the Missing Length or Width | Level 1
Direct children of grade 4 and grade 5 to figure out the area and the length or width in each word problem. Rearrange the rectangle formula, making the missing dimension the subject, substitute and solve for length or width.
Find the Missing Length or Width | Level 2
Level up with this set of printable worksheets featuring word problems to solve for the unknown dimension using the area and the dimension given. The word problems involve either unit conversion or determining the cost to provide ample practice for 5th grade and 6th grade children.
Area of Rectangular Paths | Word Problems
Read the scenario and draw the rectangular path, decompose the path into non-overlapping rectangles and find the area of each individual rectangle, add the areas to determine the area of the rectangular paths in this set of pdf worksheets.
Related Worksheets
» Area of Squares
» Area of Parallelograms
» Area of Polygons
» Rectangles
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## Area and Perimeter Problem Solving
Subject: Mathematics
Age range: 11-14
Resource type: Lesson (complete)
Last updated
19 December 2014
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## alannaboylan
Thank you.Excellent resource.
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## Rachel Garbett
Fantastic resource, excellent link to other mathematical areas. Thank you for sharing!
## bhavintoprani
great resource. Thanks.
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## Not quite what you were looking for? Search by keyword to find the right resource:
#### IMAGES
1. Year 5 Problem Solving with Place Value Maths Challenge
2. Worksheet Grade 5 Math Real Life Word Problem
3. 😍 Problem solving area and perimeter. Fourth grade Lesson Master those Area and Perimeter Word
4. Area and Perimeter Problem Solving Activity for most able Year 5- 6
5. Year 6 Problem Solving: Area and Perimeter Puzzle (One Worksheet and Answer)
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1. Area and Perimeter KS2
Age 7 to 11 Challenge Level Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume? Numerically Equal Age 7 to 11 Challenge Level Can you draw a square in which the perimeter is numerically equal to the area? Fitted Age 7 to 11 Challenge Level
2. Area and Perimeter Ages 9
Year 5 Measure the Perimeter of Composite Rectilinear Shapes Differentiated Worksheets 4.2 (16 reviews) KS2 Area and Perimeter Games 4.2 (20 reviews) Solving Perimeter Problems Differentiated Worksheets 4.8 (10 reviews)
3. Area and perimeter: Year 5: Planning tool
Area and perimeter: Year 5: Planning tool Home / Planning tool / 5 / Measurement / Area and perimeter Planning tool Year levels F 1 2 3 4 5 6 7 8 9 10 Strands Number Algebra Measurement Space Probability Statistics Topics Metric units and using instruments Area and perimeter Time and duration Angles and parallel lines Expected level of development
4. PDF Year 5 Area of Rectangles Reasoning and Problem Solving
Reasoning and Problem Solving - Area of Rectangles - Year 5 Greater Depth Reasoning and Problem Solving Area of Rectangles Developing 1a. Hafsa is correct because 5cm x 5cm = 25cm2 so 25 tiles are needed. 2a. Various answers, for example: W=3cm and L=4cm, W=1cm and L=12cm 3a.
Area worksheets for grade 5 are easy and flexible, students can download these worksheets in PDF format for free. Explore more topics at Cuemath's Math Worksheets. 5th Grade Area Worksheets - Worksheets aid in improving the problem-solving skills of students in turn guiding the kids to learn and understand the patterns as well as the logic of ...
6. Autumn Maths Year 5 Perimeter and Area
Step 1 Convert Units of Length Additional Supporting Step → Step 2 Find Unknown Lengths Additional Supporting Step → 5 Autumn Block 5 Consolidation Pack Additional Supporting Step → Differentiated maths resources for Autumn Block 5 (Perimeter and Area) in small steps for KS2 children in Year 5.
7. KS2/KS3 Area/Perimeter Problems
Age range: 7-11 Resource type: Other File previews docx, 102.51 KB docx, 70.56 KB docx, 196.47 KB jpg, 962.88 KB jpg, 1.4 MB jpg, 1004.11 KB Having looked online at various area and perimeter questions, i felt they were all too similar, with very little problem solving/investigation skills being used. So, I decided to create these.
8. Area Word Problems Worksheet
Perfect for developing children's fluency, reasoning and problem-solving skills. More area word problems to expand teaching. ... Interactive PDF: White Rose Maths Supporting Year 6: Spring Block 5 Perimeter, Area and Volume: Area of a Triangle (3) Comparing Area Differentiated Year 4 Maths Worksheets. What is Area? Year 4 Differentiated Maths ...
9. Area and perimeter problem solving
Area and perimeter problem solving. Subject: Mathematics. Age range: 5-7. Resource type: Lesson (complete) dianatany. 4.48 89 reviews. ... 3 years ago. report. 4. Really great challenging problems that have helped me get better on the are and perimeter topic.
10. PlanIt Maths Year 5 Measurement Lesson Pack 5: Area of ...
Teach year 5 children to calculate the area of rectangles (including squares) by using the formula of length × width. As well as using given measurements to calculate area, children are also taught how to find missing side lengths from the information provided.
11. White Rose Maths Compatible Y5 Step 4 Area of Rectangles
This maths mastery teaching pack will deepen year 5 children's understanding of how to calculate the area of rectangles (including squares) by using the formula of length × width, which progresses on from their learning in year 4 of using a grid to calculate area.
12. Area and Perimeter Word Problems
teaching resource Area and Perimeter Word Problems Updated: 20 Nov 2023 Get students to solve area and perimeter word problems with this set of three worksheets. Editable: Google Slides Non-Editable: PDF Pages: 4 Pages Curriculum: AUS V8, AUS V9 Years: 4 - 5 Download Preview File Get inspired!
13. Browse Printable Area Worksheets
Area of a Triangle #1. Worksheet. Calculating Surface Area #1. Interactive Worksheet. Find the Area of a Rectangle: Level 2. Interactive Worksheet. Quilting Coordinates: Coordinate Plane Performance Task. Worksheet. Shady Shapes: An Area Activity.
14. KS2 Area and Perimeter Worksheet including problem solving & money
pdf, 596.59 KB. KS2 area and perimeter worksheet - the first part requires children to calculate the area/perimeter of different fields. The second part of the task focuses on problem solving, requiring children to calculate costs for total perimeter fence needed, how many sheep can fit into the fields and the total cost of the sheep.
15. Area Word Problems Worksheet
area problem solving . area problems . area and perimeter year 5 . year 4 area . area year 4 . area and perimeter ... Interactive PDF: White Rose Maths Supporting Year 6: Spring Block 5 Perimeter, Area and Volume: Area and Perimeter. Area of Composite Shapes Differentiated Maths Activity Sheets. Multi-Step Word Problems Worksheet.
