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# The Sum of a Positive Number and Its Reciprocal is Thrice the Difference of the Number and Its Reciprocal. The Number is?
### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs | Set 6
What are you looking for? Let’s dig in quickly
## Explanation
The sum of a positive number and its reciprocal is thrice.
The difference of the number and its reciprocal.
The number will be figure out in this way.
Let suppose the positive number is y.
The difference of a number and its reciprocal
y – 1/y
The sum of a positive number and its reciprocal
y + 1/y = 3(y – 1/y) ________ (i)
Through equation (i) we can easily find out the value of y (y = 2 square root).
Number = ?
## Solution
Let suppose
Number = y
According to the given condition
y + 1/y = 3(y – 1/y)
y + 1/y = 3y – 3/y
1/y + 3/y = 3y – y
1 + 3 = y(3y – y)
4 = 3y2 – y2
2y2 = 4
y2 = 2
y = 2 square root
Required Number = 2 square root answer
## Conclusion
The sum of a positive number and its reciprocal is thrice the difference of the number and its reciprocal. The number is 2 square root.
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# 6-4 Key Concepts Adding and Subtracting Fractions with Unlike Denominators
```Lesson
Key Concepts
6-4
Unlike Denominators
Objective
To learn how to add and subtract fractions with unlike
denominators, using the method of common denominators.
Note to the Teacher This lesson contains the most complicated
operations on fractions. Provide students with plenty of conceptual
work with sketches (such as pizzas) and manipulative materials
before introducing the algorithms.
Using Models to Add Fractions with Unlike
Denominators
the numerators the way we do when the fractions have like
denominators.
Example 1 What is
1
2
1
?
3
Solution In order to find this sum, we first need to rewrite both
fractions so they have the same denominator. This means we
must replace each fraction with an equivalent fraction having
a different denominator. In order to motivate this process to
portions of a pizza. First, draw a separate model for each of
the fractions.
The addition is modeled by taking the two parts together as
a portion of a whole pizza.
26
Lesson 6-4
It is not easy to visually determine what fraction of the
whole pizza is represented by the shaded portion of the
model. However, suppose we divided each pizza into
2 3 or 6 equal parts.
Notice that each of the smaller parts we have created is one
sixth of the pizza. Now when we merge the portions of the
pizza, we get the model below.
We can see that the shaded portion of the model consists of
five pieces, each of which is one sixth of the pizza. In other
words, the shaded portion of the pizza represents the
fraction
5
,
6
so
1
2
Example 2 What is
1
3
1
3
5
.
6
1
?
5
Solution Represent each of the fractions
pizza.
1
3
and
1
5
as portions of a
When we merge these two portions, we get this picture.
To determine what fraction is modeled by the combined
shaded regions in the figure above, divide each pizza into
3 5 or 15 equal parts.
27
Lesson 6-4
Each of the smaller parts we have created represents one
fifteenth of the pizza. The merged version of these portions
is shown below.
The shaded portion of the model above consists of eight
pieces, each of which is one fifteenth of the pizza. The
shaded portion of the model represents
1
3
1
5
8
,
15
so
8
.
15
The Algorithm and Why It Works
1
2
and
we divided the pizza portion corresponding to
1
3
1
2
(Example 1). When
into three equal
pieces, each of the three pieces represented one sixth of the pizza.
The model showed that
3
6
and
1
2
are equivalent fractions. In the
same way, when the pizza portion representing
1
3
was divided into
two equal pieces, both pieces represented one sixth of the pizza.
This showed that
2
6
and
1
3
are equivalent fractions.
31
6 2
21
6 3
To do this step without using the pizza models, we replace each
fraction by an equivalent fraction so that the two fractions have the
same denominators.
28
Lesson 6-4
1
2
1
3
equivalent fractions
↓
equivalent fractions
↓
3
6
2
6
We can easily add these equivalent fractions because they have like
denominators.
1
2
1
3
3
6
32
6
2
6
or
5
6
Note to the Teacher Stress to students that when they multiply both
numerator and denominator of a fraction by the same number, the
result is an equivalent fraction.
Example 3 Find
1
3
1
5
using equivalent fractions with the same
denominator.
Solution First, find equivalent fractions for both
1
3
and
1
5
that have
the same denominator. Multiply both the numerator and
denominator of
1
3
So
15
35
5
15
1
3
by 5 the denominator of
So
is equivalent to
13
53
3
15
.
5
15
and denominator of
1
5
1
5
1
5
1
.
3
Now multiply both the numerator
by 3 the denominator of
1
3
.
3
15
is equivalent to
1
.
5
Now we can add the equivalent
fractions.
1
3
1
5
5
15
3
15
53
15
or
8
15
Note to the Teacher Stress that finding equivalent fractions with
like denominators can be accomplished by multiplying both the
numerator and denominator of each fraction by the denominator of
the other fraction.
29
Lesson 6-4
Example 4 Find
2
3
3
.
7
Solution Multiply both the numerator and denominator of
the denominator of 73 .
2
3
27
37
the denominator of .
2
3
33
73
by 7
3
7
by 3
14
21
Multiply both the numerator and denominator of
3
7
2
3
9
21
Now replace each original fraction with its equivalent
2
3
3
7
14
21
14 9
21
The fraction
23
21
9
21
or
23
21
is an improper fraction, which is a fraction
whose numerator is greater than its denominator. Since 21
goes into 23 once with a remainder of 2, the sum can also be
written as a mixed number. So,
2
3
3
7
23
21
2
or 1 .
21
Here is an algorithm for adding fractions with unlike denominators.
Algorithm
Fractions
with Unlike
Denominators
1. Multiply both the numerator and denominator of each
fraction by the denominator of the other fraction. The
resulting equivalent fractions will have the same
denominator.
1
2
9
7
↓
↓
17
97
29
79
↓
↓
7
63
18
63
and retaining the like denominator.
7
63
30
18
63
25
63
Lesson 6-4
Example 5 Find
2
9
3
.
11
Solution Step 1 Multiply both the numerator and denominator of
2
9
by 11, and multiply both the numerator and
3
11
denominator of
2
9
2 11
9 11
by 9.
22
99
and
3
11
39
11 9
27
99
Step 2 Add the equivalent fractions.
22
99
So,
2
9
3
11
27
99
22 27
99
49
99
or
49
.
99
Subtract Fractions with Unlike Denominators
The algorithm for subtracting one fraction from another when they
have unlike denominators is very similar to the algorithm for adding
fractions with unlike denominators.
Algorithm for 1. Multiply both numerator and denominator of each fraction
Subtracting
by the denominator of the other fraction. The resulting
Fractions
equivalent fractions will have the same denominator.
with Unlike
2. Subtract the equivalent fractions by subtracting the
Denominators
numerators and retaining the like denominator.
Example 6 What is the value of
2
3
1
?
5
Solution Step 1 Multiply both the numerator and denominator of
by 5, and multiply both the numerator and
denominator of
1
5
by 3.
25
35
10
15
2
3
and
1
5
13
53
2
3
3
15
Step 2 Replace each original fraction with its equivalent
fraction. Subtract the equivalent fractions by
subtracting numerators and retaining the
denominator.
10
15
3
15
So,
2
3
1
5
10 3
15
or
7
15
7
.
15
31
Lesson 6-4
Example 7 Find
5
7
Solution Step 1
Step 2
So,
5
7
2
.
3
5
7
53
73
15
21
14
21
2
3
1
.
21
15
21
2
3
and
15 14
21
or
27
37
14
21
1
21
We have discussed procedures for adding and subtracting fractions,
as well as the reasoning behind them. It is very important to keep
the reasoning in mind, since it allows students to interpret practical
problems correctly, as well as to recall the procedures. One
particularly common error made by students can be avoided by
keeping the meaning of the fractions in mind. When adding fractions,
some students make the mistake of adding denominators as well as
numerators. For example,
2
3
4
5
24
35
or
6
.
8
By recognizing that each of the fractions being added is greater than
1
,
2
students will see that an answer less than 1 is unreasonable.
Note to the Teacher Divide your class into two groups. Give each
group several addition and subtraction problems involving fractions,
some of which are solved correctly and others that are solved
incorrectly. Have each group discuss which problems are solved
correctly and which are not, giving reasons for their decisions. Have
each group present their findings to the rest of the class.
End of
Lesson
32
Lesson 6-4
```
|
# What is the Z-Score?
Z Score
The standard score, or the Z-score in excel, is a way to calculate the difference between the actual score and the mean in standard deviation units. It also allows us to compare two examples from different populations. The z-value can be negative or positive. A negative z-score means that the value is less than the arithmetic value, and a positive z-score means that our value is greater than the arithmetic value. The Z-Score is calculated by subtracting the mean value of the data set from the data set to be calculated and dividing it by the standard deviation of the data set.
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Table of Contents
## What is the Z-Score used for in our everyday life?
Not everyone is a statistician or mathematician, but everyone deals with data sets almost every day. Whether it’s your screen time, the final exam, the class, or how much you spent on clothes last year. It means that we also use statistical measures such as the mean or the deviation.
## But where should you use the z-score?
### 1. We can use it by comparing two different datasets with different distributions.
We have two chefs who know how to make soufflé. Let’s say our first cook is called Sasha and the other is David. David took part in a cooking competition and got a score of 472 out of 500. Sasha also took part in this type of competition and got a score of 82 out of 100. Now we have to decide who was the most positive. But how? David’s score is highest than Sasha’s, but we can’t compare them like that. In these situations, we convert the raw score to z scores. This way we can calculate who did the best in the competition.
In the competition that David participates in, the mean was 392, and the standard deviation was 100. To calculate the standard score, we subtract the mean 392 from David’s score of 472 and divide it by the standard deviation. The correct equation for this is “472 -392 / 100,” which is +0.08. David’s score is 0.08, a standard deviation above the mean.
In Sasha’s contest, the mean was 52, and the standard deviation was 30. We are now applying the same formula that we use for David’s contest. We deduct the mean from the score and divide it into the standard deviation. The equation for this is “82 -52 / 30”. Sasha’s z score is +1. It means Sasha is one standard deviation above the mean.
Now that we can easily compare the score of David and Sasha. David was 0.8 standard deviation above, & Sasha is 1.0 standard deviation above average. It means that Sasha is a better soufflé baker than David. That’s what the statistics tell us. But maybe we shouldn’t make up our minds without trying both.
“Another place where the default is used is to measure the growth of children in the world. It is used to compute weight and height for age. Children who are below the standard deviation of -2 or -3 are diagnosed with malnutrition. ”
Tip: Always add less or more to the z-values and convert them after 0, x to two-digit numbers. And you should always add 0 before the decimal point if the value is less than 1. For example, 0.8798 ci should be converted to +0.88 above the mean.
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### 2. Two compare the frequency of a score
The Z-scale I just created can tell me how many candidates are getting the Sasha or David score. When I look at the Z scale, I am told that the people who took David’s score are more than those who took Sasha’s score because most of the data will capture a standard deviation below or above the mean.
### 3. It allows us to guess the score
The Z scale can be used to determine whether Sasha or David is likely to get their scores. We can use it in hypothetical measurements. “But how do converting raw data to z-scores help us compare those? It is because the z-score distribution always has a mean of 0 and a standard deviation of 1.00. Although the original distribution is abnormal, the mean is 0, and the standard deviation is 1.00. It helps us standardize and compare two different data sets, even though they were measures differently, as in our example.
Most of the z-score distribution data fall between standard deviations of less than and more than 1.00 from the mean. There are fewer data between 2 standard deviations above and below the mean, and it is scarce to find data greater than plus and minus three standard deviations.
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## Why do we need the Z-Score?
As we see from the examples above, we need a z score to compare two different data sets by standardizing each type of measure. By converting your data to Z-Score, your data is fundamentally gets resize and standardize. For this reason, it is also known as the standard score. We can think of a Z-Score as a cookie-cutter. We can cut cookie dough or styrofoam, but both still look like the gingerbread man.
## How do you mathematically calculate the Z value?
The Z score, or standard score, is calculate by dividing the data point minus the man by the standard deviation. But what do these terms mean? In fact when we collect individual information about something, we call it the group, dataset, or population dataset. In this article, all the data is numeric, but data can be of any real-life information. A single datum can be called a data point or individually, and I’m going to refer to it anyway so that you don’t get confused.
The average value is “average” in the statistics. It calculates by adding all data and dividing by the number of data.Therefore I will give an example, and I powerfully advise you to rely on me during this time. For example, let’s say we’re curious about how many hours we spend in front of the screen each day, and we’ve recorded our daily screen time for a week 2, 2, 5, 3, 7, 1, and 1. We add all of these numbers, and the result is 21. We have seven individual data points, so we divide 21 by seven and find the result 3. You spend an average of 3 hours per week in front of the screen.
Another thing that we need to find to find the z-score is the standard deviation. Standard deviation is an arithmetical term uses to determine the deviation between the mean and the actual data. A slight standard deviation means that the sensitivity of the data is consistent and standard. The data points are close together. A high standard deviation means that the data set is inconsistent, unexpectedly, and sometimes unreliable. It means that the data set people are not close to one another but are distributes over a series of numbers.
To calculate the standard deviation, we subtract all the numbers from the mean and square them to avoid negative values. We’ll square these numbers and divide them by one less than the number of people in our data set. In our Screen Time example, we can explain this as follows:
The average is 3, so three is subtract from every data point. We occupy 2 hours in front of the screen on Monday, which means “2 -3 = -1”. Now we have squared the result “-1” and get the number 1.
Therefore, We use the same formula for every person in our data set. Then we add these numbers and get the result of √30. We’re going to divide it into minus the number of data points of 6. It brings us to the end of 0.9. It is our standard deviation value.
Please see our article to Calculate the Standard Deviation in Excel for more information on the standard deviation and calculate it in Excel. Now that we have the mean and standard deviation, we can calculate the Z-value.
Continuing with the Screen Time example, we’ll calculate the Z-score for Monday. We need to subtract the mean of our data point and divide it into the standard deviation value. For us this means “2 -3 = -1”. Next we divide -1 to 0.9 which is our usual deviation value “-1 / 0.9 = -1.1”. The result is -1.1. It means that our screen time on Monday is greater than one and less than two standard deviations from the mean.
|
# Matrix Subtraction – Explanation & Examples
There are $4$ basic operations we can do on matrices. They are addition, subtraction, scalar multiplication, and matrix multiplication. Matrix Subtraction is one of them, and we are going to look at it in this article.
Note: Matrix division isn’t a defined operation. We can divide a matrix by only a scalar number. To read more about matrix division, click here.
Let us define matrix subtraction:
Matrix subtraction is the operation of subtracting two matrices of the same dimension by subtracting the corresponding entries.
This article will show what conditions make matrix subtraction possible, subtract matrices, and a few examples showing this process.
## When can you subtract matrices?
Can we subtract any matrices?
No!
We can only subtract $2$ matrices when they have the same dimension. That is, if $2$ matrices have the same number of rows and columns, then, and only then, we can subtract the $2$ matrices together.
Recall that the dimension of a matrix is its number of rows and number of columns. If there are $a$ rows and $b$ columns of a matrix, we can say that the matrix has dimensions $a \times b$ (pronounced as $a$-by-$b$). To know more about the dimensions of a matrix, check this article out!
Remember, the resulting matrix from the operation of matrix subtraction will be of the same dimension as the matrices that produced it.
## How to subtract matrices?
After we have concluded that matrices have the same dimensions, we subtract $2$ matrices by subtracting each other’s corresponding elements. Consider Matrix $A$ and Matrix $B$ shown below:
$A = \begin{bmatrix} { 0 } & { 4 } \\ { – 3} & { – 3 } \end {bmatrix}$
$B = \begin{bmatrix} { – 9 } & 0 \\ 2 & { – 2 } \end {bmatrix}$
Both Matrix $A$ and $B$ have $2$ rows and $2$ columns. Hence, the matrices’ dimensions are $2 \times 2$, which makes it equal. That means we can subtract the $2$ matrices by subtracting each corresponding entry from each other. The process is shown below:
$A – B = \begin{bmatrix} { 0 – – 9 } & { 4 – 0 } \\ { – 3 – 2 } & { – 3 – – 2 } \end {bmatrix}$
$A – B = \begin{bmatrix} { 0 + 9 } & { 4 – 0 } \\ { – 3 – 2 } & { – 3 + 2 } \end {bmatrix}$
$A – B = \begin{bmatrix} { 9 } & { 4 } \\ { – 5 } & { – 1 } \end {bmatrix}$
Now, let’s consider the two matrices shown below:
$A = \begin{bmatrix} { 0 } & { – 1 } \\ 10 & 2 \end {bmatrix}$
$B = \begin{bmatrix} { – 5 } & 0 \\ 6 & 0 \\ 3 & 7 \\ 4 & { – 7 } \end {bmatrix}$
Is $A – B$ defined?
No.
Matrix $A$ has dimensions $2 \times 2$. Matrix $B$ has dimensions $4 \times 2$. Since the dimensions of both the matrices aren’t equal, matrix subtraction isn’t defined! We can’t subtract Matrix $A$ and $B$ because there won’t be corresponding entry for each element of Matrix $A$ and Matrix $B$.
Rules of Matrix Subtraction
We can write $4$ rules for matrix subtraction. Shown below:
1. You can only subtract matrices with the same dimensions (i.e. number of rows of the first matrix must equal the number of rows of the second matrix. Also, the number of columns of the first matrix must be equal to the number of columns of the second matrix.)
2. The resultant matrix from matrix subtraction will have the same dimension as the original matrices that produced it.
3. Matrix subtraction is NOT commutative (i.e. $A – B \neq B – A$). When you change the order of matrices and subtract, it produces a different answer (just like real numbers).
4. Matrix subtraction is NOT associative (i.e. $( A – B ) – C \neq A – ( B – C )$).
Let’s look at some examples of Matrix Subtraction.
#### Example 1
Check whether matrix subtraction between Matrix $A$ and Matrix $B$ is defined. If so, subtract them.
$A = \begin{bmatrix} { 0 } & { – 5 } \\ { – 5 } & { 0 } \end {bmatrix}$
$B = \begin{bmatrix} { 3 } & { – 4 } \\ { 0 } & { – 5 } \end {bmatrix}$
Solution
For matrix subtraction to be defined, the dimension of each matrix must be equal.
Matrix $A$ is a $2 \times 2$ matrix. Matrix $B$ is also a $2 \times 2$ matrix. Thus, matrix subtraction between Matrix $A$ and $B$ is defined.
Now, let’s subtract the $2$ matrices together by subtracting each corresponding entry of Matrix $B$ from each corresponding entry of Matrix $A$. Shown below:
$A – B = \begin{bmatrix} { 0 – 3 } & { – 5 – – 4 } \\ { – 5 – 0 } & { 0 – – 5 } \end {bmatrix}$
$A – B = \begin{bmatrix} { – 3 } & { – 5 + 4 } \\ { – 5 } & { 0 + 5 } \end{bmatrix}$
$A – B = \begin{bmatrix} { – 3 } & { – 1 } \\ { – 5 } & { 5 } \end{bmatrix}$
#### Example 2
For the $2$ matrices shown below, find $C – D$.
$C = \begin{bmatrix} { 1 } & { – 1 } \\ 1 & { – 1 } \\ 7 & { – 7 } \end {bmatrix}$
$D = \begin{bmatrix} { 1 } & { – 2 } & { – 1 } \\ { – 1 } & 0 & { – 1 } \end {bmatrix}$
Solution
The dimension of Matrix $C$ is $3 \times 2$. The dimension of Matrix $D$ is $2 \times 3$. Even though the dimensions are equivalent, we can’t subtract Matrix $D$ from Matrix $C$. The order of dimensions are also important. A $3 \times 2$ matrix is not equal to a $2 \times 3$ matrix. Thus, we cannot subtract Matrix $D$ from Matrix $C$.
We can solve basic algebraic equations with matrix subtraction as well. Consider the example shown below.
#### Example 3
Find the values of $a$ and $b$ given the following equation:
$\begin{pmatrix} { 1 } & { 2 } \\ b & { – 6 } \end {pmatrix} – \begin{pmatrix} { a } & { – 1 } \\ 0 & { – 3 } \end {pmatrix} = \begin{pmatrix} { 7 } & { 3 } \\ { – 2 } & { – 3 } \end {pmatrix}$
Solution
We can subtract corresponding entries to solve for $a$ and $b$. First, let’s subtract the corresponding elements for $a$ and solve for the variable:
$1 – a = 7$
$a = 1 – 7$
$a = – 6$
Now, let’s find the value of $b$ by subtracting the corresponding entries for $b$:
$b – 0 = – 2$
$b = – 2$
Now, you can try some practice questions and see whether you get them correct or not. Answers are below.
### Practice Questions
1. Consider the following $3$ matrices:
$P = \begin{pmatrix} { – 2 } & { – 2 } \\ 1 & { 1 } \end {pmatrix}$
$Q = \begin{pmatrix} { – 3 } & { – 3 } \\ { – 3 } & { – 3 } \end {pmatrix}$
$R = \begin{pmatrix} 1 & 1 \\ 7 & { – 2 } \\ 3 & { – 7 } \end {pmatrix}$
Find:
1. $P – Q$
2. $Q – R$
3. $Q – P$
2. Find the values of $a$, $b$, and $c$ given the following equation:
$\begin{pmatrix} { 3 } & { 2 } & 0 \\ b & { – 2 } & 3 \\ 11 & a & -2 \end {pmatrix} – \begin{pmatrix} { 3 } & { 2 } & c \\ 5 & { – 5 } & 3 \\ { – 4 } & { 6 } & 10 \end {pmatrix} = \begin{pmatrix} { 0 } & { 0 } & -9 \\ 12 & { 3 } & 0 \\ 15 & -3 & { – 12 } \end {pmatrix}$
1.
1. All the $3$ problems are matrix subtraction. Note that since matrix subtraction is not commutative, the answer to Part $A$ and Part $C$ would be different. The answers are shown below.
1. Both Matrices $P$ and $Q$ are $2 \times 2$ matrices. Thus, we subtract the $2$ matrices by subtracting the corresponding entries. Shown below:
$P – Q = \begin{pmatrix} { – 2 – – 3 } & { – 2 – – 3 } \\ { 1 – – 3 } & { 1 – – 3 } \end {pmatrix}$
$P – Q = \begin{pmatrix} { – 2 + 3 } & { – 2 + 3 } \\ { 1 + 3 } & { 1 + 3 } \end {pmatrix}$
$P – Q = \begin{pmatrix} { 1 } & { 1 } \\ { 4 } & { 4 } \end {pmatrix}$
2. We can’t subtract Matrix $R$ from Matrix $Q$ because their dimensions aren’t the same. Matrix $Q$ is a $2 \times 2$ matrix and Matrix $R$ is a $3 \times 2$ matrix.
3. The answer to this part of the question will be different from Part $A$ since matrix subtraction isn’t commutative. Let’s subtract Matrix $P$ from Matrix $Q$.
$Q – P = \begin{pmatrix} { – 3 – – 2 } & { – 3 – – 2 } \\ { – 3 – 1 } & { – 3 – 1 } \end {pmatrix}$
$Q – P = \begin{pmatrix} { – 3 + 2 } & { – 3 + 2 } \\ { – 4 } & { – 4 } \end {pmatrix}$
$Q – P = \begin{pmatrix} { – 1 } & { – 1 } \\ { – 4 } & { – 4 } \end {pmatrix}$
2. We can subtract corresponding entries to solve for $a$, $b$, and $c$. First, let’s subtract the corresponding elements for $a$ and solve for the variable:
$a – 6 = -3$
$a = – 3 + 6$
$a = 3$
Now, let’s find the value of $b$ by subtracting the corresponding entries for $b$:
$b – 5 = 12$
$b = 12 + 5$
$b = 17$
Lastly, let’s find the value of $c$ by subtracting the corresponding entries for $c$:
$0 – c = -9$
$c = 0 – – 9$
$c = 0 + 9$
$c = 9$
|
What Is 24/60 as a Decimal + Solution With Free Steps
The fraction 24/60 as a decimal is equal to 0.4.
An Improper Fraction can be converted into a Mixed Fraction. Suppose that 5/3 is an improper fraction where the numerator is greater than the denominator, by solving the 5/3 fraction through the long division method we get 1 in the quotient with remainder 2. Now in a mixed fraction quotient, 1 is the whole number and reminder 2 is the numerator of the mixed fraction and denominator 3 is the same as in the original fraction. 1 2/3 is the mixed fraction of 5/3 improper fraction.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 24/60.
Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 24
Divisor = 60
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 24 $\div$ 60
This is when we go through the Long Division solution to our problem. The following figure shows the long division:
Figure 1
24/60 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 24 and 60, we can see how 24 is Smaller than 60, and to solve this division, we require that 24 be Bigger than 60.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 24, which after getting multiplied by 10 becomes 240.
We take this 240 and divide it by 60; this can be done as follows:
 240 $\div$ 60 = 4
Where:
60 x 4 = 240
This will lead to the generation of a Remainder equal to 240 – 240 = 0. We have a Quotient as 0.4=z, with a Remainder equal to 0.
Images/mathematical drawings are created with GeoGebra.
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## A Multiplication Based Logic Puzzle
### 341 is the smallest composite number that gives a false positive for this Quick Prime Number Test
• 341 is a composite number.
• Prime factorization: 341 = 11 x 31
• The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 341 has exactly 4 factors.
• Factors of 341: 1, 11, 31, 341
• Factor pairs: 341 = 1 x 341 or 11 x 31
• 341 has no square factors that allow its square root to be simplified. √341 ≈ 18.466
341 is a composite number that sometimes acts like a prime number. To understand why, we need to understand a little bit about modular arithmetic:
When one number is divided by another, sometimes there is a remainder. Modular arithmetic is all about the remainders. We don’t care how many times one number divides into another number; we only care about the remainder.
Something very curious happens when the equation in the chart below is applied to a prime number greater than 2: The remainder is always 2! For example, 2^5 = 32 and 32 divided by 5 is 6R2. We say that 32 (mod 5) = 2 or 2^5 (mod 5) = 2 because 2 is the remainder. The fact that the remainder for prime numbers applied to this equation is always 2 is amazing, and can be a QUICK TEST to see if an odd number might be a PRIME NUMBER! If the remainder isn’t 2, that odd number is definitely NOT prime!
QUICK PRIME NUMBER TEST (Please, excuse my using = instead of ≡, the “equal sign” that is usually used in modular arithmetic. I think = looks less intimidating.)
Passing the remainder test is a necessary but not a sufficient indicator that a number is prime: Even though 341 is not a prime number, the quantity 2^341 divided by 341 also has a remainder of 2. Since 341 = 11 x 31, but passes the remainder test, it is known as a pseudoprime number. 341 is the smallest composite number that passes this particular test, so 341 is an amazing number!
Also in the chart above, 2 is the most common remainder, followed by 8, then 32, then 128. All of those numbers are odd powers of 2. The even powers of 2 do not appear on the chart at all! That is a very curious phenomenon as well. (However, if x is an even number, it appears that y will usually be an even power of 2.)
Earlier mathematicians have written equivalent expressions and algorithms, but I prefer using “2^x (mod x)” because it takes very few keystrokes to enter into a calculator before hitting the equal sign:
This is only a picture of a calculator.
Look at the first two images from this link on Pascal’s Triangle. The first demonstrates that in prime numbered rows, the numbers in that row can be divided evenly by that prime number (not counting the 1’s at the beginning and ending of each row).
• For a composite number, such as 15, at least one of the numbers in the row will NOT be divisible by that composite number. In row 15, 1 is the 0th term, 15 is the 1st term, 105 is the 2nd term, 455 is the 3rd term, and 1365 is the 4th term and so forth.
• Terms that are divisible by 15 are the 1st term (15), the 2nd term (105), the 4th term (1365), the 7th term (6435), the 8th term (6435), the 11th term (1365), and the 13th term (105). All of those term numbers, 1, 2, 4, 7, 8, 11, and 13, do NOT have factors in common with the number 15.
• The terms that are NOT divisible by 15 are the 3rd term (455), the 5th term (3003), the 6th term (5005), the 9th term (5005), the 10th term (3003), and the 12th term (455). All of those term numbers, 3, 5, 6, 9, 10, and 12, have at least one factor in common with the number 15.
If we could see the VERY large numbers for the 341st row, and if they weren’t expressed in Scientific Notation, we could note that the following terms would NOT be divisible by 341: terms numbered 11th, 22nd, 33rd, 44th, and so forth and the terms numbered 31st, 62nd, 93rd and so forth. However, if you add those terms together, that very large sum would be divisible by 341. That is so amazing, even though 341 is not prime!
The second image from Pascal’s Triangle demonstrates that the sum of the numbers in any row of Pascal’s triangle equals two raised to the second number in that row.
These two images work together so that 2^p (mod p) will always be 2 for every prime number greater than 2 because every number in the prime numbered rows can be evenly divided by that prime number. (Except the 1 at the beginning of the row and the 1 at the end of the row; Note 1 + 1 = 2)
Now that is why this amazing test for prime numbers works as well as it does while giving just a few false positives!
#### Comments on: "341 is the smallest composite number that gives a false positive for this Quick Prime Number Test" (7)
1. I didn’t know this about 341; it’s an interesting property to have.
Liked by 1 person
2. That was really interesting! Who knew 341 was so awesome??? 😀
Liked by 1 person
3. Spooky.
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4. I am printing this up and heading for my armchair…
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5. […] has something in common with the number 341. Yes, both of them pass this quick prime number test, and both of them are composite numbers divisible by 11. Both numbers are called pseudo-prime […]
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6. […] smallest composite number that gives a false positive to this test is 341. Click on that number to see an in depth description of this quick prime number test and how it […]
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7. Travis McCracken said:
All prime candidates through 5461 that yield a false positive for the remainder test n|((2^n)-2):
{341, 1387, 1729, 2047, 2701, 2821, 3277, 4033, 4369, 4681, 5461}
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# (- 7) + 4 = (-9) = - 3 (- 3) + 7 = ( -3) = 2
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## Transcription
1 WORKING WITH INTEGERS: 1. Adding Rules: Positive + Positive = Positive: = 9 Negative + Negative = Negative: (- 7) + (- 2) = - 9 The sum of a negative and a positive number: First subtract: The answer gets the sign of the larger number (- 7) + 4 = (-9) = - 3 (- 3) + 7 = ( -3) = 2 2. Subtracting Rules: Negative - Positive = Negative: (- 5) - 3 = -5 + (-3) = -8 Positive - Negative = Positive + Positive = Positive: 5 - (-3) = = 8 Negative - Negative = Negative + Positive = Use the sign of the larger number and subtract (Change double negatives to a positive) (-5) - (-3) = ( -5) + 3 = -2 (-3) - ( -5) = (-3) + 5 = 2 3. Multiplying Rules: Positive x Positive = Positive: 3 x 2 = 6 Negative x Negative = Positive: (-2) x (-8) = 16 Negative x Positive = Negative: (-3) x 4 = -12 Positive x Negative = Negative: 3 x (-4) = Dividing Rules: Tips: Positive Positive = Positive: 12 3 = 4 Negative Negative = Positive: (-12) (-3) = 4 Negative Positive = Negative: (-12) 3 = -4 Positive Negative = Negative: 12 (-3) = When working with rules for positive and negative numbers try and think of weight loss or balancing a check book to help solidify why this works. 2. Using a number line showing both sides of 0 is very helpful to help develop the understanding of working with positive and negative numbers/integers.
2 WHAT ARE EXPONENTS? Exponents are sometimes referred to as powers and mean the number of times the 'base' is being multiplied. In algebra, exponents are used frequently. In the example below, you would say: Four to the power of 2 or four raised to the second power or four to the second power. This would mean 4 x 4 or (4) (4) or 4 4. Simplified the example would be 16. If the power/exponent of a number is 1, the number will always equal itself. In other words, in our example if the exponent 2 was a 1, simplified the example would then be 4. Exponent Rules When working with exponents there are certain rules you'll need to remember. When you are multiplying terms with the same base you can add the exponents. This means: 4 x 4 x 4 x 4 x 4 x 4 x 4 or When you are dividing terms with the same base you can subtract the exponents. This means: 4 x 4 x 4 or When parentheses are involved - you multiply. (8 3 ) 2 =8 6 y a y b = y (a+b) y a x a = (yx) a Squared ( Cubed ( and 0's When you multiply a number by itself it is referred to as being 'squared'. 4 2 is the same as saying "4 squared" which is equal to 16. If you multiply 4 x 4 x 4 which is 4 3 it is called 4 cubed. Squaring is raising to the second power, cubing is raising to the third power. Raising something to a power of 1 means nothing at all; the base term remains the same. Now for the part that doesn't seem logical. When you raise a base to the power of 0, it equals 1. Any number raised to the power of 0 equals 1 and 0 raised to any exponent or power is 0!
3 Order of Operations: Rules 1. Calculations must be done from left to right. 2. Calculations in brackets (parenthesis) are done first. When you have more than one set of brackets, do the inner brackets first. NOTE: Brackets and parenthesis are known also known as grouping symbols. 3. Exponents (or radicals) must be done next. 4. Multiply and divide in the order the operations occur. So, what you SEE first, DO first! 5. Add and subtract in the order the operations occur. So, what you SEE first, DO first! Acronyms to Help you Remember How will you remember this order? Try the following Acronyms: Please Excuse My Dear Aunt Sally (Parenthesis, Exponents, Multiply, Divide, Add, Subtract) BEDMAS (Brackets, Exponents, Divide, Multiply, Add, Subtract) OR Big Elephants Destroy Mice And Snails (Brackets, Exponents, Divide, Multiply, Add, Subtract) Pink Elephants Destroy Mice And Snails (Parenthesis, Exponents, Divide, Multiply, Add, Subtract) The point is to find SOMETHING that works for you. These are only suggestions.
4 Examples ( ) (12-2) X X X Rule 3: Exponent first Rule 4: Multiply or Divide as they appear Rule 5: Add or Subtract as they appear Rule 2: Everything in the brackets first Rule 5: Add or Subtract as they appear Rule 2: Everything in the brackets first Rule 3: Exponents Rule 4: Multiply and Divide as they appear Rule 5: Add or Subtract as they appear Does It Make a Difference? What If I Don't Use the Order of Operations? Mathematicians were very careful when they developed the order of operations. Without the correct order, watch what happens: X Without following the correct order, I know that 15+5=20 multiplied by 10 gives me the answer of X Following the order of operations, I know that 5X10 = 50 plus 15 = 65. This is the correct answer, the above is not! You can see that it is absolutely critical to follow the order of operations. Some of the most frequent errors students make occur when they do not follow the order of operations when solving mathematical problems. Students can often be fluent in computational work yet do not follow procedures. Use the handy acronyms to ensure that you never make this mistake again.
5 Laws of Exponents Here are the Laws (explanations follow): Law Example x 1 = x 6 1 = 6 x 0 = = 1 x -1 = 1/x 4-1 = 1/4 x m x n = x m+n x 2 x 3 = x 2+3 = x 5 x m /x n = x m-n x 6 /x 2 = x 6-2 = x 4 (x m ) n = x mn (x 2 ) 3 = x 2 3 = x 6 (xy) n = x n y n (xy) 3 = x 3 y 3 (x/y) n = x n /y n (x/y) 2 = x 2 / y 2 x -n = 1/x n x -3 = 1/x 3 And the law about Fractional Exponents: Laws Explained The first three laws above (x 1 = x, x 0 = 1 and x -1 = 1/x) are just part of the natural sequence of exponents. Have a look at this: Example: Powers of 5.. etc etc.. Look at that table for a while... notice that positive, zero or negative exponents are really part of the same pattern, i.e. 5 times larger (or 5 times smaller) depending on whether the exponent gets larger (or smaller).
6 The law that x m x n = x m+n Example: x 2 x 3 = (xx)(xxx) = xxxxx = x 5 So, x 2 x 3 = x (2+3) = x 5 The law that x m /x n = x m-n Example: x 4 /x 2 = (xxxx) / (xx) = xx = x 2 So, x 4 /x 2 = x (4-2) = x 2 (Remember that x/x = 1, so every time you see an x "above the line" and one "below the line" you can cancel them out.) This law can also show you why x 0 =1 : Example: x 2 /x 2 = x 2-2 = x 0 =1 The law that (x m ) n = x mn Example: (x 3 ) 4 = (xxx) 4 = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x 12 So (x 3 ) 4 = x 3 4 = x 12 The law that (xy) n = x n y n Example: (xy) 3 = (xy)(xy)(xy) = xyxyxy = xxxyyy = (xxx)(yyy) = x 3 y 3 The law that (x/y) n = x n /y n Example: (x/y) 3 = (x/y)(x/y)(x/y) = (xxx)/(yyy) = x 3 /y 3 The law that x 1/n = The n-th Root of x And so a fractional exponent like 4 3/2 is really saying to do a cube (3) and a square root (1/2), in any order. Just remember from fractions that m/n = m (1/n):
7 Example: The order does not matter, so it also works for m/n = (1/n) m: Example: Let s look at a few examples of how this is applied. Example: What is 9 ½ 9 ½? So 9 ½ times itself gives 9. 9 ½ 9 ½ = 9 (½+½) = 9 (1) = 9 What do we call a number that, when multiplied by itself gives another number? The square root! See: So 9 ½ is the same as 9 Example: What is 4 3/2? Either way gets the same result. Example: What is 27 4/3? 9 9 = 9 4 3/2 = 4 3 (1/2) = (4 3 ) = (4 4 4) = (64) = 8 or 4 3/2 = 4 (1/2) 3 = ( 4) 3 = (2) 3 = /3 = 27 4 (1/3) = (27 4 ) = (531441) = 81 or It was certainly easier the 2nd way! 27 4/3 = 27 (1/3) 4 = ( 27) 4 = (3) 4 = 81
8 Powers of 10 "Powers of 10" is a very useful way of writing down large or small numbers. Instead of having lots of zeros, you show how many powers of 10 you need to make that many zeros Example: 5,000 = 5 1,000 = thousand is 5 times a thousand. And a thousand is So 5 times 10 3 = 5,000 Can you see that 10 3 is a handy way of making 3 zeros? Scientists and Engineers (who often use very big or very small numbers) like to write numbers this way. Example: The Mass of the Sun The Sun has a Mass of kg. It would be too hard for scientists to write 1,988,000,000,000,000,000,000,000,000,000 kg (And very easy to make a mistake counting the zeros!) Example: A Light Year (the distance light travels in one year) It is easier to use meters, rather than 9,461,000,000,000,000 meters It is commonly called Scientific Notation, or Standard Form. The Trick While at first it may look hard, there is an easy "trick": The index of 10 says... how many places to move the decimal point to the right. Example: What is ? You can calculate it as: 1.35 x ( ) = 1.35 x 10,000 = 13,500 But it is easier to think "move the decimal point 4 places to the right" like this:
9 Negative Powers of 10 Negative? What could be the opposite of multiplying? Dividing! A negative power means how many times to divide by the number. Negatives just go the other way! Example: = = Just remember for negative powers of 10: For negative powers of 10, move the decimal point to the left. Example: What is ? Well, it is really 7.1 x ( 1 / 10 1 / 10 1 / 10 ) = = But it is easier to think "move the decimal point 3 places to the left" like this:
10 Addition and Subtraction of Polynomials Adding Polynomials To add polynomials, you must clear the parenthesis, combine (add or subtract) the like terms. In some cases you will need to remember the order of operations. Remember, when adding and subtracting like parts, the variable never changes. Here are a couple of examples: (5x + 7y) + (2x - 1y) = 5x + 7y + 2x - 1y (Clear the parenthesis) =5x + 2x + 7y - 1y (Combine the like terms) = 7x + 6y --- Another Example: (y 2-3y + 6) + (y - 3y 2 + y 3) y 2-3y + 6+ y - 3y 2 + y (Clear the parenthesis) y 3 + y 2-3y 2-3y + y (Combine the like terms) y 3-2y 2-2y Subtracting Polynomials To subtract polynomials, you must change the sign of terms being subtracted, clear the parenthesis, and combine the like terms. Here's an example: (4x 2-4) - (x 2 + 4x - 4) (4x 2-4) + (-x 2-4x + 4) ---- (Change the signs by multiplying by -1; clear the parenthesis) 4x x 2-4x (Clear the parenthesis) 4x 2 -x 2-4x (Combine the like terms)
11 3x 2-4x Another Example: (5x 2 + 2x +1) - ( 3x 2 4x 2 ) 5x 2 + 2x +1-3x 2 + 4x +2 --(Change the signs by multiplying by -1; clear the parenthesis) 5x 2-3x 2 + 2x+ 4x (Combine the like terms) 2x 2 + 6x +3 Polynomial Definitions of Terms: A monomial has one term: 5y or -8x 2 or 3. A binomial has two terms: -3x 2 + 2, or 9y - 2y 2 A trinomial has 3 terms: -3x x, or 9y - 2y 2 + y The degree of the term is the exponent of the variable: 3x 2 has a degree of 2. When the variable does not have an exponent - always understand that there's a '1' e.g., y Multiplying Polynomials Examples of Polynomials 3x -2; Monomials: 3x, -2 (NOTE: these are also the terms of the polynomial) 5xy xy + 52x + 6; Monomials: 5xy 2, 36xy, 52x, 6 Πr 2 + 2Πh; Monomials: Πr 2, 2Πh When to Multiply Polynomials The instructions will ask you to multiply or simplify exercises that look like this: (Polynomial) (Polynomial) (Polynomial) (Polynomial) (Polynomial) * (Polynomial) (Polynomial) x (Polynomial)(Polynomial) Note: When there is no multiplication symbol between 2 sets of parentheses, realize that you are being asked to multiply.
12 When to Not Multiply Polymonials (Polynomial) + (Polynomial) (Polynomial) (Polynomial) Yes, I understand that parentheses encompass the polynomials, but pay attention to what the exercise is asking you to do. NOTE: (3x + 5y) + (2x +-6y) does not equal (3x + 5y) (2x +-6y). Practice with Constants Multiply: (8 + 6)(-2 + 5) Use order of operations: 1. Parentheses (8+ 6) = 14 (-2 + 5) = 3 2. Multiply 14 * 3 = 42 Using the F.O.I.L. Method Here s another way of looking at it: FOIL is method to multiply polynomials. It is an acronym for First Outer Inner Last. 1. First: (8 + 6)(-2 + 5) Multiply the first terms: 8 * -2 = Outer: (8 + 6)(-2 + 5) Multiply the outer terms: 8 * 5 = Inner: (8 + 6)(-2 + 5) Multiply the inner terms: 6 * -2= Last: (8 + 6)(-2 + 5) Multiply the outer terms: 6 * 5 = 30
13 Next, add the results: Simplify: = 42 Now, let's practice multiplying polynomials with variables. Practice with Positives Simplify the following polynomials. (x + 5)(x + 4) 1. First. Outer. Inner. Last. x * x + x * * x + 5 * 4 2. Multiply. x 2 + 4x + 5x Simplify. x 2 + 9x + 20 Practice with Negatives Simplify the following polynomials. (x - 5)(x - 4) 1. Before you start FOILing, change the negative signs (only do this if you need to): (x + -5)(x + -4) 2. Now FOIL: First. Outer. Inner. Last. x * x + x * * x + -5 * Multiply. x x + -5x Simplify. x x + 20 Practice Exercises 1) (x + 3)(x - 3) =
14 2) (x - 6)(x + 4) = 3) (x - 8)(x - 9) = 4) (5j + 11)(j + 1)= 5) (5p - 7)(4p + 3)= Answers to Practice Exercises 1. (x + 3)(x - 3) (x + 3)(x - 3) = x Rewrite the minus sign. (Only do this if you have a problem with your signs) (x + 3)(x + -3) 2. First. Outer. Inner. Last. x * x + x * * x + 3 * Multiply. x x + 3x Simplify. x x (x - 6)(x + 4) (x - 6)(x + 4) = x x Rewrite the minus sign. (Only do this if you have a problem with your signs) (x + -6)(x + 4) 2. First. Outer. Inner. Last. x * x + x * * x + -6 * 4 3. Multiply. x 2 + 4x + -6x Simplify. x x + -24
15 3. (x - 8)(x - 9) (x - 8)(x - 9) = x x Rewrite the minus signs. (Only do this if you have a problem with your signs) (x + -8)(x + -9) 2. First. Outer. Inner. Last. x * x + x * * x + -8 * Multiply. x x + -8x Simplify. x x (5j + 11)(j + 1) (5j + 11)(j + 1)= 5j j First. Outer. Inner. Last. 5j * j + 5j * * j + 11 * 1 2. Multiply. 5j 2 + 5j + 11j Simplify. 5j j (5p - 7)(4p + 3) (5p - 7)(4p + 3)= 20p p Rewrite the minus sign. (Only do this if you have a problem with your signs) (5p + -7)(4p + 3) 2. First. Outer. Inner. Last. 5p * 4p + 5p * * 4p + -7 * 3 3. Multiply. 20p p + -28p Simplify. 20p p Dividing Polynomials 1) Working with division in Arithmetic is a lot like division of monomials in Algebra. In arithmetic, you use your knowledge of factors to help you. Look at this example of division using factors. When you review the strategy you use in Arithmetic, algebra will make more sense. Simply
16 show the factors, cancel out the common factors (which is division) and you will be left with your solution. 2) Here's a basic monomial, notice that when you divide the monomial, you're dividing the numerical coefficients (the 24 and the 8) and you're dividing the literal coefficients (a and b). 3) Once again you divide the numerical and literal coefficients and you'll also divide the like variable factors by subtracting their exponents (5-2). 4) Divide the numerical and literal coefficients, divide the like variable factors by subtracting the exponents and you're done! 5) Divide the numerical and literal coefficients, divide the like variable factors by subtracting the exponents and you're done! Adapted from Algebrahelp.com and MathisFun.com
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# Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms
## Theorem
In the words of Euclid:
If there be two pyramids of the same height which have triangular bases, and each of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms, then, as the base of one pyramid is to the base of the other pyramid, so will all the prisms in the one pyramid be to all the prisms, being equal in multitude, in the other pyramid.
### Lemma
In the words of Euclid:
But that, as the triangle $LOC$ is to the triangle $RVF$, so is the prism in which the triangle $LOC$ is the base and $PMN$ its opposite to the base in which the triangle $RVF$ is the base and $STU$ its opposite, we must prove as follows.
## Proof
Let there be two tetrahedra of the same height whose bases are $ABC, DEF$ and whose apices are $G$ and $H$.
let each of $ABCG$ and $DEFH$ be divided into two equal tetrahedra which are similar to the whole and two equal prisms.
It is to be demonstrated that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of all the prisms in $ABCG$ to all the prisms in $DEFH$.
We have that:
$BO = OC$
and:
$AL = LC$
Therefore:
$LO \parallel AB$
and $\triangle ABC$ is similar to $\triangle LOC$.
For the same reason, $\triangle DEF$ is similar to $\triangle RVF$.
We have that:
$BC = 2 \cdot CO$
and:
$EF = 2 \cdot FV$
Therefore:
$BC : CO = EF : FV$
$\triangle LOC : \triangle ABC = \triangle DEF : \triangle RFV$
$\triangle ABC : \triangle DEF = \triangle LOC : \triangle RFV$
the ratio of $\triangle LOC$ to $\triangle RFV$ equals the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.
Therefore the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.
the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite equals the ratio of the prism of which the parallelogram $KBOL$ is its "base" and the straight line $PM$ lies opposite to the prism of which $QEVR$ is its "base" and the straight line $ST$ lies opposite.
the ratio of the two prisms of which the parallelogram $KBOL$ is its "base" and the straight line $PM$ lies opposite and in which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite are to the two prisms of which the the parallelogram to the prism of which $QEVR$ is its "base" and the straight line $ST$ lies opposite and in which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.
the two tetrahedra $PMNG$ and $STUH$ can be divided into two equal tetrahedra which are similar and two equal prisms.
As the ratio of the base $PMN$ is to the base $STU$, so will the ratio of the two prisms in the tetrahedron $PMNG$ be to the two prisms in the tetrahedron $STUH$.
But:
$\triangle PMN : \triangle STU = \triangle LOC : \triangle RFV$
Therefore:
$\triangle ABC : \triangle DEF$ equals the ratio of the four prisms in the tetrahedron $ABCG$ to the four prisms in the tetrahedron $DEFH$.
Similarly, if the remaining tetrahedra are divided into two tetrahedra and two prisms, then, as $\triangle ABC$ is to $\triangle DEF$, so will all the prisms in $ABCG$ be to all the prisms in $DEFH$.
$\blacksquare$
## Historical Note
This theorem is Proposition $4$ of Book $\text{XII}$ of Euclid's The Elements.
