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# Gaussian Elimination
• Introduction
• Resolution Method: Gaussian Elimination and the Rouché-Capelli theorem.
• 10 Resolved Systems by Gaussian Elimination
## Introduction
A system of equations (linear) is a group of (linear) equations with various unknown factors. Generally speaking, the unknown factors appear in various equations.
What an equation with various unknown factors does is relates them amongst each other.
Solving a system consists in finding the value for the unknown factors in a way that verifies all the equations that make up the system.
• If there is a single solution (one value for each unknown factor) we will say that the system is Consistent Independent System (CIS).
• If there are various solutions (the system has infinitely many solutions), we say that the system is a Consistent Dependent System (CDS). .
• If there is no solution, and this will happen if there are two or more equations that can't be verified at the same time, we say it's an Inconsistent System (IS). For example, the following system of equations
$$\begin{cases} \begin{array}{lcl} y & = & 0 \\ 2x + y & = & 0 \\ 2x + y & = & 2 \end{array} \end{cases}$$
is inconsistent because of we obtain the solution x = 0 from the second equation and, from the third, x = 1.
In this section we are going to solve systems using the Gaussian Elimination method, which consists in simply doing elemental operations in row or column of the augmented matrix to obtain its echelon form or its reduced echelon form (Gauss-Jordan).
## Resolution Method
1. We apply the Gauss-Jordan Elimination method: we obtain the reduced row echelon form from the augmented matrix of the equation system by performing elemental operations in rows (or columns).
2. Once we have the matrix, we apply the Rouché-Capelli theorem to determine the type of system and to obtain the solution(s), that are as:
Let A·X = B be a system of m linear equations with n unknown factors, m and n being natural numbers (not zero):
• AX = B is consistent if, and only if,
$$rank(A)=rank(A|B)$$
• AX = B is consistent independent if, and only if,
$$rank(A)=n=rank(A|B)$$
Note: The elemental operations in rows or columns allow us to obtain equivalent systems to the initial one, but with a form that simplifies obtaining the solutions (if there are). Also, there are quicker tools to work out the solutions in the CIS, like Cramer's rule.
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# What Is 25 percent of 200 + Solution with Free Steps
The outcome of 25% of 200 is the number 50. This solution may be obtained by multiplying the component 0.25 by 200.
The twenty five percent of two hundred is very useful calculation for daily life. For example, suppose you were to buy a product on an online platform worth 200 US dollars. All of a sudden there is an end of season sale on the product worth 25% on the listed price. Since you know that 25 percent of 200 is equal to 50, you can clearly say that you will get a discount of 50 US dollars on the product. In short, you will have to pay only 150 US dollars instead of 200 US dollars.
In this article, the method for calculating a percentage of a given number is described. We’ll explain the procedure for figuring out what percentage of 200 is equal to 25%.
## What Is 25 percent of 200?
The result of 25 percent of 200 is 50. This result may be obtained by multiplying the component 0.25 by 200.
The solution to the question “25% of 200” may be found by multiplying 200 by a fraction of 25/100 and then reducing the result to a fractional number.
## How To Calculate 25 percent of 200?
The following easy mathematical procedures can help you find 25% of 200.
### Step 1
Writing 25 percent of 200 in mathematical form:
25 percent of 200 = 25% x 200
### Step 2
Substitute the % symbol with the fraction 1/100:
25 percent of 200 = ( 25 x 1/100 ) x 200
### Step 3
Rearranging the above equation gives:
25 percent of 200 = ( 25 x 200 ) / 100
### Step 4
Multiplying 25 with 200:
25 percent of 200 = ( 5000 ) / 100
### Step 5
Dividing 5000 by 100:
25 percent of 200 = 50
Therefore, the 25 percent of 200 is equivalent to 50
We can visualize that 25 percent of 200 is equal to 50 by using the following pie chart.
Figure 1: Pie Chart for 25 percent of 200
The orange portion of the preceding figure, which equals 50, represents 25 percent of 200. The figure’s overall area corresponds to the value of 200 in its entirety. The remaining 150, or 75% of the total 200, are represented by the green region.
The percentage is a technique in which physical quantities are converted on a scale of zero to a hundred for comparison.
All the Mathematical drawings/images are created using GeoGebra.
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# Sequences
. Introduction
. Terminology
. Some Common Sequences
. Linear Sequences
. Position-Position Rule
. Position-Term Rule
. Times Tables As Sequences
. A Matchstick Pattern Application
. Quadratic Sequences
. An Example In Using A Quadratic Sequence
. Geometric Sequences
. Fibonacci Sequence
## Introduction
Simply put, a sequence is a (usually) a list of numbers. These are separated by a comma and may or may not continue forever. Usually we hope there is a rule which links the numbers to each other. However it is possible to have a random sequence in which there is, of course, no rule by definition.
## Terminology
There is some terminology associated with sequences which we need to know.
Here is a typical(?) sequence: 3, 7, 11, 15, ...
Each number is called a term. So 3 is a term, 7 is a term and so on. Actually 3 is the first term (1st term), 7 is the second term (2nd term) and so on.
Notice that the terms are separated by commas. At the end we have the typographical symbol called an ellipsis, the three dots, which is used to indicate that the terms continue on and on forever.
Each term occupies a position in the sequence, so we can see that the three occupies the first position or position one, the seven occupies the second position or position two etc.
One thing we can investigate to our advantage, especially for the simpler sequences we encounter, is the difference between the terms. Actually, we specifically look at the term in the higher position and subtract the previous term. So in our sequence above we get 7 - 3 = 4. So the first difference between the first two terms is +4. In fact we will find this is the difference between all the successive terms for this sequence.
Note: We can also add the terms together. In this case we have what is called a series.
## Some Common And Special Sequences
A constant sequence: 3, 3, 3, 3, ... The same number all the time The even numbers: 2, 4, 6, 8, 10, ... Go up in twos The odd numbers 1, 3, 5, 7, 9, 11, ... Go up in twos, but starts at 1 The prime numbers 2, 3, 5, 7, 11, 13, 17,... What rule links these? The square numbers 1, 4, 9, 16, 25, 36, ... Each term is its position squared A linear sequence 3, 7, 11, 15, ... The difference between the terms is constant (=4 in this case) A geometric sequence 1, 2, 4, 8, 16, 32, ... Each term is a constant multiple of the previous term - this is the doubling sequence. A quadratic sequence 4, 5, 8, 13, 20, 29, ... Discussed later, but look at the differences. The triangular numbers 1, 3, 6, 10, 15, 21, ... Actually a quadratic type of sequence The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, ... Each term is the sum of the previous two terms.
There are many other types of sequence: exponential, logarithmic, power, trigonometric, Taylor, etc.
## Linear Sequences
The feature that defines a sequence as being a linear sequence is that the difference between successive terms is always the same value, a constant. For instance:
1, 2, 3, 4, 5, ... the difference between successive terms is +1.
4, 7, 10, 13, ... the difference between successive terms is + 3
10, 9.5, 9.0, 8.5, 8.0, ... the difference between successive terms is -0.5
One of our main aims in working with sequences is to try and determine the value of any term at any position we state. Another aim might be to decide if a particular value is actually in a given sequence.
## Term To Term Rule
There are two ways we can work out our terms. The first one we’ll look at is to work out the term from the previous term or terms.
For instance suppose we are given the sequence
4, 7, 10, 13, ...
We can see that the difference is +3. So our rule to get the next term is to add 3 to the previous term.
Here is another one:
10, 6, 2, -2, ...
The next term is the previous term subtract 4.
We need now to use a more mathematical notation.
We can call the term at position n, Tn This is just read: “tee of en or tee at en”.
Using this notation we can call the term at position 1 T1, the term at position 2 T2 and so on.
The term before Tn is therefore Tn-1, and the one after Tn is Tn+1
So for the sequence 7, 12, 17, 22, ... we could say that
Tn = Tn-1 + 5
With our term to term rule, we of course have to specify at least one term in the sequence, usually the first one T1.
One of the problems with the term to term rule for linear sequences is that if I give you the first term and the rule and then ask you to find the 900th term, you are going to have to do quite a bit of work (unless you do a little thinking!).
## Position To Term Rule
Another approach is to work out a position - term rule instead of a term - term rule. With this all I need to do is to provide a position and the rule will work out the value of the term for me.
To do this we need to do a little preparatory work first. So let's investigate the sequences in our ordinary times tables.
## Times Tables As Sequences
The 2× table: 2, 4, 6, 8, 10, ...
Note that the difference between successive terms is 2. We, of course, expect this since the two times table is the sequence of numbers that goes up in 2's.
We can say that the term is simply the position times by 2. So that the term at the first position is 1 x 2 = 2, the term at the second position is 2 x 2 = 4, the term at the third position is 3 x 2 and so on. With this rule we can easily work out the term at any position. For example, the term at position 27 is 27 x 2 = 54. Easy.
Now let's write this using a more mathematical notation. Let Tp be the term at position p and p be the position.
For our 2× table T = p x 2, but we write this in algebra by putting the number first to get T = 2 x p or Tp = 2p
Let's look at the 7 time table. 7, 14, 21, 28, 35, ...
The difference between the terms is 7 and the formula is the position × 7 or
Tp = 7p.
This applies to any times table. The 3× table is given by the rule or formula Tp = 3p, the four times table by the formula Tp = 4p, and so on.
Let's consider again the 3× table:
3, 6, 9, 12, 15, ...
The rule is Tp = 3p.
What if we add 1 to each term? Well we would get the sequence:
4, 7, 10, 13, 16, ...
and surely the formula is Tp = 3p + 1
Here, we can see the difference between terms is still 3.
We are now in a position where we can work out the formula for any linear sequence.
Consider: 5, 7, 9, 11, 13, ...
The difference is 2, so we suggest that this sequence must be based upon the 2× table, i.e. Tp = 2p
2, 4, 6, 8, 10, ...
We now need to ask what we do to each term in this sequence to get our desired sequence of: 5, 7, 9, 11, 13, ...
If we look at each term in turn we can see that the sequence terms are 3 more than the 2× table terms. So I have added 2 to the 2× table. Using our formula we can say that our sequence is Tp = 2p + 3
Here is another example: 6, 10, 14, 18, 22, ...
The difference is 4 so we must have Tp = 4p, the 4× table 4, 8, 12, 16, ..
And we are adding 2 to each term to get our sequence, so we must have the formula Tp = 4p + 2
If I have this formula, how do I use it? Well suppose I want to know what the value of the term at position 37 is? I can just substitute the p by 42 in the formula so T = 4x37 + 2. (Remember the multiplication is done first) This equals 150. This is the term at position 37.
## A Matchstick Pattern Application
Quite often (certainly at the lower levels) we have to work with matchstick or other patterns. Here is a complete example of how such a question might work.
Here we have the first three patterns in a sequence of patterns. Each pattern is made up of matchsticks, and we can see that the pattern grows in the same way for each new pattern. If we study the patterns carefully we can see that the first pattern has three matchsticks, makes one triangle and is pattern one in the sequence. The second pattern is obtained by adding two matchstick to the first patterns so it has five matchsticks and comprises two triangles. The third pattern has another two matchsticks and has seven matchsticks and makes three triangles.
We could easily define a term to term rule for this sequence. In words we could say that we add two matchsticks to the previous pattern. In symbolic language we could say that Mn+1 = Mn + 2, where Mn is the number of matchsticks at position n.
However, it is arguably better to use the position - term idea to get a formula for the number of matchsticks. Let's call the number of matchsticks M and let's call the position by the letter P.
The actual sequence is 3, 5, 7, 9, ...
We know that the pattern goes up in twos, so is based upon the 2× table, so
M = 2P, gives 2, 4, 6, 8, ...
and each term is 1 more than the two times table: M = 2P + 1
With this formula we can now work out how many matchsticks in any pattern. For instance how many matchsticks are there in pattern 23?
M = 2 x 23 + 1 = 47
We can also work backwards. What pattern could we make with 70 matchsticks. To do this we reverse the formula.
70 = 2P + 1,
So take off 1, leaves 69. Divide by 2, to get 34.5, so we can make pattern 34 (note we don't have enough matchsticks to make the next pattern - don't round up your answer.)
It is also useful to try and see how the formula relates to the original situation.
The 2P obviously refers to the fact that each pattern has two extra matchsticks. So pattern 1 has 1 lot of 2, patterns 2 has 2 lots of 2, pattern 3 has 3 lots of 2. Each pattern also has the one matchstick at the beginning, which is what the +1 relates to.
## Quadratic Sequences
Not all sequences are linear. For instance consider this sequence:
1, 4, 9, 16, 25, 36, ...
This is the sequence of square numbers. i.e. 1×1, 2×2, 3×3 etc.
If we work out the difference between the terms we get:
To try and describe this rule in words is a little tricky. We can say that the next term is obtained by adding to the previous term the previous difference add 2. This is a clumsy sentence. Or we could say that the difference goes up by two each time and we add this to the previous term.
To try and write this in symbols is also tricky. Tn+1 = Tn + ?
However, what do we put for the question mark, since it changes for each position.
For these types of sequences we find it easier to work with position - term definitions.
In position 1 we have T1 = 1 x 1 = 1
In position 2 we have T2 = 2 x 2 = 4
In position 3 we have T3 = 3 x 3 = 9
So in position n we have Tn = n x n = n2, so our formula is Tn = n2
If we look at our sequence we might have spotted that if we take the difference between the differences, what is called the second difference, we get 2 which is a constant.
So here we have a 2nd difference of 2 which comes from having a sequence of n2
Surely if I double the terms of the original sequence to get 2, 8, 18, 32, 50, ... I will be doubling the 2nd difference value to get 4.
The formula for this sequence is 2n2 and its second difference is 4.
In general then the coefficient of n2 is half of the second difference. (Whereas for the linear sequence the coefficient of n was the first difference.)
If the second difference = 1 then the sequence is based upon
If we find that the second difference is a constant, then we know that the sequence is based upon n2 terms, which we call a quadratic sequence.
We can also see that in general a quadratic sequence has the form
Tn = an2 + bn + c
This is made up of two parts, the first part is Tn = an2 and the other part is the linear sequence Tn = bn + c.
So if we know that we have a quadratic sequence, because our second difference is a constant, then we can take away the an2 part from our sequence to leave a linear sequence. We can then find out what this is and by putting the two parts together we can work out our formula.
Here is an example:
5, 11, 19, 29, 41, ...
We can work out our differences
We have a 2nd difference of 2 so our sequence is based upon n2.
If we take away the terms of this we shall end up with a linear sequence
We can now work out the formula for this linear sequence. it goes up in 3's so is based on the 3x table and each term is 1 more than the 3× table. The formula is therefore Tn = 3n + 1
If we now put our two parts together we get:
Tn = n2 + 3n + 1
## An Example In Using A Quadratic Sequence
A pond builder! builds square ponds and surrounds them with square tiles. The builder wants to know how many tiles are required for a particular size pond. (How contrived is this!)
Here are some sketches she made:
We can see that for a 1 by 1 pond we need 8 tiles, for a 2 by 2 pond we need 12 tiles, for a 3 by 3 pond we need 18 tiles and so on.
We have a sequence: 8, 12, 18, 26, 36, ...
The first difference is 4, then 6 then 8 and so on, the second difference is always 2.
So we know we have a quadratic and it contains n2 tiles. If we take off the n2 tiles we end up with 7, 8, 9, 10 etc. This is a linear sequence: n+6.
So our original sequence is Tn = n2 + n + 6
If the builder wants to build a 5 by 5 pond she will need 36 tiles.
## Geometric Sequences
A geometric sequence is one where each term is increased by multiplying by a fixed amount (called the common ratio).
A simple example is the doubling sequence, 1, 2, 4, 8, 16, 32, 64 etc. Here the common ratio is ×2.
Note how quickly the sequence increases.
If the common ratio is ½ then the sequence will keep decreasing, e.g.
160, 80, 40, 20, 10, 5, 2.5, …
## Fibonacci Sequences
Fibonacci sequences are named after Leonardo of Pisa (c 1200) who was the son of a certain Bonaccio hence Fi Bonaccio.
These sequences have the characteristic that each term is based upon some combination of two or more previous terms.
The original Fibonacci sequence goes 1, 1, 2, 3, 5, 8, ... Each term is the sum of the two previous terms.
One interesting fact about the Fibonacci terms is that if we work out the ratio between successive terms we find that the ratio converges to a very important number called the Golden Ratio or Divine Proportion, given the symbol Φ . It’s value is 1.618… This number is irrational.
There are some important results to do with Φ. For instance:
We can also show that:
=
This strange thing on the right is called a continued fraction. Maths is full of interesting links and relationships…if you search for them.
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# Combinations and Permutations
## What's the Difference?
In English we use the word "combination" loosely, without thinking if theorder of things is important. In other words: "My fruit salad is a combination of apples, grapes and bananas"We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad. "The combination to the safe was 472". Now we do care about the order. "724" would not work, nor would "247". It has to be exactly 4-7-2. So, in Mathematics we use more precise language: If the order doesn't matter, it is a Combination. If the order does matter it is a Permutation.
## So, we should really call this a "Permutation Lock"!
In other words: A Permutation is an ordered Combination. To help you to remember, think "Permutation ...Position"
Permutations
There are basically two types of permutation: 1. Repetition is Allowed: such as the lock above. It could be "333". 2. No Repetition: for example the first three people in a running race. You can't be first and second.
## 1. Permutations with Repetition
These are the easiest to calculate. When you have n things to choose from ... you have n choices each time!
When choosing r of them, the permutations are: n n ... (r times) (In other words, there are n possibilities for the first choice, THEN there are npossibilites for the second choice, and so on, multplying each time.) Which is easier to write down using an exponent of r: n n ... (r times) = nr Example: in the lock above, there are 10 numbers to choose from (0,1,..9) and you choose 3 of them: 10 10 ... (3 times) = 103 = 1,000 permutations So, the formula is simply: nr where n is the number of things to choose from, and you choose rof them (Repetition allowed, order matters)
## 2. Permutations without Repetition
In this case, you have to reduce the number of available choices each time. For example, what order could 16 pool balls be in? After choosing, say, number "14" you can't choose it again. So, your first choice would have 16 possibilites, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be: 16 15 14 13 ... = 20,922,789,888,000 But maybe you don't want to choose them all, just 3 of them, so that would be only: 16 15 14 = 3,360 In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls. But how do we write that mathematically? Answer: we use the "factorial function " The factorial function (symbol: !) just means to multiply a series of descending natural numbers.Examples: 4! = 4 3 2 1 = 24 7! = 7 6 5 4 3 2 1 = 5,040 1! = 1
Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations.
So, if you wanted to select all of the billiard balls the permutations would be: 16! = 20,922,789,888,000 But if you wanted to select just 3, then you have to stop the multiplying after 14. How do you do that? There is a neat trick ... you divide by 13! ... 16 15 14 13 12 ... 13 12 ... Do you see? 16! / 13! = 16 15 14 The formula is written: = 16 15 14 = 3,360
where n is the number of things to choose from, and you choose rof them (No repetition, order matters)
Examples :
Our "order of 3 out of 16 pool balls example" would be: 16! (16-3)! = 16! = 20,922,789,888,000 = 3,360
13! 6,227,020,800 (which is just the same as: 16 15 14 = 3,360) How many ways can first and second place be awarded to 10 people? 10! (10-2)! = 10! = 3,628,800 = 90
## 8! 40,320 (which is just the same as: 10 9 = 90)
Notation
Instead of writing the whole formula, people use different notations such as these:
Example: P(10,2) = 90
Combinations
There are also two types of combinations (remember the order does not matter now): 1. Repetition is Allowed: such as coins in your pocket (5,5,5,10,10) 2. No Repetition: such as lottery numbers (2,14,15,27,30,33)
## 1. Combinations with Repetition
Actually, these are the hardest to explain, so I will come back to this later.
## 2. Combinations without Repetition
This is how lotteries work. The numbers are drawn one at a time, and if you have the lucky numbers (no matter what order) you win! The easiest way to explain it is to:
assume that the order does matter (ie permutations), then alter it so the order does not matter. Going back to our pool ball example, let us say that you just want to know which 3 pool balls were chosen, not the order. We already know that 3 out of 16 gave us 3,360 permutations. But many of those will be the same to us now, because we don't care what order! For example, let us say balls 1, 2 and 3 were chosen. These are the possibilites: Order does matter Order doesn't matter
123 123 132 213 231 312 321 So, the permutations will have 6 times as many possibilites. In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is: 3! = 3 2 1 = 6 (Another example: 4 things can be placed in 4! = 4 3 2 1 = 24 different ways, try it for yourself!) So, all we need to do is adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in the order any more):
That formula is so important it is often just written in big parentheses like this:
where n is the number of things to choose from, and you choose rof them (No repetition, order doesn't matter) It is often called "n choose r" (such as "16 choose 3")
## And is also known as the "Binomial Coefficient"
Notation
As well as the "big parentheses", people also use these notations:
Example
So, our pool ball example (now without order) is: 16! = 16! 3!13! = 20,922,789,888,000 66,227,020,800 = 560
## 3!(16-3)! Or you could do it this way:
161514 321
3360 6
= 560
So remember, do the permutation, then reduce by a further "r!" ... or better still ... Remember the Formula! It is interesting to also note how this formula is nice and symmetrical:
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# Finding the Centre and Radius from the Equation of a Circle(KS3, Year 8)
homesitemapgraphs and coordinatesfinding the center and radius from the equation of a circle
We can find the centre and the radius of a circle from the equation of a circle. Imagine we wanted to find the centre and radius of the circle with the equation:
## How to Find the Centre and Radius from the Equation of a Circle
Finding the centre and radius from the equation of a circle is easy.
## How to Find the Centre from the Equation of a Circle
Find the Cartesian coordinates of the centre of the circle from the equation.
## Question
What is the centre of the circle with the equation (x − 1)2 + (y + 2)2 = 9?
## 1
Find the brackets with the x in it.
## 2
Look at the number in these brackets and the sign in front of it.
In our example, it is − 1.
## 3
Change the sign of the answer (−1) to find the x-coordinate of the centre of the circle.
−1 → 1
The x-coordinate of the centre of the circle is 1.
## 4
Find the brackets with the y in it.
## 5
Look at the number in these brackets and the sign in front of it.
In our example, it is + 2.
## 6
Change the sign of the answer (+2) to find the y-coordinate of the centre of the circle.
+ 2 → − 2
The y-coordinate of the centre of the circle is −2.
The centre of the circle with the equation (x − 1)2 + (y + 2)2 = 9 is:
## How to Find the Radius from the Equation of a Circle
Find the radius of the circle from the equation.
## Question
What is the radius of the circle with the equation (x − 1)2 + (y + 2)2 = 9?
## 1
Find the number not in brackets. It is usually written to the right of the equals sign (=).
In our example, it is 9.
## 2
Find the square root of the answer (9).
√9 = 3
The radius of the circle with the equation (x − 1)2 + (y + 2)2 = 9 is 3. The circle with the equation (x − 1)2 + (y + 2)2 = 9 has centre (1, −2) and radius 3.
## Lesson Slides
The slider below shows another real example of how to find the centre and radius from the equation of a circle.
## The Center of a Circle
The equation of a circle is:
The center is (a, b).
• The number being subtracted from the x in the brackets is the x-coordinate of the center.
• The number being subtracted from the y in the brackets is the y-coordinate of the center.
What if a coordinate of the center is negative? Imagine the center of the circle is (−1, 2). The equation will start:
(x − −1)2 + ...
Remember, that subtracting a negative number is the same as adding the positive number:
(x − −1)2 = (x + 1)2
A negative coordinate will have a + sign in front of it. A positive coordinate will have a sign in front of it. With practise, it is straightforward to read off the center of the circle.
## Equations That Don't Quite Look Right
Don't be confused if you see an equation which looks like this:
(x − 1)2 + (y − 3)2 − 49 = 0
This is still an equation of a circle, as can be seen with a little rearranging:
(x − 1)2 + (y − 3)2 = 49
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# Grouping of Data
## What is Grouping Data?
Grouping data plays a significant role when we have to deal with large data. This information can also be displayed using a pictograph or a bar graph. Data formed by arranging individual observations of a variable into groups, so that a frequency distribution table of these groups provides a convenient way of summarizing or analyzing the data. This is how we define grouped data.
In mathematics in the topic grouping data ,we basically learn to define grouped data mathematically. When the number of observations is very large,we may condense the data into several groups, by the concept of grouping of data. We record the frequency of observations falling in each of the groups.Presentation of data in groups along with the frequency of each group is called the frequency distribution of the grouped data.
### What are The Advantages of Grouping Data?
The Advantages of grouping data in statistics are-
• It helps to focus on important subpopulations and ignores irrelevant ones.
• Grouping of data improves the accuracy/efficiency of estimation.
### Frequency Distribution Table for Grouped Data
To analyse the frequency distribution table for grouped data when the collected data is large, then we can follow this approach to analyse it easily.
Example
Consider the marks of 50 students of class VII obtained in an examination. The maximum marks of the exam is 50.
23, 8, 13, 18, 32, 44, 19, 8, 25, 27, 10, 30, 22, 40, 39, 17, 25, 9, 15, 20, 30, 24, 29, 19, 16, 33, 38, 46, 43, 22, 37, 27, 17, 11, 34, 41, 35, 45, 31, 26, 42, 18, 28, 30, 22, 20, 33, 39, 40, 32
If we create a frequency distribution table for each and every observation, then it will form a large table. So for easy understanding, we can make a table with a group of observations say 0 to 10, 10 to 20 etc.
The distribution obtained in the above table is known as the grouped frequency distribution. This helps us to bring various significant inferences like:
(i) Many students have secured between 20-40, i.e. 20-30 and 30-40.
(ii) 8 students have secured higher than 40 marks, i.e. they got more than 80% in the examination.
In the above-obtained table, the groups 0-10, 10-20, 20-30,… are known as class intervals (or classes). It is observed that 10 appears in both intervals, such as 0-10 and 10-20. Similarly, 20 appears in both the intervals, such as as10-20 and 20-30. But it is not feasible that an observation either 10 or 20 can belong to two classes concurrently. To avoid this inconsistency, we choose the rule that the general conclusion will belong to the higher class. It means that 10 belongs to the class interval 10-20 but not to 0-10. Similarly, 20 belongs to 20-30 but not to 10-20, etc.
Consider a class say 10-20, where 10 is the lower class interval and 20 is the upper class interval. The difference between upper and lower class limits is called class height or class size or class width of the class interval.
This is how we create a frequency distribution table for grouped data as shown above.
### Histogram
We can show the above frequency distribution table graphically using a histogram. We need to consider class intervals on the horizontal axis and we need to consider the frequency on the vertical axis.
Let’s See A Few Grouped Data Examples in Detailed Step-by-Step Explanations.
Example 1. The marks obtained by forty students of class VIII in an examination are listed below:
16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13, 21, 13, 15, 19, 24, 16, 2, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23
Divide the data into five groups, namely, 0-5, 5-10, 10-15, 15-20 and 20-25, where 0-5 means marks greater than or equal to 0 but less than 5 and similarly 5-10 means marks greater than or equal to 5 but less than 10, and so on. Prepare a grouped frequency table for the grouped data.
Solution: We need to arrange the given observations in ascending order. After arranging them in ascending order we get them as
0, 1, 2, 2, 3, 3, 5, 5, 5, 6, 7, 7, 7, 8, 8, 9, 10, 10, 12, 12, 12, 13, 13, 13, 15, 16, 16,
16, 17, 18, 18, 18, 19, 20, 21, 21, 23, 23, 23, 24
Thus, the frequency distribution of the data may be given as follows:
Note: Here, each of the groups that is 0-5, 5-10, 10-15, 15-20 and 20-25 is known as a class interval. In the class interval 10-15, the number 10 is known as the lower limit and 15 is known as the upper limit of the class interval and the difference between the upper limit and the lower limit of any given class interval is known as the class size.
Thus, the class size in the above frequency distribution is equal to 5.
The mid value of a class is known to be its class mark and the class mark is obtained by adding its upper and lower class limits and dividing the sum by 2.
Thus, the class mark of 0-5 range is equal to (0 + 5)/2 = 2.5
And the class mark of 5-10 range is equal to (5 + 10)/2 = 7.5, etc.
### Questions to be Solved:
Question 1)The weights (in kg) of 35 persons are given below:
43, 51, 62,47, 48, 40, 50, 62, 53, 56, 40, 48, 56, 53, 50, 42, 55, 52, 48, 46, 45, 54, 52, 50, 47, 44, 54, 55, 60, 63, 58, 55, 60, 53,58
Prepare a frequency distribution table taking equal to the class size. One such class is the 40-45 class (where 45 is not included).
Solution) We may represent the data as given below:
1. What is a group of data called?
Grouped data is a statistical term used in data analysis. Raw data can be organized by grouping together similar measurements in a table. This grouped frequency table is also called grouped data.
2. What is grouped and ungrouped data?
Data is often described as ungrouped or grouped. Ungrouped data is the data given as indi- vidual data points. Grouped data is data given in intervals whereas Ungrouped data without a frequency distribution.
3. How can we convert ungrouped data to grouped data?
The first step is to determine how many classes you want to have. Next, you subtract the lowest value in the data set from the highest value in the data set and then you divide by the number of classes that you want to have.
4. What is grouped data examples?
Grouped data is data that has been bundled together in categories. Frequency tables and histograms can be used to show this type of data: 1) Relative frequency histogram showing book sales for a certain day, sorted by price. 2) A grouped frequency table showing grouped data by height.
These are the few grouped data examples from many other examples out there.
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# IRR Internal Rate of Return
The internal rate of return (IRR) is the discount rate which will give a net present value (NPV) of zero when applied to a series of cash flows.
## IRR Calculation
Consider as an example, the following cash flow diagram. At the start of year 1 (today) there is a cash out flow of 2,000 representing an investment in a project. For simplicity, with no further investment, the amount of 2,500 is returned in 2 years time at the end of year 2.
The business decides that the appropriate discount rate to use is 8%. The discount rate is the value the business places on its money.
The net present value of this project is the sum of the present values of each of the cash flows. Further details on the calculation of NPV can be found in our net present value tutorial.
```NPV = -2,000 + 2,500 /(1 + 8%)2
NPV = -2,000 + 2,143.35
NPV = 143.35
```
The NPV is greater than zero so this project must have a rate of return greater than 8%.
If the same calculation is carried at at a discount rate of 14% the net present value is:
```NPV = -2,000 + 2,500 /(1 + 14%)2
NPV = -2,000 + 1,923.67
NPV = -76.33
```
The NPV is less than zero, so this project must have a rate of return less than 14%.
If we vary the discount rate, at a given value the NPV will be equal to zero, and this is the rate of return for that project, this rate is referred to as the internal rate of return or IRR.
In this simple case to solve for the value of the discount rate we would need to solve the equation
```NPV = 0 = -2,000 + 2,500 /(1 + i)2
(1 + i)2 = 2,500/2,000
1 + 2i + i2 = 5 / 4
Solving this quadratic equation for i and taking the only positive value
i = 11.80%
```
At a rate of 11.80% the NPV of this project will be zero, and we can say that the internal rate of return or IRR of the project is 11.80%.
## IRR Calculation With Multiple Cash Flows
In the simple example above it was possible to solve for i, in cases where there are multiple cash flows at later stages in the projects life, it is not possible to solve directly for i and iterative techniques need to be applied.
Consider a project with the following cash flow diagram. In this case the initial cash is paid at at the start of period 1 (today). This is followed by a further payment at the end of period 1, and then receipts at the end of period 2 and the end of period 3.
Calculating the NPV at a discount rate of 10% we get
```NPV = -5,000 + 3,000 / (1 + 10%)2 + 7,000 / (1 + 10%)3
NPV = -5,000 + 2,479.34 + 5,259.20
NPV = 2,738.54
```
The net present value of the project at 10% is positive, so the return must be greater than the 10%.
The same calculation at 40% gives the following:
```NPV = -5,000 + 3,000 / (1 + 40%)2 + 7,000 / (1 + 40%)3
NPV = -5,000 - 1,530.61 + 2,551.02
NPV = -918.37
```
The net present value of the project at 40% is negative, so the return must be less than the 40%.
This process can now be repeated at values between 10% and 40% until a solution is reached. At 29.62% the calculations show the following:
```NPV = -5,000 + 3,000 / (1 + 29.62%)2 + 7,000 / (1 + 29.62%)3
NPV = -5,000 - 1,785.62 + 3,214.39
NPV = -0.00
```
The iterative process can also be carried out using a financial calculator or the Excel IRR function
## Using the IRR
Once the IRR has been calculated it can used used together with a few simple rules as follows:
1. If the IRR of the project is greater than the required rate of return (discount rate), it should be accepted.
2. If the IRR of the project is less than the required rate of return (discount rate), it should be rejected.
3. If only one project can be chosen, then the project with the highest IRR should be accepted.
## Limitations and Disadvantages of IRR
The IRR calculation formula will evaluate a project correctly is there is an initial payment (negative cash flow) followed by cash receipts (positive cash flows). If the cash flows are reversed so that the project has a cash receipt (positive cash flow) followed by cash payments (negative cash flow) then the IRR method can be less accurate and care needs to be taken with the solutions produced.
In the event that the project has multiple cash flows which change sign throughout the project, then internal rate of return equation will produce multiple IRRs for the project. The method does not distinguish between the answers, and there is no way of deciding which is the correct solution.
Finally, calculating the internal rate of return tells an investor nothing about the absolute size of a project, it will give the same rate of return irrespective of whether the project is small or large in scale.
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# GRE Math : How to find the common factors of squares
## Example Questions
← Previous 1
### Example Question #1 : Arithmetic
Simplify:
Explanation:
When simplifying the square root of a number that may not have a whole number root, it's helpful to approach the problem by finding common factors of the number inside the radicand. In this case, the number is 24,300.
What are the factors of 24,300?
24,300 can be factored into:
When there are factors that appear twice, they may be pulled out of the radicand. For instance, 100 is a multiple of 24,300. When 100 is further factored, it is (or 10x10). However, 100 wouldn't be pulled out of the radicand, but the square root of 100 because the square root of 24,300 is being taken. The 100 is part of the24,300. This means that the problem would be rewritten as:
But 243 can also be factored:
Following the same principle as for the 100, the problem would become
because there is only one factor of 3 left in the radicand. If there were another, the radicand would be lost and it would be 9*10*3.
9 and 10 may be multiplied together, yielding the final simplified answer of
### Example Question #1 : Basic Squaring / Square Roots
Explanation:
To solve the equation , we can first factor the numbers under the square roots.
When a factor appears twice, we can take it out of the square root.
Now the numbers can be added directly because the expressions under the square roots match.
### Example Question #1 : Arithmetic
Simplify.
Explanation:
To simplify, we must try to find factors which are perfect squares. In this case 16 is a factor of 624 and is also a perfect square.
Therefore we can rewrite the square root of 624 as:
### Example Question #1 : How To Find The Common Factors Of Squares
Reduce to its simplest form.
Explanation:
To simplify, we must try to find factors which are perfect squares. In this case 20 is a factor of 400 and is also a perfect square.
Thus we can rewrite the problem as:
Note:
### Example Question #1 : Arithmetic
Simplify.
Explanation:
Use the following steps to reduce this square root.
To simplify, we must try to find factors which are perfect squares. In this case 144 is a factor of 720 and is also a perfect square.
Thus we can rewrite the problem as follows.
### Example Question #1 : Arithmetic
Find the square root of .
Explanation:
Use the following steps to find the square root of
To simplify, we must try to find factors which are perfect squares. In this case 900 is a factor of 1800 and is also a perfect square.
Thus we can rewrite the problem as follows.
### Example Question #1 : Simplifying Square Roots
Simplify.
Explanation:
To simplify, we must try to find factors which are perfect squares. In this case 9 is a factor of 54 and is also a perfect square.
To reduce this expression, use the following steps:
### Example Question #2 : Simplifying Square Roots
Reduce.
Explanation:
To simplify, we must try to find factors which are perfect squares. In this case 36 is a factor of 72 and is also a perfect square.
To reduce this expression, use the following arithmetic steps:
### Example Question #11 : Arithmetic
Which quantity is greater: or ?
Not enough information to determine the relationship between these two quantities.
Explanation:
To simplify, we must try to find factors which are perfect squares. In this case 30 is a factor of 900 and is also a perfect square.
The square root of is equal to:
However,
Thus,
Reduce.
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# How do we solve for the mean?
How did you solve for the mean
You can find the mean, or average, of a data set in two simple steps: Find the sum of the values by adding them all up. Divide the sum by the number of values in the data set.
What is the formula used for mean
The mean formula is given as the average of all the observations. It is expressed as mean = (sum of observations) ÷ (total number of observations).
How would you solve the mean of the problem
The mean can be calculated only for numeric variables, no matter if they are discrete or continuous. It's obtained by simply dividing the sum of all values in a data set by the number of values. The calculation can be done from raw data or for data aggregated in a frequency table.
How can we solve mean median mode
To find the mean, add up the values in the data set and then divide by the number of values that you added. To find the median, list the values of the data set in numerical order and identify which value appears in the middle of the list. To find the mode, identify which value in the data set occurs most often.
How do you solve sample and mean
Mean. And as we increase the sample size n. The sample mean will approach the population mean so let's say if we increase the sample size from 100 to a thousand.
How do you find the mean of the data
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest.
What is the mean sample formula
The general formula for calculating the sample mean is given by x̄ = ( Σ xi ) / n. Here, x̄ represents the sample mean, xi refers all X sample values and n stands for the number of sample terms in the data set.
What is the mean mode formula
How to Calculate the Mean Using Mean Median Mode Formula If the set of 'n' number of observations is given then the mean can be easily calculated by using a general mean median mode formula that is, Mean = {Sum of Observations} ÷ {Total number of Observations}.
How do you find the mean in statistics
Now. You may have to use a calculator if you have lots of different data points but that's okay. Well. I added everything up I got a total of 74 divided by 9. Now.
How do you solve for mean and frequency
A frequency table shows how many times a value occurs. To find the sum of all the values, multiply each category by its frequency and then find the total of the results. The sum of all the frequencies shows how many values there are. The mean is the sum of all the values divided by how many values there are.
How did you find the mean in math
By eight now we can calculate the sum of those numbers. And it's going to be 548 and then we divide. By 8 to get our mean 548 divided by 8 gives us a mean of 68.
How do you solve mode
Five shows up three times. So it looks like one. And five both show up the most number of times. So I'm going to say my mode is one.
How do you find the mean of a sample
The following steps will show you how to calculate the sample mean of a data set:Add up the sample items.Divide sum by the number of samples.The result is the mean.Use the mean to find the variance.Use the variance to find the standard deviation.
What is the formula for the mean of grouped data
The mean formula to find the mean of a grouped set of data can be given as, x̄ = Σfx/Σf, where, x̄ is the mean value of the set of given data, f is the frequency of each class and x is the mid-interval value of each class.
What is mean in math statistics
In mathematics and statistics, the mean refers to the average of a set of values. The mean can be computed in a number of ways, including the simple arithmetic mean (add up the numbers and divide the total by the number of observations), the geometric mean, and the harmonic mean.
What is the formula of sample mean and population mean
Calculation. The formula used for the sample mean evaluation is: x̄ =∑xi /n. The formula used for the population mean calculation is: μ=∑X / N.
How do you find the mean and median of grouped data
SummaryFor grouped data, we cannot find the exact Mean, Median and Mode, we can only give estimates.To estimate the Mean use the midpoints of the class intervals: Estimated Mean = Sum of (Midpoint × Frequency)Sum of Frequency.To estimate the Median use: Estimated Median = L + (n/2) − BG × w.To estimate the Mode use:
What is the mean in in math
A mean in math is the average of a data set, found by adding all numbers together and then dividing the sum of the numbers by the number of numbers. For example, with the data set: 8, 9, 5, 6, 7, the mean is 7, as 8 + 9 + 5 + 6 + 7 = 35, 35/5 = 7.
Why calculate the mean of the means
The mean of means is simply the mean of all of the means of several samples. By calculating the mean of the sample means, you have a single value that can help summarize a lot of data. The mean of means, notated here as μ x ¯ , is actually a pretty straightforward calculation.
How do you find the sample mean formula
Mean. And as we increase the sample size n. The sample mean will approach the population mean so let's say if we increase the sample size from 100 to a thousand.
Do you find the mean in math
By eight now we can calculate the sum of those numbers. And it's going to be 548 and then we divide. By 8 to get our mean 548 divided by 8 gives us a mean of 68.
How do I find the mean in statistics
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set.
What is the formula of mean method in statistics
The formula to find the mean of an ungrouped data is given below: Suppose x1, x2, x3,….., xn be n observations of a data set, then the mean of these values is: x ― = ∑ x i n. Here, xi = ith observation, 1 ≤ i ≤ n.
How do you know what the mean is
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest.
What is the mean formula in statistics table
Mean from a frequency table is when we find the mean average from a data set which has been organised into a frequency table. To calculate the mean we find the total of the values and divide the total by the number of values. The number of values is the total frequency. This can be abbreviated to n.
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Real Numbers Ex-1.3 Ch-1 [Solutions] NCERT Maths Class 10th - Study Cbse Notes
# Real Numbers Ex-1.3 Ch-1 [Solutions] NCERT Maths Class 10th
Exercise 1.3
Q1. Prove that √5 is irrational.
Answer: Let take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get
b√5 = a
To remove root, Squaring on both sides, we get
5b2 = a2 … (i)
Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also.
That means 5 will divide a. So we can write
a = 5c
Putting value of a in equation (i) we get
5b2 = (5c)2
5b2 = 25c2
Divide by 25 we get
b2/5 = c2
Similarly, we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it contradicts.
Hence √5 is not a rational number, it is irrational.
Q2. Prove that 3 + 2√5 is irrational.
Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a-3b)/b
Now divide by 2, we get
√5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.
Hence, 3 + 2√5 is a irrational number.
3. Prove that the following are irrationals:
(i) 1/√2 (ii) 7√5 (iii) 6 + √2
(i) Let take that 1/√2 is a rational number.
So we can write this number as
1/√2 = a/b
Here a and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get
1 = (a√2)/b
Now multiply by b
b = a√2
divide by a we get
b/a = √2
Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts.
Hence, 1/√2 is a irrational number
(ii) Let take that 7√5 is a rational number.
So we can write this number as
7√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get
√5 = a/(7b)
Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts.
Hence, 7√5 is a irrational number.
(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
√2 = a/b – 6
√2 = (a-6b)/b
Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number.
But √2 is a irrational number so it contradicts.
Hence, 6 + √2 is a irrational number.
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# x,yWhat is the value of x and y if the point (x, y) is equidistant from (3, 6) and (4,8)?
sciencesolve | Certified Educator
You need to use the of equations that give the coordinates x,y midpoint of segment, such that:
`{(x = (x_1 + x_2)/2),(y = (y_1 + y_2)/2):}`
Considering `(x_1,y_1) = (3,6)` and `(x_2,y_2) = (4,8)` yields:
`x = (3 + 4)/2 => x = 7/2 => x = 3.5`
`y = (6 + 8)/2 => y = 7`
Hence, evaluating the coordinates of the midpoint, under the given conditions, yields `x = 3.5, y = 7.`
giorgiana1976 | Student
Since the point (x,y) is equidistant from the endpoints A(3, 6) and B(4,8), that means that the point is located on the midperpendicular of the segment AB.
The slope of the midperpendicular multiplied by the slope of the segment AB is -1.
mAB*m = -1
We'll write the equation of the segment line AB.
(yB - yA)/(y-yA) = (xB-xA)/(x-xA)
We'll substitute the coordinates of A and B:
(8-6)/(y-6) = (4-3)/(x-3)
2/(y-6) = 1/(x-3)
We'll cross multiply and we'll get:
y - 6 = 2(x-3)
The slope of the segment AB is mAB = 2.
mAB*m = -1
2*m = -1
m = -1/2
The equation of the midperpendicular is:
y - yM = (-1/2)(x-xM)
We'll calculate the coordinates of th midpoint of the segment AB:
xM = (xA+xB)/2
xM = (3+4)/2
xM = 7/2
yM = (yA+yB)/2
yM = (6+8)/2
yM = 7
y - 7 = (-1/2)(x - 7/2)
y - 7 = -x/2 + 7/4
4y - 28 = -2x + 7
2x - 4y - 35 = 0
The equation of the midperpendicular is: 2x - 4y - 35 = 0, so that the coordinates that verify the equation of the midperpendicular are equidistant from the endpoints A and B.
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The line of symmetry is the axis or imaginary line that passes through the center of an object and divides it into identical halves. If we cut an equilateral triangle into two halves, then it forms two right-angled triangles. Similarly, rectangle, square, circle are examples of a line of symmetry. The line of symmetry is also called an axis of symmetry. Also, it is named a mirror line where it forms two reflections of an image. A basic definition of a line of symmetry is it divides an object into two halves.
## Types of Lines of Symmetry
There are mainly two types considered in Lines of Symmetry concepts. They are
1. Vertical Line of Symmetry
2. Horizontal Line of Symmetry
Vertical Line of Symmetry: If the axis of the shape cuts it into two equal halves vertically then it is called a Vertical Line of Symmetry. The mirror image of the one half appears in a vertical or straight standing position. Examples for vertical Line of Symmetry are H, M, A, U, O, W, V, Y, T.
Horizontal Line of Symmetry: If the axis of the shape cuts it into two equal halves horizontally, then it is called as Horizontal Line of Symmetry. The mirror image of the one half appears as the other similar half. Examples of Horizontal Line of Symmetry are C, B, H, E.
Three Lines of Symmetry: An equilateral triangle is an example of three lines of symmetry. This is symmetrical along its three medians.
Four Lines of Symmetry: A square is an example of Four Lines of Symmetry. The symmetrical lines are two along the diagonals and two along with the midpoints of the opposite sides.
Five Lines of Symmetry: A regular pentagon is an example of Five Lines of Symmetry. The symmetrical lines are joining a vertex to the mid-point of the opposite side.
Six Lines of Symmetry: A regular hexagon is an example of Six Lines of Symmetry. The symmetrical lines are 3 joining the opposite vertices and 3 joining the mid-points of the opposite sides.
Infinite Lines of Symmetry: A circle is an example of Infinite Lines of Symmetry. It has infinite or no lines of symmetry. It is symmetrical along all its diameters.
### Line of Symmetry Examples
Check out some of the examples of Line of Symmetry and learn completely with clear details.
1. Line segment:
From the figure, there is one line of symmetry. line of symmetry of a Line segment passes through its center. There may infinite line passes through the line segment and forms different angles. But we only consider a line as a line of symmetry that cuts the line segment into two equal halves. The Line segment AB is symmetric along the perpendicular bisector l.
2. An angle:
From the figure, there is one line of symmetry. An angle measures the amount of ‘turning’ between two straight lines that meet at a point. The figure is symmetric along the angle bisector OC.
3. An isosceles triangle:
From the figure, there is one line of symmetry. The isosceles triangle figure is symmetric along the bisector of the vertical angle. The Median PL. If the isosceles triangle is also an equilateral triangle, then it has three lines of symmetry. An isosceles triangle has exactly two sides of equal length. Therefore, it has only 1 line of symmetry that passes from the vertex between the two sides of equal length to the midpoint of the side opposite that vertex.
4. Semi-circle:
From the figure, there is one line of symmetry. The Semi-circle figure is symmetric along the perpendicular bisector l. of the diameter AB. A semi-circle does not have any rotational symmetry.
5. Kite:
From the figure, there is one line of symmetry. The Kite is symmetric along with the diagonal BS. A kite is a quadrilateral with two different pairs of adjacent sides that are equal in length and also have only one line of symmetry.
6. Isosceles trapezium:
From the figure, there is one line of symmetry. The Isosceles trapezium figure is symmetric along the line l joining the midpoints of two parallel sides AB and DC. The isosceles trapezium is a convex quadrilateral consists a pair of non-parallel sides that are equal and another pair of sides is parallel but not equal.
7. Rectangle:
From the figure, there are two lines of symmetry. The Rectangle figure is symmetric along the lines l and m joining the midpoints of opposite sides. There are 2 symmetry lines of a rectangle which are from its length and breadth. They cut the rectangle into two equal halves. They appear mirror to each other.
8. Rhombus:
From the figure, there are two lines of symmetry. The Rhombus figure is symmetric along the diagonals AC and BD of the figure. Both the lines of symmetry in a rhombus are from its diagonals. So, it can also say the rhombus lines of symmetry are both diagonals.
### Lines of Symmetry in Alphabets
Have a look at the letters that have the line of symmetry.
One Line of symmetry: The letters consist of One line of symmetry are A B C D E K M T U V W Y.
Vertical Line of symmetry: The letters consist of a Vertical line of symmetry is A M T U V W Y.
Horizontal Line of symmetry: The letters consist of a Horizontal line of symmetry is B C D E K.
Two Lines of Symmetry: The letters consist of Two lines of symmetry are H I X. These are having both horizontal and vertical lines of symmetry.
No Lines of Symmetry: The letters consist of No lines of symmetry are F G J L N P Q R S Z. These have neither horizontal nor vertical lines of symmetry.
Infinite Lines of Symmetry: The letter having Infinite lines of symmetry is O.
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# 2/10/13
## 6.5.5.2. Numerical Example
6. Process or Product Monitoring and Control 6.5. Tutorials 6.5.5. Principal Components
## 6.5.5.2. Numerical Example
Calculation of principal components example Sample data set A numerical example may clarify the mechanics of principal component analysis.
Let us analyze the following 3-variate dataset with 10 observations. Each observation consists of 3 measurements on a wafer: thickness, horizontal displacement and vertical displacement.
## Solve for the roots of R
Next solve for the roots of R, using software value proportion 1 1.769 2 .927 3 .304 .590 .899 1.000
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## 6.5.5.2. Numerical Example
Notice that Each eigenvalue satisfies |R- I| = 0. The sum of the eigenvalues = 3 = p, which is equal to the trace of R (i.e., the sum of the main diagonal elements). The determinant of R is the product of the eigenvalues. The product is 1 x 2 x 3 = .499. Compute the first column of the V matrix Substituting the first eigenvalue of 1.769 and R in the appropriate equation we obtain
This is the matrix expression for 3 homogeneous equations with 3 unknowns and yields the first column of V: .64 .69 -.34 (again, a computerized solution is indispensable). Compute the remaining columns of the V matrix Repeating this procedure for the other 2 eigenvalues yields the matrix V
Notice that if you multiply V by its transpose, the result is an identity matrix, V'V=I. Compute the L1/2 matrix Now form the matrix L1/2 , which is a diagonal matrix whose elements are the square roots of the eigenvalues of R. Then obtain S, the factor structure, using S = V L1/2
So, for example, .91 is the correlation between variable 2 and the first principal component. Compute the communality Next compute the communality, using the first two eigenvalues only
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## Diagonal elements report how much of the variability is explained
Communality consists of the diagonal elements. var 1 .8662 2 .8420 3 .9876 This means that the first two principal components "explain" 86.62% of the first variable, 84.20 % of the second variable, and 98.76% of the third.
## Compute the coefficient matrix
The coefficient matrix, B, is formed using the reciprocals of the diagonals of L1/2
## Compute the principal factors
Finally, we can compute the factor scores from ZB, where Z is X converted to standard score form. These columns are the principal factors.
## Principal factors control chart
These factors can be plotted against the indices, which could be times. If time is used, the resulting plot is an example of a principal factors control chart.
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# Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 chapter 1 - Linear Equations in Two Variables [Latest edition]
Textbook page
## Chapter 1: Linear Equations in Two Variables
Practice Set 1.1Practice Set 1.2Practice Set 1.3Practice Set 1.4Practice Set 1.5Problem Set - 1
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 Chapter 1 Linear Equations in Two VariablesExercise Practice Set 1.1 [Pages 4 - 5]
Practice Set 1.1 | Q 1 | Page 4
Complete the following activity to solve the simultaneous equations.
5x + 3y = 9 -----(I)
2x + 3y = 12 ----- (II)
Practice Set 1.1 | Q 2.1 | Page 5
Solve the following simultaneous equation.
3a + 5b = 26; a + 5b = 22
Practice Set 1.1 | Q 2.2 | Page 5
Solve the following simultaneous equation.
x + 7y = 10; 3x – 2y = 7
Practice Set 1.1 | Q 2.3 | Page 5
Solve the following simultaneous equation.
2x – 3y = 9; 2x + y = 13
Practice Set 1.1 | Q 2.4 | Page 5
Solve the following simultaneous equation.
5m – 3n = 19; m – 6n = –7
Practice Set 1.1 | Q 2.5 | Page 5
Solve the following simultaneous equation.
5x + 2y = –3; x + 5y = 4
Practice Set 1.1 | Q 2.6 | Page 5
Solve the following simultaneous equation.
$\frac{1}{3}x + y = \frac{10}{3}; 2x + \frac{1}{4}y = \frac{11}{4}$
Practice Set 1.1 | Q 2.7 | Page 5
Solve the following simultaneous equation.
99x + 101y = 499; 101x + 99y = 501
Practice Set 1.1 | Q 2.8 | Page 5
Solve the following simultaneous equation.
49x – 57y = 172; 57x – 49y = 252
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 Chapter 1 Linear Equations in Two VariablesExercise Practice Set 1.2 [Page 8]
Practice Set 1.2 | Q 1 | Page 8
Complete the following table to draw graph of the equations - (I) x + y = 3 (II) x – y = 4
x + y = 3
x 3 y 5 3 (x,y) (3,0) (0,3)
x – y = 4
x -1 0 y 0 -4 (x,y) (0,-4)
Practice Set 1.2 | Q 2.1 | Page 8
Solve the following simultaneous equations graphically.
= 6 ; – = 4
Practice Set 1.2 | Q 2.2 | Page 8
Solve the following simultaneous equations graphically.
x + y = 5 ; x – y = 3
Practice Set 1.2 | Q 2.3 | Page 8
Solve the following simultaneous equations graphically.
x + y = 0 ; 2x – y = 9
Practice Set 1.2 | Q 2.4 | Page 8
Solve the following simultaneous equations graphically.
3x – y = 2 ; 2x – y = 3
Practice Set 1.2 | Q 2.5 | Page 8
Solve the Following Simultaneous Equations Graphically.
3x – 4y = –7 ; 5x – 2y = 0
Practice Set 1.2 | Q 2.6 | Page 8
Solve the following simultaneous equations graphically.
2x – 3y = 4 ; 3y – x = 4
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 Chapter 1 Linear Equations in Two VariablesExercise Practice Set 1.3 [Page 16]
Practice Set 1.3 | Q 1 | Page 16
Fill in the blanks with correct number$\begin{vmatrix}3 & 2 \\ 4 & 5\end{vmatrix}$ = 3 x ____ - ____ x 4 = ____ - 8 = ____
Practice Set 1.3 | Q 2.1 | Page 16
Find the values of following determinant.
$\begin{vmatrix}- 1 & 7 \\ 2 & 4\end{vmatrix}$
Practice Set 1.3 | Q 2.2 | Page 16
Find the values of following determinant.
$\begin{vmatrix}5 & 3 \\ - 7 & 0\end{vmatrix}$
Practice Set 1.3 | Q 2.3 | Page 16
Find the values of following determinant.
$\begin{vmatrix}\frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2}\end{vmatrix}$
Practice Set 1.3 | Q 3.1 | Page 16
Solve the following simultaneous equations using Cramer’s rule.
3x – 4y = 10 ; 4x + 3y = 5
Practice Set 1.3 | Q 3.2 | Page 16
Solve the following simultaneous equations using Cramer’s rule.
4x + 3y – 4 = 0 ; 6x = 8 – 5y
Practice Set 1.3 | Q 3.3 | Page 16
Solve the following simultaneous equations using Cramer’s rule.
x + 2y = –1 ; 2x – 3y = 12
Practice Set 1.3 | Q 3.4 | Page 16
Solve the following simultaneous equations using Cramer’s rule.
6x – 4y = –12 ; 8x – 3y = –2
Practice Set 1.3 | Q 3.5 | Page 16
Solve the following simultaneous equations using Cramer’s rule.
4m + 6n = 54 ; 3m + 2n = 28
Practice Set 1.3 | Q 3.6 | Page 16
Solve the following simultaneous equations using Cramer’s rule.
$2x + 3y = 2 ; x - \frac{y}{2} = \frac{1}{2}$
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 Chapter 1 Linear Equations in Two VariablesExercise Practice Set 1.4 [Page 19]
Practice Set 1.4 | Q 1.1 | Page 19
Solve the following simultaneous equations.
$\frac{2}{x} - \frac{3}{y} = 15; \frac{8}{x} + \frac{5}{y} = 77$
Practice Set 1.4 | Q 1.2 | Page 19
Solve the following simultaneous equations.
$\frac{10}{x + y} + \frac{2}{x - y} = 4; \frac{15}{x + y} - \frac{5}{x - y} = - 2$
Practice Set 1.4 | Q 1.3 | Page 19
Solve the following simultaneous equations.
$\frac{27}{x - 2} + \frac{31}{y + 3} = 85; \frac{31}{x - 2} + \frac{27}{y + 3} = 89$
Practice Set 1.4 | Q 1.4 | Page 19
Solve the following simultaneous equations.
$\frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4}; \frac{1}{2\left( 3x + y \right)} - \frac{1}{2\left( 3x - y \right)} = - \frac{1}{8}$
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 Chapter 1 Linear Equations in Two VariablesExercise Practice Set 1.5 [Page 26]
Practice Set 1.5 | Q 1 | Page 26
Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Practice Set 1.5 | Q 2 | Page 26
Complete the following.
Practice Set 1.5 | Q 3 | Page 26
The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Practice Set 1.5 | Q 4 | Page 26
The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Practice Set 1.5 | Q 5 | Page 26
Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Practice Set 1.5 | Q 6 | Page 26
Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 Chapter 1 Linear Equations in Two VariablesExercise Problem Set - 1 [Pages 27 - 29]
Problem Set - 1 | Q 1.1 | Page 27
Choose correct alternative for the following question.
To draw graph of 4x + 5y = 19, Find y when x = 1.
• 4
• 3
• 2
• -3
Problem Set - 1 | Q 1.2 | Page 27
Choose correct alternative for the following question.
For simultaneous equations in variables x and y, D= 49, Dy = –63, D = 7 then what is x ?
• 7
• -7
• 1/7
• (-1)/7
Problem Set - 1 | Q 1.3 | Page 27
Choose correct alternative for the following question.
Find the value of $\begin{vmatrix}5 & 3 \\ - 7 & - 4\end{vmatrix}$
• –1
• –41
• 41
• 1
Problem Set - 1 | Q 1.4 | Page 27
Choose correct alternative for the following question.
To solve x + y = 3 ; 3x – 2y – 4 = 0 by determinant method find D.
(A) 5
(B) 1
(C) -5
(D) -1
Problem Set - 1 | Q 1.5 | Page 27
Choose correct alternative for the following question.
ax + by = c and mx + ny = d and an ≠ bm then these simultaneous equations have -
• Only one common solution.
• No solution.
• Infinite number of solutions.
• Only two solution.
Problem Set - 1 | Q 2 | Page 27
Complete the following table to draw the graph of 2x – 6y = 3
x -5 x y x 0 (x,y) (-5,x) (x,0)
Problem Set - 1 | Q 3.1 | Page 27
Solve the following simultaneous equation graphically.
2x + 3y = 12 ; x – y = 1
Problem Set - 1 | Q 3.2 | Page 27
Solve the following simultaneous equation graphically.
x – 3y = 1 ; 3x – 2y + 4 = 0
Problem Set - 1 | Q 3.3 | Page 27
Solve the following simultaneous equation graphically.
5x – 6y + 30 = 0 ; 5x + 4y – 20 = 0
Problem Set - 1 | Q 3.4 | Page 27
Solve the following simultaneous equation graphically.
3x – y – 2 = 0 ; 2x + y = 8
Problem Set - 1 | Q 3.5 | Page 27
Solve the following simultaneous equation graphically.
3x + y = 10 ; x – y = 2
Problem Set - 1 | Q 4.1 | Page 27
Find the value of the following determinant.
$\begin{vmatrix}4 & 3 \\ 2 & 7\end{vmatrix}$
Problem Set - 1 | Q 4.2 | Page 27
Find the value of the following determinant.
$\begin{vmatrix}5 & - 2 \\ - 3 & 1\end{vmatrix}$
Problem Set - 1 | Q 4.3 | Page 27
Find the value of the following determinant.
$\begin{vmatrix}3 & - 1 \\ 1 & 4\end{vmatrix}$
Problem Set - 1 | Q 5.1 | Page 28
Solve the following equations by Cramer’s method.
6x – 3y = –10 ; 3x + 5y – 8 = 0
Problem Set - 1 | Q 5.2 | Page 28
Solve the following equation by Cramer’s method.
4m – 2n = –4 ; 4m + 3n = 16
Problem Set - 1 | Q 5.3 | Page 28
Solve the following equations by Cramer’s method.
3x – 2= $\frac{5}{2}$ ; $\frac{1}{3}x + 3y = - \frac{4}{3}$
Problem Set - 1 | Q 5.4 | Page 28
Solve the following equations by Cramer’s method.
7x + 3y = 15 ; 12y – 5x = 39
Problem Set - 1 | Q 5.5 | Page 28
Solve the following equations by Cramer’s method.
$\frac{x + y - 8}{2} = \frac{x + 2y - 14}{3} = \frac{3x - y}{4}$
Problem Set - 1 | Q 6.1 | Page 28
Solve the following simultaneous equations.
$\frac{2}{x} + \frac{2}{3y} = \frac{1}{6} ; \frac{3}{x} + \frac{2}{y} = 0$
Problem Set - 1 | Q 6.2 | Page 28
Solve the following simultaneous equations.
$\frac{7}{2x + 1} + \frac{13}{y + 2} = 27 ; \frac{13}{2x + 1} + \frac{7}{y + 2} = 33$
Problem Set - 1 | Q 6.3 | Page 28
Solve the following simultaneous equations.
$\frac{148}{x} + \frac{231}{y} = \frac{527}{xy} ; \frac{231}{x} + \frac{148}{y} = \frac{610}{xy}$
Problem Set - 1 | Q 6.4 | Page 28
Solve the following simultaneous equations.
$\frac{7x - 2y}{xy} = 5 ; \frac{8x + 7y}{xy} = 15$
Problem Set - 1 | Q 6.5 | Page 28
Solve the following simultaneous equations.
$\frac{1}{2\left( 3x + 4y \right)} + \frac{1}{5\left( 2x - 3y \right)} = \frac{1}{4} ; \frac{5}{\left( 3x + 4y \right)} - \frac{2}{\left( 2x - 3y \right)} = - \frac{3}{2}$
Problem Set - 1 | Q 7.1 | Page 28
Solve the following word problem.
A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.
Problem Set - 1 | Q 7.2 | Page 29
Solve the following word problem.
Kantabai bought $1\frac{1}{2}$ kg tea and 5 kg sugar from a shop. She paid Rs 50 as return fare for rickshaw. Total expense was Rs 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid Rs 880 for that. Find the rate of sugar and tea per kg.
Problem Set - 1 | Q 7.3 | Page 29
Solve the following word problem.
To find number of notes that Anushka had, complete the following activity.
Problem Set - 1 | Q 7.4 | Page 29
Solve the Following Word Problem.
Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Problem Set - 1 | Q 7.5 | Page 29
Solve the Following Word Problem.
In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is Rs 720. Find daily wages of skilled and unskilled workers.
Problem Set - 1 | Q 7.6 | Page 29
Solve the Following Word Problem.
Places A and B are 30 km apart and they are on a st raight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
## Chapter 1: Linear Equations in Two Variables
Practice Set 1.1Practice Set 1.2Practice Set 1.3Practice Set 1.4Practice Set 1.5Problem Set - 1
## Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 chapter 1 - Linear Equations in Two Variables
Balbharati solutions for Textbook for SSC Class 10 Mathematics 1 chapter 1 (Linear Equations in Two Variables) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board Textbook for SSC Class 10 Mathematics 1 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Balbharati textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Textbook for SSC Class 10 Mathematics 1 chapter 1 Linear Equations in Two Variables are Linear Equations in Two Variables Applications, Cramer'S Rule, Cross - Multiplication Method, Substitution Method, Elimination Method, Equations Reducible to a Pair of Linear Equations in Two Variables, Simple Situational Problems, Inconsistency of Pair of Linear Equations, Consistency of Pair of Linear Equations, Introduction of System of Linear Equations in Two Variables, Graphical Method of Solution of a Pair of Linear Equations, Determinant of Order Two, Pair of Linear Equations in Two Variables.
Using Balbharati Class 10th Board Exam solutions Linear Equations in Two Variables exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Balbharati Solutions are important questions that can be asked in the final exam. Maximum students of Maharashtra State Board Class 10th Board Exam prefer Balbharati Textbook Solutions to score more in exam.
Get the free view of chapter 1 Linear Equations in Two Variables Class 10th Board Exam extra questions for Textbook for SSC Class 10 Mathematics 1 and can use Shaalaa.com to keep it handy for your exam preparation
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# [Smart Math] Arithmetic Problem 24
Here’s and example of a SMART MATH problem for ARITHMETIC.
### Problem
The sum of 5 successive number is 100. What is the product of the first and last number?
1. 246
2. 282
3. 396
4. 484
5. 555
### The Usual Method
For 5 consecutive numbers, to have their sum = 100, we have the numbers as:
$(x-2)$, $(x-1)$, $x$, $(x+1)$ and $(x+2)$
$\therefore (x-2)+(x-1)+x+(x+1)+(x+2)=100$
$\therefore 5x=100$
$\therefore x=20$
Hence the numbers are:
$(x-2)$ = 18
$(x-1)$ = 19
$x$ = 20
$(x+1)$ = 21
$(x+2)$ = 22
Hence product of the first and last number is 18 x 22 = 396
(Ans: 3)
Estimated Time to arrive at the answer = 60 seconds.
### Using Technique
One can intuitively know that to have sum of 5 consecutive numbers to be 100, we need $\frac{100}{5}=20$ as the middle number. We also know that in Arithmetic Progression, the product of the two consecutive numbers equidistant from a central number is always less than the square of the central number. Since 20 is the central number, the product of the first and last number (equidistant from the central number) will be less than $20^{2}=400$. Also since the numbers are close to each other (being consecutive), the product will not be ‘too’ less than 400. Only 396 satisfies this condition.
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# Introduction to Proofs: An Active Exploration of Mathematical Language
## Section4.2Mathematical Induction
In this section we learn a new proof technique, mathematical induction. This technique is useful for proving statements about the positive (or nonnegative) integers. It is based on the following principle.
### Principle of Mathematical Induction.
Let $$P(n)$$ be a property defined for integers $$n\text{,}$$ and let $$a$$ be a fixed integer. Suppose the following are true:
1. $$P(a)$$ is true;
2. For all $$k\in\mathbb{Z}$$ with $$k\geq a\text{,}$$ if $$P(k)$$ is true than $$P(k+1)$$ is true;
then for all $$n\geq a\text{,}$$ $$P(n)$$ is true.
The way to think about the Principle of Mathematical Induction is that if you know the statement is true for some starting value, ($$P(a)$$ is true), and if you can show that knowing the statement is true for some value allows you to know it is true for the next value ($$P(k)\rightarrow P(k+1)$$), then you know it for all values greater than or equal to $$a\text{.}$$
So why should this work? Suppose you know two things:
1. $$P(1)$$ is true, and
2. $$P(k)\rightarrow P(k+1)\text{.}$$
Note, you do not know $$P(k)$$ is true, just that if it is true, then $$P(k+1)$$ will be true. Now since $$P(1)$$ is true, by (2), $$P(2)$$ must be true. Since $$P(2)$$ is true, $$P(3)$$ must be true, etc. In this way we can see that $$P(n)$$ must be true for all $$n\geq 1\text{.}$$
The Principle of Mathematical Induction lets us skip all the intermediate steps, $$P(1)\rightarrow P(2), P(2)\rightarrow P(3),$$ and conclude $$P(n)$$ once we have (1) and (2).
### Structure of a Mathematical Induction Proof.
To prove $$P(n), \forall n\geq 1\text{,}$$
• Base Step.
Show $$P(1)$$ is true.
• Induction Step.
Assume $$P(k)$$ for some $$k\in\mathbb{Z}\text{,}$$ show $$P(k+1)\text{.}$$
Conclude $$P(n)$$ is true for all $$n\geq 1$$
In the above structure we used $$a=1$$ for simplicitiy, but an induction proof could have a base step starting at a different $$a\text{.}$$ Most commonly, the base step starts with 0 or 1.
When writing induction proofs, make sure you use the actual statement you are proving rather than the notation $$P(1), P(k), P(n)\text{.}$$
Since the induction step in mathematical induction connects a statement about $$k$$ to a statement about $$k+1\text{,}$$ we need to be comfortable with the relationship between sums of $$k$$ terms and sums of $$k+1$$ terms.
#### (a)
Write out the summation for $$\sum_{i=1}^{k}i\text{.}$$
#### (b)
Write out the summation for $$\sum_{i=1}^{k+1}i\text{.}$$
#### (c)
How do (a) and (b) differ?