16. Problem-solving Maths Investigations for Year 5
Hamilton provide an extensive suite of problem-solving maths investigations for Year 5 to facilitate mathematical confidence, investigative inquiry and the development of maths meta skills in 'low floor - high ceiling' activities for all. Explore all our in-depth problem solving investigations for Year 5.
17. Calculating Areas: Year 5
These calculating area worksheets provide extra challenge for Y5 children. A variety of problems are spread across three sections, enabling you to use the whole sheet during a single lesson or to select specific problems for different teaching sessions, or home learning. Y5 maths objective: calculate and compare the area of rectangles ...
18. Area and Perimeter Worksheets
Area and Perimeter worksheets with Word Problems for Grade 5 where you calculate with the given dimensions of the shape. This is an excellent resource for Grade 5 to understand real life problems. Subscribe to www.grade1to6.com for \$25 / Rs 2000 a year or buy this Grade 5 math workbook for \$6 / Rs 450 here. Download Now. Learning made easy ...
19. Year 5 Area of Rectangles Lesson
This Year 5 Area of Rectangles lesson covers the prior learning of counting squares, before moving onto the main skill of finding the area of rectangles. ... powerpoint can be used to model the questions that the children will complete on the Varied Fluency and Reasoning & Problem Solving worksheets as part of this lesson. Login to download.
20. Area and perimeter
Unit test. Test your understanding of Area and perimeter with these % (num)s questions. Area and perimeter help us measure the size of 2D shapes. We'll start with the area and perimeter of rectangles. From there, we'll tackle trickier shapes, such as triangles and circles.
21. Area of a Rectangle Word Problems
Direct children of grade 4 and grade 5 to figure out the area and the length or width in each word problem. Rearrange the rectangle formula, making the missing dimension the subject, substitute and solve for length or width. Find the Missing Length or Width | Level 2
22. PDF Year 5 Area of Compound Shapes Reasoning and Problem Solving
1a. Various answers, for example: Accept any compound shapes with an area of 16cm2. Each shape should have 4 squares shaded. 2a. A. 9 x 5 = 45cm2; B. 5 x 5 = 25cm2 3a.
23. Area and Perimeter Problem Solving
Age range: 11-14 Resource type: Lesson (complete) File previews pptx, 312.58 KB Some reasoning questions around the topic of area and perimeter. Lots of opportunities for class discussion about methods and problem solving, and linking in to solving algebraic equations and simplifying expressions! Creative Commons "NoDerivatives" |
# How do you solve log _(2) (log _(2) (log _(2) x))) = 2?
Dec 17, 2015
Make repeated use of the fact that ${2}^{{\log}_{2} \left(x\right)} = x$ to find that
$x = 65536$
#### Explanation:
Applying the property of logarithms that
${a}^{{\log}_{a} \left(x\right)} = x$
we have
${\log}_{2} \left({\log}_{2} \left({\log}_{2} \left(x\right)\right)\right) = 2$
$= {2}^{{\log}_{2} \left({\log}_{2} \left({\log}_{2} \left(x\right)\right)\right)} = {2}^{2}$
$\implies {\log}_{2} \left({\log}_{2} \left(x\right)\right) = 4$
$\implies {2}^{{\log}_{2} \left({\log}_{2} \left(x\right)\right)} = {2}^{4}$
$\implies {\log}_{2} \left(x\right) = 16$
$\implies {2}^{{\log}_{2} \left(x\right)} = {2}^{16}$
$\implies x = 65536$
Dec 17, 2015
$x = {2}^{16} = 65536$
#### Explanation:
Start from this point: if you know that $a = b$, then it must be ${2}^{a} = {2}^{b}$. So, start from your equation, and deduce that
${2}^{{\log}_{2} \left({\log}_{2} \left({\log}_{2} \left(x\right)\right)\right)} = {2}^{2}$
Now, by definition, ${2}^{{\log}_{2} \left(z\right)} = z$, while of course ${2}^{2} = 4$. So, the equation becomes
${\log}_{2} \left({\log}_{2} \left(x\right)\right) = 4$
And now we're in the same situation as before, so we can apply the same logic:
${2}^{{\log}_{2} \left({\log}_{2} \left(x\right)\right)} = {2}^{4}$
Which means
${\log}_{2} \left(x\right) = 16$
One last iteration gives us
$x = {2}^{16} = 65536$
${\log}_{2} \left({\log}_{2} \left({\log}_{2} \left({2}^{16}\right)\right)\right) = {\log}_{2} \left({\log}_{2} \left(16\right)\right) = {\log}_{2} \left(4\right) = 2$ |
```Solving
Linear
Systems
Solving
Linear
Systems
3-6
3-6 in Three Variables
in Three Variables
Warm Up
Lesson Presentation
Lesson Quiz
HoltMcDougal
Algebra 2Algebra 2
Holt
3-6
Solving Linear Systems
in Three Variables
Warm Up
Solve each system of equations algebraically.
1.
x = 4y + 10
4x + 2y = 4
(2, –2) 2.
6x – 5y = 9
2x – y =1
(–1,–3)
Classify each system and determine the
number of solutions.
3x – y = 8
x = 3y – 1
4.
3.
6x – 2y = 2
6x – 12y = –4
inconsistent; none
Holt McDougal Algebra 2
consistent, independent; one
3-6
Solving Linear Systems
in Three Variables
Objectives
Represent solutions to systems of
equations in three dimensions
graphically.
Solve systems of equations in three
dimensions algebraically.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Systems of three equations with three
variables are often called 3-by-3 systems.
In general, to find a single solution to any
system of equations, you need as many
equations as you have variables.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Recall from Lesson 3-5 that the graph of a
linear equation in three variables is a plane.
When you graph a system of three linear
equations in three dimensions, the result is
three planes that may or may not intersect.
The solution to the system is the set of points
where all three planes intersect. These
systems may have one, infinitely many, or no
solution.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Identifying the exact solution from a
graph of a 3-by-3 system can be very
difficult. However, you can use the
methods of elimination and substitution to
reduce a 3-by-3 system to a 2-by-2
system and then use the methods that
you learned in Lesson 3-2.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 1: Solving a Linear System in Three
Variables
Use elimination to solve the system of equations.
5x – 2y – 3z = –7
1
2x – 3y + z = –16
2
3x + 4y – 2z = 7
3
Step 1 Eliminate one variable.