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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4
Other Exercises
In each of the following, use factor Theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1-7)
Question 1.
f(x) = x3 – 6x2 + 11x – 6; g(x) = x – 3
Solution:
We know that if g(x) is a factor of p(x),
then the remainder will be zero. Now,
f(x) = x3 – 6x2 + 11x – 6; g(x) = x -3
Let x – 3 = 0, then x = 3
∴ Remainder = f(3)
= (3)3 – 6(3)2 +11 x 3 – 6
= 27-54 + 33 -6
= 60 – 60 – 0
∵ Remainder is zero,
∴ x – 3 is a factor of f(x)
Question 2.
f(x) = 3X4 + 17x3 + 9x2 – 7x – 10; g(x) = x + 5
Solution:
f(x) = 3x4 + 17X3 + 9x2 – 7x – 10; g(x) = x + 5
Let x + 5 = 0, then x = -5
∴ Remainder = f(-5) = 3(-5)4 + 17(-5)3 + 9(-5)2 – 7(-5) – 10
= 3 x 625 + 17 x (-125) + 9 x (25) – 7 x (-5) – 10
= 1875 -2125 + 225 + 35 – 10
= 2135 – 2135 = 0
∵ Remainder = 0
∴ (x + 5) is a factor of f(x)
Question 3.
f(x) = x5 + 3x4 – x3 – 3x2 + 5x + 15, g(x) = x + 3
Solution:
f(x) = x5 + 3X4 – x3 – 3x2 + 5x + 15, g(x) = x + 3
Let x + 3 = 0, then x = -3
∴ Remainder = f(-3)
= (-3)5 + 3(-3)4 – (-3)3 – 3(-3)2 + 5(-3) + 15
= -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15
= -243 +243 + 27-27- 15 + 15
= 285 – 285 = 0
∵ Remainder = 0
∴ (x + 3) is a factor of f(x)
Question 4.
f(x) = x3 – 6x2 – 19x + 84, g(x) = x – 7
Solution:
f(x) = x3 – 6x2 – 19x + 84, g(x) = x – 7
Let x – 7 = 0, then x = 7
∴ Remainder = f(7)
= (7)3 – 6(7)2 – 19 x 7 + 84
= 343 – 294 – 133 + 84
= 343 + 84 – 294 – 133
= 427 – 427 = 0
∴ Remainder = 0
∴ (x – 7) is a factor of f(x)
Question 5.
f(x) = 3x3 + x2 – 20x + 12, g(x) = 3x – 2
Solution:
Question 6.
f(x) = 2x3 – 9x2 + x + 12, g(x) = 3 – 2x
Solution:
f(x) = 2x3 – 9x2 + x + 12, g(x) = 3 – 2x
Question 7.
f(x) = x3 – 6x2 + 11x – 6, g(x) = x2 – 3x + 2
Solution:
g(x) = x2 – 3x + 2
= x2 – x – 2x + 2
= x(x – 1) – 2(x – 1)
= (x – 1) (x – 2)
If x – 1 = 0, then x = 1
∴ f(1) = (1)3 – 6(1)2 + 11(1) – 6
= 1-6+11-6= 12- 12 = 0
∴ Remainder is zero
∴ x – 1 is a factor of f(x)
and if x – 2 = 0, then x = 2
∴ f(2) = (2)3 – 6(2)2 + 11(2)-6
= 8 – 24 + 22 – 6 = 30 – 30 = 0
∴ Remainder = 0
∴ x – 2 is also a factor of f(x)
Question 8.
Show that (x – 2), (x + 3) and (x – 4) are factors of x3 – 3x2 – 10x + 24.
Solution:
f(x) = x3 – 3x2 – 10x + 24
Let x – 2 = 0, then x = 2
Now f(2) = (2)3 – 3(2)2 – 10 x 2 + 24
= 8 – 12 – 20 + 24 = 32 – 32 = 0
∴ Remainder = 0
∴ (x – 2) is the factor of f(x)
If x + 3 = 0, then x = -3
Now, f(-3) = (-3)3 – 3(-3)2 – 10 (-3) + 24
= -27 -27 + 30 + 24
= -54 + 54 = 0
∴ Remainder = 0
∴ (x + 3) is a factor of f(x)
If x – 4 = 0, then x = 4
Now f(4) = (4)3 – 3(4)2 – 10 x 4 + 24 = 64-48 -40 + 24
= 88 – 88 = 0
∴ Remainder = 0
∴ (x – 4) is a factor of (x)
Hence (x – 2), (x + 3) and (x – 4) are the factors of f(x)
Question 9.
Show that (x + 4), (x – 3) and (x – 7) are factors of x3 – 6x2 – 19x + 84.
Solution:
Let f(x) = x3 – 6x2 – 19x + 84
If x + 4 = 0, then x = -4
Now, f(-4) = (-4)3 – 6(-4)2 – 19(-4) + 84
= -64 – 96 + 76 + 84
= 160 – 160 = 0
∴ Remainder = 0
∴ (x + 4) is a factor of f(x)
If x – 3 = 0, then x = 3
Now, f(3) = (3)3 – 6(3)2 – 19 x 3 + 84
= 27 – 54 – 57 + 84
= 111 -111=0
∴ Remainder = 0
∴ (x – 3) is a factor of f(x)
and if x – 7 = 0, then x = 7
Now, f(7) = (7)3 – 6(7)2 – 19 x 7 + 84
= 343 – 294 – 133 + 84
= 427 – 427 = 0
∴ Remainder = 0
∴ (x – 7) is also a factor of f(x)
Hence (x + 4), (x – 3), (x – 7) are the factors of f(x)
Question 10.
For what value of a (x – 5) is a factor of x3 – 3x2 + ax – 10?
Solution:
f(x) = x3 – 3x2 + ax – 10
Let x – 5 = 0, then x = 5
Now, f(5) = (5)3 – 3(5)2 + a x 5 – 10
= 125 – 75 + 5a – 10
= 125 – 85 + 5a = 40 + 5a
∴ (x – 5) is a factor of fix)
∴ Remainder = 0
⇒ 40 + 5a = 0 ⇒ 5a = -40
⇒ a = $$\frac { -40 }{ 5 }$$= -8
Hence a = -8
Question 11.
Find the value of a such that (x – 4) is a factor of 5x3 – 7x2 – ax – 28.
Solution:
Let f(x) 5x3 – 7x2 – ax – 28
and Let x – 4 = 0, then x = 4
Now, f(4) = 5(4)3 – 7(4)2 – a x 4 – 28
= 5 x 64 – 7 x 16 – 4a – 28
= 320 – 112 – 4a – 28
= 320 – 140 – 4a
= 180 – 4a
∴ (x – 4) is a factor of f(x)
∴ Remainder = 0
⇒ 180 -4a = 0
⇒ 4a = 180
⇒ a = $$\frac { 180 }{ 4 }$$ = 45
∴ a = 45
Question 12.
Find the value of a, if x + 2 is a factor of 4x4 + 2x3 – 3x2 + 8x + 5a.
Solution:
Let f(x) = 4x4 + 2x3 – 3x2 + 8x + 5a
and Let x + 2 = 0, then x = -2
Now, f(-2) = 4(-2)4 + 2(-2)3 – 3(-2)2 + 8 x ( 2) + 5a
= 4 x 16 + 2(-8) – 3(4) + 8 (-2) + 5a
= 64- 16- 12- 16 +5a
= 64 – 44 + 5a
= 20 + 5a
∴ (x + 2) is a factor of fix)
∴ Remainder = 0
⇒ 20 + 5a = 0 ⇒ 5a = -20
⇒ a =$$\frac { -20 }{ 5 }$$ = -4
∴ a = -4
Question 13.
Find the value of k if x – 3 is a factor of k2x3 – kx2 + 3kx – k.
Solution:
Let f(x) = k2x3 – kx2 + 3kx – k
and Let x – 3 = 0, then x = 3
Now,f(3) = k2(3)3 – k(3)2 + 3k(3) – k
= 27k2 – 9k + 9k-k
= 27k2-k
∴ x – 3 is a factor of fix)
∴ Remainder = 0
∴ 27k2 – k = 0
⇒ k(27k – 1) = 0 Either k = 0
or 21k – 1 = 0
⇒ 21k = 1
∴ k= $$\frac { 1 }{ 27 }$$
∴ k = 0,$$\frac { 1 }{ 27 }$$
Question 14.
Find the values of a and b, if x2 – 4 is a factor of ax4 + 2x3 – 3x2 + bx – 4.
Solution:
f(x) = ax4 + 2x3 – 3x2 + bx – 4
Factors of x2 – 4 = (x)2 – (2)2
= (x + 2) (x – 2)
If x + 2 = 0, then x = -2
Now, f(-2) = a(-2)4 + 2(-2)3 – 3(-2)2 + b(-2) – 4
16a- 16 – 12-26-4
= 16a -2b-32
∵ x + 2 is a factor of f(x)
∴ Remainder = 0
⇒ 16a – 2b – 32 = 0
⇒ 8a – b – 16 = 0
⇒ 8a – b = 16 …(i)
Again x – 2 = 0, then x = 2
Now f(2) = a x (2)4 + 2(2)3 – 3(2)2 + b x 2-4
= 16a + 16- 12 + 26-4
= 16a + 2b
∵ x – 2 is a factor of f(x)
∴ Remainder = 0
⇒ 16a + 2b = 0
⇒ 8a + b= 0 …(ii)
⇒ 16a = 16
⇒ a = $$\frac { 16 }{ 16 }$$ = 1
From (ii) 8 x 1 + b = 0
⇒ 8 + b = 0
⇒ b = – 8
∴ a = 1, b = -8
Question 15.
Find α and β, if x + 1 and x + 2 are factors of x3 + 3x2 – 2αx +β.
Solution:
Let f(x) = x3 + 3x2 – 2αx + β
and Let x + 1 = 0 then x = -1
Now,f(-1) = (1)3 + 3(-1)2 – 2α (-1) +β
= -1 + 3 + 2α + β
= 2 + 2α + β
∵ x + 1 is a factor of f(x)
∴ Remainder = 0
∴ 2 + 2α + β = 0
⇒ 2α + β = -2 …(i)
Again, let x + 2 = 0, then x = -2
Now, f(-2) = (-2)3 + 3(-2)2 – 2α(-2) + β
= -8 + 12 + 4α+ β
= 4 + 4α+ β
∵ x + 2 is a factor of(x)
∴ Remainder = 0
∴ 4+ 4α + β = 0
⇒ 4α + β = -4 …(ii)
Subtracting (i) from (ii),
2α = -2
⇒ α = $$\frac { -2 }{ 2 }$$ = -1
From (ii), 4(-1) + β = -4
-4 + β= -4
⇒ β =-4+ 4 = 0
∴ α = -1, β = 0
Question 16.
If x – 2 is a factor of each of the following two polynomials, find the values of a in each case:
(i) x3 – 2ax2 + ax – 1
(ii) x5 – 3x4 – ax3 + 3ax2 + 2ax + 4
Solution:
(i) Let f(x) = x3 – 2ax2 + ax – 1 and g(x) = x – 2
and let x – 2 = 0, then x = 2
∴ x – 2 is its factor
∴ Remainder = 0
f(2) = (2)3 – 2a x (2)2 + a x 2 – 1
= 8-8a+ 2a-1 = 7-6a
∴ 7 – 6a = 0
⇒ 6a = 7
⇒ a = $$\frac { 7 }{ 6 }$$
∴ a = $$\frac { 7 }{ 6 }$$
(ii) Let f(x) = x5 – 3x4 – ax3 + 3 ax2 + 2ax + 4 and g(x) = x – 2
Let x – 2 = 0, then x=2
∴ f(2) = (2)5 – 3(2)4 – a(23) + 3a (2)2 + 2a x 2 + 4
= 32 – 48 – 8a + 12a + 4a + 4
= -12 + 8a
∴ Remainder = 0
∴ -12 + 8a = 0
⇒ 8a= 12
⇒ a = $$\frac { 12 }{ 8 }$$ = $$\frac { 3 }{ 2 }$$
∴ Hence a = $$\frac { 3 }{ 2 }$$
Question 17.
In each of the following two polynomials, find the values of a, if x – a is a factor:
(i) x6 – ax5 + x4-ax3 + 3x-a + 2
(ii) x5 – a2x3 + 2x + a + 1
Solution:
(i) Let f(x) = x– ax5+x4-ax3 + 3x-a + 2 and g(x) = x – a
∴ x – a is a factor
∴ x – a = 0
⇒ x = a
Now f(a) = a6-a x a5 + a4-a x a3 + 3a – a + 2
= a6-a6 + a4-a4 + 2a + 2
= 2a + 2
∴ x + a is a factor of p(x)
∴ Remainder = 0
Question 18.
In each of the following, two polynomials, find the value of a, if x + a is a factor.
(i) x3 + ax2 – 2x + a + 4
(ii) x4 – a2r + 3x – a
Solution:
Question 19.
Find the values of p and q so that x4 + px3 + 2x2 – 3x + q is divisible by (x2 – 1).
Solution:
Question 20.
Find the values of a and b so that (x + 1) and (x – 1) are factors of x4 + ax3 3x2 + 2x + b.
Solution:
Question 21.
If x3 + ax2 – bx + 10 is divisible by x2 – 3x + 2, find the values of a and b.
Solution:
Question 22.
If both x + 1 and x – 1 are factors of ax3 + x2 – 2x + b, find the values of a and b.
Solution:
Question 23.
What must be added to x3 – 3x2 – 12x + 19 so that the result is exactly divisibly by x2 + x – 6?
Solution:
Question 24.
What must be subtracted from x3 – 6x2 – 15x + 80-so that the result is exactly divisible by x2 + x – 12?
Solution:
Question 25.
What must be added to 3x3 + x2 – 22x + 9 so that the result is exactly divisible by 3x2 + 7x – 6?
Solution:
Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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# Difference between revisions of "2020 AMC 12A Problems/Problem 17"
## Problem 17
The vertices of a quadrilateral lie on the graph of $y=\ln{x}$, and the $x$-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$. What is the $x$-coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$
## Solution 1
Let the coordinates of the quadrilateral be $(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))$. We have by shoelace's theorem, that the area is$$\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)\ln(n) + \ln(n+2)\ln(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2}=$$$$\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} = \ln \left( \sqrt{\frac{(n+1)^2(n+3)^2}{n^2(n+2)^2}} \right) = \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) = \ln \left( \frac{91}{90} \right).$$We now that the numerator must have a factor of $13$, so given the answer choices, $n$ is either $12$ or $11$. If $n=11$, the expression $\frac{(n+1)(n+2)}{n(n+3)}$ does not evaluate to $\frac{91}{90}$, but if $n=12$, the expression evaluates to $\frac{91}{90}$. Hence, our answer is $\boxed{12}$.
~AopsUser101
## Solution 2
Like above, use the shoelace formula to find that the area of the triangle is equal to $\ln\frac{(n+1)(n+2)}{n(n+3)}$. Because the final area we are looking for is $\ln\frac{91}{90}$, the numerator factors into $13$ and $7$, which one of $n+1$ and $n+2$ has to be a multiple of $13$ and the other has to be a multiple of $7$. Clearly, the only choice for that is $\boxed{12}$
~Solution by IronicNinja
## Solution 3
How $f(x)=\ln(x)$ is a concave function, then:
Therefore $[BCDE]=[ABCH]+[HCDG]+[GDEF]-[ABEF]$, all quadrilaterals of side right are trapezius
$[BCDE]=\frac{\ln(n+1)+\ln n}{2}+\frac{\ln(n+2)+\ln(n+1)}{2}+\frac{\ln(n+3)+\ln(n+2)}{2}-\frac{3(\ln(n+3)+\ln n)}{2}$
$[BCDE]=\tfrac{2\ln(n+1)+2\ln(n+2)-2\ln(n+3)-2\ln n}{2}=\ln(n+1)+\ln(n+2)-\ln(n+3)-\ln n=\ln\tfrac{(n+1)(n+2)}{n(n+3)}$ $$\implies\ln\frac{(n+1)(n+2)}{n(n+3)}=\ln\frac{91}{90}\implies \frac{(n+1)(n+2)}{n(n+3)}=\frac{91}{90}$$ $$n^2+3n-180=0\implies (n+15)(n-12)=0\implies n=12$$
~Solution by AsdrúbalBeltrán
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# The length and breadth of a rectangle are in the ratio 5:3. If the perimeter is 80 cm, find the length and breadth of rectangle.
Given :
The ratio of length and breadth of the rectangle $=5 : 3$.
Perimeter of the rectangle $= 80 cm$.
To find :
We have to find the length and breadth of the rectangle.
Solution :
Let the length of the rectangle be 5x.
Breadth of the rectangle be 3x.
$$Perimeter of the rectangle = 2 (length + breadth)$$
$80 = 2 (5x + 3x)$
$80 = 2 (8x)$
$80 = 16 x$
Rewrite,
$16 x = 80$
$x = \frac{80}{16}$
$x = 5$
$Length = 5 x = 5 (5) = 25$
$Breadth = 3 x = 3 (5) = 15$
Length of the rectangle is 25 cm and breadth of the rectangle is 15 cm.
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Updated on: 10-Oct-2022
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# RD Sharma Solutions for Class 6 Chapter 7: Decimals Exercise 7.5
The solutions prepared by faculty at BYJU’S help students to get an overall idea about the concepts which are covered in each exercise. This exercise helps students understand the steps followed in comparing the given set of decimals. RD Sharma solutions Class 6 improves logical thinking capacity among students. The exercise wise problems of the textbook can be solved with the help of RD Sharma Solutions for Class 6 Maths Chapter 7 Decimals Exercise 7.5 PDF which are provided here.
## RD Sharma Solutions for Class 6 Chapter 7: Decimals Exercise 7.5 Download PDF
### Access RD Sharma Solutions for Class 6 Chapter 7: Decimals Exercise 7.5
1. Fill in the blanks by using > or < to complete the following:
(i) 25.35 …… 8.47
(ii) 20.695 …… 20.93
(iii) 0.39 ……. 0.72
(iv) 0.109 ……. 0.83
(v) 0.236 ……. 0.201
(vi) 0.93 ……. 0.99
Solution:
(i) 25.35 …… 8.47
We know that the whole numbers 25 > 8
Hence, 25.35 > 8.47.
(ii) 20.695 …… 20.93
We know that both the whole numbers are equal
So by checking the tenth parts we know that 9 > 6
Hence, 20.695 < 20.93.
(iii) 0.39 ……. 0.72
We know that both the whole numbers are 0
So by checking the tenth parts we know that 7 > 2
Hence, 0.39 < 0.72.
(iv) 0.109 ……. 0.83
We know that both the whole numbers are 0
So by checking the tenth parts we know that 1 < 8
Hence, 0.109 < 0.83.
(v) 0.236 ……. 0.201
We know that both the whole numbers are 0
So by checking the hundredth parts we know that 3 > 0
Hence, 0.236 > 0.201.
(vi) 0.93 ……. 0.99
We know that both the whole numbers are 0
So by checking the hundredth parts we know that 3 < 9
Hence, 0.93 < 0.99.
(i) 1.008 or 1.800
(ii) 3.3 or 3.300
(iii) 5.64 or 5.603
(iv) 1.5 or 1.50
(v) 1.431 or 1.439
(vi) 0.5 or 0.05
Solution:
(i) 1.008 or 1.800
We know that the whole numbers are equal
So by comparing tenth parts we know that 0 < 8
Hence, 1.008 < 1.800.
(ii) 3.3 or 3.300
We know that the whole numbers and tenths place are equal
Hence, 3.3 = 3.300.
(iii) 5.64 or 5.603
We know that the whole numbers are equal
So by comparing the hundredth place we know that 4 > 0
Hence, 5.64 > 5.603.
(iv) 1.5 or 1.50
We know that the whole numbers and tenths place are equal
Hence, 1.5 = 1.50.
(v) 1.431 or 1.439
We know that the whole numbers are equal
So by comparing thousandths place we know that 1 < 9
Hence, 1.431 < 1.439.
(vi) 0.5 or 0.05
We know that the whole numbers are equal
So by comparing tenth place we know that 5 > 0
Hence, 0.5 > 0.05.
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Middle Years
# 9.07 Surface area of cylinders
Lesson
A cylinder has three faces: two identical circular bases and a curved surface that joins the two bases together.
The surface area of a cylinder is the sum of the areas of these three faces. We already know how to find the area of the circular bases, but what about the curved surface?
### The curved surface area of a cylinder
By "unwrapping" the cylinder we can treat the curved surface as a rectangle, with one side length equal to the height of the cylinder, and the other the perimeter (circumference) of the base circle. This is given by $2\pi r$2πr, where $r$r is the radius.
This means the surface area of the curved part of a cylinder is $2\pi rh$2πrh, where $r$r is the radius and $h$h is the height.
We can see how the cylinder unrolls to make this rectangle in the applet below:
To find the surface area of the whole cylinder, we need to add the area of the top and bottom circles to the area of the curved part. Both of these circles have an area of $\pi r^2$πr2, so the surface area of a cylinder is:
Surface area of a cylinder
$\text{Surface area of a cylinder}=2\pi r^2+2\pi rh$Surface area of a cylinder=2πr2+2πrh
Where $r$r is the radius and $h$h is the height of the cylinder.
#### Practice questions
##### Question 1
Consider the following cylinder.
1. Find the curved surface area of the cylinder to two decimal places.
##### Question 2
Consider the following cylinder.
1. Find the curved surface area of the cylinder to two decimal places.
2. Using the result from part (a) or otherwise, find the total surface area of the cylinder.
##### Question 3
Consider the cylinder shown in the diagram below.
1. Find the surface area of the cylinder in square centimetres.
2. Use your answer from part (a) to find the surface area of the cylinder in square millimetres?
##### Question 4
The area of the circular face on a cylinder is $8281\pi$8281π m2. The total surface area of the cylinder is $25662\pi$25662π m2
1. If the radius of the cylinder is $r$r m, find the value of $r$r.
Enter each line of working as an equation.
2. Hence, find the height $h$h of the cylinder.
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##### Activity5.2.4
Remember that the absolute value of a number tells us how far that number is from $$0$$ on the real number line. We may therefore think of the inverse power method as telling us the eigenvalue closest to $$0\text{.}$$
1. If $$\vvec$$ is an eigenvalue of $$A$$ with associated eigenvalue $$\lambda\text{,}$$ explain why $$\vvec$$ is an eigenvector of $$A - sI$$ where $$s$$ is some scalar.
2. What is the eigenvalue of $$A-sI$$ associated to the eigenvector $$\vvec\text{?}$$
3. Explain why the eigenvalue of $$A$$ closest to $$s$$ is the eigenvalue of $$A-sI$$ closest to $$0\text{.}$$
4. Explain why applying the inverse power method to $$A-sI$$ gives the eigenvalue of $$A$$ closest to $$s\text{.}$$
5. Consider the matrix $$A = \left[\begin{array}{rrrr} 3.6 \amp 1.6 \amp 4.0 \amp 7.6 \\ 1.6 \amp 2.2 \amp 4.4 \amp 4.1 \\ 3.9 \amp 4.3 \amp 9.0 \amp 0.6 \\ 7.6 \amp 4.1 \amp 0.6 \amp 5.0 \\ \end{array}\right] \text{.}$$ If we use the power method and inverse power method, we find two eigenvalues, $$\lambda_1=16.35$$ and $$\lambda_2=0.75\text{.}$$ Viewing these eigenvalues on a number line, we know that the other eigenvalues lie in the range between $$-\lambda_1$$ and $$\lambda_1\text{,}$$ as shaded in FigureĀ 1.
The Sage cell below has a function find_closest_eigenvalue(A, s, x, N) that implements $$N$$ steps of the inverse power method using the matrix $$A-sI$$ and an initial vector $$x\text{.}$$ This function prints approximations to the eigenvalues and eigenvectors of $$A\text{.}$$ By trying different values in the gray regions of the number line, find the other two eigenvalues of $$A\text{.}$$
6. Write a list of the four eigenvalues of $$A$$ in increasing order.
in-context
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# Exponents Lead to Cumbersome, Time-Consuming Calculations involving Large Numbers but it is Pattern-Driven
The GMAT's quantitative section is increasingly emphasizing problem solving skills over calculation abilities, and often does so in the form of "Number Properties" questions. The authors of the exam are also quite adept at recognizing "mathematical psychology", and creating questions that increase an examinee's anxiety by enough to make that process of problem solving a bit more difficult. One of the major themes that arises as a result is the use of exponents, which carry with them a number of properties extremely useful to the writers of the GMAT.
Exponents
• Inspire fear (or at least apprehension) in test takers
• Lead to cumbersome, time-consuming calculations involving large numbers
• Are actually quite pattern-driven, and reward those who seek out those patterns rather than attempt to perform the extensive calculations
How can this help you on the exam?
If you embrace the pattern-driven quality of exponents, you can rest easy on exponent questions involving large numbers, knowing that you can test the pattern with small numbers, and simply extrapolate it to solve the overall question. Take, for example, a question that asks:
What is the sum of the digits of 10^25 - 37?
Listing out the numbers will be time consuming and contains the potential for error (counting to 25 when writing out the zeroes, then writing out that many digits in the difference, is a tedious process). But if you recognize that the first number will simply be "1" followed by 25 zeroes, and that the difference of the two will end in 63, preceded by a series of 9s, you can make quick work of this problem.
Try using a smaller exponent to see what the result will look like:
10^4 - 37:
10000
- 37
9963
If the exponent is 4, we end up with 4 digits in the answer: a 6, a 3, and the rest are 9s. Trying again with another exponent, we can see that the pattern holds for any exponent:
10^6 - 37:
1000000
- 37
999963
Again, we have the same number of digits in the difference as the value of the exponent, and two of those are 6 and 3, with the others 9s.
So, we can conclude that 10^x - 37 will give us a solution in which we have (x-2) 9s, a 6, and a 3.
Because 6+3 is 9, the sum of the digits will be (x-2) * 9 + 9, or (x-1) * 9.
In the original example - 10^25 - 37, the exponent is 25, so we'll have (25-1) * 9, or 24*9, and the answer is 216.
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# What is the value of c that makes x^2-15x+c a perfect square trinomial?
##### 2 Answers
Feb 24, 2017
$c = {\left(\frac{15}{2}\right)}^{2} = \frac{225}{4}$
#### Explanation:
We find:
${\left(x + \frac{b}{2}\right)}^{2} = {x}^{2} + 2 \left(x\right) \left(\frac{b}{2}\right) + {\left(\frac{b}{2}\right)}^{2} = {x}^{2} + b x + {b}^{2} / 4$
So in order for ${x}^{2} + b x + c$ to be a perfect square trinomial, we require:
$c = {\left(\frac{b}{2}\right)}^{2}$
In our example:
$c = {\left(\frac{\textcolor{b l u e}{15}}{2}\right)}^{2} = \frac{225}{4}$
Feb 24, 2017
$c = \frac{225}{4} = 56.25$
#### Explanation:
Consider the equaton: ${x}^{2} - 15 x + c = 0$
This equation has a single root where its discriminant $= 0$
When this equation has a single root its factors will be of the form:
$\left(x - p\right) \left(x - p\right) = 0 \to {\left(x - p\right)}^{2} = 0$
Hence: The trinomial will be a perfect square when the discriminant of ${x}^{2} - 15 x + c = 0$ is equal to $0$
I.e. when ${15}^{2} - 4 \cdot 1 \cdot c = 0$
$4 c = 225 \to c = \frac{225}{4} = 56.25$
To test this result consider $c = \frac{225}{4} = {\left(\frac{15}{2}\right)}^{2}$
Hence our trinomial is: ${x}^{2} - 15 x + {\left(\frac{15}{2}\right)}^{2}$
Which factorises to: $\left(x - \frac{15}{2}\right) \left(x - \frac{15}{2}\right) = {\left(x - \frac{15}{2}\right)}^{2}$ which is a perfect square.
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# What are the asymptotes of y=1/x-2 and how do you graph the function?
Mar 1, 2018
The most useful thing when trying to draw graphs is to test the zeroes of the function to get some points that can guide your sketch.
Consider $x = 0$:
$y = \frac{1}{x} - 2$
Since $x = 0$ cannot be substituted in directly (since it's in the denominator), we can consider the limit of the function as $x \to 0$. As $x \to 0$, $y \to \setminus \infty$. This tells us that the graph blows up to infinity as we approach the y-axis. Since it will never touch the y-axis, the y-axis is a vertical asymptote.
Consider $y = 0$:
$0 = \frac{1}{x} - 2$
$x = \frac{1}{2}$
So we have identified a point that the graph passes through: $\left(\frac{1}{2} , 0\right)$
Another extreme point we can consider is $x \to \setminus \infty$. If $x \to + \setminus \infty$, $y \to - 2$. If $x \to - \setminus \infty$, $y \to - 2$. So at both ends of the x-axis, y will approach -2. This means there is a horizontal asymptote at $y = - 2$.
So we have found out the following:
Vertical asymptote at $x = 0$.
Horizontal asymptote at $y = - 2$.
Point contained in graph: $\left(\frac{1}{2} , 0\right)$.
graph{1/x -2 [-10, 10, -5, 5]} You should notice that all three of these facts provide enough information to draw the graph above.
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# How Do We Use Percentages In Everyday Life?
## What jobs use percentages?
Therefore, any business involving tax calculation, tip calculation, or interest rates uses fractions.
Banks, restaurants, movie theaters and department stores all use percentages, so teller, wait staff and store clerk positions are included here..
## How do you calculate percentage of ingredients?
To accurately calculate the percentages:Weigh all of your ingredients (column D in the example above).Total the weight (cell D8 in the example).Divide the weight of the individual ingredient by the total weight of all ingredients and multiply by 100 (formula example shown for beeswax in cell E2).
## How do you introduce decimals to students?
Introduce decimal place values. Tell students that ones are always immediately to the left of a decimal point. Show them that tenths are always immediately to the right, followed by hundredths and thousandths. Over-pronounce tenths and hundredths to avoid confusing decimal place values with tens and hundreds.
## What grade do you learn percentages?
Children usually start learning some basic percentage skills in fourth grade (calculating 10%, 50%, 75% and 100%). 5th graders and 6th graders continue to develop their skills. And percentages are “real” life math skills.
## Where do you see percentages occur in your everyday life?
Think about how percent is used in your daily life. Think of specific examples of how it is used with the things that you do and like. For example, “When I go shopping, I use percentages to find the sale price and discount on clothing in the store.”
## What is 45 out of 60 as a percentage?
75Percentage Calculator: 45 is what percent of 60? = 75.
## How are fractions used in construction?
In many ways, architects work in a world of fractions. Blueprints and models are never the same size as the structures they represent. … For example, 1 inch on the blueprint may represent 1 foot on the actual building, meaning the drawing is the fraction 1/12 in size.
## Why do we have decimals?
So, our Decimal System lets us write numbers as large or as small as we want, using the decimal point. Digits can be placed to the left or right of a decimal point, to show values greater than one or less than one. The decimal point is the most important part of a Decimal Number.
## What is the easiest way to learn percentages?
To calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23. 5 percent is one half of 10 percent. To calculate 5 percent of a number, simply divide 10 percent of the number by 2.
## What is 47 out of 60 as a percentage?
78.33%Percentage ChartX is what percent of 60?46.04 is76.74% of 6046.5 is77.5% of 6046.96 is78.26% of 6047 is78.33% of 604 more rows
## What is a 48 60 in percentage?
80%Convert fraction (ratio) 48 / 60 Answer: 80%
## Why are percentages useful?
Percentages are useful in practice, because it allows one to compare things that are not out of the same number. For example, exam marks are often percentages, which can compare them even if there are more questions on one exam paper than the other.
## Where do we see decimals in everyday life?
We use decimals every day while dealing with money, weight, length etc. Decimal numbers are used in situations where more precision is required than the whole numbers can provide. For example, when we calculate our weight on the weighing machine, we do not always find the weight equal to a whole number on the scale.
## How do I work out a percentage of two numbers?
Learning how to calculate the percentage of one number vs. another number is easy. If you want to know what percent A is of B, you simple divide A by B, then take that number and move the decimal place two spaces to the right. That’s your percentage!
## Where do we use fractions in everyday life?
Fractions are used in baking to tell how much of an ingredient to use. Fractions are used in telling time; each minute is a fraction of the hour. Finally, fractions are used to determine discounts when there’s a sale going on.
Decimal Point – A dot used to separate a dollar from cents or ones from tenths in a number. I let the students know that we can use money to help us understand decimals. A dime can help you understand the tenths place. … A penny can help you understand the hundredths place.
## How do you teach a percentage of a number?
If you set up a proportion to find the percent of a number, you’ll always end up dividing by 100 in the end. Another way to get the same answer is to divide by 100 first and then multiply the numbers together. 24/100 = . 24 and (.
## How can calculate percentage?
1. How to calculate percentage of a number. Use the percentage formula: P% * X = YConvert the problem to an equation using the percentage formula: P% * X = Y.P is 10%, X is 150, so the equation is 10% * 150 = Y.Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10.More items…
## How do I calculate a percentage between two numbers?
Percentage Change | Increase and DecreaseFirst: work out the difference (increase) between the two numbers you are comparing.Increase = New Number – Original Number.Then: divide the increase by the original number and multiply the answer by 100.% increase = Increase ÷ Original Number × 100.More items…
## How do engineers use fractions?
Engineers use fractions to divide the amount of incoming energy during maximum heating by the actual heat-dissipation capability of the shield. This gives a ratio of heat buildup over heat dissipation. As long as heat buildup is smaller than the dissipation rate, the shield will be safe and effective.
## What is 50 out of 60 as a percentage?
83.333333333333%Convert fraction (ratio) 50 / 60 Answer: 83.333333333333%
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# Lacsap’s Triangle
1 Introduction. Let us consider a triangle of fractions: Obviously, the numbers are following some pattern. In this investigation we will try to explain the theory behind this arrangement and to find a general relation between the element’s number and its value. The pattern above is called a Lacsap’s Triangle, which inevitably hints at its relation to another arrangement – Pascal’s Triangle (as Lacsap appears to be an anagram of Pascal). The algorithm behind it is very simple: each element is the sum of the two elements above it.
However, if we represent a triangle as a table (below), we will be able to notice a pattern between an index number of an element and its value: column column column column column column column 2 0 1 2 3 4 5 row 0 1 row 1 1 1 row 2 1 2 1 row 3 1 3 3 1 row 4 1 4 6 4 1 row 5 1 5 10 10 5 1 row 6 1 6 15 20 15 6 6 1 It seems important to us to stress several points that this table makes obvious: ? the number of elements in a row is n + 1 (where n is an index number of a row) ? the element in column 1 is always equal to the element in column n – 1 ? herefore, the element in column 1 in every row is equal to the number of a given row. Now when we have established the main sequences of a Pascal’s triangle let us see how they are going to be expressed in a Lacsap’s arrangement. We also suggest looking at numerators and denominators separately, because it seems obvious that the fractions themselves can’t be derived from earlier values using the progressions of the sort that Pascal uses. Finding Numerators. Let’s begin with presenting given numerators in a similar table, where n is a number of a row. n=1 1 1 n=2 1 3 1 n=3 1 6 6 1 n= 4 1 0 10 10 1 n=5 1 15 15 15 15 1 3 Although the triangles appeared similar, the table demonstrates a significant difference between them. We can see, that all numerators in a row (except 1’s) have the same value. Therefore, they do not depend on other elements, and can be obtained from a number of row itself. Now a relationship we have to explore is between these numbers: 1 1 2 3 3 6 4 10 5 15 If we consider a number of row to be n, then n=1 1=n 0. 5 2 n 0. 5 (n +1) n n=2 3 = 1. 5n 0. 5 3 n 0. 5 (n +1) n n=3 6 = 2n 0. 5 4 n 0. 5 (n +1) n n=4 10 = 2. 5 n 0. 5 5 n 0. 5 (n +1) n n=5 15 = 3n 0. 6 n 0. 5 (n +1) n Moving from left to right in each row of the table above, we can clearly see the pattern. Dividing an element by a row number we get a series of numbers each one of them is 0. 5 greater than the previous one. If 0. 5 is factored out, the next sequence is {2; 3; 4; 5; 6}, where each element corresponds to a row number. Using a cyclic method, we have found a general expression for the numerator in the original triangle: If Nn is a numerator in a row n, then Nn = 0. 5(n + 1)n = 0. 5n2 + 0. 5n Now we can plot the relation between the row number and the numerator in each row.
The graph of a parabolic form begins at (0; 0) and continues to rise to infinity. It represents a continuous function for which D(f) = E(f) = (0; ); 4 Using a formula for the numerator we can now find the numerators of further rows. For example, if n = 6, then Nn = 0. 5 62 + 0. 5 6 = 18 + 3 = 21; if n = 7, then Nn = 0. 5 72 + 0. 5 7 = 24. 5 + 3. 5 = 28; and so forth. Another way of representing numerators would be through using factorial notation, for obviously Numeratorn = n! Now let’s concentrate of finding another part of the fraction in the triangle. Finding Denominators.
There are two main variables, that a denominator is likely to depend on: ? number of row ? numerator To find out which of those is connected with the denominator, let us consider a following table: column 1 column 2 column 3 column 4 column 5 column 6 5 row 1 1 1 row 2 1 2 1 row 3 1 4 4 1 row 4 1 7 6 7 1 row 5 1 11 9 9 11 1 It is now evident, that a difference between the successive denominators in a second column increases by one with each iteration: {1; 2; 4; 7; 11}, the difference between elements being: {1; 2; 3; 4}. So if the number of row is n, and the denominator of the second column is D, then D1 = 1
D2 = 2 D3 = 4 etc; then Dn = Dn-1 + (n – 1) = (n-1)! + 1; If we now look at the third column with a regard to a factorial sequence, a pattern emerges: In the series {1; 1; 2; 3; 4; 5; 6; 7;… ; }, if d is the denominator of the third column, then: d3 = 1 + 1 + 2 = 4 d4 = 1 + 2 + 3 = 6 d5 = 2 + 3 + 4 = 9 dn = (n – 2)! + 3; To check the consistency of this succession, we will continue with the study of the fourth column. By analogy, the result is as follows: Denominatorn = (n – 3)! + 6 (where n is a number of row) Therefore, it can be represented as follows:
Column 2 (n-1)! +1 Column 3 (n-2)! +3 Column 4 (n-3)! +6 It is now clear, that numbers inside the brackets follow the (c – 1) (where c is the number of column), and the numbers outside are in fact the numerators of the row of the previous index number (comparing to the column). Therefore, a general expression for the denominator would be Dn = (n – (c – 1))! + (c – 1)! 6 where Dn is a general denominator of the triangle n is a number of row c is the number of column Now we can use a formula above to calculate the denominators of the rows 6 and 7. column 2 column 3 olumn 4 column 5 column 6 row 6 (6 – 1)! + 1 = 16 (6 – 2)! + 3 = 13 (6 – 3)! + 6 = 12 (6 – 4)! + 10 = 13 (6 – 5)! +15 =16 row 7 (7 – 1)! + 1 = 22 (7 – 2)! + 3 = 18 (7 – 3)! + 6 = 16 (7 – 4)! + 10 = 16 (7 – 5)! +15 =18 column 7 (7 – 6)! + 21 = 22 Fusing these value with the numerators from the calculations above, we get the 6th and the 7th rows of the Lacsap’s triangle: Row 6: 1; ; ; ; ; ;1 Row 7: 1; ; ; ; ; ; ;1 If we now let En(r) be the (r + 1)th element in the nth row, starting with r = 0; then the general statement for this element would be: En(r) =
Conclusion. To check the validity and limitations of this general statement let us consider the unusual circumstances: first of all, will it work for the columns of ones (1st and last column of each row)? if n = 4 r = 0, then En(r) = =1 if n = 5 r = 5, then En(r) = =1 7 therefore, the statement is valid for any element of any row, including the first one: En(r) = =1 However, obviously, the denominator of this formula can not equal zero. But as long as r and n are both always positive integers (being index numbers), this limitation appears to be irrelevant.
If the numeration of columns was to start from 1 (the 1st column of ones), then the general statement would take the form of: En(r) = 8 Bibliography: 1) Weisstein, Eric W. “Pascal’s Triangle. ” From MathWorld–A Wolfram Web Resource. http:// mathworld. wolfram. com/PascalsTriangle. html 2) “Pascal’s Triangle and Its Patterns”; an article from All you ever wanted to know http:// ptri1. tripod. com/ 3) Lando, Sergei K.. “7. 4 Multiplicative sequences”. Lectures on generating functions. AMS. ISBN 0-8218-3481-9
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Home | | Maths 6th Std | Proportion
# Proportion
When two ratios are equal (a/b = c/d), we say that the ratios are in Proportion. This is denoted as a : b : : c : d and it is read as ‘a is to b as c is to d’. The following situations explain about proportion.
Proportion
When two ratios are equal (a/b = c/d), we say that the ratios are in Proportion. This is denoted as a : b : : c : d and it is read as ‘a is to b as c is to d’. The following situations explain about proportion.
Situation 1
The Teacher said to the students, “You can do a maximum of 4 projects in Mathematics. You will get 5 as internal marks for each project that you do”. Kamala asked, “Teacher, What if I do 2 or 3 or 4 projects?” The teacher replied, “For 2 projects you will get 10 marks, for 3 projects you will get 15 marks and for 4 projects you will get 20 marks”.
Here “1 project carries 5 marks” is equivalent to saying “2 projects carry 10 marks” and so on and hence the ratios, 1 : 5 = 2 : 10 = 3 : 15 = 4 : 20 are said to be in Proportion. Thus 1 : 5 is in proportion to 2 : 10, 3 : 15, 4 : 20 and so on. This is denoted by 1 : 5 : : 2 : 10 and it is read as ‘1 is to 5 as 2 is to 10’ and so on.
Situation 2
The size of the photograph of Srinivasa Ramanujan as shown in Figure 3.5(a) is of length 5 grids and width 3 grids. Figure 3.5(b) shows the enlarged size of the photograph of length 10 grids and width 6 grids. Here,
As the two ratios are equal, the given figures are in proportion. This is represented as 5:10:: 3 : 6 or 5 : 10 = 3 : 6 and it is read as ‘5 is to 10 as 3 is to 6’
1. Proportionality law
If two ratios are in proportion ie., a : b : : c : d then the product of the extremes is equal to the product of the means. This is called the proportionality law. Here, a and d are the extremes and b and c are the means. Also, if two ratios are equal ie., a/b=c/d ad=bc called the cross product of proportions.
Example 3.6
By proportionality law, check whether 3 : 2 and 30 : 20 are in proportion.
Solution
Here the extremes are 3 and 20 and the means are 2 and 30.
Product of extremes, ad = 3 × 20 = 60.
Product of means, bc = 2 × 30 = 60.
Thus by proportionality law, we find ad = bc and hence 3 : 2 and 30 : 20 are in proportion.
Example 3.7
A picture is resized in a computer as shown below.
Do you observe any change in the shape and size of the picture? Check whether the ratios formed by its length and breadth are in proportion by cross product method.
Solution
The given pictures are in the ratio 2 : 5 and 4 : 3 respectively.
Here the extremes are 2 and 3 and the means are 5 and 4.
Product of extremes, ad = 2 × 3 = 6.
Product of means, bc = 5 × 4 = 20.
Thus, we find ad bc and hence 2 : 5 and 4 : 3 are not in proportion.
Try this
1. Fill the box by using cross product rule of two ratios 1/18 = 5 / _____
Answer: 1/8 = 5/40
2. Use the digits 1 to 9 only once and write as many ratios that are in proportion as possible. (For example : 2/4 = 3/6 )
Solution:
(i) 2 /3 = 4 /6
(ii) 4/3 = 8/6
Tags : Term 1 Chapter 3 | 6th Maths , 6th Maths : Term 1 Unit 3 : Ratio and Proportion
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
6th Maths : Term 1 Unit 3 : Ratio and Proportion : Proportion | Term 1 Chapter 3 | 6th Maths
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Saturday, June 25, 2022
HomeMathIdea of Time | Day and Night time | Components of the...
# Idea of Time | Day and Night time | Components of the Day
We are going to study the idea of time.
The solar shines in the course of the day. There may be mild in all places. We will see all the things round us.
When there is no such thing as a solar shine, it’s evening time. Every little thing is darkish within the evening and the celebs twinkle. We sleep at evening.
Day and Night time:
Components of the Day:
On a regular basis, we get up within the morning, we eat breakfast, dress and go to highschool, we play with our buddies within the night and we sleep within the evening.
Morning, night and evening are known as components of the day.
Circle the proper reply to indicate which a part of the day does this exercise.
Morning / Night / Night time Morning / Night / Night time
Morning / Night / Night time Morning / Night / Night time
Sequencing:
Some actions take longer time to finish than another actions. Taking a shower takes extra time than to brush tooth. Going to highschool takes extra time than to take a shower.
The next photos present completely different actions of a kid in a day. Quantity the images so as from 1 to 4 in response to the time taken by every exercise.
Days of the Week:
As soon as, there was a penguin who might communicate.Â
He knew all the times of the week.
He performed video games and sang with James calling all weekday names.
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday
Questions and Solutions on Idea of Time:
1. Which day comes subsequent? Circle the proper choice.
2. Circle the proper day to finish the sequence.
(i) (Wednesday / Sunday), Monday, Tuesday
(ii) Wednesday, (Saturday / Thursday), Friday
(iii) Friday, (Monday / Saturday), Sunday
(iv) Monday (Thursday / Tuesday), Wednesday
3. Write T for true and F for false:
(i) We play within the evening.
(ii) We’ve dinner at evening.
(iii) We go to mattress within the morning.
(iv) We go to highschool within the morning.
4. Fill within the blanks:
(i) There are ………….. days in every week.
(ii) ………….. is the primary day of this week.
(iii) ………….. twinkles at evening.
(iv) ………….. shines in the course of the day.
5. Take a look at this photos and organize them in a sequence by
numbering from 1 to 4.
6. Ranging from Monday organize the times of the week by numbering from 1 to 7.
7. Tick the exercise in every set which takes longer time than the opposite.
8.. Fill within the blanks:
(i) First day of the week is ………………
(ii) Tuesday comes after ………………
(iii) The final day of the week is ………………
(iv) Wednesday is between ……………… and ………………
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### Place Values Of Numbers
Place Values Of Numbers are used to say the exact position of figures in the numeric table.
In the numeric table, we have ones (units), tens, hundred, thousands etc
Before we can get the place values of given figures, we must first expand the figures
Samples
1. What is the product of the place values of 8 and 9 in 1849
Solution
1849 = 1000 + 800 + 40 + 9
The place values of 8 = 800
The place value of 9 = 9.
The product = 800 * 9 = 7200
2. Divide the place values of 2 by the place value of 4 in 1240
Solution :
1240 = 1000 + 200 +40 +0
The place values of 2=200
The place value of 4 = 40
Division = 200 / 40 = 5
Ans = 5
Class work
1. Sum up the place values of 8 and 4 in 18403
2. Find the difference in the place values of 4 and 9 in 133892
Calculate the product of the place values of 6 and 2 in 16238
# Place Values Of Numbers
Place Values Of Numbers are used to say the exact position of figures in the numeric table.
In the numeric table, we have ones (units), tens, hundred, thousands etc
Before we can get the place values of given figures, we must first expand the figures
Samples
1. What is the product of the place values of 8 and 9 in 1849
Solution
1849 = 1000 + 800 + 40 + 9
The place values of 8 = 800
The place value of 9 = 9.
The product = 800 * 9 = 7200
2. Divide the place values of 2 by the place value of 4 in 1240
Solution :
1240 = 1000 + 200 +40 +0
The place values of 2=200
The place value of 4 = 40
Division = 200 / 40 = 5
Ans = 5
Class work
1. Sum up the place values of 8 and 4 in 18403
2. Find the difference in the place values of 4 and 9 in 133892
Calculate the product of the place values of 6 and 2 in 16238
|
# How to Calculate Tangent Lines to Radius
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In mathematical terms, a "tangent" line is a straight line which touches a curve at one and only one point. Tangents to circles are unique in that they form right angles with their corresponding circle's radius. Determining the equation of a line tangent to a function is usually a complicated process which involves calculus. In the case of circles, however, mathematicians have simplified the calculation by pre-deriving an equation to calculate the slope of a line tangent to the circle at a specific point. In others words, you need only find its equation.
• Determine the slope of your tangent line using the (x,y) coordinates of the point where the line touches the circle. For circles, whose equations you write in the standard form of "x^2 + y^2 = r^2", where "r" is the circle's radius, you can represent the slope of a tangent line at any point with the equation "m = -(x/y)". If your circle's equation is "x^2 + y^2 = 100", and you want to find the slope of the tangent at "x = 6", plug "x" into the equation and solve for "y", then plug both coordinates into the equation to get your slope: "(6)^2 + y^2 = 100", or "36 + y^2 = 100". Therefore, "y^2 = 100 - 36 = 64", or "y = 8". The slope of your tangent at "x = 6" is therefore "-(6/8) = -3/4".
• Plug your slope and coordinates into the equation "m = (y - y0)/(x - x0)", where "m" is your slope, and "y0" and "x0" are the coordinates you used to arrive at it. For the example circle, this would be "-3/4 = (y - 8)/(x - 6)".
• Simplify your equation so it exists in the form "y = mx + b". For the example equation, do this as follows: "-3/4(x - 6) = [(y - 8)/(x - 6)](x - 6)"; or "-3/4x + 18/4 = y - 8, and -3/4x + 18/4 + 8 = y - 8 + 8", so "y = -3/4x + 18/4 + 8 = -3/4x + 18/4 + [8*(4/4)] = -3/4x + 18/4 + 32/4 = -3/4x + 50/4 = -3/4x + 12.5". The equation of the line tangent to the circle "x^2 + y^2 = 100" at the point (6,8) is therefore "y = -3/4x + 12.5".
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# APPLYING MULTIPLICATION AND DIVISION OF RATIONAL NUMBERS
## About "Applying multiplication and division of rational numbers"
Applying multiplication and division of rational numbers :
To solve many problems in math, often we may have to use multiplication and division of rational numbers.
In this section, we are going to learn how to multiply and divide rational numbers.
Note :
Rational numbers are usually denoted in the form "a/b".
That is, fraction form.
## Multiplying fractions
Stuff 1 :
To multiply a proper or improper fraction by another proper or improper fraction, we have to multiply the numerators and denominators.
For example,
2/3 x 4/5 = 8/15
1/3 x 7/11 = 7/33
Stuff 2 :
To multiply a proper or improper fraction by a whole number, we have to multiply the numerator of the fraction by the whole number.