Prove $$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$$ for $$n\geq 1\text{.}$$
Base Step: Let $$n=1\text{.}$$ Then
\begin{gather*} \sum_{i=1}^{n}i=\sum_{i=1}^{1}i=1\\ \frac{n(n+1)}{2}=\frac{1(1+1)}{2}=1 \end{gather*}
Since the left hand side of the equation equals the right hand side, $$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$$ for $$n=1\text{.}$$
Induction Step:
Assume $$\sum_{i=1}^{k}i=\frac{k(k+1)}{2}$$ for some $$k\geq 1\text{.}$$
Show $$\sum_{i=1}^{k+1}i=\frac{(k+1)(k+2)}{2}\text{.}$$
Proof of induction step:
\begin{align*} \sum_{i=1}^{k}i&=\frac{k(k+1)}{2}\ \ \textrm{ by the induction assumption}\\ \sum_{i=1}^{k+1}i&= \sum_{i=1}^{k}i+(k+1)\ \ \textrm{ add the } k+1 \textrm{ term of the sum}\\ &=\frac{k(k+1)}{2}+(k+1)\\ &=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}\\ &=\frac{k(k+1)+2(k+1)}{2}\ \ \textrm{ factor out }(k+1) \textrm{ on the top}\\ &=\frac{(k+1)(k+2)}{2}. \end{align*}
Thus, $$\sum_{i=1}^{k+1}i=\frac{(k+1)(k+2)}{2}\text{.}$$
Hence, by induction $$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$$ for $$n\geq 1\text{.}$$
Note, in the base step we looked at each side of what we wanted to show separately. We can refer to this as a “left hand side/ right hand side proof”, or in short hand, a LHS/RHS proof. If we just start with the equation we want to show, then we are assuming what we are trying to prove. To avoid this, it is best, when trying to show two things are equal in the base step, to do a LHS/ RHS proof.
Example 4.2.1 now gives us a useful formula for finding the sum of 1 through $$n\text{:}$$
$$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}.\tag{4.2.1}$$
### Activity4.2.2.Adding $$k+1$$ Terms.
Before the next example, we practice the relationship between sums of $$k$$ terms and sums of $$k+1$$ terms.
#### (a)
Write out the summation for $$\sum_{i=0}^{k}r^i\text{.}$$
#### (b)
Write out the summation for $$\sum_{i=0}^{k+1}r^i\text{.}$$
#### (c)
How do (a) and (b) differ?
Prove $$\sum_{i=0}^{n}2^i=2^{n+1}-1$$ for $$n\geq 0\text{.}$$
Base Step: Let $$n=0\text{.}$$ Then
\begin{gather*} \sum_{i=0}^{0}2^i=2^0=1\\ 2^{0+1}-1=2-1=1 \end{gather*}
Since the left hand side of the equation equals the right hand side, $$\sum_{i=0}^{n}2^i=2^{n+1}-1$$ for $$n=0\text{.}$$
Induction Step:
Assume $$\sum_{i=0}^{k}2^i=2^{k+1}-1$$ for some $$k\geq 0\text{.}$$
Show $$\sum_{i=0}^{k+1}2^i=2^{k+2}-1\text{.}$$
Proof of induction step:
\begin{align*} \sum_{i=0}^{k}2^i&=2^{k+1}-1 \textrm{ by the induction assumption}\\ \sum_{i=0}^{k+1}2^i&= \sum_{i=0}^{k}2^i+(2^{k+1}) \textrm{ add the } k+1 \textrm{ term of the sum}\\ &=2^{k+1}-1+2^{k+1}\\ &=2^{k+1}+2^{k+1}-1 \\ &=2^{k+1}(1+1)-1 \\ &=2^{k+1}(2)-1 \\ &=2^{k+2}-1. \end{align*}
Thus, $$\sum_{i=0}^{k+1}2^i=2^{k+2}-1\text{.}$$
Hence, by induction $$\sum_{i=0}^{n}2^i=2^{n+1}-1$$ for $$n\geq 0\text{.}$$
Example 4.2.2 can be generalized to the sum $$\sum_{i=0}^{n}r^i\text{.}$$
Prove $$\sum_{i=0}^{n}r^i=\frac{r^{n+1}-1}{r-1}$$ for $$n\geq 0\text{.}$$
Base Step: Let $$n=0\text{.}$$ Then
\begin{gather*} \sum_{i=0}^{0}r^i=r^0=1\\ \frac{r^{0+1}-1}{r-1}=\frac{r-1}{r-1}=1 \end{gather*}
Since the left hand side of the equation equals the right hand side, $$\sum_{i=0}^{n}r^i=\frac{r^{n+1}-1}{r-1}$$ for $$n=0\text{.}$$
Induction Step:
Assume $$\sum_{i=0}^{k}r^i=\frac{r^{k+1}-1}{r-1}$$ for some $$k\geq 0\text{.}$$
Show $$\sum_{i=0}^{k+1}r^i=\frac{r^{k+2}-1}{r-1}\text{.}$$
Proof of induction step:
\begin{align*} \sum_{i=0}^{k}r^i&=\frac{r^{k+1}-1}{r-1} \textrm{ by the induction assumption}\\ \sum_{i=0}^{k+1}r^i&= \sum_{i=0}^{k}r^i+(r^{k+1}) \textrm{ add the } k+1 \textrm{ term of the sum}\\ &=\frac{r^{k+1}-1}{r-1}+(r^{k+1})\\ &=\frac{r^{k+1}-1}{r-1}+\frac{(r-1)(r^{k+1})}{r-1}\\ &=\frac{r^{k+1}-1+(r-1)(r^{k+1})}{r-1} \textrm{ factor out } r^{k+1} \textrm{ on the top}\\ &=\frac{r^{k+1}(1+r-1)-1}{r-1} \\ &=\frac{r^{k+1}(r)-1}{r-1} \\ &=\frac{r^{k+2}-1}{r-1}. \end{align*}
Thus, $$\sum_{i=0}^{k+1}r^i=\frac{r^{k+2}-1}{r-1}\text{.}$$
Hence, by induction $$\sum_{i=0}^{n}r^i=\frac{r^{n+1}-1}{r-1}$$ for $$n\geq 0\text{.}$$
The sum in Example 4.2.3 is called a geometric sum. Now we have a useful formula for the sum of $$r^i$$ for various values of $$r>1\text{:}$$
$$\sum_{i=0}^{n}r^i=\frac{r^{n+1}-1}{r-1}.\tag{4.2.2}$$
### Activity4.2.3.Practice with Induction.
Suppose you want to prove $$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$$ for all $$n\geq 1$$ by induction.
#### (a)
What would you want to prove in the base step?
#### (b)
What should you assume in the induction step?
#### (c)
What should you show in the induction step?
#### (d)
Now try to put this all together in the form of an induction proof. If you are unable to prove it, where do you get stuck?
The closed form of a sum is the computational formula for the sum. For example, the closed form of $$\sum_{i=0}^{n}2^i$$ is $$2^{n+1}-1\text{,}$$ as proved in Example 4.2.2.
### ExercisesExercises
#### 1.
Use mathematical induction to prove for all integers $$n\geq 1\text{,}$$
\begin{equation*} \sum_{i=1}^n 2i=n^2+n. \end{equation*}
#### 2.
Use mathematical induction to prove for all integers $$n\geq 1\text{,}$$
\begin{equation*} \sum_{i=1}^ni^3=\Bigl[\frac{n(n+1)}{2}\Bigr]^2. \end{equation*}
#### 3.
Use mathematical induction to prove for all integers $$n\geq 1\text{,}$$
\begin{equation*} \sum_{i=1}^{n}i(i!)=(n+1)!-1. \end{equation*}
#### 4.
Use Example 4.2.3 to find a formula for $$\sum_{i=0}^n 3^i\text{.}$$ Prove your formula directly using induction.
#### 5.
Find a formula for $$\sum_{i=0}^n 2i-1\text{.}$$ Prove your formula using induction.
#### 6.
Observe that
\begin{align*} \frac{1}{1\cdot 3}\amp =\frac{1}{3}\\ \frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}\amp =\frac{2}{5}\\ \frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}\amp =\frac{3}{7}\\ \frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}\amp =\frac{4}{9} \end{align*}
Conjecture a general formula for $$\sum_{i=1}^{n}\frac{1}{(2i-1)(2i+1)}\text{,}$$ and prove your conjecture by mathematical induction.
#### 7.
Evaluate the sum $$\sum_{i=1}^{n}\frac{i}{(i+1)!}$$ for all $$n=1, 2, 3, 4, 5\text{.}$$ Make a conjecture for a formula for the sum for a general $$n\text{,}$$ and prove your conjecture by induction.
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# How do you solve and write the following in interval notation: 8x -3x + 2< 2(x + 7)?
Jun 2, 2016
$\left(- \infty , 4\right)$
#### Explanation:
We have the inequality
$8 x - 3 x + 2 < 2 \left(x + 7\right)$
On the left hand side, the $8 x$ and $- 3 x$ can be combined:
${\overbrace{8 x - 3 x}}^{8 x - 3 x = 5 x} + 2 < 2 \left(x + 7\right)$
$5 x + 2 < 2 \left(x + 7\right)$
Next, on the right hand side, distribute the $2$:
$5 x + 2 < {\overbrace{2 \left(x + 7\right)}}^{2 \left(x\right) + 2 \left(7\right)}$
$5 x + 2 < 2 x + 14$
Subtract $2 x$ from both sides.
${\overbrace{5 x - 2 x}}^{3 x} + 2 < {\overbrace{2 x - 2 x}}^{0} + 14$
$3 x + 2 < 14$
Subtract $2$ from both sides.
$3 x + {\overbrace{2 - 2}}^{0} < {\overbrace{14 - 2}}^{12}$
$3 x < 12$
Divide both sides by $3$.
$\frac{3 x}{3} < \frac{12}{3}$
$x < 4$
Now, we need to express $x < 4$ in interval notation. Since $x$ has to be less than $4$, we know that $4$ is the upper bound of the interval.
Since there is no lower bound, $x$ can extend all the way down, from $4$, to $0$, into the negative numbers, working its way left on the number line all the way to negative infinity.
So, the interval is $\left(- \infty , 4\right)$. Note that ( and ) are used instead of [ and ] because $x < 4$, not $x \le 4$. Additionally, ( and ) are always used with $\infty$ because $x$ can never equal $\infty$.
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Quick Math Homework Help
Master the 7 pillars of school success that I have learned from 25 years of teaching.
# What is an outlier in Math?
An outlier is a number in a data set that is much smaller or larger than the other numbers in the data set.
90,86,15,86,92
15 would be an outlier in this data set.
90,15,25,18,24
In this data set, 90 would be considered an outlier
The measure of center may be affected by an outlier.
Determine how the outlier affects the median and the mean in the
following scores.
35, 90, 88, 93, 96
The outlier in this data set is 35
Determine how the outlier affects the median by finding the median without the outlier included.
The median without the outlier = The average of 88 + 93 = 90.5
How much did the median change? 90.5 – 88 = 2.5
Next, determine how the outlier affects the mean by finding the mean
without the outlier included.
Mean without the outlier = 90 + 88 + 93 + 96 = 91.75
The mean changed by 11.35 91.75-80.4 = 11.35
The mean was changed the most by the outlier.
In conclusion, when comparing the measure of center, and you have an outlier it is best to use the median.
If you have don’t have an outlier then use the mean when finding the measure of center.
You may also enjoy..
Middle School Math
## Identify the outlier in the following data sets.
Keywords: outlier in Math | | 6th Grade Math | Measure of center | Impact of outlier on measure of center
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Easy Class 11 Notes for Motion In A Straight Line
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Motion in a straight line is one of the fundamental topics in the field of physics, particularly in the early stages of learning. Understanding the concept of motion in a straight line lays the foundation for more complex topics in the realm of mechanics. In Class 11, students are introduced to the basic principles and equations governing motion in a straight line. In this comprehensive guide, we will delve into the various aspects of motion in a straight line and provide easy-to-understand notes for Class 11 students.
Introduction to Motion in a Straight Line
Motion in a straight line, also known as rectilinear motion, refers to the movement of an object along a straight path. The key parameters that describe the motion of an object in a straight line include displacement, velocity, and acceleration.
Displacement
• Definition: Displacement is the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction.
• Formula: Displacement (( S )) is given by the equation:
[ S = \text{Final position} - \text{Initial position} ]
Velocity
• Definition: Velocity is the rate of change of displacement with respect to time. It is a vector quantity.
• Formula: Average velocity (( \overline{v} )) is calculated as:
[ \overline{v} = \frac{\text{Total displacement}}{\text{Total time taken}} ]
Acceleration
• Definition: Acceleration is the rate of change of velocity with respect to time. It is a vector quantity.
• Formula: Average acceleration (( \overline{a} )) is given by:
[ \overline{a} = \frac{\text{Change in velocity}}{\text{Time taken for the change}} ]
Equations of Motion
In Class 11, students are introduced to the equations of motion that govern the motion of an object in a straight line under constant acceleration.
1. First Equation of Motion:
[ v = u + at ]
where:
( v ) = final velocity,
( u ) = initial velocity,
( a ) = acceleration,
( t ) = time taken.
1. Second Equation of Motion:
[ s = ut + \frac{1}{2}at^2 ]
where:
( s ) = displacement,
( u ) = initial velocity,
( a ) = acceleration,
( t ) = time taken.
1. Third Equation of Motion:
[ v^2 = u^2 + 2as ]
where:
( v ) = final velocity,
( u ) = initial velocity,
( a ) = acceleration,
( s ) = displacement.
Graphical Representation of Motion
Graphs are essential tools for visualizing and analyzing motion in a straight line. In Class 11, students learn about the graphical representation of motion using distance-time and velocity-time graphs.
Distance-Time Graph
• A distance-time graph shows how the distance traveled by an object changes with time.
• The slope of a distance-time graph represents the speed of the object.
• A horizontal line on a distance-time graph indicates that the object is at rest.
Velocity-Time Graph
• A velocity-time graph illustrates how the velocity of an object changes with time.
• The area under a velocity-time graph represents the displacement of the object.
• Different segments of the graph indicate different types of motion, such as uniform motion, accelerated motion, or decelerated motion.
Key Concepts in Motion In A Straight Line
Uniform Motion
• An object is said to be in uniform motion when it covers equal distances in equal intervals of time.
• In uniform motion, the velocity of the object remains constant.
Non-Uniform Motion
• In non-uniform motion, an object does not cover equal distances in equal intervals of time.
• The velocity of the object changes over time in non-uniform motion.
Relative Motion
• Relative motion refers to the calculation of the motion of one object with respect to another moving or stationary object.
• Understanding relative motion is crucial in scenarios where multiple objects are in motion.
Acceleration Due to Gravity
• Near the surface of the Earth, all objects experience a constant acceleration due to gravity denoted as ( g ).
• The standard value of acceleration due to gravity is approximately ( 9.81 \, \text{m/s}^2 ) downward.
1. What is the difference between distance and displacement?
• Distance is the total path length traveled by an object, irrespective of direction. It is a scalar quantity.
• Displacement is the change in position of an object in a specific direction. It is a vector quantity.
2. How is velocity different from speed?
• Velocity is a vector quantity that includes both the speed of an object and its direction.
• Speed is a scalar quantity that only depicts how fast an object is moving, without considering direction.
3. What is the significance of the slope of a distance-time graph?
• The slope of a distance-time graph represents the speed of the object. A steeper slope indicates higher speed, while a horizontal line indicates the object is at rest.
4. Can an object have zero velocity and non-zero acceleration simultaneously?
• Yes, an object can have zero velocity and non-zero acceleration if its speed is constant but it changes direction. This occurs in cases of circular motion.
5. How do you calculate the area under a velocity-time graph?
• The area under a velocity-time graph represents the displacement of the object. To calculate the displacement, find the area enclosed by the graph and the time axis.
In conclusion, mastering the concepts of motion in a straight line is essential for building a strong foundation in physics. By understanding the fundamental principles, equations, and graphical representations of motion, Class 11 students can navigate more complex topics in mechanics with confidence and clarity.
Diya Patel
Diya Patеl is an еxpеriеncеd tеch writеr and AI еagеr to focus on natural languagе procеssing and machinе lеarning. With a background in computational linguistics and machinе lеarning algorithms, Diya has contributеd to growing NLP applications.
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# Thread: Determing Antiderivatives from a Word Problem
1. ## Determing Antiderivatives from a Word Problem
"A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/ $s^{2}$. What is the distance covered before the car comes to a stop?"
I know that the antiderivative of the acceleration is the velocity, and the antiderivative of the velocity is its distance, but I need help on how to derive the equations from this word problem.
2. Well, we can simply use this one formula:
$a = c$
Meaning the acceleration is constant, in this case, it is $-22 \frac{ft}{s^2}$
So, we simply take the antiderivative to find the velocity function:
$\int -22dx$
$= -22x + C$
Well, we know that this equals the velocity, but it's in $\frac{mi}{h}$ and not $\frac{ft}{s}$, so we have to convert it.
$50 \frac{mi}{h} = \frac{220}{3} \frac{ft}{s}$
Now we set the constant equal to the initial velocity, seeing as it fits the equation to find the final equation:
$v_0 = \frac{220}{3} = C$
And we know it slows down to 0, so:
$v_f = 0$
$0 = -22x + \frac{220}{3}$
We can generalize a formula from this:
$v_f = v_0 + at$
$-22t = at$
$C = v_0$
$0 = v_f$
Now we take the antiderivative of the right side of that function:
$\int [v_0 + at]dt$
$= v_0t + \frac{1}{2}at^2 + D$
This is the position function, and we can assume that since the initial velocity was the C value, that the initial position value would be the D value:
$D = h_0$
So, this function is a function of position, which is commonly called s(t):
$s(t) = h_0 + v_0t + \frac{1}{2}at^2$
Now we plug in what we know (Assuming we started at position = 0):
$s(t) = \frac{220}{3}t - 11t^2$
3. ## Re: Determing Antiderivatives from a Word Problem
I would have written it slightly differently but basically the same work as Aryth did:
"A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/ s^{2}. What is the distance covered before the car comes to a stop?"
One thing you have to be careful about is that speed is give in "miles per hour" while acceleration is given in "feet per second per second". We must change to the same units. There are 5280 feet per mile so 50 miles per hour is 50(5280)= 264000 feet per hour. There are 60 minutes per hour and 60 seconds per minute so there are 3600 seconds per hour. 264000 feet per hour is 26400/3600= 73 and 1/3 feet per second.
Acceleration is the derivative (rate of change) of velocity so $\frac{dv}{dt}= -22$ (negative because this is deceleration).
Integrating, v= -22t+ C and since it had an initial (t= 0) speed of 73 and 1/3 feet per second, v= -22t+ 73 1/3.
v= dx/dt= -22t+ 73 1/3 so $x= -11t^2+ (73 1/3)t+ C$. Since we want to know how far the car travels before it stops, we can take x= 0 as the initial position and so C= 0. The car will have stopped when v= 0 so solve -22t+ 73 1/3= 0 for t and put that value of t into $x= -11t^2+ (73 1/3)t$.
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# Division of Fraction by a Whole Number – Definition, Examples | How to Divide a Fraction with a Whole Number?
Here you can learn completely about the concept of dividing fractions with whole numbers. By going through this entire article you will be able to know what is meant by a fraction and what is a whole number, and the procedure to divide fraction numbers and mixed fractions numbers with whole numbers with few solved examples on this concept with clear explanation.
Also, refer
## What is meant by Division of Fractions?
Dividing fractions can be obtained by multiplying the fractions in reverse of one of its two fraction numbers which means by writing the reciprocal of one fraction, reciprocal means if a fraction is given as $$\frac {x}{y}$$ then the reciprocal of it will $$\frac {y}{x}$$. This means we are simply interchanging the position of numerator and denominator with each other.
Division of fractions requires equivalent fractions to solve them. So first of all we have to make sure that the fractions are equivalent and then we need to follow all the steps in dividing the fractions. So generally when we are dividing fractions in the direct method it requires more effort, but don’t worry we are providing a simple and easy method here.
## Dividing Fraction with a Whole Number
Division of Fraction number by a whole number can be done easily by following the below-mentioned step-by-step procedure.
As we already know, a whole number is nothing but the real numbers, including zero and positive and negative integers.
1. We have to convert the whole numbers into a fraction, this can be obtained by simply adding 1 as its denominator.
2. After converting the given whole number into a fraction number we have to find the reciprocal of the given number.
3. Now, we need to multiply the converted fraction with the given fraction number.
4. After doing so we need to simplify the equation to get its lowest terms.
### Division of Fractions with Whole Numbers Examples
Example 1:
Solve the equation divide a fraction number, $$\frac { 6 }{ 5 }$$ with a whole number 10?
Solution:
According to our steps first, we need to covert our given whole number 10 into a fractional number by adding 1 as the denominator. So our whole number becomes $$\frac { 10 }{ 1 }$$
Now after converting take the reciprocal of the obtained fractional number, which gives $$\frac { 1 }{ 10 }$$
Now we need to multiply obtained fraction number and the given fraction number. $$\frac { 6 }{ 5 } * \frac { 1 }{ 10 }$$
To simplify we nedd to muilpty numerators and denominators $$\frac { 6 * 1 }{ 5 * 10 }$$
The result will be $$\frac { 6 }{ 50 }$$.
Hence, the result obtained by dividing factional number $$\frac { 6 }{ 5 }$$ and the whole number 10 is $$\frac { 6 }{ 50 }$$.
This can further simplifed as $$\frac { 3 }{ 25 }$$ because both 6 and 50 can be divided by 2.
Answer: $$\frac { 3 }{ 25 }$$
Example 2:
Solve the equation divide a fraction number, $$\frac { 4 }{ 3 }$$ with a whole number 8?
Solution:
According to our steps first, we need to covert our given whole number 8 into a fractional number by adding 1 as the denominator. So our whole number becomes $$\frac { 8 }{ 1 }$$
Now after converting take the reciprocal of the obtained fractional number, which gives $$\frac { 1 }{ 8 }$$
Now we need to multiply obtained fraction number and the given fraction number $$\frac { 4 }{ 3 } * \frac { 1 }{ 8 }$$
To simplify we nedd to muilpty numerators and denominators $$\frac { 4 * 1 }{ 3 * 8 }$$
The result will be $$\frac { 4 }{ 24 }$$.
Hence, the result obtained by dividing factional number $$\frac { 4 }{ 3 }$$ and the whole number 8 is $$\frac { 4 }{ 24 }$$.
This can further simplifed as $$\frac { 1 }{ 6 }$$ because both 4 and 24 can be divided by 2.
Answer: $$\frac { 1 }{ 6 }$$
## Dividing Mixed Fraction with a Whole Number
To divide the given mixed fractions with a whole number initially, we need to convert the given mixed fraction to a normal fraction(improper fraction), by doing this we can solve the given problem with the same method we used for solving division problems on fraction with the whole number.
We need to follow the below-mentioned steps
1. First, we need to convert the given mixed fraction number into a fraction number
2. We have to convert the whole numbers into a fraction, this can be obtained by simply adding 1 as its denominator.
3. After converting the given while number into a fraction number we have to find the reciprocal of the given number.
4. Now, we need to multiply the converted fraction with the given fraction number.
5. After doing so we need to simplify the equation to get its lowest terms.
### Division of Mixed Fractions with Whole Numbers Examples
Example 1:
Solve the equation divide the mixed fraction 3$$\frac { 1 }{ 2 }$$ with a whole number 7?
Solution:
First, we need to convert 3$$\frac { 1 }{ 2 }$$ to a simple fraction, which gives $$\frac { 7 }{ 2 }$$.
Now we need to covert our given whole number 7 into a fractional number by adding 1 as the denominator. So our whole number becomes $$\frac { 7 }{ 1 }$$
Now after converting we need to find the reciprocal of the obtained fractional number, which gives $$\frac { 1 }{ 7 }$$
Now we need to multiply obtained fraction number and the given fraction number $$\frac { 7 }{ 2 } * \frac { 1 }{ 7 }$$
To simplify we nedd to muilpty numerators and denominators $$\frac { 7 * 1 }{ 2 * 7 }$$
The result will be $$\frac { 7 }{ 14 }$$.
Hence, the result obtained by dividing mixed factional number 3$$\frac { 1 }{ 2 }$$ and the whole number 7 is $$\frac { 7 }{ 14 }$$
This can further simplifed as $$\frac { 1 }{ 2 }$$ because both 7 and 14 can be divided by 7.
$$\frac { 1 }{ 2 }$$
Example 2:
Solve the equation divide the mixed fraction 2$$\frac { 2 }{ 5 }$$ with a whole number 2?
Solution:
First, we need to convert 2$$\frac { 2 }{ 5 }$$ to a simple fraction, which gives $$\frac { 12 }{ 5 }$$.
Now we need to covert our given whole number 2 into a fractional number by adding 1 as the denominator. So our whole number becomes $$\frac { 2 }{ 1 }$$
Now after converting we need to find the reciprocal of the obtained fractional number, which gives $$\frac { 1 }{ 2 }$$
Now we need to multiply obtained fraction number and the given fraction number $$\frac { 12 }{ 5 } * \frac { 1 }{ 2 }$$
To simplify we nedd to muilpty numerators and denominators $$\frac { 12 * 1 }{ 5 * 2 }$$
The result will be $$\frac { 12 }{ 10 }$$.
Hence, the result obtained by dividing mixed factional number 2$$\frac { 2 }{ 5 }$$ with a whole number 2 is $$\frac { 12 }{ 10 }$$
This can further simplifed as $$\frac { 6 }{ 5 }$$ because both 12 and 10 can be divided by 2.
$$\frac { 6 }{ 5 }$$
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# Percentages and Decimals
## Definition
Decimals are used to indicate the fractional component of a number. For example, $\frac{5}{2}$ in decimal form is $2.5$.
There are three types of decimals:
• A terminating decimal does not go on forever. Example: $13.5982$.
• A repeating decimal goes on forever but the digits repeat. Example: $0.352525252\dots$, which can be rewritten more succinctly as $0.3\overline{52}$ by putting a line over the repeating digits.
• An irrational decimal goes on forever but the digits have no repeating pattern. Example: $\pi = 3.14159\dots$.
A percentage is a number displayed as a fraction of 100, often denoted with a percent sign %, such as $98\%$, which is 98 parts of 100.
## Technique
You will often have to manipulate numbers between their decimal, percentage, and fraction forms.
To convert between percentage and decimal, you simply move the decimal point left or right two places.
\begin{aligned} 17\% &= 0.17 \\ 189\% &= 1.89 \\ 0.4\% &= 0.004 \end{aligned}
Percentages can be rewritten as fractions by placing them over a denominator of $100$.
\begin{aligned} 17\% &= \frac{17}{100} \\ 189\% &= \frac{189}{100} \\ 0.4\% &= \frac{0.4}{100} = \frac{4}{1000} = \frac{2}{500} \end{aligned}
To convert a terminating decimal into a fraction, first count the number of decimal places. Then divide the decimal's digits over 1 followed by the number of zeroes equal to the number of decimal places.
\begin{aligned} 0.125 &= \frac{125}{1000} = \frac{1}{8} \\ 3.72 &= \frac{372}{100} = \frac{93}{25} \end{aligned}
Decimals that are irrational cannot be converted into fractions, but repeating decimals can. Let's look at a technique for converting repeating decimals into fractions:
### Express $0.3\overline{52}$ as a fraction.
Let's have $0.3\overline{52}$ be equal to the fraction $x$.
There are two repeating digits in this fraction, so we multiply $x$ twice by $10$ to get $100x$. Now we have the equations
$\begin{array}{rcr} 100x &=& 35.2\overline{52} \\ x &=& 0.3\overline{52} \end{array}$
If we subtract them, we get $99x = 34.9$. Solving for $x$ gives us the answer:
$x = \frac{34.9}{99} = \frac{349}{990} \quad _\square$
Note that the number of 10's you multiply with depends on the number of repeating digits. For example, if $x = 0.\overline{12345}$, you would multiply by $10$ five times to get $100000x$. The rest of the steps remain the same.
## Application and Extensions
### Susan and Sally have just finished dinner at a nice sushi restaurant. Their waiter gives them their bill and they see that their (pre-tax and pre-tip total) is $50. If tax is 5% and they decide to give a 19% tip, how much money should Susan and Sally pay? Tax and tip are calculated based on the original price, so tax is $\50 \times 0.05 = \2.50$ and tip is $\50 \times 0.19 = \9.50$. Their total bill is therefore $\50 + \2.50 + \9.50 = \62$. $_\square$ ### A clothing store is having a clearance sale. Their advertisement says "Take 50% off already already reduced prices!" You decide to buy a clearance jacket with a 15%-off label, and pay$34 for it. What was the jacket's original price?
Let $P$ be the original price of the jacket. A 15%-off sale means that 15% of the original price is subtracted from the original price, or 100%-15% = 85% of the original price. Clearance takes another 50% off the sale price. So,
$(P \times 0.85) \times 0.5 = 34$
Solving for $P$ gives us the original price of $\80$. $_\square$
Note by Arron Kau
7 years, 2 months ago
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# RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise.
Board RBSE Te×tbook SIERT, Rajasthan Class Class 10 Subject Maths Chapter Chapter 3 Chapter Name Polynomials E×ercise Miscellaneous Exercise Number of Questions Solved 23 Category RBSE Solutions
## Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise
Multiple Choice Questions
RBSE Solutions For Class 10 Maths Chapter 3 Miscellaneous Question 1.
If one zero of polynomial f(x) = 5x2 + 13x + k is inverse of (RBSESolutions.com) other, then k will be :
(A) 0
(B) $$\frac { 1 }{ 5 }$$
(C) 5
(D) 6
Solution
RBSE Class 10 Maths Chapter 3 Miscellaneous Question 2.
Zeros of (RBSESolutions.com) polynomial x2 – x – 6 are :
(A) 1, 6
(B) 2, -3
(C) 3, -2
(D) 1, -6
Solution
f(x) = x2 – x – 6
= x2 – 3x + 2x – 6
= x(x – 3) + 2(x – 3)
= (x – 3)(x + 2)
for zeros f(x) = 0
⇒ (x – 3)(x + 2) = 0
⇒ x = 3 or x = -2
Hence, option (C) is correct.
RBSE Solutions For Class 10 Maths Chapter 3 Question 3.
If 3 is a zero of polynomial 2x2 + x + k, then (RBSESolutions.com) value of k will be
(A) 12
(B) 21
(C) 24
(D) -21
Solution
Let f(x) = 2x2 + x + k
One zero is 3
f(3) = 0
⇒ 2(3)2 + 3 + k = 0
⇒ 2 × 9 + 3 + k = 0
⇒ 18 + 3 + k = 0
⇒ k = -21
Hence, option (D) is correct.
Miscellaneous Exercise 3 Class 10 Question 4.
If α, β are zeros of (RBSESolutions.com) polynomial x2 – p(x + 1) – c such that (α + 1)(β + 1) = 0, then c will
(A) 0
(B) -1
(C) 1
(D) 2
Solution
RBSE Solution Class 10 Maths Chapter 3 Question 5.
If roots of quadratic (RBSESolutions.com) equation x2 – kx + 4 = 0 then k will be
(A) 2
(B) 1
(C) 4
(D) 3
Solution
Given equation x2 – kx + 4 = 0
Comparing it with ax2 + bx + c = 0
We get a = 1, b= -k and c = 4
D = b2 – 4 ac = (-k)2 – 4 × 1 × 4 = k2 – 16
Roots are same
D = 0
⇒ k2 – 16 = 0
⇒ k2 = 16
⇒ k = ±√16
⇒ k = ± 4
Hence, option (C) is correct.
Class 10 Maths RBSE Solution Chapter 3 Question 6.
If x = 1, is common root of (RBSESolutions.com) equation ax2 + ax + 3 = 0 and x2 + x + b = 0, then ab will be:
(A) 1
(B) 3.5
(C) 6
(D) 3
Solution
Putting x = 1 in equation
ax2 + ax + 3 = 0
⇒ a(1)2 + a(1) + 3 = 0
⇒ a + a + 3 = 0
⇒ 2a = – 3
⇒ a = $$\frac { -3 }{ 2 }$$
and putting x = 1 in equation
x2 + x + b = 0
⇒ (1)2 + (1) + b = 0
⇒ 1 + 1 + b = 0
⇒ b = -2
ab = $$\frac { -3 }{ 2 }$$ × -2 = 3
Hence, option (D) is correct.
Class 10 RBSE Maths Solution Ch 3 Question 7.
Discriminant of quadratic (RBSESolutions.com) equation 3√3 x2 + 10x + √3 = 0
(A) 10
(B) 64
(C) 46
(D) 30
Solution
Comparing 3√3x2 + 10x + √3 by ax2 + bx + c = 0,
a = √3, b = 10 and c = √3
Discriminant(D) = b2 – 4ac
= (10)2 – 4 × 3√3 × √3
= 100 – 4 × 3 × 3
= 100 – 36
= 64
Hence, option (B) is correct.
RBSE Solutions For Class 10 Maths Chapter 3 Miscellaneous Question 8.
Nature of roots of quadratic (RBSESolutions.com) equation 4x2 – 12x – 9 = 0 is :
(A) Real and same
(B) Real and distinct
(C) Imgiary and same
(D) Imaginary and distinct
Solution
Given equations 4x2 – 12x – 9 = 0
Where a = 4, b = -12, c = -9
Discriminant (D) = b2 – 4ac
= (-12)2 – 4 × 4 × (-9)
= 144 + 144
= 288 > 0
Hence, option (B) is correct.
RBSE Solutions For Class 10 Maths Chapter 3 Question 9.
H.C.F. of (RBSESolutions.com) expressions 8a2b2c and 20ab3c2 is:
(A) 4ab2c
(B) 4abc
(C) 40a2b3c2
(D) 40abc
Solution
8a2b2c = 2 × 2 × 2 × a2 × b2 × c = 23 × a2 × b× c
and 20ab3c3 = 2 × 2 × 5 × a × b3 × c3 = 22 × 5 × a × b3 × c3
Product of common least power = 22 × a × b2 × c = 4ab2c
Hence, option (A) is correct.
RBSE Class 10 Maths Chapter 3 Miscellaneous Question 10.
Find L.C. M. of expressions x2 – 1 and x2 + 2x + 1
(A) x + 1
(B) (x2 – 1) (x – 1)
(C) (x – 1) (x + 1)2
(D) (x2 – 1) (x + 1)
Solution
x2 – 1 = (x – 1) (x + 1)
and x2 + 2x + 1 = (x + 1) (x + 1) = (x + 1)2
Product of highest powers = (x – 1)(x + 1)2
Hence, option (C) is correct.
RBSE Class 10 Maths Chapter 3 Miscellaneous Question 11.
L.C.M. of (RBSESolutions.com) expressions 6x2y4 and 10xy2, then 30x2y4 H.C.F. will be:
(A) 6x2y2
(B) 2xy2
(C) 10x2y2
(D) 60x3y6
Solution :
RBSE Solutions For Class 10 Maths Chapter 3 Miscellaneous Question 12.
To find roots of quadratic (RBSESolutions.com) equation ax2 + bx + c = 0 write Shridhar Acharya formula.
Solution
RBSE Class 10 Maths Chapter 3 Miscellaneous Exercise Question 13.
Write nature of roots by general form of discriminant of equation ax2 + bx + c = 0.
Solution
Nature of roots of quadratic equation ax2 + bx + c = 0, a 0 depends on its discriminant value, D = b2 – 4ac
(i) If D = b2 – 4ac > 0, then roots will be real and distinct.
If α and β are roots of the equation then
Miscellaneous Exercise Class 10 Question 14.
Find zeros of quadratic (RBSESolutions.com) equation 2x2 – 8x + 6 and test the relationship between zeros and coefficient.
Solution
Given polynamial
f(x) = 2x2 – 8x + 6
= 2x2 – 6x – 2x + 6
= 2x(x – 3) – 2(x – 3)
= (x – 3) (2x – 2)
To find zeros of polynomial f(x), f(x) = 0.
⇒ f(x) = 0
⇒ (x – 3)(2x – 2) = 0
⇒ x – 3 = 0 or 2x – 2 = 0
⇒ x – 3 or x = 1
Zeros of polynomial 2x2 – 8x + 6 are 1 and 3.
Relation between zeros and coefficient:
Sum of zeros = 1 + 3 = 4
Product of zeros = 1 × 3 = 3
comparing (RBSESolutions.com) polynomial by ax2 + bx + c = 0
a = 2, b = -8, c = 6
Thus there is a relationship between zeros and coefficient.
RBSE Solutions For Class 10 Maths Chapter 3 Miscellaneous Question 15.
If α and β are zeros of quadratic (RBSESolutions.com) equation f(x) = x2 – px + q, then find the values of the following:
(a) α2 + β2
(b) $$\frac { 1 }{ \alpha } +\frac { 1 }{ \beta }$$
Solution
Given α and β are zeros quadratic equation f(x) = x2 – px + q
RBSE Maths Solution Class 10 Chapter 3 Question 16.
If polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided (RBSESolutions.com) by another polynamial x2 – 2x + k and remainder obtained x + a, then find k and a.
Solution
Given x4 – 6x3 + 16x2 – 25x + 10 divided by x2 – 2x + k obtained remainder is x + a
RBSE Solutions For Class 10 Maths Chapter 12 Miscellaneous Question 17.
The area of a rectangular (RBSESolutions.com) plot is 528 m2. Length of a plot (in m) is 1 more than twice of breadth. By forming quadratic equation, find the length and breadth of the plot.
Solution
Let breadth of a plot is x m.
According to question,
Length of plot = (2 × breadth) + l = (2 × x + 1) = (2x + 1) m.
Area of rectangular plot = l × b = (2x + 1) × x = (2x2 + x) sq. m.
Given : Area of plot = 528 sq. m.
2x2 + x = 528
⇒ 2x2 + x – 528 = 0
Required quadratic (RBSESolutions.com) equation is :
2x2 + x – 528 = 0
⇒ 2x2 + 33x – 32x – 528 = 0
⇒ x(2x + 33) – 16(2x + 33) = 0
⇒ (2x + 33) (x – 16) = 0
⇒ x – 16 = 0 or 2x + 33 = 0
⇒ x = 16 or x = $$\frac { –33 }{ 2 }$$ (impossible)
Hence, Length of plot is 2x + 1 = 2 × 16 + 1 = 33 m
Exercise 3.5 Class 10 RBSE Question 18.
Solve quadratic (RBSESolutions.com) equation x2 + 4x – 5 = 0 by completing the square method.
Solution
x2 + 4x – 5 = 0
⇒ x2 + 4x = 5
Adding square of half of coefficient of x on both sides
x2 + 2 × 2x + (2)2 = 5 + (2)2
⇒ (x + 2)2 = 5 + 4
⇒ (x + 2)2 = 9
⇒ x + 2 = ±√9
⇒ x + 2 = ± 3
Taking + ve sign x = 3 – 2 = 1
Taking – ve sign x = -3 – 2 = – 5
Hence, solution of given quadratic equation are 1 and – 5.
RBSE Solutions For Class 10 Maths Chapter 3 Miscellaneous Question 19.
Solve the following (RBSESolutions.com) equations by factorisation method.
Solution
RBSE Class 10 Maths Chapter 4 Miscellaneous Solutions Question 20.
If -5 is a root of quadratic (RBSESolutions.com) equation 2x2 + px – 15 = 0 and roots of quadratic equation p(x2 + x) + k = 0 are same, then find k.
Solution
Given
One root of quadratic equation 2x2 + px – 15 = 0 is -5.
2(-5)2 + p(-5) – 15 = 0
⇒ 2 × 25 – 5p – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 5p = 35
⇒ p = 7
Putting this value in (RBSESolutions.com) quadratic equation p(x2 + x) + k = 0
7 (x2 + x) + k = 0
⇒ 7x2 + 7x + k = 0
Comparing by ax2 + bx + c = 0
a = 7, b = 7 c = k
root of equation are equal
Discriminant (D) = 0
⇒ b2 – 4ac = (7)2 – 4 × 7 × k
⇒ 0 = 49 – 28k
⇒ 28k = 49
⇒ k = $$\frac { 49 }{ 28 }$$
Thus, k = $$\frac { 7 }{ 4 }$$
Chapter 3 Maths Class 10 Question 21.
Using, Shridhar Acharya (RBSESolutions.com) formula, solve the following quadratic equations : p2x2 + (p2 – q2) x – q2 = 0
Solution
Given equation p2x2 + (p2 – q2) x – q2 = 0
Comparing by ax2 + bx + c = 0, we get
a = p2, b = (p2 – q2) and c = -q2
By Shridhar Acharya formula
Class 10 Maths Chapter 3 Question 22.
L.C.M. and H.C.F. of two quadratic (RBSESolutions.com) expression are respectively x3 – 7x + 6 and (x – 1). Find the expression.
Solution
Least common multipile (L.C.M.) = x3 – 7x + 6
Highest common factor (H.C.F.) = (x – 1)
Factorising expression x3 – 7x + 6
putting x = 1 in this expression, we get = (1)3 – 7(1) + 6 = 1 – 7 + 6 = 0
At x = 0, expression = 0
(x – 1) is a factor
(x3 – 7x + 6)
= (x – 1) (x2 + x – 6)
= (x – 1) [x2 + 3x – 2x – 6]
= (x – 1)[x(x + 3) – 2(x + 3)]
= (x – 1)(x + 3)(x – 2)
Now L.C.M. = (x – 1) (x – 2) (x + 3)
and H.C.F. = (x – 1)
factror (x – 1) will be (RBSESolutions.com) common in two expressions
first expression = (x – 1) (x – 2) = x2 – 3x + 2
and second expression = (x – 1)(x + 3) = x2 + 2x – 3
Miscellaneous Exercise 3 Class 10 Question 23.
L.C.M. of two polynomials is x3 – 6x2 + 3x + 10 and H.C.F. is (x + 1) if one polynomial is x2 – 4x – 5, then find the other.
Solution
L.C.M. = x3 – 6x2 + 3x + 10
H.C.F. = (x + 1)
Putting x = 1 in x3 – 6x2 + 3x + 10
x = 1 putting
= (1)3 – 6(1)2 + 3(1) + 10
= 1 – 6 + 3 + 10 ≠ 0
putting x = -1
= (-1)3 – 6(-1)2 + 3(-1) + 10
= -1 – 6 – 3 + 10 = 0
at x = – 1, expression = 0
(x + 1) is a factor of expression
We hope the given RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise, drop a comment below and we will get back to you at the earliest.
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Home » College Entrance ExamLETPMA Entrance ExamReviewersUPCAT » Sequence and Series
# Sequence and Series
Patterns and rules surround us.
For instance, the dry and wet seasons are alternating throughout the year; the stripes in the petals of a flower are arranged systematically; funds in a regular savings account earn interest monthly, and even your daily morning routine follows a well-defined pattern.
Quantities and numbers also follow the rules and patterns. For instance, counting by 2s is one of the ways to define a pattern for numbers.
This review will focus on sequence – a set of numbers following a certain pattern. Once we’ve unraveled its essential concepts, we will deal with series or the sum of these patterned numbers.
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## What is a Sequence?
Look at this given set of numbers:
5, 10, 15, 20, 25, 30, ____.
What number must be placed in the blank?
Yes, it should be 35.
It’s pretty easy to guess the missing number. By observing the numbers above, you can deduce that the next number can be obtained by adding 5 to 30. I’m sure you can do this because you know that the numbers follow a particular pattern (i.e., adding 5 to the previous number to get the next one).
The numbers above (5, 10, 15, 20, 25, 30, 35) exemplify a sequence.
A sequence of real numbers is an ordered list of real numbers. We will be most interested in sequences of numbers with a specific pattern. As we have mentioned earlier, the pattern of the sequence above is that it adds 5 to the previous number to get the succeeding one (5 + 5 = 10; 10 + 5 = 15; 15 + 5 = 20 …)
Sample Problem: What is the missing number in 72, 63, 54, 45, ______?
Solution: The sequence above follows a specific pattern. It subtracts 9 from the previous number to obtain the succeeding one (72 – 9 = 63; 63 – 9 = 54 …). Therefore, to get the missing number: 45 – 9 = 36.
### Terms of a Sequence
Each element in a sequence is called a term. Thus, in the sequence 5, 10, 15, 20, 25, 30, 35, each number is a sequence term. Moreover, 5 is the first term, 10 is the second term, 15 is the third term, and so on.
### Notations in Sequence
Generally, the first term of a sequence can be expressed as a1. The subscript “1” indicates that it is the first term of a particular sequence.
For example, in the sequence 5, 10, 15, 20, 25, 30, 35, the first term is 5. Therefore, a1 = 5.
Now, if the first term of a sequence is expressed as a1, the second term must be a2, the third term must be a3, the fourth term must be a4, and so on. Thus, the nth term of a sequence is written as an
Hence, in general, a sequence can be written as:
a1, a2, a3, a4 …, an
Sample Problem: Determine a2 and a3 of 2, 5, 8, 11, 14, 17.
Solution: a2 and a3 refer to the sequence’s second and third terms, respectively. Since the second term of the sequence is 5 while the third term is 8, then a2 = 5 and a3 = 8.
### Rule of a Sequence
We have mentioned earlier that the pattern or rule of this sequence is that it adds 5 to a previous term to obtain the next one:
Using a formula or an equation, we can provide a more precise description of the sequence rule.
For instance, the rule of the sequence 5, 10, 15, 20, 25, 30, 35 can be expressed as an = 5n where an is the nth term of the sequence. Note that the nth term means any term of the given sequence.
If we let n = 1, then by substitution: a(1) = 5(1) = 5. This means that the first term of the sequence defined by the rule an = 5n is 5.
If n = 2, then a(2) = 5(2) = 10. This means that the second term of the sequence defined by the rule an = 5n is 10.
Similarly, if n = 5, then a(5) = 5(5) = 25. This means that the fifth term of the sequence defined by the rule an = 5n is 25.
Sample Problem 1: The sequence 5, 10, 15, 20, 25, 30, 35 is defined by the equation an = 5n. What is the 100th term of the sequence?
Solution: To find the 100th term of the sequence, which is defined by the equation an = 5n, we let n = 100:
a100 = 5(100) = 500
Therefore, the 100th term of the sequence is 500.
Sample Problem 2: A sequence is defined by the rule an = 7n – 1. What is the first term of this sequence? How about its 20th term?
Solution: To find the first term of the sequence defined by the rule an = 7n – 1, we let n = 1:
a1 = 7(1) – 1 = 6
Therefore, the first term of the sequence is 6.
On the other hand, to find the 20th term, we let n = 20:
a20 = 7(20) – 1 = 140 – 1 = 139
Thus, the 20th term of the sequence is 139.
Sample Problem 3: Given the sequence with a rule defined by the equation an = 5 – 2n, what is the 15th term?
Solution: To find the 15th term, we let n = 15:
a15 = 5 – 2(15) = 5 – 30 = -25
Thus, the 15th term of the sequence is -25.
### Types of Sequence
#### 1. Arithmetic Sequence
An arithmetic sequence is a sequence wherein every succeeding term is obtained by adding a fixed number to the previous term.
Our previous example (i.e., 5, 10, 15, 20, 25, 30, 35) is an example of an arithmetic sequence because its succeeding term is derived by adding a fixed number (5) to the previous term.
The common difference is the fixed number added to every term to get the succeeding one.
For instance, the sequence 5, 10, 15, 20, 25, 30, and 35 has a common difference of 5.
Sample Problem 1: Which of the following is/are arithmetic sequence(s)?
a. 8, 10, 12, 14, 16, 18,…
b. 15, 12, 9, 6, 3,…
c. 2, 4, 8, 16, 32, …
Solution:
a. Every succeeding sequence term is obtained by adding 2 to the previous term.
Therefore, this sequence is an arithmetic sequence with a common difference of 2.
b. This sequence works this way: subtracting 3 from the previous term to obtain the next one. Subtracting 3 from a number also means adding -3 to a number.
Since we are still adding a fixed number (-3) to every term in the sequence, this is an arithmetic sequence. Thus, this is an example of an arithmetic sequence with a common difference of -3.
c. This is not an arithmetic sequence because the number added to each term to obtain the next one is not fixed or constant.
Therefore, only sequences in a and b are arithmetic.
Sample Problem 2: Determine the common difference of the following arithmetic sequence.
a. 11, 22, 33, 44, 55,…
b. -8, -5, -2, 1, 4,…
c. ⅔, 1, 4/3, 5/3,…
Solution: To find the common difference (d) of an arithmetic sequence, we pick any term in the given sequence and subtract from it the term that precedes it. Mathematically,
d = an – an – 1
Where d is a common difference, an is any term of the sequence (nth term), and an – 1 is the term that precedes an
a. To identify the common difference in this arithmetic sequence, we can subtract from any of its terms the term that precedes it
Let’s try 33; the term preceding it is 22. Hence, the common difference is:
d = 33 – 22 = 11
We can still get a common difference if we try another term. Let’s try 44; the term preceding it is 33:
d = 44 – 33 = 11
Hence, the common difference is 11.
b. Using the same method we used in the previous item, we can pick any term and subtract from it the term preceding it:
d = -2 – (-5) = -2 + 5 = 3
Hence, the common difference is 3.
c. To find the common difference, we pick any term and subtract from it the term that precedes it:
d = 5/3 – 4/3 = ⅓
Therefore, the common difference is ⅓.
##### Finding the nth Term of an Arithmetic Sequence
Suppose we have this arithmetic sequence: 6, 9, 12, 15, 18,… and want to find its 25th term. What will you do to find this term?
If you’re planning to add 3 to each previous term until you reach the 25th term, then you’ll waste a lot of your precious time. Perhaps you’re good at arithmetic and can reach the 25th term in a few seconds. However, what if you’re looking for the 100th, 250th, or 500th term? Enumerating the terms one by one becomes inconvenient and time-consuming in this case.
Luckily, mathematicians have done the heavy work and given us a general formula that we can use to find any term of an arithmetic sequence without listing each term manually. This formula is presented in the image below:
Here is the meaning of the variables in the formula above
• an is any term of the sequence; this is the term you’re looking for
• a1 is the first term of the given sequence
• n is the number of the term you’re looking for (e.g., if you’re looking for the 25th term, then n = 25). If the last term of an arithmetic sequence is specified, n also represents the total number of terms in the sequence.
• d is the common difference of the arithmetic sequence
We will not discuss how this formula is derived because it’s quite long and complex.
For now, let’s answer some examples to understand better how to use this formula.
Sample Problem 1: Determine the 25th term of the sequence 6, 9, 12, 15, 18,…
Solution: The formula to find any term of an arithmetic sequence is defined as:
an = a1 + (n – 1)d
• The first term of the sequence is 6, so we have a1 = 6
• Since we are looking for the 25th term, then we have n = 25
• The common difference (d) of the given sequence is 3 (e.g., 9 – 6 = 3)
Substituting the values we have derived:
an = a1 + (n – 1)d
a25 = 6 + (25 – 1)3
a25 = 6 + (24)3
a25 = 6 + 72
a25 = 78
Thus, the 25th term of the sequence is 78.
Let us verify if the formula works by listing each term of the given sequence manually until the 25th term:
6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78
The list above shows that 78 is the 25th term of the given sequence.
Sample Problem 2: Identify the 54th term of the sequence 17, 28, 39, 40, 51, 62,…
Solution: The formula to find any term of an arithmetic sequence is defined as:
an = a1 + (n – 1)d
• The first term of the sequence is 17, so we have a1 = 17
• Since we are looking for the 54th term, then we have n = 54
• The common difference (d) of the given sequence is 11 (28 – 17 = 11)
Substituting the values we have derived:
a54 = a1 + (n – 1)d
a54 = 17 + (54 – 1)11
a54 = 17 + (53)11
a54 = 17 + 583
a54 = 600
Thus, the 54th term of the sequence is 600.
Sample Problem 3: Selena plans to save money for 30 days. On the first day, she saved ₱20. On the second day, she saved ₱22. On the third day, she saved ₱24. On the fourth day, she saved ₱26. How much will Selena save on the 30th day?
Solution: How Selena saves money follows an arithmetic sequence of
20, 22, 24, 26,… This sequence has a common difference of 2 (d = 22 – 20 = 2).
To find how much Selena will save on the 30th day, we need to find the 30th term of the given sequence. The formula to find any term of an arithmetic sequence is defined as:
an = a1 + (n – 1)d
• The first term of the sequence is 20, so we have a1 = 20
• Since we are looking for the 30th term, we have n = 30
• The common difference (d) of the given sequence is 2
Substituting the values we have derived:
a30 = a1 + (n – 1)d
a30 = 20 + (30 – 1)2
a30 = 20 + (29)2
a30 = 20 + 58
a30 = 78
Thus, the 30th term of the sequence is 78; Selena will save ₱78 on the 30th day.
Sample Problem 4: How many numbers from 1 to 100 are divisible by 4?
Solution: At first glance, the given problem seems unrelated to the concept of the arithmetic sequence. However, this problem is solvable by applying what we have learned about this type of sequence.
If a number is divisible by 4, it is also a multiple of 4. For instance, 16 is a multiple of 4, so it also means that it is also divisible by 4.
To identify the numbers that are multiple (and divisible) by four between 1 to 100, we count by 4s:
4, 8, 12, 16, 20, 24, …
Note that the list of numbers above is an arithmetic sequence with a common difference of 4.
Thus, to find how many numbers from 1 – 100 are divisible by 4, we need to identify how many terms there are in the sequence.
However, we cannot identify the number of terms in the sequence without figuring out its last term. Note that the last number between 1 to 100 that is divisible by 4 is 96. This also means that the last term of our sequence should be 96.
Hence, the arithmetic sequence corresponding to all numbers between 1 to 100 divisible by 4 is 4, 8, 12, 16, 20, 24, …, 96. We can now identify how many terms there are in this sequence:
The formula for identifying any term of a given arithmetic sequence is
an = a1 + (n – 1)d
However, this time we are not looking for an but n. Note that if the last term of a sequence is specified, then n may represent the total number of terms in a sequence. Since the sequence 4, 8, 12, 16, 20, 24, …, 96 ends with 96, we have a sequence with a specified last term. So, n in this problem serves as the total number of terms in the sequence (the total number of numbers from 1 – 100 that are divisible by 4).
• Since 4 is the first number in the sequence, then we have a1 = 4
• Since we are interested in finding the number of terms in 4, 8, 12, 16, 20, …, 96 or the value of “n,” the value of an should be the last term which is 96.
• The common difference (d) is 4
Substituting these values in the formula, we can now solve for n:
an = a1 + (n – 1)d
96 = 4 + (n – 1)4
Let us now do some algebra to solve the value of n:
Since n represents the number of terms in the sequence 4, 8, 12, 16, 20, 24, … 96, there are 24 terms in this sequence. This also implies that 24 numbers between 1 to 100 are divisible by 4.
Sample Problem 5: The first term of an arithmetic sequence is -9, while its 12th term is 1. What is the common difference between this arithmetic sequence?
Solution: We have a1 = -9 since -9 is the first term of the sequence. On the other hand, a12 (the an in this case) is 1. Furthermore, we have n = 12. Our goal is to solve for the common difference or d.
Thus, the common difference in the sequence in this problem is 10/11.
#### 2. Geometric Sequence
In an arithmetic sequence, there’s a fixed number added to each term to obtain the succeeding term (called the common difference). Meanwhile, a geometric sequence has each term multiplied by a fixed number to obtain the next one.
Look at this sequence: 3, 6, 12, 24, 48, …
Unlike in an arithmetic sequence, there is no common difference. Instead, each term is multiplied by 2 to obtain the next one.
Since this sequence is characterized by multiplying a term by 2 to get the next one, this is an example of a geometric sequence.
In a geometric sequence, each term is multiplied by a fixed number to get the next term; the fixed number is called the common ratio (denoted by the letter r).
This means that the common ratio of the sequence 3, 6, 12, 24, 48, … is 2.
Sample Problem 1: Which of the following is/are geometric sequence(s)?
a. 1, 4, 7, 10, 13, …
b. 2, 6, 18, 54, …
c. 9, 11, 14, 18, …
Solution:
a. This sequence has a common difference of 3. Thus, it is not a geometric sequence.
b. This sequence has a common ratio of 3. This means we multiply the previous term by 3 (e.g., 2 x 3 = 6; 6 x 3 = 18) to obtain the next term. Hence, this is a geometric sequence.
c. This sequence is neither arithmetic nor geometric sequence. It has no common difference or common ratio.
Sample Problem 2: Identify the common ratio of the following sequence.
a. 1, 5, 25, 125, 625, …
b. 3, -6, 12, -24, 48, …
Solution:
Here’s a simple trick to identify the common ratio of a geometric sequence. Pick any term from the sequence, then divide it by the preceding term. Let us apply this trick to find the common ratio of the sequence:
a. Let us pick 25. The term that precedes it is 5. Hence, the common ratio is:
r = 25/5 = 5
b. Let us pick -24. The term that precedes it is 12. Thus, the common ratio is:
r = -24/12 = -2
Again, we can find the common ratio of a geometric sequence by picking any term from the sequence (except the first term), then dividing it by the term that precedes it. Mathematically,
Where r is the common ratio, an is the nth term (any sequence term), and an-1 is the term that precedes an
##### Finding the nth Term of a Geometric Sequence
Like in an arithmetic sequence, we can also determine any term of a geometric sequence using a formula. Suppose we want to identify the 10th term of 3, 6, 12, 24, … we can use the formula below to find it without listing the terms.
Where:
• an is any term of the sequence; this is the term you’re looking for
• a1 is the first term of the given sequence
• n is the number of the term you’re looking for (For example, if you’re looking for the 25th term, n = 25)
• r is the common ratio of the geometric sequence
Sample Problem 1: Find the 10th term of the geometric sequence 3, 6, 12, 24, …
Solution: The first term of the sequence is 3. Thus, we have a1 = 3. Since we are looking for the 10th term or a10, we have n = 10.
The common ratio of the sequence is r = 6/3 = 2.
Using the formula for the nth term of a geometric sequence: an = a1rn – 1
Therefore, the 10th term of the geometric sequence is 1,536.
Sample Problem 2: What is the 8th term of the geometric sequence 1/16, ⅛, ¼, ½, …
Solution: The first term of the sequence is 1/16. Thus, we have a1 = 1/16. Since we are looking for the 8th term or a8, we have n = 8.
The common ratio of the sequence is
You can review our module about how to divide fractions to learn how we can obtain the answer above.
Using the formula for the nth term of a geometric sequence: an = a1rn – 1
Therefore, the 8th term of the geometric sequence is 8.
#### 3. Harmonic Sequence
Before discussing a harmonic sequence, look at this arithmetic sequence: 4, 7, 10, 13, 16, 19, …
Let us take the reciprocal of each term of the arithmetic sequence above:
The set of fractions above is an example of a harmonic sequence.
A harmonic sequence is derived from the reciprocals of the terms of an arithmetic sequence. This also means that the reciprocal of the terms of a harmonic sequence forms an arithmetic sequence.
Going back to our example of a harmonic sequence above, taking the reciprocal of each term forms an arithmetic sequence with a common difference of 3:
This type of sequence is called “harmonic” because it is said that ancient Greek mathematicians like Pythagoras used this particular type of numerical pattern to devise a “mathematical” theory that could explain musical harmony.
Sample Problem: Is 1/9, 1/7, ⅕, ⅓, … a harmonic sequence?
Solution: If we take the reciprocal of each term of the sequence above, then:
9, 7, 5, 3, …
The sequence above is an arithmetic sequence because it has a common difference of -2. Therefore, the original sequence is a harmonic sequence.
#### 4. Fibonacci Sequence
A Fibonacci sequence is a sequence wherein the succeeding term is obtained by adding two previous terms. It’s named after Leonardo “Fibonacci” of Pisa, an Italian mathematician who first discussed it during the early 1200s.
Here’s an example of a Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, …
In the sequence above, any term (except for the first and second terms) is obtained by adding two previous terms. For instance, to obtain 2, we add two terms preceding it which are 1 and 1 (i.e., 1 + 1 = 2). Similarly, to get 8, we add the two terms before it (i.e., 3 + 5 = 8).
Therefore, the nth term of a Fibonacci sequence can be expressed mathematically as follows:
an = an-1 + an – 2
where an is any sequence term, with an – 1 and an-2 corresponding to the two terms that precede an
Sample Problem: What is the missing number in this sequence: 3, 3, 6, 9, 15, 24, ____?
Solution: The sequence above is a Fibonacci sequence because each succeeding term is obtained by adding two previous terms.
Hence, we add 15 and 24: 15 + 24 = 39 to find the missing number.
Thus, the missing number is 39.
## What is a Series?
In mathematics, a series is a sum of all terms of a particular sequence.
For instance, consider this sequence: 3, 5, 7, 9, 11, …
If we add all of the terms of the sequence above, then it becomes a series: 3 + 5 + 7 + 9 + 11 + …
Hence, if a1, a2, a3, …, an represents the terms of a sequence, then a series S is defined as:
S = a1 + a2 + a3 + … + an
Sample Problem: Transform the sequence -4, -2, 0, 2, 4, 6, … into a series.
Solution: To turn the sequence above into a series, we express it as a sum of its terms:
(-4) + (-2) + 0 + 2 + 4 + 6 + …
### Finite Sum and Infinite Series
A mathematical series can be a finite or an infinite sum (series).
For example, if we are just interested in adding the first ten terms of the sequence 3, 5, 7, 9, …., the series 3 + 5 + 7 + 9 … + 21 is a partial sum of the sequence (21 is the 10th term of the given sequence).
Thus, a finite series refers to the sum of the first n terms of an arithmetic or geometric sequence.
On the other hand, if we are adding a limitless number of terms in a sequence, then we’re now dealing with an infinite series. For instance, adding all terms of the sequence 3, 5, 7, 9, … up to infinity is an example of an infinite series.
An infinite series is a series where an infinite number of terms of a sequence are added.
You might find it ridiculous why someone would add the terms of a sequence like 3, 5, 7, 9, … up to infinity. We cannot estimate an exact sum because we are dealing with infinite addends. The sum will get larger and larger without reaching a definite end.
However, some infinite series “converge” to an exact result whether you believe it or not. That is, it is possible to obtain an exact sum by adding an endless number of terms for some sequences. For instance, if you add all terms of the sequence ½, ¼, ⅛, 1/16, .. up to infinity, the sum will be 1.
I know this thing seems counterintuitive and mind-boggling. Don’t worry; we will shed light on this confusing thing later as we explain the type of infinite series with an “exact” sum.
### Types of Series
#### 1. Arithmetic Series
An arithmetic series is the sum of the terms in an arithmetic sequence.
For instance, consider the arithmetic sequence: 4, 5, 7, 9, … 21.
Let S be the sum of the terms of this arithmetic sequence. Therefore, we have
S = 4 + 5 + 7 + 9 + … + 21
From here, we can see that S is an example of an arithmetic series.
Note that an arithmetic series must have a definite number of terms. In other words, the addends must be finite. In our previous example, S = 4 + 5 + 7 + 9 + … + 21, we add the terms from 4 to 21. This allows us to have an “exact” value of S.
If we don’t specify a definite number of terms, we add an infinite number. Suppose that instead of S = 4 + 5 + 7 + 9 + … + 21, we just write it as S2 = 4 + 5 + 7 + 9 + … .
Since S2 has no definite end, we’re adding an infinite number of terms, and the sum will be larger and larger without bounds.
In short, an arithmetic series must have a definite number of addends.
So, to formally define an arithmetic series, we express it mathematically as follows:
Sn = a1 + a2 + a3 + … + an
Such that a1, a2, a3, …, an are terms of an arithmetic sequence and an represents the end of the sequence.
Sample Problem: Which of the following is/are arithmetic serie(s)?
1. 7 + 11 + 16 + 22 + … + 320
2. 6 + 12 + 18 + 24 + … + 120
3. 9 + 8 + 7 + 6 + … + (-1)
Solution:
1. This is not an arithmetic series. Note that the addends do not satisfy the condition to become an arithmetic sequence because they have no common difference.
• 11 – 7 = 4
• 16 – 11 = 5
• 22 – 16 = 6
1. This is an arithmetic series since the addends are terms of an arithmetic sequence with a common difference of 6.
1. This is an arithmetic series since the addends are terms of an arithmetic sequence with a common difference of -1.
##### Formula for the Sum of an Arithmetic Series
We can use two formulas depending on the given situation to find the sum of an arithmetic series.
Such that:
• a1 is the first term
• n is the number of terms being added to the series
• an is the nth term of the sequence. If the sequence has a specified last term, then an is also the last term.
• d is the common difference between the terms or addends in the series
We use the first formula when the sequence has no specified last term. For example, if we want to find the sum of the first 50 terms of 2 + 4 + 6 + 8 + …, then we use the first formula. On the other hand, the second formula is used when the last term of the sequence is specified. For example, we can use the second formula to compute the sum of 1 + 2 + 3 + … + 50.
It’s easy to derive the formulas above. However, we will not discuss this anymore because it is beyond the scope of our reviewer.
Let us solve some sample questions to understand further how to use these formulas.
Sample Problem 1: What is the sum of all whole numbers from 1 to 100?
Solution: By writing the problem as a mathematical sentence, we’ll obtain the following:
1 + 2 + 3 + 4 + … + 100
It’s clear here that we are dealing with an arithmetic series since the terms have a common difference of 1. Since the sequence has a specified last term (which is 100), then it’s advisable to use the second formula:
• Since the first term of the series is 1, we have a1 = 1
• The last term of the series is 100, so an = 100
• There are 100 whole numbers from 1 to 100, so we have n = 100
Let us insert these values into the formula for the sum of an arithmetic series.
Therefore, the sum of numbers from 1 to 100 is 5050. If you have a lot of spare time, you can verify this manually by doing the addition process.
Sample Problem 2: What is the sum of 4 + 6 + 8 + 10 + … + 32?
Solution: Since the given arithmetic series above has a definite last term (which is 32), then we will use formula #2.
The first term of the given series is 4. So we have a1 = 4. Meanwhile, the last term of the sequence is 32, so we have an = 32.
For the value of n, we need to determine the number of terms in the series 4 + 6 + 8 + 10 + … + 32. We can do this by returning to our formula for the nth term of an arithmetic sequence. You may manually count the number of terms if you don’t want to perform some algebra.
Let us do the algebra stuff by using the formula:
an = a1 + (n – 1)d
We have a1 = 4, an = 32, and d = 2. Let us insert these values into the formula above and then solve for the value of n:
Since we have computed that n = 15, there are 15 terms from 4 to 32 in the series 4 + 6 + 8 + 10 + … + 32.
Our variables for the formula to identify the sum of the arithmetic series are complete.
We have a1 = 4, an = 32, n = 15, and d = 2. Let us substitute these values in the formula:
Hence, the sum of the arithmetic series 4 + 6 + 8 + 10 + … + 32 is 270.
Sample Problem 3: Bea saved ₱100 on February 1 in her savings account. On February 2, she saved ₱105. On February 3, she saved ₱110, and so on. How much money did Bea save at the end of February (assuming it is not a leap year)?
Solution: How Bea saved money during February follows an arithmetic sequence with a common difference of 5.
• February 1 = 100
• February 2 = 105
• February 3 = 110
To determine the total amount of money that Bea saved for the entire month of February, add the daily amount of money saved:
Total amount that Bea saved in February = 100 + 105 + 110 + …
The addition sentence above is an arithmetic series. We shall use the first formula since the last term is not specified.
Since 100 is the first term of the series, we have a1 = 100
Note that we are looking for the total amount of money Bea saved in February. Since it is not a leap year, Bea saved money for 28 days. Therefore, we have n = 28.
Lastly, the common difference is 5. So we have d = 5.
Let us input these values into the formula for the sum of an arithmetic series.
Therefore, Bea saved a total amount of ₱4,690 for the entire month of February.
#### 2. Geometric Series
A geometric series is the sum of the terms of a geometric sequence.
Consider this geometric sequence: 1, 3, 9, 27, … Suppose that S represents the sum of the terms of this sequence such that S = 1 + 3 + 9 + 27 + … Since S is the sum of the terms of a geometric sequence, then S is a geometric series.
Unlike with arithmetic series, the terms in a geometric series can be finite or infinite. A finite geometric series has a definite number of terms. This means that a finite geometric series has a “last” term. It is also the sum of the first n terms of a geometric sequence.
For instance, the geometric series 2 + 4 + 8 + … + 256 is a finite geometric series because it has a definite number of terms. Note that there is the last term in this series which is 256.
On the other hand, an infinite geometric series has an infinite number of terms being added. This means it has no “last” term because it has endless addends.
However, not every infinite geometric series has an “exact” sum. For instance, the infinite geometric series 2 + 4 + 8 + 16 + … doesn’t have an exact sum. The sum that we will obtain will get larger and larger without bounds.
An infinite geometric series will have an “exact” sum if it satisfies a special condition. That is, the absolute value of the common ratio (r) must be between 0 to 1:
0 < | r | < 1
For instance, the geometric series ½ + ¼ + ⅛ + … will have an “exact” sum since its common ratio is ½ (¼ ÷ ½ = ½). Note that the absolute value of ½ is between 0 to 1.
Thus, in general, a finite geometric series can be expressed mathematically as:
Sn = a1 + a2 + a3 + a4 + … + an
where a1, a2, a3, …, an are terms of a geometric sequence and an is the “last” addend of the series.
If the absolute value of the common ratio of the terms is between 0 and 1 (0 < |r| < 1), then the infinite geometric series can be expressed as:
S = a1 + a2 + a3 + a4 + …
where a1, a2, a3, a4, … are terms of a geometric sequence with no definite “last” term.
Sample Problem 1: Which of the following is a geometric series?
a. 3 + (-6) + 12 + … + (-96)
b. 7 + 9 + 11 + … 25
c. 2 + 1 + 1/2 + …
Solution: The series in A and C are geometric series with a common ratio of -2 and ½, respectively. Meanwhile, the series in B is an arithmetic series with a common difference of 2.
