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# Section5.2Finding eigenvectors numerically¶ permalink
When presented with a square matrix $A\text{,}$ we have typically found its eigenvalues as the roots of the characteristic polynomial $\det(A-\lambda I) = 0$ and the associated eigenvectors as the null space $\nul(A-\lambda I)\text{.}$ Unfortunately, this approach is not practical when we are working with large matrices. First, finding the charactertic polynomial of a large matrix requires considerable computation, as does finding the roots of that polynomial. Second, finding the null space of a singular matrix is plagued by numerical problems, as we will see in the preview activity.
In this section, we will explore a technique called the power method that finds numerical approximations to the eigenvalues and eigenvectors of a square matrix. Generally speaking, this method is how eigenvectors are found in practical computing applications.
##### Preview Activity5.2.1
Let's recall some earlier observations about eigenvalues and eigenvectors.
1. How are the eigenvalues and associated eigenvectors of $A$ related to those of $A^{-1}\text{?}$
2. How are the eigenvalues and associated eigenvectors of $A$ related to those of $A-3I\text{?}$
3. If $\lambda$ is an eigenvalue of $A\text{,}$ what can we say about the pivot positions of $A-\lambda I\text{?}$
4. Suppose that $A = \left[\begin{array}{rr} 0.8 \amp 0.4 \\ 0.2 \amp 0.6 \\ \end{array}\right] \text{.}$ Explain how we know that $1$ is an eigenvalue of $A$ and then explain why the following Sage computation is incorrect.
5. Suppose that $\xvec_0 = \twovec{1}{0}\text{,}$ and we define a sequence $\xvec_{k+1} = A\xvec_k\text{;}$ in other words, $\xvec_{k} = A^k\xvec_0\text{.}$ What happens to $\xvec_k$ as $k$ grows increasingly large?
6. Explain how the eigenvalues of $A$ are responsible for the behavior noted in the previous question.
# Subsection5.2.1The power method
Our goal is to find a technique that produces numerical approximations to the eigenvalues and associated eigenvectors of a matrix $A\text{.}$ We begin by searching for the eigenvalue having the largest absolute value. We call this the dominant eigenvalue.
Let's begin with the positive stochastic matrix $A=\left[\begin{array}{rr} 0.8 \amp 0.4 \\ 0.2 \amp 0.6 \\ \end{array}\right] \text{.}$ We spent quite a bit of time studying this type of matrix in Section 4.5; in particular, we saw that any Markov chain will converge to the unique steady state vector. Let's rephrase this statement in terms of the eigenvectors of $A\text{.}$
In this case, we have eigenvalues $\lambda_1 = 1$ and $\lambda_2 =0.4\text{,}$ and associated eigenvectors $\vvec_1 = \twovec{2}{1}$ and $\vvec_2 = \twovec{-1}{1}\text{.}$
Suppose we begin with the vector
\begin{equation*} \xvec_0 = \twovec{1}{0} = \frac13 \vvec_1 - \frac13 \vvec_2 \end{equation*}
and find
\begin{equation*} \begin{aligned} \xvec_1 \amp {}={} A\xvec_0 = \frac13 \vvec_1 - \frac13(0.4) \vvec_2 \\ \xvec_2 \amp {}={} A^2\xvec_0 = \frac13 \vvec_1 - \frac13(0.4)^2 \vvec_2 \\ \xvec_3 \amp {}={} A^3\xvec_0 = \frac13 \vvec_1 - \frac13(0.4)^3 \vvec_2 \\ \amp \vdots \\ \xvec_k \amp {}={} A^k\xvec_0 = \frac13 \vvec_1 - \frac13(0.4)^k \vvec_2 \\ \end{aligned} \end{equation*}
and so forth. The point is that the powers $0.4^k$ become increasingly small as $k$ grows so that $\xvec_k\approx \frac13\vvec_1$ when $k$ is large. As $k$ grows large, the contribution from the eigenvector $\vvec_2$ to the vectors $\xvec_k$ becomes increasingly insignificant. Therefore, the vectors $\xvec_k$ become increasingly close to a vector in the eigenspace $E_1\text{.}$ If we did not know the eigenvector $\vvec_1\text{,}$ we could find a basis vector for $E_1$ in this way.
Let's now look at the matrix $A = \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 2 \\ \end{array}\right] \text{,}$ which has eigenvalues $\lambda_1=3$ and $\lambda_2 = 1$ and associated eigenvectors $\vvec_1 = \twovec{1}{1}$ and $\vvec_{2} = \twovec{-1}{1}\text{.}$ Once again, begin with the vector $\xvec_0 = \twovec{1}{0}=\frac12 \vvec_1 - \frac12 \vvec_2$ so that
\begin{equation*} \begin{aligned} \xvec_1 \amp {}={} A\xvec_0 = 3\frac12 \vvec_1 - \frac12 \vvec_2 \\ \xvec_2 \amp {}={} A^2\xvec_0 = 3^2\frac13 \vvec_1 - \frac12 \vvec_2 \\ \xvec_3 \amp {}={} A^3\xvec_0 = 3^3\frac13 \vvec_1 - \frac12 \vvec_2 \\ \amp \vdots \\ \xvec_k \amp {}={} A^k\xvec_0 = 3^k\frac13 \vvec_1 - \frac12 \vvec_2\text{.} \\ \end{aligned} \end{equation*}
As the figure shows, the vectors $\xvec_k$ are stretched by a factor of $3$ in the $\vvec_1$ direction and not at all in the $\vvec_2$ direction. Consequently, the vectors $\xvec_k$ become increasingly long, but their direction gets closer to the direction of the eigenvector $\vvec_1=\twovec{1}{1}$ associated to the dominant eigenvalue.
To find an eigenvector associated to the dominant eigenvalue, we will prevent the length of the vectors $\xvec_k$ from growing arbitrarily large by multiplying by an appropriate normalizing constant. There are several ways to do this; we will describe one simple way. Given the vector $\xvec_k\text{,}$ we identify its component having the largest absolute value and call it $m_k\text{.}$ We then define $\overline{\xvec}_k = \frac{1}{m_k} \xvec_k\text{,}$ which means that the component of $\overline{\xvec}_k$ having the largest absolute value is $1\text{.}$
For example, beginning with $\xvec_0 = \twovec{1}{0}\text{,}$ we find $\xvec_1 = A\xvec_{0} = \twovec{2}{1}\text{.}$ The component of $\xvec_1$ having the largest absolute value is $m_1=2$ so we multiply by $\frac{1}{m_1} = \frac12$ to obtain $\overline{\xvec}_1 = \twovec{1}{\frac12}\text{.}$ Then $\xvec_2 = A\overline{\xvec}_1 = \twovec{\frac52}{2}\text{.}$ Now the component having the largest absolute value is $m_2=\frac52$ so we multiply by $\frac25$ to obtain $\overline{\xvec}_2 = \twovec{1}{\frac45}\text{.}$
The resulting sequence of vectors $\overline{\xvec}_k$ is shown in the figure. Notice how the vectors $\overline{\xvec}_k$ now approach the eigenvector $\vvec_1\text{.}$ In this way, we find the eigenvector $\vvec=\twovec{1}{1}$ of the matrix $A\text{.}$ This is the power method for finding an eigenvector associated to the dominant eigenvalue of a matrix.
##### Activity5.2.2
Let's begin by considering the matrix $A = \left[\begin{array}{rr} 0.5 \amp 0.2 \\ 0.4 \amp 0.7 \\ \end{array}\right]$ and the initial vector $\xvec_0 = \twovec{1}{0}\text{.}$
1. Compute the vector $\xvec_1 = A\xvec_0\text{.}$
2. Find $m_1\text{,}$ the component of $\xvec_1$ that has the largest absolute value. Then form $\overline{\xvec}_1 = \frac 1{m_1} \xvec_1\text{.}$ Notice that the component having the largest absolute value of $\overline{\xvec}_1$ is $1\text{.}$
3. Find the vector $\xvec_2 = A\overline{\xvec}_1\text{.}$ Identify the component $m_2$ of $\xvec_2$ having the largest absolute value. Then form $\overline{\xvec}_2 = \frac1{m_2}\overline{\xvec}_1$ to obtain a vector in which the component with the largest absolute value is $1\text{.}$
4. The Sage cell below defines a function that implements the power method. Define the matrix $A$ and initial vector $\xvec_0$ below. The command power(A, x0, N) will print out the multiplier $m$ and the vectors $\overline{\xvec}_k$ for $N$ steps of the power method.
How does this computation identify an eigenvector of the matrix $A\text{?}$
5. What is the corresponding eigenvalue of this eigenvector?
6. How do the values of the multipliers $m_k$ tell us the eigenvalue associated to the eigenvector we have found?
7. Consider now the matrix $A=\left[\begin{array}{rr} -5.1 \amp 5.7 \\ -3.8 \amp 4.4 \\ \end{array}\right] \text{.}$ Use the power method to find the dominant eigenvalue of $A$ and an associated eigenvector.
Notice that the power method gives us not only an eigenvector $\vvec$ but also its associated eigenvalue. As in the activity, consider the matrix $A=\left[\begin{array}{rr} -5.1 \amp 5.7 \\ -3.8 \amp 4.4 \\ \end{array}\right] \text{,}$ which has eigenvector $\vvec=\twovec{3}{2}\text{.}$ The first component has the largest absolute value so we multiply by $\frac13$ to obtain $\overline{\vvec}=\twovec{1}{\frac23}\text{.}$ When we multiply by $A\text{,}$ we have $A\overline{\vvec} = \twovec{-1.30}{-0.86}\text{.}$ Notice that the first component still has the largest absolute value so that the multiplier $m=-1.3$ is the eigenvalue $\lambda$ corresponding to the eigenvector. This demonstrates the fact that the multipliers $m_k$ approach the eigenvalue $\lambda$ having the largest absolute value.
Notice that the power method requires us to choose an initial vector $\xvec_0\text{.}$ For most choices, this method will find the eigenvalue having the largest absolute value. However, an unfortunate choice of $\xvec_0$ may not. For instance, if we had chosen $\xvec_0 = \vvec_2$ in our example above, the vectors in the sequence $\xvec_k = A^k\xvec_0=\lambda_2^k\vvec_2$ will not detect the eigenvector $\vvec_1\text{.}$ However, it usually happens that our initial guess $\xvec_0$ has some contribution from $\vvec_1$ that enables us to find it.
The power method, as presented here, will fail for certain unlucky matrices. This is examined in Exercise 5.2.4.5 along with a means to improve the power method to work for all matrices.
# Subsection5.2.2Finding other eigenvalues
The power method gives a technique for finding the dominant eigenvalue of a matrix. We can modify the method to find the other eigenvalues as well.
##### Activity5.2.3
The key to finding the eigenvalue of $A$ having the smallest absolute value is to note that the eigenvectors of $A$ are the same as those of $A^{-1}\text{.}$
1. If $\vvec$ is an eigenvector of $A$ with associated eigenvector $\lambda\text{,}$ explain why $\vvec$ is an eigenvector of $A^{-1}$ with associated eigenvalue $\lambda^{-1}\text{.}$
2. Explain why the eigenvalue of $A$ having the smallest absolute value is the reciprocal of the dominant eigenvalue of $A^{-1}\text{.}$
3. Explain how to use the power method applied to $A^{-1}$ to find the eigenvalue of $A$ having the smallest absolute value.
4. If we apply the power method to $A^{-1}\text{,}$ we begin with an intial vector $\xvec_0$ and generate the sequence $\xvec_{k+1} = A^{-1}\xvec_k\text{.}$ It is not computationally efficient to compute $A^{-1}\text{,}$ however, so instead we solve the equation $A\xvec_{k+1} = \xvec_k\text{.}$ Explain why an $LU$ factorization of $A$ is useful for implementing the power method applied to $A^{-1}\text{.}$
5. The following Sage cell defines a command called inverse_power that applies the power method to $A^{-1}\text{.}$ That is, inverse_power(A, x0, N) prints the vectors $\xvec_k\text{,}$ where $\xvec_{k+1} = A^{-1}\xvec_k\text{,}$ and multipliers $\frac{1}{m_k}\text{,}$ which approximate the eigenvalue of $A\text{.}$ Use it to find the eigenvalue of $A=\left[\begin{array}{rr} -5.1 \amp 5.7 \\ -3.8 \amp 4.4 \\ \end{array}\right]$ having the smallest absolute value.
6. The inverse power method only works if $A$ is invertible. If $A$ is not invertible, what is its eigenvalue having the smallest absolute value?
7. Use the power method and the inverse power method to find the eigenvalues and associated eigenvectors of the matrix $A = \left[\begin{array}{rr} -0.23 \amp -2.33 \\ -1.16 \amp 1.08 \\ \end{array}\right] \text{.}$
With the power method and the inverse power method, we can now find the eigenvalues of a matrix $A$ having the largest and smallest absolute values. With one more modification, we can find all the eigenvalues of $A\text{.}$
##### 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 Figure 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.
There are clearly restrictions on the matrices to which this technique applies. We have been making the mild assumption that the eigenvalues of $A$ are real and distinct. If $A$ has repeated or complex eigenvalues, some other technique will need to be used.
# Subsection5.2.3Summary
We have explored the power method as a tool for numerically approximating the eigenvalues and eigenvectors of a matrix.
• After choosing an initial vector $\xvec_0\text{,}$ we define the sequence $\xvec_{k+1}=A\xvec_k\text{.}$ As $k$ becomes increasingly large, the direction of the vectors $\xvec_k$ increasingly closely approximates the direction of the eigenspace corresponding to the eigenvalue $\lambda_1$ having the largest absolute value.
• We normalize the vectors $\xvec_k$ by multiplying by $\frac{1}{m_k}\text{,}$ where $m_k$ is the component having the largest absolute value. In this way, the vectors $\overline{\xvec}_k$ approach an eigenvector associated to $\lambda_1\text{.}$ The multipliers $m_k$ approach the eigenvalue $\lambda_1\text{.}$
• To find the eigenvalue having the smallest absolute value, we apply the power method using the matrix $A^{-1}\text{.}$
• To find the eigenvalue closest to some number $s\text{,}$ we apply the power method using the matrix $(A-sI)^{-1}\text{.}$
# Subsection5.2.4Exercises
This Sage cell has the commands power, inverse_power, and find_closest_eigenvalue that we have developed in this section. After evaluating this cell, these commands will be available in any other cell on this page.
##### 1
Suppose that $A$ is a matrix having eigenvalues $-3\text{,}$ $-0.2\text{,}$ $1\text{,}$ and $4\text{.}$
1. What are the eigenvalues of $A^{-1}\text{?}$
2. What are the eigenvalues of $A+7I\text{?}$
##### 2
Use the commands power, inverse_power, and find_closest_eigenvalue to approximate the eigenvalues and associated eigenvectors of the following matrices.
1. $A= \left[\begin{array}{rr} -2 \amp -2 \\ -8 \amp -2 \\ \end{array}\right] \text{.}$
2. $A= \left[\begin{array}{rr} 0.6 \amp 0.7 \\ 0.5 \amp 0.2 \\ \end{array}\right] \text{.}$
3. $A= \left[\begin{array}{rrrr} 1.9 \amp -16.0 \amp -13.0 \amp 27.0 \\ -2.4 \amp 20.3 \amp 4.6 \amp -17.7 \\ -0.51 \amp -11.7 \amp -1.4 \amp 13.1 \\ -2.1 \amp 15.3 \amp 6.9 \amp -20.5 \\ \end{array}\right] \text{.}$
##### 3
Use the techniques we have seen in this section to find the eigenvalues of the matrix
\begin{equation*} A= \left[\begin{array}{rrrrr} -14.6 \amp 9.0 \amp -14.1 \amp 5.8 \amp 13.0 \\ 27.8 \amp -4.2 \amp 16.0 \amp 0.9 \amp -21.3 \\ -5.5 \amp 3.4 \amp 3.4 \amp 3.3 \amp 1.1 \\ -25.4 \amp 11.3 \amp -15.4 \amp 4.7 \amp 20.3 \\ -33.7 \amp 14.8 \amp -22.5 \amp 9.7 \amp 26.6 \\ \end{array}\right] \text{.} \end{equation*}
##### 4
Consider the matrix $A = \left[\begin{array}{rr} 0 \amp -1 \\ -4 \amp 0 \\ \end{array}\right] \text{.}$
1. Describe what happens if we apply the power method and the inverse power method using the initial vector $\xvec_0 = \twovec{1}{0}\text{.}$
2. Find the eigenvalues of this matrix and explain this observed behavior.
3. How can we apply the techniques of this section to find the eigenvalues of $A\text{?}$
##### 5
We have seen that the matrix $A = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right]$ has eigenvalues $\lambda_1 = 3$ and $\lambda_2=-1$ and associated eigenvectors $\vvec_1 = \twovec{1}{1}$ and $\vvec_2=\twovec{-1}{1}\text{.}$
1. Describe what happens when we apply the inverse power method using the initial vector $\xvec_0 = \twovec{1}{0}\text{.}$
2. Explain why this is happening and provide a contrast with how the power method usually works.
3. How can we modify the power method to give the dominant eigenvalue in this case?
##### 6
Suppose that $A$ is a $2\times2$ matrix with eigenvalues $4$ and $-3$ and that $B$ is a $2\times2$ matrix with eigenvalues $4$ and $1\text{.}$ If we apply the power method to find the dominant eigenvalue of these matrices to the same degree of accuracy, which matrix will require more steps in the algorithm? Explain your response.
##### 7
Suppose that we apply the power method to the matrix $A$ with an initial vector $\xvec_0$ and find the eigenvalue $\lambda=3$ and eigenvector $\vvec\text{.}$ Suppose that we then apply the power method again with a different initial vector and find the same eigenvalue $\lambda=3$ but a different eigenvector $\wvec\text{.}$ What can we conclude about the matrix $A$ in this case?
##### 8
The power method we have developed only works if the matrix has real eigenvalues. Suppose that $A$ is a $2\times2$ matrix that has a complex eigenvalue $\lambda = 2+3i\text{.}$ What would happen if we apply the power method to $A\text{?}$
##### 9
Consider the matrix $A=\left[\begin{array}{rr} 1 \amp 1 \\ 0 \amp 1 \\ \end{array}\right] \text{.}$
1. Find the eigenvalues and associated eigenvectors of $A\text{.}$
2. Make a prediction about what happens if we apply the power method and the inverse power method to find eigenvalues of $A\text{.}$
3. Verify your prediction using Sage. |
# FAQ: What Are The Kinds Of Sets In Math?
## How many types of sets are there in mathematics?
Answer: There are various kinds of sets like – finite and infinite sets, equal and equivalent sets, a null set. Further, there are a subset and proper subset, power set, universal set in addition to the disjoint sets with the help of examples. Question 4: What are the properties of sets?
## What is set and its types?
The different types of sets are explained below with examples. Empty Set or Null Set: A set which does not contain any element is called an empty set, or the null set or the void set and it is denoted by ∅ and is read as phi. For example: (a) The set of whole numbers less than 0.
## What are the sets in math?
A set in mathematics is a collection of well defined and distinct objects, considered as an object in its own right. Sets are one of the most fundamental concepts in mathematics.
## What is set 7 math?
Sets. A set is a collection of unique objects i.e. no two objects can be the same. Objects that belong in a set are called members or elements. Elements of set can be anything you desire – numbers, animals, sport teams. Representing Sets.
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## What are the symbols of sets?
Symbol Meaning Example
{ } Set: a collection of elements {1, 2, 3, 4}
A ∪ B Union: in A or B (or both) C ∪ D = {1, 2, 3, 4, 5}
A ∩ B Intersection: in both A and B C ∩ D = {3, 4}
A ⊆ B Subset: every element of A is in B. {3, 4, 5} ⊆ D
## What is AUB in math?
The union of the sets A and B, denoted by A U B, is the set that contains those elements that are either in A or in B, or in both. The intersection of the sets A and B, denoted by A n B, is the set containing those elements in both A and B. A n B = 1x | x ∈ A < x ∈ Bl.
## What are the 2 types of set?
Types of a Set
• Finite Set. A set which contains a definite number of elements is called a finite set.
• Infinite Set. A set which contains infinite number of elements is called an infinite set.
• Subset.
• Proper Subset.
• Universal Set.
• Empty Set or Null Set.
• Singleton Set or Unit Set.
• Equal Set.
## What is called set?
Georg Cantor, one of the founders of set theory, gave the following definition at the beginning of his Beiträge zur Begründung der transfiniten Mengenlehre: A set is a gathering together into a whole of definite, distinct objects of our perception or our thought—which are called elements of the set.
## What is basic set?
Sets are well-determined collections that are completely characterized by their elements. Thus, two sets are equal if and only if they have exactly the same elements. The basic relation in set theory is that of elementhood, or membership.
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## How do you name sets in math?
The following conventions are used with sets:
1. Capital letters are used to denote sets.
2. Lowercase letters are used to denote elements of sets.
3. Curly braces { } denote a list of elements in a set.
## What is proper set example?
A proper subset of a set A is a subset of A that is not equal to A. In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least one element that is not in B. For example, if A={1,3,5} then B={1,5} is a proper subset of A.
## How do you write a set?
Notation: A set is usually denoted by capital letters, i.e. A,B,C,…,X,Y,Z,… etc., and the elements are denoted by small letters, i.e. a,b,c,…,x,y,z,… etc. If A is any set and a is the element of set A, then we write a∈A, read as a belongs to A.
## What is the symbol for empty set?
Empty Set: The empty set (or null set) is a set that has no members. Notation: The symbol ∅ is used to represent the empty set, { }.
## What are subsets Math 7th grade?
A set A is a subset of another set B if all elements of the set A are elements of the set B. In other words, the set A is contained inside the set B. The subset relationship is denoted as A⊂B. Since B contains elements not in A, we can say that A is a proper subset of B.
## What is roster method?
The roster method is defined as a way to show the elements of a set by listing the elements inside of brackets. An example of the roster method is to write the set of numbers from 1 to 10 as {1,2,3,4,5,6,7,8,9 and 10}. An example of the roster method is to write the seasons as {summer, fall, winter and spring}. |
# Class 9 Physics Chapter 7 Numerical Problems
Class 9 Physics Chapter 7 numerical problems are according to the new syllabus of the Punjab Board. Chapter 7 is related to the properties of matter in which you will study the working principles of how matter is converted into different substances using multiple phenomenons.
## Class 9 Physics Chapter 7 Numerical Problems
The problems have been solved with easy-to-use methods using the following formulae:
Density = mass/volume
P = F / A
D = wx ρ / (w1 – w2
A = 2πR2
F1 / a = F2 / A
Y = FL / A△L
### Numerical Problems Chapter 7 – Properties of Matter
#### 7.1 A wooden block measuring 40 cm x 10 cm x 5 cm has a mass 850 g. Find the density of 3 wood. (425 kgm – 3)
Solution:
Volume of wooden block = V = 40 cm x 10 cm x 5cm = 2000 cm3
V = 2000 x 1/1000000 m3
V = 0.002 m3
Mass = m = 850 g = = 0.85 kg
Density of wood = ρ =?
Density = mass/volume
ρ = 0.85 / 0.02
ρ = 425 kgm– 3
#### 7.2 How much would be the volume of ice formed by freezing 1 litre of water? (1.09 litre)
Solution:
Volume of water = 1 litre
Volume of ice = ?
1 litre of water = 1 kg mass and density = 1000 kg – 3
Since the density of ice is 0.92 times of liquid water, therefore,
Volume of ice = mass/density
Volume = 1000 / 920
V = 1.09 litre
#### (iii) A gold bar of mass 0.2 kg. The density of gold is 19300 kgm – 3.(1.04×10 – 5 m3)
Solution:
Mass of iron sphere = m = 5 kg
Density of iron = ρ = 8200 kgm – 3
Volume of iron sphere = V =?
Volume = Mass / Density
Volume = 5 / 8200
Volume = 0.00060975 = 6.0975 x 10 – 4
Volume = 6.1 x 10 – 4m3
Mass of lead shot = m = 200 g = kg = 0.2 kg
Density of lead = r = 11300 kgm – 3
Volume of lead shot = V =?
Volume = Mass / Density
Volume = 0.2 / 11300
Volume = 0.000017699 = 1.76699 x 10 – 5
Volume = 1.77 x 10 – 5m3
Mass of gold bar = m = 0.2 kg
Density of gold = ρ = 19300 kgm – 3
Volume of gold bar = V =?
Volume = Mass / Density
Volume = 0.2 / 19300
Volume = 0.000010362 = 1.0362 x 10 – 5
Volume = 1.04 x 10 – 5m3
#### 7.4 The density of air is 1.3 kgm – 3. Find the mass of air in a room measuring 8m x 5m x 4m. (208 kg)
Solution:
Density of air = ρ = 1.3 kgm – 3
Volume of room = V = 8 m x 5 m x 4 m = 160 m3
Mass of air = m =?
Mass of air = Density of air x volume of the room
Mass of air = 1.3 x 160
Mass of air = 208 kg
#### 7.5 A student presses her palm by her thumb with a force of 75 N. How much would be the pressure under her thumb having contact area 1.5 cm2 ? (5×105 Nm – 2)
Solution:
Force = F = 75 N
Contact Area A = 1.5 cm2 = 1.5 x x m2 = 1.5 x 10 – 4 m2
Pressure under the thumb = P =?
P = F / A
P = 75 / 1.5 x 104
P = 5 x 105 Nm – 2
#### 7.6 The head of a pin is a square of side 10 mm. Find the pressure on it due to a force of 20 N. (2×105 Nm – 2)
Solution:
Force = F = 20 N
Area of head of a pin = A = 10mm x 10mm = 10/10 cm x 10/10 cm
A = 1/100 m x 1/100 m
A = 10 – 4 m2
Pressure under the thumb = P =?
= F/A
= 20 / 1x 10-4
P = 2 x 105 Nm – 2
#### (ii) Density of the wood. (1778 Nm – 2 , 889 kgm – 3)
Solution:
Length of the smallest side of the block = 7.5 cm
Mass of the block = m = 1000g = 1kg
Pressure exerted by the block = P =?
Density of wood = ρ =?
Since the smallest edge of the block is rested on the horizontal surface. Therefore, the area of the block will be:
Area = A = 7.5 cm x 7.5 cm = 56.25 cm2
A = 56.25 x 1/100 x 1/100 m2 = 56.25 x 10 – 4 m2
Pressure under the thumb = P =?
P = F/A = mg/A
P = 1x 10 / 56.25 x 10-4
P = 0.1778 x 104
P = 1778 Nm – 2
Volume = V = 20 cm x 7.5 cm x 7.5 cm = 1125 cm3
V = 1125 x 1/100 m x 1/100 m x 1/100 m
V = 1125 x 10 – 6 m3
V = 1.125 x 10 – 3 m3
Density = Mass / Volume
ρ = 1/ 1.125 x 10-3 = 0.8888 x 103 = 888.8 kgm – 3
ρ = 889 kgm – 3
#### 7.8 A cube of glass of 5 cm side and mass 306 g, has a cavity inside it. If the density of glass is 2.55 gcm – 3. Find the volume of the cavity. (5 cm3)
Solution:
Size of the cube = 7.5 cm
Mass of the cube = m = 306 g
Density of glass = ρ = 2.55 kgm – 3
Volume of the cavity = V =?
Volume of the whole cube = 5 cm x 5 cm x 5 cm = 125 cm3
Volume of the glass = Mass / Density
Volume = 306 / 2.55
V = 120 cm3
Volume of the cavity = 125 cm3 – 120 cm3
V = 5 cm3
#### 7.9 An object has weight 18 N in air. Its weight is found to be 11.4 N when immersed in water. Calculate its density. Can you guess the material of the object? (2727 kgm – 3, Aluminium)
Solution:
Weight of object in air = w1 = 18 N
Weight of object immersed in water = w2 = 11.4 N
Density of glass = ρ = 1000 kgm – 3
Density of the object = D =?
Nature of the material =?
D = wx ρ / (w1 – w2
D = 18 / 18-11.4 x 1000
D = 18/6.6 x 1000 = 2.727 x 103 = 2727 kgm – 3
The density of aluminium is 2700 kgm – 3, and the calculated value of density is 2727 kgm– 3 nearest to the density of aluminium, so the material of the object is aluminium.
#### 7.10 A solid block of wood of density 0.6 gcm – 3 weighs 3.06 N in air. Determine (a) volume of the block (b) the volume of the block immersed when placed freely in a liquid of density 0.9 gcm – 3 ?(510 cm3 , 340 cm3)
Solution:
Density of wood = D = 0.6 gcm – 3
Weight of the wooden block = w= 3.06 N
Since w = mg
m = w / g
m = 3.06 / 10
m = 0.306 kg = 306 g
Density of liquid = D = 0.9 gcm – 3
Volume of the block V =?
Volume of the block immersed in a liquid V =?
Density = Mass / Volume
Volume = Mass / Volume
V = 306 / 0.6 = 510 cm3
Volume = Mass / Density
V = 306 / 0.9 = 340 cm3
#### 7.11 The diameter of the piston of a hydraulic press is 30 cm. How much force is required to lift a car weighing 20 000 N on its piston if the diameter of the piston of the pump is 3 cm? (200 N)
Solution: Diameter = D = 30 cm
Radius of the piston = R = D / 2 = 30 / 2 = 15 cm = 15 / 100 m = 0.15 m
Area of the piston = A = 2πR2 = 2 x 3.14 x (0.15)2
A = 0.1413 m2
Weight of the car = w = F2 = 20000 N
Diameter of the piston = d = 3 cm
Radius of the piston = R = D/2 = 3/2 = 1.5 cm = 1.5/100 m = 0.015 m
Area of the piston = A = 2πR2 = 2 x 3.14 x (0.015)2
A = 1.413 x 10 – 3 m2
Force = F1 = ?
F1 / a = F2 / A
F= F2 x a / A
F= 200000 N x 1.413 x 10 – 3 / 0.1413
F= 200000 N x 0.01
F= 200 N
#### 7.12 A steel wire of cross-sectional area 2×10 – 5 m2 is stretched through 2 mm by a force of 4000 N. Find Young’s modulus of the wire. The length of the wire is 2 m. (2×1011 Nm – 2)
Solution:
Cross-sectional area = A = 2 x 10– 5 m2
Extension = △L = 2 mm = 2 x 10-3 m = 0.002 m
Force = F = 4000 N
Length of the wire = L = 1m
Y = FL / A△L
Y = 4000×2 / 2×10– 3 x 0.002 = 8000 / 0.004 x 10 – 5
Y = 8000 / 0.004 x 10 – 5
Y = 2,000,000 x 10 – 5 = 2 x 1011 Nm – 2
Class 9 Physics Full Book Numerical Problems with Solution
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A Cyclic Inequality in Three Variables XXI
Solution 1
Note that $\displaystyle \sqrt{\frac{ab}{7}}\le\frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}.\,$ Indeed, that is equivalent to $(a-b)^2(7a^2-ab+7b^2)\ge 0,\,$ which is true, with equality for $a=b.\,$
Taking the product of three such inequalities we obtain the required one. Equality when $a=b=c.$
Solution 2
Observe that
$\displaystyle a^2-ab+b^2=\frac{3}{4}(a-b)^2+\frac{1}{4}(a+b)^2\ge\frac{1}{4}(a+b)^2,\text{and}\\ \displaystyle a^2+5ab+b^2=-\frac{3}{4}(a-b)^2+\frac{7}{4}(a+b)^2\le\frac{7}{4}(a+b)^2.$
Also, for $a,b,c\gt 0,\,$ $(a+b)(b+c)(c+a)\ge 8abc.\,$ Combining that
\displaystyle \begin{align} \prod_{cycl}\frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}&\ge\frac{\displaystyle \left(\frac{1}{4}\right)^3\prod_{cycl}(a+b)^2}{\displaystyle \left(\frac{\sqrt{7}}{2}\right)^3\prod_{cycl}(a+b)}\\ &=\frac{1}{56\sqrt{7}}(a+b)(+c)(c+a)\\ &\ge\frac{abc}{7\sqrt{7}}. \end{align}
Solution 3
$\displaystyle a^2-ab+b^2=\frac{3}{4}(a-b)^2+\frac{1}{4}(a+b)^2\ge\frac{1}{4}(a+b)^2.\,$ Similarly, $\displaystyle b^2-bc+c^2\ge\frac{(b+c)^2}{4}\,$ and $\displaystyle c^2-ca+a^2\ge\frac{(c+a)^2}{4}.\,$ Thus,
\displaystyle \begin{align} \prod_{cycl}(a^2-ab+b^2) &\ge \frac{(a+b)^2(b+c)^2(c+a)^2}{64}\\ &=\frac{\displaystyle \prod_{cycl}(a+b)\cdot\prod_{cycl}(a+b)}{64}\\ &\ge\frac{8abc}{64}(a+b)(b+c)(c+a)\\ &=\frac{(a+b)(b+c)(c+a)}{8}. \end{align}
Suffice it to prove that
$\displaystyle \frac{\displaystyle abc\prod_{cycl}(a+b)}{8}\ge\frac{abc}{7\sqrt{7}}\prod_{cycl}\sqrt{a^2+5ab+b^2}$
which is the same as
(a)
$\displaystyle \prod_{cycl}\frac{\sqrt{7}(a+b)}{2}\ge \prod_{cycl}\sqrt{a^2+5ab+b^2}$
Suffice it to prove that $\displaystyle \frac{\sqrt{7}(a+b)}{2}\ge\sqrt{a^2+5ab+b^2},\,$ which is equivalent to
\displaystyle \begin{align} &7(a^2+b^2+2ab)\ge 4(a^2+5ab+b^2)\,\Longleftrightarrow\\ &3(a-b)^2\ge 0, \end{align}
which is true.
Solution 4
The required inequality is equivalent to
$\displaystyle 7^3\prod_{cycl}(a^2-ab+b^2)^2\ge (abc)^2\prod_{cycl}(a^2+5ab+b^2).$
Now,
\begin{align} 7(a^2-ab+b^2)^2&-\,ab(a^2+5ab+b^2)\\ &=7(a^2+b^2)^2-15ab(a^2+b^2)+2(ab)^2\\ &=(a-b)^2(7a^2+7b^2-ab)\ge 0, \end{align}
so $7(a^2-ab+b^2)^2\ge ab(a^2+5ab+b^2).\,$ Rotating $a,b,c\,$ and taking the product yields the required inequality.
Solution 5
Consider
\begin{align} 7(a^2-ab+b^2)^2&-\,ab(a^2+5ab+b^2)\\ &=7(a^2-ab+b^2)^2-\,ab(a^2-ab+b^2)-6a^2b^2\\ &=7(a^2-ab+b^2)^2-\,7ab(a^2-ab+b^2)\\ &\qquad\qquad\qquad\qquad+\,6ab(a^2-ab+b^2)-6a^2b^2\\ &=7(a^2-ab+b^2)(a^2-ab+b^2-ab)\\ &\qquad\qquad\qquad\qquad+\,6ab(a^2-ab+b^2-ab)\\ &=(a-b)^2[(a^2-ab+b^2)+6(a^2+b^2)]\ge 0. \end{align}
Hence, $\displaystyle \frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}\ge\frac{\sqrt{ab}}{\sqrt{7}}.\,$ Rotating $a,b,c\,$ and taking the product yields the required inequality.
Solution 6
First note that $\displaystyle \frac{x^2-x+1}{\sqrt{x^2+5x+1}}\ge\frac{1}{2\sqrt{7}}(x+1),\,$ $(xyz=1)\,$ the RHS being the tangent of the LHS:
With this in mind, set $\displaystyle \frac{a}{b}=x,\,$ $\displaystyle \frac{b}{c}=y,\,$ $\displaystyle \frac{c}{a}=z.\,$ The inequality reduces to
$\displaystyle \frac{x^2-x+1}{\sqrt{x^+5x+1}}\ge\frac{1}{7\sqrt{7}}.$
We have
\displaystyle\begin{align} LHS &\ge \frac{x+1}{2\sqrt{7}}\cdot\frac{y+1}{2\sqrt{7}}\cdot\frac{z+1}{2\sqrt{7}}\\ &=\frac{(x+1)(y+1)(z+1)}{8\cdot 7\sqrt{7}}\\ &\ge\frac{2\sqrt{x}\cdot 2\sqrt{y}\cdot 2\sqrt{z}}{8\cdot 7\sqrt{7}}\\ &=\frac{1}{7\sqrt{7}}. \end{align}
Solution 7
Let's find $k\gt \,$ for which $k(a^2+5ab+b^2)\le a^2+b^2,\,$ i.e.,
$(k-1)a^2-5ab+(k-1)b^2\ge 0.$
The discriminant of the quadratic form is $5^2-4(k-1)^2,\,$ which is not positive for $2|k-1|\le 5.\,$ The simplest suitable value is $\displaystyle k=\frac{7}{2}.$
Now we have
\displaystyle \begin{align} &\prod_{cycl}\sqrt{a^2+5ab+b^2}\le\left(\frac{7}{2}\right)^{\frac{3}{2}}\prod_{cycl}\sqrt{a^2+b^2}\\ &\prod_{cycl}(a^2-ab+b^2)\ge\prod_{cycl}\left(\frac{a^2+b^2}{2}\right). \end{align}
Dividing one by the other and taking into account that $a^2+b^2\ge 2ab,\,$ we obtain the required inequality.
Acknowledgment
This problem from the Romanian Mathematical Magazine, has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. Solution 1 is by Leo Giugiuc; Solution 2 is by Kevin Soto Palacios; Solution 3 by Soumava Chakraborty; Solution 4 by Diego Alvariz; Solution 5 by Ravi Prakash; Solution 6 is by Sladjan Stankovik; Solution 7 is by Srinivas Vemuri. |
## Calculus 10th Edition
a) The graph is on the figure. b) The least rapid increase coincides with the least slope and that happened in the year of 2008. c) The average change rate is the increase of $2.8$ million people per year. d) The prediction is that there would be $337.8$ million people living in US in $2020$.
a) The plot is given on the figure and for technical reason is done on computer. To draw it by hand just draw points $(t,y)$ from the table putting $t$ horizontally and $y$ vertically as in the plot given down. Then connect the points (red points on the figure) drawing a straight segment between two neighboring points. b) To find the slope of the line between two points divide the difference in their $y$ coordinate with the difference in their $t$ coordinate. We will do that for each pair of neighboring points and find which slope is the lowest and that means that during that year the change was the least rapid: $$m_1 = \frac{295.8-293}{5-4} = 2.8;$$ $$m_2 = \frac{298.6-295.8}{6-5} = 2.8;$$ $$m_3 = \frac{301.6-298.6}{7-6} = 3;$$ $$m_4 = \frac{304.4-301.6}{8-7} =2.8;$$ $$m_5=\frac{307.0-304.4}{9-8} = 2.6.$$ Here we see that $m_5$ which corresponds to the segment from $t=8$ to $t=9$ is the lowest which means that during the year of $2008$ the increase was the least rapid. c) The average rate of change is just the total difference in $y$ from $t=9$ to $t=4$ divided by the difference in $t$ which is: $$m_{avg} = \frac{307.0-293.0}{5} = 2.8.$$ This tells us that the average rate of change in the US population during the period of $2004-2009$ is $2.8$ million people per year. d) If the population keeps changing in this rate of $2.8$ million people per year then it will increase on average for $2.8$ million people every year from $2009$ to $2020$ which is the difference of $11$ years so the population in $2020$ would be that in the $2009$ plus the total increase: $$y=307.0+2.8*11 = 337.8$$ in millions. |
# Thread: Solve in two different ways
1. ## Solve in two different ways
Solve in two different ways $\frac{dy}{dx}=xy^3-xy$.
-------------------------------------------------------------------------
Separable:
$\frac{dy}{dx}=x(y^3-y)$
$\frac{dy}{y^3-y}=xdx$
$\frac{dy}{y(y+1)(y-1)}=xdx$
Let $\frac{1}{y(y+1)(y-1)}=\frac{A}{y}+\frac{B}{y+1}+\frac{C}{y-1}$
$1=A(0+1)(0-1)$
$A=-1$
$1=B(-1)(-1-1)$
$B=\frac{1}{2}$
$1=C(1+1)$
$C=\frac{1}{2}$
$\int \left[\frac{-1}{y}+\frac{1}{2(y+1)}+\frac{1}{2(y-1)}\right]dy=\int xdx$
$-\ln y+\frac{1}{2}\ln (y+1)+\frac{1}{2}\ln (y-1)=\frac{x^2}{2}+C'$
$-2\ln y+\ln (y+1)+\ln (y-1)=x^2+C$, where $C=2C'$.
$\ln \frac{(y+1)(y-1)}{y^2}=x^2+C$
Bernoulli:
$\frac{dy}{dx}+xy=xy^3$
Let $v=y^{-2}$.
$y=v^{-1/2}$
$\frac{dy}{dx}=-\frac{1}{2}v^{-3/2}\frac{dv}{dx}$
$-\frac{1}{2}v^{-3/2}\frac{dv}{dx}+xv^{-1/2}=xv^{-3/2}$
Multiplying throughout by $-2v^{3/2}$,
$\frac{dv}{dx}-2xv=-2x$
The integrating factor is $e^{\int -2xdx}=e^{-x^2}$.
$e^{-x^2}\frac{dv}{dx}-2xe^{-x^2}v=-2xe^{-x^2}$
$\frac{d}{dx}(ve^{-x^2})=-2xe^{-x^2}$
$ve^{-x^2}=e^{-x^2}+C$
$y^{-2}e^{-x^2}=e^{-x^2}+C$
$y^{-2}=Ce^{x^2}+1$
$y=(Ce^{x^2}+1)^{-1/2}$
-------------------------------------------------------------------------
Where have I gone wrong?
2. ## Re: Solve in two different ways
Originally Posted by alexmahone
Solve in two different ways $\frac{dy}{dx}=xy^3-xy$.
-------------------------------------------------------------------------
Separable:
$\frac{dy}{dx}=x(y^3-y)$
$\frac{dy}{y^3-y}=xdx$
$\frac{dy}{y(y+1)(y-1)}=xdx$
Let $\frac{1}{y(y+1)(y-1)}=\frac{A}{y}+\frac{B}{y+1}+\frac{C}{y-1}$
$1=A(0+1)(0-1)$
$A=-1$
$1=B(-1)(-1-1)$
$B=\frac{1}{2}$
$1=C(1+1)$
$C=\frac{1}{2}$
$\int \left[\frac{-1}{y}+\frac{1}{2(y+1)}+\frac{1}{2(y-1)}\right]dy=\int xdx$
$-\ln y+\frac{1}{2}\ln (y+1)+\frac{1}{2}\ln (y-1)=\frac{x^2}{2}+C'$
$-2\ln y+\ln (y+1)+\ln (y-1)=x^2+C$, where $C=2C'$.
$\ln \frac{(y+1)(y-1)}{y^2}=x^2+C$
Bernoulli:
$\frac{dy}{dx}+xy=xy^3$
Let $v=y^{-2}$.
$y=v^{-1/2}$
$\frac{dy}{dx}=-\frac{1}{2}v^{-3/2}\frac{dv}{dx}$
$-\frac{1}{2}v^{-3/2}\frac{dv}{dx}+xv^{-1/2}=xv^{-3/2}$
Multiplying throughout by $-2v^{3/2}$,
$\frac{dv}{dx}-2xv=-2x$
The integrating factor is $e^{\int -2xdx}=e^{-x^2}$.
$e^{-x^2}\frac{dv}{dx}-2xe^{-x^2}v=-2xe^{-x^2}$
$\frac{d}{dx}(ve^{-x^2})=-2xe^{-x^2}$
$ve^{-x^2}=e^{-x^2}+C$
$y^{-2}e^{-x^2}=e^{-x^2}+C$
$y^{-2}=Ce^{x^2}+1$
$y=(Ce^{x^2}+1)^{-1/2}$
-------------------------------------------------------------------------
Where have I gone wrong?
What makes you think there's something wrong with either of them (except for leaving off the modulus signs in your logarithms)? If you try to write your separable equation as y in terms of x, you should get something that is correct up to an integration constant.
3. ## Re: Solve in two different ways
Originally Posted by Prove It
What makes you think there's something wrong with either of them (except for leaving off the modulus signs in your logarithms)? If you try to write your separable equation as y in terms of x, you should get something that is correct up to an integration constant.
$\ln \left|\frac{(y+1)(y-1)}{y^2}\right|=x^2+C$
$\frac{(y+1)(y-1)}{y^2}=\pm e^{(x^2+C)}=\pm e^Ce^{x^2}$
$\frac{(y+1)(y-1)}{y^2}=C'e^{x^2}$, where $C'=\pm e^C$
$y^2-1=C'e^{x^2}y^2$
$y^2(1-C'e^{x^2})=1$
$y^2=(1-C'e^{x^2})^{-1}$
$y=(1-C'e^{x^2})^{-1/2}$, where $C'$ is the negative of the constant in the Bernoulli method. |
# Discovering Decimals Part 3: Multiplication and Division
This is the last part of number operations dealing with decimals: multiplication and division in a concrete and pictorial method. This is actually not part of the KG-5th grade OAS standards, but it is addressed in 6th grade and for those of you utilizing the CCSS, you will find multiplication and division with decimals starts in 5th grade. Or you may have advanced students who are ready to explore this concept. There are a couple of freebies included in this post. Read on to find them!
Multiplying Decimals:
Typically we teach our students that when you multiply 2 numbers together, the product is larger than the 2 factors. And when we divide two numbers, the quotient is smaller than the dividend.
Be careful about stating this generalization: This remains true when multiplying whole numbers (or even a combination of whole and decimal such as 5.2 x 6.4 = 33.28 in which the product is larger than either factor), but NOT with decimals or fractions less than 1 (example: .7 x .2 = .14 in which .14 is less than either factor .7 or .2). With division of decimals less than 1, the answer is often a whole number larger than either the divisor or the dividend (such as 5 divided by 1/2 = 10). This is a difficult concept, but modeling and practicing with concrete and pictorial models helps to see the reasoning. So, be careful not to say, “When you multiply two numbers together you always get a bigger number.” because it’s not always true.
To further understand this principal, let’s look at the use of the times (x) sign with whole numbers. The problem 5 x 2 could be shown in an array as 5 rows with 2 in each row (phrased as 5 “rows of” 2). It can be shown in a set model such as 5 groups and each group has 2 (phrased as 5 “groups of” 2) . It can also be shown in an area model (box) as a shape divided equally into 5 rows and 2 columns (phrased as 5 “by” 2). Continue reading
# Multiplication Strategies Part 4: Doubling and Halving (and Lattice)
Doubling and . . .
I hope you have had a chance to look at Parts 1, 2, and 3 of my multiplication strategy posts. These strategies are especially helpful with 3rd – 5th grade students (and beyond). I have been reading a book by Dr. Nicki Newton called “Guided Math in Action.” She discusses five components to being mathematically proficient. One of them is strategic competence. What is strategic competence? The National Research Council defines it as “the ability to formulate mathematical problems, represent them, and solve them.” The first process std. in Common Core (Make Sense of Problems and Persevere in Solving Them) emphasizes strategic competence in this way: “they try special cases and simpler
Halving Multip. Strategy
forms of the original problem in order to gain insight into its solution, . . . students check their answers to problems using a different method, and they continually ask themselves, Does this make sense? . . . and, they can understand the approaches of others to solving complex problems and identify correspondences between different approaches.”
With all of that said, I would like to show two other multiplication strategies: 1) Doubling and Halving, and 2) the Lattice Method. All of the strategies picture boards I have shown are available on this free PDF (It’s 10 pages): Multiplication Strategies PDF
Doubling and Halving
Doubling and halving is supported by the associative property. It also enables the student to use mental math strategies. Here is an example: Original Problem: 25 x 12. When changed to 50 x 6, I can solve it mentally which equals 300. From what I have researched and applied, here are some tips:
• Use when a problem has one even and one odd factor (or two even factors). It won’t work with 2 odd factors because you can’t break an odd number in half and still get a whole number.
• Double the odd factor and halve the even factor. Do this 1-3 times until you get two factors which can be multiplied mentally.
• Most useful when the odd factor has a 5 in the one’s place (because 5 doubled is 10, so the result will be a multiple of 10 which can usually be taken care of mentally).
How is the above problem connected to the associative property?
• Think of 25 x12 as 25 x (2 x 6).
• If I apply the associative property, I get (25 x 2) x 6 = 50 x 6 = 300.
• I can also go another step further and show that 50 x 6 = 50 x (2 x 3) = (50 x 2) x 3 = 100 x 3 = 300.
• Here is another one: Think of 15 x 24 as 15 x (2 x 12) = (15 x 2) x 12 = 30 x 12 = 360.
• See my pictures below of how that looks in array form with the problems 3 x 4 and 5 x 12. |
# If both of the zeros of ax^2+bx+c are positive or both negative then what can you say about the signs of a, b and c ?
Aug 6, 2017
See explanation...
#### Explanation:
$a {x}^{2} + b x + c$
• If the zeros are both positive then $a$ and $c$ have the same sign and $b$ has the opposite sign.
• If the zeros are both negative then $a$, $b$ and $c$ have the same sign.
To see these, suppose the zeros are $x = \alpha$ and $x = \beta$.
We have:
$a {x}^{2} + b x + c = a \left(x - \alpha\right) \left(x - \beta\right)$
$\textcolor{w h i t e}{a {x}^{2} + b x + c} = a \left({x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta\right)$
$\textcolor{w h i t e}{a {x}^{2} + b x + c} = a {x}^{2} - \left(\alpha + \beta\right) a x + \alpha \beta a$
Matching the coefficients, we find:
$\left\{\begin{matrix}b = - \left(\alpha + \beta\right) a \\ c = \alpha \beta a\end{matrix}\right.$
If $\alpha > 0$ and $\beta > 0$ then $- \left(\alpha + \beta\right) < 0$ and $\alpha \beta > 0$. Hence $b$ has the opposite sign to $a$ and $c$ has the same sign as $a$.
If $\alpha < 0$ and $\beta < 0$ then $- \left(\alpha + \beta\right) > 0$ and $\alpha \beta > 0$. Hence $a$, $b$ and $c$ all have the same sign.
$\textcolor{w h i t e}{}$
Note
The converses of these two results are not generally true.
There are quadratic equations $a {x}^{2} + b x + c$ with $a , c$ having the same sign and $b$ having the same or opposite sign which have no real zeros.
For example:
${x}^{2} + x + 1$
has no real zeros.
It still has two zeros, but they are so called "imaginary" a.k.a. non-real complex numbers. They do not lie on the real number line, but somewhere else in the complex plane. The real number line forms the $x$ axis of the complex plane, the $y$ axis being called the imaginary axis.
All complex numbers can be expressed in the form $x + y i$, where $x , y$ are real numbers and $i$ is the imaginary unit, satisfying:
${i}^{2} = - 1$
For example, the zeros of ${x}^{2} + x + 1$ are:
$- \frac{1}{2} + \frac{\sqrt{3}}{2} i \text{ }$ and $\text{ } - \frac{1}{2} - \frac{\sqrt{3}}{2} i$
It is unfortunate that the term imaginary is used of such numbers - coined when such numbers were poorly understood. They are just as real as Real numbers and better behaved in some respects.
Aug 10, 2017
Some more explanation...
#### Explanation:
Attempting to make this a little easier to follow, let us start with the simpler case where the leading coefficient is $1$
${x}^{2} + p x + q$
If $\alpha$ and $\beta$ are the two zeros, then both $\left(x - \alpha\right)$ and $\left(x - \beta\right)$ must be factors.
Using FOIL to help multiply, we find:
$\left(x - \alpha\right) \left(x - \beta\right) = {\overbrace{\left(x \cdot x\right)}}^{\text{First"+overbrace(x * (-beta))^"Outside"+overbrace((-alpha) * x)^"Inside"+overbrace((-alpha) *(-beta))^"Last}}$
$\textcolor{w h i t e}{\left(x - \alpha\right) \left(x - \beta\right)} = {x}^{2} - \beta x - \alpha x + \alpha \beta$
$\textcolor{w h i t e}{\left(x - \alpha\right) \left(x - \beta\right)} = {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta$
Comparing this with the expression ${x}^{2} + p x + q$ we find:
• If $\alpha > 0$ and $\beta > 0$ then $- \left(\alpha + \beta\right) < 0$, so $p < 0$.
Also $\alpha \beta > 0$, so $q > 0$.$\textcolor{w h i t e}{\frac{0}{0}}$
In summary, the pattern of the signs of the coefficients of ${x}^{2} + p x + q$ is $+ - +$.$\textcolor{w h i t e}{\frac{0}{0}}$
For example ${x}^{2} - 3 x + 2 = \left(x - 1\right) \left(x - 2\right)$ has zeros $1$ and $2$.$\textcolor{w h i t e}{\frac{0}{0}}$
• If $\alpha < 0$ and $\beta < 0$ then $- \left(\alpha + \beta\right) > 0$, so $p > 0$.$\textcolor{w h i t e}{\frac{0}{0}}$
Also $\alpha \beta > 0$, so $q > 0$.$\textcolor{w h i t e}{\frac{0}{0}}$
In summary, the pattern of the signs of the coefficients of ${x}^{2} + p x + q$ is $+ + +$.$\textcolor{w h i t e}{\frac{0}{0}}$
For example ${x}^{2} + 3 x + 2 = \left(x + 1\right) \left(x + 2\right)$ has zeros $- 1$ and $- 2$.
The more general case of $a {x}^{2} + b x + c$ is similar, but we have to deal with that initial coefficient $a$, which can be positive or negative.
We have:
$a {x}^{2} + b x + c = a \left({x}^{2} + \frac{b}{a} x + \frac{c}{a}\right) = a \left(x - \alpha\right) \left(x - \beta\right)$
• If $\alpha > 0$ and $\beta > 0$ then the coefficients $1$, $\frac{b}{a}$ and $\frac{c}{a}$ have signs in the pattern $+ - +$.$\textcolor{w h i t e}{\frac{0}{0}}$
Hence if $a > 0$ then the pattern of the signs of the coefficients of $a {x}^{2} + b x + c$ is also $+ - +$.$\textcolor{w h i t e}{\frac{0}{0}}$
If instead $a < 0$ then the pattern of the signed of the coefficients of $a {x}^{2} + b x + c$ is $- + -$.$\textcolor{w h i t e}{\frac{0}{0}}$
• If $\alpha < 0$ and $\beta < 0$ then the coefficients $1$, $\frac{b}{a}$ and $\frac{c}{a}$ have signs in the pattern $+ + +$.$\textcolor{w h i t e}{\frac{0}{0}}$
Hence if $a > 0$ then the pattern of the signs of the coefficients of $a {x}^{2} + b x + c$ is also $+ + +$.$\textcolor{w h i t e}{\frac{0}{0}}$
If instead $a < 0$ then the pattern of the signs of the coefficients of $a {x}^{2} + b x + c$ is $- - -$. |
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# If the lines ${a_1}x + {b_1} y + {c_1} = 0$ and ${a_2}x + {b_2} y + {c_2} = 0$ cut the coordinate axis in concyclic points, thenA ${a_1} {a_2} = {b_1} {b_2}$B $\dfrac{{{a_1}}} {{{a_2}}} = \dfrac{{{b_1}}} {{{b_2}}}$C ${a_1} + {a_2} = {b_1} + {b_2}$D ${a_1} {b_1} = {a_2} {b_2}$
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Hint: In this question we have been given equations of two lines which cut the coordinate axis in concyclic points which gives us a very important relation. Let suppose the first line intersect at the points A and B and similarly the second line at the points C and D. AB and CD are chords around the x and the y axis with origin O. Therefore, the relation between the points A, B, C, D, as they are concyclic would be $OA \times OB = OC \times OD$. We would use this relation to solve the equation further and find the relation between ${a_1}, {a_2}, {b_1}, {b_2}$.
We have been provided with two equation ${a_1}x + {b_1} y + {c_1} = 0$ and ${a_2}x + {b_2} y + {c_2} = 0$. So, we will be finding the coordinates of both the equations one by one.
The first equation is ${a_1}x + {b_1}y + {c_1} = 0$ , so the coordinates of this equation will be $A\left( {\dfrac{{ - {c_1}}}{{{a_1}}},0} \right)$ and $B\left( {0,\dfrac{{ - {c_1}}}{{{b_1}}}} \right)$
Similarly, for the second equation ${a_2}x + {b_2}y + {c_2} = 0$, the coordinates would be $C\left( {\dfrac{{ - {c_2}}}{{{a_2}}},0} \right)$ and $D\left( {0,\dfrac{{ - {c_2}}}{{{b_2}}}} \right)$
And since these points A, B, C, D, are concyclic the following relation would be true for them $OA \times OB = OC \times OD$.
The values will be $\left( {\dfrac{{ - {c_1}}}{{{a_1}}}} \right) \times \dfrac{{ - {c_2}}}{{{a_2}}} = \left( {\dfrac{{ - {c_1}}}{{{b_1}}}} \right) \times \dfrac{{ - {c_2}}}{{{b_2}}}$
Now solving this equation using cross multiplication method the final relation comes out to be ${a_1} {a_2} = {b_1} {b_2}$
So, the answer comes out to be ${a_1} {a_2} = {b_1} {b_2}$ which is your option (a). |
# What Is 47/64 as a Decimal + Solution With Free Steps
The fraction 47/64 as a decimal is equal to 0.7343.
Fraction is defined as the parts of the whole number. Fraction expression is written in p/q form. p is the value of the parts of the whole number and q is the whole number value. 47/64 is the fraction expression where p=47 is the value of parts of the whole number and q=64 is the whole number value.
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 47/64.
## 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 = 47
Divisor = 64
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 = 47 $\div$ 64
This is when we go through the Long Division solution to our problem. The following figure shows the long division:
Figure 1
## 47/64 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 47 and 64, we can see how 47 is Smaller than 64, and to solve this division, we require that 47 be Bigger than 64.
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 47, which after getting multiplied by 10 becomes 470.
We take this 470 and divide it by 64; this can be done as follows:
 470 $\div$ 64 $\approx$ 7
Where:
64 x 7 = 448
This will lead to the generation of a Remainder equal to 470 – 448 = 22. Now this means we have to repeat the process by Converting the 22 into 220 and solving for that:
220 $\div$ 64 $\approx$ 3Â
Where:
64 x 3 = 192
This, therefore, produces another Remainder which is equal to 220 – 192 = 28. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 280.
280 $\div$ 64 $\approx$ 4Â
Where:
64 x 4 = 256
Finally, we have a Quotient generated after combining the three pieces of it as 0.734=z, with a Remainder equal to 24.
Images/mathematical drawings are created with GeoGebra. |
# How do you solve using the completing the square method x^2+8x=9?
Feb 28, 2016
$x = 1$ or $x = 9$
(see below for method of completing the square).
#### Explanation:
A general squared binomial has the relation:
$\textcolor{w h i t e}{\text{XXX}} {\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$
If ${x}^{2} + 8 x$ are the first two terms of an expanded squared binomial
then $a = 1$ and $b = 4$
and the third term needed to complete the square would be ${b}^{2} = 16$.
If we add $16$ to the left side to complete the square there
we will need to add $16$ to the right side to maintain the equality.
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 8 x + 16 = 9 + 16$
Re-writing the left side as a squared binomial and simplifying the right side:
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 4\right)}^{2} = 25$
If we take the square root of both sides:
$\textcolor{w h i t e}{\text{XXX}} \left(x + 4\right) = \pm 5$
which implies:
either $x = 1$ or $x = - 9$ |
# Degree of Differential equations
The exponent of the highest derivative of a function in a non-radical and non-fractional differential equation is called the degree of a differential equation.
## Introduction
In differential equations, one or more derivatives of a function are involved as coefficients in the terms and the power of the highest derivative of the function is considered as the degree of the differential equations.
$4\dfrac{d^2y}{dx^2}-y+9 = 0$
It is a differential equation, where $y$ represents a function in $x$. In this case, the function $y$ is differentiated twice. So, we have to evaluate the exponent of the highest derivative of the function.
$\implies$ $4\Bigg(\dfrac{d^2y}{dx^2}\Bigg)^1-y+9 = 0$
Therefore, the degree of this differential equation is one and it is called the differential equation of first degree.
In some cases, the differential equations may contain fractions and radicals. So, the radicals and fractions should be eliminated from the differential equations firstly for evaluating the degree of any differential equation.
### Examples
Let’s learn the concept of degree of differential equations from some more understandable examples.
$(1)\,\,\,$ $\dfrac{du}{d\theta} \,=\, 8u+7$
In this example, there is only one derivative and we have to evaluate its index to find the degree of differential equation.
$\implies$ $\Bigg(\dfrac{du}{d\theta}\Bigg)^1 \,=\, 8u+7$
Therefore, the degree of this differential equation is $1$ and it is called the differential equation of first degree.
$(2)\,\,\,$ $7\Bigg(\dfrac{d^2l}{dx^2}\Bigg)^{\Large \frac{5}{2}}+8\dfrac{dl}{dx}+9 = 0$
The variable $l$ represents a function in $x$ in the given differential equation. The function $l$ is differentiated two times in the first term and one time in the second term. The degree of the differential equation is defined by considering highest derivative but its exponent is a fraction. In fact, a degree of an equation cannot be a fraction. By eliminating the fraction form, the differential equation can be written as follows.
$\implies$ $49\Bigg(\dfrac{d^2l}{dx^2}\Bigg)^5$ $-$ $64\Bigg(\dfrac{dl}{dx}\Bigg)^2$ $-$ $144\dfrac{dl}{dx}$ $-$ $81$ $\,=\,$ $0$
In this example, the second derivative is highest derivative and its exponent is $5$. So, the degree of this differential equation is $5$. Hence, it is called the differential equation of fifth degree.
$(3)\,\,\,$ $\sqrt[\displaystyle 3]{\dfrac{d^9z}{dy^9}+5\dfrac{d^4z}{dy^4}-2\dfrac{dz}{dy}+16}$ $\,=\,$ $3$
In this example, the differential equation is in radical form. It is difficult for beginners to evaluate the degree of the differential equation. So, eliminate the root form and it is written as follows.
$\implies$ $\dfrac{d^9z}{dy^9}+5\dfrac{d^4z}{dy^4}-2\dfrac{dz}{dy}-11$ $\,=\,$ $0$
The ninth derivative is highest derivative in this case and its exponent is $1$. Therefore, the degree of this equation is one and it is called as the differential equation of first degree.
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# Vector Multiplication by Number Calculator
## This calculator multiplies a vector by a number and gives a detailed solution to all stages of the calculation.
Dimension of space
Specify the form of the vector
Enter the coordinates of the vector
a̅ = { ; }
Enter the value of the number q by which you want to multiply the vector
q =
How to Multiply a Vector by a Number in 2d and 3d Space
Example #1
Let's multiply the vector in two-dimensional space by the number q. Vector coordinates are given by points.
Coordinates of point A of vector AB: (5 ; 9)
Coordinates of point B of vector AB: (-2 ; 11)
The numbers q by which you need to multiply the vector AB = 12
Step by step solution:
In order to multiply a vector by a number, it is necessary to multiply each coordinate of the vector by a given number.
Let's calculate the coordinates of the first vector at two points A and B:
AB = {xB - xA ; yB - yA} = {-2 - 5 ; 11 - 9} = {-7 ; 2}
AB โ
q = {ABx โ
q ; ABy โ
q} = {-7 โ
12 ; 2 โ
12} = {-84 ; 24}
Example #2
Let's multiply a vector in three-dimensional space by q.
Vector coordinates a: (5 ; 9 ; -2)
The number q by which you need to multiply the vector a = 2.6
Step by step solution:
In order to multiply a vector by a number, it is necessary to multiply each coordinate of the vector by a given number.
a โ
q = {ax โ
q ; ay โ
q ; az โ
q} = {5 โ
2.6 ; 9 โ
2.6 ; -2 โ
2.6} = {13 ; 117/5 ; -26/5} = {13 ; 23.4 ; -5.2}
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# Surface area of a cone.
The surface of a cone can be derived from the area of a square pyramid
Start with a pyramid square and just keep increasing the number of sides of the base. After a very large number of sides, you can see that the figure will be finally look like a cone.This is illustrated below:
This observation is important because we can use the formula of the surface of a pyramid square to conclude that a cone l is the height of the slope.
The area of the square is s2
The area of a triangle is (s × l) / 2
Since there are 4 triangles, the region is 4 × (s × l) / 2 = 2 × s × l
Therefore, the area, call it her is:
SA = s2 + 2 × s × l:
In General, to find the area of a regular pyramid whose base is a, the perimeter is p and the height of the slope is l, we use the following formula:
S = A + 1/2 (P × l)
Once again a is the area of the base. For a figure with 4 sides, A = s2 with s = length on one side
Where the 1/2 (P × l) come from?
Either s the length of the base of a regular pyramid. Then, the area of a triangle is
(s × l) / 2
For n triangles and this also means that the base of the pyramid has n sides, we get
(n × s × l) / 2
Now P = n × s. When n = 4, of course, P = 4 × s as it has already shown.
We have, after replaning × s n by P, S = A + 1/2 (P × l)
Let us now use it to derive the formula for the surface of a cone
For a cone, the base is a circle, A = p × r2
P = 2 × p × r
To find the tilt, height, l, just use the Pythagorean theorem
l = r2 + h2
l = v (r2 + h2)
Putting all together, we get:
S = A + 1/2 (P × l)
S = p × r2 + 1/2 (2 × p × r × v (r2 + h2))
S = p × r2 + p × r × v (r2 + h2)
Example # 1:
Find the surface of a cone with a radius of 4 cm and 8 cm high
S = p × r2 + p × r × v (r2 + h2)
S = 3.14 × 42 + 3.14 × 4 × v (42 + 82)
S = 3.14 × 16 + 12.56 × v (16 + 64)
S = 50.24 + 12.56 × v (80)
S = 50.24 + 12.56 × 8.94
S = 50.24 + 112.28
S = cm2 162.52
Example # 2:
Find the surface of a cone with a radius of 9 cm and 12 cm in height
S = p × r2 + p × r × v (r2 + h2)
S = 3.14 × 92 + 3.14 × 9 × v (92 + 122)
S = 3.14 × 81 + 28.26 × v (81 + 144)
S = 254.34 + 28,26 × v (225)
S = 254.24 + 28.26 × 15
S = 254.24 + 423.9
S = cm2 678.14 |
Free Algebra Tutorials!
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
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Equ. #2:
Equ. #3:
Equ. #4:
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Equ. #8:
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Number of inequalities to solve: 23456789
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Example
The solutions of a quadratic equation are 5 and
Work backwards to find a quadratic equation with these solutions.
Solution
Begin by writing the two solutions. To clear the fraction in the second solution, multiply both sides of the second equation by 3. x = 5 or 3x = -2 Move the constant to the left side of each equation. x - 5 = 0 or 3x + 2 = 0 Since each binomial is equal to 0, their product is 0. (x - 5)(3x + 2) = 0 Multiply the binomials. Combine like terms. 3x2 + 2x - 15x - 10 = 0 3x2 - 13x - 10 = 0
The quadratic equation 3x2 - 13x - 10 = 0 has the given solutions, 5 and
Note:
3x2 - 13x - 10 = 0
Any nonzero multiple of this equation also has solutions 5 and
Here are some examples:
6x2 - 26x - 20 = 0
15x2 - 65x - 50 = 0
The quadratic formula states that the solutions of a quadratic equation, ax2 + bx + c = 0, are
Let’s see what happens when we combine the solutions first by addition and then by multiplication:
• When we add the solutions, the radicals are eliminated. Add the numerators to form one fraction. Combine like terms. The square root terms add to zero. Cancel the common factor, 2.
Thus, for a quadratic equation, ax2 + bx + c = 0, the sum of the solutions is
• When we multiply the solutions, the radicals are eliminated.
Multiply the numerators and multiply the denominators.
Combine like terms.
Cancel the common factor, 4a.
Thus, for a quadratic equation, ax2 + bx + c = 0, the product of the solutions is |
# If A=begin{bmatrix}1 & 2&0 1 & 1&01&4&0 end{bmatrix}, B=begin{bmatrix}1 & 2&3 1 & 1&-12&2&2 end{bmatrix} text{ and } C=begin{bmatrix}1 & 2&3 1 & 1&-11&1&1end{bmatrix}. Show that AB=AC text{ but } B neq C
If . Show that
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odgovoreh
Step 1
Given data:
The first matrix is $A=\left[\begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 1& 4& 0\end{array}\right]$
The second matrix is $B=\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& -1\\ 2& 2& 2\end{array}\right]$
The third matrix is $C=\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& -1\\ 1& 1& 1\end{array}\right]$
Step 2
The multiplication of AB matrices is,
$AB=\left[\begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 1& 4& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& -1\\ 2& 2& 2\end{array}\right]$
$=\left[\begin{array}{ccc}\left(1+2\right)& \left(2+2\right)& \left(3-2\right)\\ \left(1+1\right)& \left(2+1\right)& \left(3-1\right)\\ \left(1+4\right)& \left(2+4\right)& \left(3-4\right)\end{array}\right]$
$=\left[\begin{array}{ccc}3& 4& 1\\ 2& 3& 2\\ 5& 6& -1\end{array}\right]$
The multiplication of AC matrices is,
$AC=\left[\begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 1& 4& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& -1\\ 1& 1& 1\end{array}\right]$
$=\left[\begin{array}{ccc}\left(1+2\right)& \left(2+2\right)& \left(3-2\right)\\ \left(1+1\right)& \left(2+1\right)& \left(3-1\right)\\ \left(1+4\right)& \left(2+4\right)& \left(3-4\right)\end{array}\right]$
$=\left[\begin{array}{ccc}3& 4& 1\\ 2& 3& 2\\ 5& 6& -1\end{array}\right]$
The multiplication of AB and AC is equal.
Thus, AB=AC, but B not equal to C. |
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# Class 12 NCERT Solutions – Mathematics Part I – Chapter 4 Determinants – Exercise 4.6 | Set 2
• Difficulty Level : Expert
• Last Updated : 05 Apr, 2021
### 3y – 5z = 9
Solution:
Matrix form of the given equation is AX = B
i.e.
∴ |A| =
∴ Solution is unique.
Now, X = A-1B = (adj.A)B
Therefore, x=1, y=1/2, z=3/2
### x + y + z = 2
Solution:
Matrix form of the given equation is AX = B
i.e
∴ |A| =
∴ Solution is unique.
Now, X = A-1B = (adj.A)B
Therefore, x = 2, y = -1, z = 1
### 3x – y – 2z = 3
Solution:
Matrix form of given equation is AX = B
i.e.
∴ |A| =
∴ Solution is unique.
Now, X = A-1B = (adj.A)B
Therefore, x = 1, y = 2, z = -1
### 2x – y + 3z = 12
Solution:
Matrix form of given equation is AX = B
i.e.
∴ |A| =
∴ Solution is unique.
Now, X = A-1B = (adj.A)B
Therefore, x = 2, y = 1, z = 3
### x + y – 2z = – 3
Solution:
Given: A=
Now, |A|=
∴ |A|=
Means, A-1 exists.
Now,
From eq. (1),
A-1=
Now, Matrix form of given equation is AX = B
i.e.
∵ Solution is unique.
∴ X=A-1B
⇒
Therefore, x = 1, y = 2, z = 3
### Question 16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is 60 rupees. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is 90 rupees. The cost of 6 kg onion 2 kg wheat and 3 kg rice is 70 rupees. Find cost of each item per kg by matrix method.
Solution:
Let Rs x, Rs y, Rs z per kg be the prices of onion, wheat and rice respectively.
A.T.Q.
4x+3y+2z=60
2x+4y+6z=90
6x+2y+3z=70
Matrix form of equation is AX = B
where, A=,B=and X=
=>
Now, |A|=
∴ Solution is unique
Now, |
### MATLAB Lecture #2
```MATLAB Lecture #2 EGR 271 – Circuit Theory I
Linear Systems
An important use of matrices is in the solution of systems of linear equations,
or linear systems. Linear systems occur in numerous areas of engineering and
mathematics, including electric circuits and electric systems.
A linear system might be described by the following equations:
a11x1 a12 x 2 a13 x 3 b1
a 21x1 a 22 x 2 a 23 x 3 b 2
a 31x1 a 32 x 2 a 33 x 3 b3
These equations could be written in matrix form as:
a 11
a
21
a 31
a 12
a 22
a 32
a 13 x1 b1
a 23 x 2 b 2
a 33 x 3 b 3
The matrix equation could be written as: Ax = b
1
MATLAB Lecture #2 EGR 271 – Circuit Theory I
Several methods can be used to solve linear systems in the form Ax = b,
including:
1) Using the inverse matrix, A-1
2) Gaussian elimination
3) Gauss-Jordan reduction
2
MATLAB Lecture #2 EGR 271 – Circuit Theory I
3
Solving Linear Systems using A-1:
Recall that when multiplying matrices that AB BA.
As a result, if both sides of a matrix equation are multiplied by another matrix,
there is a difference between pre-multiplying and post-multiplying.