In this system, z is a reasonable choice to eliminate
first because the coefficient of z in the second
equation is 1 and z is easy to eliminate from the
other equations.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
1
2
5x – 2y – 3z = –7
5x – 2y – 3z = –7
3(2x –3y + z = –16)
Multiply equation
6x – 9y + 3z = –48 to equation 1 .
11x – 11y
Use equations
in x and y.
3
and
2
= –55
4
to create a second equation
1
3
2
3x + 4y – 2z = 7
3x + 4y – 2z = 7
Multiply equation
2(2x –3y + z = –16) 4x – 6y + 2z = –32 -2 by 2, and add
7x – 2y
Holt McDougal Algebra 2
= –25
to equation
5
3
.
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
You now have a 2-by-2 system.
Holt McDougal Algebra 2
11x – 11y = –55
4
7x – 2y = –25
5
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
Step 2 Eliminate another variable. Then solve for
the remaining variable.
You can eliminate y by using methods from
Lesson 3-2.
4
5
equation
–2(11x – 11y = –55) –22x + 22y = 110 1 Multiply
-4 by –2, and
11(7x – 2y = –25) 77x – 22y = –275 equation -5 by 11
55x
1
= –165
x = –3
Holt McDougal Algebra 2
Solve for x.
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
Step 3 Use one of the equations in your 2-by-2
system to solve for y.
4
11x – 11y = –55
1
11(–3) – 11y = –55
Substitute –3 for x.
1
y=2
Holt McDougal Algebra 2
Solve for y.
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
Step 4 Substitute for x and y in one of the original
equations to solve for z.
2
2x – 3y + z = –16
2(–3) – 3(2) + z = –16
z = –4
The solution is (–3, 2, –4).
Holt McDougal Algebra 2
Substitute
1 –3 for x and
2 for y.
1Solve for y.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1
Use elimination to solve the system of equations.
–x + y + 2z = 7
1
2x + 3y + z = 1
2
–3x – 4y + z = 4
3
Step 1 Eliminate one variable.
In this system, z is a reasonable choice to eliminate
first because the coefficient of z in the second
equation is 1.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1 Continued
1
2
–x + y + 2z = 7 Multiply equation
–x + y + 2z = 7
–2(2x + 3y + z = 1) –4x – 6y – 2z = –2 -2 by –2, and add
1
–5x – 5y
Use equations
in x and y.
1
and
3
=5
to equation
.
4
to create a second equation
1
1
3
–x + y + 2z = 7
–2(–3x – 4y + z = 4)
–x + y + 2z = 7 Multiply equation
6x + 8y – 2z = –8 -3 by –2, and add
to equation
5x + 9y
Holt McDougal Algebra 2
= –1
5
1
.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1 Continued
You now have a 2-by-2 system.
–5x – 5y = 5
5x + 9y = –1
Holt McDougal Algebra 2
4
5
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1 Continued
Step 2 Eliminate another variable. Then solve for
the remaining variable.
You can eliminate x by using methods from
Lesson 3-2.
4
5
–5x – 5y = 5
5x + 9y = –1
4y = 4
y=1
Holt McDougal Algebra 2
Solve for y.
5
1
to equation
4
.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1
Step 3 Use one of the equations in your 2-by-2
system to solve for x.
4
–5x – 5y = 5
Substitute 1 for y.
–5x – 5(1) = 5
1
–5x – 5 = 5
–5x = 10
x = –2
Holt McDougal Algebra 2
Solve for x.
1
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1
Step 4 Substitute for x and y in one of the original
equations to solve for z.
2
2x +3y + z = 1
2(–2) +3(1) + z = 1
–4 + 3 + z = 1
z=2
Substitute –2 for x and
1 for y.
Solve for 1z.
1
The solution is (–2, 1, 2).
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
You can also use substitution to solve a
3-by-3 system. Again, the first step is to
reduce the 3-by-3 system to a 2-by-2
system.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
The table shows the number of each type of
ticket sold and the total sales amount for each
night of the school play. Find the price of each
type of ticket.
Orchestra
Mezzanine Balcony
Total Sales
Fri
200
30
40
\$1470
Sat
250
60
50
\$1950
Sun
150
30
0
\$1050
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 2 Continued
Step 1 Let x represent the price of an orchestra seat,
y represent the price of a mezzanine seat, and z
represent the present of a balcony seat.
Write a system of equations to represent the data in
the table.
200x + 30y + 40z = 1470
1
250x + 60y + 50z = 1950
2
Saturday’s sales.
150x + 30y = 1050
3
Sunday’s sales.
Friday’s sales.
A variable is “missing” in the last equation; however,
the same solution methods apply. Elimination is a good
choice because eliminating z is straightforward.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 2 Continued
Step 2 Eliminate z.
Multiply equation
1
2
1
by 5 and equation
5(200x + 30y + 40z = 1470)
–4(250x + 60y + 50z = 1950)
2
1000x + 150y + 200z = 7350
–1000x – 240y – 200z = –7800
y
=5
By eliminating z, due to the coefficients of x, you also
eliminated x providing a solution for y.
Holt McDougal Algebra 2
Solving Linear Systems
in Three Variables
3-6
Example 2 Continued
Step 3 Use equation
150x + 30y = 1050
150x + 30(5) = 1050
3
x=6
Holt McDougal Algebra 2
3
to solve for x.
Substitute 5 for y.
Solve for x.
3-6
Solving Linear Systems
in Three Variables
Example 2 Continued
Step 4 Use equations
1
or
2
to solve for z.
1
1
200x + 30y + 40z = 1470
200(6) + 30(5) + 40z = 1470
Substitute 6 for x and 5 for y.
Solve for x.
z=3
The solution to the system is (6, 5, 3). So, the
cost of an orchestra seat is \$6, the cost of a
mezzanine seat is \$5, and the cost of a balcony
seat is \$3.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 2
Jada’s chili won first place at the winter fair.
The table shows the results of the voting.
How many points are first-, second-, and
Winter Fair Chili Cook-off
Name
1st
Place
2nd
Place
3rd
Place
Total
Points
3
1
4
15
Maria
2
4
0
14
Al
2
2
3
13
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 2 Continued
Step 1 Let x represent first-place points, y represent
second-place points, and z represent thirdplace points.
Write a system of equations to represent the data in
the table.
3x + y + 4z = 15
1
2x + 4y = 14
2
Maria’s points.
2x + 2y + 3z = 13
3
Al’s points.
A variable is “missing” in one equation; however, the
same solution methods apply. Elimination is a good
choice because eliminating z is straightforward.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 2 Continued
Step 2 Eliminate z.
