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# Sets: Definition, Elements and Notation of Sets
In day-to-day life, we often discuss the collection of objects of a specific kind, like, collection of coins, collection of books, players of a cricket team, etc.
So, can we say that this is a collection of good cricket players or a set of good cricket players? The answer will vary from person to person. Hence this is not a well defined collection.
Similarly, we come across some collections in mathematics as well. For example: collection of irrational numbers, lines, composite numbers, etc. Let’s understand when we can call a collection of things a set.
• Definition of Set
• Elements of a set
• Notation of Sets
• Cardinality of a Set
• Practice Problems
• FAQs
## Definition of Set
A set is a well-defined collection of elements or objects. A set is always denoted by a capital letter. Some examples of sets:
• R: Set of all real numbers
• N: Set of all natural numbers
• The collection of all states of India
Note: A collection of good students is not a set as the term “good” is vague i.e. not well defined.
## Elements of a Set
The objects in a set are called its members/elements. Elements can be present in any order and in general, we won't repeat elements in a set.
If a is an element of set A then we write a A and is read as “a belongs to A”.
If a is not an element of set ‘A’ then we write a A and is read as “a does not belong to A”.
## Notation of Sets
A set is represented within curly braces { }. There are two common ways of representing the set:
• Roster or Tabular Notation
In Roster form, all the elements of a set are listed, separated by commas and enclosed within braces.
Example: A= the set of even natural numbers will be represented in Roster/Tabular form as A = {2,4,6,8,....}
• Set Builder Notation
In the set-builder form, property or properties which are satisfied by all the members of the set are listed. We write, {x:x satisties property P}, which is read as “ set of all those such that (represented by | or :) , each x has property P
Example: A= The set of even natural numbers will be represented in set builder form as $A=\left\{x:x=2n,n\in N\right\}$
## Cardinality of a Set
The number of elements present in a given set is known as the cardinality or order of that set.The order/cardinality can be finite or infinite depending on the number of elements present in the set. The cardinality of a given set A is denoted by
Example: Let a set A= {1,2,3,4} then cardinality or order of set A is n(A) = 4.
## Practice Problems of Sets
Example : Write the set $setD=t|{t}^{3}=t,t\in R$ in the roster form.
Here, $setD=t|{t}^{3}=t,t\in R$
Now, $setD=t|{t}^{3}=t,t\in R$
Hence, the roster form is $D=\left\{-1,0,1\right\}$
Example : If Y={1,2,3,....10} and a represents any element of Y, write the set containing all the elements satisfying the condition, . Also state the cardinality of the obtained set.
Here,Y={1,2,3,....10} and aY
Since
Hence, the required set is {4,5,6,7,8,9,10} & since the required set contains seven elements,
Therefore, n(A) = 7
Example : Let A be the set of all real numbers between 4 and 5, including 4 but not 5. Describe A in set builder notation, using the variable name as x.
In the set builder form, we could write
Example : Let $B=2m+5n|m,n\in N.$. Is 10∈ B? Is 13 ∈ B? Explain.
For different values of mand n, we can see B={7, 9, 11, 12,13, 14, 15, 16 ...}. From this list we can say that 10 ∉ B, as no natural numbers mand n, satisfies 2m+5n=10. However,
13 ∈ B, as for m=4and n=1, 2m+5n= 13
Example : Describe sets A and Bin roster form, where
A={n : nZ and n2≤4}and B={x : xR and x2-3x+2=0}
## FAQs of Sets
Question 1. Can the cardinality of a set be equal to zero?
Answer: If the set is empty i.e. it contains no elements then the cardinality of that set is zero.
Question 2. Can we represent sets using diagrams?
Question 3. Can small letters be used to represent a set?
Answer: A set is always represented by a capital letter while its elements are written in small letters
## NCERT Class 11 Maths Chapters
Sets Relations and Functions Triginometric Functions Mathematical Induction Numbers and Quadriatic Equations Linear Inequalities Premutations and Combinations Binomial Theorem Sequence and Series Straight Lines Conic Sections 3 D Geometry Limits and Derivatives Mathematical Reasoning Statistics Probability
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Cheyanne Leigh
2021-01-19
Polynomial equation with real coefficients that has roots
broliY
Given:) $-1,4-2i$
Used Formula:
$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
Calculation:
If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 4 — 2i must have another root as 4 + 2i.
Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial.
Hence, below mentioned equation is written as.
$\left[x-\left(-1\right)\right]\left[x-\left(4-2i\right)\right]\left[x-\left(4+2i\right)\right]=0$
$\left(x+1\right)\left(x-4+2i\right)\left(x-4-2i\right)=0$
Further use arithmetic rule.
$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
Here, $a=x-4,b=2i$
Now, the polynomial equation is,
$\left(x+1\right)\left[\left(x-4{\right)}^{2}-\left(2i{\right)}^{2}\right]=0$
Use arithmetic rule.
$\left(a-b{\right)}^{2}={a}^{2}-2ab+{b}^{2}$
And ${i}^{2}=-1$
Now, the polynomial equation is,
$\left(x+1\right)\left({x}^{2}-8x+16+4\right)=0$
$\left(x+1\right)\left({x}^{2}-8x+20\right)=0$
$\left({x}^{3}+{x}^{2}-8{x}^{2}-8x+20x+20\right)=0$
${x}^{3}-7{x}^{2}+12x+20=0$
Hence, the polynomial equation of giveb roots $-1,4-2iis{x}^{3}-7{x}^{2}+12x+20=0$
Do you have a similar question?
|
# Ex.1.2 Q6 Real Numbers Solution - NCERT Maths Class 10
Go back to 'Ex.1.2'
## Question
Explain why $$7 × 11 × 13 + 13$$ and $$7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$$ are composite numbers.
Video Solution
Real Numbers
Ex 1.2 | Question 6
## Text Solution
What is unknown?
Whether $$7{{ }} \times {{ }}11{{ }} \times {{ }}13{{ }} + {{ }}13$$ and $$7{{ }} \times {{ }}6{{ }} \times {{ }}5{{ }} \times {{ }}4{{ }} \times {{ }}3{{ }} \times {{ }}2{{ }} \times {{ }}1{{ }} + {{ }}5$$ are composite numbers?
Reasoning:
To solve this question, recall that:
• Prime numbers are whole numbers whose only factors are $$1$$ and itself.
• Composite number are the positive integers which has factors other than $$1$$ and itself.
Now, simplify $$7{{ }} \times {{ }}11{{ }} \times {{ }}13{{ }} + {{ }}13$$ and $$7{{ }} \times {{ }}6{{ }} \times {{ }}5{{ }} \times {{ }}4{{ }} \times {{ }}3{{ }} \times {{ }}2{{ }} \times {{ }}1{{ }} + {{ }}5.$$ On simplifying them, you will find that both the numbers have more than two factors. So, if the number has more than two factors, it will be composite.
Steps:
It can be observed that,
\begin{align} 7\times 11\times &13+13\,\\&=13\left( 7\times 11+1 \right) \\ \,\,&=13\left( 77+1 \right) \\ &=13\,\,\times \,78 \\ &=13\times 13\times 6\times 1 \\&=13\times13\times2\times3\times1\end{align}
The given number has $$2,3,13$$ and $$1$$ as its factors. Therefore, it is a composite number.
\begin{align}7\times &6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \\&= {5 \times \left( {7\!\times\! 6 \!\times\! 4 \!\times\! 3 \!\times\! 2 \!\times\! 1 \!+\!1} \right)}\\&= {5 \times \left( {1008 + 1} \right)}\\ &={ 5 \times 1009 \times 1}\end{align}
$$1009$$ cannot be factorised further. Therefore, the given expression has $$5,1009$$ and $$1$$ as its factors. Hence, it is a composite number.