Sample Problem 2: Which of the following infinite geometric series satisfies the condition to have an exact sum?
a. 9 + 27 + 81 + …
b. -1 + 5 + (-25) + 125 + …
c. 16 + 4 + 1 + …
Solution: A and B have a common ratio of 3 and -5, respectively. Note that the respective absolute values of these numbers don’t fall within the 0-1 range. Hence, they do not satisfy the condition of having an exact sum. On the other hand, the common ratio in C is ¼, which falls within the 0 -1 range. Therefore, this series will have an exact sum.
##### Formula To Find the Sum of a Finite Geometric Series
Shown below is the formula for the sum of a finite geometric series:
Such that:
• a1 is the first term (or addend) of the geometric series
• r is the common ratio
• n is the number of terms being added
Sample Problem 1: Compute the sum of the first 15 terms of the geometric series 1 + 2 + 4 + 8 + …
Solution:
• The first term of the sequence is 1. So we have a1 = 1
• The common ratio is 2 (2 ÷ 1 = 2). So we have r = 2
• We are looking for the sum of the first 15 terms. Thus, we have n = 15.
Let us substitute these values for the formula for the sum of a finite geometric series.
Based on our calculation above, the sum is 32,767.
Sample Problem 2: Compute the sum of the first six terms of the sequence -3 + 12 + (-48) + …
Solution:
• The first term of the sequence is -3. So we have a1 = -3.
• The common ratio is -4 (12 (-3) = -4). So we have r = -4
• We are looking for the sum of the first 6 terms. Thus, we have n = 6.
Let us insert these values into the formula for a finite geometric series sum.
Based on our calculation above, the sum is 2,457.
##### Formula To Find the Sum of an Infinite Geometric Series
Shown below is the formula for the sum of an infinite geometric series.
Such that:
• a1 is the first term (or addend) of the geometric series
• r is the common ratio such that 0 < | r | < 1
Sample Problem: What is the value of ½ + ¼ + ⅛ + …?
Solution: The first term of the sequence is ½. So we have a1 = ½.
Meanwhile, the common ratio is ½ (¼ ÷ ½ = ½)
By substitution:
Thus, ½ + ¼ + ⅛ + … = 1
##### BONUS: A Geometric Interpretation on Infinite Geometric Series
We have learned earlier that using the formula for infinite geometric series, the sum of ½ + ¼ + ⅛ + 1/16 + .. is equal to 1. Although this formula is mathematically sound and valid, I know that this is still too abstract and difficult to grasp.
Using geometry, we can provide a more intuitive explanation of why ½ + ¼ + ⅛ + 1/16 + .. = 1.
Imagine we have a ½ sheet of colored paper placed on a table.
Now, let us place a ¼ sheet of colored paper near the ½ sheet of colored paper.
Add a ⅛ sheet of colored paper.
Add a 1/16 sheet of colored paper. We also perform this step for 1/32, 1/64, 1/128, etc.
Doing this method multiple times allows us to obtain a total amount of paper equivalent to 1 whole sheet of colored paper.
This shows that ½ + ¼ + ⅛ + 1/16 + .. = 1.
Next topic: Using Algebra to Solve Word Problems
Previous topic: Cartesian Coordinate System
## Test Yourself!
### 3. Math Mock Exam + Answer Key
Written by Jewel Kyle Fabula
Last Updated
### Jewel Kyle Fabula
Jewel Kyle Fabula is a Bachelor of Science in Economics student at the University of the Philippines Diliman. His passion for learning mathematics developed as he competed in some mathematics competitions during his Junior High School years. He loves cats, playing video games, and listening to music.
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The perimeter of a triangle right angled at C is 70, and the in-radius is 6. Then |a - b| equals:
This question was previously asked in
DSSSB TGT Maths Female Subject Concerned - 18 Nov 2018 Shift 3
View all DSSSB TGT Papers >
1. 2
2. 9
3. 8
4. 1
Option 4 : 1
Detailed Solution
Given:
Perimeter = 70
Right angled at C
Concept:
Semiperimeter (s) = $$\dfrac{1}{2}$$(a + b + c)
Inradius (r) = Area of triangle / Semiperimeter (s)
Area of triangle = $$\dfrac{1}{2}$$ × base × height = $$\dfrac{1}{2}$$ab
Pythagoras Theorem:
Hypotenuse2 = Height2 + Base2
c2 = a2 + b2
Solution:
Semiperimeter (s) = $$\dfrac{1}{2}$$(a + b + c)
As, a + b + c = 70
Semiperimeter (s) = 70 / 2 = 35
Area = $$\dfrac{1}{2}$$ab
Inradius (r) = Area / semiperimeter
So,
6 = ($$\dfrac{1}{2}$$ab) / 35
ab = 420
From Pythagorus Theorem, c2 = a2 + b2
⇒ a2 + b2 = c2
⇒ a2 + b2 + 2ab - 2ab = c2
Using identity: (a + b)2 = a2 + b2 + 2ab
= (a + b)2 - 2ab = c2
As, a + b + c = 70 ⇒ a + b = 70 - c
Substituting, ab = 420
=
=
Substituting,
= $$\sqrt{(70-c)^2 - 4\times420}$$
= $$\sqrt{(70-29)^2 - 1680}$$
= $$\sqrt{(41)^2 - 1680}$$
= $$\sqrt{1681 - 1680}$$
= $$\sqrt{1}$$ = 1 unit
Hence, |a - b| = 1 unit
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# Expansion Method of Multiplication
We will discuss here about the expansion method of multiplication. Now we will learn to multiply a 2-digit number and a 3-digit number by a 1-digit number using the method of expansion.
Solved examples for using the expand method of multiplication:
1. Multiply 82 × 7.
= (80 + 2) × 7, (breaking 82 as 8 × 10 + 2 × 1)
= 7 × 80 + 7 × 2
= 560 + 14
= 574
2. Find the value of 432 × 8.
(400 + 30 + 2) × 8, (breaking 432 as 4 × 100 + 3 × 10 + 2 × 1)
= 400 × 8 + 30 × 8 + 2 × 8
= 3200 + 240 + 16
= 3456
Similarly we can multiply a 4-digit number by a 1-digit number.
3. Find the product of 1023 and 3.
1023 × 3
(1000 + 20 + 3) × 3
= 3 × 1000 + 3 × 20 + 3 × 3
= 3000 + 60 + 9
= 3069
4. Find the product of 1378 and 5.
1378 × 5
(1000 + 300 + 70 + 8) × 5
= 5 × 1000 + 5 × 300 + 5 × 70 + 5 × 8
= 5000 + 1500 + 350 + 40
= 6890
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# How do you write y = -1/3x - 9 in standard form?
Mar 1, 2017
$\textcolor{red}{1} x + \textcolor{b l u e}{3} y = \textcolor{g r e e n}{- 27}$
#### Explanation:
The standard form of a linear equation is: $\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$
Where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1
To transform this equation to standard form, first, multiply each side of the equation by $\textcolor{red}{3}$ to eliminate the fractions:
$\textcolor{red}{3} \times y = \textcolor{red}{3} \left(- \frac{1}{3} x - 9\right)$
$3 y = \left(\textcolor{red}{3} \times - \frac{1}{3} x\right) - \left(\textcolor{red}{3} \times 9\right)$
$3 y = \left(\cancel{\textcolor{red}{3}} \times - \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} x\right) - 27$
$3 y = - 1 x - 27$
Next, add $\textcolor{red}{1 x}$ to each side of the equation to place the $x$ and $y$ variables on the left side of the equation as the standard form requires:
$\textcolor{red}{1 x} + 3 y = \textcolor{red}{1 x} - 1 x - 27$
$1 x + 3 y = 0 - 27$
$\textcolor{red}{1} x + \textcolor{b l u e}{3} y = \textcolor{g r e e n}{- 27}$
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# Solving Percent Problems Using Proportions
## Presentation on theme: "Solving Percent Problems Using Proportions"— Presentation transcript:
Solving Percent Problems Using Proportions
COURSE 2 LESSON 6-5 117 is 45% of what number? The model shows the relationship. = 117 n 45 100 Write the proportion. Write the cross products. 45n = 117(100) Divide each side by 45. 45n 45 117(100) = n = 260 Simplify. 117 is 45% of 260. 6-5
Solving Percent Problems Using Proportions
COURSE 2 LESSON 6-5 Use a proportion to solve: 96 is 15% of what number? Model the relationship. The model shows the relationship. = 96 n 15 100 Write the proportion. Write the cross products. 15n = 96(100) Divide each side by 15. 15n 15 96(100) = n = 640 Simplify. Check 15% of 640 = 0.15 • 640 = 96 6-5
Solving Percent Problems Using Proportions
COURSE 2 LESSON 6-5 Use a proportion to solve. 1. What percent of 240 is 60? is what percent of 120? 3. 75% of what number is 66? is 96% of what number? 25% 75% 88 25 6-5
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# NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1
## Chapter 14 Ex.14.1 Question 1
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in $$20$$ houses in a locality. Find the mean number of plants per house.
Number of plants $$0 - 2$$ $$2 - 4$$ $$4 - 6$$ $$6 - 8$$ $$8 - 10$$ $$10 -12$$ $$12 - 14$$ Number of houses $$1$$ $$2$$ $$1$$ $$5$$ $$6$$ $$2$$ $$3$$
Which method did you use for finding the mean, and why?
### Solution
What is known?
The number of plants in $$20$$ houses in a locality.
What is unknown?
The mean number of plants per house and the method used for finding the mean.
Reasoning:
We can solve this question by any method of finding mean but here we will use direct method to solve this question because the data given is small.
The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations.
We know that if $$\ x_1, x_2, \ldots, x_n$$ are observations with respective frequencies $$f_{1}, f_{2}, \ldots, f_{n}$$ then this means observation $$\ x_1 \text { occurs } f_1 \text { times, } x_2 \text { occurs } f_2$$ times, and so on.
$$\ x$$ is the class mark for each interval, you can find the value of $$\ x$$ by using
class mark
\begin{align}\left(x_i\right) =\frac{\text { upper limit + lower limit }}{2}\end{align}
Now, the sum of the values of all the observations $$=\ f_1 \ x_1+\ f_2 x_2+\ldots+\ f_n \ x_n$$ and the number of observations $$=f_1+f_2+\ldots+f_n$$.
So, the mean of the data is given by
\begin{align}\ \overline x\!=\!\frac{f_{1} x_{1}\!+\!f_{2} x_{2}\!+\!\ \ldots \ldots\!+\!f_{n} x_{n}}{f_{1}\!+\!f_{2}\!+\!\ldots \ldots\!+\!f_{n}}\end{align}
\begin{align}\overline x=\frac{\sum f_{i} x_{i}}{\Sigma f_{i}}\end{align} where $$i$$ varies from $$1$$ to $$n$$
Steps:
Number of plants Number of houses$$\ (f_i)$$ $$\ x_i$$ $$\ f_ix_i$$
$$0 - 2$$ $$1$$ $$1$$ $$1$$
$$2 - 4$$ $$2$$ $$3$$ $$6$$
$$4 - 6$$ $$1$$ $$5$$ $$5$$
$$6 - 8$$ $$5$$ $$7$$ $$35$$
$$8 - 10$$ $$6$$ $$9$$ $$54$$
$$10 - 12$$ $$2$$ $$11$$ $$22$$
$$12 - 14$$ $$3$$ $$13$$ $$39$$
$$\Sigma f_i = 20$$ $$\Sigma f_ix_i = 162$$
From the table it can be observed that,
\begin{align}\Sigma f_i=20, \\ \Sigma f_ix_i =162\end{align}
\begin{align} \text { Mean } \overline x &=\frac{\Sigma f_i x_i}{\Sigma f_i} \\ &=\frac{162}{20} \\ &=8.1 \end{align}
Thus, the mean number of plants each house has $$8.1.$$
Here, we have used the direct method because the value of $$x_i$$ and $$f_i$$ are small.
## Chapter 14 Ex.14.1 Question 2
Consider the following distribution of daily wages of $$50$$ workers of a factory.
Daily wages (in Rs) $$500 –520$$ $$520 – 540$$ $$540 –560$$ $$560 – 580$$ $$580 – 600$$ Number of workers $$12$$ $$14$$ $$8$$ $$6$$ $$10$$
Find the mean daily wages of the workers of the factory by using an appropriate method.
### Solution
What is known?
Distribution of daily wages of $$50$$ workers of a factory is given-
What is unknown?
The mean daily wages of the workers of the factory.
Reasoning:
We will use Assumed Mean Method to solve this questionbecause the data given is large.
Sometimes when the numerical values of $$x_i$$ and $$f_i$$ are large, finding the product of $$x_i$$ and $$f_i$$ becomes tedious.We can do nothing with the $$f_i’s$$, but we can change each xi to a smaller number so that our calculations become easy.Now we have to subtract a fixed number from each of these $$x_i’s$$.
The first step is to choose one among the $$x_i’s$$ as the assumed mean, and denote it by $$‘a’.$$ Also, to further reduce our calculation work, we may take ‘a’ to be that $$x_i$$ which lies in the centre of \begin{align}{x} _1, {x} _ 2, \ldots, {x_n}\end{align}. So, we can choose $$a.$$
The next step is to find the difference $$d_i$$ between a and each of the $$x_i$$’s, that is, the deviation of ‘a’ from each of the $$x_i$$'s i.e., $$d_i = x_i – a$$
The third step is to find the product of $$d_i$$ with the corresponding$$f_i$$, and take the sum of all the $${f}_{{i}} {d}_{{i}}^{\prime} {s}$$
Now put the values in the below formula
\begin{align}\operatorname{Mean} \,\,(\overline{{x}})= {a}+\left(\frac{\Sigma {f}_{{i}} {u}_{{i}}}{\Sigma {f}_{{i}}}\right) \end{align}
Steps:
We know that,
class mark ($$x_i$$) \begin{align}={\frac{\text { upper limit+lower limit }}{2}} \end{align}
Taking assumed mean, a = 550
Daily wages (in Rs) No of workers $$(f_i)$$ $$(X_i)$$ $$d_i = x_i -150$$ $$f_iu_i$$ $$500-520$$ $$12$$ $$510$$ $$- 40$$ $$- 480$$ $$520-540$$ $$14$$ $$530$$ $$- 20$$ $$- 280$$ $$540-560$$ $$8$$ $$550 (a)$$ $$0$$ $$0$$ $$560-580$$ $$6$$ $$570$$ $$20$$ $$120$$ $$580-600$$ $$10$$ $$590$$ $$40$$ $$400$$ $$\Sigma f_i=50$$ $$\Sigma f_id_i= -240$$
It can be observed from the table,
\begin{align} \Sigma f_{i} &=50 \\ \Sigma f_{i} u_{i} &=-240 \end{align}
\begin{align}{{{\rm Mean }}\,\,(\overline {{x}} ) } &={{{a}} + \left( {\frac{{{{ }}\Sigma {{f_iu_i}}}}{{\Sigma {{f_i}}}}} \right){{h}}}\\{}&{ = 550 + \left( {\frac{{ - 240}}{{50}}} \right)}\\{}&{ = 550 - \frac{{24}}{5}}\\&=550-4.8\\ {}&{ = 545.2}\end{align}
Thus, the mean daily wages of the workers of the factory is $$\rm Rs.545.20$$
## Chapter 14 Ex.14.1 Question 3
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs $$18$$. Find the missing frequency $$f$$.
Daily pocket allowance (in Rs) $$11 – 13$$ $$13 – 15$$ $$15 – 17$$ $$17 – 19$$ $$19 – 21$$ $$21 – 23$$ $$23 – 25$$ Number of children $$7$$ $$6$$ $$9$$ $$13$$ $$f$$ $$5$$ $$4$$
### Solution
What is known?
The mean pocket allowance is Rs $$18$$.
What is unknown?
The missing frequency $$f$$.
Reasoning:
We will use assumed mean method to solve this question
\begin{align} \operatorname{Mean,}\,\,\,(\overline{{x}})={a}+\left(\frac{\Sigma {f}_{{i}} {d}_{{i}}}{\Sigma {f}_{{i}}}\right) \end{align}
Solution:
We know that,
Class mark
{{x}_{i}}=\,\,\frac{\left[ \begin{align} & \text{Upper}\ \text{class}\ \text{limit}\text{+} \\ & \text{Lower}\ \text{class}\ \text{limit} \\ \end{align} \right]}{2}
Taking assumed mean$$, a = 18$$
Steps:
Daily pocket allowance (in Rs) No of children $$(f_i)$$ $$X_i$$ $$d_i = x_i -a$$ $$f_id_i$$ $$11 – 13$$ $$7$$ $$12$$ $$-6$$ $$-42$$ $$13 – 15$$ $$6$$ $$14$$ $$-4$$ $$-24$$ $$15 – 17$$ $$9$$ $$16$$ $$-2$$ $$-18$$ $$17 – 19$$ $$13$$ $$18 (a)$$ $$0$$ $$0$$ $$19 – 21$$ $$f$$ $$20$$ $$2$$ $$2f$$ $$21 – 23$$ $$5$$ $$22$$ $$4$$ $$20$$ $$23 – 25$$ $$4$$ $$24$$ $$6$$ $$24$$ $$\Sigma f_i=40 +f$$ $$\Sigma f_id_i=2f-40$$
From the table, we obtain
$$\Sigma {f}_{i} = 40 + f$$
$$\Sigma {f}_{i} {d}_{i} = 2f - 40$$
\begin{align} \text { Mean, }\,\,(\overline{{x}}) &={a}+\left(\frac{\Sigma {f}_{{i}} {d}_{{i}}}{\Sigma {f}_{{i}}}\right) \\ 18 &=18+\left(\frac{-40+2 {f}}{40+f}\right) \\ 18-18 &=\frac{2 {f}-40}{40+f} \\ 2f - 40 &= 0 \\ {f} &=20 \end{align}
Hence, the missing frequency $$f$$ = $$20$$.
## Chapter 14 Ex.14.1 Question 4
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heart beats per minute $$65 – 68$$ $$68 – 71$$ $$71 – 74$$ $$74 – 77$$ $$77 – 80$$ $$80 – 83$$ $$83 – 86$$ Number of women $$2$$ $$4$$ $$3$$ $$8$$ $$7$$ $$4$$ $$2$$
### Solution
What is known?
The heart beats per minute of $$30$$ women.
What is unknown?
The mean heart beats per minute for these women.
Reasoning:
We will use Step-deviation Method to solve this question because the data given is large and will be convenient to apply to all the $$d_i$$ have a common factor.
Sometimes when the numerical values of $$x_i$$ and $$f_i$$ are large, finding the product of $$x_i$$ and $$f_i$$becomes tedious. We can do nothing with the $$f_i,$$ but we can change each $$x_i$$ to a smaller number so that our calculations become easy. Now we have to subtract a fixed number from each of these $$x_i.$$
The first step is to choose one among the $$x_i$$as the assumed mean and denote it by $$‘a’$$. Also, to further reduce our calculation work, we may take $$‘a’$$ to be that $$x_i$$ which lies in the centre of $$x_1, x_2, . . ., x_n.$$ So, we can choose $$a.$$
The next step is to find the difference $$‘d_i’$$ between a and each of the $$x_i,$$ that is, the deviation of $$‘a’$$ from each of the $$x_i.$$ i.e.,$$d_i = x_i - a$$
The third step is to find $$‘u_i’$$ by dividing $$d_i$$ and class size $$h$$ for each of the $$x_i.$$ i.e.,$$u_i = \frac{{d_i}}{h}$$
The next step is to find the product of $$u_i$$ with the corresponding $$f_i,$$ and take the sum of all the $$f_iu_i$$
The step-deviation method will be convenient to apply if all the $$d_i$$have a common factor.
Now put the values in the below formula
Mean, $$\overline x = a + (\frac{{\Sigma f_i u_i}}{\Sigma f_i}) \times h$$
Steps:
We know that,
Class mark, $$x_i = \frac{{\text{Upper class limit + Lower class limit}}}{2}$$
Class size, $$h=3$$
Taking assumed mean,$$a= 75.5$$
Number of heart beats per minute No of women $$(f_i)$$ $$X_i$$ $$d_i = x_i -a$$ \begin{align} {u_i}=\frac{ {x_i}-{a}}{h}\end{align} $$f_iu_i$$ $$65-68$$ $$2$$ $$66.5$$ $$-9$$ $$-3$$ $$-6$$ $$68-71$$ $$4$$ $$69.5$$ $$-6$$ $$-2$$ $$-8$$ $$71 -74$$ $$3$$ $$72.5$$ $$-3$$ $$-1$$ $$-3$$ $$74-77$$ $$8$$ $$75.5(a)$$ $$0$$ $$0$$ $$0$$ $$77 – 80$$ $$7$$ $$78.5$$ $$3$$ $$1$$ $$7$$ $$80 – 83$$ $$4$$ $$81.5$$ $$6$$ $$2$$ $$8$$ $$83 – 86$$ $$2$$ $$84.5$$ $$9$$ $$3$$ $$6$$ $$\Sigma f_i=30$$ $$\Sigma f_iu_i=4$$
From the table, we obtain
$$\Sigma f_i=30$$
$$\Sigma f_i u_i=4$$
\begin{align} \operatorname{Mean}\,\,(\overline{{x}}) &={a}+\left(\frac{\Sigma {f}_{{i}} {u}_{{i}}}{\Sigma {f}_{{i}}}\right) {h} \\ \overline{{x}} &=75.5+\left(\frac{4}{30}\right) 3 \\ \overline{{x}} &=75.5-\frac{{2}}{5} \\\overline{{x}} &=75.5-0.4 \\ \overline {{x}}&=75.9 \end{align}
Hence, the mean heartbeat per minute for these women is $$75.9$$
## Chapter 14 Ex.14.1 Question 5
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes $$50 – 52$$ $$53 – 55$$ $$56 – 58$$ $$59 – 61$$ $$62 – 64$$ Number of boxes $$15$$ $$110$$ $$135$$ $$115$$ $$25$$
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
### Solution
What is known?
The distribution of mangoes according to the number of boxes.
What is unknown?
The mean number of mangoes kept in a packing box.
Reasoning:
We solve this question by step deviation method.
Hence, the given class interval is not continuous. First we have to make it continuous. There is a gap of $$1$$ between two class interval. Therefore, $$0.5$$ has to be added to the upper class limit and $$0.5$$ has to be subtracted from the lower class limit of each interval.
$\text{Mean,}\; \overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$
Solution:
We know that,
Class mark,
$${x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}$$
Class size, $$h = 3$$
Taking assumed mean, $$a = 57$$
Class interval No of house hold $$(f_i)$$ $$x_i$$ $$d_i = x_i -a$$ \begin{align}{u_i}=\frac{ d_i}{h}\end{align} $$F_iu_i$$ $$49.5-52.5$$ $$15$$ $$51$$ $$-6$$ $$-2$$ $$-30$$ $$52.5-55.5$$ $$110$$ $$54$$ $$-3$$ $$1$$ $$-110$$ $$55.5-58.5$$ $$135$$ $$57(a)$$ $$0$$ $$0$$ $$0$$ $$58.5-61.5$$ $$115$$ $$60$$ $$3$$ $$1$$ $$115$$ $$61.5-64.5$$ $$25$$ $$63$$ $$6$$ $$2$$ $$50$$ $$\Sigma f_i=400$$ $$\Sigma f_iu_i=25$$
From the table, we obtain
\begin{align}\sum {{f_i} = 400} \\\sum {{f_i}{u_i}} = 25\end{align}
Mean, $$\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$$
\begin{align}& = 57 + \left( {\frac{{25}}{{400}}} \right) \times 3\\ &= 57 + \frac{1}{{16}} \times 3\\ &= 57 + \frac{3}{{16}}\\ &= 57 + 0.19\\ &= 57.19\end{align}
The mean number of mangoes kept in a packing box are $$57$$.$$19$$
## Chapter 14 Ex.14.1 Question 6
The table below shows the daily expenditure on food of $$25$$ households in a locality.
Daily expenditure (in Rs) $$100 – 150$$ $$150 – 200$$ $$200 – 250$$ $$250 – 300$$ $$300 – 350$$ Number of households $$4$$ $$5$$ $$12$$ $$2$$ $$2$$
Find the mean daily expenditure on food by a suitable method.
### Solution
What is known?
The daily expenditure on food of $$25$$ households in a locality.
What is unknown?
The mean daily expenditure on food.
Reasoning:
We will solve this question by step deviation method.
Mean, $$\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$$
Steps:
We know that,
Class mark, $${x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}$$
Class size, $$h = 50$$
Taking assumed mean, $$a = 225$$
Daily expenditure in Rs. Number of house holds $$(f_i)$$ $$x_i$$ $$d_i = x_i -a$$ \begin{align}{u_i}=\frac{ d_i}{h}\end{align} $$f_iu_i$$ $$100 –150$$ $$4$$ $$125$$ $$-100$$ $$-2$$ $$-8$$ $$150 – 200$$ $$5$$ $$175$$ $$-50$$ $$-1$$ $$-5$$ $$200 – 250$$ $$12$$ $$225 (a)$$ $$0$$ $$0$$ $$0$$ $$250 – 300$$ $$2$$ $$275$$ $$50$$ $$1$$ $$2$$ $$300 –350$$ $$2$$ $$325$$ $$100$$ $$2$$ $$4$$ $$\Sigma f_i=25$$ $$\Sigma f_iu_i=-7$$
From the table, we obtain
$\begin{array}{l}\sum {{f_i} = 25} \\\sum {{f_i}{u_i}} = - 7\end{array}$
\begin{align}{\text { Mean }(\overline{{x}})}&={{a}+\left(\frac{\sum { f_iu_i }}{\sum {f} _{i}}\right) {h}} \\ {\overline{{x}}}&={225+\left(\frac{-7}{25}\right) 50} \\ {\overline{{x}}}&={225-14} \\ {\overline{{x}}}&={211}\end{align}
Thus, the mean daily expenditure on food is Rs $$211$$.
## Chapter 14 Ex.14.1 Question 7
To find out the concentration of $$SO_2$$ in the air (in parts per million, i.e., ppm), the data was collected for $$30$$ localities in a certain city and is presented below:
Concentration of $$SO_2$$ (in ppm) Frequency $$0.00 - 0.04$$ $$4$$ $$0.04 - 0.08$$ $$9$$ $$0.08 - 0.12$$ $$9$$ $$0.12 - 0.16$$ $$2$$ $$0.16 - 0.20$$ $$4$$ $$0.20 - 0.24$$ $$2$$
Find the mean concentration of $$SO_2$$ in the air.
### Solution
What is known?
The Concentration of $$SO_2$$ in the air (in parts per million, i.e., ppm) for $$30$$ localities in a certain city:
What is unknown?
The mean concentration of $$SO_2$$in the air
Reasoning:
We solve this question by step deviation method.
Mean, $$\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$$
Steps:
We know that,
Class mark,$${x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}$$
Class size$$, h = 0.04$$
Taking assumed mean$$,a = 0.14$$
Content of SO2 Frequency $$(f_i)$$ $$x_i$$ $$d_i = x_i -a$$ \begin{align}{u_i}=\frac{ d_i}{h}\end{align} $$f_iu_i$$ $$0.00-0.04$$ $$4$$ $$0.02$$ $$-0.12$$ $$-3$$ $$-12$$ $$0.04-0.08$$ $$9$$ $$0.06$$ $$-0.08$$ $$-2$$ $$-18$$ $$0.08-0.12$$ $$9$$ $$0.10$$ $$-0.04$$ $$-1$$ $$-9$$ $$0.12-0.16$$ $$2$$ $$0.14(a)$$ $$0$$ $$0$$ $$0$$ $$0.16-0.20$$ $$4$$ $$0.18$$ $$0.04$$ $$1$$ $$4$$ $$0.20-0.24$$ $$2$$ $$0.22$$ $$0.08$$ $$2$$ $$4$$ $$\Sigma f_i=30$$ $$\Sigma f_iu_i=-31$$
From the table, we obtain
$\begin{array}{l}\sum {{f_i} = 30} \\\sum {{f_i}{u_i}} = - 31\end{array}$
\begin{align} \text { Mean }(\overline{{x}}) &={a}+\left(\frac{\sum {f}_{{i}} {u}_{{i}}}{{f}_{{i}}}\right) {h} \\ \overline{{x}} &=0.14+\left(\frac{-31}{30}\right) 0.04 \\ \overline{{x}} &=0.14-| 0.04133 \\ \overline{{x}} &=0.099 \end{align}
The mean concentration of $$SO_2$$ in the air is $$0.099$$.
## Chapter 14 Ex.14.1 Question 8
A class teacher has the following absentee record of $$40$$ students of a class for the whole term. Find the mean number of days a student was absent.
Number of days $$0 – 6$$ $$6 – 10$$ $$10 – 14$$ $$14 – 20$$ $$20 – 28$$ $$28 – 38$$ $$38 – 40$$ Number of students $$11$$ $$10$$ $$7$$ $$4$$ $$4$$ $$3$$ $$1$$
### Solution
What is known?
The Absentee record of $$40$$ students of a class for the whole term.
What is unknown?
The mean number of days a student was absent.
Reasoning:
We solve this question by assumed mean method.
Mean, $$\overline x = a + \left( {\frac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}} \right)$$
Steps:
We know that,
Class mark,$${x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}$$
Taking assumed mean,$$a=17$$
Number of days Number of students $$f_i$$ $$x_i$$ $$d_i = x_i -a$$ $$f_id_i$$ $$0 – 6$$ $$11$$ $$3$$ $$-14$$ $$-154$$ $$6 – 10$$ $$10$$ $$8$$ $$-9$$ $$-90$$ $$10 – 14$$ $$7$$ $$12$$ $$-5$$ $$-35$$ $$14 – 20$$ $$4$$ $$17(a)$$ $$0$$ $$0$$ $$20 – 28$$ $$4$$ $$24$$ $$7$$ $$28$$ $$28 – 38$$ $$3$$ $$33$$ $$18$$ $$48$$ $$38 – 40$$ $$1$$ $$39$$ $$22$$ $$22$$ $$\Sigma f_i=40$$ $$\Sigma f_id_i=-181$$
From the table,we obtain
$\begin{array}{l}\sum {{f_i} = 40} \\\sum {{f_i}{d_i}} = - 181\end{array}$
\begin{align} \operatorname{Mean}(\overline{{x}}) &={a}+\left(\frac{\Sigma f_{1} d_{i}}{\Sigma {f}_{{i}}}\right) \\ \overline{{x}} &=17+\left(\frac{-181}{40}\right) \\ \overline{{x}} &=17-\frac{181}{40} \\ \overline{{x}} &=17-4.525 \\ \overline{{x}} &=12.475 \\ \overline{{x}} & =12.48 \end{align}
Thus, the mean number of days a student was absent is $$12.48$$.
## Chapter 14 Ex.14.1 Question 9
The following table gives the literacy rate (in percentage) of $$35$$ cities. Find the mean literacy rate.
Literacy rate (in %) $$45 – 55$$ $$55 – 65$$ $$65 – 75$$ $$75 – 85$$ $$85 – 95$$ Number of cities $$3$$ $$10$$ $$11$$ $$8$$ $$3$$
### Solution
What is known?
The Literacy rate (in percentage) of $$35$$ cities
What is unknown?
The Mean literacy rate.
Reasoning:
We solve this question by assumed mean method.
Mean, $$\overline x = a + \left( {\frac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}} \right)$$
Steps:
We know that,
Class mark, $${x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}$$
Taking assumed mean$$, a=70$$
Literacy rate No of cities$$(f_i)$$ $$X_i$$ $$d_i = x_i -a$$ $$f_id_i$$ $$45 – 55$$ $$3$$ $$50$$ $$-20$$ $$-60$$ $$55 – 65$$ $$10$$ $$60$$ $$-10$$ $$-100$$ $$65 – 75$$ $$11$$ $$70(a)$$ $$0$$ $$0$$ $$75 – 85$$ $$8$$ $$80$$ $$10$$ $$80$$ $$85 – 95$$ $$3$$ $$90$$ $$20$$ $$60$$ $$\Sigma f_i=35$$ $$\Sigma f_id_i=-20$$
From the table,we obtain
$\begin{array}{l}\sum {{f_i} = 35} \\\sum {{f_i}{d_i}} = - 20\end{array}$
\begin{align} \text { Mean }(\overline{{x}}) &={a}+\left(\frac{\Sigma f_{i} d_{i}}{\Sigma {f}_{{i}}}\right) \\ \overline{{x}} &=70+\left(\frac{-20}{35}\right) \\ \overline{{x}} &=70-\frac{4}{7} \\ \overline{{x}} &=70-0.57 \\ \overline{{x}} &=69.43 \end{align}
Thus, the mean literacy rate is $$69.43$$%
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# Mixed Number and Fraction Estimation
## Estimating sums and differences
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Mixed Number and Fraction Estimation
License: CC BY-NC 3.0
Travis was practicing for the long jump. His first jump was 834\begin{align*}8 \frac{3}{4}\end{align*} feet. After practicing for a month, he was able to add 213\begin{align*}2 \frac{1}{3}\end{align*} feet to his record. About how many feet is Travis able to jump? Estimate the distance.
In this concept, you will learn how to estimate the sums and differences of fractions and mixed numbers.
### Estimating Mixed Numbers and Fractions
A sum is the answer in addition problem. You can also estimate the sum of fractions. To estimate a sum of two fractions, round the fractions to the nearest half. Here is an addition problem.
38+17=\begin{align*}\frac{3}{8} + \frac{1}{7} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
First, round each fraction to the nearest half.
Let’s start with 38\begin{align*}\frac{3}{8}\end{align*}. For a whole divided into 8 parts, 12=48\begin{align*}\frac{1}{2}=\frac{4}{8}\end{align*}. 38\begin{align*}\frac{3}{8}\end{align*} is rounded to 12\begin{align*}\frac{1}{2}\end{align*}.
For a whole divided into 7 parts, one-half is between 37\begin{align*}\frac{3}{7}\end{align*} and 47\begin{align*}\frac{4}{7}\end{align*}17\begin{align*}\frac{1}{7}\end{align*} is rounded 0.
3812120\begin{align*}\frac{3}{8}& \approx \frac{1}{2}\\ \frac{1}{2} & \approx 0\end{align*}Then, rewrite the problem and find the sum.
12+0=12\begin{align*}\frac{1}{2} + 0 = \frac{1}{2}\end{align*}
The estimate sum is 12\begin{align*}\frac{1}{2}\end{align*}.
When estimating the sum of mixed numbers, round to the nearest whole number instead of the nearest half. Here is an addition problem involving mixed numbers.
345+219=\begin{align*}3 \frac{4}{5} + 2 \frac{1}{9} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
First, round each mixed number to the nearest whole number.
345\begin{align*}3 \frac{4}{5}\end{align*} will either round down to 3 or up to 4. Look at the fraction part of the mixed number. 45\begin{align*}\frac{4}{5}\end{align*} is greater than 12\begin{align*}\frac{1}{2}\end{align*}. Round up to 4.
219\begin{align*}2 \frac{1}{9}\end{align*} will either round down to 2 or up to 3. One-ninth is a very small fraction and less than 12\begin{align*}\frac{1}{2}\end{align*}. Round down to 2.
34521942\begin{align*}3\frac{4}{5} & \approx 4 \\ 2\frac{1}{9} & \approx 2\end{align*}
Then, rewrite the problem and find the sum.
4+2=6\begin{align*}4+2=6\end{align*}
The estimate sum is 6.
When talking about a difference, you are talking about subtraction. Estimating the difference of fractions and mixed numbers is similar to estimating sums. Round each fraction or mixed number and then subtract to find the estimate.
4537=\begin{align*}\frac{4}{5} - \frac{3}{7} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
First, round each fraction to the nearest half.
For a whole divided into 5 parts, one-half is between 25\begin{align*}\frac{2}{5}\end{align*} and 35\begin{align*}\frac{3}{5}\end{align*}45\begin{align*}\frac{4}{5}\end{align*} is rounded to 1.
For a whole divided into 7 parts, is one-half is between 37\begin{align*}\frac{3}{7}\end{align*} and 47\begin{align*}\frac{4}{7}\end{align*}. 37\begin{align*}\frac{3}{7}\end{align*} is rounded to 12\begin{align*}\frac{1}{2}\end{align*}.
4537112\begin{align*}\frac{4}{5} & \approx 1 \\ \frac{3}{7} & \approx \frac{1}{2}\end{align*}
Then, rewrite the problem and subtract.
112=12\begin{align*} 1 - \frac{1}{2} = \frac{1}{2}\end{align*}
The estimate difference is 12\begin{align*}\frac{1}{2}\end{align*}.
### Examples
#### Example 1
Earlier, you were given a problem about Travis practicing the long jump.
Travis started at 834\begin{align*}8\frac{3}{4}\end{align*} feet and was able to add 213\begin{align*}2\frac{1}{3}\end{align*} feet to his long jump record. Round the numbers and add the numbers to find an estimate of his current long jump record.
834+213\begin{align*}8\frac{3}{4} + 2\frac {1}{3} \end{align*}
First, round the mixed numbers to the nearest whole number.
83492132\begin{align*}8\frac{3}{4} \approx 9 \\ 2\frac {1}{3} \approx 2\end{align*}
Then, rewrite the problem and add.
9+2=11\begin{align*}9+2=11\end{align*}
Travis can now jump approximately 11 feet.
#### Example 2
Estimate the difference:
634218=\begin{align*}6 \frac{3}{4} - 2 \frac{1}{8} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}.
First, round each mixed number to the nearest whole number. Look at the fraction part of each mixed number.
34\begin{align*} \frac{3}{4}\end{align*} is greater than 12\begin{align*}\frac{1}{2}\end{align*}. 634\begin{align*}6\frac{3}{4}\end{align*} rounds up to 7.
18\begin{align*}\frac{1}{8}\end{align*} is less than 12\begin{align*}\frac{1}{2}\end{align*}. 218\begin{align*}2 \frac{1}{8}\end{align*} rounds down to 2.
Then, rewrite the problem and subtract.
72=5\begin{align*}7-2=5\end{align*}
The estimate sum is 5.
#### Example 3
Estimate the sum: 49+78=\begin{align*}\frac{4}{9} + \frac{7}{8} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}.
First, round each fraction to the nearest half.
4978121\begin{align*}\frac{4}{9}&\approx \frac{1}{2}\\\frac{7}{8}&\approx 1\end{align*}
Then, rewrite the problem and add.
12+1=112\begin{align*}\frac{1}{2} + 1 = 1\frac{1}{2}\end{align*}
The estimate sum is 112\begin{align*}1 \frac{1}{2}\end{align*}.
#### Example 4
Estimate the sum: 67+111=\begin{align*}\frac{6}{7} + \frac{1}{11} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}.
First, round each fraction to the nearest half.
\begin{align*}\frac{6}{7} & \approx 1 \\ \frac{1}{11} & \approx 0\end{align*}
Then, rewrite the problem and add.
\begin{align*}1+0=1\end{align*}
The estimate sum is \begin{align*}1\end{align*}.
#### Example 5
Estimate the difference: \begin{align*}5 \frac{1}{3} - 2 \frac{3}{4} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}.
First, round the mixed numbers to the nearest whole number.
\begin{align*}5 \frac{1}{3} & \approx 5 \\ 2 \frac{3}{4} & \approx 3\end{align*}
Then, rewrite the problem and find the difference.
\begin{align*}5-3=2\end{align*}
The estimate difference is \begin{align*}2\end{align*}.
### Review
Estimate the following sums.
1. \begin{align*}\frac{1}{5} + \frac{4}{5} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
2. \begin{align*}\frac{8}{9} + \frac{4}{6} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
3. \begin{align*}\frac{2}{9} + \frac{4}{5} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
4. \begin{align*}\frac{3}{6} + \frac{2}{3} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
5. \begin{align*}\frac{5}{6} + \frac{2}{3} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
6. \begin{align*}\frac{1}{12} + \frac{9}{11} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
7. \begin{align*}\frac{6}{12} + \frac{10}{11} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
8. \begin{align*}1 \frac{1}{10} + 2 \frac{1}{2} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
9. \begin{align*}4 \frac{2}{3} + 5 \frac{4}{5} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
10. \begin{align*}7 \frac{1}{9} + 8 \frac{1}{5} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
11. \begin{align*}14 \frac{5}{9} + 8 \frac{4}{5} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
12. \begin{align*}4 \frac{2}{3} + 7 \frac{1}{7} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
13. \begin{align*}18 \frac{1}{13} + 7 \frac{2}{10} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
14. \begin{align*}11 \frac{12}{13} + 4 \frac{1}{10} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
15. \begin{align*}22 \frac{5}{7} + 11 \frac{1}{5} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
Estimate each difference.
1. \begin{align*}\frac{4}{5} - \frac{1}{4} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
2. \begin{align*}\frac{4}{5} - \frac{3}{4} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
3. \begin{align*}\frac{9}{10} - \frac{3}{6} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
4. \begin{align*}\frac{11}{12} - \frac{1}{7} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
5. \begin{align*}\frac{10}{13} - \frac{1}{10} = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
To see the Review answers, open this PDF file and look for section 6.3.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Difference The result of a subtraction operation is called a difference.
Estimate To estimate is to find an approximate answer that is reasonable or makes sense given the problem.
fraction A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
Mixed Number A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
Sum The sum is the result after two or more amounts have been added together.
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# Solving Compound Inequalities
Objective Learn how to solve compound inequalities.
In this lesson, we are dealing with compound inequalities, expressions in which more than one inequality applies to the same variable. When inequalities are multiplied by a negative number, the inequality symbol must be reversed.
# Compound Inequalities
Let's talk a little about the notation and the terminology.
A compound inequality is an expression like 3 3x + 4 < 9 or 1x 2 7. The two inequality symbols indicate that the variable is to satisfy each part. So, for example, the inequality 2 < 2x - 5 < 4 is read “two is less than 2x - 5 and 2x - 5 is less than four” or “2x - 5 is between 2 and 4”.
## Solving Compound Inequalities
First, remember that solving an inequality (compound or otherwise) means describing its solution set, since there is more than one solution. Compound inequalities can be solved using the same methods that are used to solve single inequalities, the Addition and Multiplication Properties of Inequalities.
Example 1
Solve 3 2 x + 1 8.
Solution
First, subtract 1 from each part so that no constants appear in the expression that contains the variable.
3 2x + 1 < 8 2 2x < 7 Subtract 1 from each part. 1 x < Divide each part by 2.
The solution set is
Why is this considered solving the inequality if, after all, we still have an inequality? Why is this one any better than the original inequality?
The answer is that the inequalities now apply directly to x, not to an algebraic expression involving x. This means that it is much easier to visualize the solution set, as the points to the right of and including 1 and to the left of . Solving compound inequalities means isolating x in the inequalities.
Example 2
Solve -1 3 - x 5.
Solution
First, subtract 3 from each part to remove any constants from the expression that contains x.
-1 3 - x 5 -4 -x 2 Subtract 3 from each side.
Next, multiply each part of the inequality by -1 in order to isolate x in the middle. Since -1 is negative, reverse the inequality symbols.
4 x -2
The solution set is { x | 4 x -2 }.
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# RS Aggarwal Class 6 Solutions Chapter 7 - Decimals Ex 7A (7.1)
## RS Aggarwal Class 6 Chapter 7 - Decimals Ex 7A (7.1) Solutions Free PDF
Convert each of the following into a fraction in its simplest form:
(1) 0.9
We have:
= 0.9 = $\frac {9} {10}$
(2) 0.6
We have:
= 0.6 = $\frac {6} {10} = \frac {3} {5}$
(3) 0.08
We have:
= 0.08 = $\frac {8} {100} = \frac {4} {50} = \frac {2} {25}$
(4) 0.15
We have:
= 0.15 = $\frac {15} {100} = \frac {3} {20}$
(5) 0.48
We have:
= 0.48 = $\frac {48} {100} = \frac {12} {25}$
(6) 0.053
We have:
= 0.053 = $\frac {53} {1000}$
(7) 0.125
We have:
= 0.125 = $\frac {125} {1000} = \frac {25} {200} = \frac {5} {40} = \frac {1} {8}$
(8) 0.224
We have:
= 0.224 = $\frac {224} {1000} = \frac {56} {250} = \frac {28} {125}$
Convert each of the following as a mixed fraction:
(9) 6.4
We have:
= 6.4 = $\frac {64} {10} = \frac {32} {5} = 6 \frac {2} {5}$
(10) 16.5
We have:
= 16.5 = $\frac {165} {10} = \frac {33} {2} = 16 \frac {1} {2}$
(11) 8.36
We have:
= 8.36 = $\frac {836} {100} = \frac {209} {25} = 8 \frac {9} {25}$
(12) 4.275
We have:
= 4.275 = $\frac {4275} {1000} = \frac {171} {40} = 4 \frac {11} {40}$
(13) 25.06
We have:
= 25.06 = $\frac {2506} {100} = \frac {1253} {50} = 25 \frac {3} {50}$
(14) 7.004
We have:
= 7.004 = $\frac {7004} {1000} = \frac {1751} {250} = 7 \frac {1} {250}$
(15) 2.052
We have:
= 2.052 = $\frac{2052}{1000} = \frac{513}{250} = 2\frac{13}{250}$
(16) 3.108
We have:
= 3.108 = $\frac {3108} {1000} = \frac {777} {250} = 3 \frac {27} {250}$
Convert each of the following into decimal:
(17) $\frac {23} {10}$
We have:
$\frac {23} {10}$ = $2 \frac {3} {10}$
= 2 + 0.3 = 2.3
(18) $\frac {167} {100}$
We have:
$\frac {167} {100}$ = $1 \frac {67} {100}$ = 1 + 0.67 = 1.67
(19) $\frac {1589} {100}$
We have:
$\frac {1589} {100}$ = $15 \frac {89} {100}$ = 15 + 0.89 = 15.89
(20) $\frac {5413} {1000}$
We have:
$\frac {5413} {1000}$ = $5 \frac {413} {1000}$ = 5 + 0.413 = 5.413
(21) $\frac {21415} {1000}$
We have:
$\frac {21415} {1000}$ = $21 \frac {415} {1000}$ = 21 + 0.415 = 21.415
(22) $\frac {25} {4}$
We have:
$\frac {25} {4}$ = $6 \frac {1} {4}$ = 6 + 0.25 = 6.25
(23) $3 \frac {3} {5}$
= $\frac {18} {5}$
We have:
$3 \frac {3} {5}$ = 3 + 0.6 = 3.6
(24) $1 \frac {4} {25}$
= $1 \frac {4} {25}$ = $\frac {29} {25}$
We have:
$1 \frac {4} {25}$ = 1 + 0.16 = 1.16
(25) $5 \frac {17} {50}$
= $5 \frac {17} {50}$ = $\frac {267} {50}$
We have:
$5 \frac {17} {50}$ = 5 + 0.34 = 5.34
(26) $12 \frac {3} {8}$
$12 \frac {3} {8}$ = $\frac {99} {8}$
We have:
$12 \frac {3} {8}$ = 12 + 0.375 = 12.375
(27) $2 \frac {19} {40}$
$2 \frac {19} {40}$ = $\frac {99} {40}$
We have:
$2 \frac {19} {40}$ = 2 + 0.475 = 2.475
(28) $\frac {19} {20}$
$\frac {19} {20}$
We have:
$\frac {19} {20}$ = 0.95
(29) $\frac {37} {50}$
$\frac {37} {50}$
We have:
$\frac {37} {50}$ = 0.74
(30) $\frac {107} {250}$
$\frac {107} {250}$
We have:
$\frac {107} {250}$ = 0.428
(31) $\frac {3} {40}$
$\frac {3} {40}$
We have:
$\frac {3} {40}$ = 0.075
(32) $\frac {7} {8}$
$\frac {7} {8}$
We have:
$\frac {7} {8}$ = 0.875
(33) Using decimals, express
(i) 8 kg 640 g in kilograms
8 kg + 640 gm = 8 kg + $\frac {640} {1000} kg$
8 kg + 0.640 kg = 8.640 kg
(ii) 9 kg 37 g in kilograms
9 kg + 37 gm = 9 kg + $\frac {37} {1000} kg$
9 kg + 0.037 kg = 9.037 kg
(iii) 6 kg 8 g in kilograms
6 kg + 8 gm = 6kg + $\frac {8} {1000} kg$
6 kg + 0.008 kg = 6.008 kg
(34) Using decimals, express
(i) 4 km 365 m in in kilometers
4 km 365 m = 4 km + $\frac {365} {1000} \; km \; \left [ Since\; 1\; km\; =\; 1000\; m \right ]$
4 km + 0.365 km = 4.365 km
(ii) 5 km 87 m in kilometers
5 km 87 m = 5 km + $\frac {87} {1000} \; km \; \left [ Since\; 1\; km\; =\; 1000\; m \right ]$
5 km + 0.087 km = 5.087 km
(iii) 3 km 6 m in kilometers
3 km 6 m = 3 km + $\frac {6} {1000} \; km \; \left [ Since\; 1\; km\; =\; 1000\; m \right ]$
3 km + 0.006 km = 3.006 km
(iv) 270 m in kilometers
$\frac {270} {1000} \; km \; \; = 0.270\; km \; \left [ Since\; 1\; km\; =\; 1000\; m \right ]$
(v) 35 m in kilometers
$\frac {35} {1000} \; km \; = 0.035\; km \; \left [ Since\; 1\; km\; =\; 1000\; m \right ]$
(vi) 6 m in kilometers
$\frac {6} {1000} \; km \; = 0.006\; km \; \left [ Since\; 1\; km\; =\; 1000\; m \right ]$
(35) Using decimals, express
(i) 15 kg 850 g in kilograms
15 kg + 850 gm = 15 kg + $\frac {850} {1000} \; kg \; \left [ Since\; 1\; kg\; =\; 1000\; gm \right ]$
15 kg + 0.850 kg = 15.850 kg
(ii) 8 kg 96 g in kilograms
8 kg + 96 gm = 8 kg + $\frac {96} {1000} \; kg \; \left [ Since\; 1\; kg\; =\; 1000\; gm \right ]$
8 kg + 0.096 kg = 8.096 kg
(iii) 540 g in kilograms
540 gm = $\frac {540} {1000} = 0.540 \; kg \; \left [ Since\; 1\; kg\; =\; 1000\; gm \right ]$
(iv) 8 g in kilograms
8 gm = $\frac {8} {1000} = 0.008 \; kg \; \left [ Since\; 1\; kg\; =\; 1000\; gm \right ]$
(36) Using decimals, express:
(i) Rs 18 and 25 paise in rupees
Rs 18 + 25 paise = Rs 18 + $Rs\; \frac {25} {100} \; \left [ Since\;Re\; 1\;\; =\; 100\; paise \right ]$
Rs 18 + Rs 0.25 = Rs 18.25
(ii) Rs 9 and 8 paise in rupees
Rs 9 and 8 paise = Rs 9 + $Rs\; \frac {8} {100} \; \left [ Since\;Re\; 1\;\; =\; 100\; paise \right ]$
Rs 9 + Rs 0.08 = Rs 9.08
(iii) 32 paise in rupees
32 paise = Rs $Rs\; \frac {32} {100} = Rs\; 0.32 \; \left [ Since\;Re\; 1\;\; =\; 100\; paise \right ]$
(iv) 5 paise in rupees
5 paise = $Rs\; \frac {5} {100} = Rs\; 0.05 \; \left [ Since\;Re\; 1\;\; =\; 100\; paise \right ]$
Exercise 7C
(1) 9.6, 14.8, 37 and 5.9
9.6, 14.8, 37 and 5.9
Converting the decimals into like decimals:
9.6, 14.8, 37.0 and 5.9
Let us write the given numbers in the column form:
Hence, the sum of the given numbers is 67.3
(2) 23.7, 106.94, 68.9 and 29.5
23.7, 106.94, 68.9 and 29.5
Converting the decimals into like decimals:
23.70, 106.94, 68.90 and 29.50
Let us write the given numbers in the column form:
Hence, the sum of the given numbers is 229.04
(3) 72.8, 7.68, 16.23 and 0.7
72.8, 7.68, 16.23 and 0.7
Converting the decimals into like decimals:
72.80, 7.68, 16.23 and 0.70
Let us write the given numbers in the column form:
Hence, the sum of the given numbers is 97.41.
(4) 18.6, 84.75, 8.345 and 9.7
18.6, 84.75, 8.345 and 9.7
Converting the decimals into like decimals:
18.600, 84.750, 8.345 and 9.700
Let us write the given numbers in the column form:
Hence, the sum of the given numbers is 121.395
(5) 8.236, 16.064, 63.8 and 27.53
8.236, 16.064, 63.8 and 27.53
Converting the decimals into like decimals:
8.236, 16.064, 63.800 and 27.530
Let us write the given numbers in the column form:
Hence, the sum of the given numbers is 115.630.
(6) 28.9, 19.64, 123.697 and 0.354
28.9, 19.64, 123.697 and 0.354
Converting the decimals into like decimals:
28.900, 19.640, 123.697 and 0.354
Let us write the given numbers in the column form:
Hence, the sum of the given numbers is 172.591
(7) 4.37, 9.638, 17.007 and 6.8
4.37, 9.638, 17.007 and 6.8
Converting the decimals into like decimals:
4.370, 9.638, 17.007 and 6.800
Let us write the given numbers in the column form:
Hence, the sum of the given numbers is 37.815.
(8) 14.5, 0.038, 118.573 and 6.84
14.5, 0.038, 118.573 and 6.84
Converting the decimals into like decimals:
14.500, 0.038, 118.573 and 6.840
Let us write the given numbers in the column form:
Hence, the sum of the given numbers is 139.951
9. During three days of week, a rickshaw puller earns Rs 32.60, Rs 56.80 and Rs 72 respectively. What is his total earning during these days?
Earning on the 1st day of the week = Rs 32.60
Earning on the 2nd day of the week = Rs 56.80
Earning on the 3rd day of the week = Rs 72.00
Total earning = Rs 161.40
10. A man purchases an almirah for Rs 11025, gives Rs 172.50 as its cartage and spends Rs 64.80 on its repair. How much does the almirah cost him?
Cost of the almirah = Rs 11025.00
Money spend on cartage = Rs 172.50
Money spent on repair = Rs 64.800
Total cost of the almirah = Rs 11262.3
11. Ramesh covers 36 km 235 m by taxi, 4 km 85 m by rickshaw and 1 km 80 m on foot. What is the total distance covered by him?
Distance covered by the taxi = 36 km 235 m
Distance covered by the rickshaw = 4 km 085 m
Distance covered on foot = 1 km 080 m
Total distance covered = 41km 400m
12. A bag contains 45 kg 80 g of sugar and the mass of the empty bag is 950 g. What is the mass of the bag containing this much of sugar?
Weight of sugar in the bag = 45 kg 080 g
Weight of the empty bag = 0 kg 950 g
Total weight of the bag = 46kg 30g
13. Ramu bought 2 m 70 cm cloth for his shirt and 2 m 60 cm cloth of his pyjamas. Find the total length of cloth bought by him.
Length of the cloth for his shirt = 2 m 70 cm
Length of the cloth for his pyjamas = 2 m 60 cm
The total length of the cloth bought = 5m 30 cm
14. Radhika bought 2 m 5 cm cloth for her salwar and 3 m 35 cm cloth for her shirt. Find the total length of cloth bought by her.
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# how can I solve for $x$?
I am having a bit of trouble solving for x when trying to find $f^{-1}$. I have $$y=\frac{x+5}{x-4}$$ How can I get x on one side? I tried multiplying both sides by the denominator but I am still left with an $x$ on both sides...
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That was a good start! After that move the terms containing an $x$ to the other side. Remember to swap signs when appropriate. – Jyrki Lahtonen Sep 3 '12 at 6:36
You just need to expand out the brackets and re-factor:
$$\begin{array}{rcl} y & = & \frac{x+5}{x-4}\\ y(x-4) & = & x+5\\ yx-4y&=&x+5\\ yx-x &=&4y+5\\ x(y-1)&=&4y+5\\ x&=&\frac{4y+5}{y-1} \end{array}$$
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Note that $x+5 = 1\cdot(x-4) + 9$. Just as with numbers, this means that $x-4$ divides $x+5$ once with a remainder of 9. So you can write $(x+5)/(x-4)$ as $1+\frac{9}{x-4}$. It should be easy to get $x$ on one side now.
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For doing $x+5=1.(x-4)+9$, assuming $x\neq 4$ is essential. :-) – Babak S. Sep 3 '12 at 6:38
Just as with numbers, one needs to be careful when dividing by zero. Simply by paying careful attention to the statement of the problem we can see that $x=4$ could be problematic. – shoda Sep 3 '12 at 17:18
Usually when we are working with rational functions; we delete the roots of its denominator of our domain. So, what you did here is someway legal. – Babak S. Sep 3 '12 at 17:35
First multiply out to get rid of the fraction: $y(x-4)=x+5$, or $yx-4y=x+5$. Now collect the $x$ terms on one side and everything else on the other: $yx-x=5+4y$. Factor the lefthand side: $(y-1)x=5+4y$. Now just divide both sides by $y-1$, and you’ll have $x$ in terms of $y$.
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Multiply with the denominator:
$$y(x-4)=x+5$$ $$xy -4y=x+5$$ Take all x and bring them to the left side, all other terms to the right side. $$xy -x = 5+4y$$ Single out x: $$x(y-1)=5+4y$$ Divide by the factor: $$x=\frac{5+4y}{y-1}$$
Note: $x\ne4$, because of the first equation.
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– Cross multiply the fractions. 4 * 15 = 10 * x.
– Solve the equation for x. x = (4 * 15) / 10.
– Simplify for x. x = 6.
The solution of an equation is the set of all values that, when substituted for unknowns, make an equation true. For equations having one unknown, raised to a single power, two fundamental rules of algebra, including the additive property and the multiplicative property, are used to determine its solutions.
Subsequently, What is 5/6 minus 2/3 as a fraction?
56 – 23 is 16.
Also, How do you add and subtract fractions?
Fractions – Adding and Subtracting fractions – First Glance. Like fractions are fractions with the same denominator. You can add and subtract like fractions easily – simply add or subtract the numerators and write the sum over the common denominator.
How do you solve for X on one side of a fraction?
Last Review : 14 days ago.
## How do you add and subtract fractions with different denominators?
Make their denominators equal using the concept of least common multiple. Then subtract their numerators accordingly. Rewrite each fraction to its equivalent fraction with a denominator equal to the LCM = 30, then subtract their numerators. Make sure to reduce your answer to the lowest term.
## What is the first step in solving for x?
The first step to solving for ​x​ is going to be getting ​x​ alone on one side of the equation and everything else on the other side. Remember the algebraic golden rule: What you do to one side of the equation, you must do to the other side. That’s how the equation stays equal!
## How do I subtract fractions?
– Make sure the bottom numbers (the denominators) are the same.
– Subtract the top numbers (the numerators). Put the answer over the same denominator.
– Simplify the fraction (if needed).
## What are the steps for adding and subtracting fractions?
– Check to see if the fractions have the same denominator.
– If they don’t have the same denominator, then convert them to equivalent fractions with the same denominator.
– Once they have the same denominator, add or subtract the numbers in the numerator.
## How do you solve for x in a fraction?
How to Solve for x in Fractions. Solve for x by cross multiplying and simplifying the equation to find x. Example: Given the equation 4/10 = x/15 solve for x. Since 60 = 60 is true, you can be sure that x = 6 is the correct answer.
## How do you solve equations with fractions?
– Find the least common denominator of all the fractions in the equation.
– Multiply both sides of the equation by that LCD. …
– Isolate the variable terms on one side, and the constant terms on the other side.
– Simplify both sides.
## How do you add and subtract fractions step by step?
Step 1: Find the Lowest Common Multiple (LCM) between the denominators. Step 2: Multiply the numerator and denominator of each fraction by a number so that they have the LCM as their new denominator. Step 3: Add or subtract the numerators and keep the denominator the same.
## What are the steps to subtracting fractions?
– Make sure the bottom numbers (the denominators) are the same.
– Subtract the top numbers (the numerators). Put the answer over the same denominator.
– Simplify the fraction (if needed).
## What are the steps to adding fractions?
To add fractions there are Three Simple Steps: Step 1: Make sure the bottom numbers (the denominators) are the same. Step 2: Add the top numbers (the numerators), put that answer over the denominator. Step 3: Simplify the fraction (if needed)
## How do you add fractions with different denominators?
If the denominators are not the same, then you have to use equivalent fractions which do have a common denominator . To do this, you need to find the least common multiple (LCM) of the two denominators. To add fractions with unlike denominators, rename the fractions with a common denominator. Then add and simplify.
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# Probability of independent events
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0/1
##### Intros
###### Lessons
1. Differences between independent events and dependent events
0/9
##### Examples
###### Lessons
1. Sam is hunting a treasure. He is equally likely to travel along any pathway. His trip ends at 1, 2, 3, 4, 5, or 6.
1. What is the probability that Sam finds the treasure?
2. What is the probability that Sam does not find the treasure?
2. Claire rolls two 6-sided dice.
1. If Claire rolls two threes or two fives, then she will buy herself a donut. How many favorable outcomes are there?
2. What is the probability of getting at least one odd number?
3. If she rolls the two dice and adds the numbers together, how many times will the sum be 7?
3. Anna has 18 clothes in her closet. She has 6 shirts, 5 tees, 4 tanks and 3 sweaters. If she is taking 1 top from the closet randomly, what is the probability that:
1. She will take out a tee?
2. She will take out a shirt or a sweater?
3. She will not take out a shirt or a tank?
4. She will take out a sweater and a tee, if she takes out 1 item of cloth, puts it back and takes out a second one?
###### Topic Notes
Similar to the section about organizing outcomes, in this section we practice determining the outcomes of two independent events along with organizing these outcomes using tables and tree diagrams. In this section, we are also going to learn how to determine the number of possible outcomes in two independent events. In addition to finding the number of possible outcomes, in this section we practice finding the number of favorable outcomes in two independent events. Finally, in this section we use tables and tree diagrams to solve probability problems.
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# Rationalisation (mathematics)
(Redirected from Rationalization (mathematics))
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For other uses, see Rationalization.
In elementary algebra, root rationalisation is a process by which surds in the denominator of an irrational fraction are eliminated.
These surds may be monomials or binomials involving square roots, in simple examples. There are wide extensions to the technique.
## Rationalisation of a monomial square root and cube root
For the fundamental technique, the numerator and denominator must be multiplied by the same factor.
Example 1:
$\frac{10}{\sqrt{a}}$
To rationalise this kind of monomial, bring in the factor $\sqrt{a}$:
$\frac{10}{\sqrt{a}} = \frac{10}{\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{{10\sqrt{a}}}{\sqrt{a}^2}$
The square root disappears from the denominator, because it is squared:
$\frac{{10\sqrt{a}}}{\sqrt{a}^2} = \frac{10\sqrt{a}}{a}$
This gives the result, after simplification:
$\frac{{10\sqrt{a}}}{{a}}$
Example 2:
$\frac{10}{\sqrt[3]{b}}$
To rationalise this radical, bring in the factor $\sqrt[3]{b}^2$:
$\frac{10}{\sqrt[3]{b}} = \frac{10}{\sqrt[3]{b}} \cdot \frac{\sqrt[3]{b}^2}{\sqrt[3]{b}^2} = \frac{{10\sqrt[3]{b}^2}}{\sqrt[3]{b}^3}$
The cube root disappears from the denominator, because it is cubed:
$\frac{{10\sqrt[3]{b}^2}}{\sqrt[3]{b}^3} = \frac{10\sqrt[3]{b}^2}{b}$
This gives the result, after simplification:
$\frac{{10\sqrt[3]{b}^2}}{{b}}$
## Dealing with more square roots
For a denominator that is:
$\sqrt{2}+\sqrt{3}\,$
Rationalisation can be achieved by multiplying by the Conjugate:
$\sqrt{2}-\sqrt{3}\,$
and applying the difference of two squares identity, which here will yield −1. To get this result, the entire fraction should be multiplied by
$\frac{ \sqrt{2}-\sqrt{3} }{\sqrt{2}-\sqrt{3}} = 1.$
This technique works much more generally. It can easily be adapted to remove one square root at a time, i.e. to rationalise
$x +\sqrt{y}\,$
by multiplication by
$x -\sqrt{y}$
Example:
$\frac{3}{\sqrt{3}+\sqrt{5}}$
The fraction must be multiplied by a quotient containing ${\sqrt{3}-\sqrt{5}}$.
$\frac{3}{\sqrt{3}+\sqrt{5}} \cdot \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} = \frac{3(\sqrt{3}-\sqrt{5})}{\sqrt{3}^2 - \sqrt{5}^2}$
Now, we can proceed to remove the square roots in the denominator:
$\frac{{3(\sqrt{3}-\sqrt{5}) }}{\sqrt{3}^2 - \sqrt{5}^2} = \frac{ 3 (\sqrt{3} - \sqrt{5} ) }{ 3 - 5 } = \frac{ 3( \sqrt{3}-\sqrt{5} ) }{-2}$
## Generalisations
Rationalisation can be extended to all algebraic numbers and algebraic functions (as an application of norm forms). For example, to rationalise a cube root, two linear factors involving cube roots of unity should be used, or equivalently a quadratic factor.
## References
This material is carried in classic algebra texts. For example:
• George Chrystal, Introduction to Algebra: For the Use of Secondary Schools and Technical Colleges is a nineteenth-century text, first edition 1889, in print (ISBN 1402159072); a trinomial example with square roots is on p. 256, while a general theory of rationalising factors for surds is on pp. 189–199.
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## Calculus (Part-2)-geometric explanation of differentiation
Online maths tutors like the following concept very much
Geometric explanation of differentiation- If we find the derivative of function f(x) at x = x0 then it is equal to the slope of the tangent to the graph of given function f(x) at the given point [(x0, f(x0))].
But what is a tangent line?
It is not merely a simple line that joins the graph of given function at one point.
It is actually the limit of the secant lines joining points P = [(x0, f(x0)] and Q on the graph of f(x) as Q moves very much close to P.
The tangent line contacts the graph of given function at given point [(x0, f(x0)] the slope of the tangent line matches the direction of the graph at that point. The tangent line is the straight line that best approximates the graph at that point.
As we are given the graph of given function, we can draw the tangent to this graph easily. Still, we’ll like to make calculations involving the tangent line and so will require a calculative method to explore the tangent line.
We can easily calculate the equation of the tangent line by using the slope-point form of the line. We slope of a line is m and its passing through a point (x0,y0) then its equation will be
y − y0 = m(x − x0)
So now we have the formula for the equation of the tangent line. It’s clear that to get an actual equation for the tangent line, we should know the exact coordinates of point P. If we have the value of x0 with us we calculate y0 as
y = f(x0)
The second thing we must know is the slope of line
m = f’(x0)
Which we call the derivative of given function f(x).
Definition:
The derivative f’(x0) of given function f at x0 is equal to the slope of the tangent line to
y = f(x) at the point P = (x0, f(x0).
Differentiation Using Formulas- We can use derivatives of different types of functions to solve our problems :
(ix) D (secx) = secx . tanx (x) D (cosecx) = – cosecx . cotx
(xii) D (constant) = 0 where D =
These formulas are result of differetiation by first principle
Inverse Functions And Their Derivatives :
Theorems On Derivatives: If u and v are derivable function of x, then,
(i) (ii) where K is any constant
(iii) known as “ Product Rule ” (iv)
known as “Quotient Rule ”
(v) If y = f(u) & u = g(x) then “Chain Rule ”
Logarithmic Differentiation: To find the derivative of : (i) a function which is the product or quotient of a number of functions OR
(ii) a function of the form where f & g are both differentiable, it will be found convenient to take the logarithm of the function first & then differentiate. This is called Logarithmic Differentiation.
Implicit Differentiation: (i) In order to find dy/dx, in the case of implicit functions, we differentiate each term w.r.t. x regarding y as a function of x & then collect terms in dy/dx together on one side to finally find dy/dx.
(ii) In answers of dy/dx in the case of implicit functions, both x & y are present.
Parametric Differentiation: If y = f(q) & x = g(q) where q is a parameter, then
Derivative Of A Function w.r.t. Another Function-: Let y = f(x) & z = g(x)
Derivatives Of Order Two & Three: Let a function y = f(x) be defined on an open interval (a, b). It’s derivative, if it exists on(a, b) is a certain function f'(x) [or (dy/dx) or y’ ] & is called the first derivative of y w.r.t. x. If it happens that the first derivative has a derivative on (a, b) then this derivative is called the second derivative of y w. r. t. x & is denoted by f”(x) or (d2y/dx2) or y”. Similarly, the 3rd order derivative of y w. r. t. X, if it exists, is defined by It is also denoted by f”'(x) or y”’.
All online maths tutors suggest solving fair amount of questions based on these concepts
Example-Find the tangent line to the following function at z=3 for the given function
Solution
We can find the derivative of the given function using basic differentiation as discussed in the previous post
We are already given that z=3 so
Equation of tangent line is
y − y0 = m(x − x0) here y0=R(3)=√7
Putting these values we get equation of tangent line
In my online maths tutors series, I will discuss Application of Derivatives
Example- Differentiate the following function
Ans: We can apply quotient rule in this questions
## IB Mathematics Tutors-Part 1 (Calculus)
In my IB Mathematics Tutors series, I will explain different topics taught at HL and SL levels of IB Mathematics. Calculus is the first one.
If we want to understand the importance of Calculus in IB Diploma Programme, We should have a look at the number of teaching hours recommended for it. It’s 40 hours in SL(out of 150 total hours) and 48 (out of 240 total hours) hours in HL. This makes calculus the most important topic for IB Mathematics Tutors as well as for IB students.