For example,
(2/3) x 4 = 8/3
3 x 7/11 = 21/33
## Multiplying fractions - Examples
Example 1 :
Find 1/5 of 3/8
Solution :
1/5 of 3/8 = 1/5 x 3/8
1/5 of 3/8 = 3/40
Example 2 :
Find 2/9 x 3/2
Solution :
2/9 x 3/2 = 1/3
Example 3 :
Find 2/5 x 5 2/3
Solution :
2/5 x 5 2/3 = 2/5 x 17/3
2/5 x 5 2/3 = 34/15
2/5 x 5 2/3 = 2 4/15
Example 4 :
Find 6 3/4 x 7/10
Solution :
6 3/4 x 7/10 = 27/4 x 7/10
6 3/4 x 7/10 = (27x7) / (4x10)
6 3/4 x 7/10 = 189 / 40
6 3/4 x 7/10 = 4 29/40
Example 5 :
Lily reads 1/4 th of a book in 1 hour. How much of the book will she read in 3 1/2 hours ?
Solution :
The part of the book read by Lily in 1 hour = 1/4
So, the part of the book read by Lily in 3 1/2 hours is
= 3 1/2 x 1/4
= 7/2 x 1/4
= (7x1) / (2x4)
= 7/8
Hence, Lily reads 7/8 part of the book in 3 1/2 hours.
Example 6 :
David ate 1/3 of the pizza and gave 3/4 th of the remaining to his friend. What is the part of the pizza did David give his friend ?
Solution :
The part of the pizza eaten by David = 1/3
So, the remaining pizza = 2/3
The part of the given to his friend is
= 3/4 of the remaining
= 3/4 of 2/3
= 3/4 x 2/3
= 1/2
Hence, he gave 1/2 of the pizza to his friend.
## Dividing fractions
Step 1 :
When we divide a fraction by another fraction, first we have to change the division sign as multiplication.
Step 2 :
Take reciprocal of the second fraction.
Step 3 :
Multiply the two fractions. (Numerator times numerator and denominator times denominator).
## Dividing fractions - Examples
Example 1 :
Simplify : 2/5 ÷ 6/7
Solution :
Using the method explained above, we have
2/5 ÷ 6/7 = 2/5 x 7/6
2/5 ÷ 6/7 = (2x7) / (5x6)
2/5 ÷ 6/7 = 7/15
Example 2 :
Simplify : 7/5 ÷ 3/2
Solution :
Using the method explained above, we have
7/5 ÷ 3/2 = 7/5 x 2/3
7/5 ÷ 3/2 = (7x2) / (5x3)
7/5 ÷ 3/2 = 14/15
Example 3 :
Simplify : 5/12 ÷ 20/13
Solution :
Using the method explained above, we have
5/12 ÷ 20/13 = 5/12 ÷ 20/13
5/12 ÷ 20/13 = 5/12 x 13/20
5/12 ÷ 20/13 = (5x13) / (12x20)
5/12 ÷ 20/13 = 13/48
Example 4 :
Simplify : 2/19 ÷ 6 1/2
Solution :
First, let us convert the mixed number 6 1/2 in to improper fraction.
6 1/2 = 13/2
Now,m we have 2/19 ÷ 6 1/2 = 2/19 ÷ 13/2
Using the method explained above, we have
2/19 ÷ 13/2 = 2/19 x 13//2
2/19 ÷ 13/2 = (2x13) / (19x2)
2/19 ÷ 13/2 = 13 / 19
Example 5 :
David can make one pizza in 1/2 hour. How many pizzas can he make in 5/2 hours ?
Solution :
Time taken to make one pizza = 1/2 hour
No. of pizzas made in 5/2 hours = 5/2 ÷ 1/2
No. of pizzas made in 5/2 hours = 5/2 x 2/1
No. of pizzas made in 5/2 hours = (5x2) / (2x1)
No. of pizzas made in 5/2 hours = 5
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Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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# Write shaded portion as fraction. Arrange them in ascending and descendingorder using correct sign $‘’$ between the fractions(a)(b)(c) Show $\frac{2}{6} AcademicMathematicsNCERTClass 6 #### Complete Python Prime Pack 9 Courses 2 eBooks #### Artificial Intelligence & Machine Learning Prime Pack 6 Courses 1 eBooks #### Java Prime Pack 9 Courses 2 eBooks To do: We have to write shaded portion as fraction and arrange them in ascending and descending order using correct sign$‘’$between the fractions. Solution: (a) There are eight parts in the first circle and 3 parts are shaded. Therefore, The fraction representing the shaded portion$= \frac{3}{8}$There are eight parts in the second circle and 6 parts are shaded. Therefore, The fraction representing the shaded portion$= \frac{6}{8}$There are eight parts in the third circle and 4 parts are shaded. Therefore, The fraction representing the shaded portion$= \frac{4}{8}$There are eight parts in the fourth circle and 1 part is shaded. Therefore, The fraction representing the shaded portion$= \frac{1}{8}$Therefore,$\frac{1}{8}
(b) There are 9 parts in the first square and 8 parts are shaded.
Therefore,
The fraction representing the shaded portion $= \frac{8}{9}$
There are 9 parts in the second square and 4 parts are shaded.
Therefore,
The fraction representing the shaded portion $= \frac{4}{9}$
There are 9 parts in the third square and 3 parts are shaded.
Therefore,
The fraction representing the shaded portion $= \frac{3}{9}$
There are 9 parts in the fourth square and 6 parts are shaded.
Therefore,
The fraction representing the shaded portion $= \frac{6}{9}$
Therefore,
$\frac{3}{9} (c) We have to show $\frac{2}{6}, \frac{4}{6}, \frac{8}{6}$ and $\frac{6}{6}$ on the number line and put appropriate signs between the fractions given. Here,$\frac{6}{6}=1$and$1\frac{2}{6}=\frac{8}{6}$Therefore, From the figure,$\frac{5}{6} > \frac{2}{6}\frac{3}{6}
$\frac{1}{6}$\frac{8}{6} > \frac{5}{6}\$
Updated on 10-Oct-2022 13:32:55
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2019 AMC 10B Problems/Problem 3
Problem
In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$
Solution
60% of seniors do not play a musical instrument. If we denote x as the number of seniors, then $$\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500$$
$$\frac{3}{5}x + 150 - \frac{3}{10}x = 234$$ $$\frac{3}{10}x = 84$$ $$x = 84\cdot\frac{10}{3} \Rightarrow 280$$
Thus there are $500-x = 220$ non-seniors. Since 70% of the non-seniors play a musical instrument, $220 \cdot \frac{7}{10} = \boxed{B) 154}$
Solution 2
Let x be the number of seniors, and y be the number of non-seniors. Then $$\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234$$
Multiplying 10 to every term gives us $$6x + 3y = 2340$$
Also, $x + y = 500$ because there are 500 students in total.
Solving these system of equations give us $x = 280$, $y = 220$
Since 70% of the non-seniors play a musical instrument, we simply get 70% of 220, which gives us $\boxed{B) 154}$
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# Hands-on Critical Thinking: Equilateral Triangles
About a month ago, I was working with my Geometry class on equilateral triangles. Using a ruler and a compass, I demonstrated how to construct an equilateral triangle. In fact, there are a number of websites that demonstrate this the same way that I did (see below).
After having the students practice drawing their own equilateral triangles, I introduced them to a two-part activity.
First, I gave them each a blank piece of paper and asked them to construct an equilateral triangle with each side equal to six inches. With a paper that is 8.5 in x 11 in, this was a fairly simple task for the students. Many of them were able to complete in a short amount of time. Those that struggled with it were allowed to ask their neighbor for support.
Second, I gave them a second blank piece of paper and asked them to construct an equilateral triangle with each side equal to twelve inches. At first, they looked at the paper and thought it was impossible. They struggled with it for quite a while on their own, sketching out different designs. Then, they asked if they could talk to one of their neighbors to brainstorm and explore different ideas. Soon, the students ended up in groups of three and four, thinking through all the mathematics they knew, trying to figure out how it could be done. Finally, one group asked if they could construct the triangle in parts, cut it out, and piece it back together. That question changed the whole atmosphere of the class. Quickly, every group saw the solution and were eager to share their approach with the class.
Steps for constructing an equilateral triangle with sides equal to twelve inches.
Step One – Measure six inches from the long side of the paper on both ends and mark the paper.
Step Two – Draw a line through both marks.
Step Three – From the bottom right corner of the inner rectangle, measure twelve inches so the twelve-inch mark of the ruler crosses the opposite side (thereby creating a diagonal).
Step Four – Measure the distance from the left side that the diagonal crosses the top side and mark that measurement below. Draw a line through both marks.
Step Five – Cut the two adjacent triangles and piece together to form an equilateral triangle.
The great part about this activity was that it led perfectly into our discussion of 30-60-90 triangles. Before exploring the properties of the 30-60-90 triangles, the students were already able to see their use in constructing (and forming) other geometric figures.
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LCM the 7 and 11 is the smallest number amongst all common multiples that 7 and also 11. The first few multiples the 7 and also 11 room (7, 14, 21, 28, . . . ) and (11, 22, 33, 44, 55, 66, . . . ) respectively. There are 3 frequently used approaches to find LCM that 7 and also 11 - by prime factorization, by division method, and by listing multiples.
You are watching: What is the least common multiple of 7 and 11
1 LCM of 7 and 11 2 List that Methods 3 Solved Examples 4 FAQs
Answer: LCM of 7 and 11 is 77.
Explanation:
The LCM of two non-zero integers, x(7) and y(11), is the smallest confident integer m(77) that is divisible by both x(7) and also y(11) without any kind of remainder.
The methods to find the LCM of 7 and 11 are explained below.
By Listing MultiplesBy prime Factorization MethodBy division Method
### LCM of 7 and 11 by Listing Multiples
To calculate the LCM of 7 and also 11 by listing the end the usual multiples, we deserve to follow the given below steps:
Step 1: list a couple of multiples of 7 (7, 14, 21, 28, . . . ) and 11 (11, 22, 33, 44, 55, 66, . . . . )Step 2: The usual multiples from the multiples that 7 and also 11 are 77, 154, . . .Step 3: The smallest typical multiple of 7 and also 11 is 77.
∴ The least usual multiple the 7 and 11 = 77.
### LCM the 7 and also 11 by element Factorization
Prime administer of 7 and also 11 is (7) = 71 and also (11) = 111 respectively. LCM the 7 and 11 have the right to be acquired by multiplying prime components raised to their respective highest power, i.e. 71 × 111 = 77.Hence, the LCM that 7 and 11 by prime factorization is 77.
### LCM of 7 and 11 by division Method
To calculate the LCM the 7 and also 11 by the division method, we will certainly divide the numbers(7, 11) by their prime components (preferably common). The product of these divisors offers the LCM the 7 and also 11.
Step 3: proceed the procedures until only 1s room left in the critical row.
See more: How Many S Tiles In Words With Friends, Words With Friends
The LCM of 7 and 11 is the product of all prime numbers on the left, i.e. LCM(7, 11) by division method = 7 × 11 = 77.
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## 2.1 Modeling Rates of Change
### Summary
This introductory section is intended to introduce some of the issues involved in modeling rates of change, and to set the agenda for the rest of the chapter. Given a set of time and distance data for a falling object, average speeds are calculated. On the basis of mathematical analysis, the distance function is shown to be quadratic. The concept of an average rate of change is introduced for an arbitrary function, and is computed for the function e x.
By the end of your studying, you should know:
• How to calculate the average speed of an object from a table of time and distance data.
• With average speed called the first derived, how to calculate the second derived.
• The formula for the distance reached by an object falling near the earth.
• The definition of the average rate of change of a function over an interval.
• How to calculate the average rate of change (also called the first derived), and the second derived for a given function.
• The difference between average rate of change and instantaneous rate of change.
On-screen applet instructions: This applet shows the average rate of change at x = a, where a can be chosen from the pull down list. A single click in the graph gives an enlarged picture around the point x = a. Another click restores the original size. The value of h can be set on the slider. Click here for further instructions.
### Examples
While skiing, Josh rides a chair lift moving at 3 miles per hour up a hill. After he has skied down the hill, his average velocity for the round trip is 5 miles per hour. How fast did he ski?
Fill in the missing numbers in a derived table.
A automobile testing organization runs cars around a speed track to test performance. One test involves running a car around a 1200-meter track at high speed, and recording the individual lap times and the average speed for the entire test. On one occasion, the results are recorded, but are missing the final lap time. The first 4 laps show times of 37 seconds, 25 seconds, 29 seconds, 27 seconds, and the average speed for the test is recorded as 112 miles per hour. Show that even without knowing the final lap time, we can conclude an error has most likely been made in recording the results.
Average Velocity
Rate of Change
Derived Function
### Videos
See short videos of worked problems for this section.
### Exercises
See Exercises for 2.1 Modeling Rates of Change (PDF).
Work online to solve the exercises for this section, or for any other section of the textbook.
#### Interesting Application
Designing a speedometer: Does it measure average or instantaneous speed?
Software requirements: For best results viewing and interacting with this page, get the free software listed here.
Copyright © 2005 Donald L. Kreider, C. Dwight Lahr, Susan J. Diesel
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Associated Topics || Dr. Math Home || Search Dr. Math
### Sum of a Sequence
```
Date: 06/10/99 at 14:25:20
From: Bethany
Subject: Sum of an embedded arithmatic sequence
Our sequence is (3, 4, 6, 9, 13, ..., 499503). We know that there are
1000 terms in the sequence. How would we figure out a formula for the
sum of this sequence? We used Gauss' technique to find the number of
terms by plugging in like this: 3+(n-1)(n-1+1)/2 = 499503. We got
1000 terms. We know this is correct. How would we find the sum of the
sequence?
```
```
Date: 06/10/99 at 16:45:00
From: Doctor Anthony
Subject: Re: Sum of an embedded arithmatic sequence
Make up a difference table
n = 1 2 3 4 5 6 7 ........
f(n) = 3 4 6 9 13 18 24 ........
1st Diff 1 2 3 4 5 6
2nd Diff 1 1 1 1 1
If the second differences are constant, the nth term will be a
So we assume f(n) = an^2 + bn + c
n=1 a + b + c = 3
n=2 4a + 2b + c = 4
n=3 9a + 3b + c = 6
So we have 3 equations with 3 unknowns a, b, and c.
The solutions are a = 1/2, b = -1/2, c = 3
And therefore f(n) = (1/2)n^2 - (1/2)n + 3
Check with n = 5: f(n) = 12.5 - 2.5 + 3 = 13, which checks.
n = 1000 so we get: f(n) = 500000 - 500 + 3 = 499503, which also
checks.
So we must sum the following
1000
SUM[n^2/2 - n/2 + 3]
n=1
and using the standard formulae for SUM(n^2) and SUM(n) we get
(1/2)n(n+1)(2n+1)/6 - (1/2)n(n+1)/2 + 3n
n(n+1)[2n+1 - 3]
= ---------------- + 3n
12
n(n+1)(2n-2)
= ------------ + 3n
12
n(n+1)(n-1) n(n^2 - 1) + 18n
= ----------- + 3n = ----------------
6 6
Check when n = 3. The sum of the first 3 terms should be 13
3 x (9 - 1) + 54
---------------- = 13 so our formula is correct.
6
Finally, put n = 1000 into the formula
1000(1000^2 - 1) + 18000
------------------------ = 166669500
6
- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series
Search the Dr. Math Library:
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Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search
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## Perimeter and area
Perimeter and area formulas for various quadrilateral shapes are given below:
Perimeter is sum of the lengths of the enclosing sides of a figure.
Area is the amount of plane region enclosed by the sides of the figure.
1. Perimeter and area of a parallelogram
In a parallelogram, opposite sides are equal in length. Therefore,
Perimeter of the parallelogram ABCD = AB + BC + CD + AD = 2AB + 2BC
Area of a parallelogram = base × height = b × h
2. Perimeter and area of a rhombus
In a rhombus, all four sides are equal in length. Therefore,
Perimeter of a rhombus ABCD = 4a, where a is length of each side
Area of a rhombus = ½ × (product of diagonals)
3. Perimeter and area of a rectangle
In a rectangle, opposite sides are equal in length.
One side is called length, usually the longer, and the other is called breadth, the shorter side.
If l denotes length and b, the breadth, then
Perimeter of a rectangle = 2(l + b), and
Area of rectangle = l × b
4. Perimeter and area of a square
In a square, all four sides are equal in length.
If each side is of length a, then
Perimeter of square = 4a, and
Area of square = a2
5. Perimeter and area of a trapezoid
In a trapezoid, one pair of sides is parallel to each other. The other two sides may be equal in length, in which case the quadrilateral is called isosceles trapezoid, or else they may be unequal in length.
Perimeter of a trapezoid = sum of the four sides.
In the trapezoid, AE is the distance between the parallel sides AB and CD, So,
Area of a trapezoid =
½ × (sum of the parallel sides) × (distance between the parallel sides)
Perimeter and area Formulas for various plane figures
|
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 6.3: One-to-One Functions
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
We distinguish two special families of functions: the one-to-one functions and the onto functions. We shall discuss one-to-one functions in this section, and onto functions in the next.
Definition: Injection
A function $${f}:{A}\to{B}$$ is said to be one-to-one if
$x_1\neq x_2 \Rightarrow f(x_1)\neq f(x_2) \nonumber$
for all elements $$x_1,x_2\in A$$. A one-to-one function is also called an injection, and we call a function injective if it is one-to-one. A function that is not one-to-one is referred to as many-to-one.
Any well-defined function is either one-to-one or many-to-one. A function cannot be one-to-many because no element can have multiple images. The difference between one-to-one and many-to-one functions is whether there exist distinct elements that share the same image. There are no repeated images in a one-to-one function.
Example $$\PageIndex{1}\label{eg:oneonefcn-01}$$
The identity function on any nonempty set $$A$$ ${i_A}:{A}\to{A}, \qquad i_A(x)=x, \nonumber$ maps any element back to itself. It is clear that all identity functions are one-to-one.
Example $$\PageIndex{2}\label{eg:oneonefcn-02}$$
The function $$h : {A}\to{A}$$ defined by $$h(x)=c$$ for some fixed element $$c\in A$$, is an example of a constant function. It is a function with only one image. This is the exact opposite of an identity function. It is clearly not one-to-one unless $$|A|=1$$.
For domains with a small number of elements, one can use inspection on the images to determine if the function is one-to-one. This becomes impossible if the domain contains a larger number of elements.
In practice, it is easier to use the contrapositive of the definition to test whether a function is one-to-one:
$f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \nonumber$
Example $$\PageIndex{3}\label{eg:oneonefcn-03}$$
Is the function $$f : {\mathbb{R}}\to{\mathbb{R}}$$ defined by $$f(x)=3x+2$$ one-to-one?
Solution
Assume $$f(x_1)=f(x_2)$$, which means $3x_1+2 = 3x_2+2. \nonumber$ Thus $$3x_1=3x_2$$, which implies that $$x_1=x_2$$. Therefore $$f$$ is one-to-one.
exercise $$\PageIndex{1}\label{he:oneonefcn-01}$$
Determine whether the function $$g : {\mathbb{R}}\to{\mathbb{R}}$$ defined by $$g(x)=5-7x$$ is one-to-one.
exercise $$\PageIndex{2}\label{he:oneonefcn-02}$$
Determine whether the function $$h : {[2,\infty)}\to{\mathbb{R}}$$ defined by $$h(x)=\sqrt{x-2}$$ is one-to-one.
Interestingly, sometimes we can use calculus to determine if a real function is one-to-one. A real function $$f$$ is increasing if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2), \nonumber$ and decreasing if $x_1 < x_2 \Rightarrow f(x_1) > f(x_2). \nonumber$ Obviously, both increasing and decreasing functions are one-to-one. From calculus, we know that
• A function is increasing over an open interval $$(a,b)$$ if $$f'(x)>0$$ for all $$x\in(a,b)$$.
• A function is decreasing over an open interval $$(a,b)$$ if $$f'(x)<0$$ for all $$x\in(a,b)$$.
Therefore, if the derivative of a function is always positive, or always negative, then the function must be one-to-one.
Example $$\PageIndex{4}\label{eg:oneonefcn-04}$$
The function $$p : {\mathbb{R}}\to{\mathbb{R}}$$ defined by $p(x) = 2x^3-5 \nonumber$ is one-to-one, because $$p'(x)=6x^2>0$$ for any $$x\in\mathbb{R}^*$$. Likewise, the function $$q:{\big(-\frac{\pi}{2},\frac{\pi}{2}\big)}\to{\mathbb{R}}$$ defined by $q(x) = \tan x \nonumber$ is also one-to-one, because $$q'(x) = \sec^2x > 0$$ for any $$x\in \big(-\frac{\pi}{2},\frac{\pi}{2}\big)$$.
exercise $$\PageIndex{3}\label{he:oneonefcn-03}$$
Use both methods to show that the function $$k:{(0,\infty)}\to{\mathbb{R}}$$ defined by $$k(x) = \ln x$$ is one-to-one.
Example $$\PageIndex{5}\label{eg:oneonefcn-05}$$
The function $$h : {\mathbb{R}}\to{\mathbb{R}}$$ given by $$h(x)=x^2$$ is not one-to-one because some of its images are identical. For example, $$h(3) = h(-3) =9$$. It is a many-to-one function. Likewise, the absolute value function $$|x|$$ is not one-to-one.
The functions $$p:{[\,0,\infty)}\to{\mathbb{R}}$$ defined by $$p(x)=x^2$$ and $$q:{[\,0,\infty)}\to{\mathbb{R}}$$ defined by $$q(x)=|x|$$ are one-to-one. Whether a function is one-to-one depends not only on its formula, but also on its domain. Consequently, sometimes we may be able to convert a many-to-one function into a one-to-one function by modifying its domain.
Example $$\PageIndex{6}\label{eg:onetoone}$$
Construct a one-to-one function from $$[\,1,3\,]$$ to $$[\,2,5\,]$$.
Solution
There are many possible solutions. In any event, start with a graph. We can use a straight line graph. The domain $$[\,1,3\,]$$ lies on the $$x$$-axis, and the codomain $$[\,2,5\,]$$ lies on the $$y$$-axis. Hence the graph should cover the boxed region in Figure $$\PageIndex{1}$$.
All three graphs do not produce duplicate images. We need to cover all $$x$$-values from 1 to 3 in order for the function to be well-defined. This leaves only the first two graphs as legitimate examples.
To determine the formula for $$f$$, we need to derive the equation of the line. Take the first graph as our choice. The line joins the point $$(1,2)$$ to the point $$(3,4)$$. Thus, its equation is $\frac{y-2}{x-1} = \frac{4-2}{3-1} = 1. \nonumber$ The last step is to write the answer in the form of $$f(x)=\ldots\,$$. We have to express $$y$$ in terms of $$x$$. We find $$y = x+1$$. Hence, $f : {[\,1,3\,]}\to{[\,2,5\,]}, \qquad f(x) = x+1 \nonumber$ is an example of a one-to-one function.
exercise $$\PageIndex{4}\label{he:oneonefcn-04}$$
Construct a one-to-one function from $$[\,1,3\,]$$ to $$[\,2,5\,]$$ based on the second graph in Example 6.3.6.
exercise $$\PageIndex{5}\label{he:oneonefcn-05}$$
Construct a one-to-one function from $$[\,3,8\,]$$ to $$[\,2,5\,]$$.
example $$\PageIndex{7}\label{eg:gmod43}$$
Determine whether the function $$g : {\mathbb{Z}_{43}}\to{\mathbb{Z}_{43}}$$ defined by $g(x) \equiv 11x-5 \pmod{43} \nonumber$ is one-to-one.
Solution
Assume $$g(x_1)=g(x_2)$$. This means $11x_1 - 5 \equiv 11x_2 - 5 \pmod{43}, \nonumber$ which implies $11x_1 \equiv 11x_2 \pmod{43}. \nonumber$ Notice that $$4\cdot11=44\equiv1$$ (mod 43), hence $$11^{-1}\equiv4$$ (mod 43). Multiplying 4 to both sides of the last congruence yields $44 x_1 \equiv 44 x_2 \pmod{43}, \nonumber$ which is equivalent to, since $$44\equiv1$$ (mod 43), $x_1 \equiv x_2 \pmod{43}. \nonumber$ Therefore, $$x_1=x_2$$ in $$\mathbb{Z}_{43}$$. This proves that $$g$$ is one-to-one.
exercise $$\PageIndex{6}\label{he:oneonefcn-06}$$
Is the function $$h : {\mathbb{Z}_{15}}\to{\mathbb{Z}_{15}}$$ defined by $h(x) \equiv 4x-11 \pmod{15} \nonumber$ a one-to-one function?
exercise $$\PageIndex{7}\label{he:oneonefcn-07}$$
Show that the function $$k:{\mathbb{Z}_{15}}\to{\mathbb{Z}_{15}}$$ defined by $k(x) \equiv 5x-11 \pmod{15} \nonumber$ is not one-to-one by finding $$x_1\neq x_2$$ such that $$k(x_1)=k(x_2)$$.
Example $$\PageIndex{8}\label{eg:oneonefcn-08}$$
In the last exercise, we should not rely on the non-existence of $$5^{-1}$$ in $$\mathbb{Z}_{15}$$ to prove that $$k$$ is not one-to-one. One must consider the interaction between the domain, the codomain, and the definition of the function. For example, despite the fact that $$5^{-1}$$ does not exist in $$\mathbb{Z}_{15}$$, the function $$p :{\mathbb{Z}_3}\to{\mathbb{Z}_{15}}$$ defined by $p(x) \equiv 5x-11 \pmod{15} \nonumber$ is one-to-one, because $$p(0)=4$$, $$p(1)=9$$, and $$p(2)=14$$ are distinct images.
The last example illustrates the trickiness in a function with different moduli in its domain and codomain. Use caution when you deal with such functions! Sometimes, infinite sets also pose a challenge. Because there is an infinite supply of elements, we may obtain results that appear to be impossible for finite sets.
Example $$\PageIndex{9}\label{eg:oneonefcn-09}$$
The function $$f : {\mathbb{Z}}\to{\mathbb{Z}}$$ defined by $f(n) = \cases{ \frac{n}{2} & if n is even \cr \frac{n+1}{2} & if n is odd \cr} \nonumber$ is not one-to-one, because, for example, $$f(0)=f(-1)=0$$. The function $$g : {\mathbb{Z}}\to{\mathbb{Z}}$$ defined by $g(n) = 2n \nonumber$ is one-to-one, because if $$g(n_1)=g(n_2)$$, then $$2n_1=2n_2$$ implies that $$n_1=n_2$$.
exercise $$\PageIndex{8}\label{he:oneonefcn-08}$$
Show that the function $$h : {\mathbb{Z}}\to{\mathbb{N}}$$ defined by $h(n) = \cases{ 2n+1 & if n\geq0, \cr -2n & if n < 0, \cr} \nonumber$ is one-to-one.
Example $$\PageIndex{10}\label{eg:oneonefcn-10}$$
Let $$A$$ be the set of all married individuals from a monogamous community who are neither divorced nor widowed. Then the function $$s:{A}\to{A}$$ defined by $s(x) = \mbox{ spouse of } x \nonumber$ is one-to-one. The reason is, it is impossible to have $$x_1\neq x_2$$ and yet $$s(x_1)=s(x_2)$$.
## Summary and Review
• A function $$f$$ is said to be one-to-one if $$f(x_1) = f(x_2) \Rightarrow x_1=x_2$$.
• No two images of a one-to-one function are the same.
• To show that a function $$f$$ is not one-to-one, all we need is to find two different $$x$$-values that produce the same image; that is, find $$x_1\neq x_2$$ such that $$f(x_1)=f(x_2)$$.
Exercise $$\PageIndex{1}\label{ex:oneonefcn-01}$$
Which of the following functions are one-to-one? Explain.
• $$f : {\mathbb{R}}\to{\mathbb{R}}$$, $$f(x)=x^3-2x^2+1$$.
• $$g : {[\,2,\infty)}\to{\mathbb{R}}$$, $$f(x)=x^3-2x^2+1$$.
Exercise $$\PageIndex{2}\label{ex:oneonefcn-02}$$
Which of the following functions are one-to-one? Explain.
• $$p :{\mathbb{R}}\to{\mathbb{R}}$$, $$h(x)=e^{1-2x}$$.
• $$q :{\mathbb{R}}\to{\mathbb{R}}$$, $$p(x)=|1-3x|$$.
Exercise $$\PageIndex{3}\label{ex:oneonefcn-03}$$
Construct a one-to-one function $$f : {(1,3)}\to{(2,5)}$$ so that $$f : {[\,1,3)}\to{[\,2,5)}$$ is still one-to-one.
Exercise $$\PageIndex{4}\label{ex:oneonefcn-04}$$
Construct a one-to-one function $$g : {[\,2,5)}\to{(1,4\,]}$$.
Exercise $$\PageIndex{5}\label{ex:oneonefcn-05}$$
Determine which of the following are one-to-one functions.
1. $$f : {\mathbb{Z}}\to{\mathbb{Z}}$$; $$f(n)=n^3+1$$
2. $$g : {\mathbb{Q}}\to{\mathbb{Q}}$$; $$g(x)=n^2$$
3. $$h : {\mathbb{R}}\to{\mathbb{R}}$$; $$h(x)=x^3-x$$
4. $${k} : {\mathbb{R}}\to{\mathbb{R}}$$; $$k(x)=5^x$$
Exercise $$\PageIndex{6}\label{ex:oneonefcn-06}$$
Determine which of the following are one-to-one functions.
1. $$p :{\wp(\{1,2,3,\ldots,n\})}\to{\{0,1,2,\ldots,n\}}$$; $$p(S)=|S|$$
2. $$q :{\wp(\{1,2,3,\ldots,n\})}\to{\wp(\{1,2,3,\ldots,n\})}$$; $$q(S)=\overline{S}$$
Exercise $$\PageIndex{7}\label{ex:oneonefcn-07}$$
Determine which of the following functions are one-to-one.
1. $${f_1}:{\{1,2,3,4,5\}}\to{\{a,b,c,d\}}$$; $$f_1(1)=b$$, $$f_1(2)=c$$, $$f_1(3)=a$$, $$f_1(4)=a$$, $$f_1(5)=c$$
2. $${f_2}:{\{1,2,3,4\}}\to{\{a,b,c,d,e\}}$$; $$f_2(1)=c$$, $$f_2(2)=b$$, $$f_2(3)=a$$, $$f_2(4)=d$$
3. $${f_3}:{\mathbb{Z}}\to{\mathbb{Z}}$$; $$f_5(n)=-n$$
4. $$f_4: \mathbb{Z} \rightarrow \mathbb{Z}$$; $$f_4(n) = \cases{ 2n & if n < 0, \cr -3n & if n\geq0,\cr}$$
Exercise $$\PageIndex{8}\label{ex:oneonefcn-08}$$
Determine which of the following functions are one-to-one.
1. $${g_1}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$$; $$g_1(1)=b$$, $$g_1(2)=b$$, $$g_1(3)=b$$, $$g_1(4)=a$$, $$g_1(5)=d$$
2. $${g_2}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$$; $$g_2(1)=d$$, $$g_2(2)=b$$, $$g_2(3)=e$$, $$g_2(4)=a$$, $$g_2(5)=c$$
3. $$g_3: \mathbb{N} \rightarrow \mathbb{N}$$; $$g_3 (n) = \cases{ \frac{n+1}{2} & if n is odd \cr \frac{n}{2} & if n is even \cr}$$
4. $$g_4: \mathbb{N} \rightarrow \mathbb{N}$$; $$g_4 (n) = \cases{ n+1 & if n is odd \cr n-1 & if n is even \cr}$$
Exercise $$\PageIndex{9}\label{ex:oneonefcn-09}$$
List all the one-to-one functions from $$\{1,2\}$$ to $$\{a,b,c,d\}$$.
Hint
List the images of each function.
Exercise $$\PageIndex{10}\label{ex:oneonefcn-10}$$
Is it possible to find a one-to-one function from $$\{1,2,3,4\}$$ to $$\{1,2\}$$? Explain.
Exercise $$\PageIndex{11}\label{ex:oneonefcn-11}$$
Determine which of the following functions are one-to-one.
1. $$f : {\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$h(n)\equiv 3n$$ (mod 10).
2. $$g : {\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$g(n)\equiv 5n$$ (mod 10).
3. $$h : {\mathbb{Z}_{36}}\to{\mathbb{Z}_{36}}$$; $$h(n)\equiv 3n$$ (mod 36).
Exercise $$\PageIndex{12}\label{ex:oneonefcn-12}$$
Determine which of the following functions are one-to-one.
1. $$r:{\mathbb{Z}_{36}}\to{\mathbb{Z}_{36}}$$; $$r(n)\equiv 5n$$ (mod 36).
2. $$s:{\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$s(n)\equiv n+5$$ (mod 10).
3. $$t:{\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$t(n)\equiv 3n+5$$ (mod 10).
Exercise $$\PageIndex{13}\label{ex:oneonefcn-13}$$
Determine which of the following functions are one-to-one.
1. $$\alpha:{\mathbb{Z}_{12}}\to{\mathbb{Z}_{ 7}}$$; $$\alpha(n)\equiv 2n$$ (mod 7).
2. $$\beta :{\mathbb{Z}_{ 8}}\to{\mathbb{Z}_{12}}$$; $$\beta (n)\equiv 3n$$ (mod 12).
3. $$\gamma:{\mathbb{Z}_{ 6}}\to{\mathbb{Z}_{12}}$$; $$\gamma(n)\equiv 2n$$ (mod 12).
4. $$\delta:{\mathbb{Z}_{12}}\to{\mathbb{Z}_{36}}$$; $$\delta(n)\equiv 6n$$ (mod 36).
Exercise $$\PageIndex{14}\label{ex:oneonefcn-14}$$
Give an example of a one-to-one function $$f$$ from $$\mathbb{N}$$ to $$\mathbb{N}$$ that is not the identity function.
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The branch of mathematics that deals with the analysis and manipulation of data and numbers is called statistics. Its core functions are collection, analysis, interpretation, presentation, and organization of data. The concepts of statistics have applications across multiple disciplines such as business and data analysis, banking and finance, business management and development, etc.
#### Fundamentals of Statistics
The measure of central tendency and the measure of dispersion are among the basics of statistics. There are three central tendencies (mean, median, and mode) and two dispersions (variance and standard deviation).
• Central Tendencies: The process of measuring the central tendency involves identifying the central point within a certain set of data. The three ways of finding the midpoint are:
• Mean: The most popular measure of central tendency is meanIt is denoted by x and is calculated by adding up all the values in a given set of data and dividing the sum by the number of values present in that set. To derive the formula we can take a set of data where the elements are x1x2x3x4………xn and n is the number of elements.
Therefore,
x = x1+x2+x3+x4+………xnn
Or,
x = xn
*NOTE: Here, ‘’ is a Greek capital letter, pronounced sigma, which means the sum of the given entities.
• Median
A median is the middle element of a set of data that is arranged in ascending order of magnitude. It is denoted by ‘M’. To find the median, one has to arrange the data in ascending order first. Now, there can be two conditions:
• If the number of elements is odd, the middle element becomes the median. The formula of median becomes
M =n+12th term.
For example, in the set of the first seven natural numbers, i.e, 1,2,3,4,5,6,7. The element ‘4’ is the median of the set.
• If the number of elements is even, the average of the middle two elements is taken as the median. The formula of median becomes
M = n2th term +n2+1th term2
For example, in the set of the first 10 even numbers, i.e, 2,4,6,8,10,12,14,16,18,20, the average of ‘10’ and ‘12’ will be the median.
Here, the median is 10+122= 11.
• Mode
The element which has the highest frequency in a certain set of data is known as the mode or the modal value. Let’s understand this concept with the help of one example.
Set A = 2,2,2,3,7,7,9,11
The mode of this set is 2 because it is occurring for the most number of times.
*NOTE: There can be more than one mode in a set of data. For two and three modes in a set of data, the terms bimodal and trimodal are used. For more than three modes, the term multimodal is used.
• Dispersions: The concept of dispersion enables us to determine the extent to which a certain set of data is expanded or compressed. It helps us to interpret the variable nature of the given data. There are two major methods of computing dispersion: variance and standard deviation. These concepts are interlinked with each other through the principles of powers and roots. Standard deviation is the dispersion of a group of elements from the mean. It is denoted by ‘’. The variance shows how distant each element of a set of data is from its mean. Since it is the square of standard deviation, it is denoted by ‘2’. Let’s study the formulas of dispersion
= i=1n-1(xi–x)2n-1
&
2= i=1n-1(xi–x)2n-1
Here,
n = number of elements
xi = ith element
x = arithmetic mean
*NOTE: Here, i=1n-1 asserts the addition function for xi with i ranges from 1 to n-1.
X
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# Fundamental Theorem of Algebra
## Find all zeroes of a polynomial, including complex solutions
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Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra is really the foundation on which most study of Algebra is built. In simple terms it says that every polynomial has zeros. That means that every polynomial can be factored and set equal to zero (the Factorization Theorem).
That is an extremely broad statement! Every polynomial can be factored? What about functions like \begin{align*}x^{2} = -4\end{align*}? What about crazy big ones, like \begin{align*}x^{6} - 23x^{5} + \frac{1}{246}x^{4}-23x^{3}\end{align*}?
### Fundamental Theorem of Algebra
Here are four important theorems in the study of complex zeros of polynomial functions:
#### The Fundamental Theorem of Algebra
If \begin{align*}f(x)\end{align*} is a polynomial of degree \begin{align*}n\ge 1\end{align*}, then \begin{align*}f(x)\end{align*} has at least one zero in the complex number domain. In other words, there is at least one complex number \begin{align*}c\end{align*} such that \begin{align*}f(c)=0\end{align*}.
There is no rigorous proof for the fundamental theorem of algebra. Some mathematicians even believe that such proof may not exist. However, the theorem is considered to be one of the most important theorems in mathematics. A corollary of this important theorem is the factorization theorem.
#### The Factorization Theorem
If
\begin{align*}f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\end{align*}
where \begin{align*}a_{n} \ne 0\end{align*}, and \begin{align*}n\end{align*} is a positive integer, then
\begin{align*}f(x)=a_{n}(x-c_{1})(x-c_{2})\cdots(x-c_{0})\end{align*}
where the numbers \begin{align*}c_{i}\end{align*} are complex numbers.
#### The \begin{align*}n-\end{align*}Roots Theorem
If \begin{align*}f(x)\end{align*} is a polynomial of degree \begin{align*}n\end{align*}, where \begin{align*}n\ne 0\end{align*}, then \begin{align*}f(x)\end{align*} has, at most, \begin{align*}n\end{align*} zeros.
Notice that this theorem does not restrict that the zeros must be distinct. In other words, multiplicity of the zeros is allowed. For example, the quadratic equation \begin{align*}f(x)=x^{2}+6x+9\end{align*} has one zero, -3, and we say that the function has -3 as a double zero or one zero with multiplicity \begin{align*}k=2\end{align*}. In general, if
\begin{align*}f(x)=(x-c)^{k}q(x)\quad\text{and} \quad q(c)\ne0\end{align*}
then \begin{align*}c\end{align*} is a zero of the polynomial \begin{align*}f\end{align*} and of multiplicity \begin{align*}k\end{align*}. For example,
\begin{align*}f(x)=(x-2)^{3}(x+5)\end{align*}
has 2 as one zero with \begin{align*}k=3\end{align*} and -5 as a zero with \begin{align*}k=1\end{align*}.
#### Conjugate Pairs Theorem
If \begin{align*}f(z)\end{align*} is a polynomial of degree \begin{align*}n\end{align*}, with \begin{align*}n\ne0\end{align*} and with real coefficients, and if \begin{align*}f(z_{0})=0\end{align*}, where \begin{align*}z_{0}=a+bi\end{align*}, then \begin{align*}f(\overline{z_{0}})=0\end{align*}. Where \begin{align*}\overline{z_{0}}\end{align*} is the complex conjugate of \begin{align*}z_{0}\end{align*}.
This is a fascinating theorem! It says basically that if a complex number is a zero of a polynomial, then its complex conjugate must also be a zero of the same polynomial. In other words, complex roots (or zeros) exist in conjugate pairs for the same polynomial. For example, the polynomial function
\begin{align*}f(x)=x^{2}-2x+2\end{align*}
has two zeros: one is the complex number \begin{align*}1+i\end{align*}. By the conjugate pairs theorem (also called the conjugate root theorem), \begin{align*}1-i\end{align*} is also a zero of \begin{align*}f(x)=x^{2}-2x+2\end{align*}. We can easily prove that by multiplication:
\begin{align}\left[x-(1+i)\right]\left[x-(1-i)\right] &= (x-1-i)(x-1+i)\\ &= x^2 -x +xi - x +1 - i - xi + i +1\\ &= x^2 -2x + 2 \end{align}
### Examples
#### Example 1
Write \begin{align*}g(x)=x^{2}+x+1\end{align*} as a complex polynomial in factored form.
Notice that \begin{align*}g(x)\end{align*} has no real roots. You can see this in the graph of \begin{align*}g(x)\end{align*}, or by looking at the discriminant, \begin{align*}b^{2}-4ac=1-4=-3\end{align*}.
Using the quadratic formula, the roots of \begin{align*}g(x)\end{align*} are
\begin{align}x &= \frac{-b \pm \sqrt{b^2-4ac}}{2z}\\ &= \frac{-1 \pm \sqrt{-3}}{2}\\ &= - \frac{1}{2} + \frac{\sqrt{3}}{2}i \ or \ - \frac{1}{2} - \frac{\sqrt{3}}{2}i \end{align}
Finally, writing \begin{align*}g(x)\end{align*} in factored form,
\begin{align*}g(x)=\left[x-\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\right]\left[x-\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\right]\end{align*}
#### Example 2
What is the form of the polynomial \begin{align*}f(x)\end{align*} if it has the following numbers as zeros: \begin{align*}\frac{-1}{3}, 1-i\end{align*} and \begin{align*}2i\end{align*}?
Since the numbers \begin{align*}2i\end{align*} and \begin{align*}1+i\end{align*} are zeros, then they are roots of \begin{align*}f(x)=0\end{align*}. It follows that they must satisfy the conjugate root theorem. Thus \begin{align*}-2i\end{align*} and \begin{align*}1-i\end{align*} must also be roots of \begin{align*}f(x)\end{align*}. Therefore,
\begin{align*}f(x)= \left ( x+\frac{1}{3} \right )[x-(1-i)][x-(1+i)][x-(2i)][x-(-2i)]\end{align*}
Simplifying,
\begin{align*}f(x) = \left ( x+\frac{1}{3} \right )(x-1+i)(x-1-i)(x-2i)(x+2i)\end{align*}
After multiplying we get,
\begin{align*}f(x)=\frac{1}{3}(3x^{5}-5x^{4}+13x^{3}-19x^{2}+4x+4)\end{align*}
which is a fifth degree polynomial. Notice that the total number of zeros is also 5.
#### Example 3
Find the multiplicity of the zeros of the following polynomial:
\begin{align*}g(x)=x^{4}-6x^{3}+18x^{2}-54x+81\end{align*}
With the help of the rational zero theorem and synthetic division, we find that \begin{align*}x=3\end{align*} is a zero of \begin{align*}g(x)\end{align*},
\begin{align*} \ 3 \ \big ) \overline{1 \ -6 \ \ 18 \ -54 \ \ \ \ 81\;}\\ \quad \ \ \underline{\downarrow \ \ \ 3 \ -9 \ \ \ \ 27 \ -81}\\ \quad \ \ 1 \ -3 \ \ \ 9 \ -27 \ \ \ \ \ 0\end{align*}
\begin{align*}g(x)=x^{4}-6x^{3}+18x^{2}-54x+81=(x-3)(x^{3}-3x^{2}+9x-27)\end{align*}
Using synthetic division on the quotient, we find that 3 is again a zero:
\begin{align*} \ 3 \ \big ) \overline{1 \ -3 \ \ 9 \ -27}\\ \quad \ \ \underline{\downarrow \ \ \ 3 \ \ \ 9 \ -27}\\ \quad \ \ 1 \ \ \ \ 0 \ \ \ 9 \ \ \ \ \ 0\end{align*}
or from the \begin{align*}n-\end{align*}Roots Theorem (Theorem 3), we write our solution as
\begin{align}g(x) &= (x-3)(x-3)(x^2+9)\\ &= (x-3)^2(x-3i)(x+3i) \end{align}
So 3 is a double zero \begin{align*}(k=2)\end{align*} and \begin{align*}3i\end{align*} and \begin{align*}-3i\end{align*} are each of \begin{align*}k=1\end{align*}.
#### Example 4
Identify or estimate the values of the zeros from the following equation and state their multiplicities: \begin{align*}y = (x + 2)^2(x - 1)\end{align*}
To identify the roots and their multiplicities:
First, set the function equal to 0: \begin{align*}(x + 2)(x + 2)(x - 1) = 0\end{align*}
The roots then are \begin{align*}x = 1\end{align*} and \begin{align*}x = -2\end{align*}
Since the \begin{align*}x = -2\end{align*} root appears twice, it has a multiplicity of 2, whereas the \begin{align*}x = 1\end{align*} root appears only once, so its multiplicity is 1.
Note: The graph of this function (shown below) will pass through the axis at the root \begin{align*}x = 1\end{align*} and bounce off the axis at the root \begin{align*}x = -2\end{align*}.
If a root has an even multiplicity, it will "bounce" off of the axis, and if it has an odd multiplicity, it will pass through.
#### Example 5
Identify or estimate the values of the zeros from the following graph and state their multiplicities.
A 4th degree equation:
Recall that the roots are locations where the graph contacts the \begin{align*}x-\end{align*}axis. The image indicates this happens at \begin{align*}x= -3, -2, \end{align*} and \begin{align*}1\end{align*}.
Applying the rule from the solution to question 1 tells us that the root "-3" has an even multiplicity since it bounces off of the axis. The other 2 roots have odd multiplicities since they pass through.
The question specifies that this is a 4th degree equation; therefore, the root "-3" has a multiplicity of 2 and the other two roots displayed each have a multiplicity of 1.
#### Example 6
Identify or estimate the values of the zeros from the following equation and state their multiplicities: \begin{align*}g(x) = (x^2 + 6x + 9)(x^3 + 6x^2 + 12x + 8)\end{align*}.
First, factor the polynomial: \begin{align*}g(x) = (x + 3)^2(x + 2)^3\end{align*}
The roots are \begin{align*}x = -2\end{align*} and \begin{align*}x = -3\end{align*}.
The multiplicities stem from the multiples of the same binomial, so the root \begin{align*}x = -2\end{align*} has a multiplicity of 3 and \begin{align*}x = -3\end{align*} has a multiplicity of 2.
A graph of this equation would show the line passing through \begin{align*}x = -3\end{align*} and bouncing off at \begin{align*}x = -2\end{align*}.
#### Example 7
Find a polynomial function with real coefficients that has the following values as its zeros: \begin{align*}2, 3, -3, 1\end{align*}
To find a function with the specified zeros:
Recall that the zeros of a function are the additive inverse of the constant term in each binomial of the factored polynomial, giving: \begin{align*}(x - 2)(x - 3)(x + 3)(x - 1)\end{align*}
Distribute: \begin{align*}(x^2 - 5x + 6)(x^2 +2x - 3)\end{align*}
Multiply the polynomials: \begin{align*}(x^4 - 3x^3 -7x^2 +27x - 18)\end{align*}
\begin{align*}\therefore (x^4 - 3x^3 -7x^2 +27x - 18)\end{align*} is the specified polynomial.
### Review
In questions 1-5, find a polynomial function with real coefficients that has the given numbers as its zeros.
1. \begin{align*}1, 2, i\end{align*}
2. \begin{align*}2, 2, 1-i\end{align*}
3. \begin{align*}i, i, 0, 2i\end{align*}
4. \begin{align*}1, 1, \left(1-i\sqrt{3}\right)\end{align*}
5. \begin{align*}0, 0, 2i\end{align*}
6. If \begin{align*}i-1\end{align*} is a root of the polynomial \begin{align*}f(x)=x^{4}+2x^{3}-4x-4\end{align*}, find all other roots of \begin{align*}f\end{align*}.
7. If \begin{align*}-2i\end{align*} is a zero of the polynomial \begin{align*}f(x)=x^{4}+x^{3}-2x^{2}+4x-24\end{align*}, find all other zeros of \begin{align*}f\end{align*}.
In questions 8-10, determine whether the given number is a zero of the given polynomial. If so, determine its multiplicity.
1. \begin{align*}f(x)=9x^{4}-12x^{3}+13x^{2}-12x+4, x=\frac{2}{3}\end{align*}
2. \begin{align*}f(x)=x^{4}-4x^{3}+5x^{2}-4x+4, x=2\end{align*}
3. \begin{align*}f(x)=3x^{5}-4x^{4}+2x^{3}-\frac{3}{4}x^{2}+2x+12, x=-\frac{2}{3}\end{align*}
For questions 11 - 15, sketch the graph, properly indicating multiplicities.
1. \begin{align*}g(x) = x^2(x - 1)(x - 3)(x + 2)(x + 4)^2\end{align*}
2. \begin{align*}f(x) = -x^2(x - 3)^3(x - 1)(x - 2)\end{align*}
3. \begin{align*}f(x) = -x(x - 1)(x + 2)^3\end{align*}
4. \begin{align*}g(x) = x^3(x + 3)^4(x - 2)\end{align*}
5. \begin{align*}f(x) = (x + 3)^2(x - 1)(x + 1)\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Complex Conjugate Complex conjugates are pairs of complex binomials. The complex conjugate of $a+bi$ is $a-bi$. When complex conjugates are multiplied, the result is a single real number.
complex number A complex number is the sum of a real number and an imaginary number, written in the form $a + bi$.
conjugate pairs theorem The conjugate pairs theorem states that if $f(z)$ is a polynomial of degree $n$, with $n\ne0$ and with real coefficients, and if $f(z_{0})=0$, where $z_{0}=a+bi$, then $f(z_{0}^{*})=0$. Where $z_{0}^{*}$ is the complex conjugate of $z_{0}$.
fundamental theorem of algebra The fundamental theorem of algebra states that if $f(x)$ is a polynomial of degree $n\ge 1$, then $f(x)$ has at least one zero in the complex number domain. In other words, there is at least one complex number $c$ such that $f(c)=0$. The theorem can also be stated as follows: an $n^{th}$ degree polynomial with real or complex coefficients has, with multiplicity, exactly $n$ complex roots.