Example:
B=C
(original equation)
AB = AC (pre-multiplying the original equation by matrix A)
BA = CA (post-multiplying the original equation by matrix A)
But since AB BA, the result of pre-multiplying and post-multiplying is
clearly different. This needs to be kept in mind when solving linear equations
in the form Ax = b.
Ax = b
(system of linear equations)
A-1Ax = A-1b (pre-multiply by A-1)
So linear systems
can be solved using
Ix = A-1b
(since A-1A = I)
x = A-1b
x = A-1b
(since Ix = x)
x = inv(A)*b in MATLAB
MATLAB Lecture #2 EGR 271 – Circuit Theory I
4
Example: Solve the system of equations below using x = A-1b with MATLAB.
x1 2x2 3x3 9
x1 3x2 4x3 11
x1 4x2 3x3 7
Solution:
Note: MATLAB may give warnings about using
this method to solve equations and may recommend
a different method. We will discuss this later.
MATLAB Lecture #2 EGR 271 – Circuit Theory I
Gaussian Elimination
Which system of equations below is easier to solve?
x yz7
3x 2y z 11
4x - 2y 2z 8
x y z7
- y - 2z - 10
10z 40
The system on the right is easier because we can easily:
• Solve for z in the 3rd equation
• Substitute the value of z into the 2nd equation and solve for y
• Substitute the values of y and z into the 1st equation and solve for x
The right set of equations is easier to solve for because it is in row-echelon
form where we can easily use back substitution.
It may be hard to recognize, but the two systems of equations are equivalent!
5
MATLAB Lecture #2 EGR 271 – Circuit Theory I
Gaussian Elimination
x yz7
3x 2y z 11
4x - 2y 2z 8
Original equations
1 1 1 7
3 2 1 11
4 2 2 8
Original equations in
augmented matrix form
x y z7
- y - 2z - 10
10z 40
Equations in row-echelon form
1
7
1 1
0 1 2 10
0 0 10 40
Augmented matrix manipulated
into row-echelon form
The matrix was manipulated using elementary row operations.
The process is called Gaussian elimination.
6
MATLAB Lecture #2 EGR 271 – Circuit Theory I
7
Manipulating augmented matrices is similar to how we manipulate equations.
Example:
2x 3 y 4z 5
4 x 6 y 8z 10
Multiply both sides
of an equation by 2
2
a
21
a 31
3
4
a 22
a 32
a 23
a 33
5 2R 1 4
a
b 2
21
a 31
b 3
6
8
a 22
a 32
a 23
a 33
10
b 2
b 3
This was a type of
elementary row
operation
Example:
x 3y 2z 4
0 x 11y 4 z 9
2 x 5 y 8 z 1 Add 2 times Eq 1 to Eq 2
to form a new equation
1
- 2
a 31
3
2
5
a 32
8
a 33
4
1
1 R 2 2 R 1 0
a 31
b 3
3
2
11
a 32
4
a 33
This was another type of
4
elementary row operation.
9
Note that Eq 2 was
b 3
replaced by the new
equation.
MATLAB Lecture #2 EGR 271 – Circuit Theory I
Elementary Row Operations
There are three types of elementary row operations:
1) Multiply a row by a non-zero constant
Notation: 3R1
(multiply row 1 by 3)
2) Interchange two rows
Notation: R2,3
(interchange row 2 and row 3)
3) Add a multiple of one row to another row
Notation: R2 + (3)R1
(add 3 times row 1 to row 2)
Note: Elementary column operations cannot be used to solve systems of
equations, but they could perhaps be used in other applications not covered in
this course.
8
MATLAB Lecture #2 EGR 271 – Circuit Theory I
9
Example 1: Solve the following three equations using Gaussian elimination.
x yz7
Pivot element
3x 2y z 11
Pivot column
4x - 2y 2z 8
1 1 1 7
Solution: Form the augmented matrix: 3 2 1 11 R 2 - 3R 1
4 2 2 8 R 3 - 4 R 1
Perform elementary row operations
using column 1 as the pivot column:
1
7
1 1
0 1 2 10
0 6 2 20 R 3 - 6R 2
Perform elementary row operations
using column 2 as the pivot column:
1
7
1 1
0 1 2 10
0 0 10 40
Back substitute to solve
for x, y, and z.
Row 3 : 10z 40,
so z 4
Row 2 : - y - 2(4) - 10, so y 2
Row 1 : x 2 4 7,
so x 1
The matrix is now
in row-echelon
form
MATLAB Lecture #2 EGR 271 – Circuit Theory I
Note: Although it is not required, it is common to adjust each column so that
For the last example, we could continue as follows:
1
7
1 1
1 1 1 7
0 1 2 10 1R 0 1 2 10
2
0 0 10 40 0.1R 3 0 0 1 4
Now the back substitution is even easier:
Row 3 : z 4
Row 2 : y 2(4) 10, so y 2
Row 1 : x 2 4 7,
so x 1
Checking results: It is a good idea to check your results by substituting the
answers back into the original equations. Try this for the problem above:
x yz7
3x 2y z 11
4x - 2y 2z 8
10
MATLAB Lecture #2 EGR 271 – Circuit Theory I
11
Rearranging rows: When performing Gaussian reduction, the pivot element
must be non-zero. If there is a zero in the pivot element position, it is useful to
rearrange the rows (one of the three elementary row operations).
Example: Solve the following system of equations.
y 3z 4
- x 2y 3
2x 3y 4z 1
1 3 4 R 1,2 1 2 0 3 (1)R 1 1 2 0 3
0
1 2 0 3
0
0 1 3 4
1
3
4
2 3 4 1
2 3 4 1
2 3 4 1
1 2 0 3
1 2 0 3
1 2 0 3
0 1 3 4
0 1 3 4
0 1 3 4
2 3 4 1 R 3 (2)R 1 0 1 4 7 R 3 (1)R 2 0 0 1 3
Row 3 : z 3
Row 2 : y 3(3) 4, so y - 5
Row 1 : x - 2(-5) - 3,
so x - 13
(Sub intooriginalequationsto check)
MATLAB Lecture #2 EGR 271 – Circuit Theory I
12
Gauss-Jordan Elimination: In using Gauss-Jordan elimination (or GaussJordan reduction), we continue where Gaussian elimination left off and use
additional elementary row operations until the augmented matrix is in reduced
row-echelon form. This will eliminate the need for back substitution.
x 2y 3z 9
- x 3y
-4
2x - 5y 5z 17
Elementary row
9
1
operations
1 2 3
1 3 0 4
2 5 5 17
Gaussian
elimination
0
0
2
1
0
Elementary row
3 9 operations 1
1
0 1 0 1
3 5
Gauss-Jordan
1 2
0 0 1 2
elimination
Row-echelon
form
See next slide for
step-by-step details
Use back
substitution to
find results
0 0
Reduced
row-echelon
form
directly: x = 1,
y = -1, z = 2
MATLAB Lecture #2 EGR 271 – Circuit Theory I
13
Gauss-Jordan Elimination: Solve the system of equations below using
Gauss-Jordan reduction (same example as on previous slide but detail added).
x 2y 3z 9
- x 3y
-4
2x - 5y 5z 17
Row-echelon
form
Solution:
1 2 3 9
1 2 3 9
1 2 3 9
1 2 3 9
1 3 0 4 R R 0 1
0 1 3 5
0 1 3 5
3
5
2
1
2 5 5 17 R 3 (2)R 1 0 1 1 1 R 3 R 2 0 0 2 4 (1 / 2)R 3 0 0 1 2
1 2 3 9 R 1 (2)R 2 1 0 9 19 R 1 (9)R 3 1 0 0 1
0 1 3 5
0 1 3 5 R (3)R 0 1 0 1
2
3
0 0 1 2
0 0 1 2
0 0 1 2
Pivot element
Pivot column
Reduced
row-echelon
form
Results: x = 1, y = -1, z = 2
MATLAB Lecture #2 EGR 271 – Circuit Theory I
14
Example: Solve the system of equations below using Gauss-Jordan
reduction.
2x 4 y - 2
x 2y 2z 7
3x 3y z 11
Results: x = 3, y = -2, z = 4
MATLAB Lecture #2 EGR 271 – Circuit Theory I
rref( ) - a useful function in MATLAB for reducing
an augmented matrix into reduced row echelon form
Example: Use rref( ) to solve the following systems
of equations (both from earlier examples).
System 1:
x 2y 3z 9
- x 3y
-4
2x - 5y 5z 17
System 2:
x yz7
3x 2y z 11
4x - 2y 2z 8
15
MATLAB Lecture #2 EGR 271 – Circuit Theory I
Left Division in MATLAB - It is recommended that a system of linear
equations in the form
Ax = b
be solved using left division
x = A\b
x = inv(A)*b
In general, x = A\b is more stable and faster. Why? Some reasons include:
• inv(A) may not exist
• x = A\b uses Gaussian elimination and:
• Scales matrix entries to minimize errors
• Uses faster algorithms for special matrices, such as sparse, symmetrical
or banded matrices.
16
MATLAB Lecture #2 EGR 271 – Circuit Theory I
17
Example: - The following represents an ill-conditioned linear system. The
error in the result depends highly on the number of significant digits used
unless the equations are scaled.
• Note the warning associated with
using x1 = inv(A)*b.
• Note that the results are different.
MATLAB Lecture #2 EGR 271 – Circuit Theory I
18
Example: A) Solve the following equations using Gauss-Jordan reduction
B) Solve the equations in MATLAB using three methods:
KVL, meshes 1-3 yields:
• x = inv(A)*b
• rref( )
• x = A\b
Answers(tocheck your results): I1 22.46A, I2 13.85A, I3 8.31A
``` |
# Parabolas Lesson 6 Episode 1 (Teachers)
### Making Sense
Sasha and Keoni use the equation y = x2/(4p) to plot a parabola for p = 1/4. They make a conjecture for how the shape of the parabola will change as gets larger.
### Episode Supports
Students’ Conceptual Challenges
After using the equation to plot points on the parabola y = x2 with a p-value of 1/4, Keoni struggles to locate coordinates of the focus of the parabola [5:06-5:24]. He first places the focus at (0,1) [5:35-5:59]. Keoni states that (0,1) is a “general place” to put the focus.
➤Sasha and Keoni notice a conflict when asked to state the p-value for the parabola when the focus is one unit above the origin. They restate that p is the distance between the focus and the vertex. Keoni notices that they are currently working with a p-value of 1/4. Consequently, Sasha and Keoni adjust the focus location [6:02-6:19].
Focus Questions
For use in a classroom, pause the video and ask these questions:
1. [Pause video at 5:05]. What are some other points that you know are on the parabola because of the geometry of the parabola?
2. [Pause video at 9:16]. What are the coordinates of the red point that Keoni says is on the parabola?
Supporting Dialogue
Invite students attend to the reasoning of others while reflecting on multiple strategies:
1. Stop the video at [10:00]. Ask one student to present one method for checking to see if the point (½, ¼) is on the parabola. Ask a second student to use the first student’s method to check a different point, say (1,1).
2. Repeat the process for a new method of checking.
Math Extensions
1. What happens when the focus is below the vertex? Graph the parabola with a focus at (0, –¼) and vertex at (0, 0). Label the focus, the directrix, and several points on the parabola. |
# GRE Arithmetic | Decimals
The value of a digit can be determined by using its digit, position of digit in the number and base of the given number system. Decimal number system has base 10 as it uses 10 digits from 0 to 9 and the successive positions to the left of the decimal point are as: 100, 101, 102, 103, … and so on. Successive positions to the right of the decimal point are as: 10-1, 10-2, 10-3, … and so on. Example: For example, the digits of the number 4321.1234 have following place values: 4321.1234 can be written as,
```= (4 x 1000)+(3 x 100)+(2 x 10)+(1 x l)
+ (1 x 0.1)+(2 x 0.01)+(3 x 0.001)+(4 x 0.0001)
= (4 x 103)+(3 x 102)+(2 x 101)+(1 x 100)
+ (1 x 10-1)+(2 x 10-2)+(3 x 10-3)+(4 x 10-4)
= 4000 + 300 + 20 + 1 + 0.1 + 0.02 + 0.003 + 0.0004
= 4321.1234 ```
Since each place value is a power of 10, every decimal can be converted to an integer divided by a power of 10. We can convert a fractional decimal number to its equivalent fraction with integers in numerator and denominator. Example-1:
`4.5 = 4 + 5/10 = 40/10 + 5/10 = 45/10`
Example-2:
`75.18 = 75 + 18/100 = 7500/100 + 18/100 = 7518/100`
Example-3:
`0.369 = 0 + 369/1000 = 369/1000`
We can also convert an Integer in numerator and denominator to its equivalent decimal fraction by dividing numerator by denominator. The result will either terminate or repeat without end. Repeating part of a decimal can be indicate using bar over repeating digits. Example-4:
`3/4 = 0.75`
Example-5:
`134/20 = 6.7`
Example-6:
`30/28 = 1.0714285`
Note:
• Every Rational number can be expressed as a terminating or repeating decimal and converse is also true.
• Irrational numbers can not be expressed in numerator/denominatr form because these numbers neither have repeating fraction (i.e., √2 = 1.41421356237) nor has finite fraction (0.03033033303333…).
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# Difference between revisions of "2021 AIME II Problems/Problem 15"
## Problem
Let $f(n)$ and $g(n)$ be functions satisfying $$f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}$$and $$g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}$$for positive integers $n$. Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$.
## Solution 1
Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2 for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$. Note that this formula also returns the correct value when $n=(k+1)^2$, but not when $n=k^2$. Thus $f(n)=k^2+3k+2-n$ for $k^2.
If $2 \mid (k+1)^2-n$, $g(n)$ returns the same value as $f(n)$. This is because the recursion once again stops at $(k+1)^2$. We seek a case in which $f(n), so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$.
Write $7f(n)=4g(n)$, which simplifies to $3k^2+k-10=3n$. Notice that we want the $LHS$ expression to be divisible by 3; as a result, $k \equiv 1 \pmod{3}$. We also want n to be strictly greater than $k^2$, so $k-10>0, k>10$. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is $k=16$, giving $n=258$.
Indeed - if we check our answer, it works. Therefore, the answer is $\boxed{258}$.
-Ross Gao
## Solution 2 (More Variables)
We restrict $n$ in which $k^2 for some positive integer $k,$ or $$n=(k+1)^2-p\hspace{15mm}(1)$$ for some nonnegative integer $p.$ By observations, we get \begin{align*} f\left((k+1)^2\right)&=k+1, \\ f\left((k+1)^2-1\right)&=k+2, \\ f\left((k+1)^2-2\right)&=k+3, \\ &\cdots \\ f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{15mm}(2) \\ \end{align*} If $n$ and $(k+1)^2$ have the same parity, then starting with $g\left((k+1)^2\right)=k+1,$ we will get $g(n)=f(n)$ by a similar process. This contradicts the precondition $\frac{f(n)}{g(n)} = \frac{4}{7}.$ Therefore, $n$ and $(k+1)^2$ have different parities, from which $n$ and $(k+2)^2$ have the same parity.
Along with the earlier restriction, note that $k^2 or $$n=(k+2)^2-2q\hspace{15mm}(3)$$ for some positive integer $q.$ By observations, we get \begin{align*} g\left((k+2)^2\right)&=k+2, \\ g\left((k+2)^2-2\right)&=k+4, \\ g\left((k+2)^2-4\right)&=k+6, \\ &\cdots \\ g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15mm}(4) \\ \end{align*}
~MRENTHUSIASM |
# How to Identify the Parts of an Algebraic Expression
Related Topics:
More Lessons for Grade 7 Math
Math Worksheets
Videos, solutions, worksheets, games and activities to help Algebra 1 or grade 7 students learn how to identify the parts of an algebraic expression. Vocabulary for Algebraic Expressions. How to write and evaluate expressions?
Parts of an expression (terms, coefficients and variables) and evaluating expressions
What is a term?
A term is any coefficient and it's variable.
Examples:
1. Identify the terms in the expression 4x - 2y + 5
2. Identify the coefficient and variable in the term.
3. Evaluate an expression.
4. If x = 4 and y = 2 evaluate 3x + 5y
5. Evaluate the expression 3 + a - 2b if a = 7 and b = 2.
6. Jim and Jane like to go running in the morning. Jim runs twice as far as Jane. What is the combined distance they travel? Let J be the distance Jane runs.
Vocabulary for Algebraic Expressions
In an algebraic expression, terms are things that are separated by either a + or -.
The official definition is: A term is a number, a variable, or a product or quotient of numbers and variables raised to powers.
Example:
Identify the terms in the following expression:
2x4 + 3x3 - 5x2 - 8x + 12
Each of the terms is made up of a coefficient and a variable. The coefficient is the number in front of the variable
Example:
1. Name the coefficient in the term 2x4
2. Name the coefficient in the term x
3. Name the coefficient in the term -x
When we just have a number, we call that a constant.
Example:
Name the constant in the following expression.
2x4 + 3x3 - 5x2 - 8x + 12
When an algebraic expression is written properly, we call it descending order.
The first term is called the leading term and the number in front of the variable is called the leading coefficient.
Variables and Algebraic
Variable is a symbol, usually a letter, that represents one or more numbers.
Algebraic expression is a mathematical phrase that can include numbers, variables, and operation symbols.
Examples:
1. Write an algebraic expression for each phrase
a) the sum of n and 8
b) t minus 15
2. Define a variable and write an algebraic expression for each phrase.
a) ten more than twice a number
b) three times a number minus 6
c) 9 less than a number
d) the sum of twice a number and 31
An equation is a mathematical sentence that uses an equal sign.
An open sentence is an equation that contains one or more variables.
Example:
1. Write an equation to show the total incomes from selling tickets to a school play for \$5 each.
2. Write an equation for the data in the table.
How to Identify Parts of an Expression?
Example:
Identify the terms, like terms, coefficients, and constants in the expression
8m - 6m - 4 + m -7.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
## Engage NY Eureka Math 7th Grade Module 4 Lesson 13 Answer Key
### Eureka Math Grade 7 Module 4 Lesson 13 Example Answer Key
Example 1.
The scale factor from Drawing 1 to Drawing 2 is 60%. Find the scale factor from Drawing 2 to Drawing 1. Explain your reasoning.
The scale drawing from Drawing 2 to Drawing 1 is an enlargement. Drawing 1 is represented by 100%, and Drawing 2, a reduction of Drawing 1, is represented by 60%. A length in Drawing 2 is the whole, so the scale factor from Drawing 2 to 1 is length in Drawing 1 = percent × length in Drawing 2.
100% = percent × 60%
$$\frac{100 \%}{60 \%}$$ = $$\frac{1}{0.60}$$ = $$\frac{1}{\frac{3}{5}}$$ = $$\frac{5}{3}$$ = 166 $$\frac{2}{3}$$%
Example 2.
A regular octagon is an eight-sided polygon with side lengths that are all equal. All three octagons are scale drawings of each other. Use the chart and the side lengths to compute each scale factor as a percent. How can we check our answers?
To check our answers, we can start with 10 (the length of the original Drawing 1) and multiply by the scale factors we found to see whether we get the corresponding lengths in Drawings 2 and 3.
Drawing 1 to 2: 10(1.20) = 12
Drawing 2 to 3: 12($$\frac{2}{3}$$) = 8
Example 3.
The scale factor from Drawing 1 to Drawing 2 is 112%, and the scale factor from Drawing 1 to Drawing 3 is 84%. Drawing 2 is also a scale drawing of Drawing 3. Is Drawing 2 a reduction or an enlargement of Drawing 3? Justify your answer using the scale factor. The drawing is not necessarily drawn to scale.
First, I needed to find the scale factor of Drawing 3 to Drawing 2 by using the relationship
Quantity = Percent × Whole.
Drawing 3 is the whole. Therefore,
Drawing 2 = Percent × Drawing 3
112% = Percent × 84%
$$\frac{1.12}{0.84}$$ = $$\frac{112}{84}$$ = $$\frac{4}{3}$$ = 133 $$\frac{1}{3}$$%
Since the scale factor is greater than 100%, Drawing 2 is an enlargement of Drawing 3.
Explain how you could use the scale factors from Drawing 1 to Drawing 2 (112%) and from Drawing 2 to Drawing 3 (75%) to show that the scale factor from Drawing 1 to Drawing 3 is 84%.
The scale factor from Drawing 1 to Drawing 2 is 112%, and the scale factor from Drawing 2 to Drawing 3 is 75%; therefore, I must find 75% of 112% to get from Drawing 2 to Drawing 3. (0.75)(1.12) = 0.84. Comparing this answer to the original problem, the resulting scale factor is indeed what was given as the scale factor from Drawing 1 to
Drawing 3.
### Eureka Math Grade 7 Module 4 Lesson 13 Exercise Answer Key
Opening Exercise
Scale factor: $$\frac{\text { length in SCALE drawing }}{\text { Corresponding length in ORIGINAL drawing }}$$
Describe, using percentages, the difference between a reduction and an enlargement.
A scale drawing is a reduction of the original drawing when the lengths of the scale drawing are smaller than the lengths in the original drawing. The scale factor is less than 100%.
A scale drawing is an enlargement of the original drawing when the lengths of the scale drawing are greater than the lengths in the original drawing. The scale factor is greater than 100%.
Use the two drawings below to complete the chart. Calculate the first row (Drawing 1 to Drawing 2) only.
Compare Drawing 2 to Drawing 1. Using the completed work in the first row, make a conjecture (statement) about what the second row of the chart will be. Justify your conjecture without computing the second row.
Drawing 1 will be a reduction of Drawing 2. I know this because the corresponding lengths in Drawing 1 are smaller than the corresponding lengths in Drawing 2. Therefore, the scale factor from Drawing 2 to Drawing 1 would be less than 100%.
Compute the second row of the chart. Was your conjecture proven true? Explain how you know.
The conjecture was true because the calculated scale factor from Drawing 2 to Drawing 1 was 62.5%. Since the scale factor is less than 100%, the scale drawing is indeed a reduction.
### Eureka Math Grade 7 Module 4 Lesson 13 Problem Set Answer Key
Question 1.
The scale factor from Drawing 1 to Drawing 2 is 41 $$\frac{2}{3}$$%. Justify why Drawing 1 is a scale drawing of Drawing 2 and why it is an enlargement of Drawing 2. Include the scale factor in your justification.
Quantity = Percent × Whole
Length in Drawing 1 = Percent × Length in Drawing 2
100% = Percent × 41 $$\frac{2}{3}$$%
$$\frac{100 \%}{41 \frac{2}{3} \%}$$ = $$\frac{100 \cdot 3}{41 \frac{2}{3} \cdot 3}$$ = $$\frac{300}{125}$$ = $$\frac{12}{5}$$ = 2.40 = 240%
Drawing 1 is a scale drawing of Drawing 2 because the lengths of Drawing 1 would be larger than the corresponding lengths of Drawing 2.
Since the scale factor is greater than 100%, the scale drawing is an enlargement of the original drawing.
Question 2.
The scale factor from Drawing 1 to Drawing 2 is 40%, and the scale factor from Drawing 2 to Drawing 3 is 37.5%. What is the scale factor from Drawing 1 to Drawing 3? Explain your reasoning, and check your answer using an example.
To find the scale factor from Drawing 1 to 3, I needed to find 37.5% of 40%, so (0.375)(0.40) = 0.15. The scale factor from Drawing 1 to Drawing 3 would be 15%.
Check: Assume the length of Drawing 1 is 10. Then, using the scale factor for Drawing 2, the corresponding length of Drawing 2 would be 4. Then, applying the scale factor to Drawing 3, Drawing 3 would be 4(0.375) = 1.5. To go directly from Drawing 1 to Drawing 3, which was found to have a scale factor of 15%, then 10(0.15) = 1.5.
Question 3.
Traci took a photograph and printed it to be a size of 4 units by 4 units as indicated in the diagram. She wanted to enlarge the original photograph to a size of 5 units by 5 units and 10 units by 10 units.
a. Sketch the different sizes of photographs.
b. What was the scale factor from the original photo to the photo that is 5 units by 5 units?
The scale factor from the original to the 5 by 5 enlargement is $$\frac{5}{4}$$ = 1.25 = 125%.
c. What was the scale factor from the original photo to the photo that is 10 units by 10 units?
The scale factor from the original to the 10 by 10 photo is $$\frac{10}{4}$$ = 2.5 = 250%.
d. What was the scale factor from the 5 × 5 photo to the 10 × 10 photo?
The scale factor from the 5 × 5 photo to the 10 × 10 photo is $$\frac{10}{5}$$ = 2 = 200%.
e. Write an equation to verify how the scale factor from the original photo to the enlarged 10 × 10 photo can be calculated using the scale factors from the original to the 5 × 5 and then from the 5 × 5 to the 10 × 10.
Scale factor original to 5 × 5: (125%)
Scale factor 5 × 5 to 10 × 10: (200%)
4(1.25) = 5
5(2.00) = 10
Original to 10 × 10, scale factor = 250%
4(2.50) = 10
The true equation 4(1.25)(2.00) = 4(2.50) verifies that a single scale factor of 250% is equivalent to a scale factor of 125% followed by a scale factor of 200%.
Question 4.
The scale factor from Drawing 1 to Drawing 2 is 30%, and the scale factor from Drawing 1 to Drawing 3 is 175%. What are the scale factors of each given relationship? Then, answer the question that follows. Drawings are not to scale.
a. Drawing 2 to Drawing 3
The scale factor from Drawing 2 to Drawing 3 is
$$\frac{175 \%}{30 \%}$$ = $$\frac{1.75}{0.30}$$ = $$\frac{175}{30}$$ = $$\frac{35}{6}$$ = 5 $$\frac{5}{6}$$ = 583 $$\frac{1}{3}$$%.
b. Drawing 3 to Drawing 1
The scale factor from Drawing 3 to Drawing 1 is
$$\frac{1}{1.75}$$ = $$\frac{100}{175}$$ = $$\frac{4}{7}$$ ≈ 57.14%.
c. Drawing 3 to Drawing 2
The scale factor from Drawing 3 to Drawing 2 is
$$\frac{0.3}{1.75}$$ = $$\frac{30}{175}$$ = $$\frac{6}{35}$$ ≈ 17.14%.
To check my answers, I can work backwards and multiply the scale factor from Drawing 1 to Drawing 3 of 175% to the scale factor from Drawing 3 to Drawing 2, and I should get the scale factor from Drawing 1 to Drawing 2.
(1.75)(0.1714) ≈ 0.29995 ≈ 0.30 = 30%
### Eureka Math Grade 7 Module 4 Lesson 13 Exit Ticket Answer Key
Question 1.
Compute the scale factor, as a percent, for each given relationship. When necessary, round your answer to the nearest tenth of a percent.
a. Drawing 1 to Drawing 2
Drawing 2 = Percent × Drawing 1
3.36 = Percent × 1.60
$$\frac{3.36}{1.60}$$ = 2.10 = 210%
b. Drawing 2 to Drawing 1
Drawing 1 = Percent × Drawing 2
1.60 = Percent × 3.36
$$\frac{1.60}{3.36}$$ = $$\frac{1}{2.10}$$ ≈ 0.476190476 ≈ 47.6%
c. Write two different equations that illustrate how each scale factor relates to the lengths in the diagram.
Drawing 1 to Drawing 2:
1.60(2.10) = 3.36
Drawing 2 to Drawing 1:
3.36(0.476) = 1.60
Question 2.
Drawings 2 and 3 are scale drawings of Drawing 1. The scale factor from Drawing 1 to Drawing 2 is 75%, and the scale factor from Drawing 2 to Drawing 3 is 50%. Find the scale factor from Drawing 1 to Drawing 3. |
# Inter 1st Year Maths 1A Products of Vectors Important Questions
Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Products of Vectors Important Questions which are most likely to be asked in the exam.
## Intermediate 1st Year Maths 1A Products of Vectors Important Questions
Question 1.
If $$\bar{a}$$ = 6$$\bar{i}$$ + 2$$\bar{j}$$ + 3$$\bar{k}$$ and $$\bar{b}$$ = 2$$\bar{i}$$ – 9$$\bar{j}$$ + 6$$\bar{k}$$, then find $$\bar{a}$$ . $$\bar{b}$$ and the angle between $$\bar{a}$$ and $$\bar{b}$$.
Solution:
Question 2.
If a = i + 2j – 3k, b = 3i – j + 2k then show that a + b and a – b are perpendicular to each other. (A.P) (Mar. ’15, May ’11)
Solution:
a + b = (1 + 3)i + (2 – 1)j+ (-3 + 2) k
= 4i + j – k
Similarly a – b = – 2i + 3j – 5k
(a + b). (a – b) = 4(-2) + 1(3) + (-1)(-5) = 0
Hence (a + b) and (a – b) are mutually perpendicular.
Question 3.
Let $$\bar{a}$$ and $$\bar{b}$$ be non- zero, non-collinear vectors. If |$$\bar{a}$$ + $$\bar{b}$$ | = | $$\bar{a}$$ – $$\bar{b}$$ |, then find the angle between $$\bar{a}$$ and $$\bar{b}$$.
Solution:
| $$\bar{a}$$ + $$\bar{b}$$ |2 = | a – b |2.
⇒ |$$\bar{a}$$|2 + |$$\bar{b}$$|2 + 2 ($$\bar{a}$$. $$\bar{b}$$)
⇒ |$$\bar{a}$$|2 + |$$\bar{b}$$|2 – 2 ($$\bar{a}$$. $$\bar{b}$$)
⇒ 4($$\bar{a}$$ . $$\bar{b}$$) = 0 ⇒ $$\bar{a}$$ . $$\bar{b}$$ = 0
⇒ Angle between $$\bar{a}$$ and $$\bar{b}$$ is 90°.
Question 4.
If |$$\bar{a}$$| = 11, |$$\bar{b}$$| = 23 and | $$\bar{a}$$ – $$\bar{b}$$ | = 30, then find the angle between the vectors $$\bar{a}$$, $$\bar{b}$$ and also find | $$\bar{a}$$ + $$\bar{b}$$ |.
Solution:
Given that | $$\bar{a}$$ | = 11, | $$\bar{b}$$ | = 23 and | $$\bar{a}$$ – $$\bar{b}$$ | = 30, Let ($$\bar{a}$$, $$\bar{b}$$) = θ.
Question 5.
If $$\bar{a}$$ = $$\bar{i}$$ – $$\bar{j}$$ – $$\bar{k}$$ and $$\bar{b}$$ = 2$$\bar{i}$$ – 3$$\bar{j}$$ + $$\bar{k}$$, then find the projection vector of $$\bar{b}$$ on $$\bar{a}$$ and its magnitude.
Solution:
Question 6.
If P, Q, R and S are points whose position vectors are $$\bar{i}$$ – $$\bar{k}$$, –$$\bar{i}$$ +2$$\bar{j}$$, 2$$\bar{i}$$ – 3$$\bar{k}$$ and 3$$\bar{i}$$ – 2$$\bar{j}$$ – $$\bar{k}$$ respectively, then find the component of $$\overline{\mathbf{R S}}$$ on $$\overline{\mathbf{P Q}}$$.
Solution:
Let ‘O’ be the origin.
$$\overline{\mathrm{OP}}$$ = $$\overline{\mathbf{i}}$$ – $$\overline{\mathbf{k}}$$, $$\overline{\mathrm{OQ}}$$ = –$$\overline{\mathbf{i}}$$ + 2$$\overline{\mathbf{j}}$$, $$\overline{\mathrm{OR}}$$ = 2$$\overline{\mathbf{i}}$$ – 3$$\overline{\mathbf{k}}$$ and $$\overline{\mathrm{OR}}$$ = 3$$\overline{\mathbf{i}}$$ – 2$$\overline{\mathbf{j}}$$ – $$\overline{\mathbf{k}}$$
The component of $$\overline{\mathrm{RS}}$$ and $$\overline{\mathrm{PQ}}$$
Question 7.
If the vectors λ$$\overline{\mathbf{i}}$$ – 3$$\overline{\mathbf{j}}$$ + 5$$\overline{\mathbf{k}}$$ and 2 λ$$\overline{\mathbf{i}}$$ – λ$$\overline{\mathbf{j}}$$ + $$\overline{\mathbf{k}}$$ are perpendicular to each other, find λ. (T.S) (Mar. ’16)
Solution:
Since the vectors are perpendicular
⇒ (λ$$\overline{\mathbf{i}}$$ – 3$$\overline{\mathbf{j}}$$ + 5$$\overline{\mathbf{k}}$$) . (2λ$$\overline{\mathbf{i}}$$ – λ$$\overline{\mathbf{j}}$$ – $$\overline{\mathbf{k}}$$) = 0
⇒ λ(2λ) + (-3)(-λ) + 5(-1) = 0
⇒ 2λ2 + 3λ – 5 = 0
⇒ (2λ + 5)(λ – 1) = 0
∴ λ = 1 (or) $$\frac{-5}{2}$$
Question 8.
Let $$\overline{\mathbf{a}}$$ = 2$$\overline{\mathbf{i}}$$ + 3$$\overline{\mathbf{j}}$$ + $$\overline{\mathbf{k}}$$, $$\overline{\mathbf{b}}$$ = 4$$\overline{\mathbf{i}}$$ + $$\overline{\mathbf{j}}$$ and $$\overline{\mathbf{c}}$$ = $$\overline{\mathbf{i}}$$ – 3$$\overline{\mathbf{j}}$$ – 7$$\overline{\mathbf{k}}$$. Find the vector $$\overline{\mathbf{r}}$$ such that $$\overline{\mathbf{r}}$$ . $$\overline{\mathbf{a}}$$ = 9, $$\overline{\mathbf{r}}$$ . $$\overline{\mathbf{b}}$$ = 7 and $$\overline{\mathbf{r}}$$ . $$\overline{\mathbf{c}}$$ = 6.
Solution:
Let $$\overline{\mathbf{r}}$$ = x$$\overline{\mathbf{i}}$$ + y$$\overline{\mathbf{j}}$$ + z$$\overline{\mathbf{k}}$$
∴ $$\overline{\mathbf{r}}$$ . $$\overline{\mathbf{a}}$$ = 9 ⇒ 2x + 3y + z = 9
$$\overline{\mathbf{r}}$$ . $$\overline{\mathbf{b}}$$ = 7 ⇒ 4x + y = 7 and $$\overline{\mathbf{r}}$$ . $$\overline{\mathbf{c}}$$ = 6 ⇒ x – 3y – 7z = 6
By solving these equations, we get x = 1, y = 3 and z = -2
∴ $$\overline{\mathbf{r}}$$ = $$\overline{\mathbf{i}}$$ + 3$$\overline{\mathbf{j}}$$ – 2$$\overline{\mathbf{k}}$$
Question 9.
Show that the points 2$$\overline{\mathbf{i}}$$ – $$\overline{\mathbf{j}}$$ + $$\overline{\mathbf{k}}$$, $$\overline{\mathbf{i}}$$ – 3$$\overline{\mathbf{j}}$$ – 5$$\overline{\mathbf{k}}$$ and 3$$\overline{\mathbf{i}}$$ – 4$$\overline{\mathbf{j}}$$ – 4$$\overline{\mathbf{k}}$$ are the vertices of a right angle triangle. Also, find the other angles.
Solution:
Let ‘O’ be the origin and A, B, C be the given points, then
Question 10.
Prove that the angle θ between any two diagonals of a cube is given by cos θ = $$\frac{1}{3}$$. (May ’11)
Solution:
Let us suppose that the cube is a unit cube
Question 11.
Let $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$, $$\overline{\mathbf{c}}$$ be non—zero mutually orthogonal vectors, if x $$\overline{\mathbf{a}}$$ + y$$\overline{\mathbf{b}}$$ + z = 0, then x = y = z = 0.
Solution:
x$$\overline{\mathbf{a}}$$ + y$$\overline{\mathbf{b}}$$ + z$$\overline{\mathbf{c}}$$ = 0
⇒ $$\overline{\mathbf{a}}$$ • (x$$\overline{\mathbf{a}}$$ + y$$\overline{\mathbf{b}}$$ + z$$\overline{\mathbf{c}}$$) = 0
⇒ x($$\overline{\mathbf{a}}$$ • $$\overline{\mathbf{a}}$$) = 0
⇒ x = 0 (∵ $$\overline{\mathbf{a}}$$ • $$\overline{\mathbf{a}}$$ ≠ 0)
Similarly y = 0, z = 0.
Question 12.
Let $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$ and $$\overline{\mathbf{c}}$$ be mutually orthogonal vectors of equal magnitudes. Prove that the vector $$\overline{\mathbf{a}}$$ + $$\overline{\mathbf{b}}$$ + $$\overline{\mathbf{c}}$$ is equally inclined to each of $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$ and $$\overline{\mathbf{c}}$$, the angle of inclination being cos-1$$\left(\frac{1}{\sqrt{3}}\right)$$.
Solution:
Similarly, it can be proved that $$\overline{\mathbf{a}}$$ + $$\overline{\mathbf{b}}$$ + $$\overline{\mathbf{c}}$$ inclines at an angle of cos-1$$\left(\frac{1}{\sqrt{3}}\right)$$ with $$\overline{\mathbf{b}}$$ and $$\overline{\mathbf{c}}$$.
Question 13.
The vectors $$\overline{\mathbf{A B}}$$ = 3$$\bar{i}$$ – 2$$\bar{j}$$ + 2$$\bar{k}$$ and $$\overline{\mathbf{A D}}$$ = $$\bar{i}$$ – 2$$\bar{k}$$ represent the adjacent sides of a parallelogram ABCD. Find the angle between the diagonals.
Solution:
Question 14.
For any two vectors a and b, show that
i) |a • b| (Cauchy – Schwartz inequality)
ii) || a + b | ≤ ||a|| + ||b| (triangle inequality
Solution:
i)
If a = 0 or b = 0, the inequalities hold trivially.
ii)
Question 15.
Find the cartesian equation of the plane passing through the point (-2, 1, 3) and perpendicular to the vector 3$$\bar{i}$$ + $$\bar{j}$$ + 5$$\bar{k}$$.
Solution:
Let A = -2$$\bar{i}$$ + $$\bar{j}$$ + 3$$\bar{k}$$ and $$\bar{r}$$ = x$$\bar{i}$$ + y$$\bar{j}$$ + z$$\bar{k}$$ be any point in the plane
∴ $$\overline{\mathrm{AP}}$$ = (x + 2)$$\bar{i}$$ + (y – 1)$$\bar{j}$$ + (z – 3)$$\bar{k}$$
∴ $$\overline{\mathrm{AP}}$$ is perpendicular to 3$$\bar{i}$$ + $$\bar{j}$$ + 5$$\bar{k}$$
⇒ (x + 2)3 + (y – 1)1 + (z – 3)5 = 0
⇒ 3x + y + z – 10 = 0
Question 16.
Find the cartesian equation of the plane through the point A (2, -1, -4) and parallel to the plane 4x – 12y – 3z – 7 = 0.
Solution:
The equation of the plane parallel to
4x – 12y – 3z – 7 = 0 be
4x -12y – 3z = p
∴ If passes through the point A(2, -1, -4)
⇒ 4(2) – 12(-1) – 3(-4) = p
⇒ p = 32
∴ The equation of the required plane is
4x – 12y – 3z = 32
Question 17.
Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4.
Solution:
Equation to the planes is 2x – 3y – 6z = 5
⇒ $$\overline{\mathbf{r}}$$ .(2$$\overline{\mathbf{i}}$$ – 3$$\overline{\mathbf{j}}$$ – 6$$\overline{\mathbf{k}}$$) = 5 ———- (1)
and 6x + 2y – 9z = 4 —— (2)
∴ The angle between the planes $$\overline{\mathbf{r}} \cdot \overline{\mathbf{n}}_{1}$$ = p, and $$\overline{\mathbf{r}} \cdot \overline{\mathbf{n}}_{2}$$ = q is θ, then
Question 18.
Find unit vector orthogonal to the vector 3$$\bar{i}$$ + 2$$\bar{j}$$ +6$$\bar{k}$$ and coplanar with thevectors 2$$\bar{i}$$ + $$\bar{j}$$ + $$\bar{k}$$ and $$\bar{i}$$ – $$\bar{j}$$ + $$\bar{k}$$.
Solution:
⇒ (2x + y)3 + (x – y)2 + (x + y)6 = 0
⇒ 6x + 3y + 2x – 2y + 6x + 6y = 0
⇒ 14x + 7y = 0
⇒ y = -2x —— (1)
∵ |$$\overline{\mathbf{r}}$$| = 1
⇒ (2x + y)2 + (x – y)2 + (x + y)2 = 1
⇒ 4x2 + 4xy + 9 + 2(x2 + y2) = 1
⇒ 6x2 + 4x (-2x) + 3(-2x)2 = 1 (By using (1))
⇒ 10x2 = 1
Question 19.
If $$\overline{\mathbf{a}}$$ = 2$$\overline{\mathbf{i}}$$ – 3$$\overline{\mathbf{j}}$$ + 5$$\overline{\mathbf{k}}$$, $$\overline{\mathbf{b}}$$ = –$$\overline{\mathbf{i}}$$ + 4$$\overline{\mathbf{j}}$$ + 2$$\overline{\mathbf{k}}$$, then find $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$ and unit vector perpendicular to both $$\overline{\mathbf{a}}$$ and $$\overline{\mathbf{b}}$$.
Solution:
$$\overline{\mathbf{a}}$$ = 2$$\overline{\mathbf{i}}$$ – 3$$\overline{\mathbf{j}}$$ + 5$$\overline{\mathbf{k}}$$ and $$\overline{\mathbf{b}}$$ = –$$\overline{\mathbf{i}}$$ + 4$$\overline{\mathbf{j}}$$ + 2$$\overline{\mathbf{k}}$$
Question 20.
If $$\bar{a}$$ = 2$$\bar{i}$$ – 3$$\bar{j}$$ + 5$$\bar{k}$$, $$\bar{b}$$ = –$$\bar{i}$$ + 4$$\bar{j}$$ + 2$$\bar{k}$$, then find ($$\bar{a}$$ + $$\bar{b}$$) × ($$\bar{a}$$ – $$\bar{b}$$) and unit vector perpendicular to both $$\bar{a}$$ + $$\bar{b}$$ and $$\bar{a}$$ – $$\bar{b}$$.
Solution:
Question 21.
Find the area of the parallelogram for which the vectors $$\bar{a}$$ = 2$$\bar{i}$$ – 3$$\bar{j}$$ and $$\bar{b}$$ = 3$$\bar{i}$$ – $$\bar{k}$$ are adjacent sides. (Mar. ; May ‘08)
Solution:
The vector area of the given parallelogram
Question 22.
If $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$, $$\overline{\mathbf{c}}$$ and $$\overline{\mathbf{d}}$$ are vectors such that $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$ = $$\overline{\mathbf{c}}$$ × $$\overline{\mathbf{d}}$$ and $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{c}}$$ = $$\overline{\mathbf{b}}$$ × $$\overline{\mathbf{d}}$$ then show that the vectors $$\overline{\mathbf{a}}$$ – $$\overline{\mathbf{d}}$$ and $$\overline{\mathbf{b}}$$ – $$\overline{\mathbf{c}}$$ are parallel.
Solution:
Question 23.
If $$\overline{\mathbf{a}}$$ = $$\overline{\mathbf{i}}$$ + 2$$\overline{\mathbf{j}}$$ + 3$$\overline{\mathbf{k}}$$ and $$\overline{\mathbf{b}}$$ = 3$$\overline{\mathbf{i}}$$ + 5$$\overline{\mathbf{j}}$$ – $$\overline{\mathbf{k}}$$ are two sides of a triangle, then find
its area.
Solution:
Area of the triangle = $$\frac{1}{2}|\bar{a} \times \bar{b}|$$
Question 24.
In Δ ABC, if $$\overline{\mathrm{BC}}$$ = $$\bar{a}$$, $$\overline{\mathrm{CA}}$$ = $$\bar{b}$$, $$\overline{\mathrm{AB}}$$ = $$\bar{c}$$ then show that $$\bar{a}$$ × $$\bar{b}$$ = $$\bar{b}$$ × $$\bar{c}$$ = $$\bar{c}$$ × $$\bar{a}$$.
Solution:
Question 25.
Let $$\bar{a}$$ = 2$$\bar{i}$$ – $$\bar{j}$$ + $$\bar{k}$$ and $$\bar{b}$$ = 3$$\bar{i}$$ + 4$$\bar{j}$$ – $$\bar{k}$$. If θ is the angle between $$\bar{a}$$ and $$\bar{b}$$ then find sin θ.
Solution:
Question 26.
Let $$\bar{a}$$, $$\bar{b}$$ and $$\bar{c}$$ be such that $$\bar{c}$$ ≠ 0,
$$\bar{a}$$ × $$\bar{b}$$ = $$\bar{c}$$, $$\bar{b}$$ × $$\bar{c}$$ = $$\bar{a}$$. Show that $$\bar{a}$$, $$\bar{b}$$, $$\bar{c}$$ are pair wise orthogonal vectors and |$$\bar{b}$$| = 1, |$$\bar{c}$$| = |$$\bar{a}$$|.
Solution:
Question 27.
Let $$\overline{\mathbf{a}}$$ = 2$$\overline{\mathbf{i}}$$ + $$\overline{\mathbf{j}}$$ – 2$$\overline{\mathbf{k}}$$ ; $$\overline{\mathbf{b}}$$ = $$\overline{\mathbf{i}}$$ + $$\overline{\mathbf{j}}$$. If $$\overline{\mathbf{c}}$$ is a vector such that $$\overline{\mathbf{a}}$$ . $$\overline{\mathbf{c}}$$ = |$$\overline{\mathbf{c}}$$|, |$$\overline{\mathbf{c}}$$ – $$\overline{\mathbf{a}}$$| = $$2 \sqrt{2}$$ and the angle between $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$ = $$2 \sqrt{2}$$ and the angle between $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$ and $$\overline{\mathbf{c}}$$ is 30°, then find the value of |($$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$) × $$\overline{\mathbf{c}}$$|
Solution:
Question 28.
Let $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$ be two non-collinear unit vectors. If $$\bar{\alpha}$$ = $$\overline{\mathbf{a}}$$ – ($$\overline{\mathbf{a}}$$ • $$\overline{\mathbf{b}}$$)$$\overline{\mathbf{b}}$$ and $$\bar{\beta}$$ =
= $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$, then show that |$$\overline{\boldsymbol{\beta}}$$| = |$$\bar{\alpha}$$|.
Solution:
Question 29.
A non-zero vector $$\bar{a}$$ is parallel to the line of intersection of the plane determined by the vectors $$\bar{i}$$, $$\bar{i}$$ + $$\bar{j}$$ and the plane determined by the vectors $$\bar{i}$$ – $$\bar{j}$$, $$\bar{i}$$ + $$\bar{k}$$. Find the angle between $$\bar{a}$$ and the vector $$\bar{i}$$ – 2$$\bar{j}$$ + 2$$\bar{k}$$.
Solution:
Let l be the line of intersection of the planes determined by the pairs $$\bar{i}$$, $$\bar{i}$$ + $$\bar{j}$$ and $$\bar{i}$$ – $$\bar{j}$$, $$\bar{i}$$ + $$\bar{k}$$
∴ $$\bar{n}_{1}$$ is perpendicular to l and $$\bar{n}_{2}$$ is also perpendicular to l.
∴ $$\bar{a}$$ is parallel to the line l, follows that $$\bar{a}$$ is perpendicularto both $$\bar{n}_{1}$$, and $$\bar{n}_{2}$$
Question 30.
Let $$\overline{\mathbf{a}}$$ = 4$$\overline{\mathbf{i}}$$ + 5$$\overline{\mathbf{j}}$$ – $$\overline{\mathbf{k}}$$, $$\overline{\mathbf{b}}$$ = $$\overline{\mathbf{i}}$$ – 4$$\overline{\mathbf{j}}$$ + 5$$\overline{\mathbf{k}}$$ and $$\overline{\mathbf{c}}$$ = 3$$\overline{\mathbf{i}}$$ + $$\overline{\mathbf{j}}$$ – $$\overline{\mathbf{k}}$$. Find the vector $$\bar{\alpha}$$ which is perpendicular to both $$\overline{\mathbf{a}}$$ and $$\overline{\mathbf{b}}$$ and $$\bar{\alpha} \cdot \mathbf{c}$$ = 21.
Solution:
Question 31.
For any vector a, show that |a × i|2 + |a × j|2 + |a × k|2 = 2|a|2.
Solution:
Let a = x i + y j + z k.
Then a × i = -yk + zj.
∴ |a × i|2 = y2 + z2
Similarly |a × j|2 = z2 + x2 and |a × k|2 = x2
+ y2
∴ |a × i|2 + |a × j|2 + |a × k|2 = 2(x2 + y2 + z2) = 2|a|2
Question 32.
If $$\overline{\mathbf{a}}$$ is a non-zero vector and $$\overline{\mathbf{b}}$$, $$\overline{\mathbf{c}}$$ are two vector such that $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$ = $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{c}}$$ and $$\overline{\mathbf{a}}$$ • $$\overline{\mathbf{b}}$$ = $$\overline{\mathbf{a}}$$ • $$\overline{\mathbf{c}}$$, then prove that $$\overline{\mathbf{b}}$$ = $$\overline{\mathbf{c}}$$.
Solution:
Question 33.
Prove that the vectors $$\bar{a}$$ = 2$$\bar{i}$$ – $$\bar{j}$$ + $$\bar{k}$$, $$\bar{b}$$ = $$\bar{i}$$ – 3$$\bar{j}$$ – 5$$\bar{k}$$ and $$\bar{c}$$ = 3$$\bar{i}$$ – 4$$\bar{j}$$ – 4$$\bar{k}$$ are coplanar.
Solution:
Question 34.
Find the volume of the parallelopiped whose cotersminus edges are represented by the vectors 2$$\bar{i}$$ – 3$$\bar{j}$$ + $$\bar{k}$$, $$\bar{i}$$ – $$\bar{j}$$ + 2$$\bar{k}$$ and 2$$\bar{i}$$ + $$\bar{j}$$ – $$\bar{k}$$.
Solution:
Let $$\bar{a}$$ = 2$$\bar{i}$$ – 3$$\bar{j}$$ + $$\bar{k}$$
Question 35.
If the vectors $$\bar{a}$$ = 2$$\bar{i}$$ – $$\bar{j}$$ + $$\bar{k}$$, $$\bar{b}$$ = $$\bar{i}$$ + 2$$\bar{j}$$ – 3$$\bar{k}$$ and $$\bar{c}$$ = 3$$\bar{i}$$ + p$$\bar{j}$$ + 5$$\bar{k}$$ are coplanar, then find p.
Solution:
Question 36.
Show that
$$\overline{\mathbf{i}}$$ × ($$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{i}}$$) + $$\overline{\mathbf{j}}$$ × ($$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{j}}$$) + k × ($$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{k}}$$) = 2$$\overline{\mathbf{a}}$$ for any vector $$\overline{\mathbf{a}}$$.
Solution:
Question 37.
Prove that for any three vectors $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$, $$\overline{\mathbf{c}}$$, [$$\overline{\mathbf{b}}$$ + $$\overline{\mathbf{c}}$$ $$\overline{\mathbf{c}}$$ + $$\overline{\mathbf{a}}$$ $$\overline{\mathbf{a}}$$ + $$\overline{\mathbf{b}}$$] = 2[$$\overline{\mathbf{a}}$$ $$\overline{\mathbf{b}}$$ $$\overline{\mathbf{c}}$$]
Solution:
Question 38
For any three vectors $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$, $$\overline{\mathbf{c}}$$ prove that [$$\overline{\mathbf{b}}$$ × $$\overline{\mathbf{c}}$$ $$\overline{\mathbf{c}}$$ × $$\overline{\mathbf{a}}$$ $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$] = [$$\overline{\mathbf{a}}$$ $$\overline{\mathbf{b}}$$ $$\overline{\mathbf{c}}$$]2
Solution:
Question 39.
Let $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$ and $$\overline{\mathbf{c}}$$ be unit vectors such that $$\overline{\mathbf{b}}$$ is not parallel to $$\overline{\mathbf{c}}$$ and $$\overline{\mathbf{a}}$$ × ($$\overline{\mathbf{b}}$$ × $$\overline{\mathbf{c}}$$) = $$\frac{1}{2} \bar{b}$$. Find the angles made by $$\overline{\mathbf{a}}$$ with each of $$\overline{\mathbf{b}}$$ and $$\overline{\mathbf{c}}$$.
Solution:
Since $$\overline{\mathbf{b}}$$ and $$\overline{\mathbf{c}}$$ are non—collinear vectors. Equating corresponding coeffs. on both sides
Question 40.
Let $$\overline{\mathbf{a}}$$ = $$\overline{\mathbf{i}}$$ + $$\overline{\mathbf{j}}$$ + $$\overline{\mathbf{k}}$$, $$\overline{\mathbf{b}}$$ = 2$$\overline{\mathbf{i}}$$ – $$\overline{\mathbf{j}}$$ + 3$$\overline{\mathbf{k}}$$, $$\overline{\mathbf{c}}$$ = $$\overline{\mathbf{i}}$$ – $$\overline{\mathbf{j}}$$ and $$\overline{\mathbf{d}}$$ = 6$$\overline{\mathbf{i}}$$ + 2$$\overline{\mathbf{j}}$$ + 3$$\overline{\mathbf{k}}$$. Express $$\overline{\mathbf{d}}$$ interms of $$\overline{\mathbf{b}}$$ × $$\overline{\mathbf{c}}$$, $$\overline{\mathbf{c}}$$ × $$\overline{\mathbf{a}}$$ and $$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$ (May. ’12)
Solution:
Then from the above problem,
Question 41.
For any four vectors $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$, $$\overline{\mathbf{c}}$$ and $$\overline{\mathbf{d}}$$ prove that
($$\overline{\mathbf{b}}$$ × $$\overline{\mathbf{c}}$$) . ($$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{d}}$$) + ($$\overline{\mathbf{c}}$$ × $$\overline{\mathbf{a}}$$) . ($$\overline{\mathbf{b}}$$ × $$\overline{\mathbf{d}}$$) + ($$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$) . ($$\overline{\mathbf{c}}$$ × $$\overline{\mathbf{d}}$$) = 0.
Solution:
Question 42.
Find the equation of the plane passing through the points A = (2, 3, -1), B = (4, 5, 2) and C = (3, 6, 5).
Solution:
Let ‘O’ be the origin
Let P be any point on the plane passing through the points A, B, C.
Hence, equation of the required plane is
Question 43.
Find the equation of the plane passing through the point A (3, -2, -1) and parallel to the vector $$\bar{b}$$ = $$\bar{i}$$ – 2$$\bar{j}$$ + 4$$\bar{k}$$ and $$\bar{c}$$ = 3$$\bar{i}$$ + 2$$\bar{j}$$ – 5$$\bar{k}$$.
Solution:
The required plane passes through A = (3, -2, -1) and is parallel to the vectors
$$\bar{b}$$ = $$\bar{i}$$ – 2$$\bar{j}$$ + 4$$\bar{k}$$ and $$\bar{c}$$ = 3$$\bar{i}$$ + 2$$\bar{j}$$ – 5$$\bar{k}$$. Hence if P is a point in the plane, then AP, b and C are coplanar and [AP b c] = 0.
i.e., 2x + 17y + 8z + 36 = 0 is the equation of the required plane.
Question 44.
Find the vector equation of the plane passing through the intersection of the planes r. (i + j + k) = 6 and r. (2i + 3j+ 4k) = -5 and the point (1, 1, 1).
Solution:
Substituting this value of λ in equation (1), we get
Question 45.
Find the distance of a point (2, 5, -3) from the plane r. (6i – 3j + 2k) = 4.
Solution:
Here a = 2i + 5j – 3k, N = 6i – 3j + 2k and d = 4.
∴ The distance of the point (2, 5, -3) from the given plane is
Question 46.
Find the angle between the line $$\frac{x+1}{2}$$ = $$\frac{y}{3}$$ = $$\frac{z-3}{6}$$ and the plane 10x + 2y – 11z = 3.
Solution:
Let φ be the angle between the given line and the normal to the plane.
Converting the given equations into vector form, we have
r = (-i + 3k) + λ(2i + 3j + 6k)
and r. (10i + 2j – 11k) = 3
Here b = 2i + 3j + 6k and n = 10i + 2j – 11k
Question 47.
For any four vectors a, b, c and d, (a × b) × (c × d) = [a c d]b – [b c d]a and (a × b) × (c × d) = [a b d]c – [a b c]d.
Solution:
Let m = c × d
∴ (a × b) × (c × d) = (a × b) × m
= (a.m)b – (b.m)a
= (a. (c × d))b – (b. (c × d))a
= [a c d]b – [b c d]a.
Again Let a × b = n.
Then(a × b) × (c × d) = n × (c × d)
= (n .d)c – (n.c)d
= ((a × b). d) c – ((a × b). c)d
= [a b d]c – [a b c]d.
Question 48.
Find the shortest distance between the skew line r = (6i + 2j + 2k) + t(i – 2j + 2k) and r = (-4i – k) + s (3i – 2j – 2k).
Solution:
The first line passes through the point
A(6, 2, 2) and is parallel to the vector b = i – 2j + 2k. Second line passes through the point C(-4, 0, -1) and is parallel to the vector d = 3i – 2j – 2k
Question 49.
The angle ¡n a semi circle is a right angle. (May ’08)
Solution:
Let O be the centre and AOB be the diameter of the given šemi circle. Let P be a point on the semi circle. With reference to O as the origin, if the position vectors of A and P are taken as $$\overline{\mathrm{a}}$$ and $$\overline{\mathrm{p}}$$,then
Question 50.
The altitude of a triangle are concurrent.
ii) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:
i) In the given triangle ABC, let the altitudes from A, B drawn to BC. CA respectively intersect them at D, E. Assume that AD and BE intersect at O, join CO and produce it to meet AB at F. With reference to O, let the position vectors of A, B and C be a, b and c respectively.
From Fig. we have
ii) In the given ABC, let the mid-points of BC, CA and AB be D, E and F respectively. Let the perpendicular bisectors drawn to BC and CA at D and E meet at O. Join OF. With respect to O, let the position vectors of A, B and C be $$\bar{a}$$, $$\bar{b}$$ and $$\bar{c}$$ respectively.
On adding equations (1) and (2), we obtain
Question 51.
Show that the vector area of the quadrilateral ABCD having diagonals $$\overline{\mathbf{A C}}$$, $$\overline{\mathbf{B D}}$$ is $$\frac{1}{2}(\overline{A C} \times \overline{B D})$$
Solution:
If the point of intersection of the diagonals AC, BD of the given quadrilateral ABCD is assumed as Q, the vector area of the quadrilateral ABCD.
= Sum of the vector areas of ΔQAB, ΔQBC, ΔQCD and ΔQDA.
Question 52.
For any vectors a, b and c (a × b) × c = (a.c) b – (b.c) a (AP) (Mar. 15 May ‘06; June ’04)
Solution:
Equation (1) is evidently true if a and b are parallel.
Now suppose that a and b are non-parallel.
Let O denote the origin. Choose points A and B such that $$\overline{\mathrm{OA}}$$ = $$\overline{\mathbf{a}}$$ and $$\overline{\mathrm{OB}}$$ = $$\overline{\mathbf{b}}$$. Since $$\overline{\mathbf{a}}$$ and $$\overline{\mathbf{b}}$$ are non-parallel, the points O, A and B are non-collinear. Hence they determine a plane.
Let i denote the unit vector along $$\overline{\mathrm{OA}}$$.
Let j be a unit vector in the OAB plane perpendicular to i. Let k = i × j. Then
Question 53.
For any four vectors a, b, c and d(a × b). (c × d) = $$\left|\begin{array}{cc} a \cdot c & a \cdot d \\ b . c & b . d \end{array}\right|$$ and in particular (a × b)2 = a2b2 – (a. b)2.
Solution:
Question 54.
In the above fou rmula if $$\overline{\mathbf{a}}$$, $$\overline{\mathbf{b}}$$, $$\overline{\mathbf{c}}$$ and $$\overline{\mathbf{d}}$$($$\overline{\mathbf{a}}$$ × $$\overline{\mathbf{b}}$$). ($$\overline{\mathbf{c}}$$ × $$\overline{\mathbf{d}}$$) = ($$\overline{\mathbf{a}}$$ . $$\overline{\mathbf{c}}$$)($$\overline{\mathbf{b}}$$ . $$\overline{\mathbf{d}}$$) – ($$\overline{\mathbf{a}}$$ . $$\overline{\mathbf{d}}$$)($$\overline{\mathbf{b}}$$ . $$\overline{\mathbf{c}}$$) = $$\left|\begin{array}{ll} \overline{\mathbf{a}} \cdot \overline{\mathbf{c}} & \overline{\mathbf{a}} \cdot \overline{\mathbf{d}} \\ \overline{\mathbf{b}} \cdot \overline{\mathbf{c}} & \bar{b} \cdot \overline{\mathbf{d}} \end{array}\right|$$.
Solution:
Since this is a scalar, this is also called the scalar product of four vectors. |
This standard deviation example questions can help you to calculate mean, variance, SD easily. To find the standard deviation, first write the computational formula for the standard deviation of the sample. Solution: Question 2: A health researcher read that a 200-pound male can burn an average of 524 calories per hour playing tennis. Conclusion – Standard Deviation Examples. Example: Standard deviation in a normal distribution You administer a memory recall test to a group of students. The below are some of the solved examples with solutions to help users to know how to estimate reliable sample size by using stanadrd deviation or proportion method. Population Standard Deviation. A slightly more complex calculation is called sample standard deviation. The standard deviation for discrete series can be calculated by approaches stated below: Direct method. Example 8.14. Keep reading for standard deviation examples and the different ways it appears in daily life. Population standard deviation looks at the square root of the variance of the set of numbers. of the mean. A small standard deviation can be a goal in certain situations where the results are restricted, for example, in product manufacturing and quality control. N is the number of terms in the population. This number is the standard deviation of the sample. Evaluate the standard deviation. The data follows a normal distribution with a mean score of 50 and a standard deviation of 10. Scroll down the page for more examples and solutions on using the normal distribution formula. Sample and population standard deviation . The sample mean is 5.343 with a sample standard deviation of 0.397. Example 1: Let us consider a data sample : 10,13,7,9,6 . The normal distribution is described by two parameters: the mean, μ, and the standard deviation, σ. The probability distribution is: x-152 154 156 158 160 162 164 P (x-) 1 16 2 16 3 16 4 16 3 16 2 16 1 16. The standard deviation measures the dispersion of the dataset which is relative to its mean. The mean and standard deviation of 15 observations are found to be 10 and 5 respectively. Variance and Standard Deviation for Grouped Data Calculator. For example, the mean of the following two is the same: 15, 15, 15, 14, 16 and 2, 7, 14, 22, 30. Check that this is a valid PDF and calculate the standard deviation of X.. Figure 14. To verify that f(x) is a valid PDF, we must check that it is everywhere nonnegative and that it integrates to 1.. We see that 2(1-x) = 2 - 2x ≥ 0 precisely when x ≤ 1; thus f(x) is everywhere nonnegative. In Note 6.5 "Example 1" in Section 6.1 "The Mean and Standard Deviation of the Sample Mean" we constructed the probability distribution of the sample mean for samples of size two drawn from the population of four rowers. Population Standard Deviation Formula . If a set has a low standard deviation, the values are not spread out too much. Khan Academy is a 501(c)(3) nonprofit organization. This means that we can report what proportion the standard deviation is of the mean. However, as we are often presented with data from a sample only, we can estimate the population standard deviation from a sample standard deviation. As in discrete series another column of frequency gets added, the formula for calculation of standard deviation using direct approach is altered to incorporate frequency is stated below: Standard deviation(σ)= √(∑fD²)/N) Standard Deviation Problems with Solutions. For example, if we have a standard deviation of 1.5 and a mean of 5, the ratio of the standard deviation to the mean is 0.3. Suppose random samples of size n are drawn from a population with mean μ and standard deviation σ. Standard Deviation formula to calculate the value of standard deviation is given below: (Image to be added soon) Standard Deviation Formulas For Both Sample and Population. It is symbolized by ${s}$ . If you are studying the post metric syllabus of the stats then you are most probably going to come across this measure and it will form the significant part of your exams as well. A normal probability plot for Example 9. Sample standard deviation and bias. The coefficient of variation can be reported as a percentage. Find solutions for your homework or get textbooks Search. Standard Deviation Example. Formula : Mean : Mean = Sum of X values / N(Number of Values) Variance : … Our mission is to provide … However, the second is clearly more spread out. Let the assumed mean, A = 35, c = 10 . Around 99.7% of scores are between 20 and 80. The standard deviation measures how concentrated the data are around the mean; the more concentrated, the smaller the standard deviation. It is calculated as the square root of variance. Example: 3, 8, 14, 18, 25, 22, 15, 9, 5 The sample size is small and we don’t know anything about the distribution of the population, so we examine a normal probability plot. home / study / math / statistics and probability / statistics and probability solutions manuals / Statistics for the Behavioral Sciences / 7th edition / chapter 5 / problem 15P. A low standard deviation means that the data is very closely related to the average, thus very reliable. The mean profit earning for a sample of 41 businesses is 19, and the S.D. The distribution looks normal so we will continue with our test. Home . Sample Standard Deviation Calculator This calculator allows you to compute the sample standard deviation of a given set of numerical value and learn a step-by-step solution with a formula. This is the currently selected item. It is given by the formula: The capital Greek letter sigma is commonly used in mathematics to represent a summation of all the numbers in a grouping. σ = $\sqrt{\sum (X-\mu)^{2/n}}$ Sample Standard Deviation Formula. Standard deviation (by mean method) σ = If d i = x i – are the deviations, then . Up Next. The sum of all variances gives a, which is the square of the standard deviation. 37 males were randomly selected and the mean number of calories burned per hour playing squash was 534.8 with a standard deviation of 45.9 calories. Sample Standard Deviation. We will reject… Statistics for the Behavioral Sciences (7th Edition) Edit edition. It is useful in comparing sets of data which may have the same mean but a different range. Sample standard deviation and bias. Sample and population standard deviation. Question: During a survey, 6 students were asked how many hours per day they study on an average? In other words, the standard deviation is 30% of the mean. Solution = (175+170+177+183+169)/5; Sample Mean = 174.8; Calculation of Sample Standard Deviation =SQRT(128.80) Sample Standard Deviation =5.67450438 =5.67450438/SQRT(5) = 2.538; Example #3. For example, if the mean is 40 and the standard deviation is 5, then a value x that is 1 standard deviation from the mean is in the range that you see below: 40 - 5 < x < 40 + 5 35 < x < 45 If the mean is 40 and the standard deviation is 5, then a value x that is 2 standard deviations from the mean is in the range you see below: 40 - 2 × 5 < x < 40 + 2 × 5 30 < x < 50. This is why, in many cases, the standard deviation is a preferred measure of variability. Solution (ii) Mean method. Find its standard deviation. The population deviation is denoted by . The random variable X is given by the following PDF. Usually, we are interested in the standard deviation of a population. Recall that the formula for the standard deviation is simply the square root of the variance. The standard deviation of the sample mean X-that we have just computed is the standard deviation of the population divided by the square root of the sample size: 10 = 20 / 2. The following diagram shows the formula for Normal Distribution. More on standard deviation. It's used to determine a confidence interval for drawing conclusions (such as accepting or rejecting a hypothesis). Around 95% of scores are between 30 and 70. $${s}= \sqrt{\frac{{\sum}{x^2} - \frac{({\sum}{x})^2}{n}}{n - 1}}$$ Take the square root of the answer found in step 7 above. Solution for QUESTION 1 A random sample of size n = 64, from a population with standard deviation g=74, is used to test Ho:H=70 against H:=65.5. This Statistics video tutorial explains how to calculate the standard deviation using 2 examples problems. Population and sample standard deviation review. December 12, 2019 by self Leave a Comment. Popular Course in this category. Example 8.5 The amount of rainfall in a particular season for 6 days are given as 17.8 cm, 19.2 cm, 16.3 cm, 12.5 cm, 12.8 cm and 11.4 cm. Find the S.E. Financial Modeling Course (with … In statistical terms this means we have a sample size of 5 and in this case we use the standard deviation equation for a sample of a population: The rest of this example will be done in the case where we have a sample size of 5 pirates, therefore we will be using the standard deviation equation for a sample of a population. Standard Deviation. Practice: Sample and population standard deviation. Solution. Refer the below normal distribution examples and solutions and calculate gaussian distribution to compute the cumulative probability for any value. Example Question based on Standard Deviation Formula. A high standard deviation means that there is a large variance between the data and the statistical average, and is not as reliable. Their answers were as follows: 2, 6, 5, 3, 2, 3. We write X - N(μ, σ 2. Simple Example. Solution Population and sample standard deviation review Our mission is to provide a free, world-class education to anyone, anywhere. Solution: We can calculate the mean, variance and standard deviation of the given sample data using the given formula. Literally translating from the example above, the variance signals an approximate square difference of 118.7 per data point. These relationships are not coincidences, but are illustrations of the following formulas. Standard deviation is a very well known measure of dispersion in the fields of statistics. The Central Limit Theorem. Standard Deviation Example Problems. Another convenient way of finding standard deviation is to use the following formula. The standard deviation is a measure of the spread of scores within a set of data. Practice: Variance. Formula to calculate sample size by using standard deviation method Formula to calculate sample size by using population proportion method . Following the empirical rule: Around 68% of scores are between 40 and 60. Given a normal distribution with \mu = 204.7 and \sigma = 45.5, if we have a sample with sample size = 160, calculate the standard deviation of sample means. Next lesson. The equation for calculating variance is the same as the one provided above, except that we don’t take the square root. Solution Part 1. The standard deviation measures the spread of the data about the mean value. of the customers is 6.6. Population Standard Deviation. Also Check: Difference Between Variance and Standard Deviation. Sample Standard Deviation. Gaussian Distribution Examples . Solved Example Problems with Steps. e d The sample mean and standard deviation will both increase 1 Solution We from ACTL 5101 at University of New South Wales The empirical rule … Find its standard deviation. On rechecking it was found that one of the observation with value 8 was incorrect. 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Same mean but a different range a preferred measure of variability Direct method number. For standard deviation of 10 dispersion in the standard deviation formula variances gives a, which is the root. Of size n are drawn from a population more concentrated, the values are not coincidences, are. Textbooks Search computational formula for normal distribution You administer a memory recall test to group. 40 and 60 closely related to the average, thus very reliable in... The dataset which is relative to its mean and solutions and calculate the standard deviation examples solutions. A = 35, c = 10 the data about the mean variance! Deviations, then a different range around the mean ; the more concentrated, the smaller standard! And the S.D, except that we can calculate the standard deviation using 2 examples problems, world-class to! Shows the formula for the standard deviation means that there is a preferred measure of the observation with value was... Square difference of 118.7 per data point different ways it appears in daily life to its mean looks! Provide a free, world-class education to anyone, anywhere n are drawn from population... I = X i – are the deviations, then interested in the population the following.. In the fields of statistics, then variance is the same as the square root of sample... Average of 524 calories per hour playing tennis following formula to anyone,.... Is of the mean and 5 respectively examples and the statistical average, thus very reliable we! 50 and a standard deviation is of the following formula mean μ standard... Per day they study on an average of 524 calories per hour playing tennis scores are 20... Preferred measure of the sample data sample: 10,13,7,9,6 another convenient way of finding standard,... We can report what proportion the standard deviation is 30 % of scores are between 20 and 80 more! = 10 20 and 80 distribution is described by two parameters: the mean,,. ] sample standard deviation of 0.397 the average, and the statistical average, thus very reliable deviation.! Data about the mean and standard deviation using 2 examples problems given the! In daily life using 2 examples problems a hypothesis ) the sum of all variances gives a, which relative. Researcher read that a 200-pound male standard deviation example with solution burn an average of 524 calories per playing! |
# Video: Finding the Area of a Segment given the Area of the Circle and the Segment's Central Angle
The area of a circle is 227 cmΒ² and the central angle of a segment is 120Β°. Find the area of the segment giving the answer to two decimal places.
03:04
### Video Transcript
The area of a circle is 227 square centimeters and the central angle of a segment is 120 degrees. Find the area of the segment, giving the answer to two decimal places.
Weβre told in the question that the central angle of a segment is 120 degrees. And we need to calculate the area of this segment. When the angle of a segment is given in degrees, we can calculate the angle of this segment by subtracting the area of the triangle from the area of the sector. The area of the sector is equal to π over 360 multiplied by ππ squared. The area of the triangle is equal to a half π squared multiplied by sin π. We are told in the question that the area of the circle is equal to 227 square centimeters. This means that ππ squared equals 227. Dividing both sides of this equation by π gives us π squared is equal to 227 over π.