What is Calculus- Calculus is an ancient Latin word. It means ‘small stones’ used for counting. In every branch of Mathematics, we study of something specific, like in Geometry, we study about shapes, in Algebra, we study about arithmetic operations, in coordinate, we study about locating a point. In calculus, we do the mathematical study of continuous change. It mainly has two branches- Read more
## How to find sum and product of zeros of equations
In my previous post on IB Mathematics, I have discussed how to solve a quadratic polynomial using Quadratic formula. Here I will tell you about different relationships based on sum and product of quadratic polynomial, cubic polynomial and bi-quadratic polynomials.
ax2 + bx + c = 0
Sum of the roots = −b/a
Product of the roots = c/a
If we know the sum and product of the roots/zeros of a quadratic polynomial, then we can find that polynomial using this formula
x2 − (sum of the roots)x + (product of the roots) = 0
## Cubic:
Now let us look at a Cubic (one degree higher than Quadratic):
ax3 + bx2 + cx + d=0
if α, β and γ are the zeros of this cubic polynomial then
α+β+γ=-b/a
αβ+βγ+γα=c/a
αβγ=-d/a
If we know these relationships of polynomials then we cal also calculate the polynomial using this formula:
x³-(α+β+γ)x²+(αβ+βγ+γα)x-(α+β+γ)=0
If we are given a bi-quadratic polynomial with degree 4 like:
ax 4+bx³+cx²+dx+e=0
and its roots/zeros are α, β, γ, and δ then
α+β+γ+δ=-b/a
αβ+αγ+αδ+βγ+βδ+γδ=c/a
αβγ+βγδ+γδα+αβδ=-d/a
αβγδ=e/a
using these formulas of sum and product of zeros of polynomials, we can find a lot of relationships in zeros of polynomials. Usually, we are asked to find these types of relationships in zeros.
Question: If α and β are the zeros of polynomial x²-px+q=0 then find the following relationships.
i) 1/α+1/β ii) α²β+αβ² iii) α²+β² iv) α/β+β/α v) α³+β³
Ans: To find these relationships we convert every value either in sum (α+β) or in the product (αβ) of zeros. For this conversion, we use following Mathematical Tricks
1)Try to take common
2) Try to take L.C.M
3) Try to make a Perfect Square
4) Use algebraic identities wherever required
If we use above steps properly, in most of the cases we are able to convert everything either in sum or in product of zeros
If we compare the given equation with std. form ax²+bx+c=0 then
a=1, b=-p and c=q
sum of zeros α+β=-b/a=-(-p/a)=p
product of zeros αβ=c/a=q/1=q (i) 1/α+1/β=β+α/αβ [By L.C.M]
=p/q
(ii) α²β+αβ² = αβ(α+β) [By common]
=q.p
(iii) α²+β² = (α²+β²+2αβ)-2αβ [add and subtract 2αβ, make it a perfect square ]
=(α+β)²-2αβ
=p²-2q
(iv) α/β+β/α = β²+α²/βα [By taking L.C.M]
we have already found β²+α² that is p²-2q so
α/β+β/α=(p²-2q)/q
(v) (α+β)³=(α+β)³-3αβ(α+β) [Direct algebraic identity]
=p³-3qp
You can further read about quadratics in PDF(quadratics ) given here. There are a lot of practise questions given in this PDF
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THE SAT MATH: ADDITIONAL TOPICS - McGraw-Hill Education SAT 2017 Edition (Mcgraw Hill's Sat) (2016)
## McGraw-Hill Education SAT 2017 Edition (Mcgraw Hill's Sat) (2016)
### THE SAT MATH: ADDITIONAL TOPICS
1. Understanding Geometric Relationships
2. Understanding Basic Trigonometry
3. Understanding Complex Numbers
What other special topics are included on the SAT Math test?
About 10% (6 out of 58 points) of the SAT Math questions are “Additional Topics” questions. These include topics like
• analyzing triangles using the Pythagorean Theorem
• graphing circles and other figures in thexy -plane
• analyzing areas, circumferences, chords, and sectors of circles
• measuring angles and arcs in radians
• working with area and volume and their formulas
• using the theorems of congruence and similarity
• working with basic trigonometric relationships including cofunction identities
• calculating with imaginary and complex numbers
Why are these topics important?
These topics from geometry, trigonometry, and advanced analysis are crucial to work in engineering, physics, architecture, and even design. Although they are not essential to every college major, they do provide tools for understanding and analyzing advanced concepts across the curriculum.
Sound intimidating? It”s not.
Some of you have already spent some time in math class studying these topics. If not, the three skills described in these 12 lessons will give you the knowledge and practice you need to master them.
Skill 1: Understanding Geometric Relationships
Lesson 1: Intersecting and parallel lines
In the figure above, ABCD is a parallelogram, and point B lies on . If x = 40, what is the value of y ?
1. A) 40
2. B) 50
3. C) 60
4. D) 70
(Medium ) Since ABCD is a parallelogram, we can take advantage of the Parallel Lines Theorem.
The Intersecting Lines Theorem
When two lines cross, four angles are formed. The vertical angles are congruent and adjacent angles are supplementary (that is, they have a sum of 180°).
The Parallel Lines Theorem
When two parallel lines are crossed by a third line, eight angles are formed. If the third line is perpendicular to one of the parallel lines, then it”s perpendicular to the other and all eight angles are right angles. Otherwise, all four acute angles are congruent, all four obtuse angles are congruent, and any acute angle is supplementary to any obtuse angle.
When dealing with parallel lines, especially in complicated figures, we can simplify things by considering angles in pairs. The important pairs form one of four letters: F, Z, C, or U.
First, let”s mark up the diagram with what we know from the Parallel Lines Theorem.
Since the pairs of opposite sides are parallel, the consecutive angles in the parallelogram must be supplementary (that is, have a sum of 180°). Notice that these pairs of consecutive angles form “U”s or “C”s as mentioned in the previous Helpful Tip. This implies that opposite angles are congruent in a parallelogram.
Since is a straight (180°) angle:
y + x + x + x = 180
Substitute x = 40 and simplify:
y + 120 = 180
Subtract 120:
y = 60
Therefore, the correct answer is (C).
In the figure above, lines l and m are parallel. What is the value of x ?
1. A) 43
2. B) 79
3. C) 86
4. D) 101
(Hard ) Although our diagram includes parallel lines, it doesn”t seem to show any of the parallel line “letter pairs” that we discussed above, because no line directly connects the parallel lines. We can fix this problem by drawing an extra line that”s parallel to l and m through the vertex of the angle.
Now we have two “Z” pairs of angles (otherwise known as “alternate interior” pairs) that show that the middle angle is actually the sum of two smaller angles of 36° and 43°, and therefore, x = 36 + 43 = 79, and the correct answer is (B).
Lesson 2: Triangles
Angle Sum Theorem
The sum of the measures of the angles in any triangle is 180°.
We can prove this with the “draw an extra line” trick. If we take any triangle, pick any of its vertices, and draw a line through that vertex that is parallel to the opposite side, we get a picture like the one above. Since the line we”ve drawn is a 180° angle, and since the “Z” angle pairs must be congruent, we”ve proven that a + b + c = 180.
Side-Angle Theorem
The largest angle in a triangle is always across from the largest side, and the smallest angle is always across from the smallest side.
Isosceles Triangle Theorem
If two sides in a triangle are congruent, the two angles across from those sides are also congruent. Conversely, if two angles in a triangle are congruent, the two sides across from them are also congruent.
Exterior Angle Theorem
If the side of a triangle is extended beyond a vertex, it makes an exterior angle with the adjacent side. The measure of this exterior angle is equal to the sum of the two remote interior angles.
The Triangle Inequality
The sum of any two sides of a triangle must always be greater than the third side.
In the figure above, if AD = DB = DC , what is the value of x + y ?
1. A) 72
2. B) 90
3. C) 96
4. D) 108
(Medium ) Since angle ADB and angle BDC are supplementary and AD = DB = DC , we can take advantage of the Isosceles Triangle Theorem to mark up the diagram.
Now let”s look at triangle ABC . Since its interior angles must have a sum of 180°, x + x + y + y = 180, and therefore, 2x + 2y = 180 and x + y = 90. So the correct answer is (B). Notice that this fact is independent of the measures of the other two (108° and 72°) angles. As long as AD = DB = DC , this relationship will hold. We can see these angle relationships if we notice that these three segments could all be radii of a circle centered at D .
You may remember from studying geometry that any “inscribed” angle (an angle inside a circle with a vertex on the circle) intercepts an arc on the circle that is twice its measure. Since angle ABC is an inscribed angle that intercepts a 180° arc, it must have a measure of 90° and therefore, x + y = 90.
The figure above shows three intersecting lines. What is the value of c in terms of a and b ?
1. A) 180 −ab
2. B) 180 −a+ b
3. C) 90 +ba
4. D) a+b
(Easy ) First, we should notice that two of the angles are “vertical” to two interior angles of the triangle, and the other is an exterior angle.
Since the c ° angle is an exterior angle to the triangle, the Exterior Angle Theorem tells us that c = a + b , so the correct answer is (D).
Alternately, we could just choose reasonable values for a and b , like a = 50 and b = 90, and then analyze the diagram in terms of these values. This would imply that the interior angles of the triangle are 50°, 90°, and 40°, and c ° would then be the measure of the supplement of 40°, which is 140°. If we then plug these values for a and b into all of the choices, the only one that yields 140 is D.
Lesson 3: The xy -plane
Note: Figure not drawn to scale.
In the xy -plane above, points A and B lie on the graph of the line y = 6. If OB has a slope of and AB = 5, what is the slope of ?
(Medium-hard ) To analyze this diagram, we must recall the definition of slope from Chapter 7 , Lesson 5.
First, let”s drop two perpendicular segments from A and B to points C and D , respectively, on the x -axis. Since A and B lie on the line y = 6, they are both 6 units from the x -axis, and so AC = BD = 6. Then, since the slope of OBis ½, BD /OD = ½, and therefore, OD = 12. Since AB = 5, CD = 5 also, and therefore, OC = 12 – 5 = 7. (Don”t worry that looks shorter than in the diagram. Remember, the figure is not drawn to scale!) This gives us everything we need to find the slope of , which connects (0, 0) to (7, 6). By the slope formula from Chapter 7 , Lesson 5, slope = (6 – 0)/(7 – 0) = 6/7 = 0.857.
Working in the Coordinate Plane
Finding Segment Midpoints. To find the coordinates of a midpoint , just average the coordinates of the endpoints .
Finding Slopes. To find the slope of a line in the xy -plane from any two points on the line, use the slope formula .
Finding Areas. Remember that the area of a figure is just the number of unit squares that fit inside it. You don”t always need to use a special formula to find the area of a figure. Even for very complicated shapes, you can sometimes find the area just by counting squares.
In the figure above, point M (not shown) is the midpoint of and point N (not shown) is the midpoint of . What is the slope of ?
(Medium ) To find the midpoint of a segment, we just need to take the average of the endpoints. Point M , the midpoint of , therefore has coordinates , and point N , the midpoint of , has coordinates . By the Slope Formula, then, the slope of is .
Lesson 4: The Pythagorean Theorem and the Distance Formula
What is the perimeter of quadrilateral ABCD in the figure above?
(Medium ) The perimeter of a figure is the distance around its edges. It”s easy to find the lengths of and because they are horizontal. The length of a horizontal segment is just the difference between the x -coordinates of its endpoints. The length of is 15 – 2 = 13, and the length of is 24 – (–3) = 27. To find the lengths of and , we can drop two vertical lines from points A and B to the bottom edge. This shows that and are hypotenuses of two right triangles as shown in the figure below.
(Take a minute to confirm the lengths of all the segments for yourself.) With this information, we can find AD and BC by the Pythagorean Theorem.
The Pythagorean Theorem
If a , b , and c represent the sides of a right triangle in which c is the longest side (the hypotenuse),
a2 + b2 = c2
Special Right Triangles
The SAT Math test expects you to be familiar with four families of special right triangles: 45°-45°-90° triangles, 30°-60°-90° triangles, 3-4-5 triangles , and 5-12-13 triangles . Take some time to familiarize yourself with these particular side-side relationships and side-angle relationships so that you can use these relationships when you recognize these triangles in SAT Math questions.
(3x )2 + (4x )2 = (5x )2
9x 2 + 16x 2 = 25x 2
(5y )2 + (12y )2 = (13y )2
25y 2 + 144y 2 = 169y 2
So, according to our diagram:
AD 2 = 52 + 122 = 169
BC 2 = 92 + 122 = 225
Notice that triangle on the left is a 5-12-13 special right triangle, and the triangle on the right is a 3-4-5 special right triangle. Noticing these relationships provides a shortcut to using the Pythagorean Theorem.
Take the square root:
BC = 15
Therefore, the perimeter of ABCD is 13 + 15 + 27 + 13 = 68.
The Distance Formula
We can generalize the technique we used in the previous problem to find the distance between any two points in the xy -plane. Just think of this distance as the length of the hypotenuse of a right triangle, as in the figure below. In other words, the Pythagorean Theorem and the Distance Formula are one and the same.
The 3-D Distance Formula
If we need to find the distance between two points in three-dimensional xyz -space, we just need to use a modified version of the distance formula that includes the extra z -dimension. You can see where this formula comes from if you imagine trying to find the length of the longest diagonal through a rectangular box.
The length of this diagonal, AC , is also the hypotenuse of right triangle ABC , and so its length is given by the Pythagorean Theorem.
Exercise Set 1: Geometry (No Calculator)
1
In the figure above, MNOP is a square and Q is the midpoint of . If , what is the area of square MNOP ?
2
Lines l and m are parallel in the figure above. What is the value of x ?
3
In the figure above, what is the value of a + b + c ?
4
Lines l and m are parallel in the figure above. Which of the following expresses the value of y in terms of x ?
1. A) 95 – 2x
2. B) 165 – 2x
3. C) 175 – 2x
4. D) 185 – 2x
5
In the figure above, what is the distance between the midpoints (not shown) of the two line segments?
1. A)
2. B)
3. C)
4. D)
6
What is the perimeter of an equilateral triangle inscribed in a circle with circumference 24π?
1. A)
2. B)
3. C)
4. D)
Exercise Set 1: Geometry (Calculator)
Questions 7–9 are based on the figure below.
Note: Figure not drawn to scale.
7
In the figure above, what is the perimeter of quadrilateral ABCO , to the nearest integer?
8
In the figure above, what is the area, in square units, of ABCO ?
9
In the figure above, point K (not shown) is the midpoint of , and point M (not shown) is the midpoint of . What is the slope of ?
10
In the xy -plane, point H has coordinates (2, 1) and point J has coordinates (11, 13). If is parallel to the x -axis and is parallel to the y -axis, what it the perimeter of triangle HJK ?
11
Note: Figure not drawn to scale.
In the figure above, what is the value of x ?
1. A)
2. B)
3. C)
4. D)
Questions 12–15 are based on the situation described below.
In the xy -plane, ABCD is a square. Point A has coordinates (–1, 2) and point B has coordinates (3, 5).
12
Which of the following could be the coordinates of C ?
1. A) (0, 9)
2. B) (6, 0)
3. C) (2, –2)
4. D) (–4, 6)
13
What is the area of square ABCD ?
1. A) 25
2. B) 28
3. C) 30
4. D) 32
14
What is the slope of ?
1. A)
2. B)
3. C)
4. D)
15
What is the distance between C and the midpoint of ?
1. A)
2. B)
3. C)
4. D)
EXERCISE SET 1: GEOMETRY ANSWER KEY
No Calculator
1 . 16/9 or 1.77 or 1.78 If we define x as the length of , then the length of one side of the square is 2x , and so the area of square MNOP is (2x )(2x ) = 4x 2 . To find this value, we can apply the Pythagorean Theorem to right triangle QNO :
Simplify:
Divide by 5:
Multiply by 4:
2 . 133 The key is to notice simple relationships between angles until we get around to x .
3 . 210 Draw three lines as shown:
Since the polygon divides into 3 triangles, the sum of its internal angles is (3)(180°) = 540°. Therefore a + b + c + 240 + 90 = 540, and so a + b + c = 210.
4 . C Using the Crossed Lines Theorem and the Parallel Lines Theorem, we can mark up the diagram like this:
This shows that x + y + x + 5 = 180, and so y = 175 – 2x .
5 . B The midpoint of the top segment is , and the midpoint of the bottom segment is , therefore, the distance between them is
6 . C To solve this problem we must draw a diagram and find the relationship between the radius of the circle and the sides of the triangle. By the Isosceles Triangle Theorem, if all three sides of a triangle are congruent, then all three angles must be congruent. Since these angles also must have a sum of 180°, they must each be 60°. If we draw the bisectors of each of these angles, we divide the triangle into six smaller triangles. These smaller triangles are congruent 30°-60°-90 ° triangles, as shown here:
Since the circumference of the circle (2πr ) is 24π, its radius is 12. Since each of the hypotenuses of our right triangles is also a radius of the circle, we can find all of the sides of these triangles using the 30°-60°-90° relationships. Each side of the equilateral triangle is therefore , and its perimeter is therefore .
Calculator
7 . 43 Using the distance formula, we can calculate the lengths of each segment. , , and . Therefore, the perimeter is approximately 15.30 + 7.61 + 10 + 10.20 = 43.11, which rounds to 43.
8 . 107 Since we do not have a formula that directly calculates the area of such an odd-shaped quadrilateral, we must analyze its area in terms of simpler shapes. The simplest way to do this is by drawing a box around it. This turns the area of interest into a rectangle minus three right triangles, all of which have areas that can be easily calculated.
9 . 6/5 or 1.2 The midpoint of is (1.5, 7.5) and the midpoint of is (6.5, 13.5); therefore, the slope of the segment between them is 6/5.
10 . 36 If point K is on the same horizontal line as (2, 1), it must have a y -coordinate of 1, and if it is on the same vertical line as (11, 13), it must have an x -coordinate of 11. Therefore, K is the point (11, 1), and so HK = 9, JK = 12, and . Notice that it is a 3-4-5 triangle!
11 . C Since the sum of the interior angles of any triangle is 180°, y + y + 2y = 4y = 180, and therefore y = 45. Therefore, this is a 45°-45°-90° right triangle. Since two angles are equal, the two opposite sides must also be equal, so 3m = 2m + 5 and so m = 5 and the two legs each have measure 15. Using the Pythagorean Theorem or the 45°-45°-90° shortcut, we can see that .
12 . A The key to questions 12 through 15 is a good diagram in the xy -plane that represents the given information:
If ABCD is a square, then the points A , B , C , and D must appear in that order around the square. Notice that to get from point A to point B , we must move 4 units to the right and 3 units up. This means that, in order to get to point C along a perpendicular of the same length, we must go either 3 units right and 4 units down , or 3 units left and 4 units up . This puts us either at (6, 1) or (0, 9).
13 . A The diagram shows that AB is the length of the hypotenuse of a right triangle with legs 3 and 4. You should recognize this as the special 3-4-5 right triangle. If AB = 5, then the area of the square is 52 = 25.
14 . A Notice that the slope of is the same regardless of which option we choose for C . In either case, the slope formula tells us that the slope is −4/3.
15 . D The midpoint of (point M above) is (1, 3.5). We can use the distance formula to find the distance between this point and either of the possible locations of C . (Notice that the distance is the same either way.) Alternately, we might notice that MC is the hypotenuse of a right triangle with legs 5 and 2.5. Either way, we get a value of .
Lesson 5: Circles
Which of the following equations represents a circle in the xy -plane that passes through the point (1, 5) and has a center of (3, 2)?
1. A)
2. B) (x– 3)2+ (y – 2)2 = 13
3. C) (x– 1)2+ (y – 5)2 = 13
4. D) (x– 3)2+ (y – 2)2 = 25
Equations of Circles
(xh )2 + (yk )2 = r 2
A circle is defined as the set of all points in a plane that are a fixed distance, r , from a fixed point, (h , k ). The distance r is called the radius and (h , k ) is the center .
Therefore, by the Distance Formula, any point (x , y ) on the circle must satisfy the equation
or (xh )2 + (yk )2 = r 2
(Easy ) Since our circle has a center at (3, 2), its equation must have the form (x – 3)2 + (y – 2)2 = r 2 , which eliminates choice (C). To find r , the radius, we simply have to find the distance between the center and any point on the circle. By the distance formula, this is , and therefore, . The correct answer is (B). If you chose (A), keep in mind that the equation for a circle has r 2 on the right side, not r .
What is the area, in square centimeters, of a circle with circumference of 16π centimeters?
1. A) 8π
2. B) 16π
3. C) 32π
4. D) 64π
(Easy ) If the circumference of the
circle is 16π centimeters:
r = 16π
Divide by 2π:
r = 8
Therefore, by the area formula:
Area = πr 2 = π(8)2 = 64π
So the correct answer is (D).
Circumference of a Circle
circumference = π d = r
The number π (≈ 3.14159 …) is defined as the number of diameters it takes to get around a circle. Put another way, π is the ratio of the circumference of any circle to its diameter. Since any diameter is twice the radius,
circumference = 2πr
Area of a Circle
A = πr 2
If we cut any circle into tiny enough sectors, and reassemble them as shown below, we can create a parallelogram-like shape that has a height of r and a length that is half of the circumference, or π r .
Since the area of any parallelogram is equal to its base times its height, the area of a circle is (πr )(r ) = πr 2 .
Tangents to Circles
A tangent line to a curve is a line that touches the curve without crossing it. A tangent line to any circle is perpendicular to the radius at the point of tangency.
In the figure above, is tangent to the circle at point P , and . If the circle has an area of 100π, what is the area of triangle MOQ ?
(Hard ) The first thing we should do is draw radius . Since this is a radius to the point of tangency, it is perpendicular to the tangent. We should also write in the given measures.
The area of the circle is 100π:
π(OP )2 = 100π
Divide by π:
(OP )2 = 100
Take square root:
OP = 10
Notice that is the height of triangle MOQ if is taken as its base. If we can find the length of base , we can simply use the triangle angle formula . To find MQ , we can use the Pythagorean Theorem to find MPand PQ and just add them together.
Pythagorean Theorem for triangle OPM :
Pythagorean Theorem for triangle OPQ :
Simplify:
100 + (MP )2 = 269
100 + (PQ )2 = 244
Subtract 100:
(MP )2 = 169
(PQ )2 = 144
Take square root:
MP = 13
PQ = 12
Therefore MQ =
MP + PQ =
13 + 12 = 25, so:
Area of
Lesson 6: Radians, chords, arcs, and sectors
What is the degree measure of an angle that measures 4.5 radians?
1. A) 4.5π°
2. B)
3. C)
4. D)
(Medium ) Although many students will get this question wrong, it is very simple if you know how to convert radians to degrees. All we need to do to convert any radian measure to a degree measure is to multiply it by the conversion factor (as explained below). Therefore, 4.5 radians = 4.5 radians and the correct answer is (C).
A radian is simply the radius of a circle used as a “measuring stick” for an arc on the circle and for its corresponding central angle.
Because circumference = 2πr , a full rotation of 360° equals 2π radians, and 180° equals π radians.
Therefore, we may use as a conversion factor to convert a degree measure to radians, and as a conversion factor to convert a radian measure to degrees.
Take some time to memorize the radian measures of the common degree measures above. Put them on flashcards and study them until you”ve mastered them.
The circle above has an area of 100π square centimeters. If chord is 8 centimeters long, how far, in centimeters, is from the center of the circle?
1. A) 6
2. B) 8
3. C)
4. D)
(Medium ) First, let”s draw three extra line segments:
Since πr 2 = 100π, r = 10. If we draw a perpendicular from the center to the chord, the length of this segment is the distance from the center to the chord. This segment also bisects the chord, dividing it into two equal segments of 4 centimeters each. This allows us to use the Pythagorean Theorem to find this distance:
42 + x 2 = 102
Simplify:
16 + x 2 = 100
Subtract 16:
x 2 = 84
Take square root:
Therefore, the correct answer is (D).
Chords
A chord is any line segment that connects two points on a circle. The longest chord in a circle is its diameter , which passes through the center.
The perpendicular segment from the center of the circle to a chord always bisects that chord.
Note: Figure not drawn to scale.
In the figure above, AC is a diameter of the circle with center O , OB = 12, and the length of arc AB is 7π. What is the value of x ?
1. A) 60
2. B) 72
3. C) 75
4. D) 78
(Medium ) Since the circle has a radius of 12, its circumference is 2π(12) = 24π. Since AC is a diameter, then the measure of arc AC is half the circumference, or 12π. If the length of arc AB is 7π, then the length of arc BC is 12π – 7π = 5π. Since the central angle of x ° is the same fraction of 360° as its arc BC is to the
entire circumference,
Cross multiply:
24πx = 1,800π
Divide by 24π:
x = 75
Therefore, the correct answer is (C).
Arcs and Sectors
An arc is a continuous part of the circumference of a circle. Every arc has a corresponding central angle. The ratio of an arc length to the circumference is equal to the ratio of its central angle to 360 ° (or, in radians, 2π).
A sector is a part of the interior of a circle bounded by an arc and two radii. The ratio of a sector area to the area of the circle is equal to the ratio of its central angle to 360 ° (or, in radians, 2π).
Note: Figure not drawn to scale.
In the figure above, AC is a diameter of the circle with center O , OB = 7, and the measure of ACB is 20°. What is the area of the shaded sector?
1. A)
2. B)
3. C)
4. D)
(Medium -Hard ) Since , , and are all radii, triangles AOB and BOC are isosceles. Therefore, we can analyze the diagram with the Isosceles Triangle Theorem and the Angle Sum Theorem:
Since the central angle of the sector is 40°, the area of the sector is 40°/360° = 1/9 the area of the circle. Since the area of the circle is π(7)2 = 49π, the area of the sector is 49π/9 square units. Therefore, the correct answer is (D).
Lesson 7: Areas and volumes
Reference Information
Every SAT Math section will include the following Reference Information . Take some time to familiarize yourself with these area and volume formulas.
The number of degrees of arc in a circle is 360.
The number of radians of arc in a circle is 2π.
The sum of the measures in degrees of the angles of a triangle is 180.
The figure above shows a wooden cylindrical tube with a length of 10 centimeters and a diameter of 4 centimeters with a cylindrical hole with a diameter of 2 centimeters that extends 40% of the length of the tube. The tube is closed on the end opposite to the hole. The density of the wood is 4.2 grams per cubic centimeter. What is the mass of this tube, to the nearest gram? (Recall that mass = density × volume )
1. A) 151 grams
2. B) 343 grams
3. C) 468 grams
4. D) 475 grams
(Medium ) To find the mass of the tube, we must multiply its density by its volume. To find its volume, we must subtract the volume of cylindrical hole from the volume of the wooden cylinder. The large cylinder has a radius of 2 (remember, the diameter is 4 and so the radius is 4 ÷ 2 = 2) and a length of 10, so its volume is π(2)2 (10) = 40π. The cylindrical hole has a radius of 1 (because its diameter is 2) and a length of (0.40)(10) = 4, so the volume of the hole is π(1)2 (4) = 4π. Therefore the total volume of the closed tube is 40π – 4π = 36π ≈ 113.1. Since the mass is equal to the volume times the density, its mass is (113.1)(4.2) = 475 grams, so the correct answer is (D).
What is the area, in square units, of quadrilateral ABCD above?
(Medium ) You might remember this figure from Lesson 4, in which we found its perimeter. Now we are asked to find its area. Unfortunately, we are not given any formula for calculating the area of this kind of quadrilateral. (You might remember that its technical name is a trapezoid , but in fact we don”t need to know anything special about trapezoids to solve this problem.) In such situations, it helps to remember the Strange Area Rule .
Strange Area Rule
When asked to find a “strange area,” that is, the area of a region for which you do not have a simple formula, try to analyze the region into the sum or the difference of simpler shapes .
In this case, we can look at this area in two different ways: as a rectangle plus two right triangles, or as a bigger rectangle minus two right triangles:
We should get the same result from either method. With the first method, the area of the trapezoid is the area of the rectangle plus the areas of two right triangles. This gives us a total area of (12)(13) + (1/2)(5)(12) + (1/2)(9)(12) = 156 + 30 + 54 = 240. With the second method, the area of the trapezoid is the area of the large rectangle minus the areas of the two right triangles. This gives us a total area of (12)(27) – (1/2)(5)(12) – (1/2)(9)(12) = 324 – 30 – 54 = 240. Bingo!
Lesson 8: Similar figures
In the figure above, and are line segments that intersect at point P . What is the value of m ?
(Medium ) The key to this question is noticing that the two triangles are similar . That is, that they are the same shape, although they may be different sizes.
Similarity
In geometry, similar really just means the same shape . Two figures are similar if and only if all pairs of corresponding angles are congruent, and all pairs of corresponding sides are proportional.
Always be on the lookout for similar triangles in SAT Math diagrams!
AA (Angle-Angle) Similarity Theorem
If two triangles have two pairs of congruent corresponding angles, then (1) the remaining pair of corresponding angles must be congruent, and (2) the triangles must be similar.
SAS (Side-Angle-Side) Similarity Theorem
If two triangles have two pairs of proportional corresponding sides, and the corresponding angles between those sides are also congruent, then the triangles must be similar.
Perimeters, Areas, and Volumes of Similar Figures
If two polygons are similar with corresponding sides in ratio of a :b , then the corresponding perimeters of those figures have a ratio of a :b and their corresponding areas have a ratio of a 2 :b 2 . If two solids have corresponding lengths in a ratio of a :b , then their volumes have a ratio of a 3 :b 3 .
Coming back to our diagram, if and intersect at point P , the Crossed Lines Theorem tells us that ∠APC and ∠BPD must be congruent, and so, by the AA Theorem, ΔAPC ∼ ΔBPD . Therefore, the corresponding sides are proportional:
Cross multiply:
8m = 70
Divide by 8:
m = 70/8 = 8.75
Exercise Set 2: Geometry (No Calculator)
1
A cereal company sells oatmeal in two sizes of cylindrical containers. The radius of the larger container is twice that of the smaller, and the height of the larger container is 50% greater than the smaller. If the smaller container holds 10 ounces of oatmeal, how many ounces can the larger container hold?
2
Note: Figure not drawn to scale.
In the figure above, is tangent to both circles, which are tangent to each other. If the smaller circle has a circumference of 4π and the larger circle has a circumference of 4π, what is the length of ?
3
What is the area, in square inches, of a circle with diameter 6π2 inches?
1. A) 9π4
2. B) 9π5
3. C) 36π4
4. D) 36π5
4
What is the length of the longest line segment that connects two vertices of a rectangular box that is 6 units wide, 4 units long, and 2 units tall?
1. A)
2. B)
3. C)
4. D)
5
Which of the following equations represents a circle in the xy -plane that intersects the x -axis at (3, 0) and (9, 0)?
1. A) (x– 6)2+ (y – 4)2 = 25
2. B) (x– 3)2+ (y – 9)2 = 25
3. C) (x– 6)2+ (y – 4)2 = 36
4. D) (x– 3)2+ (y – 9)2 = 36
6
In the figure above, P and N are the centers of the circles and PN = 6. What is the area of the shaded region?
1. A)
2. B)
3. C)
4. D)
7
The diagram above shows a hexagon with all sides congruent and all angles congruent. What is the value of k ?
1. A)
2. B)
3. C)
4. D)
Exercise Set 2: Geometry (Calculator)
8
What is the area, in square units, of the quadrilateral above?
9
What is the degree measure, to the nearest whole degree, of an angle that measures 5.6 radians?
10
In the figure above, arc has a length of 11.5. To the nearest integer, what is the value of x ?
11
The Great Pyramid in Giza, Egypt, has a height of 140 meters and a volume of 2.6 million cubic meters. If a scale model of the Great Pyramid is to be built that is 2 meters high, what will be the volume, in cubic meters, of this model?
12
Which of the following equations defines a circle that is tangent to the y -axis?
1. A) (x– 2)2+ (y + 3)2 = 2
2. B) (x– 2)2+ (y + 3)2 = 3
3. C) (x– 2)2+ (y + 3)2 = 4
4. D) (x– 2)2+ (y + 3)2 = 9
Questions 13 and 14 refer to the diagram below.
The figure above shows two rectangles that share a common vertex, and is a line segment that passes through C .
13
What is the ratio of the area of rectangle ABCD to the area of rectangle AEFG ?
1. A) 3:5
2. B) 9:25
3. C) 5:8
4. D) 25:64
14
If CD = 9, what is the perimeter of rectangle AEFG ?
1. A) 67.2
2. B) 72.6
3. C) 76.2
4. D) 78.6
15
Point O is the center of the circle above. What is the area of the shaded region?
1. A)
2. B)
3. C)
4. D)
EXERCISE SET 2: GEOMETRY ANSWER KEY
No Calculator
1 . 60 If the smaller cylinder has a radius of r and a height of h , its volume is πr 2 h . The larger cylinder therefore must have a radius of 2r and a height of 1.5h , and a volume of π(2r )2 (1.5h ) = 6πr 2 h . Since this is 6 times the volume of the smaller cylinder, it must hold 10 × 6 = 60 ounces of oatmeal.
2 . 8 First, let”s draw the radii to the points of tangency, the segment joining the centers, and the segment from the center of the smaller circle that is perpendicular to the radius of the larger circle. Since the tangent segment is perpendicular to the radii, these segments form a rectangle and a right triangle.
Since the circumference of the smaller circle is 4π, its radius is 2, and since the circumference of the larger circle is 16π, its radius is 8. The hypotenuse of the right triangle is the sum of the two radii: 2 + 8 = 10. One of the legs of the right triangle is the difference of the two radii: 8 – 2 = 6.
Pythagorean Theorem:
x 2 + 62 = 102
Simplify:
x 2 + 36 = 100
Subtract 36:
x 2 = 64
Take square root:
x = 8
3 . B Diameter = 2r :
2r = 6π2
Divide by 2:
r = 3π2
Area formula:
πr 2 = π(3π2 )2
Simplify:
πr 2 = π(9π4 )
Simplify:
πr 2 = 9π5
4 . C From the 3-D Distance Formula back in Lesson 4, the length of the diagonal is
5 . A All of the equations are clearly equations of circles, so our only task is to verify that one of these equations is satisfied by both point (3, 0) and point (9, 0). Simply by plugging these coordinates into the equations, we can verify that only the equation in (A) is true for both points: (3 – 6)2 + (0 – 4)2 = 25 and (9 – 6)2 + (0 – 4)2 = 25.
6 . C In this problem, we have to take advantage of the Strange Area Rule from Lesson 7. First we should draw the segments from P and N to the points of intersection. Since each of these segments is a radius, they have equal measure (6), and form two equilateral 60°-60°60 ° triangles.
The shaded region is composed of two circle “segments,” each of which is a sector minus a triangle, as shown in the figure above. The sector, since it has a 120° central angle, has an area 1/3 of the whole circle, or (1/3) (π(6)2 ) = 12π and the triangle has area . Therefore, the shaded region has an area of .
7 . B Each side of the hexagon has length 9 – 3 = 6. Each interior angle of a regular hexagon has measure (6 – 2)(180°)/6 = 120°, so the segments shown form two 30°-60°-90° triangles with lengths shown below.
Therefore, .
Calculator
8 . 142 First, let”s draw a rectangle around the figure as shown.
This shows that the area we want is the area of the rectangle minus the areas of the three triangles: (16)(14) – (1/2)(2)(16) – (1/2)(12)(6) – (1/2)(10)(6) = 224 – 16 – 36 – 30 = 142.
9 . 321 To convert any angle from radians to degrees, we just multiply by the conversion factor (180°)/(π radians). 5.6 × 180°/π = 320.86 ≈ 321°.
10 . 66 In a circle with radius 10, and arc of length 11.5 has a radian measure of 11.5/10 = 1.15 radians. In degrees, this equals 1.15 × 180°/π = 65.89°≈ 66°.
11 . 7.58 If two similar solids have sides in ratio of a:b , then their volumes are in a ratio of a 3 :b 3 . The ratio of the heights is 140:2 = 70:1, so the ratio of volumes is 703 :13 = 343,000:1. This means that the volume of the model is 2,600,000 ÷ 343,000 ≈ 7.58 cubic meters.
12 . C As a quick sketch will verify, in order for a circle to be tangent to the y -axis, its radius must equal the absolute value of the x -coordinate of its center. Since the center of each square is (2, –3), the radius must be 2. The only circle with a radius of 2 is (C).
13 . D By the AA Theorem, triangle ACD is similar to triangle AFG , and so rectangle ABCD is similar to rectangle AEFG . The ratio of the corresponding sides is equal to the ratio of their diagonals, which is 15:24 = 5:8. Therefore, the ratio of their areas is 52 :82 = 25:64
14 . A If CD = 9, we can find AD by the Pythagorean Theorem. (AD )2 + (CD )2 = (AC )2
Substitute:
(AD )2 + 92 = 152
Simplify:
(AD )2 + 81 = 225
Subtract 81:
Take square root:
This means that the perimeter of ABCD is 12 + 9 + 12 + 9 = 42. Since the ratio of the perimeters of similar figures equals the ratio of corresponding sides,
Cross multiply:
5p = 336
Divide by 5:
p = 67.2
15 . A The two radii and the chord form an isosceles triangle.
x + x + 4x = 180
Simplify:
6x = 180
Divide by 6:
x = 30
Therefore, the diagram should look like this:
As we saw in question 6, this portion of the circle is called a “segment,” and we find its area by taking the area of the sector minus the area of the triangle. The sector has area (120/360)(π82 ) = 64π/3, and the triangle has area , so the segment has an area of .
Skill 2: Understanding Basic Trigonometry
Lesson 9: The basic trigonometric functions
Which of these is equivalent to (No calculator)
1. A)
2. B)
3. C)
4. D)
The Basic Trigonometric Functions
Any of the three basic trigonometric functions, like all functions, takes an input number and transforms it into an output number. A trigonometric function takes an angle, θ , as an input and constructs a right triangle with θ as one of its acute angles. The output is then the ratio of two sides of that triangle as defined by the mnemonic SOH-CAH-TOA.
But these definitions are limited, because they only work when θ is an acute angle. What if it”s a larger angle, like 135°, or even a negative angle, like –20°? To find the trigonometric ratios for these angles, we use the unit circle.
The Unit Circle
The unit circle is just a circle with radius 1 centered at the origin of the x y-plane. When using the unit circle, all angles must be in standard position, that is, with vertex at the origin and measured counterclockwise from the positive x -axis (just like the angle θ in the following figure).
When an angle, θ , is in standard position, its terminal ray intersects the unit circle in the point (x 1 , y 1 ). If we drop a vertical line segment from this point to the x -axis, we form a right triangle with legs of length x 1 and y 1 and a hypotenuse of length 1 (as shown above).
So now let”s go back to the definitions of the basic trigonometric functions. In terms of this right triangle, what are the sine, cosine, and tangent of θ ?
This suggests three important theorems:
• Thesine of any angle is the y -coordinate of its corresponding point on the unit circle.
• Thecosine of any angle is the x -coordinate of its corresponding point on the unit circle.
• Thetangent of any angle is the ratio of the y -coordinate to the x -coordinate of its corresponding point on the unit circle.
(Medium-hard ) First, you may find it useful to convert the angles to degree measures using the conversion factor (180°/π radians). This gives us π/4 radians = 45° and π/6 radians = 30°. We should recognize these as angles in two Special Right Triangles:
Using the definitions for sine and cosine above, these triangles show us that and sin . Therefore, and the correct answer is (C).
If sin x = a and sin z = −a , where x and z are in radians, and , which of the following could be the value of z in terms of x ?
1. A) π –x
2. B) x– π
3. C) 2π +x
4. D)
(Hard ) The statement indicates that x is an angle in quadrant II, where the sine (the y -value of the points on the unit circle) is positive. Let”s draw this situation on the unit circle so we can visualize it. (We don”t want to confuse the angles called x and y in the problem with the x -coordinates and y -coordinates in the xy -plane. For this reason, let”s label the terminal rays for the angles “angle x ” and “angle z .”) Recall that the sine of any angle is the y-coordinate of the point on the unit circle that corresponds to that angle. If sin x = a , then a is the y -coordinate of the point on the unit circle that corresponds to “angle x ,” as shown in the diagram. If sin z = –a , then –a is the y -coordinate of the point on the unit circle that corresponds to “angle z .” There are two possible locations for “angle z ” as shown in the diagram.
At this point, it may be easiest to simply pick a value for “angle x ” that is between π/2 (≈ 1.57) and π (≈ 3.14), such as x = 2. Since sin 2 ≈ 0.909 (remember to put your calculator into “radian mode”), a = 0.909. Now we just need to find which angle among the choices has a sine of − 0.909
1. A) sin(π –2) = 0.909
2. B) sin(2 – π) = –0.909
3. C) sin(2π + π) = 0.0909
4. D) sin(2 – π/2) = 0.416
Therefore, the correct answer is (B).
The Pythagorean Identity
sin2 x + cos2 x = 1 for all values of x
An identity is an algebraic equation that is true for all values of the unknown, and not just for particular values. We can prove the Pythagorean Identity by just applying the Pythagorean Theorem to our right triangle and thinking about the trigonometric ratios.
Apply Pythagorean Theorem:
Divide by (hypotenuse)2 :
Simplify using trig definitions:
sin2 θ + cos2 θ = 1
If b is an angle measure such that sin , what is the value of (sin b – cos b )2 ?
1. A)
2. B)
3. C)
4. D)
(Medium ) The expression we are trying to evaluate includes squared trigonometric ratios, so we will probably have to take advantage of the Pythagorean Identity.
Expression to evaluate:
(sin b – cos b )2
FOIL:
sin2 b – 2sin b cos b + cos2 b
Rearrange using Commutative Law of Addition:
sin2 b + cos2 b – 2sin b cos b
Substitute sin2 b + cos2 b = 1:
1 – 2sin b cos b
Now we”ll have to find the value of sin b cos b , which we can find with the given equation.
Given equation:
Multiply by cos b :
Substitute sin b cos b =
into original expression:
1 – 2sin b cosb
So the correct answer is (A).
Lesson 10: The trigonometry of complementary angles
If sin and , which of the following is equal to sin
1. A)
2. B)
3. C)
4. D)
Trigonometry of Complementary Angles
The two acute angles in a right triangle are complements of one another, that is, they have a sum of 90° (or, in radians, ). So, if one of the angles has a radian measure of x , the other has a measure of – x .
If we look at the trigonometric ratios for this new angle, we see that these ratios are related to the trigonometric ratios of its complement by the following rule: The trigonometric ratio of any angle equals the cofunction of its complement .
(Medium-hard ) Let”s start by drawing a picture of this situation. Since y is the measure of an acute angle, we can imagine it as the interior angle of a right triangle. Since its sine is equal to a/b , we can say that the opposite side has measure a and the hypotenuse has measure b .
Now we can find the length of the remaining leg (let”s call it k ) in terms of a and b using the Pythagorean
Theorem:
k 2 + a 2 = b 2
Subtract a 2 :
k 2 = b 2a 2
Take the square root:
Also, we know that the other acute angle has a measure of , so let”s complete the picture:
Now, finding the value of is just a matter of using the definition of sine: SOH.
So the correct answer is (D).
Exercise Set 3: Trigonometry (No Calculator)
1
What is the greatest possible value of f if
2
If cos , what is the value of
3
If (sin x – cos x )2 = 0.83, what is the value of (sin x + cos x )2 ?
4
Which of the following is equivalent to
1. A)
2. B)
3. C)
4. D) 1
5
If sin >θ < 0 and sin θ cos θ < 0, then θ must be in which quadrant of the figure above?
1. A) I
2. B) II
3. C) III
4. D) IV
6
If sin and , which of the following expressions is equal to
1. A)
2. B)
3. C) 1 − sin2x
4. D)
7
If sin b = a , which of the following could be the value of cos (b + π)?
1. A)
2. B) a2– 1
3. C)
4. D) 1 –a2
8
If and , what is the value of cos x ?
1. A)
2. B)
3. C)
4. D)
EXERCISE SET 3: TRIGONOMETRY ANSWER KEY
No Calculator
1 . 7/2 or 3.5 The discussion in Lesson 9 about the definition of the sine function and the unit circle made it clear that the value of the sine function ranges from –1 to 1. Therefore, the maximum value of is or 3.5.
2 . 1/36 or .027 or .028 An radian measure of π/3 is equivalent to 60°. If you haven”t memorized the fact that cos(60 °) = ½, you can derive it from the Reference Information at the beginning of every SAT Math section, which includes the 30°-60°-90° special right triangle. Since a = ½, (a /3)2 = (1/6)2 = 1/36.
3 . 1.17
(sin x – cos x )2 = 0.83
FOIL:
sin2 x – 2sin x cos x + cos2 x = 0.83
Regroup:
sin2 x + cos2 x – 2sin x cos x = 0.83
Simplify:
1 – 2sin x cos x = 0.83
Subtract 1:
–2sin x cos x = –0.17
Multiply by –1:
2sin x cos x = 0.17
Evaluate this expression:
(sin x + cos x )2
FOIL:
sin2 x + 2sin x cos x + cos2 x
Regroup:
sin2 x + cos2 x + 2sin x cos x
Substitute:
1 + 0.17 = 1.17
4 . D sin(π/6) = ½ and cos(π/3) = ½, so sin(π/6)/cos(π/3) = 1.
5 . D If sin θ < 0, then θ must be either in quadrant III or in quadrant IV. (Remember that sine corresponds to the y -coordinates on the unit circle, so it is negative in those quadrants where the y -coordinates are negative.) If sin θcos θ < 0, then cos θ must be positive (because a negative times a positive is a negative). Since cos θ is only positive in quadrants I and IV (because cosine corresponds to the x -coordinates on the unit circle), θ must be in quadrant IV
6 . B First, notice that a/b and b/a are reciprocals. Next, we can use the identity in Lesson 10 that to see that choice (B) is just the reciprocal of sin x . Alternately, we can just choose a value of x , like x = 1, and evaluate sin 1 = 0.841. The correct answer is the expression that gives a value equal to the reciprocal of 0.841, which is 1/0.841 = 1.19. Plugging in x = 1 gives (A) 0.841, (B) 1.19, (C) 0.292, (D) 0.540.
7 . C Recall from the Pythagorean Identity that . Substituting sin b = a gives . The angle b + π is the reflection of angle b through the origin, so cos(b + π) is the opposite of cos b , which means that .
8 . D Recall from the Pythagorean Identity that cos2 x = 1 − sin2 x
Substitute cos2 x = 1 – sin2 x :
Cancel common factor:
Reciprocate:
Skill 3: Understanding Complex Numbers
Lesson 11: Understanding the imaginary number i and the complex plane
Imaginary Numbers
The imaginary number i is defined as the principal square root of –1.
The square root of a negative number is not on the real number line, because the square of a real number cannot be negative. Therefore, the square roots of negative numbers must reside on their own number line, which we call the “imaginary axis” , which is perpendicular to the real axis, intersecting it at the origin. The plane defined by the real axis and the imaginary axis is called the complex plane .
Given that , which of the following is equal to
1. A) i
2. B) −i
3. C) 1
4. D) −1
(Medium-hard ) To answer this question, we just need to know the basic exponent rules and the definition of i .
Original expression:
Factor:
Substitute i 2 = –1
Multiply by i/i :
Substitute i 2 = –1:
–(–i ) = i
Therefore, the correct answer is (A).
Lesson 12: Adding, multiplying, dividing, and simplifying complex numbers
Complex Numbers
The sum of a real number and an imaginary number is called a complex number . All complex numbers can be expressed in the form a + bi where a and b are real numbers and .
Every complex number a + bi corresponds to the point (a , b ) on the complex plane.
To add complex numbers, just combine “like” terms.
Original expression:
(3 – 2i ) + (2 + 6i )
Regroup with Commutative and Associative Laws of Addition:
(3 + 2) + (–2i + 6i )
Simplify:
5 + 4i
Multiplying Complex Numbers
To multiply complex numbers, just FOIL and combine like terms.
Original expression:
(3 – 2i )(2 + 6i )
FOIL:
(3)(2) + (3)(6i ) + (–2i )(2) + (–2i )(6i )
Simplify:
6 + 18i – 4i –12i 2
Substitute i 2 = –1:
6 + 18i – 4i – 12(–1)
Combine like terms:
18 + 14i
Dividing Complex Numbers
To divide complex numbers, express the quotient as a fraction, multiply numerator and denominator by the complex conjugate of the denominator, and simplify. The complex conjugate of a + bi is abi .
Original expression:
Multiply numerator and
denominator
by conjugate of denominator:
FOIL:
Simplify:
Distribute division:
Powers of i
The successive powers of i (i 1 , i 2 , i 3 , i 4 , i 5 , i 6 , i 7 . . .) cycle counterclockwise around the unit circle in the complex plane.
We can verify this by expanding any positive integer power of i .
Expression to be evaluated:
i 13
Expand:
(i )(i )(i )(i )(i )(i )(i )(i )(i )(i )(i )(i )(i )
Group in pairs:
(i × i )(i × i )(i × i )(i × i )(i × i )(i × i )(i )
Substitute i 2 = –1:
(–1)(–1)(–1)(–1)(–1)(–1)(i )
Simplify:
i
This implies that i n = 1 if n is a multiple of 4. (That is, i 4 = 1, i 8 = 1, i 12 = 1, etc.) This gives us a convenient way to simplify large powers of i : just replace the exponent with the remainder when it is divided by 4 . For instance, i 39 = i 3 = –i , because 3 is the remainder when 39 is divided by 4.
If , where , which of the following is equal to K 2 ?
1. A) 2i
2. B) 4i
3. C) 4 +i
4. D) 4
Therefore, the correct answer is (D).
Which of the following is NOT equal to i 6i 2 ?
1. A) i5i
2. B) i4
3. C) 2i3+ 2i
4. D) 1 +i6
(Medium ) Here, we have to use our knowledge about powers of i . Since i 6 = (i × i )(i × i )(i × i ) = (–1)(–1)(–1) = –1, and i 2 = –1, the given expression, i 6i 2 , is equal to (–1) – (–1) = 0. Simplifying each choice gives us
1. A) i5i = ii = 0
2. B) i4= 1
3. C) 2i3+ 2i = –2i + 2i = 0
4. D) 1 +i6 = 1 + (–1) = 0
Therefore, the correct answer is (B).
Exercise Set 4: Complex Numbers (No Calculator)
1
If a + bi = (1 + 2i )(3 – 4i ), where a and b are constants and , what is the value of a + b ?
2
If , where a and b are constants and , what is the value of a ?
3
For what value of b does (b + i )2 = 80 + 18i ?
4
The solutions of the equation x 2 – 2x + 15 = 0 are and , where a and b are positive numbers. What is the value of a + b ?
5
Given that , which of the following is equal to
1. A)
2. B)
3. C)
4. D)
6
Which of the following expressions is equal to (2 + 2i )2 ?
1. A) 0
2. B) 4i
3. C) 8i
4. D) 4 – 4i
7
If B (3 + i ) = 3 – i , what is the value of B ?
1. A)
2. B)
3. C)
4. D)
8
x 2 + kx = –6
If one of the solutions to the equation above is , what is the value of k ?
1. A) –4
2. B) –2
3. C) 2
4. D) 4
9
If i m = –i , which of the following CANNOT be the value of m ?
1. A) 15
2. B) 18
3. C) 19
4. D) 27
EXERCISE SET 4: COMPLEX NUMBERS ANSWER KEY
No Calculator
1 . 13
(1 + 2i )(3 – 4i )
FOIL:
(1)(3) + (1)(–4i ) + (2i )(3) + (2i )(–4i )
Simplify:
3 – 4i + 6i – 8i 2
Substitute i 2 = –1:
3 – 4i + 6i –8(–1)
Combine like terms:
11 + 2i
Therefore, a = 11 and b = 2, so a + b = 13.
2 . 7/5 or 1.4
Multiply conjugate:
FOIL:
Substitute i 2 = –1:
Combine like terms:
Distribute division:
3 . 9
(b + i )2
FOIL:
(b + i )(b + i ) = b 2 + bi + bi + i 2
Substitute i 2 = –1:
b 2 + bi + bi – 1
Combine like terms:
(b 2 – 1) + 2bi
Since this must equal 80 + 18i , we can find b by solving either b 2 – 1 = 80 or 2b = 18. The solution to both equations is b = 9.
4 . 15 The equation we are given is a quadratic equation in which a = 1, b = –2, and c = 15. Therefore, we can use the quadratic formula:
Substitute:
Simplify:
Simplify:
Distribute division:
Therefore, a = 1 and b = 14, so a + b = 15.
5 . B
FOIL:
Substitute i 2 = –1:
Simplify:
Multiply by i/i :
Substitute i 2 = -1:
6 . C
(2 + 2i )2
FOIL:
(2 + 2i )(2 + 2i ) = 4 + 4i + 4i + 4i 2
Substitute i 2 = –1:
4 + 8i – 4 = 8i
7 . D
B (3 + i ) = 3 – i
Divide by 3 + i :
FOIL:
Substitute i 2 = –1:
Simplify:
Distribute division:
8 . B
x 2 + kx = –6
x 2 + kx + 6 = 0
Substitute :
FOIL:
Simplify:
Distribute:
Collect terms:
Therefore, both 2 + k = 0 and . Solving either equation gives k = –2.
9 . B As we discussed in Lesson 10, the powers of i are “cyclical,” and i m = –i if and only if m is 3 more than a multiple of 4. The only number among the choices that is not 3 more than a multiple of 4 is (B) 18.
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# What does l l mean in math?
What does l l mean in math? This can certainly be one of the most common questions asked by people who are trying to learn math. The question is-what does l l mean in math? _ l _ is used in an expression to represent raising any number to the power of l. For example, if a=4, then 4^l=4 X 4 X 4 X 4 X 4 = 256.
#### What does ll stand for in math?
L is the letter for the number 1 and I is the number 1. Together they represent a vector, a quantity that has both direction and magnitude. | is the notation used to represent a vector that points in the positive direction of the x-axis. There are two types of vectors. Scalar vectors are vectors that can be represented with only one number. A vector is a vector that requires two or more numbers to describe. For example, the vector with components (1, 1) is represented by ax + by + c*z.
#### What does || mean in calculus?
|| is the absolute value function. In math, we can say that if a and b are numbers, then |a| = |b|. It means that if a and b have the same sign, then |a|=|b|, and if they have different signs, then |a| ≠ |b|. Absolute value is the distance of a number from zero. |2| = 2, |-2| = 2, and |-2| ≠ -2, but -2 ≠ -2. So you can see, it is the distance of a number from zero. For example, if a = -2 and b = 2, then |a||b| = 2. If a = -2 and b = -2, then |a||b| = 4.
#### What does II mean in geometry?
Geometry refers to the various branches of mathematics that study properties of space, properties of geometric figures, their symmetry and relations between them. It includes the study of lines, circles, triangles, squares, rectangles and so on. Geometry is based on three fundamental elements: points, lines and planes. A point is the smallest, indivisible and unextendable entity that has no volume, size or length. A line is the straight edge consisting of infinite points and a plane is the flat surface which consists of infinite lines.
#### What is the L symbol in math?
It is known as Laplace operator. It is a type of differential operator and is denoted by the letter L. It is also known as Laplacian operator. It is used a lot in the study of diffusion and heat transfer. For example, if a function is continuous and differentiable in an open region of the space, then the Fourier transform of this function is related to the Laplace operator. This operator is particularly important in the study of harmonic functions on the real line, which are solutions of Laplace’s equation.
#### What does || || mean in linear algebra?
The symbol “||” is called the “orthant” symbol. It is used to denote an orthant in a normed vector space. An orthant is the set of all points that are equidistant from some fixed point and some fixed direction. The symbol “||” is frequently used in the “||x||” metric in linear algebra, which is the Euclidian norm, the square root of the sum of the squares of the elements of a vector.
#### What does double || mean in math?
In mathematics, the double pipe sign || is used to denote “or”. It is often confused with the single pipe character | (called the “vertical pipe”) which is used in computer programming languages to denote the “or” operator. The double pipe is actually the English letter P, which is the traditional abbreviation for “or”. The double pipe is used in mathematics to indicate a range of values. For example, 1 â?¦ n
#### What is upside down L?
The upside down L is a psychological trend that is common among all the innovative products of Apple Inc. It is an illusory symbol of the apple logo. The logo for Apple products is composed of a normal white apple on top of a blue box. When people see the logo, they often have a natural tendency to look for the reflection of the apple in the blue box.
However, obviously there is no apple in the box, just the color of the blue box. When people see the blue color, they tend to see a white shape instead in the same shape of the apple, but upside down. This is the illusion of the apple. This has been a very effective marketing strategy. You can see that almost everyone else is using a green box on the Apple’s iTune logo. This is because they don’t want to let the Apple’s blue color take the center of attention.
#### How do you type a reverse L?
A reverse L is a type of cursive letter, where the letter crosses over itself. I have just learned how to write this letter and thought I could help others learn how to write it as well. To write a reverse L first make a capital L, then put your pen on the bottom part of the L and move it towards the left. Keep going until you get to the top part of the L. Note that you want to keep the bottom part of the L straight, so be careful that you don’t start to make a capital I. It takes some practice to get used to writing the reverse L, but you will get good at it in no time!
#### What is this symbol called in math?
The symbol is called the definite integral, and is represented by the letter “i” with a line over it. There are many ways to define a definite integral, and most definitions fall into one of the following categories: a) The integral of a function from a to b is the limit of the Riemann sums of the function on the interval [a, b] as lengths of the subintervals approach zero; b) The integral of a function from a to b is the limit of the area of rectangles of height f(x) and width dx as lengths of the subintervals approach zero; c) The integral of a function from a to b is the limit of the net area (positive if the function is increasing, negative if the function is decreasing, and zero if the function is constant) under the curve y = f(x) between x = a and x = b as the width of rectangles considered approaches zero. The definite integral is a fundamental concept in calculus and is used for many applications in a variety of fields, including physics, engineering, economics, and finance.
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# Difference between revisions of "2015 AMC 10A Problems/Problem 22"
The following problem is from both the 2015 AMC 12A #17 and 2015 AMC 10A #22, so both problems redirect to this page.
## Problem
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256}$
## Solutions
### Solution 2
We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by $2^8 = 256$ at the end. We casework on how many people are standing.
Case $1:$ $0$ people are standing. This yields $1$ arrangement.
Case $2:$ $1$ person is standing. This yields $8$ arrangements.
Case $3:$ $2$ people are standing. This yields $\dbinom{8}{2} - 8 = 20$ arrangements, because the two people cannot be next to each other.
Case $4:$ $4$ people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding $2$ possible arrangements.
More difficult is:
Case $5:$ $3$ people are standing. First, choose the location of the first person standing ($8$ choices). Next, choose $2$ of the remaining people in the remaining $5$ legal seats to stand, amounting to $6$ arrangements considering that these two people cannot stand next to each other. However, we have to divide by $3,$ because there are $3$ ways to choose the first person given any three. This yields $\dfrac{8 \cdot 6}{3} = 16$ arrangements for Case $5.$
Alternate Case $5:$ Use complementary counting. Total number of ways to choose 3 people from 8 which is $\dbinom{8}{3}$. Sub-case $1:$ three people are next to each other which is $\dbinom{8}{1}$. Sub-case $2:$ two people are next to each other and the third person is not $\dbinom{8}{1}$ $\dbinom{4}{1}$. This yields $\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16$
Summing gives $1 + 8 + 20 + 2 + 16 = 47,$ and so our probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$.
### Solution 3
We will count how many valid standing arrangements there are counting rotations as distinct and divide by $256$ at the end. Line up all $8$ people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires $2$ spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.
If there are $4$ standing, there are ${4 \choose 4}=1$ ways to place them. For $3,$ there are ${3+2 \choose 3}=10$ ways. etc. Summing, we get ${4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34$ ways.
Now we consider that the far right person can be standing as well, so we have ${3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13$ ways
Together we have $34+13=47$, and so our probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$.
### Solution 5
We will count the number of valid arrangements and then divide by $2^8$ at the end. We proceed with casework on how many people are standing.
Case $1:$ $0$ people are standing. This yields $1$ arrangement.
Case $2:$ $1$ person is standing. This yields $8$ arrangements.
Case $3:$ $2$ people are standing. To do this, we imagine having 6 people with tails in a line first. Notate "tails" with $T$. Thus, we have $TTTTTT$. Now, we look to distribute the 2 $H$'s into the 7 gaps made by the $T$'s. We can do this in ${7 \choose 2}$ ways. However, note one way does not work, because we have two H's at the end, and the problem states we have a table, not a line. So, we have ${7 \choose 2}-1=20$ arrangements.
Case $4:$ $3$ people are standing. Similarly, we imagine 5 $T$'s. Thus, we have $TTTTT$. We distribute 3 $H$'s into the gaps, which can be done ${6 \choose 3}$ ways. However, 4 arrangements will not work. (See this by putting the H's at the ends, and then choosing one of the remaining 4 gaps: ${4 \choose 1}=4$) Thus, we have ${6 \choose 3}-4=16$ arrangements.
Case $5:$ $4$ people are standing. This can clearly be done in 2 ways: $HTHTHTHT$ or $THTHTHTH$. This yields $2$ arrangements.
Summing the cases, we get $1+8+20+16+2=47$ arrangements. Thus, the probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$
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# Unit 1 Study Guide
## Number Sets
Natural Numbers: Any number that's not 0, a fraction, a decimal, or a negative.
Ex: 1,2,3,4,5,6
Whole Numbers: Any number that's not a fraction, decimal, or a negative.
Ex: 0,1,2,3,4,5,6
Integers: Any number that's not a fraction or a decimal.
Ex: -1,0,1,2
Rational Numbers: Any number that's not irrational.
Ex: 1.5, 7/4, -2, 0, 5
Irrational Numbers: A number that goes on forever, but never repeats.
Ex: 3.14 or Pi
## Fraction To Decimals
To turn A fraction to a decimal, simply divide the numerator by the denominator.
Ex: 2/4 4(2.0 4x5=20 Answer: .5
## Operations With All Rational Numbers
ADDITION: Integers (positive + Positive= Positive) (Positive+Negative=negative) If the negative number is Bigger
Ex: 9+ -12= -3 Ex: 12+ -9= 3
Decimals: negative + negative = negative) (positive + negative= negative) (+ + + = +)
Ex: 9.2+10.8= 20.0 Ex: -9.5+ -7.0 = -16.5 Ex: -9.0 - 3.0 = -12 Ex: -9.5 - (-9.5) = 0
Fractions:
• Find common Denominator
• Turn to mixed fraction
• Turn to proper fraction
SUBTRACTION: (negative-negative=positive) (negative-Positive=negative)
Positive-negative=positive) (positive-positive=negative)
Integers: Ex: -18 - (-12)= -6 Ex: 18 - 6= 12 Ex: -20 - (-40)= 20 EX: 2-2=0
Decimals: Ex: -4.7 + -5.3= -10 EX: -5.1 - 5 = -10.1 Ex: 5 - (-10) = 15 -4.7 - (-4.7) = 0
Fractions:
• Find common Denominator
• Turn to mixed fraction
• Subtract straight across
• Turn to proper fraction
MULTIPLICATION: (Positive x Positive= positive) (Negative x Negative= positive)
(Negative x Positive= Negative)
Integers: 2*-11= -22 3*3= 9 -9*-10= 90
Decimals: EX: 5.4 x 13= 54 x 130 = 7020 = 70.20
• 2.4 x 7 = 24 x 70
• 2.44 x 7 = 244 x 700
Fractions:
• Convert all mixed numbers to improper fractions.
• Flip second fraction EX: 2/4 x 5/4 = 2/4 x 4/5
• Multiply the numerators.
• Multiply the denominators.
DIVISION: (negative/negative=positive) (positive/negative=negative) (posititve/+=+)
Decimals: 4.8/0.02= 480/2 = 240
Integers: (Same rules as multiplying integers) 16/-4= -4
Fractions: 4/5 / 1/2 = 4/5 x 2/1
• flip second fraction
• Multiply numerator
• multiply denominators
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# Factors of 871: Prime Factorization, Methods, and Examples
The factors of a given number are those integers that when divided by that number give a zero remainder. the integers can be both positive and negative.
 871 has a total of 4 factors. Details are provided in the article.
### Factors of 871
Here are the factors of number 871.
Factors of 871: 1, 13, 67, and 871
### Negative Factors of 871
The negative factors of 871 are similar to its positive aspects, just with a negative sign.
Negative Factors of 871: -1, -13, -67, and -871
### Prime Factorization of 871
The prime factorization of 871 is the way of expressing its prime factors in the product form.
Â
Prime Factorization: 13 x 67
In this article, we will learn about the factors of 871 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 871?
The factors of 871 are 1, 13, 67, and 871. These numbers are the factors as they do not leave any remainder when divided by 871.
The factors of 871 are classified as prime numbers and composite numbers. The prime factors of the number 871 can be determined using the prime factorization technique.
## How To Find the Factors of 871?
You can find the factors of 871 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 871, create a list containing the numbers that are exactly divisible by 871 with zero remainders. One important thing to note is that 1 and 871 are the 871’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 871 are determined as follows:
$\dfrac{871}{1} = 871$
$\dfrac{871}{13} = 67$
$\dfrac{871}{67} = 13$
$\dfrac{871}{871} = 1$
Therefore, 1, 13, 67, and 871 are the factors of 871.
### Total Number of Factors of 871
For 871, there are 4Â positive factors and 4Â negative ones. So in total, there are 8 factors of 871.Â
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 871 is given as:
Factorization of 871 is 1 x 13 x 67.
The exponent of 1, 13, and 67 is 1.
Adding 1 to each and multiplying them together results in 8.
Therefore, the total number of factors of 871 is 8. 4 are positive, and 4 factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 871 by Prime Factorization
The number 871 is a composite number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Â
Before finding the factors of 871 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 871, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 871 can be expressed as:
871 = 13 x 67
## Factors of 871 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
Â
For 871, the factor pairs can be found as:
1 x 871 = 871
13 x 67 = 871
The possible factor pairs of 871 are given as (1, 871) and (13, 67).
All these numbers in pairs, when multiplied, give 871 as the product.
The negative factor pairs of 871 are given as:
-1 x -871 = 871Â
-13 x -67 = 871
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -13, -67, and -871 are called negative factors of 871.
The list of all the factors of 871, including positive as well as negative numbers, is given below.
Factor list of 871: 1, -1, 13, -13, 67, -67, 871, and -871
## Factors of 871 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 871 are there?
### Solution
The total number of Factors of 871 is 8.
Factors of 871 are 1, 13, 67, and 871.
### Example 2
Find the factors of 871 using prime factorization.
### Solution
The prime factorization of 871 is given as:
871 $\div$ 13 = 67Â
67 $\div$ 67 = 1Â
So the prime factorization of 871 can be written as:
13 x 67 = 871
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Algebra and Trigonometry 2e
# 10.5Polar Form of Complex Numbers
Algebra and Trigonometry 2e10.5 Polar Form of Complex Numbers
## Learning Objectives
In this section, you will:
• Plot complex numbers in the complex plane.
• Find the absolute value of a complex number.
• Write complex numbers in polar form.
• Convert a complex number from polar to rectangular form.
• Find products of complex numbers in polar form.
• Find quotients of complex numbers in polar form.
• Find powers of complex numbers in polar form.
• Find roots of complex numbers in polar form.
“God made the integers; all else is the work of man.” This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science.
We first encountered complex numbers in Complex Numbers. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem.
## Plotting Complex Numbers in the Complex Plane
Plotting a complex number $a+bi a+bi$ is similar to plotting a real number, except that the horizontal axis represents the real part of the number, $a, a,$ and the vertical axis represents the imaginary part of the number, $bi. bi.$
## How To
Given a complex number $a+bi, a+bi,$ plot it in the complex plane.
1. Label the horizontal axis as the real axis and the vertical axis as the imaginary axis.
2. Plot the point in the complex plane by moving $a a$ units in the horizontal direction and $b b$ units in the vertical direction.
## Example 1
### Plotting a Complex Number in the Complex Plane
Plot the complex number $2−3i 2−3i$ in the complex plane.
## Try It #1
Plot the point $1+5i 1+5i$ in the complex plane.
## Finding the Absolute Value of a Complex Number
The first step toward working with a complex number in polar form is to find the absolute value. The absolute value of a complex number is the same as its magnitude, or $| z |. | z |.$ It measures the distance from the origin to a point in the plane. For example, the graph of $z=2+4i, z=2+4i,$ in Figure 2, shows $| z |. | z |.$
Figure 2
## Absolute Value of a Complex Number
Given $z=x+yi, z=x+yi,$ a complex number, the absolute value of $z z$ is defined as
$| z |= x 2 + y 2 | z |= x 2 + y 2$
It is the distance from the origin to the point $( x,y ). ( x,y ).$
Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, $( 0,0 ). ( 0,0 ).$
## Example 2
### Finding the Absolute Value of a Complex Number with a Radical
Find the absolute value of $z= 5 −i. z= 5 −i.$
## Try It #2
Find the absolute value of the complex number $z=12−5i. z=12−5i.$
## Example 3
### Finding the Absolute Value of a Complex Number
Given $z=3−4i, z=3−4i,$ find $| z |. | z |.$
## Try It #3
Given $z=1−7i, z=1−7i,$ find $| z |. | z |.$
## Writing Complex Numbers in Polar Form
The polar form of a complex number expresses a number in terms of an angle $θ θ$ and its distance from the origin $r. r.$ Given a complex number in rectangular form expressed as $z=x+yi, z=x+yi,$ we use the same conversion formulas as we do to write the number in trigonometric form:
$x=rcosθ y=rsinθ r= x 2 + y 2 x=rcosθ y=rsinθ r= x 2 + y 2$
We review these relationships in Figure 5.