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# Nonlinear Systems Warm Up Solve each quadratic equation
• Slides: 20
Nonlinear Systems Warm Up Solve each quadratic equation by factoring. Check your answer. 5, -2 -2 Find the number of real solutions of each equation using the discriminant. 3. 25 x 2 - 10 x + 1 = 0 one 1. x 2 - 3 x - 10 = 0 2. -3 x 2 - 12 x = 12 4. 2 x 2 + 7 x + 2 = 0 two 5. 3 x 2 + x + 2 = 0 none Holt Mc. Dougal Algebra 1
Nonlinear Systems Recall that a system of linear equations is a set of two or more linear equations. A solution of a system is an ordered pair that satisfies each equation in the system. Points where the graphs of the equations intersect represent solutions of the system. A nonlinear system of equations is a system in which at least one of the equations is nonlinear. For example, a system that contains one quadratic equation and one linear equation is a nonlinear system. Holt Mc. Dougal Algebra 1
Nonlinear Systems A system made up of a linear equation and a quadratic equation can have no solution, one solution, or two solutions, as shown below. Holt Mc. Dougal Algebra 1
Nonlinear Systems Example 1: Solving a Nonlinear System by Graphing Solve the system by graphing. Check your answer. y = x 2 + 4 x + 3 y=x+3 Holt Mc. Dougal Algebra 1
Nonlinear Systems Check It Out! Example 1 1. Solve the system by graphing. Check your answer. y = x 2 - 4 x + 5 y=x+1 Holt Mc. Dougal Algebra 1
Nonlinear Systems Remember! The substitution method is a good choice when either equation is solved for a variable, both equations are solved for the same variable, or a variable in either equation has a coefficient of 1 or -1. Holt Mc. Dougal Algebra 1
Nonlinear Systems Example 2: Solving a Nonlinear system by substitution. Solve the system by substitution. y = x 2 - x - 5 y = -3 x + 3 Holt Mc. Dougal Algebra 1
Nonlinear Systems Example 2: Continued The solutions are (4, 15) and (2, – 3). Holt Mc. Dougal Algebra 1
Nonlinear Systems Check It Out! Example 2 1. Solve the system by substitution. Check your answer. y = 3 x 2 - 3 x + 1 y = -3 x + 4 Holt Mc. Dougal Algebra 1
Nonlinear Systems Check It Out! Example 2 Continued The solutions are ( – 1, 7) and (1, 1). Holt Mc. Dougal Algebra 1
Nonlinear Systems Remember! The elimination method is a good choice when both equations have the same variable term with the same or opposite coefficients or when a variable term in one equation is a multiple of the corresponding variable term in the other equation. Holt Mc. Dougal Algebra 1
Nonlinear Systems Example 3 : Solving a Nonlinear System by Elimination. Solve each system by elimination. A 3 x - y = 1 y = x 2 + 4 x - 7 Holt Mc. Dougal Algebra 1
Nonlinear Systems Example 3 : Continued The solution is (– 3, – 10 ) and (2, 5). Holt Mc. Dougal Algebra 1
Nonlinear Systems Example 3 : Continued B y = 2 x 2 + x - 1 x - 2 y = 6 Holt Mc. Dougal Algebra 1
Nonlinear Systems Example 3 : Continued - 1 ± √– 63 x= 8 Holt Mc. Dougal Algebra 1 Since the discriminant is negative, there are no real solutions
Nonlinear Systems Check It Out! Example 3 1. Solve each system by elimination. Check your answers. . a 2 x - y = 2 y = x 2 - 5 Holt Mc. Dougal Algebra 1
Nonlinear Systems Check It Out! Example 3 Continued The solution is (3, 4) and (– 1, – 4). Holt Mc. Dougal Algebra 1
Nonlinear Systems Remember! The elimination method is a good choice when both equations have the same variable term with the same or opposite coefficients or when a variable term in one equation is a multiple of the corresponding variable term in the other equation. Holt Mc. Dougal Algebra 1
Nonlinear Systems Lesson Quiz: Part-1 Solve each system by the indicated method. 1. Graphing: 2. Substitution: Holt Mc. Dougal Algebra 1 y = x 2 - 4 x + 3 y=x-1 y = 2 x 2 - 9 x - 5 y = -3 x + 3 (1, 0), (4, 3) (-1, 6), (4, -9)
Nonlinear Systems Lesson Quiz: Part-2 3. Elimination: y = x 2 + 2 x - 3 x-y=5 no solution 4. Elimination: y = x 2 - 7 x + 10 2 x - y = 8 (3, -2), (6, 4) Holt Mc. Dougal Algebra 1
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# Cancel Fractions when Multiplying
In this worksheet, students will practise cancelling fractions in multiplications sums so that the numbers are lower and, therefore, easier to calculate with.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: Pearson Edexcel, OCR, Eduqas, AQA,
Curriculum topic: Number, Fractions, Decimals and Percentages
Curriculum subtopic: Structure and Calculation Fractions
Difficulty level:
#### Worksheet Overview
Simplifying is one of the most useful skills when working with fractions.
It can make the numbers in our fractions much easier to deal with.
When cancelling a fraction:
You need to find the highest or greatest common factor (HCF) of the two numbers and then divide both the numerator and denominator by this.
e.g. Cancel the fraction:
48 60
For this, we need to find the largest number you can divide both 48 and 60 by.
In this case, it's 12.
48 60
=
48 ÷ 12 60 ÷ 12
=
4 5
Cancelling when multiplying:
This situation is a bit harder to get your head around.
Without going into the reasons too deeply, we can cancel fractions before we multiply them by looking for common factors in the numbers.
This means we need to find a number we can divide two (of the four) numbers by, where one of the numbers is a denominator and one is a denominator.
Huh? Sounds tricky?
Let's look at a few examples to clarify.
e.g. Calculate:
3 4
x
5 9
You should notice here that there is 3 on the top of the left-hand fraction and a 9 on the bottom of the right-hand one.
This means we can divide both of these numbers by 3 and we will still get the same answer when we multiply the fractions.
31 4
x
5 = 1 x 5 = 5 93 4 3 12
e.g. Calculate:
1 2
x
4 5
You should notice here that there is a 2 on the bottom of the left-hand fraction and a 4 on the top of the right-hand one.
This means we can divide both of these by 2 and still get the same answer when we multiply the fractions.
1 21
x
42 = 1 x 2 = 2 5 1 5 5
Are these the only scenarios to be aware of?
There could be a combination of both the examples we have shown you.
Don't forget that you could also cancel a fraction if the numerator and the denominator have a common factor within the same fraction.
In this activity, we will practise applying the process shown above to cancel fractions in multiplications sums so that the numbers are lower and, therefore, easier to calculate with.
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# How to Graph Absolute Value Inequalities?
To graph an absolute value inequality, first, you need to solve the absolute value inequality.
The absolute value of a number describes its distance from zero. To graph absolute value inequality you must first solve the inequality and then represent absolute value inequalities on a number line.
## Step-by-step guide to graph absolute value inequalities
• Isolate the absolute value expression.
• Write the equivalent compound inequality.
• Find two values.
• Draw a number line and Place dots or open dots on the two points corresponding to the solutions. if the symbol is ($$≥$$ or $$≤$$) then you fill in the dot if the symbol is ($$>$$ or $$<$$) then you do not fill in the dot.
• Sketch a line and show all numbers the variable can be.
### Graphing Absolute Value Inequalities– Example 1:
Solve and graph $$|3x-3|<9$$.
Solution:
Use absolute properties: if $$|a|<b$$, $$b > 0$$ then: $$-b<a<b → -9 < 3x-3 < 9$$
Add $$3$$ to each side of inequality: $$-9+3 < 3x-3+3 < 9+3 → -6 < 3x < 12$$
Divide each side of inequality by $$3$$: $$-2 < x < 4$$
Graph it on a number line.
### Graphing Absolute Value Inequalities– Example 2:
Solve and graph $$|x+2| ≥ 3$$.
Solution:
Split into two inequalities: $$x+2 ≥ 3$$ or $$x+2 ≤ -3$$.
Subtract $$2$$ from each side of each inequality:
$$x+2-2 ≥ 3-2$$ → $$x ≥ 1$$
or
$$x+2-2 ≤ -3-2$$ → $$x ≤ -5$$
$$x ≥ 1$$ or $$x ≤ -5$$.
Graph the numbers that satisfy both conditions:
## Exercises for Graph Absolute Value Inequalities
### Solve and graph the following inequalities.
• $$\color{blue}{4|x| ≥ 12}$$
• $$\color{blue}{|x-2| < 1}$$
• $$\color{blue}{|x+2| ≥ 2}$$
• $$\color{blue}{5|x+2| ≥ 15}$$
• $$\color{blue}{4|x| ≥ 12}$$
• $$\color{blue}{|x-2| < 1}$$
• $$\color{blue}{|x+2| ≥ 2}$$
• $$\color{blue}{5|x+2| ≥ 15}$$
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# How do you divide (5x+15)/(12x-6) div(10x^2)/(18)?
Aug 30, 2015
$\frac{3}{2} \cdot \frac{x + 3}{x - 2} \cdot \frac{1}{x} ^ 2$
#### Explanation:
Your starting expression looks like this
$\frac{5 x + 15}{12 x - 6} \cdot \frac{18}{10 {x}^{2}}$
Factor the numerator and the denominator of the first fraction to get
$\frac{5 \left(x + 3\right)}{6 \left(x - 2\right)}$
The expression will now become
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} \left(x + 3\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} \left(x - 2\right)} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{18}}} 3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}} 2 {x}^{2}}$
$\frac{x + 3}{x - 2} \cdot \frac{3}{2 {x}^{2}} = \textcolor{g r e e n}{\frac{3}{2} \cdot \frac{x + 3}{x - 2} \cdot \frac{1}{x} ^ 2}$
|
# If $x^{2}+\frac{1}{x^{2}}=18$, find the values of $x+\frac{1}{x}$ and $x-\frac{1}{x}$.
Given:
$x^2 + \frac{1}{x^2} = 18$
To do:
We have to find the values of $x + \frac{1}{x}$ and $x - \frac{1}{x}$.