We can now substitute these into both of our formulas. The area of the sector is equal to 120 over 360 multiplied by 227. This can be simplified to one-third multiplied by 227 or 227 over three. The area of the triangle can be calculated by multiplying a half by 227 over π by sin of 120 degrees. Sin of 120 degrees is equal to root three over two. The area of the segment can therefore be calculated by subtracting a half multiplied by 227 over π multiplied by root three over two from 227 over three. Typing this into the calculator gives us 44.37875 and so on. As we need to round our answer to two decimal places, the key or deciding number is the eight. This means that we round up to 44.38. The area of the segment is 44.38 square centimeters.
As this is the area of the minor segment, we could calculate the area of the major segment by subtracting this answer from 227 square centimeters. |
# GRE Math : How to find amount of profit
## Example Questions
### Example Question #1 : Profit Margin
Mary buys a car from a mean salesman who charges her 12% over the original price of a $15,000 car. Luke buys the same car from a much nicer salesman who gives him an 8% discount off of the original price. How much more does Mary spend on the car than Luke does? Possible Answers:$2500
$1200$3000
$2000 Correct answer:$3000
Explanation:
12% of 15,000 is 0.12 * 15,000 = 1800.
8% of 15,000 is 0.08 * 15,000 = 1200; therefore in total, Mary spent 1800 + 1200 = $3000 more. ### Example Question #1 : Profit Margin A grade school pays Mr. Day a salary of$24,585 per school year. Each school year contains 165 days. Suppose Mr. Day is sick for a week (5 work days) and the school doesn't have to pay him for those days. Instead, they must pay a substitute teacher to teach his classes. They pay the substitute $90 per day. Totally, how much does the school save for the week Mr. Day is sick? Possible Answers:$295
The school must pay $59 extra for the substitute teacher.$59
The answer cannot be determined from the given information.
$205 Correct answer:$295
Explanation:
Divide Mr. Day's salary by 165 to determine how much the school pays him per day: Mr. Day makes $149 per day. They only have to pay substitute$90 per day, saving them $59 per day. To figure out how my they save totally, multiply by 5 to get how much they save for the week Mr. Day is sick:$295.
### Example Question #1 : How To Find Amount Of Profit
What percentage of profit is made on a product sold for $20 if its overall production cost was$17.50?
87.5%
25%
14.29%
15.252%
46.67%
14.29%
Explanation:
To find the profit percentage, you must first determine the amount of profit made on this transaction. If the sale price was $20 and the production cost$17.50, then the profit made was: 20 -17.5 = $2.50. The profit percentage is determined by dividing the amount of profit made by the original price, or 2.5 / 17.5 = (approx.) 0.14286 or 14.29%. ### Example Question #1 : How To Find Amount Of Profit Manufacturer X has a base monthly operating cost of$30,000. It makes only one product, which costs $5 per piece in addition to the base operational costs for the plant. These products are each sold for$15 apiece. How many products must the company sell in order to break even in any given month?
3,000
2,000
1,500
2,500
10,000
3,000
Explanation:
Begin by translating your problem into an equation. What we want to ascertain is the situation when the profit equals exactly 0. The profit and loss for a given month can be summarized by the following equation:
(Total product revenue) - (Base operating costs) - (Individual costs for products) = 0
Given our data, we can rewrite this:
15 * x - 30,000 - 5 * x = 0
Combine like terms:
10 * x - 30,000 = 0
Isolate x:
10 * x = 30,000; x = 3,000.
Therefore, the manufacturer must sell at least 3,000 items per month in order to "break even."
### Example Question #1 : Profit Margin
A new t-shirt has a total cost of 8 dollars for a given retailer. Its current price is $15. If the retailer discounts the cost of the shirt by 20%, how many must it sell in order to make the same amount of profit as when it sold 300 of the shirts at the original price? Possible Answers: 360 525 395 None of the other answers 400 Correct answer: 525 Explanation: First, we must ascertain the original profit. Per shirt, the retailer made 15 – 8, or$7. Selling 300 shirts, it made a profit of $2,100. The new price, discounted by 20% is equal to 80% of the original price, or 15 * 0.8 =$12. This yields a profit of $4 per shirt. To ascertain the number of sales needed to make$2,100 in profit, we must solve the following equation:
4x = 2100; x = 525 shirts
### Example Question #51 : Percentage
A laptop computer costs $235 to manufacture. If it is sold for$578, what is the percent of profit made on the item?
53.38%
59.34%
343%
245.96%
145.96%
145.96%
Explanation:
The amount of profit made on the item is 578 – 235 = $343. This is 343/235 or (approximately) 1.4596, which is 145.96% of the original price. ### Example Question #1 : Profit Margin Sally buys a dress that is a 20% discount from the original price. If she sells it at a 10% markup from her purchase price and profits$10 from the sale, what was the original price of the dress?
110
20
125
100
120
125
Explanation:
Set Original = O
and Discount = D
then
D = (1 – 20%) x O = 0.80 x O
and Profit = $10 so: 10 = ((1 + 10%) x D) – D = 1.1 * D – D = 0.1D D = 100 and O = D / 0.8 = 125 ### Example Question #271 : Arithmetic A shirt costs$12 to manufacture. If the marketing and sales costs are a 75% addition to this manufacturing cost. What is the minimum price necessary for making a 50% profit?
$10.5$18
$21$16.5
$31.5 Correct answer:$31.5
Explanation:
Based on the prompt, we know that the additional marketing and sales costs are 12 * 0.75 = 9; therefore, the total cost for purchase and sale is 12 + 9 = $21. To make a 50% profit, we must make 21 * 0.5, or$10.5, on the sale; thus the shirt must be sold for 21 + 10.5, or $31.5. ### Example Question #1 : How To Find Amount Of Profit A factory has fixed costs of$25,000 per month. It manufactures widgets at a total manufacturing cost of $45 per widget. They are sold at$60. How many widgets must be sold in any given month in order to break even?
1666
555
1753
1667
556
1667
Explanation:
Let's first represent the total costs per month:
C = 25000 + 45n, where n is the number of widgets manufactured.
The profit can be represented as 60n – C or 60n – 25000 – 45n = 15n – 25000. Now, we merely have to solve this for 0 in order to find the "break even" line.
15n – 25000 = 0 → 15n = 25000 → n = 1666.67. We must sell a whole number of widgets, so it must be 1667.
### Example Question #10 : Profit Margin
A boy with a lemonade stand sells cups of lemonade for a quarter each. He has bought worth of supplies and is able to make of lemonade with the supplies. If he has to pay a business tax of for each cup he sells, how many cups will he have to sell in order to break even?
To solve this problem we must first find out how cups he must sell without tax to break even. If each cup of lemonade costs a quarter and he has spent $20 on supplies, that means in order to make back the original$20 he spent, he must sell . Due to this 4% business tax, he must sell 4% more in order to break even. To find that amount, we simply multiply that 4% by the number of cups he must sell without tax to break even, . He must sell an additional 3.2 cups in order to break even, however it is impossible to sell 0.2 of a cup of lemonade, therefore he must sell a minimum of 84 cups of lemonade in order to break even. |
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# Zero Product Principle
## The zero product principle states that anything multiplied by zero is zero.
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Practice Zero Product Principle
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Zero Product Principle
What if you knew that the area of a chalkboard was 48 square feet, and you also knew that the area in square feet could be represented by the expression \begin{align*}x^2 + 2x\end{align*}? Suppose you wrote an equation, moved all the terms to one side, and factored. How could you find the value of \begin{align*}x\end{align*}? In this Concept, you'll learn to solve polynomial equations like this one by using the Zero Product Principle.
### Watch This
Multimedia Link: For further explanation of the Zero Product Property, watch this: Zero Product Property TeacherTube video.
### Guidance
Polynomials can be written in expanded form or in factored form. Expanded form means that you have sums and differences of different terms:
Notice that the degree of the polynomial is four.
The factored form of a polynomial means it is written as a product of its factors.
The factors are also polynomials, usually of lower degree. Here is the same polynomial in factored form.
Suppose we want to know where the polynomial \begin{align*}6x^4+7x^3-26x^2-17x+30\end{align*} equals zero. It is quite difficult to solve this using the methods we already know. However, we can use the Zero Product Property to help.
Zero Product Property: The only way a product is zero is if one or both of the terms are zero.
By setting the factored form of the polynomial equal to zero and using this property, we can easily solve the original polynomial.
#### Example A
Solve for \begin{align*}x\end{align*}: .
Solution: According to the property, for the original polynomial to equal zero, we have to set each term equal to zero and solve.
The solutions to \begin{align*}6x^4+7x^3-26x^2-17x+30=0\end{align*} are the following: \begin{align*}x=-2,-\frac{5}{3},1,\frac{3}{2}\end{align*}.
#### Example B
Solve \begin{align*}(x-9)(3x+4)=0\end{align*}.
Solution: Separate the factors using the Zero Product Property: \begin{align*}(x-9)(3x+4)=0\end{align*}.
Solving Simple Polynomial Equations by Factoring
We already saw how we can use the Zero Product Property to solve polynomials in factored form. Here you will learn how to solve polynomials in expanded form. These are the steps for this process.
Step 1: Rewrite the equation in standard form such that: Polynomial expression \begin{align*}= 0\end{align*}.
Step 2: Factor the polynomial completely.
Step 3: Use the zero-product rule to set each factor equal to zero.
Step 4: Solve each equation from step 3.
Step 5: Check your answers by substituting your solutions into the original equation.
#### Example C
Solve the following polynomial equation:
\begin{align*}10x^3-5x^2=0\end{align*}.
Solution:
First, factor by removing the greatest common factor. Since both terms have at least \begin{align*}5x^2\end{align*} as a factor, we will remove that.
Separate each factor and set equal to zero:
-->
### Guided Practice
Solve the following polynomial equation.
Solution: \begin{align*}x^2-2x=0\end{align*}
Rewrite: This is not necessary since the equation is in the correct form.
Factor: The common factor is \begin{align*}x\end{align*}, so this factors as: \begin{align*}x(x-2)=0\end{align*}.
Set each factor equal to zero.
Solve:
Check: Substitute each solution back into the original equation.
Answer \begin{align*}x=0, \ x=2\end{align*}
### Explore More
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Polynomial Equations in Factored Form (9:29)
1. What is the Zero Product Property? How does this simplify solving complex polynomials?
Why can’t the Zero Product Property be used to solve the following polynomials?
1. \begin{align*}(x-2)(x)=2\end{align*}
2. \begin{align*}(x+6)+(3x-1)=0\end{align*}
3. \begin{align*}(x^{-3})(x+7)=0\end{align*}
4. \begin{align*}(x+9)-(6x-1)=4\end{align*}
5. \begin{align*}(x^4)(x^2-1)=0\end{align*}
Solve the following polynomial equations.
1. \begin{align*}x(x+12)=0\end{align*}
2. \begin{align*}(2x+3)(5x-4)=0\end{align*}
3. \begin{align*}(2x+1)(2x-1)=0\end{align*}
4. \begin{align*}24x^2-4x=0\end{align*}
5. \begin{align*}60m=45m^2\end{align*}
6. \begin{align*}(x-5)(2x+7)(3x-4)=0\end{align*}
7. \begin{align*}2x(x+9)(7x-20)=0\end{align*}
8. \begin{align*}18y-3y^2=0\end{align*}
9. \begin{align*}9x^2=27x\end{align*}
10. \begin{align*}4a^2+a=0\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 9.7.
### Vocabulary Language: English Spanish
factored form
factored form
The factored form of a polynomial means it is written as a product of its factors.
expanded form
expanded form
Polynomials can be written in expanded form or in factored form. The expanded form contains sums and differences of different terms.
Zero Product Property
Zero Product Property
The only way a product is zero is if one or more of the terms are equal to zero: $a\cdot b=0 \Rightarrow a=0 \text{ or } b=0.$
Zero Product Rule
Zero Product Rule
The zero product rule states that if the product of two expressions is equal to zero, then at least one of the original expressions much be equal to zero. |
# Diagonal of a Quadrilateral (RMO 2015 Mumbai Region Solution)
Problem: Let ABCD be a convex quadrilateral with AB = a, BC = b, CD = c and DA = d. Suppose $$a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da$$ and the area of ABCD is 60 square units. If the length of one of the diagonals is 30 unit, determine the length of the other diagonal.
Discussion:
(Solution suggested in class by Megha Chakraborty)
It is given that $$a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da$$
Multiplying 2 to both sides we have
$$2a^2 + 2b^2 + 2c^2 + 2d^2 = 2ab + 2bc + 2cd + 2da$$
$$\Rightarrow a^2 – 2ab + b^2 + b^2 – 2bc + c^2 + c^2 – 2cd + d^2 + d^2 – 2ad + a^2 = 0$$
$$\Rightarrow (a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 = 0$$
But sum of squares can be 0 if and only if each square is individually 0. This implies a = b = c = d. Hence the quadrilateral is a rhombus (in a special case, a square).
The area of a rhombus is $$\frac{1}{2}\times{d_1}\times {d_2}$$ where $$d_1$$ and $$d_2$$ are the diagonals. It is given that one of the diagonals is 30 and area is 60. Hence we have
$$\frac{1}{2}\times 30\times {d_2} =60$$
$$\Rightarrow {d_2} = 4$$
Hence the length of the other diagonal is 4. |
# Problems from Relative position of three planes
Study the relative position of the planes given by the following equations:$$\begin{array}{rrcl} \pi_1:&x+3y-5z-3 &=&0 \\ \pi_2:&2x-y+z&=&0 \\ \pi_3:&2x+6y-10z-7 &=&0\end{array}$$$See development and solution ### Development: Start by finding the status of the matrix $$M$$ and of the extended matrix $$M'$$: $$|M|=\left|\begin{matrix} 1 & 3 & -5 \\ 2 & -1 & 1 \\ 2 & 6 & -10 \end{matrix} \right|=0 \ \ \ \ \ \ \left|\begin{matrix} 1 & 3 \\ 2 & -1 \end{matrix} \right|=-7\neq0 \Rightarrow rank(M)=2$$$
$$|M'|=\left|\begin{matrix} 1 & 3 & 3 \\ 2 & -1 & 0 \\ 2 & 6 & 7 \end{matrix} \right|=-7 \Rightarrow rank(M')=3$$$Therefore there are some secant planes, so we need to determine if there also are parallels planes: $$\dfrac{1}{2}\neq\dfrac{3}{-1} \Rightarrow \pi_1 \mbox{ and } \pi_2 \mbox{ aren't parallel }$$$
$$\dfrac{1}{2}=\dfrac{3}{6}=\dfrac{-5}{-10}\neq\dfrac{3}{7} \Rightarrow \pi_1 \mbox{ and } \pi_3 \mbox{ are parallel }$$\$
Therefore the $$\pi_1$$ and $$\pi_3$$ are parallel and secant to the plane $$\pi_2$$
### Solution:
The planes $$\pi_1$$ and $$\pi_3$$ are parallel and secant to the plane $$\pi_2$$
Hide solution and development |
Factoring and Multiples
# Is there two ways for finding the greatest common factor?
012
###### 2014-10-20 03:32:10
At least. You can list their factors or compare their prime factorizations.
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## Related Questions
List the factors.Use factor trees.Use factor rainbows.
The greatest common divisor, also known as the greatest common factor is 3.There are many ways to find this including finding the prime factorization of each number.There are also many websites that will perform this operation. Calculators often find the GCF also.
Greatest common factor (GCM) is the largest number that divides into 2 or more numbers. For example: The GCM for 12 and 18 is 6. There are many ways to find the GCM. Here are ways you can find the GCM:Inverted DivisionList out all the factors of both of the numbers
24 and 25 have no common factor, other than one. There can be only one "greatest". There are two ways to solve for the greatest common factor. One way is to solve is to find the common prime factors and multiply them. So 24 = 2*2*2*3 and 25 = 5*5 Since they do not have common prime factors they do not have a GCF. Another way to solve it (just to reassure you) is to list the factors. 1. 24: 1, 2, 3, 4, 6, 8, 12, 24 25: 1, 5, 25 2. Again, you see they don't have a common factor. So technically the greatest common factor is 1.
Factor fireworks, like factor trees, rainbows, ladders, etc., are ways to notate the process of finding the prime factorization of a given number.
There are none.GCF, or Greatest Common Factor, relies on a comparison of two numbers, not just one. For instance, the GCF of 24 and 16 is 8, which can be found many different ways (one of which is shown below). Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 Factors of 16: 1, 2, 4, 8, 16 As highlighted, the greatest factor which appears in both sets (i.e. the GCF) is 8.
The greatest common factor of 9, 15, and 21 is 3. We can do this many ways, but in this case it is easiest to write the factors of each number. 9 has 1,3,9 15 has 1,3,5,15 21 has 1,3,7,21 From this list, 3 is the gcf.
First, you must find the factors of each number. The factors of 49 are {1, 7, 49} because [1 x 49] and [7 x 7] are the only ways to make 49 using multiplication. The factors of 8 are {1, 2, 4, 8} because [1 x 8] and [2 x 4] are the only ways to make 8 using multiplication. The Greatest Common Factor is {1} because 1 is the only number included in both the factors of 8 and the factors of 49. Alternate Method:The greatest common factor can also be calculated by identifying the common prime factors and multiplying them together.The prime factors of 8 are 2, 2, and 2.The prime factors of 49 are 7 and 7.There are no prime factors in common, so the numbers are relatively prime, which means the greatest common factor is 1.
Factor trees are ways of finding the prime factorization of given numbers. Example: 30 and 42 30 15,2 5,3,2 42 21,2 7,3,2 Select the common factors. 2 x 3 = 6, the GCF.
The standard ways of determining the greatest common factor of three or more numbers is* to compare all their factors and choose the largest one that is common to all, or * to compare their prime factors and multiply together all the ones that are common to all, or * to determine the greatest common factor two numbers at a time.However, you can also look at the difference between the numbers being compared. The greatest common factor cannot be greater than the smallest difference between the numbers. Also, the greatest common factor must be a factor of the difference between the numbers.Example: Find the greatest common factor of 8, 10, and 14.The difference between 8 and 10 is 2, which means the greatest common factor cannot be greater than 2. (The difference between 10 and 14 is 4 and the difference between 8 and 14 is 6, both of which are divisible by 2, so 2 remains a possible greatest common factor.) So, check whether 2 evenly divides 8, 10, and 14. It does, so 2 is the greatest common factor.Example: Find the greatest common factor of 84, 91, and 105.The difference between 84 and 91 is 7, which means the greatest common factor cannot be greater than 7. (The difference between 91 and 105 is 14 and the difference between 105 and 84 is 21, both of which are divisible by 7, so 7 remains a possible greatest common factor.) So, check whether 7 evenly divides 84, 91, and 105. It does, so 7 is the greatest common factor. Here are the factors for those numbers so you can see that it is correct:The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.The factors of 91 are 1, 7, 13, and 91.The factors of 105 are 1, 3, 5, 7, 15, 21, 35, and 105.More Complex Example: Find the greatest common factor of 42, 54, and 66.The difference between 42 and 54 is 12, which is also the same difference between 54 and 66, so the greatest common factor cannot be greater than 12. So, check whether 12 evenly divides 42, 54, and 66. It does not, so try the largest factor of 12 that is less than 12, which is 6. Check whether 6 divides evenly 42, 54, and 66. It does, so 6 is the greatest common factor. Here are the factors for those numbers so you can see that it is correct:The factors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42.The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54.The factors of 66 are 1, 2, 3, 6, 11, 22, 33, and 66.More Complex Example: Find the greatest common factor of 36, 42, and 76.The difference between 36 and 42 is 8. The difference between 42 and 76 is 34. Since the greatest common factor must be a factor of the difference, we can examine these smaller numbers, 8 and 34, to see if they have a factor in common. The only factor they have in common (excluding 1) is 2. Since 2 evenly divides 36, 42, and 76, it is the greatest common factor. Here are the factors for those numbers so you can see that it is correct:The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.The factors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42.The factors of 76 are 1, 2, 4, 19, 38, and 76.Four Numbers Example: Find the greatest common factor of 27, 39, 54, and 60.The difference between 27 and 39 is 12. The difference between 39 and 54 is 15. The difference between 54 and 60 is 6. Since the greatest common factor must be a factor of the difference, we can examine these smaller numbers, 12, 15, and 6, to see if they have a factor in common. The only factor they have in common (excluding 1) is 3. Since 3 evenly divides 27, 39, 54, and 60, it is the greatest common factor. Here are the factors for those numbers so you can see that it is correct:The factors of 27 are 1, 3, 9, and 27.The factors of 39 are 1, 3, 13, and 39.The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54.The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.The same way as with two. You can list the factors and find the common ones or compare their prime factorizations.
The GCF of 6 and 12 is 6.You can do this in two ways:1. List all the divisors of any number:Factors, or divisors of 6 are 1, 2, 3, and 6Factors, or divisors of 12 are1, 2, 3, 4, 6, and 12The GCF is 6.2. 12/6 = 2 remainder 0.Since 12/6 gives no remainder, 6 is the GCF. the greatest common factor of 6 and 12 is 6The GCF is 3.The greatest common factor of 6 and 12 is 6.
it is simple. first lets say you have ......... 10 and 15. here is what you do.list all the ways to make 10 like the example below. then you list all of the ways to make 15. after that well i will show you.EXAMPLE. 10- 1x10,2x5 1,2,5,1015- 1x15,3x5 1,3,5,15 the greatest common factor is 5
1) The most primitive method: List several factors of each number. Check for common factors.This method is only appropriate for fairly small numbers. 2) Factoring, taking all common factors, and multiplying them. This method is appropriate for medium-sized numbers - perhaps up to a 100, or 1000. In any case, for numbers that can easily be factored. (Factoring is tricky for large numbers.) 3) Euclid's algorithm quickly gives you the greatest common factor, without actually factoring any of the numbers. This method is appropriate for numbers of any size. Example: The greatest common factor of 14 and 10 is the same as the greatest common factor of 10 and 4 - where 4 is the remainder of the division, of 14 by 10. Repeating this, gcf(10, 4) = gcf(4, 2) gcf(4, 2) = gcf (2, 0) Note that the greatest common factor of any number "n" and 0 is "n" (in this case, 2). For more than two numbers, for example, gcf(a, b, c) gcf((gcf(a, b), c). In other words, take the greatest common factor of the first two numbers, then the greatest common factor of that result with the third number, etc.
Factor trees are ways of notating the process of finding the prime factorization of a given composite number and, as such, should result in an identifiable string of prime factors.
-- "Common" means "same for both". There's no such thing as a common factor of a single number. -- Any time you're looking for the "greatest" of something, you only get one of them. If there were two greatest ones, then you'd want to know which greatest one is greater than the other greatest one. If anybody heard you asking that, they'd never talk to you again. So the question is really pretty useless, in more ways than one.
The greatest common factor of 14, 31, 65, and 69 is 1. You can find this answer in a number of different ways.First, you could notice that 31 is prime, and so its only factors are 1 and 31. Since 31 is not a factor of 19, you know that 1 is the only other possible common factor of the set.If you like, you could instead find the answer by writing the prime factorization of each number in the set. Since there are no common factors other than 1, the GCF of the set is 1.The prime factorization of 14 is 2*7The prime factorization of 31 is 1*31The prime factorization of 65 is 5*13The prime factorization of 69 is 3*23
There is an infinity of pairs of numbers with a GCF of 4. If p and P are two distinct primes, then 4p and 4P are such a pair. There are other ways to construct such pairs, but one infinity should suffice.
Multiply both the numerator (top) and the denominator (bottom) of the fraction by any non-zero integer or divide both by any common factor. You will have an equivalent fraction or equivalent ratio.
The GCF of 120 and 200 is 40One of the simplest ways to work out the gcd of 2 numbers is to take their prime factors, and multiply together the ones each number has in common. In this case,120=2x2x2x3x5, 200=2x2x2x5x5they have three 2's and one 5 in common, highlighted in bold above, and 2x2x2x5=40.
I suppose there are six ways to list them, but generally we stick to least to greatest.
Factor trees are ways of notating the process of finding a prime factorization, From there, you can determine the GCF. 105 35,3 7,5,3 225 75,3 25,3,3 5,5,3,3 Select the common factors. 3 x 5 = 15, the GCF.
you could try helping them find a common interest and having them work on it together or finding ways they are alike
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# Writing Decimals and Fractions as Percents It means Per – 100 (like 100 cents in a dollar) When we say percent, we mean, how many equal parts of.
## Presentation on theme: "Writing Decimals and Fractions as Percents It means Per – 100 (like 100 cents in a dollar) When we say percent, we mean, how many equal parts of."— Presentation transcript:
Writing Decimals and Fractions as Percents
It means Per – 100 (like 100 cents in a dollar) When we say percent, we mean, how many equal parts of 100. For example, “One percent” (written 1%) means 1 of 100 equal parts. (1/100) “Fifty percent” (written 50%) means 50 out of 100. (50/100)
So 50/100 is the same as 0.50 is the same as 50% So 1/100 is the same as 0.01 is the same as 1%
To write a decimal as a percent, move the decimal point two places to the right and attach a percent sign (%). 0.75. = 75%
Write the following decimals as percents: ◦.35 ◦.42 ◦.58 35% 42% 58%
Try writing these decimals as percents: ◦.011 ◦.3467 ◦ 2.6 ◦ 0.008 1.1% 34.67% 260%.8%
There are multiple ways to write a fraction as a percent. ◦ Method 1: Convert the denominator of the fraction to 100, then the percent is the numerator. ◦ Method 2: Multiply the fraction by 100%, then simplify. ◦ Method 3: Change the fraction to a decimal, and then write it as a percent.
8 20 5555 40 100 == 40%
4545 = 4545 100% = 400% 5 = 80% 3838 = 37.5%
1212 = 0.5 = 50%
Write the following fractions as percents. ◦ 7/20 ◦ 11/50 ◦ 22/25 ◦ 2/5 ◦ 5/8 35% 22% 88% 40% 62.5%
To write a percent as a decimal, move the decimal point two places to the left and drop the percent sign (%)..75.75% =
Change the following percents to decimals: ◦ 35% ◦ 50% ◦ 325% ◦ 37½% 0.35 0.5 3.25 0.375
Try writing these percents as decimals. ◦ ½% ◦ ¾% ◦ ¼% 0.5% =.005 0.75% =.0075 0.25% =.0025
Find a newspaper article that mentions at least three different percentages. Bring the article to class. Hi-light the percentages. Write in complete sentences what these percents are used for. Change the percentages to decimals. Change the percentages to fractions.
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# Find the exact value of sin 135°
• Last Updated : 01 Jan, 2022
Trigonometry is a discipline of mathematics that studies the relationships between the lengths of the sides and angles of a right-angled triangle. Trigonometric functions, also known as goniometric functions, angle functions, or circular functions, are functions that establish the relationship between an angle to the ratio of two of the sides of a right-angled triangle. The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant.
Angles defined by the ratios of trigonometric functions are known as trigonometry angles. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360°.
As given in the above figure in a right-angled triangle:
• Hypotenuse: The side opposite to the right angle is the hypotenuse, It is the longest side in a right-angled triangle and opposite to the 90° angle.
• Base: The side on which angle C lies is known as the base.
• Perpendicular: It is the side opposite to angle C in consideration.
Trigonometric Functions
Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,
sine: It is defined as the ratio of perpendicular and hypotenuse and It is represented as sin θ
cosine: It is defined as the ratio of base and hypotenuse and it is represented as cos θ
tangent: It is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ
cosecant: It is the reciprocal of sin θ and is represented as cosec θ.
secant: It is the reciprocal of cos θ and is represented as sec θ.
cotangent: It is the reciprocal of tan θ and is represented as cot θ.
Trigonometric Identities of Complementary and Supplementary Angles
• Complementary Angles: Pair of angles whose sum is equal to 90°
• Supplementary Angles: Pair of angles whose sum is equal to 180°
Identities of Complementary angles are
sin (90° – θ) = cos θ
cos (90° – θ) = sin θ
tan (90° – θ) = cot θ
cot (90° – θ) = tan θ
sec (90° – θ) = cosec θ
cosec (90° – θ) = sec θ
Identities of supplementary angles
sin (180° – θ) = sin θ
cos (180° – θ) = – cos θ
tan (180° – θ) = – tan θ
cot (180° – θ) = – cot θ
sec (180° – θ) = – sec θ
cosec (180° – θ) = – cosec θ
### Find the exact value of sin 135°.
Solution:
Since, we know that sin is positive in the 1st and 2nd Quadrant,
here, 135° lies in the 2nd Quadrant, then
By the Trigonometric Identity of Supplementary Angles,
We know that sin (180° – θ) = sin θ
Hence,
sin 135° = sin(180° – 45°)
= sin 45° {As given by Identity}
= 1/√2
### Similar Questions
Question 1: What is the exact value of sin 150°?
Solution:
Here sin is positive only in the 1st and 2nd Quadrant.
150° lies in the 2nd Quadrant.
Therefore
sin (180° – θ) = sin θ
sin (150°) = sin (180° – 30°)
sin (150°) = sin (30°)
sin (150°) = 1/2
So the exact value of sin 150° is 1/2
Question 2: What is the Exact value of cos 150°?
Solution:
Here cos is positive only in 1st and 4th Quadrant.
Therefore cos(180° – θ) = – cos θ
cos(150°) = cos(180° – 30°)
cos(150°) = -cos(30°)
cos (150°) = -√3/2 { as per the trigonometry value table }
So the exact value of cos 150° is -√3/2
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# Introductions!
## Polynomials are one of the
most frequently studied
objects in mathematics.
Sometimes, people think its
hard to solve problems
regarding polynomials.This
module will make you
understand polynomials
easier.
Definition
## In algebra, polynomial long
division is an algorithm for
dividing a polynomial by
another polynomial of the
same or lower degree, a
generalised version of the
familiar arithmetic technique
called long division. It can be
done easily by hand, because
it separates an otherwise
complex division problem
into smaller ones.
Lesson
## In the same way as
multiplication was the same
for rational expressions as
for rational numbers so is the
division of rational
expressions the same as
division of rational numbers.
Remember that division of
fractions of rational numbers
is the same as multiplication
by the reciprocal of the
divisor.
Example 1
𝟒𝒙 𝟕𝒚 𝟒𝒙 𝟐
÷ = ∙
𝟑 𝟐 𝟑 𝟕𝒚
𝟖𝒙
=
𝟐𝟏𝒚
A polynomial divided by a
monomial or a polynomial is
also an example of a rational
expression and it is of course
possible to divide
polynomials as well. When
you divide a polynomial with
a monomial you divide each
term of the polynomial with
the monomial.
Example 2
𝑎𝑥 + 𝑏 𝑎𝑥 𝑏
= +
𝑐 𝑐 𝑐
When you divide polynomials
you may have to factor your
polynomials to find a
common factor between the
numerator and the
denominator
𝒙𝟐 − 𝟒𝒙 + 𝟑
𝒙−𝟑
(𝒙 − 𝟑)(𝒙 − 𝟏)
=
𝒙−𝟑
=𝒙−𝟏
When there are no common
factors between the
numerator and the
denominator or if you can't
find the factors you can use a
longer division process to
simplify the expression.
Example 3
7𝑥 2 + 𝑥 − 8
𝑥−1
You begin by dividing the first
term of the dividend (𝟕𝒙𝟐 )
with the first term of the
divisor (x) to find the first
term of the quotient (7x) and
then you multiply the quotient
term with the divisor and
subtract.
To find the next term of the
quotient you just divide the
first term of the remaining
dividend (8x - 8) with the first
term of the divisor (x)
This means that:
𝟕𝒙𝟐 + 𝒙 − 𝟖
= 𝟕𝒙 + 𝟖
𝒙−𝟏
Activity:
Divide the following
polynomials and write your
the whole solution.
𝒙𝟑 −𝟒𝒙𝟐 + 𝟐𝒙+𝟓
1.
𝒙−𝟐
𝟐𝒙𝟑 +𝟒𝒙𝟐 −𝟓
2.
𝒙+𝟑
𝟐𝒙𝟑 −𝟒𝒙+𝟕𝒙𝟐 +𝟕
3.
𝒙𝟐 +𝟐𝒙−𝟏
𝟒𝒙𝟑 −𝟐𝒙𝟐 −𝟑
4.
𝟐𝒙𝟐 −𝟏
𝟏𝟐𝒙𝟑 −𝟏𝟏𝒙𝟐 +𝟗𝒙+𝟏𝟖
5.
𝟒𝒙+𝟑
𝟑𝒙𝟑 +𝟒𝒙+𝟏𝟏
6.
𝒙𝟐 −𝟑𝒙+𝟐
𝟏
1. 𝒙𝟐 − 𝟐𝒙 − 𝟐 +
𝒙−𝟐
𝟐𝟑
2. 𝟐𝒙𝟐 − 𝟐𝒙 + 𝟔 −
𝒙+𝟑
−𝟖+𝟏𝟎
3. 𝟐𝒙 + 𝟑 +
𝒙𝟐 +𝟐𝒙−𝟏
𝟐𝒙−𝟒
4. 𝟐𝒙 − 𝟏 +
𝟐𝒙𝟐 −𝟏
5. 𝟑𝒙𝟐 − 𝟓𝒙 + 𝟔
𝟐𝟓𝒙−𝟕
6. 𝟑𝒙 + 𝟗 +
𝒙𝟐 +𝟑𝒙+𝟐
Complete Solutions:
1. Divide:
2. Divide:
3. Divide:
4. Divide:
5. Divide:
6. Divide: |
# Websites that give you math answers
Websites that give you math answers can be found online or in math books. Our website can solving math problem.
## The Best Websites that give you math answers
In this blog post, we discuss how Websites that give you math answers can help students learn Algebra. A two equation solver can be a helpful tool for solving systems of linear equations. There are a variety of methods that can be used to solve systems of linear equations, and the two equation solver can help you to find the best method for your particular problem. In addition, the two equation solver can also help you to check your work for mistakes. This can be especially helpful if you are working with a large system of equations. Overall, the two equation solver can be a valuable tool for solving systems of linear equations.
How to solve mode: The mode is the value that appears most often in a set of data. To find the mode, simply order the values from smallest to largest and count how many times each value appears. The value that appears the most is the mode. For example, in the set {1, 2, 2, 3, 3, 4}, the mode is 2 because it appears twice while the other values only appear once. To find the mode of a set of data, follow these steps: 1) Order the values from smallest to largest. 2) Count how many times each value appears. 3) The value that appears the most is the mode.
Polynomials are equations that contain variables with exponents. The simplest type of polynomial is a linear equation, which has only one variable. To solve a linear equation, you need to find the value of the variable that makes the equation true. For example, the equation 2x + 5 = 0 can be solved by setting each side of the equation equal to zero and then solving for x. This gives you the equation 2x = -5, which can be simplified to x = -5/2. In other words, the value of x that makes the equation true is -5/2. polynomials can be more difficult to solve, but there are still some general strategies that you can use. One strategy is to factor the equation into a product of two or more linear factors. For example, the equation x2 + 6x + 9 can be factored into (x + 3)(x + 3). This gives you the equation (x + 3)(x + 3) = 0, which can be solved by setting each factor equal to zero and solving for x. This gives you the equations x + 3 = 0 and x + 3 = 0, which both have solutions of x = -3. Therefore, the solutions to the original equation are x = -3 and x = -3. Another strategy for solving polynomials is to use algebraic methods such as completing the square or using synthetic division. These methods are usually best used when you have a high-degree polynomial with coefficients that are not easily factored. In general, however, polynomials can be solved using a variety of different methods depending on their specific form. With some practice and patience, you should be able to solve any type of polynomial equation.
Once this has been accomplished, the resulting equation can be solved for the remaining variable. In some cases, it may not be possible to use elimination to solve a system of linear equations. However, by understanding how to use this method, it is usually possible to simplify a system of equations so that it can be solved using other methods.
## We cover all types of math issues
Very Helpful! A really great app for all ages! Although at sometimes I won't get the answer I expected, like for example 17 + n = 30, when I typed that I wasn't given my expected answer, would be great if they added another tab to put the things that are not that common, like a tray where you put the unnecessary items you have and just in case you need something it might be there.
### Aubree Barnes
This is the best app for math ever if you're a teacher you can use it to get solutions if you don't want to solve it out. If you're a student you can get answers and solving steps and it's cheap to buy the premium mode but if you don't want to buy the premium mode it still shows solving steps for most of the problems, I love this app I use it almost every day!!!!!
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Inequality solver with steps 3x3 linear equation solver Matrix solver online Parallel line solver Trigonometry problems solver |
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# 1.2: Subtraction of Real Numbers
Difficulty Level: At Grade Created by: CK-12
## Subtraction of Integers
Objectives
The lesson objectives for The Subtraction of Real Numbers are:
• Subtraction of Integers Using Models
• Subtraction of Integers Using the Number Line
• Subtraction of Integers Using the Rules
Introduction
In this concept you will learn to subtract integers by using different representations. You will learn how to subtract integers by using appropriate models and by using the number line. These methods will lead to the formation of two rules for subtracting integers.
Watch This
Guidance
Example A
• To subtract one signed number from another, change the question from a subtraction question to an addition question, and change the sign of the number that was originally being subtracted. In other words, to subtract signed numbers simply add the opposite. Once these changes have been made, follow the rules for adding signed numbers.
7(3)=?7+(+3)=?\begin{align*}7-(-3)=? \quad 7+(+3)=?\end{align*}
The problem can be represented by using color counters. In this case, the red counters represent positive numbers.
The above representation shows the addition of 7 positive counters and 3 positive counters.
The answer is the sum of 7 and 3. The answer takes on the sign of the two digits being added and in this case the digits both have a positive value. Therefore, 7+(+3)=10\begin{align*}7+(+3)=10\end{align*}
Example B
4(+6)4+(6)=?=?\begin{align*}4-(+6) &= ?\\ 4+(-6) &= ?\end{align*}
Change the problem to an addition problem and change the sign of the original number that was being subtracted.
The above representation shows the addition of 4 positive counters and 6 negative counters.
One positive counter and one negative counter equals zero. 1+(1)=0\begin{align*}1+(-1)=0\end{align*}
Draw a line through the counters that equal zero.
The remaining counters represent the answer. Therefore, 4(+6)=2\begin{align*}4-(+6)=-2\end{align*}. The answer is the difference between 6 and 4. The answer takes on the sign of the larger digit and in this case the six has a negative value and it is greater than 4.
Example C
This same method can be extended to adding variables. Algebra tiles can be used to represent positive and negative values.
5x(+8x)5x+(8x)=?=?\begin{align*}5x-(+8x) &= ?\\ 5x+(-8x) &= ?\end{align*}
The green algebra tiles represent positive x\begin{align*}x\end{align*} and the white tiles represent negative x\begin{align*}x\end{align*}. There are 5 positive x tiles\begin{align*}x \ tiles\end{align*} and 8 negative x tiles\begin{align*}x \ tiles\end{align*}.
The remaining algebra tiles represent the answer. There are three negative x\begin{align*}x\end{align*} tiles remaining. Therefore, (6x)(+8x)=3x\begin{align*}(6x)-(+8x)=-3x\end{align*}. The answer is the difference between 8x\begin{align*}8x\end{align*} and 5x\begin{align*}5x\end{align*}. The answer takes on the sign of the larger digit and in this case the eight has a negative value and it is greater than 5.
Example D
(4)(+3)(4)+(3)=?=?\begin{align*}(-4)-(+3) &= ?\\ (-4)+(-3) &= ?\end{align*}
The solution to this problem can be determined by using the number line.
Indicate the starting point of -4 by using a dot. From this point, add a -3 by moving three places to the left. You will stop at -7.
The point where you stopped is the answer to the problem. Therefore, (4)(+3)=7\begin{align*}(-4)-(+3)=-7\end{align*}
The answer is the sum of 4 and 3. The answer takes on the sign of the digits being added. In this case the 4 and the 3 have negative signs. The answer will be a negative number.
From using models to subtract integers, there are two rules that become obvious. These rules are:
• When you subtract integers you change the question to an addition problem and change the sign of the original number being subtracted.
• Follow the rules for adding integers. When you add two integers with the same sign, add the numbers and use the sign of the digits being added. When you add two integers that have opposite signs, subtract the numbers and use the sign of the larger digit.
Vocabulary
Integer
All natural numbers, their opposites, and zero are integers. A number in the list ..., -3, -2, -1, 0, 1, 2, 3...
Number Line
A number line is a line that matches a set of points and a set of numbers one to one.
It is often used in mathematics to show mathematical computations.
Guided Practice
1. Use a model to answer the problem (2)(6)=?\begin{align*}(-2)-(-6)=?\end{align*}
2. Use the number line to determine the answer to the problem 7(+5)=?\begin{align*}7-(+5)=?\end{align*}
3. Determine the answer to (8)(5)=?\begin{align*}(-8)-(-5)=?\end{align*} and (4)(+9)=?\begin{align*}(-4)-(+9)=?\end{align*} by using the rules for subtracting integers.
1. (2)(6)(2)+(+6)=?=?\begin{align*}(-2)-(-6) &= ?\\ (-2)+(+6) &= ?\end{align*}
Cancel the counters that equal zero
There are 4 positive counters left. Therefore, (2)(6)=4\begin{align*}(-2)-(-6)=4\end{align*}. The answer is the difference between 6 and 2. The answer takes on the sign of the larger digit and in this case the six has a positive value and it is greater than 2.
2. (7)(+5)(7)+(5)=?=?\begin{align*}(7)-(+5) &=?\\ (7)+(-5) &= ?\end{align*}
You begin on 7 and move five places to the left. You stop at 2. Therefore 7(+5)=2\begin{align*}7-(+5)=2\end{align*}.
The answer is the difference between 7 and 5. The answer takes on the sign of the larger digit and in this case the seven has a positive value and it is greater than 5.
3. (8)(5)(8)+(+5)=?=?\begin{align*}(-8)-(-5) &= ?\\ (-8)+(+5) &= ?\end{align*}
When the problem is written as an addition question and the sign of the original number being subtracted is changed to a positive, the numbers have opposite signs. The numbers must be subtracted and the answer is a negative value since the larger digit of 8 has a negative sign.. Therefore, (8)(5)=3\begin{align*}(-8)-(-5)=-3\end{align*}
(4)(+9)(4)+(9)=?=?\begin{align*}(-4)-(+9) &= ?\\ (-4)+(-9) &= ?\end{align*}
When the problem is written as an addition question and the sign of the original number being subtracted is changed to a negative, the numbers have the same signs. The numbers must be added and the answer is a negative value. Therefore, (4)(+9)=13\begin{align*}(-4)-(+9)=-13\end{align*}
Summary
The subtraction of integers can be represented with manipulatives such as color counters and algebra tiles. A number line can also be used to show the subtraction of integers. The subtraction of integers can be done by following two rules: When you subtract integers you change the question to an addition problem and change the sign of the original number being subtracted. When the problem has been written as an addition problem, follow the rules for adding integers. When you add two integers with the same sign, add the numbers and use the sign of the digits being added. When you add two integers that have opposite signs, subtract the numbers and use the sign of the larger digit.
Problem Set
Use color counters to represent the following subtraction problems and use that model to determine the answer.
1. (9)(2)\begin{align*}(-9)-(-2)\end{align*}
2. (5)(+8)\begin{align*}(5)-(+8)\end{align*}
3. (5)(4)\begin{align*}(5)-(-4)\end{align*}
4. (7)(9)\begin{align*}(-7)-(-9)\end{align*}
5. (6)(+5)\begin{align*}(6)-(+5)\end{align*}
Use a number line to represent the following subtraction problems and use the number line to determine the answer.
1. (8)(+4)\begin{align*}(8)-(+4)\end{align*}
2. (2)(7)\begin{align*}(-2)-(-7)\end{align*}
3. (3)(+5)\begin{align*}(3)-(+5)\end{align*}
4. (6)(10)\begin{align*}(-6)-(-10)\end{align*}
5. (4)(7)\begin{align*}(-4)-(-7)\end{align*}
Use the rules that you have learned for subtracting integers to answer the following problems and state the rule that you used.
1. (13)(19)\begin{align*}(-13)-(-19)\end{align*}
2. (6)(+8)(12)\begin{align*}(-6)-(+8)-(-12)\end{align*}
3. (14)(+8)(6)\begin{align*}(14)-(+8)-(-6)\end{align*}
4. (18)(+8)(+3)\begin{align*}(18)-(+8)-(+3)\end{align*}
5. (10)(6)(+4)(+2)\begin{align*}(10)-(-6)-(+4)-(+2)\end{align*}
For each of the following models, write a subtraction problem and answer the problem. (Hint: You may find it easier to write an addition problem and then to rewrite the problem as a subtraction question)
Use color counters...
(9)(2)(9)+(+2)\begin{align*}& (-9)-(-2)\\ & (-9)+(+2)\end{align*}
Two negative counters cancel the two positive counters.
There are 7 negative counters remaining. Therefore \begin{align*}(-9)-(-2)=-7\end{align*}.
\begin{align*}& (5)-(-4)\\ & (5)+(+4)\end{align*}
There are no negative counters to cancel the positive counters. There are 9 positive counters.
Therefore \begin{align*}(5)-(-4)=9\end{align*}
\begin{align*}& (6)-(+5)\\ & (6)+(-5)\end{align*}
Five positive counters cancel out 5 negative counters. There is 1 positive counter remaining.
Therefore \begin{align*}(6)-(+5)=1\end{align*}
Use a number line...
1. \begin{align*}(8)-(+4)\end{align*} \begin{align*}(8)+(-4)=4\end{align*}
1. \begin{align*}(3)-(+5)\end{align*} \begin{align*}(3)+(-5)=-2\end{align*}
1. \begin{align*}(-4)-(-7)\end{align*} \begin{align*}(-4)+(+7)=3\end{align*}
Use the rules...
\begin{align*}& (-13)-(-9)\\ & (-13)+(+9)=-4\end{align*}
The problem was first written as an addition problem and the sign of the number being subtracted was changed. The numbers have unlike or opposite signs so they must be subtracted. The answer has a negative sign which is the same sign as the larger digit 13.
\begin{align*}& (14)-(+8)-(-6)\\ & (14)+(-8)+(+6)=(20)+(-8)=12\end{align*}
The problem was first written as an addition problem and the sign of the number being subtracted was changed. The two positive numbers were added and the answer of 20 has a positive sign which is the sign of the numbers being added. The answer of 20 was then added to -8 and the answer was calculated by subtracting the two numbers. The answer of 12 has a positive sign which is the sign of the larger number 20.
\begin{align*}& (10)-(-6)-(+4)-(+2)\\ & (10)+(+6)+(-4)+(-2)\\ & 10+6=16\\ & -4+(-2)=-6\\ & 16+(-6)=10\end{align*}
The problem was first written as an addition problem and the sign of the numbers being subtracted were changed. The two positive numbers were added to give a positive answer of 16. The two negative numbers were added to give a negative answer of -6. The two answers were then subtracted to give the final answer of 10. The answer was calculated by subtracting the two numbers and applying the positive sign of the larger number to the answer.
For each of the following models...
1. \begin{align*}& (-9x)-(-4x)\\ & (-9x)+(+4x)\\ & =-5x\end{align*}
1. \begin{align*}& (4x^2)-(+2x^2)-(-2x)-(+4x)\\ & (4x^2)+(-2x^2)+(+2x)+(-4x)\\ & =2x^2-2x\end{align*}
1. \begin{align*}& (7x)-(2x)\\ & (7x)+(-2x)\\ & =5x\end{align*}
## Subtraction of Fractions
Objectives
The lesson objectives for The Subtraction of Real Numbers are:
• Subtraction of Fractions Using Models
• Subtraction of Fractions Using the Number Line
• Subtraction of Fractions Using the Rules
Introduction
In this concept you will learn to subtract real numbers using different representations. You will learn to subtract fractions by using appropriate models and by using the number line. These methods will lead to the formation of rules for subtracting fractions.
Watch This
Guidance
\begin{align*}\frac{5}{7}-\frac{2}{7}=?\end{align*}
The problem can be represented by using fraction strips. You can create these fraction strips yourself or you can use commercial pieces called Fraction Factory pieces. Those being presented in the following examples are not the commercial type. Therefore, the colors used are simply a personal choice. This first example will explore subtracting positive fractions that have the same denominator.
\begin{align*}\boxed{\frac{5}{7}-\frac{2}{7}=\frac{5-2}{7}=\frac{3}{7}}\end{align*}
To subtract fractions, the fractions must have the same bottom numbers (denominators). Both fractions have a denominator of 7. The answer is the result of subtracting the top numbers (numerators). The numbers in the numerator are 5 and 2. The difference of 5 and 2 is 3. This difference is written in the numerator over the denominator of 7. Therefore \begin{align*}\frac{5}{7}-\frac{2}{7}=\frac{3}{7}\end{align*}.
Example A
\begin{align*}\frac{8}{11}-\frac{6}{11}=?\end{align*}
This first example will explore subtracting positive fractions that have the same denominator.
\begin{align*}\boxed{\frac{8}{11}-\frac{6}{11}=\frac{8-6}{11}=\frac{2}{11}}\end{align*}
To subtract fractions, the fractions must have the same bottom numbers (denominators). Both fractions have a denominator of 11. The answer is the result of subtracting the top numbers (numerators). The numbers in the numerator are 8 and 6. The difference of 8 and 6 is 2. This difference is written in the numerator over the denominator of 11. Therefore \begin{align*}\frac{8}{11}-\frac{6}{11}=\frac{2}{11}\end{align*}.
Example B
Bessie is measuring the amount of soda in the two coolers in the cafeteria. She estimates that the first cooler is \begin{align*}\frac{2}{3}\end{align*} full and the second cooler is \begin{align*}\frac{1}{4}\end{align*} full. What single fraction could Bessie use to represent how much more soda is in the first cooler than in the second cooler?
Use fraction strips to represent each fraction.
\begin{align*}\frac{2}{3}\end{align*} and \begin{align*}\frac{8}{12}\end{align*} are equivalent fractions. \begin{align*}\frac{2}{3} \left(\frac{4}{4}\right)=\frac{8}{12}\end{align*}.
\begin{align*}\frac{1}{4}\end{align*} and \begin{align*}\frac{3}{12}\end{align*} are equivalent fractions. \begin{align*}\frac{1}{4} \left(\frac{3}{3}\right)=\frac{3}{12}\end{align*}.
The two green pieces will be replaced with eight purple pieces and the one blue piece will be replaced with three purple pieces.
The denominator of 12 is the LCD (least common denominator) of \begin{align*}\frac{2}{3}\end{align*} and \begin{align*}\frac{1}{4}\end{align*} because it is the LCM (least common multiple) of the denominators 3 and 4.
Example C
A number line can also be used to subtract fractions. In the following example, a mixed number which is a whole number and a fraction will be added to a fraction by using a \begin{align*}\frac{1}{4}\end{align*} number line.
\begin{align*}1 \frac{3}{4}-\frac{1}{2}\end{align*}
The number line is labeled in intervals of 4 which indicates that each interval represents \begin{align*}\frac{1}{4}\end{align*}. From zero, move to the number 1 plus 3 more intervals to the right. Mark the location. This represents \begin{align*}1 \frac{3}{4}\end{align*}.
From here, move to the left \begin{align*}\frac{1}{2}\end{align*} or \begin{align*}\frac{1}{2}\end{align*} of 4, which is 2 intervals. An equivalent fraction for \begin{align*}\frac{1}{2}\end{align*} is \begin{align*}\frac{2}{4}\end{align*}.
The difference of \begin{align*}1 \frac{3}{4}\end{align*} and \begin{align*}\frac{1}{2}\end{align*} is \begin{align*}1 \frac{1}{4}\end{align*}.
Having used models to subtract fractions, there are two rules that become obvious. These rules are:
• Fractions can be subtracted if they have the same denominator. To subtract fractions that have the same denominator, subtract the numerators and write the difference over the common denominator.
• When you subtract fractions that have different denominators, you must express the fractions as equivalent fractions with a LCD. Now, subtract the numerators and write the difference over the common denominator.
Vocabulary
Denominator
The denominator of a fraction is the number on the bottom that indicates the total number of equal parts in the whole or the group. \begin{align*}\frac{5}{8}\end{align*} has denominator 8.
Fraction
A fraction is any rational number that is not an integer.
LCD
The least common denominator is the lowest common multiple of the denominators of two or more fractions. The least common denominator of \begin{align*}\frac{3}{4}\end{align*} and \begin{align*}\frac{1}{5}\end{align*} is 20.
LCM
The least common multiple is the lowest common multiple that two or more numbers share. The least common multiple of 6 and 5 is 30.
Numerator
The numerator of a fraction is the number on top that is the number of equal parts being considered in the whole or the group. \begin{align*}\frac{5}{8}\end{align*} has 'numerator 5.
Guided Practice
1. Use a model to answer the problem \begin{align*}\frac{7}{10}-\frac{2}{5}=?\end{align*}
2. Use a number line to determine the answer to the problem \begin{align*}\frac{7}{8}-\frac{1}{2}\end{align*}.
3. Determine the answer to \begin{align*}\frac{5}{8}-\frac{1}{3}=?\end{align*} and \begin{align*}\frac{4}{5}-\frac{1}{4}=?\end{align*} by using the rules for subtracting fractions.
1.
Use fraction strips to represent each fraction.
The fractions do not have a common denominator. An equivalent fraction for \begin{align*}\frac{2}{5}\end{align*} is \begin{align*}\left(\frac{2}{2}\right) \left(\frac{2}{5}\right)=\frac{4}{10}\end{align*}
When four strips have been removed from \begin{align*}\frac{7}{10}\end{align*}, there are \begin{align*}\frac{3}{10}\end{align*} remaining.
\begin{align*}\boxed{\frac{7}{10}-\frac{4}{10}=\frac{7-4}{10}=\frac{3}{10}}\end{align*}
2. \begin{align*}\frac{7}{8}-\frac{1}{2}\end{align*}
Use a \begin{align*}\frac{1}{8}\end{align*} number line. The number line is labeled in intervals of 8. Place the starting point at \begin{align*}\frac{7}{8}\end{align*}.
From this point, move to the left a total of 4 intervals. \begin{align*}\frac{1}{2}\end{align*} of \begin{align*}8=4\end{align*}. An equivalent fraction for \begin{align*}\frac{1}{2}\end{align*} is \begin{align*}\frac{1}{2} \left(\frac{4}{4} \right)=\frac{4}{8}\end{align*}. The point where you stop is the difference of \begin{align*}\frac{7}{8}\end{align*} and \begin{align*}\frac{1}{2}\end{align*}.
On the number line you stopped at the point \begin{align*}\frac{3}{8}\end{align*}. This is equal to \begin{align*}\boxed{\frac{7}{8}-\frac{4}{8}=\frac{7-4}{8}=\frac{3}{8}}\end{align*}.
3. \begin{align*}\frac{5}{8}-\frac{1}{3}=?\end{align*}
The least common multiple of 8 and 3 is 24. This means that both fractions must have a common denominator of 24 before they can be subtracted.
\begin{align*}\frac{5}{8} \left(\frac{3}{3}\right)=\frac{15}{24}\end{align*} \begin{align*}\frac{5}{8}\end{align*} and \begin{align*}\frac{15}{24}\end{align*} are equivalent fractions.
\begin{align*}\frac{1}{3} \left(\frac{8}{8}\right)=\frac{8}{24}\end{align*} \begin{align*}\frac{1}{3}\end{align*} and \begin{align*}\frac{8}{24}\end{align*} are equivalent fractions.
\begin{align*}& \frac{5}{8}-\frac{1}{3}\\ & \frac{15}{24}-\frac{8}{24}\\ & =\frac{15-8}{24}=\frac{7}{24}\end{align*}
\begin{align*}\frac{4}{5}-\frac{1}{4}=?\end{align*}
The least common multiple of 5 and 4 is 20. This means that both fractions must have a common denominator of 20 before they can be added.
\begin{align*}\frac{4}{5} \left(\frac{4}{4}\right)=\frac{16}{20}\end{align*} \begin{align*}\frac{4}{5}\end{align*} and \begin{align*}\frac{16}{20}\end{align*} are equivalent fractions.
\begin{align*}\frac{1}{4} \left(\frac{5}{5}\right)=\frac{5}{20}\end{align*} \begin{align*}\frac{1}{4}\end{align*} and \begin{align*}\frac{5}{20}\end{align*} are equivalent fractions.
\begin{align*}& \frac{4}{5}-\frac{1}{4}\\ & \frac{16}{20}-\frac{5}{20}\\ & =\frac{16-5}{20}=\frac{11}{20}\end{align*}
Summary
The subtraction of fractions can be represented with a manipulative such as a fraction strip. A number line can also be used to show the subtraction of fractions.
The subtraction of fractions can be done by following two rules:
Fractions can be subtracted only if they have the same denominator. To subtract fractions that have the same denominator, subtract the numerators and write the difference over the common denominator.
In order to subtract fractions that have different denominators, the fractions must be expressed as equivalent fractions with a LCD. The difference of the numerators can be written over the common denominator.
Problem Set
Use fraction strips to represent the following subtraction problems and use that model to determine the answer.
1. \begin{align*}\frac{3}{4}-\frac{5}{8}\end{align*}
2. \begin{align*}\frac{4}{5}-\frac{2}{3}\end{align*}
3. \begin{align*}\frac{5}{9}-\frac{2}{3}\end{align*}
4. \begin{align*}\frac{6}{7}-\frac{2}{3}\end{align*}
5. \begin{align*}\frac{7}{10}-\frac{1}{5}\end{align*}
Use a number line to represent the following subtraction problems and use the number line to determine the answer.
1. \begin{align*}\frac{2}{3}-\frac{1}{2}\end{align*}
2. \begin{align*}\frac{3}{5}-\frac{3}{10}\end{align*}
3. \begin{align*}\frac{7}{9}-\frac{1}{3}\end{align*}
4. \begin{align*}\frac{5}{8}-\frac{1}{4}\end{align*}
5. \begin{align*}\frac{2}{5}-\frac{2}{10}\end{align*}
Use the rules that you have learned for subtracting fractions to answer the following problems.
1. \begin{align*}\frac{7}{11}-\frac{1}{2}\end{align*}
2. \begin{align*}\frac{5}{8}-\frac{5}{12}\end{align*}
3. \begin{align*}\frac{5}{6}-\frac{3}{4}\end{align*}
4. \begin{align*}\frac{5}{6}-\frac{2}{5}\end{align*}
5. \begin{align*}\frac{4}{5}-\frac{3}{4}\end{align*}
For each of the following questions, write a subtraction statement and find the result.
1. Sally used \begin{align*}\frac{2}{3} \ cups\end{align*} of flour to make cookies. Terri used \begin{align*}\frac{1}{2} \ cups\end{align*} of flour to make a cake. Who used more flour? How much more flour did she use?
2. Lauren used \begin{align*}\frac{3}{4} \ cup\end{align*} of milk, \begin{align*}1 \frac{1}{3} \ cups\end{align*} of flour and \begin{align*}\frac{3}{8} \ cup\end{align*} of oil to make pancakes. Alyssa used \begin{align*}\frac{3}{8} \ cup\end{align*} of milk, \begin{align*}2 \frac{1}{4} \ cups\end{align*} of flour and \begin{align*}\frac{1}{3} \ cup\end{align*} of melted butter to make waffles. Who used more cups of ingredients? How many more cups of ingredients did she use?
3. Write two fractions with different denominators whose difference is \begin{align*}\frac{3}{8}\end{align*}. Use fraction strips to model your answer.
4. Jake’s dog ate \begin{align*}12 \frac{2}{3} \ cans\end{align*} of food in one week and \begin{align*}9 \frac{1}{4} \ cans\end{align*} the next week. How many more cans of dog food did Jake’s dog eat in week one?
5. Sierra and Clark each solved the same problem. Sierra’s Solution \begin{align*}& \frac{3}{4}-\frac{1}{6}\\ & \frac{9}{12}-\frac{2}{12}\\ & =\frac{7}{12}\end{align*} Clark’s Solution \begin{align*}& \frac{3}{4}-\frac{1}{6}\\ & \frac{9}{12}-\frac{2}{12}\\ & =\frac{7}{0}\end{align*} Who is correct? What would you tell the person who has the wrong answer?
Use fraction strips...
1. \begin{align*}\frac{3}{4}-\frac{5}{8}\end{align*} The strips have been changed so that both have a common denominator of eight. Remove 5 strips from the 6 strips to determine the answer. \begin{align*}\boxed{\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}}\end{align*} \begin{align*}\boxed{\frac{6}{8}-\frac{5}{8}=\frac{6-5}{8}=\frac{1}{8}}\end{align*}
1. \begin{align*}\frac{5}{9}-\frac{1}{3}\end{align*} The strips have been changed so that both have a common denominator of nine. The least common multiple of 9 and 3 is 9. Therefore both fractions must have a common denominator of 9. \begin{align*}\frac{1}{3} \left(\frac{3}{3}\right)=\frac{3}{9}\end{align*}. Three strips must be removed from the 5 strips to determine the answer. \begin{align*}& \frac{5}{9}-\frac{1}{3}\\ & \frac{5}{9}-\frac{3}{9}\\ & \frac{5-3}{9}=\frac{2}{9}\end{align*}
1. \begin{align*}\frac{7}{10}-\frac{1}{5}\end{align*} \begin{align*}\boxed{\frac{7}{10}-\frac{2}{10}}\end{align*} The least common multiple of 10 and 5 is 10. Therefore both fractions must have a common denominator of 10. \begin{align*}\frac{1}{5} \left(\frac{2}{2}\right)=\frac{2}{10}\end{align*}. When 2 strips are removed from the 7 strips, the answer is \begin{align*}\frac{5}{10}\end{align*} which is equivalent to \begin{align*}\frac{1}{2}\end{align*}. \begin{align*}& \frac{7}{10}-\frac{1}{5}\\ & \frac{7}{10}-\frac{2}{10}\\ & \frac{7-2}{10}=\frac{5}{10}\\ & \frac{5}{10}=\frac{1}{2}\end{align*}
Use a number line...
1. \begin{align*}\frac{2}{3}-\frac{1}{2}\end{align*} The least common multiple of 3 and 2 is 6. Use a number line that is marked in intervals of 6. \begin{align*}\frac{2}{3} & \rightarrow \frac{2}{3} \left(\frac{2}{2}\right)=\frac{4}{6}\\ \frac{1}{2} & \rightarrow \frac{1}{2} \left(\frac{3}{3}\right)=\frac{3}{6}\end{align*} Begin by placing a dot above the \begin{align*}4^{th}\end{align*} interval and then move 3 places to the left. You will be at your answer. \begin{align*}& \frac{2}{3}-\frac{1}{2}\\ & \frac{4}{6}-\frac{3}{6}\\ & \frac{4-3}{6}=\frac{1}{6}\end{align*}
1. \begin{align*}\frac{7}{9}-\frac{1}{3}\end{align*} Use a \begin{align*}\frac{1}{9}\end{align*} number line since the least common multiple of 9 and 3 is 9. \begin{align*}\frac{2}{3} \rightarrow \frac{1}{3} \left(\frac{3}{3}\right)=\frac{3}{9}\end{align*} \begin{align*}& \frac{7}{9}-\frac{1}{3}\\ & \frac{7}{9}-\frac{3}{9}\\ & \frac{7-3}{9}=\frac{4}{9}\end{align*}
1. \begin{align*}\frac{2}{5}-\frac{2}{10}\end{align*} The least common multiple of 5 and 10 is 10. Use a number line that is labeled in intervals of 10. \begin{align*}\frac{2}{5} \rightarrow \frac{2}{5} \left(\frac{2}{2}\right)=\frac{4}{10}\end{align*} \begin{align*}& \frac{2}{5}-\frac{2}{10}\\ & \frac{4}{10}-\frac{2}{10}\\ & \frac{2}{10}=\frac{1}{5}\end{align*}
Use the rules...
1. \begin{align*}\frac{7}{11}-\frac{1}{2}\end{align*} The least common multiple of 11 and 2 is 22. \begin{align*}& \frac{7}{11} \left(\frac{2}{2}\right)-\frac{1}{2} \left(\frac{11}{11}\right)\\ & \frac{14}{22}-\frac{11}{22}\\ & =\frac{3}{22}\end{align*}
1. \begin{align*}\frac{5}{6}-\frac{3}{4}\end{align*} The least common multiple of 4 and 6 is 12. \begin{align*}& \frac{5}{6} \left(\frac{2}{2}\right)-\frac{3}{4} \left(\frac{3}{3}\right)\\ & \frac{10}{12}-\frac{9}{12}\\ & =\frac{1}{12}\end{align*}
1. \begin{align*}\frac{4}{5}-\frac{3}{4}\end{align*} The least common multiple of 5 and 4 is 20. \begin{align*}& \frac{4}{5} \left(\frac{4}{4}\right)-\frac{3}{4} \left(\frac{5}{5}\right)\\ & \frac{16}{20}-\frac{15}{20}\\ & =\frac{1}{20}\end{align*}
For each of the following questions...
1. \begin{align*}\frac{2}{3}\end{align*} and \begin{align*}\frac{1}{2}\end{align*} The least common multiple of 3 and 2 is 6. Change both fractions to the same denominator to determine who used the most flour. \begin{align*}\frac{2}{3} \left(\frac{2}{2}\right)&= \frac{4}{6}\\ \frac{1}{2} \left(\frac{3}{3}\right) &= \frac{3}{6}\end{align*} Sally used more flour than Terri since \begin{align*}& \frac{4}{6} > \frac{3}{6}\\ & \frac{4}{6}-\frac{3}{6}\\ & \frac{4-3}{6}\\ & =\frac{1}{6}\end{align*}
Sally used \begin{align*}\frac{1}{6}\end{align*} more cups of flour than Terri.
1. To obtain an answer of \begin{align*}\frac{3}{8}\end{align*} by subtracting two fractions, the fractions with the denominator of 8 could be \begin{align*}\frac{4}{8}\end{align*} and \begin{align*}\frac{1}{8}\end{align*}. Two fractions with different denominators whose difference is \begin{align*}\frac{3}{8}\end{align*} are \begin{align*}\frac{1}{2}\end{align*} and \begin{align*}\frac{1}{8}\end{align*}.
1. Sierra’s Solution \begin{align*}& \frac{3}{4} - \frac{1}{6}\\ & \frac{9}{12}-\frac{2}{12}\\ & =\frac{7}{12}\end{align*} Clark’s Solution \begin{align*}& \frac{3}{4}-\frac{1}{6}\\ & \frac{9}{12}-\frac{2}{12}\\ & =\frac{7}{0}\end{align*} Sierra’s solution is correct. I would tell Clark that when fractions are subtracted, the fractions must have a common denominator. The difference of the numerators is placed over the common denominator. The denominators are NOT subtracted.
## Subtraction of Decimal Numbers
Objectives
The lesson objectives for The Subtraction of Real Numbers are:
• Subtraction of Positive Decimal Numbers.
• Subtraction of Positive and Negative Decimal Numbers
• Subtraction of Decimals Using the Rules
Introduction
In this concept you will learn to subtract decimal numbers. You will learn first to subtract decimal numbers that are positive values by applying the vertical alignment method. Then, you will subtract decimal numbers that are both negative and positive values. You will again apply the rules for adding integers since subtracting decimals that are signed numbers is the same as adding the opposite. Mastering these concepts will lead to the formation of rules for subtracting decimal numbers.
Watch This
Guidance
Jeremy and his family are driving to visit his grandparents. On the first day they drove 234.8 miles and on the second day they drove 251.6 miles. How many more miles did they drive on the second day?
To answer this question, the decimal numbers should be written using the vertical alignment method. The decimal number of greater magnitude should be placed above the number of smaller magnitude. Magnitude is simply the size of the number without respect to its sign or direction. The number -42.8 has a direction to the left and a magnitude of 42.8. The decimal points must be kept directly under each other as well as the digits must be kept in the same place value in line with each other. This means that digits in the ones place must be directly below digits in the ones place, digits in the tenths place must be in the tenths column, digits in the hundredths place must be in the hundredths column and so on. Once the numbers have been correctly aligned, the subtraction process is the same as subtracting whole numbers.
The decimal number 251.6 is of greater magnitude than 234.8. The numbers must be vertically aligned with the larger one above the smaller one. Now the numbers can be subtracted.
They drove 16.8 miles more on the second day.
Example A
Subtract: \begin{align*}57.62 - 6.18\end{align*}
Subtracting decimals is similar to subtracting whole numbers. We line up the decimal points so that we can subtract corresponding place value digits (e.g. tenths from tenths, hundredths from hundredths, and so on). As with whole numbers, we start from the right and work toward the left remembering to borrow when it is necessary.
To answer this question, the decimal numbers should be written using the vertical alignment method. Don’t forget to put the number of greater magnitude above the smaller number.
\begin{align*}& \quad 57. \cancel{\overset{5}{6}} \ ^1 2\\ & \underline{ \; \; -6.1 \; \; \; 8}\\ & \quad 51.4 \ \ 4\end{align*}
Example B
\begin{align*}(98.04)-(32.801)\end{align*}
Begin by writing the question using the vertical alignment method.
\begin{align*}& \quad 98.04\\ & \underline{-32.801}\\ & \end{align*}
The decimal points must be kept directly under each other as well as the digits must be kept in the same place value in line with each other. This means that digits in the ones place must be directly below digits in the ones place, digits in the tenths place must be in the tenths column, digits in the hundredths place must be in the hundredths column and so on. To ensure that the digits are aligned correctly, add zero to 98.04.
\begin{align*}& \ \ 98.04 {\color{blue}0}\\ & \underline{-32.801}\\ & \end{align*}
Subtract the numbers.
\begin{align*}& \ \ 9 \overset{7}{\cancel{8}}.^1 0 \overset{3}{\cancel{4}} \ ^1 {\color{blue}0}\\ & \underline{-32. \; 80 \; \; \; 1}\\ & \ \ 65. \ 23 \quad 9\end{align*}
Example C
\begin{align*}(67.65)-(-25.43)\end{align*}
The first step is to write the problem as an addition problem and to change the sign of the original number being subtracted. In other words, add the opposite.
\begin{align*}(67.65)+(+25.43)\end{align*}
Now, write the problem using the vertical alignment method. The two decimal numbers that are being added have positive signs. Apply the same rule that you used when adding integers that had same signs – add the numbers and use the sign of numbers in the answer.
\begin{align*}& \ \quad 67.65\\ & \underline{ \;\; +25.43}\\ & +93.08\end{align*}
Example D
\begin{align*}(137.4)-(+259.687)\end{align*}
The first step is to write the problem as an addition problem and to change the sign of the original number being subtracted. In other words, add the opposite.
\begin{align*}(137.4)+(-259.687)\end{align*}
Now write the problem using the vertical alignment method. Remember to put 259.687 above 137.4 because 259.687 is the number of greater magnitude. The two decimal numbers that are being added have opposite signs. Apply the same rule that you used when adding integers that had opposite signs – subtract the numbers and use the sign of the larger number in the answer.
\begin{align*}&-259.687\\ & \underline{ \; +137.4 \;\;\;}\\ & \end{align*}
To ensure that the digits are aligned correctly, add zeros to 137.4.
\begin{align*}& -259.687\\ & \underline{ \; +137.4 {\color{blue}00}}\\ & \end{align*}
Subtract the numbers.
\begin{align*}& \ -259.687\\ & \underline{ \;\; +137.4 {\color{blue}00}}\\ & -122.287\end{align*}
The decimal numbers being added have opposite signs. This means that the sign of the answer will be the same sign as that of the number of greater magnitude. In this problem the answer will have a negative sign.
Vocabulary
Decimal Number
A decimal number is a fraction whose denominator is 10 or some multiple of 10.
Decimal Point
A decimal point is the place marker in a decimal number that separates the whole number and the fraction part. The decimal number 326.45 has the decimal point between the six and the four.
Magnitude
A magnitude is the size of a number without respect to its direction. The number -35.6 has a direction to the left and a magnitude of 35.6.
Guided Practice
1. Subtract these decimal numbers: \begin{align*}(243.67)-(196.3579)\end{align*}
2. \begin{align*}(32.47)-(-28.8)-(19.645)\end{align*}
3. Josie has $59.27 in her bank account. She went to the grocery store and wrote a cheque for$62.18 to pay for the groceries. Describe Josie’s balance in her bank account now.