Figure 5
We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point $( x,y ). ( x,y ).$ The modulus, then, is the same as $r, r,$ the radius in polar form. We use $θ θ$ to indicate the angle of direction (just as with polar coordinates). Substituting, we have
$z=x+yi z=rcosθ+( rsinθ )i z=r( cosθ+isinθ ) z=x+yi z=rcosθ+( rsinθ )i z=r( cosθ+isinθ )$
## Polar Form of a Complex Number
Writing a complex number in polar form involves the following conversion formulas:
$x=rcosθ y=rsinθ r= x 2 + y 2 x=rcosθ y=rsinθ r= x 2 + y 2$
Making a direct substitution, we have
$z=x+yi z=( rcosθ )+i( rsinθ ) z=r( cosθ+isinθ ) z=x+yi z=( rcosθ )+i( rsinθ ) z=r( cosθ+isinθ )$
where $r r$ is the modulus and $θ θ$ is the argument. We often use the abbreviation $rcisθ rcisθ$ to represent $r( cosθ+isinθ ). r( cosθ+isinθ ).$
## Example 4
### Expressing a Complex Number Using Polar Coordinates
Express the complex number $4i 4i$ using polar coordinates.
## Try It #4
Express $z=3i z=3i$ as $rcisθ rcisθ$ in polar form.
## Example 5
### Finding the Polar Form of a Complex Number
Find the polar form of $−4+4i. −4+4i.$
## Try It #5
Write $z= 3 +i z= 3 +i$ in polar form.
## Converting a Complex Number from Polar to Rectangular Form
Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given $z=r( cosθ+isinθ ), z=r( cosθ+isinθ ),$ first evaluate the trigonometric functions $cosθ cosθ$ and $sinθ. sinθ.$ Then, multiply through by $r. r.$
## Example 6
### Converting from Polar to Rectangular Form
Convert the polar form of the given complex number to rectangular form:
$z=12( cos( π 6 )+isin( π 6 ) ) z=12( cos( π 6 )+isin( π 6 ) )$
## Example 7
### Finding the Rectangular Form of a Complex Number
Find the rectangular form of the complex number given $r=13 r=13$ and $tanθ= 5 12 . tanθ= 5 12 .$
## Try It #6
Convert the complex number to rectangular form:
$z=4( cos 11π 6 +isin 11π 6 ) z=4( cos 11π 6 +isin 11π 6 )$
## Finding Products of Complex Numbers in Polar Form
Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham De Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.
## Products of Complex Numbers in Polar Form
If $z 1 = r 1 (cos θ 1 +isin θ 1 ) z 1 = r 1 (cos θ 1 +isin θ 1 )$ and $z 2 = r 2 (cos θ 2 +isin θ 2 ), z 2 = r 2 (cos θ 2 +isin θ 2 ),$ then the product of these numbers is given as:
$z 1 z 2 = r 1 r 2 [ cos( θ 1 + θ 2 )+isin( θ 1 + θ 2 ) ] z 1 z 2 = r 1 r 2 cis( θ 1 + θ 2 ) z 1 z 2 = r 1 r 2 [ cos( θ 1 + θ 2 )+isin( θ 1 + θ 2 ) ] z 1 z 2 = r 1 r 2 cis( θ 1 + θ 2 )$
Notice that the product calls for multiplying the moduli and adding the angles.
## Example 8
### Finding the Product of Two Complex Numbers in Polar Form
Find the product of $z 1 z 2 , z 1 z 2 ,$ given $z 1 =4(cos(80°)+isin(80°)) z 1 =4(cos(80°)+isin(80°))$ and $z 2 =2(cos(145°)+isin(145°)). z 2 =2(cos(145°)+isin(145°)).$
## Finding Quotients of Complex Numbers in Polar Form
The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.
## Quotients of Complex Numbers in Polar Form
If $z 1 = r 1 (cos θ 1 +isin θ 1 ) z 1 = r 1 (cos θ 1 +isin θ 1 )$ and $z 2 = r 2 (cos θ 2 +isin θ 2 ), z 2 = r 2 (cos θ 2 +isin θ 2 ),$ then the quotient of these numbers is
$z 1 z 2 = r 1 r 2 [ cos( θ 1 − θ 2 )+isin( θ 1 − θ 2 ) ], z 2 ≠0 z 1 z 2 = r 1 r 2 cis( θ 1 − θ 2 ), z 2 ≠0 z 1 z 2 = r 1 r 2 [ cos( θ 1 − θ 2 )+isin( θ 1 − θ 2 ) ], z 2 ≠0 z 1 z 2 = r 1 r 2 cis( θ 1 − θ 2 ), z 2 ≠0$
Notice that the moduli are divided, and the angles are subtracted.
## How To
Given two complex numbers in polar form, find the quotient.
1. Divide $r 1 r 2 . r 1 r 2 .$
2. Find $θ 1 − θ 2 . θ 1 − θ 2 .$
3. Substitute the results into the formula: $z=r( cosθ+isinθ ). z=r( cosθ+isinθ ).$ Replace $r r$ with $r 1 r 2 , r 1 r 2 ,$ and replace $θ θ$ with $θ 1 − θ 2 . θ 1 − θ 2 .$
4. Calculate the new trigonometric expressions and multiply through by $r. r.$
## Example 9
### Finding the Quotient of Two Complex Numbers
Find the quotient of $z 1 =2(cos(213°)+isin(213°)) z 1 =2(cos(213°)+isin(213°))$ and $z 2 =4(cos(33°)+isin(33°)). z 2 =4(cos(33°)+isin(33°)).$
## Try It #7
Find the product and the quotient of $z 1 =2 3 (cos(150°)+isin(150°)) z 1 =2 3 (cos(150°)+isin(150°))$ and $z 2 =2(cos(30°)+isin(30°)). z 2 =2(cos(30°)+isin(30°)).$
## Finding Powers of Complex Numbers in Polar Form
Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. It states that, for a positive integer $n, z n n, z n$ is found by raising the modulus to the $nth nth$ power and multiplying the argument by $n. n.$ It is the standard method used in modern mathematics.
## De Moivre’s Theorem
If $z=r( cosθ+isinθ ) z=r( cosθ+isinθ )$ is a complex number, then
$z n = r n [ cos( nθ )+isin( nθ ) ] z n = r n cis( nθ ) z n = r n [ cos( nθ )+isin( nθ ) ] z n = r n cis( nθ )$
where $n n$ is a positive integer.
## Example 10
### Evaluating an Expression Using De Moivre’s Theorem
Evaluate the expression $( 1+i ) 5 ( 1+i ) 5$ using De Moivre’s Theorem.
## Finding Roots of Complex Numbers in Polar Form
To find the nth root of a complex number in polar form, we use the $nth nth$ Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding $nth nth$ roots of complex numbers in polar form.
## The nth Root Theorem
To find the $nth nth$ root of a complex number in polar form, use the formula given as
$z 1 n = r 1 n [ cos( θ n + 2kπ n )+isin( θ n + 2kπ n ) ] z 1 n = r 1 n [ cos( θ n + 2kπ n )+isin( θ n + 2kπ n ) ]$
where $k=0,1,2,3,...,n−1. k=0,1,2,3,...,n−1.$ We add $2kπ n 2kπ n$ to $θ n θ n$ in order to obtain the periodic roots.
## Example 11
### Finding the nth Root of a Complex Number
Evaluate the cube roots of $z=8( cos( 2π 3 )+isin( 2π 3 ) ). z=8( cos( 2π 3 )+isin( 2π 3 ) ).$
## Try It #8
Find the four fourth roots of $16(cos(120°)+isin(120°)). 16(cos(120°)+isin(120°)).$
## Media
Access these online resources for additional instruction and practice with polar forms of complex numbers.
## 10.5 Section Exercises
### Verbal
1.
A complex number is $a+bi. a+bi.$ Explain each part.
2.
What does the absolute value of a complex number represent?
3.
How is a complex number converted to polar form?
4.
How do we find the product of two complex numbers?
5.
What is De Moivre’s Theorem and what is it used for?
### Algebraic
For the following exercises, find the absolute value of the given complex number.
6.
$5+3i 5+3i$
7.
$−7+i −7+i$
8.
$−3−3i −3−3i$
9.
$2 −6i 2 −6i$
10.
$2i 2i$
11.
$2.2−3.1i 2.2−3.1i$
For the following exercises, write the complex number in polar form.
12.
$2+2i 2+2i$
13.
$8−4i 8−4i$
14.
$− 1 2 − 1 2 i − 1 2 − 1 2 i$
15.
$3 +i 3 +i$
16.
$3i 3i$
For the following exercises, convert the complex number from polar to rectangular form.
17.
$z=7cis( π 6 ) z=7cis( π 6 )$
18.
$z=2cis( π 3 ) z=2cis( π 3 )$
19.
$z=4cis( 7π 6 ) z=4cis( 7π 6 )$
20.
$z=7cis( 25° ) z=7cis( 25° )$
21.
$z=3cis( 240° ) z=3cis( 240° )$
22.
$z= 2 cis( 100° ) z= 2 cis( 100° )$
For the following exercises, find $z 1 z 2 z 1 z 2$ in polar form.
23.
$z 1 =2 3 cis( 116° ); z 2 =2cis( 82° ) z 1 =2 3 cis( 116° ); z 2 =2cis( 82° )$
24.
$z 1 = 2 cis( 205° ); z 2 =2 2 cis( 118° ) z 1 = 2 cis( 205° ); z 2 =2 2 cis( 118° )$
25.
$z 1 =3cis( 120° ); z 2 = 1 4 cis( 60° ) z 1 =3cis( 120° ); z 2 = 1 4 cis( 60° )$
26.
$z 1 =3cis( π 4 ); z 2 =5cis( π 6 ) z 1 =3cis( π 4 ); z 2 =5cis( π 6 )$
27.
$z 1 = 5 cis( 5π 8 ); z 2 = 15 cis( π 12 ) z 1 = 5 cis( 5π 8 ); z 2 = 15 cis( π 12 )$
28.
$z 1 =4cis( π 2 ); z 2 =2cis( π 4 ) z 1 =4cis( π 2 ); z 2 =2cis( π 4 )$
For the following exercises, find $z 1 z 2 z 1 z 2$ in polar form.
29.
$z 1 =21cis( 135° ); z 2 =3cis( 65° ) z 1 =21cis( 135° ); z 2 =3cis( 65° )$
30.
$z 1 = 2 cis( 90° ); z 2 =2cis( 60° ) z 1 = 2 cis( 90° ); z 2 =2cis( 60° )$
31.
$z 1 =15cis( 120° ); z 2 =3cis( 40° ) z 1 =15cis( 120° ); z 2 =3cis( 40° )$
32.
$z 1 =6cis( π 3 ); z 2 =2cis( π 4 ) z 1 =6cis( π 3 ); z 2 =2cis( π 4 )$
33.
$z 1 =5 2 cis( π ); z 2 = 2 cis( 2π 3 ) z 1 =5 2 cis( π ); z 2 = 2 cis( 2π 3 )$
34.
$z 1 =2cis( 3π 5 ); z 2 =3cis( π 4 ) z 1 =2cis( 3π 5 ); z 2 =3cis( π 4 )$
For the following exercises, find the powers of each complex number in polar form.
35.
Find $z 3 z 3$ when $z=5cis( 45° ). z=5cis( 45° ).$
36.
Find $z 4 z 4$ when $z=2cis( 70° ). z=2cis( 70° ).$
37.
Find $z 2 z 2$ when $z=3cis( 120° ). z=3cis( 120° ).$
38.
Find $z 2 z 2$ when $z=4cis( π 4 ). z=4cis( π 4 ).$
39.
Find $z 4 z 4$ when $z=cis( 3π 16 ). z=cis( 3π 16 ).$
40.
Find $z 3 z 3$ when $z=3cis( 5π 3 ). z=3cis( 5π 3 ).$
For the following exercises, evaluate each root.
41.
Evaluate the cube root of $z z$ when $z=27cis( 240° ). z=27cis( 240° ).$
42.
Evaluate the square root of $z z$ when $z=16cis( 100° ). z=16cis( 100° ).$
43.
Evaluate the cube root of $z z$ when $z=32cis( 2π 3 ). z=32cis( 2π 3 ).$
44.
Evaluate the square root of $z z$ when $z=32cis( π ). z=32cis( π ).$
45.
Evaluate the square root of $z z$ when $z=8cis( 7π 4 ). z=8cis( 7π 4 ).$
### Graphical
For the following exercises, plot the complex number in the complex plane.
46.
$2+4i 2+4i$
47.
$−3−3i −3−3i$
48.
$5−4i 5−4i$
49.
$−1−5i −1−5i$
50.
$3+2i 3+2i$
51.
$2i 2i$
52.
$−4 −4$
53.
$6−2i 6−2i$
54.
$−2+i −2+i$
55.
$1−4i 1−4i$
### Technology
For the following exercises, find all answers rounded to the nearest hundredth.
56.
Use the rectangular to polar feature on the graphing calculator to change $5+5i 5+5i$ to polar form.
57.
Use the rectangular to polar feature on the graphing calculator to change $3−2i 3−2i$ to polar form.
58.
Use the rectangular to polar feature on the graphing calculator to change $−3−8i −3−8i$ to polar form.
59.
Use the polar to rectangular feature on the graphing calculator to change $4cis( 120° ) 4cis( 120° )$ to rectangular form.
60.
Use the polar to rectangular feature on the graphing calculator to change $2cis( 45° ) 2cis( 45° )$ to rectangular form.
61.
Use the polar to rectangular feature on the graphing calculator to change $5cis( 210° ) 5cis( 210° )$ to rectangular form.
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## Intermediate Algebra (12th Edition)
$\bf{\text{Solution Outline:}}$ Substitute the given value for $x$ in the given equation, $\sqrt{8x-3}-2x=0 .$ If the left side of the equation becomes equal to the right side of the equation, then the given value of $x$ is a solution to the equation. $\bf{\text{Solution Details:}}$ a) Substituting $x$ with $\dfrac{3}{2}$ in the given equation results to \begin{array}{l}\require{cancel} \sqrt{8\left( \dfrac{3}{2} \right)-3}-2\left( \dfrac{3}{2} \right)=0 \\\\ \sqrt{12-3}-3=0 \\\\ \sqrt{9}-3=0 \\\\ 3-3=0 \\\\ 0=0 \text{ (TRUE)} .\end{array} Hence, $x= \dfrac{3}{2}$ is a solution to the given equation. b) Substituting $x$ with $\dfrac{1}{2}$ in the given equation results to \begin{array}{l}\require{cancel} \sqrt{8\left( \dfrac{1}{2} \right)-3}-2\left( \dfrac{1}{2} \right)=0 \\\\ \sqrt{4-3}-1=0 \\\\ \sqrt{1}-1=0 \\\\ 1-1=0 \\\\ 0=0 \text{ (TRUE)} .\end{array} Hence, $x= \dfrac{1}{2}$ is a solution to the given equation.
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# mathSolve x=(x+6)^1/2.
justaguide | Certified Educator
The equation to be solved is x=(x+6)^1/2
x=(x+6)^1/2
square both the sides
=> x^2 = x + 6
=> x^2 - x - 6 = 0
=> x^2 - 3x + 2x - 6 = 0
=> x(x - 3) + 2(x - 3) = 0
=> (x + 2)(x - 3) = 0
=> x = 3 and x = -2
The solution of the equation is x = 3 and x = -2
tonys538 | Student
The equation x=(x+6)^1/2 has to be solved.
x=(x+6)^1/2
Take the square of both the sides
x^2 = x + 6
x^2 - x - 6 = 0
x^2 - 3x + 2x - 6 = 0
x(x - 3) + 2(x - 3) = 0
(x + 2)(x - 3) = 0
x = -2 and x = 3
Both of these solutions satisfy the given equation as the square root of a positive number can be negative,
giorgiana1976 | Student
We'll eliminate the square root from the right side. For this reason, we'll square both sides.
x^2 = [(x+6)^1/2]^2
x^2 = x + 6
We'll subtract x + 6 both sides:
x^2 - x - 6 = 0
We'll factorize and we'll get:
(x - 3)(x + 2) = 0
x - 3 = 0
x = 3
x + 2 = 0
x = -2
We'll verify the values of x in the equation:
x = 3
3 = sqrt(3+6)
3 = sqrt9
3 = 3
x = -2
-2 = sqrt(-2+6)
-2 = sqrt4
-2 is different from 2.
So, we'll reject the solution x = -2.
The only solution of the equation is x = 3
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# How do you use the tangent line approximation to approximate the value of ln(1003) ?
Aug 25, 2014
The answer is $3 \ln \left(10\right) + .003$
Another term for tangent line approximation is linear approximation. The linear approximation function is:
$L \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \left(x - a\right)$
So we need to find the derivative:
$f \left(x\right) = \ln \left(x\right)$
$f ' \left(x\right) = \frac{1}{x}$
Now, we need to pick an $a$ that is easy to compute. Unfortunately, with $\ln$ there isn't an easy way to compute a decimal value. So we will go with $a = 1000 = {10}^{3}$:
$f \left(a\right) = f \left({10}^{3}\right) = \ln \left({10}^{3}\right) = 3 \ln \left(10\right)$
$f ' \left(a\right) = f ' \left(1000\right) = \frac{1}{1000}$
So our linear approximation is:
$L \left(x\right) \approx 3 \ln \left(10\right) + \frac{1}{1000} \left(x - 1000\right)$
$L \left(1003\right) \approx 3 \ln \left(10\right) + \frac{1}{1000} \left(1003 - 1000\right)$
$\approx 3 \ln \left(10\right) + \frac{3}{1000}$
$\approx 3 \ln \left(10\right) + .003$
We should leave this as the answer since it's supposed to be mental math. But let's look at how accurate this is: $\ln \left(1003\right) = 6.910750788$ and $L \left(1003\right) \approx 6.910755279$ which is accurate to 5 decimal places; so that's pretty good.
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HARD
### EXPLANATION
To see how to solve this problem, let’s take an example:
5 3
10100
01100
10011
00010
00100
3 5
5 1
5 3
First, let’s pretend all lights can be pressed.
Note that the order in which the lights are pressed doesn’t matter, and pressing a light twice is the same as not pressing it at all. Therefore a solution consists of a set of lights.
A solution is determined by which lights in the top row are pressed: Since the order doesn’t matter, we can press those first, and then “chase the lights”: For every light in the top row that is on, we have to press the light below it, and we can’t press any other lights in the second row. Then, for every light in the second row that is on, we have to press the light below it. The same for the third row, and so on.
If we press lights on the top row, not corresponding to a solution, and then chase the lights, only the bottom row will have lights on. What’s more, the effect of pushing some lights at the top, say k of them, and chasing the lights, then pressing another light at
the top and chasing the lights again, is the same as first pressing all k+1 lights (some of them possibly the same) and then chasing the lights. Also, whether a light is pressed an odd or even number of times doesn’t change.
To solve the puzzle, first chase the lights, to reduce the problem to the case where only the bottom row has any lights on. In our example, we get
00000
00000
00000
00000
10001
Now, for each light on the top row, start with all lights off, press that light, chase the lights, and see what the effect on the bottom row is. For instance
11000
10000
00000
00000
00000
becomes
00000
00000
00000
00000
01101
So, pressing (1,1) and chasing the lights will toggle lights 2, 3 and 5 on the bottom row. We want a set of lights in the top row, such that the combined effect on the bottom row is to toggle lights 1 and 5. In other words we have to solve five equations in five unknowns, where arithmetic is performed modulo 2. In Matrix notation: Ax=b, where:
| 0 1 1 0 1 | | 1 |
| 1 1 1 0 0 | | 0 |
A = | 1 1 0 1 1 | and b = | 0 |
| 0 0 1 1 1 | | 0 |
| 1 0 1 1 0 | | 1 |
Column j of A is the bottom row after pressing (1,j) on a blank grid and chasing the lights. b is the bottom row after chasing the lights in the original configuration.
But some lights can’t be pressed. This adds some additional constraints to the system of equations. Starting with a blank grid, pressing one button at the top, and chasing the lights, we look at which faulty lights are pressed in the process. For instance, starting with (1,1), no faulty lights are pressed. Starting with (1,3), we press (3,5) and (5,3). Going through all five lights at the top, we have the constraints
| 0 0 1 0 1 | | 0 |
| 0 0 0 0 1 | x = | 1 |
| 0 0 1 0 0 | | 1 |
The rows correspond to (3,5), (5,1) and (5,3), in that order. The column on the right corresponds to faulty buttons that are pressed in the first step of the solution, when we chase the lights to clear everything except the last row. Putting these equations together, we must solve
| 0 1 1 0 1 | | 1 |
| 1 1 1 0 0 | | 0 |
| 1 1 0 1 1 | | 0 |
| 0 0 1 1 1 | x = | 0 |
| 1 0 1 1 0 | | 1 |
| 0 0 1 0 1 | | 0 |
| 0 0 0 0 1 | | 1 |
| 0 0 1 0 0 | | 1 |
Using Gauss elimination, this reduces to
| 1 0 0 0 0 | | 0 |
| 0 1 0 0 0 | | 1 |
| 0 0 1 0 0 | | 1 |
| 0 0 0 1 0 | x = | 0 |
| 0 0 0 0 1 | | 1 |
| 0 0 0 0 0 | | 0 |
| 0 0 0 0 0 | | 0 |
| 0 0 0 0 0 | | 0 |
So there is a unique solution, which is to press lights 2, 3, and 5 in the top row, and chase the lights, giving the solution set {(1,2),(1,3),(1,5),(2,3),(2,5),(3,2),(3,3),(4,3)}.
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# Derivative of tan x: Proof by Quotient, Chain & First Principle
In this article, you will learn what is the derivative of tan x as well as prove the derivative of tan x by chain rule, first principal rule, and quotient rule.
I have already discussed the derivative of sin x, cos x, and sec x in a previous article you can check from here.
So, without wasting time let's get started.
## What is tan x?
tan x is a trigonometric function which is equal to the (sin x)/(cos x).
This trigonometric function is mostly used in right-angle triangles.
## Derivative of tan x
The derivative of tan x is equal to the sec²x.
We can prove the derivative of tan x in three ways first by using the quotient rule and second by using the first principle rule and the last chain rule.
### Derivative of tan x Proof by Quotient Rule
The formula of the quotient rule is,
dy/dx = {v (du/dx) - u (dv/dx)}/v²
Where,
dy/dx = derivative of y with respect to x
v = variable v
du/dx = derivative of u with respect to x
u = variable u
dv/dx = derivative of v with respect to x
v = variable v
As we know,
tan x = sin x/cos x
So,
Let us
y = tan x
and,
u = sin x
v = cos x
Now putting these values on the quotient rule formula, we will get
dy/dx = [ cos x × d/dx (sin x) - sin x × d/dx (cos x)] / (cos²x)
= [cos x · cos x - sin x (-sin x)] / (cos²x)
= [cos²x + sin²x] / (cos²x)
So, from the Pythagoras theorem, we know that cos²x + sin²x = 1
So,
dy/dx = 1 / cos²x
So, 1 / cos²x = sec²x
Hence,
d/dx(tan x) = sec²x
Thus, we proved the derivative of tan x will be equal to sec²x using the quotient rule method.
### Derivative of tan x Proof by First Principle Rule
According to the first principle rule, the derivative limit of a function can be determined by computing the formula:
For a differentiable function y = f (x)
We define its derivative w.r.t x as :
dy/dx = f ' (x) = limₕ→₀ [f(x+h) - f(x)]/h
f'(x) = limₕ→₀ [f(x+h) - f(x)]/h
This limit is used to represent the instantaneous rate of change of the function f(x).
Let,
f (x) = tan x
So,
f(x + h) = tan (x + h)
Putting these values on the above first principle rules equation.
f'(x) = limₕ→₀ [tan(x + h) - tan x] / h
Since, we know tan x = sin x/cos x
So, putting these values
f'(x) = limₕ→₀ [ [sin (x + h) / cos (x + h)] - [sin x / cos x] ] / h
= limₕ→₀ [ [sin (x + h ) cos x - cos (x + h) sin x] / [cos x · cos(x + h)] ]/ h
As we know,
sin (a - b) = sin a cos b - cos a sin b
So,
f'(x) = limₕ→₀ [ sin (x + h - x) ] / [ h cos x × cos(x + h)]
= limₕ→₀ [ sin h ] / [ h cos x × cos(x + h)]
= limₕ→₀ (sin h)/ h limₕ→₀ 1 / [cos x × cos(x + h)]
So, after applying the limit,
limₕ→₀ (sin h)/ h = 1.
f'(x) = 1 [ 1 / (cos x × cos(x + 0))]
f'(x) = 1/cos² x
As we know,
1/cos x = sec x
So,
f'(x) = sec²x.
Thus, we proved the derivative of tan x will be equal to sec²x using the first principle rule method.
### Derivative of tan x Proof by Chain Rule
Let us,
y = tan x
As we know,
tan x = 1/cot x
So,
y = 1 / (cot x)
= (cot x)⁻¹
By using the chain rule,
The formula of chain rule is,
dy/dx = (dy/du) × (du/dx)
Where,
dy/dx = derivative of y with respect to x
dy/du = derivative of y with respect to u
du/dx = derivative of u with respect to x
After putting these values we can find,
dy/dx = -1 (cot x)⁻² × d/dx (cot x)
Since,
d/dx (cot x) = (-cosec²x) and a⁻ⁿ = 1/aⁿ
So,
dy/dx = -1/(cot² x) × (-cosec²x)
As we know,
1/(cot² x) = tan² x
So,
dy/dx = (tan² x) × (cosec²x)
Now we know that the trigonometric formula
tan x = sin x/cos x and cosec x = 1/sin x
So after putting these values we will get,
dy/dx = (sin² x)/(cos² x) × (1/sin² x)
= 1/cos² x
As we know,
1/cos x = sec x
d/dx (tan x) = sec²x
Thus, we proved the derivative of tan x will be equal to sec²x using the chain rule method.
## FAQ Related to Derivative of tan x
### What is a derivative of tan x?
The derivative of tan x is equal to the sec²x.
### Where is tan equal to 1?
At value tan 45° is equal to 1.
### What is tan Infinity?
At value tan 90° is equal to infinity.
### What is the differentiation of tan x?
The differentiation of tan x is equal to the sec²x.
So friends here I discussed all aspects related to the derivative of tan x
I hope you enjoy this topic If you have any doubt then you can ask me through comments or direct mail.
Thank You.
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Statistics - Ti 83 Exponential Regression
Ti 83 Exponential Regression is used to compute an equation which best fits the co-relation between sets of indisciriminate variables.
Formula
${ y = a \times b^x}$
Where −
• ${a, b}$ = coefficients for the exponential.
Example
Problem Statement:
Calculate Exponential Regression Equation(y) for the following data points.
Time (min), Ti Temperature (°F), Te 0 5 10 15 140 129 119 112
Solution:
Let consider a and b as coefficients for the exponential Regression.
Step 1
${ b = e^{ \frac{n \times \sum Ti log(Te) - \sum (Ti) \times \sum log(Te) } {n \times \sum (Ti)^2 - \times (Ti) \times \sum (Ti) }} }$
Where −
• ${n}$ = total number of items.
${ \sum Ti log(Te) = 0 \times log(140) + 5 \times log(129) + 10 \times log(119) + 15 \times log(112) = 62.0466 \\[7pt] \sum log(L2) = log(140) + log(129) + log(119) + log(112) = 8.3814 \\[7pt] \sum Ti = (0 + 5 + 10 + 15) = 30 \\[7pt] \sum Ti^2 = (0^2 + 5^2 + 10^2 + 15^2) = 350 \\[7pt] \implies b = e^{\frac {4 \times 62.0466 - 30 \times 8.3814} {4 \times 350 - 30 \times 30}} \\[7pt] = e^{-0.0065112} \\[7pt] = 0.9935 }$
Step 2
${ a = e^{ \frac{\sum log(Te) - \sum (Ti) \times log(b)}{n} } \\[7pt] = e^{\frac{8.3814 - 30 \times log(0.9935)}{4}} \\[7pt] = e^2.116590964 \\[7pt] = 8.3028 }$
Step 3
Putting the value of a and b in Exponential Regression Equation(y), we get.
${ y = a \times b^x \\[7pt] = 8.3028 \times 0.9935^x }$
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# 2005 AMC 12B Problems/Problem 25
## Problem
Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?
$\mathrm{(A)}\ \frac {5}{256} \qquad\mathrm{(B)}\ \frac {21}{1024} \qquad\mathrm{(C)}\ \frac {11}{512} \qquad\mathrm{(D)}\ \frac {23}{1024} \qquad\mathrm{(E)}\ \frac {3}{128}$
## Solution
### Solution 1
We approach this problem by counting the number of ways ants can do their desired migration, and then multiple this number by the probability that each case occurs.
Let the octahedron be $ABCDEF$, with points $B,C,D,E$ coplanar. Then the ant from $A$ and the ant from $F$ must move to plane $BCDE$. Suppose, without loss of generality, that the ant from $A$ moved to point $B$. Then, we must consider three cases.
• Case 1: Ant from point $F$ moved to point $C$
On the plane, points $B$ and $C$ are taken. The ant that moves to $D$ can come from either $E$ or $C$. The ant that moves to $E$ can come from either $B$ or $D$. Once these two ants are fixed, the other two ants must migrate to the "poles" of the octahedron, points $A$ and $F$. Thus, there are two degrees of freedom in deciding which ant moves to $D$, two degrees of freedom in deciding which ant moves to $E$, and two degrees of freedom in deciding which ant moves to $A$. Hence, there are $2 \times 2 \times 2=8$ ways the ants can move to different points.
• Case 2: Ant from point $F$ moved to point $D$
On the plane, points $B$ and $D$ are taken. The ant that moves to $C$ must be from $B$ or $D$, but the ant that moves to $E$ must also be from $B$ or $D$. The other two ants, originating from points $C$ and $E$, must move to the poles. Therefore, there are two degrees of freedom in deciding which ant moves to $C$ and two degrees of freedom in choosing which ant moves to $A$. Hence, there are $2 \times 2=4$ ways the ants can move to different points.
• Case 3: Ant from point $F$ moved to point $E$
By symmetry to Case 1, there are $8$ ways the ants can move to different points.
Given a point $B$, there is a total of $8+4+8=20$ ways the ants can move to different points. We oriented the square so that point $B$ was defined as the point to which the ant from point $A$ moved. Since the ant from point $A$ can actually move to four different points, there is a total of $4 \times 20=80$ ways the ants can move to different points.
Each ant acts independently, having four different points to choose from. Hence, each ant has probability $1/4$ of moving to the desired location. Since there are six ants, the probability of each case occuring is $\frac{1}{4^6} = \frac{1}{4096}$. Thus, the desired answer is $\frac{80}{4096}= \boxed{\frac{5}{256}} \Rightarrow \mathrm{(A)}$.
### Solution 2
Let $f(n)$ be the number of cycles of length $n$ the can be walked among the vertices of an octahedron. For example, $f(3)$ would represent the number of ways in which an ant could navigate $2$ vertices and then return back to the original spot. Since an ant cannot stay still, $f(1) = 0$. We also easily see that $f(2) = 1, f(3) = 2$.
Now consider any four vertices of the octahedron. All four vertices will be connected by edges except for one pair. Let’s think of this as a square with one diagonal (from top left to bottom right). EDIT: This part is wrong as if you choose the 4 vertices that have a cross section as a square, there exists no connecting diagonal.
$[asy] size(30); defaultpen(0.6); pair A = (0,0), B=(5,0), C=(5,5), D=(0,5); draw(A--B--C--D--cycle); draw(B--D); [/asy]$
Suppose an ant moved across this diagonal; then the ant at the other end can only move across the diagonal (which creates 2-cycle, bad) or it can move to another vertex, but then the ant at that vertex must move to the spot of the original ant (which creates 3-cycle, bad). Thus none of the ants can navigate the diagonal and can either shift clockwise or counterclockwise, and so $f(4) = 2$.
For $f(6)$, consider an ant at the top of the octahedron. It has four choices. Afterwards, it can either travel directly to the bottom, and then it has $2$ ways back up, or it can travel along the sides and then go to the bottom, of which simple counting gives us $6$ ways back up. Hence, this totals $4 \times (2+6) = 32$.
Now, the number of possible ways is given by the sum of all possible cycles, $$a \cdot f(2) \cdot f(2) \cdot f(2) + b \cdot f(2) \cdot f(4) + c \cdot f(3) \cdot f(3) + d \cdot f(6)$$
where the coefficients represent the number of ways we can configure these cycles. To find $a$, fix any face, there are $4$ adjacent faces to select from to complete the cycle. From the four remaining faces there are only $2$ ways to create cycles, hence $a = 8$.
To find $b$, each cycle of $2$ faces is distinguished by their common edge, and there are $12$ edges, so $b = 12$.
To find $c$, each three-cycle is distinguished by the vertex, and there are $8$ edges. However, since the two three-cycles are indistinguishable, $c = 8/2 = 4$.
Clearly $d = 1$. Finally,
$$8(1)(1)(1) + 12(1)(2) + 4(2)(2) + (32) = 80$$
Each bug has $4$ possibilities to choose from, so the probability is $\frac{80}{4^6} = \frac{5}{256}$.
## Solution 3
We use the idea of cycles.
Case 1: Two 3-cycles. We have 4 ways to divide the faces into cycles. There are 8 faces, and once a face has been chosen, the other three vertices must form the other cycle of length 3. Then there are $2^2$ ways to choose the direction of the cycle. There are $4 \cdot 4 = 16.$
Case 2: Three 2-cycles. There are $4 \cdot 2 = 8$ ways to do so.
Case 3: One 4-cycle, one 2-cycle. There are $12 \cdot 2 = 24$ ways here because there are 12 different edges for the 2-cycle and we can choose the direction of the 4-cycle.
Case 4: One 6-cycle. Visualizing the octahedron as a 4-sided pyramid pointing up above one pointing down, we look at the ways an ant can start from the top vertex and visit all vertices without revisiting one along the way and returning to the top. Because the first step must be from the top to the middle square ring, we choose one vertex to move to and will multiply the final result by four.
We divide the paths into two sub-cases.
Sub-case 1: The ant continues to the bottom vertex. In this case, the ant has visited three corners of a square, but can not next visit the fourth corner or there will be no way to connect the remaining two vertices of the octahedron. Thus, the ant must next visit one of the other two vertices, and that selection decides the remainder of the path. Thus, this sub-case has 2 possible paths.
Sub-case 2: The ant continues along the "equator" square ring. By symmetry, there are two choices. Choose one, and we will multiply the final result by two. From that second point on the equator, there are two choices. If it continues to a third point on the equator, there is only one path to complete the cycle. If it next moves to the bottom vertex, there are two ways to complete the cycle. This gives a total of three possible paths. Multiplying by 2 gives 6 for this sub-case.
There are $(2 + 6) \cdot 4 = 32$ ways here.
Overall, there are $80$ cases. Answer is $80/2^{12} = 5/256$
~Williamgolly + Dr. 17
## Solution 4
Let $ACEDFB$ be the octahedron such that points $CDEF$ are coplanar, $C/D$ are opposite each other and $E/F$ are opposite each other.
The set of points that the ants from $A/B$ can travel to is $\{C,D,E,F\}$, the set of points that the ants from $C/D$ can travel to is $\{A,B,E,F\}$ and the set of points that the ants from $E/F$ can travel to is $\{A,B,C,D\}$. So the problem is analogous to finding the number of ways to take 2 elements from each of these sets to make the set $\{A,B,C,D,E,F\}$.
• Case 1: Ants from $E/F$ travel to $A/B$
If this occurs then the ants from $C/D$ must travel to $E/F$ and therefore the ants from $E/F$ must travel to $C/D$. So this gives 1 case.
• Case 2: Ants from $E/F$ travel to $A/C$
If this happens then one of the ants from $C/D$ must go to $B$ and one of the ants from $A/B$ must go to $D$ and the remaining ant from $C/D$ can choose to go to $E$ or $F$ and then the remaining ant from $A/B$ has only 1 choice. So this gives 2 cases.
The ants from $E/F$ traveling to $A/B$ is equivalent to them traveling to $C/D$ (both remove 2 elements from another set). And the ants from $E/F$ traveling to $A/C$ is equivalent to them traveling to $A/D$ or $B/C$ or $B/D$ (both remove 1 element from the other 2 sets). Also, for each of the 3 pairs of ants there are 2 ways the pair of ants can go to 2 points (for example, $E\mapsto A$ and $F\mapsto B$ or $E\mapsto B$ and $F\mapsto A$) so there are $2^{3}(2\cdot1 + 4\cdot2) = 80$ possibilties. There are a total of $2^{12}$ ways the ants can move so the answer is $\frac{80}{2^{12}} = \boxed{\textbf{(A) } \frac{5}{256}}$. ~LuisFonseca123
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# Bundling in Maths
Last updated date: 17th Apr 2024
Total views: 151.2k
Views today: 1.51k
## Introduction to Bundling
For young elementary students, understanding place value is an important maths concept. They need to understand that a digit's value is determined by where it is in a number. A digit's place value in a number is its value in that specific place. For example- the place value of the digit 1 in the number 152 is the hundredth. The number 5 and the number 2 have five at tens and two at one place, respectively. Students can better understand number properties and how larger numbers can be divided into smaller numbers by learning how to bundle numbers. In this article, we will learn about bundling in maths by understanding place value.
## Define Bundling
Bundling is also known as a grouping. This method of grouping numbers by combining smaller units to get a bigger one. For example, adding ten ones gives one ten. Adding 10 tens equals one hundred.
## Place Value Using the Bundle of 10
We know that the hundreds, tens, and ones are represented by the three digits of a three-digit number.
Consider the following examples:
The numbers 100, 200, 300, 400, 500, 600, 700, 800, and 900 denote one, two, three, four, five, six, seven, eight, or nine hundred. A bundle of tens, or a hundred, can be considered a group of tens (with 0 tens and 0 ones).
Bundling
Let’s see the product bundling examples: A hundred is framed as a collection of 10 tens or simply a hundred in the above image. Here, a hundred have been constructed from 10 tens bundles.
## 3-Digit Number Using Bundling
Use bundles of one hundred with no tens or ones to represent numbers such as 100, 200, 300,... 900.
Let’s see an example of price bundling: Let us assume 1 stick costs 10 rupees. In one bundle, there are 100 sticks. Therefore, 200 is equal to 2 bundles of 10 tens. 3 bundles of 10 tens equal 300.
3-digit Number Bundling
Here we can see that there are 3 bundles of 100, which mean 300, 5 bundles of 10, that is 50, and 6 one.
Therefore the number is 356.
Hence, the cost will become 3560 rupees.
Let us see another example of bundling 451. We know that 4 is at a hundred places, 5 is at tens place, and 9 is at ones. So, 400 is equal to 4 bundles of 10 tens, 50 is equal to 5 bundles of tens, and 1 is equal to one single stick.
Bundling of 451
## Solved Examples
Q 1. Represent number 279 using bundles.
Ans: We have given 279, of which 2 is at hundred places, 7 is at tens place, and 9 is at one’s place. Therefore, 100 is equal to 10 bundles of 10 tens, 70 is equal to 7 bundles of tens, and 9 is equal to nine single sticks.
179 Using Bundles of Ten
Q 2. Represent number 93 using bundles of ten.
Ans: We have given a two-digit number 93, of which 9 is at tens place, and 2 is at one’s place. Therefore, 90 is equal to 9 bundles of tens, and 3 is equal to three single sticks.
93 Using Bundles of Ten
## Practice Problem
Q 1. Count the number of sticks and write the number.
Bundling of Sticks
Ans: 16
Q 2. Count the number of sticks and write the number.
Bundling of Sticks
Ans: 57
## Summary
In this article, we have learned about the place value of a number. With the help of bundling, we have learned how easily we can understand the place values of the numbers. And as we know, bundling in simple terms is the grouping of the numbers and finding the values using it. With the help of sticks, we have portrayed this article to make it easy to understand the concept of place value. Place value is the value of a digit according to its position in the number such as ones, tens, hundreds, and so on. Some solved examples and practice questions are given in this article to make your understanding of numbers easier.
## FAQs on Bundling in Maths
1. How does bundling help kids learn to count?
We know that illustrations always help while learning anything. So in the case of bundling, we will have bundles that will help the kids to learn more frequently and easily.
2. What exactly is place value?
Every digit in a number in mathematics has a place value. The value a digit in a number represents based on where it is in the number is known as place value. For instance, the place value of 7 in 3,743 is 700, which is 7 hundred. Even though the digits are the same in both numbers, we can see that their place value changes due to their position.
3. How is place - value chart helpful?
Place Value Chart is a very helpful table format that enables us to determine the place value of each digit based on where it appears in a number. Making a place-value model with actual objects, such as place-value blocks or bundles of sticks, writing the digits in the chart, and then writing the number in the usual or standard form, are excellent ways to see the place-value relationships in numbers.
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# Trigonometric Identities (Revision : 1.
4)
1 Trigonometric Identities you must remember
The big three trigonometric identities are sin2 t + cos2 t = 1 sin(A + B) = sin A cos B + cos A sin B cos(A + B) = cos A cos B sin A sin B (1) (2) (3)
Using these we can derive many other identities. Even if we commit the other useful identities to memory, these three will help be sure that our signs are correct, etc.
## 2 Two more easy identities
From equation (1) we can generate two more identities. First, divide each term in (1) by cos2 t (assuming it is not zero) to obtain tan2 t + 1 = sec2 t. When we divide by sin t (again assuming it is not zero) we get 1 + cot2 t = csc2 t. (5)
2
(4)
## 3 Identities involving the difference of two angles
From equations (2) and (3) we can get several useful identities. First, recall that cos(t) = cos t, From (2) we see that sin(A B) = sin(A + (B)) = sin A cos(B) + cos A sin(B) which, using the relationships in (6), reduces to sin(A B) = sin A cos B cos A sin B. In a similar way, we can use equation (3) to nd cos(A B) = cos(A + (B)) = cos A cos(B) sin A sin(B) which simplies to cos(A B) = cos A cos B + sin A sin B. (8) Notice that by remembering the identities (2) and (3) you can easily work out the signs in these last two identities. 1 (7) sin(t) = sin t. (6)
## 4 Identities involving products of sines and cosines
If we now add equation (2) to equation (7) sin(A B) = sin A cos B cos A sin B +(sin(A + B) = sin A cos B + cos A sin B) we nd sin(A B) + sin(A + B) = 2 sin A cos B and dividing both sides by 2 we obtain the identity sin A cos B = 1 1 sin(A B) + sin(A + B). 2 2 (9)
In the same way we can add equations (3) and (8) cos(A B) = cos A cos B + sin A sin B +(cos(A + B) = cos A cos B sin A sin B) to get cos(A B) + cos(A + B) = 2 cos A cos B which can be rearranged to yield the identity cos A cos B = 1 1 cos(A B) + cos(A + B). 2 2 (10)
Suppose we wanted an identity involving sin A sin B. We can nd one by slightly modifying the last thing we did. Rather than adding equations (3) and (8), all we need to do is subtract equation (3) from equation (8): cos(A B) = cos A cos B + sin A sin B (cos(A + B) = cos A cos B sin A sin B) This gives cos(A B) cos(A + B) = 2 sin A sin B or, in the form we prefer, sin A sin B = 1 1 cos(A B) cos(A + B). 2 2 (11)
## Double angle identities
Now a couple of easy ones. If we let A = B in equations (2) and (3) we get the two identities sin 2A = 2 sin A cos A, cos 2A = cos2 A sin2 A. 2 (12) (13)
## 6 Identities for sine squared and cosine squared
If we have A = B in equation (10) then we nd cos A cos B = 1 1 cos(A A) + cos(A + A) 2 2 1 1 2 cos A = cos 0 + cos 2A. 2 2
Simplifying this and doing the same with equation (11) we nd the two identities cos2 A = 1 (1 + cos 2A), 2 1 sin2 A = (1 cos 2A). 2 (14) (15)
## 7 Identities involving tangent
Finally, from equations (2) and (3) we can obtain an identity for tan(A + B): tan(A + B) = sin(A + B) sin A cos B + cos A sin B = . cos(A + B) cos A cos B sin A sin B
Now divide numerator and denominator by cos A cos B to obtain the identity we wanted: tan(A + B) = tan A + tan B . 1 tan A tan B (16)
We can get the identity for tan(A B) by replacing B in (16) by B and noting that tangent is an odd function: tan(A B) = tan A tan B . 1 + tan A tan B (17)
8 Summary
There are many other identities that can be generated this way. In fact, the derivations above are not unique many trigonometric identities can be obtained many different ways. The idea here is to be very familiar with a small number of identities so that you are comfortable manipulating and combining them to obtain whatever identity you need to.
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Combinatorics - Combinatorial Analysis
Variations or permuted combinations (permutations without repetition)
Variations with repetition (or permuted combinations with repetition)
Variations or permuted combinations (permutations without repetition)
The variations of size r chosen from a set of n different objects are the permutations of combinations of r.
The number of variations of size r chosen from n objects equals the number of combinations of size r multiplied by the r! permutations,
Example: Find the number of variations of size 3 that can be made from digits 1, 2, 3, 4 and write them out.
Solution: Since, n = 4 and r = 3 then
Notice that there are 4 combinations of size 3 chosen from the given 4 digits, each of which gives six permutations as is shown below.
The variations are, 1 2 3 1 2 4 1 3 4 2 3 4
1 3 2 1 4 2 1 4 3 2 4 3
2 1 3 2 1 4 3 1 4 3 2 4
2 3 1 2 4 1 3 4 1 3 4 2
3 1 2 4 1 2 4 1 3 4 2 3
3 2 1 4 2 1 4 3 1 4 3 2.
Variations with repetition (or permuted combinations with repetition)
The number of ways to choose r objects from a set of n different objects when order is important and one object can be chosen more than once
Example: Find the number of variations with repetition of size 4 that can be made from digits 0, 1, 2 and write them out.
Solution: Since, n = 3 and r = 4 then the total number of the variations with repetition is
As in the above example, first we should select all combinations with repetition of size 4 from the 3 given digits, and then write the permutations of each of them.
Let calculate the number of combinations
The 15 distinct combinations are divided into four groups depending on the number of permutations each group yields,
a b c d 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 2 1 1 1 1 0 0 0 2 0 0 2 2 1 1 0 2 2 2 2 2 1 1 1 0 1 1 2 2 2 2 0 1 1 1 1 2 2 2 2 0 2 2 2 1
The 3 combinations from the group a can not be rearranged.
Each combination from the group b, c and d, gives,
b) c) d)
as they are the permutations of the 4 objects some of which are the same.
Thus, the permuted combinations (or the variations) of the group b are,
0 0 0 1 0 0 0 2 1 1 1 0 1 1 1 2 2 2 2 0 2 2 2 1
0 0 1 0 0 0 2 0 1 1 0 1 1 1 2 1 2 2 0 2 2 2 1 2
0 1 0 0 0 2 0 0 1 0 1 1 1 2 1 1 2 0 2 2 2 1 2 2
1 0 0 0 2 0 0 0 0 1 1 1 2 1 1 1 0 2 2 2 1 2 2 2 ( 6 ´ 4 = 24 variations)
the variations of the group c are, the variations of the group d are,
0 0 1 1 0 0 2 2 1 1 2 2 0 0 1 2 1 1 0 2 2 2 0 1
0 1 0 1 0 2 0 2 1 2 1 2 0 0 2 1 1 1 2 0 2 2 1 0
0 1 1 0 0 2 2 0 1 2 2 1 0 1 0 2 1 0 1 2 2 0 2 1
1 1 0 0 2 2 0 0 2 2 1 1 0 1 2 0 1 0 2 1 2 0 1 2
1 0 1 0 2 0 2 0 2 1 2 1 0 2 0 1 1 2 1 0 2 1 2 0
1 0 0 1 2 0 0 2 2 1 1 2 0 2 1 0 1 2 0 1 2 1 0 2
( 3 ´ 6 = 18 variations) 1 0 0 2 0 1 1 2 0 2 2 1
1 0 2 0 0 1 2 1 0 2 1 2
1 2 0 0 0 2 1 1 0 1 2 2
2 0 0 1 2 1 1 0 1 2 2 0
2 0 1 0 2 1 0 1 1 2 0 2
2 1 0 0 2 0 1 1 1 0 2 2
( 3 ´ 12 = 36 variations)
Therefore, the total number of the variations of size 4 with repetition chosen from the given 3 digits are,
V(3, 4) = 3 + 6 ´ 4 + 3 ´ 6 + 3 ´ 12 = 3 · (1 + 8 + 6 + 12) = 3 · 27 = 3 · 33 = 34 = 81.
Example: Find the number of variations with repetition of size 8 that can be made from the binary digits 0, 1.
Solution: Since, n = 2 and r = 8 then the total number of the variations with repetition is
V(n, r) = nr => V(2, 8) = 28 = 256.
First we should select the combinations of size 8 that can be made from the 2 binary digits, then examine the number of ways each combination can be rearranged or permuted to prove the total number of variations.
So, the number of the combinations is
The 9 combinations are,
1)
0 0 0 0 0 0 0 0
this combination can not be rearranged or permuted
2)
1 1 1 1 1 1 1 1
this combination can not be rearranged or permuted
3)
0 0 0 0 0 0 0 1
4)
0 0 0 0 0 0 1 1
5)
0 0 0 0 0 1 1 1
6)
0 0 0 0 1 1 1 1
7)
0 0 0 1 1 1 1 1
8)
0 0 1 1 1 1 1 1
9)
0 1 1 1 1 1 1 1
Therefore, the total number of the variations of size 8 with repetition chosen from the given 2 digits are,
V(2, 8) = 2 + 2 ´ 8 + 2 ´ 28 + 2 ´ 56 + 70 = 2 · (1 + 8 + 28 + 56 + 35) = 2 · 128 = 2 · 27 = 28
V(2, 8) = 28 = 256.
Both numerical and non numerical data can be processed by the computer as all letters digits and special characters are coded (represented as a unique sequence of binary digits 0 and 1) using binary variations.
Combinatorics and probability contents
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# Archimedes Principle Explanation
## Introduction
Archimedes’ Principle is an important principle of fluid mechanics. It is discovered by a Greek mathematician and scientist Archimedes. This principle is utilized to estimate the volume of an object that has an irregular shape, density, and distinct gravity of an object.
## What is Archimedes' Principle?
A body placed on a fluid (liquid or gas) will experience an upward buoyant force. This force is equal to the amount of fluid displaced by the body. In the case of a completely immersed body or partially immersed body, the volume of the fluid displaced by the body will be equal to the volume of that body.
## The statement of Archimedes' Principle
If a body is totally or partly plunged in a liquid at rest, few parts of its weight appear to lose. This apparent loss of weight in the body is equal to the weight of the fluid displaced by that body.
## Archimedes' Principle Formula
Archimedes’ law of fluid states that the buoyancy force on a body due to the fluid is equal to the weight of the fluid displaced by the body. The mathematical expression of Archimedes’ law can be written as,
$$\mathrm{F=\rho Vg}$$
Here,
F= buoyant force act on immersed bogy
ρ=density of the fluid
V= volume of the displaced fluid
## Archimedes' Principle Derivation
Let us consider, a cube is immersed in a liquid of density ρ. The depth of the ABCD and EFGH surfaces of the cube from the upper surface of the liquid be $\mathrm{h_2}$ and $\mathrm{h_1,}$ respectively. The length of each side of the cube is l. The liquid exerts a normal thrust on each surface of the cube. The lateral thrust acting horizontally on the mutually opposite vertical faces ABHE and CDFG of the cube is equal and opposite.
Similarly, the lateral thrust acting on the faces AEFD and BCGH balances each other. So, in the horizontal direction, the resultant lateral force is zero. But, an unequal thrust acts on the surfaces ABCD and EFGH due to the difference in depths of these surfaces from the upper side of the liquid.
The downward pressure of the liquid at any point on $\mathrm{ABCD=\rho gh_1}$
The downward thrust on $\mathrm{ABCD=l^2\:\rho gh_1}$
The upward pressure of the liquid at any point on $\mathrm{EFGH=\rho\:gh_2}$
Then, The upward thrust on $\mathrm{EFGH=l^2 \rho gh_2}$
Since $\mathrm{h_2 > h_1}$, the upward thrust acting on the cube is greater than the downward thrust.
Then,
net upward thrust $\mathrm{l^2\:\rho gh_2-l^2 ρgh_1=l^2 \rho g(h_2-h_1)=l^3 \rho g }$
Since, $\mathrm{h_2-h_1=l }$
But, $\mathrm{l^3=V (volume\:of\:the\:cube)}$
Hence,
$$\mathrm{F = l^3\:\rho\:g=V\:\rho g}$$
So, the buoyancy force on the cube = weight of the liquid displaced by the cube.
## Laws of Floating
There are three laws of flotation. So, let W is the weight of the body and w is the buoyant force.
• If W>w, i.e., the weight of the body is greater than buoyant force then the body will go down to the bottom of the liquid.
• If W<w, i.e., the weight of is less than buoyant force then the body will float partially submerged in the liquid
• If W=w, i.e., then the body will stay afloat in liquid if its whole volume is just immersed in liquid.
## Archimedes' Principle Examples
Example 1: A piece of wood of volume 20.5 cm3 is tied to a piece of lead of volume 1 cm3. State whether the combination will float or sink in water. [Specific gravity of wood and lead are respectively 0.5 and 11.4].
Ans: $\mathrm{Weight\:of\:lead=(1×11.4)gm×g=11.4gm×g}$
$\mathrm{Weight\:of\:the\:piece\:of\:wood\: =(20.5×0.5)gm×g=10.25gm×g}$
$$\mathrm{Total\:weight=21.65\:gm×g}$$
$\mathrm{Total\:Volume=(20.5+1)cm^3=21.5Cm^3}$
So, the weight of displaced water by the combination=21.5gm×g
Hence, the weight of the combination >weight of displaced water
Therefore, the combination will sink into the water.
## Archimedes' Principle Experiment
According to Archimedes’ principle,
$$\mathrm{Apparent\:weight=weight\:of\:the \:body-buoyant\:force }$$
Let’s have a look at an experiment,
• Let us take a container filled with water. The container is connected with a bowl such that if the water overflows it will displace the bowl through a pipe.
• Now take any solid object, and measure its weight by employing a spring balance in the air. Note down the value, it is 6kg.
• Next, attach the object to the spring balance and immerse it in water. The spring balance must be out of water.
• Now, as the weight is lowered some water will be enter into the bowl. The weight of the displaced water is 2 kg.
• The spring balance will display less value than previously. Note down the value which is 4 kg.
• We found that when we subtract the new weight of the body from the first measured value, it will come out equal to the weight of the displaced water.
MikeRun, Archimedes-principle, CC BY-SA 4.0
## Application of Archimedes' Principle
### The upward Motion of the Balloon
The upward motion of a balloon depends on the thrust of the air. The net weight of the balloon is much less compared to the air displaced by it. Hence, the buoyant force acting on the balloon is greater than its weight. So, the balloon undergoes a resultant upward force and moves up.
### Submarine
A submarine can be submerged in water and float in water too. A submarine is comprised of counterbalance tanks and these tanks can be loaded up with air or water. At the point when the tanks are loaded up with air, the submarine floats on water in light of the fact that the heaviness of the submarine becomes lower than the heaviness of the water dislodged by it. At the point when water is permitted to enter the tanks, the heaviness of the submarine becomes more noteworthy than the heaviness of the water uprooted by it and consequently, the submarine lowers in water.
### Life Belts
A life belt is an air-filled bag. The weight of the life belt along with the weight of the person is less than the weight of the water displaced. Hence, by wearing a life belt a person can float.
## Conclusion
A body partially or fully plunged in a liquid will experience a loss of weight. This loss of weight is equivalent to the weight of the liquid shift by the body. This is known as Archimedes' Principle. This rule does not apply to a body that is in a weightless condition such as in artificial satellites or free-fall. Archimedes' Principle acts in the case of submarines, life belts, balloons, swimming, ships, etc. With the help of this Principle volume of a solid of any shape, the density of a substance, etc can be determined.
## FAQs
Q1. What is Buoyancy?
Ans. The ability of a fluid to exert an upward force at rest condition on a body immersed in that fluid is called buoyancy.
Q2. Explain whether Archimedes' Principle is applicable in the case of a free-falling body under gravity.
Ans. We know that, free-falling body is weightless. So, this Principle isn’t applicable in this case. Because a free-falling body is weightless.
Q3. An egg sinks in freshwater, but it floats in salt water. Why?
Ans. The density of an egg is higher than fresh water. So that, it sinks into the water. But, when a suitable amount of salt is added to water, the density of salt water becomes higher than an egg, and the egg floats on saline water.
Q4. Will Archimedes' Principle be applicable within an artificial satellite revolving around the sun?
Ans. All bodies become weightless inside an artificial satellite revolving around the sun. Hence, the weight of the object and the fluid displaced by the object both will be zero and the object will not experience any buoyant force. Therefore, Archimedes' Principle will not be applicable within an artificial satellite revolving around the sun.
Q5. A solid sphere and hollow sphere having the same mass and external radius are immersed in the same liquid. Which one will feel heavier?
Ans. The radius and volume of both solid and hollow spheres are equal. Hence, the liquid displaced by them will be the same. Therefore, the loss of weight in both cases will be equal and the apparent weight of both spheres will be the same.
Updated on: 18-Apr-2023
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2220hw4sol
# 2220hw4sol - Math 2220 Problem Set 4 Solutions Spring 2010...
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Unformatted text preview: Math 2220 Problem Set 4 Solutions Spring 2010 Section 4.3 : 4. Use Lagrange multipliers to find the critical points of the function f ( x, y ) = xy subject to the constraint 2 x − 3 y = 6. Solution. The critical points are solutions of the system y = λ (2) x = λ ( − 3) 2 x − 3 y = 6 The first two equations give 2 x = − 3 y . Substituting into the third equation, we obtain x = 3 / 2 and y = − 1, so (3 / 2 , − 1) is the only critical point. 8. Use Lagrange multipliers to find the critical points of the function f ( x, y, z ) = x + y + z subject to the constraints y 2 − x 2 = 1 and x + 2 z = 1. Solution. The critical points are solutions of the system 1 = λ ( − 2 x ) + μ (1) 1 = λ (2 y ) + μ (0) 1 = λ (0) + μ (2) y 2 − x 2 = 1 x + 2 z = 1 The third equation gives μ = 1 / 2 and then the first two equations simplify to λx = − 1 / 4 and λy = 1 / 2. These equations imply in particular that x and y are nonzero so we can solve them for λ to get λ = − 1 / 4 x = 1 / 2 y and hence y = − 2 x . Plugging this into the fourth equation gives x = ± radicalbig 1 / 3. Using the fifth equation to obtain the z values, we thus have two critical points ( x, y, z ) = ( radicalbig 1 / 3 , − 2 radicalbig 1 / 3 , (1 − radicalbig 1 / 3) / 2) and ( − radicalbig 1 / 3 , 2 radicalbig 1 / 3 , (1 + radicalbig 1 / 3) / 2) 18. Find the maximum and minimum values of f ( x, y, z ) = x + y − z on the sphere x 2 + y 2 + z 2 = 81. Explain how you know there must be both a maximum and a minimum attained. Solution. The critical points are solutions of the system 1 = λ (2 x ) 1 = λ (2 y ) − 1 = λ (2 z ) x 2 + y 2 + z 2 = 81 1 Math 2220 Problem Set 4 Solutions Spring 2010 The first three equations give that x = y = − z . Substituting into the last equation we get 3 x 2 = 81, i.e., x = 3 √ 3. Thus we have 2 critical points (3 √ 3 , 3 √ 3 , − 3 √ 3) and ( − 3 √ 3 , − 3 √ 3 , 3 √ 3). The values of the function f at these two points are 9 √ 3 and − 9 √ 3....
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# Equations and Inequalities
Scheme of work: Year 12 A-Level: Pure 1: Equations and Inequalities
#### Prerequisite Knowledge
• Be able to use set notation to represent equalities.
• Represent linear equalities and their solutions graphically.
• Solve a quadratic equation in the form x2 + bx + c = 0
• Solve a pair of simultaneous equations by elimination.
#### Success Criteria
• Solve linear inequalities, e.g. ax + b > cx + d
• Solve quadratic inequalities, e.g. ax2 + bx + c > 0
• Solve simultaneous linear equations by substitution.
• Use the substitution method to solve simultaneous equations where one equation is linear and the other is quadratic.
• Give solutions of inequalities using set notation.
• Represent quadratic inequalities and their solutions graphically.
#### Key Concepts
• Linear simultaneous equations can be solved by elimination or substitution.
• When using graphs with solving simultaneous quadratic and linear equations the number of intersections is equal to the number of solutions.
• When you multiply or divide an inequality by a negative number, you need to change the inequality sign to its opposite.
• To solve a quadratic inequality
• find the roots and intercept of the quadratic equation
• sketch the graph of the quadratic function, then
• use the sketch to find the required set of values.
#### Common Misconceptions
• Students often forget to change the inequality sign when multiplying or dividing by a negative.
• Students can get confused with the set notation for solving quadratic inequalities. Encourage them to sketch a graph and marked on the desired range of values. For instance, x2 – 7x + 12 < 0 can have the incorrect solution of x < 3 and x > 4 rather than 3 < x < 4.
• When solving simultaneous equations students often forget to find both the x and y solutions after finding one.
## Equations and Inequalities Resources
### Mr Mathematics Blog
#### Estimating Solutions by Rounding to a Significant Figure
Explore key concepts, FAQs, and applications of estimating solutions for Key Stage 3, GCSE and IGCSE mathematics.
#### Understanding Equivalent Fractions
Explore key concepts, FAQs, and applications of equivalent fractions in Key Stage 3 mathematics.
#### Transforming Graphs Using Function Notation
Guide for teaching how to transform graphs using function notation for A-Level mathematics.
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## Area of Compound Shapes
A compound shape is a shape made up of different shapes.
Here is a compound shape:
To work out the area of this shape we can split it into two rectangles.
There are two ways of splitting the shape into two rectangles. We can split the shape like this:
To find the area of the whole shape we will work out the area of the two rectangles and add them together.
To find the area of a rectangle we need to multiply the length by the width.
We know the length and the width of the bigger rectangle. We need to find a missing length of the smaller rectangle.
The height of the whole shape is 9 cm and the height of the bottom shape is 5 cm. To find the height of the top shape we need to take 5 away from 9
9 - 5 = 4
(The smaller rectangle is a square)
We now have all the information we need to calculate the areas.
The area of the large rectangle is 5 × 10 = 50 cm2
The area of the square is 4 × 4 = 16 cm2
The area of the whole shape is 16 + 50
16 + 50 = 66 cm2
We could have also split the shape like this:
We need to find the base of the rectangle on the left. The base of the whole shape is 20 cm and the base of the shape on the right is 4 cm.
We need to calculate: 10 - 4
10 - 4 = 6 cm
The area of the rectangle on the left is 5 × 6 = 30 cm2
The area of the rectangle on the right is 4 × 9 = 36 cm2
The area of the whole shape is 30 + 36
30 + 36 = 66 cm2
Here is another compound shape:
To find the area of the compound shape we need to split it into a rectangle and a triangle.
We know the length and width of the rectangle.
The area of a triangle = 12 × base × height
We need to work out the base and the height of the triangle.
The base is 12 - 6 = 6 cm
The height is 11 - 4 = 7 cm
The area of the rectangle = 12 × 4 = 48 cm2
The area of the triangle = 12 × 6 × 7
12 × 6 × 7 = 3 × 7 = 21 cm2
The total area = 48 + 21 = 69 cm2
Try these:
Find the area of the shape:
Find the area of the shape:
In this shape we have a parallelogram cut out of a rectangle:
To find the area of this shape we need to take away the area of the parallelogram from the area of the rectangle.
The area of a rectangle = length × width
Area of rectangle = 11 × 7 = 77 cm2
The area of a parallelogram = base × height
Area of parallelogram = 2 × 5 = 10 cm2
The area of the shape = 77 - 10 = 67 cm2
In this shape we have a circle cut out of a triangle:
To find the area of this shape we need to take away the area of the circle from the area of the triangle.
The area of a triangle = 12 × base × height
The triangle's area = 12 × 9 × 11 = 992 = 49.5 cm2
The area of a circle = π × radius2
The radius of a circle is half of its diameter.
42 = 2 cm
The circle's area = π × 22
The circle's area = π × 4
We can write this as 4π cm2
We can leave our answer in terms of pi (if we have a calculator we can convert it to a decimal).
The area of the shape = 49.5 - 4π cm2
We can type this into a calculator to convert the answer to a decimal, this would give an answer of 36.9 cm2 to one decimal place.
Try this:
Find the area of the shape:
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Let’s understand the concept of per unit system by solving an example. In the one-line diagram below, the impedance of various components in a power system, typically derived from their nameplates, are presented. The task now is to normalize these values using a common base.
Figure 1: Oneline Diagram Of A Power System
Now that you have carefully examined the system and its parameters, the equivalent impedance diagram for the above system would look something like the following.
Figure 2: Impedance Diagram Of A Power System
Resistive impedance for most components have been ignored. Rotating machines have been replaced with a voltage source behind their internal reactance. Capacitive effects between lines and to ground are ignored as well.
To obtain the new normalized per unit impedances, first we need to figure out the base values (Sbase, Vbase, Zbase) in the power system. Following steps will lead you through the process.
### Step 1: Assume a system base
Assume a system wide $S_{base}$ of 100MVA. This is a random assumption and chosen to make calculations easy when calculating the per unit impedances.
So, $S_{base}$ = 100MVA
### Step 2: Identify the voltage base
Voltage base in the system is determined by the transformer. For example, with a 22/220kV voltage rating of T1 transformer, the $V_{base}$ on the primary side of T1 is 22kV while the secondary side is 220kV. It does not matter what the voltage rating of the other components are that are encompassed by the $V_{base}$ zone.
See figure below for the voltage bases in the system.
Figure 3: Voltage Base In The Power System
### Step 3: Calculate the base impedance
The base impedance is calculated using the following formula:
$Z_{base}=\frac{{kV_{base}}^2}{S_{base MVA}}$ Ohms…………………………………………………………………..(1)
For T-Line 1: $Z_{base}=\frac{(220)^2}{100}$= 484 Ohms
For T-Line 2: $Z_{base}=\frac{(110)^2}{100}$= 121 Ohms
For 3-phase load: $Z_{base}=\frac{(11)^2}{100}$= 1.21 Ohms
### Step 4: Calculate the per unit impedance
The per unit impedance is calculated using the following formulas:
$Z_{p.u.}=\frac{Z_{actual}}{Z_{base}}$ ……………………………………………………………………………..(2)
$Z_{p.u._{new}}=Z_{p.u._{old}}(\frac{S_{base_{new}}}{S_{base_{old}}})(\frac{V_{base_{old}}}{V_{base_{new}}})^2$ ……………………………….(3)
The voltage ratio in equation (3) is not equivalent to transformers voltage ratio. It is the ratio of the transformer’s voltage rating on the primary or secondary side to the system nominal voltage on the same side.
For T-line 1 using equation (2): $X_{l1_{p.u.}}=\frac{48.4}{484}$= 0.1 pu
For T-line 2 using equation (2): $X_{l2_{p.u.}}=\frac{65.43}{121}$= 0.5 pu
Power Factor: $\cos^{-1}(0.6)=\angle{53.13}$
Thus, $S_{3\phi}(load)=57\angle{53.13}$
$Z_{act}=\frac{(V_{rated})^2}{\overline{S}^*}= \frac{10.45^2}{57\angle{-53.13}}$= 1.1495+j1.53267 Ohms
Per unit impedance of 3-phase load using equation (2)= $\frac{1.1495+j1.5326}{1.21}$ = 0.95+j1.2667 pu
For generator, the new per unit reactance using equation (3)
$X_{sg}= 0.18(\frac{100}{90})(\frac{22}{22})^2$ = 0.2 pu
For transformer T1: $X_{t1}= 0.1(\frac{100}{50})(\frac{22}{22})^2$ = 0.2 pu
For transformer T2: $X_{t2}= 0.06(\frac{100}{40})(\frac{220}{220})^2$ = 0.15 pu
For transformer T3: $X_{t3}= 0.064(\frac{100}{40})(\frac{22}{22})^2$0.16 pu
For transformer T4: $X_{t4}= 0.08(\frac{100}{40})(\frac{110}{110})^2$0.2 pu
For Motor, $X_{sm}= 0.185(\frac{100}{66.5})(\frac{10.45}{11})^2$0.25 pu
The equivalent impedance network with all the impedances normalized to a common system base and the appropriate voltage base is provided below.
Per Unit Impedance Diagram
### Summary:
1. Assume a Sbase for the entire system.
2. The Vbase is defined by the transformer and any off-nominal tap setting it may have.
3. Zbase is derived from the Sbase and Vbase.
4. The new per unit impedance is obtained by converting the old per unit impedance on old base values to new ones. See equations (2) and (3).
* * * * *
### 53 Responses to Per Unit System – Practice Problem Solved For Easy Understanding
1. Tom says:
Can someone help please how can i calculate R and X from p.u. to base units?