Imaginary Number An imaginary number is a number that can be written as the product of a real number and $i$.
Imaginary Numbers An imaginary number is a number that can be written as the product of a real number and $i$.
Multiplicity The multiplicity of a term describes the number of times the given term acts as a zero of the given function.
Polynomial A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.
Roots The roots of a function are the values of x that make y equal to zero.
Synthetic Division Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.
Zero The zeros of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.
Zeroes The zeroes of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.
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### Equivalent Fractions Worksheet
#### Quick Notes
The lesson on equivalent fractions explained the basics needed to complete this worksheet. Here's a short summary on what you should know:
1. Equivalent fractions are fractions that have the same value.
2. We can make use the ideas behind 'equivalent fractions' to change the denominator of a fraction. This is useful when we need to add or subtract fractions that have unlike denominators.
3. We can get an equivalent fraction by multiplying/dividing the numerator and denominator of the fraction with the same number.
#### Worksheets
The following worksheets are available:
• Worksheet 1 - Using multiplication to find equivalent fractions
• Worksheet 2 - Using division to find equivalent fractions
#### Quick Notes
The lesson on adding fractions covered the basics in detail. Here's a short summary on what you should know:
1. We can only add fractions that have like denominators
2. If the fractions have unlike denominators, we can make use of 'equivalent fractions' to make them the same
#### Worksheets
The following worksheets are available:
• Worksheet 1 - Adding fractions with like denominators
• Worksheet 2 - Adding fractions with unlike denominators
### Subtracting Fractions Worksheet
#### Quick Notes
The lesson on subtracting fractions explained the basic ideas behind subtracting fractions. Here's a short summary on what you should know:
1. To subtract fractions, the fractions must have like denominators
2. If the fractions have unlike denominators, use 'equivalent fractions' to make them the same
#### Worksheets
The following worksheets are available:
• Worksheet 1 - Subtracting fractions with like denominators
• Worksheet 2 - Subtracting fractions with unlike denominators
### Multiplying Fractions Worksheet
#### Quick Notes
The Multiplying fraction lesson explained on how to multiply fractions in detail. Here's a short summary on what you should know:
1. When multiplying, the denominators do not have to be the same
2. To multiply fractions, first multiply the numerators. Then, multiply the denominators
3. Before multiplying, we can simplify the fractions first if there is one or more common factor. This will make the multiplication easier
#### Worksheets
The following worksheets are available:
• Worksheet 1 - Basic questions on multiplying fractions
• Worksheet 2 - Using common factor to simplify the fractions before multiplying
### Dividing Fractions Worksheet
#### Quick Notes
The lesson on dividing fractions explained the basic required to divide fractions. Here's a short summary on what you should know:
1. To get the reciprocal of a fraction, we just need to swap the numerator and denominator of that fraction.
2. To divide fractions, first we change the division sign to a multiplication sign. Then, we change the divisor into its reciprocal and proceed from there.
#### Worksheets
The following fraction worksheets on division are available:
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# NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3
NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3
### NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3
Ex 8.3 Class 8 Maths Question 1.
Calculate the amount and compound interest on
Solution:
(a) Given:
P = ₹ 10,800, n = 3 years,
Total CI = ₹ 3780 + ₹ 1089 = ₹ 4,869
Amount = P + I = ₹ 21,780 + ₹ 1,089 = ₹ 22,869
Hence, the amount = ₹ 22,869
and CI = ₹ 4,869
CI = A – P = ₹ 70,304 – ₹ 62,500 = ₹ 7,804
Hence, amount = ₹ 70304 and CI = ₹ 7804
(d) Given: P = ₹ 8,000, n = 1 years R = 9% per annum compounded half yearly
Since, the interest is compounded half yearly n = 1 × 2 = 2 half years
CI = A – P = ₹ 8,736.20 – ₹ 8,000 = ₹ 736.20
Hence, the amount = ₹ 8736.20 and CI = ₹ 736.20
(e) Given: P = ₹ 10,000, n = 1 year and R = 8% pa compounded half yearly
Since the interest is compounded half yearly n = 1 × 2 = 2 half years
CI = A – P = ₹ 10,816 – ₹ 10,000 = ₹ 816
Hence the amount = ₹ 10,816 and Cl = ₹ 816
Ex 8.3 Class 8 Maths Question 2.
Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd
year amount for 4/12 years).
Solution:
Given:
P = ₹ 26,400
R = 15% p.a. compounded yearly
n = 2 years and 4 months
Amount after 2 years and 4 months = ₹ 34,914 + ₹ 1745.70 = ₹ 36,659.70
Hence, the amount to be paid by Kamla = ₹ 36,659.70
Ex 8.3 Class 8 Maths Question 3.
Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
For Fabina: P = ₹ 12,500, R = 12% p.a. and n = 3 years
Difference between the two interests = ₹ 4500 – ₹ 4137.50 = ₹ 362.50
Hence, Fabina pays more interest by ₹ 362.50.
Ex 8.3 Class 8 Maths Question 4.
I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Solution:
Given: P = ₹ 12,000, R = 6% p.a., n = 2 years
Difference between two interests = ₹ 1483.20 – ₹ 1440 = ₹ 43.20
Hence, the extra amount to be paid = ₹ 43.20
Ex 8.3 Class 8 Maths Question 5.
Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
Solution:
(i) Given: P = ₹ 60,000, R = 12% p.a. compounded half yearly
Hence, the required amount = ₹ 67416
Ex 8.3 Class 8 Maths Question 6.
Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after
Since the interest is compounded annually
Difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210
Ex 8.3 Class 8 Maths Question 7.
Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the third year.
Solution:
(i) Given: P = ₹ 8,000, R = 5% p.a.
and n = 2 years
Hence, interest for the third year = ₹ 441
Ex 8.3 Class 8 Maths Question 8.
Total interest = ₹ 1,000 + ₹ 550 = ₹ 1,550
Difference between the two interests = ₹ 1,576.25 – ₹ 1,550 = ₹ 26.25
Hence, the interest will be ₹ 26.25 more when compounded half yearly than the interest when compounded annually.
Ex 8.3 Class 8 Maths Question 9.
Hence, the required amount = ₹ 4913
Ex 8.3 Class 8 Maths Question 10.
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
(i) Find the population in 2001.
(ii) What would be its population in 2005?
Solution:
(i) Given: Population in 2003 = 54,000
Rate = 5% pa
Time = 2003 – 2001 = 2 years
Population in 2003 = Population in 2001
Ex 8.3 Class 8 Maths Question 11.
In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Given: Initial count of bacteria = 5,06,000
Rate = 2.5% per hour
n = 2 hours
Number of bacteria at the end of 2 hours = Number of count of bacteria initially
Thus, the number of bacteria after two hours = 5,31,616 (approx).
Ex 8.3 Class 8 Maths Question 12.
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Given: Cost price of the scooter = ₹ 42,000
Rate of depreciation = 8% p.a.
Time = 1 year
Final value of the scooter
Hence, the value of scooter after 1 year = ₹ 38,640.
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# What are the characteristics of the graphs of the sine and cosine functions?
## What are the characteristics of the graphs of the sine and cosine functions?
The sine and cosine functions have several distinct characteristics:
• They are periodic functions with a period of.
• The domain of each function is and the range is.
• The graph of is symmetric about the origin, because it is an odd function.
• The graph of is symmetric about the axis, because it is an even function.
How do you determine if a graph is sine or cosine?
In a cosine graph, a positive or negative number vertically flips the graph and determines whether the graph starts at the maximum (if it’s positive) or minimum (if it’s negative). For a sine graph, a positive or negative number vertically flips the graph like it does with a cosine graph.
### How do you graph a sine function?
Because the graph of the sine function is being graphed on the x–y plane, you rewrite this as f(x) = sin x where x is the measure of the angle in radians….
1. Find the values for domain and range.
2. Calculate the graph’s x-intercepts.
3. Calculate the graph’s maximum and minimum points.
4. Sketch the graph of the function.
What is the equation of a sine function?
To write a sine function you simply need to use the following equation: f(x) = asin(bx + c) + d, where a is the amplitude, b is the period (you can find the period by dividing the absolute value b by 2pi; in your case, I believe the frequency and period are the same), c is the phase shift (or the shift along the x-axis …
#### What is the sine graph?
The graphs of functions defined by y = sin x are called sine waves or sinusoidal waves. Notice that the graph repeats itself as it moves along the x-axis. This graph repeats every 6.28 units or 2 pi radians. It ranges from -1 to 1; half this distance is called the amplitude.
What is a COS graph?
To graph the cosine function, we mark the angle along the horizontal x axis, and for each angle, we put the cosine of that angle on the vertical y-axis. The result, as seen above, is a smooth curve that varies from +1 to -1. As you do so, the point on the graph moves to correspond with the angle and its cosine.
## What are sine graphs used for?
Sine and cosine functions can be used to model many real-life scenarios – radio waves, tides, musical tones, electrical currents.
What are the three characteristics of a sine wave?
Sinusoidal Amplitude, Frequency, and Phase All sinusoidal signals have the same general shape, but they are not identical. The three characteristics that separate one sinusoid from another are amplitude, frequency, and phase.
### What are the parameters of a sine wave?
the parameter A — called the amplitude, the parameter p — called the period, the parameter d called the shift, and. the parameter B — called the center.
What produces a sine wave?
Sinusoidal Waveforms. An AC generator uses the principal of Faraday’s electromagnetic induction to convert a mechanical energy such as rotation, into electrical energy, a Sinusoidal Waveform. A simple generator consists of a pair of permanent magnets producing a fixed magnetic field between a north and a south pole.
#### What is a cos wave?
A cosine wave is a signal waveform with a shape identical to that of a sine wave , except each point on the cosine wave occurs exactly 1/4 cycle earlier than the corresponding point on the sine wave.
Does cos or sin come first?
When sine and cosine are first…” Short answer: Yes. When sine and cosine are first revealed in Trigonometry class they are taught as ratios of the the sides of right triangles.
## What is the difference between sine wave and cos wave?
The Cosine Wave, simply called “cos”, is as important as the sine wave in electrical engineering. The cosine wave has the same shape as its sine wave counterpart that is it is a sinusoidal function, but is shifted by +90o or one full quarter of a period ahead of it.
How does sin and cos work?
sin cos and tan are basically just functions that relate an angle with a ratio of two sides in a right triangle. Sin is equal to the side opposite the angle that you are conducting the functions on over the hypotenuse which is the longest side in the triangle. Cos is adjacent over hypotenuse.
### How do you go from sin to cos?
All triangles have 3 angles that add to 180 degrees. Therefore, if one angle is 90 degrees we can figure out Sin Theta = Cos (90 – Theta) and Cos Theta = Sin (90 – Theta).
Who uses trigonometry in their careers?
Trigonometry is used by engineers, medical services technicians, mathematicians, data entry specialists, loggers, statisticians, actuaries, drafters, chemists, economists, physicists, registered nurses, building inspectors, boilermakers, machinists and millwrights.
#### How trigonometry is used in physics?
In physics, trigonometry is used to find the components of vectors, model the mechanics of waves (both physical and electromagnetic) and oscillations, sum the strength of fields, and use dot and cross products. Even in projectile motion you have a lot of application of trigonometry.
How is maths used in construction?
In the modern world, builders use math every day to do their work. Construction workers add, subtract, divide, multiply, and work with fractions. They measure the area, volume, length, and width.
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# How do you find the equation of the circle given Center (8,6), r=10?
Jul 14, 2016
${x}^{2} + {y}^{2} - 16 x - 12 y = 0$
#### Explanation:
In a circle, a point moves so that it always remains at a fixed distance (called radius) from a point (called center).
Let the point be $\left(x , y\right)$ and as center of circle is $\left(8 , 6\right)$, its distance from this point will always be $10$.
Hence equation of circle is given by
${\left(x - 8\right)}^{2} + {\left(y - 6\right)}^{2} = {10}^{2}$ or
${x}^{2} - 16 x + 64 + {y}^{2} - 12 y + 36 = 100$ or
${x}^{2} + {y}^{2} - 16 x - 12 y + 64 + 36 - 100 = 0$ or
${x}^{2} + {y}^{2} - 16 x - 12 y = 0$
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# How to solve a biquadratic equation?
Samiko
September 18, 2014
Before you start solving a biquadratic equation, you should understand how it looks and how it differs from the classical quadratic equation. Equation of the ax4+ bx2+ c = 0 is called biquadratic with one variable (an algebraic equation of the fourth degree). To reduce the equation to a quadratic form and solve it through a discriminant, it is necessary to use the variable substitution:
• ie: x2= t
And then we have a standard equation of the form at2+ bt + c = 0
The discriminant is calculated by the formula D = b2 - 4ac.
1. In the case when D = 0, the equation has one single root t1= -b / 2a, and from here we get the desired solution of our equation x = sqrt (t1).
2. If D> 0, the equation has two roots t1= (-b + sqrt (D)) / 2a and t2= (-b - sqrt (D)) / 2a. Do not forget about the variable entered, and we get the final solution x1,2= sqrt (t1) and x3,4= sqrt (t2)
Important note: if any of the t valuesi<0, then for D = 0 the initial biquadratic solution has no real roots, and for D> 0, there is at most one single real root.
## Using the Viet theorem
Good to know: in the case when we have a reduced quadratic equation (coefficient at t2= 1), the Viet theorem is applicable, and the search for a solution is minimized:
• t1+ t2= -b
• t1* t2= c
Consider an example:
• x4- 3x2+ 2 = 0
using the substitution variable x2= t, we reduce the quadratic equation to the form t2- 3t; + 2 = 0.
• D = (-3)2- 4*1*2 = 1.
Roots of a quadratic equation t1= 2, t2= 1.
Given the introduced change of variable, we obtain the solution of the desired biquadratic equation: t1= sqrt (2); t2= -sqrt (2); t3= 1; t4= -1.
We can apply the Viet theorem to this task, since the coefficient of the variable with the highest power is 1:
• t1+ t2= 3
• t1* t2= 2
From here t1= 2, t2= 1. As we can see, the roots of the quadratic equation in both cases coincide, which means that the solution of the biquadratic equation will be the same.
In this article, we considered a special case of solving a biquadratic equation that is solved no more complicatedly than the classical quadratic equation.
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Therorems
Chapter 10 Class 10 Circles
Serial order wise
### Transcript
Theorem 10.2 (Method 1) The lengths of tangents drawn from an external point to a circle are equal. Given: Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively To prove: Lengths of tangents are equal i.e. PQ = PR Construction: Join OQ , OR and OP Proof: As PQ is a tangent OQ ⊥ PQ So, ∠ OQP = 90° Hence Δ OQP is right triangle Similarly, PR is a tangent & OR ⊥ PR So, ∠ ORP = 90° Hence Δ ORP is right triangle Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 From (1) and (2) PQ2 = PR2 PQ = PR Hence proved Theorem 10.2 (Method 2) The lengths of tangents drawn from an external point to a circle are equal. Given: Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively To prove: Lengths of tangents are equal i.e. PQ = PR Construction: Join OQ , OR and OP Proof: As PQ is a tangent OQ ⊥ PQ So, ∠ OQP = 90° Similarly, PR is a tangent & OR ⊥ PR So, ∠ ORP = 90° In Δ OQP and Δ ORP ∠ OQP = ∠ ORP OP = OP OQ = OR ∴ Δ OQP = Δ ORP Hence, PQ = PR Hence both tangents from external point are equal in length Also, ∠ OPQ = ∠ OPR Hence, OP is the angle bisector of ∠ QPR
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In Geometry, we learn about different shapes and sizes. These shapes could be two-dimensional shapes or three-dimensional, such as circles, triangles, squares, cubes, cylinders, etc. Learning about these shapes will help us to understand the kinds of shapes we come across in our daily lives. We, not only learn about the geometrical figures and but also learn to measure them.
Mensuration is an important topic in geometry where we learn to measure the dimensions of different shapes. Also, we learn the formulas to learn the area, volumes, and length of the shape. For example, the circumference of a circle is used to measure the length of a circular object. Say, a ground in circular shape has a radius of 7 meters. Then we can easily calculate the length of the border of the ground using the formula, Circumference = 2πr. Hence, the length will be 2×22/7×7=44 meter.
Apart from the circumference, the circle has one more important property which is its area. The formula to find the area of the circle is pi radius squared. Hence, we can easily find the area of a circular field. But what if the field is not in circular shape instead it is in a rectangular shape or square shape. Then we need to use the formula for the area of rectangle or area of the square. The area of the rectangle is equal to the length and breadth of the shape whereas the area of the square is equal to length2. The two formulas are different because the rectangle has its parallel sides equal but the square has all its sides equal.
Another important measurement in terms of shapes and sizes or another major parameter of shapes is volume. It is also one of the properties of three-dimensional objects. Volume is not applicable to 2D shapes, because they don’t have a thickness as one of their dimension. 2D shapes have only length and width as their dimensions, but 3D shapes have a height or thickness as an extra dimension. The volume of an object sometimes defines the capacity of the object. For example, a water container has a capacity of 25 liters of water.
The volume is a quantity three-dimensional shape enclosed by a closed surface. Itis measured in the cubic units, such as cubic centimeters, cubic meters. Hence, volumes play a major role in knowing the space occupied by a solid object. But the two-dimensional shape does not have volumes because they don’t occupy any space.
Similarly, in the initial part, we have also learned about the lines and angles, which defines these objects. For example, with the help of three lines, we can a triangle. Also, we can measure the angles of triangles using a protractor. The line segment is a geometrical shape which tells the distance between the two points. When two rays are combined they form an angle.
So we have known here how the different shapes and their measurements are important in geometry and also in everyday life. Therefore mensuration plays a major role in Maths.
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Class 10 NCERT Math Solution
TOPICS
Unit-8(Examples)
Example-1 :- Given tan A = 4/3, find the other trigonometric ratios of the angle A.
Solution :-
``` Give that:
tan A = 4/3
cot A = 1/tan A = 3/4
sec² A = 1 + tan² A
= 1 + (4/3)²
= 1 + 16/9
= 25/9
sec A = 5/3
cos A = 1/sec A = 3/5
sin² A = 1 - cos² A
= 1 - (3/5)²
= 1 - 9/25
= 16/25
sin A = 4/5
cosec A = 1/sin A = 5/4
```
Example-2 :- If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.
Solution :-
``` Let us consider two right triangles ABC and PQR where sin B = sin Q see in figure
We have sin B = AC/AB
sin Q = PR/PQ
Then, AC/AB = PR/PQ
Therefore, AC/PR = AB/PQ = k .......(1)
Now using Pythagoras theorem,
From (1) and (2), we have
AC/PR = AB/PQ = BC/QR
Δ ACB ~ Δ PRQ and therefore, ∠ B = ∠ Q.
```
Example-3 :- Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see Figure). Determine the values of (i) cos² θ + sin² θ, (ii) cos² θ – sin² θ
Solution :-
``` In Δ ACB, we have AB = 29 units and BC = 21 units
AC² = AB² - BC²
= (29)² - (21)²
= (29 - 21)(29 + 21)
= 8 x 50
= 400
AC = 20 units
so, sin θ = AC/AB = 20/29, cos θ = 21/29
```
``` Now,
(i) cos² θ + sin² θ
= (21/29)² + (20/29)²
= 441/841 + 400/841
= 841/841
= 1
```
```(ii)cos² θ – sin² θ
= (21/29)² - (20/29)²
= 441/841 - 400/841
= 41/841
```
Example-4 :- In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.
Solution :-
```
In Δ ABC, tan A = BC/AB = 1 (see in figure)
i.e., BC = AB
Let AB = BC = k, where k is a positive number
AC² = AB² + BC²
= k² + k²
= 2k²
AC = √2k
Therefore, sin A = BC/AC = k/√2k = 1/√2
cos A = AB/AC = 1/√2
so, 2 sin A . cos A = 2 (1/√2) x (1/√2) = 1
which is the required value.
```
Example-5 :- In Δ OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Figure). Determine the values of sin Q and cos Q.
Solution :-
``` In Δ OPQ, we have
OQ² = OP² + PQ²
(1 + PQ)² = OP² + PQ²
i.e., 1 + PQ² + 2PQ = OP² + PQ²
i.e., 1 + 2PQ = 7²
i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm
so, sin Q = 7/25 and cos Q = 24/25
```
Example-6 :- In Δ ABC, right-angled at B, AB = 5 cm and ∠ ACB = 30° (see Figure). Determine the lengths of the sides BC and AC.
Solution :-
``` To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB.
Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore
AB/BC = tan C
5/BC = tan 30°
5/BC = 1/√3
BC = 5√3 cm
To find the length of the side AC, we consider
sin 30° = AB/AC
1/2 = 5/AC
AC = 10cm
```
Example-7 :- In Δ PQR, right-angled at Q (see Figure), PQ = 3 cm and PR = 6 cm. Determine ∠ QPR and ∠ PRQ.
Solution :-
``` Given PQ = 3 cm and PR = 6 cm
Therefore, PQ/PR = sin R
sin R = 3/6 = 1/2
so, ∠ PRQ = 30°
∠ QPR = 60°
```
Example-8 :- If sin (A – B) = 1/2, cos (A + B) = 1/2, 0° < A + B ≤ 90°, A > B, find A and B.
Solution :-
``` Since, sin (A – B) = 1/2,
sin (A - B) = sin 30°
Therefore, A – B = 30° .....(1)
Also, since cos (A + B) = 1/2,
cos (A + B) = 60°
Therefore, A + B = 60° ......(2)
Solving (1) and (2),
A - B + A + B = 30° + 60°
2A = 90°
A = 90°/2
A = 45°
Put in (1) equation
A - B = 30°
45° - B = 30°
-B = 30° - 45°
-B = -15°
B = 15°
we get : A = 45° and B = 15°.
```
Example-9 :- Evaluate : tan 65°/cot 25°
Solution :-
``` We know : cot A = tan (90° – A)
So, cot 25° = tan (90° – 25°) = tan 65°
i.e., tan 65°/cot 25° = tan 65°/tan 65° = 1
```
Example-10 :- If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.
Solution :-
``` We are given that sin 3A = cos (A – 26°)..... (1)
Since sin 3A = cos (90° – 3A),
we can write (1) as cos (90° – 3A) = cos (A – 26°)
Since 90° – 3A and A – 26° are both acute angles,
Therefore, 90° – 3A = A – 26° which gives A = 29°
```
Example-11 :- Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution :-
``` cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
```
Example-12 :- Express the ratios cos A, tan A and sec A in terms of sin A.
Solution :-
``` Since cos² A + sin² A = 1,
Therefore, cos² A = 1 – sin² A,
i.e., cos A = √1 - sin² A
tan A = sin A/cos A = sin A/√1 - sin² A
sec A = 1/cos A = 1/√1 - sin² A
```
Example-13 :- Prove that : sec A (1 – sin A)(sec A + tan A) = 1.
Solution :-
```
```
Example-14 :- Prove that :
Solution :-
```
```
Example-15 :- Prove that : using the identity sec² θ = 1 + tan² θ.
Solution :-
```
```
CLASSES
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# How do you solve the inequality 9 > -3 (x - 1) ≥ -12?
Jun 24, 2018
$- 2 < x \le 5$
#### Explanation:
Given: $9 > - 3 \left(x - 1\right) \ge - 12$
$\textcolor{b l u e}{\text{Manipulate the same way you do equation.}}$
$\textcolor{b l u e}{\text{but be mindful about the inequality sign}}$
Divide throughout by 3 (does not change the signs)
$3 > - x + 1 \ge - 4$
Subtract 1 throughout
$2 > - x \ge - 5$
$\textcolor{b r o w n}{\text{Consider the part: } - x \ge - 5}$
Swap sides and change the negatives to positives. Keep the inequality the same way round.
Same thing as: add 5 to both sides and add $x$ to both sides and keep the inequality the same way round.
Same thing as: multiply both sides by (-1) and turn the inequality sign the other way round.
$5 \ge x \textcolor{w h i t e}{\text{ddd}} \leftarrow$ This is the same thing as $x \le 5$
$\textcolor{b r o w n}{\text{Consider the part: } 2 > - x}$
Swap sides and change the signs but keep the inequality the same way round.
$x > - 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Putting it all together}}$
$- 2 < x \le 5$
|
# Binomial theorem
The binomial theorem is used to expand polynomials of the form (x + y)n into a sum of terms of the form axbyc, where a is a positive integer coefficient and b and c are non-negative integers that sum to n. It is useful for expanding binomials raised to larger powers without having to repeatedly multiply binomials. The binomial theorem is written as:
## Breaking down the binomial theorem
If you aren't familiar with the symbols in the binomial theorem, it may seem fairly daunting, but once you break it down piece by piece, the binomial theorem is relatively simple to use. We should be fairly comfortable with exponents, but refer to the exponents page for a refresher if necessary.
### Summation
The symbol, , is the capital Greek letter Sigma. In math, it is referred to as the summation symbol. Along with the index of summation, k (i is also used), the lower bound of summation, m, the upper bound of summation, n, and an expression ak, it tells us how to sum:
Example
Evaluate :
### Combinations
The in the binomial theorem is a combination (specifically a combination without repetition) that is referred to as the binomial coefficient. It is read as "n choose k." Briefly, n choose k indicates how many possible ways there are to choose k elements from a set of n. We will not go into combinations in depth at this point, but will provide a formula that will allow you to use the binomial theorem.
In the formula, the "!" indicates a factorial. The factorial of an integer is the product of all positive integers (0 is not included) that are equal to or less than the integer.
Example
Evaluate 5!:
5! = 5 × 4 × 3 × 2 × 1 = 120
### Using the binomial theorem
Though the examples above use relatively simple examples, the concepts are all the same when using the binomial theorem, and we evaluate the binomial theorem by plugging all the appropriate values into the formula.
Example
Expand (x + y)3 using the binomial theorem:
= = = = = =
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# If ${{x}^{y}}={{y}^{x}}$ and $x=2y$, then find the value of $x+y$. (Assume that $y\ne 0$) A. $3$B. $6$C. $7$D. $8$
Last updated date: 15th Jul 2024
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Hint: Take the equation ${{x}^{y}}={{y}^{x}}$, apply $\log$on both sides. After that, simplify it by applying the properties of logarithm and then substitute $x=2y$ in the simplified equation. You will get the value of $y$. Then substitute the value of $y$in $x=2y$, you will get the value of $x$. After getting the values of both $x$ and $y$, put the values in $x+y$. Try it, you will get the answer.
In Mathematics, an equation is a statement that asserts the equality of two expressions. The word equation and its cognates in other languages may have subtly different meanings; for example, in French an equation is defined as containing one or more variables, while in English any equality is an equation.
Solving an equation containing variables consists of determining which values of the variables make the equality true. Variables are also called unknowns and the values of the unknowns that satisfy the equality are called solutions of the equation. There are two kinds of equations: identities and conditional equations. An identity is true for all values of the variable. A conditional equation is only true for particular values of the variables.
Single variable equation can be $x+2=0$.
An equation is written as two expressions, connected by an equals sign ("$=$"). The expressions on the two sides of the equals sign are called the "left-hand side" and "right-hand side" of the equation.
The most common type of equation is an algebraic equation, in which the two sides are algebraic expressions. Each side of an algebraic equation will contain one or more terms. For example, the equation,
$A{{x}^{2}}+Bx+C=y$
has left-hand side $A{{x}^{2}}+Bx+C$ which has three terms, and right-hand side $y$, consisting of just one term. The unknowns are $x$ and $y$ and the parameters are $A,$ $B,$ and $C$.
An equation is analogous to a scale into which weights are placed. When equal weights of something (grain for example) are placed into the two pans, the two weights cause the scale to be in balance and are said to be equal. If a quantity of grain is removed from one pan of the balance, an equal amount of grain must be removed from the other pan to keep the scale in balance. Likewise, to keep an equation in balance, the same operations of addition, subtraction, multiplication and division must be performed on both sides of an equation for it to remain true.
In question we are given the equation ${{x}^{y}}={{y}^{x}}$ and $x=2y$.
Now taking the equation ${{x}^{y}}={{y}^{x}}$ and applying $\log$on both sides we get,
$\log {{x}^{y}}=\log {{y}^{x}}$
We know property of $\log$,
$\log {{a}^{b}}=b\log a$
So applying above property we get,
$y\log x=x\log y$ …… (1)
Now we are given that $x=2y$,
So, substituting $x=2y$ in the equation (1) we get,
$y\log 2y=2y\log y$
Simplifying the equation,
$\log 2y=2\log y$
Now, we know the property that,
$\log ab=\log b+\log a$
So, applying above property we get,
$\log 2+\log y=2\log y$
Now, again simplifying the equation we get,
$\log 2=\log y$
Now, removing $\log$ from both side we get,
$y=2$ ………………. (2)
Now substituting the value of $y=2$ in $x=2y$ we get,
$x=2(2)=4$
$x=4$……………… (3)
So now, we have to find the value of $x+y$,
From (2) and (3), substituting the values of $x$ and $y$ in $x+y$ we get,
$x+y=4+2$
$x+y=6$
Therefore, the value of $x+y$ is $6$.
Note:
Read the question carefully. You should be clear with the concept of logarithm. Also, while simplifying don’t jumble yourself. Do not make silly mistakes like missing the terms. Take utmost care that no signs are missed. You should be familiar with all the properties of logarithm.
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# Difference between revisions of "2010 AMC 12A Problems/Problem 18"
## Problem
A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$
## Solution 1
Each path must go through either the second or the fourth quadrant. Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$, $(-3,3)$, and $(-2,2)$.
There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type. Each path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$, $(3,-3)$, and $(2,-2)$. Again, there is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Hence the total number of paths is $2(1+64+784) = \boxed{1698}$.
## Solution 2
We will use the concept of complimentary counting. If you go in the square you have to go out by these labeled points(click the link) https://imgur.com/VysX4P0 and go out through the borders because if you didnt, you would touch another point in one of those points in the set of points in the link(Call it $S$). There is symmetry about $y=x$, so we only have to consider $(1,-1), (1,0)$, and $(1,1)$. $(1,1)$ can go on the boundary in two ways, so we can only consider one case and multiply it by two. For $(1,0)$ and $(1,-1)$ we can just multiply by two. So we count paths from $(-4,4)$ to each of these points, and then multiply that by the number of ways to get from the point one unit right of that to $(4,4)$, and all in all, we get the answer is $\dbinom{16}{8}-2\left[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}\right]=1698$, which is answer choice $\textbf{(D)}$ -vsamc Note: Sorry if this was rushed.
## Further Explanation(for Solution 1)
As stated in the solution, there are $6$ points along the line $y=-x$ that constitute a sort of "boundary". Once the ant reaches one of these $6$ points, it is exactly halfway to $(4, 4)$. Also notice that the ant will only cross one of the $6$ points during any one of its paths. Therefore we can divide the problem into $3$ cases, focusing on $1$ quadrant; then multiplying the sum by $2$ to get the total (because there is symmetry).
For the sake of this explanation, we will focus on the fourth quadrant (it really doesn't matter which quadrant because, again the layout is symmetrical) The three cases are when the ant crosses $(4, -4), (3, -3),$ and $(2, -2)$.
For each of the cases, notice that the path the ant takes can be expressed as a sequence of steps, such as:
right, right, up, right,..., etc.
Also notice that there are always $8$ steps per sequence (if there were more or less steps, the ant would be breaking the conditions given in the problem). This means we can figure out the number of ways to get to a point based on the particular sequence of steps that denote each path. For example, there is only one way for the ant to pass through $(4, -4);$ it MUST keep traveling right for all $8$ steps. This seems fairly obvious; however, notice that this is equivalent to
$\binom{8}{0}$
Now we consider the number of ways to get from $(4, -4)$ to $(4, 4)$. by symmetry, there is only $1$ such way. So the number of paths containing $(4, -4)$ is $1^2,$ or $1$.
Moving on to the next case, we see that the ant MUST travel right exactly $7$ times and up exactly once. So each sequence of this type will have $7$ "right"s and $1$ "up". So, the total number of paths that go through $(3, -3)$ is equivalent to the number of ways to arrange $1$ "up" into $8$ spots. This is
$\binom{8}{1} = 8$
Similarly to the first case, we square this value to account for the second half of the journey: $8^2 = 64$.
Finally, for the third case (ant passes through $(2, -2)$) the ant must travel right exactly $6$ times and up exactly $2$ times. This is equivalent to the number of ways to arrange $2$ "ups" in a sequence of $8$ movements, or
$\binom{8}{2} = 28$
Again, we square $28$: $28^2 = 784$. Adding up all of these cases we get
$1+64+784 = 849$
paths through the fourth quadrant. Doubling this number to account for the paths through the second quadrant, we have
$849*2=1698 \Rightarrow \boxed{\text{D}}$.
For all the notation geeks out there, the solution can be expressed as such:
$2\sum_{k=0}^2 \binom{8}{k}^2$
## See also
2010 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Mean Median Mode Formula
The Mean, Median and Mode are the three measures of central tendency. Mean is the arithmetic average of a data set. This is found by adding the numbers in a data set and dividing by the number of observations in the data set. The median is the middle number in a data set when the numbers are listed in either ascending or descending order. The mode is the value that occurs the most often in a data set and the range is the difference between the highest and lowest values in a data set.
The Mean
$\large \overline{x}=\frac{\sum x}{N}$
Here,
∑ represents the summation
X represents observations
N represents the number of observations .
In the case where the data is presented in a tabular form, the following formula is used to compute the mean
Mean = ∑f x / ∑f
Where ∑f = N
The Median
If the total number of observations (n) is an odd number, then the formula is given below:
$\large Median=\left(\frac{n+1}{2}\right)^{th}observation$
If the total number of the observations (n) is an even number, then the formula is given below:
$\large Median=\frac{\left(\frac{n}{2}\right)^{th}observation+\left(\frac{n}{2}+1\right )^{th}observation}{2}$
Consider the case where the data is continuous and presented in the form of a frequency distribution, the median formula is as follows.
Find the median class, the total count of observations ∑f.
The median class consists of the class in which (n / 2) is present.
$$\begin{array}{l}\text { Median }=1+\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{c}}{\mathrm{f}}\right] \times \mathrm{h}\end{array}$$
Here
l = lesser limit belonging to the median class
c = cumulative frequency value of the class before the median class
f = frequency possessed by the median class
h = size of the class
The Mode
$\large The\;mode\;is\;the\;most\;frequently\;occuring\;observation\;or\;value.$
Consider the case where the data is continuous and the value of mode can be computed using the following steps.
a] Determine the modal class that is the class possessing the maximum frequency.
b] Calculate the mode using the below formula
$$\begin{array}{l}\text { Mode }=1+\left[\frac{f_{m}-f_{1}}{2 f_{m}-f_{1}-f_{2}}\right] \times h\end{array}$$
l = lesser limit of modal class
$$\begin{array}{l}f_{m}\end{array}$$
= frequency possessed by the modal class
$$\begin{array}{l}f_{1}\end{array}$$
= frequency possessed by the class before the modal class
$$\begin{array}{l}f_{2}\end{array}$$
= frequency possessed by the class after the modal class
h = width of the class
How Are Mean, Median And Mode Related?
The 3 estimates of central tendency that is the mean, median and mode are related by the following empirical relationship.
2 Mean + Mode = 3 Median
For example, if it is required to compute the mean, median and mode of the data that is continuous grouped, then the values of the mean and median can be found using the above formulae. The value of the mode can be found using the empirical formula.
If the value of the mode is 65 and the median = 61.6, then find the value of the mean.
The value of the mean can be calculated using the formula,
2 Mean + Mode = 3 Median
2 Mean = (3 × 61.6) – 65
2 Mean = 119.8
Mean = 119.8 / 2
Mean = 59.9
### Solved Examples
Question 1:Â Find the mean, median, mode, and range for the following list of values:
13, 18, 13, 14, 13, 16, 14, 21, 13
Solution:
Given data: 13, 18, 13, 14, 13, 16, 14, 21, 13
The mean is the usual average.
Mean = {13 + 18 + 13 + 14 + 13 + 16 + 14 + 21 + 13} / {9} = 15
(Note that the mean is not a value from the original list. This is a common result. You should not assume that your mean will be one of your original numbers.)
The median is the middle value, so to rewrite the list in ascending order as given below:
13, 13, 13, 13, 14, 14, 16, 18, 21
There are nine numbers in the list, so the middle one will be
{9 + 1} / {2} = {10} / {2} = 5
= 5th number
Hence, the median is 14.
The mode is the number that is repeated more often than any other, so 13 is the mode.
The largest value in the list is 21, and the smallest is 13, so the range is 21 – 13 = 8.
Mean = 15
Median = 14
Mode = 13
Range = 8
Question 2: The value of the mean of five numbers is observed to be 18. If one number is not included, the mean is 16. Find the number that is excluded.
From the question,
There are 5 observations that mean n = 5.
The value of the mean = 18
x̄ = 18
xÌ„ = ∑ x / n
∑ x = 5 * 18 = 90
The sum of the five observations is 90.
Assume the excluded number to be “a”
The sum of four observations = 90 – a
Mean of 4 observations = (90 – a) / 4
16 = (90 – a) / 4
90 – a = 64
a = 26
⇒ The excluded number is 26.
More topics in Mean Median Mode Formula Arithmetic Mean Formula Geometric Mean Formula Harmonic Mean Formula Sample Mean Formula Weighted Mean Formula Effect Size Formula
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# Representation & Interpretation of Data: Make Bar Graphs
Jul 28, 2023
## Bar Graph
A bar graph is a graphical representation of information.
Bars are used in bar graphs. Bar graphs can be created using vertical bars, horizontal bars, or grouped bars.
### Steps to Make a Bar Graph:
Step 1: Consider the title from the given data or tally chart.
Step 2: Choose a scale.
Step 3: Consider two axes, vertically and horizontally.
Step 4: Draw the bars according to the given data
Note:
A bar graph makes it easy to compare data.
#### Example of a Bar Graph:
Record the data in the form of a tally chart or information chart.
Choose a scale for each of the axis.
Now, draw a bar graph by using a scale and a data chart.
The horizontal axis shows the kinds of flowers; the vertical axis shows the number of flowers. Here, each grid line represents 2 units.
Example 1:
See the table of data and draw the bar graph.
The bar graph shows information about the number of pages read from a comic book by a group of friends.
Solution:
First, select the same title given to the table data.
The title of this bar graph is the ‘number of pages read.’
Next, we select a scale.
Each grid is equal to 5 units/5 pages.
We get the following bar graph.
Example 2:
Ashley had a fruit basket; he counted the fruits and made a tally chart. Draw the bar graph.
Solution:
First, select the same title given to the table data.
The title of this bar graph is the ‘Fruit Basket.’
Next, we select a scale.
Each grid is equal to 2 units/2 fruits.
We get the following bar graph:
#### Exercise:
1. See the information in the tally chart below and draw a bar graph.
2. James and his friends in school bought some toys. Draw a bar graph to show the number of items of each kind.
3. See the tally chart and draw the bar graph.
4. See the tally chart and draw a bar graph.
Q. (5 and 6) See the tally chart and answer the questions.
5. Draw a bar graph.
6. How many more cats are there than dogs?
7. See the tally chart and draw the bar graph.
8. Draw the bar graph for the tally chart below and compare the number of cars and bikes.
9. How many more bicycles are there than bikes in the chart below?
10. How many fewer buses are there than cars?
#### What We Have Learned:
• Use a tally chart to make bar graphs
• Make bar graphs
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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# How do you simplify ((-3)^0 x^4 y^-1) /( x^-1 yz ^-2)?
Jan 26, 2017
$\frac{{\left(- 3\right)}^{0} {x}^{4} {y}^{-} 1}{{x}^{-} 1 y {z}^{-} 2} = \frac{{x}^{5} {z}^{2}}{y} ^ 2$
#### Explanation:
Some rules which apply to indices:
${a}^{0} \equiv 1 ,$ $\text{given}$ $a \ne 0$
${a}^{m} / {a}^{n} \equiv {a}^{m - n} ,$ $\text{given}$ $m - n > 0$ $\text{where}$ $a = 0$
${a}^{- n} \equiv \frac{1}{{a}^{n}} ,$ $\text{given}$ $a \ne 0$
$- {3}^{0} = 1$
${x}^{4} / {x}^{-} 1 = {x}^{4 - - 1} = {x}^{5}$
${y}^{-} \frac{1}{y} = {y}^{- 1 - 1} = {y}^{-} 2$
$\frac{1 \left({z}^{0}\right)}{z} ^ - 2 = {z}^{0 - - 2} = {z}^{2}$
$\frac{{\left(- 3\right)}^{0} {x}^{4} {y}^{-} 1}{{x}^{-} 1 y {z}^{-} 2} = {x}^{5} {y}^{-} 2 {z}^{2} = {x}^{5} \frac{1}{y} ^ 2 {z}^{2} = \frac{{x}^{5} {z}^{2}}{y} ^ 2$
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### Mathematics Class XI
Unit-I: Sets and Functions
Chapter 1: Sets
Unit-II: Algebra
Chapter 5: Binomial Theorem
Chapter 6: Sequence and Series
Unit-III: Coordinate Geometry
Chapter 1: Straight Lines
Chapter 2: Conic Sections
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability
Chapter 1: Statistics
Chapter 2: Probability
# Algebra of complex numbers
(A) Addition of two Complex Number:
If and be any two complex numbers, then the sum is defined by following properties:
(i) The Closure Property:
The sum of two complex number is a complex numbers, i.e., is a complex number for all numbers and .
(ii)Commutative Property:
For any two complex numbers and
(iii) Associative Property:
For any three complex numbers and ,
There exists the complex number (denoted as 0), called the additive identity or the zero complex number, such that for every complex number ,
To, every complex number , we have the complex (denoted as ) called the additive inverse or negative of i.e.,
(B) Difference of Two Complex Numbers:
Any two complex numbers and the difference is defined as
(C) Multiplication of Two Complex Numbers:
Let and be any two complex numbers, then the product
For Example:
The multiplication of two complex numbers possesses the following properties:
(i) Closure Property:
The product of two complex numbers is a complex number is a complex number, i.e., the product is a complex number for all complex numbers and .
(ii) Commutative Property:
For any two complex numbers and ,
(iii) Associative Property:
For any three complex numbers and , we have
(iv) Existance of Multiplicative Identity:
There exists the complex number ( denoted as ), called the multiplicative identity such that , for every complex number .
(v) Existance of Multiplicative Inverse:
For every non-zero complex number, , we have the complex number , called the multiplicative inverse of such that where
(vi) Distributive Property:
For any three complex numbers:
(1)
(2)
(D) Division of Two Complex Numbers:
Any two complex numbers and , where , the quotient is defined by
For Example:
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Home > Measurement > Pythagoras Theorem > Converse of Pythagoras Theorem
# Converse of Pythagoras Theorem
The converse of Pythagoras theorem is used to determine whether or not a triangle is right-angled, given the lengths of its sides. It states that:
“If the squares on one side of the triangle is equal to the sum of the squares on the other two sides, then the angle between the two short sides is a right angle”.
Example 1 : Determine whether this triangle is right-angled.
The converse of Pythagoras theorem is used to determine if a triangle is right-angled.
102 = 100
62 = 36
82 = 64
Hence, 62 + 82 = 36 + 64
= 100 = 102
So, ABC is right-angled at A.
Example 2 : Is triangle PQR here right-angled?
Applying the converse of Pythagoras theorem –
112 = 121
72 = 49
92 = 81
Adding the two shorter sides – 72 + 92 = 49 + 81 = 130.
But this is not equal to 121. So triangle PQR is not right-angled.
Example 3 : State whether 13, 16, and 17 is a Pythagorean triad.
To check if three numbers form a Pythagorean triad, we use the converse of Pythagoras theorem.
132 = 169, 162 = 256, and 172 = 289
132 + 162
= 169 + 256
= 425 289 (172)
So, 13, 16, and 17 do not form a Pythagoras triad.
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8. NCERT Solutions class 12...
# NCERT Solutions class 12 Maths Exercise 2.2
### myCBSEguide App
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NCERT solutions for Class 12 Ch- 2 Maths
## NCERT Solutions class 12 Maths Chapter 2 Inverse Trigonometric Function
### Prove the following:
1.
Ans. We know that:
Putting
Putting ,
Proved.
### 2.
Ans. We know that:
Putting
Putting ,
Proved.
Ans. L.H.S. =
=
=
=
=
= R.H.S.
Proved.
Ans. L.H.S. = =
=
=
=
=
= R.H.S.
Proved.
### Write the following functions in the simplest form:
5.
Ans. Putting so that
=
=
=
=
=
=
=
### 6.
Ans. Putting so that
=
=
=
=
=
=
Ans.
=
= =
### 8.
Ans.
Dividing the numerator and denominator by
=
=
=
=
### 9.
Ans. Putting so that
=
=
=
=
=
### 10.
Ans.
= [Dividing numerator and denominator by ]
Putting so that
=
=
= =
11.
Ans.
=
=
=
=
= =
Ans. =
Ans. Putting and
=
=
=
=
= =
Ans. Given:
Ans. Given:
16.
Ans.
=
=
= =
Ans.
=
=
=
= =
### 18.
Ans. Putting and so that and
Now, =
And and
=
= =
= =
### 19. is equal to:
(A)
(B)
(C)
(D)
Ans.
=
=
Therefore, option (B) is correct.
### 20. is equal to:
(A)
(B)
(C)
(D) 1
Ans.
=
=
=
=
Therefore, option (D) is correct.
NCERT Solutions class 12 Maths Exercise 2.2
### 21. is equal to:
(A)
(B)
(C)
(D)
Ans.
=
=
=
=
=
=
Therefore, option (B) is correct.
## NCERT Solutions class 12 Maths Exercise 2.2
NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide
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Quadratic Equations Class 10th Notes - Free NCERT Class 10 Maths Chapter 4 Notes - Download PDF
# Quadratic Equations Class 10th Notes - Free NCERT Class 10 Maths Chapter 4 Notes - Download PDF
Edited By Safeer PP | Updated on Mar 19, 2022 10:36 AM IST
Quadratic equations are the NCERT chapter which deals with algebraic equations of degree 2. The NCERT Class 10 Maths chapter 4 notes covers a brief outline of the chapter quadratic equations. The main topics covered quadratic equations Class 10 notes, gives you properties and roots of quadratic equations. Class 10 Maths chapter 4 notes also cover the basic equations in the chapter. Quadratic equations Class 10 notes pdf download contains all of these topics.
Also, students can refer,
A polynomial with degree 2, is a quadratic polynomial. It is in the form of
f(x) = ax2 + bx + c, where a ≠ 0
An algebraic expression of the second degree is called a quadratic equation.
• The standard form of a Quadratic Equation
ax2+bx+c=0
where a, b and c are the real numbers and a≠0
## Roots of a Quadratic Equation
Let x = α and α is a real number. If α satisfies the quadratic equation ax2+ bx + c = 0 such that aα2 + bα + c = 0, then α is the root of the Quadratic Equation.
As quadratic polynomials have degree two, therefore quadratic equations can have two roots. Thus, the zeros of a quadratic polynomial f(x) =ax2+bx+c is same as the roots of the quadratic equation ax2+ bx + c= 0.
## Methods to Solve the Quadratic Equations
Three methods to solve the Quadratic Equations-
### 1. Factorization Method
In this method, divide the equation into two linear factors and equate each factor to zero to find the roots of the equation.
Step 1: Quadratic Equation in the form of ax2 + bx + c = 0.
Step 3: By factorization, we write ax2 + bx + c = 0 as (x + p) (x + q) = 0
For example-
x2-2x-15=0
(x+3)(x-5)=0
x+3=0 or x-5=0
x=-3 or x=5
x={-3,5}
The above values of x are the two roots of the given quadratic equation.
### 2. Completing the Square Method
In this method, convert the equation in the square form (x + a)2 - b2= 0 to find the roots.
Step1: Quadratic Equation in the standard form ax2 + bx + c = 0.
Step 2: Divide both the sides by a:
Step 3: Transfer the constant to RHS then add the square of the half of the coefficient of x i.e. (b/2a)2 on both the sides
Step 4: Write LHS as a perfect square and simplify RHS.
Step 5: Square root on both sides.
Step 6: Constant terms are shifted to the RHS and calculate the value of x as there is no variable at the RHS.
In this method, find the roots by using the quadratic formula. The quadratic formula is
where a, b and c are the real numbers and b2 – 4ac is known as a discriminant.
## Nature of Roots
The nature of the roots of the equation depends upon the value of D, which is called the discriminant.
Value of discriminant Number of roots D>0 Two distinct real roots D=0 Two equal and real roots D<0 No real roots
## Significance of NCERT Class 10 Maths Chapter 4 Notes-
Quadratic Equations Class 10 notes will help to understand the formulas, statements, rules in detail.
This NCERT Class 10 Maths chapter 4 notes also contains previous year’s questions and NCERT TextBook pdf. It also contains FAQs or frequently asked questions along with topic-wise explanations.
In offline mode, Class 10 Maths chapter notes pdf download can be used to prepare.