1. \begin{align*}(243.67)-(196.3579)\end{align*}
Write the decimal numbers using the vertical alignment method.
\begin{align*}& \quad 243.67\\ & \underline{-196.3579}\\ & \end{align*}
To ensure that the digits are properly aligned, add zeros to 243.67
\begin{align*}& \quad 243.67 {\color{blue}00}\\ & \underline{-196.3579}\\ & \end{align*}
Subtract the numbers. Work from right to left and borrow when it is necessary.
\begin{align*}& \ \ \overset{1}{\cancel{2}} \ \overset{13}{\cancel{4}} \ {^1} 3.6 \overset{6}{\cancel{7}} \ \overset{9}{^1 {\color{blue}\bcancel{0}}} \ {^1} {\color{blue}0}\\ & \underline{ -1 \; 9 \; \;\; 6.35 \;\; 7 \;\;\; 9\;\;}\\ & \quad \ \ 4 \ \ 7.31 \ \ 2 \ \ 1\end{align*}
2. \begin{align*}(32.47)-(-28.8)-(19.645)\end{align*}
Write the question as an addition problem and change the sign of the original number being subtracted.
\begin{align*}(32.47)+(+28.8)+(-19.645)\end{align*}
\begin{align*}& \ \ 32.47 \qquad \text{Add a zero to} \ 28.8\\ & \underline{ +28.8 {\color{blue}0}}\\ & \end{align*}
\begin{align*}& \ \ 32.47\\ & \underline{ +28.8 {\color{blue}0}}\\ & \ \ 61.27\end{align*}
The numbers being added are both positive so the answer will also be positive.
\begin{align*}(+61.27)+(-19.645)\end{align*}
Write the problem using the vertical alignment method.
\begin{align*}& \quad 61.27 {\color{blue}0}\\ & \underline{-19.645}\\ & \end{align*}
The number of greater magnitude was written above the smaller number. A zero was added to 61.27. The numbers have opposite signs so they will be subtracted and the answer will have the same sign as the larger number – positive.
\begin{align*}& \quad 61.27 {\color{blue}0}\\ & \underline{-19.645}\\ & \ \ 41.625\end{align*}
3. Amount in her bank account - $59.27 Amount of the written cheque -$62.18
The amount of the cheque is greater than the amount of money in the account.
The account will have a negative value. This means that her account is overdrawn.
Summary
The subtraction of decimal numbers is simply the subtraction of whole numbers and like fraction parts. To make this process simpler, the decimal numbers are written using the vertical alignment method with the number of greater magnitude being written above the number of lesser magnitude. The decimal points are aligned as well as the numbers are aligned according to their place value. The numbers in each vertical column are then subtracted by starting right and working to the left. If the decimal numbers are signed numbers, the rules for subtracting integers are applied to the problem.
Problem Set
Subtract the following decimal numbers:
1. \begin{align*}42.37-15.32\end{align*}
2. \begin{align*}37.891-7.2827\end{align*}
3. \begin{align*}579.237-45.68\end{align*}
4. \begin{align*}4.2935-0.327\end{align*}
5. \begin{align*}16.074-7.58\end{align*}
Subtract the following signed decimal numbers:
1. \begin{align*}(-17.39)-(-49.68)\end{align*}
2. \begin{align*}(92.75)+(-106.682)\end{align*}
3. \begin{align*}(-72.5)-(-77.57)-(31.724)\end{align*}
4. \begin{align*}(-82.456)-(279.83)+(-567.3)\end{align*}
5. \begin{align*}(-57.76)-(-85.9)-(33.84)\end{align*}
Determine the answer to the following problems.
1. The diameter of No. 12 bare copper wire is 0.08081 in., and the diameter of No. 15 bare copper wire is 0.05707 in. How much larger is No.12 wire than No. 15 wire?
2. The resistance of an armature while it is cold is 0.208 ohm. After running for several minutes, the resistance increases to 1.340 ohms. Find the increase in resistance of the armature.
3. The highest temperature recorded in Canada this year was \begin{align*}114.8^\circ F\end{align*}. The lowest temperature of \begin{align*}-62.9^\circ F\end{align*} was recorded in February this year. Find the difference between the highest and lowest temperatures recorded in Canada this year.
4. The temperature in Alaska was recorded as \begin{align*}-78.64^\circ F\end{align*} in January of 2010 and as \begin{align*}-59.8^\circ F\end{align*} on the same date in 2011. What is the difference between the two recorded temperatures?
5. Laurie has a balance of -32.16 in her bank account. Write a problem that could represent this balance. Answers Subtract the following decimal numbers: 1. \begin{align*}42.37-15.32\end{align*} Use the vertical alignment method. \begin{align*}& \quad \overset{3}{\cancel{4}} \ {^1} 2.37\\ & \underline{ \; -1 \; \; 5.32}\\ & \quad 2 \ \ 7.05\end{align*} Subtract right to left and borrow when it is necessary. 1. \begin{align*}579.237-45.68\end{align*} Use the vertical alignment method. \begin{align*}& \ 57 \overset{8}{\cancel{9}}.\overset{11}{\cancel{2}} \ ^1 37\\ & \underline{-45. 6 \quad 8{\color{blue}0}}\\ & \ 533. 5 \quad 57\end{align*} Add a zero to 45.68 and subtract the numbers. 1. \begin{align*}16.074-7.58\end{align*} Use the vertical alignment method. \begin{align*}& \overset{0}{\cancel{1}} \ \overset{15}{\cancel{6}}. \ \overset{9}{^1\bcancel{0}} \ {^1} 74\\ & \underline{ - \; 7. \; \; 5 \; \; 8{\color{blue}0}}\\ & \quad \ 8. \ 4 \ \ 94\end{align*} Add a zero to 7.58 and subtract the numbers. Subtract the following signed decimal numbers: \begin{align*}&(-17.39)-(-49.68)\\ &(-17.39)+(+49.68)\end{align*} Write the problem as an addition problem and change the sign of the original number being subtracted. \begin{align*}& \quad 49.68\\ & \underline{ - 17.39\;}\end{align*} Write the problem using the vertical alignment method. Place the number of greater magnitude above the number of lesser magnitude. Follow the rules for adding integers. \begin{align*}& \quad 49. \overset{5}{\cancel{6}} \ {^1} 8\\ & \underline{ - 17. 3 \quad 9}\\ & \quad 32.2 \ \ 9\end{align*} 1. \begin{align*}(-72.5)-(-77.57)-(31.724)\end{align*} Write the problem as an addition problem and change the sign of the original numbers being subtracted. \begin{align*}(-72.5)+(+77.57)+(-31.724)\end{align*} \begin{align*}& \ \ -72.5 {\color{blue}00}\\ & \underline{ \quad -31. 724}\\ & -104.224\end{align*} Write the problem using the vertical alignment method. Add zeros to -72.5 and follow the rules for the addition of integers. Follow the rules for adding integers. Numbers with the same sign must be added and the sign of the numbers being added will be the sign of the answer. \begin{align*}(-104.224)+(+77.57)\end{align*} The numbers being added have opposite signs. Place the number of greater magnitude on top. Subtract the numbers and apply the sign of the larger number to the answer. \begin{align*}& - \overset{9}{\bcancel{10}} \ \overset{13}{\bcancel{4}}. \overset{11}{\bcancel{2}} \ {^1} 24\\ & \ \underline{+ \; \;\; 7 \; 7. \;5 \;\;\; 7{\color{blue}0}}\\ & - \ \ 2 \ 6. \; 6 \quad 54\end{align*} \begin{align*}& (-57.76)-(-85.9)-(33.84)\\ & (-57.76)+(+85.9)+(-33.84)\end{align*} \begin{align*}& -57.76 \quad \ \ -91.60\\ & \underline{ \; -33.84} \qquad \underline{ \; +85.90}\\ & -91.60 \qquad - 5.70\end{align*} Determine the answer to the following problems. 1. \begin{align*}0.08081-0.05707\end{align*} \begin{align*}& \quad 0.0 \ \overset{7}{\bcancel{8}} \ {^1} 0 \ \overset{7}{\bcancel{8}} \ {^1} 1\\ & \underline{ \; -0.0 \; 5 \; \; 7 \; 0 \; \;\; 7\;\;}\\ & \quad 0.0 \ 2 \ \ 3 \ 7 \ \ 4\end{align*} The No. 12 wire is 0.02374 in. thicker than No. 15 copper wire. \begin{align*}& 114.8-(-62.9)\\ & 114.8+(+62.9)\end{align*} \begin{align*}& \ 114.8\\ & \underline{+62.9}\\ & \ 177.7\end{align*} The difference between the lowest and highest recorded temperatures is \begin{align*}177.7^\circ F\end{align*}. 1. -32.16 is the balance of the bank account. Laurie had a balance of $243.39 in her bank account. Her car payment of$275.55 was taken from the same account. This transaction left Laurie with a negative balance of \$32.16.
## Summary
In this lesson you have learned how to subtract real numbers by using a variety of models. The real numbers that you subtracted were integers, fractions and decimal numbers. The models that were used to subtract integers were color counters, algebra tiles and a number line. By using these models, you learned that subtracting integers was actually adding the opposite. The subtraction problems were written as addition problems and the sign of the original number being subtracted was changed. To calculate the answer, the two rules for adding integers were applied. After subtracting integers, you then learned how to subtract fractions by using fraction strips and a number line. You learned that fractions can only be subtracted if they have a common denominator. The difference of the numerators of the fractions being subtracted is placed over the common denominator. The last real numbers that were subtracted were decimal numbers. You learned that a number has both direction and magnitude. The direction of a positive number is to the right and that of a negative number is to the left. This direction is determined by the numbers location with respect to zero on the number line. The magnitude of a number is simply its size with no regard to its sign. When subtracting decimal numbers that were signed numbers, the rules for the addition of integers were applied to the problems.
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# Bridge over troubled water
Q. Please clarify the meanings of "Y-1 bridge through 10" when adding single-digit numbers, and "partition with 5 and a bit" when adding 6, 7, 8 and 9, and suggest good ways to teach these concepts. This is the first time I have taught this subject and I want to make sure I know exactly what they are.
A. Partitioning and "bridging through 10" are examples of methods suggested in the National Numeracy Strategy to help pupils become proficient in mental calculations and to lay foundations for later, more formal, pencil-and-paper methods. In the Strategy document you will find a description of the stages that help one understand this process. Page 32 states that, as outcomes, Year 1 pupils should: "Begin to partition and recombine by breaking units of 6, 7, 8 or 9 into '5 and a bit'. For example, work out mentally that: 5 + 8 = 5 plus (5 and 35 + 5 + 3 10 + 3 = 13."
The numbers are partitioned into 5s and hence easily into 10s.
Page 40 of the document says pupils should be taught to:
"Add or subtract a pair of numbers mentally I by bridging through 10 or 100, or a multiple of 10 or 100, and adjusting. As outcomes, Year 1 pupils should: Begin to add a pair of single-digit numbers, crossing 10.
* Use two steps and cross 10 as a middle stage. For example, work out mentally that: 6 + 7 = 13 and explain that: 6 + 7 = 6 + 4 + 3 10 + 3 = 13" The use of tallying into groups of five, using matchsticks or straws, helps pupils to develop a feel for breaking the number into "five and a bit".
Consider the strategy's example of 6 + 7; using the tally of matchsticks this becomes 5 + 1 + 5 + 2 = 10 + 3 = 13.
Helping pupils to grasp the numbers through this tallying process is extremely useful. You can then introduce the number line, so they have a visual representation of the process. In Cambridge-Hitachi's Mult-e-Maths, lesson 7 from Year 3 of the Addition and Subtraction strand contains a useful tool for showing a number line representation of bridging through 10.
Add any two-digit and single-digit numbers and they are represented by a "loop" on the number line, eg 16 + 7 becomes 16 + 4 + 3 = 20 + 3 = 23.
Dragging the right-hand end of the loop splits it into two parts whose sizes can be adjusted to demonstrate the usefulness of adding 7 to 16 by adding 4 and then 3. Changes to the number line are reflected in the lower addition statement on the screen.
The total can then be discussed and revealed as answers in the written statements and as a label on the number line. Partitioning is also used with larger numbers: bridging through multiples of 10.
The addition can also take place through compensation. This is quite an advanced approach and not necessarily one that pupils would naturally attempt. This is where the numbers are "made up" to the next easiest number and then that amount is subtracted as compensation. eg 17 + 9 = 17 + 10 - 1 26 26 + 11 = 26 + 10 + 1 = 37 You may find the following worth reading: * Ian Thompson's paper on mental strategieswww.m-a.org.ukdocslibrary 2076.pdf
* How Do We Want Children to be Numerate by Mike Askew and Margaret Brown of King's College London www.bera.ac.ukpublicationspdfs520668_Num.pdf
* For more information on Mult-e-Maths (the bridging through 10 tool is included as part of the demo): www.cambridge-hitachi.com productsprimarymultemathsstrands.htmNoNo
* In the online game 'Add it up', you select numbers to clear the board, providing fun and mental practice: www.flasharcade.comgame.php?additup2
* The following could be used for practice or to set problems for the class to discuss in a lesson starter: www.amblesideprimary.comamblewebmentalmathsnumberbond.html
* A selection of activities isavailable at: www.coxhoe.durham.sch.ukCurriculumNumeracy.htm
You need to visit the site first: I couldn't get all of the activities to work, but there are plenty to choose from.
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## 3 Ways to Find the Slope of an Equation
To solve the point-slope form for a given straight line to find the equation of the given line, we can follow these steps: Step 1: Note down the slope, m m of the straight line and
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## Slope formula (equation for slope)
Calculating slope from a graph using the slope formula The slope formula is used to find the slope of a line that joins two points (x1,y1) ( x 1, y 1) and (x2,y2) ( x 2, y 2). sing this
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## Finding the Slope of a Line from an Equation
The slope is represented mathematically as: m = y 2 - y 1 x 2 - x 1 In the equation above, y2 - y1 = Δy, or vertical change, while x2 - x1 = Δx, or horizontal change, as shown in the graph provided.
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## How to Find the Slope of an Equation
Calculate slope, m using the formula for slope: Slope Formula m = (y2 − y1) (x2 − x1) m = rise run = Δy Δx = y2 − y1 x2 − x1 Here you need to know the coordinates of 2 points on a line, (x 1, y 1)
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What is the value of x in the equation 3x + 5 = 17? The value of x in the equation 3x + 5 = 17 is 12.
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## 3 Ways to Find the Slope of a Line
Step 1: Identify the values of , , , and . [Show solution] Step 2: Plug in these values to the slope formula to find the slope. [Show solution] Step 3: Gut check. Make sure this slope makes
## How to Use the Formula and Calculate Slope
The slope formula is used to find the slope of a line that joins two points (x₁, y₁) and (x₂, y₂). Using this formula, the slope of the line is, m = (y₂ - y₁) / (x₂ - x₁). We can use the same formula to find
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# 9.5: Rare Events, the Sample, Decision and Conclusion
Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test.
## Rare Events
Suppose you make an assumption about a property of the population (this assumption is the null hypothesis). Then you gather sample data randomly. If the sample has properties that would be very unlikely to occur if the assumption is true, then you would conclude that your assumption about the population is probably incorrect. (Remember that your assumption is just an assumption—it is not a fact and it may or may not be true. But your sample data are real and the data are showing you a fact that seems to contradict your assumption.)
For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a$100 bill. The probability of this happening is $$\frac{1}{200} = 0.005$$. Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A "rare event" has occurred (Didi getting the$100 bill) so Ali doubts the assumption about only one \$100 bill being in the basket.
Using the Sample to Test the Null Hypothesis
Use the sample data to calculate the actual probability of getting the test result, called the $$p$$-value. The $$p$$-value is the probability that, if the null hypothesis is true, the results from another randomly selected sample will be as extreme or more extreme as the results obtained from the given sample.
A large $$p$$-value calculated from the data indicates that we should not reject the null hypothesis. The smaller the $$p$$-value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it.
Draw a graph that shows the $$p$$-value. The hypothesis test is easier to perform if you use a graph because you see the problem more clearly.
Example $$\PageIndex{1}$$
Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm. and the distribution of heights is normal.
• The null hypothesis could be $$H_{0}: \mu \leq 15$$
• The alternate hypothesis is $$H_{a}: \mu > 15$$
The words "is more than" translates as a "$$>$$" so "$$\mu > 15$$" goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis.
Since $$\sigma$$ is known ($$\sigma = 0.5 cm.$$), the distribution for the population is known to be normal with mean $$μ = 15$$ and standard deviation
$\dfrac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{10}} = 0.16. \nonumber$
Suppose the null hypothesis is true (the mean height of the loaves is no more than 15 cm). Then is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test works by asking the question how unlikely the sample mean would be if the null hypothesis were true. The graph shows how far out the sample mean is on the normal curve. The p-value is the probability that, if we were to take other samples, any other sample mean would fall at least as far out as 17 cm.
The $$p$$-value, then, is the probability that a sample mean is the same or greater than 17 cm. when the population mean is, in fact, 15 cm. We can calculate this probability using the normal distribution for means.
$$p\text{-value} = P(\bar{x} > 17)$$ which is approximately zero.
A $$p$$-value of approximately zero tells us that it is highly unlikely that a loaf of bread rises no more than 15 cm, on average. That is, almost 0% of all loaves of bread would be at least as high as 17 cm. purely by CHANCE had the population mean height really been 15 cm. Because the outcome of 17 cm. is so unlikely (meaning it is happening NOT by chance alone), we conclude that the evidence is strongly against the null hypothesis (the mean height is at most 15 cm.). There is sufficient evidence that the true mean height for the population of the baker's loaves of bread is greater than 15 cm.
Exercise $$\PageIndex{1}$$
A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5.
• $$H_{0}: \mu leq 12$$
• $$H_{a}: \mu > 12$$
The $$p$$-value is 0.0013
Draw a graph that shows the $$p$$-value.
$$p\text{-value} = 0.0013$$
## Decision and Conclusion
A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the $$p$$-value and a preset or preconceived $$\alpha$$ (also called a "significance level"). A preset $$\alpha$$ is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem.
When you make a decision to reject or not reject $$H_{0}$$, do as follows:
• If $$\alpha > p\text{-value}$$, reject $$H_{0}$$. The results of the sample data are significant. There is sufficient evidence to conclude that $$H_{0}$$ is an incorrect belief and that the alternative hypothesis, $$H_{a}$$, may be correct.
• If $$\alpha \leq p\text{-value}$$, do not reject $$H_{0}$$. The results of the sample data are not significant.There is not sufficient evidence to conclude that the alternative hypothesis,$$H_{a}$$, may be correct.
When you "do not reject $$H_{0}$$", it does not mean that you should believe that H0 is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of $$H_{0}$$.
Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem.
Example $$\PageIndex{2}$$
When using the $$p$$-value to evaluate a hypothesis test, it is sometimes useful to use the following memory device
• If the $$p$$-value is low, the null must go.
• If the $$p$$-value is high, the null must fly.
This memory aid relates a $$p$$-value less than the established alpha (the $$p$$ is low) as rejecting the null hypothesis and, likewise, relates a $$p$$-value higher than the established alpha (the $$p$$ is high) as not rejecting the null hypothesis.
Fill in the blanks.
Reject the null hypothesis when ______________________________________.
The results of the sample data _____________________________________.
Do not reject the null when hypothesis when __________________________________________.
The results of the sample data ____________________________________________.
Reject the null hypothesis when the $$p$$-value is less than the established alpha value. The results of the sample data support the alternative hypothesis.
Do not reject the null hypothesis when the $$p$$-value is greater than the established alpha value. The results of the sample data do not support the alternative hypothesis.
Exercise $$\PageIndex{2}$$
It’s a Boy Genetics Labs claim their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows:
• $$H_{0}: p = 0.50, H_{a}: p > 0.50$$
• $$\alpha = 0.01$$
• $$p\text{-value} = 0.025$$
Interpret the results and state a conclusion in simple, non-technical terms.
Since the $$p$$-value is greater than the established alpha value (the $$p$$-value is high), we do not reject the null hypothesis. There is not enough evidence to support It’s a Boy Genetics Labs' stated claim that their procedures improve the chances of a boy being born.
## Review
When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p-value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind:
• $$\alpha > p-value$$, reject the null hypothesis
• $$\alpha \leq p-value$$, do not reject the null hypothesis
## Glossary
Level of Significance of the Test
probability of a Type I error (reject the null hypothesis when it is true). Notation: $$\alpha$$. In hypothesis testing, the Level of Significance is called the preconceived $$\alpha$$ or the preset $$\alpha$$.
$$p$$-value
the probability that an event will happen purely by chance assuming the null hypothesis is true. The smaller the $$p$$-value, the stronger the evidence is against the null hypothesis. |
# How to solve circumference of a circle
We will explore How to solve circumference of a circle can help students understand and learn algebra. We will also look at some example problems and how to approach them.
## How can we solve circumference of a circle
The solver will provide step-by-step instructions on How to solve circumference of a circle. Solving for a side in a right triangle can be done using the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. Using this theorem, it is possible to solve for any side in a right triangle given the length of the other two sides. For example, if the length of one side is 3 and the length of the other side is 4, then the hypotenuse must be 5, since 3^2 + 4^2 = 25. In order to solve for a side, all you need is the lengths of the other two sides and a calculator. However, it is also possible to estimate the length of a side without using a calculator. For example, if you know that one side is 10 and the other side is 8, you can estimate that the hypotenuse is 12 since 8^2 + 10^2 = approximately 144. Solving for a side in a right triangle is a simple matter as long as you know the Pythagorean theorem.
How to solve radicals can be a tricky process, but there are a few steps that can help. First, rationalize the denominator by multiplying by an accessory root. This will eliminate any fractions in the denominator. Next, extract any perfect square roots from the radical. For example, if the radical is 4√5, you would take out the 2√5. Finally, simplify the radical by using absolute value signs and grouping like terms. How to solve radicals may seem complicated at first, but with some practice it can become second nature.
Imagine being able to simply take a picture of a math word problem and have the answer pop up on your screen almost instantaneously. That's what one new app promises to do. The app, called PhotoMath, uses the camera on your smartphone or tablet to take a picture of a math problem and then displays the answer. Just point your camera at a problem and PhotoMath will do the rest. The app can solve problems ranging from simple addition and subtraction to more complex equations involving fractions and decimals. It can even handle problems that require multiple steps, such as long division. And if you're not satisfied with the answer it gives you, PhotoMath also provides step-by-step instructions for how to solve the problem. PhotoMath is still in its early stages, so it doesn't always get things right. But it shows promise as a tool that could one day make solving math problems a breeze. So if you're struggling with a math problem, why not give PhotoMath a try? It just might be the answer you're looking for.
Domain and range are two important concepts in mathematics. Domain refers to the set of all possible input values for a function, while range refers to the set of all possible output values. Both concepts can be difficult to grasp, but there are a few simple steps that can help. First, it is important to understand what inputs and outputs are. Inputs are the values that are fed into a function, while outputs are the values that the function produces. Once this is understood, it is fairly easy to identify the domain and range of a given function. To do this, simply list all of the possible input values and then identify the corresponding output values. In some cases, it may also be helpful to graph the function to visualize the relationship between inputs and outputs. By understanding these basic concepts, it is possible to solve domain and range problems with ease.
Solving by square roots Solving by square roots Solving by square roots Solving by square Solving by square Solving Solving by Solving Solving Solving Solving Solvingsolving solving Equation Assume the given equation is of the form: ax^2 + bx + c = 0. Then, the solution to the equation can be found using the following steps: 1) Determine the value of a, b, and c. 2) Find the discriminant, which is equal to b^2 - 4ac. 3) If the discriminant is negative, then there are no real solutions to the equation. 4) If the discriminant is equal to zero, then there is one real solution to the equation. 5) If the discriminant is positive, then there are two real solutions to the equation. 6) Use the quadratic formula to find the value of x that solves the equation. The quadratic formula is as follows: x = (-b +/-sqrt(b^2-4ac))/2a. |
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# Airspeed of an aeroplane is 400km/hr. The wind is blowing at 200km/hr towards the east direction. In what direction should the pilot try to fly the plane to move exactly towards the northeast.
Last updated date: 24th Feb 2024
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Hint: When the plane is moving and the wind is blowing due east in this situation when the plane steers towards northeast then the effect of wind blowing will also affect the angle turned by the aeroplane it will turn more than it was steered to turn. So, if we want to steer at a particular angle then we need to steer at an angle lower than desired because the rest of the turn will be due to the effect of wind.
Complete step by step solution:
Consider the following diagram in this case: the plane is flying and the wind is blowing due east. Now here it is said that the plane needs to be steered exactly northeast. Now as we can see that the exact northeast will be at $45^\circ$.
So if we need to travel at $45^\circ$ we will take a steer at an angle more than $45^\circ$ so the wind will show its effect and the resultant of wind and aeroplane will be at $45^\circ$.
Now in this figure, the red line is the actual path the plane needs to turn and the blue line is the path we will turn to compensate for the effect of wind blowing due east.
Now resolving the components of ${V_a}$ we get ${V_a}\operatorname{Cos} \theta$ along AB.
And ${V_a}\operatorname{Sin} \theta$ perpendicular to AB
Similarly,
Now we will resolve ${V_w}$ for AB
So, we will get ${V_w}\operatorname{Cos} 45^\circ$ along AB
And ${V_w}\operatorname{Sin} 45^\circ$ perpendicular to that
Now in the figure, we can see that the components ${V_a}\operatorname{Cos} \theta$ and ${V_a}\operatorname{Cos} 45^\circ$ will steer the plane along with AB
And ${V_w}\operatorname{Sin} 45^\circ$, ${V_a}\operatorname{Sin} \theta$ acting in opposite directions will be equal and cancel out each other.
So the net resultant force will be along AB and plane will steer exactly at $45^\circ$
On equating we get ${V_a}\operatorname{Sin} \theta = {V_w}\operatorname{Sin} 45^\circ$
Substituting the values, we get
${V_a}$ is the velocity of the aeroplane $= 400m{s^{ - 1}}$
${V_w}$ is the velocity of the wind $= 200m{s^{ - 1}}$
$\Rightarrow 400\operatorname{Sin} \theta = 200\operatorname{Sin} 45^\circ$
$\Rightarrow \operatorname{Sin} \theta = \dfrac{{200}}{{400}}\operatorname{Sin} 45^\circ$
$\Rightarrow \operatorname{Sin} \theta = \dfrac{1}{2}\operatorname{Sin} 45^\circ$
$\Rightarrow \theta = 20.70^\circ$
Final answer: The pilot needs to turn an angle of $20.70^\circ$ to move exactly northeast.
Note: The resultant of both the motion will give us the actual motion.
When a plane is flying and it takes a turn and there is wind flowing then the effect of the wind will also be considered in steering the plane.
The effect of wind generally increases or decreases the angle the pilot needs to steer the plan to get into a particular direction. |
# How to Calculate Area of Triangle?
The simplest way to find the area of a triangle is to take half of its base times the height. However, there are various other formulas to find the area of a triangle, depending upon the information we have. Furthermore, it is possible to calculate the area of a triangle just from the sides and angles of a triangle without knowing its height. In this article, we will see different methods of how to calculate area of triangle.
## 3 Methods to Calculate Area of Triangle
• Method 1: Using base and height
• Method 2: Using side lengths
• Method 3: Using trigonometry
### Method 1: Using base and height
1. #### By finding the base and height of the triangle
In a triangle, the base is one side of the triangle and height is the measure of the tallest point on a triangle. Moreover, we can find it by drawing a line perpendicular from the base to the opposite vertex. Such as, you have a triangle whose base is 5cm long and its height is 3cm long.
1. #### Form a formula for the area of the triangle
So, the formula here is $$Area = \frac{1}{2}(bh)$$. Also, here b is the base of the triangle and h is the height of the triangle.
1. #### Now put the values in the formula
Firstly, multiple the value of base with height then multiplies the product by $$\frac{1}{2}$$. In addition, this will give the area of the triangle in square units. In the above example the solution will be:
$$Area = \frac{1}{2}(bh)$$
$$Area = \frac{1}{2}(5)(3)$$
$$Area = \frac{1}{2}(15)$$
$$Area = 7.5cm^{2}$$
Hence, the area of a triangle is $$7.5cm^{2}$$
1. #### How to find the area of a right triangle
In a right triangle, two sides are perpendicular to each other. So, one will be the height of the triangle and the other will be the base of the triangle. Thus, even, if the height or base is unstated, you are given them if you know the side lengths. So, you can use the $$Area = \frac{1}{2}(bh)$$ formula to find the area.
Moreover, you can also use this formula if you know one side, plus the length of the hypotenuse. Furthermore, the hypotenuse is the longest side of the right triangle and is opposite to the right triangle using the Pythagorean Theorem ($$a^{2} + b^{2} = c^{2}$$.
Suppose, if the hypotenuse of a triangle is side c, and the height and base would be the other two sides (a and b). Moreover, if you knew that the hypotenuse is 5cm and the base 4cm. so you can find the height by Pythagorean Theorem.
$$a^{2} + b^{2} = c^{2}$$
$$a^{2} + 4^{2} = 5{2}$$
$$a^{2} + 16 = 25$$
$$a^{2} + 16 – 16 = 25 – 16$$
$$a^{2} = 9$$
$$a = 3$$
So put the value in the area formula, and use two perpendicular side (a and b) to substitute the base and height:
$$Area = \frac{1}{2}(bh)$$
$$Area = \frac{1}{2}(4)(3)$$
$$Area = \frac{1}{2}(12)$$
$$Area = 6$$
So, the area of a triangle is $$6cm^{2}$$.
### Method 2: Using side lengths
1. #### Calculate the semi-perimeter of the triangle
The semi-perimeter is equal to the half of its perimeter. So, to find the semi-perimeter we need to find its perimeter of the by adding up the length of its three sides then multiplying the product by $$\frac{1}{2}$$.
E.g. suppose if a triangle has three sides that are 5cm, 4cm and 3cm long, then the semi-perimeter will be:
$$s = \frac{1}{2}(3 + 4 + 5)$$
$$s = \frac{1}{2}(12)$$ = 6
1. #### Apply Heron’s formula
Heron’s formula is $$Area = \sqrt{s(s-a)(s-b)(s-c)}$$. Here ‘s’ is the semi-perimeter of a triangle and a, b, and c are the side lengths of a triangle.
1. #### Put values in the formula and calculate it
While putting the values in the formula, make sure you substitute the semi-perimeter for each instance of s in the formula
$$Area = \sqrt{s(s-a)(s-b)(s-c)}$$
$$Area = \sqrt{6(6-3)(6-4)(6-5)}$$
$$Area = \sqrt{6(3)(2)(1)}$$
$$Area = \sqrt{6(6)}$$
$$Area = \sqrt{36}$$ = 6
Therefore, the area of the triangle is $$6cm^{2}$$.
### Method 3: Using trigonometry
1. #### Find the length of two adjacent sides and included angle
Firstly, the adjacent sides refer to those sides that meet at the vortex. Also, the angle is the angle between these two sides.
E.g. a triangle with two adjacent sides that measure 150cm and 231cm in length. And the angle between them is 123 degrees.
1. #### Now use the trigonometry formula for the area of a triangle
The trigonometry formula is $$Area = \frac{bc}{2}sin A$$. Here b and c are the adjacent sides and A is the angle between them.
1. #### Put values in the formula
Firstly, validate that you substitute for the variables b and c. Also, multiply their values and divide them by 2.
E.g. $$Area = \frac{bc}{2}sin A$$
$$Area = \frac{(150)(231)}{2}sin A$$
$$Area = \frac{34650}{2}sin A$$
$$Area = 17325sin A$$
1. #### Plug the values of sine in formula and multiply them
Calculate the value of sine using scientific calculator. After that, put the value into the formula.
The sine of 123 degree angle is 0.83867.
$$Area = 17325sin A$$
$$Area = 17325(0.83687)$$ =14529.96
Hence the area of the triangle is $$14530cm^{2}$$.
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Home » Blog » Deduce properties of the solutions of a given differential equation
# Deduce properties of the solutions of a given differential equation
Let be a solution for all of the second order differential equation
Without attempting to solve the equation answer the following questions.
1. If and has an extremum at , prove that this extremum must be a minimum.
2. If has an extremum at 0 decide whether this extremum is a maximum or minimum and prove your assertion.
3. If the solution satisfies the conditions
find the minimum value for the constant such that
Incomplete.
1. Evangelos says:
a.) We know from our given information that c cannot be zero. And we know that at the point where x=c, we have a local extremum where f'(c) = 0. So our differential equation becomes
And since c is not zero, we can divide both sides, giving us
We now have two cases
1.) c is greater than zero
If c is greater than zero, then
Thus,
And by Theorem 4.9, since f”(c) is greater than 0, f'(c) must be a local minimum.
2.) c is less than zero
If c is less than zero, then
And by Theorem 4.9, since f”(c) is greater than 0, f(c) must be a local minimum.
As such, for all c not equal to zero, the point f(c) must be a local minimum.
b.) From our given information, we know that if f has an extremum at x=0, then f'(0) = 0. To determine if this is a local maximum, minimum, or neither, we must compute the value of f”(0). If we were to divide both sides by x and take the limit as x approaches zero
We would get the indeterminate form of 0/0
But if this is the case, then we can use L’Hopital’s Rule and take the derivative with respect to x of the numerator and denominator, and then take the limit as x approaches zero.
And since this value is greater than zero, the point f(0) is a local minimum.
c.) If we can assume that f(0) = f'(0) = 0, we can use the result from part (b.) to show the following:
But with our initial conditions and the result of part (b.), we know that
So the smallest value of A that fulfills the condition
And thus, the conditions
for all x greater than or equal to zero, must be
Looks like that’s all for now, thanks to RoRi for opening up the comments sections, and thanks to the readers for… reading :)
• Evangelos says:
So, the latex didn’t render properly for my cases, it should be as follows
Case 1.) c > 0
Case 2.) c < 0
• Evangelos says:
D’oh! |
# Basic Algebra
### Also on IntMath
Related algebra chapters:
This chapter contains elementary algebra tutorials on the following topics:
1. Adding and Subtracting Algebraic Expressions, shows you how to do problems like: Simplify: -2[-3(x − 2y) + 4y].
2. Multiplication of algebra expressions, has examples like:
Expand (2x + 3)(x2x − 5).
3. Division of algebraic expressions, for example: (12a2b) ÷ (3ab2)
4. Solving Equations, like this one: 5 − (x + 2) = 5x.
5. Formulas and Literal Equations, which shows how to solve an equation for a particular variable.
6. Applied Verbal Problems shows why we are doing all this.
## What is Algebra?
Algebra is the branch of mathematics that uses letters in place of some unknown numbers.
You've been using algebra since your early schooling, when you learned formulas like the area of a rectangle, with width w, height h:
A = w × h
We used letters to stand for numbers. Once we knew the width and height, we could substitute them into the formula and find our area.
Another one you may have seen is the area of a circle, with radius r:
A = πr2
As soon as we know the length of the sides, we can find the area.
Literal numbers (the letters used in algebra) can either stand for variables (the value of the letter can change, like the w, h and r in the examples of the area of a rectangle and the area of a circle) or constants (where the value does not change), for example:
π (the ratio circumference/diameter of a circle, value 3.141592....)
g (the accelaration due to gravity, 9.8 m/s2),
e (which has a constant value of 2.781828...).
And as my students constantly ask...
## Why Do We Have to do This?
Algebra is a powerful tool for problem solving in science, engineering, economics, finance, architecture, ship-building and many other day-to-day tasks.
If we didn't use letters in place of numbers (and used words instead), we would be writing many pages for each problem and it would be much more confusing.
This elementary algebra chapter follows on from the earlier chapter on Numbers.
### If you find this chapter difficult...
If you struggle with this chapter, it may be a good idea to go back and remind yourself about basic number properties first, since that's important background.
### On with the show
OK, let's move on and learn some basic algebra tips:
top
### Online Algebra Solver
This algebra solver can solve a wide range of math problems.
### Algebra Lessons on DVD
Easy to understand algebra lessons on DVD. See samples before you commit. |
## Engage NY Eureka Math 5th Grade Module 4 Lesson 22 Answer Key
### Eureka Math Grade 5 Module 4 Lesson 22 Problem Set Answer Key
Question 1.
Solve for the unknown. Rewrite each phrase as a multiplication sentence. Circle the scaling factor and put a box around the number of meters.
a. $$\frac{1}{2}$$ as long as 8 meters = ______ meter(s)
4 meters.
Explanation:
Given that $$\frac{1}{2}$$ as long as 8 meters which is $$\frac{1}{2}$$ × 8 = 4 meters.
b. 8 times as long as $$\frac{1}{2}$$ meter = _______ meter(s)
4 meters.
Explanation:
Given that $$\frac{1}{2}$$ meter is 8 times as long, so 8 × $$\frac{1}{2}$$ which is 4 meters.
Question 2.
Draw a tape diagram to model each situation in Problem 1, and describe what happened to the number of meters when it was multiplied by the scaling factor.
a.
The scaling factor is less than 1, so the number of meters decreases.
Explanation:
The scaling factor is less than 1, so the number of meters decreases.
b.
The scaling factor is greater than 1, so the number of meters increased.
Explanation:
The scaling factor is greater than 1, so the number of meters increased.
Question 3.
Fill in the blank with a numerator or denominator to make the number sentence true.
a. 7 × $$\frac{}{4}$$ < 7
7 × $$\frac{}{4}$$ < 7.
Explanation:
Given that 7 × $$\frac{}{4}$$ < 7, so here in the numerator we will place a number that is less than 4. So we will place 3 in the numerator which will be 7 × $$\frac{3}{4}$$ < 7.
b. $$\frac{7}{}$$ × 15 > 15
$$\frac{7}{2}$$ × 15 > 15
Explanation:
Given that $$\frac{7}{2}$$ × 15 > 15, so here in the denominator we will place a number that is less than 7. So we will place 2 in the denominator which will be $$\frac{7}{2}$$ × 15 > 15.
c. 3 × $$\frac{}{5}$$ = 3
3 × $$\frac{5}{5}$$ = 3
Explanation:
Given that 3 × $$\frac{}{5}$$ = 3, so to justify the answer we will place 5 in the numerator which is 3 × $$\frac{5}{5}$$ = 3.
Question 4.
Look at the inequalities in each box. Choose a single fraction to write in all three blanks that would make all three number sentences true. Explain how you know.
a.
$$\frac{5}{2}$$.
Explanation:
Multiplying by a fraction greater than 1 will make the product larger than the other factor.
b.
$$\frac{1}{2}$$.
Explanation:
Multiplying by a fraction less than 1 will make the product less than the other factor.
Question 5.
Johnny says multiplication always makes numbers bigger. Explain to Johnny why this isn’t true. Give more than one example to help him understand.
4 times 0.5 equals 2.
Explanation:
Given that Johnny says multiplication always makes numbers bigger which is not true because if we multiply any number by a decimal number, we will make it smaller. Because if we multiply a number by something less than one, we will get something less than itself. This also works if you multiply a number by a fraction. For example, 4 times 0.5 equals 2 because you are getting half of the original number which is 4.
Question 6.
A company uses a sketch to plan an advertisement on the side of a building. The lettering on the sketch is $$\frac{3}{4}$$ inch tall. In the actual advertisement, the letters must be 34 times as tall. How tall will the letters be on the building?
The letters on the building would be 25 $$\frac{1}{2}$$ inch.
Explanation:
Given that a company uses a sketch to plan an advertisement on the side of a building and lettering on the sketch is $$\frac{3}{4}$$ inch tall and the letters must be 34 times as tall. So to find the height of the building, we will multiply 34 × $$\frac{3}{4}$$ inch and the height of the letters on the building which is 34 × $$\frac{3}{4}$$ inch
= $$\frac{104}{4}$$ inch
= $$\frac{51}{2}$$ inch
= 25 $$\frac{1}{2}$$ inch.
Therefore the letters on the building would be 25 $$\frac{1}{2}$$ inch.
Question 7.
Jason is drawing the floor plan of his bedroom. He is drawing everything with dimensions that are $$\frac{1}{12}$$ of the actual size. His bed measures 6 ft by 3 ft, and the room measures 14 ft by 16 ft. What are the dimensions of his bed and room in his drawing?
The dimensions of his room in his drawing are 1 $$\frac{1}{3}$$ by1 $$\frac{1}{6}$$ ft,
The dimensions of his bed in his drawing are $$\frac{1}{2}$$ ft by $$\frac{1}{4}$$ ft.
Explanation:
Given that Jason is drawing the floor plan of his bedroom and he is drawing everything with dimensions that are $$\frac{1}{12}$$ of the actual size and his bed measures 6 ft by 3 ft, and the room measures 14 ft by 16 ft. So the dimensions of his room in his drawing are $$\frac{1}{12}$$ of 16 ft and $$\frac{1}{12}$$ of 14 ft which is
= $$\frac{1}{12}$$ × 16
= $$\frac{4}{3}$$
= 1 $$\frac{1}{3}$$
$$\frac{1}{12}$$ of 14 ft
= $$\frac{1}{12}$$ × 14 ft
= $$\frac{7}{6}$$
= 1 $$\frac{1}{6}$$ ft
So the dimensions of his room in his drawing are 1 $$\frac{1}{3}$$ by1 $$\frac{1}{6}$$ ft.
For his bed in his drawing are $$\frac{1}{12}$$ of 6 ft by $$\frac{1}{12}$$ of 3 which is
= $$\frac{1}{12}$$ × 6
= $$\frac{1}{2}$$ ft
$$\frac{1}{12}$$ of 3
= $$\frac{1}{12}$$ × 3
= $$\frac{1}{4}$$ ft.
So the dimensions of his bed in his drawing are $$\frac{1}{2}$$ ft by $$\frac{1}{4}$$ ft.
### Eureka Math Grade 5 Module 4 Lesson 22 Exit Ticket Answer Key
Fill in the blank to make the number sentences true. Explain how you know.
a. $$\frac{}{3}$$ × 11 ˃ 11
$$\frac{4}{3}$$ × 11 ˃ 11.
Explanation:
Given that $$\frac{}{3}$$ × 11 ˃ 11, so here in the numerator we will place a number that is greater than 3. So we will place 4 in the numerator which will be $$\frac{4}{3}$$ × 11 ˃ 11.
b. 5 × $$\frac{}{8}$$ ˂ 5
5 × $$\frac{5}{8}$$ ˂ 5.
Explanation:
Given that 5 × $$\frac{}{8}$$ ˂ 5, so here in the numerator we will place a number that is less than 8. So we will place 5 in the numerator which will be 5 × $$\frac{5}{8}$$ ˂ 5.
c. 6 × $$\frac{2}{}$$ = 6
6 × $$\frac{2}{2}$$ = 6
Explanation:
Given that 6 × $$\frac{2}{2}$$ = 6, so to justify the answer we will place 2 in the numerator which is 6 × $$\frac{2}{2}$$ = 6
### Eureka Math Grade 5 Module 4 Lesson 22 Homework Answer Key
Question 1.
Solve for the unknown. Rewrite each phrase as a multiplication sentence. Circle the scaling factor and put a box around the number of meters.
a. $$\frac{1}{3}$$ as long as 6 meters = ______ meter(s)
2 meters.
Explanation:
Given that $$\frac{1}{3}$$ as long as 6 meters which is $$\frac{1}{3}$$ × 6 = 2 meters.
b. 6 times as long as $$\frac{1}{3}$$ meter = ______ meter(s)
2 meters.
Explanation:
Given that $$\frac{1}{3}$$ meter is 6 times as long, so 6 × $$\frac{1}{3}$$ which is2 meters.
Question 2.
Draw a tape diagram to model each situation in Problem 1, and describe what happened to the number of meters when it was multiplied by the scaling factor.
a.
The scaling factor is less than 1, so the number of meters decreases.
Explanation:
The scaling factor is less than 1, so the number of meters decreases.
b.
The scaling factor is greater than 1, so the number of meters increased.
Explanation:
The scaling factor is greater than 1, so the number of meters increased.
Question 3.
Fill in the blank with a numerator or denominator to make the number sentence true.
a. 5 × $$\frac{}{3}$$ ˃ 5
5 × $$\frac{2}{3}$$ ˃ 5.
Explanation:
Given that 5 × $$\frac{4}{3}$$ ˃ 5, so here in the numerator, we will place a number that is greater than 3. So we will place 4 in the numerator which will be 5 × $$\frac{4}{3}$$ ˃ 5.
b. $$\frac{6}{}$$ × 12 ˂ 12
$$\frac{6}{}$$ × 12 ˂ 12.
Explanation:
Given that $$\frac{6}{7}$$ × 12 ˂ 12, so here in the numerator, we will place a number that is greater than 6. So we will place 7 in the numerator which will be $$\frac{6}{7}$$ × 12 ˂ 12.
c. 4 × $$\frac{}{5}$$ = 4
4 × $$\frac{5}{5}$$ = 4
Explanation:
Given that 4 × $$\frac{5}{5}$$ = 4, so to justify the answer we will place 5 in the numerator which is 4 × $$\frac{5}{5}$$ = 4.
Question 4.
Look at the inequalities in each box. Choose a single fraction to write in all three blanks that would make all three number sentences true. Explain how you know.
a.
$$\frac{5}{4}$$.
Explanation:
Multiplying by a fraction greater than 1 will make the product larger than the other factor.
b.
$$\frac{1}{2}$$.
Explanation:
Multiplying by a fraction less than 1 will make the product less than the other factor.
Question 5.
Write a number in the blank that will make the number sentence true.
a. 3 × _____ ˂ 1
3 × $$\frac{1}{4}$$ < 1.
Explanation:
To make the number sentence true we will place the number which is less than $$\frac{1}{3}$$, so we will place $$\frac{1}{4}$$ which will be less than 1. So the expression will be 3 × $$\frac{1}{4}$$ < 1.
b. Explain how multiplying by a whole number can result in a product less than 1.
When a positive whole number is multiplied by a fraction between 0 and 1, the product is less than the whole number. When a number greater than 1 is multiplied by a number greater than 1, the product is greater than both numbers.
Question 6.
In a sketch, a fountain is drawn $$\frac{1}{4}$$ yard tall. The actual fountain will be 68 times as tall. How tall will the
fountain be?
The actual height of the fountain is 17 yards.
Explanation:
Given that a fountain is drawn $$\frac{1}{4}$$ yard tall and the actual fountain will be 68 times as tall. So the actual height of the fountain is $$\frac{1}{4}$$ × 68 which is 17 yards.
Question 7.
In blueprints, an architect’s firm drew everything $$\frac{1}{24}$$ of the actual size. The windows will actually measure 4 ft by 6 ft and doors measure 12 ft by 8 ft. What are the dimensions of the windows and the doors in the drawing?
Given that an architect’s firm drew everything $$\frac{1}{24}$$ of the actual size and the windows will actually measure 4 ft by 6 ft, so the dimensions of the length of the windows are $$\frac{1}{24}$$ × 4 which is $$\frac{1}{6}$$ ft, so in inch, it will be $$\frac{1}{6}$$ × 12 which is 2 in. And the width of the windows is $$\frac{1}{24}$$ × 6 which is $$\frac{1}{4}$$, so in inch, it will be $$\frac{1}{4}$$ × 12 which is 3 in. So the dimensions of the windows are 2 in by 3 in. Given that the measures of the door are 12 ft by 8 ft, so the dimensions of the length of the doors are $$\frac{1}{24}$$ × 12 which is $$\frac{1}{2}$$ ft, so in inch, it will be $$\frac{1}{2}$$ × 12 which is 6 in. And the width of the windows is $$\frac{1}{24}$$ × 8 which is $$\frac{1}{3}$$, so in inch, it will be $$\frac{1}{3}$$ × 12 which is 4 in. |
# How do you find the dimensions of a rectangular box that has the largest volume and surface area of 56 square units?
##### 1 Answer
Mar 17, 2015
The box should be a cube and each dimension should be $\setminus \frac{2 \setminus \sqrt{21}}{3} \setminus \approx 3.05$ units. The maximum volume is $\setminus \frac{56 \setminus \sqrt{21}}{9} \setminus \approx 28.5$ cubic units.
Let $x$, $y$, and $z$ be the dimensions of the box. Then $V = x y z$ and $S = 2 x y + 2 x z + 2 y z$ are the volume and surface areas of the box.
It might make some intuitive sense to you that the answer is a cube so that $x = y = z$. If it does, then the surface area being 56 square units implies that $S = 6 {x}^{2} = 56$ so that ${x}^{2} = {y}^{2} = {z}^{2} = \setminus \frac{56}{6} = \setminus \frac{28}{3}$ and $x = y = z = \setminus \sqrt{\setminus \frac{28}{3}} = \setminus \frac{2 \setminus \sqrt{7}}{\setminus \sqrt{3}} = \setminus \frac{2 \setminus \sqrt{21}}{3}$ and $V = x y z = {\left(\setminus \frac{2 \setminus \sqrt{21}}{3}\right)}^{3} = \setminus \frac{8 \setminus \cdot 21 \setminus \sqrt{21}}{27} = \setminus \frac{56 \setminus \sqrt{21}}{9}$
More rigorously, we can use calculus:
Since $S = 56$, we can say that $2 z \left(x + y\right) = 56 - 2 x y$ so that $z = \setminus \frac{28 - x y}{x + y}$. This means we can write the volume as a function of two variables:
$V = \setminus \frac{x y \left(28 - x y\right)}{x + y} = \setminus \frac{28 x y - {x}^{2} {y}^{2}}{x + y} .$
The partial derivatives of this function are:
$\setminus \frac{\setminus \partial V}{\setminus \partial x} = \setminus \frac{\left(x + y\right) \left(28 y - 2 x {y}^{2}\right) - \left(28 x y - {x}^{2} {y}^{2}\right) \setminus \cdot 1}{{\left(x + y\right)}^{2}} = \setminus \frac{{y}^{2} \left(28 - {x}^{2} - 2 x y\right)}{{\left(x + y\right)}^{2}}$
and
$\setminus \frac{\setminus \partial V}{\setminus \partial y} = \setminus \frac{\left(x + y\right) \left(28 x - 2 {x}^{2} y\right) - \left(28 x y - {x}^{2} {y}^{2}\right) \setminus \cdot 1}{{\left(x + y\right)}^{2}} = \setminus \frac{{x}^{2} \left(28 - {y}^{2} - 2 x y\right)}{{\left(x + y\right)}^{2}} .$
Setting these both equal to zero results in the system of equations ${x}^{2} + 2 x y = 28 , {y}^{2} + 2 x y = 28$. The first of these equations can be solved for $y$ to get $y = \setminus \frac{28 - {x}^{2}}{2 x}$.
Plugging this into the second equation leads to ${\left(\setminus \frac{28 - {x}^{2}}{2 x}\right)}^{2} + 2 x \setminus \cdot \setminus \frac{28 - {x}^{2}}{2 x} = 28$. Multiplying both sides of this last equation by $4 {x}^{2}$ gives ${\left(28 - {x}^{2}\right)}^{2} + 112 {x}^{2} - 4 {x}^{4} = 112 {x}^{2}$, which reduces to $3 {x}^{4} + 56 {x}^{2} - 784 = 0$.
Using the quadratic formula, ${x}^{2} = \setminus \frac{- 56 \setminus \pm \setminus \sqrt{{56}^{2} - 4 \setminus \cdot 3 \setminus \cdot \left(- 784\right)}}{2} = \setminus \frac{- 56 \setminus \pm \setminus \sqrt{12544}}{6} = \setminus \frac{- 56 \setminus \pm 112}{6}$. The positive one is ${x}^{2} = \setminus \frac{56}{6} = \setminus \frac{28}{3}$ so that $x = \setminus \sqrt{\setminus \frac{28}{3}} = \setminus \frac{2 \setminus \sqrt{7}}{\setminus \sqrt{3}} = \setminus \frac{2 \setminus \sqrt{21}}{3} \setminus \approx 3.05$. The symmetry in the original equations implies that $y = \setminus \frac{2 \setminus \sqrt{21}}{3} \setminus \approx 3.05$ as well. If you use the equation $z = \setminus \frac{28 - x y}{x + y}$ you'll get $z = \setminus \frac{2 \setminus \sqrt{21}}{3} \setminus \approx 3.05$ as well. That this gives a maximum value can be checked with the 2nd derivative test. |
## What are the characteristics of a linear function?
Linear functions are those whose graph is a straight line. A linear function has one independent variable and one dependent variable. The independent variable is x and the dependent variable is y. a is the constant term or the y intercept.
## What are the 4 types of linear functions?
Summary. Students learn about four forms of equations: direct variation, slope-intercept form, standard form and point-slope form.
## What is a linear function 7th grade?
A linear function is a function that represents a straight line on the coordinate plane. For example, y = 3x – 2 represents a straight line on a coordinate plane and hence it represents a linear function.
## What are the types of linear functions?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form.
## What are the 5 examples of linear equation?
Some of the examples of linear equations are 2x – 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, 3x – y + z = 3.
## What linear function means?
Definition of linear function
1 : a mathematical function in which the variables appear only in the first degree, are multiplied by constants, and are combined only by addition and subtraction.
## How do you find a linear function?
The linear function formulas are: y = mx + b (slope-intercept form)
1. (x, y) in every equation is a general point on the line.
2. (x1,y1) ( x 1 , y 1 ) is any fixed point on the line.
3. m is the slope of the line. …
4. (a, 0) and (0, b) are the x-intercept and y-intercept respectively.
5. A, B, and C are constants.
## How do you write a linear function?
A linear function is represented by the equation y = mx + b where:
1. y is the y-coordinate.
2. m is the slope of the line, or how steep it is.
3. x is the x-coordinate.
4. b is the y-intercept, or where the line crosses the y-axis on a graph.
## How do you show a function is linear?
A linear function must satisfy f(cx)=cf(x) for any number c. The other requirement for a linear function is that applying f to the sum of two inputs x and y is the same thing as adding the results from being applied to the inputs individually, i.e., f(x+y)=f(x)+f(y).
## How do you know if a function is linear?
So linear functions, the way to tell them is for any given change in x, is the change in y always going to be the same value. For example, for any one-step change in x, is the change in y always going to be 3? Is it always going to be 5? If it’s always going to be the same value, you’re dealing with a linear function.
## How do you find a linear function?
The linear function formulas are: y = mx + b (slope-intercept form)
## What is not a linear function?
Non-linear means the graph is not a straight line. The graph of a non-linear function is a curved line. A curved line is a line whose direction constantly changes. A cautionary note: Economists are accustomed to designate all lines in graphs as curves – both straight lines and lines which are actually curved.
## How do you find linear function?
A linear function is represented by the equation y = mx + b where:
1. y is the y-coordinate.
2. m is the slope of the line, or how steep it is.
3. x is the x-coordinate.
4. b is the y-intercept, or where the line crosses the y-axis on a graph.
## Which equation represent a linear function?
A linear function is a function of the form f(x) = ax + b, where a and b are real numbers. Here, a represents the gradient of the line, and b represents the y-axis intercept (which is sometimes called the vertical intercept).
## How do you write a linear function from a word problem?
Writing Systems of Linear Equations from Word Problems
1. Understand the problem. Understand all the words used in stating the problem. Understand what you are asked to find. …
2. Translate the problem to an equation. Assign a variable (or variables) to represent the unknown. …
3. Carry out the plan and solve the problem.
## Why is f called a linear function?
A linear function is a function of the form f(x) = mx + b, where m and b are constants. We call these functions linear because there graphs are lines in the plane. |
# FINDING LCM AND HCF OF POLYNOMIALS AND RELATIONSHIP WORKSHEET
Finding LCM and HCF of Polynomials and Relationship Worksheet :
Here we are going to see some practice questions on finding HCF and LCM.
To find GCD or LCM, first we have to find the factors of the given expression or polynomial.
• If we have coefficients of x or y term, then we have to decompose the coefficients as much as possible.
• If we have quadratic or cubic expression, then we have to factorize using suitable algebraic identities.
• To get GCD, multiply the common factors
• To get LCM, multiply highest factors.
## Finding LCM and HCF of Polynomials and Relationship Worksheet - Questions
(1) Find the LCM and GCD for the following and verify that f (x) × g(x) = LCM × GCD
(i) 21x2y, 35xy2 Solution
(ii) (x3 −1)(x +1), (x3 +1) Solution
(iii) (x2y + xy2), (x2 + xy) Solution
(2) Find the LCM of each pair of the following polynomials
(i) a2 + 4a −12, a2 −5a + 6 whose GCD is a -2 Solution
(ii) x 4 -27a3x, (x -3a)2 whose GCD is (x -3a) Solution
(3) Find the GCD of each pair of the following polynomials
(i) 12(x4 -x3), 8(x4 −3x3 +2x2) whose LCM is 24x3(x -1)(x -2)
(ii) (x3 + y3), (x4 + x2y2 + y4) whose LCM is (x3 + y3)(x2 + xy + y2)
(4) Given the LCM and GCD of the two polynomials p(x) and q(x) find the unknown polynomial in the following table
(i)LCM= a3 −10a2 +11a + 70GCD = a - 7p(x) = a2 −12a + 35find q(x) (ii)LCM = (x2 +y2)(x4 +x2y2+y4)GCD = (x2 -y2) (x4 −y4)q(x) = (x2 +y2 −xy)find p(x)
After having gone through the stuff given above, we hope that the students would have understood, how to find HCF and LCM of polynomials.
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Word problems on simple equations
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Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
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Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Beginners Guide to Arithmetic Operations on Korn Shell Variables
Place an integer expression in two pairs of parentheses ((…)) to invoke an Arithmetic evaluation. For readability, place a space before and after any arithmetic operator within the double pair of parentheses. The table below lists the standard arithmetic operations.
Operator Operation Example Result
+ Addition ((x = 24 + 25)) 49
- Subtraction ((x = 100 - 25)) 75
* Multiplication ((x = 4 * 5)) 20
/ Division ((x = 10 / 3)) 3
% Integer remainder ((x = 10 % 3)) 1
All arithmetic operations are performed using integer arithmetic, which can cause minor surprises:
``````\$ x=15.38
\$ y=15.72
\$ ((z = x + y))
\$ echo \$z
30
``````
Everything is integer arithmetic. Note what happens when division is performed in a calculation. For example, when computing the percentage of the eligible voters who actually voted, the following statement would most likely assign 0 to prct:
``````((prct = voted / eligible_voters * 100))
``````
The following algebraically equivalent statement is the correct way to compute the percentage of eligible voters in the Korn shell:
``````(((prct = voted * 100 / eligible_voters))
``````
## Arithmetic Precedence
• Expressions (operations) within parentheses are evaluated first. Use a single pair of parentheses around an expression to force it to be evaluated first.
• Multiplication (*) and division (% and /) have greater precedence than addition (+) and subtraction (-).
• Arithmetic expressions are evaluated from left to right.
• When there is the slightest doubt, use parentheses to force the evaluation order.
## Bit-wise Operations
The six bit-wise operators are listed in the table below:
Operator Operation Example Result
# base 2#1101010 or 16#6A 10#106
<< Shift bits left ((x = 2#11 << 3)) 2#11000
» Shift bits right ((x = 2#1001 » 2)) 2#10
& Bit-wise AND ((x = 2#101 & 2#110)) 2#100
Bit-wise OR ((x = 2#101
^ Bit-wise exclusive OR ((x = 2#101 ^ 2#110)) 2#11
The # operator designates the base of the value that follows it. For example, 2#101 means that the value 101 is a base 2 (binary) value (2#101 would be 5 in the decimal base).
The << operator performs a binary shift left by as many bits as are indicated by the number that follows the operator. The expression 2#10 << 1 yields the value 2#100. The expression 2#10100<< 2 yields the value 2#1010000.
Vacated positions are padded with 0 or 1 based on whether the number is positive or negative, respectively. The » operator performs a binary shift right by as many bits as are indicated by the number that follows the operator. The expression 2#10 » 1 yields the value 2#1. The expression 2#10100 » 2 yields the value 2#101.
The & operator ANDs two binary numbers together. This means that the AND operation is performed on the corresponding digit of 0 or 1 in each number, resulting in a 1 if both digits are 1; otherwise, the result is 0.
The | operator ORs two binary numbers together. This means that the OR operation is performed on the corresponding digit of 0 or 1 in each number, resulting in a 1 if either digit is a 1; otherwise, the result is 0.
The ^ operator performs an exclusive OR on two binary numbers together. This means that an exclusive OR is performed on the corresponding digit of 0 or 1 in each number, resulting in 1 if only one of the digits is a 1. If both digits are 0 or both are 1, then the result is 0. |
Triangles in Seventh Grade Common Core Math - dummies
# Triangles in Seventh Grade Common Core Math
Viewed on their own, the seventh grade standards about angles and triangles can seem a little odd. However, when you see these standards in context, you’ll understand why seventh graders study what they do in Common Core math.
The road from elementary school geometry to high school geometry has been rocky. Traditionally, children have spent a great deal of time recognizing and identifying shapes in the primary grades but have encountered a huge gap and advance little in their geometry knowledge until tenth grade, when they’re expected to prove geometry theorems.
The Common Core standards seek to fill that gap by slowly making the geometry more challenging at every grade, which means for seventh graders that they’re taking on a small part of what is traditionally in a high school geometry course. Seventh graders play around with drawing triangles in an attempt to determine precisely what conditions are necessary in order to know that two triangles are identical to each other (the math term for identical is congruent).
If the triangles have three pairs of congruent sides and three pairs of congruent angles, then the two triangles are congruent, as shown in the figure.
Congruent triangles.
But the three angles of a triangle add up to 180 degrees, so if you only know that two of the three pairs of angles are congruent, the third pair must be, which means that you only need three pairs of sides and two pairs of angles. Seventh graders build this kind of argument as they investigate which conditions must give congruent triangles and which may not.
Studying the properties of angles supports some of this work. A bunch of vocabulary goes along with it. Two angles are supplementary if their measures add to 180 degrees. Two angles are complementary if their measures add to 90 degrees. Two angles are vertical if they result from two intersecting lines but have only a vertex in common. Two angles are adjacent if they share a side. |
# Second Term 05/061 Roots of Equations Bracketing Methods.
## Presentation on theme: "Second Term 05/061 Roots of Equations Bracketing Methods."— Presentation transcript:
Second Term 05/061 Roots of Equations Bracketing Methods
Second Term 05/062 Root We are given f(x), a function of x, and we want to find α such that f(α) = 0 α is called the root of the equation f(x) = 0, or the zero of the function f(x)
Second Term 05/063 Example: Interest Rate Suppose you want to buy an electronic appliance from a shop and you can either pay an amount of 12,000 or have a monthly payment of 1,065 for 12 months. What is the corresponding interest rate? A is the monthly payment P is the loan amount x is the interest rate per period of time n is the loan period To find the yearly interest rate, x, you have to find the zero of We know the payment formulae is:
Second Term 05/064 Finding Roots Graphically Not accurate However, graphical view can provide useful info about a function. –Multiple roots? –Continuous? –etc.
Second Term 05/065 The following root finding methods will be introduced: A. Bracketing Methods A.1. Bisection Method A.2. Regula Falsi B. Open Methods B.1. Fixed Point Iteration B.2. Newton Raphson's Method B.3. Secant Method
Second Term 05/066 Bracketing Methods By Mean Value Theorem, we know that if a function f(x) is continuous in the interval [a, b] and f(a)f(b) < 0, then the equation f(x) = 0 has at least one real root in the interval (a, b).
Second Term 05/067 Usually f(a)f(b) > 0 implies zero or even number of roots –[figure (a) and (c)] f(a)f(b) < 0 implies odd number of roots –[figure (b) and (d)]
Second Term 05/068 Exceptional Cases Multiple roots –Roots that overlap at one point. –e.g.: f(x) = (x-1)(x-1)(x-2) has a multiple root at x=1. Functions that discontinue within the interval
Second Term 05/069 Algorithm for bracketing methods Step 1: Choose two points x l and x u such that f(x l )f(x u ) < 0 Step 2: Estimate the root x r (note: x l < x r < x u ) Step 3: Determine which subinterval the root lies: if f(x l )f(x r ) < 0 // the root lies in the lower subinterval set x u to x r and goto step 2 if f(x l )f(x r ) > 0 // the root lies in the upper subinterval set x l to x r and goto step 2 if f(x l )f(x r ) = 0 x r is the root
Second Term 05/0610 How to select x r in step 2? 1.Bisection Method Guess without considering the characteristics of f(x) in (xl, xu) 2.False Position Method (Regula Falsi) Use "average slope" to predict the root
Second Term 05/0611 A.1. Bisection Method Each guess reduce the search interval by half
Second Term 05/0612 Bisection Method – Example Find the root of f(x) = 0 with an approximated error below 0.5%. (True root: α=14.7802 )
Second Term 05/0613 Example (continue) nxlxl xrxr xuxu f(xl)f(xl)f(xr)f(xr)f(xu)f(xu)f(xl)f(xu)f(xl)f(xu)εaεa 012166.067-2.269 11214166.0671.569-2.269> 0 21415161.569-0.425-2.269< 06.667% 31414.5151.5690.552-0.425> 03.448% 414.514.75150.5520.0590-0.425> 01.695% 514.7514.875150.0590-0.184-0.425< 00.840% 614.7514.812514.8750.0590-0.0629-0.184< 00.422%
Second Term 05/0614 Error Bounds The true root, α, must lie between x l and x u. xlxl xuxu xrxr x l (1) x u (1) After the 1 st iteration, the solution, x r (1), should be within an accuracy of x r (1) Let x r (n) denotes x r in the n th iteration
Second Term 05/0615 Error Bounds x l (2) x u (1) x u (2) Suppose the root lies in the lower subinterval. x r (2) After the 2 nd iteration, the solution, x r (2), should be within an accuracy of
Second Term 05/0616 Error Bounds In general, after the n th iteration, the solution, x r (n), should be within an accuracy of If we want to achieve an absolute error of no more than E α
Second Term 05/0617 Implementation Issues The condition f(x l )f(x r ) = 0 (in step 3) is difficult to achieve due to errors. We should repeat until x r is close enough to the root, but we don't know what the root is! Therefore, we have to estimate the error as and repeat until e a < e s (acceptable error)
Second Term 05/0618 Bisection Method (as C function) // xl, xu: Lower and upper bound of the interval // es: Acceptable relative percentage error // xr: Estimated root in zero iteration (can simply // take xl or xu) // iter_max: Maximum # of iterations double Bisect(double xl, double xu, double es, double xr, int iter_max) { double xr_old; // Est. root in the previous step double ea; // Est. error int iter = 0; // Keep track of # of iterations do { iter++; xr_old = xr;
Second Term 05/0619 xr = (xl + xu) / 2; // Estimate root if (xr != 0) ea = fabs((xr – xr_old) / xr) * 100; test = f(xl) * f(xr); if (test < 0) xu = xr; else if (test > 0) xl = xr; else ea = 0; } while (ea > es && iter < iter_max); return xr; }
Second Term 05/0620 Additional Implementation Issues Function call is a relatively slow operation. In the previous example, function f() is called twice in each iteration. Is it necessary? –We only need to update one of the bounds (see step 3 in the algorithm for the bracketing method).
Second Term 05/0621 Revised Bisection Method (as C function) double Bisect(double xl, double xu, double es, double xr, int iter_max) { double xr_old; // Est. root in the previous step double ea; // Est. error int iter = 0; // Keep track of # of iterations double fl, fr; // Save values of f(xl) and f(xr) fl = f(xl); do { iter++; xr_old = xr; xr = (xl + xu) / 2; // Estimate root fr = f(xr);
Second Term 05/0622 if (xr != 0) ea = fabs((xr – xr_old) / xr) * 100; test = fl * fr; if (test < 0) xu = xr; else if (test > 0) { xl = xr; fl = fr; } else ea = 0; } while (ea > es && iter < iter_max); return xr; }
Second Term 05/0623 Comments on Bisection Method The method is guaranteed to converge. However, the convergence is slow as we gain only one binary digit in accuracy in each iteration.
Second Term 05/0624 A.2. Regula Falsi Method Also known as the false-position method, or linear interpolation method. Unlike the bisection method which divides the search interval by half, regula falsi interpolates f(x u ) and f(x l ) by a straight line and the intersection of this line with the x-axis Is used as the new search position. The slope of the line connecting f(x u ) and f(x l ) represents the "average slope" (i.e., the value of f'(x)) of the points in [x l, x u ].
Second Term 05/0625
Second Term 05/0626 False-position vs Bisection False position in general performs better than bisection method. Exceptional Cases: –(Usually) When the deviation of f'(x) is high and the end points of the interval are selected poorly. –For example,
Second Term 05/0627 Iteration xlxl xuxu xrxr ε a (%)ε t (%) 101.30.6535 20.651.30.97533.325 30.9751.31.137514.313.8 40.9751.13751.056257.75.6 50.9751.056251.0156254.01.6 Iteration xlxl xuxu xrxr ε a (%)ε t (%) 101.30.0943090.6 20.094301.30.1817648.181.8 30.181761.30.2628730.973.7 40.262871.30.3381122.366.2 50.338111.30.4078817.159.2 Bisection Method (Converge quicker) False-position Method
Second Term 05/0628
Second Term 05/0629 Summary Bracketing Methods –f(x) has the be continuous in the interval [ x l, x u ] and f(x l )f(x u ) < 0 –Always converge –Usually slower than open methods Bisection Method –Slow but guarantee the best worst-case convergent rate. False-position method –In general performs better than bisection method (with some exceptions). |
+0
# 23MT
0
78
1
+929
HHow do you do this?
Nov 10, 2019
#1
+2551
+2
This is hard, dgfgrafgdfge111 hang on for a ride that is very confusing and is probably wrong.
FIND $$SQ$$
Based on pythagorean thoerem, we can see that it is $$25$$
Next, we find the altitude of $$\Delta{SPQ}$$.
We FIRST do that by finding the area, which is 20*15 = 300 / 2 = 150
Using the triangle area formula $$\frac{1}{2}ab=150$$, with A being the altitude, and B being the base, we can plug in and solve for altitude.
The base is 25, which is SQ.
So plugging in:
$$\frac{1}{2}a(25)=150$$
$$\frac{1}{2}a=6$$
$$a=12$$
So we find the altitude is 12. SRQ is congruent to SPQ, so SRQ also has an altitude of 12.
To help us visualize better, we unfold the two visible faces of the cube with triangles SPQ and SRQ from three dimensions into 2 dimensions.
If folded back into 3 dimensions, the length of the orange line will be the answer. (try to imagine it, I know its hard)
The orange, red, and half the top blue line form a right triangle in 3 dimensions.
While the red, green, and half the bottom blue form a right triangle in 2 dimensions.
The length of the blue line is 24, because we know the altitude is 12, because 12 * 2 = 24.
We need to know the length of the green line. So we can use pythagorean theorem to find the length of the red line.
We can use some geoemtric proofs to prove that SQ Is trisected. And the green part is 1/3 of SQ. (ask for the proof if you need it there is a lot to write.)
So since it is proved that SQ Is trisected, we know that the green length is $$\frac{25}{3}$$.
So, after millions of years of calculations, I think i got (c) as the answer
I am not so sure, so please check!
Oof my brain got fried
Nov 10, 2019
edited by CalculatorUser Nov 10, 2019 |
# SOLVING ONE STEP EQUATIONS
## About "Solving one step equations"
"Solving one step equations" is nothing but the initial stuff of learning algebra in math.
A one-step equation is as straightforward as it sounds. We just have to perform one step in order to solve the equation.
We have to isolate the variable which comes in the equation.
That is, we have to get rid of the number which is added to the variable or subtracted from the variable or multiplied by the variable or divides the variable.
## Solving one step equations - Examples
Example 1 :
Solve : 5 + x = 3
Solution :
Here 5 is added to the variable "x". To get rid of 5, we have to take "negative 5" on both sides and solve the equation as explained below.
Hence, the vale of "x" is "-2"
Let us look at the next example on "Solving one step equations"
Example 2 :
Solve : p - 7 = 3
Solution :
Here 7 is subtracted from the variable "p". To get rid of "-7", we have to take "positive 7" on both sides and solve the equation as explained below.
Hence, the vale of "p" is "10"
Let us look at the next example on "Solving one step equations"
Example 3 :
Solve : 2r = 6
Solution :
Here "r" is multiplied by 2. To get rid of 2, we have to divide by 2 on both sides and solve the equation as explained below.
Hence, the vale of "r" is "3"
Let us look at the next example on "Solving one step equations"
Example 4 :
Solve : (1/4)m = 3
Solution :
Here "m" is divided by 4. To get rid of 4, we have to divide by 4 on both sides and solve the equation as explained below.
Hence, the vale of "m" is "12"
Example 5 :
Solve : 8 - p = 12
Solution :
Here "p" is having negative sign.
In this problem, first we have to make "p" to be positive.
For that, we have to add "p" on both sides. When we do so, we will have "12 + p" on the right side of the equation.
Then, to get rid of "12" on the right side, we have to subtract 12 on both sides.
Therefore, we have to add "p" and subtract "12" on both sides and solve the equation as explained below.
Hence, the vale of "p" is "-4"
## Solving one step equations - Practice problems
Problem 1 :
Solve : m - 10 = -15
Solution :
Here "10" is subtracted from "m". To get rid of 10, we have to add 10 on both sides and solve the equation as explained below.
(m - 10) + 10 = (-15) + 10
m = -5
Hence, the value of "m" is "-5
Problem 2 :
Solve : v + 5 / 3 = - 1/ 3
Solution :
Here "5/3" is added to "v". To get rid of 5/3, we have to subtract 5/3 on both sides and solve the equation as explained below.
(v + 5/3) - 5/3 = (-1/3) - 5/3
v = -2
Hence, the value of "v" is "-2"
Let us look at the next problem on "Solving one step equations"
Problem 3 :
Solve : 38 - m = -44
Solution :
Here "m" is having negative sign.