Solution:
The given expression is $x^2 + \frac{1}{x^2} = 18$. Here, we have to find the values of $x + \frac{1}{x}$ and $x - \frac{1}{x}$. So, by using the identities $(a+b)^2=a^2+2ab+b^2$...................(i) and $(a-b)^2=a^2-2ab+b^2$.............(ii), we can find the required values.
Now,
$x^2 + \frac{1}{x^2} = 18$
Adding $2$ on both sides, we get,
$x^2 + \frac{1}{x^2} + 2 = 18+2$
$x^2 + \frac{1}{x^2} + 2 \times x \times \frac{1}{x} = 20$ (Since $2\times x \times \frac{1}{x}=2$)
$(x+\frac{1}{x})^2=20$ [Using (i)]
Taking square root on both sides, we get,
$x+\frac{1}{x}=\sqrt{20}$
Now,
$x^2 + \frac{1}{x^2} = 18$
Subtracting $2$ on both sides, we get,
$x^2 + \frac{1}{x^2} - 2 = 18-2$
$x^2 + \frac{1}{x^2} - 2 \times x \times \frac{1}{x} = 16$ (Since $2\times x \times \frac{1}{x}=2$)
$(x-\frac{1}{x})^2=16$ [Using (ii)]
Taking square root on both sides, we get,
$x-\frac{1}{x}=\sqrt{16}$
$x-\frac{1}{x}=4$
Hence, the value of $x+\frac{1}{x}$ is $\sqrt{20}$ and the value of $x-\frac{1}{x}$ is $4$.
Updated on: 01-Apr-2023
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# Further Mathematical Methods (Linear Algebra)
## Solutions For Problem Sheet 8
In this sheet, we looked at orthogonal projections and how to analyse sets of data using least squares
fits. These latter questions were all fairly easy as they just involved number crunching.
1. Let X be the subspace of R3 spanned by the vectors [1, 2, 3]t and [1, 1, 1]t . We are asked to find
a matrix P such that Px is the orthogonal projection of x R3 onto X. To do this, we recall from
the lectures that:
## If A is an m n matrix of rank n, then the orthogonal projection onto R(A) is given by
P = A(At A)1 At .
and so if we can find a matrix A such that R(A) = X, then the matrix which we seek will be
P = A(At A)1 At .
## So, since R(A) = CS(A), a suitable matrix would be
1 1
A = 2 1 ,
3 1
as this has rank two (since the vectors [1, 2, 3]t and [1, 1, 1]t are linearly independent) and R(A) = X.
Thus, the desired orthogonal projection is given by
1 1
t 1 2 3 14 0 t 1 1 3 0
AA= 2 1 = = (A A) = ,
1 1 1 0 3 42 0 14
3 1
and so,
t 1 t 1 3 0 1 2 3 1 3 6 9
(A A) A = = .
42 0 14 1 1 1 42 14 14 14
Thus, the matrix
1 1 17 20 5
1 3 6 9 1
P = A(At A)1 At = 2 1 = 20 26 4 ,
42 14 14 14 42
3 1 5 4 41
## will give us the sought after orthogonal projection onto X.
Note: We can check that this matrix does represent an orthogonal projection since it is clearly
symmetric (i.e. Pt = P) and it is also idempotent as
17 20 5 17 20 5 17 20 5
1 1
P2 = PP = 2 20 26 4 20 26 4 = 20 26 4 = P.
42 42
5 4 41 5 4 41 5 4 41
Indeed, if you are really keen, you can check that this matrix represents an orthogonal projection
onto X by noting that for any vector x R3 we have
17 20 5 x1 17 20 5
42Px = 20 26 4 x2 = x1 20 + x2 26 + x3 4 ,
5 4 41 x3 5 4 41
and showing that the set of vectors Lin{[17, 20, 5]t , [20, 26, 4]t , [5, 4, 41]t } = X, as this will imply
that the vector Px is in X. (This can be done, for instance, by using these vectors to form the rows
1
of a matrix and then performing row operations on this matrix until its rows give you a set of vectors
that clearly spans X.)1
2. We are given a real n n matrix A which is orthogonally diagonalisable, i.e. there exists an
orthogonal matrix P such that
Pt AP = D,
where D is a diagonal matrix made up from the eigenvalues of A and we are told that all of the
eigenvalues and eigenvectors of this matrix are real. The orthogonal matrix P has column vectors
given by the orthonormal set of vectors S = {x1 , x2 , . . . , xn } where xi is the eigenvector corresponding
to the eigenvalue i , i.e. Axi = i xi . As such, we have PPt = I and so, writing this out in full we
have:
xt1
xt2
I= x
1 x 2 x
n .. ,
.
xtn
which can be multiplied out to give (see the aside at the end of this question),
I = x1 xt1 + x2 xt2 + + xn xtn .
(Notice, that this is a spectral decomposition of the identity matrix.)2 Then, multiplying both sides
of this expression by A, we get
A = Ax1 xt1 + Ax2 xt2 + + Axn xtn ,
which gives
A = 1 x1 xt1 + 2 x2 xt2 + + n xn xtn .
Thus, defining the n n matrix Ei to be such that Ei = xi xti we get
A = 1 E1 + 2 E2 + + n En ,
which is a spectral decomposition of A.
## To prove that the matrices Ei are such that
(
Ei if i = j
Ei Ej =
0 if i =
6 j
we note that, by definition,
Ei Ej = xi xti xj xtj = hxi , xj ixi xtj ,
where we have used our convention. Thus, as S is an orthonormal set, we have
(
1 if i = j
hxi , xj i =
0 if i 6= j
and so, (
1 xi xti if i = j
Ei Ej =
0 xi xtj if i =
6 j
which means that, (
Ei if i = j
Ei Ej =
0 if i =
6 j
where 0 is the n n zero matrix (as required).
To show that the matrix Ei represents an orthogonal projection we note that since:
1
But, life is too short.
2
The identity matrix has = 1 (multiplicity n) as its eigenvalues. Also, each of the xi will be an eigenvector of the
identity matrix corresponding to this eigenvalue since Ixi = 1 xi .
2
E2i = Ei Ei = Ei (from above), the matrix Ei is idempotent and as such it represents a projection.
Eti = (xi xti )t = xi xti = Ei , the matrix Ei is symmetric and so it represents an orthogonal
projection.
Thus, Ei represents an orthogonal projection (as required). Now, we know from the lectures that S
is a basis for Rn and so, any vector x Rn can be written as
n
X
x= j xj ,
j=1
which means that as Ei = xi xti , and the vectors in S are orthonormal, we have
n
X n
X n
X n
X
Ei x = j Ei xj = j (xi xti )xj = j xi (xti xj ) = j hxi , xj ixi = i xi ,
j=1 j=1 j=1 j=1
i.e. Ei orthogonally projects any vector in Rn onto Lin{xi } (as required). Consequently, we can see
that
Xn n
X
Ax = i Ei x = i i xi ,
i=1 i=1
and so the linear transformation represented by A takes the component of x in the direction of each
eigenvector (i.e. i xi for each eigenvector xi ) and multiplies it by a factor given by the corresponding
eigenvalue. (For example, if i > 1, then the vector i xi is stretched [or dilated] by a factor of i .)
So, basically, the spectral decomposition allows us to describe linear transformations in terms of the
sum of the components of a vector in the direction of each eigenvector scaled by the appropriate
eigenvalue. A simple illustration of this is given in Figure 1.
2 2 x2 Ax
x x
2 x2
x2
0 x1 1 x1 0 1 1 x1
Figure 1: This figure illustrates the results of Question 2 in R2 . Notice that the eigenvectors x1 and
x2 are orthogonal (i.e. perpendicular) and have the same [i.e. unit] length. In the left-hand diagram
we see that the vector x = 1 x1 + 2 x2 can be decomposed into components 1 x1 and 2 x2 in the
directions of the eigenvectors x1 and x2 respectively. Indeed, these two components are given by
E1 x and E2 x [respectively]. In the right-hand diagram, the two components have been multiplied
by the relevant eigenvalue (notice that 0 1 1 and 2 1) and the sum of these new vectors
gives us the vector Ax as expected from the theory above.