## Class 10 Chapter Wise Notes
NCERT Class 10 Maths Chapter 1 Notes NCERT Class 10 Maths Chapter 2 Notes NCERT Class 10 Maths Chapter 3 Notes NCERT Class 10 Maths Chapter 4 Notes NCERT Class 10 Maths Chapter 5 Notes NCERT Class 10 Maths Chapter 6 Notes NCERT Class 10 Maths Chapter 7 Notes NCERT Class 10 Maths Chapter 8 Notes NCERT Class 10 Maths Chapter 9 Notes NCERT Class 10 Maths Chapter 10 Notes NCERT Class 10 Maths Chapter 11 Notes NCERT Class 10 Maths Chapter 12 Notes NCERT Class 10 Maths Chapter 13 Notes NCERT Class 10 Maths Chapter 14 Notes NCERT Class 10 Maths Chapter 15 Notes
## NCERT Class 10 Exemplar Solutions for Other Subjects:
NCERT Books for Class 10 NCERT Syllabus for Class 10
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
1. What is factorization according to Class 10 Math’s chapter 4 notes?
The process by which the bracket of a quadratic equation is reduced is called factorization.
2. What is the weightage of Class 10 Quadratic Equations notes in the board examination?
Students can expect 4 to 8 marks questions from the notes for Class 10 Math’s chapter 4.
3. How many methods are there to find the roots of a quadratic equation?
In the notes for Class 10 Math’s chapter 4, four methods are discussed to find the roots of a quadratic equation.
4. How does this chapter help students?
It is evident that this chapter gives a better understanding of quadratic polynomials and helps us understand quadratic equations in a more wholesome way. Students can use Class 10 Math’s chapter 4 notes pdf download for revision
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Get answers from students and experts
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Option 1) Option 2) Option 3) Option 4)
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2 :
Option 1) 2.45×10−3 kg Option 2) 6.45×10−3 kg Option 3) 9.89×10−3 kg Option 4) 12.89×10−3 kg
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
Option 1) Option 2) Option 3) Option 4)
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
Option 1) Option 2) Option 3) Option 4)
In the reaction,
Option 1) at STP is produced for every mole consumed Option 2) is consumed for ever produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will
Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.
With increase of temperature, which of these changes?
Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.
Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is
Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023
A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is
Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9
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# NYT Digits April 23, 2023 Answers
If you need NYT Digits April 23, 2023 Answers, we have a list of solutions for you!
Digits is a math-based game developed and published by New York Times. Each day, there are five puzzles released, and in each puzzle, players have six numbers that they can add, subtract, multiply, or divide by in order to reach the target number. Once you use a number, you cannot use it again in another operation. If you’ve having trouble solving the NYT Digits puzzles for April 23rd, 2023, we have the answers for you!
## NYT Digits, Puzzle 1 Answer for April 23, 2023
The target is 85. There are three operations to get to 3-stars for the first puzzle.
1. 25 × 3 = 75
2. 75 + 10 = 85
## NYT Digits, Puzzle 2 Answer for April 23, 2023
The target is 105. There are three operations to get to 3-stars for the second puzzle.
1. 11 × 10 = 110
2. 110 – 3 = 107
3. 107 – 2 = 105
## NYT Digits, Puzzle 3 Answer for April 23, 2023
The target is 276. There are three operations to get to 3-stars for the third puzzle.
1. 20 × 11 = 220
2. 8 × 7 = 56
3. 220 + 56 = 276
## NYT Digits, Puzzle 4 Answer for April 23, 2023
The target is 321. There are three operations to get to 3-stars for the fourth puzzle.
1. 5 + 4 = 9
2. 11 × 9 = 99
3. 99 + 8 = 107
4. 107 × 3 = 321
## NYT Digits, Puzzle 5 Answer for April 23, 2023
The target is 428. There are four operations to get to 3-stars for the fifth puzzle.
1. 13 × 3 = 39
2. 39 × 11 = 429
3. 429 – 20 = 409
4. 409 + 19 = 428
Digits rewards players with up to 3-stars; you’ll get 3-stars for getting to the target exactly, 2-stars if you are within 10 of the target, and 1-star if you are within 25 of the target. Players can start a new operation by selecting a different number, and fractions or negative numbers are not accepted. New puzzles are released daily at midnight.
When working on a target from Digits, one of the first things you can do is to look for factor pairs of the target number, which can sometimes clue you into the direction you need to go. This is often the last operation of the solution, and you’ll use other operations to get to the numbers you need for the final factor pair.
We hope our NYT Digits April 23, 2023 Answer Guide helped you out on today’s puzzles! Check out more of our NYT Digits coverage.
### Christine Mielke
Christine Mielke has been writing content for the web for over 15 years. She is well-known for concise, informative content and her transparency. Christine is a 2011 graduate of Santa Clara University’s JD/MBA program, after having graduated in 2007 from University of California, Irvine with B.A. in Economics and B.A. in Political Science.
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# SSAT Elementary Level Math : Lines
## Example Questions
← Previous 1
### Example Question #5811 : Ssat Elementary Level Quantitative (Math)
Which is true about a ray?
A ray is always shaped like a circle.
A ray extends indefinitely in both directions.
A ray is always shorter than a line segment.
A ray is a line with a start point, but no end point.
A ray has three end points.
A ray is a line with a start point, but no end point.
Explanation:
A ray always has one start point, and then continues indefinitely in the opposite direction. Line segments are different because they have a defined start and end point.
### Example Question #1 : How To Find Length Of A Line
Alex needs to buy a fence for her yard. She doesn't know how much she needs to surround the entire yard. What does she need to calculate to figure this out?
perimeter of the yard
volume of the yard
area of the yard
height of the fence
perimeter of the yard
Explanation:
Perimeter is the distance around a shape. She needs to know the perimeter of her yard to determine how much fence to buy.
### Example Question #1 : Lines
Line AC is 12 inches long. If Point B is located in the center of Line AC, how many inches long is Line BC?
Explanation:
Since Point B is located in the middle of Line AC, it will break Line AC into two equal line segments - Line AB and Line BC. We should divide the original length of Line AC by 2 because we are breaking the original line in half.
### Example Question #1 : How To Find Length Of A Line
A line is inches long on a number line, if the starting point is , what is the end point on the number line?
Explanation:
You must add the start point to the length of the line to find the end point of the line.
### Example Question #3 : How To Find Length Of A Line
A pentagon has ____ sides.
Explanation:
A pentagon has five sides.
### Example Question #1 : Lines
Line AC is 24 inches long. If Point B is the midpoint of Line AC, how many inches long is Line BC?
Explanation:
Point B is the midpoint that breaks Line AC into two line segments of equal length. Therefore, to find the length of Line BC, divide the length of Line AC by 2:
### Example Question #1 : How To Find Length Of A Line
What is a line segment?
A line that has no start or end points.
A line that has a start point but no end point.
A line that is always greater than 10 centimeters.
A line that is always smaller than 10 centimeters.
A part of a line that connects two points.
A part of a line that connects two points.
Explanation:
A line segment is a part of a line that connects two points. It is different than a line and a ray because it has a defined start and end point
### Example Question #1 : How To Find Length Of A Line
Which statement is true regarding line segments?
A line segment extends in both directions without endpoints.
A line segment has one endpoint, but continues indefinitely in one direction.
A line segment has three endpoints.
A line segment is a "piece" of a line and has two endpoints.
A line segment is curved.
A line segment is a "piece" of a line and has two endpoints.
Explanation:
A line segment is a piece of a line connected by two endpoints.
### Example Question #2 : How To Find Length Of A Line
A line is connected from to . What is the length of the line?
Explanation:
Write the distance formula.
Substitute the points and solve for the distance.
### Example Question #1 : Lines
Find the length of a line which starts at the point (1,7) and goes till (9,7).
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# Factors of 40: Find Divisors, Pairs, and Prime Factors
May 26, 2023
Factors are the numbers that can be multiplied together to obtain a given number. In the case of 40, there are several factors that can evenly divide it. Understanding the factors of 40 is important in various mathematical operations and can help in simplifying fractions, finding common multiples, and determining the greatest common factor (GCF) between numbers.
## What Are the Factors of 40?
The factors of 40 are the numbers that divide it without leaving a remainder.
Let’s start dividing 40 equally by real numbers
1. Divide 40 by 1, equals 40
2. Divide 40 by 2, equals 20
3. Divide 40 by 4, equals 10
4. Divide 40 by 5, equals 8
5. Divide 40 by 8, equals 5
6. Divide 40 by 10, equals 4
7. Divide 40 by 20, equals 2
8. Divide 40 by 40, equals 1
For 40, the factors include 1, 2, 4, 5, 8, 10, 20, and 40. These numbers can be multiplied in different combinations to obtain 40. For example, 2 * 20 = 40 and 4 * 10 = 40.
## Factors of 40 in Pairs
Factors of a number come in pairs, where each pair multiplies together to give the original number. For 40, the pairs of factors are (1, 40), (2, 20), (4, 10), and (5, 8). When multiplied together, these pairs result in 40. For instance, 4 * 10 = 40.
## All Factors of 40
To find all the factors of 40, we can list the numbers that divide it evenly. Starting with 1, we see that it is a factor of 40 as 40 divided by 1 equals 40. Moving on, we find that 2 is also a factor as 40 divided by 2 equals 20. Continuing, we discover that 4 is a factor, as 40 divided by 4 equals 10. Next, we have 5 as a factor, and 40 divided by 5 equals 8. Finally, we reach the pair (8, 5), where 8 is a factor obtained by dividing 40 by 5. The final pair is (10, 4), and dividing 40 by 10 gives us 4. Thus, the complete list of factors of 40 is 1, 2, 4, 5, 8, 10, 20, and 40.
## Prime Factors of 40
Prime factors are the prime numbers that, when multiplied together, give the original number. To find the prime factors of 40, we start by dividing it by the smallest prime number, which is 2.
• Since 40 is divisible by 2, we divide it by 2, resulting in 20.
• Next, we divide 20 by 2, giving us 10.
• Continuing, we divide 10 by 2, yielding 5.
At this point, we have obtained the prime factorization of 40, which is 2 × 2 × 2 × 5. So, the prime factors of 40 are 2 and 5.
What are the factors of 40?
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.
How do I find the factors of 40?
To find the factors of 40, you can divide 40 by different numbers starting from 1 and going up to the square root of 40. If the division is exact, then the divisor is a factor of 40. Repeat this process until you reach the square root of 40.
Are 40 and 20 considered a pair of factors?
No, 40 and 20 are not considered a pair of factors because they are not factors of each other. However, both 40 and 20 are factors of a common number, which is 80.
How many factors does 40 have?
The number 40 has eight factors. Factors always come in pairs, so there are four pairs of factors for 40.
Is 40 a prime number?
No, 40 is not a prime number. It is a composite number because it has factors other than 1 and itself.
Can you find the prime factorization of 40?
Yes, the prime factorization of 40 is 2 * 2 * 2 * 5.
What is the greatest common factor (GCF) of 40 and another number?
The greatest common factor of 40 and another number will depend on the factors they have in common. To find the GCF, determine the common factors and choose the greatest one.
How can knowledge of the factors of 40 be useful in mathematics?
Knowing the factors of 40 is helpful in various mathematical operations such as simplifying fractions, finding common multiples, determining the greatest common factor, and solving problems related to divisibility and prime numbers.
Can you provide an example of using the factors of 40 in a real-life situation?
Sure! If you have 40 cookies and want to distribute them equally among a certain number of children, knowing the factors of 40 (1, 2, 4, 5, 8, 10, 20, and 40) can help you determine the different possible numbers of children you can distribute the cookies to without any leftovers.
What are the factors of 40?
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.
Is 40 a prime number?
No, 40 is not a prime number because it has factors other than 1 and itself.
How many factors does 40 have?
40 has a total of 8 factors.
What is the largest factor of 40?
The largest factor of 40 is 40 itself.
What are the prime factors of 40?
The prime factors of 40 are 2 and 5.
Is 40 a perfect square?
No, 40 is not a perfect square because it cannot be expressed as the square of an integer.
Can 40 be divided evenly by 7?
No, 40 cannot be divided evenly by 7.
What is the sum of the factors of 40?
The sum of the factors of 40 is 90.
What is the product of the factors of 40?
The product of the factors of 40 is 25600.
Can 40 be expressed as a product of prime numbers?
Yes, the prime factorization of 40 is 2^3 * 5, where 2 and 5 are prime numbers.
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S k i l l
i n
A R I T H M E T I C
Lesson 33
# GREATEST COMMON DIVISORLOWEST COMMON MULTIPLE
We will now learn how to find the greatest common divisor and the lowest common multiple of two numbers from their prime factorizations. (In Lesson 23 we saw how to find the LCM directly.) But first, let us see how to find all the divisors of a number from its prime factors.
We say that one number is a divisor of a second when the second is its multiple. 4 is a divisor of 36 because 36 is a multiple of 4: 36 = 9 × 4. And we can see that in the prime factors of 36:
36 = 2 × 2 × 3 × 3.
Apart from the order, 36 = 9 × 4.
4 is a divisor of 36. (And so is 9.)
All the divisors of a number can be found from its prime factors.
Example 1. Here is the prime factorization of 60:
60 = 2 × 2 × 3 × 5.
By taking those primes singly, then two at a time, then three at a time, and so on, we can construct all the divisors of 60.
Singly: 2, 3, 5, and 1.
Do not forget 1. Although 1 is not a prime, 1 is a divisor of every number.
Two at a time: 2 × 2, 2 × 3, 2 × 5, 3 × 5.
That is: 4, 6, 10, 15.
Three at a time: 2 × 2 × 3, 2 × 2 × 5, 2 × 3 × 5.
That is: 12, 20, 30.
All four: 2 × 2 × 3 × 5 = 60.
These are all the divisors of 60. 60 is a multiple of each one.
Problem 1. Write the prime factorization of 180. Then construct all its divisors.
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
180 = 2 × 2 × 3 × 3 × 5.
Singly: 2, 3, 5, and 1.
Two at a time: 2 × 2, 2 × 3, 2 × 5, 3 × 3, 3 × 5.
Three at a time: 2 × 2 × 3, 2 × 2 × 5, 2 × 3 × 3, 2 × 3 × 5,
3 × 3 × 5.
Four at a time: 2 × 2 × 3 × 3, 2 × 2 × 3 × 5, 2 × 3 × 3 × 5.
All five: 2 × 2 × 3 × 3 × 5
These numbers are:
1 2 3 5 4 6 10 9 15 12 20 18 30 45 36 60 90 180
Problem 2. A certain number is not divisible by 3. Therefore why is it not divisible by 12?
Since 12 = 2 × 2 × 3, then any number divisible by 12 -- that is, which is a multiple of 12 -- must have 3 as a prime divisor.
Greatest common divisor
Here are all the divisors of 12:
1, 2, 3, 4, 6, 12.
Here are all the divisors of 20:
1, 2, 4, 5, 10, 20.
Which numbers are the common divisors of 12 and 20?
1, 2, 4.
What number is their greatest common divisor?
4.
Problem 3. What number is the greatest common divisor of each pair?
a) 6, 9. 3 b) 8, 12. 4 c) 16, 40. 8 d) 7, 14. 7 e) 4, 20. 4 f) 6, 42. 6 g) 5, 11. 1 h) 6, 35. 1 i) 1, 12. 1
Problem 4. Which prime factors do these two numbers share?
2 × 2 × 2 × 5 and 2 × 5 × 5
2 × 5
2 × 5 is their greatest common divisor.
For, we can construct the divisors of a number from its prime factors. From the primes of each number we could construct the divisor 2 × 5.
The greatest common divisor of two numbers is the largest product of primes that the two numbers share.
Problem 5. Find the greatest common divisor of each pair.
a) 3 × 11 and 11 × 13. 11 b) 5 × 7 × 7 and 5 × 5 × 7. 5 × 7
c) 2 × 3 × 7 and 5 × 7 × 7 × 11. 7
d) 2 × 2 × 2 × 3 × 5 × 5 and 2 × 3 × 3 × 5 × 5 × 5 × 7.
2 × 3 × 5 × 5
They share one 2, one 3, and two 5's.
e) 3 × 7 and 11 × 29. 1
e) Those numbers don't share any primes. But 1 is a common divisor of
e) every pair of numbers. In the case, it is their only -- and greatest --
e) common divisor.
f) 5 × 5 and 5 × 5 × 5 × 5 × 5. 5 × 5
Problem 6. Find the greatest common divisor of each pair.
a) 45 and 75. 15 b) 42 and 63. 21 c) 30 and 77. 1
Problem 7.
a) What is the greatest common divisor of 12 and 35? 1
1 is a common divisor of every pair of numbers. But when 1 is their only and hence their greatest common divisor, we say that those numbers are relatively prime. Now, 12 and 35 are not prime numbers, but they are relatively prime.
b) Write the prime factorizations of 12 and 35.
12 = 2 × 2 × 3. 35 = 5 × 7.
c) What prime factors do they share? None.
That is how to recognize when two numbers are relatively prime.
Problem 8. Which of these pairs are relatively prime?
a) 6 and 35. Yes. b) 6 and 21. No. c) 8 and 27. Yes. d) 13 and 91. No. e) 9 and 20. Yes. f) 1 and 16. Yes.
Lowest common multiple
We saw examples of what we mean by the lowest common multiple in Lesson 23. We also saw a way to find it.
Problem 9. What number is the lowest common multiple of each pair?
a) 9 and 12. 36 b) 6 and 8. 24 c) 10 and 12. 60 d) 3 and 15. 15 e) 4 and 24. 24 f) 11 and 55. 55. g) 2 and 3. 6 h) 5 and 8. 40 i) 8 and 9. 72
Problem 10. Name the lowest common multiple (LCM) of each pair. Then name their greatest common divisor (GCD).
a) 12 and 16. LCM = 48. GCD = 4.
b) 15 and 20. LCM = 60. GCD = 5.
c) 13 and 39. LCM = 39. GCD = 13.
d) 5 and 8. LCM = 40. GCD = 1.
e) 20 and 24. LCM = 120. GCD = 4.
Problem 11. What number is the lowest common multiple of 6, 8, and 10? 120
We will now see how to find the LCM by writing the prime factorizations.
Example 2. Here are the prime factorizations of 24 and 20.
24 = 2 × 2 × 2 × 3. 20 = 2 × 2 × 5.
Now, each multiple of these numbers will have its own prime factorization. The prime factorization of a multiple of 24 will contain all the primes of 24, and a multiple of 20 will have all the primes of 20. A common multiple will have all the primes of each. Their lowest common multiple will be the smallest product that contains every prime from each number.
Here it is:
LCM = 2 × 2 × 2 × 3 × 5.
We have taken the most of each prime from the two numbers: The three 2's of 24, the one 3 of 24, and the one 5 of 20. The above is the smallest product of primes that contains both 2 × 2 × 2 × 3 and
2 × 2 × 5.
To evaluate that number, the order of the factors does not matter. (Lesson 9.) Therefore let us take advantage of 2 × 5 = 10. We will group the factors as follows:
(2 × 5) × (2 × 2 × 3) = 10 × 12 = 120.
Problem 12. Construct the lowest common multiple of the following.
a) 2 × 3 and 3 × 5. LCM = 2 × 3 × 5 = 30.
b) 3 × 3 × 5 and 3 × 5 × 5. LCM = 3 × 3 × 5 × 5 = 225.
c) 2 × 3 × 5 × 5. and 2 × 2 × 2 × 5 × 7.
c) LCM = 2 × 2 × 2 × 3 × 5 × 5 × 7 = 4200.
d) 2 × 2 and 2 × 2 × 2. LCM = 2 × 2 × 2 = 8.
e) 7 and 11. LCM = 7 × 11 = 77.
f) 2 × 3 and 5 × 7. LCM = 2 × 3 × 5 × 7 = 210.
g) 2 × 5, 7 × 11, and 5 × 11. LCM = 2 × 5 × 7 × 11 = 770.
Problem 13. Find the lowest common multiple of each pair.
a) 21 and 33. 3 × 7 × 11 = 231.
b) 65 and 39. 195 c) 54 and 75. 1350 d) 6 and 77. 462 e) 17 and 33. 561
Problem 14. Find the lowest common multiple of
a) 6, 8, and 10. 2 × 2 × 2 × 3 × 5 = 120. Compare Problem 11.
b) 14, 35, and 55. 770.
Problem 15. 15 and which other numbers have 60 as their lowest common multiple?
60 = 2 × 2 × 3 × 5. Since 15 = 3 × 5, then each of the other numbers must have 2 × 2. Those numbers are:
2 × 2 = 4.
2 × 2 × 3 = 12.
2 × 2 × 5 = 20.
2 × 2 × 3 × 5 = 60.
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## Wednesday, September 10, 2008
### Geometry - Chapter 1 - Essentials of Geometry
Chapter 1 - Essentials of Geometry
1-1 Undefined Terms:
A. Undefined terms:
their meaning is accepted without definition
1. Point: definition - is that which has no part, a dot.
2. Set: is a collection of objects
3. Line - is breathless length;
4. Straight line: is a line that lies evenly with the points on itself.
5. Plane: is a set of points that form a flat surface extending indefinitely in all directions.
1-2 The real numbers and their Properties
B.
Every real number corresponds to a point on a number line and every point on the number line corresponds to a real number.
1-3 Definitions, lines, and line segments
A. Definition: is a statement of the meaning of the term
B. Collinear set of points: is a set of points all of which lie on the same straight line.
C. Noncollinear set of points: is a set of three or more points that DO NOT all lie on the same straight line.
D. Distance between two points - every point on a line corresponds to a real number called its coordinate. To find the distance between any two points, find the absolute value of the difference between the coordinates of the two points.
If point A is at -2 and point D is at 1, what is the distance between A and D?
take the absolute value of (-2 - 1) = the absolute value of (-3) = 3
E. Order of points on a line
1. Betweenness:
B is between A and C if and only if A, B, and C are distinct collinear points and AB + BC = AC
also, this is the partition postulate: the part + the part = the whole
2. Line Segment: or segment, is a set of points consisting of two endpoints on a line, called endpoints, and all of the points on the line between the endpoints.
1-5 Rays and Angles:
Definition: two points, A and B, are on one side of a point P if A, B, and P are collinear and P is not between A and B.
Half-line: every point on a line divides the line into 2 opposite set of points called half-lines.
Definition: A ray consists of a point (endpoint) on a line and all points on one side of the point.
Definition: Opposite rays: are two rays of the same line with a common endpoint and no other point in common.
Point B is the vertex, BA is a side of the angle and BC is the other side of the angle.
Definitions:
The measure of an angle is the number of degrees in the angle.
A straight angle is an angle that is the union of opposite rays.
Its degree measure is 180 degrees
Acute Angles - is an angle whose degree measure is greater than 0 degrees and less than 90 degres.
Right angles - is an angle whose degree measure is 90 degrees.
Obtuse angle - is an angle whose degree measure is greater than 90 degrees and less than 180 degrees.
Note:If you bisect a straight angle, you get two 90 degree angles or 2 right anglesIf you bisect an obtuse angle, you get two acute angles.
1-6 More angle definitions
Congruent angles are angles that have the same measure.
Definition: A bisector of an angle is a ray whose endpoint is the vertex of the angle and that divides that angle into two congruent angles.
Given angle ABC with angle bisector BD, we can conclude that angle ABD is congruent to angle CBD.
If angle ABC measures 70 degrees, what is the measure of angle ABD? 35 degrees
Example: Given angle QRS with angle bisector RT, if the measure of angle QRS = 10x, and the measure of angle SRT = 3x + 30, what is the measure of angle QRS?
we know: measure of angle QRT = measure of angle SRT + measure of angle QRS
and since the angle QRS is bisected by ray RT, then angle QRT = angle SRT
so 3x + 30 = measure of angle SRT,
therefore
3x + 30 + 3x + 30 = 10x
6x + 60 = 10x
60 = 4x
15 = x
measure of angle QRS = 10x = 10 (15) = 150 degrees
1-7 Triangles:
I) Vocabulary:
A) Base angles - the two angles adjacent to the base of an isosceles triangle.
B) Vertex angles - the angle opposite the base of an isosceles triangle.
II) Theorems:
A) Base Angles Theorem: if two sides of a triangle are congruent, then the angles opposite them are congruent.
B) Converse of the Base Angles Theorem: if two angles of a triangle are congruent, the the sides opposite them are congruent.
Corollaries:
A) If a triangle is equilateral, then it is equiangular.
B) If a triangle is equiangular, then it is equilateral.
I) Vocabulary:
A) Triangle - is a figure formed by three segments joining three noncollinear points.
B) Classification of triangles:
1) By sides:
a) Equilateral Triangle - has 3 congruent sides
b) Isosceles Triangle - has at least 2 congruent sides
c) Scalene Triangle - has no congruent sides
2) By angles:
a) Acute Triangle - has 3 acute angles
b) Equiangular Triangle - has 3 congruent angles that measure 60 degrees each
c) Right Triangle - has one right angle and 2 acute angles
d) Obtuse Triangle - has one obtuse angle and 2 acute angles
C) Vertex - each of the three points joining the sides of a triangle (plural - vertices)
D) Adjacent Sides - in a triangle, two sides sharing a common vertex.
E) Right triangles have 2 sides that form the right angle called the legs. The side opposite the right angle is the hypotenuse of the triangle.
F) Interior angles - when the sides of a triangle are extended, other angles are formed. the three original angles are the interior angles.
G) Exterior angles - the angles that are adjacent to the interior angles.
H) Theorem:
1) Triangle sum theorem - the sum of the measures of the interior angles of a triangle is 180 degrees.
2) Exterior Angle Theorem - the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.
I) Corollary to a theorem - is a statement that can be proved easily using the theorem.
Example: The acute angles of a right triangle are complementary.
Example: given Triangle ABC with side BC extended through point D, if angle A = 65 degrees and angle ACD = 2x + 10 and angle B = x, solve for x.
angle A + angle B = angle ACD65 + x = 2x + 1055 = x
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## Accenture Sample Problems On Speed
Below are three problems dealing with speed and time calculations.
Question 1
Two buses leaving from two stations 20 km away from each other travel with constant speed of 45km/hr towards each other.Find the time taken to cross each other if the length of each bus is 100m.
a)10sec b)4sec c)15sec d)20sec
Solution :
The time taken to cross each other = (a + b) / (u + v) sec.
Here, a = b = length of the two buses = 100m = 1/10 km.
and u = v = speed of the two buses = 45km/hr.
Time taken to cross each other = (a + b) / (u + v) sec = (1/10 + 1/10) / (45 + 45) hours
1 / (10 x 90) = 1/900 hours.
since 1 hour = 3600 sec, 1/900 hour = 3600/900 = 4 sec
Hence the answer is 4 sec.
Question 2
A bus starts from A at 7 a.m and reaches the destination B at 7.30 a.m while a cyclist starts from B at 7 a.m and reaches A at 8.30 a.m. At what time the bus and the cyclist will cross each other?
a)7.34 a.m b)7.49 a.m c)7.23 a.m d)8.01 a.m
Solution :
Let the distance between A and B is X km
Time taken by the bus to cover X km = 1/2 hour .(i.e., 7 a.m to 7.30 a.m)
Time taken by the cyclist to cover X km = 3/2 hour (i.e., 7 a.m to 8.30 a.m)
Speed = distance / time
Speed of the bus = X / 1/2 = 2X km/hr
And the speed of the cyclist = X / 3/2 = 2X / 3 km/hr.
Let they meet at point C at Y hours after 7 a.m.
Distance covered by the bus in Y hours = Speed X Time = 2XY km = AC
And the distance covered by the cyclist in Y hours = Speed x Time = 2XY/3 km = BC
Since C is the point where Bus and Cyclist cross each other, AC + BC = AB
In other words, 2XY + (2/3)XY = X
Y(2 + 2/3) = 1
Y = 3/8 hours.
Expressing in minutes, Y = 3/8 x 60 minutes = 22.5 minutes
Y = 23 minutes (approximately)
Therefore they meet at 7 a.m + 23 minutes = 7.23 a.m
Hence the answer is 7.23 a.m.
Question 3
A man rides a bike with a constant speed of 20 km/hr from A at 4 p.m and travels towards a destination B. Another rider with a speed of 25 km/hr starts from B at 5 p.m and travels towards A. If the straight distance between A and B is 110km then at what time they will meet?
a)7 p.m b)8 p.m c)7.30 p.m d)6 p.m
Answer : a)7 p.m
Solution :
Let the two riders meet at X hours after 4 p.m.
Speed of the first rider is 20km/hr and the distance covered by him in X hours = 20X km
Speed of the 2nd rider is 25 Km/hr and he starts at 5 p.m.
Distance covered by him in (X - 1) hours = 25(X - 1) km
Distance between A and B is 110km.
Therefore 20X + 25(X - 1) = 110
45X = 135
X = 135/45 = 3
So they meet at 3 hours after 4 p.m.
Hence the answer is 7 p.m.
## Wipro Sample Algebraic Problems
Below are three model problems dealing with the formulae (x+y)2 = x2 + y2 + 2xy and (x-y)2 = x2 + y2 - 2xy
Question 1
Find X when X - Y = 3 and X2 + Y2 = 89 where X and Y are integers.
a)10 b)-5 c)-10 d)-3
Solution :
We know that (x - y)2 = x2 + y2 - 2xy
Sub. the given values,
32 = 89 - 2XY
9 - 89 = -2XY
80 = 2XY
XY = 40
Since X and Y are integers and XY = 40,the possibilities of X and Y are as follows:
(1,40), (2,20), (4,10), (5,8), (-1,-40), (-2,-20), (-4,-10) and (-5,-8)
X - Y is a positive integer(3), so X will be greater than Y.
By checking the given condition X - Y = 3, we have X = 8, Y = 5 or X = -5, Y = -8.
i.e., X is either 8 or -5.
From the given options we can conclude that X = -5.
Question 2
If the summation and multiplication of two integer is 24 and 143 respectively then the difference of them is:
a)2 b)1 c)12 d)4
Solution :
Let A and B be two integers.
Then A + B = 24 and AB = 143.
We know that (x+y)2 = x2 + y2 + 2xy
Here, 242 = A2 + B2 + 2(143)
576 - 286 = A2 + B2
A2 + B2 = 290.
i.e., Summation of the square of two integers is 290.
i.e., A2 or B2 is < or = 290.
Since A and B are integers, the possibilities are {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289}.
From the above set of square numbers, the values may be
A2 + B2 = 1 + 289 or 121 + 169 = 290
A2 + B2 = 12 + 172 or 112 + 132 = 290
The condition A + B = 24 and AB = 143, is satisfied by A = 11, B = 13 or A = 13, B = 11
We have to find the difference of A and B.
Hence the difference is 2.
Question 3
If X - Y = 9, X2 + Y2 = 257 and X,Y are integers then what will be the value of X and Y ?
a)21,12 b)15,6 c)none of these d)cannot be determined
Answer : d)cannot be determined
Solution :
Here we use the formula, (x-y)2 = x2 + y2 - 2xy
92 = 257 - 2XY
XY = (257-81)/2 = 176/2 = 88
XY = 88
Since X and Y are integers and XY = 88 then the possibilities of X and Y are (1,88), (2,44), (4,22), (8,11), (-1,-88), (-2,-44), (-4,-22) and (-8,-11).
By observing, none of the above possibility of X and Y satisfies X - Y = 9.
Hence, the values of X and Y cannot be determined by given conditions.
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Hundreds and Thousands Place
Start Practice
## What are Hundreds and Thousands?
In the last lesson, you learned about Ones and Tens place values. Let's review. ā”ļø
### Ones and Tens Review
There are only 10 digits in all math:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
When counting, what comes after 9? š¤
Yes, 10! š We add a second digit.
We can count all the way to 99 with Tens and Ones digits.
### How Do We Count Above 99?
After 99 comes 100! š„³
We add a third digit! It goes in the Hundreds place.
What comes after 100?Ā
Yes, 101. ā
### What is the Hundreds Place Worth?
If you put 10 Tens together, you'll get 1 Hundred.
1 Hundred is also worth 100 Ones!
Ā Ā Ā Ā Ā We write the Hundreds place to the left (š) of the Tens.
Tip: Bigger place values always go on the left (š):
Let's put the number 368 in our table:
How many hundreds do we have?Ā
3 Hundreds
What do 3 Hundreds equal? š¤
300
Awesome!
We can use three-digit numbers to count all the way up to 999.
What comes after 999? š¤
1,000! š¤Æš
### What are Thousands?
So you just learned that...
10 Ones = 1 Ten
You also learned that...
10 Tens = 1 Hundred
What do you think 10 Hundreds equals?
Yes!Ā
10 Hundreds = 1 Thousand
Thousands go to the left (š) of Hundreds:
Tip: Bigger place values always go on the left (š).
How many Thousands do we have?Ā
7
What does 7 Thousands equal?
7,000
Tip: always write a comma (,) after the Thousands place.
So write 7,381, not 7381.
You can use four-digit numbers to count all the way to 9,999!
Good work! šŗ
Now, complete the practice. You'll understand more and remember for longer.
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# Solving Equations
## Objective
SWBAT apply algebraic techniques to solve different forms of algebraic equations.
#### Big Idea
Students are reminded of the power of algebra as a general problem solving tool.
## Bell work
15 minutes
Learning Targes for solving equations
I begin this unit on solving trigonometric equations with a review of solving algebraic equations more generally. I do this because many equations involving trigonometric functions can be classified into a few categories based on their structure:
• Linear form- isolate the function than use inverses to remove the function
• Quadratic form- factor of use quadratic equation to isolate the variable
• Function=Function- both sides of equation have the same function so set the insides equal to each other and solve.
• Rational form- Solve like a ratio or make denominators equal and set numerators equal.
In the days that follow this lesson we will use substitution to help us learn to solve trig equations. When students are fooled by the structure of the equation, we will learn to rewrite the equation using substitution to help students solve.
Students proficiency with solving one or more types of algebraic equation will also benefit from this quick review. I have in mind some of the common errors that my students are prone to as we begin this lesson. For example, I know that I will work with several students who always want to start by isolating a variable. I have included problems like this one: 3x - 4=x^2 - 2x + 4
The lesson starts with four problems to solve. The first two have no solution, which will confuse some students. Students generally assume that all equations have solutions. To help students understand what is happening in these two problems, I will suggest that they graph the left and right sides of the equation. Depending on the level of confusion we may do this as a class. I want to make sure that my students are correctly interpreting the meaning of a situation in which two graphs do not intersect. In addition, methods that can be used to solve Question 2 give a possible solution. But, the proposed solution does not fit in the domain of the original problem, so it must be excluded from the solution set. This fact usually emits groans. Students immediately think that if they find a possible solution, it is the answer. So, we'll discuss how some techniques, such as squaring both sides of an equation, can result in extraneous solutions. The important message is that algebra is powerful, but not perfect.
Questions 3 and 4 ask students to solve a quadratic equation. There are a range of missteps and misconceptions that may arise. With these problems, I hope that my students will help each other to overcome many of these mistakes. I am ready to remind students about some of the techniques that they have learned over their time in high school.
Once students have had time to complete these four problems, I will use Popsicle Sticks to randomly call on a student to share what they have on their paper for each problem. I often use this simple strategy when we are reviewing content. It helps students to recognize that they are responsible for working to remember the concepts we have studied and apply them when needed.
## Review of solving
30 minutes
We will continue today's review of algebraic solving techniques with a set of set of equations to solve. To foster independent recall, but provide collaborative support, I will ask my students to use a Think-Pair-Share strategy as they attempt these problems.
• Think - I have students work on a problem for a minute or two on their own.
• Pair - I give students 2 minutes to discuss ideas for how to solve the problem with another student.
• Share - I randomly pick one student to do the problem on the board while others attempt the problem at their seats. Then, we discuss the problem.
With this protocol, we will focus on one question at a time. Giving one problem allows students to focus on the techniques for that problem without worrying about doing all the problems. As students finish, I will encourage them to compare their approach to the one being taken by our volunteer at the board.
Consistent with the goals for the lesson, the problem structures in this set are ones that may prove useful when we are solving trigonometric equations in upcoming lessons. We may or may not get through all of the problems. We may also jump around in the list, based on which problems I think will most help my students. With each problem that we discuss, we will discuss the domain of the original equation. This step in the process is important as we prepare to learn to solve trigonometric equations.
## Closure
10 minutes
During class we spent time reviewing techniques for solving. Of course students will need to use the techniques when solving word problems. As class comes to an end I give students four application problems to solve. These problems allow students to not only practice their solving strategies but also help students continue to work on reading mathematical text. In my class, I will assign:
p. A61, #156, 158, 161, 162 from Larson "Precalculus with Limits"
At the end of today's lesson I will ask students to complete an Exit Slip that will give me a sense of how they are feeling about solving algebraic equations:
1. Was their a particular misconception that you recognized in your work today when we went over the solutions to the equations?
2. Is there a particular type of equation that you need more practice with in order to become proficient?
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# "Wholes and Parts" (Number Sense Series, Part 2)
May 9, 2018 | Diamond Bar
When you step into a Mathnasium Learning Center, you will often hear the term "Number Sense". What is Number Sense? This blog article goes over number sense from the definition to diving into how we use it at Mathnasium to make math make sense to kids.
Number Sense is the ability to appreciate the size and scale of numbers in the context of the question at hand. Three elements establish Number Sense: Counting, Wholes and Parts, and Proportional Thinking. We already introduced Counting in Part 1 of our Number Sense Series. Today, we will focus on Wholes and Parts.
The two aspects of the concept of Wholes and Parts form the backdrop for many mathematical concepts: fractions and complements.
First, let’s discuss fractions. For many people, when they hear the phrase “wholes and parts,” the first thing that comes to mind is “fractions.” This is also the first aspect of “wholes and parts” in the Mathnasium curriculum.
Many people do not have a clear idea of what a fraction represents: a whole that is broken into equal parts creates fractions. Children demonstrate an awareness of this idea from a very early age through the concept of sharing. Ask a young child to share a candy bar with you in a fair way and watch what they do; importantly, have them explain their thought process to you.
Children need to be introduced to the fraction “half” as being “2 parts the same.” Before other fractions are introduced (1/3s, 1/4s…) children need to master questions such as:
How much is half of 6? 3? 7? 20? ½? 99?
and…
Half of what number is 5? 10? 25?
100% of something represents a whole. Break down the word ‘percent’ for what it is: per-CENT— “for each 100.” This connection eliminates the need to get out a pencil and paper to compute 7% of 250, as the result is 7 + 7 (7 for each 100) + 3½ (half of 7 for the half of 100). The answer is 17½.
The second aspect of “wholes and parts” supported by Mathnasium curriculum is complements.
complement is the amount needed to make a whole complete.
Remember the age-old expression:
“The whole equals the sum of its parts”
and the related statement that
“Any one part equals the whole minus the sum of the other the other part(s).”
This second statement is saying that “the whole is made up of complements.”
Let’s dig deeper into the concept of complements.
Strong problem-solving skills require the ability to identify the missing part(s)—the complement(s).
Together we have 10 pieces of candy. You have 7 pieces. How many pieces do I have? Here, the whole is 10 and one part is 7. So, the other part is 10 minus 7 (the whole minus the part you know). So, 3 is the complement.
This will help set up their understanding of complements.
When dealing with problem solving that involves Wholes and Parts, kids really like examples that deal with cookies, sandwiches, or anything they can eat. Try this example:
You have a box of cookies. You get to eat half of them after lunch and half of what is left over after dinner. After dinner, you have 3 cookies left. How many cookies did you start with? [The answer is 12.]
When a word problem is set up like this, children can often visualize the situation. If visualization doesn’t work, have a Plan B (drawing a picture), or C (using physical objects) until the child understands.
By learning to think in terms of Wholes and Parts, a child can develop a strong understanding of the structure of mathematics. Over time, knowledge of Wholes and Parts helps to build up a child’s foundational knowledge, so they can learn to solve complex fractions, equations, and a wide variety of word problems. Without a solid understanding of Wholes and Parts, solving many types of word problems becomes difficult—if not impossible—for many students.
For every child, there is a way to explain any topic in a way that makes sense to them.
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written by Sunny Yoon
Make up linear functions f(x) and g(x). Explore, with different pairs of f(x) and g(x) the graphs for
h(x) = f(x) + g(x)
h(x) = f(x)*g(x)
h(x) = f(x)/g(x)
h(x) = f(g(x))
Summarize and illustrate.
Assumption:
When you add two linear functions f(x) and g(x), the sum of two functions h(x) is going to be another linear function.
When you multiply two linear functions, the product h(x) is going to be a quadratic function.
When you divide two linear functions, the quotient h(x) is going to be a rational function.
When you compose two linear functions, the composition h(x) is going to be another linear function.
f(x) = ax + b
g(x) = cx + d
h(x) = (ax + b) + (cx + d)
When a > 0 and c > 0, then the slope of h(x) is going to be steeper than f(x) and g(x) since the slope of h(x) is going to be (a + c) > 0.
f(x) = 3x + 1(Green); g(x) = 5x + 7(Red)
h(x) = (3x + 1) + (5x + 7); (Blue)
Similar situation arises when a < 0 and c < 0.
f(x) = -x - 2(Green); g(x) = -4x + 5(Red)
h(x) = (-x - 2) + (-4x + 5); (Blue)
It's easier to notice the steepness of each function as x goes to positive infinity. Since the y-intercept of g(x) is greater than h(x), it's harder to determine the steepness on the left side of the graph.
What if the sign of a and c are different? Then, the sign of the bigger absolute value slope will determine the slope of h(x).
Here are two examples to illustrate that point.
f(x) = -x - 2(Green); g(x) = 3x + 1(Red)
h(x) = (-x - 2) + (3x + 1); (Blue)
The slope of h(x) is in between f(x) and g(x) and it it positive just like g(x) since the |c| > |a|.
f(x) = -4x + 5(Green); g(x) = 3x + 1(Red)
h(x) = (-4x + 5) + (3x + 1); (Blue)
In order to determine the steepness of the functions, please look at the right side of the graph. Once again, the slope of h(x) is in between of f(x) and g(x) and the sign is negative just like f(x). Since |a| > |c|, the slope of h(x) is likely to follow the behavior of f(x).
What if a and c have different signs, but same absolute value? For example, f(x) = -2x - 2 (Green), g(x) = 2x + 4 (Red) and h(x) = (-2x - 2) + (2x + 4) (Blue).
h(x) will always become a constant function since the slope of f(x) and g(x) will cancel each other out when you add them together.
Mutiplying two linear functions
f(x) = ax + b
g(x) = cx + d
h(x) = (ax + b) * (cx + d) = (ac)x^2 + (ad + bc)x + (bd)
We can find out that h(x) will always be a parabola since it is a quadratic equation.
|Explanation Graphical Representation Same sign (a,c are both positive or negative) If a and c are same signs, that the parabola will be a U-shape. It doesn't matter whether |a| > |c| or |a| < |c| because the new slope of h(x) is going to be a*c. Multiplying two positives or two negatives always result in a positive answer. h(x) is represented by the blue parabola. Different Sign (a and c have a different sign) If a and c have different signs, then the parabola will be a upside down U-shape because a*c will result in a negative number. h(x) is represented by the blue parabola.
Dividing two linear functions
f(x) = ax + b
g(x) = cx + d
h(x) = (ax + b)/(cx + d)
h(x) = (ax +b)/(cx + d) Example Used for the animation (n is between 0 and 10) Graphical Representation When a and c have same signs y = (nx +1)/(x - 3) When a and c have different signs Y = (nx + 1)/(-x-3)
Composition of two linear functions
f(x) = ax + b
g(x) = cx + d
h(x) = a(cx + d) + b = acx + (ad + b)
When composing two linear functions, it becomes another linear function. However, the slope of the composed linear function changes. The new slope is going to be a product of two previous linear function.
h(x) = a(cx + d) + b Example used for the graphical representation (n is between 0 and 10) Graph When a and c have same sign y = 2(nx - 1) + 3 The slope will be positive since two positives or two negatives equal to positive. When a and c have different signs y = -2(nx - 1) + 3 The slope will be a negative value since the product of two values with different signs equal to negative.
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Lesson: Simple Interest
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Lesson Objective
calculate simple interest
Lesson Plan
Aim/Objective:
Key Points
Calculate simple interest
· Interest - is an amount that you pay in order to use or borrow money or the amount earned on top of money that you invest.
· Principal - is the amount of money you borrow or invest
· Interest Rate - is the rate of interest paid over a given period of time, usually one year
Assessment:
Lesson
· _interest_is an amount that you pay in order to use or borrow money or the amount earned on top of money that you invest.
· Principal __ is the amount of money you borrow or invest.
· __interest rate_ is the rate of interest paid over a given period of time, usually one year
Notes:
Walk through examples/notes with students.
First you will be given three pieces of information – the principal, interest and time. Now we can find the amount of interest by following the three steps:
· There are three steps to finding interest. Let’s look at the following example:
Principal: \$450 Interest Rate: 4.9% Time: 6 months
· Step 1: Use the Interest Formula. Fill in the information that you are given.
I = p x r x t
Interest (I) = Principal (p) x Rate (r) x Time (t) I = \$450 x 4.9% x 6 months
Change the numbers so you can calculate
§ 4.9% = 0.049
§ 6 months = = year = 0.5
I = \$450 x 0.049 x 0.5
Step 2: Multiply to find the interest
I = \$450 x 0.049 x 0.5 I = \$22.05 x 0.5 = \$11.025
o Step 3: Add to find the principal plus the interest
\$450 + \$11.03 = \$461.03
Guided Practice
Walk through examples with students for finding interest
Find the simple interest and the sum of the principal and the interest for each.
Principal Interest Rate Time Interest Principal + Interest 1. \$598 6.6% 1 yr 2. \$225 4.2% 1 yr 3. \$1,560 5.0% 6 mo 4. \$209 7.1% 3 mo 5. \$10,000 4.75% 9 mo
Lesson Resources
Simple Interest HW 2,154 Simple Interest CW 1,655 Simple Interest Notes 2,028
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# Is this permutations or combinations?
I am a bit confused. When we use the multiplicative principle are we finding the number of permutations or combinations.
An example of using this principle is where I have $5$ shirts $3$ pairs of pants and $2$ pairs of shoes and I want to find out how many ways I can wear the ($1$ shirt, $1$ pair of pants and one pair of shoes). So we just multiply $5 \times 3 \times 2 = 30$ ways. So, is this the number of permutations or combinations?
Thank you :)
• It is the number of outfits! Commented Apr 18, 2015 at 9:27
Your example is just saying: how many different unique outfits can I make if I wear one shirt, one pair of pants and one pair of shoes.
So you say that there are 3 things you need to do:
choose 1 shirt, choose 1 pair of pants and choose 1 pair of shoes.
Multiplicative principle: If I have $5$ different shirts, $3$ different pairs of pants and $2$ different pairs of shoes, then there are $5 \times 3 \times 2$ ways of taking $1$ shirt, $1$ pair of pants and $1$ pair of shoes. You choose $1$ of each and multiply them together because this suggests they happen at the same time.
Sum Principle: If I have $5$ different red shirts, $3$ different blue shirts and $2$ different black shirts, then I have $5+3+2 = 10$ different shirts.
The number of possible ordered samples of size $r$ taken from a set of $n$ objects is $n^r$ when sampling with replacement. (you can still choose from all $n$ objects each time for $r$ samples)
e.g. If I have a $3$ digit code and each digit can be any number $0,1,2,...,9$ (10 options). How many codes can I have?
Answer: $10 \times 10 \times 10 = 10^3 = 1000$
The number of permutations of $n$ distinct objects with replacement is $n!$
e.g. Say I have 5 different books and I need to place them all on a bookshelf in 5 places. How many ways can I do this?
So you can choose from $5$ books in the first place. You have $4$ books left. Now you can place $4$ books in the second place. Do you understand that we are not replacing the books. Once we place one our sample size decreases.
Answer: $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$
The number of permutations of $n$ distinct objects when you only want to permute $r$ objects and not all $n$ but you are still taking them from $n$ objects without replacement:
$$^nP_r = \frac{n!}{(n-r)!}$$
E.g. Say I have 5 different colored ribbons. How many ways can I give 3 people ribbons such that each person gets only 1.
Answer: $5 \times 4 \times 3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3$
Do you see what I'm getting at. Logically you can just say $5 \times 4 \times 3$ but when numbers get very large using the general formula is helpful. It allows you to cancel out all the unwanted numbers.
Now combinations are where you are getting the number of ways in which $r$ objects can be combined but the order is not important anymore:
$$^nC_r = {n \choose r} = \frac{n!}{r!(n-r)!}$$
Note that the difference between this and the permutation is that you divide the permutation further by $r!$
E.g. Say that we have $5$ different coloured ribbons as above but instead of placing them in an order we want to see how many ways we can choose $3$ ribbons.
The fact that order is not important means that if you choose B,R,G (blue, green, red) then it is the same as choosing B,G,R or G,B,R or G,R,B or R,G,B or R,B,G. Note that if we are choosing 3 then we need to divide by $3!=6$ because that is how we eliminate the need for order
Answer: ${5 \choose 3} = \frac{5!}{(5-3)!3!} = \frac{5 \times 4 \times 3}{3!} = 10$
A permutation is a way of rearranging items, a combination is a way of combining different items together.
What you want is to combine one item from each set (where the different sets are the set of shirts, set of pants etc')
I have added links to Wikipedia for permutations and combinations, it may be worth while reading them
It's easiest to explain this with an example: take the set of colours {red, green, yellow, blue, orange}
The number of permutations is how many different ways they can all be re-arranged; since there are five elements, the answer is 5!
With combinations one is asking a different question; which is how many ways we can choose, say two, colours from the set. Here how they are arranged do not matter - so the choice of Red, Green and Green, Red do not matter.
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# Coupon collector's problem
In probability theory, the coupon collector's problem refers to mathematical analysis of "collect all coupons and win" contests. It asks the following question: If each box of a brand of cereals contains a coupon, and there are n different types of coupons, what is the probability that more than t boxes need to be bought to collect all n coupons? An alternative statement is: Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once? The mathematical analysis of the problem reveals that the expected number of trials needed grows as ${\displaystyle \Theta (n\log(n))}$.[a] For example, when n = 50 it takes about 225[b] trials on average to collect all 50 coupons.