In this problem, first we have to make "m" to be positive.
For that, we have to add "m" on both sides. When we do so, we will have "- 44 + m" on the right side of the equation.
Then, to get rid of "-44" on the right side, we have to add 44 on both sides.
Therefore, we have to add "m" and "44" on both sides and solve the equation as explained below.
(38 - m) + m + 44 = (-44) + m + 44
82 = m
Hence, the value of "m" is "82"
Problem 4 :
Solve : 11.7 = 1.1 + n
Solution :
Here "1.1" is added to "n". To get rid of 1.1, we have to subtract 1.1 on both sides and solve the equation as explained below.
(11.7) -1.1 = (1.1 + n) - 1.1
10.6 = n
Hence, the value of "n" is "10.6"
Let us look at the next problem on "Solving one step equations"
Problem 5 :
Solve : -25.7 - v = -40.3
Solution :
Here "v" is having negative sign.
In this problem, first we have to make "v" to be positive.
For that, we have to add "m" on both sides. When we do so, we will have "- 40.3 + v" on the right side of the equation.
Then, to get rid of "-40.3" on the right side, we have to add 40.3 on both sides.
Therefore, we have to add "v" and "40.3" on both sides and solve the equation as explained below.
(-25.7 - v) + v + 40.3 = (-40.3) + v + 40.3
14.6 = v
Hence, the value of "v" is "40.3"
## Solving one step equations - Word problems
Problem 1 :
When 7 is added to a number, we get 25. Find the number.
Solution :
Let "x' be the required number.
According to the question, we have
x + 7 = 25
Here "7" is added to "x". To get rid of 7, we have to subtract 7 on both sides and solve the equation as explained below.
(x + 7) - 7 = (25) - 7
x = 18
Hence, the required number is "18".
Let us look at the next word problem on "Solving one step equations"
Problem 2 :
When we multiply a number by 4, we get 124. Find the number.
Solution :
Let "x' be the required number.
According to the question, we have
4x = 124
Here "x" is multiplied by "4". To get rid of 4, we have to divide by 4 on both sides and solve the equation as explained below.
4x / 4 = 124 / 4
x = 31
Hence, the required number is "31".
Let us look at the next word problem on "Solving one step equations"
Problem 3 :
When we divide a number by 7, we get 14. Find the number.
Solution :
Let "m' be the required number.
According to the question, we have
m / 7 = 14
Here "m" is divided by "7". To get rid of 7, we have to multiply by 7 on both sides and solve the equation as explained below.
(m/7) x 7 = 14 x 7
m = 98
Hence, the required number is "98".
Let us look at the next word problem on "Solving one step equations"
Problem 4 :
John had some candies. He gave 5 candies to his friend and now he has 18 candies. How many candies did John initially have ?
Solution :
Let "m' be the no. of candies that John initially had.
According to the question, we have
m - 5 = 18
Here "5" is subtracted from "m". To get rid of 5, we have to add 5 on both sides and solve the equation as explained below.
(m - 5) + 5 = 18 + 5
m = 23
Hence, John initially had 23 candies.
Let us look at the next word problem on "Solving one step equations"
Problem 5 :
Alex borrowed some money from Jose. After 3 years, Alex returned 2 times of borrowed money to Jose. If the returned money is \$226, how much money did Alex borrow from Jose ?
Solution :
Let "x' be the borrowed money.
According to the question, we have
2x = 226
Here "x" is multiplied by 2. To get rid of 2, we have to divide by 2 on both sides and solve the equation as explained below.
2x / 2 = 226 / 2
x = 113
Hence, the borrowed money is \$113.
## Solving one step equations - Worksheets
Worksheet -1
Worksheet -2
Worksheet -3
Worksheet -4
Worksheet -5
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Lecture 1: The Real Numbers
1.1 What Are They?
This course will deal with multivariate calculus, basically the analytical study of regions of n-dimensional real space. Before considering this, we should review what we mean by the term 'real numbers'.
The basic idea is that the real numbers are the elements of the set of real numbers. However, a set is just a set; something about which we can take subsets and do other set theoretic operations. In order to understand this, we need to consider the set of real numbers as something with more structure. The set of real numbers is actually a complete ordered field.
Let us pause to describe what this means:
Definition 1: A field is a triple (F, +, *) where F is a set and + and * are binary operators (called addition and multiplication) which satisfy the following axioms:
1. Both addition and multiplication are commutative and associative and multiplication is distributive over addition. In symbols, this means that for all a, b, and c in F, one has:
1. a + b = b + a, a*b = b*a
2. (a + b) + c = a + (b + c), (a*b)*c = a*(b*c)
3. a*(b + c) = a*b + a*c
2. (Identities) There are two different elements 0 and 1 (called zero and one) in F such that for all a in F: a + 0 = a and a*1 = a.
3. (Inverses) For every a in F, there is at least one b in F such that a + b = 0. Further, if a is non-zero, there is at least one c in F such that a*b = 1. The elements b (respectively c) are called the additive (respectively the multiplicative) inverses of a and are denoted -a (respectively 1/a).
Definition 2: An ordered field is a pair ((F, +, *), <) where (F, +, *) is a field and < is binary relation on F such that:
1. (Transitivity) For every a, b, c in F, if a < b and b < c, then a < c.
2. (Trichotomy) For every a and b in F, exactly one of the following holds:
1. a < b
2. a = b
3. b < a
3. For all a, b, and c in F, if a < b, then a + c < b + c. Further, if c > 0, then a*c < b*c.
It is assumed that you have probably already had a course in which you showed that most of the standard rules of algebra follow from the assumption that one has an ordered field. (If you haven't ever done this before, then take home and work through Landau's Foundations of Analysis; he will show you in a weekend read how to derive the basic properties of the real numbers starting from Peano postulates. Just for practice, however, you might try to show:
Exercise 1 In any field, additive and multiplicative identities and inverses are unique. In an ordered field, if a < b and c < 0, then b*c < a*c.
Definition 3: A subset S of an ordered field F is bounded above by an element a in F if b < a for all b in S. The upper bound a is called a least upper bound for S if every other upper bound of S is larger than a.
Definition 4: An ordered field F is said to be complete if every subset S of F which has an upper bound, also has a least upper bound.
Exercise 2: Show that least upper bounds are unique. Mimic the above definition to define lower bounds and greatest lower bounds. Show that in complete fields, every set S with a lower bound has a greatest lower bound.
It is useful to have the notion of complete ordered field in order to sort out some basic properties of real numbers. But the real value is that these assumptions alone are enough to completely characterize the real numbers. The result is expressed in terms of:
Definition 5: An order isomorphism between two ordered fields ((F,+,*),<) and ((G,+,*),<) is a 1-1 and onto map f:F ->G such that for all a and b in f, one has:
1. f(a + b) = f(a) + f(b)
2. f(a*b) = f(a)*f(b)
3. If a < b, then f(a) < f(b).
So basically, two ordered fields are order isomorphic if we cannot distinguish between them by using their addition and multiplication operations and their order relation. In these terms the basic result is:
Theorem 1: There is a complete ordered field and any two complete ordered fields are order isomorphic.
This result will not be proved here, but is found in algebra books. Here are some of the ideas of a proof:
1. Using the elements 0 and 1, one can build up the integers through addition, and then one can take quotients to get the rational numbers Q.
2. Define a sequence of elements of an ordered field to be a Cauchy sequence if for every , there is an n such that for any i, j greater than n, one has . We say that the sequence has limit x if for every , there is an n such that for every i > n, one has .
3. Two sequences and are said to be equivalent if the sequence has limit 0. You can show that this is an equivalence relation.
4. If and are Cauchy sequences, then adding or multiplying corresponding terms gives new Cauchy sequences. Similarly, taking additive inverses gives a Cauchy sequence and so does taking multiplicative inverses if the original sequence has a non-zero limit and none of the elements are zero.
5. So that equivalence is preserved under addition and multiplication of Cauchy sequences.
6. Take the set S of Cauchy sequences with elements in the rational numbers Q. The set of equivalence classes of such sequences can be shown to be a complete ordered field.
1.2 The Cauchy-Schwartz Inequality
Henceforth, let use let denote the real numbers, i.e. the set part of a complete ordered field. By taking n-tuples of real numbers we get n-dimensional real space, denoted ; its elements are called points. The points are demoted or simply (x^i). Points can be added componentwise and we can multiply a real number by a point to get another point whose coordinates are just the original coordinates multiplied by the real number. Of course, this makes into an n-dimensional vector space. But, we can also define a dot product:
As usual, this allows us to define the length of a point as the square root of the dot product of the point with itself:
Now the length function behaves like you would expect. For example:
Proposition 1 (Triangle Inequality) For any two points x and y in , one has
Proof: Squaring both sides, we see this amounts to showing that
is at most
Comparing terms, we see that the triangle inequality is equivalent to:
Proposition 2 (Cauchy-Schwartz Inequality) For any two points x and y in , one has , i.e.
Exercise 3: Draw a picture in the case n = 2 and identify various factors of the left hand side divided by the right to convince yourself that the result is basically equivalent to the addition formula for cosines.
Recall from your calculus course, that the dot product was shown to be:
where is the angle between the vectors from the origin to the points x and y respectively. This was probably only shown in the case n = 2; but it is another way of understanding why Proposition 2 should be true. In our case, we will actually do things in the opposite direction, i.e. prove that Proposition 2 is true and use it to define the angle .
The main use of the dot product in Calculus was to give you the means of calculating the projection of one vector on another. Using this interpretation, can you see why Proposition 2 should be true?
Given two vectors (i.e. two points) x and y, if we subtract from x the vector projection p of x on y, then we get a vector perpendicular to y. Now, geometrically, this vector should be smallest vector amongst all vectors of the form for all possible real . Geometrically, we have `solved' the problem: Minimize .
Ahh! we are finally doing calculus!
So let's do calculus: To minimize this, we should minimize the square,
If x or y is zero, this is easy (What is the answer?); so assume both are non-zero. Now, if think of x and y as being constants, then this is just a parabola. So, either from your knowledge of parabolas or from elementary calculus, it is easy to see when this is minimized: Setting the derivative to zero gives . Substituting back shows that the minimum value is . Since this minimum value is non-negative, we have just proved Proposition 2.
A slight variant: We know that the expression on the right side of the displayed formula must be positive unless x is a multiple of y. But, if it is positive for all values of , then the roots must both be complex and so the discriminant is negative. But this is precisely the Cauchy-Schwartz inequality.
Remark: Make sure you sort out all the geometry from the proof to be sure we really have a proof here and not just heuristic hand waving. When you have done this, compare to the proof in Spivak -- hopefully, then you will understand his very slick proof.
Another approach entirely Let's go back to the notion of angle between two vectors. Along with the dot product, you had a vector cross product in the case of . It gave you a vector that was perpendicular to the original two vectors, say v and w, and its length was precisely . It could also be calculated as the value of the determinant of a strange looking matrix in which the top row had the unit vectors i, j, and k in the coordinate axis directions and the other two rows had the coordinates of v and w respectively. Remember the formula:
(If not, then it is time to prove it -- yes, the Pythagorean Theorem in its many guises, including trigonometry, is your best friend.)
Identities are mysterious and wonderful. If you write the above in terms of all the coordinates, then you simply have an algebraic identity.
Exercise 4: Use the above identity to prove the Cauchy-Schwartz inequality. Now generalize it to n dimensions and get another proof of Proposition 2. |
In this lesson, you will learn how to calculate the probability of a permutation by analyzing a real-world example in which the order of the events does matter. We’ll also review what a factorial is. We will then go over some examples for practice.
## Probability
A probability is the likelihood of an event occurring.
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When calculating the probability of an event, use the formula number of favorable outcomes over the number of total outcomes. We can calculate the probability of any event occurring by examining the choices that are available. Let’s look at an example to see how simple it is to find the probability.
Jimmy bought a bag full of gumballs at the store. In the bag there were 20 red, 15 blue, and 12 yellow gumballs. If Jimmy reaches in the bag and selects a gumball at random, what is the probability that he will select a red gumball?The first thing that Jimmy knows is that if he adds the number of blue, red, and yellow gumballs together, he has a total of 47 gumballs in the bag. Jimmy knows that to calculate the probability of selecting a red gumball, he must use the equation the number of favorable outcomes over the number of total outcomes.Since there were 20 red gumballs, the number of favorable outcomes is 20. Also, there were 47 total gumballs, so the number of total outcomes is 47.
We can see that the probability of Jimmy selecting a red gumball from the bag is 20/47.To finish calculating the probability, we need to always check and see that our fraction is in its simplest form. Fortunately for Jimmy, his fraction is in simplest form. So the probability of Jimmy drawing a red gumball from the bag at random is 20/47.
## Permutations
To calculate the total number of outcomes, you occasionally have to use a permutation. A permutation is a method to calculate the number of events occurring where order matters. To calculate a permutation, you will need to use the formula nPr = n! / (nr)!. In this equation, n represents the number of items to choose from and r represents how many items are being chosen.Another function of this equation is the use of the ! (exclamation point). This symbol stands for ‘factorial.
‘ A factorial is the product of all positive integers equal to and less than your number. A factorial is written as the number followed by an exclamation point. For example, to write the factorial of 5, you would write 5!.
To calculate the factorial of 5, you would multiply the positive integers equal to and less than 5. 5! = 5 x 4 x 3 x 2 x 1. By multiplying these numbers together, we can find that 5! = 120. Let’s look at another example: how would we write and solve the factorial of 10? The factorial of 10 would be written as 10!. To calculate: 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800.Let’s look at an example where we must calculate the permutation.
At Rockland University, it is time for the student government election. There are 20 students who have submitted the necessary paperwork to appear on the ballot. How many different ways can the student body select a President, Vice President, Secretary, and Treasurer?First of all, this example is a permutation because the order in which the students are selected for the positions does matter. We know that there are 20 students who have been chosen to run for office – this will represent the n variable in our equation. We also know that the student government consists of only 4 positions, which is our r term. So in this permutation, we have 20 items to choose from for the 4 positions. The equation for our permutation would look like 20P4 = 20! / (20 – 4)!.
To begin solving this permutation, we need to subtract the denominator. (20 – 4)! = 16! Next, let’s expand each factorial. As we work this equation, we’re going to cancel out common terms in both the numerator and denominator. We can see that there is a 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1 on both the top and bottom.
These terms can be cancelled out, leaving 20 x 19 x 18 x 17 on top and nothing on bottom, which becomes an understood 1.To solve this equation, we will now multiply the top and bottom of our fraction. 20 x 19 x 18 x 17 = 116,280. The denominator has a term of 1, so our fraction is 116,280/1. We can now see that there are 116,280 ways that the student government can be selected from those 20 students.
## Calculating Probability of Permutations
When calculating the probability of a permutation, we will be combining these two skills.
The use of permutations will give us the total number of outcomes and then we will need to calculate the total number of favorable outcomes. Let’s look at an example to see how these two skills can be combined.Jim was given a new smartphone with a 5-digit passcode. The passcode could contain any single digit number (0-9), but each number cannot be repeated. Unfortunately, Jim has forgotten his passcode and must try to guess the number. What is the probability of Jim guessing the correct 5-digit passcode?This problem will involve two processes: first of all, the use of permutations to calculate the number of passcodes that Jim could choose from, and secondly, the probabilities to calculate the chances of Jim selecting the correct passcode.
First, we need to calculate the permutation of Jim’s total outcomes. There are 10 items, and 5 of them are being chosen. This permutation would look like 10P5 when written in notation form. Next, we will need to plug in our values into the formula for a permutation. Remember, the equation to calculate a permutation is nPr = n! / (nr)!.
Let’s plug in our values. For all of the n variables we will plug in a 10 and for the r variables a 5. Our equation now looks like 10P5 = 10! / (10 – 5)!. To work this equation, we need to subtract on the bottom: (10 – 5) = 5!.
We now have 10!/5!. Next, let’s expand our factorials. On top, 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, and on bottom, 5! = 5 x 4 x 3 x 2 x 1.
As we continue working this equation, we will cancel out common terms in both the numerator and denominator. Since there is 5 x 4 x 3 x 2 x 1 on both the top and bottom, we can cancel out these terms. As these terms are canceled out, they disappear. Our equation now looks like 10 x 9 x 8 x 7 x 6 /1. As we multiply, 10 x 9 x 8 x 7 x 6 = 30,240. This means that there are 30,240 different outcomes for passcodes that Jim could use.To calculate the probability that Jim selects the correct passcode, we will use the equation total number of favorable outcomes over the total number of outcomes.
Since there is only one possible passcode that will unlock the phone, the number of favorable outcomes is 1. After calculating the permutation of events, we know that there are 30,240 total outcomes. So the probability that Jim selects the correct passcode is 1/30,240.
## Lesson Summary
In this video, we’ve looked at several examples in which we calculate the probability of a permutation. To find the probability of an event occurring, we will use the equation number of favorable outcomes over the number of total outcomes.When calculating the total number of outcomes, we sometimes need to use a permutation. A permutation is a method to calculate the number of events occurring where order matters. To calculate a permutation, we will need to use the formula nPr = n! / (nr)!. In this equation, n represents the number of items to choose from and r represents how many items are being chosen.
Once you have found the permutation, you will plug it in as the total number of outcomes. Remember after you find the probability that all probabilities must be in simplest form.
## Learning Outcome
After viewing the lesson, you should be able to:
• Define what permutations are
• Solve for probability
• Calculate a probability problem using permutations
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2.4 Solving Quadratic Inequalities There are several ways to solve quadratic inequalities. The first method examines each factor of the inequality and how various values affect them when substituted in for the variable. METHOD 1 Example 1. Solve Start by getting zero on one side of the inequality and everything else on the other side. (Why is it incorrect to say “set the inequality equal to zero”?) answer The inequality is easier to solve if the quadratic term is positive. The reason we get everything on one side of the inequality and zero on the other is because now we only have to worry about the sign of the expression on the left side. As long as x2 2x 3 is a negative number, when a value is substituted in for x, the inequality will be true. Notice on the number line: Any number less than zero is a negative number and any number greater than zero is a positive number. Zero is neither positive nor negative. Therefore, the symbolism is referring to all negative numbers and the symbolism is referring to all positive numbers. Instead of just randomly filling numbers in for x to find which values make the inequality true (which would take a lot of time), it is easier to first factor the expression and then see how each factor is affected. factor There are now two quantities being multiplied together. For the inequality to be true, their product must be a negative number when values are substituted in for x. The only way the product of two numbers can be negative is if one of the factors is positive and one is negative. Here are the possibilities for this problem: We will now look to see how every number affects each of the factors. Start by finding the numbers that when filled in for x will make each factor equal to zero. These numbers, 3 and -1, are called “critical values”. The factors are neither positive nor negative at these numbers because they, the factors, equal zero. For the factor x-3, at x = 3 the factor equals zero. For any number greater than 3 that is filled in for x, the factor will end up as a positive number. For any number less than 3 that is filled in for x, the factor will end up as a negative number. Try filling in some numbers for x and seeing for yourself that this is true. Example 2. For ; For the factor x+1, at x = -1 the factor equals zero. For any number greater than -1 that is filled in for x, the factor will end up as a positive number. For any number less than -1 that is filled in for x, the factor will end up as a negative number. Try filling in some numbers for x and seeing for yourself that this is true. Example 3. For ; Remember, we are looking for all the numbers that will make one factor a negative number and the other factor a positive number. We will set the two number lines up one above the other and take a look: Example 4. If we extend a line from each of the critical values points to the other number line, it will be easier to look at the given information: Example 5. Now we can see that any number to the left of -1 will cause each of the factors to be negative, any number between -1 and 3 will cause one factor to be negative and one to be positive and any number to the right of 3 will cause each factor to be positive. When two negative numbers are multiplied together, the product is positive. When two positive numbers are multiplied together, the product is positive. When a negative and a positive are multiplied together, the product is negative therefore we want the numbers between -1 and 3 so that when filled into both of the factors in the problem, the product will end up negative and less than zero. will be true when numbers between -1 and 3 are filled in for x. Example 6. Let x = 2, which is a number between -1 and 3. Then becomes which is < 0 and therefore makes the inequality true. Using inequality notation to write the solution: Using interval notation to write the solution: (-1,3) What happens if -1 or 3 is substituted in for x? We know that each of these numbers will cause one of the factors to equal zero. When one of the factors is zero the product will then be zero. Example 7. For let x = 3, one of the critical numbers The inequality then becomes We do not want the product to equal zero so the critical numbers are not included in the solution. NOTE: Critical numbers will always cause the product to equal zero when substituted in for the variable. Example 8. Solve Get everything on one side and zero on the other (keep the x2 term positive) Factor This time we want the product to be positive ( ), or equal to zero. Find the critical numbers (the values that make each factor equal zero) Set up the number line sign chart for each factor and the product. Look for the values that cause the product to be positive. NOTICE: This time there are two intervals that contain values that cause the product to be positive. All the values in each of these intervals will be solutions. Check the critical values to see if they will be part of the solution For the critical value, x = 6, is when 6 is substituted in for x ? Is ? Is ? Is ? YES! So 6 is a solution For the critical value, x = -2, is when -2 is substituted in for x ? Is ? Is ? Is ? YES! So -2 is a solution Write the solution Using inequality notation to write the solution: Using interval notation to write the solution: The second method for solving quadratic inequalities is a shortcut for the first method. Method 2 The first three steps of this method are identical to the first three steps in Method 1 1. Get everything on one side of the inequality sign and zero on the other. Be sure to keep the “x2” term positive. 2. Factor 3. Find the critical numbers Example 9. Solve Get everything on one side and zero on the other (keep the x2 term positive) Factor Find the critical values Now this is where Method 2 becomes a shortcut for Method 1. Instead of setting up separate number lines for each of the factors, both critical numbers will be marked on the same number line. The number line has now been divided into three intervals. Look back at Example 5. Notice how after a line was extended from each of the critical values, the number line became divided into three intervals and notice that all the values in each interval affected the product in the same way either all the values in the interval caused the product to be positive or all the values in the interval caused the product to be negative. Because we know that every value in each interval affects the product in the same way, in Method 2, after marking the critical values on the number line, we will pick a test value from each interval to represent that particular interval and see how the product is affected. Let’s label the intervals 1, 2 and 3 to make them easier to refer to. From Interval 1, let x be the test value -2. Remember, it doesn’t matter which value is picked as the test value because all the values in this interval will affect the product in the same way, from Interval 2, let x be the test value 0 (my favorite test value) and from Interval 3, let x = 4. Now fill each one of these test values into the original inequality - not the factored inequality - the original inequality, and see whether the inequality is then true or not. For Interval 1, if x = -2 then becomes Write the word “false” above Interval 1. For Interval 2, if x = 0 then becomes This inequality IS true therefore, all the values in Interval 2 will be solutions for the inequality. Write the word “true” above Interval 2. For Interval 3, if x = 4 then becomes Write the word “false” above Interval 3. If you want, you can now test the critical numbers by substituting them in for the variable one at a time to see how they affect the inequality. They will, however, always make one side of the inequality equal to the other side. For the critical number -1, if x = -1 then becomes This inequality is NOT true therefore, 1 is not one of the solutions for the inequality. Put a parenthesis (which means “not included”) on this number on the number line. For the critical number 3, if x = 3 then becomes This inequality is NOT true therefore, 3 is not one of the solutions either. Put a parenthesis also on this number on the number line. The correct solution for this inequality is all the numbers in Interval 2, the interval where the word “true” is written. Remember, however, that the critical numbers are not part of the solution. In inequality notation, the solution is -1 < x < 3. In interval notation, the solution is (-1,3). Example 10. Solve Get everything on one side and zero on the other (keep the x2 term positive) Factor Find the critical numbers for each factor For x-6, x = 6; for x+2, x = -2 Mark the critical numbers on the number line to create intervals Pick a test value from each interval to represent that interval. Fill it into the original inequality for the variable and see if it is a solution or not. For Interval 1, let x = -3 then becomes This inequality IS true therefore, all the values in Interval 1 will be solutions for the inequality. Write the word “true” above Interval 1. For Interval 2, let x = 0 then becomes This inequality is NOT true therefore, all the values in Interval 2 will not be solutions for the inequality. Write the word “false” above Interval 2. For Interval 3, let x = 7 then becomes This inequality IS true therefore, all the values in Interval 1 will be solutions for the inequality. Write the word “true” above Interval 3. Check the critical numbers to see if they are solutions also by filling them into the original inequality for the variable. For the critical number -2, if x = -2 then becomes This inequality IS true therefore, -2 is one of the solutions. Put a bracket (which means “included”) on this number on the number line. For the critical number 6, if x = 6 then becomes This inequality IS true therefore, 6 is also one of the solutions. Put a bracket (which means “included”) on this number on the number line. The correct solution for this inequality is all the numbers in Interval 1 and Interval 3, the intervals where the word “true” is written. In this example, the critical numbers are also part of the solution. In inequality notation, the solution is . In interval notation, the solution is . Method 3 is a graphical approach to solving the quadratic inequality Let y1 = x2 and y2 = 2x + 3 The graph of y1 = x2 is a parabola because it is a quadratic equation. Y2 = 2x + 3 is a linear equation and so its graph is a straight line with a slope of 2 and a y-intercept at (0,3). Now plot these two graphs on the same set of axes. When solving the inequality , you are actually solving the inequality where stands for the values (or outputs) of the equation after various values are filled in for x (or the input) and stands for the values (or outputs) of the equation after various values are filled in for x (or the input). When solving the inequality , all you need to do is look at the graphs of the two equations and find where the values are less than (or below) the values. With the above two graphs, it is rather easy to see that the graphs intersect at the points (-1,1) and (3,9). At these points, actually equals . Between the points (-1,1) and (3,9), the graph of is below the graph of and so between the points (-1,1) and (3,9), all the values are less than the values. The only thing left to do now to solve the inequality is to determine which x’s were used to draw this part of the graph. Look at the points listed on the following graphs. Notice that for each point, the x of the point that was used to get the y is the x on the x-axis that lies directly below the point. Can you see on the graphs below, then, that the x’s that cause the graph of to be less than (or below) the graph of are the x’s in between x = -1 and x = 3. These are the x’s that are the solutions to the inequality . In inequality notation, the solution is . In interval notation, the solution is (-1,3). Example 11. Solve using Method 3 Let y1 = x2 and y2 = 4x+12 Graph these two equations on the same set of axes You must now find where the two graphs intersect When the graphs intersect each other, the y values of the graphs are identical at these points: y1 = y2 . So set the equations equal to each other and solve for x. To find the y values that correspond to these x’s just plug the x’s into either equation. It doesn’t matter which equation because they will both end up equaling the same number. Test it for yourself. At x = 6, y1 = x2 becomes y1 = (6)2 y2 = 4x+12 becomes y2 = 4(6)+12 = 36 = 24+12 = 36 See, identical answers. At x = -2 y1 = (-2)2 = 4 The points of intersection for the graphs are at (-2,4) and (6,36). Now determine where or y1 = x2 is above y2 = 4x+12. Looking at the above graphs, it can be seen that y1 is above y2 to the left of the point (-2,4) and to the right of the point (6,36). Since in the original problem, you are trying to find the x’s that cause x2 to be greater than or equal to 4x+12, or the x’s used to construct the graphs that cause y1 to be greater than or equal to y2, the solution to the problem in inequality notation is . In interval notation, the solution is . solutions solutions |
Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
5.1 Math IC Algebra Strategies 5.2 Writing Equations 5.3 Manipulating Equations 5.4 Zero Product 5.5 Absolute Value 5.6 Inequalities 5.7 Systems of Equations
5.8 Common Word Problems 5.9 Logarithms 5.10 Polynomials 5.11 Key Formulas 5.12 Review Questions 5.13 Explanations
Inequalities
Before you get too comfortable with expressions and equations, we should introduce inequalities. An inequality is like an equation, but instead of relating equal quantities, it specifies exactly how two quantities are not equal. There are four types of inequalities:
1. x > y: “x is greater than y.”
2. x < y: “x is less than y.”
3. x ≥ y: “x is greater than or equal to y.”
4. x ≤ y: x is less than or equal to y.”
Solving inequalities is exactly like solving equations except for one very important difference: when both sides of an inequality are multiplied or divided by a negative number, the direction of the inequality switches.
Here are a few examples:
Solve for x in the inequality – 3 < 2y.
Solve for x in the inequality ≥ –2.
Notice that in the last example, the inequality had to be reversed. Another way to express the solution is x ≥ –2. To help remember that multiplication or division by a negative number reverses the direction of the inequality, remember that if x > y, then –x < –y, just as 5 > 4 and –5 < –4. Intuitively, this idea makes sense, and it might help you remember this special rule of inequalities.
Absolute Value and Inequalities
When absolute values are included in inequalities, the solutions come in two varieties.
1. If the absolute value is less than a given quantity, then the solution is a single range, with a lower and an upper bound. For example,
Solve for x in the inequality |2x – 4| ≤ 6.
• First, solve for the upper bound:
• Second, solve for the lower bound:
• Now, combine the two bounds into a range of values for x. –1 ≤ x ≤ 5 is the solution.
1. The other solution for an absolute value inequality involves two disjoint ranges: one whose lower bound is negative infinity and whose upper bound is a real number, and one whose lower bound is a real number and whose upper bound is infinity. This occurs when the absolute value is greater than a given quantity. For example,
Solve for x in the inequality |3x + 4| > 16.
• First, solve for the upper range:
• Then, solve for the lower range:
• Now combine the two ranges to form the solution, which is two disjoint ranges: –∞ < x < –203 or 4 < x < ∞.
When working with absolute values, it is important to first isolate the expression within absolute value brackets. Then, and only then, should you solve separately for the cases in which the quantity is positive and negative.
Ranges
Inequalities are also used to express the range of values that a variable can take. a < x < b means that the value of x is greater than a and less than b. Consider the following word-problem example:
A very complicated board game has the following recommendation on the box: “This game is only appropriate for people older than 40 but no older than 65.” What is the range of the age of people for which the board game is appropriate?
Let a be the age of people for which the board game is appropriate. The lower bound of a is 40, and the upper bound is 65. The range of a does not include its lower bound (it is appropriate for people “older than 40”), but it does include its upper bound (“no older than 65”, i.e., 65 is appropriate, but 66 is not). Therefore, the range of the age of people for which the board game is appropriate can be expressed by the inequality:
Here is another example:
A company manufactures car parts. As is the case with any system of mass production, small errors occur on virtually every part. The key for this company to succeed in making viable car parts is to keep the errors within a specific range. The company knows that a particular piece they manufacture will not work if it weighs less than 98% of its target weight or more than 102% of its target weight. If the target weight of this piece is 21.5 grams, in what range of weights must the piece measure for it to function?
The boundary weights of this car part are .98 21.5 = 21.07 and 1.02 21.5 = 21.93 grams. The problem states that the piece cannot weigh less than the minimum weight or more than the maximum weight in order for it to work. This means that the part will function at boundary weights themselves, and the lower and upper bounds are included. The answer to the problem is 21.07 ≤ x ≤ 21.93, where x is the weight of the part in grams.
Finding the range of a particular variable is essentially an exercise in close reading. Every time you come across a question involving ranges, you should carefully peruse the problem to pick out whether a particular variable’s range includes its bounds or not. This inclusion is the difference between “less than or equal to” and simply “less than.”
Operations on Ranges
Operations like addition, subtraction, and multiplication can be performed on ranges just like they can be performed on variables. For example:
If 4 < x < 7, what is the range of 2x + 3?
To solve this problem, simply manipulate the range like an inequality until you have a solution. Begin with the original range:
Then multiply the inequality by 2:
There is one crucial rule you need to know about multiplying ranges: if you multiply a range by a negative number, you must flip the greater-than or less-than signs. For instance, if you multiply the range 2 < x < 8 by –1, the new range will be –2 > x > –8. Math IC questions that ask you to perform operations on ranges of one variable will often test your alertness by making you multiply the range by a negative number.
Some range problems on the Math IC will be made slightly more difficult by the inclusion of more than one variable. In general, the same basic procedures for dealing with one-variable ranges applies to adding, subtracting, and multiplying two-variable ranges.
Addition with Ranges of Two or More Variables
If –2 < x < 8 and 0 < y < 5, what is the range of x + y?
Simply add the ranges. The lower bound is –2 + 0 = –2. The upper bound is 8 + 5 = 13. Therefore, –2 < x + y < 13.
Subtraction with Ranges of Two or More Variables
Suppose 4 < s < 7 and –3 < t < –1. What is the range of s – t?
In this case, you have to find the range of –t. By multiplying the range of t by –1 and reversing the direction of the inequalities, we find that 1 < –t < 3. Now we can simply add the ranges again to find the range of s – t. 4 + 1 = 5, and 7 + 3 = 10. Therefore, 5 < s – t < 10.
In general, to subtract ranges, find the range of the opposite of the variable being subtracted, and then add the ranges as usual.
Multiplication with Ranges of Two or More Variables
If –1 < j < 4 and 6 < k < 12, what is the range of jk?
First, multiply the lower bound of one variable by the lower and upper bounds of the other variable:
Then, multiply the upper bound of one variable with both bounds of the other variable:
The least of these four products becomes the lower bound, and the greatest is the upper bound. Therefore, –12 < jk < 48.
Let’s try one more example of performing operations on ranges:
If 3 ≤ x < 7 and , what is the range of 2(x + y)?
The first step is to find the range of x + y. Notice that the range of y is written backward, with the upper bound to the left of the variable. Rewrite it first:
Next add the ranges to find the range of x + y:
We have our bounds for the range of x + y, but are they included in the range? In other words, is the range 0 < x + y < 11, 0 ≤ x + y ≤ 11, or some combination of these two?
The rule to answer this question is the following: if either of the bounds that are being added, subtracted, or multiplied is non-inclusive (< or >), then the resulting bound is non-inclusive. Only when both bounds being added, subtracted, or multiplied are inclusive (≤ or ≥) is the resulting bound also inclusive.
The range of x includes its lower bound, 3, but not its upper bound, 7. The range of y includes both its bounds. Therefore, the range of x + y is 0 ≤ x + y < 11, and the range of 2(x + y) is 0 ≤ 2(x + y) < 22.
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# Reduced Row Echelon Form of Homogeneous Systems of Equations
5. In general, a matrix's row echelon form can vary a bit. A matrix's reduced row echelon
form is always unique. In other words, there is only one specific reduced row echelon form
matrix associated with each matrix.
(a) Consider the following homogeneous system:
2x1 ¡ x2 + x4 + 4x5 = 0
2x1 ¡ 2x2 + x3 + 4x4 ¡ 3x5 = 0
2x1 ¡ 4x2 + x3 + 6x4 + 6x5 = 0
Write this system as an augmented matrix, use your calculator and ¯nd the row echelon
form, and the reduced row echelon form.
(b) Consider the following homogeneous system:
2x1 ¡ 2x2 + x3 + 4x4 ¡ 3x5 = 0
2x1 ¡ x2 + x4 + 4x5 = 0
2x1 ¡ 4x2 + x3 + 6x4 + 6x5 = 0
Write this system as an augmented matrix. Use your calculator and ¯nd the row echelon
form and the reduced row echelon form. How is this system di®erent than the ¯rst? What
row operation would move the augmented matrix from part (a) to this one?
(c) Consider the following homogeneous system:
2x1 - 4x2 + x3 + 6x4 + 6x5 = 0
2x1 - x2 + x4 + 4x5 = 0
2x1 - 2x2 + x3 + 4x4 - 3x5 = 0
Write this system as an augmented matrix. Use your calculator and find the row echelon
form and the reduced row echelon form. How is this system different than the first? What
row operation would move the augmented matrix from part (a) to this one?
(d) Explain why the row echelon forms found in parts (b) and (c) are alternative forms of
the row echelon form matrix found in part (a).
(e) Comparing the three different row echelon forms, how are they different? What is the
same about them? What does this suggest about row echelon form of any matrix?
(f) How are the reduced row echelon forms different for the above augmented matrices?
What does this suggest about the reduced row echelon form on any matrix?
(g) While we are here, solve the homogeneous system.
#### Solution Preview
Please see the attached file for the complete solution.
Thanks for using BrainMass.
(a) The augmented matrix of the system is:
The REF form of a matrix is not unique.
The RREF form is unique and is:
1 0 0 0 -0.25 0
0 1 0 -1 -4.5 0
0 0 1 2 -11.5 0
(b) The augmented matrix of the system is:
The REF form of a matrix is not unique.
The RREF form is unique and is:
1 0 0 0 ...
#### Solution Summary
Reduced Row Echelon Form of Homogeneous Systems of Equations is investigated.
\$2.19 |
Basic Properties of Complex Numbers Review
# Basic Properties of Complex Numbers Review
We will now review some of the recent material regarding complex numbers.
• Recall from the The Set of Complex Numbers page that the Imaginary Unit is defined to be $i = \sqrt{-1}$. An Imaginary Number is a number of the form $bi$ where $b \in \mathbb{R}$. For example, $2i$ is an imaginary number. A Complex Number is a number of the form $a + bi$ where $a, b \in \mathbb{R}$. For example, $3 + 2i$ is a complex number. The set of complex numbers is denoted by $\mathbb{C}$.
• If $z = a + bi \in \mathbb{C}$ then we defined the Real Part of $z$ as $\mathrm{Re} (z) = a$, and we defined the Imaginary Part of $z$ as $\mathrm{Im} (z) = b$. Furthermore, $z$ can be represented as an ordered pair or vector $(a, b)$ in the complex plane where the complex plane is $\mathbb{R}^2$ with the $x$-axis labelled as the Real Axis and the $y$-axis labelled as the Imaginary Axis.
(1)
\begin{align} \quad z + w = (a + c) + (b + d)i \end{align}
• We defined Multiplication between $z = a + bi, w = c + di \in \mathbb{C}$ to obtain the Product $z \cdot w$ by:
(2)
\begin{align} \quad z \cdot w = (ac - bd) + (ad + bc)i \end{align}
• On the Division of Complex Numbers page we defined Division of two complex numbers. First we noted that if $w = c + di \in \mathbb{C}$ and $w \neq 0$ then the inverse of $w$ is:
(3)
\begin{align} \quad w^{-1} = \frac{1}{w} = \frac{c}{c^2 + d^2} - \frac{d}{c^2 + d^2}i \end{align}
• So if $z = a + bi, w = c + di \in \mathbb{C}$ and $w \neq 0$ then we can define the Quotient $\displaystyle{\frac{z}{w}}$ as:
(4)
\begin{align} \quad zw^{-1} = \frac{z}{w} = \frac{ac + bd}{c^2 + d^2} - \frac{ad - bc}{c^2 + d^2}i \end{align}
• On the The Set of Complex Numbers is a Field page we then noted that the set of complex numbers $\mathbb{C}$ with the operations of addition $+$ and multiplication $\cdot$ defined above make $(\mathbb{C}, +, \cdot)$ an algebraic field (similarly to that of the real numbers with the usually defined addition and multiplication).
• On the The Conjugate of a Complex Number page we then defined the Complex Conjugate of $z = a + bi$ to be the complex number $\overline{z} = a - bi$. We then proved some basic properties regarding the complex conjugate. We saw that $z = \overline{z}$ if and only if $z \in \mathbb{R}$, $z + \overline{z} = 2a$, $z - \overline{z} = 2bi$, and that $z \cdot \overline{z} = a^2 + b^2$.
(5)
\begin{align} \quad \mid z \mid = \sqrt{a^2 + b^2} \end{align}
• Geometrically, the absolute value of a complex number $z = a + bi$ gives us the length of the position vector $(a, b)$ in the complex plane.
• On the Square Roots of Complex Numbers page we proved a very important result which said that if $z \in \mathbb{C}$ then there exists a $w \in \mathbb{C}$ such that $z = w^2$. In other words, every complex number has a square root. More generally, every complex number has precisely two square roots (allowing repeated roots).
• We saw that the square roots of a complex number $z$ are both real if and only if $z$ is real and positive. We saw that the square roots of $z$ are both imaginary if and only if $z$ is real and negative. We saw that the square roots of $z$ are the same if and only if $z = 0$. |
# what is 4/5 of 10
## How do you find 4/5 of a number?
To find one fifth of a number we divide the number by five. Then, to find four fifths of a number, we first find one fifth of that number and then multiply this by four.
## What is 4 out of 5 as a fraction?
Answer: 4 out of 5 can be written as 4/5 and is equal to 80%. Let’s understand the conversion of a fraction to a percentage. Explanation: 4 out of 5 can be written as 4/5.
## What can 4 5 be written?
0.8
Answer: 4/5 as a decimal is 0.8.
## How do you solve this math problem?
Here are four steps to help solve any math problems easily:
1. Read carefully, understand, and identify the type of problem. …
2. Draw and review your problem. …
3. Develop the plan to solve it. …
4. Solve the problem.
## What is 4/5 as a percentage?
Answer: 4/5 as a percent is 80%.
Let’s look into the solution. To get 4/5 as a percent, we multiply it by 100.
## What is 4/10 as a percentage?
Fraction to percent conversion table
Fraction Percent
2/10 20%
3/10 30%
4/10 40%
5/10 50%
## What is 10 as a fraction?
1/10
Example Values
Percent Decimal Fraction
10% 0.1 1/10
12½% 0.125 1/8
20% 0.2 1/5
25% 0.25 1/4
## How do you write 4/10 as a decimal?
Answer: The fraction 4/10 written as a decimal number is equal to 0.4.
## What is the reduced fraction 4 10?
410 is not in lowest terms. Since both the numerator and denominator are even numbers, they can be divided by 2 .
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## How do you write 4/5 as a decimal?
To convert a fraction into a decimal, divide the top of the fraction (numerator) by the bottom of the fraction (denominator). In this case, the fraction is 45 , so divide 4 by 5 . This equals 0.8 .
## What is fraction math?
fraction, In arithmetic, a number expressed as a quotient, in which a numerator is divided by a denominator. In a simple fraction, both are integers. A complex fraction has a fraction in the numerator or denominator. In a proper fraction, the numerator is less than the denominator.
## What are fractions for kids?
Fractions are used to represent smaller pieces (or parts) of a whole.
• The parts might make up one thing, or more than one thing. …
• It’s important to note that a whole can mean more than one thing. …
• You learn that when you count up, the numbers have more value. …
• Positive whole numbers (like 1, 2 or 65) are simple.
## Who created math?
Archimedes is known as the Father of Mathematics. Mathematics is one of the ancient sciences developed in time immemorial.
## Is 1/4th a quarter?
One fourth is equivalent to the fraction: 1/4. Therefore, it is a quarter of an amount.
## What’s 1/10 as a percentage?
10%
To convert 1/10 into a percentage we can multiply the numerator and denominators by 10 to get 10/100, which is 10%. We can also divide 1 by 10 to get 0.1 as a decimal and then multiply by 100 to get 10%. Both methods give the same answer. The fraction 10/100 is 10% because percent means out of 100.
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## How do you find out the percentage?
Percentage can be calculated by dividing the value by the total value, and then multiplying the result by 100. The formula used to calculate percentage is: (value/total value)×100%.
## What is a 3 out of 10 grade?
30%
Answer: 3/10 can be written as 30% as a percentage.
## How do you turn 3 5 into a percent?
1. “Percent” or “%” means “out of 100” or “per 100”, Therefore x% can be written as x100 .
2. 35=x100.
3. 100⋅35=100⋅x100.
4. 3005=100 ⋅x100.
5. 35=60100=60%
## How do you write 0.5 as a percentage?
Multiply 0.5 by 100/100. Since 100/100 = 1, we are only multiplying by 1 and not changing the value of our number. 50/100 is 50 over 100 and means 50 per 100. 50 “per 100” means 50 “percent” or 50%.
## What is 0.4 as a fraction?
2/5
Answer: 0.4 can be written in a fraction as 4/10 or 2/5. Explanation: First, we need to remove the decimal.
## How do multiply fractions?
There are 3 simple steps to multiply fractions
1. Multiply the top numbers (the numerators).
2. Multiply the bottom numbers (the denominators).
3. Simplify the fraction if needed.
## How do you simplify fractions?
You can simplify a fraction if the numerator (top number) and denominator (bottom number) can both be divided by the same number. Six twelfths can be simplified to one half, or 1 over 2 because both numbers are divisible by 6. 6 goes into 6 once and 6 goes into 12 twice.
## What is 0.8 as a number?
The decimal 0.8 is a rational number. It is equivalent to the fraction 8/10.
READ: see when a website was last updated
## What is 0.01 as a fraction?
1100
Hence in fractions, 0.01 is 1100 .
## What is 80 in a fraction?
4/5
Answer: 80% as a fraction in simplest form is 4/5.
## What is 5 over 10 as a decimal?
Fraction to decimal conversion table
Fraction Decimal
5/10 0.5
6/10 0.6
7/10 0.7
8/10 0.8
## What is 4 hundredths as a decimal?
Decimal Numbers Hundredths
0.04 4
0.4
4
40
## doremi phép thuật thần kỳ (phần 1) (tập 4) (5/10) Ba người không phải là quá nhiều đâu
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## What is 43 percent of 69?
43 percent of 69 is 29.67
# Percentage calculators
Use these percentage calculators to work out percentages quickly. If you want to know how to work them out yourself, follow the examples below to learn how.
## Calculate what percent one value is of another
out of is how many percent?
## Calculate the percentage of a value
What is percent of ?
# The calculations explained
## How do I calculate what percent 25 is of 200?
To calculate 25 of 200 as a percentage, first calculate what one percent of 200 is. To do so, divide 200 by 100.
200 ÷ 100 = 2
Now divide 25 by 2 to get the answer.
25 ÷ 2 = 12%
## How do I calculate what 50% of 200 is?
First calculate what one percent of 200 is by dividing it by 100.
200 ÷ 100 = 2
Now multiply 50 by 2 to get the answer.
50 × 2 = 100
## Exercises
### Question
There is a total of 40 pens. India holds 13. What percent of all the pens does India have?
India has 13 pens. There are 40 pens in total. 1% of 40 is 40 ÷ 100 = 0.4. Now, divide the amount India has by one percent. 13 ÷ 0.4 = 32.5% . India has 32.5% of the pens.
### Question
There is a collection with 50 pencils from which Abigail helps themself to 48. What percentage of pencils does Abigail have?
Abigail has 48 pencils. There are 50 pencils as a whole. One % of 50 is 50 ÷ 100 = 0.5. Next, divide the amount Abigail has by one percent. 48 ÷ 0.5 = 96% . Abigail has 96% of the pencils.
### Question
A basket contains 500 sticks. Aaron is given 435. How many percent of the 500 sticks does Aaron have?
Aaron has 435 sticks. There are 500 sticks as a whole. One % of 500 is 500 ÷ 100 = 5. So, divide the amount Aaron has by one percent. 435 ÷ 5 = 87% . Aaron has 87% of the sticks.
### Question
Out of 220 sticks, if Cuthbert takes 147, what percent of the sticks does Cuthbert have?
Cuthbert has 147 sticks. There are 220 sticks in total. One % of 220 is 220 ÷ 100 = 2.2. Now, divide the amount Cuthbert has by one percent. 147 ÷ 2.2 = 66.82% . Cuthbert has 66.82% of the sticks.
### Question
Out of 88 pens, if Candida takes 63, what percent of the pens does Candida have?
Candida has 63 pens. There are 88 pens in total. One % of 88 is 88 ÷ 100 = 0.88. So, divide the amount Candida has by one percent. 63 ÷ 0.88 = 71.59% . Candida has 71.59% of the pens.
### Question
Find 35 percent of 93.
To get the 35% of 93, first divide 93 by 100 to work out one%: 93 ÷ 100 = 0.93. Now multiply that by 100 to get the answer. 35 × 0.93 = 32.55
### Question
There is a bag with 20 sweets from which Lex helps themself to 15. What percentage of sweets does Lex have?
Lex has 15 sweets. There are 20 sweets as a whole. 1% of 20 is 20 ÷ 100 = 0.2. So, divide the amount Lex has by one percent. 15 ÷ 0.2 = 75% . Lex has 75% of the sweets.
### Question
There is a bag with 400 pies from which Didier helps themself to 120. What percentage of pies does Didier have?
Didier has 120 pies. There are 400 pies as a whole. 1% of 400 is 400 ÷ 100 = 4. Next, divide the amount Didier has by one percent. 120 ÷ 4 = 30% . Didier has 30% of the pies.
### Question
A drawer contains 12 cakes. Aaron is given 8. How many percent of the 12 cakes does Aaron have?
Aaron has 8 cakes. There are 12 cakes combined. One percent of 12 is 12 ÷ 100 = 0.12. So, divide the amount Aaron has by one percent. 8 ÷ 0.12 = 66.67% . Aaron has 66.67% of the cakes.
### Question
What is 83 percent of 413?
To work out 83% of 413, first divide 413 by 100 to work out one%: 413 ÷ 100 = 4.13. Now times that by 100 to get the answer. 83 × 4.13 = 342.79
### Question
94% of 455 is what?
To calculate 94% of 455, first divide 455 by 100 to get one percent: 455 ÷ 100 = 4.55. Now multiply that by 100 to get the answer. 94 × 4.55 = 427.7
### Question
If Bill takes 6 fidget-spinners from a packet of 250 fidget-spinners, what percent of the fidget-spinners does Bill have?
Bill has 6 fidget-spinners. There are 250 fidget-spinners as a whole. One % of 250 is 250 ÷ 100 = 2.5. So, divide the amount Bill has by one percent. 6 ÷ 2.5 = 2.4% . Bill has 2.4% of the fidget-spinners.
### Question
84% of 169 is what?
To figure out 84 percent of 169, first divide 169 by 100 to work out 1 percent: 169 ÷ 100 = 1.69. Now times that by 100 to get the answer. 84 × 1.69 = 141.96
### Question
Find 35 percent of 181.
To figure out 35% of 181, first divide 181 by 100 to work out one percent: 181 ÷ 100 = 1.81. Now multiply that by 100 to get the answer. 35 × 1.81 = 63.35
### Question
What is 15 percent of 167?
To work out 15 percent of 167, first divide 167 by 100 to get one%: 167 ÷ 100 = 1.67. Now times that by 100 to get the answer. 15 × 1.67 = 25.05
### Question
43 percent of 288 is what?
To work out 43 percent of 288, first divide 288 by 100 to work out 1%: 288 ÷ 100 = 2.88. Now multiply that by 100 to get the answer. 43 × 2.88 = 123.84
### Question
What is 60% of 162? |
Chapter 3 - McGraw Hill Higher Education
```Describing Data:
Numerical Measures
Chapter 3
McGraw-Hill/Irwin
GOALS
1. Calculate the arithmetic mean, weighted mean,
median, mode, and geometric mean.
2. Explain the characteristics, uses, advantages, and
disadvantages of each measure of location.
3. Identify the position of the mean, median, and mode
for both symmetric and skewed distributions.
4. Compute and interpret the range, mean deviation,
variance, and standard deviation.
5. Understand the characteristics, uses, advantages,
and disadvantages of each measure of dispersion.
6. Understand Chebyshev’s theorem and the Empirical
Rule as they relate to a set of observations.
3-2
Parameter Versus Statistics
PARAMETER A measurable characteristic
of a population.
STATISTIC A measurable characteristic of a
sample.
3-3
Population Mean
For ungrouped data, the population mean is the sum of all the population values divided by the total number of
population values. The sample mean is the sum of all the sample values divided by the total number of sample
values.
EXAMPLE:
3-4
The Median
MEDIAN The midpoint of the values after they have been ordered from the smallest to the largest, or the largest to
the smallest.
1.
2.
3.
4.
PROPERTIES OF THE MEDIAN
There is a unique median for each data set.
It is not affected by extremely large or small values and is therefore a valuable measure of central tendency
when such values occur.
It can be computed for ratio-level, interval-level, and ordinal-level data.
It can be computed for an open-ended frequency distribution if the median does not lie in an open-ended
class.
EXAMPLES:
The ages for a sample of five college students are:
21, 25, 19, 20, 22
The heights of four basketball players, in inches, are:
76, 73, 80, 75
Arranging the data in ascending order gives:
Arranging the data in ascending order gives:
73, 75, 76, 80.
19, 20, 21, 22, 25.
Thus the median is 21.
3-5
Thus the median is 75.5
The Mode
MODE The value of the observation that appears most frequently.
3-6
The Relative Positions of the Mean,
Median and the Mode
3-7
The Geometric Mean
Useful in finding the average change of percentages, ratios, indexes, or growth rates over time.
It has a wide application in business and economics because we are often interested in finding the
percentage changes in sales, salaries, or economic figures, such as the GDP, which compound or
build on each other.
The geometric mean will always be less than or equal to the arithmetic mean.
The formula for the geometric mean is written:
EXAMPLE:
Suppose you receive a 5 percent increase in salary this year and a 15 percent
increase next year. The average annual percent increase is 9.886, not 10.0. Why is
this so? We begin by calculating the geometric mean.
GM
3-8
( 1.05 )( 1.15 ) 1.09886
Measures of Dispersion
A measure of location, such as the mean or the median, only describes the center of the data. It is valuable from
that standpoint, but it does not tell us anything about the spread of the data.
For example, if your nature guide told you that the river ahead averaged 3 feet in depth, would you want to wade
across on foot without additional information? Probably not. You would want to know something about the variation
in the depth.
A second reason for studying the dispersion in a set of data is to compare the spread in two or more distributions.
RANGE
MEAN DEVIATION
VARIANCE AND STANDARD DEVIATION
3-9
EXAMPLE – Mean Deviation
EXAMPLE:
The number of cappuccinos sold at the Starbucks location in the Orange Country
Airport between 4 and 7 p.m. for a sample of 5 days last year were 20, 40, 50, 60,
and 80. Determine the mean deviation for the number of cappuccinos sold.
Step 1: Compute the mean
x
n
x
20 40 50 60 80
50
5
Step 2: Subtract the mean (50) from each of the observations, convert to positive if difference
is negative
Step 3: Sum the absolute differences found in step 2 then divide by the number of
observations
3-10
Variance and Standard Deviation
VARIANCE The arithmetic mean of the squared deviations from the mean.
STANDARD DEVIATION The square root of the variance.
3-11
The variance and standard deviations are nonnegative and are zero only
if all observations are the same.
For populations whose values are near the mean, the variance and
standard deviation will be small.
For populations whose values are dispersed from the mean, the
population variance and standard deviation will be large.
The variance overcomes the weakness of the range by using all the
values in the population
EXAMPLE – Population Variance and
Population Standard Deviation
The number of traffic citations issued during the last five months in Beaufort County, South Carolina, is
reported below:
What is the population variance?
Step 1: Find the mean.
N
x
19 17 ... 34 10
12
348
29
12
Step 2: Find the difference between each observation and the mean, and square that difference.
Step 3: Sum all the squared differences found in step 3
Step 4: Divide the sum of the squared differences by the number of items in the population.
2
(X
)
N
3-12
2
1, 488
12
124
Sample Variance and
Standard Deviation
Where :
2
s is the sample variance
X is the value of each observatio n in the sample
X is the mean of the sample
n is the number
EXAMPLE
The hourly wages for a sample of part-time
employees at Home Depot are: \$12, \$20,
\$16, \$18, and \$19.
What is the sample variance?
3-13
of observatio ns in the sample
Chebyshev’s Theorem and Empirical Rule
The arithmetic mean biweekly amount
contributed by the Dupree Paint
employees to the company’s profitsharing plan is \$51.54, and the standard
deviation is \$7.51. At least what percent
of the contributions lie within plus 3.5
standard deviations and minus 3.5
standard deviations of the mean?
3-14
The Arithmetic Mean and Standard
Deviation of Grouped Data
EXAMPLE:
Determine the arithmetic mean vehicle
selling price given in the frequency
table below.
3-15
EXAMPLE
Compute the standard deviation of the vehicle
selling prices in the frequency table below.
``` |
Mathematics is the science that builds the world. As a scientist, and a simple person - no one can do without it. First, young children are taught to count, then add, subtract, multiply and divide, secondary symbols are used for secondary school, and in the older one they can not do without them.
But today we will talk about what all the known mathematics are built on. About the community of numbers called "limits of sequences".
## What are sequences and where is their limit?
The meaning of the word "sequence" is not difficult to interpret. It is the construction of things where someone or something is located in a certain order or line. For example, the queue for tickets to the zoo is a sequence. And it can only be one! If, for example, look at the queue in the store - this is one sequence. And if one person from this queue suddenly leaves, then this is another line, another order.
The word "limit" is also easily interpreted - this is the end of something. However, in mathematics, the limits of sequences are those on the number line to which the sequence of numbers tends. Why does it seek, not end? It's simple, the number line has no end, and most of the sequences, like rays, have only a beginning and look like this:
Hence the definition of a sequence is a function of the natural argument. In simpler words, this is the series of terms of a certain set.
## How is the numerical sequence constructed?
The simplest example of a numerical sequence can look like this: 1, 2, 3, 4, ... n ...
In most cases, for practical purposes, sequences are constructed from numbers, and each subsequent member of the series, denoted by X, has a name. For example:
x1 - the first member of the sequence;
x2 The second term of the sequence;
x3 - the third term;
xn - Enne member.
In practical methods, a sequence is given by a general formula in which there is some variable. For example:
Xn = 3n, then the series of numbers will look like this:
It is worth remembering that for general recording of sequences you can use any Latin letters, not just X. For example: y, z, k, etc.
## Arithmetic progression as part of sequences
Before looking for the limits of sequences, it is advisable to delve deeper into the very notion of such a numerical series, which all faced, being in middle classes. Arithmetic progression is a series of numbers in which the difference between neighboring members is constant.
The problem: "Let a1 = 15, and the step of progression of the numerical series d = 4. Build the first 4 members of this series "
Solution: a1 = 15 (by condition) - the first member of the progression (numerical series).
a2 = 15 + 4 = 19 - the second member of the progression.
a3 = 19 + 4 = 23 - the third term.
a4 = 23 + 4 = 27 is the fourth term.
## Sequence types
Most of the sequences are infinite, it's worth remembering for life. There are two interesting types of numerical series. The first is given by the formula an = (-1) n. Mathematicians often call this sequence a flasher. Why? Let us check its number series.
-1, 1, -1. 1, -1, 1, etc. On such an example it becomes clear that the numbers in the sequences can easily be repeated.
The factorial sequence. It is easy to guess - in the formula defining the sequence, there is a factorial. For example: an = (n + 1)!
Then the sequence will look like this:
a3 = 1x2x3x4 = 24, and so on.
There is even a sequence consisting of the same number. Thus, an = 6 consists of an infinite set of sixes.
## Determination of the limit
Limits of sequences have long existed in mathematics. Of course, they deserve their own competent design. So, it's time to learn the definition of the limits of sequences. To begin with, consider in detail the limit for a linear function:
1. All limits are abbreviated lim.
2. A limit entry consists of a shortening of lim, some variable tending to a certain number, zero or infinity, and also from the function itself.
It is easy to understand that the definition of the limit of a sequence can be formulated as follows: it is a number to which all the terms of the sequence approach infinitely. A simple example: ax = 4x + 1. Then the sequence itself will look like this.
5, 9, 13, 17, 21 ... x ...
Thus, this sequence will increase indefinitely, and, therefore, its limit is equal to infinity as x → ∞, and it should be written like this:
If we take a similar sequence, but x will tend to 1, we get:
A number of numbers will be: 1.4, 1.8, 4.6, 4.944, etc. Each time you need to substitute the number more and more to the unit (0.1, 0.2, 0.9, 0.986). From this series it is clear that the limit of a function is five.
From this part it is worth remembering what is the limit of a numerical sequence, the definition and method of solving simple tasks.
## General designation of the limit of sequences
Having analyzed the limit of a numerical sequence, its definition and examples, it is possible to start a more complex topic. Absolutely all the limits of sequences can be formulated with a single formula, which is usually disassembled in the first semester.
So, what does this set of letters, modules and signs of inequalities mean?
∀ is the universal quantifier replacing the phrases "for all", "for everything", etc.
∃ - existence quantifier, in this case means that there is a certain value of N, which belongs to the set of natural numbers.
A long vertical stick, following N, means that the given set N is "such that". In practice, it can mean "such, that", "such, that," and so on.
Next comes the module. Obviously, the module is the distance, which by definition can not be negative. Hence the modulus of the difference is strictly less than the "epsilon."
To fix the material, read the formula aloud.
## Uncertainty and definiteness of the limit
The method of finding the limit of sequences, which was considered above, even though simple in application, but not so rational in practice. Try to find the limit for this function:
If we substitute the various values of "X" (each time increasing: 10, 100, 1000, etc.), then we get ∞ in the numerator, but ∞ in the denominator. It turns out a rather strange fraction:But is it really? To calculate the limit of a numerical sequence in this case, it seems easy enough. You could leave everything as is, because the answer is ready, and received it on reasonable terms, however, there is another method specifically for such cases.
To begin with, we find the highest degree in the numerator of the fraction - this is 1, since x can be represented as x 1.
Now we find the highest degree in the denominator. Also 1.
We divide both the numerator and the denominator into a variable in the highest degree. In this case, we divide the fraction by x 1.
Next, we find to what value each term that contains the variable tends. In this case, fractions are considered. As x → ∞, the value of each of the fractions tends to zero. When writing in writing, it is worth making such footnotes:
The following expression is obtained:
Of course, the fractions containing x did not become zeros! But their value is so small that it is completely allowed not to take it into account in the calculations. In fact, x will never be 0 in this case, because you can not divide by zero.
## What is the neighborhood?
Suppose, at the disposal of the professor, a complex sequence, presumably given by an equally complex formula. The professor found the answer, but is it suitable? After all, all people are wrong.
Auguste Cauchy in his time came up with a great way to prove the limits of sequences. His method was called operating with neighborhoods.
Suppose that there is a point a, its neighborhood in both directions on the number line is equal to ε ("epsilon"). Since the last variable is distance, its value is always positive.
Now we give a sequence xn and assume that the tenth member of the sequence (x10 ) occurs in a neighborhood of a. How to write this fact in mathematical language?
Assume that x10 is to the right of a, then the distance x10 -a \u0026 lt; ε, however, if "tenth" is located to the left of the point a, then the distance is negative, and this is impossible, therefore, the left side of the inequality must be added to the module. It turns out that | x10 -a | \u0026 lt; ε.
Now it's time to clarify in practice the formula, which was mentioned above. A certain number a is said to be a finite point of the sequence if for any of its limits the inequality ε\u003e 0 holds, and the entire neighborhood has its natural number N such that all terms of the sequence with more significant numbers appear inside the sequence | xn - a | \u0026 lt; ε.
With such knowledge it is easy to implement the decision of the limits of the sequence, to prove or disprove the ready answer.
Theorems on the limits of sequences are an important part of the theory, without which practice is impossible. There are only four main theorems, remembering which, one can simplify the course of the decision or evidence at times:
1. Uniqueness of the limit of a sequence. The limit for any sequence can be only one or not at all. The same example with a queue that can have only one end.
2. If the series of numbers has a limit, then the sequence of these numbers is bounded.
3. The limit of the sum (difference, product) of sequences is equal to the sum (difference, product) of their limits.
4. The limit of the quotient of the division of two sequences is equal to the quotient limit if and only if the denominator does not vanish.
## Proof of Sequences
Sometimes it is required to solve the inverse problem, to prove the given limit of a numerical sequence. Consider the example.
Prove that the limit of the sequence given by the formula is zero.
By the rule considered above, for any sequence the inequality | xn - a | \u0026 lt; ε. We substitute the given value and the reference point. We get:
We express n through "epsilon" to show the existence of a certain number and to prove the existence of a limit of sequence.
At this stage, it is important to recall that "epsilon" and "en" are positive numbers and not equal to zero. Now we can continue the further transformations, using knowledge about the inequalities obtained in the secondary school.
Here such a convenient method can prove the limit of a numerical sequence, no matter how complex it may seem at first glance. The main thing is not to panic when seeing the task.
## Maybe it's not there?
The existence of a sequence limit is not necessary in practice. It is easy to find such series of numbers that really have no end. For example, the same "flasher" xn = (-1) n. it is obvious that a sequence consisting of only two digits, cyclically repeating, can not have a limit.
The same story is repeated with sequences consisting of one number, fractional, having in the course of computations an uncertainty of any order (0/0, ∞ / ∞, ∞ / 0, etc.). However, it should be remembered that an incorrect calculation also takes place. Sometimes the limit of the followers will be found by rechecking your own decision.
## The monotone sequence
Above we considered several examples of sequences, methods for solving them, and now we try to take a more specific case and call it a "monotonous sequence".
Definition: any sequence is validly called monotonically increasing if it satisfies the strict inequality xn \u0026 lt; xn+1. Also, any sequence is said to be monotone decreasing if it satisfies the inequality xn \u0026 gt; xn+1.
Along with these two conditions, there are also similar non-strict inequalities. Correspondingly, xn ≤ xn+1 (nondecreasing sequence) and xn ≥ xn+1 (nonincreasing sequence).
But it is easier to understand this by examples.
The sequence given by the formula xn = 2 + n, forms the following series of numbers: 4, 5, 6, etc. This is a monotonically increasing sequence.
## Limit of convergent and bounded sequence
A bounded sequence is a sequence having a limit. A convergent sequence is a series of numbers having an infinitesimal limit.
Thus, the limit of a bounded sequence is any real or complex number. Remember that there can be only one limit.
The limit of a convergent sequence is an infinitesimal (real or complex) value. If you draw a sequence diagram, then at a certain point it will seem to converge, strive to address a certain amount. Hence the name is a convergent sequence.
## The limit of a monotone sequence
The limit of such a sequence may or may not be. At first it is useful to understand when it is, hence it is possible to pounce on the proof of the absence of a limit.
Among the monotonous sequences, the convergent and divergent are distinguished. Convergent is a sequence that is formed by the set x and has a real or complex limit in the given set. Divergent is a sequence that does not have a limit in its set (neither real nor complex).
And the sequence converges, if the geometric image of its upper and lower limits converge.
The limit of a convergent sequence in many cases can be equal to zero, since any infinitesimal sequence has a known limit (zero).
What a convergent sequence they take, they are all limited, but not all limited sequences converge.
The sum, the difference, the product of two convergent sequences is also a convergent sequence. However, the quotient can also be convergent if it is defined!
## Various actions with limits
Sequence limits are the same essential (in most cases) value, as well as numbers and numbers: 1, 2, 15, 24, 362, etc. It turns out that some operations can be performed with limits.
First, like numbers, limits of all sequences you can add and subtract. On the basis of the third theorem about limits of sequences, fairly following equality: limit of the sum of the sequence equal to the sum of their limits.
Secondly, based on the fourth theorem on the limits of sequences, the following equality holds: the limit of the product of the n-th number of sequences is equal to the product of their limits. The same is true for division: the limit of a particular two sequences is equal to their particular limits, provided that the limit is not zero. After all, if the limit of sequences is zero, then we get a division by zero, which is impossible.
## Properties of Sequence Values
1. The sum of any number of arbitrarily small quantities will also be small.
2. The sum of any number of large quantities will be infinitely large.
3. The product of arbitrarily small quantities is infinitesimal.
4. The product of arbitrarily large numbers is infinitely large.
5. If the original sequence tends to an infinitely large number, then the inverse of it will be infinitesimal and tend to zero.
In fact, computing the limit of a sequence is not such a difficult task if you know a simple algorithm. But the limits of sequences are a topic requiring maximum attention and perseverance. Of course, it's enough just to grasp the essence of the solution of such expressions. Beginning with small, in time you can reach large peaks. |
Collins Maths Solutions Class 7 Chapter 8
Collins Maths Solutions Class 7 Chapter 8 Profit and Loss
Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Collins Maths Class 7 Mathematics, Chapter 8, Profit and Loss. Here students can easily find chapter 7 solutions with exercise wise explanations. Students will find proper solutions for Exercise 8.1, 8.2 and 8.3 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Profit and Loss Exercise 8.1 Solution :
Question no – (1)
Solution :
Cost Price = 65/25
= 2.60 Rs
Sell = 3.50 Rs
Selling Price,
= (3.50 × 25)
= 87.5 Rs
Profit,
= (87.5 – 65)
= 22.5 Rs
So, shopkeeper will profit 22.5 Rs.
Question no – (2)
Solution :
Cost Price of pen drive = 210 Rs.
Selling Price of pen drive = 250 Rs.
Profit = (250 – 210)
= 40 Rs.
Hence, the profit will be 40 Rs.
Question no – (3)
Solution :
Selling price of cell phone = 8200 Rs.
Make a profit of = 540 Rs.
Cost Price will be = (selling price – profit)
= (8200 – 540)
= 7660 Rs.
So, the cost price of the cell phone will be 7660 Rs.
Question no – (4)
Solution :
Cost price of the bat = 990 Rs.
Make a loss = 77 Rs
Selling price = (cost price – loss)
= (990 – 77)
= 913 Rs.
Therefore, the selling price of the bat will be 913 Rs.
Question no – (5)
Solution :
= 30/3
= 10 pieces.
Selling price of total,
= 24/2
= 12 Rs.
Now, profit by selling 1 chocolate
= (12 – 10) Rs
= 2 Rs.
Profit and Loss Exercise 8.2 Solution :
Question no – (1)
Solution :
Profit percentage,
= (186 – 150/150 × 100)
= (36/150 × 100)
= 24%
Therefore, the profit percentage will be 24%.
Question no – (2)
Solution :
Cost price of banana,
= 48/12
= 4 Rs.
Selling price of banana,
= (3.80 × 12)
= 45 Rs
So, loss percent,
= (3/48 × 100)
= 6.25%
Therefore, the loss percent will be 6.25%.
Question no – (3)
Solution :
Cost price of the dress = 600 Rs
Selling price of the dress = 690 Rs.
So, profit percent,
= (690 – 600/600 × 100)
= (90/600 × 100)
= 15%
Hence, the profit percent will be 15%.
Question no – (4)
Solution :
Selling price of cosmetics = 1550 Rs.
Cost price of cosmetics,
= (1550 + 310)
= 1860 Rs.
Loss percent,
= (310/1860 × 100)
= 16.67%
Thus, the loss percent will be 16.67%.
Profit and Loss Exercise 8.3 Solution :
Question no – (1)
Solution :
(a) Cost price = Rs 750, profit = 10%
Selling Price,
= (750 × 110/100)
= 825 Rs.
Hence, the selling price will be 825 Rs.
(b) Cost price = Rs 600, loss = 9%
Selling Price,
= (600 × 91/100)
= 546 Rs.
So, the selling price will be 546 Rs.
(c) Cost price = Rs 1000, loss = 2%
Selling Price,
= (1000 × 98/100)
= 980 Rs.
Hence, the selling price will be 980 Rs.
(d) Cost price = Rs 350, overhead expenses = Rs 50, profit = 15%
Total cost price,
= (350 + 500)
= 400 Rs
Now, Selling Price,
= (400 × 115/100)
= 460 Rs.
Thus, the selling price will be 460 Rs.
Question no – (2)
Solution :
Cost price of the television set = 15,000 Rs
Make a profit = 12%
Selling Price,
= (1500 × 112/100)
= 16800 Rs.
Therefore, the selling price of the television set will be 16800 Rs
Question no – (3)
Solution :
Selling price at a loss of 6%
= (500000 × 94/100)
= 470,000 Rs
Now, selling price at a profit of 6%
= (500000 × 106/100)
= 530,000 Rs.
Therefore, the selling price of the car will be 530, 000 Rs.
Next Chapter Solution :
Updated: June 15, 2023 — 8:21 am |
# Math Expressions Grade 5 Unit 1 Lesson 5 Answer Key Fractions Greater Than One
## Math Expressions Common Core Grade 5 Unit 1 Lesson 5 Answer Key Fractions Greater Than One
Math Expressions Grade 5 Unit 1 Lesson 5 Homework
Name the mixed number shown by the shaded parts.
Fractions Greater Than One Math Expressions Grade 5 Unit 1 Lesson 5 Question 1.
_____
3(1/2)
The mixed number for the above shaded parts is 3(1/2).
Explanation:
In the above image we can observe four objects. In that three objects are completely shaded and one object is not completely shaded. So the mixed number for the above shaded parts is 3(1/2).
Grade 5 Unit 1 Lesson 5 Answer Key Math Expressions Fractions Greater Than One Question 2.
______
2(2/3)
The mixed number for the above shaded parts is 2(2/3).
Explanation:
In the above image we can observe three flags. In that two flags are completely shaded and one flag is not completely shaded. So the mixed number for the above shaded parts is 2(2/3).
Answer Key Fractions Greater Than One Math Expressions Grade 5 Unit 1 Lesson 5 Question 3.
______
1(3/7)
The mixed number for the above shaded parts is 1(3/7).
Explanation:
In the above image we can observe two objects. In that one object is completely shaded and one object is not completely shaded. So the mixed number for the above shaded parts is 1(3/7).
Write the mixed number as a fraction.
Question 4.
2$$\frac{1}{3}$$ = ____
2(1/3) = (7/3)
Explanation:
First multiply 3 with 2 then the product is 6. Add 6 to 1 then the sum is 7. The mixed number 2(1/3) in fraction form as (7/3).