Aside: You may be surprised that the matrix product PPt can be multiplied out to give:
xt1
xt2
PP =
t
x1 x2 xn
.. = x1 xt1 + x2 xt2 + + xn xtn ,
.
t
xn
3
a fact which we have just used in both the lectures and the previous question. I will not justify this
in any general way, but I will show why it holds in the case where P is a 3 3 matrix. To see this,
suppose that the vectors x1 , x2 , x3 which constitute the columns of the matrix P are given by,
x11 x21 x31
x1 = x12 , x2 = x22 , and x3 = x32 .
x13 x23 x33
## That is, the matrix product PPt is given by,
x11 x21 x31 x11 x12 x13
t
PP = x12 x22 x32 x21 x22 x23 ,
x13 x23 x33 x31 x32 x33
## and multiplying these two matrices together yields,
P P P
P xi1 xi1 P xi1 xi2 P xi1 xi3
PPt = P xi2 xi1 P xi2 xi2 P xi2 xi3 ,
xi3 xi1 xi3 xi2 xi3 xi3
where all of these summations run from i = 1 to 3. However, as we can add matrices by adding their
corresponding elements, this is the same as
x11 x11 x11 x12 x11 x13 x21 x21 x21 x22 x21 x23 x31 x31 x31 x32 x31 x33
t
PP = x12 x11 x12 x12 x12 x13 + x22 x21 x22 x22 x22 x23 + x32 x31 x32 x32 x32 x33 ,
x13 x11 x13 x12 x13 x13 x23 x21 x23 x22 x23 x23 x33 x31 x33 x32 x33 x33
## which, you will notice, is just
x11 x21 x31
PPt = x12 x11 x12 x13 + x22 x21 x22 x23 + x32 x31 x32 x33 ,
x13 x23 x33
i.e. we have,
PPt = x1 xt1 + x2 xt2 + x3 xt3 ,
if P is a 3 3 matrix, as desired.
Note: In the solutions to Questions 3 to 7 we start by considering the system of equations which the
data would ideally satisfy. That is, if there were no errors in the data, then given a rule with certain
parameters, we would expect to get values for these parameters which were solutions to this system
of equations. However, as there are errors in the data, this system of equations will be inconsistent,
and hence we will be unable to solve them for the parameters. This is why we look for a least squares
fit! The values of the parameters that we find using this analysis are the ones that minimise the
least squares error between the rule and the data. (Notice that, in general, these parameters will not
satisfy any of the equations that the data would ideally satisfy.)3
3. The quantities x and y are related by a rule of the form y = ax + b for some constants a and b.
We are given some data, i.e.
x 1 2 3 4
y 5 3 2 1
3
Indeed, this raises an interesting question: Do we need to check that each system of equations is actually inconsistent
before we apply this method? The answer is, of course, no. In the unlikely event (unlikely because surely no-one
would set a least squares question that can be solved without using a least squares analysis) that the equations are
consistent (that is, in the case where the error terms are all zero), the least squares solution will still give the correct
answer. This is because, in such a situation, the least squares solution x = (At A)1 At b will give the unique solution
to the set of equations that the data satisfy. To see why, look at Question 8. (For a further illustration of this idea, see
the remarks following Questions 5 and 7.
4
and asked to find the least squares estimate of the parameters a and b in the rule above. To do this,
we note that ideally4 a and b would satisfy the system of equations
a+b=5
2a + b = 3
3a + b = 2
4a + b = 1
and writing them in matrix form, i.e. setting
1 1 5
2 1 a 3
A=
3 1 , x = b and b =
,
2
4 1 1
we let Ax = b, so that we can use the fact that x = (At A)1 At b gives a least squares solution to
this system. Then, as
1 1
1 2 3 4 2 1
= 30 10 1 4 10
At A = t 1
= (A A) = ,
1 1 1 1 3 1 10 4 20 10 30
4 1
and
5
1 2 3 4 3 1 1
At b = = 21
= x =
4 10 21
=
13
,
1 1 1 1 2 11 20 10 30 11 10 60
1
the required least squares estimates are a = 1.3 and b = 6.5
## 4. The quantities x and y are known to be related by a rule of the form
m
y= + c,
x
for some constants m and c. We are given some data, and as we need to fit this data to a curve
containing a 1/x term, it is convenient to supplement the table of data with an extra row, i.e.
x 1/5 1/4 1/3 1/2 1
1/x 5 4 3 2 1
y 4 3 2 2 1
Now, to find the least squares estimate of m and c in the rule above, we note that ideally6 these
parameters would satisfy the system of equations
5m + c = 4
4m + c = 3
3m + c = 2
2m + c = 2
m+c=1
4
As noted above, I say ideally as these equations are inconsistent. (Again, this is why we are looking for a least
squares solution!) We can see that they are inconsistent as subtracting the first two equations gives a = 2, and
hence b = 7. But this solution satisfies neither the third equation (as 6 + 7 = 1 6= 2), nor the fourth equation (as
8 + 7 = 1 6= 1).
5
Thus, putting these values into the rule, we see that
y = 1.3x + 6,
is the curve which minimises the least square error between the rule and the data. Notice that this line has a negative
gradient as one might expect from the data (i.e. as x increases, y decreases!).
6
As noted above, I say ideally as these equations are inconsistent. We can see that this is the case because the third
and fourth equations imply that m = 0, and hence c = 2, but this solution fails to satisfy any of the three remaining
equations.
5
So, writing them in matrix form, i.e. setting
5 1 4
4 1 3
A= 3 1 , x = m and b = 2 ,
c
2 1 2
1 1 1
and letting Ax = b, we can use the fact that x = (At A)1 At b gives a least squares solution to this
system. Then, as
5 1
4 1
5 4 3 2 1 1
t
AA= 3 1 = 55 15 t 1
= (A A) =
5 15
,
1 1 1 1 1 15 5 50 15 55
2 1
1 1
and
4
3
5 4 3 2 1 1 1
Ab= t 2 = 43
= x =
5 15 43
=
35
,
1 1 1 1 1 12 50 15 55 12 50 15
2
1
## the required least squares estimates are m = 0.7 and c = 0.3.7
We are also asked why it would be wrong to suppose that this was equivalent to the problem of
fitting a curve of the form
z = xy = cx + m,
through the data points (xy, x). The reason why this supposition would be wrong is that in the
problem which we solved, we effectively introduced an error term r which made the matrix equation
Ax = b consistent. That is, the system of equations represented by the matrix equation
Ax = b r,
would have solutions that corresponded to the values of the parameters for the data in question.
However, as we are ignorant of the values which the components of r take, we have to content
ourselves with finding the values of the parameters which minimise the least squares error given by
krk. Due to the way that this is set up, each component of the vector r is the vertical distance
between a data point and a curve given by the rule. Thus, when we find the curve that minimises
the least squares error, we are therefore just minimising the sum of the squares of these vertical
distances. Now, when we adopt the rule
z = xy = cx + m,
through the data points (xy, x), although this is algebraically the same, the errors that we introduce
will not be the vertical distances between the data points and the curve y = m x + c, but those between
the data points (z, x) and the curve z = cx + m. So, as this transformation of the data points alters
the distances that we are trying to minimise, it will alter the least squares fit that we find.
7
Thus, putting these values into the rule, we see that
7 1 3
y= + ,
10 x 10
is the curve which minimises the least square error between the rule and the data.
6
5. We are asked to find the least squares fit of the form y = m x + c through the data points
(x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ). To do this, we use the matrix approach developed in the lectures to
show that the parameters m and c are given by
P P P P P P P
n xi yi ( xi )( yi ) ( yi )( x2i ) ( xi )( xi yi )
m = P P and c = P P ,
n x2i ( xi )2 n x2i ( xi )2
where all summations in these formulae run from i = 1 to n. In this general case, the parameters
would ideally8 satisfy the system of equations
c + x1 m = y1
c + x2 m = y2
..
.
c + xn m = yn
## and writing them in matrix form, i.e. setting
1 x1 y1
1 x2
c y2
A = . . , x = and b = .. ,
.. .. m .
1 xn yn
and letting Ax = b, we can use the fact that x = (At A)1 At b gives a least squares solution to this
system. So, noting that all of the summations below will run from i = 1 to n, we get
1 x1
P
t 1 1 1 1 x2 n xi
AA= = P P 2
x1 x2 xn ... ... xi xi
1 xn
P 2 P
1
= (At A)1 = P P Pxi xi ,
n x2i ( xi )2 xi n
and
y1
P
t 1 1 1 y2 yi
Ab= .. = P
x1 x2 xn . xi yi
yn
P 2 P P
1 x xi y i
= x = P 2 P P i P
n xi ( xi )2 xi n xi yi
P P P P
c 1 x2iP( yiP) ( xi )P
( xi yi )
x = = P 2 P ,
m n xi ( xi )2 ( xi ) ( yi ) + n xi yi
## which gives the desired result.
Remark: As we have mentioned above, if the data was free from error, and as such, the data
points were all on the curve in question, then the above system of equations would be consistent.
Consequently, it would be easy to solve them for the parameters m and c. But, just for the sake of
completeness, we will now show that the above result will also give this answer. To do this, let us
8
As this is a least squares fit question we assume that the data points are such that these equations are inconsistent.
(But, also see the remark below.)
7
assume that we have solved the [now] consistent set of equations and found the parameters to be m
and c. This means that each data point (xi , yi ) would satisfy the equation yi = mxi + c, and as such,
our result becomes
P 2 P P P
1 xiP( {mx i + c}) ( xi )P
( xi {mxi + c})
x = P P P
n x2i ( xi )2 ( xi ) ( {mxi + c}) + n xi {mxi + c}
P P P 2 P P P P
1 m x2i ( xi ) + c xi ( 1) m ( xi ) x2i c ( xi )2
= P 2 P P P P P 2 P
n xi ( xi )2 m ( xi )2 c ( xi ) ( 1) + nm xi + nc xi
P P 2
1 cn P x2i c( xP
i) P
= P 2 P :Using the fact that 1=n
n xi ( xi )2 m( xi )2 + nm x2i
c
= x = ,
m
as expected.