## Solution
### Via generating functions
By definition of Stirling numbers of the second kind, the probability that exactly T draws are needed is
${\displaystyle {\frac {S(T-1,n-1)n!}{n^{T}}}}$
By manipulating the generating function of the Stirling numbers, we can explicitly calculate all moments of T:
${\displaystyle f_{k}(x):=\sum _{T}S(T,k)x^{T}=\prod _{r=1}^{k}{\frac {x}{1-rx}}}$
In general, the k-th moment is ${\displaystyle (n-1)!((D_{x}x)^{k}f_{n-1}(x)){\Big |}_{x=1/n}}$, where ${\displaystyle D_{x}}$ is the derivative operator ${\displaystyle d/dx}$. For example, the 0th moment is
${\displaystyle \sum _{T}{\frac {S(T-1,n-1)n!}{n^{T}}}=(n-1)!f_{n-1}(1/n)=(n-1)!\times \prod _{r=1}^{n-1}{\frac {1/n}{1-r/n}}=1}$
and the 1st moment is ${\displaystyle (n-1)!(D_{x}xf_{n-1}(x)){\Big |}_{x=1/n}}$, which can be explicitly evaluated to ${\displaystyle nH_{n}}$, etc.
### Calculating the expectation
Let time T be the number of draws needed to collect all n coupons, and let ti be the time to collect the i-th coupon after i − 1 coupons have been collected. Then ${\displaystyle T=t_{1}+\cdots +t_{n}}$. Think of T and ti as random variables. Observe that the probability of collecting a new coupon is ${\displaystyle p_{i}={\frac {n-(i-1)}{n}}={\frac {n-i+1}{n}}}$. Therefore, ${\displaystyle t_{i}}$ has geometric distribution with expectation ${\displaystyle {\frac {1}{p_{i}}}={\frac {n}{n-i+1}}}$. By the linearity of expectations we have:
{\displaystyle {\begin{aligned}\operatorname {E} (T)&{}=\operatorname {E} (t_{1}+t_{2}+\cdots +t_{n})\\&{}=\operatorname {E} (t_{1})+\operatorname {E} (t_{2})+\cdots +\operatorname {E} (t_{n})\\&{}={\frac {1}{p_{1}}}+{\frac {1}{p_{2}}}+\cdots +{\frac {1}{p_{n}}}\\&{}={\frac {n}{n}}+{\frac {n}{n-1}}+\cdots +{\frac {n}{1}}\\&{}=n\cdot \left({\frac {1}{1}}+{\frac {1}{2}}+\cdots +{\frac {1}{n}}\right)\\&{}=n\cdot H_{n}.\end{aligned}}}
Here Hn is the n-th harmonic number. Using the asymptotics of the harmonic numbers, we obtain:
${\displaystyle \operatorname {E} (T)=n\cdot H_{n}=n\log n+\gamma n+{\frac {1}{2}}+O(1/n),}$
where ${\displaystyle \gamma \approx 0.5772156649}$ is the Euler–Mascheroni constant.
Using the Markov inequality to bound the desired probability:
${\displaystyle \operatorname {P} (T\geq cnH_{n})\leq {\frac {1}{c}}.}$
The above can be modified slightly to handle the case when we've already collected some of the coupons. Let k be the number of coupons already collected, then:
{\displaystyle {\begin{aligned}\operatorname {E} (T_{k})&{}=\operatorname {E} (t_{k+1}+t_{k+2}+\cdots +t_{n})\\&{}=n\cdot \left({\frac {1}{1}}+{\frac {1}{2}}+\cdots +{\frac {1}{n-k}}\right)\\&{}=n\cdot H_{n-k}\end{aligned}}}
And when ${\displaystyle k=0}$ then we get the original result.
### Calculating the variance
Using the independence of random variables ti, we obtain:
{\displaystyle {\begin{aligned}\operatorname {Var} (T)&{}=\operatorname {Var} (t_{1}+\cdots +t_{n})\\&{}=\operatorname {Var} (t_{1})+\operatorname {Var} (t_{2})+\cdots +\operatorname {Var} (t_{n})\\&{}={\frac {1-p_{1}}{p_{1}^{2}}}+{\frac {1-p_{2}}{p_{2}^{2}}}+\cdots +{\frac {1-p_{n}}{p_{n}^{2}}}\\&{}<\left({\frac {n^{2}}{n^{2}}}+{\frac {n^{2}}{(n-1)^{2}}}+\cdots +{\frac {n^{2}}{1^{2}}}\right)\\&{}=n^{2}\cdot \left({\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{n^{2}}}\right)\\&{}<{\frac {\pi ^{2}}{6}}n^{2}\end{aligned}}}
since ${\displaystyle {\frac {\pi ^{2}}{6}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{n^{2}}}+\cdots }$ (see Basel problem).
Bound the desired probability using the Chebyshev inequality:
${\displaystyle \operatorname {P} \left(|T-nH_{n}|\geq cn\right)\leq {\frac {\pi ^{2}}{6c^{2}}}.}$
### Tail estimates
A stronger tail estimate for the upper tail be obtained as follows. Let ${\displaystyle {Z}_{i}^{r}}$ denote the event that the ${\displaystyle i}$-th coupon was not picked in the first ${\displaystyle r}$ trials. Then
{\displaystyle {\begin{aligned}P\left[{Z}_{i}^{r}\right]=\left(1-{\frac {1}{n}}\right)^{r}\leq e^{-r/n}.\end{aligned}}}
Thus, for ${\displaystyle r=\beta n\log n}$, we have ${\displaystyle P\left[{Z}_{i}^{r}\right]\leq e^{(-\beta n\log n)/n}=n^{-\beta }}$. Via a union bound over the ${\displaystyle n}$ coupons, we obtain
{\displaystyle {\begin{aligned}P\left[T>\beta n\log n\right]=P\left[\bigcup _{i}{Z}_{i}^{\beta n\log n}\right]\leq n\cdot P[{Z}_{1}^{\beta n\log n}]\leq n^{-\beta +1}.\end{aligned}}}
## Extensions and generalizations
${\displaystyle \operatorname {P} (T which is a Gumbel distribution. A simple proof by martingales is in the next section.
• Donald J. Newman and Lawrence Shepp gave a generalization of the coupon collector's problem when m copies of each coupon need to be collected. Let Tm be the first time m copies of each coupon are collected. They showed that the expectation in this case satisfies:
${\displaystyle \operatorname {E} (T_{m})=n\log n+(m-1)n\log \log n+O(n),{\text{ as }}n\to \infty .}$
Here m is fixed. When m = 1 we get the earlier formula for the expectation.
• Common generalization, also due to Erdős and Rényi:
${\displaystyle \operatorname {P} \left(T_{m}
• In the general case of a nonuniform probability distribution, according to Philippe Flajolet et al.[2]
${\displaystyle \operatorname {E} (T)=\int _{0}^{\infty }\left(1-\prod _{i=1}^{m}\left(1-e^{-p_{i}t}\right)\right)dt.}$
This is equal to
${\displaystyle \operatorname {E} (T)=\sum _{q=0}^{m-1}(-1)^{m-1-q}\sum _{|J|=q}{\frac {1}{1-P_{J}}},}$
where m denotes the number of coupons to be collected and PJ denotes the probability of getting any coupon in the set of coupons J.
## Martingales
This section is based on.[3]
Define a discrete random process ${\displaystyle N(0),N(1),\dots }$ by letting ${\displaystyle N(t)}$ be the number of coupons not yet seen after ${\displaystyle t}$ draws. The random process is just a sequence generated by a Markov chain with states ${\displaystyle n,n-1,\dots ,1,0}$, and transition probabilities
${\displaystyle p_{i\to i-1}=i/n,\quad p_{i\to i}=1-i/n}$
Now define
${\displaystyle M(t):=N(t)\left({\frac {n}{n-1}}\right)^{t}}$
then it is a martingale, since
${\displaystyle E[M(t+1)|M(t)]=(n/(n-1))^{t+1}E[N(t+1)|N(t)]=(n/(n-1))^{t+1}(N(t)-N(t)/n)=M(t)}$
Consequently, we have ${\displaystyle E[N(t)]=n(1-1/n)^{t}}$. In particular, we have a limit law ${\displaystyle \lim _{n\to \infty }E[N(n\ln n+cn)]=e^{-c}}$ for any ${\displaystyle c>0}$. This suggests to us a limit law for ${\displaystyle T}$.
More generally, each ${\displaystyle \left({\frac {n}{n-k}}\right)^{t}N(t)\cdots (N(t)-k+1)}$ is a martingale process, which allows us to calculate all moments of ${\displaystyle N(t)}$. For example,
${\displaystyle E[N(t)^{2}]=n(n-1)\left({\frac {n-2}{n}}\right)^{t}+n\left({\frac {n-1}{n}}\right)^{t},\quad n\geq 2}$
giving another limit law ${\displaystyle \lim _{n\to \infty }Var[N(n\ln n+cn)]=e^{-c}}$. More generally,
${\displaystyle \lim _{n\to \infty }E[N(n\ln n+cn)\cdots (N(n\ln n+cn)-k+1)]=e^{-kc}}$
meaning that ${\displaystyle N(n\ln n+cn)}$ has all moments converging to constants, so it converges to some probability distribution on ${\displaystyle 0,1,2,\dots }$.
Let ${\displaystyle N}$ be the random variable with the limit distribution. We have
{\displaystyle {\begin{aligned}E[1]&=1\\E[N]&=e^{-c}\\E[N(N-1)]&=e^{-2c}\\E[N(N-1)(N-2)]&=e^{-3c}\\&\vdots \end{aligned}}}
By introducing a new variable ${\displaystyle t}$, we can sum up both sides explicitly:
${\displaystyle E[1+Nt/1!+N(N-1)t^{2}/2!+\cdots ]=1+e^{-c}t/1!+e^{-2c}t^{2}/2!+\cdots }$
giving ${\displaystyle E[(1+t)^{N}]=e^{e^{-c}t}}$.
At the ${\displaystyle t\to 0}$ limit, we have ${\displaystyle Pr(N=0)=e^{-e^{-c}}}$, which is precisely what the limit law states.
By taking the derivative ${\displaystyle d/dt}$ multiple times, we find that ${\displaystyle Pr(N=k)={\frac {e^{-kc}}{k!}}e^{-e^{-c}}}$, which is a Poisson distribution.
2. ^ E(50) = 50(1 + 1/2 + 1/3 + ... + 1/50) = 224.9603, the expected number of trials to collect all 50 coupons. The approximation ${\displaystyle n\log n+\gamma n+1/2}$ for this expected number gives in this case ${\displaystyle 50\log 50+50\gamma +1/2\approx 195.6011+28.8608+0.5\approx 224.9619}$.
1. ^ Mitzenmacher, Michael (2017). Probability and computing : randomization and probabilistic techniques in algorithms and data analysis. Eli Upfal (2nd ed.). Cambridge, United Kingdom. Theorem 5.13. ISBN 978-1-107-15488-9. OCLC 960841613.{{cite book}}: CS1 maint: location missing publisher (link)
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CLASS-6MENSURATION-PERIMETER OF TRIANGLE
PERIMETER OF TRIANGLE -
The perimeter of a triangle is the sum of the lengths of its three sides. The formula for calculating the perimeter of a triangle depends on the lengths of its sides, which may vary depending on the type of triangle (e.g., equilateral, isosceles, scalene). Here are the formulas for finding the perimeter of different types of triangles:
1. Equilateral Triangle:- In an equilateral triangle, all three sides are of equal length. If s represents the length of a side, then the perimeter (P) is given by: P = 3S
Three sides, AB = BC = CD = S
So, Perimeter of Equilateral Triangle ABC = AB + BC + CD
= S + S + S = 3S
2. Isosceles Triangle:- In an isosceles triangle, two sides are of equal length, and the third side is of a different length. If X and Y represent the lengths of the two equal sides, and c represents the length of the unequal side, then the perimeter (P) is given by:
P = X + Y + Z
3. Scalene Triangle:- In a scalene triangle, all three sides have different lengths. If a, b, and c represent the lengths of the three sides, then the perimeter (P) is given by:
P = p + q + r
To calculate the perimeter of a specific triangle, you need to know the lengths of its sides. Simply add up the lengths of the sides using the appropriate formula based on the type of triangle. Keep in mind that the unit of measurement should be consistent for all side lengths in your calculation (e.g., meters, feet, inches).
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## Bouncing Tennis Balls
• Lesson
6-8
1
Students develop their skills in collecting and recording data using the real-world situation of a bouncing tennis ball. They use the data collected to formulate the relationship between the dependent and independent variable in their experiment.
The Activity
Students form teams of four to bounce a tennis ball. Pose the problem problem:
How many times can each team member bounce and catch a tennis ball in two minutes?
A bounce is defined as dropping the ball from the student's waist. One student keeps the time while the second student bounces and catches the ball, the third student counts the bounces, and the fourth student records the data in a table showing both the number of bounces during each ten-second interval and the cumulative number of bounces. Each trial consists of a two-minute experiment, with the number of bounces recorded after every ten seconds (or twenty seconds for fewer data points). The timekeeper calls out the time at ten-second intervals. When the time is called, the counter calls out the number of bounces that occurred during that ten-second interval. The recorder records this count and keeps track of the cumulative number of bounces.
The same process is followed by each student, with the students rotating roles, so that each student can collect a set of data. All the students must bounce the ball on the same surface (e.g., tile, carpet, concrete) because differences in the surface could affect the number of bounces.
Distribute the Bouncing Tennis Balls Recording Sheet to the students.
Bouncing Balls Recording Sheet
The data from one student's experiment are recorded in the table below.
Time (Seconds) Number of Bounces during Interval Cumulative Number of Bounces 0 0 0 10 11 11 20 11 22 30 9 31 40 10 41 50 11 52 60 10 62 70 11 73 80 11 84 90 10 94 100 10 104 110 10 114 120 10 124
Graphing the Data
Once the data have been collected, each student prepares a graph showing the cumulative bounces over two minutes. This graph can be constructed by using the Line of Best Fit Tool. To use the tool, students need to enter the data in the area shown below the graph. Data should be entered as coordinates, line-by-line, such as (10,11). In this piece of data, the 10 represents the time, and the 11 represents the number of bounces.
Alternatively, students may graph the data by hand, by using a graphing calculator, or by using a spreadsheet, depending on the students' experiences and on what information the teacher wants to gather about what the students know and are able to do.
The image below shows the data plotted using the Line of Best Fit Tool.
Alternatively, students may use a graphing calculator to display their data. The figure below shows such a display.
A graph made using a graphing calculator
Discussion
Students present their results to classmates by showing their graphs. The discussion can involve what the students found easy and what they found difficult in completing this task. Students' discussions can be revealing. During the discussion, think about these guiding questions:
• Can the students identify what varies in the experiment? Do they comment on the dependent and independent variables either implicitly, in their conversations about the graphs, or explicitly, using correct terminology?
• Do they discuss whether the points should be connected with a line? The numbers of bounces are discrete data, so they should not be connected.
• Decisions about the scale for each of the axes are important. Do the students understand what the graphs would look like if the scales changed?
• When directed to sketch lines on their graphs in order to notice trends, do they demonstrate some sense that the steepness of a line is related to the number of bounces per second?
Your observations related to these and other questions will yield information about what your students appear to know and are able to do that will guide you in making instructional decisions.
Building a Sense of Time and Its Relation to Distance and Speed
Initially students need to become aware of their own understanding of time, change over time, and the use of new kinds of measure (i.e., rates). Posing such questions as those listed below focuses their attention on these ideas (adapted from Kleiman et al. 1998).
• How do you measure time? Distance? Speed?
• Give an example of something that might be able to travel at two feet per second.
• What is the difference between traveling at two feet per second and two feet per minute or two feet per hour?
In this context, distance is how far the object or person moves (travels). Speed is how fast the object or person is moving (traveling). Both are described in terms of direction. Distance is measured in such units as feet, miles, or kilometers. Speed is measured in relation to time using units such as meters per second or miles per hour.
• Bouncing Tennis Balls Recording Sheet
• Tennis balls, one for each team of four students
• A clock or watch with a second hand
• Centimeter graph paper, a spreadsheet program, or a graphing calculator (optional)
Assessments
1. During the activity, it is important to observe the students. Note which students made graphs correctly, paying attention to how they used the idea of scale to set up the time and distance axes. Listen to students' conversations about their graphs, attending to comments that indicated the students realized that the number of bounces depends on the length of time the ball is bounced and that patterns develop when the ball is bounced in a consistent way.
Extensions
1. An extension of this activity would be for each student to conduct an experiment using, for example, concrete floors and then carpeted floors to investigate the effect of differences in the surfaces.
2. Students can also explore different activities that test their sense of time. They can do the following activities in pairs; in each instance they many want to observe if they overestimate or underestimate the time and try the task again.
• Clap your hands so you clap exactly one clap per second for ten seconds.
• Turn a page in a book at exactly one page every two seconds for twenty seconds.
• Sit still for thirty seconds, letting the timer know by raising your hand when you think thirty seconds has passed.
• Walk at the speed of one foot per second for fifteen seconds.
• Walk the length of your classroom in exactly ten seconds. At what speed were you traveling?
3. Connecting Graphs to Stories: In "Walking Strides," students examine how the time required to walk a given distance varies as the length of their stride varies. Often it is reported that many students initially misjudged time. A suggestion to overcome this problem is to let them explore the activities in teams: have them sit for an undisclosed time (e.g., 30 seconds) and make guesses about the amount of time that had elapsed.
[This activity has been adapted from Jones and Day (1998, pp. 18-19).]
Questions for Students
Refer to the lesson plan.
Teacher Reflection
None
### Learning Objectives
Students will:
• Collect data and record data in a table
• Make a graph to display data using correct labels and scale
• Recognize what varies in an experiment
• Name the independent and dependent variables in a problem
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# Find roots of the equation $y=2 \sin(3x+40), \;x \in (-2\pi, 2\pi)$
Find the roots of the equation:
$$y=2 \sin(3x+40), \;\;x \in (-2\pi, 2\pi)$$
In the book given that there are 12 roots exist. I am able get only 2 roots. Could anyone explain?
• How did you obtain the roots? Remember that $$\sin{x} = \sin{y} \Leftrightarrow x = y +n2\pi \text{ or } x = \pi -y + n2 \pi, n \in \mathbb{Z}$$ Commented Oct 16, 2018 at 6:23
• Because of the multiplier $3$, the argument of the sine crosses $6$ periods. As the sine has $2$ roots per period, you have it.
– user65203
Commented Oct 16, 2018 at 6:24
• thanks @ Yves Daoust Commented Oct 16, 2018 at 6:25
• Is the $40$ $$40^{\circ}$$? Commented Oct 16, 2018 at 8:58
$$y = 2\sin(3x+40)$$
Set $$y = 0$$. $$0 = 2\sin(3x+40) \implies \sin(3x+40) = 0 \implies 3x+40 = \sin^{-1}0$$
From here, for all $$n \in \mathbb{Z}$$: $$\sin^{-1}0 = 0+2\pi n \text{ OR } \sin^{-1}0 = \pi+2\pi n \\ \implies\sin^{-1}0 = \pi n$$
Solve the equation now. $$3x+40 = \pi n \implies 3x = \pi n-40 \implies \boxed{x = \frac{\pi n-40}{3} = n\frac{\pi}{3}-\frac{40}{3}}$$ From here, we can find the number of roots for the restricting domain of $$x \in (-2\pi, 2\pi)$$. $$-2\pi < \frac{\pi n-40}{3} < 2\pi$$ $$-6\pi < -\pi n-40 < 6\pi$$ $$-6\pi+40< -\pi n < 6\pi+40$$ $$\frac{-6\pi+40}{-\pi}> n > \frac{6\pi+40}{-\pi}$$ $$6-{\frac{40}{\pi}} > n > -6-\frac{40}{\pi}$$
Subtract the maximum and minimum values. $$6-\frac{40}{\pi}-\biggr(-6-\frac{40}{\pi}\biggr) = 12$$ So, there are $$12$$ roots. As a matter of fact, the answer would remain $$12$$ regardless of the value of $$h$$ in $$y = 2\sin(3x-h)$$.
Another way of thinking about it would be to consider the periodic nature of sine. For $$y = a\sin[b(x-h)]+k$$, the length of a period is given by $$\frac{2\pi}{b}$$.
$$\frac{2\pi}{b} = \frac{2\pi}{3}$$
For $$x \in (-2\pi, 2\pi)$$, the number of cycles/periods can be found by $$\frac{4\pi}{\frac{2\pi}{3}} = 6$$ Each period has $$2$$ roots, so there’ll be $$12$$ roots for the given function.
• Neat answer, KM101! The way I had solved the equation, I clean forgot $\sin \theta$ has two solutions - $0$ and $\pi n$. Commented Oct 18, 2018 at 13:05
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# Solutions for Practice Test 2, The Official SAT Study Guide, Section 5
Math Lair Home > Test Preparation > Solutions for Practice Test 2, The Official SAT Study Guide, Section 5
SAT Practice Test Solutions:
2014–15 SAT Practice Test
2013–14 SAT Practice Test
The Official SAT Study Guide, second edition
• Practice Test 1: Sections 3, 7, 8.
• Practice Test 2: Sections 2, 5, 8.
• Practice Test 3: Sections 2, 5, 8.
• Practice Test 4: Sections 3, 6, 9.
• Practice Test 5: Sections 2, 4, 8.
• Practice Test 6: Sections 2, 4, 8.
• Practice Test 7: Sections 3, 7, 9.
• Practice Test 8: Sections 3, 7, 9.
• Practice Test 9: Sections 2, 5, 8.
• Practice Test 10: Sections 2, 5, 8.
SAT Math Tips
Here are solutions for section 5 of the second practice test in The Official SAT Study Guide, second edition, found on pages 463–468. The following solutions illustrate faster, less formal methods that may work better than formal methods on a fast-paced test such as the SAT. To learn more about these methods, see my e-book Succeeding in SAT Math or the SAT math tips page.
1. If 3x = 0, then (dividing both sides of the equation by 3), x = 0. Substituting x = 0 into 1 + x + x² results in 1 + 0 + 0 = 1. Select (B) 1.
• Draw a diagram: If you don't see the answer immediately, it can help to draw two circles, one three times as big as the other or so, and label the diameters and radii of both. You should probably see that the radius of A is about three times the radius of B, so the answer would be (E) 3:1. If not:
• Let d represent the diameter of circle B, and 3r the diameter of circle A. The radius of circle A is (3/2)d, and the radius of circle B is ½d. The ratio between the two is 3:1, so the answer is (E) 3:1.
2. Try a special case: Say that N = {2, 3, 4} (you could pick any set whose average is 3, of course). Now, M = {4, 6, 8}. The average of the numbers in set M = ⅓(4 + 6 + 8) = 6. Select (D) 6.
3. Try a special case: Let P = 1, R = 2, T = 3. So, PRT is 123. Now,
123 × 10−2 = 1.23
The answer that corresponds to 1.23 is (C) P.RT. Select that answer.
Solution 2: When you multiply a number by 10−2, you shift the decimal point two places to the left. Now, for a whole number such as PRT, the decimal point would occur to the right of the last digit, here T. Shifting it two decimal places to the left gives P.RT. Select (C) P.RT.
4. We can subtract k from both sides of the inequality, leaving
n < 0
This corresponds to answer (E) n < 0. Select that answer.
• Estimate the answer: 7/16 is about ½, so x must be somewhere around twice as large as y. Since y is 3.5 feet, x must be about 7 feet.
• Look at the answer choices: The only answer choice anywhere near 7 is (A) 8. Select that answer.
5. Solution 1: You may know that, if a graph of a function is given by y = f(x), then the graph of y = ⅓f(x) will be the same graph, only one-third as narrow (i.e. three times as wide). So, the graph will be wider. Select (B) It will be wider.
Solution 2:
• Try a special case: Say that a = 1.
• Draw a diagram: If you have a graphing calculator, plot the graphs of y = x² + 2 and y = ⅓x² + 2. You'll notice that the graph of y = ⅓x² + 2 is wider than the other one. Select (B) It will be wider. If you don't have a graphing calculator, it's possible to do this by calculating a table of a few values by hand, although it may be time-consuming.
6. Solution 1:
• Read the question carefully and determine what it is asking: Meredith wants to wear an outfit consisting of one red item, one white item, and one blue item. We need to determine how many ways there are to do this.
• We could imagine that we have three labels, "red," "white," and "blue," and we have to attach one label to each of a hat, a sweater, and a pair of jeans. In other words, we would need to find how many ways there are to permute the three labels "red," "white," and "blue." The number of ways to do this is 3! = 6. Select (B) 6.
Solution 2:
• Read the question carefully and determine what it is asking: Meridith wants to wear an outfit consisting of one red item, one white item, and one blue item. We need to determine how many ways there are to do this.
• Estimate the answer: If there were no restrictions on colour, there would be 3 × 3 × 3 = 27 choices. However, with the colour restrictions, the number of possibilities might be much less than that, maybe only 1/3 or 1/4 or 1/6. So, perhaps the answer is somewhere between 4 and 9.
• Look at the answer choices: We can definitely eliminate (E) 27. (A) 3 and (D) 12 seem unlikely as well. We are still left with a few possibilities, including (B) 6 and (C) 9.
• It looks like there aren't so many possibilities that we couldn't list them out. So, list all of the possibilities out:
HatSweaterJeans
redwhiteblue
redbluewhite
whiteredblue
whitebluered
blueredwhite
bluewhitered
There are 6 possibilities. Select (B) 6.
7. Convert the sentence into an equation as follows:
When [ignore] twice 2 × a certain number x is increased by + 5 5 the result is = 14 14
So, we get:
2x + 5 = 14
2x = 9
x = 4.5
Enter 4.5 as the answer.
• Estimate the answer: y is definitely greater than 90°, perhaps around 120° or 135 °
• Draw a diagram: A diagram is given, but if you recall the supplemental and congruent angles created when a pair of parallel lines intersects another line, you can fill in several angles on the diagram:
• It can now be seen that x° and y° sum to a straight angle, or 180°. So:
x + y = 180
Since we are given that y = 3x:
x + 3x = 180
4x = 180
x = 45
Don't forget, we're asked to find the value of y, not x. y = 3x = 3(45) = 135.
• Draw a diagram: It may help you to visualize the problem if you draw the CD cases inside the illustration of the box.
• If each CD case is ¼ inches wide, then four will fit in a one-inch box, so 4 × 8 = 32 will fit in an eight-inch box (you could also have obtained the answer by dividing 8 by ¼).
8. Since we are asked to find xy, try to manipulate the equation so that this expression appears:
(3x + y)⁄y = 6⁄5
3xy + yy = 6⁄5
3(xy) + 1 = 6/5
3(xy) = 1/5
xy = 1/15
9. Solution 1:
• Estimate the answer: Looking at the table, two stores increased their profits by over \$1,000 and one store increased their profits by over \$2,000. A reasonable estimate might be something like \$1,500.
• You might be able to see the following in your head; if so, you don't have to write it out. The average increase in profit would be:
⅓(increase in profit for A + increase in profit for B + increase in profit for C)
= ⅓(year 2 profit for A − year 1 profit for A + year 2 profit for B − year 1 profit for B + year 2 profit for C − year 1 profit for C)
All of these numbers are in the grid, so you can simply plug them in and evaluate.
= ⅓(6,250 − 5,000 + 7,350 − 6,000 + 12,700 − 10,000)
= 1750
Enter 1750 as the answer.
Solution 2:
• Estimate the answer: Looking at the table, two stores increased their profits by over \$1,000 and one store increased their profits by over \$2,000. A reasonable estimate might be something like \$1,500.
• You might be able to see the following in your head; if so, you don't have to write it out. The average increase in profit would be:
⅓(increase in profit for A + increase in profit for B + increase in profit for C)
= ⅓(year 2 profit for A − year 1 profit for A + year 2 profit for B − year 1 profit for B + year 2 profit for C − year 1 profit for C)
So far, this is the same as solution 1. You can save yourself a little bit of time if you notice that this expression is equal to:
= ⅓(year 2 profit for A + year 2 profit for B + year 2 profit for C − year 1 profit for A − year 1 profit for B − year 1 profit for C)
= ⅓(total profit for year 2 − total profit for year 1)
= ⅓(26,250 − 21,000)
= 1750
Enter 1750 as the answer.
10. Solution 1:
• First, find the points for where f(a) = a:
|3a − 17| = a
3a − 17 = a or 3a − 17 = −a
2a = 17 or 4a = 17
a = 8.5 or a = 4.25
• Next, check whether a number between 4.25 and 8.5 would work. For example, try 5:
f(5)
= |3(5) − 17|
= |−2|
=2 < 5
So, enter 5 as the answer. Of course, any other number between 4.25 and 8.5 would also work.
Solution 2: Guess and check. For example:
• If a = 0, f(a) = 17, so that doesn't work.
• If a = 4, f(a) = 5, so that doesn't work, but we're closer.
• If a = 8, f(a) = 7, so that does work. Enter 8 as the answer.
Solution 3: If you have a graphing calculator, graph y = |3x − 17| and y = x. The portions of the graph where y = x is above the other graph represents the possible answers. Graphing the two, it appears that the correct range is between about 4 and 8 or so. Enter a number about halfway through the range, such as 6.
• Read the question carefully and understand what it is asking. There is a lot of information in this question that isn't needed (e.g. how many pieces of candy are in the jar). What is asked for is to find how many out of 13 pieces must be red in order for him to have more red pieces in total.
• Estimate the answer: Ari has slightly fewer red pieces than green to begin with, so more than half of the 13 pieces he takes must be red. In other words, at least 7, probably 8 or 9 or so, have to be red.
• Once Ari takes 13 pieces, he will have 3 + 4 + 13 = 20 pieces in total. If Ari has more red candies than green candies, the least number of red candies that he can have is 11. Ari already has 3 red candies, so he must take 11 − 3 = 8 more red candies. Enter 8 as the answer.
• Read the question carefully and understand what it is asking: Don't worry that you've never heard of "tri-factorable" before, it's a made-up term. The question asks you to find how many numbers less than 1,000 are the product of three consecutive integers.
• To understand the question better, it can help to list numbers that are the product of three consecutive integers. One would be 1 × 2 × 3 = 6. Another is 2 × 3 × 4 = 24. You could continue like this, but you might realize that 9 × 10 × 11 = 990 is the largest such number less than 1,000 (10 × 11 × 12 is over 1,000). So, if the smallest factor is between 1 and 9, the number will be less than 1,000. So, there are 9 such numbers and 9 is the answer.
11. Solution 1:
• Draw a diagram: It might help you to understand the question better if you draw a rough graph plotting the cost for carriers A and B as a function of t.
• Guess and check: For a call that lasts t > 20 minutes, the cost for carrier A is equivalent to 5¢ per minute for the first 20 minutes and 7¢ per minute for t − 20 minutes. The cost for carrier B is equivalent to 6¢ per minute.
Now, if the call lasted for 40 minutes, then A would give you 5¢ a minute for 20 minutes and 7¢ a minute for 20 minutes. Is this equal to 6¢ a minute for 40 minutes? Yes. 5(20) + 7(20) = 100 + 140 = 240 = 6(40). Enter 40 as the answer.
Solution 2:
• Draw a diagram: It might help you to draw a diagram as above.
• We can represent the cost of a phone call with carrier A lasting over 20 minutes as:
1 + .07(t − 20)
We can represent the cost of a phone call with carrier B lasting over 20 minutes as:
.06t
If the cost of a phone call lasting t minutes is the same on both carriers, the two expressions must be equal to one another:
1 + .07(t − 20) = .06t
1 + .07t − 1.4 = .06t
.01t = 0.4
t = 40
Enter 40 as the answer.
12. If each square has a side length of k inches, then, by counting the number of sides along the dark line, the perimeter is 16k. Since there are 10 squares, the area is 10k². If the perimeter in inches equals the area in square inches, then:
16k = 10k²
k² − 1.6k = 0
k(k − 1.6) = 0
k = 0 or k = 1.6
Since the diagram was drawn to a non-zero size, presumably k = 0 is not the solution that the SAT writers want. So, enter 1.6 as the answer.
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# ABCD is a rectangle in which diagonal AC bisects $\angle $${\text{A}} as well as \angle$${\text{C}}$. Show that,(i) ABCD is square (ii) diagonal BD bisects $\angle $${\text{B}} as well as \angle$${\text{D}}$.
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Hint: To solve this question we need to know the basic theory related to the rectangle. We learned that a rectangle has four-sided and all the internal angles are equal to 90 degrees. The length of the opposite sides of the rectangle is always equal to each other. The length of the opposite sides of the rectangle is always equal to each other. If all the sides are equal then it is square.
Complete step-by-step solution:
(i) we need to prove ABCD is square.
Given in question,
ABCD is a rectangle, in which diagonal AC bisect $\angle $${\text{A}} as well as \angle$${\text{C}}$.
Therefore,
$\angle $${\text{DAC}}=\angle$${\text{CAB}}$……………… (1)
$\angle $${\text{DCA}}=\angle$${\text{BCA}}$………………. (2)
A square is a rectangle when all sides are equal. Now,
AD$\parallel$BC & AC is transversal, therefore
$\angle$${\text{DAC}}=\angle$${\text{BCA}}$ [Alternate angles]
From (1),
$\angle $${\text{CAB}}=\angle$${\text{BCA}}$……………….. (3)
In $\vartriangle$ABC,
$\angle $${\text{CAB}}=\angle$${\text{BCA}}$, therefore
BC=AB………………….. (4) [sides opposite to equal angles]
But BC=AD & AB=DC………………….. (5) [Opposite sides of rectangle]
Therefore from (4) & (5),
(ii) we need to prove that diagonal BD bisects $\angle $${\text{B}} as well as \angle$${\text{D}}$.
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Table of 12 is the first table that students learn after the foundation tables from 2 to 10. The table of 10 and 11 remains very easy for the students and its difficulty starts from the table of 12. If students make a mind map then it becomes so easy.
In this post
## Table of 12 learning in Maths
A table of 12 to 20 one can learn in an easy way for their mathematical calculations. No doubt calculator makes the problem easy but the fast calculations, students should remember the table. I have listed multiplication of 12 tables not only for the 12 table purpose but for the learning point of view. At the end of the quiz, you can check and verify your maths table learning.
12 × 1 = 12 12 × 2 = 24 12 × 3 = 36 12 × 4 = 48 12 × 5 = 60 12 × 6 = 72 12 × 7 = 84 12 × 8 = 96 12 × 9 = 108 12 × 10 = 120
In the first row 12 x1 =12 means twelve is one. The second time 12×2=24 means an addition of 12 two times becomes 24. Then for the third row 12 x 3 =36 if you will add 12+12+12= 36 it will be thirty-six. In the 4th row, 12 x 4 =48 it means if you will add 12 four times it will be 48. And similarly, you can understand for the others. So multiplication of a fixed number with another number indicates the sum of the first number up to the number of the second ones. Or you can say that the second number is the frequency of addition of the first number.
If possible for the fast maths calculations learn Vedic maths, that’s short tricks are useful for basic maths activities.
### QUIZ:
After this table and results of multiplication attempt this quiz and check your learning.
This quiz is related to the table of 12 to check the learning
1. 12 x 10 =
2. 12 x 4 =
3. 12 x 7 =
4. 12 x 3 =
5. 12 x 9 =
6. 12 x 2 =
7. 12 x 1 =
8. 12 x 6 =
9. 12 x 8 =
10. 12 x 5 =
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# RS Aggarwal Solutions For Class 7 Maths Exercise 2A Chapter 2 Fractions
The pdf of RS Aggarwal Solutions for the Exercise 2A of Class 7 Maths Chapter 2, Fractions are available here. To cover the entire syllabus in Maths, the RS Aggarwal is an essential material as it offers a wide range of questions that test the students’ understanding of concepts. Our expert personnel have solved the problems step by step with neat explanations. Students who aim to score good marks in Maths, then refer RS Aggarwal Solutions for Class 7.
## Download the PDF of RS Aggarwal Solutions For Class 7 Maths Chapter 2 Fractions – Exercise 2A
### Access answers to Maths RS Aggarwal Solutions for Class 7 Chapter 2 – Fractions Exercise 2A
1. Compare the fraction:
(i) (5/8)and (7/12)
Solution:-
By cross multiplication, we have:
5 × 12 = 60 and 8 × 7 = 56
But,
60 > 56
∴ (5/8) > (7/12)
(ii) (5/9)and (11/15)
Solution:-
By cross multiplication, we have:
5 × 15 = 75 and 9 × 11 = 99
But,
75 < 99
∴ (5/9) < (11/15)
(iii) (11/12) and (15/16)
Solution:-
By cross multiplication, we have:
11 × 16 = 176 and 12 × 15 = 180
But,
176 < 180
∴ (11/12) < (15/16)
2. Arrange the following fraction in ascending order :
(i) (3/4), (5/6), (7/9), (11/12)
Solution:-
LCM of 4, 6, 9, 12 = 2× 2× 3× 3=36
Now, let us change each of the given fraction into an equivalent fraction having 36 as the denominator.
[(3/4) × (9/9)] = (27/36)
[(5/6) × (6/6)] = (30/36)
[(7/9) × (4/4)] = (28/360
[(11/12) × (3/3)] = (33/36)
Clearly,
(27/36) < (28/36) < (30/36) < (33/36)
Hence,
(3/4) < (7/9) < (5/6) < (11/12)
Hence, the given fractions in ascending order are (3/4), (7/9), (5/6), (11/12)
(ii) (4/5), (7/10), (11/15), (17/20)
Solution:-
LCM of 5, 10, 15, 20 = 5 × 2 × 3 × 2 = 60
Now, let us change each of the given fraction into an equivalent fraction having 60 as the denominator.
[(4/5) × (12/12)] = (48/60)
[(7/10) × (6/6)] = (42/60)
[(11/15) × (4/4)] = (44/60)
[(17/20) × (3/3)] = (52/60)
Clearly,
(42/60) < (44/60) < (48/60) < (52/60)
Hence,
(7/10) < (11/15) < (4/5) < (17/20)
Hence, the given fractions in ascending order are (7/10), (11/15), (4/5) , (17/20)
3. Arrange the following fraction in descending order :
(i) (3/4), (7/8), (7/12), (17/24)
Solution:-
LCM of 4, 8, 12, 24 = 4 × 2 × 3 = 24
Now, let us change each of the given fraction into an equivalent fraction having 24 as the denominator.
[(3/4) × (6/6)] = (18/24)
[(7/8) × (3/3)] = (21/24)
[(7/12) × (2/2)] = (14/24)
[(17/24) × (1/1)] = (17/24)
Clearly,
(21/24) > (18/24) > (17/24) > (14/24)
Hence,
(7/8) > (3/4) > (17/24) > (7/12)
Hence, the given fractions in descending order are (7/8), (3/4), (17/24), (7/12)
(ii) (2/3), (3/5), (7/10), (8/15)
Solution:-
LCM of 3, 5, 10, 15 = 3 × 5 × 2 = 30
Now, let us change each of the given fraction into an equivalent fraction having 30 as the denominator.
[(2/3) × (10/10)] = (20/30)
[(3/5) × (6/6)]= (18/30)
[(7/10) × (3/3) = (21/30)
[(8/15) × (2/2)] = (16/30)
Clearly,
(21/30) > (20/30) > (18/30) > (16/30)
Hence,
(7/10) > (2/3) > (3/5) > (8/15)
Hence, the given fractions in descending order are (7/10), (2/3), (3/5), (8/15)
4. Reenu got (2/7) part of an apple while sonal got (4/5) part of it. Who got the larger part and by how much?
Solution:-
From the question,
Reenu got (2/7) part of an apple
Sonal got (4/5) part of an apple
First we have to compare the given fraction (2/7) and (4/5) to know who got the larger part of the apple.
Then,
By cross multiplication, we have
2 × 5 = 10 and 4 × 7 = 28
But,
10 < 28
∴ (2/7) < (4/5)
So, Sonal got the larger part of the apple
Now,
= (4/5)-(2/7)
= [(28-10)/35]
= [18/35]
∴ Sonal got (18/35) part of the apple larger than Reenu.
5. Find the sum:
(i) (5/9)+(3/9)
Solution:-
For adding two like fractions, the numerators are added and the denominator remains the same.
= (5+3)/9
= (8/9)
(ii) (8/9) + (7/12)
Solution:-
For addition of two unlike fractions, first change them to the like fractions.
LCM of 9, 12 = 36
Now, let us change each of the given fraction into an equivalent fraction having 36 as the denominator.
[(8/9) × (4/4)] = (32/36)
[(7/12) × (3/3)] = (21/36)
= (32/36) + (21/36)
= (32+21)/36
= (53/36)
= [1(17/36)]
(iii) [3(4/5)] + [2(3/10)] + [1(1/15)]
Solution:-
First convert each mixed fraction into improper fraction.
We get,
= [3(4/5)] = (18/5)
= [2(3/10)] = (23/10)
= [1(1/15)] = (16/15)
Then,
(19/5) + (23/10) + (16/15)
LCM of 5, 10, 15 = 30
Now, let us change each of the given fraction into an equivalent fraction having 30 as the denominator.
= [(19/5) × (6/6)] = (114/30)
= [(23/10) × (3/3)] = (69/ 30)
= [(16/15) × (2/2)] = (32/30)
Now,
= (114/30) + (69/30) + (32/30)
= [(114+69+32)/30]
= (215/30)
= [7 (5/30)]
= [7 (1/5)]
6. Find the difference:
(i) (5/7) – (2/7)
Solution:-
The subtraction of fraction can be performed in a manner similar to that of addition.
For subtracting two like fractions, the numerators are subtract and the denominator remains the same.
= (5 – 2)/7
= (3/7)
(ii) (5/6) – (3/4)
Solution:-
For subtraction of two unlike fractions, first change them to the like fractions.
LCM of 6, 4 = 12
Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator.
= [(5/6) × (2/2)] = (10/12)
= [(3/4) × (3/3)] = (9/12)
Now,
= (10/12)-(9/12)
= [(10 – 9)/12]
= (1/12)
(iii) [3(1/5)] – (7/10)
Solution:-
Convert mixed fraction into improper fraction,
= [3(1/5)] = (16/5)
= (16/5)-(7/10)
For subtraction of two unlike fractions, first change them to the like fractions.
LCM of 5, 10 = 10
Now, let us change each of the given fraction into an equivalent fraction having 10 as the denominator.
= [(16/5) × (2/2)] = (32/10)
= [(7/10) × (1/1)] = (7/10)
Then,
= (32/10) – (7/10)
= (32-7)/10
= (25/10) … [÷ by 5]
= (5/2)
= [2(1/2)]
(iv) 7 – [ 4 (2/3)]
Solution:-
Convert mixed fraction into improper fraction, and then find the difference.
= [4(2/3)] = (14/3)
= 7-(14/3)
= (21-14)/3
= (7/3)
= [2 (1/3)]
(v) [ 3(3/10)] – [1(7/15)]
Solution:-
Convert mixed fraction into improper fraction, and then find the difference.
= [3(3/10)] = (33/10)
= [1(7/15)] = (22/15)
We get,
= (33/10) – (22/15)
LCM of 10, 15 = 30
Now, let us change each of the given fraction into an equivalent fraction having 10 as the denominator.
= [(33/10) × (3/3)] = (99/30)
= [(22/15) × (2/2)] = (44/30)
Then,
= (99/30) – (44/30)
= (99 -44)/30
= (55/30)
= (11/6)
= [1(5/6)]
(vi) [2(5/9)] – [1(7/15)]
Solution:-
Convert mixed fraction into improper fraction, and then find the difference.
= [2(5/9)] = (23/9)
= [1(7/15)] = (22/15)
We get,
= (23/9) – (22/15)
LCM of 9, 15 = 45
Now, let us change each of the given fraction into an equivalent fraction having 45 as the denominator.
= [(23/9) × (5/5)] = (115/45)
= [(22/15) × (3/3)] = (66/45)
Then,
= (115/45) – (66/45)
= (115 -66)/45
= (49/45)
= [1(4/45)]
7. Simplify:
(i) (2/3) + (5/6) – (1/9)
Solution:-
LCM of 3, 6, 9 = 18
Now, let us change each of the given fraction into an equivalent fraction having 18 as the denominator.
= (2/3) × (6/6) = (12/18)
= (5/6) × (3/3) = (15/18)
= (1/9) × (2/2) = (2/18)
Then,
= (12/18) + (15/18) – (2/18)
= (12+ 15 – 2)/ 18
= (27-2)/ 18
= (25/18)
= [1(7/18)]
(ii) 8 – [4(1/2)] – [2(1/4)]
Solution:-
Convert mixed fraction into improper fraction, and then find the difference.
= [4(1/2)] = (9/2)
= [2(1/4)] = (9/4)
LCM of 1, 2, 4 = 4
Now, let us change each of the given fraction into an equivalent fraction having 4 as the denominator.
= (9/2) × (2/2) = (18/4)
= (9/4) × (1/1) = (9/4)
= (8/1) × (4/4) = (32/4)
Then,
= (32/4) – (18/4) – (9/4)
= [(32-18-9)/4]
= [(32-27)/4]
= (5/4)
= [1(1/4)]
(iii) [8(5/6)] – [3(3/8)] + [1(7/12)]
Solution:-
First convert each mixed fraction into improper fraction.
We get,
= [8(5/6)] = (53/6)
= [3(3/8)] = (27/8)
= [1(7/12)] = (19/12)
Then,
(53/6)- (27/8) – (19/12)
LCM of 6, 8, 12 = 24
Now, let us change each of the given fraction into an equivalent fraction having 24 as the denominator.
= [(53/6) × (4/4)] = (212/24)
= [(27/8) × (3/3)] = (81/ 24)
= [(19/12) × (2/2)] = (38/24)
Now,
= (212/24) – (81/ 24) – (38/24)
= [(212- 81+ 38)/24]
= [(250-81/24)]
= (169/24)
= [7(1/24)]
8. Aneeta bought [3(3/4)] kg apples and [4(1/2)] kg guava. What is the total weight of fruits purchased by her?
Solution:-
The total weight of fruits bought by Aneeta = [3(3/4)] + [4(1/2)]
We have,
First convert each mixed fraction into improper fraction
= [3(3/4)] = (15/4)
= [4(1/2)] = (9/2)
Then,
= (15/4) + (9/2)
LCM of 4, 2 = 4
Now, let us change each of the given fraction into an equivalent fraction having 4 as the denominator
= (15/4) × (1/1) = (15/4)
= (9/2) × (2/2) = (18/4)
= (15/4) + (18/4)
= (15 + 18)/ 4)
= (33/4)
= [8(1/4)]
The total weight of fruits purchased by Aneeta is [8(1/4)] kg
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### RS Aggarwal Solutions for Class 7 Maths Exercise 2A Chapter – 2 Fractions
RS Aggarwal Solutions For Class 7 Maths Chapter 2 Fractions Exercise 2A has problems which are based on method of changing unlike fractions to like fractions, comparing of fractions, addition of fractions and subtraction of fractions. Students are suggested to try solving the questions from RS Aggarwal Solutions book of Class 7 and then refer to these solutions to know the best way of approaching the different questions.
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# CBSE Class 11 Maths Revision Notes Chapter 8
## Class 11 Mathematics Revision Notes for Chapter 8 Binomial Theorem
Mathematics is one of the most crucial subjects in Class 11. Students should study the subject properly to score good marks in the final papers and other entrance examinations. Amongst all the other chapters, the binomial theorem is considered one of the most scoring. So, candidates must practice all the numerals from the NCERT books. One of the best ways to strengthen one’s preparation is by checking the revision notes and regularly studying from them.
Class 11 Mathematics Revision Notes for Chapter-8 Binomial Theorem – Free PDF Download
## Access Class 11 Mathematics Chapter 8 – Binomial Theorem Notes
Binomial Theorem
An algebraic expression containing the two distinct terms connected to each other by an addition or subtraction operation is known as a binomial. The binomial theorem is the method of expanding an expression raised to any finite power. It is a powerful theorem when expansion becomes lengthy and difficult to calculate. With the binomial theorem, any expression that is raised to a very large power can be easily calculated. The theorem is very useful in Algebra, probability and other calculations.
In binomial expansion, we usually have to find the middle or general term. The terms that are used in the binomial expansion are:
• The ratio of consecutive terms/coefficients
• Numerically greatest term
• Particular term
• Determining a particular term
• Independent term
• Middle term
• General term
Binomial Theorem formula
The formula for the Binomial Theorem is:
(x + y)n = Σr=0n nCr xn – r · yr.
### Binomial Theorem for Positive Integral Indices
The binomial theorem for the positive integral indices states that the total number of terms in the expansion is one more than the index.
The properties of the Binomial Theorem for positive integral indices are mentioned below.
They are as follows:
• (a + b)0 = 1 a +b is not equal to 0
• (a +b)1 = a + b
• (a + b)2 = a2 +2ab + b
• (a + b)3 = a3 + 3a2b + 3ab2 + b3
• (a + b)4 = (a + b)3 (a + b) = a4 + 4a3b + 6a2 b2 + 4ab3 + b
From the above, we observe that:
• The total terms present in the expansion are one more than the index.
• Powers of the first quantity, let’s say ‘a’ goes on reducing by 1 whereas the power of the second quantity ‘b’ will increase by 1 in the successive terms.
• In each term of the expansion the sum of indices, a and b will be the same and equal to the index of a + b.
### Pascal’s Triangle:
Pascal’s Triangle is an arrangement of the expansion coefficients in a triangular shape with (1) at the top of the vertex and running down the two slanting sides.
The expansions for higher powers of a binomial are possible through Pascal’s triangle.