Question 5.
4$$\frac{2}{5}$$ = ____
4(2/5) = (22/5)
Explanation:
First multiply 5 with 4 then the product is 20. Add 20 with 2 then the sum is 22. The mixed number 4(2/5) in fraction form as (22/5).
Question 6.
3$$\frac{3}{4}$$ = ____
3(3/4) = (15/4)
Explanation:
First multiply 4 with 3 then the product is 12. Add 12 with 3 then the sum is 15. The mixed number 3(3/4) in fraction form as (15/4).
Question 7.
1$$\frac{5}{8}$$ = ____
1(5/8) = (13/8)
Explanation:
First multiply 8 with 1 then the product is 8. Add 8 with 5 then the sum is 13. The mixed number 1(5/8) in fraction form as (13/8).
Write the fraction as a mixed number.
Question 8.
$$\frac{7}{6}$$ = _____
(7/6) = 1(1/6)
Explanation:
The fraction (7/6) can be written in mixed number as 1(1/6).
Question 9.
$$\frac{8}{3}$$ = _____
(8/3) = 2(2/3)
Explanation:
The fraction (8/3) can be written in mixed number as 2(2/3).
Question 10.
$$\frac{9}{2}$$ = _____
(9/2) = 4(1/2)
Explanation:
The fraction (9/2) can be written in mixed number as 4(1/2).
Question 11.
$$\frac{10}{7}$$ = _____
(10/7) = 1(3/7)
Explanation:
The fraction (10/7) can be written in mixed number as 1(3/7).
Complete. Give the answer as a mixed number.
Question 12.
$$\frac{3}{5}$$ + $$\frac{4}{5}$$ = _____
(3/5) + (4/5) = (7/5)
(3/5) + (4/5) = 1(2/5)
Explanation:
Add these two fraction (3/5) and (4/5) then the sum is (7/5). The fraction (7/5) can be written in mixed number as 1(2/5).
Question 13.
$$\frac{6}{4}$$ + $$\frac{3}{4}$$ = _____
(6/4) + (3/4) = (9/4)
(6/4) + (3/4) = 2(1/4)
Explanation:
Add these two fraction (6/4) and (3/4) then the sum is (9/4). The fraction (9/4) can be written in mixed number as 2(1/4).
Question 14.
$$\frac{2}{9}$$ + $$\frac{8}{9}$$ = _____
(2/9) + (8/9) = (10/9)
(2/9) + (8/9) = 1(1/9)
Explanation:
Add these two fraction (2/9) and (8/9) then the sum is (10/9). The fraction (10/9) can be written in mixed number as 1(1/9).
Question 15.
7 + $$\frac{2}{3}$$ = _____
7 + (2/3) = (23/3)
7 + (2/3) = 7(2/3)
Explanation:
Add these number and fraction 7 and (2/3) then the sum is (23/3). The fraction (23/3) can be written in mixed number as 7(2/3).
Question 16.
Alicia walked $$\frac{7}{8}$$ mile on Saturday and $$\frac{6}{8}$$ mile on Sunday. How far did she walk over the weekend? Give the answer as a mixed number.
Alicia walked (7/8) mile on Saturday.
Alicia walked (6/8) mile on Sunday.
(7/8) + (6/8) = (13/8)
(7/8) + (6/8) = 1(5/8)
Alicia walked 1(5/8) over the weekend.
Explanation:
Alicia walked (7/8) mile on Saturday and (6/8) mile on Sunday. Add these two miles (7/8) and (6/8) then the sum is (13/8). The mixed fraction of (13/8) is 1(5/8). Alicia walked 1(5/8) over the weekend.
Question 17.
The dark chain is $$\frac{5}{12}$$ yard long. The light one is $$\frac{9}{12}$$ yard long. How long will they be if they are joined?
The dark chain is (5/12) yard long.
The light one is (9/12) yard long.
(5/12) + (9/12) = (14/12)
(5/12) + (9/12) = 1(2/12)
1(2/12) yard long if both the chains are joined.
Explanation:
The dark chain is (5/12) yard long and the light one is (9/12) yard long. Add these two chains lengths (5/12) and (9/12) then the sum is (14/12). The mixed fraction of (14/12) is 1(2/12). 1(2/12) yard long if both the chains are joined.
Math Expressions Grade 5 Unit 1 Lesson 5 Remembering
Solve.
Question 1.
The dog has gone $$\frac{5}{8}$$ of the way across the yard. How much farther does it have to go to reach the gate?
The dog has gone (5/8) of the way across the yard.
(8/8) – (5/8) = (3/8)
The dog has to go (3/8) to reach the gate.
Explanation:
The dog has gone (5/8) of the way across the yard. By adding (5/8) with (3/8) then the sum is (8/8). So the dog has to go (3/8) to reach the gate.
Question 2.
The cat has gone $$\frac{7}{16}$$ of the way across the yard. How much farther does it have to go to reach the gate?
The cat has gone (7/16) of the way across the yard.
(16/16) – (7/16) = (9/16)
The cat has to go (9/16) to reach the gate.
Explanation:
The cat has gone (7/16) of the way across the yard. By adding (7/16) with (9/16) then the sum is (16/16). So the cat has to go (9/16) to reach the gate.
Question 3.
I cleaned $$\frac{6}{9}$$ of my room, and my friend cleaned $$\frac{2}{9}$$ of my room. How much of my room do we still have to clean?
I cleaned (6/9) of my room.
My friend cleaned (2/9) of my room.
(6/9) + (2/9) = (8/9)
I and my friend cleaned (8/9) of my room.
(9/9) – (8/9) =(1/9)
We still have to clean (1/9) my room.
Explanation:
I cleaned (6/9) of my room and my friend cleaned (2/9) of my room. Add (6/9) and (2/9) then the sum is (8/9). I and my friend cleaned (8/9) of my room. Subtract (8/9) from (9/9) then the difference is (1/9). We still have to clean (1/9) my room.
Question 4.
Mrs. Spencer’s class is signing up to play sports. $$\frac{8}{26}$$ of the students want to play soccer and $$\frac{12}{26}$$ want to play basketball. The rest of the students want to play baseball. What fraction of the students wants to play baseball?
The students want to play soccer are (8/26).
The students want to play basketball are (12/26).
(8/26) + (12/26) = (20/26)
The students wants to play both soccer and basketball are (20/26).
(26/26) – (20/26) = (6/26)
The students wants to play baseball are (6/26).
Explanation:
Mrs. Spencer’s class is signing up to play sports. The students want to play soccer are (8/26). The students want to play basketball are (12/26). Add (8/26) and (12/26) then the sum is (20/26). The students wants to play both soccer and basketball are (20/26). Subtract (20/26) from (26/26) then the difference is (6/26). The students wants to play baseball are (6/26).
Compare.
Question 5.
$$\frac{2}{6}$$ $$\frac{1}{6}$$
(2/6) > (1/6)
Explanation:
By comparing these two numbers (2/6) and (1/6). (2/6) is greater than (1/6).
Question 6.
$$\frac{4}{9}$$ $$\frac{4}{10}$$
(4/9) >(4/10)
Explanation:
By comparing these two numbers (4/9) and (4/10). (4/9) is greater than (4/10).
Question 7.
$$\frac{7}{12}$$ $$\frac{13}{24}$$
(7/12) > (13/24)
Explanation:
By comparing these two numbers (7/12) and (13/24). (7/12) is greater than (13/24).
Question 8.
$$\frac{3}{5}$$ $$\frac{1}{3}$$
(3/5) > (1/3)
Explanation:
By comparing these two numbers (3/5) and (1/3). (3/5) is greater than (1/3).
Question 9.
$$\frac{4}{6}$$ $$\frac{6}{9}$$
(4/6) = (6/9)
Explanation:
By comparing these two numbers (4/6) and (6/9). (4/6) is equal to (6/9).
Question 10.
$$\frac{4}{5}$$ $$\frac{5}{6}$$
(4/5) < (5/6)
Explanation:
By comparing these two numbers (4/5) and (5/6). (4/5) is less than (5/6).
Question 11.
$$\frac{7}{12}$$ $$\frac{3}{4}$$
(7/12) < (3/4)
Explanation:
By comparing these two numbers (7/12) and (3/4). (7/12) is less than (3/4).
Question 12.
$$\frac{3}{5}$$ $$\frac{4}{9}$$
(3/5) > (4/9)
Explanation:
By comparing these two numbers (3/5) and (4/9). (3/5) is greater than (4/9).
Question 13.
$$\frac{7}{9}$$ $$\frac{7}{8}$$
Stretch Your Thinking Find two fractions that are between $$\frac{3}{5}$$ and $$\frac{4}{5}$$. |
## Precalculus (6th Edition) Blitzer
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix} 2 & -4 & 1 \\ 1 & -3 & 1 \\ 3 & -7 & 2 \\ \end{matrix} \right|\begin{matrix} 3 \\ 5 \\ 12 \\ \end{matrix} \right]$ Solve this matrix as below to get: $\left[ \left. \begin{matrix} 2 & -4 & 1 \\ 1 & -3 & 1 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} 3 \\ 5 \\ 3 \\ \end{matrix} \right]$ $By,\ 3{{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$ $\left[ \left. \begin{matrix} 2 & -4 & 1 \\ 0 & -2 & 1 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} 3 \\ 7 \\ 3 \\ \end{matrix} \right]$ $By,\ 2{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$ $\left[ \left. \begin{matrix} 2 & -4 & 1 \\ 0 & -2 & 1 \\ 0 & 0 & 0 \\ \end{matrix} \right|\begin{matrix} 3 \\ 7 \\ 4 \\ \end{matrix} \right]$ $By,\ {{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$ Convert the last row in equation form, \begin{align} 0x+0y+0z=4 & \\ 0=4 & \\ \end{align} And this statement is not true, so there is no solution of this system. Hence, there is no solution for this system of equations. |
Related Articles
Algebraic Operations on Complex Numbers | Class 11 Maths
• Last Updated : 04 Dec, 2020
A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation i² = −1. For example, 5+6i is a complex number, where 5 is a real number and 6i is an imaginary number. Therefore, the combination of both the real number and imaginary number is a complex number. There can be four types of algebraic operations on complex numbers which are mentioned below. The four operations on the complex numbers include:
• Subtraction
• Multiplication
• Division
To add two complex numbers, just add the corresponding real and imaginary parts.
(a + bi) + (c + di) = (a + c) + (b + d)i
Examples:
• (7 + 8i) + (6 + 3i) = (7 + 6) + (8 + 3)i = 13 + 11i
• (2 + 5i) + (13 + 7i) = (2 + 13) + (7 + 5)i = 15 + 12i
• (-3 – 6i) + (-4 + 14i) = (-3 – 4) + (-6 + 14)i = -7 + 8i
• (4 – 3i ) + ( 6 + 3i) = (4+6) + (-3+3)i = 10
• (6 + 11i) + (4 + 3i) = (4 + 6) + (11 + 3)i = 10 + 14i
### Subtraction of Complex Numbers
To subtract two complex numbers, just subtract the corresponding real and imaginary parts.
(a + bi) − (c + di) = (a − c) + (b − d)i
Examples:
• (6 + 8i) – (3 + 4i) = (6 – 3) + (8 – 4)i = 3 + 4i
• (7 + 15i) – (2 + 5i) = (7 – 2) + (15 – 5)i = 5 + 10i
• (-3 + 5i) – (6 + 9i) = (-3 – 6) + (5 – 9)i = -9 – 4i
• (14 – 3i) – (-7 + 2i) = (14 – (-7)) + (-3 – 2)i = 21 – 5i
• (-2 + 6i) – (4 + 13i) = (-2 – 4) + (6 – 13)i = -6 – 7i
### Multiplication of Two Complex Numbers
Multiplication of two complex numbers is the same as the multiplication of two binomials. Let us suppose that we have to multiply a + bi and c + di. We will multiply them term by term.
(a + bi) ∗ (c + di) = (a + bi) ∗ c + (a + bi) ∗ di
= (a ∗ c + (b ∗ c)i)+((a ∗ d)i + b ∗ d ∗ −1)
= (a ∗ c − b ∗ d + i(b ∗ c + a ∗ d))
Example 1: Multiply (1 + 4i) and (3 + 5i).
(1 + 4i) ∗ (3 + 5i) = (3 + 12i) + (5i + 20i2)
= 3 + 17i − 20
= −17 + 17i
Note: Multiplication of complex numbers with real numbers or purely imaginary can be done in the same manner.
Example 2: Multiply 5 and (4 + 7i).
5 ∗ (4+7i) can be viewed as (5 + 0i) ∗ (4 + 7i)
= 5 ∗ (4 + 7i)
= 20 + 35i
Example 3: Multiply 3i and (2 + 6i).
3i ∗ (2 + 6i) can be viewed as (0 + 3i) ∗ (2 + 6i)
= 3i ∗ (2 + 6i)
= 6i + 18i2
= 6i − 18
= −18 + 6i
Example 4: Multiply (5 + 3i) and (3 + 4i).
(5+3i) ∗ (3+4i) = (5 + 3i) ∗ 3 + (5 + 3i) ∗ 4i
= (15 + 9i) + (20i + 12i2)
= (15 − 12) + (20 + 9)i
= 3 + 29i
### Review of Complex Numbers Addition, Subtraction and Multiplication
1. (a + bi) + (c + di) = (a + c) + (b + d)i
2. (a + bi) − (c + di) = (a − c) + (b − d)i
3. (a + bi) ∗ (c + di) = ((a ∗ c − b ∗ d) + (b ∗ c + a ∗ d)i)
## Conjugate of a Complex Number
In any two complex numbers, if only the sign of the imaginary part differs then, they are known as a complex conjugate of each other. Thus conjugate of a complex number a + bi would be a – bi.
### What’s the use of a complex conjugate?
Thus we can observe that multiplying a complex number with its conjugate gives us a real number. Thus the division of complex numbers is possible by multiplying both numerator and denominator with the complex conjugate of the denominator.
Property 1:
Property 2:
Property 3:
Property 4:
Property 5:
### Division of Two Complex Numbers
Division of complex numbers is done by multiplying both numerator and denominator with the complex conjugate of the denominator.
Example 1:
Example 2:
Example 3:
Example 4:
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
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Finding the Length of a Rolled-up Carpet
One of the more frequent questions we had on Ask Dr. Math was about how to find the length of material (carpet, paper, wire, etc.) on a roll, knowing only the inner and outer diameters and something else: the thickness of the material, or the number of turns, or the original size of the roll. Here we’ll look at several derivations of a formula, and variations of it.
Approximating a spiral
We’ll start with this reverse question from 1995, to establish the concept:
Calculating the Diameter of a Carpet Roll
How do you calculate the diameter of a carpet roll when you have the length and the thickness?
Doctor Andrew answered with some useful ideas:
Hi! Great question. For a darn good approximation you could find the cross-sectional area (the area of the circular end of the roll) needed for the roll, assume that the roll is a perfect cylinder, and then calculate the diameter required for that cross-sectional area. If a carpet has length L and thickness t, the cross-sectional area of the roll is the same as the cross-sectional area of the edge of the carpet when it is lying on the floor, which is the very long rectangle with length L and thickness t. See if you can take it from here to find the diameter.
Here are the dimensions he has in mind for such a perfect cylinder:
It has two diameters (the core, C, and the outer diameter, D), thickness t, and a length when rolled out, L:
(Not to scale!)
Neither of us has ever seen a carpet roll shaped perfectly cylindrically since it is a spiral with an edge that usually sticks out off the rest of the roll. I imagine that trying to describe the shape of the spiral could be pretty tough. You could probably come up with a number of approximations, the most accurate of which would take into account how the spiral begins and what kind of space would be left in the center of the roll due to physical limitations on how the carpet can bend. You might notice that the approximation in the above paragraph becomes less and less accurate as the ratio of L to t (the fraction L/t) becomes smaller and smaller. If you think more about it, you'll realize that when a carpet bends it is already changing its shape (stretching on the outside, compressing on the inside) in a way that makes describing it exactly with mathematics very difficult.
Here is a slightly better image of what the roll might look like, though the bend would not really be this smooth:
The more layers (and the thinner the carper), the closer we can approximate it with a cylinder.
In what follows, we’ll find the length given the dimensions of the roll, reversing what was said here. Most of the answers will be based on the area idea. Next time we’ll consider the compression issue.
Derivation from cross-section area
Next, from 1996, we get an actual formula:
Length of Material on a Cylindrical Roll
Is there a formula for calculating the length of material on a roll knowing the roll diameter, the core diameter, and the material thickness?
I recently tried using a formula I saw in a newsletter, but it was not even close when I tried a real life example.
The formula was something like (64.54*(D*D)-(C*C))/T*1000 where D is the roll diameter, C is the core diameter, and T is the thickness, all in inches. The answer was supposedly in feet.
His formula, which is not quite written right, can be written correctly as $$\frac{64.54(D^2-C^2)}{1000T}$$
Doctor Anthony derived that formula, using the area approach and then converting to inches by dividing by 12:
The area of cross-section of the roll can be expressed in two ways. When stretched out in a straight line it is a rectangle of area L*T where L is the length in inches, and T the thickness viewed edge on. When rolled on the drum, this same cross-section will be pi*R^2 - pi*r^2 where R = outer radius (=D/2), and r = inner radius (=C/2)
So we have L*T = pi{D^2/4 - C^2/4}
L = pi/(4T){D^2 - C^2} inches
Length = pi/(48T){D^2 - C^2} feet
= (0.06545/T){D^2 - C^2) feet (T,D and C in inches)
Try this in real life. It will be sensitive to the value of T you use, so measure that carefully.
That’s exactly the formula the anonymous questioner had seen. Without the unit conversion (so that the length is in whatever units you use for the other measurements), the formula is $$L=\frac{\pi(D^2-C^2)}{4T}$$
The last few answers here will look at substitutes for the thickness, making it easier to be accurate.
Derivation by summing the layers
For an alternative derivation of the same formula, consider this 2001 question:
Fabric Left on a Roll
I am going to write a program for my TI-83+ (I can also use a TI89 if necessary) to calculate how many feet of material are left on a roll. I know the outside diameter of the core (this is the tube around which the film is wrapped, I know the outside diameter of the roll, and I know the thickness (mil thickness) of the material. What I don't know is the formula to use to calculate the feet of material left on the roll.
What I thought I could do is start out with the distance around the core. In this particular case, that would be pi*3.75 (the outside diameter). Then each additional layer of material would be the previous outside diameter plus twice the thickness (1 mil = 0.001"). I'd have a field in which I'd keep a running total of these inches. I'd set up another field to keep track of the total number of "wraps" of material, which would be a calculated field (outside diameter of the roll/2 - 3.75/2 [this removes the core's dimension] divided by the mil thickness of the film). I'd put a loop in the program to keep adding to the running total of inches until the number of wraps = 0.
This would seem to be the hard way, but as I said, I don't know what formula to use. I know there has to be a formula I can use for this calculation to make things a lot easier. Does anybody out there know the formula I'm looking for?
Doctor Robert answered, using the suggested method and finding the same formula we’ve seen:
I'll take a whack at this one. No guarantee of success.
Let me define the diameter of the tube as Do and the diameter of the roll as D. Let t be the thickness of the material. Then I figure that the number of wraps, n, on the tube, is:
n = (D-Do)/(2*t)
and we need to find the total length of these n wraps.
I'll figure the length of a wrap to be the circumference of a circle whose radius is from the center of the tube to the center of the wrap I'm dealing with. In other words, the radius of the first wrap is Do/2 + (1/2)t; the radius of the second wrap is Do/2 + (3/2)t; the radius of the third wrap is Do/2 + (5/2)t.
Each layer add t to the radius, and thus $$2t$$ to the diameter, so we divide the difference in diameters by $$2t$$ to count the layers.
He is using the radius to the middle of each layer (figuring that each layer is compressed on the inside and stretched on the outside), so he takes the first layer as having radius $$\frac{t}{2}$$ more than the core radius, and adds t to the radius for each layer. This makes an arithmetic series:
Now the circumference is 2pi times the radius and we need to sum a bunch of circumferences. Generalizing on the formula above, the radius of the kth wrap is:
radius of kth wrap = Do/2 + t(2k-1)/2 where k= 1,2,3,...
Length of material
= sum from k=1 to n of 2*pi*[Do/2 + t*(2k-1)/2]
= n*2*pi*Do/2 + sum from k=1 to n of 2*pi*t(2k-1)/2)
= n*pi*Do + pi*t times sum from k=1 to n of (2k-1)
Now the sum of 1+3+5 + ...+2n-1 = n^2
So, length = n*pi*Do + pi*t*n^2
But length = n*pi(Do + n*t)
But n = (D-Do)/(2t) so the formula is
Length = (D-Do)*pi/(2t)*[Do + (D-Do)/2]
= (D-Do)*pi/(4t)[Do+D]
= (D^2-Do^2)*pi/(4t)
I hope that this formula is correct and serves you well. There is no guarantee.
The summation is $$\sum_{k=1}^n 2\pi\left[\frac{D_o}{2}+\frac{t(2k-1)}{2}\right]=n\pi D_o+\pi t\sum_{k=1}^n (2k-1)\\=n\pi D_o+\pi tn^2=\pi n(D_o+nt)$$ This uses the fact that the sum of n consecutive odd numbers is $$n^2$$, which we proved in Inductive Proofs: Four Examples.
So the formula becomes $$L=\pi\cdot\frac{(D-D_o)}{2t}\cdot\left[D_o+\frac{D-D_o}{2}\right]\\=\frac{\pi(D-D_o)}{2t}\cdot\frac{D+D_o}{2}=\frac{\pi(D^2-D_o^2)}{4t}$$
Note that this gives the same formula as before, by adding all the layers in our idealized cylinder.
Derivation from average length of a layer
This question from 2003 provides a slightly different approach equivalent to the same idea:
Determining Length of Material Remaining on a Roll
Many apologies if this is not what you do, but I'm at a loss for where to find this answer.
I need to know the mathematical formula to determine the length of material as it is on a roll. Specifically; my company supplies rubber and tape, both in rolls on a core. Without unrolling the roll, we'd like to determine the length of the material with the following information: Outside diameter of the core, outside diameter of the whole roll and thickness of the material (determined by a micrometer).
Do you have any such formula?
We do, indeed!
Hi, Laura.
We get this question a lot; here are some of the explanations in our archive:
Length of Material on a Cylindrical Roll
http://mathforum.org/library/drmath/view/51723.html
How Much Carpet is Left?
http://mathforum.org/library/drmath/view/56481.html
Fabric Left on a Roll
http://mathforum.org/library/drmath/view/54357.html
Briefly, the length of the material will be the product of the number of layers and the average length of one layer in the roll. Both are easy to find: Number of layers = (Do-Di)/(2t) [total thickness/one layer] Average layer = pi(Do+Di)/2 [circumference at average diameter] where Do and Di are the outer and inner diameters, and t is the thickness of the material. So the length is L = pi(Do+Di)(Do-Di)/(4t) = pi/4 (Do^2 - Di^2)/t
We get the same formula; the idea is that the sum of an arithmetic series is the number of terms times the average of all the terms, which is the average of the first and last.
A small error in the thickness measurement can make a big difference; I would recommend measuring it not with a micrometer, but by measuring the outside diameter of a roll of known length and calculating t from this formula in reverse. That will ensure that the number you use reflects the way the material lies on the roll. But try both measurements to see how they compare.
This last comment has great practical importance. Not only is it hard to accurately measure a small thickness, but that may be changed by compression or by loose wrapping. Measuring the roll itself makes sure we use the real thickness. So let’s do that!
Given the size of a full roll
We’ll turn that last idea into a formula with this 1999 question:
How Much Carpet is Left?
We use rolled carpet in our business and would like to know how much is left on a roll given the remaining diameter.
Given: Beginning OD, core OD, number of layers, length of roll, New OD
Find: Length of carpet used or remaining.
There’s more information here than we need. We won’t be using the number of layers, but we’ll do that in the next answer.
Hi, Zene, welcome to Ask Dr. Math.
I enjoy solving problems from the working world - students often ask us, "Will I ever have a use for this stuff?" The answer is, you'd be surprised how often math comes in handy at work.
Here is a method that will give an approximate length of the carpet roll; the thinner the carpet relative to the diameter of the core, the better the approximation.
Consider a cross-section of the roll. It is an annulus (like a washer - a disk with a hole in the middle). Its area is the area of the outer circle minus the area of the inner circle (the core).
Here is an annulus (yellow), whose area is $$\pi R^2-\pi r^2$$:
Let C be the core diameter, F the diameter of a full roll, and L the length of carpet on a full roll. Let's also define T, the thickness of the carpet.
The cross-section area of the full carpet is
Area = (pi/4)(F^2 - C^2)
(Read F^2 as F squared.) When you unroll the carpet, its cross-sectional area is LT (L times T). This area should be (approximately) the same as the area of the roll, so we can find the thickness T:
T = (pi/4)(F^2 - C^2)/L
You can check the accuracy of the calculation so far by counting the layers of carpet on the full roll; this should be
n = (F - C)/(2T)
= (F - C)L/((pi/2)(F^2 - C^2))
= 2L/(pi(F + C))
You can also think of this calculation as dividing the length by the average circumference of a turn.
Now, if the diameter of a partial roll is D, we can use the same area calculations as above, but use the known T to find the length P of the carpet on the partial roll:
P = Area/T
= (pi/4)(D^2 - C^2)/((pi/4)(F^2 - C^2)/L)
= L(D^2 - C^2)/(F^2 - C^2)
So there you have it. The length of carpet on a partial roll is the length on a full roll times the difference between the diameter squared and the core diameter squared, divided by the same calculation for the full roll.
This is $$P=\frac{L(D^2-C^2)}{F^2-C^2}$$ That’s nice: Divide the current difference of squared diameters, by the original, and you get the fraction of the roll that is left.
Given the number of turns
But there’s another way to avoid the sensitivity to the measured thickness, seen in this 2003 question:
Length of Coiled Belt
Working with conveyor belts we use a formula that gives a very close approximation of the length of a coil of belt, but I would like to know the reasoning of it. The outside and inside diameters of the coil (in inches) are added and then multiplied by the number of wraps, this is then multiplied by a constant (.1309), and the result is to be read in feet.
It works, but I would like to know why. I tried to reason it out but I am stumped.
This turns out to be a valid formula: $$0.1309N(D+d)$$
I answered, starting with the formula we’ve seen:
Hi, Manny.
We have a number of answers to questions about the length on a roll of some material; one of them may give a formula like this based on the number of turns.
The cross sectional area of the coil is the difference between the circle formed by the outer diameter, D, and that formed by the inner diameter, d:
A = pi (D/2)^2 - pi (d/2)^2 = pi/4 (D^2 - d^2)
The cross-sectional area of the belt straightened out is
A = TL
where T is the thickness and L is the length. Setting these equal,
TL = pi/4 (D^2 - d^2)
This is our basic formula.
But T can be found by dividing (D-d)/2 by the number of turns N:
(D-d)/(2N) L = pi/4 (D^2 - d^2)
Solving for L,
L = pi/4 2N (D^2-d^2)/(D-d)
Since D^2 - d^2 = (D-d)(D+d), this is
L = pi/2 N (D+d)
That is your formula; to convert inches to feet we have to divide by 12, and (pi/2)/12 = 0.1309.
The calculation of T here is equivalent to the calculation of N in the previous answer. The new formula is $$L=\frac{\pi}{2}N(D+d)$$ using consistent units, and $$L=\frac{\pi}{2}N(D+d)\cdot\frac{1\text{ foot}}{12\text{ inches}}\approx0.1309N(D+d)$$ to get feet of length given diameters in inches.
Next time we’ll look at further variations in the question, focusing on how reality can change the formula.
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Question:
# Ivy spends $5.85 for 3.2 pounds of fruit and$9.00 for 2.7 pounds of vegetables. What is the average cost per pound Ivy spent on the fruits and vegetables?
A $2.50 explanation Step 1: Interpret the question To find the average in this scenario, we need to take the ratio of the total cost of the fruits and vegetables to the total number of pounds of them. Therefore, we have a division relationship between the total cost and the total number of pounds. The total cost of the fruits and vegetables will be the sum of their costs. Total cost: 5.85 + 9 The total number of pounds of fruits and vegetables will be the sum of their weights. The total number of pounds: 3.2 + 2.7 To get the average cost per pound for the fruits and vegetables, we divide the 2 totals. Average cost per pound= (Total cost) ÷ (Total number of pounds) Step 2: Round our numbers We have to round some of these numbers before we add them together. The numbers 5.85 and 2.7 will need to be rounded. 5.85 will be rounded to the nearest one's place. Let’s round 5.85. We underline the place value we are rounding to: 5.85 When we look to the right of the 5 we see an 8, which tells us the underlined digit goes up by one. So, 5 goes to 6. Everything to the right of 6 turns to 0. 6.00 Since the zero after the decimal has no value we can get rid of it. 5.85 rounds to 6. 2.7 will also be rounded to the nearest one's place. Let’s round 2.7. We underline the place value we are rounding to: 2.7 When we look to the right of the 2 we see a 7, which tells us the underlined digit goes up by one. So, 2 goes to 3. Everything to the right of 3 turns to 0. 3.0 Since the zero after the decimal has no value we can get rid of it. 2.7 rounds to 3. Step 3: Solve First, find the total cost of the fruits and vegetables. 6 + 9 = 15 So, the total cost of the fruits and vegetables is$15.
Next, find the total weight of the fruits and vegetables.
3 + 3 = 6
So, the total weight of the fruits and vegetables is 6 pounds.
Finally, we get the average cost per pound of the fruits and vegetables by dividing the total cost by the number of pounds.
15 ÷ 6 = 2.5
So, the average cost per pound for the fruits and vegetables is \$2.50.
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# What is 1/474 as a decimal?
## Solution and how to convert 1 / 474 into a decimal
1 / 474 = 0.002
1/474 or 0.002 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Both represent numbers between integers, in some cases defining portions of whole numbers The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. So let’s dive into how and why you can convert 1/474 into a decimal.
## 1/474 is 1 divided by 474
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We use this as our equation: numerator(1) / denominator (474) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's how our equation is set up:
### Numerator: 1
• Numerators are the portion of total parts, showed at the top of the fraction. With a value of 1, you will have less complexity to the equation; however, it may not make converting any easier. 1 is an odd number so it might be harder to convert without a calculator. Values like 1 doesn't make it easier because they're small. Let's take a look below the vinculum at 474.
### Denominator: 474
• Denominators represent the total parts, located at the bottom of the fraction. 474 is one of the largest two-digit numbers to deal with. And it is nice having an even denominator like 474. It simplifies some equations for us. Overall, two-digit denominators are no problem with long division. So without a calculator, let's convert 1/474 from a fraction to a decimal.
## Converting 1/474 to 0.002
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 474 \enclose{longdiv}{ 1 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 474 \enclose{longdiv}{ 1.0 }$$
Uh oh. 474 cannot be divided into 1. So that means we must add a decimal point and extend our equation with a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 474 into 1 + 0 or 10.
### Step 3: Solve for how many whole groups you can divide 474 into 10
$$\require{enclose} 00.0 \\ 474 \enclose{longdiv}{ 1.0 }$$
Now that we've extended the equation, we can divide 474 into 10 and return our first potential solution! Multiply by the left of our equation (474) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 474 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. Same goes for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Here are examples of when we should use each.
### When you should convert 1/474 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.002 to 1/474 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 1/474 = 0.002 what would it be as a percentage?
• What is 1 + 1/474 in decimal form?
• What is 1 - 1/474 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.002 + 1/2? |
New Zealand
Level 7 - NCEA Level 2
# Tangent Ratios (proofs and equivalences)
Lesson
Imagine a right-angled triangle with one of the acute angles labelled $\phi$ϕ. If the side opposite the angle $\phi$ϕ has length $a$a and the side adjacent to the angle $\phi$ϕ has length $b$b, then we write $\tan\phi=\frac{a}{b}$tanϕ=ab. But this is the same as the equivalent fraction $\frac{\frac{a}{c}}{\frac{b}{c}}$acbc obtained by dividing the numerator and the denominator by the number $c$c.
If we now let $c$c be the hypotenuse of the triangle, we see that the complicated looking fraction is just $\frac{\sin\phi}{\cos\phi}$sinϕcosϕ. Hence, we can write the identity
$\tan\phi\equiv\frac{\sin\phi}{\cos\phi}$tanϕsinϕcosϕ
(This is an identity in the sense that it is true for every value of $\phi$ϕ.)
Another very useful identity is obtained in a similar way. In the same triangle, we have the statement of Pythagoras's theorem $a^2+b^2\equiv c^2$a2+b2c2, which remains true for a given $c$c whatever the values of $a$a and $b$b, provided the triangle continues to be right-angled.
On dividing both sides of this equivalence by $c^2$c2, we obtain $\frac{a^2}{c^2}+\frac{b^2}{c^2}\equiv1$a2c2+b2c21. We recognise this as just
$\sin^2\phi+\cos^2\phi\equiv1$sin2ϕ+cos2ϕ1.
This can now be divided through by $\cos^2\phi$cos2ϕ to obtain a further identity involving the tangent function:
$\tan^2\phi+1\equiv\frac{1}{\cos^2\phi}=\sec^2\phi$tan2ϕ+11cos2ϕ=sec2ϕ
#### Examples
##### Example 1
Find a simpler way of writing $\frac{\sin^2A+\sin A\cos A}{\cos^2A+\sin A\cos A}$sin2A+sinAcosAcos2A+sinAcosA.
Taking a common factor out of the numerator and out of the denominator, we can write the expression as
$\frac{\sin A\left(\sin A+\cos A\right)}{\cos A\left(\cos A+\sin A\right)}$sinA(sinA+cosA)cosA(cosA+sinA).
On cancelling the common term from the numerator and the denominator, we have $\frac{\sin A}{\cos A}=\tan A$sinAcosA=tanA.
Often, students are asked to verify an identity. This is usually done by rearranging or simplifying one side of the identity until it looks identical to the other side. The working above essentially proves the identity
$\frac{\sin^2A+\sin A\cos A}{\cos^2A+\sin A\cos A}\equiv\tan A$sin2A+sinAcosAcos2A+sinAcosAtanA.
##### Example 2
If $5\sin x=13\cos x$5sinx=13cosx and $x$x is between $0$0 and $2\pi$2π, what are the possible values of $x$x?
From $5\sin x=13\cos x$5sinx=13cosx we obtain $\frac{\sin x}{\cos x}=\frac{13}{5}=2.6$sinxcosx=135=2.6
That is, $\tan x=2.6$tanx=2.6. Therefore, in the first quadrant, $x=\arctan2.6\approx1.2036$x=arctan2.61.2036 or $\frac{\pi}{2.6101}$π2.6101. There should also be a third-quadrant solution since $\tan x=\tan\left(\pi+x\right)$tanx=tan(π+x). That is, the second solution is$\pi+\frac{\pi}{2.6101}=\frac{3.6101\times\pi}{2.6101}$π+π2.6101=3.6101×π2.6101.
#### Worked Examples
##### Question 1
By finding the ratio represented by $\sin\theta$sinθ, $\cos\theta$cosθ and $\tan\theta$tanθ in the given figure, we want to prove that $\frac{\sin\theta}{\cos\theta}=\tan\theta$sinθcosθ=tanθ.
1. Write down the expression for $\sin\theta$sinθ.
2. Write down the expression for $\cos\theta$cosθ.
3. Hence, form an expression for $\frac{\sin\theta}{\cos\theta}$sinθcosθ.
4. Write down the expression for $\tan\theta$tanθ.
5. Does $\frac{\sin\theta}{\cos\theta}=\tan\theta$sinθcosθ=tanθ?
Yes
A
No
B
Yes
A
No
B
##### Question 2
.
Given that $\sin x=\frac{4}{5}$sinx=45 and $\cos x=\frac{3}{5}$cosx=35, find $\tan x$tanx.
##### Question 3
Find the value of $x$x when
$8\sin x=15\cos x$8sinx=15cosx. |
$$\DeclareMathOperator{cosec}{cosec}$$
Number and Algebra
Furthermore
Official Guidance, clarification and syllabus links:
Counting principles may be used in the development of the theorem.
$$^nC_r$$ should be found using both the formula and technology.
Example: Find $$r$$ when $$^6C_r=20$$, using a table of values generated with technology.
Formula Booklet:
Binomial theorem $$n \in \mathbb{N}$$ $$(a + b)^n = a^n +\;^nC_1 a^{n-1}b + \ldots + \;^nC_r a^{n-r}b^r + \ldots + b^n$$ $$^nC_r = \frac{n!}{r!(n-r)!}$$
Example
To expand $$(2x + 3y)^5$$ using the binomial theorem, we can use the formula:
$$(a + b)^n = a^n + nC_1 a^{n-1}b + \ldots + nC_r a^{n-r}b^r + \ldots + b^n$$
Substituting $$a = 2x$$, $$b = 3y$$, and $$n = 5$$ into the formula, we get:
$$(2x + 3y)^5 = (2x)^5 + 5C_1 (2x)^4(3y) + 5C_2 (2x)^3(3y)^2 + 5C_3 (2x)^2(3y)^3 + 5C_4 (2x)(3y)^4 + (3y)^5$$
Simplifying further:
$$(2x + 3y)^5 = 32x^5 + 160x^4 \cdot 3y + 80x^3 \cdot 9y^2 + 40x^2 \cdot 27y^3 + 10x \cdot 81y^4 + 243y^5$$
Which results in:
$$(2x + 3y)^5 = 32x^5 + 480x^4y + 720x^3y^2 + 1080x^2y^3 + 810xy^4 + 243y^5$$
This video on the Binomial Theorem is from Revision Village and is aimed at students taking the IB Maths AA Standard level course
How do you teach this topic? Do you have any tips or suggestions for other teachers? It is always useful to receive feedback and helps make these free resources even more useful for Maths teachers anywhere in the world. Click here to enter your comments.
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# Momentum
## Linear momentum
Momentum is equal to mass times velocity.
${\displaystyle {\vec {p}}=m{\vec {v}}}$
## Angular momentum
Angular momentum of an object revolving around an external axis ${\displaystyle O}$ is equal to the cross-product of the position vector with respect to ${\displaystyle O}$ and its linear momentum.
${\displaystyle {\vec {L}}={\vec {r}}\times {\vec {p}}={\vec {r}}\times m{\vec {v}}}$
Angular momentum of a rotating object is equal to the moment of inertia times angular velocity.
${\displaystyle {\vec {L}}=I{\vec {\omega }}}$
## Force and linear momentum, torque and angular momentum
Net force is equal to the change in linear momentum over the change in time.
${\displaystyle {\vec {F}}={\frac {\Delta {\vec {p}}}{\Delta t}}}$
Net torque is equal to the change in angular momentum over the change in time.
${\displaystyle {\vec {\tau }}={\frac {\Delta {\vec {L}}}{\Delta t}}}$
## Conservation of momentum
${\displaystyle {\begin{matrix}{\vec {p}}_{i}={\vec {p}}_{f}\\{\vec {L}}_{i}={\vec {L}}_{f}\end{matrix}}}$
Let us prove this law.
We'll take two particles ${\displaystyle a,b}$ . Their momentums are ${\displaystyle {\vec {p}}_{a},{\vec {p}}_{b}}$ . They are moving opposite to each other along the ${\displaystyle x}$-axis and they collide. Now force is given by:
${\displaystyle {\vec {F}}={\frac {d{\vec {p}}}{\mathrm {d} t}}}$
According to Newton's third law, the forces on each particle are equal and opposite.So,
${\displaystyle {\frac {d{\vec {p}}_{a}}{dt}}=-{\frac {d{\vec {p}}_{b}}{dt}}}$
Rearranging,
${\displaystyle {\frac {d{\vec {p}}}{dt}}+{\frac {d{\vec {p}}_{b}}{dt}}=0}$
This means that the sum of the momentums does not change with time. Therefore, the law is proved.
## Variables
p: momentum, (kg·m/s) m: mass, (kg) v: velocity (m/s) L: angular momentum, (kg·m2/s) I: moment of inertia, (kg·m2) ω: angular velocity (rad/s) α: angular acceleration (rad/s2) F: force (N) t: time (s) r: position vector (m) Bold denotes a vector quantity. Italics denotes a scalar quantity.
## Definition of terms
Momentum (p): Mass times velocity. (kg·m/s) Mass (m) : A quantity that describes how much material exists, or how the material responds in a gravitational field. Mass is a measure of inertia. (kg) Velocity (v): Displacement divided by time (m/s) Angular momentum (L): A vector quantity that represents the tendency of an object in circular or rotational motion to remain in this motion. (kg·m2/s) Moment of inertia (I): A scalar property of a rotating object. This quantity depends on the mass of the object and how it is distributed. The equation that defines this is different for differently shaped objects. (kg·m2) Angular speed (ω): A scalar measure of the rotation of an object. Instantaneous velocity divided by radius of motion (rad/s) Angular velocity (ω): A vector measure of the rotation of an object. Instantaneous velocity divided by radius of motion, in the direction of the axis of rotation. (rad/s) Force (F): mass times acceleration, a vector. Units: newtons (N) Time (t) : (s) Isolated system: A system in which there are no external forces acting on the system. Position vector (r): a vector from a specific origin with a magnitude of the distance from the origin to the position being measured in the direction of that position. (m)
## Calculus-based Momentum
Force is equal to the derivative of linear momentum with respect to time.
${\displaystyle {\vec {F}}={\frac {d{\vec {p}}}{dt}}}$
Torque is equal to the derivative of angular momentum with respect to time.
${\displaystyle {\vec {\tau }}={\frac {d{\vec {L}}}{dt}}}$ |
# Graphing Inequalities
## What does it mean?
### Definitions:
When graphing linear inequalities, we use a lot of the same concepts that we used when we were solving linear equations. The difference: the solution to the linear equation is only one number, but when a variable is less than or greater than a number, there are an infinite number of values that would be a part of the answer.
### Inequality Signs and Interval Notation:
#### Inequality
x > a x is greater than a
x < a x is less than a
x > 2
x < 2
(a, )
(-∞, a)
(-∞, 2)
#### Inequality
x ≥ a x is greater than or equal to a
x ≤ a x is less than or equal to a
x ≥ 2
x ≤ 2
[a, )
(-∞, a]
(-∞, 2]
### Multiplication/Division Properties for Inequalities:
Multiplying or dividing the same NEGATIVE number to both sides of an inequality reverses the sign of the inequality.
1. If a < b and c is negative, then ac > bc
2. If a < b and c is negative, then a/c > b/c
## What does it look like?
$$-2 < 5 \space multiplied \space by \space (-3)$$ $$(-3)(-2) > 5(-3)$$ $$6 > (-15)$$ $$-2 < 5 \space divided \space by \space (-3)$$ $$\frac{-2}{-3} > \frac{5}{-3}$$ $$\frac{2}{3} > \frac{-5}{3}$$ Graphing Linear Inequalities: y > mx + b & y ≥ mx + b
• Graph the solution to y < 2x + 3
The first step is to find the "equals" part. For two-variable linear inequalities, the "equals" part is the line $$y = 2x + 3$$
Now we are ready to do the "y less than" part. In other words, this is where we need to shade one side of the line or the other. We need y LESS THAN the line, so we want all the points below the line.
• Graph the solution to 2x – 3y ≤ 6
First, solve for y: $$2x – 3y ≤ 6$$ $$–3y ≤ –2x + 6$$ $$y ≥ \frac{2}{3}x – 2$$ Now, find the "equals" part, which is the line $$y = \frac{2}{3}x – 2$$
We have inequality where "y is greater than". The notation for a strict inequality is a dashed line. So the border of our solution region actually looks like this:
By using a dashed line, we know where the border is but we also know that the border isn't included in the solution. Since this is a "y greater than" inequality, we want to shade above the line.
## You'll use it...
Temperature conversion, exchange rates for currency, plans for cell phones, travel, in cooking recipes, marketing and much more.
## Video
### Graphing Lines 1
Watch a Khan Academy Video »
Length: 9:50
## Video
### Interpreting Linear Graphs
Watch a Khan Academy Video »
Length: 5:05
## Video
### Graphing Inequalities
Watch a Khan Academy Video »
Length: 8:04
## Video
### Graphing Linear Inequalities in Two Variables
Watch a Khan Academy Video »
Length: 2:42 |
# How do you find the vertex and intercepts for f(x) = 2x^2 - 4x + 3?
Jun 14, 2018
$\text{see explanation}$
#### Explanation:
$\text{express the quadratic in "color(blue)"vertex form}$
•color(white)(x)y=a(x-h)^2+k
$\text{where "(h,k)" are the coordinates of the vertex and a is}$
$\text{a multiplier}$
$\text{using the method of "color(blue)"completing the square}$
$f \left(x\right) = 2 \left({x}^{2} - 2 x + \frac{3}{2}\right)$
$\textcolor{w h i t e}{f \left(x\right)} = 2 \left({x}^{2} + 2 \left(- 1\right) x + 1 - 1 + \frac{3}{2}\right)$
$\textcolor{w h i t e}{f \left(x\right)} = 2 {\left(x - 1\right)}^{2} + 1$
$\text{vertex } = \left(1 , 1\right)$
$\text{for y-intercept set x = 0}$
$f \left(0\right) = 3 \Rightarrow \left(0 , 9\right)$
$\text{for x-intercepts set y = 0}$
$2 {\left(x - 1\right)}^{2} + 1 = 0$
${\left(x - 1\right)}^{2} = - \frac{1}{2}$
$\text{this has no real solutions hence no x-intercepts}$
graph{2x^2-4x+3 [-10, 10, -5, 5]} |
# 2016 AMC 8 Problems/Problem 11
## Problem
Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$
## Solutions
### Solution 1
We can write the two digit number in the form of $10a+b$; reverse of $10a+b$ is $10b+a$. The sum of those numbers is: $$(10a+b)+(10b+a)=132$$$$11a+11b=132$$$$a+b=12$$ We can use brute force to find order pairs $(a,b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$. Thus, our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$; or $\boxed{\textbf{(B)} 7}$ ordered pairs.
### Solution 2 -SweetMango77
Since the numbers are “mirror images,” their average has to be $\frac{132}{2}=66$. The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$ and $6-3=3$, so our lowest tens digit is $3$. The numbers between $9$ and $3$ inclusive is $9-3+1=\boxed{\text{(B)}\;7}$ total possibilities.
~savannahsolver |
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### Section 5.5 : Partial Fractions
7. Determine the partial fraction decomposition of each of the following expression.
$\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}}$
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Start Solution
The first step is to determine the form of the partial fraction decomposition. For this problem the partial fraction decomposition is,
$\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{A}{{2x + 3}} + \frac{B}{{x - 2}} + \frac{C}{{{{\left( {x - 2} \right)}^2}}}$ Show Step 2
The LCD for this expression is $$\left( {2x + 3} \right){\left( {x - 2} \right)^2}$$. Adding the terms back up gives,
$\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{{A{{\left( {x - 2} \right)}^2} + B\left( {2x + 3} \right)\left( {x - 2} \right) + C\left( {2x + 3} \right)}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}}$ Show Step 3
Setting the numerators equal gives,
$4{x^2} - 22x + 7 = A{\left( {x - 2} \right)^2} + B\left( {2x + 3} \right)\left( {x - 2} \right) + C\left( {2x + 3} \right)$ Show Step 4
For this problem we can pick “good” values of $$x$$ to determine only two of the three constants. Here is that work.
\begin{array}{l}{\displaystyle x = - \frac{3}{2}:}\\{x = 2:}\end{array}\hspace{0.25in}\begin{aligned}49 & = A{{\left( { - \frac{7}{2}} \right)}^2} = \frac{{49}}{4}A\\ - 21 & = 7C\end{aligned}\hspace{0.25in} \to \hspace{0.25in}\begin{array}{l}{A = 4}\\{C = - 3}\end{array} Show Step 5
To get the remaining constant we can use any value of $$x$$ and plug that along with the values of $$A$$ and $$C$$ we found in the previous step into the equation from Step 3.
It really doesn’t matter what value of $$x$$ we pick as long as it isn’t $$x = - \frac{3}{2}$$ or $$x = 2$$ since we used those in the previous step. The idea here is to pick a value of $$x$$ that won’t create “large” or “messy” numbers, if possible. Good choices are often $$x = 0$$ or $$x = 1$$, provided they weren’t used in the previous step of course.
For this problem $$x = 0$$ seems to be a good choice. Here is the work for this step.
\begin{align*}7 & = \left( 4 \right){\left( { - 2} \right)^2} + B\left( 3 \right)\left( { - 2} \right) + \left( { - 3} \right)\left( 3 \right)\\ 7 & = 7 - 6B\\ 0 & = - 6B\hspace{0.25in} \Rightarrow \hspace{0.25in}B = 0\end{align*}
In this case one of the constants ended up being zero. This happens on occasion but there is no way, in general, to know ahead of time that was going to happen so don’t worry about it. If it turns out one of the constants is zero then we’ll figure that out when we do the work.
Show Step 6
The partial fraction decomposition is then,
$\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{4}{{2x + 3}} - \frac{3}{{{{\left( {x - 2} \right)}^2}}}}}$ |
Session 8, Part B:
Inverse Proportions (60 minutes)
In This Part: Splitting a Prize | Ms. Anwar's Backyard | Another Inverse Porportion
Imagine that the teachers in your school decide to play the lottery together. If they win, the prize is \$800,000. Problem B1 is a table that shows how much each teacher will get, depending on how many of them contribute to buy the tickets. Note 4
Problem B1
Fill in the rest of this table: Note 5
Number of teachers Amount each will receive 1 800,000 2 400,000 3 4 80,000 53,333.33
Number of teachers Amount each will receive 1 800,000 2 400,000 3 266,666.66 4 200,000 10 80,000 15 53,333.33 20 40,000 100 8,000
Problem B2 Describe a rule that fits the table. Try to find more than one rule. Explain why your rule will work if the table is continued.
Problem B3 Graph the rule that shows how much each teacher will receive. Describe how this graph is different from other functions you've seen.
Make sure you distinguish this function from all the other functions you've seen -- linear, exponential, quadratic, and cyclic. Close Tip Make sure you distinguish this function from all the other functions you've seen -- linear, exponential, quadratic, and cyclic.
Video Segment In this video segment, the class discusses the shape of the graph for the lottery problem. Then they discuss whether or not the graph crosses the line y = 0. Watch this segment after you've completed Problem B3. If you get stuck on the problem, you can watch the video segment to help you. What would a value of y = 0 mean in terms of the lottery problem? There is an important difference between this type of function -- a decreasing curve -- and a decreasing line. You can find this segment on the session video, approximately 10 minutes and 15 seconds after the Annenberg Media logo.
Functions like this are called inverse proportions. The relationship between the two variables in this function is called inverse variation. Inverse proportions are another example of nonlinear functions.
You can think of inverse proportions in two ways:
• output = some constant / input, or • input * output = some constant
There are many applications of these kinds of functions. If you're getting paid a fixed amount of money to do a job, for example, your hourly rate depends on how quickly you complete the work. The shorter the time, the larger the hourly rate.
Session 8: Index | Notes | Solutions | Video |
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• ## Related Books
### Angles
Main definitions and formulas:
• Degrees are a unit of measurement by which a circle is divided into 360 equal parts, denoted 360° .
• Radians are a unit of measurement by which a circle is divided into 2π parts, denoted 2πR.
• Since the circumference of a circle is 2πr, this means that a 1R angle cuts off an arc whose length is exactly equal to the radius. (It is [1/(2π )] of the whole circle.)
• Since 2π≈ 6.28..., this means that 1R is about one sixth of a circle. But we seldom use whole numbers of radians. Instead we use multiples of π . For example, (π /2)R is exactly one fourth of a circle.
• degree measure × π180 = radian measure
• radian measure × 180 π = degree measure
• Coterminal angles are angles that differ from each other by adding or subtracting multiples of 2πR (i.e. 360° ). If you graph them in the coordinate plane starting at the x-axis, they terminate at the same place.
• Complementary angles add to (π /2)R(i.e. 90° ).
• Supplementary angles add to πR (i.e. 180° ).
Example 1:
If a circle is divided into 18 equal angles, how big is each one, in degrees and radians?
Example 2:
2. Convert (5π /12)R into degrees.
Example 3:
For each of the following angles, determine which quadrant it is in and find a coterminal angle between 0 and 360° or between 0 and 2πR.
1. 1000°
2. − (19π /6)R
3. -586°
4. (22π /7)R
Example 4:
Convert the following "common values" from degrees to radians: 0, 30, 45, 60, 90. Find the complementary and supplementary angles for each one, in both degrees and radians.
Example 5:
For each of the following angles, determine which quadrant it is in and find a coterminal angle between 0 and 360° or between 0 and 2πR.
1. − (5π /4)R
2. 735°
3. − (7π /3)R
4. -510°
### Angles
A circle is divided into 12 equal angles. Calculate the measure of each angle in degrees and radians.
• In order to calculate the degrees and radians, recall that a circle is 360°
• Calculate the measure of each angle in degrees first
• [360/12]° = 30°
• In radians, we know that 360° is 2π. So, we can calculate the measure of each angle in radians
• [(2π)/12] = [(π)/6]
30° and [(π)/6]
• Recall the equation for converting degrees to radians
• degree measure ×[(π)/(180°)] = radian measure
• 185° ×[(π)/(180°)] =
• [(185°π)/(180°)]
[(37π)/36]
Convert [(7π)/10] into degrees
• Recall the equation for converting radians to degrees
• radian measure ×[(180°)/(π)] = degree measure
• [(7π)/10] ×[(180° )/(π)] =
• 7 ×18°
126°
What is the degree measure of an arc whose measure is [(3π)/5] radians?
• radian measure ×[(180°)/(π)]= degree measure
• [(3π)/5] ×[(180°)/(π)] =
• [(540°π)/(5π)]
• The radians cancel so now just divide
108°
What is the radian measure of an arc whose measure is 76 °?
• Recall the formula for converting degrees to radians
• degree measure ×[(π)/(180°)] = radian measure
• 76°× [(π)/(180° )] =
• [(76° π)/(180° )]
• The degrees will cancel. Simplify your fraction
[(19π)/45]
Determine which quadrant the following angle is in and find a coterminal angle between 0° and 360° :
460°
• Notice that 460° is larger than 360°, so we must subtract 360° from our given angle until we reach an angle that is between 0° and 360°
• 460° - 360° = 100° which is an angle that is between 0° and 360° and it is coterminal to 450°
• Now we can determine which quadrant our angle is in by using the following:
Quadrant I has angles between 0° and 90°
Quadrant II has angles between 90° and 180°
Quadrant III has angles between 180° and 270°
Quadrant IV has angles between 270° and 360°
Determine which quadrant the following angle is in and find a coterminal angle between 0° and 360° :
[( − 13π)/15]
• Notice that [( − 13π)/15] is smaller than 0°, so we must add 360° or 2π to our given angle until we reach an angle that is between 0 and 360° (i.e. between 0 and 2π)
• [( − 13π)/15] + 2π = [(17π)/15] which is an angle that is between 0 and 2π and it is coterminal to [( − 13π)/15]
• Now we can determine which quadrant our angle is in by using the following:
Quadrant I has angles between 0 and [(π)/2]
Quadrant II has angles between [(π)/2] and π
Quadrant III has angles between π and [(3π)/2]
Quadrant IV has angles between [(3π)/2]and 2π
Find the complementary angle for each of the following angles:
a. 47°
b. [(π)/12]
• Recall that complementary angles add to 90° or [(π)/2]
a. 90° − 47° = 43°
b. [(π)/2] − [(π)/12] = [(5π)/12]
Find the supplementary angle for each of the following angles:
a. 114°
b. [(4π)/5]
• Recall that supplementary angles add to 180° or π
• a. 180° − 114° = 66°
b. π− [(4π)/5] = [(π)/5]
a. 66°
b. [(π)/5]
For each of the following angles, determine which quadrant it is in and find a coterminal angle between 0° and 360° or between 0 and 2π
a. [( − 6π)/11]
b. 623°
c. [( − 27π)/13]
d. 1572°
• a. [( − 6π)/11]is smaller than 0 so you have to add 2π to find a coterminal angle
• [( − 6π)/11] + 2π = [(16π)/11] which is in quadrant III
• b. 623° is larger than 360° so you have to subtract 360° to find a coterminal angle
• 623° − 360° = 263° which is in quadrant III
• c. [( − 27π)/13] is smaller than 0 so you have to add 2p to find a coterminal angle
• [( − 27π)/13] + 2π = [( − π)/13] which is still smaller than 0 so keep adding 2p
• [( − π)/13] + 2π = [(25π)/13] which is in quadrant IV
• d. 1572° is larger than 360° so you have to subtract 360° to find the coterminal angle
• 1572° − 360° = 1212° which is still larger than 360° so keep subtracting by 360°
• 1212° − 360° = 852°
852° − 360° = 492°
492° − 360° = 132° which is in quadrant II
a. [(16π)/11]; III
b. 263°; III
c. [(25π)/13]; IV
d. 132°; II
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Angles
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Degrees 0:22
• Circle is 360 Degrees
• Splitting a Circle
• Circle is 2 Pi Radians
• Half-Circle and Right Angle
• Converting Between Degrees and Radians 6:24
• Formulas for Degrees and Radians
• Coterminal, Complementary, Supplementary Angles 7:23
• Coterminal Angles
• Complementary Angles
• Supplementary Angles
• Example 1: Dividing a Circle 10:38
• Example 2: Converting Between Degrees and Radians 11:56
• Example 3: Quadrants and Coterminal Angles 14:18
• Extra Example 1: Common Angle Conversions
• Extra Example 2: Quadrants and Coterminal Angles
### Transcription: Angles
Hi we are here to do some extra examples on measuring angles and converting back and forth between degrees and radians.0000
I hope you had a chance to try this out on your own a little bit.0008
There are common values that you use a lot in all kinds of trigonometric functions and situations.0012
It is worth at least working them out once on your own and memorize them after that.0020
The common values are 0 degrees, 30 degrees, 45degrees, 60degrees, and 90 degrees.0027
What we are going to do is find the complementary and supplementary angles for each one in both degrees and radians.0037
The reason these angles are so important is because 90 degrees is a right angle0047
What we are doing is chopping a right angle up into either two equal pieces which gives us 45 degrees or three equal pieces which gives us the 30 degrees and 60 degrees angles.0054
Those are very common ones that come up very often.0065
It is worth knowing what these are in both degrees and radians.0068
Knowing their complements and supplements and knowing what the complements and supplements are in degrees and radians as well.0073
Let me make a little chart here, we are starting out in degrees.0082
We have the 0 degrees angle, 30 degrees, 45 degrees, 60 degrees, and 90 degrees.0087
Let me convert those into radians first.0098
The 0 degrees angle is still 0 in radians.0102
Well remember that a 90 degrees angle is pi over 2 radians.0107
A 30 degrees angle is 1/3 of that, it’s pi over 2 divided by 3, that is pi over 6.0114
45 degrees is 90 degrees divided by 2, that is pi over 2 divided by 2 which is pi over4.0123
60 degrees is twice 30 degrees, 60 degrees is 2 x pi over 6, 2 pi over 6 is pi over 3.0131
Those are pretty convenient fraction if you write them in radians.0143
The complementary angles we will do it in terms of degrees first.0147
Remember complementary angles add up to 90 degrees.0154
If you know what the angle is, you will do 90 degrees minus that number to get the complementary angle.0158
If you start with the 0 degrees angle, the complementary angle is 90 degrees because 90 minus 0 is 90.0165
30 degrees angle the complementary angle is 60 because those add up to 90.0173
45 degrees angle is its own complement because 45 and 45 is 90.0178
60 degrees angle its complement is 30.0185
A 90 angle its complement is 0.0187
Let us do those in terms of radiance0192
The 0 radiant angle its complement is going to be pi over 2 because those add up to pi over 2.0198
Remember that is the same as a 90 degree angle.0207
Pi over 6 + pi over 3 is pi over 2.0210
You can work out the fractions there or you can remember that 30 + 60 equals 90.0216
45 degree angle or pi over 4 is its own complement.0222
Pi over 3 we already figured out that its complement is pi over 6.0228
Pi over 2 its complement is 0 because those add up to pi over 2 or 90 degree angle.0235
Let us figure out the supplementary angles.0243
Remember supplementary angle means that they add up to 180 degrees.0246
In each case we are looking for what adds up to pi over 2 or 90.0251
45 a 180 - 45 is 135.0265
180 - 60 is 120.0271
180-90 is 90 itself.0275
Finally, if we find the supplementary angles in terms of radians.0280
Remember we are looking here for, before they add it up to a 180 degrees in terms of radians they should add it up to pi.0286
0 + pi adds up to pi.0295
Pi over 6 + 5 pi over 6 adds up to pi.0299
Pi over 4 + 3 pi over 4 adds up to pi.0306
Pi over 3 + 2 pi over 3 adds up to pi.0313
Pi over 2 + pi over 2 adds up to pi.0320
All of these are common values.0326
These are all conversions back and forth between degrees and radians that you should now very well as a trigonometry student just because these angles come up so often.0328
It is probably worth understanding the pictures behind each of these numbers that I have written down.0337
For example, when we look at complementary angles.0344
Here is a 30 degree angle and its complement is a 60 degree angle.0349
The 60 degree angle is pi over 3 radians.0362
Add those together, you will get pi over 3 + pi over 6 adds up to pi over 2.0367
Here is another right angle and I’m dividing it in 2 into 45 degrees, which is pi over 4 radians.0377
We see that that angle is its own complement 45 degrees is pi over 4 radians.0385
I will do the supplements in blue.0394
If we start out with a 30 degree angle or pi over 6 radians.0399
Then its supplement is a 150 degree angle which is 5 pi over 6.0411
If we start out with a 60 degree angle which is pi over 3 radians then its supplement.0423
Remember to put them together and they are supposed to make 180 degree or pi radians is 120 degrees which is 2 pi over 3 radians.0436
Finally if you start out with a 45 degree angle which is pi over 4 radians then its supplement is 135 degrees which is 3 pi over 4 radians.0452
All of those are angles that you should know very well both in terms of degrees and radians because we will be seeing a lot of them in our trigonometry lessons.0475
Finally, our example here is for each of the following angles.0000
I want to find out what quadrant it is in.0005
I want to find the coterminal angle between 0 and 360 degrees over 0 and 2 pi radians.0009
We got 4 angles here and I’m going to give you in both degrees and radians.0018
We are going to start out with -5 pi over 4 radians and that will do 735 degrees, -7 pi/3 radians and -510 degrees.0027
If you want you can try those on your own.0047
Ok, that is not between 0 and 2 pi radians.0057
What we are going to is add a 2 pi to it, 2 pi is 8 pi/4, (8 pi/4 – 5 pi/4) is 3 pi/4.0060
Let us graph that, 0 pi/2 pi, 3 pi/2, and 2 pi, 3 pi/4 is between pi/2 and pi.0076
It is right there, that is in the second quadrant.0094
Our answer here is 3 pi/4 and it is in the second quadrant.0107
The way you can understand how that came from the original -5 pi/4.0115
If you went in the other direction 5 pi/4 in the other direction from the traditional direction because it is negative. You will end up at that angle.0121
Its coterminal angle is 3 pi over 4 in quadrant 2.0131
735 degrees that is way bigger than 360.0136
Let us drop down multiples of 360, we can actually take out two multiples of 360 right away.0141
I’m going to subtract 720 degrees and we get 15 degrees.0148
If you write things in terms of degrees here, 0 is 0 degrees, Pi/2 is 90 degrees, Pi is 180 degrees, 3 pi/2 is 270 degrees, 2pi is 360 degrees.0154
15 degrees is between 0 and 90. That is right about there0174
What that means is 735 would actually go out in circle twice and end up right at the same place that 15 degrees ended up.0182
So, the coterminal angle is 15 degrees and that is in the first quadrant.0192
Let us look at -7pi/ 3, that is less than 0 so I’m going to add 2 pi to that, plus 2 pi, well 2 pi is 6 pi/3.0203
That would give us – pi/3, that is still less than 0, let me add another 2 pi.0217
That is again 6 pi/3 – pi/3, would give us 5 pi/3, that is between 0 and 2 pi.0227
Now, we just have to figure what out quadrant it is in.0238
5 pi/3 is a little bit bigger than 3 pi/2, 3 pi/2 is 1 ½, 5 pi/3 is 1.67.0241
So, 5 pi/3 is a little bit bigger than 3 pi/2.0252
Let me erase some of these extra angles, they are getting in our way.0260
I will show you where 5 pi over 3 is.0273
If you go between pi and 2 pi, there is 4 pi/3, there is 5 pi/3.0277
And so, 5 pi/3 is right there, that is clearly in the fourth quadrant.0283
Finally, -510 degrees, where does that one end up? That is definitely less than 0.0295
So, I will add 360 degrees and we will end up with -150 degrees, still less than 0 there so I will add another 360 degrees, that gives us 210 degrees.0303
That is between 0 and 360 degrees, we know we found our coterminal angle, 210 degrees is just past 180 degrees.0322
In fact, it is 30 degrees past 180 degrees, it is right there, that is in the third quadrant.0333
Just to recap here, finding these coterminal angles and finding out what quadrant they are in.0346
To find the coterminal angle, you take the angle that you are given in degrees or radians and you add or subtract multiples of either 2 pi radians or 360 degrees.0351
Then you add or subtract these multiples until you get them in to the range that you want, 0 to 2 pi radians, 0 to 360 degrees.0357
Once you get it into those ranges then you can break it down finally and ask are you between 0, pi over 2, pi, 3 pi over 2, or 2 pi.0380
That tells you which quadrant you are in or in terms of degrees that would be 0, 90 degrees, 180 degrees, 270 degrees, or 360 degrees.0394
That will tell you what quadrant you are in if you are in terms of degrees.0407
That is the first or our lessons on trigonometry for www.educator.com.0412
Here we talked about angles, we have not really got in to the trigonometric functions yet.0417
In the next lessons we will start talking about sine and cosine, all the different identities and relationships,how you use them in triangles?0421
We will talk about tangents and secants.0429
That is all coming in the later lectures on www.educator.com.0431
Hi, This is Will Murray and I'm going to be giving the trigonometry lectures for educator.com.0000
We're very excited about the trigonometry series.0006
In particular, for me, trigonometry is the class that got me excited about math.0008
I'm really looking forward to working with you on learning some trigonometry.0013
We're going to start right away here, learning about angles.0018
The first thing you have to understand is that there's two different ways to measure angles.0021
People use degrees which you probably already heard of, and radians which you may not hear about until you start to take your first trigonometry class.0029
They're just two different ways in measuring.0037
You can use either one but you really need to know how to use both, and convert back and forth.0041
That's what I'll be covering in this first lecture.0044
Degrees are unit of measurement in which a circle gets divided into 360 degrees0050
If you have a full circle, the whole thing is 360 degrees.0059
That's 360 degrees.0066
Then if you have just a piece of a circle, then it gets broken up into smaller chunks.0067
For example, an angle that's half of a circle here that's 180 degrees, because that's half of 360, a quarter of a circle which is a right angle, that would be 90 degrees, then so on.0072
You can take a 90-degree angle and break it up into two equal pieces.0093
Then each one of those pieces would be a 45-degree angle.0098
Or you could break up a 90-degree angle into three equal pieces, and each one of those would be a 30-degree angle.0104
We'll be studying trigonometric functions of these different angles.0114
In the meantime, it's important just to get comfortable with measuring angles in terms of degrees.0118
The second unit of measurement we're going to use to measure angles is called radians.0124
That's a little bit more complicated.0129
The idea is that, you take a circle, and remember that the circumference of a circle is equal to 2Π times the radius, that's 2Π r, it's one of those formulas that you learned in geometry.0135
What you do with radians is, you break the circle up, and you say the entire circle is 2Π radians.0151
What that means is that a one-radian angle, well, if the entire circle is 2Π radians, then 1 radian, use a little r to specify the radians, cuts off an arc that is 1 over 2Π of the whole circle.0161
One radian, an angle that is 1 radian cuts of a fraction of the circle that is 1 over 2Π.0184
Since the radius is 2Π times r, sorry, the circumference is 2Π times r, if you have 1 over 2Π of the whole circumference, what you get is exactly the length of the radius.0194
That's why they're called radians is because if you take one-radian angle, it cuts of an arc length that is exactly equal to the radius.0214
It takes a little bit of getting used to.0228
What you have to remember that's important is that the whole circle is 2Π radians.0230
That means a half circle, a 180-degree angle, is Π radians.0240
A right angle, a 90-degree angle, or a quarter circle is Π over 2 radians, and so on.0250
Then you can break that down into the even smaller angles like we talked about before.0262
If you take a right angle and you cut it in half, so that was a 45-degree angle before, that's Π over 4 radians because it's half of Π over 2.0268
If you take a right angle and you cut it into three equal pieces, so those are 30-degree angles before, in terms of radians, that's Π over 2 divided by 3, so that's Π over 6 radians.0281
You want to practice going back and fort between degrees and radians, and kind of getting and into the feel of how big angles are in terms of degrees and radians.0299
We'll practice some of that here in this lecture.0307
That means we're breaking an entire circle up in the 2Π radians, so the circle gets broken up into about 6.28 radians.0317
What I've shown up here is pretty accurate that one radian is about one-sixth of a circle.0332
It's about 60 degrees but it's not exact there because it's not exactly 6 it's 6.28 something.0339
People almost always talk about radians in multiples of Π the same way I was doing here, where I said the circle is 2Π radians, the half circle is Π radians, the right angle is Π over 2.0355
People almost always talk about radians in multiples of Π and degrees in terms of whole numbers.0366
Sometimes, they don't even bother to write the little r.0374
It's just understood that if you're using a multiple of Π , then you're probably talking about radians.0376
Let's practice going back and forth between degrees and radians.0385
Remember that 360 degrees is a whole circle.0389
What that means is that Π radians is equal to 360 degrees over 2, which is 180 degrees.0398
That gives you the formula to convert back and forth between degree measurement and radian measurement.0415
If you know the measurement in degrees, you multiply by Π over 180 and that tells you the measurement in radians.0421
If you know the measurement in radians, you just multiply by 180 over Π , and that tells you the measurement in degrees.0429
We'll practice that in some of the examples later on.0439
We got a few more definitions here.0442
Coterminal angles, what that means is that their angles that differ from each other by a multiple of 2Π radians,0446
remember that's a whole circle, or if you think about it in degrees, 360 degrees.0456
For example, if you take a 45-degree angle, and then you add on 360 degrees, that would count as a 360 plus 45 is 405 degrees.0462
Forty-five and 405 degrees are coterminal.0484
In the language of radians, 45 degrees is Π over 4.0491
If you add on 2Π radians, if you add on a whole circle to that, you would get...0497
This should end up here.0511
If you add on 2Π plus Π over 4, well 2Π is 8Π over 4, so you get 9Π over 4.0512
Then Π over 4 and 9Π over 4 are coterminal angles.0522
The reason they're called coterminal angles is because we often draw angles starting with one side on the positive x-axis.0528
I'll draw this in blue.0544
Then we draw the other side of the angle just wherever it ends up.0545
Coterminal angles are angles that will end up at the same place, that's why they're called coterminal.0550
If they differ from each other, if one is 2Π more than the other one, or 360 degrees more than the other one, or maybe 720 degrees more than the other one, then we call them coterminal because they really end up on the same terminal line here.0561
Couple other definitions we need to learn.0580
Complementary angles are angles that add up to being a right angle, in other words, 90 degrees or Π over 2.0583
If you have two angles, like this two angles right here , that add up to being a right angle, 90 degrees or Π over 2, those are complementary.0591
Supplementary angles are angles that add up to being a straight line, in other words, Π radians or 180 degrees.0608
Those two angles right there are supplementary.0620
That's all the vocabulary that you need to learn about angles, but we'll go through and we'll do some examples of each one to give you some practice.0629
Here's our first example.0640
If a circle is divided into 18 equal angles, how big is each one in degrees and radians?0642
Let me try drawing this.0648
We've got this circle and it's divided into a whole bunch of little angles but each one is the same.0653
We want to figure out how big each one is, in terms of degrees and radians.0664
Let's solve this in degrees first.0668
Remember that a circle is 360 degrees.0671
If it's divided into 18 parts, then each part will be 20 degrees.0676
Each one of those angles will be 20 degrees.0682
Remember that an entire circle is 2Π radians.0691
If we divide that by 18, then we get Π over 9 radians will be the size of each one of those little angles.0695
You can measure this angle either way, we say 20 degrees is equal to Π over 9 radians.0706
Second example here, we want to convert back and forth between degrees and radians.0718
Let's practice that.0721
We want to convert 27 degrees into radians.0724
Well, let's remember the formula here, the conversion formula, is Π over 180.0727
So we do 27 times Π over 180.0733
That's our conversion formula from degrees into radians.0738
I'm just going to leave the Π because it doesn't really cancelled anything.0743
The 27 over 180 does simplify.0747
I could take a 9 out of each ones.0752
That would be 3.0753
If we take a 9 out of 180, then there'd be 20.0755
What we end up with is 3Π over 20 radians, as our answer there.0761
Converting back and forth between degrees and radians is just a matter of remembering this conversion factor, Π over 180 gets you from degrees into radians.0777
For the second part of this example, we're given a radian angle measurement, 5Π over 12.0789
We want to convert that into degrees, 5Π over 12 radians.0794
We just multiply by the opposite conversion factor, 180 over Π.0804
Let's see here.0813
The pis cancel.0814
One hundred eighty over 12 is 15.0818
That's 5 times 15 degrees.0826
That gives us 75 degrees.0833
The same angle that you would measure, in radians is being 5Π over 12, will come out to be a 75-degree angle.0838
Converting back and forth there is just a matter of remembering the Π and the 180, and multiplying by one over the other to convert back and forth.0848
Third example is some practice with coterminal angles.0859
In each case, what we want to do is, we're given an angle and we want to find out what quadrant it's in.0863
That's assuming that all the angles are drawn in the standard position with their starting side on the positive x-axis.0872
We want to start on the positive x-axis.0882
We want to see which one of the four quadrants the angle ends up in.0885
Then we want to try to simplify these angles down by finding a coterminal angle that's between 0 and 360, or between 0 and 2Π radians.0893
Let's start out with 1000 degrees.0904
A thousand degrees is going to be, that's way bigger than 360.0908
Let me just start subtracting multiples of 360 from that.0912
If I take off 360 degrees, what I'm left with is 640 degrees.0916
That's still way bigger than 360 degrees.0926
I'll subtract off another 360 degrees and what I'm left with is 280 degrees.0928
That's between 0 and 360.0940
I found my coterminal angle there.0942
I wanna figure which quadrant it's gonna end up in.0945
Now, remember, if we start with 0 degrees being on the x-axis, that would make 90 degrees being on the positive y-axis.0947
Then over here on the negative x-axis, we'd have 180 degrees.0959
Down here is 270 degrees, because that's 180 plus 90.0964
Then 360 degrees would be back here at 0 degrees.0971
Two hundred and eighty degrees would be just past 270 degrees.0976
That's a little bit bigger than 270 degrees.0981
It would be about right there.0983
That's 280.0986
That puts it in the fourth quadrant.0990
We have -19Π over 6 radians.1005
That's one, I'll do this one red.1011
That's one that goes in the negative direction.1014
We start on the positive x-axis but now we go in the negative direction.1016
Instead of going up around past the positive y-axis, we go down in the negative direction and we go -19Π over 6.1021
If you think about it, 19Π over 6 is bigger than 2Π.1031
Let me start with 19Π over 6 and subtract off a 2Π there.1039
Well, 2Π is 12Π over 6, so that gives us 7Π over 6.1044
If we do, -19Π over 6 plus 2Π, that will give us, -19Π over 6 plus 12Π over 6 is -7Π over 6.1055
That's still not in the range that we want, because we want it to be between 0 and 2Π radians.1070
Let me add on another 2Π plus 2Π gives us positive 5Π over 6.1076
The trick with finding these coterminal angles with degrees, it was just a matter of adding or subtracting 360 degrees at a time.1084
Remember, 2Π is a whole circle.1098
We end up with 5Π over 6.1101
That is between 0 and 2Π so we're done with that part but we still have to figure out what quadrant it's in.1104
Well now 5Π over 6, where would that be?1111
Well if we map out our quadrants here, 0 is right there on the positive x-axis just as we had before, 90 degrees is Π over 2 radians, 180 degrees is Π radians, and 270 degrees is 3Π over 2 radians.1114
Then 360 degrees is 2Π radians, a full circle.1135
Where does 5Π over 6 land?1143
Well that's bigger than Π over 2, it's less than Π, so, 5Π over 6 lands about right there.1145
OK, we have another degree one.1165
Negative 586 degrees, and what are we going to do with that?1168
It's going in the negative direction so it's going down south from the x-axis.1173
Negative 586 degrees.1180
Well, 586 is way outside our range of 0 and 360.1183
Let's try adding 360 degrees to that.1187
That gives you -226 degrees, which is still outside of our range.1193
We're adding and subtracting multiples of a full circle 360 degrees.1208
That gives us positive 134 degrees.1212
Now, positive 134 degrees, that is in our allowed range between 0 and 360.1218
So we finished that part of the problem.1227
Where would that land in terms of quadrants?1229
Let me redraw my axis because those are getting a little messy.1230
That's 0, 90, 180, 270 and 360.1237
Where's 134 going to be?1246
A hundred and thirty-four is going to be between 90 and 180, almost exactly halfway between.1250
The answer to that one is that that's in the quadrant number two there.1262
Finally, we have 22Π over 7, again given in radians.1272
The question is, is that between 0 and 2Π?1281
It's not, it's too big.1284
It's bigger than 2Π.1285
Let me subtract off a multiple of 2Π.1287
I'll subtract off just 2Π, which is 14Π over 7.1291
That simplifies down to 22Π over 7 minus 14Π over 7, is 8Π over 7.1296
Now, 8Π over 7 is between 0 and 2Π.1303
We found our coterminal angle.1309
Where will that land on the axis?1311
Well, remember 0 degrees is 0 radians, 90 degrees is Π over 2 radians, 180 degrees is Π radians, and 270 degrees is 3Π over 2 radians, and finally, 360 degrees is 2Π radians.1312
Eight pi over 7 is just a little bit bigger than 1.1334
That's a little bit, or 8 over 7 is a little bigger than 1.1336
Eight pi over 7 is just a little bit bigger than Π.1341
Let's going to put it about right there which will put it in the third quadrant.1346
Let's recap how we found this coterminal angles.1360
Basically, you're given some angle and you check first whether it's in the correct range, whether it's in between 0 and 2Π radians,1363
or if it's given in degrees, whether it's between 0 and 360 degrees.1372
If it's not already in the correct range, if it's negative or if it's too big, then what you do is you add and subtract multiples of 360 degrees or 2Π radians until you'll get it into the correct range,1378
the range between 0 and 360 degrees or 0 and 2Π radians.1393
Once you get it in that range, if you want to figure out what quadrant it's in, well in degrees, it's a matter of checking 0, 90, 180, 270, and 360;1402
in radians,it's a matter of checking 0, Π over 2, Π, 3Π over 2, 2Π.1417
Which one of those ranges does it fall into?1426
That tells you what quadrant it's in.1428 |
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 1.1: Integration by parts
[ "stage:draft", "article:topic", "Integration by Parts", "authorname:thangarajahp" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Integration by parts (IBP)
$\int\,u \,dv= u\,v-\int\, v\, du.$
Note
When deciding which function should be u and dv, it is best to choose u as the function that will simplify easily.