Multiply equation
1
by 3 and equation
3
1
3(3x + y + 4z = 15)
9x + 3y + 12z = 45
3
–4(2x + 2y + 3z = 13)
–8x – 8y – 12z = –52
x – 5y
Multiply equation
4
2
4
–2(x – 5y = –7)
2x + 4y = 14
Holt McDougal Algebra 2
= –7
by –2 and add to equation
2
4
.
–2x + 10y = 14
2x + 4y = 14
y = 2 Solve for y.
Solving Linear Systems
in Three Variables
3-6
Check It Out! Example 2 Continued
Step 3 Use equation
2
2x + 4y = 14
2x + 4(2) = 14
x=3
Holt McDougal Algebra 2
2
to solve for x.
Substitute 2 for y.
Solve for x.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 2 Continued
Step 4 Substitute for x and y in one of the original
equations to solve for z.
3
2x + 2y + 3z = 13
2(3) + 2(2) + 3z = 13
6 + 4 + 3z = 13
z=1
Solve for z.
The solution to the system is (3, 2, 1). The points for
first-place is 3, the points for second-place is 2, and 1
point for third-place.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
The systems in Examples 1 and 2 have unique
solutions. However, 3-by-3 systems may have
no solution or an infinite number of solutions.
Remember!
Consistent means that the system of equations
has at least one solution.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 3: Classifying Systems with Infinite Many
Solutions or No Solutions
Classify the system as consistent or inconsistent,
and determine the number of solutions.
2x – 6y + 4z = 2
–3x + 9y – 6z = –3
5x – 15y + 10z = 5
Holt McDougal Algebra 2
1
2
3
3-6
Solving Linear Systems
in Three Variables
Example 3 Continued
The elimination method is convenient because the
numbers you need to multiply the equations are small.
First, eliminate x.
Multiply equation
1
2
1
by 3 and equation
3(2x – 6y + 4z = 2)
2(–3x + 9y – 6z = –3)
2
6x – 18y + 12z = 6
–6x + 18y – 12z = –6
0=0
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 3 Continued
1
3
Multiply equation 1 by 5 and equation 3 by –2
10x – 30y + 20z = 10
5(2x – 6y + 4z = 2)
–10x + 30y – 20z = –10
–2(5x – 15y + 10z = 5)
0 = 0
Because 0 is always equal to 0, the equation is
an identity. Therefore, the system is consistent,
dependent and has an infinite number of solutions.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3a
Classify the system, and determine the number
of solutions.
3x – y + 2z = 4
1
2x – y + 3z = 7
2
–9x + 3y – 6z = –12
Holt McDougal Algebra 2
3
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3a Continued
The elimination method is convenient because the
numbers you need to multiply the equations by are
small.
First, eliminate y.
Multiply equation
1
3
2
by –1 and add to equation
3x – y + 2z = 4
–1(2x – y + 3z = 7)
.
3x – y + 2z = 4
–2x + y – 3z = –7
x
Holt McDougal Algebra 2
1
– z = –3
4
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3a Continued
Multiply equation
2
3
2
by 3 and add to equation
3(2x – y + 3z = 7)
–9x + 3y – 6z = –12
–9x + 3y – 6z = –12
Now you have a 2-by-2 system.
–3x + 3z = 9
Holt McDougal Algebra 2
4
5
.
6x – 3y + 9z = 21
–3x
x – z = –3
3
+ 3z = 9
5
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3a Continued
Eliminate x.
4
5
3(x – z = –3)
–3x + 3z = 9
3x – 3z = –9
–3x + 3z = 9
0=0
Because 0 is always equal to 0, the equation is
an identity. Therefore, the system is consistent,
dependent, and has an infinite number of solutions.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3b
Classify the system, and determine the number
of solutions.
2x – y + 3z = 6
1
2x – 4y + 6z = 10
2
y – z = –2
3
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3b Continued
Use the substitution method. Solve for y in equation 3.
3
y – z = –2
y=z–2
Solve for y.
4
Substitute equation
4
in for y in equation
2x – y + 3z = 6
2x – (z – 2) + 3z = 6
2x – z + 2 + 3z = 6
2x + 2z = 4
Holt McDougal Algebra 2
5
1
.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3b Continued
Substitute equation
4
in for y in equation
2x – 4y + 6z = 10
2x – 4(z – 2) + 6z = 10
2x – 4z + 8 + 6z = 10
2x + 2z = 2
6
Now you have a 2-by-2 system.
2x + 2z = 4
2x + 2z = 2
Holt McDougal Algebra 2
5
6
2
.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3b Continued
Eliminate z.
5
6
2x + 2z = 4
–1(2x + 2z = 2)
02
Because 0 is never equal to 2, the equation is a
inconsistent and has no solutions.
Holt McDougal Algebra 2
3-6
Solving Linear Systems
in Three Variables
Lesson Quiz: Part I
1. At the library book sale, each type of book is
priced differently. The table shows the number of
books Joy and her friends each bought, and the
amount each person spent. Find the price of each
type of book.
Hardcover
Paper- Audio Total
back Books Spent
Hal
3
4
1
\$17
Ina
2
5
1
\$15
Joy
3
3
2
\$20
Holt McDougal Algebra 2
hardcover: \$3;
paperback: \$1;
audio books: \$4
3-6
Solving Linear Systems
in Three Variables
Lesson Quiz: Part II
Classify each system and determine the number
of solutions.
2x – y + 2z = 5
2.
–3x +y – z = –1
inconsistent; none
x – y + 3z = 2
9x – 3y + 6z = 3
3.
12x – 4y + 8z = 4
–6x + 2y – 4z = 5
Holt McDougal Algebra 2
consistent; dependent;
infinite
``` |
# Over Lesson 3–1
```Five-Minute Check (over Lesson 3–1)
CCSS
Then/Now
New Vocabulary
Key Concept: Linear Function
Example 1: Solve an Equation with One Root
Example 2: Solve an Equation with No Solution
Example 3: Real-World Example: Estimate by Graphing
Over Lesson 3–1
Determine whether y = –2x – 9 is a linear equation.
If it is, write the equation in standard form.
A. linear; y = 2x – 9
B. linear; 2x + y = –9
C. linear; 2x + y + 9 = 0
D. not linear
Over Lesson 3–1
Determine whether 3x – xy + 7 = 0 is a linear
equation. If it is, write the equation in standard
form.
A. linear; y = –3x – 7
B. linear; y = –3x + 7
C. linear; 3x – xy = –7
D. not linear
Over Lesson 3–1
Graph y = –3x + 3.
A.
B.
C.
D.