2. samantha says:
please help: Three transformers each rated 25 MVA, 38. 1 /3.81 kV are connected star-delta with a balanced load of three 0.6?, Y-connected resistors. Choose a base of 75 MVA, 66 kV for the high-voltage side of the transformer and specify the base for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then, determine the load resistance R_L i n ohms referred to the high-voltage side and the per-unit value of this resistance on the chosen base.
3. daran says:
4. Rusty says:
Shouldn’t the 3-ph load be multiplied by (100/57)?
5. Mian Faizan says:
Love it.. A job well done. 🙂
6. PAUL PEREZ says:
HELP! How do determine the equivalent three phase impedance of three single phase transformer? Example. Three each of 10 kva transformer, 7.52kv-240V, wiith an individual impedance of 4%. What is the equivalent impedance when they are connected in three phase at 30 kva.
@sayonsom
What we developed in this exercise is a positive sequence network. A transformer’s winding configuration does not affect it. The positive sequence network is good for calculating balanced load current and 3 phase faults (not involving ground).
The zero sequence network however does get affected by transformer winding configuration. Delta and ungrounded Y have a big impact in the design of zero sequence impedance network.
You will need a zero sequence network for unsymmetrical fault current analysis like L-G or L-L-G etc.
I am getting into symmetrical components with this discussion.
8. Vanjinathan P says:
good thing, to see a dedicated power system website.
9. sayonsom says:
How does the calculations change if the transformers are Y-Y or Y-Delta connected in different parts of the network?
10. Shalini says:
Awesome !!! Best explanation ever.. thank u
11. Naz says:
good job. its a comprehensive article. Appreciate your hard work
12. Jathin says:
Thanks a lot.. Nice elaboration..
13. Ashwin says:
Can someone tell me what would happen if for T2 and T4, the Primary and Secondary were swapped i.e. T2 would be 11/220 and T4 would be11/110?
I have a simialr problem and I am getting two different values for Vbase4.
14. narayan says:
IJAJ :what do you mean by….. s*-??
ans: it mean conjugate of S i.e changing sign of angle only
mike: When calculating Xtl2 using (22/22) . Vbase in T2 is 220 primary and 11 secondary, 220 come from?
ans: in all transformer we are allowed only take primary or secondary as reference. here primary is taken
15. Harsh says:
sir how to choose base kv value transformer? some time you take (22/22) nd sometime (10.45/11)
16. Aslam says:
Hi, thanks for the nice sharing,
I wana know how to model delta-grounded y transformer for load flow calculation.thanks
17. tuafi says:
nice job dude, highly comprehensible and presise, thumbs up
18. CJ says:
Good example of the method but pay attention the generator can supply only 90MVA the loads absorb 57 + 66.5 =123.5 MVA the network will in reality overload or under power the loads
19. Amey says:
if the transformer’s secondary is grounded by a neutral impedance then how to proceed with the calculations please suggest with an example
20. demis tesfaw says:
is that posible to calculate each parameter without giving a base voltage?
good job
22. IJAJ says:
what do you mean by….. s*-??
23. an says:
Can we find the short circuit current at each end?
24. samuel says:
A load of 50mw at 0.8 power factor lagging is taken from the 33kv.( taking a base MVA of 100mva), calculate the terminal voltage of the synchronous machine? (Please help me solve this question) thanks
25. Nikhil says:
very useful thanks 🙂
26. alshaia says:
How can we determine the voltage on the bus 1
27. BRian says:
Do you know how to find the voltage at the bus?
Thanks
28. tahseen says:
Hi
how we can find the voltage in bus1 in PU and in volte
29. noa says:
tanks alot save me alot of stress
30. abi says:
. Obtain the per unit impedance(reactance) diagram of the power system shown in the fig
G1 : 30MVA , 10.5KV, X?=1.6 ?
G2 : 15MVA , 6.6KV, X?=1.2 ?
G3 : 25MVA , 6.6KV, X?=0.56 ?
T1 (3 phase): 15MVA , 33/11KV , X= 15.2 ? per phase on the high tension side
T2 (3 phase): 15MVA , 33/6.2KV , X= 16 ? per phase on the high tension side
Transmission line : 20.5 ohm per phase
Load A : 15MW , 11KV , 0.9 p.f lagging
Load B : 40MW , 6.6KV , 0.85 p.f lagging
5
31. abi says:
how to convert ohms value to per unit value
32. Lee Taylor says:
Hi, great article – thanks very much! I have a similar problem to solve but I am struggling with the Zact calculation. My inputs are Vrated = 4.16kV, S = 2MVA <-36.87. Can you help?!
33. bhanu says:
awesome
34. kaushik vastarpara says:
its really bcoz by reading this my confusion abut selection of base nd other is very clear…sommust read it frend /…thank u
@Pavan @Mike: That’s a typo. Correct values are now shown in the calculations. Since the ratio of Vbase_old/Vbase_new is the same, the end result, therefore, does not change. Appreciate the feedback.
36. Pavan says:
This is really helpful. I didn’t really got it when reading through this, but when I saw the below comment by Mike, it seems like a question worth answering. However the content is really clear and understandable. Keep up the good work!
Thanks,
37. mike says:
I don’t understand one part: When calculating Xtl2 you are using (22/22) which is reflected from where? Vbase in T2 is 220 primary and 11 secondary, so where does 22 come from?
The same for Xtl4.
38. mark says:
will the impedance or p.u impedance in each line will be like in series? will the current for the PRIMARY AND SECONDARY of the transformer now be equal??? how will i find the actual line current for each line and for the whole system…
39. chris says:
A single phase ,350 kva, 1380v generator has an internal impedance Zg of j6 ohms. The generator is used to supply a load of 250kva/440v at power factor 0.78 lagging. determine: the turns ratio of the transfomer, the impedance per km if the line between the generator and the transformer is 5km, the voltage regulation of the system.
Using the ratings of the generator as base values determine the generated per unit voltage that is required to produce a full load current under short circuit condition.
CAN SOMEONE HELP ME WIT THESE CALCULATION PLZ!!!
Kam,
Once you have the impedance network, use the current division rule to determine the current flowing each line. I am not sure I understand “voltage at 3”, if bus 3 is faulted (3ph) then it is zero otherwise it should be the same as nominal voltage as seen on the secondary side of the transformer.
I will solve one for the currents in the future but for now, you will have to learn how to reduce a circuit (using KVL and KCL) to determine the currents.
41. kam says:
sorry but i didnt get my reply yet, so could you pls help me out???????? thanks a lot
42. karthik says:
very well explained but could you pls show me how to calculate voltage and current in both lines, will be very greatful , thanks a lot……….
43. kam says:
It is really well explained but could you pls show me how to calculate voltage at but 3 and current in both lines, will be very greatful , thanks a lot
44. manish says:
what if transformers are connected in star and delta connection?
45. Anayat says:
i am very new to Power side , so i really dont know abt all these concepts , what we only have T1 and T2 , and all the rating given are three phase line to line ? how we ll solve it then?
46. richa says:
very nicely explained….to the point and complete..thanks a lot 🙂
47. Sanket says:
VERY NICELY EXPLAINED THANK-YOU ………
I WILL VISIT WEBSITE AGAIN FOR FURTHER REFERENCES.
48. BABULS RAJ says:
Thank u so much…..after searching for a proper explanation for the same in so many sites, i got it finally from your site. Clear explanation with proper diagrams with multi colour…….very nice …..
Nice catch. Fixed it. Thanks.
50. Renjith M says:
Commendable work. But there is a small error. The per-unit system is the ratio of two quantities of the same units. Therefore it is unitless. Well that is what I know. So accordingly we specify the per-unit quatities as just ‘P.U’. So you need to remove the ‘Ohms’ from the text and insert ‘p.u’
51. Alfredo says:
It was very useful, but it is short because is necessary to get the complete solution, any way I liked.
52. Abdul Rauff says:
Very Good Info About PSA.Thanks Alot
53. Hilary says:
Protection engieering, i have been give the reactance as Xd’ to calculate faults on a system do i convert to Xd” how do i do this
|
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# What Is 50 Of 55?
## Introduction
When it comes to solving mathematical problems, we all have faced some difficulties in understanding the concepts. One such problem that often confuses students is finding out what is 50 of 55. However, the solution to this problem is quite simple, and in this article, we will discuss it in detail.
## Understanding the Concept
To find out what is 50 of 55, we first need to understand the concept of percentages. A percentage is a way of expressing a number as a fraction of 100. For example, 50% can be written as 50/100, which is equal to 0.5.
### Calculating 50 of 55
To calculate 50 of 55, we need to find out what percentage of 55 is 50. To do this, we can use the formula: x/100 * 55 = 50 Here, x represents the percentage we need to find out. We can solve this equation by cross-multiplying: x * 55 = 50 * 100 x = (50 * 100)/55 x = 90.91% Therefore, 50 of 55 is equal to 90.91%.
## Why is it Important to Know?
Understanding the concept of percentages and how to calculate them is essential in many real-life situations. For example, if you are shopping and see a discount of 50% on a product that costs \$55, you can quickly calculate the discounted price using the method we just discussed.
## Other Examples
Let’s take a look at some other examples to understand the concept better.
### Example 1
What is 25% of 80? To solve this problem, we can use the same formula: x/100 * 80 = 25 Cross-multiplying, we get: x * 80 = 2500 x = 31.25% Therefore, 25% of 80 is equal to 31.25.
### Example 2
What is 75% of 120? Using the formula: x/100 * 120 = 75 Cross-multiplying, we get: x * 120 = 7500 x = 62.5% Therefore, 75% of 120 is equal to 62.5.
## Conclusion
Calculating percentages may seem confusing at first, but with a little practice, it can become effortless. Understanding the concept of percentages and how to calculate them is essential in many real-life situations. We hope this article has helped you understand what is 50 of 55 and how to calculate it.
## Can Hoas Evict You?
Introduction If you are living in a community that is governed by a homeowners association …
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# Find the locus of the midpoint of the chord of the parabola ${{y}^{2}}=4ax$, which passes through the point $\left( 3b,b \right)$.
Last updated date: 13th Jul 2024
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Hint: Write the equation of chord and satisfy the given point and use formula for midpoint which is $x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$.
We are given a chord of the parabola which passes through the point $\left( 3b,b \right)$.
Here, we have to find the locus of midpoint of a given chord.
Let the midpoint of the given chord be $\left( h,k \right)$.
We know that any general point on parabola $P\left( t \right)$ is $\left( a{{t}^{2}},2at \right)$.
So, we get point $R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)$ and point $Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)$
We know that equation of any line passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is
$\left( y-{{y}_{1}} \right)=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}\left( x-{{x}_{1}} \right)$
Therefore, we get equation of chord passing through $R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)$ and point $Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)$ as,
$\left( y-2a{{t}_{1}} \right)=\dfrac{\left( 2a{{t}_{2}}-2a{{t}_{1}} \right)}{\left( at_{2}^{2}-at_{1}^{2} \right)}\left( x-at_{1}^{2} \right)$
By cancelling the like terms, we get,
$\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( t_{2}^{2}-t_{1}^{2} \right)}\left( x-at_{1}^{2} \right)$
Since we know that,
$\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$
Therefore, we get,
$\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)}\left( x-at_{1}^{2} \right)$
By cancelling the like terms, we get,
$\left( y-2a{{t}_{1}} \right)=\dfrac{2\left( x-at_{1}^{2} \right)}{\left( {{t}_{2}}+{{t}_{1}} \right)}$
After cross multiplying the above equation, we get,
$\left( y-2a{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)=2\left( x-at_{1}^{2} \right)$
Simplifying the equation, we get,
$y\left( {{t}_{2}}+{{t}_{1}} \right)-2a{{t}_{1}}{{t}_{2}}-2at_{1}^{2}=2x-2at_{1}^{2}$
Finally, we get
$y\left( {{t}_{2}}+{{t}_{1}} \right)-2a{{t}_{1}}{{t}_{2}}=2x.....\left( i \right)$
Now, we know that midpoint say $\left( x,y \right)$ of any line joining points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$is:
$x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}$ and $y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$
Therefore, we get the midpoint $\left( h,k \right)$ of chord joining $R\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)$ and point $Q\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)$ as:
$h=\dfrac{at_{1}^{2}+at_{2}^{2}}{2}....\left( ii \right)$
$k=\dfrac{2a{{t}_{1}}+2a{{t}_{2}}}{2}....\left( iii \right)$
Taking 2a common from equation $\left( iii \right)$, we get,
$k=\dfrac{2a\left( {{t}_{1}}+{{t}_{2}} \right)}{2}$.
Therefore, we get,
$\dfrac{k}{a}=\left( {{t}_{1}}+{{t}_{2}} \right)....\left( iv \right)$
Taking ‘a’ common from equation$\left( ii \right)$, we get,
$h=\dfrac{a\left( t_{1}^{2}+t_{2}^{2} \right)}{2}$
Or, $\dfrac{2h}{a}=t_{1}^{2}+t_{2}^{2}$
Since, we know that ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$
Now, we subtract 2ab from both sides. We get,
${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$
Therefore, we get,
$\dfrac{2h}{a}={{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}-2{{t}_{1}}{{t}_{2}}$
Now, we put the value of $\left( {{t}_{1}}+{{t}_{2}} \right)$ from equation (iv). We get,
$\dfrac{2h}{a}={{\left( \dfrac{k}{a} \right)}^{2}}-2{{t}_{1}}{{t}_{2}}$
Or $2{{t}_{1}}{{t}_{2}}=\dfrac{{{k}^{2}}}{{{a}^{2}}}-\dfrac{2h}{a}$
By dividing both sides by 2, we get,
${{t}_{1}}{{t}_{2}}=\dfrac{{{k}^{2}}}{2{{a}^{2}}}-\dfrac{h}{a}.....\left( v \right)$
Now, we will put the values of $\left( {{t}_{1}}+{{t}_{2}} \right)$ and $\left( {{t}_{1}}{{t}_{2}} \right)$ from equation $\left( iv \right)$and $\left( v \right)$ in equation $\left( i \right)$.
We get,
$y\left( {{t}_{2}}+{{t}_{1}} \right)-2a\left( {{t}_{1}}{{t}_{2}} \right)=2x$
$\Rightarrow y\left( \dfrac{k}{a} \right)-2a\left( \dfrac{{{k}^{2}}}{2{{a}^{2}}}-\dfrac{h}{a} \right)=2x$
$\Rightarrow \dfrac{yk}{a}-2a\left( \dfrac{{{k}^{2}}-2ha}{2{{a}^{2}}} \right)=2x$
By cancelling the like terms, we get,
$\Rightarrow \dfrac{yk}{a}-\dfrac{\left( {{k}^{2}}-2ha \right)}{a}=2x$
After simplifying and cross-multiplying above equation, we get,
$\left( yk \right)-\left( {{k}^{2}}-2ha \right)=2xa$
Now, we are given that this chord passes through point $\left( 3b,b \right)$.
Therefore, we will put $x=3b$ and $y=b$.
We get,
$bk-\left( {{k}^{2}}-2ha \right)=2\left( 3b \right)a$
$\Rightarrow bk-{{k}^{2}}+2ha=6ab$
By transposing all the terms to one side,
We get,
${{k}^{2}}-bk-2ha+6ab=0$
Now, to get the locus, we will replace h by x and k by y. We get,
${{y}^{2}}-by-2ax+6ab=0$
So, the locus of the midpoint of chord passing through $\left( 3b,b \right)$ is ${{y}^{2}}-by-2ax+6ab=0$.
Note: In these types of questions, we can directly write the equation of chord with respect to mid-point say $\left( {{x}_{1}},{{y}_{1}} \right)$ which is $\left( y-{{y}_{1}} \right)=\dfrac{2a}{{{y}_{1}}}\left( x-{{x}_{1}} \right)$ and put $\left( h,k \right)$ in place of $\left( {{x}_{1}},{{y}_{1}} \right)$ and point through which chord in passing [here (3b,b)] in place of $\left( x,y \right)$ to get locus of midpoint of chord.
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Determining the Validity of an Argument by Rules of Inference
# Determining the Validity of an Argument by Rules of Inference
Let $P$, $Q$, and $R$ be statements. So far we have discussed the following rules of inference:
• Modus Ponens: $(P \wedge (P \rightarrow Q)) \rightarrow Q$.
• Modus Tollens: $((P \rightarrow Q) \wedge \neg Q) \rightarrow \neg P$.
We will now look at an example of determining if an argument is valid by using these rules of inference. In the examples below, we will list our premises and then list our conclusion below a dotted line.
## Example 1
Determine if the following argument is valid:
(1)
\begin{matrix} \mathrm{Premise \: 1} & P\\ \mathrm{Premise \: 2} & P \rightarrow \neg Q\\ \mathrm{Premise \: 3} & \neg Q \rightarrow \neg R\\ & ---\\ \mathrm{Conclusion} & \therefore \neg R \end{matrix}
The argument is valid. Here are the steps:
(2)
\begin{align} \quad 1. & P \rightarrow \neg Q & (\mathrm{Premise \: 2}) \\ \quad 2. & \neg Q \rightarrow \neg R & (\mathrm{Premise \: 3}) \\ \quad 3. & P \rightarrow \neg R & (\mathrm{The \: Law \: of \: Syllogism \: with \: (1) \: and \: (2)}) \\ \quad 4. & P & (\mathrm{Premise \: 1}) \\ \quad 5. & \neg R & (\mathrm{Modus \: Ponens \: with \: (4) \: and \: (3)}) \end{align}
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## Using the Definition to Compute the Derivative
We have seen in the previous page how the derivative is defined: For a function f(x), its derivative at x=a is defined by
Let us give some examples.
Example 1. Let us start with the function f(x) = x2. We have
So
which means f '(a) = 2a.
What about the derivative of f(x) = xn. Similar calculations, using the binomial expansion for (x+y)n (Pascal's Triangle), yield
Example 2. Consider the function f(x)=1/x for . We have
Consequently,
Have you noticed? The algebraic trick in both of the examples above has been to factor out "h" in the numerator, so that we can cancel it with the "h" in the denominator! This is what you try to do whenever you are asked to compute a derivative using the limit definition.
You may believe that every function has a derivative. Unfortunately that is not the case.
Example 3. Let us discuss the derivative of f(x) = |x| at 0. We have
But
which implies that f '(0) does not exist.
Remark. This example is interesting. Even though the derivative at the point does not exist, the right and the left limit of the ratio do exist. In fact, if we use the slope-interpretation of the derivative we see that this means that the graph has two lines close to it at the point under consideration. They could be seen as "half-tangents". See Picture.
So let's push it a little bit more and ask whether a function always has a tangent or half-tangents at any point. That is not the case either.
Example 4. Let us consider the function for , with f(0) = 0. We have
Recall that the function has no limit when x goes to 0. So the function has no derivative and no half-derivatives as well at x=0.
What else can go wrong?
Example 5. Consider the function . Then we have
Since
then f '(0) does not exist. But observe that the graph as a geometric figure has a tangent -- albeit vertical:
In fact, the way the concept of the tangent line was introduced is based on the notion of slope. You already know that vertical lines do not have slopes. So we say that the derivative does not exist whenever the tangent line is vertical. Nevertheless keep in mind that when the limit giving the derivative is then the function has a vertical tangent line at the point.
It can be quite laborious (or impossible) to compute the derivative by hand as we have done so far. In the next pages we will show how techniques of differentiation help bypass the limit calculations and make our life much easier.
Exercise 1. Find the derivative of
Exercise 2. Discuss the differentiability of
Exercise 3. We say that the graph of f(x) has a cusp at (a,f(a)), if f(x) is continuous at a and if the following two conditions hold:
1.
as from one side (left or right);
2.
as from the other side.
Determine whether f(x) = x4/3 and g(x) = x3/5 have a cusp at (0,0).
Exercise 4. Show that if f '(a) exists, then we have
Exercise 5. A spherical balloon is being inflated. Find the rate at which its volume V is changing with respect to the radius.
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Helmut Knaust
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Browse Questions
# Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Toolbox:
• Surface Area $S=2\pi r^2+2\pi rh$
• $h=\large\frac{s-2\pi r^2}{2\pi r}$
• Volume $V=\pi r^2h$
Step 1:
Let $S$ be the given surface area of the closed cylinder whose radius is $r$ and height $h$.
Let $V$ be its volume.
Surface Area $S=2\pi r^2+2\pi rh$
$h=\large\frac{S-2\pi r^2}{2\pi r}$----(1)
Volume $V=\pi r^2h$
Substitute the value of h in the above equation
$\qquad\qquad=\pi r^2\bigg[\large\frac{S-2\pi r^2}{2\pi r}\bigg]$
$\qquad\qquad=\large\frac{1}{2}$$r(S-2\pi r) \qquad\qquad=\large\frac{1}{2}$$[Sr-2\pi r^3]$
Differentiating with respect to r we get
$\large\frac{dV}{dx}=\frac{1}{2}$$[S-6\pi r^2] Step 2: For maxima and minima \large\frac{dV}{dx}$$=0$
$S-6\pi r^2=0$
$S=6\pi r^2$
From (1)
$h=\large\frac{S-2\pi r^2}{2\pi r}$
Substitute the value os S in the above equation
$h=\large\frac{6\pi r^2-2\pi r^2}{2\pi r}$
$\;\;=\large\frac{4\pi r^2}{2\pi r}$
$\;\;=2r$
$h=2r$
Step 3:
On double differentiation of V we get,
$\large\frac{d^2V}{dr^2}=\frac{1}{2}$$(-12\pi r)$
$\qquad=-6\pi r$
$\qquad=-ve$
$\therefore V$ is maximum
Thus volume is maximum when $h=2r$
(i.e)when height of cylinder =diameter of the base.
edited Aug 19, 2013
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• Identify common factors, common multiples and prime numbers
• Number-Fractions
• Use common factors to simplify fractions; use common multiples to express fractions in the same denomination
• Compare and order fractions
• Add and subtract fractions with different denominators and mixed numbers, using the concept of equivalent fractions
• Multiply simple pairs of proper fractions, writing the answer in its simplest form
• Divide proper fractions by whole numbers
• Associate a fraction with division and calculate decimal fraction equivalents for a simple fraction
• Identify the value of each digit in numbers given to three decimal places and multiply and divide numbers by 10, 100 and 1000 giving answers up to three decimal places
• Multiply one-digit numbers with up to two decimal places by whole numbers
• Use written division methods in cases where the answer has up to two decimal places
• Solve problems which require answers to be rounded to specified degrees of accuracy
• Recall and use equivalences between simple fractions, decimals and percentages, including in different contexts.
• Ratio and Proportion
• Solve problems involving similar shapes where the scale factor is known or can be found
• Solve problems involving unequal sharing and grouping using knowledge of fractions and multiples.
• Solve problems involving the relative sizes of two quantities where missing values can be found by using integer multiplication and division facts
• Solve problems involving the calculation of percentages and the use of percentages for comparison
• Algebra
• Use simple formulae
• Generate and describe linear number sequences
• Express missing number problems algebraically
• Find pairs of numbers that satisfy an equation with two unknowns
• Enumerate possibilities of combinations of two variables
• Measurement
• Solve problems involving the calculation and conversion of units of measure, using decimal notation up to three decimal places where appropriate
• Use, read, write and convert between standard units, converting measurements of length, mass, volume and time from a smaller unit of measure to a larger unit, and vice versa, using decimal notation to up to three decimal places
• Convert between miles and kilometres
• Recognise that shapes with the same areas can have different perimeters and vice versa
• Recognise when it is possible to use formulae for area and volume of shapes
• Calculate the area of parallelograms and triangles
• Calculate, estimate and compare volume of cubes and cuboids using standard units, including cubic centimetres and cubic metres, and extending to other units
• Geometry-Properties of Shape
• Draw 2-D shapes using given dimensions and angles
• Recognise, describe and build simple 3D shapes, including making nets
• Compare and classify geometric shapes based on their properties and sizes and find unknown angles in any triangles, quadrilaterals, and regular polygons
• Illustrate and name parts of circles, including radius, diameter and circumference and know that the diameter is twice the radius
• Recognise angles where they meet at a point, are on a straight line, or are vertically opposite, and find missing angles.
• Geometry-Position and Direction
• Describe positions on the full coordinate grid (all four quadrants)
• Draw and translate simple shapes on the coordinate plane, and reflect them in the axes
• Statistics
• Interpret and construct pie charts and line graphs and use these to solve problems
• Calculate and interpret the mean as an average
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# What is the range of the function f(x)= -sqrt((x^2) -9x) ?
Jul 8, 2017
Range of $f \left(x\right) = \left(- \infty , 0\right]$
#### Explanation:
$f \left(x\right) = - \sqrt{{x}^{2} - 9 x}$
First let's consider the domain of $f \left(x\right)$
$f \left(x\right)$ is defined where ${x}^{2} - 9 x \ge 0$
Hence where $x \le 0$ and $x \ge 9$
$\therefore$ Domain of $f \left(x\right) = \left(- \infty , 0\right] \cup \left[9 , + \infty\right)$
Now consider:
${\lim}_{x \to \pm \infty} f \left(x\right) = - \infty$
Also: $f \left(0\right) = 0$ and $f \left(9\right) = 0$
Hence the range of $f \left(x\right) = \left(- \infty , 0\right]$
This can be seen by the graph of #f(x) below.
graph{-sqrt(x^2-9x) [-21.1, 24.54, -16.05, 6.74]}
Jul 8, 2017
Range : $f \left(x\right) \le 0$, in interval notation : $\left(- \infty , 0\right]$
#### Explanation:
$f \left(x\right) = - \sqrt{{x}^{2} - 9 x}$
Range: Under root should be $\ge 0$ , So $f \left(x\right) \le 0$
Range : $f \left(x\right) \le 0$, in interval notation : $\left(- \infty , 0\right]$
graph{-(x^2-9x)^0.5 [-320, 320, -160, 160]} [Ans]
|
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Class 8 Maths Chapter 1 Rational Numbers MCQs
Class 8 Maths Chapter 1-Rational Numbers MCQs (Questions and Answers) are provided here online for students. These questions are prepared as per the CBSE board syllabus (2021-2022) and NCERT guidelines. The chapter-wise objective questions are given to students to make them understand each concept and help them to score good marks in exams. Also, learn important questions for class 8 Maths here, at BYJU’S.
Practice more and test your skills on Class 8 Maths Chapter 1 Rational numbers MCQs with the given PDF here.
## MCQs on Class 8 Rational Numbers
Multiple choice questions are available for Class 8 Rational numbers with each question consisting of four answers, out of which one is correct. Students have to solve the question and choose the correct answer. They can also check their solutions here.
1. An integer can be:
A. Only Positive
B. Only Negative
C. Both positive and negative
D. None of the above
Explanation: An integer can be both positive and negative as well as zero.
i.e. …-3, -2, -1, 0, 1, 2, 3,….
2. A rational number can be represented in the form of:
A. p/q
B. pq
C. p+q
D. p-q
Explanation: A rational number can be represented in the form p/q where p and q are integers and q is not equal to zero.
3. The value of ½ x ⅗ is equal to:
A. ½
B. 3/10
C. ⅗
D. ⅖
Explanation: ½ x ⅗ = (1 x 3)/(2 x 5) = 3/10
4. The value of (½) ÷ (⅗) is equal to:
A. 3/10
B. ⅗
C. 6/5
D. ⅚
Explanation: (½) ÷ (⅗) = (½) x (5/3) = (1 x 5)/(2 x 3) = ⅚
5. The value of ½ + ¼ is equal to:
A. ¾
B. 3/2
C. ⅔
D. 1
Explanation: ½ + ¼
Making the denominator equal:
2/4 + ¼ = (2 + 1)/4 = ¾
6. The value of (5/4) – (8/3) is:
A. 17/12
B. -17/12
C. 12/17
D. -12/17
Explanation: 5/4 – 8/3
Making the denominator equal:
[(5/4) x (3/3)] – [(8/3) x (4/4)]
= (15/12) – (32/12)
= (15 – 32)/12
= -17/12
7. The associative property is applicable to:
B. Multiplication and division
D. Subtraction and Division
Explanation: As per associative property:
A + (B + C) = (A + B) + C
A × (B × C) = (A × B) × C
8. The value of (-10/3) x (-15/2) x (17/19) x 0 is:
A. 0
B. 22.66
C. 20
D. 35
Explanation: Any number multiplied by zero is equal to zero.
9. The additive identity of rational numbers is:
A. 0
B. 1
C. 2
D. -1
Explanation: Any number added to zero is equal to the number itself.
Ex: 5+0 = 5
Therefore, 0 is the additive identity of rational numbers.
10. The multiplicative identity of rational numbers is:
A. 0
B. 1
C. 2
D. -1
Explanation: Any number multiplied by 1 is equal to the number itself.
Ex: 5 x 1 = 5
Therefore, 1 is the multiplicative identity of rational numbers.
11. What is the sum of ⅔ and 4/9?
A. 6/3
B. 6/9
C. 10/9
D. 10/3
Solution: ⅔ + 4/9
⇒ ⅔ x (3/3) + 4/9
⇒ 6/9 + 4/9
⇒ 10/9
12. What is the product of 2/9 and ¾?
A. ⅙
B. ⅔
C. 1/9
D. ¼
Solution: The product of 2/9 and ¾:
⇒ 2/9 x ¾
⇒ (2 x 3)/(9 x 4)
⇒ (2 x 3)/(3 x 3 x 2 x 2)
By cancelling the common terms from numerator and denominator, we get;
⇒ 1/(3×2)
⇒ ⅙
13. What is the reciprocal of 1/9?
A. 9
B. 0
C. 1
D. None of the above
1/9 x 9 = 1
14. What is the value of 100 divided by 0?
A. 0
B. 100
C. 1
D. Undefined
100/0 = undefined
15. Which of the following is commutative for rational numbers?
C. Multiplication and division
D. Subtraction and division
16. Division of rational numbers is associative.
A. True
B. False
Explanation: Division of rational numbers is not associative.
For example,
2/5 ÷ (1/2 ÷ 1/4) = 0.2
(2/5 ÷ 1/2) ÷ 1/4 = 3.2
Hence, 2/5 ÷ (1/2 ÷ 1/4) ≠ (2/5 ÷ 1/2) ÷ 1/4
17. What is the value of ¾ + ⅚ + 2/7?
A. 10/84
B. 134/84
C. 157/84
D. 167/84
Solution: Given,
¾+⅚+2/7
LCM of 4, 6 and 7 = 84
Now, making the denominators the same and adding the given rational numbers, we get;
⇒ 63/84 + 70/84 + 24/84
⇒ 157/84
18. Find the additive inverse of 11/7?
A. 7/11
B. -7/11
C. 11/7
D.-11/7
Explanation: When a number is added to its additive inverse, then the result is zero,
11/7 – 11/7 = 0
19. How many rational numbers are there in between ¾ and 1?
A. 0
B. 1
C. 2
D. Countless
Explanation: We can write ¾ as 30/40 and 1 as 40/40.
Hence the rational numbers between them are:
31/40, 32/40, 33/40, 34/40,35/40,36/40, 37/40, 38/40, 39/40.
Note: There are countless rational numbers between any two rational numbers.
20. What should be subtracted from -⅔ to get -1?
A. ⅓
B. -⅓
C. ⅔
D. -⅔
Solution: Let x is subtracted from -⅔.
-⅔ – x = -1
-x = -1 + ⅔
-x = -⅓
x = ⅓
## Video Lesson on Rational Numbers
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# Calculation Chains
Purpose
This is a level 4 number activity from the Figure It Out series. It relates to Stage 7 of the Number Framework.
Achievement Objectives
NA4-1: Use a range of multiplicative strategies when operating on whole numbers.
NA4-1: Use a range of multiplicative strategies when operating on whole numbers.
Specific Learning Outcomes
use place value knowledge to check calculations
Description of Mathematics
Use this activity to:
• help students apply advanced additive part–whole strategies (stage 6) in addition and subtraction of large numbers
• help students apply advanced multiplicative part–whole strategies (stage 7) in multiplication and division.
Required Resource Materials
A calculator
FIO, Level 3, Number Sense and Algebraic Thinking, Book One, Calculation Chains, page 21
A classmate
Activity
In this activity, students apply number sense and their knowledge of place value to work out how a number has been transformed on a calculator. Number sense principles that may be explored through this activity include whether a number gets bigger or smaller when it’s multiplied or divided by a whole number or by a decimal number.
Question 1 is useful for students who are working at stage 6. To be able to apply appropriate number sense to questions 2 and 3, students should be making the transition to stage 7 or be already working at this level, although it is possible to benefit from the activity through trial and improvement with a calculator.
Some students may need encouragement to try a strategy beyond guessing and checking on the calculator. Possible strategies to solve 111 x = 1 221 include:
• Estimation: “I know 111 x 10 is 1 110, so it has to be a little more than that.”
• Opposite operation or working backwards: “I’m going to use the opposite operation and work out 1 221 ÷ 111.”
• Looking at the ones place: “The product in 111 x = 1221 suggests only a 1 is possible in the ones place of the box number.”
Encourage the use of these strategies by asking questions such as:
Try estimating what would go in the box before you try it on the calculator. Do you think the answer you get on your calculator is going to be more or less than your estimate? Why?
What happens to a number when you multiply it by 10? 100? 1 000?
What happens to that number when it is divided by 10? 100? 1 000?
What’s the opposite of adding, multiplying, or dividing? How might using the opposite operation help?
Could you use the answer and work backwards?
#### Extension
In question 3e, some students will be surprised that you can divide 12.5 by something and get a bigger number (25). This could start an interesting investigation:
When you multiply a number by another number, does the first number always get bigger?
When you divide a number by another number, does the first number always get smaller?
Note that Branching Out (pages 10–11 of the student book) raises the same issues.
1. a. 1
b. 909 090
c. 5 050 505
d. 3 939 393
e. 1 012
2. a. 11
b. 9
c. 5
d. 100
e. 0.5
3. a. 5
b. 50
c. 80
d. 4
e. 0.5
4. Problems will vary.
|
Definition
# What Is Median? Understanding Median & the Formula with Practical Example
## What is Median?
Median is the point in a distribution of scores that divides the distribution precisely in half when the scores are listed in numerical order. It is the number that appears midway between the highest and the lowest numbers in an array. The median is the figure in the center of an ordered, ascending order, or descending order list of numbers, although it might be more informative of that data point than just the mean.
#### Definition 2
The median represents one of the types of central tendency. Median in statistics, is the measure of central tendency which divides a group of scores in half, with half the scores falling above the median score and half below.
#### Definition 3
Median also refers to the part of the dual roadway that divides traffic flowing in different directions. There are several forms of medians, such as paint lines and sometimes even barriers.
## Understanding Median
The median is generally described in mathematics as the midpoint of a given sequence of integers. Arranging the digits yields the median. The numerals are displayed in increasing or decreasing order. The center figure is referred to as the information set's median when the values are arranged.
It is the most straightforward statistical metric to compute. The midpoint of a group of statistics is the figure in the midway or the center of such a collection. To compute the mean, sort the data at first in a sequence of lowest to biggest or biggest to the smallest number. A median will be that number that separates the upper and lower halves of a test dataset, group, or distribution function. The median varies depending on the kind of dispersal.
## Formula
• If the amount of findings is odd, then the median is calculated as follows: Median = {(n+1)/2}th term
• If the sample sizes is even, the method of calculating the median seems to be as follows: Median = [(n/2)th term + {(n/2)+1}th term]/2
Here, n = number of observations.
#### Practical Example
1. It is simple to calculate the median for just a collection with an odd set of measurements. For example, the median for the dataset- 6, 10, 14 is 10 as the observation number is odd
2. If the database is already even, then the median will be the arithmetic mean of the central two integers. For example, the median for the dataset- 3, 5, 6, and 9 will be the average of 5 and 6 which is 5.5.
3. A firm has five senior executive workers. Workers receive wages of \$2,000, \$7,000, \$10,000, \$3,000, and \$12,000. The median wage is calculated by using the formula.
• Descending order = \$2,000, \$3,000, \$7,000, \$10,000, \$12,000.
• Observations totaled = 5
• Here, n is 5 which is odd.
• So, Median =(n+1)/2nd term = [(5+1)/2]th term. = 6/3 = 3rd term= \$7,000 dollars.
#### In Sentences
• Anomalies have less impact on the median dataset than on the average.
Category: Economics
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## Related Definitions
• Median wage
Median wage is the margin between the highest paid 50...
Categories
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Common Factors
Chapter 3 Class 6 Playing with Numbers
Concept wise
## Write common factors of 4 & 18
We find factors of 4 & 18 separately
So, Common factors of 4 & 18 are 1, 2
## Write common factors of 4 & 15
We find factors of 4 & 15 separately
So, Common factors of 4 & 15 are 1
Note:
If two numbers have common factor 1
Then, they are co-prime numbers
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Factors of 4 1 × 4 = 4 2 × 2 = 4 4 × 1 = 4 We stop here as 1 & 4 have occurred earlier So factors of 4 are 1, 2, 4 Factors of 18 1 × 18 = 18 2 × 9 = 18 3 × 6 = 18 6 × 3 = 18 We stop here as 6 & 3 have occurred earlier So factors of 18 are 1, 2, 3, 6, 9, 18 Factors of 4 1 × 4 = 4 2 × 2 = 4 4 × 1 = 4 We stop here as 1 & 4 have occurred earlier So factors of 4 are 1, 2, 4 Factors of 15 1 × 15 = 15 3 × 5 = 15 5 × 3 = 15 We stop here as 3 & 5 have occurred earlier So factors of 15 are 1, 3, 5, 15
Made by
#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.
|
# Basic Geometry Practice Test for 1st Grade – [Medium]
Geometry is everywhere! From the square windows of our homes to the round wheels of our bikes, shapes make up the world around us. At Brighterly, we believe that the foundation of mathematics begins with understanding basic geometry. So, let’s dive deep into the world of shapes and see how much fun 1st grade geometry can be!
## Introduction to Shapes
When we talk about shapes in 1st grade geometry, we often start with the basics. Simple shapes that are easy to identify and draw.
• Circles: Imagine a perfectly round pizza or a coin.
• Triangles: Think of the pyramids in Egypt or a slice of pie.
• Squares: Like a window or a single piece of a checkerboard.
• Rectangles: Like a door or a chocolate bar.
These shapes are the building blocks for more complex figures you’ll learn about later. Recognizing these shapes in daily life can be a fun game. Try spotting a triangle the next time you eat dinner or finding all the rectangles in your living room.
## Understanding Edges and Corners
Now that we have identified some basic shapes, let’s understand them a little better. Shapes have edges and corners. For example, a square has 4 straight edges and 4 corners. Similarly, a triangle has 3 edges and 3 corners.
• Edges: The straight or curved lines that form the outline of the shape.
• Corners: Also known as vertices, they’re the point where two edges meet.
You can use your fingers to trace the edges of a shape or count its corners. Play a game with a friend, and see who can identify the edges and corners of different shapes faster!
## Solids vs. Flat Shapes
Geometry isn’t just about flat shapes; it’s also about solid shapes. While a circle is flat, a sphere (like a ball) is solid. A square is flat, but a cube (like a dice) is solid.
• Flat Shapes (2D): Have width and height, but no depth. Examples include circles, squares, and triangles.
• Solid Shapes (3D): Have width, height, and depth. Some primary solid shapes are spheres, cubes, and cylinders.
To understand the difference, try drawing shapes on paper and then finding solid objects around your house that match them.
## Combining Shapes
One of the most exciting parts of geometry is combining basic shapes to make new ones. By placing two triangles together, you can form a diamond. Joining many squares side by side can make a rectangle.
A fun activity is to take different colored paper cut-outs of basic shapes and see what new shapes you can create. Perhaps you can design a house using only triangles and rectangles or craft an animal with circles and squares. Let your imagination run wild!
In conclusion, 1st grade geometry is an exciting journey of discovering and understanding the shapes and patterns that make up our world. By recognizing, drawing, and combining shapes, children not only learn the fundamentals of geometry but also develop their spatial and cognitive skills. Dive into the world of shapes with Brighterly and make learning math an exciting adventure!
Basic Geometry Practice Test
Get ready for math lessons with Brighterly! Designed with care, this medium-level test is tailored to challenge and engage young minds, reinforcing their foundational knowledge in geometry.
1 / 20
Which shape has 3 sides and 3 corners?
2 / 20
How many corners does a rectangle have?
3 / 20
If you put 2 triangles together, you can make a:
4 / 20
Which shape is round and has no corners?
5 / 20
Which shape is like a can of soup?
6 / 20
A soccer ball is shaped like a:
7 / 20
How many edges does a square have?
8 / 20
Which of these is NOT a flat shape?
9 / 20
Which shape has 5 sides?
10 / 20
A dice is shaped like a:
11 / 20
A pizza slice is shaped most like which of the following?
12 / 20
Which shape rolls but does not slide?
13 / 20
Which shape has all its sides equal in length?
14 / 20
How many corners does a pentagon have?
15 / 20
A traffic cone is shaped most like:
16 / 20
If you cut a circle into 4 equal parts, each part is called a:
17 / 20
Which shape is NOT a solid shape?
18 / 20
How many long sides does a rectangle have?
19 / 20
Which of these shapes has the most sides?
20 / 20
Which shape looks like a box?
0%
Poor Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Mediocre Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Needs Improvement
Start practicing math regularly to avoid your child`s math scores dropping to C or even D.
High Potential
It's important to continue building math proficiency to make sure your child outperforms peers at school.
|
# Basic Fractions: Subtracting Fractions (Part 1)
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Recap:
All fractions have a numerator and a denominator.
The first step in subtracting 2 fractions is to check if both denominators are the same number (like denominators) or different numbers (unlike denominators).
### Subtracting Fractions With Like Denominators
Fractions whose denominators are the same are the easiest to subtract. You just need to subtract the numerators and leave the denominator unchanged.
Example: More Examples:
More:
### Subtracting Fractions With Unlike Denominators
When fractions have different denominators, we have to change them so they both have the same denominator.
We do this by finding the lowest common multiple of the 2 numbers. We use equivalent fractions.
Example:
In this example, the denominators are 3 and 2.
Step 1:We write out the first few multiples (Times Tables) of 3 and 2.We see that 6 is the smallest number that occurs in both tables (lowest common multiple).
Step 2:
We can now change both fractions so that they have the same denominator. We are changing each fraction to its equivalent fraction.
6 is the second multiple of 3 so we need to multiply both the numerator as well as the denominator by 2.
6 is the third multiple of 2 so we need to multiply both the numerator as well as denominator by 3.
Let's put it all together.
Here is an alternative way to look at it:
We cannot subtract the fractions with unlike denominators because each portion of the fraction is of a different size. We need to change them to an equal size.
After changing to equivalent fractions:
Each portion of the equivalent fractions is now of an equal size. We can easily subtract 3 portions from 4 portions.
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Review properties of equality and use them to write algebraic proofs.
1 / 22
# Review properties of equality and use them to write algebraic proofs. - PowerPoint PPT Presentation
Objectives. Review properties of equality and use them to write algebraic proofs. A proof uses logic, definitions, properties, and previously proven statements to “prove” that a conclusion is true. An important part of writing a proof is giving justifications to show that every step is valid.
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## Review properties of equality and use them to write algebraic proofs.
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Presentation Transcript
Objectives
Review properties of equality and use them to write algebraic proofs.
A proof uses logic, definitions, properties, and previously proven statements to “prove” that a conclusion is true.
• An important part of writing a proof is giving justifications to show that every step is valid.
• Possible ways to “prove” or justify each step:
• A definition
• A postulate
• A property
• Something given in the problem
Remember!
The Distributive Property states that
a(b + c) = ab + ac.
Example 1: Solving an Equation in Algebra
Solve the equation 4m – 8 = –12. Write a justification for each step.
4m – 8 = –12
+8+8
4m = –4
m = –1
t = –14
Check It Out! Example 1
AB
AB represents the length AB, so you can think of AB as a variable representing a number.
Like algebra, geometry also uses numbers, variables, and operations. For example, segment lengths and angle measures are numbers. So you can use these same properties of equality to write algebraic proofs in geometry.
Example 2: Solving an Equation in Geometry
Write a justification for each step.
NO = NM + MO
4x – 4 = 2x + (3x – 9)
4x – 4 = 5x – 9
–4 = x – 9
5 = x
Check It Out! Example 2
Write a justification for each step.
mABC = mABD + mDBC
8x°=(3x + 5)°+ (6x – 16)°
8x = 9x – 11
–x = –11
x = 11
Check It Out! Example 2
Write a justification for each step.
mABC = mABD + mDBC
8x°=(3x + 5)°+ (6x – 16)°
8x = 9x – 11
–x = –11
x = 11
Classwork:
Exit Ticket with Partner
Homework:
• 2.5 Practice B Homework Handout
• Must write out justification for each step to get credit!!
Given
z – 5 = –12
Mult. Prop. of =
z = –7
Lesson Quiz: Part I
Solve each equation. Write a justification for each step.
1.
Given
z – 5 = –12
Mult. Prop. of =
z = –7
Lesson Quiz: Part I
Solve each equation. Write a justification for each step.
1.
Given
6r – 3 = –2(r + 1)
6r – 3 = –2r – 2
Distrib. Prop.
8r – 3 = –2
8r = 1
Div. Prop. of =
Lesson Quiz: Part II
Solve each equation. Write a justification for each step.
2.6r – 3 = –2(r + 1)
WARM UP
Give an example of the following properties:
1. Reflexive Property of Equality:
2. Symmetric Property of Equality:
3. Transitive Property of Equality:
Objective:
Identify properties of equality and congruence.
Remember…..segments with equal lengths are congruent and that angles with equal measures are congruent.
So the Reflexive, Symmetric, and Transitive Properties of Equality have corresponding properties of congruence.
Remember!
Numbers are equal (=) and figures are congruent ().
Example 1: Identifying Property of Equality and Congruence
Identify the property that justifies each statement.
A. QRS QRS
B. m1 = m2 so m2 = m1
C. AB CD and CD EF, so AB EF.
D. 32° = 32°
Reflex. Prop. of .
Symm. Prop. of =
Trans. Prop of
Reflex. Prop. of =
Check It Out! Example 2
Identify the property that justifies each statement.
A. DE = GH, so GH = DE.
B. 94° = 94°
C. 0 = a, and a = x. So 0 = x.
D. A Y, so Y A
Sym. Prop. of =
Reflex. Prop. of =
Trans. Prop. of =
Sym. Prop. of
Lesson Review
Identify the property that justifies each statement.
1. x = y and y = z, so x = z.
2. DEF DEF
3. ABCD, so CDAB.
Trans. Prop. of =
Reflex. Prop. of
Sym. Prop. of
Classwork/Homework
come up with a unique example of each
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# MSBSHSE Solutions For Class 9 Maths Part 1 Chapter 4 Ratio and Proportion
MSBSHSE Solutions For Class 9 Maths Part 1 Chapter 4 Ratio and Proportion are useful for students as it helps them to score well in the class exams. BYJU’S provides the students with precise solutions, which are prepared by our dedicated specialists. All solutions from MSBSHSE Solutions For Class 9 Maths Part 1 Chapter 4 are provided here for free. These solutions are designed by subject matter experts who have assembled model questions, covering all the exercise questions from the textbook. These solutions help students to familiarize themselves with the ratio and proportion of a given number. This chapter also deals with properties of ratios, operations on equal ratios, theorem of equal ratios and Continued proportion. Solutions are available in PDF format on our website.
## Download the PDF of Maharashtra Solutions For Class 9 Maths Part 1 Chapter 4 Ratio and Proportion
### Access answers to Maths MSBSHSE Solutions For Class 9 Part 1 Chapter 4 – Ratio and Proportion
Practice set 4.1 Page no: 61
1. From the following pairs of numbers, find the reduced form of ratio of first number to second number.
i. 72, 60
ii. 38, 57
iii. 52, 78
Solution:
(i) Given 72,60
Reduced form of ratio of first number to second number:
(ii) Given 38, 57
Reduced form of ratio of first number to second number:
(iii) Given 52, 78
Reduced form of ratio of first number to second number:
2. Find the reduced form of the ratio of the first quantity to second quantity.
i. 700 Rs, 308Rs.
ii. 14Rs, 12 Rs.40 paise.
iii. 5 litre, 2500 ml
iv. 3 years 4 months, 5 years 8 months
v. 3.8 kg, 1900 gm
vi. 7 minutes 20 seconds, 5minutes 6 seconds.
(iv) Given 3 years 4 months, 5 years 8 months
3 years = 3 × 12 = 36 months
∴ 3 years 4 months = 40 months
5 years = 5 × 12 = 60
∴ 5 years 8 months = 68 months
∴ Reduced form of the ratio of 40 months and 68 months is:
(v) Given 3.8 kg, 1900 gm
3.8 kg = 3.8 × 1000 = 3800 gm
∴ Reduced form of the ratio of 3800 gm and 1900 gm is:
(vi) Given 7 minutes 20 seconds, 5minutes 6 seconds.
7 minutes = 7 × 60 = 420 seconds
∴ 7 minutes 20 seconds = 440 seconds
5minutes = 5 × 60 = 300 seconds
∴ 5 minutes 6 seconds = 306 seconds
∴ Reduced form of the ratio of 440 seconds and 306 seconds is:
3. Express the following percentages as ratios in the reduced form.
(i) 75: 100
(ii) 44: 100
(iii) 6.25%
(iv) 52: 100
(v) 0.64%
Solution:
(i) Given 75: 100
Reduced form of the ratio of 75:100 is:
(ii) Given 44:100
Reduced form of the ratio of 44:100 is:
(iii) Given 6.25%
Reduced form of 6.25% is:
(iv) Given 52:100
Reduced form of the ratio of 52:100 is:
(v) Given 0.64%
Reduced form of 0.64% is:
4. Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
Solution:
Let the persons required to build a house in 6 days be x.
Days required to build a house and number of persons are in inverse proportion.
∴ 6 × x = 8 × 3
∴ 6 x = 24
∴ x = 4
∴ 4 persons are required to build the house in 6 days.
5. Convert the following ratios into percentage.
(i) 15: 25
(ii) 47: 50
(iii) 7/10
(iv) 546/600
(v) 7/16
Solution:
(i)Â Given 15: 25
= 15/25
= ((15/25) × 100) %
= (15 × 4) %
= 60 %
(ii)Â Given 47: 50
= 47/50
= ((47/50) × 100) %
= (47 × 2) %
= 94 %
(iii)Â Given 7/10
= ((7/10) × 100) %
= (7 × 10) %
= 70 %
(iv)Â Given 546/600
= ((546/600) × 100) %
= (546/6) %
= 91 %
(v)Â Given 7/16
= ((7/16) × 100) %
= (7 × 6.25) %
= 43.75 %
6. The ratio of ages of Abha and her mother is 2: 5. At the time of Abha’s birth her mother’s age was 27 year. Find the present ages of Abha and her mother.
Solution:
Given the ratio of ages of Abha and her mother is 2 : 5.
Let the common multiple be x.
∴ Present age of Abha = 2x years and
Present age of Abha’s mother = 5x years
According to the given condition, the age of Abha’s mother at the time of Abha’s birth = 27 years
∴ 5x – 2x = 27
∴ 3x = 27
∴ x = 9
∴ Present age of Abha = 2x = 2 x 9 = 18 years
∴ Present age of Abha’s mother = 5x =5 x 9 = 45 years
The present ages of Abha and her mother are 18 years and 45 years respectively.
7. Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5: 4?
Solution:
Given present age of Vatsala = 14 years
Present age of Sara = 10 years
Let after x years, the ratio of their ages will be 5:4.
∴ Age of Vatsala after x years = (14 + x) years
Age of Sara after x years = (10 + x) years
Ratio of their ages = 5:4
On cross multiplying, we get:
56 + 4x = 50 + 5x
⇒ 5x – 4x = 56 – 50
⇒ x = 6
∴ After 6 years, their ages will be 20 years and 16 years and ratio of their ages will be 5:4.
8. The ratio of present ages of Rehana and her mother is 2: 7. After 2 years, the ratio of their ages will be 1: 3. What is Rehana’s present age?
Solution:
The ratio of present ages of Rehana and her mother is 2 : 7
Let the common multiple be x.
∴ Present age of Rehana = 2x years and Present age of Rehana’s mother = 7x years
After 2 years,
Rehana’s age = (2x + 2) years
Age of Rehana’s mother = (7x + 2) years
According to the given condition,
After 2 years, the ratio of their ages will be 1 : 3
∴ 2x+27x+2 = 13
∴ 3(2x + 2) = 1(7x + 2)
∴ 6x + 6 = 7x + 2
∴ 6 – 2 = 7x – 6x
∴ 4 = x
∴ x = 4
∴ Rehana’s present age = 2x = 2 x 4 = 8 years
∴ Rehana’s present age is 8 years.
Practice set 4.2 Page no: 63
1. Using the property , fill in the blanks substituting proper numbers in the following.
Solution:
(i) Let
∴ on comparing first two equalities, we get:
5/7 = x/28
Cross multiply and get:
7x = 28 × 5
⇒ x = 4 × 5 = 20
Now, compare the first and third equalities and get:
5/7 = 35/y
Cross multiply and get:
5y = 7 × 35
⇒ y = 7 × 7 = 49
Now, compare the first and fourth equalities and get:
5/7 = z/3.5
Cross multiply and get:
7z = 5 × 3.5
⇒ 7z = 5 × (35/10)
⇒ z = 5 × (5/10)
⇒ z = 25/10 = 2.5
∴
(ii) Let
∴ On comparing first two equalities, we get:
9/14 = 4.5/x
Cross multiply and get:
9x = 14 × 4.5
⇒ x = 14 × 0.5 = 7
Now, compare the first and third equalities and get:
9/14 = y/42
Cross multiply and get:
14y = 9 × 42
⇒ y = 9 × 3 = 27
Now, compare the first and fourth equalities and get:
9/14 = z/3.5
Cross multiply and get:
14z = 9 × 3.5
⇒ z = 9 × (3.5/14)
⇒ z = 9 × (0.25)
⇒ z = 2.25
∴
2. Find the following ratios.
(i) The ratio of radius to circumference of the circle.
(ii) The ratio of circumference of circle with radius r to its area.
(iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm.
(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area.
Solution:
(i) Let r be the radius of a circle.
Circumference of circle = 2Ï€r
Ratio of radius to circumference of circle = r/2Ï€r
= 1/2Ï€
= 1: 2Ï€
(ii) Let r be the radius of a circle.
Circumference of circle = 2Ï€r
Area of the circle =2Ï€r2
Ratio of radius to circumference of circle = 2Ï€r/Ï€r2
= 2/r
= 2: r
(iii) Side of square = 7 cm
Diagonal of square = √2 × side = 7√2 cm
Ratio of diagonal of a square to its side = 7/7√2
= 1/√2
= 1: √2
(iv) Length of rectangle = 5 cm
Breadth of rectangle = 3.5 cm
Perimeter of rectangle = 2(Length + Breadth)
= 2(5+3.5)
= 2(8.5)
= 17 cm
Area of rectangle = Length × Breadth
= 5 × 3.5
= 16.5 cm2
Ratio of Perimeter to area of rectangle
= 17/16.5
= 170/165
= 34/33
3. Compare the following pairs of ratios.
v.Â
Solution:
Therefore, both the ratios are equal, according to the ratio comparison rules.
⇒
Practice set 4.3 Page no: 70
1. If a/b = 7/3 then find the values of the following ratios.
Solution:
2. If then find the values of the following ratios.
Solution:
(i) Given:
Practice set 4.4 Page no: 73
1. Fill in the blanks of the following
Solution:
2. 5m-n=3m+4n, then find the values of the following expressions.
Solution:
(i) Given: 5m – n = 3m + 4m
⇒ 5m – 3m = 4n + n
On simplifying we get
⇒ 2m = 5n
On rearranging
⇒ m/n = 5/2
⇒ m2/n2 = 25/4
Apply componendo and dividendo:
= 29:21
(ii) Given: 5m – n = 3m + 4m
⇒ 5m – 3m = 4n + n
On simplifying we get
⇒ 2m = 5n
On rearranging
⇒ m/n = 5/2
⇒ 3m/4n = 15/8
Apply componendo and dividendo:
= 23: 7
Practice set 4.5 Page no: 77
1. Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Solution:
Let x be the number that should be subtracted from 12, 16, 21 so that the numbers remain in continued proportion.
Numbers a, b and c are said to be continued proportion if b2Â = ac.
∴ From the definition of continued proportion, we get:
⇒ (16 – x)2 = (12 – x) (21 – x)
On expanding the above equation, we get
⇒ 256 + x2 – 32x = 252 – 33x + x2
⇒-32x + 33x = 252 – 256
⇒ x= -4
∴ -4 should be subtracted from 12, 16, 21 so that the numbers remain in continued proportion.
2. If (28 – x) is the mean proportional of (23 – x) and (19 – x) then find the value of x.
Solution:
A number b is said to be mean proportional of two numbers a and c if b2Â = ac.
∴ From the definition of mean proportion, we get:
(28 – x)2Â = (23 – x) (19 – x)
On expanding the above equation, we get
⇒ 784 + x2 – 56x = 437 – 42x + x2
⇒-56x + 42x = 437 – 784
⇒ -14x= -347
⇒ x = 347/14
3. Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Solution:
Let the numbers be x, y, z.
As the numbers are in continued proportion, therefore
y2 = xz ……………… (1)
Also, the mean proportion = 12
∴ y = √xz = 12
⇒ xz = 144 …………… (2)
It is given that the sum of remaining two numbers = 26
∴ x + z = 26
⇒ x = 26 – z
Put the value of x in equation (2):
(26 – z) z = 144
Expanding and simplifying we get
⇒ 26z – z2 = 144
⇒ z2 – 26z + 144 = 0
⇒ z2 – 8z – 18z + 144 = 0
⇒ z (z – 8) – 18 (z – 8) = 0
⇒ (z – 8) (z – 18) = 0
⇒ z = 8 or z = 18
∴ x = 26 – 8 or x = 26 – 18
⇒ x = 18 or x = 8
y = 12
∴ The numbers in proportion be 8, 12, 18 or 18, 12, 8.
4. If (a + b+ c) (a –b + c) = a2 + b2 + c2, show that a, b, c are in continued proportion.
Solution:
Given: (a + b+ c) (a –b + c) = a2 + b2 + c2
On multiplying we get
⇒ a2 –ab + ac + ab – b2 + bc + ca – bc + c2 = a2 + b2 + c2
⇒ a2 –ab + ac + ab – b2 + bc + ca – bc + c2 – a2 – c2= b2 + b2
⇒ 2ac = 2b2
⇒ b2 = ac
∴ a, b, c are in continued proportion.
5. If a/b = b/c and a, b, c >0, then show that,
i. (a + b+ c) (b – c) = ab – c2
ii. (a2Â + b2) (b2Â + c2) = (ab + bc)2
iii. (a2Â + b2)/ab = (a + c)/b
Solution:
(i) Given: a/b = b/c
⇒ b2 = ac
Consider (a + b+ c) (b – c) = ab – ac + b2 – bc + cb – c2
= ab – ac + ac – c2 (∵ b2 = ac)
= ab – c2
(ii) Given a/b = b/c
⇒ b2 = ac
Consider (a2Â + b2) (b2Â + c2) = a2b2Â + a2Â c2Â + b2Â b2Â + b2c2
= a2b2 + ac(ac)+ b2(ac)+ b2c2 (∵ b2 = ac)
= a2b2 + b2(ac)+ b2(ac)+ b2c2 (∵ b2 = ac)
= a2b2Â + 2b2(ac)+ b2c2
= a2b2Â + 2ab2c+ b2c2
= (ab + bc)2
(iii) Given: a/b = b/c
⇒ b2 = ac
Consider (a2 + b2)/ab = (a2 + ac)/ab (∵ b2 = ac)
= (a + c)/b
6. Find mean proportional ofÂ
Solution:
On simplifying we get
Problem set 4 Page no: 77
1. Select the appropriate alternative answer for the following questions.
i. If 6: 5 = y: 20 then what will be the value of y?
A. 15
B. 24
C. 18
D. 22.5
Solution:
B. 24
Explanation:
Given 6:5 = y:20
⇒ 6/5 = y/20
Cross multiply and get:
5y = 6 × 20
⇒ y = 6 × 4 = 24
∴ Option B is correct.
ii. What is the ratio of 1 mm to 1 cm?
A. 1: 100
B. 10: 1
C. 1: 10
D. 100: 1
Solution:
A. 1: 100
Explanation:
We know that 1 cm = 100 mm
∴ 1mm: 1cm
⇒ 1mm: 100mm
= 1: 100
∴ Option A is correct.
iii. The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age?
A. 3: 2
B. 2: 3
C. 4: 3
D. 3: 4
Solution:
B. 2: 3
Explanation:
Given Nitin’s age = 24 years
Mohasin’s age = 36 years
∴ Ration of Nitin’s age to Mohasin’s age = 24:36
= 24/36
= 2/3
= 2:3
∴ Option B is correct.
iv. 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get?
A. 8
B. 15
C. 12
D. 9
Solution:
D. 9
Explanation:
Total bananas = 24
Ratio in which the bananas are divided = 3:5
Let number of bananas Shubham got = 3x
∴ Number of bananas Anil got = 5x
∴ 3x+5x = 24
⇒ 8x = 24
⇒ x = 3
∴ Shubham got (3 × 3) = 9 bananas.
Thus, option D is correct.
v. What is the mean proportional of 4 and 25?
A. 6
B. 8
C. 10
D. 12
Solution:
C. 10
Explanation:
Mean proportional of two numbers a and b = √(ab)
∴ Mean proportional of 4 and 25 = √ (4 × 25)
= √100 = 10
2. For the following numbers write the ratio of first number to second number in the reduced form.
i. 21, 48
ii. 36, 90
iii. 65, 117
iv. 138, 161
v. 114, 133
Solution:
(i) Ratio of 21 and 48 in the reduced form:
(To simplify, break the numbers in simpler form)
∴ Ratio of 21 and 48 in reduced form is 1:4.
(ii) Ratio of 36 and 90 in the reduced form:
(To simplify, break the numbers in simpler form)
∴ Ratio of 36 and 90 in reduced form is 2:5.
(iii) Ratio of 65 and 117 in the reduced form:
(To simplify, break the numbers in simpler form)
∴ Ratio of 65 and 117 in reduced form is 5:9.
(iv) Ratio of 138 and 161 in the reduced form:
(To simplify, break the numbers in simpler form)
∴ Ratio of 138 and 161 in reduced form is 6:7.
(v) Ratio of 114 and 133 in the reduced form:
 (To simplify, break the numbers in simpler form)
∴ Ratio of 114 and 133 in reduced form is 6:7.
3. Write the following ratios in the reduced form.
i. Radius to the diameter of a circle.
ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
iii. The ratio of perimeter to area of a square, having side 4 cm.
Solution:
(i) Let r be the radius of the circle.
Let d be the diameter of the circle.
∴ Ratio of radius to diameter in the reduced form = Radius: Diameter
∴ Ratio of radius to diameter in the reduced form = 1:2
(ii) Given: Length of rectangle = l = 4 cm
Breadth of rectangle = b = 3 cm
Diagonal of rectangle = √ (l2 + b2)
= √ (16 + 9)
= √25 = 5
∴ Diagonal of rectangle = 5 cm
Ratio of diagonal to the length of a rectangle = 4:5
(iii) Given: Side of square = 4 cm
Perimeter of square = 4 × Side = 4 × 4 = 16 cm2
Area of the square = (Side)2Â = (4)2Â = 14 cm2
The ratio of perimeter to area of a square = 16:14 = 8:7
4. Check whether the following numbers are in continued proportion.
i. 2, 4, 8
ii. 1, 2, 3
iii. 9, 12, 16
iv. 3, 5, 8
Solution:
(i) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a: b∷ b:c
or b2Â = ac
Here, a = 2, b = 4 and c = 8
∴ (4)2 = 2 × 8
⇒ 16 = 16, which holds true.
∴ 2, 4, 8 are in continued proportion.
(ii) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a: b∷ b:c
or b2Â = ac
Here, a = 1, b = 2 and c = 3
∴ (2)2 = 1 × 3
⇒ 4 = 3, which does not hold true.
∴ 1, 2, 3 are not in continued proportion.
(iii) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a: b∷ b: c
or b2Â = ac
Here, a = 9, b = 12 and c = 16
∴ (12)2 = 9 × 16
⇒ 144 = 144, which holds true.
∴ 9, 12, 16 are in continued proportion.
(iv) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion,
i.e. a: b∷ b: c
or b2Â = ac
Here, a = 3, b = 5 and c = 8
∴ (5)2 = 3 × 8
⇒ 25 = 24, which does not hold true.
∴ 3, 5, 8 are not in continued proportion.
5. a, b, c are in continued proportion. If a = 3 and c = 27 then find b.
Solution:
Given: a, b, c are in continued proportion.
Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a: b∷ b:c
or b2Â = ac
Here, a = 3, c = 27
∴ (b)2 = 3 × 27
⇒ b2 = 81
⇒ b = ±√81 = ±9
∴ b = -9 or 9
6. Convert the following ratios into percentages.
i. 37: 500
ii. 5/8
iii. 22/30
iv. 5/16
v. 144/1200
Solution:
(i)Â Given 37: 500 = 37/500
= ((37/500) × 100) %
= (37/5) %
= 7.4 %
(ii) Given 5/8 = ((5/8) × 100) %
= (5 × 12.5) %
= 62.5 %
(iii) Given 22/30 = ((22/30) × 100) %
= (220/3) %
= 73.33 %
(iv) Given 144/1200 = ((144/1200) × 100) %
= (144/12) %
= 12 %
A ratio is a mathematical expression of comparing two similar or different quantities by division. This expression can be expressed from ratio to percentage form by conversion method. It is denoted by ‘:’ symbol. For example, 2:3, 5:6, 3:7, etc. are ratios. It is also represented by ‘/’ symbol. Hence, the ratio is basically a type of division method. The concept related to Ratio and Proportion is explained in BYJU’S to help the students be thorough with the topics.
## Frequently Asked Questions on Maharashtra State Board Solutions for Class 9 Maths Part 1 Chapter 4 Ratio and Proportion
### Are these Maharashtra State Board Class 9 Maths Part 1 Chapter 4 Solutions useful to prepare for the exams?
Yes, these Maharashtra Board Class 9 Maths Part 1 Chapter 4 Solutions are very helpful. Students are recommended to practise these solutions after revising the subjects because they set the basis for the questions that could get asked often in the board exams.
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# Euler's formula explained
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• Euler's formula establishes the fundamental relationship between the trigonometric functions and the complex exponential function. Euler's formula or Euler's equation is one of the most fundamental equations in maths and engineering and has a wide range of applications
• Euler's method uses the simple formula, to construct the tangent at the point x and obtain the value of y (x+h), whose slope is, In Euler's method, you can approximate the curve of the solution by the tangent in each interval (that is, by a sequence of short line segments), at steps of h
Yes, putting Euler's Formula on that graph produces a circle: e ix produces a circle of radius 1 . And when we include a radius of r we can turn any point (such as 3 + 4i) into re ix form by finding the correct value of x and r: Example: the number 3 + 4i Euler's formula, either of two important mathematical theorems of Leonhard Euler.The first formula, used in trigonometry and also called the Euler identity, says e ix = cos x + isin x, where e is the base of the natural logarithm and i is the square root of −1 (see irrational number).When x is equal to π or 2π, the formula yields two elegant expressions relating π, e, and i: e iπ = −. We can use Euler's Formula to draw the rotation we need: Start with 1.0, which is at 0 degrees. Multiply by e i a, which rotates by a. Multiply by e i b, which rotates by b
### Euler's Formula - mathsisfun
Euler's formula relates the complex exponential to the cosine and sine functions. This formula is the most important tool in AC analysis. It is why electrical engineers need to understand complex numbers. Created by Willy McAllister Euler's formula is very simple but also very important in geometrical mathematics. It deals with the shapes called Polyhedron. A Polyhedron is a closed solid shape having flat faces and straight edges. This Euler Characteristic will help us to classify the shapes. Let us learn the Euler's Formula here Euler's formula is eⁱˣ=cos (x)+i⋅sin (x), and Euler's Identity is e^ (iπ)+1=0. See how these are obtained from the Maclaurin series of cos (x), sin (x), and eˣ. This is one of the most amazing things in all of mathematics! Created by Sal Khan
Intuition for e^(πi) = -1, using the main ideas from group theoryHelp fund future projects: https://www.patreon.com/3blue1brownAn equally valuable form of su.. EULER'S FORMULA FOR COMPLEX EXPONENTIALS According to Euler, we should regard the complex exponential eit as related to the trigonometric functions cos(t) and sin(t) via the following inspired definition:eit = cos t+i sin t where as usual in complex numbers i2 = ¡1: (1) The justification of this notation is based on the formal derivative of both sides Calvin Lin. contributed. In complex analysis, Euler's formula provides a fundamental bridge between the exponential function and the trigonometric functions. For complex numbers. x. x x, Euler's formula says that. e i x = cos x + i sin x. e^ {ix} = \cos {x} + i \sin {x}. eix = cosx +isinx. In addition to its role as a fundamental.