The priority for choosing u is:
1. $$u = ln x$$
2. $$u = x^n$$, where n= integer
3. $$u = e^{nx}$$, where n= integer
Example $$\PageIndex{1}$$:
Find $$\int x^2 \, e^{-x}\, dx$$.
Answer:
Using the priority listed above, let $$u=x^2$$ and $$dv= e^{-x} \,dx$$.
Since $$u=x^2$$ , $$du=2x$$ .
Since $$dv= e^{-x}$$, $$v= \int e^{-x} \, dx). \(= -e^{-x}$$
Using the IBP,
$$\int\,u \,dv= u\,v-\int\, v\, du$$
$$= x^2\, e^{-x}\ -\int\, e^{-x}\, 2x\, dx$$
$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{1}{2}\int\ x\, dx$$
$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{1}{2}\, \dfrac{x^2}{2}$$
$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{x^2}{4}$$
Simplifying the answer,
$$\int x \,lnx \, dx\, = \dfrac{x^2}{2}\, (ln x\, -\, \dfrac{1}{2}\, ) +\, C$$
Reminder: Make sure to write + C at the end of the answer as C is a constant.
Example $$\PageIndex{2}$$:
Find $$\int x \,lnx \, dx$$.
Answer:
Using the priority listed above, let $$u=ln x$$ and $$dv=x\,dx$$.
Since $$u=ln x$$ , $$du=\dfrac{1}{x}$$ .
Since $$dv=x$$, $$v= \dfrac{x^2}{2}$$.
Using the IBP,
$$\int\,u \,dv= u\,v-\int\, v\, du$$
$$= ln x\, \dfrac{x^2}{2}\ -\int\, \dfrac{x^2}{2}\, x\, dx$$
$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{1}{2}\int\ x\, dx$$
$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{1}{2}\, \dfrac{x^2}{2}$$
$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{x^2}{4}$$
Simplifying the answer,
$$\int x \,lnx \, dx\, = \dfrac{x^2}{2}\, (ln x\, -\, \dfrac{1}{2}\, ) +\, C$$
Reminder: Make sure to write + C at the end of the answer as C is a constant.
Example $$\PageIndex{3}$$:
Find $$\int \, lnx \, dx$$.
Answer:
Add texts here. Do not delete this text first.
Exercise $$\PageIndex{1}$$
Add exercises text here. For the automatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page.
Answer
Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. |
# How to find square perimeter
## Understanding the Concept of Perimeter
The concept of perimeter refers to the measurement of the distance around a two-dimensional shape. It is essentially the sum of all the sides of a shape or figure. Perimeter plays an important role in geometry as it helps us understand and quantify the boundary or outline of various objects, such as squares, rectangles, circles, and polygons.
When we talk about perimeter, it is essential to define a square and its properties. A square is a four-sided polygon with equal length sides and right angles at each corner. Its defining property is that all four sides are congruent in length. Understanding these properties allows us to calculate the perimeter by simply adding up all four side lengths.
To find out the length of one side of a square when given its perimeter, we can divide the total perimeter by 4 since all sides are equal in length. This gives us an easy way to determine individual side lengths based on known perimeters.
By using specific formulas for calculating perimeter, we can easily find out how long each side should be for any given square. The formula for finding the perimeter of a square is P = 4s (where P represents perimeter and s represents one side’s length). Applying this formula enables us to quickly compute perimeters without having to measure each individual side manually.
Understanding how to calculate perimeters becomes even more clear when we explore examples that involve solving different types of problems related to squares’ boundaries. By practicing with various scenarios involving different measurements and shapes, individuals can gain confidence in their ability to solve real-life situations requiring knowledge about perimeters.
Exploring different units used for measuring perimeters also broadens our understanding further. While centimeters (cm) may be commonly used for smaller shapes like squares within confined spaces, larger objects might require meters (m) or other units such as kilometers (km). Being able to convert between these units allows us flexibility in expressing measurements accurately across different contexts.
Real-life applications demonstrate the practicality of finding square perimeters. From construction projects to designing gardens, knowing the perimeter helps determine how much material or fencing is needed. It also aids in calculating distances for running tracks or determining boundary lengths on maps. Understanding this concept allows us to apply it widely and solve problems efficiently in various situations.
## Defining a Square and its Properties
A square is a geometric shape that has four equal sides and four right angles. It is a special type of rectangle, where all the sides are of the same length. The properties of a square make it unique and distinguishable from other shapes. One key property of a square is that its diagonals are congruent, meaning they have the same length. Additionally, since all four angles in a square are right angles, it can be classified as both an equilateral and equiangular polygon.
The defining characteristic of a square lies in its symmetry and balance. Each side of the square is parallel to its opposite side, creating perfect symmetry across both vertical and horizontal axes. This uniformity gives squares their aesthetic appeal and makes them ideal for various applications in art, design, architecture, and engineering.
Furthermore, squares possess certain mathematical properties that set them apart from other polygons. For instance, the perimeter of a square can be calculated by multiplying the length of one side by 4 since all sides are equal. Similarly, to find the area of a square, you simply need to multiply one side’s length by itself (squared). These formulas simplify calculations involving squares significantly.
By understanding these fundamental properties of squares – their equal sides and angles along with their symmetrical nature – we gain insight into their versatility in practical scenarios such as tiling patterns or determining optimal layouts for building structures. Whether applied in geometry problems or real-life situations like measuring floor space or planning garden beds evenly spaced with paving stones on each corner – squares play an essential role due to their well-defined properties
• A square has four equal sides and four right angles.
• The diagonals of a square are congruent.
• A square is both an equilateral and equiangular polygon.
• Each side of a square is parallel to its opposite side, creating perfect symmetry across vertical and horizontal axes.
• Squares have aesthetic appeal and are used in art, design, architecture, and engineering.
• The perimeter of a square can be calculated by multiplying the length of one side by 4.
• The area of a square can be found by multiplying the length of one side by itself (squared).
• Understanding the properties of squares helps in practical scenarios such as tiling patterns or determining optimal layouts for building structures.
## Identifying the Length of One Side of a Square
To identify the length of one side of a square, you need to understand the concept of a square and its properties. A square is a four-sided polygon with equal sides and right angles. This means that all four sides are the same length, making it easier to determine the length of one side.
One way to identify the length of one side is by using measurements provided in a problem or given information. For example, if you are told that the perimeter of a square is 20 units, you can divide this value by 4 to find that each side has a length of 5 units. Similarly, if you know the perimeter but not the individual side lengths, dividing it by 4 will give you the measurement for each side.
Another method for identifying the length of one side involves solving equations or using algebraic expressions. You may be given an equation representing either just one side or all four sides combined. By isolating and simplifying this equation, you can determine what value represents one side’s length.
By understanding these methods for identifying the length of one side in a square, you can confidently solve problems involving squares’ perimeters without relying solely on provided measurements or equations. Being able to calculate this important aspect allows for further exploration into other aspects such as area and diagonal lengths within squares.
## Calculating the Perimeter of a Square Using Side Length
To calculate the perimeter of a square using the side length, you simply need to add up all four sides. Since a square has equal sides, you can multiply the side length by 4 to find its perimeter. For example, if one side of a square measures 5 units in length, then its perimeter would be 5 + 5 + 5 + 5 = 20 units.
The formula for finding the perimeter of a square is P = 4s, where P represents the perimeter and s represents the length of one side. This formula holds true for squares because all four sides are equal in length. By substituting the given value for s into this equation, you can easily calculate the perimeter.
Calculating perimeters becomes more complex when dealing with different units of measurement. However, as long as you know how to convert between units or use appropriate conversion factors, it is still manageable. For instance, if one side of a square is measured in centimeters while another is measured in inches, you will need to convert them to either all centimeters or all inches before calculating their sum.
By understanding how to calculate the perimeter of a square using its side length and being able to apply this knowledge across various scenarios and measurements systems such as metric or imperial units, you will have acquired an essential skill that can be used in real-life situations. Whether it’s measuring fencing needed for your backyard garden or determining material requirements for building projects like flooring tiles or wallpaper borders – knowing how to find a square’s perimeter allows for accurate planning and efficient resource allocation without any unnecessary waste.
## Using the Formula for Perimeter Calculation
To calculate the perimeter of a square, you can use a simple formula. The perimeter is found by multiplying the length of one side of the square by 4. This is because a square has four equal sides, so when you add up all the sides, you get the total distance around the shape. For example, if one side of a square measures 5 units in length, then its perimeter would be 5 x 4 = 20 units.
It’s important to note that this formula works specifically for squares and not other shapes. Other polygons may have different formulas for calculating their perimeters based on their unique properties. However, with squares being regular quadrilaterals with equal sides and angles, this straightforward formula applies universally to them.
Using this formula allows for quick and efficient calculations of a square’s perimeter without needing to measure each individual side separately. It simplifies the process and provides an accurate result as long as you know the length of just one side. By understanding how to apply this formula correctly, you can easily find the perimeter of any given square in various real-life situations or mathematical problems where it may be required
## Applying the Perimeter Formula to Find the Square’s Perimeter
To apply the perimeter formula and find the perimeter of a square, we need to know the length of one side. The formula for finding the perimeter of a square is simply four times the length of one side. This means that if we know the length of one side, we can easily calculate the total distance around the square.
For example, let’s say we have a square with a side length of 5 units. To find its perimeter, we would multiply 5 by 4, which equals 20 units. So, in this case, the perimeter of our square would be 20 units.
It’s important to note that when using this formula to find a square’s perimeter, all sides must have equal lengths. If any side has a different length than others or if it is not perfectly straight, then it is not considered a perfect square and cannot be calculated using this method.
By understanding how to apply the perimeter formula and knowing how to measure or calculate one side’s length accurately, you can easily determine the total distance around any given square shape.
## Utilizing Examples to Solve Perimeter Problems
One way to understand and apply the concept of perimeter is through solving examples. Let’s consider an example where we have a square with a side length of 5 units. To find the perimeter, we simply multiply the side length by 4 since all sides of a square are equal in length. In this case, the perimeter would be 5 units multiplied by 4, which equals 20 units.
Now let’s explore another example using different units of measurement for perimeter. Suppose we have a square with a side length of 3 feet. To calculate the perimeter in inches, we need to convert feet to inches first. Since there are 12 inches in one foot, our conversion factor is 12. Thus, multiplying the side length (3 feet) by this conversion factor gives us a total perimeter of 36 inches.
In real-life applications, finding the perimeter of squares can be useful in various scenarios such as designing fences or determining how much material is needed for framing artwork or photographs. By understanding how to solve examples involving square perimeters, you can confidently tackle practical problems that involve measuring and calculating lengths around closed shapes like squares.
## Exploring Different Units of Measurement for Perimeter
One aspect of exploring the concept of perimeter involves understanding different units of measurement. While we commonly use inches, feet, or meters to measure length, these units can also be applied to measuring the perimeter of a square. For example, if one side of a square is measured in inches, then the perimeter will be expressed in terms of inches as well.
In addition to standard units like inches and feet, it’s important to note that perimeter can also be measured using other units such as centimeters or millimeters. These smaller units are useful when dealing with objects that have very small dimensions or require precise measurements. By utilizing different units for measuring perimeter, we gain flexibility in accurately describing the size and shape of various objects.
Furthermore, converting between different units of measurement for perimeter allows us to compare and analyze data more effectively. For instance, if we have the length of one side expressed in centimeters but need to calculate the total distance around a square using meters instead, conversion becomes necessary. Understanding how to convert between various unit systems enables us to work with diverse sources and solve real-life problems involving perimeters efficiently without any loss in accuracy.
## Converting Perimeter Values between Units
When working with perimeter values, it is important to be able to convert between different units of measurement. This allows for better understanding and communication of the measurements involved. To convert perimeter values between units, you need to know the conversion factors for the specific units being used.
For example, if you have a square with a side length of 5 centimeters and want to find its perimeter in meters, you would need to convert from centimeters to meters. Since there are 100 centimeters in a meter, you can divide the side length by 100 to get the equivalent length in meters. In this case, the perimeter would be 0.5 meters.
Similarly, if you have a square with a side length of 3 feet and want to find its perimeter in yards, you would need to convert from feet to yards. Since there are 3 feet in a yard, dividing the side length by 3 will give you the equivalent length in yards. So, in this case, the perimeter would be 1 yard.
Converting perimeter values between units is an essential skill when dealing with real-life applications such as construction or designing layouts. Being able to communicate measurements accurately using different units ensures that everyone involved understands and interprets them correctly. It also helps when comparing measurements across different systems or contexts where varying units may be used
## Real-life Applications of Finding Square Perimeter
Real-life applications of finding the square perimeter extend beyond the classroom and into various aspects of our daily lives. One practical application is in the field of construction, where determining the perimeter helps architects and engineers plan for fencing or enclosing a specific area. By calculating the square’s perimeter, they can accurately estimate materials needed and ensure proper security measures.
Another real-life application lies in landscaping design. When designing gardens or outdoor spaces, knowing the perimeter of a square allows landscapers to create boundaries or borders with hedges, fences, or decorative elements. This not only adds structure to the overall layout but also enhances visual appeal by creating defined areas within a larger space.
Furthermore, understanding how to find the square’s perimeter is essential in sports fields and recreational facilities. For example, when constructing an athletic track or soccer field, it is crucial to calculate its perimeter accurately. This measurement ensures that athletes compete on standardized playing surfaces while adhering to regulations governing dimensions and safety requirements.
These real-life applications highlight how finding the square’s perimeter plays a vital role in numerous industries such as construction, landscaping design, and sports facility development. By applying this mathematical concept outside academic settings, professionals can efficiently plan projects while maintaining accuracy and precision in their respective fields.
### What is the perimeter of a square?
The perimeter of a square is the total length of all its sides.
### How do you calculate the perimeter of a square?
To calculate the perimeter of a square, you can multiply the length of one side by 4.
### What is the formula for finding the perimeter of a square?
The formula for finding the perimeter of a square is P = 4s, where P represents the perimeter and s represents the length of one side.
### Can you provide an example of finding the perimeter of a square?
Sure! For example, if the length of one side of a square is 5 units, then the perimeter would be 4 times 5, which equals 20 units.
### Are there any real-life applications of finding the perimeter of a square?
Yes, there are several real-life applications. For instance, when measuring the perimeter of a garden or a fence, determining the amount of material needed for a picture frame, or calculating the distance around a square-shaped race track.
### Can the perimeter of a square be measured in different units?
Yes, the perimeter of a square can be measured in various units, such as centimeters, meters, inches, or feet.
### How can you convert the perimeter of a square from one unit to another?
To convert the perimeter of a square from one unit to another, you can use conversion factors. For example, if you want to convert the perimeter from centimeters to inches, you would multiply the value in centimeters by 0.3937 to get the equivalent length in inches.
### Are there any other formulas to find the perimeter of a square?
No, the formula P = 4s is the only formula to find the perimeter of a square.
### What happens if the length of one side of a square is unknown?
If the length of one side of a square is unknown, you cannot calculate the perimeter until you have that information.
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# INTEGRALS
```CHAPTER 8:
INTEGRALS
8.1 REVIEW: APPROXIMATING INTEGRALS WITH RIEMANN SUMS
IN 2-D
In two dimensions we have previously used Riemann sums to approximate ∫
following steps:
( )
with the
1. Divide the region a ≤ x ≤ b into intervals of width ∆x.
2. For each interval, select the height of the function that will be used for that interval.
There are various rules to choose the selected height for each interval, some of which are
presented below.
3. For each interval, approximate the area under the curve over that particular interval with
a rectangle with width ∆x and height according to the indicated rule.
4. The approximate value of ∫ ( ) can be obtained by summing the areas of the
rectangles according to the given rule.
8.2 APPROXIMATING INTEGRALS WITH RIEMANN SUMS IN 3D
VARIOUS RULES FOR RIEMANN SUMS IN 3D
To obtain the volume of a surface f (x, y) over an area A, a set of steps that are similar to those
that were used in two dimensions can be used.
1. Identify Domain of f (x, y) to be evaluated.
2. Divide the domain into rectangles with length ∆x and width ∆y.
3. For each rectangle, find the height of the surface that will be used according to the
indicated rule. There are various possible rules, a few of which are outlined below.
) can be obtained by summing the volumes of the
4. The approximate value of ∫ (
cubes as will be illustrated in the following examples.
Example Exercise 8.2.1: Approximate ∫ ∫ (
)
using rectangles of length ∆x =
2 and of width ∆y = 2 and using the height in the lower left hand corner of each division.
Solution:
1. Identify Domain of f (x, y) to be evaluated.
2. Divide the domain into rectangles with length ∆x = 2 and width ∆y = 2.
3. Place the surface over the region.
4. For each rectangle, find the height of the surface at the chosen point. In this case, find the
value of f (x, y) in the lower left corner.
5. For the rectangle, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, the height f (0, 0) in the left hand corner is z = 4 so
place the plane z = 4 above this region.
6. For the rectangle, 0 ≤ x ≤ 2, 2 ≤ y ≤ 4, the height f (0, 2) in the left hand corner is z = 6 so
place the plane z = 6 above this region.
7. For the rectangle, 2 ≤ x ≤ 4, 0 ≤ y ≤ 2, the height f (0, 2) in the left hand corner is z = 8 so
place the plane z = 8 above this region.
8. For the rectangle, 2 ≤ x ≤ 4, 2 ≤ y ≤ 4, the height f (2, 2) in the left hand corner is z = 14
so place the plane z = 14 above the region.
9. For each rectangle, find the volume of the cuboid between the rectangle in the xy plane
and the indicated plane above that region.
Rectangle
1
2
3
4
Length
2
2
2
2
Width
2
2
2
2
Height
4
6
8
10
Volume
2 x 2 x 4 = 16
2 x 2 x 6 = 24
2 x 2 x 8 = 32
2 x 2 x 10 = 40
10. Approximate the volume beneath the surface and over the region as the sum of the
volumes over the four rectangles
Volume = 16 + 24 + 32 + 40 = 112
Example Exercise 8.2.2: T he formula to represent a function f with inputs x and y and output
z f ( x, y) 0.5* x 2 y 2 4. We wish to approximate the volume below f and over the region
0 x 4,0 y 4 in the xy plane using 2 divisions in x and 2 divisions in y. The units of x,y
and z are all inches.
Solution:
1. Identify Domain of f (x, y) to be evaluated.
2. Divide x into two regions and y into two regions. This cuts the domain into four
rectangles with length ∆x = 2 and width ∆y = 2.
3. Place the surface over the region.
4. For each rectangle, find the height of the surface at the chosen point. In this case, find the
value of f (x, y) in the middle of the rectangle.
5. For the rectangle, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, the height f (1, 1) = 5.5 so place the plane z = 5.5
above this region.
6. For the rectangle 0 ≤ x ≤ 2, 2 ≤ y ≤ 4, the height in the middle of the rectangle is f (1, 3) =
13.5 so place the plane z = 13.5 above this region.
7. For the rectangle 2 ≤ x ≤ 4, 0 ≤ y ≤ 2, the height in the middle of the rectangle is f (3, 1) =
9.5 so place the plane z = 9.5 above this region.
8. For the rectangle 2 ≤ x ≤ 4, 2 ≤ y ≤ 4, the height in the middle of the rectangle is f (3, 3) =
17.5 so place the plane z = 17.5 above this region.
9. For each rectangle, find the volume of the cuboid between the rectangle in the xy plane
and the indicated plane above that region.
Rectangle
1
2
3
4
Length
2
2
2
2
Width
2
2
2
2
Height
5.5
9.5
13.5
17.5
Volume
2 x 2 x 5.5 = 22
2 x 2 x 9.5 = 38
2 x 2 x 13.5 = 54
2 x 2 x 17.5 = 70
10. Approximate the volume beneath the surface and over the region as the sum of the
volumes over the four rectangles.
Volume = 22 + 38 + 54 + 70 = 184
8.3 USING RIEMANN SUMS AND THE FUNDAMENTAL THEOREM TO
FIND PRECISE VOLUMES
If we continue the process begun in 8.2 with more divisions and the Fundamental Theorem of the
Integral Calculus, we will see that double integrals can be used to represent the precise volume
below a surface. The process to do this is shown in the following example:
Example Exercise 8.3.1: The formula to represent a function f with inputs x and y and output z
is given by z f ( x, y) 0.5* x 2 y 2 4. We wish to obtain the volume below f and over the
region 0 x 4,0 y 4 in the xy plane. The units of x,y and z are all inches.
Solution:
A. Place the surface and identify the volume we’re seeking in your 3D kit.
B. If there are two divisions in x and two divisions in y and the volume is to be approximated
using the middle value for x and y in each division, find x1 , x2 , y1 , y2 , x , and y
x1 1, x2 3, y1 1 , y2 3 , x 2 and y 2
C. For each of the four divisions on your kit, indicate the point and associated height we will use
to approximate the volume for that division.
D. Use the values obtained in part C to fill in the following table with numerical values.
Division
1
2
3
4
Length
2
2
2
2
Width
2
2
2
2
Height
5.5
9.5
13.5
17.5
Volume
2 x 2 x 5.5
2 x 2 x 9.5
2 x 2 x 13.5
2 x 2 x 17.5
E. Fill in the same table below using x1 , x2 , y1 , x , and y instead of numerical values. (Note,
the divisions should not change between the two tables.)
Division
1
2
3
4
Length
y
y
y
y
Width
x
x
x
x
Height
0.5x12 + y12 + 4
0.5x22 + y12 + 4
0.5x12 + y22 + 4
0.5x22 + y22 + 4
Volume
(0.5x1 + y12 + 4)xy
(0.5x22 + y12 + 4)xy
(0.5x12 + y22 + 4)xy
(0.5x22 + y22 + 4)xy
2
F. Express the approximate volume numerically.
Volume ≈ 2 x 2 x 5.5 + 2 x 2 x 9.5 + 2 x 2 x 13.5 + 2 x 2 x 17.5 = 184
G. Express the same volume obtained in part F using x1 , x2 , y1 , y2 x , and y
(0.5x12 + y12 + 4)xy + (0.5x22 + y12 + 4)xy + (0.5x12 + y22 + 4)xy + (0.5x22 + y22 + 4)xy
H. Express the volume obtained in part G in the form
(...)yx
(0.5x12 + y12 + 4)xy + (0.5x22 + y12 + 4)xy + (0.5x12 + y22 + 4)xy + (0.5x22 + y22 + 4)xy
= ∑
(0.5x12 y j 2 4) yx
= ∑
∑
∑
(0.5x22 y j 2 4) yx
(0.5xi 2 y j 2 4) yx
I. Use the fundamental theorem and limits to convert the approximation in part H to a precise
value.
∑ ∑ (0.5xi 2 y j 2 4) yx
∫ ∫ (0.5x 2 y 2 4)
8.4 DEFINITION OF THE INTEGRAL
Given a surface z = f (x, y) and a region in the xy plane A, the volume below the surface of z = f
) .
(x, y) and above the region A in the xy plane is defined as ∫ (
Example
If A is the rectangle 1 ≤ x ≤ 3, 0 ≤ y ≤ 4 and f (x, y) = 8.
As can be seen, the volume associated with ∫
height 8 will give us a volume of 64.
Example
If A is the region x2 + y2 ≤ 4 and f (x, y) = 5.
(
)
is a cube with length 2, width 4, and
As can be seen, the volume associated with ∫
height 5 which will give us a volume of 20πr2
(
)
is a cylinder with base radius 2 and
Example
If A is the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 4 and f (x, y) = x2 + y2.
The surface in general is the following:
The surface over the indicated range is the following:
) is not a standard volume and
As can be seen, the volume associated with ∫ (
correspondingly cannot be obtained immediately with known formulas.
8.5 IDENTIFYING VOLUMES ASSOCIATED WITH DOUBLE
INTEGRALS
) , a double integral is used to represent volumes. The
following examples, illustrate how to associate a double integral with its corresponding volume.
Example Exercise 8.5.1: Given the integral ∫ ∫ (
associated with the integral.
)
, identify the volume
Solution: Working from the outside inward, the outside integral is ∫
volume will contain values of f between x = 1 and x = 3.
indicating that our
The inside integral is ∫
indicating that for every value of x between x = 1 and x = 3, we
will accept values of y that reside between y = 1 and y = 2.
In the previous section we indicated that given a surface z = f (x, y) and a region in the xy plane
A, the volume below the surface of z = f (x, y) and above the region A in the xy plane is defined
) . The combination of the two integrals ∫ ∫
as ∫ (
defined thus far give a
rectangular region in the xy plane which corresponds to ∫
defined in the previous section.
The volume associated with ∫ ∫ (
the xy plane and below z = 2x + y + 1.
)
is given by the volume above this region in
Example Exercise 8.5.2: Given the integral ∫ ∫ (
associated with the integral.
)
Solution: Working from the outside inward, the outside integral is ∫
volume will contain values of x between x = 1 and x = 3.
, identify the volume
indicating that our
The second datum from the integral is ∫
indicating that for every value of x between x = 1
and x = 3, we will accept values of y that reside between y = 0 and y = x.
In the previous section we indicated that given a surface z = f (x, y) and a region in the xy plane
A, the volume below the surface of z = f (x, y) and above the region A in the xy plane is defined
) . The combination of the two integrals ∫ ∫
as ∫ (
defined thus far give a
trapezoidal region in the xy plane which corresponds to ∫
The volume associated with ∫ ∫ (
)
the xy plane and below the plane z = x + 3y + 2.
defined in the previous section.
is given by the volume above this region in
√
Example Exercise 8.5.3: Given the integral ∫ ∫ (
associated with the integral.
)
, identify the volume
Solution: Working from the outside inward, the first datum from the integral is ∫
indicating that our volume will contain values of y between y = 0 and y = 4.
The second datum from the integral is ∫
√
indicating that for every value of y between y = 1
and y = 4, we will accept values of x that reside between
and
√ .
In the previous section we indicated that gives a surface z = f (x, y) and a region in the xy plane A,
the volume below the surface of z = f (x, y) and above the region A in the xy plane is defined as
∫
(
)
. The combination of the two integrals ∫ ∫
region in the xy plane which corresponds to ∫
√
The volume associated with ∫ ∫ (
)
the xy plane and below the plane z = x + y + 2.
√
defined thus far give a
defined in the previous section
is given by the volume above this region in
EXERCISE PROBLEMS:
1) A. Place the volume between x = 0; x = 2, y = 0; y = 2x; z = 0; and z = x2 + y2 on your 3D kit.
B. Express the volume with an integral of the form ..dydx .
C. Express the volume with an integral of the form ..dxdy .
D. Evaluate the two integrals to verify that they are the same.
2) A. Place the volume between x = 0; x = 2, y = x2; y = 4; z = 0; and z = x2 + y2 on your 3D kit.
B. Express the volume with an integral of the form ..dydx .
C. Express the volume with an integral of the form ..dxdy .
D. Evaluate the two integrals to verify that they are the same.
3) z f x, y is represented by the following table:
y=0
y=2
y=4
x=0
x=2
x=4
1
3
3
2
3
4
3
3
5
4 4
A. Use the small magnets to construct the cuboids we will use to approximate
f ( x, y)dydx
0 0
using Riemann Sums with the Lower left corner rule with ∆x = 2 and ∆y = 2. Then find
the approximation for the volume.
4 4
B. Use the small magnets to construct the cuboids we will use to approximate
f ( x, y)dydx
0 0
using Riemann Sums with the Upper Right corner rule with ∆x = 2 and ∆y = 2. Then find
the approximation for the volume.
C. Express the volume obtained in B in the form (...)yx .
4) The formula to represent a function f with inputs x and y and output z is z = f(x, y) = 8 – x2 – y2.
4 4
A. Use the small magnets to construct the cuboids we will use to approximate f ( x, y )dydx
0 0
using Riemann Sums with the Midpoint rule with ∆x = 2 and ∆y = 2. Then find the
approximation for the volume.
B. Express the volume in A in the form (...)yx .
5) A house constructed over the rectangular region 0 ≤ x ≤ 4, 0 ≤ y ≤ 6 has a roof given by z =
f(x, y) = 2x + y +1. We wish to obtain the volume of the house.
A. Outline the region of the house on the xy plane of your kit and place its roof.
B. If there are two divisions in x and two divisions in y and the height of the roof in each
division is to be approximated using the smallest value for each variable in each division,
find x1, x2, y1 and y2 and use them to fill in the following table with numerical values.
Division
1
2
3
4
Length
Width
height
volume
C. Use the values of x1, x2, y1 and y2 to fill in the same table below using x1, x2, y1, y2, ∆x and
∆y instead of numerical values. (Note, the divisions should not change between the two
tables.)
Division
1
2
3
Length
Width
height
volume
4
D. Express the approximate volume numerically.
E. Express the volume obtained in part D in the form (...)yx .
F. Take the appropriate limits to convert the sum in part D to an integral.
6) A 32 km rectangular area that is populated with foxes represents the density of foxes with the
foxes
function f with inputs x and y (both km) and output density (
) as z = f(x, y) = xy + x and
km 2
we wish to obtain the number of foxes that live in the area 0 ≤ x ≤ 4, 0 ≤ y ≤ 8.
A. If there are two divisions in x and four divisions in y and the density in each region is to
be approximated using the middle value for each variable in each division in each
division, find x1, x2, y1, y2, y3 and y4 use them to fill in the following table with numerical
values.
Division
1
2
3
4
5
6
7
8
Length
Width
Density
Number of foxes
B. Use the values of x1, x2, y1, y2, y3 and y4 to fill in the same table below using x1, x2, y1, y2,
y3, y4, ∆x and ∆y instead of numerical values. (Note, the divisions should not change
between the two tables.)
Division
1
2
3
4
5
6
7
8
Length
Width
Density
Number of foxes
C. Express the approximate number of foxes numerically.
D. Express the mass obtained in part C in the form (...)yx .
E. Take the appropriate limits to convert the sum in part D to an integral.
``` |
Ellipse Areas
Topics: Conic section, Ellipse, Circle Pages: 4 (749 words) Published: March 15, 2013
Name: Ernest Ng
Class: 4G (23)
Date: 2-7-06
Mathematics ACE: Ellipse Areas
Before we embark on solving the problem, let us first explore the definition of ellipse.
[pic]
An ellipse is a curve that is the locus of all points in the plane the sum of whose distances [pic] and [pic] from two fixed points [pic] and [pic] (the foci) separated by a distance of [pic] is a given positive constant [pic]
[pic]
While [pic] is called the major axis, [pic] is the semi major axis, which is exactly half the distance across the ellipse. Similarly, the corresponding parameter [pic] is known as the semi minor axis.
Parallels drawn from the formula for the area of circle ([pic]) and formula for the area of an ellipse (A = [pic]ab)
Formula for the area of a circle:[pic] where [pic] is the area, and [pic] is the radius. In the case of a circle, radius a represents the semi major axis while radius b represents the semi minor axis. One can thus find the area of the circle through the formula A = [pic]ab, where a is equal to b. Hence circle, in actual fact, is a unique case of ellipse.
Proving that the area of an ellipse is πab
Procedures to take (Theory)
1. We have to let an ellipse lie along the x-axis and find the equation of the ellipse curve. 2. Upon finding the equation of the ellipse curve, we have to change the subject of the equation to y. (otherwise, we will have to do integration with respect to y if the subject of the equation is x.) 3. Next, applying what we have learnt, we can find the area bounded by the ellipse curve through definite integration between the 2 limits.
Calculations (Practical)
(Working steps are adapted from http://mathworld.wolfram.com/Ellipse.html)
Let an ellipse lie along the x-axis and find the equation of the figure where [pic]and [pic]are at [pic]and [pic]. a) Form an equation.
[pic]
b) Bring the second term to the right side and square both sides. [pic]
c)... |
# How do you use Heron's formula to find the area of a triangle with sides of lengths 2 , 5 , and 5 ?
Mar 26, 2016
$2 \sqrt{6}$
#### Explanation:
Use the Heron's formula
color(blue)(sqrt(s(s-a)(s-b)(s-c))
Where
color(red)(a,b,c=sides,s=(a+b+c)/(2)
$s$ is also defined as the Semi-perimeter of the $\triangle$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Values
color(orange)(a=2
color(orange)(b=5
color(orange)(c=5
color(orange)(s=(2+5+5)/2=12/2=6
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Start to solve it
$\rightarrow \sqrt{6 \left(6 - 2\right) \left(6 - 5\right) \left(6 - 5\right)}$
$\rightarrow \sqrt{6 \left(4\right) \left(1\right) \left(1\right)}$
$\rightarrow \sqrt{6 \left(4\right)}$
color(green)(rArrsqrt(24)=sqrt(4*6)=2sqrt6~~4.899 |
# How do I write (5/(1*2))+(5/(2*3))+(5/(3*4))+...+(5/n(n+1))+...in summation notation, and how can I tell if the series converges?
$5 \setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{k \left(k + 1\right)}$
#### Explanation:
The given series:
$\left(\frac{5}{1 \setminus \cdot 2}\right) + \left(\frac{5}{2 \setminus \cdot 3}\right) + \left(\frac{5}{3 \setminus \cdot 4}\right) + \setminus \ldots + \left(\frac{5}{n \left(n + 1\right)}\right)$
$= \setminus {\sum}_{k = 1}^{n} \setminus \frac{5}{k \left(k + 1\right)}$
$= 5 \setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{k \left(k + 1\right)}$
$= 5 \setminus {\sum}_{k = 1}^{n} \left(\frac{1}{k} - \setminus \frac{1}{k + 1}\right)$
$= 5 \left(\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \setminus \ldots + \left(\frac{1}{n} - \frac{1}{n + 1}\right)\right)$
$= 5 \left(1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \setminus \ldots + \frac{1}{n} - \frac{1}{n + 1}\right)$
$= 5 \left(1 - \frac{1}{n + 1}\right)$
$\setminus \therefore \setminus {\lim}_{n \setminus \to \setminus \infty} \setminus {\sum}_{k = 1}^{n} \setminus \frac{5}{k \left(k + 1\right)}$
$= \setminus {\lim}_{n \setminus \to \setminus \infty} 5 \left(1 - \frac{1}{n + 1}\right)$
$= 5 \left(1 - 0\right)$
$= 5$
Hence, the given series is converging
Jul 19, 2018
Please see some comlementary details below.
#### Explanation:
I added a few more details
The partial fraction decomposition is
$\frac{1}{k \left(k + 1\right)} = \frac{A}{k} + \frac{B}{k + 1}$
$= \frac{A \left(k + 1\right) + B k}{k \left(k + 1\right)}$
Compare the numerators
$1 = A \left(k + 1\right) + B k$
Let $k = 0$, $\implies$, $1 = A$
Let $k = - 1$, $\implies$, $1 = - B$
Therefore,
$\frac{1}{k \left(k + 1\right)} = \frac{1}{k} - \frac{1}{k + 1}$
Therefore,
${\sum}_{k = 1}^{n} \frac{1}{k \left(k + 1\right)} = {\sum}_{k = 1}^{n} \frac{1}{k} - {\sum}_{k = 1}^{n} \frac{1}{k + 1}$
$= 1 + {\sum}_{k = 2}^{n} \frac{1}{k} - \left({\sum}_{k = 1}^{n - 1} \frac{1}{k + 1}\right) + \frac{1}{n + 1}$
$= 1 + {\sum}_{k = 2}^{n} \frac{1}{k} - {\sum}_{k = 2}^{n} \frac{1}{k} - \frac{1}{n + 1}$
$= 1 - \frac{1}{n + 1}$
$= \frac{n}{n + 1}$
$= \frac{1}{1 + \frac{1}{n}}$
And
${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{k \left(k + 1\right)} = {\lim}_{n \to \infty} \frac{1}{1 + \frac{1}{n}}$
$= 1$
The series converges and the limit is $= 5$ |
$$\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}}$$
# 12.4: The Cross Product
Imagine a mechanic turning a wrench to tighten a bolt. The mechanic applies a force at the end of the wrench. This creates rotation, or torque, which tightens the bolt. We can use vectors to represent the force applied by the mechanic, and the distance (radius) from the bolt to the end of the wrench. Then, we can represent torque by a vector oriented along the axis of rotation. Note that the torque vector is orthogonal to both the force vector and the radius vector.
In this section, we develop an operation called the cross product, which allows us to find a vector orthogonal to two given vectors. Calculating torque is an important application of cross products, and we examine torque in more detail later in the section.
## The Cross Product and Its Properties
The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product of two vectors that generates a third vector orthogonal to the first two. Consider how we might find such a vector. Let $$u=⟨u_1,u_2,u_3⟩$$ and $$v=⟨v_1,v_2,v_3⟩$$ be nonzero vectors. We want to find a vector $$w=⟨w_1,w_2,w_3⟩$$ orthogonal to both $$u$$ and $$v$$—that is, we want to find $$w$$ such that $$u⋅w=0$$ and $$v⋅w=0$$. Therefore, $$w_1, w_2,$$ and $$w_3$$ must satisfy
$u_1w_1+u_2w_2+u_3w_3=0 \label{eq1}$
$v_1w_1+v_2w_2+v_3w_3=0. \label{eq2}$
If we multiply the top equation by $$v_3$$ and the bottom equation by $$u_3$$ and subtract, we can eliminate the variable $$w_3$$, which gives
$(u_1v_3−v_1u_3)w_1+(u_2v_3−v_2u_3)w_2=0. \nonumber$
If we select
\begin{align*} w_1 &=u_2v_3−u_3v_2 \\[5pt] w_2&=−(u_1v_3−u_3v_1), \end{align*}
we get a possible solution vector. Substituting these values back into the original equations (Equations \ref{eq1} and \ref{eq2}) gives
$w_3=u_1v_2−u_2v_1. \nonumber$
That is, vector
$w=⟨u_2v_3−u_3v_2,−(u_1v_3−u_3v_1),u_1v_2−u_2v_1⟩ \nonumber$
is orthogonal to both $$u$$ and $$v$$, which leads us to define the following operation, called the cross product.
Definition: cross product
Let $$u=⟨u_1,u_2,u_3⟩$$ and $$v=⟨v_1,v_2,v_3⟩.$$ Then, the cross product $$u×v$$ is vector
\begin{align} u×v &= (u_2v_3−u_3v_2)i−(u_1v_3−u_3v_1)j+(u_1v_2−u_2v_1)k \nonumber \\[5pt] &=⟨u_2v_3−u_3v_2,−(u_1v_3−u_3v_1),u_1v_2−u_2v_1⟩. \label{cross}\end{align}
From the way we have developed $$u×v$$, it should be clear that the cross product is orthogonal to both $$u$$ and $$v$$. However, it never hurts to check. To show that $$u×v$$ is orthogonal to $$u$$, we calculate the dot product of $$u$$ and $$u×v$$.
\begin{align*} u⋅(u×v)&=⟨u_1,u_2,u_3⟩⋅⟨u_2v_3−u_3v_2,−u_1v_3+u_3v_1,u_1v_2−u_2v_1⟩ \\[5pt] &=u_1(u_2v_3−u_3v_2)+u_2(−u_1v_3+u_3v_1)+u_3(u_1v_2−u_2v_1) \\[5pt] &=u_1u_2v_3−u_1u_3v_2−u_1u_2v_3+u_2u_3v_1+u_1u_3v_2−u_2u_3v_1\\[5pt] &=(u_1u_2v_3−u_1u_2v_3)+(−u_1u_3v_2+u_1u_3v_2)+(u_2u_3v_1−u_2u_3v_1) \\[5pt] &= 0 \end{align*}
In a similar manner, we can show that the cross product is also orthogonal to $$v$$.
Example $$\PageIndex{1}$$: Finding a Cross Product
Let $$p=⟨−1,2,5⟩$$ and $$q=⟨4,0,−3⟩$$ (Figure $$\PageIndex{1}$$). Find $$p×q$$.
Figure $$\PageIndex{1}$$: Finding a cross product to two given vectors.
Solution
Substitute the components of the vectors into Equation \ref{cross}:
\begin{align*} p×q&=⟨−1,2,5⟩×⟨4,0,−3⟩ \\[5pt] &= ⟨p_2q_3−p_3q_2,p_1q_3−p_3q_1,p_1q_2−p_2q_1⟩ \\[5pt] &= ⟨2(−3)−5(0),−(−1)(−3)+5(4),(−1)(0)−2(4)⟩ \\[5pt] &= ⟨−6,17,−8⟩.\end{align*}
Exercise $$\PageIndex{1}$$
Find $$p×q$$ for $$p=⟨5,1,2⟩$$ and $$q=⟨−2,0,1⟩.$$ Express the answer using standard unit vectors.
Hint
Use the formula $$u×v=(u_2v_3−u_3v_2)i−(u_1v_3−u_3v_1)j+(u_1v_2−u_2v_1)k.$$
$$i−9j+2k$$
Although it may not be obvious from Equation \ref{cross}, the direction of $$u×v$$ is given by the right-hand rule. If we hold the right hand out with the fingers pointing in the direction of $$u$$, then curl the fingers toward vector $$v$$, the thumb points in the direction of the cross product, as shown in Figure $$\PageIndex{2}$$.
Figure $$\PageIndex{2}$$: The direction of $$u×v$$ is determined by the right-hand rule.
Notice what this means for the direction of $$v×u$$. If we apply the right-hand rule to $$v×u$$, we start with our fingers pointed in the direction of $$v$$, then curl our fingers toward the vector $$u$$. In this case, the thumb points in the opposite direction of $$u×v$$. (Try it!)
Example $$\PageIndex{2}$$: Anticommutativity of the Cross Product
Let $$u=⟨0,2,1⟩$$ and $$v=⟨3,−1,0⟩$$. Calculate $$u×v$$ and $$v×u$$ and graph them.
Figure $$\PageIndex{3}$$: Are the cross products $$u×v$$ and $$v×u$$ in the same direction?
Solution
We have
$$u×v=⟨(0+1),−(0−3),(0−6)⟩=⟨1,3,−6⟩$$
$$v×u=⟨(−1−0),−(3−0),(6−0)⟩=⟨−1,−3,6⟩.$$
We see that, in this case, $$u×v=−(v×u)$$ (Figure $$\PageIndex{4}$$). We prove this in general later in this section.
Figure $$\PageIndex{4}$$: The cross products $$u×v$$ and $$v×u$$ are both orthogonal to $$u$$ and $$v$$, but in opposite directions.
Exercise $$\PageIndex{2}$$
Suppose vectors $$u$$ and $$v$$ lie in the xy-plane (the z-component of each vector is zero). Now suppose the x- and y-components of $$u$$ and the y-component of $$v$$ are all positive, whereas the x-component of $$v$$ is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does $$u×v$$ point?
Hint
Remember the right-hand rule (Figure $$\PageIndex{2}$$).
Up (the positive z-direction)
The cross products of the standard unit vectors $$i,j$$, and $$k$$ can be useful for simplifying some calculations, so let’s consider these cross products. A straightforward application of the definition shows that
$i×i=j×j=k×k=0.$
(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar $$0$$.) It’s up to you to verify the calculations on your own.
Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of $$i$$ and $$j$$ is parallel to $$k$$. Similarly, the vector product of $$i$$ and $$k$$ is parallel to $$j$$, and the vector product of $$j$$ and $$k$$ is parallel to $$i$$.
We can use the right-hand rule to determine the direction of each product. Then we have
\begin{align} i×j&=k \\[5pt] j×i&=−k \\[10pt] j×k&=i \\[5pt] k×j&=−i \\[10pt] k×i&=j \\[5pt] i×k&=−j. \end{align}
These formulas come in handy later.
Example $$\PageIndex{3}$$: Cross Product of Standard Unit Vectors
Find $$i×(j×k).$$
Solution
We know that $$j×k=i.$$ Therefore, $$i×(j×k)=i×i=0.$$
Exercise $$\PageIndex{3}$$:
Find $$(i×j)×(k×i).$$
Hint
Remember the right-hand rule (Figure $$\PageIndex{2}$$).
$$−i$$
As we have seen, the dot product is often called the scalar product because it results in a scalar. The cross product results in a vector, so it is sometimes called the vector product. These operations are both versions of vector multiplication, but they have very different properties and applications. Let’s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.
Properties of the Cross Product
Let $$u,v,$$ and $$w$$ be vectors in space, and let $$c$$ be a scalar.
1. Anticommutative property: $u×v=−(v×u)$
2. Distributive property: $u×(v+w)=u×v+u×w$
3. Multiplication by a constant: $c(u×v)=(cu)×v=u×(cv)$
4. Cross product of the zero vector: $u×0=0×u=0$
5. Cross product of a vector with itself: $v×v=0$
6. Scalar triple product: $u⋅(v×w)=(u×v)⋅w$
Proof
For property $$i$$., we want to show $$u×v=−(v×u).$$ We have
\begin{align*} u×v &=⟨u_1,u_2,u_3⟩×⟨v_1,v_2,v_3⟩ \\[5pt] &=⟨u_2v_3−u_3v_2,−u_1v_3+u_3v_1,u_1v_2−u_2v_1⟩ \\[5pt] &=−⟨u_3v_2−u_2v_3,−u_3v_1+u_1v_3,u_2v_1−u_1v_2⟩ \\[5pt] &=−⟨v_1,v_2,v_3⟩×⟨u_1,u_2,u_3⟩\\[5pt] &=−(v×u).\end{align*}
Unlike most operations we’ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule.
For property $$iv$$., this follows directly from the definition of the cross product. We have
$u×0=⟨u_2(0)−u_3(0),−(u_2(0)−u_3(0)),u1(0)−u_2(0)⟩=⟨0,0,0⟩=0. \nonumber$
Then, by property i., $$0×u=0$$ as well. Remember that the dot product of a vector and the zero vector is the scalar $$0$$, whereas the cross product of a vector with the zero vector is the vector $$0$$.
Property $$vi$$. looks like the associative property, but note the change in operations:
\begin{align*} u⋅(v×w)&=u⋅⟨v_2w_3−v_3w_2,−v_1w_3+v_3w_1,v_1w_2−v_2w_1⟩ \\[5pt] &= u_1(v_2w_3−v_3w_2)+u_2(−v_1w_3+v_3w_1)+u_3(v_1w_2−v_2w_1) \\[5pt] &=u_1v_2w_3−u_1v_3w_2−u_2v_1w_3+u_2v_3w_1+u_3v_1w_2−u_3v_2w_1 \\[5pt] &=(u_2v_3−u_3v_2)w_1+(u_3v_1−u_1v_3)w_2+(u_1v_2−u_2v_1)w_3 \\[5pt] &=⟨u_2v_3−u_3v_2,u_3v_1−u_1v_3,u_1v_2−u_2v_1⟩⋅⟨w_1,w_2,w_3⟩ =(u×v)⋅w.\end{align*}
$$\square$$
Example $$\PageIndex{4}$$: Using the Properties of the Cross Product
Use the cross product properties to calculate $$(2i×3j)×j.$$
Solution
\begin{align*} (2i×3j)×j&=2(i×3j)×j \\[5pt] &=2(3)(i×j)×j \\[5pt] &=(6k)×j \\[5pt] &=6(k×j) \\[5pt] &=6(−i)=−6i. \end{align*}
Exercise $$\PageIndex{4}$$:
Use the properties of the cross product to calculate $$(i×k)×(k×j).$$
Hint
$$u×v=−(v×u)$$
$$−k$$
So far in this section, we have been concerned with the direction of the vector $$u×v$$, but we have not discussed its magnitude. It turns out there is a simple expression for the magnitude of $$u×v$$ involving the magnitudes of $$u$$ and $$v$$, and the sine of the angle between them.
Magnitude of the Cross Product
Let $$u$$ and $$v$$ be vectors, and let $$θ$$ be the angle between them. Then, $$‖u×v‖=‖u‖⋅‖v‖⋅sinθ.$$
Proof
Let $$u=⟨u_1,u_2,u_3⟩$$ and $$v=⟨v_1,v_2,v_3⟩$$ be vectors, and let $$θ$$ denote the angle between them. Then
$$‖u×v‖^2=(u_2v_3−u_3v_2)^2+(u_3v_1−u_1v_3)^2+(u_1v_2−u_2v_1)^2$$
$$=u^2_2v^2_3−2u_2u_3v_2v_3+u^2_3v^2_2+u^2_3v^2_1−2u_1u_3v_1v_3+u^2_1v^2_3+u^2_1v^2_2−2u_1u_2v_1v_2+u^2_2v^2_1$$
$$=u^2_1v^2_1+u^2_1v^2_2+u^2_1v^2_3+u^2_2v^2_1+u^2_2v^2_2+u^2_2v^2_3+u^2_3v^2_1+u^2_3v^2_2+u^2_3v^2_3−(u^2_1v^2_1+u^2_2v^2_2+u^2_3v^2_3+2u_1u_2v_1v_2+2u_1u_3v_1v_3+2u_2u_3v_2v_3)$$
$$=(u^2_1+u^2_2+u^2_3)(v^2_1+v^2_2+v^2_3)−(u_1v_1+u_2v_2+u_3v_3)^2$$
$$=‖u‖^2‖v‖^2−(u⋅v)^2$$
$$=‖u‖^2‖v‖^2−‖u‖^2‖v‖^2cos^2θ$$
$$=‖u‖^2‖v‖^2(1−cos^2θ)$$
$$=‖u‖^2‖v‖^2(sin^2θ).$$
Taking square roots and noting that $$\sqrt{\sin^2θ}=\sinθ$$ for $$0≤θ≤180°,$$ we have the desired result:
$‖u×v‖=‖u‖‖v‖sinθ. \nonumber$
This definition of the cross product allows us to visualize or interpret the product geometrically. It is clear, for example, that the cross product is defined only for vectors in three dimensions, not for vectors in two dimensions. In two dimensions, it is impossible to generate a vector simultaneously orthogonal to two nonparallel vectors.
Example $$\PageIndex{5}$$: Calculating the Cross Product
Use Note to find the magnitude of the cross product of $$u=⟨0,4,0⟩$$ and $$v=⟨0,0,−3⟩$$.
Solution
We have
\begin{align*} ‖u×v‖ &= ‖u‖⋅‖v‖⋅\sinθ \\[5pt] &=\sqrt{0^2+4^2+0^2}⋅\sqrt{0^2+0^2+(−3)^2}⋅\sin\dfrac{π}{2} \\[5pt] &=4(3)(1)=12 \end{align*}
Exercise $$\PageIndex{5}$$:
Use Note to find the magnitude of $$u×v$$, where $$u=⟨−8,0,0⟩$$ and $$v=⟨0,2,0⟩$$.
Hint
Vectors $$u$$ and $$v$$ are orthogonal.
16
## Determinants and the Cross Product
Using Equation \ref{cross} to find the cross product of two vectors is straightforward, and it presents the cross product in the useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have an alternative. We can calculate the cross product of two vectors using determinant notation.
A $$2×2$$ determinant is defined by
$\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix} =a_1b_2−b_1a_2.$
For example,
$\begin{bmatrix}3&−2\\5&1\end{bmatrix} =3(1)−5(−2)=3+10=13.$
A $$3×3$$ determinant is defined in terms of $$2×2$$ determinants as follows:
$\begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=a_1\begin{bmatrix}b_2&b_3\\c_2&c_3\end{bmatrix}−a_2\begin{bmatrix}b_1&b_3\\c_1&c_3\end{bmatrix}+a_3\begin{bmatrix}b_1&b_2\\c_1&c_2\end{bmatrix}.$
Equation is referred to as the expansion of the determinant along the first row. Notice that the multipliers of each of the $$2×2$$ determinants on the right side of this expression are the entries in the first row of the $$3×3$$ determinant. Furthermore, each of the $$2×2$$ determinants contains the entries from the $$3×3$$ determinant that would remain if you crossed out the row and column containing the multiplier. Thus, for the first term on the right, $$a_1$$ is the multiplier, and the $$2×2$$ determinant contains the entries that remain if you cross out the first row and first column of the $$3×3$$ determinant. Similarly, for the second term, the multiplier is $$a_2$$, and the $$2×2$$ determinant contains the entries that remain if you cross out the first row and second column of the $$3×3$$ determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.
Example $$\PageIndex{6}$$: Using Expansion Along the First Row to Compute a $$3×3$$ Determinant
Evaluate the determinant $$\begin{bmatrix}2&5&−1\\−1&1&3\\−2&3&4\end{bmatrix}$$.
Solution
We have
\begin{align*} \begin{bmatrix}2&5&−1\\−1&1&3\\−2&3&4\end{bmatrix}=2\begin{bmatrix}1&3\\3&4\end{bmatrix}−5\begin{bmatrix}−1&3\\−2&4\end{bmatrix}−1\begin{bmatrix}−1&1\\−2&3\end{bmatrix} \\[5pt] &=2(4−9)−5(−4+6)−1(−3+2) \\[5pt] &= 2(−5)−5(2)−1(−1)=−10−10+1 \\[5pt] &=−19 \end{align*}
Exercise $$\PageIndex{6}$$
Evaluate the determinant $$\begin{bmatrix}1&−2&−1\\3&2&−3\\1&5&4\end{bmatrix}$$.
Hint
Expand along the first row. Don’t forget the second term is negative!
40
Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.
Rule: Cross Product Calculated by a Determinant
Let $$u=⟨u_1,u_2,u_3⟩$$ and $$v=⟨v_1,v_2,v_3⟩$$ be vectors. Then the cross product $$u×v$$ is given by
$u×v=\begin{bmatrix}i&j&K\\u_1&u_2&u_3\\v_1&v_2&v_3\end{bmatrix}=\begin{bmatrix}u_2&u_3\\v_2&v_3\end{bmatrix}i−\begin{bmatrix}u_1&u_3\\v_1&v_3\end{bmatrix}j+\begin{bmatrix}u_1&u_2\\v_1&v_2\end{bmatrix}k.$
Example $$\PageIndex{7}$$: Using Determinant Notation to find $$p×q$$
Let $$p=⟨−1,2,5⟩$$ and $$q=⟨4,0,−3⟩$$. Find $$p×q$$.
Solution
We set up our determinant by putting the standard unit vectors across the first row, the components of $$u$$ in the second row, and the components of $$v$$ in the third row. Then, we have
\begin{align*} p×q&=\begin{bmatrix}i&j&k\\−1&2&5\\4&0&−3\end{bmatrix}=\begin{bmatrix}2&0\\5&−3\end{bmatrix}i−\begin{bmatrix}−1&5\\4&−3\end{bmatrix}j+\begin{bmatrix}−1&2\\4&0\end{bmatrix}k \\[5pt] &= (−6−0)i−(3−20)j+(0−8)k \\[5pt] &=−6i+17j−8k.\end{align*}
Notice that this answer confirms the calculation of the cross product in Example $$\PageIndex{1}$$.
Exercise $$\PageIndex{7}$$
Use determinant notation to find $$a×b,$$ where $$a=⟨8,2,3⟩$$ and $$b=⟨−1,0,4⟩.$$
Hint
Calculate the determinant $$\begin{bmatrix}i&j&K\\8&2&3\\−1&0&4\end{bmatrix}$$.
$$8i−35j+2k$$
## Using the Cross Product
The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a parallelepiped. The following examples illustrate these calculations.
Example $$\PageIndex{8}$$: Finding a Unit Vector Orthogonal to Two Given Vectors
Let $$a=⟨5,2,−1⟩$$ and $$b=⟨0,−1,4⟩$$. Find a unit vector orthogonal to both $$a$$ and $$b$$.
Solution
The cross product $$a×b$$ is orthogonal to both vectors $$a$$ and $$b$$. We can calculate it with a determinant:
\begin{align*} a×b &=\begin{bmatrix}i&j&K\\5&2&−1\\0&−1&4\end{bmatrix}=\begin{bmatrix}2&−1\\−14\end{bmatrix}i−\begin{bmatrix}5&−1\\0&4\end{bmatrix}j+\begin{bmatrix}5&2\\0&−1\end{bmatrix}k \\[5pt] &=(8−1)i−(20−0)j+(−5−0)k \\[5pt] &=7i−20j−5k.\end{align*}
Normalize this vector to find a unit vector in the same direction:
$$∥a×b∥=\sqrt{(7)^2+(−20)^2+(−5)^2}=\sqrt{474}$$.
Thus, $$⟨\dfrac{7}{\sqrt{474}},\dfrac{−20}{\sqrt{474}},\dfrac{−5}{\sqrt{474}}⟩$$ is a unit vector orthogonal to $$a$$ and $$b$$.
Exercise $$\PageIndex{8}$$
Find a unit vector orthogonal to both $$a$$ and $$b$$, where $$a=⟨4,0,3⟩$$ and $$b=⟨1,1,4⟩.$$
Hint
Normalize the cross product.
$$⟨\dfrac{−3}{\sqrt{194}},\dfrac{−13}{\sqrt{194}},\dfrac{4}{\sqrt{194}}⟩$$
To use the cross product for calculating areas, we state and prove the following theorem.
Area of a Parallelogram
If we locate vectors $$u$$ and $$v$$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by $$‖u×v‖$$ (Figure $$\PageIndex{5}$$).
Figure $$\PageIndex{5}$$: The parallelogram with adjacent sides $$u$$ and $$v$$ has base $$‖u‖$$ and height $$‖v‖sinθ.$$
Proof
We show that the magnitude of the cross product is equal to the base times height of the parallelogram.
$\text{Area of a parallelogram}= \text{base} × \text{height} =‖u‖(‖v‖\sinθ)=‖u×v‖$
Example $$\PageIndex{9}$$: Finding the Area of a Triangle
Let $$P=(1,0,0),Q=(0,1,0),$$ and $$R=(0,0,1)$$ be the vertices of a triangle (Figure $$\PageIndex{6}$$). Find its area.
Figure $$\PageIndex{6}$$: Finding the area of a triangle by using the cross product.
Solution
We have $$\vec{PQ}=⟨0−1,1−0,0−0⟩=⟨−1,1,0⟩$$ and $$\vec{PR}=⟨0−1,0−0,1−0⟩=⟨−1,0,1⟩$$. The area of the parallelogram with adjacent sides $$\vec{PQ}$$ and $$\vec{PR}$$ is given by $$∥\vec{PQ}×\vec{PR}∥$$:
\begin{align*} \vec{PQ} \times \vec{PR} &= \begin{bmatrix}i&j&k\\−1&1&0\\−1&0&1\end{bmatrix} \\[5pt] &=(1−0)i−(−1−0)j+(0−(−1))k \\[5pt] &=i+j+k \\[10pt] ∥\vec{PQ}×\vec{PR}∥ &=∥⟨1,1,1⟩∥ \\[5pt] &=\sqrt{1^2+1^2+1^2} \\[5pt] &=\sqrt{3}. \end{align*}
The area of $$ΔPQR$$ is half the area of the parallelogram or $$\sqrt{3}/2.$$
Exercise $$\PageIndex{9}$$
Find the area of the parallelogram $$PQRS$$ with vertices $$P(1,1,0)$$, $$Q(7,1,0)$$, $$R(9,4,2)$$, and $$S(3,4,2)$$.
Hint
Sketch the parallelogram and identify two vectors that form adjacent sides of the parallelogram.
$$6\sqrt{13}$$
## The Triple Scalar Product
Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.
Definition: triple scalar product
The triple scalar product of vectors $$u, v,$$ and $$w$$ is
$u⋅(v×w).$
Calculating a Triple Scalar Product
The triple scalar product of vectors
$$u=u_1i+u_2j+u_3k, v=v_1i+v_2j+v_3k$$
and
$$w=w_1i+w_2j+w_3k$$
is the determinant of the $$3×3$$ matrix formed by the components of the vectors:
$u⋅(v×w)=\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{bmatrix}.$
Proof
The calculation is straightforward.
\begin{align*} u⋅(v×w)&=⟨u_1,u_2,u_3⟩⋅⟨v_2w_3−v_3w_2,−v_1w_3+v_3w_1,v_1w_2−v_2w_1⟩\\[5pt] &=u_1(v_2w_3−v_3w_2)+u_2(−v_1w_3+v_3w_1)+u_3(v_1w_2−v_2w_1) \\[5pt] &=u_1(v_2w_3−v_3w_2)−u_2(v_1w_3−v_3w_1)+u_3(v_1w_2−v_2w_1) \\[5pt] &=\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{bmatrix}.\end{align*}
Example $$\PageIndex{10}$$: Calculating the Triple Scalar Product
Let $$u=⟨1,3,5⟩,v=⟨2,−1,0⟩$$ and $$w=⟨−3,0,−1⟩$$. Calculate the triple scalar product $$u⋅(v×w).$$
Solution
Apply Note directly:
\begin{align*} u⋅(v×w)=\begin{bmatrix}1&3&5\\2&−1&0\\−3&0&−1\end{bmatrix}. \\[5pt] &=1\begin{bmatrix}−1&0\\0&−1\end{bmatrix}−3\begin{bmatrix}2&0\\−3&−1\end{bmatrix}+5\begin{bmatrix}2&−1\\−3&0\end{bmatrix} \\[5pt] &=(1−0)−3(−2−0)+5(0−3) \\[5pt] &=1+6−15=−8. \end{align*}
Exercise $$\PageIndex{10}$$
Calculate the triple scalar product $$a⋅(b×c),$$ where $$a=⟨2,−4,1⟩, b=⟨0,3,−1⟩$$, and $$c=⟨5,−3,3⟩.$$
Hint
Place the vectors as the rows of a $$3×3$$ matrix, then calculate the determinant.
$$17$$
When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:
$$\begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=d \begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=−d \begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=d \begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=−d$$
Verifying this fact is straightforward, but rather messy. Let’s take a look at this with an example:
\begin{align} \begin{bmatrix}1&2&1\\−2&0&3\\4&1&−1\end{bmatrix} &=\begin{bmatrix}0&3\\1&−1\end{bmatrix}−2\begin{bmatrix}−2&3\\4&−1\end{bmatrix}+\begin{bmatrix}−2&0\\4&1\end{bmatrix} \\[5pt] &=(0−3)−2(2−12)+(−2−0) \\[5pt] &=−3+20−2=15. \end{align}
Switching the top two rows we have
$$∣∣∣∣−21402131−1∣∣∣∣=−2∣∣∣211−1∣∣∣+3∣∣∣1421∣∣∣=−2(−2−1)+3(1−8)=6−21=−15.$$
Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let $$u=u_1i+u_2j+u_3k, v=v_1i+v_2j+v_3k,$$ and $$w=w_1i+w_2j+w_3k.$$ Applying Note, we have
$u⋅(v×w)=\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{bmatrix} \text{and} u⋅(w×v)=\begin{bmatrix}u_1&u_2&u_3\\w_1&w_2&w_3\\v_1&v_2&v_3\end{bmatrix}.$
We can obtain the determinant for calculating $$u⋅(w×v)$$ by switching the bottom two rows of $$u⋅(v×w).$$ Therefore, $$u⋅(v×w)=−u⋅(w×v).$$
Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:
$u⋅(v×w)=−u⋅(w×v)$
$u⋅(v×w)=v⋅(w×u)=w⋅(u×v).$
Let $$u$$ and $$v$$ be two vectors in standard position. If $$u$$ and $$v$$ are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in Note that the area of this parallelogram is $$‖u×v‖$$. Now suppose we add a third vector $$w$$ that does not lie in the same plane as $$u$$ and $$v$$ but still shares the same initial point. Then these vectors form three edges of a parallelepiped, a three-dimensional prism with six faces that are each parallelograms, as shown in Figure. The volume of this prism is the product of the figure’s height and the area of its base. The triple scalar product of $$u,v,$$ and $$w$$ provides a simple method for calculating the volume of the parallelepiped defined by these vectors.
Volume of a Parallelepiped
The volume of a parallelepiped with adjacent edges given by the vectors $$u,v$$, and $$w$$ is the absolute value of the triple scalar product (Figure $$\PageIndex{7}$$):
$V=|u⋅(v×w)|.$
Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.
Figure $$\PageIndex{7}$$: The height of the parallelepiped is given by $$∥proj_{v×w}u∥.$$
Proof
The area of the base of the parallelepiped is given by $$‖v×w‖.$$ The height of the figure is given by $$∥proj_{v×w}u∥.$$ The volume of the parallelepiped is the product of the height and the area of the base, so we have
$$V=∥proj_{v×w}u∥‖v×w‖$$
$$=∣∣\dfrac{u⋅(v×w)}{‖v×w‖}∣∣‖v×w‖$$
$$=|u⋅(v×w)|.$$
Example $$\PageIndex{11}$$: Calculating the Volume of a Parallelepiped
Let $$u=⟨−1,−2,1⟩,v=⟨4,3,2⟩,$$ and $$w=⟨0,−5,−2⟩$$. Find the volume of the parallelepiped with adjacent edges $$u,v$$, and $$w$$ (Figure $$\PageIndex{8}$$).
Figure $$\PageIndex{8}$$
Solution
We have
$$u⋅(v×w)=\begin{bmatrix}−1&−2&1\\4&3&2\\0&−5&−2\end{bmatrix} =(−1)\begin{bmatrix}3&2\\−5&−2\end{bmatrix}+2\begin{bmatrix}4&2\\0&−2\end{bmatrix}+\begin{bmatrix}4&3\\0&−5\end{bmatrix}$$
$$=(−1)(−6+10)+2(−8−0)+(−20−0)$$
$$=−4−16−20$$
$$=−40.$$
Thus, the volume of the parallelepiped is $$|−40|=40$$ units3
Exercise $$\PageIndex{11}$$
Find the volume of the parallelepiped formed by the vectors $$a=3i+4j−k, b=2i−j−k,$$ and $$c=3j+k.$$
Hint
Calculate the triple scalar product by finding a determinant.
$$8$$ units3
## Applications of the Cross Product
The cross product appears in many practical applications in mathematics, physics, and engineering. Let’s examine some of these applications here, including the idea of torque, with which we began this section. Other applications show up in later chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields (Introduction to Vector Calculus).
Example $$\PageIndex{12}$$: Using the Triple Scalar Product
Use the triple scalar product to show that vectors $$u=⟨2,0,5⟩,v=⟨2,2,4⟩$$, and $$w=⟨1,−1,3⟩$$ are coplanar—that is, show that these vectors lie in the same plane.
Solution
Start by calculating the triple scalar product to find the volume of the parallelepiped defined by $$u,v,$$ and $$w$$:
$$u⋅(v×w)=\begin{bmatrix}2&0&5\\2&2&4\\1&−1&3\end{bmatrix}$$
$$=[2(2)(3)+(0)(4)(1)+5(2)(−1)]−[5(2)(1)+(2)(4)(−1)+(0)(2)(3)]$$
$$=2−2$$
$$=0$$.