Over Lesson 3–1
Jake’s Windows uses the equation c = 5w + 15.25
to calculate the total charge c based on the number
of windows w that are washed. What will be the
charge for washing 15 windows?
A. \$75.00
B. \$85.25
C. \$87.50
D. \$90.25
Over Lesson 3–1
Which linear equation is represented by this
graph?
A. y = x – 3
B. y = 2x + 1
C. y = x + 3
D. y = 2x – 3
• Pg. 163 – 168
• Obj: Learn how to solve linear equations
by graphing and estimate solutions to a
linear equation by graphing.
• Content Standards: A.REI.10 and F.IF.7a
• Why?
– The cost of braces can vary widely. The
graph shows the balance of the cost of
treatments as payments made. This is
modeled by the function b = -85p + 5100,
where p represents the number of \$85
payments made, and b is the remaining
balance.
• If a parent has made 20 payments on her
teenager’s braces, what is the remaining
balance to be paid?
• How can you use the graph to answer the
question?
• How can a parent use the graph to find
how many payments there will be in all?
You graphed linear equations by using tables
and finding roots, zeros, and intercepts.
• Solve linear equations by graphing.
• Estimate solutions to a linear equation by
graphing.
• Linear Function – a function for which the
graph is a line
• Parent Function – the simplest function in
a family of linear functions
• Family of Graphs – a group of graphs with
one or more similar characteristics
• Root – any value that makes the equation
true
• Zeros – values of x for which f(x) = 0
Solve an Equation with One Root
A.
Method 1
Solve algebraically.
Original equation
Subtract 3 from each side.
Multiply each side by 2.
Simplify.
Solve an Equation with One Root
B.
Method 2 Solve by graphing.
Find the related function. Set the equation equal to 0.
Original equation
Subtract 2 from each side.
Simplify.
Solve an Equation with One Root
The related function is
function, make a table.
The graph intersects the x-axis
at –3.
Answer: So, the solution is –3.
To graph the
A. x = –4
B. x = –9
C. x = 4
D. x = 9
A.
x = 4;
B. x = –4;
C.
x = –3;
D. x = 3;
Solve an Equation with No Solution
A. Solve 2x + 5 = 2x + 3.
Method 1 Solve algebraically.
2x + 5 = 2x + 3
Original equation
2x + 2 = 2x
Subtract 3 from each side.
2=0
Subtract 2x from each side.
The related function is f(x) = 2. The root of the linear
equation is the value of x when f(x) = 0.
Answer: Since f(x) is always equal to 2, this function
has no solution.
Solve an Equation with No Solution
B. Solve 5x – 7 = 5x + 2.
Method 2 Solve graphically.
5x – 7 = 5x + 2
Original equation
5x – 9 = 5x
Subtract 2 from each side.
–9 = 0
Subtract 5x from each side.
Graph the related function, which is f(x) = –9. The graph
of the line does not intersect the x-axis.
solution.
A. Solve –3x + 6 = 7 – 3x algebraically.
A. x = 0
B. x = 1
C. x = –1
D. no solution
B. Solve 4 – 6x = –6x + 3 by graphing.
A.
x = –1
B. x = 1
C.
x=1
D. no solution
Estimate by Graphing
FUNDRAISING Kendra’s class is selling greeting cards to
raise money for new soccer equipment. They paid \$115 for
the cards, and they are selling each card for \$1.75. The
function y = 1.75x – 115 represents their profit y for selling
x greeting cards. Find the zero of this function. Describe what
this value means in this context.
Make a table of values.
The graph appears to intersect
the x-axis at about 65. Next,
solve algebraically to check.
Estimate by Graphing
y = 1.75x – 115
Original equation
0 = 1.75x – 115
Replace y with 0.
115 = 1.75x
65.71 ≈ x
Divide each side by 1.75.
part of a greeting card cannot be sold, they
must sell 66 greeting cards to make a profit.
TRAVEL On a trip to his friend’s house, Raphael’s average speed
was 45 miles per hour. The distance that Raphael is from his
friend’s house at a certain moment in the trip can be represented
by d = 150 – 45t, where d represents the distance in miles and t is
the time in hours. Find the zero of this function. Describe what this
value means in this context.
A. 3; Raphael will arrive at his friend’s house
in 3 hours.
B.
Raphael will arrive at his friend’s house in
3 hours 20 minutes.
C.
Raphael will arrive at his friend’s house in
3 hours 30 minutes.
D. 4; Raphael will arrive at his friend’s house in 4 hours.
A.
B.
C.
D.
A
B
C
D
``` |
# How do you solve 3/2x -15 =-12?
Jul 12, 2016
$x = 2$
#### Explanation:
$\frac{3}{2} x - 15 \textcolor{red}{+ 15} = - 12 \textcolor{red}{+ 15}$
$\textcolor{b l u e}{2 \times} \frac{3}{2} x = \textcolor{b l u e}{2 \times} 3$
$\cancel{2} \frac{3}{\cancel{2}} x = \textcolor{b l u e}{2 \times} 3$
$3 x = 6$
$x = 2$
Or we can multiply the whole equation b 2 to get rid of the denominator.
$\textcolor{b l u e}{2 \times} \frac{3}{2} x - \textcolor{b l u e}{2 \times} 15 = \textcolor{b l u e}{2 \times} \left(- 12\right)$
$3 x - 30 = - 24$
$3 x = - 24 + 30$
$3 x = 6$
$x = 2$ |
# How do you solve rational equations 2/(3x+1) = 1/x - (6x)/(3x+1)?
Dec 14, 2015
$x = \frac{1}{2}$
#### Explanation:
1) Check when the equation is defined
First of all, bear in mind that your equation is not defined if any denominator is equal to zero.
Thus, let's first compute which values for $x$ are possible:
• impossible values for the first and third demoninator: $3 x + 1 = 0 \iff x = - \frac{1}{3}$
• impossible values for the second denominator: $x = 0$
Thus, for your equation, $x \ne 0$ and $x \ne - \frac{1}{3}$.
=====================================
Now, let's solve the equation!
$\frac{2}{3 x + 1} = \frac{1}{x} - \frac{6 x}{3 x + 1}$
... to eliminate the first and the third fraction, multiply both sides of the equation with $\left(3 x + 1\right)$...
$\iff 2 = \frac{3 x + 1}{x} - 6 x$
... to eliminate the last remaining fraction, multiply both sides of the equation with $x$...
$\iff 2 x = 3 x + 1 - 6 {x}^{2}$
... bring all terms to the left side of the equation...