Other Problems.
Here are the solutions to the other questions on the analysis of data sets using least squares fits that
you might have tried.
6. An input variable and a response variable y are related by a law of the form
y = a + b cos2 ,
where a and b are constants. The observation of y is subject to error, and we are asked to use the
following data to estimate a and b using the method of least squares. As we need to fit this data to
a curve containing a cos2 term, it is convenient to supplement the table of data with a couple of
rows, i.e.
0 /6 /4
/3 /2
cos 1 3/2 1/ 2 1/2 0
cos2 1.00 0.75 0.50 0.25 0.00
y 4.1 3.4 2.7 2.1 1.6
We now note that ideally,9 the parameters a and b would satisfy the system of equations
a + 1.00b = 4.1
a + 0.75b = 3.4
a + 0.50b = 2.7
a + 0.25b = 2.1
a + 0.00b = 1.6
## and writing them in matrix form, i.e. setting
1 1.00 4.1
1 0.75 3.4
a
A= 1
0.50 , x = and b =
2.7 ,
1 b
0.25 2.1
1 0.00 1.6
9
As noted above, I say ideally as these equations are inconsistent. We can see that this is the case because the last
equation implies that a = 1.6, and so the first equation gives b = 2.5, but this solution fails to satisfy any of the three
remaining equations.
8
and letting Ax = b, we can use the fact that x = (At A)1 At b gives a least squares solution to this
system. So, as
1 1.00
1 0.75
1 1 1 1 1 5 2.50
t
AA= 1 0.50 =
1.00 0.75 0.50 0.25 0.00 1 0.25 2.50 1.875
1 0.00
t 1 1 1.875 2.50 0.6 0.8
= (A A) = = ,
3.125 2.50 5 0.8 1.6
and
4.1
3.4
1 1 1 1 1 13.9
t
Ab= 2.7 =
1.00 0.75 0.50 0.25 0.00 8.525
2.1
1.6
0.6 0.8 13.9 1.52
= x = = ,
0.8 1.6 8.525 2.52
## the required least squares estimates are a = 1.52 and b = 2.52.10
7. We are asked to find the least squares solution to the system given by x1 = a1 , a2 , . . . , an . The
notation in this question is a bit misleading (and often leads to confusion), but essentially all we
have is a single variable x1 which is found to be equal to n constants a1 , a2 , . . . , an . So, ideally,11 we
would be able to find a value of x1 such that
x1 = a1
x1 = a2
..
.
x1 = an
## and writing these in matrix form, i.e. setting
1 a1
1 a2
A = . , x = [x1 ] and b = .. ,
.. .
1 an
and letting Ax = b, we can use the fact that x = (At A)1 At b gives a least squares solution to this
system. So, as
1
t
1
t 1 1
A A = 1 1 1 . = [n] = (A A) = ,
.
. n
1
10
Thus, putting these values into the rule, we see that
## y = 1.52 + 2.52 cos2 ,
is the curve which minimises the least square error between the rule and the data.
11
As this is a least squares fit question we assume that the data points are such that these equations are inconsistent,
i.e. at least two of the ai are distinct. (But, also see the remark below.)
9
and
a1 " n # " n #
a2
X 1X
At b = 1 1 1 .. = ai = x = ai ,
. n
i=1 i=1
an
the least squares solution is
n
1X
x1 = ai ,
n
i=1
## and this is just the average of the ai for 1 i n.12
Remark: If the ai were all equal, say a1 = a2 = = an = a, then this system of equations would
be consistent, and the solution would be [unsurprisingly] x1 = a. As before, our least squares result
can also be used to get this answer, because in this case
n
1X 1
x1 = a = na = a,
n n
i=1
as expected.
## If Ax = b is consistent, then every solution of At Ax = At b also solves the original matrix
equation.
So, we are given that the matrix equation Ax = b is consistent, i.e. there are vectors x which satisfy
it. Let us take y to be such a vector, i.e. Ay = b. To show that this result is true, we have to show
that if the vector x is a solution of the matrix equation
At Ax = At b,
then x is a solution of the original matrix equation, i.e. Ax = b too. This can be done by noting
that we can write
At Ax = At b as At (Ax b) = 0,
and as we have Ay = b for some vector y (since this matrix equation is assumed to be consistent)
we have
At A(x y) = 0,
i.e. the vector x y N (At A). However, we can see that N (At A) = N (A) since13
## ut At Au = 0 = (Au)t Au = 0 = hAu, Aui = 0,
using our convention. Thus, kAuk2 = 0 and so we have Au = 0, i.e. u N (A). Consequently,
N (At A) N (A).
For any u N (A), we have Au = 0 and so, At Au = 0 too. Consequently, N (A) N (At A).
12
Incidentally, if we are trying to fit a constant function to the data, this is the one which will minimise the least
square error. (Notice that the right-hand-side of this expression is a constant for any given set of data!)
13
Compare this with the proof that N (At ) = N (AAt ) given in the lectures. (This is in the proof of (A) = (At A) =
(AAt ) for any real matrix A.) Notice that substituting At for A in this result would also have given the result needed
for this question!
10
and so we have x y N (A). Thus, for some vector z N (A) we have x y = z, and so our
solution to the matrix equation
At Ax = At b,
is x = y + z for some y and z such that Ay = b and Az = 0 respectively. Consequently, we now
note that this vector x is also a solution to the original matrix equation since
Ax = A(y + z) = Ay + Az = b + 0 = b,
as required.14
This result is important in the context of the least squares analyses considered in this problem sheet
since it guarantees that: If the matrix equation Ax = b is consistent (i.e. there are no errors in the
data), then any solution to the matrix equation
At Ax = At b,
## will have to be the unique solution given by
x = (At A)1 At b,
(as the matrix At A is assumed to be invertible in such least squares analyses), and the result that
we have just proved guarantees that this will also be a solution of Ax = b. That is, if there are no
errors, then such a least squares analysis will still give the right solution as suggested in Footnote 3.
Note: Once we have established that x y N (A) in the proof of this result, the following steps
should be obvious since we know that the solution set of the matrix equation Ax = b is just the
affine set given by
{x | Ax = b} = {y + z | Ay = b and z N (A)},
i.e. the solution set is just the translate of N (A) by the vector y. (See Figure 2.)
## Solution set of Ax=b
N(A)
0 y
z y+z
Figure 2: Any vector that differs from a solution to the matrix equation Ax = b, say y, by a vector
z N (A), is also a solution to this matrix equation. (Notice that the solution set of the matrix
equation Ax = b is just the affine set given by the translate of N (A) by y.)
14
Notice that arguing that
At Ax = At b = Ax = b,
is not sufficient to establish this result since it assumes that the matrix At is invertible and this may not be the case.
Indeed, the matrix At may not even be square!
11
Remark: A simpler proof of the result in this question can be obtained by noting that since
At (Ax b) = 0,
## N (At ) = R(A) and so, we have Ax b R(A) .
The matrix equation Ax = b is assumed to be consistent and so b R(A). But, the vector
given by Ax is in the range of A as well. So, as R(A) is closed under vector addition (since it
is a subspace), we have Ax b R(A) too.
So, as the vector Ax b is in both R(A) and R(A) , we have Ax b R(A) R(A) . Consequently,
as we know that a subspace and its orthogonal complement can be used to form a direct sum, we
have
R(A) R(A) and hence, R(A) R(A) = {0},
i.e. it must be the case that Ax b = 0. Hence, any vector x satisfying the matrix equation
At (Ax b) = 0,
## will also satisfy the matrix equation Ax = b (as required).
Harder problems
Here are the solutions for the Harder Problems. As these were not covered in class the solutions will
be a bit more detailed.
9. Let X be a subspace of the vector space V and let P denote the orthogonal projection onto X.
We are asked to show that:
## where Q(Y ) is given by
Q(Y ) = {Qy | y Y },
(i.e. it is the set of vectors which is found by multiplying each of the vectors in Y by
Q.)15
So, to establish this result, we need to use the information given above to show that
## and we can do this by noting that:
Taking any vector z Lin(X Y ), we can write z = x + y where x X and y Y . But, since
Q = I P we can write
Qy = (I P)y = y Py = y = Qy + Py,
## where Py X as P the [orthogonal] projection onto X and Qy Q(Y ). Thus, we have
z = x + Py +Qy,
| {z }
in X
and so z Lin(X Q(Y )). Consequently, Lin(X Y ) Lin(X Q(Y )).
15
This notation was also used in Question 10 on Problem Sheet 7. (Note that in the solution to this problem, we
also established that sets like Q(Y ) were subspaces of V .)
12
Taking any vector z Lin(X Q(Y )), we can write z = x + v where x X and v Q(Y ).