The volume of the parallelepiped is $$0$$ units3, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.
Exercise $$\PageIndex{12}$$
Are the vectors $$a=i+j−k, b=i−j+k,$$ and $$c=i+j+k$$ coplanar?
Hint
Calculate the triple scalar product.
No, the triple scalar product is $$−4≠0,$$ so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.
Example $$\PageIndex{13}$$: Finding an Orthogonal Vector
Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points $$P=(9,−3,−2),Q=(1,3,0),$$ and $$R=(−2,5,0).$$
Solution
The plane must contain vectors $$\vec{PQ}$$ and $$\vec{QR}$$:
$$\vec{PQ}=⟨1−9,3−(−3),0−(−2)⟩=⟨−8,6,2⟩$$
$$\vec{QR}=⟨−2−1,5−3,0−0⟩=⟨−3,2,0⟩.$$
The cross product $$\vec{PQ}×\vec{QR}$$ produces a vector orthogonal to both $$\vec{PQ}$$ and $$\vec{QR}$$. Therefore, the cross product is orthogonal to the plane that contains these two vectors:
$$\vec{PQ}×\vec{QR}=\begin{bmatrix}i&j&K\\−8&6&2\\−3&2&0\end{bmatrix}$$
$$=0i−6j−16k−(−18k+4i+0j)$$
$$=−4i−6j+2k.$$
We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the cross product to real-world situations.
Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotational effect, called torque.
Definition: Torque
Torque, $$τ$$ (the Greek letter tau), measures the tendency of a force to produce rotation about an axis of rotation. Let $$r$$ be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector $$F$$ represent the force. Then torque is equal to the cross product of $$r$$ and $$F$$:
$τ=r×F.$
See Figure $$\PageIndex{9}$$.
Figure $$\PageIndex{9}$$: Torque measures how a force causes an object to rotate.
Think about using a wrench to tighten a bolt. The torque τ applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound. Although torque is dimensionally equivalent to work (it has the same units), the two concepts are distinct. Torque is used specifically in the context of rotation, whereas work typically involves motion along a line.
Example $$\PageIndex{14}$$: Evaluating Torque
A bolt is tightened by applying a force of $$6$$ N to a 0.15-m wrench (Figure $$\PageIndex{10}$$). The angle between the wrench and the force vector is $$40°$$. Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.
Figure $$\PageIndex{10}$$: Torque describes the twisting action of the wrench.
Solution:
Substitute the given information into the equation defining torque:
$$‖τ‖=∥r×F∥=‖r‖∥F∥sinθ=(0.15m)(6N)sin40°≈0.58N⋅m.$$
Exercise $$\PageIndex{14}$$
Calculate the force required to produce $$15N⋅m$$ torque at an angle of $$30º$$ from a 150-cm rod.
Hint
$$‖τ‖=15N⋅m$$ and $$‖r‖=1.5m$$
$$20$$N
## Key Concepts
• The cross product $$u×v$$ of two vectors $$u=⟨u_1,u_2,u_3⟩$$ and $$v=⟨v_1,v_2,v_3⟩$$ is a vector orthogonal to both $$u$$ and $$v$$. Its length is given by $$‖u×v‖=‖u‖⋅‖v‖⋅sinθ,$$ where $$θ$$ is the angle between $$u$$ and $$v$$. Its direction is given by the right-hand rule.
• The algebraic formula for calculating the cross product of two vectors,
$$u=⟨u_1,u_2,u_3⟩$$ and $$v=⟨v_1,v_2,v_3⟩$$, is
$$u×v=(u_2v_3−u_3v_2)i−(u_1v_3−u_3v_1)j+(u_1v_2−u_2v_1)k.$$
• The cross product satisfies the following properties for vectors $$u,v,$$ and $$w$$, and scalar $$c:) \(u×v=−(v×u)$$
$$u×(v+w)=u×v+u×w$$
$$c(u×v)=(cu)×v=u×(cv)$$
$$u×0=0×u=0$$
$$v×v=0$$
$$u⋅(v×w)=(u×v)⋅w$$
• The cross product of vectors $$u=⟨u_1,u_2,u_3⟩$$ and $$v=⟨v_1,v_2,v_3⟩$$ is the determinant $$\begin{bmatrix}i&j&K\\u_1&u_2&u_3\\v_1&v_2&v_3\end{bmatrix}$$
• If vectors $$u$$ and $$v$$ form adjacent sides of a parallelogram, then the area of the parallelogram is given by $$‖u×v‖.$$
• The triple scalar product of vectors $$u, v,$$ and $$w$$ is $$u⋅(v×w).$$
• The volume of a parallelepiped with adjacent edges given by vectors $$u,v$$, and $$w$$ is $$V=|u⋅(v×w)|.$$
• If the triple scalar product of vectors $$u,v,$$ and $$w$$ is zero, then the vectors are coplanar. The converse is also true: If the vectors are coplanar, then their triple scalar product is zero.
• The cross product can be used to identify a vector orthogonal to two given vectors or to a plane.
• Torque $$τ$$ measures the tendency of a force to produce rotation about an axis of rotation. If force $$F$$ is acting at a distance $$r$$ from the axis, then torque is equal to the cross product of $$r$$ and $$F: τ=r×F.$$
## Key Equations
• The cross product of two vectors in terms of the unit vectors
$u×v=(u_2v_3−u_3v_2)i−(u_1v_3−u_3v_1)j+(u_1v_2−u_2v_1)k$
## Glossary
cross product
$$u×v=(u_2v_3−u_3v_2)i−(u_1v_3−u_3v_1)j+(u_1v_2−u_2v_1)k,$$ where $$u=⟨u_1,u_2,u_3⟩$$ and $$v=⟨v_1,v_2,v_3⟩$$
determinant
a real number associated with a square matrix
parallelepiped
a three-dimensional prism with six faces that are parallelograms
torque
the effect of a force that causes an object to rotate
triple scalar product
the dot product of a vector with the cross product of two other vectors: u⋅(v×w)
vector product
the cross product of two vectors |
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# HW4 - HW4 Solutions Dilip Raghavan 1 Section 5.4 Problem 48...
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Unformatted text preview: HW4 Solutions Dilip Raghavan September 28, 2008 1 Section 5.4 Problem 48. Let us first make an inventory of everything we have. The vowels: one I, one U and one O. The consonants: one N, one S, two Ts, two Rs and one C. Now, we know that the vowels must occur in alphabetical order, so let’s first put them down: I O U We also know that we must begin or end with a block of 2 Ts. We must decide whether we begin or end with them: 2 choices. TT I O U We are left with 5 letters to arrange. Let us first think about how many ways there are to distribute 5 empty slots among the letters we have arranged so far. We will then count the number of ways to arrange the letters that are left over in those empty slots. So we want to distribute 5 identical empty slots into 4 boxes: TT box 1 I box 2 O box 3 U box 4 Since we must have atleast 2 consonants between any two vowels, we must make sure that there are atleast 2 empty slots in each of boxes 2 and 3. So put two slots into each of those boxes. We have just one slot left, which we much distribute into 4 boxes: 4 ways. Now we must decide how the letters are placed in the slots: TT — I — — O — — U Since there are two identical Rs, there are 5! 2! ways to do this. So the answer is 2 × 4 × 5! 2! Problem 50. Again, let us begin by making an inventory of everything we have. The vowels: two Es, two Is. The consonants: one R, one V, one S, one T and one D. (a) Let us count the complement, which is the number of arrangements with vowels in increasing order. Firstly, the total number of arrangements is 9! 2!2! . Now if the vowels are in increasing order, let us first write them down E E I I Again, let us first distribute 5 empty slots for the 5 remaining letters into 5 boxes: parenleftBig 5 + 5 − 1 5 parenrightBig ways. Now we need to place the letter in those slots: 5! ways. So the answer is 9! 2!2! − bracketleftBiggparenleftBigg 5 + 5 − 1 5 parenrightBigg × 5! bracketrightBigg (You could also do this problem by first arranging the consonants and then distributing slots for vowels). 1 (b) Let us first arrange the vowels and then we will distribute slots for the consonants. But we must break up into four cases depending on whether the arrangement of the vowels has consecutive Es and Is or not. • Case I: no consecutive Es or Is. This means there is an I between the two Es and an E between the 2 Is. There are 2 such arrangements of the Es and Is. For example, such an arrangement might look like: E I E I Now, we are free to distribute 5 slots for the 5 consonants in any way we please into 5 boxes: parenleftBig 5 + 5 − 1 5 parenrightBig ways. Next we place the consonants in these 5 slots: 5! ways. So for this case we have 2 × parenleftBigg 5 + 5 − 1 5 parenrightBigg × 5! possibilities....
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# 2001 AMC 10 Problems/Problem 15
## Problem
A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$
## Solution 1
Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$, giving an area of $600$.
If we look at it the other way, we see the distance between the stripes is the height and the base is $50$.
$[asy] draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5),linewidth(2)); draw((2,4)--(6,2),dashed);[/asy]$
The area is still the same, so the distance between the stripes is $600/50 = \boxed{\textbf{(C)}\ 12}$.
## Solution 2
Alternatively, we could use similar triangles--the $30-40-50$ triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the $x-y-15$ triangle, where we are trying to find $y$ (the shortest distance between the two stripes). Therefore, $y$ would have to be $\boxed{\textbf{(C)}\ 12}$. |
> > > Percentage Formula
# Percentage Formula
In this lesson, students will learn about the absolute basics of Percentages. The purpose of this lesson is to help the student to get an answer one simple question related to percentage. We make a lot of use of percentage calculation in our day-to-day life. A percentage is a portion of a whole expressed as a number between the zero and 100 rather than as a fraction. Thus all of something is 100 percent. Sometimes to know the discounts on the price value the percentage formula proves much important. In this topic, we will discuss the percentage formula with examples. Let us learn it!
## Percentage Formula
### What is the Percentage?
Percent implies “for every hundred” and its sign is “%”.It is read as the percentage and therefore x% is read as x percent. Therefore, a fraction with denominator 100 is called a percent. For example, 30 % means $$\frac{30}{100}$$ (i.e. 30 parts from 100). This can also be written as 0.30.
### Formula for Percentage
To calculate p % of q, use the following formula:
$$\frac{p}{100} \times q$$
Also remember here : p % of q = q % of p
For Examples:
1. $$100% of 60 is 60 \times \frac{100}{100} = 60$$
2. $$50% of 60 is \frac{50}{100} \times 60 = 30$$
3. $$5% of 60 is \frac {5} {100} \times 60 = 3$$
Steps to find out percentage value:
1. Visualize what a percentage represents. A percentage is an expression of some part of the whole.
2. Determine the value of the whole first.
3. Then find the value that we want to turn into a percentage.
4. Then put the two values into a fraction.
5. Convert the fraction into its decimal equivalent.
6. Finally, convert the decimal into a percent.
### Some important tips are given as follows:
Basic Tip-1: If the new value of something is p times the previously given value, then the percentage increase will be $$(p-1) \times 100 %.$$
Basic Tip-2: When a quantity n is increased by k %, then the:
New quantity = $$n \times (1+ \frac { k }{ 100 })$$
Basic Tip 3: When a quantity p is decreased by k %, then the:
New quantity = $$p (1 – \frac {k} {100} )$$
Understanding the concept:
To determine the percentage value we divide the portion of the whole by the whole itself and then multiply it by 100. Therefore, if we just ate two pieces of an eight-piece pie, then we may calculate the percentage of the pie which we have consumed.
First, we will divide 2 by 8 which will be o.25. Then multiply 0.25 by 100 to get 25 percent. A percentage may also mean a portion of something but only when it has to do with numbers. When we buy furniture, then the salesman gets a percentage of what we spend.
## Solved Examples
Q.1: 60 % of some number is 360. What will be 99 % of the same number?
Solution: Let the number be x.
Given here, $$\frac{60}{100} \times x = 360$$
i.e. x = 600
Now, 99 % of 600
= $$\frac {99}{100}\times 600$$
= 594
Q.2. 50 % of some number is 360. What will be 99 % of the same number?
Solution: Let the number be y.
Given $$\frac {50}{100} \times q$$
= 360
i.e q = 720
99% of 720
$$\frac {99}{100} \times 720$$
= 712.80
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8
# THE SLOPE OF A STRAIGHT LINE
Definition of the slope
IN LESSON 33 OF ALGEBRA we discuss the equation of a straight line. Sketching the graph of the equation of a line should be a basic skill.
Consider this straight line.
The (x, y) coördinates at B have changed from the coördinates at A. By the symbol Δx ("delta x") we mean the change in the x-coördinate. That is,
Δx = x2x1.
Similarly, Δy ("delta y") signifies the resulting change in the y-coördinate.
Δy = y2y1.
Δx is the horizontal leg of that right triangle; Δy is the vertical leg.
By the slope of a straight line, then, we mean this number:
Slope = _Vertical leg_Horizontal leg = ΔyΔx =
For example,
If the value of y changes by 2 units when the value of x changes
by 3, then the slope of that line is the number 23 .
What does slope 23 mean? It indicates the rate at which a change
in the value of x produces a change in the value of y. 2 units of y per -- for every -- 3 units of x.
For every 3 units that line moves to the right, it will move up 2. That will be true between any two points on that line. Over 6 and up 4, over 15 and up 10. A straight line has one and only one slope. (Theorem 8.1 below.)
In each of those lines, the x-coördinate has increased by 1 unit. In the line on the left, however, the corresponding value of y has increased much more than in the line on the right. The line on the left has a greater slope than the line on the right. The value of y has changed at a much greater rate.
In calculus, the student will see that the slope of a tangent line to a curve --
-- is the geometrical meaning of what is called the derivative. It indicates the typically non-constant rate of change of y with respect to x -- at the point of tangency.
The slope of a straight line is very definitely a topic of precalculus.
In a typical application, the x-axis represents time and the y-axis, distance. The rate of change of distance with respect to time -- the speed -- is typically not constant. The graph is not a straight line. The slope of the tangent line gives the instantaneous speed at a specific moment of time.
Up or down?
Which line do we say is sloping "up"? And which is sloping "down"?
Since we imagine moving along the x-axis from left to right, we say that the line on the left is sloping up, and the line on the right, down.
What is more, a line that slopes up has a positive slope. While a line that slopes down has a negative slope.
For, both the x- and y-coördinates of B are greater than the coördinates of A, so that both Δx and Δy are positive. Therefore their quotient, which is the slope, is positive.
But while the x-coördinate of D is greater than the x-coördinate of C, so that Δx is positive, the y-coördinate of D is less than the y-coördinate of C, so that Δy, 5 − 8, is negative. Therefore that quotient is negative.
Problem 1. What number is the slope of each line?
Do the problem yourself first!
a) b) 34 43
c) d) − 43 − 34
Horizontal and vertical lines
What number is the slope of a horizontal line -- that is, a line parallel to the x-axis? And what is the slope of a vertical line?
A horizontal line has slope 0 because even though the value of x changes, the value of y does not. Δy = 0.
ΔyΔx = 0 Δx = 0 (Lesson 5)
For a vertical line, however, the slope is not defined. The slope tells how the y-coördinate changes when the x-coördinate changes. But the x-coördinate does not change -- Δx = 0. A vertical line does not have a slope.
ΔyΔx = Δy 0 = No value. (Lesson 5)
Problem 2.
a) Which numbered lines have a positive slope? 2 and 4.
b) Which numbered lines have a negative slope? 1 and 3.
c) What slope has the horizontal line 5? 0.
c) What slope has the vertical line 6? It does not have a slope.
Problem 3. Calculate the slope of the straight line joining the points (−1, −2) and (7, −8). And what does that slope mean?
ΔyΔx = −8 − (−2) 7 − (−1) = −8 + 2 7 + 1 = −6 8 = − 34
That slope indicates the rate of change of the y-coördinate with respect to the x-coördinate. For every 4 units the x-coördinate increases, the y-coördinate decreases by 3.
Problem 4. Let y = f(x) = x2 − 1, and let two points on its graph have these x-coördinates: x1 = 3, x2 = 5. Calculate the slope of the straight line that passes through those two points.
At each of those points, f(x) will give the value of y. At x1 = 3, y1 = 32 − 1 = 8. At x2 = 5, y2= 52 − 1 = 24. Therefore the two points are (3, 8) and (5, 24). And therefore, Δyx =
(24 − 8)/(5 − 3) = 16/2 = 8.
Problem 5.
Let y = f(x), and let its graph pass through two points P and Q whose x-coördinates are separated by a distance h. And let x be the x-coördinate of P.
a) What is the x-coördinate of Q? x + h
b) What are the x- and y-coördinates of P? (x, f(x))
c) What are the coördinates of Q? (x + h, f(x + h))
d) Write an expression for the slope of the straight line joining the
d) points P and Q.
f(x + h) − f(x) h
This is the famous difference quotient of calculus. (Topic 3.) The question calculus asks is: What is the value of that slope when h is practically 0? The answer will be the slope of the tangent line at P. That slope will be the rate of change of the function at P.
One slope
To establish the algebraic equation of any geometrical figure, we must carry over our knowledge from plane geometry; for nothing will come of nothing. With the aid of the following theorems, we will be able to prove that the equation of a straight line may take the form y = ax + b, and that a is the slope of the line. We will do that in the following Topic.
First,
Theorem 8.1. A straight line has one slope.
That is, whether we measure the slope by moving from point A to point B, or by moving from point C to point D, the number we obtain will be the same.
For, if we measure the slope from A to B, we will get BE/EA. While if we measure it between C and D, we will get DF/FC.
But FC is parallel to EA because they are both parallel to the x-axis;
and DA is a straight line that falls on two parallel straight lines;
therefore, the exterior angle DCF is equal to the opposite interior angle on the same side, angle BAE. (Euclid, I. 29)
Therefore the right triangles BEA, DFC are similar (Euclid, VI. 4), and therefore, proportionally,
BEEA = DFFC .
ΔyΔx will be the same, whether we go from A to B, or from C to D.
That is what we wanted to prove.
Problem 6. A straight line will make an angle θ with the x-axis.
Which trigonometric function of θ is its slope?
tan θ. For, a line has one slope.
"Same slope" and "parallel"
Before we state the next theorem, the student should be familiar with the logical expression if and only if.
First, if p and q are statements (sentences), then in the proposition "If p, then q," p is called the hypothesis. It is what we are given, or what we assume. There is no logic without hypotheses.
Statement q is called the conclusion. It is what we must prove on the basis of the hypothesis.
The proposition "If q, then p" is called the converse of "If p, then q." The hypothesis and conclusion are exchanged.
Finally, the proposition "p if and only if q" means: "If p, then q , and if q, then p." In other words, a proposition and its converse are both true.
For example, "A triangles is isosceles if and only if the base angles are equal." This means, "If a triangle is isosceles, then the bases angles are equal; and, conversely, if the base angles are equal, then the triangle is isosceles."
Therefore, when we prove an if and only if proposition, we must prove both a proposition and its converse.
*
The slope of a line is a number. Parallel is a geometrical property. The relationship between them is the next theorem.
Theorem 8.2. Two straight lines are parallel if and only if they have the same slope.
Let the straight lines L1, L2 be parallel. And let the slope of L1 be BC/CA; for a straight line has only one slope. (Theorem 1)
Draw the straight line ACF intersecting L2 at D, and draw FE perpendicular to DF.
Then the slope of the straight line L2 is EF/FD.
Then the slope of L2 is equal to the slope of L1. That is,
BCCA = EFFD
For, since the lines L1, L2 are parallel, and the straight line AF falls on them, the alternate angles BAC, EDF are equal (Euclid, I. 29);
therefore the right triangles BAC, EDF are similar (Euclid, VI.4),
and those sides are proportional that contain the equal angles:
BCCA = EFFD
Therefore the straight lines L1, L2 have the same slope.
Next, conversely, assume that those lines have the same slope:
BCCA = EFFD
Then the two straight lines are parallel.
For, the angles at C and F are right angles and therefore equal,
and by assumption, the sides that make them are proportional.
But if two triangles have one angle equal to one angle, and the sides that make them are proportional, then the triangles are similar,
and those angles are equal that are opposite the corresponding sides;
(Euclid, VI. 6, 4)
therefore, angle BAC is equal to angle EDF.
But angle CDG is equal to angle EDF because they are vertical angles;
(Euclid, I. 15)
therefore angle BAC is equal to angle CDG.
That is, the straight line AF falling on the straight lines L1, L2 makes the alternate angles BAC, CDG equal.
Therefore the straight line L1is parallel to the straight line L2.
(Euclid, I. 27)
Therefore two straight lines are parallel if and only if they have the same slope. Which is what we wanted to prove.
As for perpendicular lines, in Lesson 34 of Algebra we prove the following:
If two straight lines are perpendicular to one another, then the product of their slopes is −1.
That is, if the slope of one line is m, then the slope of the perpendicular line
is − 1 m .
The specification of a straight line
Finally, we can state
Theorem 8.3. A straight line may be specified by giving its slope and the coördinates of one point on it. Equivalently: Through any point there is exactly one straight line with a given slope.
Let the straight line L have slope m, and let it pass through the point P. Then L is the only straight line through P with slope m.
For, it will be impossible to draw another straight line through P parallel to L, because parallel lines do not meet.
Therefore, any two lines through P will not be parallel, and therefore they will have different slopes. (Theorem 2)
Therefore, there is only one straight line through P with a given slope.
This means that a straight line may be specified by giving its slope and the coördinates of one point on it.
Which is what we wanted to prove.
*
On the basis of this theorem, we will be able to prove, in the following Topic, that the equation of a line may take the form y = ax + b, where a is the slope of the line, and b is the y-intercept.
Next Topic: Linear functions
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# Explain the Steps Necessary to Convert a Quadratic Function – Simplifying Standard Form to Vertex Form
Converting a quadratic function from standard form to vertex form is a process that involves some algebraic manipulation.
I see this process as something like a mathematical art form, where we take the general equation of a quadratic function in standard form, which is $y = ax^2 + bx + c$, and reshape it into vertex form, expressed as $y = a(x-h)^2 + k$.
This form is remarkable because it readily gives us the coordinates of the parabola’s vertex, namely (h, k), and provides insights into the graph’s properties, such as the direction of opening and the width of the parabola.
My approach begins with understanding what these forms represent. The standard form is useful for quickly identifying the y-intercept and is usually the form in which quadratic equations are presented.
On the other hand, the vertex form unveils the maximum or minimum point of the parabola – this is the vertex – and this directly influences the graph’s shape and location on the Cartesian plane.
Getting from one form to the other enhances my understanding of quadratic functions, allowing me to fully grasp their characteristics and graph them with precision.
## Understanding the Standard Form of a Quadratic Function
When I explore quadratic functions, I often start with the standard form. This particular form is expressed as:
$$y = ax^2 + bx + c$$
Here, each letter represents a specific value: a, b, and c. They are coefficients and constants that shape the curve of the graph, known as a parabola.
• a influences the parabola’s direction and width. If a is positive, the parabola opens upwards, and if negative, it opens downwards.
• b affects the parabola’s position relative to the y-axis.
• c represents the y-intercept, the point where the parabola crosses the y-axis.
A critical point in this graph is the axis of symmetry. This is a vertical line that passes through the vertex of the parabola and divides it into two mirror-image halves. The formula for the axis of symmetry in standard form is:
$$x = -\frac{b}{2a}$$
Understanding this form lays the groundwork for transforming the quadratic equation into the vertex form. Here’s a simple breakdown of the components:
ComponentDescriptionEffect on Graph
aLeading CoefficientDirection and Width
bLinear CoefficientLateral Position
cConstant TermVertical Position
My focus on the equation’s form is crucial, as it allows me to prepare for the process of converting it to the vertex form, which can provide more information about the parabola, such as the vertex and the maximum or minimum values of the function.
## The Process of Converting a Standard Quadratic Equation to Vertex Form
When I work with quadratic functions, it’s often useful to convert them from standard form to vertex form to easily identify the vertex and the shape of the parabola. The standard form is given by $y = ax^2 + bx + c$, while the vertex form is $y = a(x – h)^2 + k$, where ( (h, k) ) represents the vertex of the parabola.
The transformation involves completing the square. Here’s how I go about it:
1. Isolate the x-terms: If ( c ) is present, I move it to the other side of the equation. $y – c = ax^2 + bx$
2. Factor out the leading coefficient (if ( a \neq 1 )) from the x-terms: $y – c = a(x^2 + \frac{b}{a}x)$
3. Complete the square inside the parentheses:
• I take $\frac{b}{2a}$, square it, and add and subtract this value inside the parentheses. $y – c + a\left(\frac{b}{2a}\right)^2 = a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2\right) – a\left(\frac{b}{2a}\right)^2$
4. Rewrite the perfect square trinomial:
• The expression inside the parentheses now forms a perfect square. $y = a\left(x + \frac{b}{2a}\right)^2 + k$
5. Determine the new k-value:
• Finally, I adjust the ( k )-value by adding the ( c ) initially moved and subtracting the value used to complete the square, multiplied by ( a ). $k = c – a\left(\frac{b}{2a}\right)^2$
6. Write the equation in vertex form:
• The equation is now in vertex form where $h = -\frac{b}{2a}$ and ( k ) is as calculated above. $y = a\left(x – \frac{b}{2a}\right)^2 + k$
StepOperationPurpose
1Isolate the x-termsPrepares equation for completion
2Factor out leading coefficientSimplifies the square completion
3-4Complete the squareCreates a perfect square trinomial
5-6Find the vertex and rewrite the equationFinalizes conversion to vertex form
By following these steps, I can find the vertex and convert my quadratic equation into a more interpretable vertex form.
## Applying the Conversion to an Example
Let’s walk through an example to convert a quadratic polynomial from standard form to vertex form. Imagine we have the quadratic polynomial $y = 2x^2 + 4x – 6$.
This represents a parabola on the graph, and we need to find its turning point, which is another term for the vertex.
First, I’ll factor out the leading coefficient from the x terms:
$$y = 2(x^2 + 2x) – 6$$
Next, to complete the square, I’ll take half of the coefficient of x, square it, and add and subtract this value inside the parentheses:
$2\left(x^2 + 2x + \left(\frac{2}{2}\right)^2 – \left(\frac{2}{2}\right)^2\right) – 62\left(x^2 + 2x + 1 – 1\right) – 6$$This simplifies to: 2\left((x+1)^2 – 1\right) – 6$$ Finally, I’ll simplify by distributing the 2 and then combine like terms: $$2(x+1)^2 – 2 – 6$$ $$2(x+1)^2 – 8$$ So the equation of our parabola in vertex form is$y = 2(x+1)^2 – 8$, and from this, I can clearly interpret that the vertex, or turning point, of the parabola is (-1, -8). With the vertex form, it’s straightforward to graph the parabola and solve for other properties like the axis of symmetry and the parabola‘s direction. This method allows us to easily solve for the vertex and provides a clearer view of the quadratic polynomial in the context of its graph. ## Additional Considerations and Tools In my experience with quadratic functions, I’ve found a few extra considerations that can really aid understanding and facilitate the conversion process from standard form, given by the equation$y = ax^2 + bx + c$, to vertex form, which is$y = a(x – h)^2 + k$. When I deal with the coefficients$a$,$b$, and$c$, I’m careful with their values as they influence the orientation and width of the parabola. If$a$is positive, the parabola opens upwards, indicating a minimum point, which is useful when I’m looking for the lowest value of the function. Conversely, a negative$a$means that the parabola opens downwards, pointing to a maximum point. Understanding the roots—where the function crosses the x-axis—is essential. Through them, I recognize the parabola’s symmetric nature, which can help with visualizing the graph before I even start the conversion. I often use the vertex calculator tools to verify my work, as they provide quick checks on my conversion. Here’s a simple checklist that I use for students or anyone learning to convert these functions: • Identify the coefficients$a$,$b$, and$c$. • Verify the leading coefficient$a$for orientation. • Find the vertex$(h, k)$using$h = -\frac{b}{2a}$and$k = c – \frac{b^2}{4a}$. • Test the function by substituting points into both forms and checking for consistent outputs. Lastly, I always keep my notes handy! They’re a compilation of solved problems and missteps from past exercises. They serve as an excellent tool for comparison and a test for the mastery of the material. Here’s a quick reference of the conversion process I’ve outlined: StepActionNote 1Write down the standard form.$y = ax^2 + bx + c$2Calculate$h$and$k$.$h = -\frac{b}{2a}$,$k = c – \frac{b^2}{4a}$3Rewrite equation into vertex form.$y = a(x – h)^2 + k$I encourage consistency in practice as it solidifies the process in your mind. With the right tools and a clear understanding, converting quadratic functions becomes more intuitive. ## Conclusion In my experience with quadratic functions, transforming the standard form$y = ax^2 + bx + c$to vertex form$y = a(x-h)^2 + k$is a manageable process once you get the hang of it. I’ve shown you how to complete the square, which is essential in revealing the vertex form. Remember, you begin by identifying a, b, and c. You then factor out the leading coefficient a from the x terms if a ≠ 1. Next, you find the value to complete the square by dividing b by 2 and squaring it, which is$\left(\frac{b}{2}\right)^2$. You add and subtract this value within the bracket to maintain the equation’s integrity. Finally, you rewrite the equation by grouping the x terms and constant, which now includes the term you subtracted. This manipulation gives you the vertex$(h, k)\$, showing the function’s peak or trough. I always ensure that the equation is simplified to its cleanest form and clearly shows the vertex.
Through practice, this method becomes intuitive. My advice is to practice with various quadratic functions to recognize patterns and build confidence in the conversion process.
Dedication to understanding this method not only helps in sketching graphs quickly but also in analyzing the properties of quadratic functions more effectively. |
# 10th Maths Chapter 2. Numbers And Sequence Exercise 2.3
10th Standard Maths Chapter 2 Exercise 2.3 Numbers And Sequences Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here
### 1. Find the least positive value of x such that
• (i) 71 ≡ x (mod 8)
• (ii) 78 + x ≡ 3 (mod 5)
• (iii) 89 ≡ (x + 3) (mod 4)
• (iv) 96 = x7 (mod 5)
• (v) 5x ≡ 4 (mod 6)
Solution:
To find the least value of x such that
(i) 71 ≡ x (mod 8)
71 ≡ 7 (mod 8)
∴ x = 7.[ ∵ 71 – 7 = 64 which is divisible by 8]
(ii) 78 + x ≡ 3 (mod 5)
⇒ 78 + x – 3 = 5n for some integer n.
75 + x = 5 n
75 + x is a multiple of 5.
75 + 5 = 80. 80 is a multiple of 5.
Therefore, the least value of x must be 5.
(iii) 89 ≡ (x + 3) (mod 4)
89 – (x + 3) = 4n for some integer n.
86 – x = 4 n
86 – x is a multiple of 4.
∴ The least value of x must be 2 then
86 – 2 = 84.
84 is a multiple of 4.
∴ x value must be 2.
(iv) 96 ≡ x7 (mod 5)
96 – x7 = 5n for some integer n.
672−x7 = 5n
672 – x = 35n.
672 – x is a multiple of 35.
∴ The least value of x must be 7 i.e. 665 is a multiple of 35.
(v) 5x ≡ 4 (mod 6)
5x – 4 = 6M for some integer n.
5x = 6n + 4
x = 6n+45
When we put 1, 6, 11, … as n values in x = 6n+45 which is divisible by 5.
When n = 1, x = 105 = 2
When n = 6, x = 36+45 = 405 = 8 and so on.
∴ The solutions are 2, 8, 14…..
∴ Least value is 2.
### 2.If x is congruent to 13 modulo 17 then 7x – 3 is congruent to which number modulo 17?
Given x ≡ 13 (mod 17) ……(1)
7x – 3 ≡ a (mod 17) ……..(2)
From (1) we get
x- 13 = 17 n (n may be any integer)
x – 13 is a multiple of 17
∴ The least value of x = 30
From (2) we get
7(30) – 3 ≡ a(mod 17)
210 – 3 ≡ a(mod 17)
207 ≡ a (mod 17)
207 ≡ 3(mod 17)
∴ The value of a = 3
### 3. Solve 5x ≡ 4 (mod 6)
Solution:
5x ≡ 4 (mod 6)
5x – 4 = 6M for some integer n.
5x = 6n + 4
x = 6n+45 where n = 1, 6, 11,…..
∴ x = 2, 8, 14,…
### 4. Solve 3x – 2 ≡ 0 (mod 11)
Solution:
3x – 2 ≡ 0 (mod 11)
3x – 2 = 11 n for some integer n.
3x = 11n + 2
### 5. What is the time 100 hours after 7 a.m.?
Solution:
100 ≡ x (mod 12) (∵7 comes in every 12 hrs)
100 ≡ 4 (mod 12) (∵ Least value of x is 4)
∴ The time 100 hrs after 7 O’ clock is 7 + 4 = 11 O’ clock i.e. 11 a.m
### 6. What is time 15 hours before 11 p.m.?
15 ≡ x (mod 12)
15 ≡ 3 (mod 12)
The value of x must be 3.
The time 15 hours before 11 o’clock is (11 – 3) 8 pm
### 7. Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Solution:
No. of days in a week = 7 days.
45 ≡ x (mod 7)
45 – x = 7n
45 – x is a multiple of 7.
∴ Value of x must be 3.
∴ Three days after Tuesday is Friday. Uncle will come on Friday.
### 8. Prove that 2n + 6 × 9n is always divisible by 7 for any positive integer n.
9 = 2 (mod 7)
9n = 2n (mod 7) and 2n = 2n (mod 7)
2n + 6 × 9n = 2n (mod 7) + 6 [2n (mod 7)]
= 2n (mod 7) + 6 × 2n (mod 7)
7 × 2n (mod 7)
It is always divisible for any positive integer n
### 9. Find the remainder when 281 is divided by 17.
Solution:
281 ≡ x (mod 17)
240 × 240 × 241 ≡ x (mod 17)
(24)10 × (24)10 × 21 ≡ x (mod 17)
(16)10 × (16)10 × 2 ≡ x(mod 17)
(165)2 × (165)2 × 2
(165) ≡ 16 (mod 17)
(165)2 ≡ 162 (mod 17)
(165)2 ≡ 256 (mod 17)
≡ 1 (mod 17) [∵ 255 is divisible by 17]
(165)2 × (165)2 × 2 ≡ 1 × 1 × 2 (mod 17)
∴ 281 ≡ 2(mod 17)
∴ x = 2
### 10.The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The aeroplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and a half hours ahead to that of London’s time, then find the time in London, when will the flight lands at London Airport.
Solution:
The duration of the flight from Chennai to London is 11 hours.
Starting time at Chennai is 23.30 hrs. = 11.30 p.m.
Travelling time = 11.00 hrs. = 22.30 hrs = 10.30 a.m.
Chennai is 412 hrs ahead to London.
= 10.30 – 4.30 = 6.00
∴ At 6 a.m. on Monday the flight will reach at London Airport. |
# Difference between revisions of "2016 AMC 10B Problems/Problem 24"
## Problem
How many four-digit integers $abcd$, with $a \not\equiv 0$, have the property that the three two-digit integers $ab form an increasing arithmetic sequence? One such number is $4692$, where $a=4$, $b=6$, $c=9$, and $d=2$.
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20$
## Solution
The numbers are $10a+b, 10b+c,$ and $10c+d$. Note that only $d$ can be zero, and that $a\le b\le c$.
To form the sequence, we need $(10c+d)-(10b+c)=(10b+c)-(10a+b)$. This can be rearranged as $10(c-2b+a)=2c-b-d$. Notice that since the left-hand side is a multiple of $10$, the right-hand side can only be $0$ or $10$. (A value of $-10$ would contradict $a\le b\le c$.) Therefore we have two cases: $a+c-2b=1$ and $a+c-2b=0$.
### Case 1
If $c=9$, then $b+d=8,\ 2b-a=8$, so $5\le b\le 8$. This gives $2593, 4692, 6791, 8890$. If $c=8$, then $b+d=6,\ 2b-a=7$, so $4\le b\le 6$. This gives $1482, 3581, 5680$. If $c=7$, then $b+d=4,\ 2b-a=6$, so $b=4$, giving $2470$. There is no solution for $c=6$.
### Case 2
This means that the digits themselves are in arithmetic sequence. This gives $1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789$.
Counting the solutions, the answer is $\textbf{(D) }17$. |
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# Simple and Compound Interest
## Use A = P(1 + r)^t to solve for A.
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Practice Simple and Compound Interest
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Simple and Compound Interest
Suppose you are re-negotiating an allowance with your parents. Currently you are given $25 per week, but it is the first of June, and you have started mowing the lawn and taking out the trash every week, and you think your allowance should be increased. You father considers the situation and makes you the following offer: "I tell you what, son. I will give you three options for your allowance, you tell me which you would like" "Option A: You keep the$25 per week"
"Option B: You take $15 this week, then$16 next week, and so on. I'll continue adding $1 per week until New Year's." "Option C: I'll give you 1 penny this week, and then double your allowance each week until the first of October, then keep it at that rate." Which option would you choose? ### Watch This Embedded Video: ### Guidance Simple interest vs. Compound Interest Simple interest is interest which accrues based only on the principal of an investment or loan. The simple interest is calculated as a percent of the principal. Simple interest: $i = p \cdot r \cdot t$ . Variable i is interest, p represents the principal amount, r represents the interest rate, and t represents the amount of time the interest has been accruing. For example, say you borrow$2,000 from a family member, and you insist on repaying with interest. You agree to pay 5% interest, and to pay the money back in 3 years.
The interest you will owe will be 2000(0.05)(3) = $300. This means that when you repay your loan, you will pay$2300. Note that the interest you pay after 3 years is not 5% of the original loan, but 15%, as you paid 5% of $2000 each year for 3 years. Now let’s consider an example in which interest is compounded. Say that you invest$2000 in a bank account, and it earns 5% interest annually. How much is in the account after 3 years?
Compound interest: $A(t) = p \cdot (1 +r)^{t}$
Here, A ( t ) is the A mount in the account after a given t ime in years, p rincipal is the initial investment, and r ate is the interest rate. Note that we use $(1 + r)$ instead of just $r$ , so we can find the entire amount in the account, not just the interest paid.
$A(t) = 2000 \cdot (1.05)^{3}$
After three years, you will have $2315.25 in the account, which means that you will have earned$315.25 in interest.
Compounding results in more interest because the principal on which the interest is calculated increased each year. Another way to look at it is that compounding creates more interest because you are earning interest on interest, and not just on the principal.
#### Example A
Use the formula for compound interest to determine the amount of money in an investment after 20 years, if you invest $2000, and the interest rate is 5% compounded annually. Solution: The investment will be worth$5306.60
A(t) = P(1 + r) t
A(20) = 2000(1.05) 20
#### Example C
What is the value of an investment after 20 years, if you invest $2000, and the interest rate is 5% compounded continuously ? Solution The more often interest is compounded, the more it increases, but there is a limit. Each time you increase the number of compoundings, you decrease the fraction of the annual interest that is applied to each compounding. Eventually, the differences become so small as to be negligible. This is known as continuous compounding . The function A ( t ) = Pe rt is the formula we use to calculate the amount of money when interest is continuously compounded, rather than interest that is compounded at discrete intervals, such as monthly or quarterly. A ( t ) = Pe rt A (20) = 2000 e .05(20) A (20) = 2000 e 1 A (20) =$5436.56
Concept question wrap-up
"Which option would you choose?"
Assume you want to make the most money possible by the end of the year. Assume also that there are 24 weeks left.
Option A = $25 \cdot 24 = 600$ total
Option B = $15 + 16 + 17... + 39 = 609$ total
Option C (assuming 16 weeks until Oct.) = $1 \cdot (2^{16}) = 655.36$ each week after Oct 1.
It is entirely possible that dear old dad didn't take exponential growth seriously enough, he may need a second job!
### Vocabulary
Simple interest is interest earned only on the initial value or principal.
Compound interest refers to interest earned on the total account at the time it is compounded, including previously earned interest.
Principal is the initial sum, before any interest is added.
Rate is the percentage at which interest accrues.
Continuous compounding refers to a loan or investment with interest that is compounded constantly, rather than on a specific schedule.
'Accrue ' means "increase in amount or value over time". If interest accrues on a bank account, you will have more money in your account. If interest accrues on a loan, you will owe more money to your lender.
### Guided Practice
1) Compare the values of the investments shown in the table. If everything else is held constant, how does the compounding influence the value of the investment?
Principal r n t
a. $4,000 .05 1 (annual) 8 b.$4,000 .05 4 (quarterly) 8
c. $4,000 .05 12 (monthly) 8 d.$4,000 .05 365 (daily) 8
e. $4,000 .05 8760 (hourly) 8 2) Determine the value of each investment. a. You invest$5000 in an account that gives 6% interest, compounded monthly. How much money do you have after 10 years?
b. You invest $10,000 in an account that gives 2.5% interest, compounded quarterly. How much money do you have after 10 years? 3) How long will it take$2000 to grow to $25,000 at a 5% interest rate? Answers 1) Use the compound interest formula. For this example, the n is the quantity that changes: $A(8) = 4000 \left (1 + \frac{.05} {n}\right )^{8n}$ Principal r n t A a.$4,000 .05 1 (annual) 8 $5909.82 b.$4,000 .05 4 (quarterly) 8 $5952.52 c.$4,000 .05 12 (monthly) 8 $5962.34 d.$4,000 .05 365 (daily) 8 $5967.14 e.$4,000 .05 8760 (hourly) 8 $5967.29 A graph of the function $f(x) = 4000 \left (1 + \frac{.05} {x}\right )^{8x}$ is shown below: The graph seems to indicate that the function has a horizontal asymptote at$6000. However, if we zoom in, we can see that the horizontal asymptote is closer to 5967.
What does this mean? This means that for the investment of $4000, at 5% interest, for 8 years, compounding more and more frequently will never result in more than about$5968.00.
2)
a. $5000, invested for 10 years at 6% interest, compounded monthly. $A(t) = P \left (1 + \frac{r} {n}\right )^{nt}$ $A(10) = 5000 \left (1 + \frac{.06} {12}\right )^{12\cdot 10}$ $A(10) = 5000 \left (1.005\right )^{120}$ $A(10) = \9096.98$ b.$10000, invested for 10 years at 2.5% interest, compounded quarterly.
Quarterly compounding means that interest is compounded four times per year. So in the equation, n = 4.
$A(t) = P \left (1 + \frac{r} {n}\right )^{nt}$
$A(10) = 6000 \left( 1 + \frac{.025} {4}\right )^{4 \cdot 10}$
$A(10) = 6000 (1.00625)^{40}$
$A(10) = \12,830.30$
In each example, the value of the investment after 10 years depends on three quantities: the principal of the investment, the number of compoundings per year, and the interest rate.
3) It will take about 50 years:
A ( t ) = Pe rt
25,000 = 2000 e .05(t)
12.5 = e .05(t) Divide both sides by 2000
ln 12.5 = ln e .05( t ) Take the ln of both sides
ln 12.5 = .05 t ln e Use the power property of logs
ln 12.5 = .05 t × 1 ln e = 1
ln 12.5 = 0.5 t Isolate t
$t = \frac{ln 12.5} {.05} \approx 50.5$
### Practice
1. What is the formula for figuring simple interest?
2. What is the formula for figuring compound interest?
3. If someone invested $4500.00, how much would they have earned after 4 years, at a simple interest rate of 2%? 4. Kyle opened up a savings account in July. He deposited$900.00. The bank pays a simple interest rate of 5% annually. What is Kyle's balance at the end of 4 years?
5. After having an account for 6 years, how much money does Roberta have in the account, if her original deposit was $11,000, and her bank's yearly simple interest rate is 8.4%? 6. Tom called his bank today to check on his savings account balance. he was surprised to find a balance of$6600, when he started the account with just $5000.00 8 years ago. Based on this data, what percentage rate has the bank been paying on the account? 7. Julie opened a 4% interest account with a bank that compounds the interest quarterly. If Julie were to deposit$3000.00 into the account at the beginning of the year, how much could she expect to have at the end of the year?
8. Susan has had a saving account for a few years now. The bank has been paying her simple interest at a rate of 5%. She has earned $45.00 on her initial deposit of$300.00. How long has she had the account?
9. What is the balance on a deposit of $818.00 earning 5% interest compounded semiannually for 5 years? 10. Karen made a decent investment. After 4 years she had$3250.00 in her account and expects to have $16,250, after another 4 years. Her savings account is a compounding interest account. How much was her original deposit? 11. What is the yearly simple interest rate that Ken earns, if after only three months he earned$16.00 on an initial $800.00 deposit? 12. Write an expression that correctly represents the balance on an account after 7 years, if the account was compounded yearly at a rate of 5%, with an initial balance of$1000.00
13. Caryl gives each of his three kids $3000.00 each, and they each use it to open up saving accounts at three different banks. Georgia, his oldest, is earning 3% annually at her bank. Kirk earns 7% annually at his bank. Lottie's bank is paying her an annual rate of 4%. At the end of 6 years show much will each of them have in their respective accounts. 14. Kathy receives an inheritance check for$3000.00 and decides to put it in a saving account so she can send her daughter to college when she gets older. After looking she finds an account that pays compounding interest annually at a rate of $14%. The balance on the account can be represented by a function, where x is the time in years. Write a function, and then use it to determine how much will be in the account at the end of 7 years. 15. Stan is late on his car payment. The finance company charges 3% interest per month it is late. His monthly payment is$300.00. What is the total amount he will owe if he pays the August first bill October first? (assuming he was able to make his September bill on - time)
Today, you get your first credit card. It charges 12.49% interest on all purchases and compounds that interest monthly. Within one day you max out the credit limit of $1,200.00. 1. If you pay the monthly accrued interest plus$50.00 towards the initial $1,200 amount every month, how much will you still owe at the end of the first 12 months? 2. How much will you have paid in total at the end of the year? You are preparing for retirement. You invest$10,000 for 5 years, in an account that compounds monthly at 12% per year. However, unless this money is in an IRA or other tax-free vehicle, with zero inflation, you also have an annual tax payment of 30% on the earned interest.
1. How much will you have in 5 years?
2. Now take into account that the money loses 3% spending value per year due to inflation, how much is what you have saved really worth at the end of the 5 years?
### Vocabulary Language: English
Accrue
Accrue
Accrue means "increase in amount or value over time." If interest accrues on a bank account, you will have more money in your account. If interest accrues on a loan, you will owe more money to your lender.
Compound interest
Compound interest
Compound interest refers to interest earned on the total amount at the time it is compounded, including previously earned interest.
Continuous compounding
Continuous compounding
Continuous compounding refers to a loan or investment with interest that is compounded constantly, rather than on a specific schedule. It is equivalent to infinitely many but infinitely small compounding periods.
Continuously compounding
Continuously compounding
Continuous compounding refers to a loan or investment with interest that is compounded constantly, rather than on a specific schedule. It is equivalent to infinitely many but infinitely small compounding periods.
Principal
Principal
The principal is the amount of the original loan or original deposit.
Rate
Rate
The rate is the percentage at which interest accrues. |
# SOLVING WORD PROBLEMS ON NUMBERS
Problem 1 :
Twice a number is 500 more than six times the number. What is the number?
Solution :
Let x be the number.
2x = 6x + 500
2x - 6x = 500
-4x = 500
x = -500/4
x = -125
So, the required number is -125.
Problem 2 :
Three-sevenths of a number is 24. Find the number.
Solution :
Let x be the number.
3/7 of x = 24
(3/7) x= 24
x = 24 (7/3)
x = 8 ⋅ 7
x = 56
So, the required number is 56.
Problem 3 :
What number increased by ¼ of itself is equal to 30?
Solution :
Let x be the required number.
x + (1/4) of x = 30
x + x/4 = 30
(4x + x)/4 = 30
5x / 4 = 30
5x = 120
x = 120/5
x = 24
So, the required number is 24.
Problem 4 :
Find a number such that ¼ of the number is 50 less than 2/3 of the number.
Solution :
Let x be the number,
1/4 of x = (2/3) of x - 50
x/4 = (2x/3) - 50
x/4 = (2x - 150)/3
Doing cross multiplication, we get
3x = 4(2x - 150)
3x = 8x - 600
3x - 8x = -600
-5x = -600
x = 600/5
x = 120
So, the required number is 120.
Problem 5 :
The denominator of a fraction exceeds the numerator of a fraction by 25. The value of the fraction is 3/8 . Find the fraction.
Solution :
Let x be the numerator.
denominator = x + 25
x/(x + 25) = 3/8
8x = 3(x + 25)
8x = 3x + 75
8x - 3x = 75
5x = 75
x = 75/5
x = 15
So, the required fraction is 15/40
Problem 6 :
If 6 times a number is decreased by 6, the result is the same as when 3 times the number is increased by 12. Find the number.
Solution :
Let x be the number,
6x - 6 = 3(x + 12)
6x - 6 = 3x + 36
6x - 3x = 36 + 6
3x = 42
x = 42/3
x = 14
So, the required number is 14.
Problem 7 :
Separate 84 into two parts such that one part will be 12 less than twice the other.
Solution :
Let x be the number. Then the other number 2x - 12
Sum of those two numbers = 84
x + 2x - 12 = 84
3x - 12 = 84
3x = 84 + 12
3x = 96
x = 96/3
x = 32
2x - 12 ==> 2(32) - 12 ==> 52
So, the required numbers are 32 and 52.
Problem 8 :
The difference between two numbers is 24. Find the numbers if their sum is 88.
Solution :
Let x and x + 24 be two numbers.
Sum of the numbers = 88
x + x + 24 = 88
2x = 88 - 24
2x = 64
x = 64/2
x = 32
x + 24 ==> 32 + 24 ==> 76
So, the required numbers are 32 and 76.
Problem 9 :
One number is 3 times another number. If 17 is added to each, the first resulting number is twice the second resulting number. Find the two numbers.
Solution :
Let x and 3x be the numbers.
x + 17 and 3x + 17
3x + 17 = 2(x + 17)
3x + 17 = 2x + 34
3x - 2x = 34 - 17
x = 17
The other number = 3x ==> 3(17) = 51
So, the two numbers are 17 and 51.
Problem 10 :
The larger of two numbers is 1 less than 3 times the smaller. If 3 times the larger is 5 more than 8 times the smaller, find the numbers.
Solution :
Let x be the smaller number
Larger number = 3x - 1
3(3x - 1) = 8x + 5
9x - 3 = 8x + 5
9x - 8x = 5 + 3
x = 8
the other number = 3x - 1==> 3(8) - 1 ==> 23
So, the required numbers are 8 and 23.
Problem 11 :
The second of three numbers is one less than the first. The third number is 5 less than twice the second. If the third number exceeds the first number by 12, find the three numbers.
Solution :
Let x be the first number.
Second number = x - 1
Third number = 2(x - 1) - 5
2x - 7 = x + 12
2x - x = 12 + 7
x = 19
x - 1 = 18
2(x - 1) - 5 = 31
So, the required numbers are 18, 19 and 31.
Problem 12 :
One number is 4 more than 5 times another number. If 6 is added to each, the first resulting number is three times the second resulting number. Find the two numbers.
Solution :
Let x be the other number
one number = 5x + 4
6 is added to each, then x + 6 and 5x + 4 + 6
5x + 10 = 3(x + 6)
5x + 10 = 3x + 18
5x - 3x = 18 - 10
2x = 8
x = 4
5x + 4 ==> 5(4) + 4
= 24
So, the required numbers are 4 and 24.
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# THE SEVEN BENCHMARK NUMBERS WHY IS IT A POWERFUL TOOL FOR PRIMARY GRADES?
Jun 1, 2023 | Red Deer
Before we talk about the seven benchmark numbers, ask your child (grade 5 or above) a simple fraction operation question like what is ½ + ½ (many kids would answer 2/4) or what is ½ of ½ (many kids would answer ‘zero’). If they don’t answer correctly, then it is highly likely their foundation of fractions is shaky. Encountering a student’s problem in fraction operation, many tutors will just teach the steps to solve this operation, without realizing that actually the root of the problem is the student’s basic understanding of fractions. This is because these tutors are not equipped with a comprehensive tool to properly assess a student’s math skills and pinpoint exactly where the problem started.
The Seven Benchmark Numbers
The seven benchmark numbers are: 0, 1, ½, 1/10, 10, 12 and 100. Using this approach is useful to make students – especially beginners – understand the interrelated, basic mathematical ideas. But why these seven numbers?
At Mathnasium, this simple seven number concept is comprehensively used to teach children critical attributes (universal properties) of the number 0, 1, 10, 12, 100, ½ and 1/10 because they can be used as models of all other numbers. This concept is a powerful tool to teach students – especially in primary grades – understand basic math concepts without overburdening them.
This approach enables us to teach the required basic mathematical knowledge in a simple and make-sense way; it is a core curriculum material for primary grades, and also the foundation of a focused remedial program for junior high, and high school.
Below is an example of how we use these benchmark numbers – in this case is to teach basic understanding of fractions.
Mathnasium Teaches for Conceptual Understanding of Fractions
Let this 4th grader show you an example of how benchmark numbers can simplify a math problem. Ignacio was asked to arrange these fractions from smallest to largest: 0, 1, ½, 5/8, 9/10, 2/5.
There are several ways to compare the values of fractions. And if you search ‘how to arrange fractions with different denominators’, most likely you will find that the first step is to make the denominators the same first. This is of course not incorrect, but for younger students, using a smarter way by using benchmark numbers can help solve this problem easily.
Instead of converting all denominators into a common name, or converting it to decimals, there is a quick and simple way to solve this. He measured these fractions against the benchmark numbers of 0, 1 and ½ .. and voilá! He was able to solve it in only a matter of seconds. (Click the picture or click here to watch).
Did you know that a fifth grader’s fraction knowledge predicts their algebra performance in high-school and overall math achievement? This 4th grader is right on track for grade 5 fraction readiness.
Ignacio demonstrates that he is expanding his number sense by growing a sense of fraction as quantity and a sense of the size of a fraction.
Understanding the size and relationship of fractions is necessary before having knowledge of solving fraction operations. Not doing this in a proper sequence could create a hard time for a student when they are in middle and high school. Remember, a fifth grader’s fraction knowledge predicts their algebra performance in high-school and overall math achievement.
Oh by the way, Ignacio can beat AI. Look at the pic below – taken from Microsoft Bing’s AI. Unlike Ignacio, it incorrectly says 5/8 is less than ½!
OK – now, to check your child’s algebra readiness, ask them: is 12/13 + 7/8 closest to 1, 2, 19 or 21?
Related Articles:
• WHOLES AND PARTS: THE BIRTH OF FRACTIONS |
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Linear Algebra
Linear Algebra. Linear algebra deals with vectors and matrices. Vectors and matrices represent collections of quantities. A single vector or matrix corresponds to many different numbers or variables. Since vectors and matrices differ by dimensionality, we shall begin with matrices.
Linear Algebra
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Presentation Transcript
1. Linear Algebra
2. Linear algebra deals with vectors and matrices. Vectors and matrices represent collections of quantities. A single vector or matrix corresponds to many different numbers or variables. Since vectors and matrices differ by dimensionality, we shall begin with matrices.
3. The collection of numbers or variables is treated as a single entity. Thus, we can perform operations on matrices and affect all of the elements within the resultant matrix.
4. One operation on matrices is scalar multiplication. When we multiply a matrix by a scalar we multiply every element in that matrix by that scalar. If then
5. Another operation on matrices is matrix addition. When we add two matrices we add them element-by-element. If then
6. Another operation on matrices is matrix multiplication. Multiplication is a little more complicated than addition. The product is taken by multiplying each row of the first by each column of the second. If then
7. Matrix multiplication is not commutative: Matrix multiplication is, howeverassociative:
8. In MATLAB, there are two types of matrix multiplication. The first is ordinary matrix multiplication as defined two slides ago. As an example,
9. In MATLAB, this matrix multiplication is performed using the * operator: >> A = [1 2; 0 1]; >> B = [2 0; 1 0]; >> C = A*B C = 4 0 1 0 (The ; within brackets separates rows. The ; at the end of statements suppresses outputs.)
10. MATLAB has another type of matrix multiplication using the .* operator: The .* operator performs element-by-element multiplication.
11. If we multiply the same two matrices as in the previous example using the MATLAB .* operator, we have >> A = [1 2; 0 1]; >> B = [2 0; 1 0]; >> C = C = A.*B C = 2 0 0 0
12. When we multiply two matrices, A and B, each row of A is multiplied by each column of B. The number of rows in the product is equal to the number of rows in A. The number of columns in the product is equal to the number of columns in B. Each row of A must have the same number of elements as each column of B. Each row times column multiplication is like a dot product.
13. Examples: Let A is 2x2 and B is 2x1 C is 2x1 A is 1x2 and B is 2x1 C is 1x1 A is 3x2 and B is 2x1 C is 3x1 A is 2x1 and B is 2x1 C is not defined: each row of A does not have enough elements for each column of B
14. An extension of matrix multiplication is matrix exponentiation. Suppose we had a matrix A. Matrix exponentiation is repeated multiplication. Since matrix multiplication is associative and all of the matrices are the same in this case, it does not matter the order in which the multiplication is performed.
15. Certain types of matrices are relatively easy to raise to a power. For example, suppose that the matrix is diagonal, in other words the matrix only has components along the diagonal:
16. It can easily be shown that
17. Another type of matrix that is relatively easy to raise to a power is one in which there are only components in an off-diagonal, e.g., Such a matrix is called a nillpotent matrix.
18. Let us see what happens when we square this matrix:
19. Let us see what happens when we cube the matrix:
20. Higher powers yield
21. It can be shown that a triagonal matrix raised to a power is still triagonal.
22. Suppose we wished to raise e (base of natural logarithms) to a matrix power: This exponentiation can be accomplished using a Taylor series:
23. If D is a diagonal matrix, then where all of the terms are diagonal and the result is diagonal.
24. So, if then
25. If T is a triagonal matrix, then where all of the terms are triagonal and the result is triagonal. However, figuring the value of eT is not as easy as figuring the value of eD.
26. A triagonal matrix can be expressed as a sum of a diagonal matrix and a nillpotent matrix.
27. The terms in. will eventually go to zero. This last result is true for matrices only when all of the diagonal elements inDare the same.
28. Example: Find eT where Solution:
29. So,
30. In MATLAB, the matrix exponentiation function is expm(). As an example, >> T = [2 1 1; 0 2 1; 0 0 2] >> T = 2 1 1 0 2 1 0 0 2 >> expm(T) ans = 7.3891 7.3891 11.0836 0 7.3891 7.3891 0 0 7.3891
31. Another operation on matrices is something like matrix division. If then A-1is called the matrix inverse of A.
32. Suppose we replace C with an identity matrixI. The identity matrix has ones along the diagonal and zeroes elsewhere We then have
33. We can transform into by performing row operations on A. Row operations are performed by multiplying the matrix by near-identity matrices. Examples are shown on the following slide.
34. So
35. So
36. The matrix inverse function in MATLAB isinv(). >> syms a b c d >> A = [a b; c d]; >> B = inv(A) B = [ d/(a*d-b*c), -b/(a*d-b*c)] [ -c/(a*d-b*c), a/(a*d-b*c)]
37. The term ad-bc appears often and is referred to as the determinant of the (2x2) matrix. In MATLAB the determinant function is det().
38. Higher-order matrix inverses can be found by the method of cofactors: each element in the matrix is replaced by its cofactor. A cofactor of an element is the determinant of the matrix consisting of the other rows and columns. As an example, consider the (3x3) matrix The cofactor for the upper-left element (5) is
39. The determinant of a higher-order matrix by multiplying any row or column by its cofactors [Odd elements are multiplied by (-1).] The inverse of a higher-order matrix is found in a series of steps.
40. Let us replace each element by its cofactor. We then multiply each odd element by –1:
41. We then take the matrix transpose: Finally, we divide the matrix by its determinant:
42. Exercise: Find the determinants and inverses of the following matrices. Check your work using MATLAB.
43. Matrices and Simultaneous Equations Suppose we had a set of equations in two independent variables x1 and x2.
More Related |
# FINDING EQUATION OF A LINE GIVEN TWO POINTS
Finding the Equation of a Line Given Two Points :
Here we are going to see some example problems on determining the equation of the line from two points.
The equation of a straight line is satisfied by the co-ordinates of every point lying on the straight line and not by any other point outside the straight line.
Equation of a line using two points on the line :
(y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁)
Here (x₁,y₁) and (x₂,y₂) are the points on the line.
## Finding the Equation of a Line Given Two Points
Question 7 :
Find the equation of the line which is passing through the points (2, -8) and (-5, 2).
Solution :
Here (x₁, y₁) = (2,-8) and (x₂, y₂) = (-5,2)
Equation of a line :
(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)
(y - (-8))/(2 - (-8)) = (x - 2)/(-5 - 2)
(y + 8)/(2 + 8) = (x - 2)/(-7)
(y + 8)/10 = (x - 2)/(-7)
-7(y + 8) = 10(x - 2)
-7y - 56 = 10 x - 20
10 x + 7y - 20 + 56 = 0
10 x + 7y + 36 = 0
Question 8 :
Find the equation of the line which is passing through the points (3,0) and (4,-1).
Solution :
Here (x₁,y₁) = (3,0) and (x₂,y₂) = (4,-1)
Equation of a line :
(y - 0)/(-1 - 0) = (x - 3)/(4 - 3)
y/(-1) = (x - 3)/1
1(y) = -1(x - 3)
y = - x + 3
x + y - 3 = 0
Question 9 :
Find the equation of the line which is passing through the points (-1,-5) and (-3,-1).
Solution :
Here (x₁,y₁) = (-1,-5) and (x₂,y₂) = (-3,-1)
Equation of a line :
(y - (-5))/(-1 - (-5)) = (x - (-1))/(-3 - (-1))
(y + 5)/(-6) = (x + 1)/(-3 + 1)
(y + 5)/(-6) = (x + 1)/(-2)
-2(y + 5) = -6(x + 1)
-2y - 10 = -6 x - 6
6 x - 2 y - 10 + 6 = 0
6 x - 2 y - 4 = 0
÷ by 2 => 3 x - y - 2 = 0
Question 10 :
Find the equation of the line which is passing through the points (0,-5) and (4,-6).
Here (x₁,y₁) = (0,-5) and (x₂,y₂) = (4,-6)
Equation of a line :
(y - (-5))/(-6 - (-5)) = (x - 0)/(4 - 0)
(y + 5)/(-6 + 5) = x/4
(y + 5)/(-1) = x/4
4(y + 5) = -1(x)
4y + 20 = - 1 x
x + 4y + 20 = 0
Question 11 :
Find the equation of the line which is passing through the points (4, -3) and (0, -2).
Here (x₁, y₁) = (4, -3) and (x₂, y₂) = (0, -2)
Equation of a line :
(y - (-3))/(-2 - (-3)) = (x - 4)/(0-4)
(y + 3)/(-2 + 3) = (x - 4)/(-4)
(y + 3)/1 = (x - 4)/(-4)
-4(y + 3) = 1(x - 4)
-4y - 12 = 1 x - 4
x + 4y + 12 - 4 = 0
x + 4y + 8 = 0
Question 12 :
Find the equation of the line which is passing through the points (2, -1) and (-1, -7).
Solution :
Here (x₁,y₁) = (2,-1) and (x₂,y₂) = (-1,-7)
Equation of a line :
(y - (-1))/(-7 - (-1)) = (x - 2)/(-1 - 2)
(y + 1)/(-7 + 1) = (x - 2)/(-3)
(y + 1)/(-6) = (x - 2)/(-3)
-3(y + 1) = -6(x - 2)
-3y - 3 = -6 x + 12
6 x - 3 y - 3 - 12 = 0
6 x - 3 y - 15 = 0
÷ by 3 => 2 x - y - 5 = 0
After having gone through the stuff given above, we hope that the students would have understood, finding the equation of a line given two points.
Apart from the stuff given in this sectionif you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
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Remainder when 2 power 256 is divided by 17
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Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Can someone please help me out with this? i am trying to solve: 54= l^2- l3
Jun 3, 2017
The solutions are $S = \left\{- 6 , 9\right\}$
#### Explanation:
$54 = {l}^{2} - l 3$
We rewrite it as
${l}^{2} - 3 l - 54 = 0$
We solve this by factorising
$\left(l + 6\right) \left(l - 9\right) = 0$
The solutions are
$l + 6 = 0$, $\implies$, $l = - 6$
and
$l - 9 = 0$, $\implies$, $l = 9$
Jun 3, 2017
$l = 9 , \mathmr{and} , l = - 6.$
#### Explanation:
${l}^{2} - 3 l = 54$
$\Rightarrow {l}^{2} - 3 l - 54 = 0.$
Now, $9 \times 6 = 54 , \mathmr{and} , 9 - 6 = 3.$
So, we split $3$ as $9 - 6 ,$ and get,
${l}^{2} - \left(9 - 6\right) l - 54 = 0.$
$\Rightarrow \underline{{l}^{2} - 9 l} + \underline{6 l - 54} = 0.$
$\Rightarrow l \left(l - 9\right) + 6 \left(l - 9\right) = 0.$
$\Rightarrow \left(l - 9\right) \left(l + 6\right) = 0.$
$\Rightarrow \left(l - 9\right) = 0 , \mathmr{and} , \left(l + 6\right) = 0.$
$\Rightarrow l = 9 , \mathmr{and} , l = - 6.$
Jun 3, 2017
$l = 9 , \mathmr{and} - 6$
#### Explanation:
There are 3 ways to solve this problem (by factoring, by completing the square and by using the quadratic formula). I'll use the quadratic formula since the other contributor solved it by factoring.
$54 = {l}^{2} - 3 l$
${l}^{2} - 3 l - 54 = 0$
$a = 1 , b = - 3 , c = - 54$
Substitute it into the quadratic formula which is this
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$\frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - \left(4\right) \left(1\right) \left(- 54\right)}}{\left(2\right) \left(1\right)}$
Simplify
$\frac{3 \pm \sqrt{225}}{2}$
$\frac{3 + 15}{2} , \frac{3 - 15}{2}$
$\frac{18}{2} , - \frac{12}{2}$
$l = 9 , \mathmr{and} - 6$ |
Factors
Factors of 150 | Prime Factorization of 150 | Factor Tree of 150
Written by Prerit Jain
Updated on: 12 Aug 2023
Factors of 150 | Prime Factorization of 150 | Factor Tree of 150
Factors of 150 are the numerals that divide the original number into equal parts. They are also whole numbers. Furthermore, factors of 150 will divide these numbers evenly. These factors can either be negative or positive. However, these factors can’t be decimals or fraction numbers.
When it comes to factors of 150, it has a total of 12 factors. They are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, and 150.
What Are the Factors of 150?
The factors of 150 are the numbers that divide the number (150) evenly without leaving any remainder. As 150 is a composite number, it means it has more than two factors, including 1 and the number itself. As it is a composite, these factors are also negative or positive. Here are the number 150 has both positive and negative factors.
What Are the Factors of 150 in Pairs?
A pair factor of 150 is a combination of two numbers, which when multiplied together gives the product of the original number (150). These pair factors are also written in positive and negative pairs as shown in the table below:
Positive Pair Factors of 150
The positive pair factors of 150 are (1, 150), (2, 75), (3, 50), (5, 30), (6, 25), and (10, 15).
Negative Pair Factors of 150
The negative pair factors of 150 are (-1, -150), (-2, -75), (-3, -50), (-5, -30), (-6, -25), and (-10, -15).
How To Find Factors of 150 Through Division Method?
Through this method, we will understand how to find factors of 150 through division in a simple way. Since 150 is a number that can be divided by 2, it means it is a factor of 150. That is how we perform the division method. Now, we will take every integer and divide it by 150, and the ones that leave no remainder are the factors of 150. Let’s see this method.
• 150 ÷ 1 = 150
• 150 ÷ 2 = 75
• 150 ÷ 3 = 25
• 150 ÷ 5 = 30
• 150 ÷ 6 = 25
• 150 ÷ 10 = 15
• 150 ÷ 15 = 10
• 150 ÷ 25 = 6
• 150 ÷ 30 = 5
• 150 ÷ 50 = 3
• 150 ÷ 75 = 2
• 150 ÷ 150 = 1
Hence, we can see that 1, 2, 3, 5, 6, 10, 15, 35, 30, 50, 70, and 150 are the factors of 150.
What Is the Prime Factorisation of 150?
Check if the given number is a prime number. If the given number is a prime number, then it has only two factors: 1 and 150. If the given number is a composite number, then it has more than two factors.
Since 150 is the composite number, we have more than two factors. Hence, we can use the prime factorization method to find the factors of 150.
The steps to find prime factors of 150 using the prime factorization method are given below:
• Step 1: Firstly, we will take the smallest prime factor to divide by 150, that is 2.
• 150 ÷ 2 = 75
• Step 2: Now, we will take another small prime number to divide 75 with as we cannot divide it with 2. So, we will take 3.
• 75 ÷ 3 = 25
• Step 3: As 25 can
• 25 ÷ 5 = 5
• 5 ÷ 5 = 1
The quotient is 1. Since 1 cannot be divided further, the prime factors of 150 are 2 X 3 X 5 X 5 or 2 X 3 X 52. We can also write it down as 2, 3, and 5, as both of them are prime numbers.
Image representation of the prime factorization of 150:
What is the Factor Tree of 150?
Students can also use the factor tree to find the prime factors of 150. In this method, we will place the number 150 at the top of the factor tree. Now, as part of the branches, we will write down any set of pair factors of the given number (150). Now, again, we will split the pair of factors into factors. This process will be continued until the branches become prime numbers. Now, circling all the prime numbers will give us the prime factors of 150.
• Step 1: Let’s place the number 150 at the top of the factor tree and write down 150 in pairs, that is, 10 and 15.
• Step 2: Since 10 is not a prime number, we will now divide 10 into 2 and 5.
• Step 3: Since 2 and 5 are prime numbers, we can no longer separate them.
• Step 4: As we can see, 15 is not a prime number, so we will split 15 into 3 and 5.
• Step 5: We can see that 3 and 5 are both prime numbers. The branches in the factor tree will now be all the prime numbers.
Hence, circling all the branches, 2 X 3 X 52 or 2 X 3 X 5 X 5= 150, which are 150 prime factors.
There are many ways in which we can write the factor tree of 150. Below is an image representation of the factor tree of 150
Solved Examples on Factors of 150
Below are some solved examples on factors of 150 for students.
What is the sum of all the factors of 150?
1, 2, 3, 5, 6, 10, 15, 25, 30, 75, and 150 are the 150 factors.
1 + 2 + 3 + 5 + 6 + 10 + 15 + 25 + 30 + 75 + 150 = 372
Hence, 372 is the sum of all 150 factors.
What’s the highest common factor between 150 and 120?
The factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75, and 150.
Factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120.
Hence, from the above, the common factors of 150 and 120 are 1, 2, 3, 5, 6, 10, 15, and 30.
The highest common factor between 150 and 120 is 30.
What’s the highest common factor between 150 and 250?
The factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75, and 150.
Factors of 250 are 1, 2, 5, 10, 25, 50, 125, and 250.
Hence, from the above, the common factors of 150 and 250 are 1, 2, 5, 10, and 25.
The highest common factor between 150 and 250 is 25.
Is 75 a factor of 150?
Yes, 75 is a factor of 150.
As 150 ÷ 75 = 2.
What is the largest factor of 150?
The largest factor of 150 is 150.
The second largest factor of 150 is 75.
What are the first 5 factors of 150?
The factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75, and 150.
The first 5 factors of 150 are 1, 2, 3, 5, and 6.
Is 125 a factor of 150?
No, 125 is not a factor of 150.
As 150 ÷ 125 = Leaves a reminder
Hence, 125 is not a factor of 150 since factors of a number (150) don’t leave a remainder after being divided.
FAQs on Factors of 150
What are the factors and prime factors of 150?
The 150 factors are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75, and 150.
The 150 prime factors are: 2 X 3 X 52 or 2 X 3 X 5 X 5.
How many 150 factors are odd?
A total of 6 odd numbers are there in factors of 150 and they are 1, 3, 5, 15, 25, and 75.
What are the factors of -150?
The factors of -150 are -1, -2, -3, -5, -6, -10, -15, -25, -30, -75, and -150.
How many pair factors are there in 150?
A total of 6 pairs are there in factors of 150.
What are the composite factors of 150?
6, 10, 15, 25, 30, 75, and 150 are the 150 composite factors. Composite factors are those that have more than 2 factors other than 1 and itself.
What are the factors of 150 in pairs?
(1, 150), (2, 75), (3, 50), (5, 30), (6, 25), and (10, 15) are the 150 factors of pairs.
What are the positive factors for 150?
1, 2, 3, 4, 6, 8,1, 2, 3, 5, 6, 10, 15, 25, 30, 75, and 150 are the positive factors for 150.
Now that you have all the necessary information regarding the factors of 150, We hope this article has been useful for you and easy to understand as well.
Calculate Factors of
The Factors are
https://wiingy.com/learn/math/factors-of-150/
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Prerit Jain
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