What does it mean to compute e^{pi i}?Full playlist: https://www.youtube.com/playlist?list=PLZHQObOWTQDP5CVelJJ1bNDouqrAhVPevHome page: https://www.3blue1bro.. 5.3 Complex-valued exponential and Euler's formula Euler's formula: eit= cost+ isint: (3) Based on this formula and that e it= cos( t)+isin( t) = cost isint: cost= eit+ e it 2; sint= e e it 2i: (4) Why? Here is a way to gain insight into this formula. Recall the Taylor series of et: et= X1 n=0 tn n!: Suppose that this series holds when the. Euler's Polyhedral Formula is a well-known formula that he devised. The rest of Euler's formula deals with complex numbers. Euler's formula states for polyhedron that that certain rules are to be followed: F + V - E = E, which is often called Euler's number and is an irrational number that rounds to 2.72. The imaginary number i is defined as the square root of -1. Pi (Π), the relationship between the diameter and circumference of a circle, is approximately 3.14 but, like e, is an irrational number. This formula is written as e (i*Π) + 1 = 0
Euler's formula about e to the i pi, explained with velocities to positions.Help fund future projects: https://www.patreon.com/3blue1brownAn equally valuable.. Euler's formula relates the complex exponential to the cosine and sine functions. This formula is the most important tool in AC analysis. It is why electrica..
### Euler's formula - Wikipedi
• Euler's identity is an equality found in mathematics that has been compared to a Shakespearean sonnet and described as the most beautiful equation.It is a special case of a foundational.
• Euler's Formula for Planar Graphs The most important formula for studying planar graphs is undoubtedly Euler's formula, first proved by Leonhard Euler, an 18th century Swiss mathematician, widely considered among the greatest mathematicians that ever lived. Until now we have discussed vertices and edges of a graph, and the way in which thes
• The result of this short calculation is referred to as Euler's formula: [4][5] eiφ = cos(φ) +isin(φ) (7) (7) e i φ = cos. . ( φ) + i sin. . ( φ) The importance of the Euler formula can hardly be overemphasised for multiple reasons: It indicates that the exponential and the trigonometric functions are closely related to each other.
• Euler's Formula explained with exponential function. Ask Question Asked 3 years, 2 months ago. Active 3 years, 2 months ago. Then your desired formula is an elementary calculation from the binomial theorem--see my answer. $\endgroup$ - alex-tang Apr 3 '18 at 3:5
### Understanding Euler's Formula
1. Euler's identity is the greatest feat of mathematics because it merges in one beautiful relation all the most important numbers of mathematics. But that's still a huge understatement, as it conceals a deeper connection between vastly different areas that Euler's identity indicates
2. imum area moment of inertia of the cross section of the column unsupported length of column column effective length facto
3. This celebrated formula links together three numbers of totally different origins: e comes from analysis, π from geometry, and i from algebra. Here is just one application of Euler's formula. The addition formulas for cos(α + β) and sin(α + β) are somewhat hard to remember, and their geometric proofs usually leave something to be desired
4. Euler's Formula Examples. Look at a polyhedron, for instance, the cube or the icosahedron above, count the number of vertices it has, and name this number V. The cube has 8 vertices, so V = 8. Next, count and name this number E for the number of edges that the polyhedron has. There are 12 edges in the cube, so E = 12 in the case of the cube
### Euler's Formula: A Complete Guide Math Vaul
1. Euler's method uses the simple formula, to construct the tangent at the point x and obtain the value of y(x+h), whose slope is, In Euler's method, you can approximate the curve of the solution by the tangent in each interval (that is, by a sequence of short line segments), at steps of h
2. (c) Explain why 2E = 6T. (Hint. All F = 2T triangles provide a total of 6T edges. But there are repeats, because each edge is common to exactly 2 triangles.) (d) Use Euler's formula V − E + F = 2 for the sphere to prove that T = 2V I +V B −2. (e) Use the fact that each simple lattice triangle has area 1 2 to deduce Pick's formula
3. The formula for compount interest is given by, A = P ( 1 + r 100. n) n. where P is the principal, r is the yearly rate of interest in percentage, n is the number of compounding periods and A is the total amount at the end of 1 year. Let, P = 1 and r = 100. If the interest is compounded annually, i.e. n = 1, then
4. Columns fail by buckling when their critical load is reached. Long columns can be analysed with the Euler column formula. F = n π 2 E I / L 2 (1) where . F = allowable load (lb, N) n = factor accounting for the end conditions. E = modulus of elastisity (lb/in 2, Pa (N/m 2)) L = length of column (in, m) I = Moment of inertia (in 4, m 4
5. Euler's constant explains so much about math and physics. We delve into the number e, or Euler's number, to find out why it's so important in mathematics
6. Euler's Method is one of the simplest of many numerical methods that now exist for solving differential equations. Euler meets Glenn? Rudy Horne, a mathematician at Morehouse College in Atlanta, was the math advisor to the movie, and it was he who suggested Euler's Method for the key blackboard scene
7. June 2007 Leonhard Euler, 1707 - 1783 Let's begin by introducing the protagonist of this story — Euler's formula: V - E + F = 2. Simple though it may look, this little formula encapsulates a fundamental property of those three-dimensional solids we call polyhedra, which have fascinated mathematicians for over 4000 years. Actually I can go further and say that Euler's formula
In this post we will explore Euler's Formula, explain what it is, where it comes from, and reveal its magic properties. Euler's Formula, coined by Leonhard Euler in the XVIIIth century, is one. Euler's formula is widely used in structural engineering calculations, but one alternative sometimes used is the Perry-Robertson theory. This is employed in the guidelines set out in Eurocode 3: Design of steel structures. The mathematical instability (i.e. buckling) of a structural member such as a beam or column is calculated for all. Euler's Method, is just another technique used to analyze a Differential Equation, which uses the idea of local linearity or linear approximation, where we use small tangent lines over a short distance to approximate the solution to an initial-value problem. Euler's Approximation. Remember This formula does not take into account the axial stress and the buckling load is given by this formula may be much more than the actual buckling load. Euler's Buckling (or crippling load) The maximum load at which the column tends to have lateral displacement or tends to buckle is known as buckling or crippling load Euler's Product Formula 1.1 The Product Formula The whole of analytic number theory rests on one marvellous formula due to Leonhard Euler (1707-1783): X n∈N, n>0 n−s = Y primes p 1−p−s −1. Informally, we can understand the formula as follows. By the Funda-mental Theorem of Arithmetic, each n≥1 is uniquely expressible in the form n.
### Euler's Formula: Definition, Formulas and Question
Euler Formula for Buckling. Euler Buckling Theory is a classical theory to find the critical buckling load for the column of any cross-section. It is usually adopted to calculate the buckling load in long columns. The derivation of Euler's formula for buckling starts from noting that bending moment in a loaded column and buckled column is. Euler's Buckling (or crippling load): The maximum load at which the column tends to have lateral displacement or tends to buckle is known as buckling or crippling load. Load columns can be analysed with the Euler's column formulas can be given as. where, E = Modulus of elasticity, l = Effective Length of column, and I = Moment of inertia of.
Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations. are solved starting at the initial condition and ending at the desired value. is the solution to the differential equation. In this problem, Starting at the initial point We continue using Euler's method until . The results. Derivation Of The Euler Equations Of Motion For A Rigid Body To derive the Euler equations of motion for a rigid body we must first set up a schematic representing the most general case of rigid body motion, as shown in the figure below. In the schematic, two coordinate systems are defined: The first coordinate system used in the Euler equations derivation is the global XYZ reference frame Monday, 14 Jun 2010. How To explain Euler's identity using triangles and spiral
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# NTS Quantitative Reasoning Practice Test 1
### NTS Quantitative Reasoning Practice Test # 1
INSTRUCTIONS:
• This Quiz is related to NTS Quantitative Reasoning Practice Test # 1 in this series
• The Quiz based on the Topic, “Quantitative Reasoning
• If you have good understanding about the subject, then you can attempt it
• There will be 10 Multiple Choice Questions (MCQs) in this test
• You have 60 Seconds to answer a single Multiple Choice Question
• Questions will be randomly changed every time you start this test
• You should practice more and more to get high marks
• Passing Criteria is 60 percent for this Quiz
• You can retake this test as many time as you like
• The progress bar at the top of screen will show your progress as well as the time remaining where timed quiz
• After the quiz, You will find Your Test Score and Grade
• If you feel that any Incorrect Answer to a Question, simply Comment us about the Question.
#### Find a positive value of the constant k so that the equation x2 + k x + 4 = 0 has only one solution.
Correct! Wrong!
For the given equation to have a solution, the left hand side must be a square. Hence x2 + k x + 4 = (x + 2)2 Expand the right hand side x2 + k x + 4 = x2 + 4x + 4 Comparing the left hand side and the right hand sides, the value of k is 4 k = 4
#### In a drawer of shirts, 8 r blue, 6 r green, and 4 r magenta. If Mason draws 2 shirts at random, what is the probability that at least one of the shirts he draws will be blue?
Correct! Wrong!
Remember that at least one is a clue, and when u see phrase, u need to find the probability of getting everything except what u want (in other words, the probability of getting any other color except blue), and then subtract that from 1. The formula for this would be 1-(the probability of getting the other colors). 1-(10/18 * 9/17)=1-5/17=12/17.
#### Solve the equation 2 - (x - 2)2 = - 18 for x.
Correct! Wrong!
Subtract 2 from both sides of the equation. - (x - 2)2 = - 20 Multiply both sides of the equation by -1. (x - 2)2 = 20 Solve by extracting the square root. x - 2 = ~+mn~ √20 = ~+mn~ 2 √5 solutions: x = 2 + 2 √5 and x = 2 - 2 √5
Correct! Wrong!
#### In the above figure above, does x = 90? (1) The length of AC is less than the length of BC. (2) The length of AB is one-fourth the circumference of the circle.
Correct! Wrong!
We need to see the fact statements. Statement (1) says that the length of AC is less than the length of BC. This clearly leads us to nowhere. So BCE. Now the fact statement (2) tells us that the length of AB is one-fourth the circumference of the circle, which clearly leads us to know that it is not the diameter, it is just a chord. So the angle subtended is not equal to 90 deg.
#### Find the area of a right isosceles triangle with hypotenuse equal to 24.
Correct! Wrong!
The two legs of a right isosceles triangle have equal lengths; let x be one of these lengths. The area A of the triangle is given by A = (1/2) x * x = (1/2)x2 We now use Pythagora's theorem to find x as follows x2 + x2 = 242 Simplify 2 x2 = 576 x2 = 288 We now calculate the area A as follows A = (1/2)x2 = (1/2) 288 = 144
#### Solve the equation (x - 1)(x + 3) = (1 - x) for x.
Correct! Wrong!
Rewrite equation with the right hand side equal to 0. (x - 1)(x + 3) - (1 - x) = 0 Rewrite as (x - 1)(x + 3) + (x - 1) = 0 Factor and solve (x - 1) [ (x + 3) + 1 ] = 0 (x - 1)(x + 4) = 0 x - 1 = 0 or x + 4 = 0 Solutions: x = 1 and x = -4
#### A number, K, is a positive integer with the special property that 3 times its unit is equal to 2 times its 10 digit. How many such numbers exist between 10 & 99?
Correct! Wrong!
Here’s another smart Question. It appears to be daunting but it’s not that tough. We start with 1 at units place. When multiplied 3 times and then divided by 2 we get 1.5. So it is ruled out. Next we try with 2. When we multiply by 3 we get 6 which when divided by 2 gives us 3.Bingo!! We get the first number 32. Similarly by trying out different numbers at unit place we get other 2 numbers as 64 and 96(which are also multiples of 32 for hint). So we get a total of 3 numbers between 10 and 99.
#### (√12 - √3)(-√12 + √3)
Correct! Wrong!
Rewrite as (√12 - √3)(-√12 + √3) = - (√12 - √3) (√12 - √3) and simplify = - (12 - 3) = -9
#### The lengths of sides BA and BC of triangle ABC are equal. Find the measure of angle x.
Correct! Wrong!
Since BA and BC have equal lengths, then the triangle is isosceles and the interior angles at A and C have equal measures which may be calculated as follows A + C + 40 = 180 or 2A = 140 or A = 70° The interior angle at A and angle x are supplementary. Hence 70 + x = 180 or x = 110°
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# A long spring is stretched by $2cm$ and its potential energy is $U$. If the spring is stretched by $10cm$ ; its potential energy will be: A) $U/5$B) $U/25$C) $5U$D) $25U$
Last updated date: 09th Apr 2024
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Hint: First, calculate the value of the spring constant using the formula for energy in a spring by putting the given values, i.e. $x = 2$ and $PE = U$ . Then use the same formula by putting $x = 10cm$ and spring constant as calculated previously to get the value of final potential energy. This will be our final answer.
Formula Used:
Potential energy in a spring, $PE = \dfrac{1}{2}k{x^2}$
Where, $k$ is the spring constant of the spring and $x$ the distance by which the string has been stretched, or extended.
Complete step by step solution:
First, we will use the formula for energy in a spring and put $x = 2$ , as given to get the value of spring constant $k$ . Once we find the value of $k$ we will move on to find the potential energy of the spring when it is stretched by $10cm$ , which will be the final answer.
We have, potential energy in a spring, $PE = \dfrac{1}{2}k{x^2}$
Where, $k$ is the spring constant of the spring and $x$ the distance by which the string has been stretched.
For $x = 2$ and $PE = U$ (given) we get $U = \dfrac{1}{2}k{(2)^2}$
$\Rightarrow k = \dfrac{{2U}}{4} = \dfrac{U}{2}$
Now we will use this value of spring constant in the formula of potential energy of the spring with $x = 10cm$ to calculate the potential energy of spring in this case.
Therefore, we get $PE = \dfrac{1}{2} \times \dfrac{U}{2} \times {(10)^2}$
On simplifying, we are left with $PE = \dfrac{U}{4} \times 100$
Which gives, $PE = 25U$
Hence, option D is the correct answer.
Note: In questions like these, make sure both the values of extension of spring are given with the same unit. If the units are different, make sure you convert the given values into the same unit before using the values in an answer. Otherwise, you may end up getting the wrong answer. A spring stores energy as potential energy and releases the stored energy as kinetic energy. The kinetic energy released is proportional to the square of the length by which the spring is compressed or stressed. Thus, the potential energy is also directly proportional to the square of the length by which the spring is compressed or stressed.
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# 1998 CEMC Gauss (Grade 7) Problems/Problem 11
## Problem
Kalyn cut rectangle R from a sheet of paper. A smaller rectangle is then cut from the large rectangle R to produce figure S. In comparing R to S,
[R is a rectangle with sides 8 and 6 cm. S is the same as R with a 4x1 rectangle cut from one of its corners.]
$\text{(A) the area and perimeter both decrease}$
$\text{(B) the area decreases and the perimeter increases}$
$\text{(C) the area and perimeter both increase}$
$\text{(D) the area increases and the perimeter decreases}$
$\text{(E) the area decreases and the perimeter stays the same}$
## Solution
Because a section of R is cut out to form S, the area of S is smaller than the area of R.
The perimeter of R is $2(6) + 2(8) = 28$ cm and the perimeter of S is $6 + 8 + 5 + 4 + 1 +4 = 28$ cm, so the perimeters of both shapes are equal.
The answer is $\boxed{\text{(E)}}.$
-edited by coolmath34
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# McGraw Hill My Math Grade 3 Chapter 5 Answer Key Understand Division
All the solutions provided in McGraw Hill My Math Grade 3 Answer Key PDF Chapter 5 Understand Division will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 3 Answer Key Chapter 5 Understand Division
Essential Question?
What does division mean?
Write two multiplication sentences for each array.
Question 1.
_____ _____
2 × 4 = 8
4 × 2 = 8
Explanation:
No of rows = 2
No of columns = 4
The two multiplication sentences are
2 × 4 = 8
4 × 2 = 8
Question 2.
_____ _____
3 × 5 = 15
5 × 3 = 15
Explanation:
Given,
No of rows = 3
No of column = 5
The two multiplication sentences are
3 × 5 = 15
5 × 3 = 15
Identify a pattern. Then find the missing numbers.
Question 3.
30, 25, 20, ___, ___, 5 Pattern: _____
15, 10, 5
Explanation:
The pattern that follows are
30 – 25 = 5
25 – 20 = 5
20 – 15 = 5
15 – 10 = 5
Question 4.
12, ___, 8, ____, 4, 2 Pattern: _____
12, 10, 8, 6, 4, 2
Explanation:
The pattern that follows are
12 – 10 = 2
10 – 8 = 2
8 – 6 = 2
6 – 4 = 2
12, 10, 8, 6, 4, 2
Question 5.
55, 45, 35, ____, 15, ____ Pattern: _____
55, 45, 35, 45, 15, 5
Explanation:
The pattern that follows are
5 × 1 = 5
5 × 3 = 15
5 × 5 = 25
5 × 7 = 35
5 × 9 = 45
55, 45, 35, 45, 15, 5
Draw the counters in the circles to make equal groups.
Question 6.
12 ÷ 2 = 6
Explanation:
No of counters = 12
No of groups = 2
12 ÷ 2 = 6 counters in each group.
Question 7.
9 ÷ 3 = 3
Explanation:
No of counters = 9
No of groups = 3
9 ÷ 3 = 3
Question 8.
Colton made 15 party invitations. His brother made 9 invitations. Write a subtraction sentence to find how many more invitations Colton made.
15 – 9 = 6 more invitations.
Explanation:
Given,
No of party invitations = 15
His brother made = 9 invitations
The subtraction sentence = 15 – 9 = 6
Hence there are 6 more invitations colton made.
Question 9.
Mrs. Jones has 21 pencils. She gives 2 pencils to Carter and 2 pencils to Mandy. Write a subtraction sentence to find how many pencils she has left.
21 – 4 = 17 pencils left.
Explanation:
Given,
No of pencils = 21
She gave 2 pencils to carter
She gave 2 pencils to Mandy.
Total of 4 pencils were given from 21.
21 – 4 = 17 pencils.
She left with 17 pencils.
My Math Words
Review Vocabulary
array equal groups pattern repeated addition
Making Connections
Use the review vocabulary to describe each example in the graphic organizer.
How are the examples similar? How are the examples different?
Given all examples are same. If we solve we get the same answer. But while solving we used different models to find the answers.
Example:
My Vocabulary Cards
Ideas for Use
• Group 2 or 3 common words. Add a word that is unrelated to the group. Then work with a friend to name the unrelated word.
• Design a crossword puzzle. Use the definition for each word as the clues.
A number that is being divided.
Circle the dividend in 12 ÷ 4 = . Then write the quotient.
Quotient = 3
Explanation:
Given,
The question is 12 ÷ 4 =
4 × 3 = 12
The answer to this question is 3.
The dividend is 12.
The number by which the dividend is being divided.
Write a division sentence in which the divisor is 5. Circle the divisor.
15 ÷ 5 = 3
Explanation:
We need to find the division sentence.
The divisor = 5
let the dividend be 15
Operations that undo each other, like multiplication and division.
Use inverse operations to write a multiplication and division sentence.
5 × 3 = 15
15 ÷ 5 = 3
Explanation:
The inverse operations to write a multiplication and division sentence.
5 × 3 = 15
15 ÷ 5 = 3
The answer to a division problem.
Write and solve a division problem. Circle the quotient.
20 ÷ 5 = 4
Explanation:
Let us solve the divison problem
Let the dividend be 20, divisor be 5
By dividing both of them the answer is 20 ÷ 5 = 4
The quotient is 4.
To separate into equal groups, to find the number of groups or the number in each group.
By dividing the snacks each friend will get equal snacks.
Explanation:
Let us take an example to solve this.
No of friends = 20
No of chocolate snacks = 5
By dividing both = 20 ÷ 5 = 4
Therefore each each will get 4 chocolate snacks.
A number sentence using numbers and the ÷ sign.
Write an example of a division sentence. Then write the sentence using words.
25 ÷ 5 = 5
25 is shared into 5 equal groups would give us 5 in each group.
Explanation:
The example of a division sentence is 25 ÷ 5 = 5.
25 is shared into 5 equal groups would give us 5 in each group.
A group of related facts using the same numbers.
Write the numbers in the fact family shown on this card.
The fact family is multiplication or division fact family.
Explanation:
The multiplication fact family is
3 × 9 = 27
9 × 3 = 27
The divison fact family is
27 ÷ 3 = 9
27 ÷ 9 = 3
To divide or “break up.”
Partition can mean “a wall that divides a room into different areas.” How does that relate to the math word?
Divide
Explanation:
The word that relate to the math word is Divide.
My Vocabulary Cards
The chart for the vocabulary cards is given in the figure below.
Explanation:
Ideas for Use
• Write an example for each card. Be sure your examples are different from what is shown on each card.
• Write the name of each lesson on the front of each blank card. Write a few study tips on the back of each card.
Subtraction of the same number over and over again.
Write a sentence comparing repeated subtraction to repeated addition.
4 + 4 + 4 = 16
Explanation:
A sentence comparing repeated subtraction to repeated addition is 4 + 4 + 4 = 16.
Basic facts using the same numbers.
Why is facts plural in this vocabulary word?
The facts are plural because we can write different facts by using the same numbers.
Explanation:
The different facts by using the same numbers is multiplication fact or division fact and addition and subtraction fact.
My Foldable
Dividend = 28, divisor = 7 and Quotient = 4
28 ÷ 7 =4
Explanation:
Foldables
Study Organizer
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 10.6: Green’s Theorem
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
## Change of variables
Note: FIXME4 lectures
In one variable, we have the familiar change of variables $\int_a^b f\bigl(g(x)\bigr) g'(x)\, dx = \int_{g(a)}^{g(b)} f(x) \, dx .$ It may be surprising that the analogue in higher dimensions is quite a bit more complicated. The first complication is orientation. If we use the definition of integral from this chapter, then we do not have the notion of $$\int_a^b$$ versus $$\int_b^a$$. We are simply integrating over an interval $$[a,b]$$. With this notation then the change of variables becomes $\int_{[a,b]} f\bigl(g(x)\bigr) \left\lvert {g'(x)} \right\rvert\, dx = \int_{g([a,b])} f(x) \, dx .$ In this section we will try to obtain an analogue in this form.
First we wish to see what plays the role of $$\left\lvert {g'(x)} \right\rvert$$. If we think about it, the $$g'(x)$$ is a scaling of $$dx$$. The integral measures volumes, so in one dimension it measures length. If our $$g$$ was linear, that is, $$g(x)=Lx$$, then $$g'(x) = L$$. Then the length of the interval $$g([a,b])$$ is simply $$\left\lvert {L} \right\rvert(b-a)$$. That is because $$g([a,b])$$ is either $$[La,Lb]$$ or $$[Lb,La]$$. This property holds in higher dimension with $$\left\lvert {L} \right\rvert$$ replaced by absolute value of the determinant.
[prop:volrectdet] Suppose that $$R \subset {\mathbb{R}}^n$$ is a rectangle and $$T \colon {\mathbb{R}}^n \to {\mathbb{R}}^n$$ is linear. Then $$T(R)$$ is Jordan measurable and $$V\bigl(T(R)\bigr) = \left\lvert {\det T} \right\rvert V(R)$$.
It is enough to prove for elementary matrices. The proof is left as an exercise.
We next notice that this result still holds if $$g$$ is not necessarily linear, by integrating the absolute value of the Jacobian. That is, we have the following lemma
Suppose $$S \subset {\mathbb{R}}^n$$ is a closed bounded Jordan measurable set, and $$S \subset U$$ for an open set $$U$$. If $$g \colon U \to {\mathbb{R}}^n$$ is a one-to-one continuously differentiable mapping such that $$J_g$$ is never zero on $$S$$. Then $V\bigl(g(S)\bigr) = \int_S \left\lvert {J_g(x)} \right\rvert \, dx .$
FIXME
The left hand side is $$\int_{R'} \chi_{g(S)}$$, where the integral is taken over a large enough rectangle $$R'$$ that contains $$g(S)$$. The right hand side is $$\int_{R} \left\lvert {J_g} \right\rvert$$ for a large enough rectangle $$R$$ that contains $$S$$. Let $$\epsilon > 0$$ be given. Divide $$R$$ into subrectangles, denote by $$R_1,R_2,\ldots,R_K$$ those subrectangles which intersect $$S$$. Suppose that the partition is fine enough such that $\epsilon + \int_S \left\lvert {J_g(x)} \right\rvert \, dx \geq \sum_{j=1}^N \Bigl(\sup_{x \in S \cap R_j} \left\lvert {J_g(x)} \right\rvert \Bigr) V(R_j)$ ... $\sum_{j=1}^N \Bigl(\sup_{x \in S \cap R_j} \left\lvert {J_g(x)} \right\rvert \Bigr) V(R_j) \geq \sum_{j=1}^N \left\lvert {J_g(x_j)} \right\rvert V(R_j) = \sum_{j=1}^N V\bigl(Dg(x_j) R_j\bigr)$ ... FIXME ... must pick $$x_j$$ correctly?
Let
FIXME
So $$\left\lvert {J_g(x)} \right\rvert$$ is the replacement of $$\left\lvert {g'(x)} \right\rvert$$ for multiple dimensions. Note that the following theorem holds in more generality, but this statement is sufficient for many uses.
Suppose that $$S \subset {\mathbb{R}}^n$$ is an open bounded Jordan measurable set, and $$g \colon S \to {\mathbb{R}}^n$$ is a one-to-one continuously differentiable mapping such that $$g(S)$$ is Jordan measurable and $$J_g$$ is never zero on $$S$$.
Suppose that $$f \colon g(S) \to {\mathbb{R}}$$ is Riemann integrable, then $$f \circ g$$ is Riemann integrable on $$S$$ and $\int_{g(S)} f(x) \, dx = \int_S f\bigl(g(x)\bigr) \left\lvert {J_g(x)} \right\rvert \, dx .$
FIXME
FIXME: change of variables for functions with compact support
FIXME4
## Exercises
Prove .
FIXME
1. If you want a funky vector space over a different field, $${\mathbb{R}}$$ is an infinite dimensional vector space over the rational numbers.
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# 2021 AMC 12B Problems/Problem 4
The following problem is from both the 2021 AMC 10B #6 and 2021 AMC 12B #4, so both problems redirect to this page.
## Problem
Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$, and the afternoon class's mean score is $70$. The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$. What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$
## Solution 1 (One Variable)
Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$. The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$. Therefore, the answer is $\boxed{\textbf{(C)} ~76}$.
~ {TSun} ~
## Solution 2 (Two Variables)
Suppose the morning class has $m$ students and the afternoon class has $a$ students. We have the following table: $$\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\ \hline & & & \\ [-2.5ex] \textbf{Morning} & m & 84 & 84m \\ \hline & & & \\ [-2.5ex] \textbf{Afternoon} & a & 70 & 70a \end{array}$$ We are also given that $\frac ma=\frac34,$ which rearranges as $m=\frac34a.$
The mean of the scores of all the students is $$\frac{84m+70a}{m+a}=\frac{84\left(\frac34a\right)+70a}{\frac34a+a}=\frac{133a}{\frac74a}=133\cdot\frac47=\boxed{\textbf{(C)} ~76}.$$ ~MRENTHUSIASM
## Solution 3 (Ratio)
Of the average, $\frac{3}{3+4}=\frac{3}{7}$ of the scores came from the morning class and $\frac{4}{7}$ came from the afternoon class. The average is $\frac{3}{7}\cdot 84+\frac{4}{7}\cdot 70=\boxed{\textbf{(C)} ~76}.$
~Kinglogic
## Solution 4 (Convenient Values)
WLOG, assume there are $3$ students in the morning class and $4$ in the afternoon class. Then the average is $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}.$
~ pi_is_3.14
## Video Solution by TheBeautyofMath
https://youtu.be/GYpAm8v1h-U (for AMC 10B)
https://youtu.be/EMzdnr1nZcE?t=608 (for AMC 12B)
~IceMatrix
~Interstigation
## Video Solution (Under 2 min!)
~Education, the Study of Everything
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## What is meaning of like fraction?
Like fractions. Fractions having the same denominator are called ‘Like fractions’ The five examples of like fractions are. 2/7,3/7,4/7,5/7 and 6/7 Unlike fractions.
## What is 6/7 as a fraction with a denominator of 42?
36⁄42
Answer: The fraction 6⁄7 can be written as 36⁄42 with denominator 42.
## What is 2/3 as a fraction with a denominator of 9?
6⁄9
Answer: The fraction 2⁄3 can be written as 6⁄9 with denominator 9.
## How do you do like fractions?
Step 1: Multiply the numerator of the first fraction by the denominator of the second fraction. Step 2: Multiply the numerator of the second fraction by the denominator of the first fraction. Step 3: Multiply the denominators of both fractions and take it as a common denominator for the results of step 1 and step 2.
## What is 26 as a fraction in simplest form?
13/50
26 percent in simplest form is the fraction 13/50.
## What is 86 percent as a fraction in simplest form?
To write the fraction in the simplest form means to eliminate the common factors in the numerator and the denominator of the fraction. Hence, the required simplified fractional form of \[86\% \] is equal to \[\dfrac{{43}}{{50}}\].
## Which set of fraction are like and which are unlike?
The fractions with the same denominators are called like fractions. In group (ii) the denominator of each fraction is different, i.e., the denominators of all the fractions are different. The fractions with different denominators are called unlike fractions. (c) (1/9, 2/7, 3/4, 2/5).
## What is like & unlike fraction?
If two or more fractions share the same denominator we call them like fractions, while on the other hand, if the denominators of two or more fractions are different, those are known as unlike fractions. For example, 1/2 and 3/2 are like fractions and 1/2 and 3/4 are unlike fractions.
## Which group of fractions are like fractions among the following?
2/7 , 3/7, 4/7 is like fraction. Since, in this fraction have all same denominator.
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Question
# Simplify the following equation$\dfrac{{\cos e{c^2}67^\circ - {{\tan }^2}23^\circ }}{{{{\sec }^2}20^\circ - {{\cot }^2}70^\circ }}$
Hint: We need to know the basic trigonometric identities and formulae to solve this problem.
The given trigonometric expression is $\dfrac{{\cos e{c^2}67^\circ - {{\tan }^2}23^\circ }}{{{{\sec }^2}20^\circ - {{\cot }^2}70^\circ }}$
We have,
$1 + {\cot ^2}\theta = \cos e{c^2}\theta$
$1 + {\tan ^2}\theta = {\sec ^2}\theta$
Using these trigonometric identities, given function can be written as
$\dfrac{{\cos e{c^2}67^\circ - {{\tan }^2}23^\circ }}{{{{\sec }^2}20^\circ - {{\cot }^2}70^\circ }} = \dfrac{{1 + {{\cot }^2}67^\circ - {{\tan }^2}23^\circ }}{{1 + {{\tan }^2}20^\circ - {{\cot }^2}70^\circ }}$
We know that, $\tan (90 - \theta ) = \cot \theta$
$\cot \left( {90 - \theta } \right) = \tan \theta$
$= \dfrac{{1 + {{\cot }^2}(90^\circ - 23^\circ ) - {{\tan }^2}23^\circ }}{{1 + {{\tan }^2}(90^\circ - 70^\circ ) - {{\cot }^2}70^\circ }}$
$= \dfrac{{1 + {{\tan }^2}23^\circ - {{\tan }^2}23^\circ }}{{1 + {{\cot }^2}70^\circ - {{\cot }^2}70^\circ }}$
$= \dfrac{1}{1} = 1$
Note:
$\cot \left( {90 - \theta } \right)\& \tan (90 - \theta )$are in the first quadrant. In the first quadrant all trigonometric functions are positive. So, tan and cot values in the first quadrant are positive.
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# Write and simplify ratios. Use proportions to solve problems.
Objectives. Write and simplify ratios. Use proportions to solve problems. Remember!. In a ratio, the denominator of the fraction cannot be zero because division by zero is undefined. A ratio compares two numbers by division. The ratio of two numbers a and b can be written as
## Write and simplify ratios. Use proportions to solve problems.
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### Presentation Transcript
1. Objectives Write and simplify ratios. Use proportions to solve problems.
2. Remember! In a ratio, the denominator of the fraction cannot be zero because division by zero is undefined. • A ratio compares two numbers by division. • The ratio of two numbers a and b can be written as • ato b • a:b • where b ≠ 0.
3. Given that two points on m are C(–2, 3) and D(6, 5), write a ratio expressing the slope of m. Substitute the given values. Simplify.
4. A ratio can involve more than two numbers. For the rectangle, the ratio of the side lengths may be written as 3:7:3:7.
5. Example 2: Using Ratios The ratio of the side lengths of a triangle is 4:7:5, and its perimeter is 96 cm. What is the length of the shortest side? Let the side lengths be 4x, 7x, and 5x. Then 4x + 7x + 5x = 96 . After like terms are combined, 16x = 96. So x = 6. The length of the shortest side is 4x = 4(6) = 24 cm.
6. Check It Out! Example 2 The ratio of the angle measures in a triangle is 1:6:13. What is the measure of each angle? x + y + z = 180° x + 6x + 13x = 180° 20x = 180° x = 9° y = 6x z = 13x y = 6(9°) z = 13(9°) y = 54° z = 117°
7. A proportion is an equation stating that two ratios are equal. In the proportion , the values a and d are the extremes. The values b and c are the means. When the proportion is written as a:b = c:d, the extremes are in the first and last positions. The means are in the two middle positions.
8. Reading Math The Cross Products Property can also be stated as, “In a proportion, the product of the extremes is equal to the product of the means.” In Algebra 1 you learned the Cross Products Property. The product of the extremes ad and the product of the means bc are called the cross products.
9. Example 3A: Solving Proportions Solve the proportion. 7(72) = x(56) Cross Products Property 504 = 56x Simplify. x = 9 Divide both sides by 56.
10. Example 3B: Solving Proportions Solve the proportion. (z – 4)2 = 5(20) Cross Products Property (z – 4)2 = 100 Simplify. (z – 4) = 10 Find the square root of both sides. Rewrite as two eqns. (z – 4) = 10 or (z – 4) = –10 Add 4 to both sides. z = 14 or z = –6
11. Check It Out! Example 3b Solve the proportion. 2y(4y) = 9(8) Cross Products Property 8y2 = 72 Simplify. y2 = 9 Divide both sides by 8. Find the square root of both sides. y = 3 y = 3 or y = –3 Rewrite as two equations.
12. Check It Out! Example 3d Solve the proportion. (x + 3)2 = 4(9) Cross Products Property (x + 3)2 = 36 Simplify. (x + 3) = 6 Find the square root of both sides. (x + 3) = 6 or (x + 3) = –6 Rewrite as two eqns. x = 3 or x = –9 Subtract 3 from both sides.
13. The following table shows equivalent forms of the Cross Products Property.
14. Example 4: Using Properties of Proportions Given that 18c = 24d, find the ratio of d to c in simplest form. 18c = 24d Divide both sides by 24c. Simplify.
15. Check It Out! Example 4 Given that 16s = 20t, find the ratio t:s in simplest form. 16s = 20t Divide both sides by 20s. Simplify.
16. 1 Marta is making a scale drawing of her bedroom. Her rectangular room is 12 feet wide and 15 feet long. On the scale drawing, the width of her room is 5 inches. What is the length? Understand the Problem Example 5: Problem-Solving Application The answer will be the length of the room on the scale drawing.
17. Make a Plan 2 Example 5 Continued Let x be the length of the room on the scale drawing. Write a proportion that compares the ratios of the width to the length.
18. 3 Solve Example 5 Continued 5(15) = x(12.5) Cross Products Property 75 = 12.5x Simplify. x = 6 Divide both sides by 12.5. The length of the room on the scale drawing is 6 inches.
19. Check the answer in the original problem. The ratio of the width to the length of the actual room is 12 :15, or 5:6. The ratio of the width to the length in the scale drawing is also 5:6. So the ratios are equal, and the answer is correct. 4 Example 5 Continued Look Back
20. 1 Understand the Problem Check It Out! Example 5 What if...?Suppose the special-effects team made a different model with a height of 9.2 m and a width of 6 m. What is the height of the actual tower? The answer will be the height of the tower.
21. Make a Plan 2 Check It Out! Example 5 Continued Let x be the height of the tower. Write a proportion that compares the ratios of the height to the width.
22. 3 Solve Check It Out! Example 5 Continued 9.2(996) = 6(x) Cross Products Property 9163.2 = 6x Simplify. 1527.2 = x Divide both sides by 6. The height of the actual tower is 1527.2 feet.
23. 4 Check It Out! Example 5 Continued Look Back Check the answer in the original problem. The ratio of the height to the width of the model is 9.2:6. The ratio of the height to the width of the tower is 1527.2:996, or 9.2:6. So the ratios are equal, and the answer is correct.
More Related
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Chapter Polynomials and Rationals Review
Chapter
Chapter 2
Section
Chapter Polynomials and Rationals Review
Sum of Rational Expressions 6 Videos
## A soft intro to Sum of Rational Expressions
A overview Introduction
## Introduction to Sum of Rational Expressions
\dfrac{A}{B} + \dfrac{C}{D} = \dfrac{AD + BC}{BD}
ex Combine and simplify
\dfrac{1}{x -1} + \dfrac{1}{2x + 1}
= \dfrac{2x + 1 + x - 1}{(x - 1)(2x + 1)}
= \dfrac{3x}{(x - 1)(2x + 1)}
2.17mins
1 Introduction to Sum of Rational Expressions
## Adding Rational Expression with more than one factors in the denominator
\dfrac{A}{BC} + \dfrac{D}{BE} = \dfrac{AE + CD}{BCE}
ex Combine and simplify.
\dfrac{1}{x^2 - 1} + \dfrac{1}{x^2 - x -2}
= \dfrac{2x - 3}{(x - 1)(x + 1)(x - 2)}
4.02mins
2 Adding Rational Expression with more than one factors in the denominator
## Subtraction of Rational Expressions
ex Subtract and simplify.
\dfrac{1}{(x -1)(x + 2)^2} - \dfrac{1}{(x -2)(x + 2)(x - 1)}
= \dfrac{-4}{(x - 1)(x + 2)^2(x - 2)}
1.39mins
3 Subtraction of Rational Expressions
## Addition of Two Rational Terms
\dfrac{2}{(x^2 -4)(x^2-4x +3)} + \frac{1}{(x^2 -x -2)(x^2 + 4x + 4)}
= \dfrac{3x^2 + 2x + 7}{(x-2)(x + 2)^2(x-1)(x + 1)(x-3)}
2.35mins
4 Addition of Two Rational Terms
ex Find the sum.
\dfrac{1}{x - 2} + \dfrac{1}{(x - 2)^2} + \dfrac{1}{(x - 2)^3}
= \dfrac{x^2 - 3x + 3}{(x - 2)^3}
2.27mins
Product of Rational Expressions 6 Videos
## Introduction to Rational Expressions
Definition Rational number is any number that can be expressed as \frac{a}{b} where a, b, are integers.
ex \dfrac{x + 1}{x - 1}, \dfrac{x^2 - 2x - 3}{(x + 2)(x - 5)}
3.50mins
Introduction to Rational Expressions
## Simplifying Rational Expressions
ex Simplify & state restrictions
\dfrac{AB}{BC} \times \dfrac{C}{D} = \dfrac{A}{D}, B \neq 0, C \neq 0
2.37mins
Simplifying Rational Expressions
## Simplifying Rational Expression in Factored Form
ex Simplify and state the restrictions.
\dfrac{(2x -1)(x + 5)}{(x -1)(x + 5)}
= \dfrac{2x - 1}{x - 1}, x \neq 1, -5
0.40mins
Simplifying Rational Expressions and its restrictions
ex Simplify and state the restrictions.
\dfrac{2x^2 -x -3}{x^2 + 2x +1}
= \dfrac{2x - 3}{x + 1}, x \neq -1
1.04mins
Simplifying Rational Expressions by Factoring and its Restrictions
## Multiplying Rational Expressions
ex Simplify and state the restrictions.
\dfrac{x^2 -4}{(x + 6)^2} \cdot \frac{x^2 + 9x + 18}{2(2 -x)}
= \dfrac{(x + 2)(x + 3)}{-2(x + 6)}, x \neq -6, 2, x ∈ \mathbb{R}
1.40mins
Multiplying Rational Expressions
## Restriction on Division of Rational Expressions
\displaystyle \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{AB}{BC}
where \displaystyle B, C, D =\neq 0
2.50mins
Restriction on Division of Rational Expressions
Solutions 59 Videos
Simplify.
(7x^2 - 2x + 1)+(9x^2 -4x + 5)-(4x^2 + 6x - 7)
0.44mins
Q1a
Simplify.
(7a^2 -4ab + 9b^2)-(-a^2+2ab + 6b^2)
0.37mins
Q1b
Determine two non-equivalent polynomials f(x) and g(x), such that f(0) =g(0) and f(1) = g(1).
2.23mins
Q2
Ms. Frizzle has three daughters: Allison, Belle, and Claire. Today, January 1, their ages are, respectively,
\displaystyle \begin{array}{cccccc} & A(n) = -(n + 30)+(2n + 5) \\ & B(n) = (7 -n)-(32 -2n) \\ & C(n) = (n - 26)-(n + 4) + (n - 3) \\ \end{array}
All ages are expressed in years, and n represents Ms. Frizzle's age.
a) Are the daughter triples? Explain.
b) Are any of them twins? Explain.
1.05mins
Q3ab
Ms. Frizzle has three daughters: Allison, Belle, and Claire. Today, January 1, their ages are, respectively,
\displaystyle \begin{array}{cccccc} & A(n) = -(n + 30)+(2n + 5) \\ & B(n) = (7 -n)-(32 -2n) \\ & C(n) = (n - 26)-(n + 4) + (n - 3) \\ \end{array}
All ages are expressed in years, and n represents Ms. Flangan's age.
(c) How old was Ms. Flanagan when Cassandra was born?
0.33mins
Q3c
Expand and simplify.
-3(7x -5)(4x - 7)
0.59mins
Q4a
Expand and simplify.
-(y^2-4y + 7)(3y^2-5y-3)
1.45mins
Q4b
Expand and simplify.
2(a +b)^3
1.21mins
Q4c
Expand and simplify.
3(x^2 -2)^2(2x - 3)^2
2.38mins
Q4d
The volume of a cone is given by \displaystyle V = \frac{1}{3}\pi r^2h. Determine the volume of the cone in simplified form if the radius is increased by x and the height is increased by 2x.
2.42mins
Q5
Simplify.
(2x^4-3x^2 - 6)+(6x^4 -x^3+4x^2 + 5)
0.38mins
Q6a
Simplify.
(x^2-4)(2x^2+ 5x - 2)
0.35mins
Q6b
Simplify.
-7x(3x^3-7x + 2)-3x(2x^2-5x+ 6)
0.43mins
Q6c
Simplify.
-2x^2(3x^3-7x+2)-x^3(5x^3+2x -8)
0.50mins
Q6d
Simplify.
-2x[5x - (2x -7)] + 6x[3x-(1 + 2x)]
0.45mins
Q6e
Simplify.
(x + 2)^2(x- 1)^2 - (x-4)^2(x + 4)^2
2.25mins
Q6f
Simplify.
(x^2+5x - 3)^2
1.21mins
Q6g
Factor.
12m^2n^3 + 18m^3n^2
0.30mins
Q7a
Factor.
x^2-9x+20
0.25mins
Q7b
Factor.
3x^2+24x + 45
0.30mins
Q7c
Factor.
50x^2-72
0.34mins
Q7d
Factor.
9x^2-6x + 1
0.11mins
Q7e
Factor.
10a^2+a - 3
0.21mins
Q7f
Factor.
2x^2y^4-6x^5y^3+8x^3y
0.33mins
Q8a
Factor.
2x(x + 4)+3(x + 4)
0.12mins
Q8b
Factor.
x^2 -3x-10
0.12mins
Q8c
Factor.
15x^2-53x +42
1.20mins
Q8d
Factor.
a^4 -16
0.25mins
Q8e
Factor.
(m - n)^2-(2m + 3n)^2
0.59mins
Q8f
Simplify. State any restrictions on the variables.
\displaystyle \frac{10a^2b + 15bc^2}{-5b}
0.49mins
Q9a
Simplify. State any restrictions on the variables.
\displaystyle \frac{130x^2y^3-20x^2z^2+50x^2}{10x^2}
0.37mins
Q9b
Simplify. State any restrictions on the variables.
\displaystyle \frac{xy - xyz}{xy}
0.23mins
Q9c
Simplify. State any restrictions on the variables.
\displaystyle \frac{16mnr-24mnp+40kmn}{8mn}
0.42mins
Q9d
Simplify. State any restrictions on the variables.
\displaystyle 8xy^2 + 12x^2y - \frac{6x^3}{2xy}
1.19mins
Q10a
Simplify. State any restrictions on the variables.
\displaystyle \frac{7a-14b}{2(a - 2b)}
0.26mins
Q10b
Simplify. State any restrictions on the variables.
\displaystyle \frac{m + 3}{m^2+10m + 21}
0.26mins
Q10c
Simplify. State any restrictions on the variables.
\displaystyle \frac{4x^2-4x -3}{4x^2-9}
0.42mins
Q10d
Simplify. State any restrictions on the variables.
\displaystyle \frac{3x^2-21x}{7x^2-28x + 21}
0.38mins
Q10e
Simplify. State any restrictions on the variables.
\displaystyle \frac{3x^2-2xy-y^2}{3x^2+4xy+y^2}
0.47mins
Q10f
Simplify. State any restrictions on the variables.
\displaystyle \frac{6x}{8y}\times \frac{2y^2}{3x}
0.31mins
Q12a
Simplify. State any restrictions on the variables.
\displaystyle \frac{10m^2}{3n}\times \frac{6mn}{20m^2}
0.26mins
Q12b
Simplify. State any restrictions on the variables.
\displaystyle \frac{2ab}{5bc} \div\frac{6ac}{10b}
1.02mins
Q12c
Simplify. State any restrictions on the variables.
\displaystyle \frac{5p}{8pq} \div \frac{3p}{12q}
0.43mins
Q12d
Simplify. State any restrictions on the variables.
\displaystyle \frac{x^2}{2xy}\times \frac{x}{2y^2} \div \frac{(3x)^2}{xy^2}
0.54mins
Q13a
Simplify. State any restrictions on the variables.
\displaystyle \frac{x^2-5x + 6}{x^2-1} \times \frac{x^2-4x -5}{x^2-4} \div \frac{x-5}{x^2+3x+2}
1.45mins
Q13b
Simplify. State any restrictions on the variables.
\displaystyle \frac{1-x^2}{1+y} \times \frac{1-y^2}{x + x^2} \div \frac{y^3-y}{x^2}
2.00mins
Q13c
Simplify. State any restrictions on the variables.
\displaystyle \frac{x^2-y^2}{4x^2-y^2}\times \frac{4x^2+8xy+3y^2}{x + y}\div \frac{2x + 3y}{2x -y}
1.47mins
Q13d
Simplify. State any restrictions on the variables.
\displaystyle \frac{4}{5x}-\frac{2}{3x}
0.25mins
Q14a
Simplify. State any restrictions on the variables.
\displaystyle \frac{5}{x + 1}- \frac{2}{x -1}
0.47mins
Q14b
Simplify. State any restrictions on the variables.
\displaystyle \frac{1}{x^2+3x -4} + \frac{1}{x^2+x-12}
0.58mins
Q14c
Simplify. State any restrictions on the variables.
\displaystyle \frac{1}{x^2-5x+6} - \frac{1}{x^2 - 9}
1.24mins
Q14d
Simplify and state any restrictions on the variables.
\displaystyle \frac{1}{2x} - \frac{7}{3x^2}+ \frac{4}{x^3}
1.11mins
Q15a
Simplify and state any restrictions on the variables.
\displaystyle \frac{3x}{x + 2} + \frac{4x}{x -6}
0.53mins
Q15b
Simplify and state any restrictions on the variables.
\displaystyle \frac{6x}{x^2 - 5x + 6} - \frac{3x}{x^2 + x - 12}
Q15c
Simplify and state any restrictions on the variables.
\displaystyle \frac{2(x-2)^2}{x^2 + 6x + 5} \times \frac{3x + 15}{(2- x)^2}
Q15d
Simplify and state any restrictions on the variables.
\displaystyle \frac{(x - 2y)^2}{x^2 -y^2} \div \frac{(x- 2y)(x + 3y)}{(x + y)^2}
2.02mins
Q15e
Simplify and state any restrictions on the variables.
\displaystyle \frac{2b -5}{b^2 -2b -15} + \frac{3b}{b^2 + b - 30} \times \frac{b^2 + 8b + 12}{b + 3}
2.02mins
Q15f
Fred’s final mark in an online course was determined entirely by two exams. The mid-term exam was out of x marks and was worth 25% of his final mark. The final exam was out of 2x marks and was worth 75% of his final mark. Fred scored 40 marks on the first exam and 60 marks on the second exam. Determine the value of x if Fred earned a final mark of 50% in the course.
3.27mins
Q16
Sam plays a game in which he selects three different numbers from 1 to n (n > 3). After he selects his numbers, four different winning numbers from 1 to n are chosen, one at a time. Sam wins if all three of his numbers are among the four winning numbers.
The first number chosen is one of Sam's! His probability of winning is now given by
\displaystyle P(n) = \frac{24}{n^3 - 3n^2 + 2n} \div \frac{3}{n}
a) Simplify P(n) and state restrictions on n.
b) What would Sam's probability of winning be if
• i) n = 5?
• ii) n = 4?
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# SOLUTION: The length of a rectangle is 5 inches longer than the width. The area of the rectangle is 50 inches squared. Find the length and the width of the rectangle.
Algebra -> -> SOLUTION: The length of a rectangle is 5 inches longer than the width. The area of the rectangle is 50 inches squared. Find the length and the width of the rectangle. Log On
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Question 572708: The length of a rectangle is 5 inches longer than the width. The area of the rectangle is 50 inches squared. Find the length and the width of the rectangle.Answer by mathsmiles(68) (Show Source): You can put this solution on YOUR website!First thing we want to do is write down what we know: Area of a rectangle: A = L x W L = 5 + W (length is 5 inches longer than the width) A = 50 (Area is 50 inches squared) Putting this all together: L x W = 50 Substituting for L with the above equation: (5+W) x W = 50 Multiplying out the paren 5W + W^2 = 50 Subtract 50 from both sides and rearrange the terms a little: W^2 + 5W - 50 = 0 Now solve: (W _ _) (W _ _) = 0 (we need to figure out the operand and the term for each) The negative whole number indicates these have different signs (W - _) (W + _) = 0 We need to find two factors of 50 whose difference (subtract them) gives 5. 50 = 25 x 2 Nope 50 = 5 x 10 Yup! Since the 5W term is positive, we need to put the 10 in the positive factor and 5 in the negative factor so ... (W - 5)(W + 10) = 0 W -5 = 0 W = 5 W + 10 = 0 W = -10 Since we're talking about the area of a rectangle, we have to assume it has positive sides or we've just entered another dimension. :-) So, W = 5 inches L = 5 + 5 (5 inches longer than the width) L = 10 inches Checking: A = L x W A = 5 x 10 50 = 50 Correct!
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# Equation set solver
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Algebra is a fundamental tool in mathematics that allows us to solve equations and understand the relationships between variables. Basic algebra problems usually involve solving for a missing variable in an equation. For example, if we are given the equation "x + 5 = 10", we can solve for "x" by subtracting 5 from both sides of the equation to get "x = 5". Algebra can be used to solve for many different types of variables, making it a very powerful tool.
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# How do I use a sign chart to solve x^2>4?
Jun 30, 2015
${x}^{2} > 4$ when $x < - 2$ and when $x > 2$.
#### Explanation:
Let's modify the inequality by subtracting $4$ from each side. Then
x^2 – 4 > 0
We start by finding the critical numbers.
Set f(x) = x^2 – 4 = 0 and solve for $x$.
$\left(x + 2\right) \left(x - 2\right) = 0$
$x + 2 = 0$ or $x - 2 = 0$
$x = - 2$ or $x = + 2$
The critical numbers are $- 2$ and $+ 2$.
Now we check for positive and negative regions.
We have three regions to consider: (a) $x < - 2$; (b) $- 2 < x < 2$; and (c) $x > 2$.
Case (a): Let $x = - 3$.
Then $f \left(- 3\right) = {\left(- 3\right)}^{2} - 4 = 9 - 4 = 5$
$f \left(x\right) > 0$ when $x < - 2$.
Case (b): Let $x = 0$.
Then $f \left(0\right) = {0}^{2} - 4 = 0 - 4 = - 4$
$f \left(x\right) < 0$ when $- 2 < x < 2$
Case (c): Let $x = 3$.
Then $f \left(3\right) = {3}^{2} - 4 = 9 - 4 = 5$
$f \left(x\right) > 0$ when $x > 2$.
If ${x}^{2} - 4 > 0$ when $x < - 2$ and when $x > 2$,
${x}^{2} > 4$ when $x < - 2$ and when $x > 2$.
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4.5: Summary of Curve Sketching
Earlier in this chapter we stated that if a function $$f$$ has a local extremum at a point $$c$$, then $$c$$ must be a critical point of f. However, a function is not guaranteed to have a local extremum at a critical point. For example, $$f(x)=x^3$$ has a critical point at $$x=0$$ since $$f'(x)=3x^2$$ is zero at $$x=0$$, but $$f$$ does not have a local extremum at $$x=0$$. Using the results from the previous section, we are now able to determine whether a critical point of a function actually corresponds to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a graph by describing whether the graph of a function curves upward or curves downward.
## The First Derivative Test
Corollary $$3$$ of the Mean Value Theorem showed that if the derivative of a function is positive over an interval $$I$$ then the function is increasing over $$I$$. On the other hand, if the derivative of the function is negative over an interval $$I$$, then the function is decreasing over $$I$$ as shown in the following figure..
Figure $$\PageIndex{1}$$: Both functions are increasing over the interval $$(a,b)$$. At each point $$x$$, the derivative $$f'(x)>0$$. Both functions are decreasing over the interval $$(a,b)$$. At each point $$x$$, the derivative $$f'(x)<0.$$
A continuous function $$f$$ has a local maximum at point $$c$$ if and only if $$f$$ switches from increasing to decreasing at point $$c$$. Similarly, $$f$$ has a local minimum at $$c$$ if and only if $$f$$ switches from decreasing to increasing at $$c$$. If $$f$$ is a continuous function over an interval $$I$$ containing $$c$$ and differentiable over $$I$$, except possibly at $$c$$, the only way $$f$$ can switch from increasing to decreasing (or vice versa) at point $$c$$ is if $$f′$$ changes sign as $$x$$ increases through $$c$$. If $$f$$ is differentiable at $$c$$, the only way that $$f′$$. can change sign as $$x$$ increases through $$c$$ is if $$f′(c)=0$$. Therefore, for a function $$f$$ that is continuous over an interval $$I$$ containing $$c$$ and differentiable over $$I$$, except possibly at $$c$$, the only way $$f$$ can switch from increasing to decreasing (or vice versa) is if $$f'(c)=0$$ or $$f′(c)$$ is undefined. Consequently, to locate local extrema for a function $$f$$, we look for points $$c$$ in the domain of $$f$$ such that $$f'(c)=0$$ or $$f′(c)$$ is undefined. Recall that such points are called critical points of $$f$$.
Note that $$f$$ need not have a local extrema at a critical point. The critical points are candidates for local extrema only. In Figure, we show that if a continuous function $$f$$ has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if $$f$$ has a local extremum at a critical point, then the sign of $$f′$$ switches as $$x$$ increases through that point.
Figure $$\PageIndex{2}$$: The function $$f$$ has four critical points: $$a,b,c$$,and $$d$$. The function $$f$$ has local maxima at $$a$$ and $$d$$, and a local minimum at $$b$$. The function $$f$$ does not have a local extremum at $$c$$. The sign of $$f'$$ changes at all local extrema.
Using Figure, we summarize the main results regarding local extrema.
• If a continuous function $$f$$ has a local extremum, it must occur at a critical point $$c$$.
• The function has a local extremum at the critical point $$c$$ if and only if the derivative $$f′$$ switches sign as $$x$$ increases through $$c$$.
• Therefore, to test whether a function has a local extremum at a critical point $$c$$, we must determine the sign of $$f′(x)$$ to the left and right of $$c$$.
This result is known as the first derivative test.
First Derivative Test
Suppose that $$f$$ is a continuous function over an interval $$I$$ containing a critical point $$c$$. If $$f$$ is differentiable over $$I$$, except possibly at point $$c$$, then $$f(c)$$ satisfies one of the following descriptions:
1. If $$f′$$ changes sign from positive when $$x<c$$ to negative when $$x>c$$, then $$f(c)$$ is a local maximum of $$f$$.
2. If $$f′$$ changes sign from negative when $$x<c$$ to positive when $$x>c$$, then $$f(c)$$ is a local minimum of $$f$$.
3. If $$f'$$ has the same sign for $$x<c$$ and $$x>c$$, then $$f(c)$$ is neither a local maximum nor a local minimum of $$f$$
Now let’s look at how to use this strategy to locate all local extrema for particular functions.
Example $$\PageIndex{1}$$: Using the First Derivative Test to Find Local Extrema
Use the first derivative test to find the location of all local extrema for $$f(x)=x^3−3x^2−9x−1.$$ Use a graphing utility to confirm your results.
Solution
Step 1. The derivative is $$f′(x)=3x^2−6x−9.$$ To find the critical points, we need to find where $$f′(x)=0.$$ Factoring the polynomial, we conclude that the critical points must satisfy
$3(x^2−2x−3)=3(x−3)(x+1)=0.$
Therefore, the critical points are x$$=3,−1.$$ Now divide the interval $$(−∞,∞)$$ into the smaller intervals $$(−∞,−1),(−1,3)$$ and $$(3,∞).$$
Step 2. Since $$f'$$ is a continuous function, to determine the sign of $$f′(x)$$ over each subinterval, it suffices to choose a point over each of the intervals $$(−∞,−1),(−1,3)$$ and $$(3,∞)$$ and determine the sign of $$f′$$ at each of these points. For example, let’s choose $$x=−2,x=0,$$ and $$x=4$$ as test points.
Interval Test Point Sign of $$f′(x)=3(x−3)(x+1)$$ at Test Point Conclusion
$$(−∞,−1)$$ $$x=−2$$ (+)(−)(−)=+ $$f$$ is increasing.
$$(−1,3)$$ $$x=0$$ (+)(−)(+)=+ $$f$$ is increasing.
$$(3,∞)$$ $$x=4$$ (+)(+)(+)=+ $$f$$ is increasing.
Step 3. Since $$f′$$ switches sign from positive to negative as $$x$$ increases through $$1,f$$ has a local maximum at $$x=−1$$. Since $$f′$$ switches sign from negative to positive as $$x$$ increases through $$3,f$$ has a local minimum at $$x=3$$. These analytical results agree with the following graph.
Figure $$\PageIndex{3}$$: The function f has a maximum at $$x=−1$$ and a minimum at $$x=3$$
Exercise $$\PageIndex{1}$$
Use the first derivative test to locate all local extrema for $$f(x)=−x^3+\frac{3}{2}x^2+18x.$$
Hint
Find all critical points of $$f$$ and determine the signs of $$f′(x)$$ over particular intervals determined by the critical points.
$$f$$ has a local minimum at $$−2$$ and a local maximum at $$3$$.
Example $$\PageIndex{2}$$: Using the First Derivative Test
Use the first derivative test to find the location of all local extrema for $$f(x)=5x^{1/3}−x^{5/3}.$$ Use a graphing utility to confirm your results.
Solution
Step 1. The derivative is
$f′(x)=\frac{5}{3}x^{−2/3}−\frac{5}{3}x^{2/3}=\frac{5}{3x^{2/3}}−\frac{5x^{2/3}}{3}=\frac{5−5x^{4/3}}{3x^{2/3}}=\frac{5(1−x^{4/3})}{3x^{2/3}}.$
The derivative $$f′(x)=0$$ when $$1−x^{4/3}=0.$$ Therefore, $$f′(x)=0$$ at $$x=±1$$. The derivative $$f′(x)$$ is undefined at $$x=0.$$ Therefore, we have three critical points: $$x=0, x=1,$$ and $$x=−1$$. Consequently, divide the interval $$(−∞,∞)$$ into the smaller intervals $$(−∞,−1),(−1,0),(0,1),$$ and $$(1,∞)$$.
Step 2: Since $$f′$$ is continuous over each subinterval, it suffices to choose a test point $$x$$ in each of the intervals from step 1 and determine the sign of $$f′$$ at each of these points. The points $$x=−2,x=−\frac{1}{2},x=\frac{1}{2},and x=2$$ are test points for these intervals.
Interval Test Point Sign of $$f′(x)=\frac{5(1−x^{4/3})}{3x^{2/3}}$$ at Test Point Conclusion
$$(−∞,−1)$$ $$x=−2$$ $$\frac{(+)(−)}{+}=−$$ $$f$$ is decreasing.
$$(−1,0)$$ $$x=−\frac{1}{2}$$ $$\frac{(+)(+)}{+}=+$$ $$f$$ is increasing.
$$(0,1)$$ $$x=\frac{1}{2}$$ $$\frac{(+)(+)}{+}=+$$ $$f$$ is increasing.
$$(1,∞)$$ $$x=2$$ $$\frac{(+)(−)}{+}=−$$ $$f$$ is decreasing.
Step 3: Since $$f$$ is decreasing over the interval $$(−∞,−1)$$ and increasing over the interval $$(−1,0), f$$ has a local minimum at $$x=−1$$. Since $$f$$ is increasing over the interval $$(−1,0)$$ and the interval $$(0,1)$$, f does not have a local extremum at $$x=0$$. Since $$f$$ is increasing over the interval $$(0,1)$$ and decreasing over the interval $$(1,∞),f$$ has a local maximum at $$x=1$$. The analytical results agree with the following graph.
Figure $$\PageIndex{4}$$: The function f has a local minimum at $$x=−1$$ and a local maximum at $$x=1$$
Exercise $$\PageIndex{2}$$
Use the first derivative test to find all local extrema for $$(x)=\frac{3}{x−1}$$.
Hint
The only critical point of $$f$$ is $$x=1.$$
$$f$$ has no local extrema because $$f′$$ does not change sign at $$x=1$$.
## Concavity and Points of Inflection
We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function.
Figure(a) shows a function $$f$$ with a graph that curves upward. As $$x$$ increases, the slope of the tangent line increases. Thus, since the derivative increases as $$x$$ increases, $$f′$$ is an increasing function. We say this function $$f$$ is concave up. Figure(b) shows a function $$f$$ that curves downward. As $$x$$ increases, the slope of the tangent line decreases. Since the derivative decreases as $$x$$ increases, $$f′$$ is a decreasing function. We say this function $$f$$ is concave down.
Definition: concavity test
Let $$f$$ be a function that is differentiable over an open interval $$I$$. If $$f′$$ is increasing over $$I$$, we say $$f$$ is concave up over $$I$$. If $$f′$$ is decreasing over $$I$$, we say $$f$$ is concave down over $$I$$.
Figure $$\PageIndex{5}$$: (a), (c) Since $$f'$$ is increasing over the interval $$(a,b)$$, we say $$f$$ is concave up over $$(a,b). (b), (d)$$ Since $$f'$$ is decreasing over the interval $$(a,b)$$, we say $$f$$ is concave down over $$(a,b).$$
In general, without having the graph of a function $$f,$$ how can we determine its concavity? By definition, a function $$f$$ is concave up if $$f′$$ is increasing. From Corollary $$3$$, we know that if $$f′$$ is a differentiable function, then $$f′$$ is increasing if its derivative $$f''(x)>0$$. Therefore, a function $$f$$ that is twice differentiable is concave up when $$f''(x)>0$$. Similarly, a function $$f$$ is concave down if $$f′$$ is decreasing. We know that a differentiable function $$f′$$ is decreasing if its derivative $$f''(x)<0$$. Therefore, a twice-differentiable function $$f$$ is concave down when $$f''(x)<0$$. Applying this logic is known as the concavity test.
Test for Concavity
Let $$f$$ be a function that is twice differentiable over an interval $$I$$.
1. If $$f''(x)>0$$ for all $$x∈I$$, then $$f$$ is concave up over $$I$$
2. If $$f''(x)<0$$ for all $$x∈I,$$ then $$f$$ is concave down over $$I$$.
We conclude that we can determine the concavity of a function $$f$$ by looking at the second derivative of $$f$$. In addition, we observe that a function $$f$$ can switch concavity (Figure). However, a continuous function can switch concavity only at a point $$x$$ if $$f''(x)=0$$ or $$f''(x)$$ is undefined. Consequently, to determine the intervals where a function $$f$$ is concave up and concave down, we look for those values of $$x$$ where $$f''(x)=0$$ or $$f''(x)$$ is undefined. When we have determined these points, we divide the domain of $$f$$ into smaller intervals and determine the sign of $$f'$$' over each of these smaller intervals. If $$f''$$ changes sign as we pass through a point $$x$$, then $$f$$ changes concavity. It is important to remember that a function $$f$$ may not change concavity at a point $$x$$ even if $$f''(x)=0$$ or $$f''(x)$$ is undefined. If, however, $$f$$ does change concavity at a point a and $$f$$ is continuous at $$a$$, we say the point $$(a,f(a))$$ is an inflection point of $$f$$.
Definition: inflection point
If $$f$$ is continuous at $$a$$ and $$f$$ changes concavity at $$a$$, the point $$(a,f(a))$$ is an inflection point of $$f$$.
Figure $$\PageIndex{6}$$: Since $$f''(x)>0$$ for $$x<a$$, the function $$f$$ is concave up over the interval $$(−∞,a)$$. Since $$f''(x)<0$$ for $$x>a$$, the function $$f$$ is concave down over the interval $$(a,∞)$$. The point $$(a,f(a))$$ is an inflection point of $$f$$.
Example $$\PageIndex{3}$$: Testing for Concavity
For the function $$f(x)=x^3−6x^2+9x+30,$$ determine all intervals where $$f$$ is concave up and all intervals where $$f$$ is concave down. List all inflection points for $$f$$. Use a graphing utility to confirm your results.
Solution
To determine concavity, we need to find the second derivative $$f''(x).$$ The first derivative is $$f'(x)=3x^2−12x+9,$$ so the second derivative is $$f''(x)=6x−12.$$ If the function changes concavity, it occurs either when $$f''(x)=0$$ or $$f''(x)$$ is undefined. Since $$f''$$ is defined for all real numbers $$x$$, we need only find where $$f''(x)=0$$. Solving the equation $$6x−12=0$$, we see that $$x=2$$ is the only place where $$f$$ could change concavity. We now test points over the intervals $$(−∞,2)$$ and $$(2,∞)$$ to determine the concavity of $$f$$. The points $$x=0$$ and $$x=3$$ are test points for these intervals.
Interval Test Point Sign of $$f''(x)=6x−12$$ at Test Point Conclusion
$$(−∞,2)$$ $$x=0$$ $$f$$ is concave down
$$(2,∞)$$ $$x=3$$ + $$f$$ is concave up
We conclude that $$f$$ is concave down over the interval $$(−∞,2)$$ and concave up over the interval $$(2,∞)$$. Since $$f$$ changes concavity at $$x=2$$, the point $$(2,f(2))=(2,32)$$ is an inflection point. Figure confirms the analytical results.
Figure $$\PageIndex{7}$$: The given function has a point of inflection at $$(2,32)$$ where the graph changes concavity.
Exercise $$\PageIndex{3}$$
For $$f(x)=−x^3+\frac{3}{2}x^2+18x$$, find all intervals where $$f$$ is concave up and all intervals where $$f$$ is concave down.
Hint
Find where $$f''(x)=0$$
$$f$$ is concave up over the interval $$(−∞,\frac{1}{2})$$ and concave down over the interval $$(\frac{1}{2},∞)$$
We now summarize, in Table, the information that the first and second derivatives of a function $$f$$ provide about the graph of $$f$$, and illustrate this information in Figure.
What Derivatives Tell Us about Graphs
Sign of $$f'$$ Sign of $$f''$$ Is $$f$$ increasing or decreasing? Concavity
Positive Positive Increasing Concave up
Positive Negative Increasing Concave down
Negative Positive Decreasing Concave up
Negative Negative Decreasing Concave sown
Figure $$\PageIndex{8}$$:Consider a twice-differentiable $$I$$ over an open interval $$I$$. If $$f'(x)>0$$ for all $$x∈I$$, the function is increasing over $$I$$. If $$f'(x)<0$$ for all $$x∈I$$, the function is decreasing over $$I$$. If $$f''(x)>0$$ for all $$x∈I$$, the function is concave up. If $$f''(x)<0$$ for all $$x∈I$$, the function is concave down on $$I$$.
## The Second Derivative Test
The first derivative test provides an analytical tool for finding local extrema, but the second derivative can also be used to locate extreme values. Using the second derivative can sometimes be a simpler method than using the first derivative.
We know that if a continuous function has a local extrema, it must occur at a critical point. However, a function need not have a local extrema at a critical point. Here we examine how the second derivative test can be used to determine whether a function has a local extremum at a critical point. Let $$f$$ be a twice-differentiable function such that $$f′(a)=0$$ and $$f''$$ is continuous over an open interval $$I$$ containing $$a$$. Suppose $$f''(a)<0$$. Since $$f''$$ is continuous over $$I, f''(x)<0$$ for all $$x∈I$$ (Figure). Then, by Corollary $$3, f′$$ is a decreasing function over $$I$$. Since $$f′(a)=0$$, we conclude that for all $$x∈I,f′(x)>0$$ if $$x<a$$ and $$f′(x)<0$$ if $$x>a$$. Therefore, by the first derivative test, $$f$$ has a local maximum at $$x=a$$. On the other hand, suppose there exists a point $$b$$ such that $$f′(b)=0$$ but $$f''(b)>0$$. Since $$f''$$ is continuous over an open interval $$I$$ containing $$b$$, then $$f''(x)>0$$ for all $$x∈I$$ (Figure). Then, by Corollary $$3,f′$$ is an increasing function over $$I$$. Since $$f′(b)=0$$, we conclude that for all $$x∈I$$, $$f′(x)<0$$ if $$x<b$$ and $$f′(x)>0$$ if $$x>b$$. Therefore, by the first derivative test, $$f$$ has a local minimum at $$x=b.$$
Figure $$\PageIndex{9}$$: Consider a twice-differentiable function $$f$$ such that $$f''$$ is continuous. Since $$f'(a)=0$$ and $$f''(a)<0$$, there is an interval $$I$$ containing $$a$$ such that for all $$x$$ in $$I$$, $$f$$ is increasing if $$x<a$$ and $$f$$ is decreasing if $$x>a$$. As a result, $$f$$ has a local maximum at $$x=a$$. Since $$f'(b)=0$$ and $$f''(b)>0$$, there is an interval $$I$$ containing $$b$$ such that for all $$x$$ in $$I$$, $$f$$ is decreasing if $$x<b$$ and $$f$$ is increasing if $$x>b$$. As a result, $$f$$ has a local minimum at $$x=b$$.