(to do so, add $- 3 x - 1 + 6 {x}^{2}$ on both sides of the equation)
$\iff 6 {x}^{2} - x - 1 = 0$
=====================================
At this point, you have a quadratic equation which can be solved e.g. with the quadratic formula.
In your case, $a = 6$, $b = - 1$ and $c = - 1$. Thus,
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{1 \pm \sqrt{1 + 24}}{12} = \frac{1 \pm 5}{12}$
The two solutions for this quadratic equation are
$x = \frac{1}{2}$ and $x = - \frac{1}{3}$.
=====================================
4) Check which solutions are possible
Since we have noted before that $x \ne - \frac{1}{3}$ needs to hold, we need to discard the second solution.
Thus, the only solution for this equation is $x = \frac{1}{2}$.
Dec 14, 2015
Another way of approaching the initial stage. It has to be the same way as given but just looks different!
#### Explanation:
To make all the denominators the same:
Let some value be k then
$x \times k = \left(3 x + 1\right)$
So $k = \left(3 + \frac{1}{x}\right)$
Thus $\frac{1}{x}$ may be written as $\frac{3 + \frac{1}{x}}{3 x + 1} \to \frac{1 \times k}{x \times k} = \frac{1}{x}$
Now we have:
$\frac{2}{3 x + 1} = \frac{3 + \frac{1}{x}}{3 x + 1} - \frac{6 x}{3 x + 1}$
As they all have the same denominator the resulting equation would still be a true if we totally removed it.
$2 = 3 + \frac{1}{x} - 6 x$
$6 x - 1 = \frac{1}{x}$
$6 {x}^{2} - x - 1 = 0$
Then solve in the normal way!
Dec 14, 2015
Just another way
#### Explanation:
$\frac{2}{3 x + 1} = \frac{1}{x} - \frac{6 x}{3 x + 1}$
Take common terms to same side
$\frac{6 x + 2}{3 x + 1} = \frac{1}{x}$
Cross multiply both sides
$6 {x}^{2} + 2 x = 3 x + 1$
$6 {x}^{2} - 3 x + 2 x - 1 = 0$
$3 x \left(2 x - 1\right) + 1 \left(2 x - 1\right) = 0$
$\left(3 x + 1\right) \left(2 x - 1\right) = 0$
So we have two solutions $x = - \frac{1}{3} , \frac{1}{2}$
x =-1/3 leads to $\infty$. Hence the solution is $x = \frac{1}{2}$ |
### Become an OU student
Succeed with maths: part 1
Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.
# 1 Fractions and decimals
You use fractions every day even if you're not really thinking about it. For example, if somebody asks you the time it could be half past 2 or a quarter to 6. The half and the quarter are parts of an hour; in other words they are fractions of a whole.
The number at the bottom of the fraction tells you how many parts the whole one has. So, has two parts in the whole one and has 4. The number at the top tells you how many of these parts there are. Hence, is 1 lot of 2 parts. You’ll be looking at this again in more detail in Week 3.
Since fractions are parts of a whole, just as decimals are, this means that you can relate fractions and decimals. Here you will only be dealing with fractions that are easily shown in a place value table.
So, say you needed to write out 2 and (three-tenths) as a decimal. The whole part is 2 and the fractional part is . Remember in order to separate the whole number from the fraction you use a decimal point, with the part of the whole number to the right. So, you would write the 2 to the left of the decimal point and the fractional part to the right of the decimal point. Thus, it would be written as 2.3. If you were dealing with on its own, you would need to show that there were no whole parts by showing a zero to the left of the decimal point. is therefore written as 0.3.
Now you’ve seen a few examples, it’s your turn to have a try yourself in the next activity. You can use a place value table if it helps you.
## Activity 1 Fractions to decimals
Timing: Allow approximately 10 minutes
a) Rewrite each of the following fractions as a decimal number.
• i.
Remember: if the number does not have a whole number part, a zero is written in the units place. This makes the number easier to read (it’s easy to overlook the decimal point).
• i.
• ii.
• ii.
• iii.
• iii.
• iv.
• iv.
b) Match the numbers below to the letter shown on the number line in Figure 1.
Figure 1 Number line exercise
Using the following two lists, match each numbered item with the correct letter.
1. B
2. C
3. A
4. D
• a.
• b.
• c.
• d. |
# Formulas for 3D Shapes
Some of the useful math geometry formulas for 3D shapes are discussed below.
(i) Area of a Triangle: Let ABC be any triangle. If AD be perpendicular to BC and BC = a, CA = b, AB = c then the area of the triangle ABC (to be denoted by ⊿) is given by,
⊿ = ¹/₂ × base × altitude.
= ¹/₂ ∙ BC ∙ AD
(b) ⊿ = √[s(s - a)(s - b)(s - c)]
Where 2x = a + b + c = perimeter of the ⊿ ABC.
(c) If the a be the length of a side of an equilateral triangle then its height = (√3/2) a and it’s area = (√3/4) a²
(ii) If a be the length and b, the breadth of a rectangle then its area = a ∙ b, length of its diagonal = √(a² + b² ) and its perimeter = 2 ( a + b).
(iii) If a be the length of a side of a square, then its area = a² length of its diagonal = a√2 and perimeter = 4a.
(iv) If the lengths of two diagonals of a rhombus be a and b respectively then its area = (1/2) ab and length of a side = (1/2) √(a² + b²)
(v) If a and b be the lengths of two parallel sides of a trapezium and h be the distance between the parallel sides then the area of the trapezium = (1/2) (a + b) ∙ h.
(vi) Area of a Regular Polygon: The area of a regular polygon of n sides = (na²/4) cot (π/n) where a is the length of a side of the polygon. In particular, if a be the length of a side of a regular hexagon then its area
= (6a²/4) ∙ cot (π/6) = (3√3/2) ∙ a²
(vii) The length of circumference of a circle of radius r is 2πr and
its area = πr²
(viii) Rectangular Parallelopiped: If a, b and c be the length, breadth and height respectively of a rectangular parallelopiped then,
(a) the area of its surfaces = 2 ( ab + bc + ca)
(b) its volume = abc and
(c) the length of diagonal = √(a² + b² + c² ).
(ix) Cube: If the length of the side of a cube be a then,
(a) the area of its surfaces = 6a²,
(b) its volume = a³ and
(c) the length of the diagonal = √3a.