But, v Q(Y ) means that there exists a y Y such that Qy = v, that is,
v = Qy = (I P)y = y Py.
But, Py X as P is the [orthogonal] projection of all vectors in V onto X. Thus, we have
z = x Py +y,
| {z }
in X
and so z Lin(X Y ) as y Y . Consequently, Lin(X Q(Y )) Lin(X Y ).
Hence, we can see that Lin(X Q(Y )) = Lin(X Y ) as required.16
To interpret this result geometrically in the case where X and Y are one-dimensional subspaces of
R3 look at Figure 2. Clearly, the beauty of this result is that it allows us to construct a subspace
Z
Qy
y
X 0
Py
Y
Q(Y)
Figure 3: This figure illustrates the result of Question 9 in R3 . Here X and Y are one-dimensional
subspaces of R3 and Z = Lin(X Y ) is the plane through the origin containing X and Y . So, by
the result above, Z = Lin(X Q(Y )) too where Q(Y ) is a subspace containing vectors that are
orthogonal to all of the vectors in X. (That is, Q(Y ) X .)
Q(Y ) Lin(X Y ) which only contains vectors that are orthogonal to every vector in X (i.e.
Q(Y ) X ) whilst allowing us to keep the same linear span.17
10. Let L and M be subspaces of the vector space V . We are asked to show that:
V = L M iff L M = {0} and Lin(L M ) = V.
(Recall that, by Theorem 2.4, if S is a set of vectors, then Lin(S) is the smallest subspace that
contains all of the vectors in S.) To do this, we use the result proved in the lectures, namely that:
V = L M iff L M = {0} and V = L + M,
i.e. we only need to show that Lin(L M ) = L + M .18 But, this is fairly obvious since if
{x1 , x2 , . . . , xk } and {y1 , y2 , . . . , yl } are bases for the subspaces L and M respectively, then
Lin(L M ) = Lin{z | z L or z M }
( k l
)
X X
= i xi + i yi | i and i are scalars
i=1 i=1
= {x + y | x L and y M }
Lin(L M ) = L + M
as required.
So, assuming that L and M are such that L M = V , we are now asked to prove that
16
Notice that this result actually holds for any projection and not just orthogonal ones. However, the nice geometric
interpretation of the result that follows does rely on P being an orthogonal projection.
17
That is, the space spanned by X Q(Y ) is the same as the space spanned by X Y .
18
Which is, incidentally, a result that we have taken to be intuitively obvious everywhere else!!
13
L M = {0}.
[Lin(L M )] = {0}.
## Firstly, to prove that
L M = {0},
we consider any vector z L M , i.e.
z L and z M .
## and so, we can see that
hz, x + yi = hz, xi + hz, yi = 0 + 0 = 0,
i.e. since V = L M , z is orthogonal to every vector x + y L + M = V . But, this can only be the
case if z = 0 and so,
L M = {0},
as required.
## Secondly, to prove that
[Lin(L M )] = {0},
we start by considering any vector z Lin(L M ), i.e. for any vectors u and v in L and M
respectively, we can write
z = u + v,
where and are any scalars. So, by the definition of orthogonal complement, it must be the case
that
hw, zi = 0,
for any vector w [Lin(L M )] . However,
## hw, zi = 0 = hw, u + vi = 0 = hw, ui + hw, vi = 0,
and so, as this must hold for any scalars and , this implies that
hw, ui = hw, vi = 0.
Thus, by the definition of orthogonal complement, it must be the case that w L and w M , i.e.
w L M . But, since V = L M , we know that L M = {0}, and so it must be the case that
w = 0. Consequently,
[Lin(L M )] = {0},
as required.
Hence, we are asked to deduce that L M = V , and to do this we use the result proved at the
beginning of the question. So, from the lectures, we note that the orthogonal complement of any
subset (and hence any subspace) of V is a subspace of V , and so we can see that
## But, as we have already established that
L M = {0},
14
to deduce the result we only need to establish that
Lin(L M ) = V.
## However, we know that
[Lin(L M )] = {0},
and so as S = S (if S is a subspace of V ), we have
## which means that
n o
Lin(L M ) = x hx, yi = 0 y {0} = {x | hx, 0i = 0}.
## Trivially, {x | hx, 0i = 0} V . (As we are working in this vector space.)
For any v V , we have hv, 0i = 0 and so v {x | hx, 0i = 0}. That is, V {x | hx, 0i = 0}.
Lin(L M ) = V,
as required.
15
|
# What are the factors 0f 25?
## What are the factors 0f 25?
So all factors of 25: 1, 5, and 25. We know that factors also include negative integers hence we can also have, list of negative factors of 25: -1, -5, and -25.
## What are the factors of the number 46?
Factors of 46
• Factors of 46: 1, 2, 23 and 46.
• Negative Factors of 46: -1, -2, -23 and -46.
• Prime Factors of 46: 2, 23.
• Prime Factorization of 46: 2 × 23 = 2 × 23.
• Sum of Factors of 46: 72.
What are the factors of 46 and 26?
For 26 and 46 those factors look like this:
• Factors for 26: 1, 2, 13, and 26.
• Factors for 46: 1, 2, 23, and 46.
### What are the common factors of 24 and 46?
What is the Greatest Common Factor?
• Factors for 24: 1, 2, 3, 4, 6, 8, 12, and 24.
• Factors for 46: 1, 2, 23, and 46.
### What is the biggest factor of 25?
Answer: The greatest factor of 25 are 5×5. The prime factor they have in common is 5. The greatest common factor is 5.
How do you find the factor of a number?
How to Find Factors of a Number?
1. Find all the numbers less than or equal to the given number.
2. Divide the given number by each of the numbers.
3. The divisors that give the remainder to be 0 are the factors of the number.
#### How many factors do 24 have?
8 factors
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Therefore, 24 has 8 factors.
#### What is the highest common factor of 46 and 26?
The GCF of 26 and 46 is 2.
What is the GCF of 33 and 77?
The GCF of 33 and 77 is 11.
## What is the GCF of 46?
To find the GCF of 46 and 69, we will find the prime factorization of the given numbers, i.e. 46 = 2 × 23; 69 = 3 × 23. ⇒ Since 23 is the only common prime factor of 46 and 69. Hence, GCF (46, 69) = 23.
## What is the LCM of 46 and 24?
The LCM of 24 and 46 is 552.
What are thefactors of 46?
Factors of 46 are all the integers that we can evenly divide 46 into specific numbers. As we know, any number divided by its factor will be equal to one of the other factors. Thus, when we divide 46 with its factor, it will be equal to another factor. We know that 46 is a composite number, that means it has more than two factors.
### What are all the factors of 25?
According to the definition of factors of 25, we know that factors of 25 are all the positive or negative integers which divide the number 25 completely. So let us simply divide the number 25 by every number which completely divides 25 in ascending order till 25. So all factors of 25: 1, 5, and 25. list of negative factors of 25: -1, -5, and -25.
### What are the positive and negative factor pairs of 46?
Such numbers are as follows: Positive factor pairs Negative factor pairs 1 × 46 = 46; (1, 46) (-1) × (-46) = 46 2 × 23 = 46; (2, 23) (-2) × (-23) = 46 23 × 2 = 46; (23, 2) (-23) × (-2) = 46 46 × 1 = 46; (46, 1) (-46) × (-1) = 46
How to find the prime factorization of 46?
By prime factorization method, we can write the prime factors of 46 as given below. The first step is to divide the number 46 with the smallest prime number, i.e. 2. Here, 2 and 23 are the prime numbers, so we cannot proceed with the division method. Thus, the prime factorisation of 46 can be written as 2 × 23.
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## How do you find the perpendicular height of a triangle?
Plug your values into the equation A=1/2bh and do the math. First multiply the base (b) by 1/2, then divide the area (A) by the product. The resulting value will be the height of your triangle!
## How do you find the angle of an oblique triangle?
First, if you know two angles and the side opposite one of them, then you can determine the side opposite the other one of them. For instance, if angle A = 30°, angle B = 45°, and side a = 16, then the law of sines says (sin 30°)/16 = (sin 45°)/b. Solving for b gives b = 16(sin 45°)/(sin 30°) = 22.6274.
## Why is it important to learn about oblique triangles?
Because of their big base, they are really important to maintain things and to support weight. This can be used in a lot of places from architecture, to any simple things anyone can do. If you know how to get advantage of trigonometry in life, you can learn how to know a lot of new things.
## What does it mean to solve an oblique triangle?
An oblique triangle is a triangle with no right angle. An oblique triangle is determined, meaning it can be solved, if a side and any two other parts are known. Three basic situations fulfill this simple requirement: when two angles and a side are given, two sides and an angle are given, or three sides are given.
## What are the different kinds of oblique triangles?
An oblique triangle does not have a right angle and can also be classified as an acute triangle or an obtuse triangle. The specialty of an oblique triangle is that it has all different angles and different lengths. To solve oblique triangles, use the laws of sine and cosine.