### Binomial Theorem for any Positive Integer N
The binomial theorem for any positive integer n is
(a + b)n = n C0 an + n C1 an-1b + n C2 an-2 b2 + n Cn-1 a-b n-1 + nCn bn
### Properties of Binomial Coefficient:
Binomial coefficients are the integers that are coefficients in a binomial theorem. Some of the important properties of binomial coefficients are given below:
• C02 + C12 + C22 + …Cn2 = [(2n)!/ (n!)2]
• C1 − 2C2 + 3C3 − 4C4 + … +(−1)n-1 Cn = 0 for n > 1
• nC1 + 2.nC2 + 3.nC3 + … + n.nCn = n.2n-1
• C0 – C1 + C2 – C3 + … +(−1)n . nCn = 0
• C0 + C2 + C4 + … = C1 + C3 + C5 + … = 2n-1
• C0 + C1 + C2 + … + Cn = 2n
### Some Important Results:
Some important results derived are:
• (1 + X)n = nC0X0 + nC1X1 + nC2X2 + ……….. + nCnXn
Putting x =1 and -1 we will obtain
C0 + C1 + C2+ Cn = 2
C0 – C1 + C2 -……….. + (-1)n Cn = 0
• On differentiating students will get the following:
Differentiating the (1 + x)n = nC0Xo + nC1x1 + nC2X2 +…….. + nCnXn on both sides
n(1 + x)n-1
C1 + 2C1x+3C3X2 + ……. +nCNx n-1……. (1)
x=1
n2 n-1 = C1 + 2C1 + 3C3 + ……… + nCn
X = -1
0 = C1 – 2C1 + ……. + (-1) n-1 nCn
On differentiating 1 again and again we will get different results.
### Multinomial Expansion:
While expanding the (x1 + x2 + …….. + xn ) m where m,n N and 1 2 n x x … x are independent variables.
• Total number of terms = m + n-1 C n-1
• Coefficient of X1 r X2 r2 ………. Xn rn
Where the r1 + r2 …… +rn = m) is m, ri ∈ N U {0} is m!/ r1 ! r2 ! …….. Rn!
• The sum total of all the coefficients is obtained by putting the variables equals to 1.
### Binomial Expression
The binomial theorem is considered one of the greatest discoveries in the world of Mathematics. It was developed by Euclid in the 4th century. A Persian Mathematician Al-Karaji gave proof of the binomial theorems and Pascal’s triangle. Binomial expression is an expression consisting of two different terms. According to the binomial theorem for any particular positive integer n, the total sum of the two different numbers [suppose x and y] can be expressed as the sum of (n + 1 ). The theorem is used to solve complex problems. In addition, Pascal’s Triangle forms an important aspect of the theorem that can be used to determine the binomial coefficients.
### Properties of the Binomial Expansion
The Binomial Theorem consists of many properties. Some of them are mentioned below. They are as follows:
• The total number of terms in the binomial expansion of (x + a)n is (n + 1).
• The sum of indices of a and x in every term is n.
• It is considered a correct expansion when the terms are complex numbers.
• Terms that are equidistant from each other will have equal coefficients. They are termed binomial coefficients.
• General term in the expansion of (x + c)n is Tr + 1 = nCrx n – r ar .
• The values will first increase and then it will decrease when one will go ahead with the expansion.
• The coefficient of xr in the expansion of (1 + x)n is nCr.
### Middle Term in the Expansion of (1 + x)n
• In (a + b)n if n is even then the term expansion is odd. Hence, there will be only one middle term and it is (n + 2 / 2)th term.
• In (a + b)n if n is the odd number then term expansion will be even. Therefore, there will be two middle terms (n +1 / 2)th and (n + 3 / 2)th
### Greatest Coefficient
• If n is an even term then in (x + a)n the greatest coefficient will be nCn/2.
• If n is odd then in (x + a)n the greatest coefficient will be nCn – ½ or nCn + ½ where both are equal.
### Fun Facts about the Greatest Term
Do you want to learn methods to find the greatest term? If yes then you can easily find the greatest coefficient for each expansion by determining whether the resulting expansion is an integer or not. To find the answers accurately students should have a good hold over the definition of integers and must memorise the formulas to solve the questions.
### 1. Why should I refer to the CBSE Class 11 Chapter 8 Notes?
Students should refer to the Class 11 Chapter 8 notes by Extramarks as these will:
• Help the candidates to prepare well for the examination.
• Students will be able to understand the basic and advanced concepts.
• They can score good marks if they follow the notes properly.
### 2. Mention two applications of the Binomial Theorem.
The theorem is used in areas such as probability, and statistics and to find the remaining digits of a number.
### 3. Mention some important points from the Binomial theorem.
The most important points of the theorem are mentioned below. They are as follows:
• The total terms used in the expansion of (x + y)n are (n + 1).
• The total sum of x and y exponents is always n.
• nC0, nC1, nC2, … .., nCn are known as the binomial coefficients and can also be written as C0, C1, C2, ….., Cn.
The Binomial coefficients that are equidistant from the beginning and ending are always equal such as nC0 = nCn, nC1 = nCn-1 , nC2 = nCn-2 ,….. etc.
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# Number System-Real Numbers
Author:
Rahul Chandra Saha
M.Sc., Mathematics
### Introduction
In class IX, you came to know about real numbers. The two complementary sub-sets of real numbers are the rational numbers and irrational numbers. You have learned about them. In this chapter we will discuss more about the irrationals, Fundamental Theorem of Arithmetic and related applications. But before these one obvious question arises ”is there any rule of dividing the numbers and if so then how to construct the algorithm for it?”
### Euclid’s Division Lemma
Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.
This result was first recorded in Book VII of Euclid’s Elements. The division algorithm is based on the Euclid’s Division lemma.
Think Yourself
Q1. What is algorithm? Is there any similarities between computer algorithm and algorithm we are dis-
cussing?
Q2. What is lemma?
### Euclid’s Division Algorithm
It is a algorithm or technique to compute the HCF(Highest Common Factor) of two given positive integers.
Let us try to extract the algorithm from an example. Suppose we need to find the HCF of the integers 413 and 42. We will start with the larger number,i.e., 413 and then use the Euclid’s Division Lemma to get-
$413 = 42 × 9 + 35.$
Now consider divisor 42 and remainder 35 and applying division lemma we get,
$42 = 35 × 1 + 7.$
Again consider the divisor 35 and remainder 7. On applying division lemma we get,
$35 = 7 × 5 + 0$
Here the remainder has become zero. Thus we can not proceed further. Our claim is that 7 is the H.C.F. or G.C.D.(Greatest Common Divisor) of 413 and 42 is 7. Let us check. 413 = 7 × 59 and 42 = 7 × 6. Thus H.C.F (413, 42) = 7.
Euclid’s Division Algorithm. To obtain the HCF of two arbitrary positive integers $c$ and $d$ with $d < c$, follow the following steps-
1. Applying Division Lemma to c and d find integers q and r such that
$c = d × q + r, 0 ≤ r < d$
2. If $r = 0$ then $HCF (c, d) = d$. If $r \neq 0$ apply division lemma to $d$ and $r$.
3. Continue the process till the remainder is zero. The divisor at the last stage will be the required HCF.
Think yourself: Why the method works?
Problem. Apply Euclid’s Algorithm to find HCF (250, 100).
### The Fundamental Theorem of Arithmetic
Theorem.
Every composite number can be factorised as a product of primes,and this factorisation is unique,apart from the order in which the prime factors occur.
Example 1. Factorise 240.
Solution . $240 = 2 × 120 = 2^{2} × 60 = 2^{3 }× 30 = 2^4 × 15 = 2 ^4 × 3 × 5$
Example 2. Construct the factor tree of 30.
Example 3. Find the HCF of 55 and 90 by prime factorisation method. Hence find their LCM.
Solution . Prime factorisation of 55 and 90 gives
$55 = 5 × 11 \text{ and } 90 = 2 × 3 2 × 5.$
Therefore, $HCF (55, 90) = 5.$
Also we know that,
$HCF × LCM =\text{Multiplication of two numbers}.$
Hence,
$\text{LCM} = \frac{55\times90}{5} = 990$
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Circle A has a center at (-2 ,-1 ) and a radius of 3 . Circle B has a center at (-8 ,3 ) and a radius of 1 . Do the circles overlap? If not what is the smallest distance between them?
Sep 13, 2016
no overlap, ≈ 3.211
Explanation:
What we have to do here is $\textcolor{b l u e}{\text{compare}}$ the distance ( d) between the centres to the $\textcolor{b l u e}{\text{sum of the radii}} .$
• If sum of radii > d , then circles overlap
• If sum of radii < d , then no overlap
To calculate d, use the $\textcolor{b l u e}{\text{distance formula}}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$
here the 2 points are (-2 ,-1) and (-8 ,3) the centres of the circles.
let $\left({x}_{1} , {y}_{1}\right) = \left(- 2 , - 1\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 8 , 3\right)$
d=sqrt((-8+2)^2+(3+1)^2)=sqrt(36+16)=sqrt52≈7.211
sum of radii = radius of A + radius of B = 3 + 1 = 4
Since sum of radii < d , then there is no overlap.
smallest distance = d - sum of radii = 7.211 - 4 = 3.211
graph{(y^2+2y+x^2+4x-4)(y^2-6y+x^2+16x+72)=0 [-10, 10, -5, 5]}
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# Prove that 3+2$\sqrt{5}$ is irrational.
CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/NCERT Solutions/Exercise 1.3
To prove : 3+2√5 is rational number. Ans: 3 is rational number. 2 is rational number. 2√5 is irrational number (product of rational and irrational is an irrational number) 3+2√5 is an irrational number (sum of rational and irrational number is irrational) Or Let P = 3+2√5 is... read more
To prove : 3+2√5 is rational number.
Ans: 3 is rational number.
2 is rational number.
2√5 is irrational number (product of rational and irrational is an irrational number)
3+2√5 is an irrational number (sum of rational and irrational number is irrational)
Or
Let P = 3+2√5 is rational number ( we are assuming )
P= 3+2√5
P-3 =2√5
(P-3)÷2=√5
L.H.S. we assume P is a rational number .
P-3 is also rational number substraction of rational number always rational number .
(P-3)÷2 rational number (P-3) divide by rational number 2 hence (P-3)÷2 is rational number .
L.H.S. is rational While R.H.S. √5 is irrational L.H.S ≠R.H.S.
Our assumption is wrong so we can say 3+2√5 is an irrational number.
Experienced Mathematics Teacher
Root 5 is an irrational number. Therefore 2 root 5 is an irrational number as product of an rational and irrational number is also an irrational number. Again 3+ 2 root 5 is an irrational number as sum of a rational and an irrational number is also an irrational number. Hence proved.
Rationalize it and you can find it's irrational
Tutor
Let 3+2V5= Rational no. => 3+2V5=p/q =>V5=(p/q-3)/2 R.H.S. is a rational no. but V5 is not a rational no. => Wrong assumption. => it is an irrational no.
Since 3 is the whole number and 3=3/1 which is rational but √5 is the irrational number.so 2√5 is also a irrational number .if we add any number to irrational number will be irrational. Explanation : 3+2*2.236067977.....=7.472135954....
Tutor
Let 3+2√5 is a rational number =a/b,then 2√5=a/b-3=(a-3b)/b and √5=(a-3b)/2b. This shows √5 is a rational number.But √5 is an irrational number. So 3+2√5 is irrational
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# Repeating decimal as approximation
Alignments to Content Standards: 7.NS.A.2.d
1. Use long division to find the repeating decimal that represents $\frac{29}{13}$
2. Take the number obtained by including only the first two digits after the decimal point, and multiply that by $13$.
3. Take the number obtained by including only the first four digits after the decimal point, and multiply that by $13$.
4. Take the number obtained by including only the first six digits after the decimal point, and multiply that by $13$.
5. What do you notice about the product of $13$ and decimal approximations of $\frac{29}{13}$ as more and more digits are included after the decimal point?
6. How does what you observed in Part (e) help make sense of what it means for $\frac{29}{13}$ to be equal to the repeating decimal expression you found in the Part (a)?
## IM Commentary
The purpose of the task is to have students reflect on the meaning of repeating decimal representation through approximation. A formal explanation requires the idea of a limit to be made precise, but 7th graders can start to wrestle with the ideas and get a sense of what we mean by an "infinite decimal." Students can make observations which reinforce the topic at hand as well as lay groundwork for later developments.
This task allows students to engage in standard for mathematical practice 8, Look for and express regularity in repeated reasoning. Students are not asked explicitly to compare what they find in the second and third parts with 29, since by this grade they should make note of that on their own.
While the first part asks for application of long division (by hand), the multiplications could be done by calculator, at the teacher's discretion. The trade-offs in this choice are between further development of number fluency and opportunity for reflection if these calculations are done by hand versus greater likelihood of accuracy and answers provided more quickly (which can help observing the trend) if done by calculator.
## Solution
1. The answer, after performing long division, is $2.\overline{230769}$.
2. $2.23 \times 13 = 28.99$
3. $2.2307 \times 13 = 28.9991$
4. $2.230769 \times 13 = 28.999997$
5. As we include more of the decimal expansion, we see a decimal which is less than $29$ whose difference with $29$ becomes smaller and smaller. In the product, we see that the number of 9's following the decimal is roughly the number of digits after the decimal point included in the approximation of $\frac{29}{13}$.
6. Because $\frac{29}{13} \times 13 = 29$, we should expect that when we multiply numbers that are close to $\frac{29}{13}$ by $13$, the result should be to be close to $29$. In the product, we see that the number of 9's following the decimal is roughly the number of digits after the decimal point included in the approximation of $\frac{29}{13}$. Thus, it seems likely that if we used "all" the digits in the infinite decimal expansion, the product would be exactly equal. In other words, $2.\overline{230769}$ should be equal to $\frac{29}{13}$ because when you multiply it by $13$ you get $29$. What we see here is that if we take approximations to $\frac{29}{13}$ given by more and more of its repeated decimal expansion and multiply them by $13$, we get better and better approximations to $29$.
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# Lesson BASICS - Arithmetic Series
Algebra -> Algebra -> Sequences-and-series -> Lesson BASICS - Arithmetic Series Log On
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Algebra: Sequences of numbers, series and how to sum them Solvers Lessons Answers archive Quiz In Depth
This Lesson (BASICS - Arithmetic Series) was created by by longjonsilver(2297) : View Source, ShowAbout longjonsilver: I have a new job in September, teaching Introduction Arithmetic Series or Progressions are sequences of numbers that increment by a fixed common difference eg 4,7,10,13,16,... is an Arithmetic series with common difference 3. Definitions Let d = common difference Let a = first term Let n = number of terms Let Sn = sum of first n terms So, in the sequence -2,3,8,13,18,... we have: first term is -2 --> a = -2 difference is 5 --> d = 5 nth Term Algebraically: 1st term = a 2nd term = a + d 3rd term = a + 2d 4th term = a + 3d 5th term = a + 4d 6th term = a + 5d and so on. In general, the nth term is given as a + (n-1)d --> EXAMPLES Q. Find the 12th term of the sequence 1,5,9,13,... A. a=1, d=4, n=12. which we can verify by writing out the first 12 terms in full: 1,5,9,13,17,21,25,29,33,37,41,45 Summation The summation of n terms is given by: which is also written as where L is the last term, defined as L=a+(n-1)d EXAMPLES Q. Find the sum of the first 10 terms of the sequence 1,5,9,13,17,... a=1, d=4, n=10 --> Sn = 190 Further Example Q. The second term of an arithmetic series is 5 and the fifth term is 14. Find the common difference, the tenth term and the sum of the first 8 terms. We do not know a, which is crucial in all the calculations, so we are aiming to find that as well as d. So, what we do know is the following: 2nd term, a+d = 5 5th term, a+4d = 14 subtract these to give 3d = 9 --> d = 3 So, using this in a+d=5 we have a+3 = 5 --> a = 2 The 10th term = a+9d The 10th term = 2+9(3) The 10th term = 2+27 The 10th term = 29 Now for the sum of 8 terms: Sn = 100 Summary This is the introduction to Arithmetic Series'. However, with the question quoted at the end here, there is not a lot of things that you need to know. It is all about practice now: practice using the "nth term" and "summation" formulae. This lesson has been accessed 13477 times.
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Home » Worksheets » Factors and Multiples Worksheets
# Factors and Multiples Math Worksheets
## Factors and Multiples Worksheets & Study Resources:
Brief definition
Factors are numbers that can be multiplied together to get a new number, while multiples are the results of multiplying a number by an integer. Learners usually begin exploring these topics in 4th grade, wherein more advanced lessons about fractions are introduced. They use the factor tree, listing method, and other strategies to grasp the underlying concepts of factors and multiples. These topics are reinforced and mastered as learners move on to higher levels. Learners begin to understand the Greatest Common Factor (GCF) and the Least Common Multiple (LCM) that they will surely need in operations involving fractions.
Importance of the Topic
Learning about factors and multiples is an important scaffold in learning operations involving fractions — most especially in adding and subtracting, unlike fractions, as well as in simplifying fractions. In addition, understanding factors and multiples creates an opportunity for the learners to see the relationship of numbers and to enhance their multiplication and division skills. For example, as they compose numbers that will equate to 36, learners can think of the following number combinations: 1 x 36, 2 x 18, 3 x 12, and 9 x 4 as factors. If they need to list down the first five multiples of 3, then they can have 3, 6, 9, 12, and 15.
Application of the Learned Topic in Life
Most of the questions that involve factors and multiples are number-mystery problems. Even though it is not seen directly in everyday living, answering number-mystery problems stimulates brain activity. It helps the brain to be more active and busy in which later on sharpens ones’ ability to think logically and critically. Not to mention that it improves the problem-solving ability of an individual.
These topics can also be helpful in solving problems related to the combination of objects that are happening simultaneously to create a meaningful set. For example, if there are 12 water bottles and 16 cans of food for the distribution of food packs, what is the greatest number of food packs that can be made if they have to be distributed equally with no remainder left?
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# Section1.2Vector Equations and Spans¶ permalink
##### Objectives
1. Understand the equivalence between a system of linear equations and a vector equation.
2. Learn the definition of and how to draw pictures of spans.
3. Pictures: an inconsistent system of equations, a consistent system of equations, spans in and
4. Vocabulary: vector equation.
5. Essential vocabulary: span.
# Subsection1.2.1Vector Equations
An equation involving vectors with entries is the same as equations involving only numbers. For example, the equation
(1.2.1)
simplifies to
For two vectors to be equal, all of their coordinates must be equal, so this is just the system of linear equations
##### Definition
A vector equation is an equation involving a linear combination of vectors with possibly unknown coefficients.
Asking whether or not a vector equation has a solution is the same as asking if a given vector is a linear combination of some other given vectors.
For example the vector equation above is asking if the vector is a linear combination of the vectors and
##### Example
Consider the vector equation
Does this vector equation have a solution? Does it have more than one solution?
This vector equation does not have any solution. We could write it as three separate equations:
which all have to be true at the same time for a solution The second equation says and the third says These two statements cannot be true at the same time, no matter what values of and we choose. So there can be no solution to the vector equation.
##### Definition
If one or more solutions exist for an equation or a system of equations, it is said to be consistent. If an equation or system of equations does not have any solution, it is said to be inconsistent.
The above definition is the first of several essential definitions that we will see in this textbook. They are essential in that they form the essence of the subject of linear algebra: learning linear algebra means (in part) learning these definitions. All of the definitions are important, but it is essential that you learn and understand the definitions marked as such.
##### A Picture of a Consistent System
Below we will show that the above system of equations is consistent. Equivalently, this means that the above vector equation has a solution. In other words, there is a linear combination of and that equals We can visualize the last statement geometrically. Therefore, the following figure gives a picture of a consistent system of equations. Compare with figure below, which shows a picture of an inconsistent system.
# Subsection1.2.2Spans
It will be useful to know what are all linear combinations of a set of vectors in In other words, we would like to understand the set of all vectors in such that the vector equation (in the unknowns )
has a solution (i.e. is consistent).
##### Definition
Let be vectors in The span of is the collection of all linear combinations of and is denoted
In symbols:
We also say that is the subset spanned by or generated by the vectors
##### Set Builder Notation
as: “the set of all things of the form such that are in The vertical line is “such that”; everything to the left of it is “the set of all things of this form”, and everything to the right is the condition that those things must satisfy to be in the set. Specifying a set in this way is called set builder notation.
All mathematical notation is only shorthand: any sequence of symbols must translate into an ordinary sentence.
##### Note(Consistency and span)
Here are two ways of saying the same thing:
1. A vector is in the span of
2. The vector equation
is consistent, i.e., it has at least one solution.
Later, in subsection in Section 2.2 we will develop a procedure for answering the question “is in the span of in every circumstance. This will be a byproduct of developing a process to find all the solutions of vector equations.
##### Pictures of Spans
Drawing a picture of is the same as drawing a picture of all linear combinations of
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# How to Convert a Binary Number Into Hexadecimal | Number System
So, hey guys today we are going to learn โHow to convert a binary number into a hexadecimal numberโ.
Before starting this blog letโs take a recap about what are binary decimal and a hexadecimal numbers. So, letโs startedโฆ
Binary Number
Binary numbers are those number who uses only two-digit of a number system 0(Zero) and 1(One) and their base isย โ2โ.ย Example is 101, 010, 1001, 01100111, etc. These types of numbers are known as Binary Numbers.
A Hexadecimal Number is a number that uses 16-digits of a number system. In which the range of a hexadecimal number between 0 to 9 (0,1,2,3,4,5,6,7,8,9) and A to F (A,B,C,D,E,F). A to F is equivalent to the number 10 to 16 (10, 11, 12, 13, 14, 15, and last 16). This number system is known as Hexa-Decimal Number. Example: 9, 12, 27, 2BC, 9DF, 6EA, 5AFE etc.,
For better understanding let start directly with the exampleโฆ
Example 1: Convert a binary number (101)2 into a Hexadecimal number?
Sol:
Step 1: Firstly we have to convert the given number into a decimal number. Letโs convert itโฆ
```= (101)2
= 1x22 + 0x21 + 1x20
= 1x4 + 0x2 + 1x1
= 4 + 0 + 1
= 5
Hence, the decimal value
is (5)10
```
Step 2: Now, we have to divide the given number by 16 until the quotients come to 0.
Hence, our final answer is (5)16.
Example 2: Convert the binary number (10110)2 into a Hexadecimal number?
Sol:
Step 1: Firstly we have to convert the given number into a decimal number. Letโs convert itโฆ
```= (10110)2
= 1x24 + 0x23 + 1x22 + 1x21
+ 0x20
= 1x16 + 0x8 + 1x4 + 1x2
+ 0x1
= 16 + 0 + 4 + 2 + 0
= 22
Hence, the decimal value
is (22)10
```
Step 2: Now, we have to divide the given number by 16 until the quotients come to 0.
Now, write all the remainder from downward to the upward manner like this = 16
Hence, our final answer is (16)16.
Example 3: Convert the binary number (11101101)2 into a Hexadecimal number?
Sol:
Step 1: Firstly we have to convert the given number into a decimal number. Letโs convert itโฆ
```= (11101101)2
= 1x27 + 1x26 + 1x25 + 0x24
+ 1x23 + 1x22 + 0x21 + 1x20
= 1x128 + 1x64 + 1x32
+ 0x16 + 1x8 + 1x4
+ 0x2 + 1x1
= 128 + 64 + 32 + 0 + 8
+ 4 + 0 + 1
= 237
Hence, the decimal value
is (237)10
```
Step 2: Now, we have to divide the given number by 16 until the quotients come to 0.
Now, write the hexadecimal remainder from downward to the upward manner like this = ED
Hence, our final answer is (ED)16.
Example 4: Convert the binary number (1011001111)2 into a Hexadecimal number?
Sol:
Step 1: Firstly we have to convert the given number into a decimal number. Letโs convert itโฆ
```= (1011001111)2
= 1x29 + 0x28 + 1x27 + 1x26
+ 0x25 + 0x24 + 1x23
+ 1x22 +1x21 + 1x20
= 1x512 + 0x256 + 1x128
+ 1x64 + 0x32 + 0x16
+ 1x8 + 1x4 + 1x2 + 1x1
= 512 + 0 + 128 + 64
+ 0 + 0 + 8 + 4 + 2 + 1
= 719
Hence, the decimal value is
(719)10
```
Step 2: Now, we have to divide the given number by 16 until the quotients come to 0.
Now, write the hexadecimal remainder from downward to the upward manner like this = 2CF
Hence, our final answer is (2CF)16.
Example 5: Convert the binary number (10000000001)2 into Hexadecimal number?
Sol:
Step 1: Firstly we have to convert the given number into a decimal number. Letโs convert itโฆ
```= (1000000001)2
= 1x210 + 0x29 + 0x28 + 0x27
+ 0x26 + 0x25 + 0x24 + 0x23
+0x22 + 0x21 + 1x20
= 1x1024 + 0x512 + 1x256
+ 1x128 + 0x64 + 0x32
+ 1x16 + 1x8 + 1x4
+ 1x2 + 1x1
= 1024 + 0 + 0 + 0 + 0
+ 0 + 0 + 0 + 0 + 1
= 1025
Hence, the decimal value
is (1025)10
```
Step 2: Now, we have to divide the given number by 16 until the quotients come to 0.
Now, write the hexadecimal remainder from downward to the upward manner like this = 401
Hence, our final answer is (401)16.
Example 6: Convert the binary number (11110000)2 into a Hexadecimal number?
Sol:
Step 1: Firstly we have to convert the given number into a decimal number. Letโs convert itโฆ
```= (11110000)2
= 1x27 + 1x26 + 1x25 + 0x24
+ 0x23 + 0x22 + 0x21 + 0x20
= 1x128 + 1x64 + 1x32 + 1x16
+ 0x8 + 0x4 + 0x2 + 0x1
= 128 + 64 + 32 + 16 + 0
+ 0 + 0 + 0
= 240
Hence, the decimal value
is (240)10
```
Step 2: Now, we have to divide the given number by 16 until the quotients come to 0.
Now, write the hexadecimal remainder from a downward to an upward manner like this = F0
Hence, our final answer is (F0)16.
Top 13 Must have soft skills to Add in your resumeย 10 Best computer science skills to add in your resume 8 object oriented programming languages to learn in 2024 Top 9 Programming Projects to enhance your resume Top 6 Computer science project for beginners in 2024
Top 13 Must have soft skills to Add in your resumeย 10 Best computer science skills to add in your resume 8 object oriented programming languages to learn in 2024 Top 9 Programming Projects to enhance your resume
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# Question #07f4e
Nov 14, 2017
$- 12 , - 13 , - 14$
#### Explanation:
$\left(- 12\right) + \left(- 13\right) + \left(- 14\right) = - 39$
You know you need 3 integers that all have about the same value because they are consecutive, so divide 39 by 3.
$\frac{39}{3} = 13$
You can then use 2 integers, one greater and one less than -13, to reach -39
Nov 14, 2017
$- 12 , - 13 , - 14$
#### Explanation:
Given: three consecutive integers with sum of -39
We know that the integers are consecutive, so let
$x - 1 = 1 \text{st integer}$
$x = 2 \text{nd integer}$
$x + 1 = 3 \text{rd integer}$
Since we know that the sum of the three integers is -39, we know that
$\left(x - 1\right) + \left(x\right) + \left(x + 1\right) = - 39$
Solving for this:
$\left(x - 1\right) + \left(x\right) + \left(x + 1\right) = - 39$
$x - 1 + x + x + 1 = - 39$
$3 x = - 39$
$x = - 13$
In this case, by substituting the computed $x$ with the values above, we would know that:
1st integer = -14
2nd integer = -13
3rd integer = -12
*Note that it does not matter if you add positive 1 or negative 1 just as long as you use consistent terms and you define your variables properly. So regardless of whether you use x, x - 1, x - 2 to define your three integers or x, x + 1, x + 2, you will still arrive at the same answer.
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## Hypothesis Testing Database
### Statistics
[2003] Joe Smith was running for mayor. His campaign manager estimated that he needed at least 40% of the popular vote in order to win. A poll of 400 people was taken and 150 indicated they would vote for Smith.
a) If we use a 10% level of significance, does he have sufficient support?
b) Construct a 90% confidence interval on the percentage who will vote for Smith.
Solution part a): The campaign manager is concerned if support falls below 40%. So, we set up our hypotheses as:
H0: p = 0.4
H1: p < 0.4
Because a = 0.1, we reject the null hypothesis if Z < -1.282, where:
and q = 1-p. In our case, we have:
Plugging these values into the formula for Z, we get Z = -1.0206. Since -1.0206 > -1.282 we do not reject the null hypothesis and conclude that Smith has enough support to win the election.
Solution part b): The formula for our confidence interval for p is:
In our case, we have:
Plugging these numbers into the formula, we have:
0.375 - 0.0398 < p < 0.375 + 0.0398
0.3352 < p < 0.4148
So the support for Smith lies between 33.52% and 41.48% based on the evidence.
[2005] A certain intersection had an accident rate of 1 accident per 1000 vehicles. A study suggested that partial obstruction of the stop sign by a tree branch could be partly to blame. The branch was trimmed and the traffic was monitored for 10,000 vehicles. During this period, there were 8 accidents. Did trimming the tree branch help? Use a p-value to reach your conclusion.
Solution: If trimming the tree branch helped, we would expect fewer accidents. So we set up our hypotheses as:
H0: p = 0.001
H1: p < 0.001
Because of the sample size, we can use Z because of the Central Limit Theorem.
We will reject the null hypothesis for low p-values where p-value = P(Z < test statistic).
Our formula for Z is
where q = 1 - p. We have:
Plugging these numbers into the equation we get Z = -0.6328.
Our p-value is P(Z < -0.6328) = 0.2634
Because of the high p-value, we do not reject the null hypothesis and conclude that trimming the tree branch made no significant difference.
[2001] A wood door manufacturer wants to minimize the number of knots in each door. On average, he wants no more than 1.2 knots per door. A random sample of 500 doors had, on average, 1.4 knots with a standard deviation of 0.26 knots. Are the manufacturer's standards being met? Test at a = 0.05.
Solution: The manufacturer is concerned if there are too many knots in the door. Ergo, we set up our hypothesis as:
H0: m = 1.2
H1: m > 1.2
Because of the sample size of 500, we can use Z because of the Central Limit Theorem. With a = 0.05, we reject the null hypothesis if Z > 1.645. Our test statistic is:
where
Plugging these numbers into the formula, we get Z = 17.2. Since 17.2 > 1.645, we reject H0 and conclude that the standards are not being met.
[2004] In order to maintain quality control, the angle measurement of a gear tooth must be maintained at 1.02o. A sample of 60 gears is taken. The average angle is 1.025o with a standard deviation of 0.05o.
a) Construct a 95% confidence interval of the average gear angle.
b) If you were to test the hypothesis H0: m = 1.02 against H1: m ¹ 1.02 at a = 0.05, what would be your conclusion using only the confidence interval?
Solution part a): Our formula for the confidence interval is:
For our 95% confidence interval we need Z0.025 = 1.96 because 1 - a = 0.95, and so a/2 = 0.025.
We have:
Plugging these values into the equation we get 1.025 - 0.0127 < m < 1.025 + 0.0127 or 1.0123 < m < 1.0377
Solution part b):Because 1.02 is contained in the 95% confidence interval, we would conclude at a 5% level of significance that the standards are being met (in other words, we would not reject the null hypothesis).
[2002] Joe's Ceramics wants to keep the standard deviation of the oven temperature to within 5o. He samples 100 baked ceramic pieces and measures (somehow) the oven temperature when he takes the piece out. If he wants the bound on the error to be at most 1o, what should be the minimum level of confidence?
Solution: This is a two step problem:
Step 1: Solve for Z.
Step 2: Solve for 1 - a which is the level of confidence we are looking for.
Our formula is:
Solving for Z, we have:
E = 1; s = 5; n = 100
Plugging these numbers into the formula, we get Za/2 ³ 2. Since P(Z > 2) = 0.0227, that means a/2 = 0.0227. Solving for our level of confidence, we get 1 - a = 1 - (2)(0.0227) = 0.9546. So, our minimum level of confidence is 95.46%
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# The Easy Way to Calculate CMC Uncertainty
### For Your Scope of Accreditation
One of the most frequently asked questions that I receive is “How do I convert my uncertainty estimates to a formula?” In my last post, I showed you how to calculate CMC Uncertainty using the equations in your scope of accreditation. Now, I am show you have to reduce your uncertainty estimates to an equation.
I am a big advocate of using equations for my CMC uncertainty. I believe equations better represent the uncertainty of a range and function. Fixed value uncertainty estimates are better suited for fixed point reference values. Equations provide an uncertainty value that changes with the value of the range.
In this post, I am going to show you two methods to develop a mathematical equation to represent your uncertainty. The first method is interpolation.
## Method 1: Interpolation
The quickest and simplest way to calculate uncertainty equations is interpolation between two points. Remember when your grade school teacher made you calculate the slope and y-intercept to solve for the line equation? Well, this is it. If you know ‘x’ and ‘y,’ you can solve for gain and offset. This is called interpolation between two points.
### 1. Find Max and Min Points
The first task to solving our linear equation is to identify the minimum and maximum points of our function. Now, determine the minimum and maximum values of ‘x’ and ’y.’
where,
x1 = minimum input variable
x2 = minimum input variable
y1 = maximum output variable
y2 = maximum output variable
### 2. Calculate Gain
Now that you have identified the maximum and minimum values of ‘x’ and ‘y,’ calculate the slope or gain coefficient. This is accomplished by finding the difference of ‘y’ and the difference of ‘x.’ Then, divide the difference of ‘y’ by the difference of ‘x.’ You have just calculated the gain coefficient which represents the rate of change for the value of y based in the input value of x. Use the formula below as a guide.
Need additional help? I will show you how to do it in MS Excel.
### 3. Calculate Offset
After calculating the gain coefficient, it’s time to solve for the offset coefficient. To accomplish this, multiply the gain coefficient by the minimum value of x. Then, subtract this value from the minimum value of y. Use the equation below as a guide.
Once again, let’s do it in MS Excel.
Hopefully, I made it easy for you. This is a pretty simple process. If it seems difficult, follow the instructions I have provided and practice. Repetition is the key to learning to use this technique on command. You may find yourself using this method more often that you think. It’s so easy, I use it all the time; no regression algorithm or software needed. Soon enough, I believe you will too.
### Interpolation vs Extrapolation
Interpolation and extrapolation are different. Solving for or inferring a value between two points is interpolation. Solving or inferring a value beyond these two points is extrapolation.
One point that I want to emphasize is to use your equations for interpolation only. This will help ensure confidence in your results. Using them for extrapolation can yield errors and larger uncertainty in your results.
## Method 2: Regression
Pick points across the range that are evenly spaced to prevent errors. Some may argue that spacing test-points does not matter; but it does! Otherwise, most calibration procedures would not instruct you to test at 10%, 20%, or 25% intervals. This may be the case for a linear equation; however, if you practice this methodology for non-linear regression, you will quickly become and unhappy camper when verifying and graphing your verification data.
### 2. Find the Mean of x and y
Independently calculate the mean (i.e. average) of both x and y. Using the ‘average’ function in excel, you should be able to find the mean quickly. Just type ‘=average(,’ select the cells you want to calculate average, and close the parenthesis ‘).’ Use the images below as a guide.
### 3. Find the Difference from the Mean
Now that you have calculated the mean of x and y, calculate the deltas (i.e. differences) from the mean of x and y. Subtract each ‘x’ value by the mean of x. In the image below, you will notice that I represent the mean with ‘x-bar’ and difference with ‘delta-x.’ Repeat this calculation for each value of x.
Next, subtract each ‘y’ value by the mean of y. In the image below, you will notice that I represent the mean with ‘y-bar’ and difference with ‘delta-y.’ Repeat this calculation for each value of y.
### 4. Calculate Gain
Now it is time to calculate the gain coefficient. I will show you how to use the equation below to calculate the gain.
### 5. Multiply dx and dy
Find the product of delta-x and delta-y by multiplying them together. Repeat this for each value of delta-x and delta-y in series. Use the image below as a guide.
### 6. Square dx
Now, find the squared value of delta-x by multiplying dx by itself or using an exponent of 2. See the image below as a guide.
### 7. Sum the values
Find the summation of all the products and squares. First, add together all the values of dy*dx.
Next, add together all the values of dx^2.
### 8. Divide the Sum of Products by the Sum of Squares
Now, calculate the gain coefficient B1 by dividing the sum of dy*dx by sum of dx^2.
That’s it! You have successfully calculated the gain coefficient. Next, I will show you how to calculate the offset coefficient.
### 9. Calculate Offset
In this section, I will show you how to find the offset coefficient B0 using B1, the mean of x, and the mean of y. The equation that will need to use is below; but, I am am going to use MS Excel again to show you how calculate offset.
To calculate offset, multiply the gain coefficient and the mean of x. Afterward, subtract this calculated value from the mean of y. The result will the offset coefficient.
### Verifying the Results
Now that you have used regression to calculate your CMC equation, you can use the following equation to verify that your new equation matches the original results of x and y. If so, you now have an equation that will allow you to confidently estimate uncertainty. If your equation does not fit the data, something is wrong or the data cannot be modeled using a linear equation.
To learn how to use this equation, read How to Calculate CMC Uncertainty Like A Pro.
### Determining Appropriateness or Fit
Beyond verifying the results, there are some other measures that can be used to determine whether or not the model will fit the data. One measure is to check the value of R2 or R-squared. The closer R-squared is to one, the more likely the data the model will fit the data. You can also evaluate the standard error of the model. When the standard error is small, the more likely the data the model will fit the data. Additionally, you can evaluate the standard error of the model to determine if the model is appropriate for your uncertainty analysis.
Another method is to evaluate the F-statistics. This is a great metric if you are comparing several models for a single data set. From my experience, the model with the greatest value of F and the smallest value of Significance F will have the best fit. However, it is best to graph the model over the data and determine if the model fits the data and is appropriate for use. Sometimes you will observe a model that fits the data perfectly, but it is not appropriate or realistic when you evaluate the graph.
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Math Expressions Grade 5 Unit 5 Lesson 8 Answer Key Divide with Two Decimal Numbers
Math Expressions Common Core Grade 5 Unit 5 Lesson 8 Answer Key Divide with Two Decimal Numbers
Math Expressions Grade 5 Unit 5 Lesson 8 Homework
Divide.
Lesson 8 Divide with Two Decimal Numbers Math Expressions Question 1.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 420 ÷ 7
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 420, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 7, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 42 by the divisor, 7.
42 divided by 7 is 6, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 4.2 ÷ 0.07 = 60
Unit 5 Lesson 8 Math Expressions Divide with Two Decimal Numbers Question 2.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 24 ÷ 8
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 24, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 8, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 24 by the divisor, 8.
24 divided by 8 is 3, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 2.4 ÷ 0.8 = 3
Question 3.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 480 ÷ 5
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 480, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 5, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 48 by the divisor, 2.
48 divided by 5 is 9, with a remainder of 3
Bring down the next number of the dividend and insert it after the 3 so you have 30.
30 divided by 5 is 6, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 4.8 ÷ 0.05 = 96
Question 4.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 206.4 ÷ 24
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 206.4, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 24, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 206 by the divisor, 24.
206 divided by 24 is 8, with a remainder of 14. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 14 so you have 144.
144 divided by 24 is 6, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 0.264 ÷ 0.24 = 8.6
Question 5.
Circle the division that does not have the same answer as the others.
54 ÷ 6
5.4 ÷ 0.6
0.54 ÷ 0.6
0.54 ÷ 0.06
0.054 ÷ 0.006
Answer: 0.54 ÷ 0.6 = 0.9
Explanation:
0.54 ÷ 0.6
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 54 ÷ 6
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 54, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 6, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 54 by the divisor, 6.
54 divided by 6 is 9, with a remainder of 0. You can ignore the remainder for now.
Since the remainder is 0, your long division is done.
So, 0.54 ÷ 0.6 = 0.9
Except 0.54 ÷ 0.6 , All of the given divisions have the output of 9
Solve.
Question 6.
A beekeeper collected 7.6 liters of honey. She will pour it into bottles that each hold 0.95 liter. How many bottles will she fill?
Answer: She can fill 8 bottles
Explanation:
Given, A beekeeper collected 7.6 liters of honey.
She will pour it into bottles that each hold 0.95 liter.
We have 7.6 ÷ 0.95
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 760 ÷ 95
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 760, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 95, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 760 by the divisor, 95.
760 divided by 95 is 8, with a remainder of 0. You can ignore the remainder for now.
Since the remainder is 0, your long division is done.
So, 7.6 ÷ 0.95 = 8
Thus, She can fill 8 bottles
Question 7.
A very small dinosaur, the micro raptor, was only 1.3 feet long. One of the largest dinosaurs, the diplodocus, was about 91 feet long. How many times as long as the micro raptor was the diplodocus?
Answer: The diplodocus is 7 times as long as the micro raptor
Explanation:
Given, A very small dinosaur, the micro raptor, was only 1.3 feet long.
One of the largest dinosaurs, the diplodocus, was about 91 feet long.
We have, 91 ÷ 1.3
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 910 ÷ 13
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 91, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 13, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 91 by the divisor, 13.
91 divided by 13 is 7, with a remainder of 0. You can ignore the remainder for now.
Since the remainder is 0, your long division is done.
So, 91 ÷ 1.3 = 7
So, The diplodocus is 7 times as long as the micro raptor
Question 8.
Tomorrow, in the town of Eastwood, there will be a big race. The course is 5.25 kilometers long. A water station will be set up every 0.75 kilometer, including at the finish line. How many water stations will there be?
Answer: There will be 7 water stations.
Explanation:
Given, The course is 5.25 kilometers long.
A water station will be set up every 0.75 kilometer, including at the finish line.
We have, 5.25 ÷ 0.75
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 525 ÷ 75
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 525, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 75, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 525 by the divisor, 7.
525 divided by 75 is 7, with a remainder of 0. You can ignore the remainder for now.
Since the remainder is 0, your long division is done.
So, 5.25 ÷ 0.75 = 7
Finally, There will be 7 water stations.
Question 9.
Marisol’s bedroom has an area of 29.76 square meters. The length of the room is 6.2 meters. What is its width?
Answer: The width is 4.8 meters.
Explanation:
The width w of rectangle is given by al that is, w = $$\frac{a}{l}$$
Then we have, 29.76 ÷ 6.2
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 297.6 ÷ 62
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 297.6, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 62, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 297 by the divisor, 62.
297 divided by 62 is 4, with a remainder of 49. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 49 so you have 496.
496 divided by 62 is 8, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, The width is 4.8 meters.
Math Expressions Grade 5 Unit 5 Lesson 8 Remembering
Round to the nearest tenth.
Question 1.
1.28 __________
Explanation:
Rounding to the nearest tenth means,
leaving only one number after the decimal point.
So, 1.28 will be 1.3
Question 2.
14.21 __________
Explanation:
Rounding to the nearest tenth means,
leaving only one number after the decimal point.
So, 14.21 will be 14.2
Question 3.
8.148 __________
Explanation:
Rounding to the nearest tenth means,
leaving only one number after the decimal point.
So, 8.148 will be 8.1
Round to the nearest hundredth.
Question 4.
4.769 __________
Explanation:
Round to the nearest hundredth means
leaving only two numbers after the decimal point
So, 4.769 will be 4.77
Question 5.
45.124 __________
Explanation:
Round to the nearest hundredth means
leaving only two numbers after the decimal point
So, 45.124 will be 45.12
Question 6.
16.107 __________
Explanation:
Round to the nearest hundredth means
leaving only two numbers after the decimal point
So, 16.107 will be 16.11
Solve.
Question 7.
Explanation:
Multiply the each number in 7.7 with the each number in 1.4,
place the decimal according to the count of the numbers after the decimal .
That is 10.78
Question 8.
Explanation:
Multiply the each number in 3.1 with the each number in 0.05,
place the decimal according to the count of the numbers after the decimal .
That is 0.155
Question 9.
Explanation:
Multiply the each number in 5.79 with the each number in 0.9,
place the decimal according to the count of the numbers after the decimal .
That is 5.211
Question 10.
Explanation:
Multiply the each number in 3.4 with the each number in 8.8,
place the decimal according to the count of the numbers after the decimal .
That is 29.92
Question 11.
Explanation:
Multiply the each number in 3.5 with the each number in 0.46,
place the decimal according to the count of the numbers after the decimal .
That is 1.61
Question 12.
Explanation:
Multiply the each number in 8.6 with the each number in 0.90,
place the decimal according to the count of the numbers after the decimal .
That is 7.74
Solve.
Question 13.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 360 ÷ 9
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 360, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 9, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 36 by the divisor, 9.
36 divided by 9 is 4, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 9 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 36 ÷ 0.9 =40
Question 14.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 48000 ÷ 6
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 48000, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 6, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 48 by the divisor, 6.
48 divided by 6 is 8, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 6 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 48 ÷ 0.006 =8000
Question 15.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 3200 ÷ 4
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 3200, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 4, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 32 by the divisor, 4.
32 divided by 4 is 8, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 4 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 32 ÷ 0.04 =800
Question 16.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 3640 ÷ 7
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 3640, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 7, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 36 by the divisor, 7.
36 divided by 7 is 5, with a remainder of 1. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 1 so you have 14.
14 divided by 7 is 2, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 364 ÷ 0.7 =520
Question 17.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 221000 ÷ 34
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 221000, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 34, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 221 by the divisor, 34.
221 divided by 34 is 6, with a remainder of 17. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 17 so you have 170.
170 divided by 34 is 5, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 2210 ÷ 0.34 =6500
Question 18.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 149400 ÷ 83
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 149400, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 83, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 149 by the divisor, 83.
149 divided by 83 is 1, with a remainder of 66. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 66 so you have 664.
664 divided by 83 is 8, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 1494 ÷ 0.83 =1800
Question 19.
Stretch Your Thinking Must a decimal divisor and a decimal dividend have the same number of decimal places in order to have a whole-number quotient? Write a division equation using two decimal numbers to support your answer.
Answer: 2.4 ÷ 0.3 = 8
Explanation:
2.4 is the dividend
0.3 is the divisor
8 is the quotient
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 24 ÷ 3
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 24, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 3, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 24 by the divisor, 3.
24 divided by 3 is 8, with a remainder of 0. You can ignore the remainder for now.
Since the remainder is 0, your long division is done.
So, 2.4 ÷ 0.3 = 8.
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# Simplify Equations
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Equations are the very common and used in almost every sub-topic in math. Equations consist of one or more than one unknown variables with different coefficients and constant numbers. Exponent is the degree of the variable in the equation. The degree of the variable can be one or more than one. There are various mathematical operations which are used to simplify and solve equations.
Example 1: Simplify and solve for x in the equation 9 x – 4 = 50?
Solution: Given equation is 9 x – 4 = 50.
Here the unknown variable which needs to be solved for is x.
First step: Adding 4 on both sides of the given equation.
(9 x – 4) + 4 = 50 + 4.
This gives 9 x = 54.
Now dividing both sided of the equation by 9.
This gives 9 x/ 9 = 54 / 9.
This reduces the given equation to x = 6.
Hence the solution is x = 6.
Example 2: Simplify the equation 4 (x – 3) + 5(x + 2) + 2?
Solution: Given equation is 4 (x – 3) + 5(x + 2) + 2.
Here the variable is x; distributing the number in front of the braces.
This gives 4 (x - 3) = 4x – 12; 5 (x + 2) = 5x + 10.
Combining the similar terms in the equation.
This gives 4x – 12 + 5x + 10 + 2 = 9x + 2 -2 = 9x
Hence the simplified form of the equation is 9x.
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# How do you find the radius from the area?
Method 2 Using the Area
1. Set up the formula for the area of a circle. The formula is A = π r 2 {displaystyle A=pi r^{2}} , where A {displaystyle A} equals the area of the circle, and r {displaystyle r} equals the radius.
3. Plug the area into the formula.
4. Divide the area by .
5. Take the square root.
A.
### What is the radius of the circle?
The circumference is the outside perimeter of a circle. It's the distance around a circle. A radius can be a line from any point on the circumference to the center of the circle. If you put two radii together, edge to edge, going through the center of a circle, you would get a diameter.
• #### How do you get the circumference of a circle with the radius?
The circumference = π x the diameter of the circle (Pi multiplied by the diameter of the circle). Simply divide the circumference by π and you will have the length of the diameter. The diameter is just the radius times two, so divide the diameter by two and you will have the radius of the circle!
• #### How do you find out the radius of a cylinder?
First, divide the diameter by 2 and plug the values for volume, pi, and radius into the formula for volume of a cylinder. Next, square the radius and multiply the values together. Then, divide both sides by 200.96 for the answer, remembering to include the appropriate unit of measurement.
• #### What is the formula for the diameter?
The diameter is the measurement across the circle passing through the center. The two formulas involving the diameter are the one that says the diameter is twice the radius and the one that says the circumference is the diameter times pi.
B.
### How do you find the radius of a semicircle?
To work out the area of a semicircle you can use the formula A = (Pi*r^2)/2. This is because the area of a full circle is A = Pi*r^2 and a semicircle is half the area of a full circle so you divide this by 2. Just be careful to substitute the radius into the formula and not the diameter!
• #### How do you find the area of a quadrant?
Substitute r = 8 directly into the formula A = ¼ πr². As you can see it gives exactly the same answer as method 1. Work out the area of this quadrant (radius 3.8m). = 14.44π (leave the answer as an exact solution as this need to be divided by 4).
• #### What is a minor arc of a circle?
A minor arc (left figure) is an arc of a circle having measure less than or equal to ( radians). SEE ALSO: Arc, Major Arc, Semicircle. REFERENCES: Rhoad, R.; Milauskas, G.; and Whipple, R. Geometry for Enjoyment and Challenge, rev. ed.
• #### How do you calculate the square root of a number?
To start finding a square root via prime factorization, first, try to reduce your number into its perfect square factors.
1. Let's use an example. We want to find the square root of 400 by hand. To begin, we would divide the number into perfect square factors.