(x) Cylinder: Let r (= OA) be the radius of the base and h (=OB) be the height of a right circular cylinder ; then
(a) area of its curved surface = perimeter of base × height = 2πrh
(b) area of the whole surface = area of its curved surface + 2 × area of circular base
= 2πrh + 2πr²
= 2πr(h + r)
(c) volume of cylinder = area of base × height
= πr²h
(xi) Cone: Let r (= OA) be the radius of the base, h (= OB), the height and I, the slant height of a right circular cone ; then
(a) l² = h² + r²
(b) area of its curved surface
= (1/2) × perimeter of the base × slant height = (1/2)∙ 2πr ∙ l = πrl
(c) area of its whole surface = area of the curved surface + area of the circular base
= πrl + πr² = πrl + πr(l + r).
(d) volume of the cone = (1/3) × area of the base × height = (1/3)πr²h
Mensuration
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## Recent Articles
1. ### Comparison of Three-digit Numbers | Arrange 3-digit Numbers |Questions
Sep 13, 24 02:48 AM
What are the rules for the comparison of three-digit numbers? (i) The numbers having less than three digits are always smaller than the numbers having three digits as:
2. ### Worksheet on Three-digit Numbers | Write the Missing Numbers | Pattern
Sep 13, 24 02:23 AM
Practice the questions given in worksheet on three-digit numbers. The questions are based on writing the missing number in the correct order, patterns, 3-digit number in words, number names in figures…
3. ### 2nd Grade Place Value | Definition | Explanation | Examples |Worksheet
Sep 13, 24 01:20 AM
The value of a digit in a given number depends on its place or position in the number. This value is called its place value.
4. ### Comparison of Two-digit Numbers | Arrange 2-digit Numbers | Examples
Sep 12, 24 03:07 PM
What are the rules for the comparison of two-digit numbers? We know that a two-digit number is always greater than a single digit number. But, when both the numbers are two-digit numbers |
## Fraction as function implicit in many lessons
After the natural numbers, 0, 1,2, etc. are introduced, the fractions are introduced. What are they? Did they already exist? Or are they an operator on natural numbers?
We can think of 1/2 as a function. Its domain is the even natural numbers. For each number 2n, it matches as output n. So the graph of the function are the ordered pairs (2n,n) for n any natural number.
Suppose that 1/2 is introduced in dividing something in half to share it. The thing divided is implicitly treated as already a multiple of 2. So if it is a donut, we cut it in half, but the two halves essentially already existed in this way of thinking of 1/2.
If we have a pizza and it has 6 slices, then 1/2 is 3 slices. The pizza is either already cut, or is first cut. The physical cutting is not what fraction means. The fraction comes into play after the cutting happens.
Something is broken in half first. Then the two halves are shared. So the mathematics part comes in after the natural number 2 already exists in the problem.
The same applies to marking 1/2 points on a ruler and so “dividing” it. The marks are drawn physically and then we take them as given and say that we have divided the ruler into 1/2 units. Marking and cutting are physical acts not mathematical ones. The math comes after the physical act. The physical act either creates a new natural number, such as two halves by cutting, or it reveals that there was a pre-existing granularity that corresponded to an inherent natural number in the problem that was capable of being divided by 2.
The point is that all these methods of teaching fractions are really fractions as operators, just covertly. In each case, there is considered to be a natural number that exists and this natural number is what is divided by 2.
So a donut is considered to have already been 2 halves. The 2 is divided by 2 to get 1, i.e. one of the halves.
A pizza of 6 slices is divided into 2 by taking 3 slices. The 6 slices pre-existed before dividing them by 2 to get 3. The dividing by 2 does not happen until we change our view to associate the number 6 with the pizza.
In the same way, we alter our perception of the donut and say it corresponds to the number 2 before we divide by 2 to get 1.
In each of these cases, the fraction as function point of view applies as soon as we change our view of the original natural of the thing as having associated with it a natural number divisible without remainder by the denominator. So the fraction as function view is being applied as soon as we change the number we associate with the object or situation. The new associated number is always one in the domain of the fraction as function. So we are always applying the fraction as function concept, even when we don’t explicitly say so.
In fraction democracy, every application or idea about fractions is presented as equally important. The student is expected to sort it out logically. They don’t. When we analyze fraction democracy, we find that all the cases presented are really a covert application of fraction as function. The application comes after a new number is associated with the problem that can be divided by the denominator.
## About New Math Done Right
Author of Pre-Algebra New Math Done Right Peano Axioms. A below college level self study book on the Peano Axioms and proofs of the associative and commutative laws of addition. President of Mathematical Finance Company. Provides economic scenario generators to financial institutions.
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## Example Questions
### Example Question #36 : Add, Subtract, Multiply, And Divide Decimals To Hundredths: Ccss.Math.Content.5.Nbt.B.7
Explanation:
We can use base ten blocks to help us solve this problem. Let's review what our base ten blocks are by using a whole number
When we put this together, we add:
To use base ten blocks to add decimal numbers, we need to think of the base ten blocks a little differently. We think of the hundreds block as one whole. The tens block as tenths because you would need ten of these to make one whole. Finally, the ones block as hundredths because you would need a hundred of these to make one whole:
Let's look at this problem:
First, we want to represent the with four tenths blocks and five hundredths blocks:
Next, we want to represent the with one tenths block and three hundredths blocks:
Now, we want to combine our blocks together:
We can see that we now have five tenths blocks and eight hundredths blocks, which means our answer is
### Example Question #37 : Add, Subtract, Multiply, And Divide Decimals To Hundredths: Ccss.Math.Content.5.Nbt.B.7
Explanation:
We can use base ten blocks to help us solve this problem. Let's review what our base ten blocks are by using a whole number
When we put this together, we add:
To use base ten blocks to add decimal numbers, we need to think of the base ten blocks a little differently. We think of the hundreds block as one whole. The tens block as tenths because you would need ten of these to make one whole. Finally, the ones block as hundredths because you would need a hundred of these to make one whole:
Let's look at this problem:
First, we want to represent the with five tenths blocks and six hundredths blocks:
Next, we want to represent the with one tenths block and seven hundredths blocks:
Now, we want to combine our blocks together:
Notice that we have more than ten hundredths blocks. This means we can take ten of the hundredths blocks and make another tenths block:
We can see that we now have seven tenths blocks and three hundredths blocks, which means our answer is
### Example Question #38 : Add, Subtract, Multiply, And Divide Decimals To Hundredths: Ccss.Math.Content.5.Nbt.B.7
Explanation:
We can use base ten blocks to help us solve this problem. Let's review what our base ten blocks are by using a whole number
When we put this together, we add:
To use base ten blocks to add decimal numbers, we need to think of the base ten blocks a little differently. We think of the hundreds block as one whole. The tens block as tenths because you would need ten of these to make one whole. Finally, the ones block as hundredths because you would need a hundred of these to make one whole:
Let's look at this problem:
First, we want to represent the with one tenths block and five hundredths blocks:
Next, we want to represent the with three tenths block and two hundredths blocks:
Now, we want to combine our blocks together:
We can see that we now have four tenths blocks and seven hundredths blocks, which means our answer is
### Example Question #1 : Add Decimals
Explanation:
When we add decimals, we can treat it like a normal addition problem, we just need to remember out decimal:
### Example Question #2 : Add Decimals
Explanation:
Adding decimals is just like adding regular numbers, you just must remember to bring down your decimal point:
Explanation:
### Example Question #2 : Add Decimals
Explanation:
You start adding on the far right which in this case is the hundredths place. . We have to carry a from the sum above the tenths place and place the other below the hundredths place.