Second Derivative Test
Suppose $$f′(c)=0,f''$$ is continuous over an interval containing $$c$$.
1. If $$f''(c)>0$$, then $$f$$ has a local minimum at $$c$$.
2. If $$f''(c)<0$$, then $$f$$ has a local maximum at $$c$$.
3. If $$f''(c)=0,$$ then the test is inconclusive.
Note that for case iii. when $$f''(c)=0$$, then $$f$$ may have a local maximum, local minimum, or neither at $$c$$. For example, the functions $$f(x)=x^3, f(x)=x^4,$$ and $$f(x)=−x^4$$ all have critical points at $$x=0$$. In each case, the second derivative is zero at $$x=0$$. However, the function $$f(x)=x^4$$ has a local minimum at $$x=0$$ whereas the function $$f(x)=−x^4$$ has a local maximum at $$x$$, and the function $$f(x)=x^3$$ does not have a local extremum at $$x=0$$.
Let’s now look at how to use the second derivative test to determine whether $$f$$ has a local maximum or local minimum at a critical point c where $$f′(c)=0.$$
Example $$\PageIndex{4}$$: Using the Second Derivative Test
Use the second derivative to find the location of all local extrema for $$f(x)=x^5−5x^3.$$
Solution
apply the second derivative test, we first need to find critical points $$c$$ where $$f′(c)=0$$. The derivative is $$f′(x)=5x^4−15x^2$$. Therefore, $$f′(x)=5x^4−15x^2=5x^2(x^2−3)=0$$ when $$x=0,±\sqrt{3}\$$.
To determine whether $$f$$ has a local extrema at any of these points, we need to evaluate the sign of $$f''$$ at these points. The second derivative is
$$f''(x)=20x^3−30x=10x(2x^2−3).$$
In the following table, we evaluate the second derivative at each of the critical points and use the second derivative test to determine whether $$f$$ has a local maximum or local minimum at any of these points.
$$x$$ $$f''(x)$$ Conclusion $$−\sqrt{3}$$ $$−30\sqrt{3}$$ Local maximum $$0$$ $$0$$ Second derivative test is inconclusive $$\sqrt{3}$$ $$30\sqrt{3}$$ Local minimum
By the second derivative test, we conclude that $$f$$ has a local maximum at $$x=−\sqrt{3}$$ and $$f$$ has a local minimum at $$x=\sqrt{3}$$. The second derivative test is inconclusive at $$x=0$$. To determine whether $$f$$ has a local extrema at $$x=0,$$ we apply the first derivative test. To evaluate the sign of $$f′(x)=5x^2(x^2−3)$$ for $$x∈(−\sqrt{3},0)$$ and $$x∈(0,\sqrt{3})$$, let $$x=−1$$ and $$x=1$$ be the two test points. Since $$f′(−1)<0$$ and $$f′(1)<0$$, we conclude that $$f$$ is decreasing on both intervals and, therefore, $$f$$ does not have a local extrema at $$x=0$$ as shown in the following graph.
Figure $$\PageIndex{10}$$:The function $$f$$ has a local maximum at $$x=−\sqrt{3}$$ and a local minimum at $$x=\sqrt{3}$$
Exercise $$\PageIndex{4}$$
Consider the function $$f(x)=x^3−(\frac{3}{2})x^2−18x$$. The points $$c=3,−2$$ satisfy $$f′(c)=0$$. Use the second derivative test to determine whether $$f$$ has a local maximum or local minimum at those points.
Hint
$$f''(x)=6x−3$$
$$f$$ has a local maximum at $$−2$$ and a local minimum at $$3$$.
We have now developed the tools we need to determine where a function is increasing and decreasing, as well as acquired an understanding of the basic shape of the graph. In the next section we discuss what happens to a function as $$→±∞.$$ At that point, we have enough tools to provide accurate graphs of a large variety of functions.
## Key Concepts
• If $$c$$ is a critical point of $$f$$ and $$f′(x)>0$$ for $$x<c$$ and $$f′(x)<0$$ for $$x>c$$, then $$f$$ has a local maximum at $$c$$.
• If $$c$$ is a critical point of $$f$$ and $$f′(x)<0$$ for $$x<c$$ and $$f′(x)>0$$ for $$x>c,$$ then $$f$$ has a local minimum at $$c$$.
• If $$f''(x)>0$$ over an interval $$I$$, then $$f$$ is concave up over $$I$$.
• If $$f''(x)<0$$ over an interval $$I$$, then $$f$$ is concave down over $$I$$.
• If $$f′(c)=0$$ and $$f''(c)>0$$, then $$f$$ has a local minimum at $$c$$.
• If $$f′(c)=0$$ and $$f''(c)<0$$, then $$f$$ has a local maximum at $$c$$.
• If $$f′(c)=0$$ and $$f''(c)=0$$, then evaluate $$f′(x)$$ at a test point $$x$$ to the left of $$c$$ and a test point $$x$$ to the right of $$c$$, to determine whether $$f$$ has a local extremum at $$c$$.
## Glossary
concave down
if $$f$$ is differentiable over an interval $$I$$ and $$f′$$ is decreasing over $$I$$, then $$f$$ is concave down over $$I$$
concave up
if $$f$$ is differentiable over an interval $$I$$ and $$f′$$ is increasing over $$I$$, then $$f$$ is concave up over $$I$$
concavity
the upward or downward curve of the graph of a function
concavity test
suppose $$f$$ is twice differentiable over an interval $$I$$; if $$f''>0$$ over $$I$$, then $$f$$ is concave up over $$I$$; if $$f''<$$ over $$I$$, then $$f$$ is concave down over $$I$$
first derivative test
let $$f$$ be a continuous function over an interval $$I$$ containing a critical point $$c$$ such that $$f$$ is differentiable over $$I$$ except possibly at $$c$$; if $$f′$$ changes sign from positive to negative as $$x$$ increases through $$c$$, then $$f$$ has a local maximum at $$c$$; if $$f′$$ changes sign from negative to positive as $$x$$ increases through $$c$$, then $$f$$ has a local minimum at $$c$$; if $$f′$$ does not change sign as $$x$$ increases through $$c$$, then f does not have a local extremum at $$c$$
inflection point
if $$f$$ is continuous at $$c$$ and $$f$$ changes concavity at $$c$$, the point $$(c,f(c))$$ is an inflection point of $$f$$
second derivative test
suppose $$f′(c)=0$$ and $$f'$$' is continuous over an interval containing $$c$$; if $$f''(c)>0$$, then $$f$$ has a local minimum at $$c$$; if $$f''(c)<0$$, then $$f$$ has a local maximum at $$c$$; if $$f''(c)=0$$, then the test is inconclusive
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# October 21, 2009
## Perimeter Greater than Area?
Why does our present title then not make any sense? Firstly we are comparing unequal things, like apples and
oranges, or like comparing distance and speed. Obviously you cannot say anything if you are asked which is
greater: distance 100 km or the speed value 200 km per hour?
But nevertheless sometimes interesting information can be gained by pursuing useless things. We are also
doing so to show how elementary questions can be asked and explored at the elementary school level (well, it
helps if children should know graphs).
Okay here is the problem: when is perimeter greater than area for a rectangle (PM being measured say in cm
units and area in sq.cm units)? When we say greater we mean numerically greater.
[Well, if you want to make “sense” of this kind of question, you may reformulate the problem thus: we have
two number crunching machines, the perimeter machine and the area machine. You feed positive numbers a
and b – into both – then at what point the output from the one will exceed the other? ]
Let us take s and t be the sides of the rectangle and PM as its perimeter and A as its area.
## 1) When s = 1: PM = 2 (length + breadth) = 2 (1 + t). Area is length x breadth = 1x t = t. Now 2 + 2t is
always greater than t. Therefore here PM > A, when s =1.
This means whenever you have one of the sides as 1, perimeter will be always (numerically) greater
than area.
2) When s = 2: again here PM > area as always 2 (2+t) = 4 + 2t > 2t. This means whenever you have s
=2, perimeter will be always (numerically) greater than area, irrespective of whatever value t takes.
3) When s = 3: Here it is bit more curious. PM > A till the other side t is less than 6, and when t > 6, A >
PM, that is area is greater than perimeter. When t is exactly equal to 6, that is when t =6, area and
perimeter are equal (numerically only, mind you). Here A = PM = 18. This is the first instance when
area is equal to perimeter for integer values of the sides (s, t = 3,6). For a given value of s, we will call
the t-value equipoint when area becomes equal to perimeter. Or the point when the relation PM > A,
becomes the other way around, that is A >PM. Again numerically please let us remember.
When s = 4: Till t reaches the value 4, PM > Area. At t = 4, PM = Area. When t > 4, A > PM. When t = 4,
Area = PM =16. Now this is the second instance when area is equal to perimeter for integer values of the sides
(s, t = 4,4). As far as we know (s,t) = (3,6) or (6,3) and (4,4) are the only instances when A = PM for integer
values of the sides. Note the equipoint is t = 4 for s =4. The other integer solutions are,
## s= 1, t = -2; s =0, t =0; s = -2, t = 1.
4) When s =5: the equipoint is t = 10/3 =3.333…, that is area = perimeter = 15 in this case. When t
>10/3, A > PM. When t < 10/3, Area < PM.
If you wondered, how we got 10/3, all we did was solve the equation when PM = Area, or: 2 (5 +t) =
5t which leads to 3t =10 or t =10/3.
5) When s = 6: we know from (3) above there is an integer equipoint at t = 3. And also therefore when t
<3, A < PM and when t > 3, A >PM.
## 9) When s= 10: equipoint is t = 2.50.
You can see the trend: as s increases, the equipoint t (or the point when the relation PM > A, becomes the
other way around, that is A >PM) decreases. Does it indefinitely decrease? Or approaches a limit beyond
which it does not go?
No t does not decrease beyond a point. It approaches a limit: when s increases, the equipoint t decreases to
(or approaches) 2.
2 (s+t) =st
## or 2s + 2t = st; or 2s = st –2t = t(s-2). From which we can write t = 2s/(s-2) or
t = 2/(1-2/s). From this it is clear when s increases towards infinity, 2/s tends to zero, or 2/s 0.
Therefore, t 2.
We show this in the graphs below, t is on the vertical axis and s is on the horizontal axis. We have plotted
t = 2s/(s-2). In the graphs (produced by Wolfram Alpha), x is our s.
The graph clearly shows as s increases towards positive infinity, it approaches 2 and as s decreases and
goes towards 2 (say from 2.5) then the equipoint t approaches infinity. The latter can be understood when
we calculate t for s = 3, 2.5, 2.2, 2.1, 2.001, 2.0001, etc. The equipoint t values are respectively 6, 10, 22,
42, 4002, 40,002, etc.
If you are further interested in this kind of ‘exciting’ stuff, as s approaches 2 from zero to 1.5, 1.75, 1.90,
the corresponding equipoint, t, approaches minus infinity, etc. Also when s takes values between 0 and
less than 2, the corresponding equipoints t are negative. But when s is negative, that is less than zero; the
corresponding equipoints are always negative.
You may also like to interpret by using your imagination like Alice in Wonderland what do the following
mean or is it pure nonsense:
For example, s = ½, t =-1, A = -½, PM = -1; or s = 3, t =-1, A = -3, PM = 2; or s= -4, t =-5, A = 20, PM = -
18.
That is, area and perimeter both negative with one parameter positive; area negative and perimeter positive
with one parameter positive; or area positive and perimeter negative when both parameters t and s are
negative.
This author cannot think of any meaningful explanation physically. But welcomes suggestions.
|
# 12.3 Surface Area of Pyramids and Cones
## Presentation on theme: "12.3 Surface Area of Pyramids and Cones"— Presentation transcript:
12.3 Surface Area of Pyramids and Cones
Geometry Mrs. Spitz Spring 2005
Objectives/Assignment
Find the surface area of a pyramid. Find the surface area of a cone. Assignment: WS 12.3A
Finding the surface area of a pyramid
A pyramid is a polyhedron in which the base is a polygon and the lateral faces are triangles with a common vertex. The intersection of two lateral faces is a lateral edge. The intersection of the base and a lateral face is a base edge. The altitude or height of a pyramid is the perpendicular distance between the base and the vertex.
More on pyramids A regular pyramid has a regular polygon for a base and its height meets the base at its center. The slant height of a regular pyramid is the altitude of any lateral face. A nonregular pyramid does not have a slant height.
Pyramid Arena
Ex. 1: Finding the Area of a Lateral Face
Architecture. The lateral faces of the Pyramid Arena in Memphis, Tennessee, are covered with steal panels. Use the diagram of the arena to find the area of each lateral face of this regular pyramid.
Hexagonal Pyramids A regular hexagonal pyramid and its net are shown at the right. Let b represent the length of a base edge, and let l represent the slant height of the pyramid. The area of each lateral face is 1/2bl and the perimeter of the base if P = 6b. So the surface area is as follows:
Hexagonal pyramid Surface Area of a Regular Pyramid
S = (Area of base) + 6(Area of lateral face) S = B + 6( ½ bl) Substitute Rewrite 6( ½ bl) as ½ (6b)l. S = B + (6b)l Substitute P for 6b S = B + Pl Surface Area of a Regular Pyramid The surface area S of a regular pyramid is: S = B + ½ Pl, where B is the area of the base, P is the perimeter of the base, and l is the slant height.
Ex. 2: Finding the surface area of a pyramid
To find the surface area of the regular pyramid shown, start by finding the area of the base. Use the formula for the area of a regular polygon, ½ (apothem)(perimeter). A diagram of the base is shown to the right.
Ex. 2: Finding the surface area of a pyramid
After substituting, the area of the base is ½ ( )(6• 6), or square meters.
Surface area Now you can find the surface area by using for the area of the base, B.
Finding the Surface Area of a Cone
A circular cone, or cone, has a circular base and a vertex that is NOT in the same plane as the base. The altitude, or height, is the perpendicular distance between the vertex and the base. In a right cone, the height meets the base at its center and the slant height is the distance between the vertex and a point on the base edge.
Finding the Surface Area of a Cone
The lateral surface of a cone consists of all segments that connect the vertex with points on the base edge. When you cut along the slant height and like the cone flat, you get the net shown at the right. In the net, the circular base has an area of r2 and the lateral surface area is the sector of a circle.
More on cones . . . You can find the area of this sector by using a proportion, as shown below. Area of sector Arc length = Set up proportion Area of circle Circumference 2r Area of sector Substitute = l2 2l 2r Multiply each side by l2 Area of sector = l2 • 2l Area of sector = rl Simplify The surface area of a cone is the sum of the base area and the lateral area, rl.
Theorem Surface Area of a Right Cone
The surface area S of a right cone is S = r2 + rl, where r is the radius of the base and l is the slant height
Ex. 3: Finding the surface area of a cone
To find the surface area of the right cone shown, use the formula for the surface area. S = r2 + rl Write formula S = 42 + (4)(6) Substitute S = 16 + 24 Simplify S = 40 Simplify The surface area is 40 square inches or about square inches.
|
# Class 6 Maths Chapter 12 Test Paper Set-3 Pdf Download CBSE
Test Paper Of Class 6 Maths Chapter 12 Ratio And Proportion
class-6-maths-chapter-12-test-paper-03
Chapter 12 Test Paper (03)
## Class 6 Maths Chapter 12 Ratio And Proportion Test Paper Set- 3 Text Form:-
PGRMS EDUCATION
SAMPLE PAPER (03)
CLASS: 6
MAX. MARKS : 30
SUBJECT: MATHEMATICS
TIME: 1:30 HOUR
CH: 12 (Ratio And Proportion)
General Instructions:
I. All questions are compulsory.
II. This question paper contains 14 questions divided into four Sections A, B, C and D.
III. Section A has 5 questions of 1 mark each. Section B has 4 questions of 2 marks each. Section C has 3 questions of 3 marks each. and Section D has 2 questions of 4 marks each.
IV. Use of Calculators is not permitted.
Section ā A
In Questions 1 to 5, there are four options, out of which one is correct. Write the correct answer.
1. The ratio of 8 books to 20 books is
(a) 2 : 5 (b) 5 : 2 (c) 4 : 5 (d) 5 : 4
2. In a box, the ratio of red marbles to blue marbles is 7:4. Which of the following could be the total number of marbles in the box?
(a) 18 (b) 19 (c) 21 (d) 22
3. The men proportion 9 and 16 is
(a) 3 (b)12 (c) 33 (d) 11
4. Find the ratio of 15 kg to 75 kg is ā¦ā¦ā¦..,
(a) 1 : 5 (b) 5 : 1 (c) 3 : 5 (d) 15 : 5.
5. 30 : 45 is equivalent ratio of
(a) 15 : 3 (b) 3 : 15 (c) 3 : 5 (d) 5 : 3
Section ā B
6. Find the ratio of the following :
(a) 81 to 108 (b) 98 to 63
7. In a year, Seema earns ā¹ 1,50,000 and saves ā¹ 50,000. Find the ratio of
(a) Money that Seema earns to the money she saves.
(b) Money that she saves to the money she spends.
8. Present age of father is 42 years and that of his son is 14 years. Find the ratio of.
(a) Age of father after 10 years to the age of son after 10 years.
(b) Age of father to the age of son when father was 30 years old.
9. Present age of father is 42 years and that of his son is 14 years. Find the ratio of:
(a) Age of father after 10 years to the age of son after 10 years.
(b) Age of father to the age of son when father was 30 years old.
Section ā C
10. Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm : 1 m and ā¹ 40 : ā¹ 160
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
(c) 2 kg : 80 kg and 25 g : 625 g
11. Of the 288 persons working in a company, 112 are men and the remaining are women. Find the ratio of the number of
(a) men to that of women.
(b) men to the total number of persons.
(c) women to the total number of persons.
12. Ekta earns ā¹ 3000 in 10 days. Find:
(i) How much will she earn in 30 days?
(ii) In how many days she will earn ā¹ 1500
(iii) What is her monthly income
Section ā D
13. Are the following statements true?
(a) 40 persons : 200 persons = ā¹ 15 : ā¹ 75
(b) 7.5 litres : 15 litres = 5 kg : 10 kg
(c) 99 kg : 45 kg = ā¹ 44 : ā¹ 20
(d) 32 m : 64 m = 6 sec : 12 sec
14 : A car travels 90 km in 2 1 2 hours.
(a) How much time is required to cover 30 km with the same speed?
(b) Find the distance covered in 2 hours with the same speed
|
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Calculus
# Leibniz’s Notation & dy/dx Meaning
11.20.2021 • 6 min read
## Rachel McLean
Subject Matter Expert
Leibniz’s notation is a fundamental type of notation for derivatives. In this article, we’ll discuss the meaning of dy/dx, how to use Leibniz’s notation, and practice some examples.
## Who Is Leibniz?
Gottfried Wilhelm Leibniz (1646 - 1716) was a 17th century German mathematician. He’s often credited with developing many of the main principles of differential and integral calculus, and is primarily recognized for what we now call Leibniz’s notation.
## What Is Leibniz’s Notation System?
Derivative notations are used to express the derivative of a function based on today’s standard definition of a derivative. The instantaneous rate of change, or derivative, of a function $f$ at $x$ is given by: $\frac{d}{{dx}}f\left( x \right) = \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{\Delta{y}}{\Delta{x}} = \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x \right)}}{\Delta{x} }$
There are many different derivative notations, but Leibniz’s notation remains one of the most popular. Given a function $f$ defined by $y = f(x)$, Leibniz's notation expresses the derivative of $f$ at $x$ as:
$\frac{dy}{dx}$
It might be tempting to think of $\frac{dy}{dx}$ as a fraction. In fact, Leibniz himself first conceptualized $\frac{dy}{dx}$ as the quotient of an infinitely small change in y by an infinitely small change in $x$, called infinitesimals. However, this understanding of Leibniz’s notation lost popularity in the 19th-century when infinitesimals were considered too imprecise to define the infrastructure of calculus. Leibniz’s original understanding of $\frac{dy}{dx}$ as a quotient has been reinterpreted to align with the modern limit-based definition of a derivative. Now, $dy$ and $dx$ are generally referred to as differentials instead of infinitesimals. While Leibniz’s notation behaves in a way similar to a fraction, it’s important that you understand the difference.
## What Is dy/dx?
We can think of $\frac{d}{dx}$ as an operator defined by the standard limit-based definition of a derivative. Suppose we have a function $f$ defined by $y = f(x)$. When we apply the operator $\frac{d}{dx}$ to $y$, we have the expression $\frac{d}{dx}y$, or $\frac{dy}{dx}$. This expression represents the derivative of $y$ with respect to $x$ (note that $y$ is the function value of $f$ at $x$). In this way, the operator $\frac{d}{dx}$ takes in one function, and outputs another! More specifically, the operator $\frac{d}{dx}$ acts on a function to produce that function’s derivative.
Let’s review some examples where Leibniz’s notation is often utilized. Consider the Chain Rule, which helps us differentiate composite functions.
Suppose we have two differentiable functions $f$ and $g$, and suppose that $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$.
Then, the composite function $h = f \circ g$, such that $h(x) = f(g(x))$ for all $x$, is differentiable at $x$.
With these conditions satisfied, the Chain Rule states:
$h’(x) = f’(g(x))g’(x)$
Suppose $y = f(u)$ and $u = g(x)$. Then, $g’(x) = \frac{du}{dx}$ and $f’(u) = \frac{dy}{du}$.
We can translate the above Chain Rule into Leibniz’s notation by writing:
Now, let's see how Leibniz’s Notation can be useful when used in the familiar Inverse Function theorem.
Suppose we have a function $f$ defined by $y = f(x)$, where $f$ is differentiable and invertible.
Suppose that $g$ is the Inverse Function theorem of $f$. Then, the Inverse Function states that $f’(x) = \frac{1}{g’(f(x))}$.
Letting $y = f(x)$ and $x = g(y)$, we can translate the Inverse Function theorem into Leibniz’s notation by writing:
$f’(x) = \frac{1}{g’(f(x))} \frac{d}{dx}f(x) = \frac{1}{g'(y)} \frac{dy}{dx} = \frac{1}{\frac{d}{dy}g(y)} \frac{dy}{dx} = \frac{1}{(\frac{dx}{dy})}$
In the above equations, we can see how Leibniz’s Notation behaves similarly to a fraction, although it must be emphasized that the derivative is not a fraction.
## Leibniz’s Notation vs. Other Notations
Leibniz’s Notation is one popular notation for differentiation, but there are several others that are also frequently used in calculus. Consider the list of derivative notations below to get an understanding of their relationship.
Note that $y’$ and $f’(x)$ are pronounced respectively as “y prime” and “f prime of x”.
Each notation has its own strengths and weaknesses in different contexts. Understanding their differences can help guide your decision on which derivative notation will work best in a given circumstance. To begin, note that Leibniz’s notation lets us easily express the derivative of a function without employing the use of another variable or function.
For example, we can express the derivative of $x^3$ simply as $\frac{d}{dx}(x^3)$.
Another benefit of Leibniz’s notation is that its notation is very suggestive. As mentioned before, Leibniz’s notation often behaves like a fraction, although it’s not one. Its appearance as a fraction suggests different ways that it can be manipulated, particularly with problems that concern the Chain Rule, the Inverse Function theorem, and integration by parts. As long as you have a solid understanding of why Leibniz’s notation is not a fraction, it’s usually okay to manipulate Leibniz’s notation as you would a fraction. That is if that helps you develop an intuition for different procedures in differential and integral calculus.
LaGrange’s notation is most popularly used for derivative problems in function notation. For example, we used LaGrange’s notation earlier to express the Chain Rule.
To give another example, if we are given $f(x) = 3x^2 - 2x$, we can easily write $f’(x) = 6x - 2$.
In problems that use function notation, LaGrange is often the preferred choice, because it’s easier to mix up functions with function values when using Leibniz’s notation.
Finally, Newton’s notation is most often used in physics, and it’s usually reserved for derivatives with respect to time, like velocity and acceleration. Newton’s notation expresses derivatives by placing a dot over the dependent variable.
## Examples of Leibniz’s Notation
Let’s work through some examples together.
### Example 1
Let $y = \sqrt{4x+2}$. Find $\frac{dy}{dx}$. We'll need to use the Chain Rule in Leibniz's notation.
Set $y = \sqrt{u}$ and $u = 4x+2$. Then, $\frac{dy}{du} = \frac{1}{2}u^{\frac{-1}{2}}$, and $\frac{du}{dx} = 4$.
Using the Chain Rule, we have:
$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$
$\frac{dy}{dx} = \frac{1}{2}u^{\frac{-1}{2}}\cdot4$
$\frac{dy}{dx} = \frac{4}{2\sqrt{u}}$
$\frac{dy}{dx} = \frac{4}{2\sqrt{4x+2}}$
$\frac{dy}{dx} = \frac{2}{\sqrt{4x+2}}$
### Example 2
Let $y = (6x+1)^2$. Find $\frac{dy}{dx}$.
We'll use the Chain Rule in Leibniz's notation.
Set $y = u^2$ and $u = 6x+1$. Then, $\frac{dy}{du}=\frac12u^{-\frac12}$, and $\frac{du}{dx} = 6$.
Using the Chain Rule, we have:
$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$
$\frac{dy}{dx} = 2u \cdot 6$
$\frac{dy}{dx} = 12(6x+1)$
$\frac{dy}{dx} = 72x+12$
### Example 3
Let $y = e^{3x+5}$. Find $\frac{dy}{dx}$.
We'll use the Chain Rule in Leibniz's notation.
Set $y = e^u$ and $u = 3x+5$. Then $\frac{dy}{du} = e^u$ and $\frac{du}{dx} = 3$.
Using the Chain Rule, we have:
$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$
$\frac{dy}{dx} = e^u \cdot 3$
$\frac{dy}{dx} = 3e^{3x+5}$
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• RD Sharma Class 11 Solutions for Maths
Class 11 RD Sharma Solutions – Chapter 20 Geometric Progressions- Exercise 20.5 | Set 1
Question 1. If a, b, c are in G.P., prove that log a, log b, log c are in A.P.
Solution:
Given: a, b and c are in G.P.
Using property of geometric mean, we get
b2 = ac
(b2)n = (ac)n
b2n = an cn
Now, use log on both the sides, we get,
log b2n = log (ancn)
log (bn)2 = log an + log cn
2 log bn = log an + log cn
Hence, proved log an, log bn, log cn are in A.P
Question 2. If a, b, c are in G.P., prove that 1/loga m, 1/logb m, 1/logc m are in A.P.
Solution:
Given: a, b and c are in GP
Using the property of geometric mean
b2 = ac
On applying log on both sides with base m, we get
logm b2 = logm ac
Using property of log
logm b2 = logm a + logm
2logm b = logm a + logm c
2/logb m = 1/loga m + 1/logc m
Hence, proved 1/loga m, 1/logb m, 1/logc m are in A.P.
Question 3. Find k such that k + 9, k – 6, and 4 forms three consecutive terms of a G.P.
Solution:
Let us considered
a = k + 9
b = k − 6
c = 4
AS we know that a, b and c are in GP, then
By using property of geometric mean, we get
b2 = ac
(k − 6)2 = 4(k + 9)
k2 – 12k + 36 = 4k + 36
k2 – 16k = 0
k = 0 or k = 16
Question 4. Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. find the numbers.
Solution:
Let us considered the first term of an A.P. = a
Common difference = d
a1 + a2 + a3 = 15
Here, the three number are: a, a + d, and a + 2d
So,
a + a + d + a + 2d = 15
3a + 3d = 15 or a + d = 5
d = 5 – a -(1)
a + 1, a + d + 3, and a + 2d + 9
They are in GP, that is:
(a + d + 3)/(a + 1) = (a + 2d + 9)/(a + d + 3)
(a + d + 3)2 = (a + 2d + 9)(a + 1)
a2 + d2 + 9 + 2ad + 6d + 6a = a2 + a + 2da + 2d + 9a + 9
(5 – a)2 – 4a + 4(5 – a) = 0
25 + a2 – 10a – 4a + 20 – 4a = 0
a2 – 18a + 45 = 0
a2 – 15a – 3a + 45 = 0
a(a – 15) – 3(a – 15) = 0
a = 3 or a = 15
d = 5 – a
d = 5 – 3 or d = 5 – 15
d = 2 or – 10
Then,
For a = 3 and d = 2, A.P is 3, 5, 7
For a = 15 and d = -10, A.P is 15, 5, -5
Hence, the numbers are 3, 5, 7 or 15, 5, – 5
Question 5. The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.
Solution:
Let us considered the first term of an A.P. be = a
Common difference = d
a1 + a2 + a3 = 21
Here, the three number are: a, a + d, and a + 2d
So,
3a + 3d = 21 or
a + d = 7.
d = 7 – a -(1)
a, a + d – 1, and a + 2d + 1
They are now in GP, that is:
(a + d – 1)/a = (a + 2d + 1)/(a + d – 1)
(a + d – 1)2 = a(a + 2d + 1)
a2 + d2 + 1 + 2ad – 2d – 2a = a2 + a + 2da
(7 – a)2 – 3a + 1 – 2(7 – a) = 0
49 + a2 – 14a – 3a + 1 – 14 + 2a = 0
a2 – 15a + 36 = 0
a2 – 12a – 3a + 36 = 0
a(a – 12) – 3(a – 12) = 0
a = 3 or a = 12
d = 7 – a
d = 7 – 3 or d = 7 – 12
d = 4 or – 5
Then,
For a = 3 and d = 4, A.P is 3, 7, 11
For a = 12 and d = -5, A.P is 12, 7, 2
Hence, the numbers are 3, 7, 11 or 12, 7, 2
Question 6. The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.
Solution:
Let us considered the first term of an A.P. = a
Common difference = d
b = a + d; c = a + 2d.
Given:
a + b + c = 18
3a + 3d = 18 or a + d = 6.
d = 6 – a -(1)
a + 4, a + d + 4, and a + 2d + 36
They are now in GP, that is:
(a + d + 4)/(a + 4) = (a + 2d + 36)/(a + d + 4)
(a + d + 4)2 = (a + 2d + 36)(a + 4)
a2 + d2 + 16 + 8a + 2ad + 8d = a2 + 4a + 2da + 36a + 144 + 8d
d2 – 32a – 128
(6 – a)2 – 32a – 128 = 0
36 + a2 – 12a – 32a – 128 = 0
a2 – 44a – 92 = 0
a2 – 46a + 2a – 92 = 0
a(a – 46) + 2(a – 46) = 0
a = – 2 or a = 46
d = 6 –a
d = 6 – (– 2) or d = 6 – 46
d = 8 or – 40
Then,
For a = -2 and d = 8, A.P is -2, 6, 14
For a = 46 and d = -40, A.P is 46, 6, -34
Hence, the numbers are – 2, 6, 14 or 46, 6, – 34
Question 7. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.
Solution:
Let us considered the three numbers = a, ar, ar2
a + ar + ar2 = 56 -(1)
Now, subtract 1, 7, 21 from the numbers, we get,
(a – 1), (ar – 7), (ar2 – 21)
The above numbers are in AP.
If three numbers are in AP,
So, according to the arithmetic mean, we can write as 2b = a + c
2 (ar – 7) = a – 1 + ar2 – 21
= (ar2 + a) – 22
2ar – 14 = (56 – ar) – 22
2ar – 14 = 34 – ar
3ar = 48
ar = 48/3
ar = 16
a = 16/r -(2)
Now, substitute the value of a in eq(1) we get,
(16 + 16r + 16r2)/r = 56
16 + 16r + 16r2 = 56r
16r2 – 40r + 16 = 0
2r2 – 5r + 2 = 0
2r2 – 4r – r + 2 = 0
2r(r – 2) – 1(r – 2) = 0
(r – 2) (2r – 1) = 0
r = 2 or 1/2
Substitute the value of r in eq(2) we get,
a = 16/r
= 16/2 or 16/(1/2)
= 8 or 32
Hence, the three numbers are (a, ar, ar2) is (8, 16, 32)
(v) (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2
Solution:
(i) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
Let LHS: a(b2 + c2)
Now, on substituting b2 = ac, we get
a(ac + c2)
a2c + ac2
c(a2 + ac)
On Substituting ac = b2 we get,
c(a2 + b2) = RHS
LHS = RHS
Hence, proved.
(ii) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
Let LHS: a2b2c2 [1/a3 + 1/b3 + 1/c3]
a2b2c2/a3 + a2b2c2/b3 + a2b2c2/c3
b2c2/a + a2c2/b + a2b2/c
(ac)c2/a + (b2)2/b + a2(ac)/c -( b2 = ac)
ac3/a + b4/b + a3c/c
c3 + b3 + a3 = RHS
LHS = RHS
Hence, proved.
(iii) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
Let LHS: (a + b + c)2 / (a2 + b2 + c2)
(a + b + c)2 / (a2 + b2 + c2) = (a + b + c)2/(a2 – b2 + c2 + 2b2)
= (a + b + c)2 / (a2 – b2 + c2 + 2ac) -(b2 = ac)
= (a + b + c)2 / (a + b + c)(a – b + c) -((a + b + c)(a – b + c) = a2 – b2 + c2 + 2ac)
= (a + b + c) / (a – b + c)
= RHS
LHS = RHS
Hence, proved.
(iv) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
Let LHS: 1/(a2 – b2) + 1/b2
On taking LCM, we get
1/(a2 – b2) + 1/b2 = (b2 + a2 – b2)/(a2 – b2)b2
= a2 / (a2b2 – b4)
= a2 / (a2b2 – (b2)2)
= a2 / (a2b2 – (ac)2) -(b2 = ac)
= a2 / (a2b2 – a2c2)
= a2 / a2(b2 – c2)
= 1/ (b2 – c2)
= RHS
LHS = RHS
Hence, proved.
(v) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
Let LHS: (a + 2b + 2c) (a – 2b + 2c)
Now, on expanding, we get
(a + 2b + 2c) (a – 2b + 2c) = a2 – 2ab + 2ac + 2ab – 4b2 + 4bc + 2ac – 4bc + 4c2
= a2 + 4ac – 4b2 + 4c2
= a2 + 4ac – 4(ac) + 4c2 -(b2 = ac)
= a2 + 4c2
= RHS
LHS = RHS
Hence, proved.
(iii) (b + c) (b + d) = (c + a) (c + d)
Solution:
(i) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
c2 = bd
Let LHS: (ab – cd) / (b2 – c2)
(ab – cd) / (b2 – c2) = (ab – cd) / (ac – bd)
= (ab – cd)b / (ac – bd)b
= (ab2 – bcd) / (ac – bd)b
= [a(ac) – c(c2)] / (ac – bd)b
= (a2c – c3) / (ac – bd)b
= [c(a2 – c2)] / (ac – bd)b
= [(a + c) (ac – c2)] / (ac – bd)b
= [(a + c) (ac – bd)] / (ac – bd)b
= (a + c) / b
= RHS
LHS = RHS
Hence, proved.
(ii) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
c2 = bd
Let RHS: (a + b)2 + 2(b + c)2 + (c + d)2
Now, on expanding, we get
(a + b)2 + 2(b + c)2 + (c + d)2 = (a + b)2 + 2 (a + b) (c + d) + (c + d)2
= a2 + b2 + 2ab + 2(c2 + b2 + 2cb) + c2 + d2 + 2cd
= a2 + b2 + c2 + d2 + 2ab + 2(c2 + b2 + 2cb) + 2cd
= a2 + b2 + c2 + d2 + 2(ab + bd + ac + cb +cd) -(c2 = bd, b2 = ac)
(a + b + c)2 + d2 + 2d(a + b + c) = {(a + b + c) + d}2
RHS = LHS
Hence, proved.
(iii) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
c2 = bd
Let LHS: (b + c) (b + d)
Now, on expanding, we get
(b + c) (b + d) = b2 + bd + cb + cd
= ac + c2 + ad + cd
= c (a + c) + d (a + c)
= (a + c) (c + d)
= RHS
LHS = RHS
Hence, proved.
(iii) a2 + b2, ab + bc, b2 + c2
Solution:
(i) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
On squaring both the sides we get,
(b2)2 = (ac)2
(b2)2 = a2c2
Hence, proved a2, b2, c2 are in G.P.
(ii) Given: a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
On squaring both the sides, we get
(b2)3 = (ac)3
(b2)3 = a3c3
(b3)2 = a3c3
Hence, proved a3, b3, c3 are in G.P.
(iii) Given: a, b, c are in GP.
Using the property of geometric mean,
b2 = ac
a2 + b2, ab + bc, b2 + c2 or (ab + bc)2 = (a2 + b2) (b2 + c2
Let LHS: (ab + bc)2
Now, on expanding, we get
(ab + bc)2 = a2b2 + 2ab2c + b2c2
= a2b2 + 2b2(b2) + b2c2 -(ac = b2)
= a2b2 + 2b4 + b2c2
= a2b2 + b4 + a2c2 + b2c2 -( b2 = ac)
= b2(b2 + a2) + c2(a2 + b2)
= (a2 + b2)(b2 + c2)
= RHS
LHS = RHS
Hence, a2 + b2, ab + bc, b2 + c2 are in GP.
(iv) (a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.
Solution:
(i) Given: a, b, c, d are in G.P.
So, a, b = ar, c = ar2, d = ar3
Now,
(b2 + c2)2 = (a2 + b2)(c2 + d2)
(a2r2 + a2r4)2 = (a2 + a2r2)(a2r4 + a2r6)
a4(r2 + r4) = a2(1 + r2)a2r4(1 + r2)
a4r4(1 + r2)2 = a4r4(1 + r2)2
L.H.S = R.H.S
⇒ (b2 + c2)2 = (a2 + b2)(c2 + d2)
Hence, proved (a2 + b2), (b2 + c2), (c2 + d2) are in G.P.
(ii) Given: a, b, c, d are in G.P.
So, a, b = ar, c = ar2, d = ar3
Now,
(b2 – c2)2 = (a2 – b2)(c2 – d2)
(a2r2 – a2r4) = (a2 – a2r2)(a2r4 – a2r6)
a4(r2 – r4)2 = a2(1 – r2) a2r4 (1 – r2)
a4r4 (1 – r2)2 = a4r4 (1 – r2)2
L.H.S = R.H.S
⇒ (b2 – c2)2 = (a2 – b2)(c2 – d2)
Hence, proved (a2 – b2), (b2 – c2), (c2 – d2) are in G.P.
(iii) Given: a, b, c, d are in G.P.
So, a, b = ar, c = ar2, d = ar3
Now,
L.H.S = R.H.S
Hence, proved are in G.P.
(iv) Given: a, b, c, d are in G.P.
So, a, b = ar, c = ar2, d = ar3
Now,
(ab + bc + cd)2 = (a2 + b2 + c2)(b2 + c2 + d2)
(a2r + a2r3 + a2r5)2 = (a2 + a2r2 + a2r4)(a2r2 + a2r4 + a2r6)
a4(r + r3 + r5)2 = a2(1 + r2 + r4) a2r2 ( 1 + r2 + r4)
a4r2(1 + r2 + r4)2 = a4r2(1 + r2 + r4)2
L.H.S = R.H.S
⇒ (ab + bc + cd)2 = (a2 + b2 + c2)(b2 + c2 + d2)
Hence, proved (a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.
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# Lesson 2: Proportional relationships
Courtesy of Wikimedia Commons
## Equations: For Students
More in this series
Lesson 1.
Lesson 3.
use our version for guardians.
## Performance Expectations (CCSS)
This lesson covers the following parts of 7.RP.A.2:
Recognize and represent proportional relationships between quantities.
1. Identify the constant of proportionality (unit rate) in … equations … of proportional relationships.
2. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn.
In the previous lesson, you looked at this proportional relationship:
Cupcakes Number Cost 1 \$3.50 2 \$7.00 3 \$10.50
You saw that the ratio was a constant \$3.50 per cupcake. If C is the cost of the cupcakes, and n is the number of cupcakes, then this ratio is represented by Cn . So
Cn = 3.50
Multiplying both sides by n yields the equation C = 3.5n. This represents the proportional relationship. For example, if n = 3, then C = 3.5(3) = 10.5. That is, the cost of ___ cupcakes is \$_________, as it says in the third row of the table.
In general, if there is a proportional relationship between a quantity y and a quantity x with a constant of proportionality k, then
yx = k, which implies y = kx
A proportional relationship between a quantity y and a quantity x that has a constant of proportionality k is represented by the equation y = kx.
If an equation in a different form can be rewritten as above, then it is a proportional relationship. If it cannot be rewritten as above, then it is not a proportional relationship.
We can take any table or verbal description of a proportional relationship and turn it into an equation. We just need to first determine the constant of proportionality or unit rate.
### Example 1
Brick machine
Time
(hours)
Number
of bricks
2 1100
5 2750
8 4400
You saw in the previous lesson that the output of this brick-making machine follows a proportional relationship. Write an equation to represent the relationship. (To print this lesson out, click on the printable version.)
Solution:
Let N = the number of bricks and t = the time in hours. The unit rate is _____ bricks per hour. The equation is N = 550t.
### Example 2
A 10-pound wheel of baby gouda cheese sells for \$64. Write an equation relating the cost to the weight of the cheese.
Solution:
The cheese sells for \$_________ per pound. Let C = the cost of the cheese in dollars and w = the weight of the cheese in pounds. Then the equation is C = 6.4w.
Remember that a proportional relationship works “both ways.” So in the above example, not only is C proportional to w, but w is also proportional to C. You can take the equation above and divide both sides by 6.4 to write the relationship the other way:
164C = w → w = 0.15625C
### Example 3
The equation K = 0.4536P represents the relationship between pounds, P, and kilograms, K. Identify the constant of proportionality and give a verbal description of the relationship. Also, give a verbal description of the reverse relationship.
Solution:
The constant of proportionality is 0.4536. This means for every 1 pound there is 0.4536 kilogram. To reverse the relationship, you need the reciprocal of the constant. In this case, it is about 2.2. So for every 1 kilogram, there are about 2.2 pounds.
### Example 4
x y 3 18 5 30 7 42
Represent y = 6x as a table, with rows for x = 3, 5 and y = 42.
Solution:
When x = 3, y = 6(3) = _____.
When x = 5, y = 6(____) = ______.
When y = 42, 42 = 6x → x = 7.
## Exercises for lesson 2
Represent each proportional relationship with an equation.
1.
Area, A
(sq. ft.)
Cost, C
90 \$81
180 \$162
360 \$324
2.
Production
Time, t (hours) Widgets, w
3 1005
5 1675
7 2345
3.
T-shirts
Number, n Cost, C
2 \$23
4 \$46
5 \$57.50
4. An artist buys unprimed canvas in rolls. Each roll has 90 square feet of canvas and costs \$54. Let C = the cost of the canvas in dollars and A = the area of the canvas in square feet.
5. A 3-pound package of ground beef sells for \$12. Let B = the cost of the ground beef in dollars and w = the weight of the ground beef in pounds.
6. A mason is building a brick wall. So far, he has laid 5 rows of brick (and cement) that reaches a height of 45 cm. Let n = the number of rows of brick and h = the height of the wall in centimeters.
For each equation, (a) represent the proportional relationship by a table with input values 1, 2, and 3, and (b) write the equation for the reverse relationship.
7. y = 4x
8. y = 0.2x
9. y = x3
10. Does each equation represent a proportional relationship?
(a) y = x + 5 (b) y = x2 (c) y = 2(x – 5) + 10
11. The equation y = 2.54x represents the relationship between inches, x, and centimeters, y. Give a verbal description of the relationship.
12. The equation y = 33.8x represents the relationship between liters, x, and fluid ounces, y. Give a verbal description of the relationship.
13. Challenge Problem:
While the constant of proportionality is usually positive in real-world problems, it could be negative. Let h = the number of hours after midnight, and T = the temperature in degrees Celsius. Explain what the relationship means in real-world terms at midnight and beyond.
NEXT LESSON: Head to Lesson 3 or look back at Lesson 1
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# Finite sum
#### MountEvariste
##### Well-known member
Find the value of $$\sum_{0 \le k \le n}(-1)^k k^n \binom{n}{k}.$$
#### lfdahl
##### Well-known member
$\sum_{k=0}^{n}(-1)^kk^n\binom{n}{k} = (-1)^nn!\;\;\;\;(1)$
In order to show the identity, we need the following lemma:
$\sum_{k=0}^{n}(-1)^kk^m\binom{n}{k} = 0 , \, \, \, \, m = 0,1,.., n-1.\;\;\; (2)$
Proof by induction:
Consider the binomial identity: $(1-x)^n =\sum_{j=0}^{n}\binom{n}{j}(-1)^jx^j \;\;\; (3)$
Case $m = 0$: Putting $x = 1$ in $(3)$ yields: $\sum_{j=0}^{n}\binom{n}{j}(-1)^j = 0$
Case $m = 1$: Differentiating $(3)$ once yields: $-n(1-x)^{n-1} =\sum_{j=0}^{n}\binom{n}{j}(-1)^jjx^{j-1}$
Again putting $x = 1$: $\sum_{j=0}^{n}\binom{n}{j}(-1)^jj = 0$.
Assume the $m$th step is OK, where $1 \leq m < n-1$. We need to show, that our lemma also holds for step $m+1$:
Differentiate $(3)$ $m+1$ times:
$(-1)^{m+1}n(n-1)...(n-m)(1-x)^{n-m-1} = \sum_{j=0}^{n}\binom{n}{j}(-1)^jj(j-1)..(j-m)x^{j-m-1}$
Evaluating at $x = 1$:
$\sum_{j=0}^{n}\binom{n}{j}(-1)^j\left ( j^{m+1}+c_mj^m+c_{m-1}j^{m-1} + ... + c_1j\right )=0 \\ \\ \sum_{j=0}^{n}\binom{n}{j}(-1)^jj^{m+1} + c_m\sum_{j=0}^{n}\binom{n}{j}(-1)^jj^m +c_{m-1}\sum_{j=0}^{n}\binom{n}{j}(-1)^jj^{m-1}+...+c_1\sum_{j=0}^{n}\binom{n}{j}(-1)^jj =0 \\ \therefore \sum_{j=0}^{n}\binom{n}{j}(-1)^jj^{m+1} = 0.$
Now we are well prepared to prove, that $(1)$ holds. This will be another proof by induction:
Let $S_n = \sum_{k=0}^{n}(-1)^kk^n\binom{n}{k}$. Then we have:
$S_0 = 1 = (-1)^00!$, and $S_1 = (-1)^00^1\binom{1}{0}+(-1)^11^1\binom{1}{1} = -1 = (-1)^11!$
Therefore, we may assume, that $(1)$ holds for some step $n > 1$: $S_n = (-1)^nn!$
$S_{n+1} = \sum_{k=0}^{n+1}(-1)^kk^{n+1}\binom{n+1}{k} = \sum_{k=1}^{n+1}(-1)^kk^n\frac{(n+1)!}{(k-1)!(n+1-k)!}\\= \sum_{j=0}^{n}(-1)^{j+1}(j+1)^n\frac{(n+1)n!}{j!(n-j)!}= -(n+1)\sum_{j=0}^{n}(-1)^j(j+1)^n\binom{n}{j} \\= -(n+1)\sum_{j=0}^{n}(-1)^j\left ( 1+\binom{n}{1}j+\binom{n}{2}j^{2}+...+\binom{n}{n-1}j^{n-1}+j^n \right )\binom{n}{j} \\=-(n+1)\left ( \sum_{m=0}^{n-1}\binom{n}{m}\sum_{j=0}^{n}(-1)^jj^{m}\binom{n}{j}+S_n\right )$
With the help of our lemma, the double sum in the parenthesis equals 0, so we are left with:
$S_{n+1} = -(n+1)S_n = (-1)^{n+1}(n+1)! \;\;\; q.e.d.$
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# Understanding Derivatives: A Comprehensive Overview
1. A-level maths topics
2. Calculus
3. Derivatives
Derivatives are an essential part of calculus and a fundamental mathematical concept that helps us understand how functions behave. They allow us to measure the rate at which a function changes, or how it “derives” from other functions. Derivatives can be used in a variety of ways, from predicting the future behavior of a system to designing new products. In this comprehensive overview, we will examine derivatives in detail, including their definition, properties, and applications. We will begin by exploring the concept of derivatives and how they can be used to measure the rate of change in a function.
Next, we will look at the different types of derivatives and their properties. Finally, we will discuss the practical applications of derivatives and how they can be used to solve complex problems. By the end of this article, you will have a thorough understanding of derivatives and how they can be applied in various areas of mathematics. The first step in understanding derivatives is to understand what they are. A derivative is a measure of how a function changes as its inputs change. For example, if the function f(x) = x2, then the derivative of f(x) is 2x.
This means that as x increases by 1 unit, f(x) will increase by 2 units.
#### Derivatives
can also be used to find the slope of a function at any point. Next, we'll discuss how to calculate derivatives. The most common method for calculating derivatives is the power rule. This rule states that the derivative of a function raised to any power is equal to the product of the power and the function raised to one less than the power.
For example, if f(x) = x3 then the derivative of f(x) is 3x2.
#### Derivatives
can also be used to find the maximum and minimum points of a function. The derivative of a function tells us whether it is increasing or decreasing at any given point. If the derivative is positive, then the function is increasing; if the derivative is negative, then the function is decreasing.
By finding the points where the derivative is equal to zero, we can determine the maximum or minimum points of a function. Finally, let's look at some applications of derivatives in calculus and other A-level maths topics.
#### Derivatives
can be used to find rates of change, which are important in many areas such as physics and economics. They can also be used to find the area under curves, which is useful for solving optimization problems. In addition, derivatives can be used to solve differential equations, which are used to model many real-world phenomena. In summary, derivatives are an important concept in calculus and other advanced mathematical topics.
They provide a measure of how a function changes as its inputs change and can be used for a variety of applications such as finding rates of change, areas under curves, and solving differential equations.
## How to Calculate Derivatives
Calculating derivatives can be a daunting task for any student learning calculus. However, it is an essential skill for understanding advanced mathematical topics. Fortunately, once you understand the power rule, calculating derivatives becomes much simpler. The power rule states that the derivative of a function raised to a power is equal to the power multiplied by the function raised to the power minus one. In other words, it states that:dy/dx = nx^(n-1)Where x is the independent variable, y is the dependent variable, and n is the power that the function is raised to. Using this rule, you can calculate derivatives more easily.
For example, if we wanted to find the derivative of x^2, we would substitute n = 2 into the equation above and get:dy/dx = 2x^(2-1) = 2x^1 = 2xTherefore, we can conclude that the derivative of x^2 is 2x. This is just one example of how to use the power rule to calculate derivatives. By understanding this rule and practicing with equations of different complexities, you can become more proficient at calculating derivatives.
## What are Derivatives?
Derivatives are an important mathematical tool used to measure the rate of change in a function's output relative to its input. They provide a powerful way to analyze how a system responds to changing conditions and how quickly it can reach a certain equilibrium. Derivatives are also used to calculate marginal changes in a system, such as profit or cost. Derivatives are used in calculus and other advanced mathematical topics.
They provide a measure of how a function changes as its inputs change. The derivative of a function is the limit of the ratio of the change in the output of the function to the change in its input. In simple terms, derivatives measure how quickly the output of a function can change when its inputs are changed. Derivatives are used to calculate various types of derivatives, such as partial derivatives and directional derivatives. These derivatives can be used to analyze the behavior of a system or to optimize it.
For example, derivatives can be used to find the maximum or minimum value of a function. Derivatives can also be used to identify areas where a system is most sensitive to changes in its inputs. Derivatives have numerous applications in engineering and economics. They can be used to calculate rates of change in prices, optimize production processes, and analyze financial investments. Derivatives can also be used to analyze the behavior of complex systems, such as computer networks and aircraft.
## Applications of Derivatives
Derivatives are an important concept in calculus and other A-level maths topics.
They have a wide range of applications, from solving equations to predicting changes in functions. In this section, we'll discuss some of the main uses of derivatives.
#### Optimization
One of the most common applications of derivatives is in optimization. This involves finding the maximum or minimum value of a function by taking its derivative and setting it to zero.
By doing this, you can find the values of the variables which give the maximum or minimum output.
#### Equation Solving
Derivatives can also be used to solve equations. This involves taking the derivative of both sides of the equation and then solving for the unknown variables. This technique is often used in physics and engineering to solve complex equations.
#### Predicting Changes
Derivatives can also be used to predict how a function will change as its inputs change. By taking the derivative of a function, you can determine whether it is increasing or decreasing, and by how much.
This can be used to make predictions about future changes in a system.
#### Calculating Rates
Another application of derivatives is in calculating rates. For example, you can use derivatives to calculate the rate at which a quantity is changing over time. This can be used to calculate things like velocity, acceleration, and other rates of change.
#### Financial Analysis
Finally, derivatives can be used to analyze financial data. By taking the derivative of a stock price or other financial data, you can gain insight into how the market is performing and make better decisions about investments. Derivatives are a powerful tool in calculus and other advanced mathematical topics that can be used to measure how a function changes as its inputs change.
Calculating derivatives can be done through a variety of methods, including differentiation rules, the chain rule, and implicit differentiation. Derivatives can also be used for a wide range of applications, such as calculating rates of change, areas under curves, and solving differential equations. With this knowledge, you should now have a better understanding of derivatives and how they can be used.
##### Shahid Lakha
Shahid Lakha is a seasoned educational consultant with a rich history in the independent education sector and EdTech. With a solid background in Physics, Shahid has cultivated a career that spans tutoring, consulting, and entrepreneurship. As an Educational Consultant at Spires Online Tutoring since October 2016, he has been instrumental in fostering educational excellence in the online tutoring space. Shahid is also the founder and director of Specialist Science Tutors, a tutoring agency based in West London, where he has successfully managed various facets of the business, including marketing, web design, and client relationships. His dedication to education is further evidenced by his role as a self-employed tutor, where he has been teaching Maths, Physics, and Engineering to students up to university level since September 2011. Shahid holds a Master of Science in Photon Science from the University of Manchester and a Bachelor of Science in Physics from the University of Bath.
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# 3.2 Vector addition and subtraction: graphical methods (Page 4/15)
Page 4 / 15
## Multiplication of vectors and scalars
If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk , or 82.5 m, in a direction $\text{66}\text{.}0\text{º}$ north of east. This is an example of multiplying a vector by a positive scalar . Notice that the magnitude changes, but the direction stays the same.
If the scalar is negative, then multiplying a vector by it changes the vector’s magnitude and gives the new vector the opposite direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector $\mathbf{A}$ is multiplied by a scalar $c$ ,
• the magnitude of the vector becomes the absolute value of $c$ $A$ ,
• if $c$ is positive, the direction of the vector does not change,
• if $c$ is negative, the direction is reversed.
In our case, $c=3$ and $A=27.5 m$ . Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1.
## Resolving a vector into components
In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x - and y -components, or the north-south and east-west components.
For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction $\text{29}\text{.0º}$ north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion , and much more when we cover forces in Dynamics: Newton’s Laws of Motion . Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components.
## Phet explorations: maze game
Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can.
## Summary
• The graphical method of adding vectors $\mathbf{A}$ and $\mathbf{B}$ involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector $\mathbf{R}$ is defined such that $\mathbf{\text{A}}+\mathbf{\text{B}}=\mathbf{\text{R}}$ . The magnitude and direction of $\mathbf{R}$ are then determined with a ruler and protractor, respectively.
• The graphical method of subtracting vector $\mathbf{B}$ from $\mathbf{A}$ involves adding the opposite of vector $\mathbf{B}$ , which is defined as $-\mathbf{B}$ . In this case, $\text{A}–\mathbf{\text{B}}=\mathbf{\text{A}}+\left(\text{–B}\right)=\text{R}$ . Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector $\mathbf{R}$ .
• Addition of vectors is commutative such that .
• The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.
• If a vector $\mathbf{A}$ is multiplied by a scalar quantity $c$ , the magnitude of the product is given by $\text{cA}$ . If $c$ is positive, the direction of the product points in the same direction as $\mathbf{A}$ ; if $c$ is negative, the direction of the product points in the opposite direction as $\mathbf{A}$ .
basically potentiometer is series circuit or parallel circuit?
What is half-life
the life in which half of the radioactive element decay
Arif
what is fluid
anything that flows is Liquid.
prakash
a substance that has no specific shape
Saleemulhaq
How submarines floats one water the same time sink in water
A submarine has the ability to float and sink. The ability to control buoyancy comes from the submarine'strim or ballast tanks which can be filled with either water or air, depending on whether the submarine needs to floator sink. When the submarine floats it means its trim tanks are filled with air
Arif
Anthony
what is work
Force times distance
Karanja
product of force and distance...
Arif
Is physics a natural science?
what is the difference between a jet engine and a rocket engine.
explain the relationship between momentum and force
A moment is equivalent multiplied by the length passing through the point of reaction and that is perpendicular to the force
Karanja
How to find Squirrel frontal area from it's surface area?
how do we arrange the electronic configuration of elements
hi guys i am an elementary student
hi
Dancan
hello
are you an elementary student too?
benedict
no bro
yes
Che
hi
Miranwa
yes
Miranwa
welcome
Miranwa
what is the four equation of motion
Miranwa
what is strain?
SAMUEL
Change in dimension per unit dimension is called strain. Ex - Change in length per unit length l/L.
ABHIJIT
strain is the ratio of extension to length..=e/l...it has no unit because both are in meters and they cancel each other
How is it possible for one to drink a cold drink from a straw?
most possible as it is for you to drink your wine from your straw
Selina
state the law of conservation of energy
energy can neither be destroy or created,but can be change from one form to another
dare
yeah
Toheeb
it can neither be created nor destroyed
Toheeb
its so sample question dude
Muhsin
what is the difference between a principle and a law?
where are from you wendy .?
ghulam
philippines
Mary
why?
Mary
you are beautiful
ghulam
are you physics student
ghulam
laws are ment to be broken
Ge
hehe ghulam where r u from?
Muhsin
yes
dare
principle are meant to be followed
dare
south Africa
dare
here Nigeria
Toheeb
principle is a rule or law of nature, or the basic idea on how the laws of nature are applied.
Ayoka
Rules are meant to be broken while principals to be followed
Karanja
principle is a rule or law of nature, or the basic idea on how the laws of nature are applied.
tathir
what is momentum?
is the mass times velocity of an object
True
it is the product of mass and velocity of an object.
The momentum possessed by a body is generally defined as the product of its mass and velocity m×v
Usman
momentum is the product of the mass of a body of its velocity
Ugbesia
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# Per Unit System – Practice Problem Solved For Easy Understanding
Let’s understand the concept of per unit system by solving an example. In the one-line diagram below, the impedance of various components in a power system, typically derived from their nameplates, are presented. The task now is to normalize these values using a common base.
Now that you have carefully examined the system and its parameters, the equivalent impedance diagram for the above system would look something like the following.
Resistive impedance for most components have been ignored. Rotating machines have been replaced with a voltage source behind their internal reactance. Capacitive effects between lines and to ground are ignored as well.
To obtain the new normalized per unit impedances, first we need to figure out the base values (Sbase, Vbase, Zbase) in the power system. Following steps will lead you through the process.
### Step 1: Assume a system base
Assume a system wide S_{base} of 100MVA. This is a random assumption and chosen to make calculations easy when calculating the per unit impedances.
So, S_{base} = 100MVA
### Step 2: Identify the voltage base
Voltage base in the system is determined by the transformer. For example, with a 22/220kV voltage rating of T1 transformer, the V_{base} on the primary side of T1 is 22kV while the secondary side is 220kV. It does not matter what the voltage rating of the other components are that are encompassed by the V_{base} zone.
See figure below for the voltage bases in the system.
### Step 3: Calculate the base impedance
The base impedance is calculated using the following formula:
Z_{base}=\frac{{kV_{base}}^2}{S_{base MVA}} Ohms…..(1)
For T-Line 1: Z_{base}=\frac{(220)^2}{100}= 484 Ohms
For T-Line 2: Z_{base}=\frac{(110)^2}{100}= 121 Ohms
For 3-phase load: Z_{base}=\frac{(11)^2}{100}= 1.21 Ohms
### Step 4: Calculate the per unit impedance
The per unit impedance is calculated using the following formulas:
Z_{p.u.}=\frac{Z_{actual}}{Z_{base}} …..(2)
Z_{p.u._{new}}=Z_{p.u._{old}}(\frac{S_{base_{new}}}{S_{base_{old}}})(\frac{V_{base_{old}}}{V_{base_{new}}})^2 …..(3)
The voltage ratio in equation (3) is not equivalent to the transformers voltage ratio. It is the ratio of the transformer’s voltage rating on the primary or secondary side to the system nominal voltage on the same side.
For T-line 1 using equation (2): X_{l1_{p.u.}}=\frac{48.4}{484}= 0.1 pu
For T-line 2 using equation (2): X_{l2_{p.u.}}=\frac{65.43}{121}= 0.5 pu
Power Factor: \cos^{-1}(0.6)=\angle{53.13}
Z_{act}=\frac{(V_{rated})^2}{\overline{S}^*}= \frac{10.45^2}{57\angle{-53.13}}
= 1.1495+j1.53267 Ohms
Per unit impedance of 3-phase load using equation (2)= \frac{1.1495+j1.5326}{1.21} = 0.95+j1.2667 pu
For generator, the new per unit reactance using equation (3)
X_{sg}= 0.18(\frac{100}{90})(\frac{22}{22})^2
= 0.2 pu
For transformer T1: X_{t1}= 0.1(\frac{100}{50})(\frac{22}{22})^2 = 0.2 pu
For transformer T2: X_{t2}= 0.06(\frac{100}{40})(\frac{220}{220})^2 = 0.15 pu
For transformer T3: X_{t3}= 0.064(\frac{100}{40})(\frac{22}{22})^20.16 pu
For transformer T4: X_{t4}= 0.08(\frac{100}{40})(\frac{110}{110})^20.2 pu
For Motor, X_{sm}= 0.185(\frac{100}{66.5})(\frac{10.45}{11})^20.25 pu
If you think you learned something today then you will love the eBook I prepared for you. It has 10 additional and unique per unit problems. Go ahead, preview it. Get the complete version for only \$4.99. Thanks for supporting this blog.
To view full load amps due to motor load and inductive load at Bus 2, see this post.
### Summary
1. Assume a Sbase for the entire system.
2. The Vbase is defined by the transformer and any off-nominal tap setting it may have.
3. Zbase is derived from the Sbase and Vbase.
4. The new per unit impedance is obtained by converting the old per unit impedance on old base values to new ones. See equations (2) and (3).
Please support this blog by sharing the article
### 62 thoughts on “Per Unit System – Practice Problem Solved For Easy Understanding”
1. please help: Three transformers each rated 25 MVA, 38. 1 /3.81 kV are connected star-delta with a balanced load of three 0.6?, Y-connected resistors. Choose a base of 75 MVA, 66 kV for the high-voltage side of the transformer and specify the base for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then, determine the load resistance R_L i n ohms referred to the high-voltage side and the per-unit value of this resistance on the chosen base.
2. HELP! How do determine the equivalent three phase impedance of three single phase transformer? Example. Three each of 10 kva transformer, 7.52kv-240V, wiith an individual impedance of 4%. What is the equivalent impedance when they are connected in three phase at 30 kva.
1. @sayonsom
What we developed in this exercise is a positive sequence network. A transformer’s winding configuration does not affect it. The positive sequence network is good for calculating balanced load current and 3 phase faults (not involving ground).
The zero sequence network however does get affected by transformer winding configuration. Delta and ungrounded Y have a big impact in the design of zero sequence impedance network.
You will need a zero sequence network for unsymmetrical fault current analysis like L-G or L-L-G etc.
I am getting into symmetrical components with this discussion.
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Home » Maths » Pythagorean Triples Formula and Lists
# Pythagorean Triples Formula and Lists
## Introduction to Pythagorean Triples
• A Pythagorean triple is made up of three positive numbers, a, b, and c, so that $a^2 + b^2 = c^2$.
• These numbers indicate the side lengths of a right-angled triangle, with ‘c’ representing the hypotenuse length.
• Pythagorean triples are an extension of Pythagoras’ Theorem, which asserts that the square of the length of the hypotenuse in a right-angled triangle equals the sum of the squares of the other two sides.
## How to generate Pythagorean Triples
There are several methods to generate Pythagorean triples, but one of the most common ways is by using Euclid’s formula
a = m² – n²
b = 2mn
c = m² + n²
Where:
• a, b, and c are the side lengths of the right-angled triangle
• m and n are any two positive integers with m > n > 0
• m and n are coprime and both should not be odd numbers
Proof of Euclid Formula
$a^2 + b^2 = (m^2 -n^2)^2 + 4m^2n^2 =m^4 + n^4 + 2m^2n^2 =(m^2 + n^2)^2= c^2$
## Important points
• if a, b , c is a Pythagorean triplet, then k.a, k.b and k.c are considered as the Pythagorean triple.
• 6, 2.5, 6.5 are the sides of the right angle triangle but this is not Pythagorean triplet as it should be positive integers only
• A primitive Pythagorean triplet has a coprime a, b, and c. (that is, they have no common divisor larger than 1)
## Examples of Generating Triplets
Question 1: Generate a Pythagorean triple using Euclid’s formula with m = 3 and n = 2.
Solution: Apply Euclid’s formula with the given values of m and n:
a = m² – n² = 3² – 2² = 9 – 4 = 5
b = 2mn = 2 × 3 × 2 = 12
c = m² + n² = 3² + 2² = 9 + 4 = 13
So, the Pythagorean triple generated is (5, 12, 13)
Question 1: Generate a Pythagorean triple using Euclid’s formula with m = 2 and n = 1.
Solution: Apply Euclid’s formula with the given values of m and n:
a = m² – n² = 2² – 1² = 3
b = 2mn = 2 × 2 × 1 = 4
c = m² + n² = 2² + 1² = 5
So, the Pythagorean triple generated is (3, 4, 5)
## How to generate Pythagorean Triples if one number is given
Case 1 : The given number is even ( m)
then
The Pythagorean Triples will be given by
m
$\frac {m^2}{2} -1$
$\frac {m^2}{2} +1$
Case 1 : The given number is odd ( m)
then
The Pythagorean Triples will be given by
m
$\frac {m^2}{2} -.5$
$\frac {m^2}{2} +.5$
Examples
If m =3, then the other will be
$\frac {m^2}{2} – .5= 4.5 -.5=4$
$\frac {m^2}{2} + .5= 4.5 +.5=5$
So triplet is (3,4,5)
## Pythagorean Triples lists
Here is Primitive Pythagorean triples till 100
1. Are all Pythagorean triples primitive?
No, not all Pythagorean triples are primitive. A primitive Pythagorean triple has side lengths that are coprime, meaning they share no common divisors other than 1. Non-primitive Pythagorean triples can be generated by multiplying all sides of a primitive triple by a common integer.
1. What are some common examples of Pythagorean triples?
Some common examples of Pythagorean triples are (3, 4, 5), (5, 12, 13), (7, 24, 25), and (8, 15, 17). These are primitive Pythagorean triples, as their side lengths share no common divisors other than 1.
1. Can a Pythagorean triple have an odd hypotenuse?
No, a Pythagorean triple cannot have an odd hypotenuse. The hypotenuse (c) will always be even since, according to Euclid’s formula, c = m² + n². Since the sum of two odd numbers is even, and the sum of two even numbers is also even, the hypotenuse will always be even.
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# How do you find the vertical, horizontal or slant asymptotes for (x+3)/(x^2-9)?
May 28, 2016
This function has a horizontal asymptote $y = 0$, a vertical asymptote $x = 3$ and a removable singularity at $x = - 3$.
#### Explanation:
$f \left(x\right) = \frac{x + 3}{{x}^{2} - 9} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x + 3}}}}{\left(x - 3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 3\right)}}}} = \frac{1}{x - 3}$
excluding $x = - 3$
As $x \to \pm \infty$, $\frac{1}{x - 3} \to 0$. So there is a horizontal asymptote $y = 0$
As $x \to {3}^{+}$, $\frac{1}{x - 3} \to + \infty$
As $x \to {3}^{-}$, $\frac{1}{x - 3} \to - \infty$
$f \left(3\right)$ is undefined since division by $0$ is undefined.
So $f \left(x\right)$ has a vertical asymptote at $x = 3$
$f \left(- 3\right)$ is undefined, since both numerator $\left(x + 3\right) = 0$ and denominator $\left({x}^{2} - 9\right) = 0$. Note however, that both left and right limits exist at $x = - 3$ and are both equal to $- \frac{1}{6}$. So there is a removable singularity at $x = - 3$ (removable by redefining $f \left(- 3\right) = - \frac{1}{6}$).
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