## What is oblique sides in maths?
Slanting. Not up-down or left-right. Angles that are not 0°, 90°, 180° or 270°
## What does an oblique line look like?
Oblique Lines Are Slanted Lines that are not parallel or perpendicular on the same plane are oblique.
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# Solving Two Step Equations Using Opposites
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This presentation shows how to solve two step linear equations using Opposite Operations and Reversing their order.
To obtain a PowerPoint format download of this presentation, go to the following page:
http://passyworldofmathematics.com/pwerpoints/
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### Solving Two Step Equations Using Opposites
1. 1. Image Source: http://www.esquire.com
2. 2. Step 1 - Work out the operations on the variable letterStep 2 - Put operations into BODMAS or PEMDAS orderStep 3 - Work out what the Opposite Operations areStep 4- Put Opposites into SAMDOB or SADMEP orderStep 5- Apply Opposites one by one to the equationStep 6 - Simplify the final answer
3. 3. There is a set order we must do Math Ops in:Brackets ( ) ParenthesisOther Things X2 ExponentsDivision / or MultiplicationMultiplication X or DivisionAddition + AdditionSubtraction Subtraction
4. 4. The Opposite operations we use for solving Equations are these ones.
5. 5. 2N + 5 = 111) Operations: + 5 and x 22) BODMAS or PEMDAS is x 2 then + 53) Opposite Operations are / 2 and – 54) SAMDOB or SADMEP order is -5 then /25) Apply Opposites one by one to both sides of the equation. (Shown on the next slide).
6. 6. From Step 4) SAMDOB or SADMEP order is - 5 then /2 2N + 5 = 11 -5 -5 2N = 6 2N = 6 2 2 N =3
7. 7. 2N + 5 = 11 Using Reversing Steps our answer was N = 3Substitute N = 3 into the original equation andcheck that it works. 2N + 5 = 11 (Sub N = 3) 2 x 3 + 5 = 11 6 + 5 = 11 11 = 11Left Side = Right Side N = 3 must be correct.
8. 8. Fill in the missing blanks in the steps below 3k - 2 = 101) Operations: - __ and x ___2) BODMAS or PEMDAS is x ___ then - ___3) Opposite Operations are / ___ and + ___4) SAMDOB or SADMEP order is + 2 then / 35) Apply Opposites one by one to both sides of the equation. (Shown on the next slide).
9. 9. From Step 4) SAMDOB or SADMEP order is + 2 then /3 3k - 2 = 10 +2 +2 3k = 12 3k = 12 3 3 k =4
10. 10. Fill in the missing blanks in the steps below n/5 + 2 = 61) Operations: __ __ and __ ___2) BODMAS or PEMDAS is __ ___ then __ ___3) Opposite Operations are x ___ and - ___4) SAMDOB or SADMEP order is - 2 then x 55) Apply Opposites one by one to both sides of the equation. (Shown on the next slide).
11. 11. From Step 4) SAMDOB or SADMEP order is - 2 then x 5 n/5 + 2 = 6 -2 -2 n/5 = 4 (Now multiply by 5) nx5 = 4x5 5 n = 20
12. 12. Fill in the missing blanks in the steps below 2(a-3) = 81) Operations: ( - __ ) and x ___2) BODMAS or PEMDAS is (- ___) then x ___3) Opposite Operations are (+ ___) and / ___4) SAMDOB or SADMEP order is /2 then (+ 3)5) Apply Opposites one by one to both sides of the equation. (Shown on the next slide).
13. 13. From Step 4) SAMDOB or SADMEP order is / 2 then (+3) 2(a-3) = 8 ( First divide by 2) 2(a-3) = 8 2 2 (a-3) = 4 (Now add 3 both sides) (a-3 + 3) = 4 + 3 a =7
14. 14. http://passyworldofmathematics.com/
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# Ogives (Cumulative Frequency Graphs)
Before viewing this page, it would be helpful to learn how to calculate the Median and Quartiles.
Cumulative Frequency is the progressive total of the frequencies. It assists us to find the median, quartiles and percentiles from large quantities of data organized into tables and graphs. To find the cumulative frequency, you add up the frequencies row by row.
An Ogive is a graph of the cumulative frequency. Interpreting quartiles, median and percentiles from a graph is more accurate than from a table.
## Example - Teenagers' CD Collections
The following table shows the number of music CDs owned by students in a Math class. Complete the table with two more columns for the cumulative frequency and cumulative percentage. Draw an ogive (a cumulative frequency graph). From the ogive, find the 1st quartile, median, 3rd quartile and 80th percentile.
NUMBER OF CDs NUMBER OF TEENAGERS 0 2 1 3 2 5 3 10 4 5
Answer:
NUMBER OF CDs NUMBER OF STUDENTS CUMULATIVE FREQUENCY CUMULATIVE PERCENTAGE 0 2 2 2⁄23 × 100 = 8.7% 1 3 5 5⁄23 × 100 = 21.7% 2 5 10 10⁄23 × 100 = 43.5% 3 10 20 20⁄23 × 100 = 87% 4 3 23 23⁄23 × 100 = 100% Σ f = 23
From Table:
n = 25
1st Quartile = (n + 1) ÷ 4 = (23 + 1) ÷ 4 = 6th student = 2 CDs Median = (n + 1) ÷ 2 = (23 + 1) ÷ 2 = 12th student = 3 CDs 3rd Quartile = (n + 1) ÷ 4 × 3 = (23 + 1) ÷ 4 × 3 = 18th student = 3 CDs 80th percentile = 80% of 25 students = 20th student = 3 CDs
From Graph (more accurate):
1st Quartile = (n + 1) ÷ 4 = (23 + 1) ÷ 4 = 6th student = 1.2 CDs Median = (n + 1) ÷ 2 = (23 + 1) ÷ 2 = 12th student = 2.2 CDs 3rd Quartile = (n + 1) ÷ 4 × 3 = (23 + 1) ÷ 4 × 3 = 18th student = 2.8 CDs 80th percentile = 80% of 25 students = 20th student = 3.0 CDs
## Did You Know That...?
Michael Jackson's "Thriller" album with sales of 110 million is the best-selling album of all time.
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# Why study math?/using manipulatives
Using manipulatives is a powerful and hands-on approach to teach elementary math concepts.[1] Manipulatives are physical objects that students can touch, move, and interact with to better understand abstract mathematical ideas. Here's a general guide on how to use manipulatives to teach various elementary math concepts:
## 1. Counting and Number Sense:
• Counters: Use small objects like buttons, coins, or cubes as counters to represent numbers. Have students group them to count and perform basic operations.
• Number Lines: Create a number line with markers or use a physical number line. Students can move along it to practice counting, addition, and subtraction.
• Base-10 Blocks: These blocks represent ones, tens, hundreds, etc. Teach addition and subtraction by physically grouping and regrouping these blocks.
• Number Bonds: Use circles or diagrams to show how numbers can be broken into parts. For example, for the number 7, you might have 3 and 4 in separate circles.
## 3. Multiplication and Division:
• Arrays: Use small objects (like beans or tiles) to create arrays for multiplication problems. For division, students can distribute objects into equal groups.
• Skip Counting: Use a number line or objects to help students understand skip counting. For example, skip counting by twos would involve physically moving two spaces at a time.
## 4. Fractions:
• Fraction Circles: Provide fraction circles or pie charts that students can manipulate to understand concepts like halves, thirds, and fourths.
• Fraction Bars: Use fraction bars to show how fractions can be added, subtracted, and compared by physically putting them together or taking them apart.
## 5. Geometry:
• Pattern Blocks: Pattern blocks, with various shapes like triangles, squares, and hexagons, are great for exploring geometric concepts. Students can use them to create patterns, explore symmetry, and understand area and perimeter.
• Geometric Solids: For 3D shapes, use physical geometric solids like cubes, spheres, and pyramids to help students visualize and identify shapes.
## 6. Measurement:
• Rulers and Measuring Cups: Use rulers to measure length and measuring cups for volume. Students can engage in activities like measuring classroom objects or ingredients for a recipe.
• Balance Scales: Explore weight and mass by using a balance scale with various objects to compare and order by weight.
## 7. Data and Probability:
• Graphs and Charts: Create simple graphs (bar graphs, pictographs) using objects like colored cubes or stickers to represent data. This helps students understand how to collect, organize, and interpret data.
• Probability Spinners: Make probability spinners for hands-on lessons on probability. Students can spin the spinner and record outcomes.
## 8. Problem Solving:
• Word Problems: Present word problems using manipulatives to help students visualize the situation and model the problem before attempting to solve it.
• Story Mats: Use story mats with visuals to represent math problems. Students can move objects on the mats to act out the problem.
## 9. Algebra:
• Variable Tiles: Introduce variables using physical tiles (e.g., letter tiles or blank squares) to represent unknown values in equations.
• Balance Scales: Use balance scales to visually demonstrate the concept of equality in equations. Add objects to both sides to maintain balance.