The decimal will be carried down and remain between the tenths place and the ones place.
The final addition portion is the ones place.
### Example Question #3 : Add Decimals
Explanation:
You start adding on the far right which in this case is the hundredths place. . We have to carry a from the sum above the tenths place and place the below the hundredths place.
The decimal will be carried down and remain between the tenths place and the ones place.
The final addition portion is the ones place.
### Example Question #4 : Add Decimals
Explanation:
You start adding on the far right which in this case is the hundredths place.
Next, add the tenths place. . We have to carry a from the sum above the ones place and place the below the tenths place.
The decimal will be carried down and remain between the tenths place and the ones place.
The final addition portion is the ones place.
Explanation: |
# PSAT Math : How to find out if a point is on a line with an equation
## Example Questions
### Example Question #1 : How To Find Out If A Point Is On A Line With An Equation
In the xy -plane, line is given by the equation 2x - 3y = 5. If line passes through the point (a ,1), what is the value of a ?
-2
5
-1
3
4
Explanation:
The equation of line l relates x -values and y -values that lie along the line. The question is asking for the x -value of a point on the line whose y -value is 1, so we are looking for the x -value on the line when the y-value is 1. In the equation of the line, plug 1 in for and solve for x:
2x - 3(1) = 5
2x - 3 = 5
2x = 8
x = 4. So the missing x-value on line l is 4.
### Example Question #2 : Other Lines
The equation of a line is: 2x + 9y = 71
Which of these points is on that line?
(4,7)
(2,7)
(-2,7)
(4,-7)
(-4,7)
(4,7)
Explanation:
Test the difference combinations out starting with the most repeated number. In this case, y = 7 appears most often in the answers. Plug in y=7 and solve for x. If the answer does not appear on the list, solve for the next most common coordinate.
2(x) + 9(7) = 71
2x + 63 = 71
2x = 8
x = 4
Therefore the answer is (4, 7)
### Example Question #3 : Other Lines
Which of the following lines contains the point (8, 9)?
Explanation:
In order to find out which of these lines is correct, we simply plug in the values and into each equation and see if it balances.
The only one for which this will work is
### Example Question #4 : Other Lines
Which point lies on this line?
Explanation:
Test the coordinates to find the ordered pair that makes the equation of the line true:
### Example Question #1 : How To Find Out If A Point Is On A Line With An Equation
Points D and E lie on the same line and have the coordinates and , respectively. Which of the following points lies on the same line as points D and E?
Explanation:
The first step is to find the equation of the line that the original points, D and E, are on. You have two points, so you can figure out the slope of the line by plugging the points into the equation
.
Therefore, you can get an equation in the line in point-slope form, which is
.
Plug in the answer options, and you will find that only the point solves the equation.
### Example Question #2 : How To Find Out If A Point Is On A Line With An Equation
Which of the following points is on the line given by the equation ?
Explanation:
In order to solve this, try each of the answer choices in the equation:
For example, when we try (3,4), we find:
This does not work. When we try all the choices, we find that only (2,4) works: |
# 1-12 Multiplication Chart
Studying multiplication right after counting, addition, and subtraction is good. Kids learn arithmetic by way of a natural progression. This progress of understanding arithmetic is truly the adhering to: counting, addition, subtraction, multiplication, lastly section. This assertion results in the question why find out arithmetic in this particular sequence? More importantly, why understand multiplication soon after counting, addition, and subtraction before department?
## The following information answer these queries:
1. Youngsters find out counting very first by associating graphic things making use of their hands and fingers. A concrete instance: How many apples are there in the basket? A lot more abstract instance is the way aged are you currently?
2. From counting amounts, another reasonable step is addition then subtraction. Addition and subtraction tables can be quite valuable instructing tools for the kids because they are visual tools making the cross over from counting less difficult.
3. Which should be discovered up coming, multiplication or section? Multiplication is shorthand for addition. At this time, youngsters use a organization understanding of addition. As a result, multiplication may be the after that logical kind of arithmetic to discover.
## Review fundamentals of multiplication. Also, assess the basic principles utilizing a multiplication table.
Let us overview a multiplication example. Utilizing a Multiplication Table, increase a number of times a few and acquire an answer twelve: 4 by 3 = 12. The intersection of row about three and line several of any Multiplication Table is twelve; twelve is definitely the answer. For youngsters beginning to find out multiplication, this is certainly effortless. They can use addition to solve the trouble as a result affirming that multiplication is shorthand for addition. Illustration: 4 x 3 = 4 4 4 = 12. It is really an excellent introduction to the Multiplication Table. The added advantage, the Multiplication Table is aesthetic and reflects back to learning addition.
## Exactly where will we start discovering multiplication making use of the Multiplication Table?
1. Initially, get knowledgeable about the table.
2. Start with multiplying by one. Start off at row # 1. Go on to line number one. The intersection of row one and line one is the perfect solution: a single.
3. Perform repeatedly these actions for multiplying by one. Multiply row a single by posts one particular by way of 12. The answers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Perform repeatedly these methods for multiplying by two. Flourish row two by columns a single via several. The solutions are 2, 4, 6, 8, and 10 respectively.
5. Let us hop ahead of time. Replicate these actions for multiplying by several. Multiply row several by posts a single by way of a dozen. The solutions are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now we will boost the quantity of issues. Recurring these steps for multiplying by about three. Increase row 3 by posts 1 by way of twelve. The responses are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. In case you are more comfortable with multiplication to date, try a test. Remedy these multiplication troubles in your mind after which evaluate your responses towards the Multiplication Table: increase six as well as 2, grow nine and three, increase 1 and 11, grow 4 and four, and grow 7 and 2. The issue responses are 12, 27, 11, 16, and 14 correspondingly.
When you acquired 4 out from five issues appropriate, design your very own multiplication assessments. Estimate the replies in your head, and look them making use of the Multiplication Table. |
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