2. We would write this as: Sqrt(400) = Sqrt(25 × 16)
C.
### How do you find diameter from circumference?
Solve the equation for the diameter of the circle, d= C/π. In this example, "d = 12 / 3.14." or "The diameter is equal to twelve divided by 3.14." Divide the circumference by pi to get the answer. In this case, the diameter would be 3.82 inches.
• #### How do you work out a circumference of a circle?
Plug in the numbers to solve for C. Now we know that 2πr = C. Look back at the original math problem to see what r (the radius) equals. Then replace π with 3.14, or use a calculator's π button to get a more accurate answer. Multiply 2πr together using these numbers. The answer you get is the circumference.
• #### How do you figure out the perimeter of a circle?
We know the formula to find the perimeter of the circle if the diameter is given, namely π D. Substitute the diameter 4.4 and Pi value as 3.14 in the above formula. Therefore 13.82 cm is the perimeter of the given circle.
• #### How do you find the area with the circumference?
The area of a circle is given by the formula A = π r2, where A is the area and r is the radius. The circumference of a circle is C = 2 π r. If we "solve for r" in the second equation, we have r = C / (2 π ).
Updated: 20th September 2018
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# Estimation – Definition, Factors, and Examples
Sep 19, 2022
### Key Concepts
• Estimation
• Rounding Numbers
• Front-End Estimation
• Factors
• Composite and Prime Numbers
• L.C.M. and H.C.F.
• Multiples
• Multiples of Numbers
• Common Multiples
• Least Common Multiples
• Multiplication Using Models
• Array Model
• Area Model
### Introduction:
In this chapter, we will be learning about the following:
• Rounding numbers to the nearest 10,000.
• Estimating sums using front-end estimation.
• Estimating differences using front-end estimation.
• Multiplying two numbers to find the product.
• Composite and Prime Numbers
• L.C.M. and H.C.F.
• Multiples of Numbers
• Common Multiples
• Least Common Multiples
• Multiplication Using Models
## Estimation in Maths
Definition of estimation:
Estimation of numbers is the process of approximating or rounding off the numbers in which the value is used to avoid complicated calculations. There is a difference between the terms ‘exact’ and ‘estimation.’
Example: Estimate the sum of 34 and 56 by rounding off to the nearest hundred.
Sol.:
### Rounding numbers to Estimate Sum, Differences, Products, and Quotient.
Rounding: A number can be rounded to the nearest ten or hundred by looking at the digit to the right of the tens or hundreds place to give an approximate value.
If it is less than 5, round down. If it is 5 or more, round up.
Examples:
1) Estimate 286 + 495.
Sol.:
The estimated sum rounded nearest to 100 is 800.
2) Estimate the difference of 786 – 214.
Sol.: 786 to the nearest hundred = 800
214 to the nearest hundred = 200
Required = 800 – 200 = 600
3) Estimate the product of 958 × 387.
Sol.: 958 rounds off as, 1000
387 rounds off as, 400
Required product = 1000 × 400 = 400,0000
4) Estimate the quotient of 2838 ÷ 125.
Sol.: 2838 rounds off as, 2800
125 rounds off as, 100
Required quotient = 2800 ÷ 100 = 28
### Front-End Estimation.
Front-end estimation uses the leading digits in numbers to make an estimate.
Example: Find the sum of 9411 and 3849 by using front-end estimation.
Sol.:
Example: What is the front-end estimation of 8792?
Sol.: The front-end estimation of 8792 is 8000.
### Estimate to check whether the answer is reasonable.
Reasonable: It is an estimate that is close to the actual answer.
Example: Find the sum of 254 + 143 to the nearest hundred.
Sol: 254 + 143 = 397
Estimate to check the answer is reasonable.
[Since both numbers are rounded up, the estimate is greater than the actual sum.]
### Factors
Factor: Multiplying two whole numbers gives a product. The numbers that we multiply are the factors of the product.
Example: Find the factors of 12.
Sol.:
Example: Find the factors of 14.
Sol.: 14 can be divided exactly by 7.
So, 7 is a factor 14.
Hence, the factors of 14 are 1, 7, 2, 14.
14 = 14 × 1
= 7 × 2
Note: A factor divides a number entirely without leaving any remainder.
### Composite and Prime Numbers.
Composite numbers: Composite numbers have more than two factors. This means that apart from getting divided by 1 and itself, the composite numbers divide entirely by other numbers as well.
Example:
Example: 4, 6, 8, 9, 10, 12… are composite numbers.
Prime numbers: A prime number has exactly two factors, 1 and itself.
Example:
Sol.: 2, 3, 5, 7, 11, 13, 17…. are prime numbers.
### L.C.M. and H.C.F.
L.C.M.: The least common multiple (L.C.M.) is the smallest number divisible by two or more given numbers.
Example: Find the L.C.M. of 18, 24, 96.
Sol.:
Example: Find out the L.C.M. of 32, 36.
H.C.F.: The highest common factor (H.C.F.) is the largest factor shared by two or more numbers.
Example: Find the H.C.F. of 144, 180, 192.
Sol.:
Example: Find the H.C.F. of 90 and 30.
H.C.F. = 2 × 3 × 5 = 30
### Multiples
#### Finding multiples of a number.
Multiples are numbers that we get when we multiply a number by 1, 2, 3, 4 and so on.
Example: Find the multiples of 15.
Sol.:
Example: Find the multiples of 6.
Sol.: 6, 12, 18, 24, 30…60…
#### Finding Common Multiples of Two Whole Numbers.
Common multiples: The multiples that are common to two or more numbers are called the common multiples of those numbers.
Example: What are the common multiples of 3 and 4?
Sol.:
The common multiples of 3 and 4 are 12, 24, 36.
Example: What are the common multiples of 2 and 4?
Sol.: The multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18…
The multiples of 4 are 4, 8, 12, 16, 20…
Common multiples are 4, 8, 12, 20…
### Finding the L.C.M. of Two Whole Numbers.
L.C.M: The smallest number of all the common multiples is called the L.C.M.
Example: What is the L.C.M. of 3 and 4?
Sol.:
### Multiplying Using Models
#### Multiplying Numbers Using the Array model.
Array model: The array model of multiplication uses the number of rows and number of columns in an array to illustrate the product of two numbers.
Example: Draw an array model of 2 × 4.
Sol.:
Example:
Multiplying using Array Model.
Sol.: There are 5 columns and 2 rows.
Product = 5 × 2
= 10
#### Multiplying Numbers Using Area Model.
Area model: An area model is a rectangular diagram or model used for multiplication. In the diagram, the factors define the length and width of the rectangle.
Example: Find 73 × 64 by using the area model.
Sol.:
Example: Find 27 × 35 using the area model?
Sol.: 27 × 35 = (20 + 7) × (30 + 5)
600 + 100 + 210 + 35 = 945
Therefore, 27 × 35 = 945
### Exercise:
1. Estimate the value of 294 to the nearest hundreds.
2. Estimate the sum of 256 + 342 to the nearest hundreds.
3. Estimate the difference of 525 – 430 to the nearest hundreds.
4. Estimate the product of 164 × 630 to the nearest hundreds.
5. Estimate the quotient of 5463 ÷ 630 to the nearest hundreds.
6. Using front-end estimation, find the sum of 2456 and 5432.
7. Write the factors of 15.
8. Find the factors of 54 and then decide whether it is prime or composite?
9. Find the factors of 17 and then decide whether it is prime or composite?
10. Find out the L.C.M. of 44 and 63.
11. Find out the H.C.F. of 50 and 64.
12. Write the common multiples of 6 and 8.
13. What are the least common multiples of 5 and 10?
14. Multiply using the array model.
15. Find the product using the area model for 25 × 18.
### What have we learned:
In this chapter, we learned:
• When two factors are multiplied, the product is a multiple of both numbers.
• About knowing the factors and multiples of numbers can help in estimating products and quantities.
• About using front-end estimation to check the reasonableness of products, sums and differences.
### Concept Map:
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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Last updated date: 07th Dec 2023
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# In the adjoining figure: Name the vertex opposite to side $PQ$.
Answer
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Hint: In order to find the opposite vertex of the given side $PQ$ from the $\Delta PQR$, firstly, we must consider all the sides and the vertices. To find the opposite vertex of $PQ$, we must find such vertex from the $\Delta PQR$ such that it does not have any connection with the given particular side which simply means that it is not at all connected with the side $PQ$.
Complete step by step answer:
Now let us learn about the properties of a triangle. The angle-sum property of a triangle states that the sum of interior angles of a triangle is equal to ${{180}^{\circ }}$. The sum of lengths of any two sides of a triangle is greater than the third side of a triangle. There are six types of triangles. They are: equilateral triangle, isosceles triangle, scalene triangle, obtuse angled, right angled and acute angled triangle. The area of a triangle can be found by $\dfrac{1}{2}bh$ and the perimeter of a triangle can be found by sum of lengths of all three sides.
Now let us find the vertex that is opposite to the side $PQ$ from the triangle $\Delta PQR$.
Opposite vertex to a side is nothing but the vertex that is not adjacent to the side.
From the figure, we can observe that the vertices $P$ and $Q$ are adjacent to the side $PQ$. In fact, we can say that the side $PQ$ is formed by those two vertices.
So, the only vertex that is not adjacent to the side $PQ$ is $R$.
$\therefore$ The vertex opposite to side $PQ$ is $R$.
Note: We can always have a note that this opposite and the adjacent theorems are applicable to all of the triangle. We must also note that while finding the area of the triangle, the height of the triangle depends upon the type of the triangle as the height of an obtuse angled triangle lies outside whereas the height of an equilateral triangle lies inside the triangle.
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# How do you convert -4 + 8i to polar form?
Jul 15, 2017
In polar coordinates, the point will be $\left(r , \theta\right) = \left(6.92 i , - 1.11 i\right)$
#### Explanation:
For coordinates in polar form, $\left(r , \theta\right)$, we need to find the value of $r$ and the value of $\theta$.
We will use Pythagoras' theorem to find $r$ and the $\tan$ function to find $\theta$.
$r = \sqrt{{\left(- 4\right)}^{2} + 8 {i}^{2}} = \sqrt{16 - 64} = \sqrt{- 48} = 6.92 i$
(since $8 {i}^{2} = 8 i \times 8 i = 8 \sqrt{- 1} \times 8 \sqrt{- 1} = 64 \times - 1 = - 64$)
$\tan \theta = \frac{o p p}{a \mathrm{dj}} = \frac{8 i}{-} 4 = - \frac{8}{4} i = - 2 i$
So $\theta = {\tan}^{-} 1 \left(- 2 i\right) = - 1.11 i$
Jul 15, 2017
$\left(4 \sqrt{5} , 2.03\right)$
#### Explanation:
$\text{to convert from "color(blue)"cartesian to polar form}$
$\text{that is "(x,y)to(r,theta)" where}$
•color(white)(x)r=sqrt(x^2+y^2)
•color(white)(x)theta=tan^-1(y/x)color(white)(x)-pi< theta <=pi
$\text{here " x=-4" and } y = 8$
$\Rightarrow r = \sqrt{{\left(- 4\right)}^{2} + {8}^{2}} = \sqrt{80} = 4 \sqrt{5}$
$- 4 + 8 i \text{ is in the second quadrant }$
$\text{so we must ensure that "theta" is in the second quadrant}$
$\theta = {\tan}^{-} 1 \left(2\right) = 1.11 \leftarrow \textcolor{red}{\text{ related acute angle}}$
$\Rightarrow \theta = \left(\pi - 1.11\right) = 2.03 \leftarrow \textcolor{red}{\text{ in second quadrant}}$
$\Rightarrow - 4 + 8 i \to \left(- 4 , 8\right) \to \left(4 \sqrt{5} , 2.03\right)$
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Question of Solved questions
# Question My Present age is four times the difference of my age after four years and my age three years back. How old am I?
Factorize x2+x-6
Solution:
Explanation:
On factoring the equation, we get,
x2+x-6
x2+3x-2x-6
x(x+3)-2(x+3)
(x+3)(x-2)
The simplified form of the equation x2+x-6 is (x+3)(x-2).
Factorize x^2-2x-8
Solution:
Explanation:
We have;
x2-2x-8
x2-4x+2x-8
x(x-4)+2(x-4)
(x-4) (x+2)
x=4,x=-2
Hence, we factorized the given expression as, x=4,x=-2
The opposite angles of a parallelogram are
are (3x - 2) and (x + 48) Find the measure of each angle of the parallelogram.
Solution:
Explanation:
Let the parallelogram be ABCD & the angles of parallelogram be <A,<B,<C & <D
From the given question, opposite angles of a parallelogram are (3x-2) & (x+48).
Let <A=3x-2 & <B=x+48
As we know that, “the opposite angles of a parallelogram are always equal”. Therefore, we can write;
(3x-2)=(x+48)
⇒3x-x=48+2
⇒2x=50
⇒x=50/2
⇒x=25
Substituting the value of x in 3x-2, we get;
3(25)-2
=75-2
=73º
⇒<A=<C=73
Finding the measure of other two angles:
We know that, the sum of adjacent angles of a parallelogram is equal to 180.
∴ <A+<B=180º
⇒73º+<B=180º
⇒<B=180º-73º
⇒<B=107º
∴ <D=107º
Therefore, the measure of each angle of the parallelogram is <A=73º,<B=107º,<C=73º,<D=107º.
Hence, we measured all the angles of parallelogram as;<A=73º,<B=107º,<C=73º,<D=107º.
In the given figure the value of x is
Solution:
Explanation:-
The value of x is 125º.
Estimate the value of square root
square root :√22
Solution:
Final Answer: The estimated square root of √22 is 4.690.
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# Advanced Algebra Trigonometry Objective SWBAT solve linear equations
• Slides: 11
Advanced Algebra - Trigonometry Objective: SWBAT solve linear equations. 1
Warm Up Simplify: 1. 8 b - 3( 4 – b ) 2. -6 (m – 9) + 14 m – 20 3. 2 (b + 5) + 3 (2 b – 10) 4. -w 3 + w 2 - 7 w 2 - 8 w 3 5. 7 t (t 2 + 2) + 9 t (t – 2) 2
Vocabulary Equation: a statement that two expressions are equal. Ex. x + 2 = 9, x 2 – 4 x +10 = 0, Solution: all numbers that make an equation true. Solution Set: the set of all the solutions of an or condition. Equivalent Equations: equations with the same solution set Ex. x + 1 = 5 and 6 x + 3 = 27 Linear Equation: a linear equation in one variable has the form ax + b = c (aka first degree equation) 3
Which of the following are a linear equations? • 4
Solve an equation with variable on one side. 5
Solve an equation with variable on both sides. Solve: 8 y – 16 = 13 y + 9 6
Solve an equation using the distributive property. Solve: 4(2 x – 9) + 5 x = -3(10 – x) 7
Solve an equation with clearing a fraction. Solve: 8
Additional Vocabulary Identity: an equation that is satisfied by every number Conditional Equation: an equation that is satisfied by some numbers but not others Contradiction: an equation that has no solution and the solution set is an empty set or null set. {0}. 9
Identify the Type of Equation. Decide whether each equation is an identity, a conditional equation, or a contradiction. Give the solution set. a. -2(x + 4) + 3 x = x – 8 b. 5 x – 4 = 11 c. 3(3 x-1) = 9 x +7 10
Homework. Page 440 – Appendix A # 2 -22 evens 11
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# Minors of a Matrix – Explained with Examples
-
The minors of a matrix are quantities that are obtained by deleting some rows and columns from a matrix and finding the determinant of the resulting submatrix. The minor M_{ij} is the determinant of the submatrix formed by deleting the i^{th} row and the j^{th} column from the given matrix.
#### Example 1:
Consider the matrix A = \begin{bmatrix} 7 & -2 & 4\\ -8 & 6 & 3\\ 5 & 0 & -1 \end{bmatrix}.
We find all the minors of this matrix by calculating the determinants of the submatrices below.
M_{11} = \left|\begin{array}{cc}6 & 3\\0 & -1\end{array}\right| = (6)\cdot (-1) - (3)\cdot (0) = -6 M_{12} = \left|\begin{array}{cc}-8 & 3\\5 & -1\end{array}\right| = (-8)\cdot (-1) - (3)\cdot (5) = -7 M_{13} = \left|\begin{array}{cc}-8 & 6\\5 & 0\end{array}\right| = -30 M_{21} = \left|\begin{array}{cc}-2 & 4\\0 & -1\end{array}\right| = 2 M_{22} = \left|\begin{array}{cc}7 & 4\\5 & -1\end{array}\right| = -27 M_{23} = \left|\begin{array}{cc}7 & -2\\5 & 0\end{array}\right| = 10 M_{31} = \left|\begin{array}{cc}-2 & 4\\6 & 3\end{array}\right| = -30 M_{32} = \left|\begin{array}{cc}7 & 4\\-8 & 3\end{array}\right| = 53 M_{33} = \left|\begin{array}{cc}7 & -2\\-8 & 6\end{array}\right| = 26
#### Example 2:
We can similarly compute minors for a 4 \times 4 matrix. If we are given the matrix,
A = \begin{bmatrix} 4 & 9 & 5 & 3 \\ -2 & -3 & 8 & 6 \\ 0 & 0 & -4 & 8 \\ 8 & 1 & 0 & 2 \end{bmatrix}
we can similarly compute the minors M_{ij} by deleting the i^{th} row and the j^{th} column and computing the determinant of the resulting submatrix. For example, we can compute M_{23} as,
M_{23} = det \left|\begin{array}{ccc}4 & 9 & 3\\0 & 0 & 8\\8 & 1 & 2\end{array}\right| = 544
The other minors are computed in a similar fashion.
#### Principal Minors of a Matrix:
When we compute minors of a matrix we delete some rows and columns. We say that a minor is principal if whenever we delete the i^{th} row we also delete the i^{th} column and vice versa. The principal minors are denoted as M^{IJ} where I is the set of deleted row indices and J is the set of deleted column indices. Note that I=J for principal minors.
For example, consider the matrix A = \begin{bmatrix} 4 & 9 & 5 & 3 \\ -2 & -3 & 8 & 6 \\ 0 & 0 & -4 & 8 \\ 8 & 1 & 0 & 2 \end{bmatrix}.
Suppose we take I=J=\{2,4\} . This means that we delete the 2^{nd} and 4^{th} rows as well as the 2^{nd} and 4^{th} columns. The principal minor in this case is computed as,
M_{IJ} = \left|\begin{array}{cc}4 & 5\\0 & -4\end{array}\right| = (4)\cdot (-4) - (5)\cdot (0) = -16
##### Leading Principal Minors of a Matrix:
The minors corresponding to the upper left portion of a matrix are known as the leading principal minors of the matrix. Considering the same example as before,
A = \begin{bmatrix} 4 & 9 & 5 & 3 \\ -2 & -3 & 8 & 6 \\ 0 & 0 & -4 & 8 \\ 8 & 1 & 0 & 2 \end{bmatrix}
The leading principal minors are,
det \begin{bmatrix} 4 \end{bmatrix} = 4 det \begin{bmatrix} 4 & 9 \\ -2 & -3 \end{bmatrix} = -30 det \begin{bmatrix} 4 & 9 & 5 \\ -2 & -3 & 8 \\ 0 & 0 & -4 \end{bmatrix} = 120 det \begin{bmatrix} 4 & 9 & 5 & 3 \\ -2 & -3 & 8 & 6 \\ 0 & 0 & -4 & 8 \\ 8 & 1 & 0 & 2 \end {bmatrix} = -5112
If all the leading principal minors of a matrix are positive, the the matrix is a positive definite matrix.
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# Adding and Subtracting Decimals Study Guide
based on 6 ratings
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Updated on Oct 4, 2011
## Introduction to Adding and Subtracting Decimals
What would life be without arithmetic, but a scene of horrors?
— SYDNEY SMITH, English writer (1771–1845)
This second decimal lesson focuses on addition and subtraction of decimals. It ends by teaching you how to add or subtract decimals and fractions together.
You have to add and subtract decimals all the time, especially when dealing with money. This lesson shows you how and gives you some word problems to demonstrate how practical this skill is in real life as well as on tests.
There is a crucial difference between adding decimals and adding whole numbers; the difference is the decimal point. The position of this point determines the accuracy of your final answer; a problem solver cannot simply ignore the point and add it in wherever it "looks" best. In order to add decimals correctly, follow these three simple rules:
1. Line the numbers up in a column so their decimal points are aligned.
2. Tack zeros onto the ends of shorter decimals to keep the digits lined up evenly.
3. Move the decimal point directly down into the answer area and add as usual.
#### Tip
The number one pitfall in adding and subtracting decimals happens when the numbers are lined up, instead of the decimals. Before you start adding or subtracting with decimals, add zeros after the last digit to the right of the decimal point to all the numbers until they each have the same amount of digits to the right of the decimal point. For a whole number, just add a decimal point and then add zeros to the right of it. Example:
Example: 3.45 + 22.1 + 0.682
1 Line up the numbers so their decimal points are even: 2 Tack zeros onto the ends of the shorter decimals to fill in the "holes": 3 Move the decimal point directly down into the answer area and add:
Look at an example that adds decimals and whole numbers together. Remember: A whole number is understood to have a decimal point to its right.
Example: 0.6 + 35 + 0.0671 + 4.36
1 Put a decimal point at the right of the whole number (35) and line up the numbers so their decimal points are aligned: 2 Tack zeros onto the ends of the shorter decimals to fill in the "holes": 3 Move the decimal point directly down into the answer area and add:
## Subtracting Decimals
When subtracting decimals, follow the same initial steps as in adding to ensure that you're adding the correct digits and that the decimal point ends up in the right place.
Example: 4.8731 – 1.7
1 Line up the numbers so their decimal points are aligned: 2 Tack zeros onto the end of the shorter decimal to fill in the "holes": 3 Move the decimal point directly down into the answer and subtract:
Subtraction is easily checked by adding the number that was subtracted to the difference (the answer). If you get back the other number in the subtraction problem, then your answer is correct. For example, let's check our last subtraction problem.
Here's the subtraction: 1. Add the number that was subtracted (1.7000) to the difference (3.1731): 2. The subtraction is correct because we got back the other number in the subtraction problem (4.8731).
Checking your subtraction is so easy that you should never pass up the opportunity!
You can check the reasonableness of your work by estimating: Round each number to the nearest whole number and subtract. Rounding 4.873 and 1.7 gives 5 and 2. Since their difference of 3 is close to your actual answer, 3.1731 is reasonable.
#### Tip
When subtracting mixed decimals gets rid of any whole numbers, write a zero in the ones place. Example: 5.67 – 4.9 = 0.77, and should not be written as .77.
150 Characters allowed
### Related Questions
#### Q:
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## How to turn decimals to fractions
Rewriting decimals as fractions: 0.15
Next, you want to multiply both the top and bottom of your new fraction by 10 for every digit to the left of the decimal point. In our example, has two digits after the decimal point, so we’ll want to multiply the entire fraction by 10 x 10, or Multiplying the fraction by To convert a decimal into a fraction, you put the numbers to the right of the decimal point in the numerator (above the fraction line). Next, you place the number 1 in the denominator, and then add as many zeroes as the numerator has digits. Reduce the fraction if needed. Decimals are nothing more than glorified fractions.
General Education. Wondering how to convert decimals to fractions? Or how to convert fractions to decimals? How do you convert a decimal to a fraction? Any decimal, even complicated-looking ones, can be converted to a fraction; you just need to follow a few steps. Below we explain how to convert both furn decimals fractikns repeating decimals to fractions.
A terminating decimal is any decimal that has a finite other of digits. In other words, it tturn an end. Examples include. Next, you want to multiply both the top and bottom of your new fraction by 10 for every digit to the left of the decimal fractiohs.
In our example. The final step is reducing the fraction to its simplest form. The simplest form of the fraction is when the top and bottom of the fraction are the smallest whole numbers they can be. Now we need to simplify. Since and are both even numbers, we know we can divide both by 2. A repeating decimal is one that has no fractiins. The decimal. Multiply by whatever value of 10 you need to get the repeating digit s on the left side of how to transport a baby grand piano decimal.
We want that fractionx on the left side of the decimal, which means moving the decimal place over one spot. So we multiply what can i do with quinces sides of the equation by 10 x 1 or In this example, with 6 as the repeating digit, you only want one 6 on the left of the decimal.
If the decimal was 0. If it helps, you can picture all repeating decimals how to clean nubuck trainers the infinity bar over them, so. Next we want to get an equation where the repeating digit is just to the right of the decimal. First, make the decimal equal to x, and determine the repeating digit s. Next, get the repeating digits on the left side of the decimal again, you only want one set of repeating digits on the left.
This involves moving the decimal three places to the right, so both sides need to be multiplied by 10 x 3 or Now get the repeating digits to the right of the decimal. The decimal needs to be moved over one space, so both sides need to be multiplied by 10 x 1. Since the numerator is larger than the denominator, this is known as an irregular fraction. Sometimes you can leave the fraction as an irregular fraction, or you may be asked to convert it to a regular fraction.
The easiest way to convert a fraction to a decimal is just to use your calculator. We explain both these methods in this section. We want the denominator, in yurn case 8, to equal a value of Next we want to get the denominator to equal 1 so we can get rid of the fraction.
Note that this method only works for frwctions fraction with a denominator that can easily be multiplied to be a value of Check out the example below. There is no number you can multiply 3 what do you need for a car loan to make it an exact multiple of 10, but you can get close.
Below is a chart with common decimal to fraction conversions. For quick estimates of decimal to fraction conversions or vice versayou can look at our chart of common conversions and see which is closest to your figure to get a ballpark idea ddcimals its conversion value. Want to know the fastest and decimqls ways to convert between Fahrenheit and Celsius?
We've got you covered! Check out decimsls guide to the best ways to convert Celsius to Fahrenheit or vice versa. Are you learning about logarithms and natural logs in math class?
We have a guide on all the natural log rules you need to know. Did you deicmals that water has a very special density? Check out our go to learn what the density of water is and how the density can change. She has taught English and biology in several countries. Our new student and parent forum, at ExpertHub. See how other students and parents are navigating high school, college, and the college admissions process.
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What ACT target score should you be aiming for? How to Get a Perfect 4. How to Write an Amazing College Essay. A Comprehensive Guide. Choose Your Test. How to Convert Decimals to Fractions Decimas do you convert a decimal to hoa fraction?
Converting a Fractons Decimal to a Fraction A terminating decimal is any decimal that has a finite other of digits. Step 1 Write the decimal divided by one. Step 2 Next, you want to multiply both the top and bottom of your new fraction by 10 for every digit to the left of the decimal point. Step 3 The final step is reducing the decimalls to its simplest form. Example Convert. Converting a Repeating Decimal to a Fraction A repeating decimal is one that has no end.
Step 2 Multiply by whatever value of 10 you need to get the repeating digit s on the feactions side of the decimal. Step 3 Next we want to get an equation where the repeating digit is just to the right of the decimal.
Step 1 We want the denominator, in this case 8, to equal ffractions value of Step 2 Next we want to get the denominator to equal 1 so we can get rid of the fraction. Common Decimal to Fraction Conversions Below is a chart with common decimal to fraction conversions.
Decimal Fraction decinals. What's Next? Have friends who also need help with test prep? Share this article! Christine Sarikas. About the Author. Search the Blog Search. Find Out How. Get the latest articles and test prep tips! Looking for Graduate School Test Prep?
Jun 28, · You can convert a decimal to a fraction by following these three easy steps. In this case, you will use the decimal as an example (see the graphic below). Step One: Rewrite the decimal number over one (as a fraction where the decimal number is numerator and the denominator is one). Convert a Repeating Decimal to a Fraction Create an equation such that x equals the decimal number. Count the number of decimal places, y. Create a second equation multiplying both sides of . Convert Decimals to Fractions – Explanation & Examples Before we learn how to convert decimals to fractions, there are a number of basic information we need to know about decimals and fractions. To start with, a decimal number is probably a number which has a dot .) between the digits, this dot is known as a decimal kristinfrey.coms:
Before we learn how to convert decimals to fractions, there are a number of basic information we need to know about decimals and fractions. To start with, a decimal number is probably a number which has a dot. Basically, decimal numbers are just fractions having a denominator expressed in power of Example of decimal numbers are: 0. The two integers a and b are referred to as the numerator and denominator respectively. There are three types of fractions namely: Proper, Improper and Mixed fraction.
We can easily convert a decimal number into a fraction by following simple steps and no calculators are required. This article has elaborated clearly all the steps of converting decimals to fractions, with some examples. Let us solve the following examples so get a better understand of how to convert a decimal into fraction. Repeating or recurring numbers are decimal numbers with the endless repeating decimal digits. Either there can be a single digit repeating or two and more digits repeating by alternating.
Examples of repeating numbers are: 0. Search for:. How to Convert Decimal to Fraction? Let us learn these steps on to convert the decimal into fractions: First, begin by counting the numbers to the right side after the decimal point. Let n be the number of digits on the right side after the decimal point.
Write the number without a decimal point as a numerator and the power of 10 n as the denominator Now the fraction can be simplified by reducing the denominator and numerator with a common factor. The simplified fraction is the required fraction from the given decimal number.
Example 1 Convert 0. Solution The number 0. Take the number as a numerator by ignoring the decimal point. Take also the power of 10 1 as the denominator.
Example 2 Convert 0. Solution First count the number of decimal places. The number of decimal places in 5. You can notice that dividing the number is the same as writing it in fractional form. By just looking at the last digits both the numerator and denominator, the numbers are even. Both the denominator and numerator have common factors, therefore simplify the fraction to its lowest form.
How to Convert a Repeating Decimal to Fraction? To convert a repeating number into a fraction, see the following example. Example 5 Convert the repeating number 0.
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# Volume
## Number of unit cubes to fill prisms, spheres, pyramids, cylinders, cones, and composite solids.
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Volume of Solids
The composite solid below is made of a cube and a square pyramid. The length of each edge of the cube is 12 feet and the overall height of the solid is 22 feet. What is the volume of the solid? Why might you want to know the volume of the solid?
#### Guidance
The volume of a solid is the number of unit cubes it takes to fill up the solid.
A prism is a solid with two congruent polygon bases that are parallel and connected by rectangles. Prisms are named by their base shape.
To find the volume of a prism, find the area of its base and multiply by its height.
A cylinder is like a prism with a circular base.
To find the volume of a cylinder, find the area of its circular base and multiply by its height.
A cone also has a circular base, but its lateral surface meets at a point called the vertex.
To find the volume of a cone, find the volume of the cylinder with the same base and divide by three.
A pyramid is similar to a cone, except it has a base that is a polygon instead of a circle. Like prisms, pyramids are named by their base shape.
To find the volume of a pyramid, find the volume of the prism with the same base and divide by three.
A sphere is the set of all points in space equidistant from a center point. The distance from the center point to the sphere is called the radius.
The volume of a sphere relies on its radius.
A composite solid is a solid made up of common geometric solids.
The volume of a composite solid is the sum of the volumes of the individual solids that make up the composite.
Example A
Find the volume of the rectangular prism below.
Solution: To find the volume of the prism, you need to find the area of the base and multiply by the height. Note that for a rectangular prism, any face can be the “base”, not just the face that appears to be on the bottom.
Example B
Find the volume of the cone below.
Solution: To find the volume of the cone, you need to find the area of the circular base, multiply by the height, and divide by three.
Example C
Find the volume of a sphere with radius 4 cm.
Solution: The formula for the volume of a sphere is .
Concept Problem Revisited
The composite solid below is made of a cube and a square pyramid. The length of each edge of the cube is 12 feet and the overall height of the solid is 22 feet.
To find the volume of the solid, find the sum of the volumes of the prism (the cube) and the pyramid. Note that since the overall height is 22 feet and the height of the cube is 12 feet, the height of the pyramid must be 10 feet.
The volume helps you to know how much the solid will hold. One cubic foot holds about 7.48 gallons of liquid. This solid would hold 16,515.84 gallons of liquid!
#### Vocabulary
The volume of a solid is the number of unit cubes it takes to fill up the solid.
prism is a solid with two congruent polygon bases that are parallel and connected by rectangles. Prisms are named by their base shape.
cylinder is like a prism with a circular base.
cone has a circular base and its lateral surface meets at a point called the vertex.
pyramid is similar to a cone, except it has a base that is a polygon instead of a circle. Pyramids are named by their base shape.
sphere is the set of all points in space equidistant from a center point. The distance from the center point to the sphere is called the radius.
composite solid is a solid made up of common geometric solids.
#### Guided Practice
1. The area of the base of the pyramid below is . The height is 5 cm. What is the volume of the pyramid?
2. The volume of a sphere is . What is the radius of the sphere?
3. The volume of a square pyramid is . The height of the pyramid is three times the length of a side of the base. What is the height of the pyramid?
1.
2.
3. . Therefore, and .
#### Practice
Find the volume of each solid or composite solid.
1.
2.
3.
4.
5. The base is an equilateral triangle.
6.
7. Explain why the formula for the volume of a prism involves the area of the base.
8. How is a cylinder related to a prism?
9. How is a pyramid related to a cone?
10. How is a sphere related to a circle?
11. If one cubic centimeter will hold 1 milliliter of water, approximately how many liters of water will the solid in #1 hold? (One liter is 1000 milliliters).
12. If one cubic centimeter will hold 1 milliliter of water, approximately how many liters of water will the solid in #3 hold? (One liter is 1000 milliliters).
13. If 231 cubic inches will hold one gallon of water, approximately how many gallons of water will the solid in #5 hold?
14. The volume of a cone is . The height is three times the length of the radius. What is the height of the cone?
15. The volume of a pentagonal prism is . The height of the prism is 3 in. What is the area of the pentagon base?
### Vocabulary Language: English
Composite
Composite
A number that has more than two factors.
Cone
Cone
A cone is a solid three-dimensional figure with a circular base and one vertex.
Cylinder
Cylinder
A cylinder is a solid figure with two parallel congruent circular bases.
Prism
Prism
A prism is a three-dimensional object with two congruent parallel bases that are polygons.
Pyramid
Pyramid
A pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex.
Sphere
Sphere
A sphere is a round, three-dimensional solid. All points on the surface of a sphere are equidistant from the center of the sphere.
Volume
Volume
Volume is the amount of space inside the bounds of a three-dimensional object.
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# $\sin{(0^°)}$ Proof
According to trigonometry, the exact value of sine of zero degrees is equal to zero.
$\sin{(0^°)} \,=\, 0$
The sine of angle zero degrees can be derived in mathematics in two different geometric approaches. Now, it is your time to learn how to derive the sine of angle zero radian value mathematically.
### Fundamental method
The sine of angle zero degrees can be derived theoretically by considering a geometric property of a side in a zero degrees right triangle. Now, imagine a right triangle that contains zero degrees angle. The $\Delta RPQ$ is an example for a right triangle with zero angle.
According to the fundamental definition of the sin function, it can be expressed in ratio form of lengths of the opposite side and hypotenuse. In fact, the sine function is written as $\sin{(0^°)}$ when the angle of a right triangle is zero grades.
$\implies$ $\sin{(0^°)}$ $\,=\,$ $\dfrac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse}$
$\implies$ $\sin{(0^°)}$ $\,=\,$ $\dfrac{QR}{PQ}$
In a zero degree right triangle, the length of opposite side is zero. It means $QR \,=\, 0$.
$\implies$ $\sin{(0^°)}$ $\,=\,$ $\dfrac{0}{d}$
$\,\,\,\therefore\,\,\,\,\,\,$ $\sin{(0^°)}$ $\,=\,$ $0$
### Experimental method
The sin of angle zero degrees is also proved practically with geometric tools by constructing a right triangle with zero degrees angle. Now, let’s start the process for constructing a right angled triangle with zero degrees angle.
1. Draw a straight line horizontally from point $A$ in the plane.
2. Take protractor and draw a zero degrees line by coinciding the point $A$ with its middle point and also coinciding the right side base line with horizontal line. In this case, the zero degree angle line is drawn over the horizontal line.
3. Set the compass to any length by a ruler. In this example, The distance between lead point and pencil point to $5 \, cm$. Now, draw an arc on zero angle line from point $A$ and it intersects the point at point $B$.
4. Use set square and draw a perpendicular line to horizontal line from point $B$ and it intersects the horizontal line at point $C$. Thus, $\Delta BAC$ is constructed geometrically by the geometric tools.
The opposite side, adjacent side and hypotenuse are $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$ respectively in $\Delta BAC$.
It is time to find the value of $\sin{(0)}$ by calculating the ratio of lengths of opposite side to hypotenuse of the right triangle $BAC$.
$\sin{(0^°)} = \dfrac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse}$
$\implies \sin{(0^°)} \,=\, \dfrac{BC}{AB}$
In this example, $BC \,=\, 0\,cm$, $AC \,=\, 5\,cm$ and $BC \,=\, 5\,cm$.
$\implies \sin{(0^°)} \,=\, \dfrac{BC}{AB}$
$\implies \sin{(0^°)} \,=\, \dfrac{0}{5}$
$\,\,\,\therefore \,\,\,\,\,\, \sin{(0^°)} \,=\, 0$
In this way, the sine of angle zero degrees equals to zero is proved in two different geometric methods.
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What is the probability of winning bingo
The chance of winning bingo varies, but generally, with 75 numbers and one card per player, it’s about 1 in 25 or a 4% chance per game.
Understanding Bingo Probabilities
Understanding the probabilities of winning in bingo requires a grasp of the basic concepts of probability and a look at the specific mathematical model that applies to bingo. This knowledge then informs the strategies players use to increase their odds of winning.
Probability Basics
Probability forms the foundation of any game of chance, and bingo is no exception. In bingo, the probability of any given number being called is equal to one divided by the number of balls remaining in the drum. As the game progresses, these odds change with each number called. For instance, if a bingo game starts with 75 numbers, the probability of the first number being called is 1 in 75. After that, if the number has not been called, it becomes 1 in 74, and so on.
Mathematical Model of Bingo
The mathematical model of bingo considers the total number of cards in play and the patterns needed to win. The chance of winning a game of bingo can be modeled using a hypergeometric distribution, which calculates the probability of a certain number of successes in a series of draws from a finite population without replacement. For example, if a player needs to cover five numbers in a row, the model will calculate the probability based on the number of ways those numbers can appear on the card.
Factors Affecting Winning Chances
Several factors affect a player’s chance of winning at bingo. The number of cards in play is a direct factor: more cards in play diminish an individual’s chance of winning, as competition is higher. Additionally, the number of players in a game, the variety of possible winning patterns, and the distribution of numbers called all play a role. A game with 100 players each holding a single card will have different odds from a game with 10 players each holding ten cards.
Strategies for Increasing Winning Odds in Bingo
To increase the odds of winning, players often employ various strategies. One such strategy is card selection. Players may choose cards that do not duplicate the numbers they already have, increasing the spread of numbers they play and potentially their chances of hitting a winning pattern. Understanding game patterns is also crucial. For instance, certain bingo games may offer a higher probability of winning with a specific pattern, which savvy players will target. However, players must also be aware of common myths, such as the gambler’s fallacy, which incorrectly assumes that past events can influence future outcomes in a purely random game.
Linking these concepts to real-world scenarios, the speed of the game and the quality of the materials used for bingo cards can also impact a player’s experience. Speed affects the time available to mark cards accurately, while the durability of cards can influence their longevity and, by extension, the cost-effectiveness for frequent players. The price of bingo cards varies, with some games offering cards for a few cents and others charging more, which can affect a player’s budget and overall expenses dedicated to the game.
Statistical Analysis of Bingo Games
In-depth statistical analysis is critical for understanding the dynamics and outcomes of bingo games. This analysis extends from data collection methods to the interpretation of results and finally, to applying findings in real-world scenarios or case studies.
Data Collection and Analysis
Effective data collection in bingo requires tracking the numbers called during a game and the patterns completed on each card. Analysts can record these details over hundreds or thousands of games to discern patterns and frequencies. For instance, cost analysis of data collection reveals that employing automated number tracking software increases efficiency, reducing human error and labor costs. The speed of data processing with such software can process a bingo game’s data in seconds, where manual tracking would take minutes.
Interpretation of Bingo Game Data
Interpreting this data involves examining the distribution of winning numbers and identifying any anomalies or expected probabilities. For instance, the time to completion of a bingo game may average at 10 minutes but may vary with the number of players and the complexity of the winning pattern. This variability can affect the duration of bingo sessions and, consequently, the venue’s operational costs and revenue.
Case Studies
Case studies in bingo statistical analysis can provide valuable insights. For example, a case study on the lifespan of a bingo card might reveal that high-quality paper or plastic cards last for an average of 5 years with weekly use, while low-quality cards may need replacement every few months. This impacts the material costs and the environmental footprint of bingo operations. Another case study could examine the value of progressive jackpot bingo games, which can accumulate prizes worth thousands of dollars over time, influencing player turnout and the financial strategies of bingo halls.
Advanced tools such as computer simulation and comparative studies with other lottery games underscore the depth of statistical analysis in enhancing the understanding of bingo probabilities. Analysis of random number generation (RNG) technology is crucial to ensuring the integrity and fairness of the game. Studies show that true RNGs must pass rigorous tests for unpredictability and uniform distribution, criteria essential for fair play and adherence to gaming laws.
Through careful analysis, stakeholders can determine the optimal balance between game attractiveness, profitability, and legal compliance. By integrating statistical data with strategic planning, bingo halls can set prices that ensure both competitiveness in the market and financial sustainability. For players, this data translates into a better grasp of the game’s value proposition, allowing them to make more informed decisions about participation and expenditure.
As with any gambling-related entertainment, the speed of play and the quality of the experience are pivotal. High-speed bingo games may increase the thrill but also raise the risk of rapid loss, affecting the perception of value and satisfaction. Conversely, games with a slower pace allow for a more social experience, potentially increasing player retention and long-term success for bingo establishments.
Exploring advanced topics in bingo probabilities takes us into the realm of computer simulations and comparative studies, where the intricacies of bingo become a field for academic inquiry and sophisticated gaming strategy development.
Computer Simulation of Bingo Games
Computer simulation allows for the modeling of millions of bingo games to forecast outcomes and test theories about number distribution and game strategy. For example, simulations may run scenarios with a fixed budget to determine the expected lifetime value of a bingo card purchase, which could be around \$50 for an average player attending weekly sessions over a year. This simulated data helps in understanding how changes in game variables, such as card price—typically ranging from \$1 to \$5 per card—can affect play frequency and overall spending.
Comparative Study of Bingo and Other Lottery Games
Comparative studies provide insight into bingo’s odds in relation to other forms of lottery games. For instance, while the odds of winning a national lottery game might be one in several million, bingo players might face odds of one in several hundred or thousand, depending on the number of cards in play and the player count. Such comparisons emphasize bingo’s relative appeal and can affect a player’s decision to choose one game over another, based on the balance of cost versus potential reward, where bingo might offer more frequent but smaller wins.
Research on Random Number Generation
Research on random number generation delves into the algorithms that ensure each bingo number call is unpredictable and fair. Rigorous testing involves examining the sequence of numbers for patterns that could compromise randomness. These tests, which can evaluate millions of number sequences, are critical in certifying the quality and integrity of bingo games. RNG research also explores the speed of number calling, which can be adjusted to maintain game tempo, affecting how quickly players reach bingo and thus the pace of turnover in a bingo hall.
Bingo Probability and Gaming Laws
Finally, a thorough grasp of bingo probabilities intersects with gaming laws and regulations, which require that games are conducted fairly and openly. Regulatory compliance might stipulate that the chance of winning should not fall below a certain threshold to prevent exploitation. For instance, regulations may mandate that bingo games provide at least a 75% return-to-player rate, ensuring that players receive a fair portion of the collective stakes over time. Legal cases, such as those challenging the fairness of a game or the disclosure of odds, also rely on statistical evidence to underpin arguments, with outcomes potentially leading to changes in industry practices.
Bingo Probability and Gaming Laws
The intersection of bingo probability with gaming laws presents a fascinating study of how legal frameworks are informed by and respond to the mathematical underpinnings of gambling games.
Regulation of Bingo Games
Regulations ensure that bingo games operate within established fairness guidelines. For example, game operators must demonstrate compliance with average payout ratios, typically set at a minimum of 75% to 80%. This means that for every dollar spent by all players, at least 75 cents are returned as winnings. Compliance with these regulations is not just a legal formality; it significantly influences the cost structure of bingo operations, including licensing fees, which can range from a few hundred to several thousand dollars annually, depending on the jurisdiction.
Fair Play and Probability Checks
Fair play is paramount in gaming, and bingo halls must regularly verify their number-drawing mechanisms to ensure compliance. Probability checks are integral to this process, requiring sophisticated statistical analysis. For instance, a bingo hall may conduct monthly audits, costing around \$500 to \$2,000 per session, to certify the randomness of their number draws. These costs reflect the value placed on maintaining a reputable gaming environment, where the speed of number calling is calibrated to align with the expected duration of a game, ensuring player engagement and turnover.
Legal Cases and Judgements
Legal cases involving bingo often revolve around disputes of fairness or false advertising claims. In such cases, the age of participants, frequency of games, and even the specific dimensions of bingo cards can become focal points of litigation. For example, judgments in such cases may result in compensatory damages or require changes to the game’s operational parameters, with potential financial implications for the operators. The cost of litigation for bingo-related cases can also be substantial, sometimes reaching tens of thousands of dollars, including legal fees and potential settlements.
Through careful consideration of these advanced topics, bingo hall operators and players alike gain an appreciation for the complexity and necessity of regulations governing the game. Not only do these laws protect the interests of players, but they also serve to uphold the integrity of the game itself, ensuring that bingo remains a fair and enjoyable form of entertainment for all participants. The balance between stringent regulation and operational efficiency remains a dynamic challenge for the industry, reflecting the broader implications of probability theory in real-world applications.
How does the number of players affect bingo odds?
With more players, your odds decrease. For example, if there are 100 players, each with one card, you have a 1% chance of winning.
What impact does the number of cards have on my winning probability?
Playing more cards improves your odds. Holding 4 out of 100 cards in play gives you a 4% chance, as opposed to 1% with one card.
Does the game type change the probability of winning?
Yes, different bingo games have various winning patterns, which can change your chances. A game with fewer combinations will have higher winning odds.
What’s the typical cost of a bingo card?
Prices vary, but they usually range from \$1 to \$5 per card, influencing the overall expense of playing.
How often does a player typically win at bingo?
The average win rate is estimated at a win fraction of 0.10, or about one win per ten games played.
What is the role of RNG in bingo?
RNGs ensure each number is equally likely to be drawn, crucial for fair play, and are tested for unpredictability.
Can strategy increase my chances of winning at bingo?
While largely luck-based, selecting non-duplicate numbers and playing multiple cards can be strategic.
What are the odds of winning a progressive jackpot in bingo?
1. It's much lower than regular games, as jackpots accumulate over time, making the chance of winning dependent on the number of games played and the players involved.
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### CFA Practice Question
Consider the following annual growth forecasts for a common stock:
Growth in years 1-2 = 50%
Growth in years 3-4 = 25%
Growth after year 4 = 10%
Assuming that the last dividend was \$0.45 per share, and the required rate of return is 20% per year, what is the value of this common stock?
A. \$13.81
B. \$11.15
C. \$8.36
Explanation: To determine the value of a common stock experiencing temporary supernormal growth, use the following equation:
{V = {[d0 * (1 + gs)^1] / k} + {[d1 * (1 + gs)^2} + ... {dn * (1 + gs)^n} + {[dn * (1 + gs)^n * (1 + gn] / (k - g)}/ (1 + k)^n}}
Where: V = the value of common stock at t0, d0 = the dividend at t0, d1 = the dividend at t1, dn = the dividend at tn, gs = the supernormal rate of growth, gn = the normal rate of growth, n = the time period "n", and k = the required rate of return.
In this example, there is a transitional growth period of two years, during which the growth rate is expected to grow at 25% annually. This period will follow the two-year supernormal growth period, and would be denoted as "g subset t" if we were to reproduce the equation illustrated above. The calculation of the value of this common stock is illustrated as follows:
{V = {[\$0.45 * (1.50)^1] / (1.20)} + {[\$0.45 * (1.50)^2] / (1.20)^2} + {[\$0.45 * (1.50)^2 * (1.25)^1] / (1.20)^3} + {[\$0.45 * (1.50)^2 * (1.25)^2] / (1.20)^4} + {{[\$0.45 * (1.50)^2 * (1.25)^2 * (1.10)^1]/ (0.20 - 0.10)}/ (1.20)^4}
which can be deduced to the following: {V = [\$0.5625 + \$0.703125 + \$0.732422 + \$0.762939 + \$8.392334] = \$11.15}
User Comment
smiley25 1.5 mins good luck.
ontrack i had that-good luck. i guessed correctly!!!
AkashKB The force is strong with ontrack
takor ...and may that strong force be with us all on the exam day!!!
Mlgraber Thank you
mpapwa22 I know exctly wht to do with this question but i took about 3 minutes. Its rough. what to do......go the ontrack way...
shiva5555 Its not that hard. Just do the main stock price. It comes out to like 8.50 from there its obvious that B is the right answer.
fenix55555 Someone know how to compute this with BAII?
MaresaJaden Calculate the cash flows and plug them into the calculator...
birdperson as MaresaJaden said, using the CF button is the best way to solve these in my opinion.
tzanchan That is why you need to budget your time effectively. There are some questions that can be answered in 30 seconds.
Nadjones How are the dividends growing...
CF1= (.45)*(1.5) or .675
CF2= (.675)*(1.5) or 1.0125
CF3= (1..125)*(1.2) or 1.2656
CF4= (1.2656)*(1.2) or 1.582
with a required rate of 20, your NPV = 2.76
From there the last divided grows @ 10% (or 1.582*1.1 = 1.74) and then you a constant constant growth formula. [{1.74/(.2-.1)} divided by (1.2^4)] = 8.39
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