Remember to scaffold your lessons gradually, starting with concrete manipulatives and then transitioning to more abstract representations as students become comfortable with the concepts. Encourage discussion and exploration while using manipulatives to deepen understanding and reinforce math concepts effectively.
1. ChatGPT generated this text responding to the prompt: "Describe how to use manipulatives to teach elementary math concepts".
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# (2x^2 - 3x+6 ) / (2x+2) division
• roger
In summary, polynomial division involves finding the coefficients of a polynomial equation by dividing it by another polynomial. This is done by setting up an equation and finding the values of the coefficients through a step-by-step process. The divisor is then distributed over the dividend and the resulting equation is solved for the coefficients. This process is similar to dividing numbers in decimal form.
roger
I don't understand how this works :
(2x^2 - 3x+6 ) / (2x+2)
I don't really understand the case using numbers either but we never divide by a+b+c etc , (ie we just divide by a single number rather than 3+5 for example)
Please explain step by step how it works. Also, why isn't the 2x and +2 separately distributive over the numerator ?
Roger
What do you know about polynomial division...?My guess is:nothing.So how about do some reading ?Okay?
Daniel.
dextercioby said:
What do you know about polynomial division...?My guess is:nothing.So how about do some reading ?Okay?
Daniel.
Your guess is wrong.
But there is a difference between understanding and simply performing a task according to a given set of rules.
Can somebody explain how it works ?
As I said, I can get the answer, but I'm not quite sure what, I'm doing..
You're dividing two polyomials,you can't be doing anything else.
Daniel.
All right, roger:
Suppose you are to perform the divison 46/5.
What that means, is that you want to find a number "a", given in decimal form, so that 46=5*a
Let us set $$a=a_{1}*10+a_{0}*1+a_{-1}*\frac{1}{10}+a_{-2}*\frac{1}{100}+++$$
We must therefore determine coeficients $$a_{i}, i=1,0,-1,-2..$$ between 0 and 9, so that the equation holds:
$$46=a_{1}*50+a_{0}*5+a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++)$$
First, we see that the only valid choice for $$a_{1}$$ between 0 and 9 is $$a_{1}=0$$
Thus, we have gained:
$$46=a_{0}*5+a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++)$$
Now, it follows that we must choose $$a_{0}=9$$ that is, we get, by subtracting 9*5 from both sides:
$$1=a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++),a_{0}=9$$
We now see that by setting $$a_{-1}=2$$ we get, by subtracting $$2*\frac{5}{10}$$ from both sides:
$$0=a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++),a_{0}=9, a_{-1}=2$$
Thus, the equation is fulfilled by setting the rest of the coefficients equal to zero, i.e.
46/5=9.2
You should think in a similar manner about polynomial division:
We want to find to find a function F(x)=P(x)+R(x), so that
(2x^2 - 3x+6 ) = (2x+2)*F(x), and
where P(x) is a polynomial (R(x) is then the "rest")
Let us set $$P(x)=a_{1}x+a_{0}$$, where $$a_{1},a_{0}$$ are real numbers we want to find.
We have:
$$2x^{2}-3x+6=a_{1}2x^{2}+(2a_{1}+2a_{0})x+2a_{0}+(2x+2)*R(x)$$
Thus, we set $$a_{1}=1$$, and subtract $$2x^{2}$$ from both sides:
$$-3x+6=(2*1+2a_{0})x+2a_{0}+(2x+2)*R(x), a_{1}=1$$
Now, we set $$2+2a_{0}=-3$$, that is, $$a_{0}=-\frac{5}{2}$$ and, subtract -3x from both sides:
$$6=-\frac{5}{2}*2+(2x+2)*R(x), a_{1}=1,a_{0}=-\frac{5}{2}$$
from which we can determine R(x):
$$R(x)=\frac{11}{2x+2}$$
Thus, we have established:
$$\frac{2x^{2}-3x+6}{2x+2}=x-\frac{5}{2}+\frac{11}{2x+2}$$
Last edited:
## 1. What is the quotient when dividing (2x^2 - 3x+6 ) by (2x+2)?
The quotient when dividing (2x^2 - 3x+6 ) by (2x+2) is x-2. This can be determined by using long division or synthetic division.
## 2. Can the division of (2x^2 - 3x+6 ) by (2x+2) be simplified further?
Yes, the division of (2x^2 - 3x+6 ) by (2x+2) can be simplified to x-2. This is the simplest form of the quotient and cannot be further simplified.
## 3. How does the division of (2x^2 - 3x+6 ) by (2x+2) relate to the concept of remainder?
The division of (2x^2 - 3x+6 ) by (2x+2) does not have a remainder because the polynomial division algorithm is used. However, if the division was done using long division, the remainder would be zero.
## 4. Is it possible to divide (2x^2 - 3x+6 ) by (2x+2) when x = -1?
Yes, it is possible to divide (2x^2 - 3x+6 ) by (2x+2) when x = -1. This would result in a quotient of -1 and a remainder of 0.
## 5. How can the division of (2x^2 - 3x+6 ) by (2x+2) be used to solve real-life problems?
The division of (2x^2 - 3x+6 ) by (2x+2) can be used to solve real-life problems involving polynomial equations, such as finding the roots of a quadratic equation. It can also be used to simplify complex algebraic expressions and solve for unknown variables.
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# How do you solve the system 1/3x-y=3 and 2x+y=25 using substitution?
Oct 2, 2017
$x = 12$ and $y = 1$
#### Explanation:
Given equations are:
(1) ------ $\frac{1}{3} x - y = 3$
(2)------$2 x + y = 25$
Multiply (1) by 3 to eliminate the fractional part,
(1) ------- $x - 3 y = 9$
$\implies x = 9 + 3 y$ -------- let this be equation (3)
Now substitute this value of $x$ from (3) in equation (2),
(2) --------- $2 \left(9 + 3 y\right) + y = 25$
$\implies 18 + 6 y + y = 25$
$\implies 7 y = 25 - 18$
$\implies y = \frac{7}{7}$
Therefore,
$y = 1$
Substituting this value of $y$ in equation (3),
$x = 9 + 3 \setminus \times 1$
$x = 9 + 3$
Therefore,
$x = 12$
So we have the values of $x$ and $y$ as,
$x = 12$ and $y = 1$
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# Middle School Curriculum Update
## Humanities
In our weekly meetings with our partner group, the Fox Class, the partnered students wrote stories and then created stop animation videos of their stories. As a follow up to their Wishes for the World, the Middle School Students held two fundraisers, a pretzel sale and a holiday breakfast and plan to distribute the proceeds to several organizations.
## Math
Pre-algebra – We have been working on extending our knowledge of working with numbers to incorporate many different aspects into our work at once, so that operations become smooth and thoughtful. These include: order of operation, arithmetic properties, decimals, fractions, and operations with negative numbers. It is incumbent on the students to truly be able to “think” with a mathematical mindset in order to understand what a problem is really asking and to develop an efficient approach to solving it. For example, grasping what the problem -1 34 ÷ 12 is asking, and to be able to estimate the answer, involves a complex understanding of numbers and mathematical concepts. Additionally, we have continued to work on a variety of mathematical puzzles and conundrums as we seek to become increasingly comfortable playing with numbers and solving ongoing complex problems.
Math 6
Math 6 students began studying Chapter 3 in the CPM math program. In this chapter, students begin by focusing on multiple representations of portions, ratios, and equivalence. It then moves into work with integers and signed rational numbers. The first half of Chapter 3 focused on equivalent fractions and ratios, and introduces the concept of multiplying by the “Giant One”, i.e., multiplying both the numerator and denominator of a fraction by the same number to get an equivalent fraction. In this section, students will also examine the connections between fraction, decimal, and percent representations for portions of a whole. We looked extensively at correlating these three numerical forms. Here they look at how the meanings of percents, fractions, and decimals are related, how to represent a quantity in each of these forms, and how to move between these representations. They also connect these ideas to the concept of ratios. Ratio and proportion are an important focus of this course and this lays the groundwork to connect this big idea to portions and operations with rational numbers.
The second half of Chapter 3 began with having students looking at motion on a number line. Integer expressions are used to represent this motion, with motion to the left being represented by adding a negative number and motion to the right represented by adding a positive number. As they work in this section, students also create their own number lines on which to represent solutions, providing practice with setting intervals and scaling one-dimensional axes in preparation for work with coordinate graphs in Lesson 3.2.4. They work to locate positive and negative numbers on the number line, which leads to finding distance using absolute value. Lastly, they connect this idea of distance to finding the length of a horizontal or vertical line segment on a coordinate graph and plotting points in all 4 quadrants. Students should be familiar with graphing coordinate points in the first quadrant from their work in previous courses.
In addition to completing Chapter 3 in the CPM program, students continued to learn new properties of algebra: exponents, monomials, binomials, and how to apply the distributive property to monomials and binomial expansions, or FOILing. Finally, students regularly complete handouts in the MathCounts program, which organizes regional and national math competitions. Students are well-established in new, rigorous materials and making good progress in their understanding of pre-algebra concepts.
## Physical Science
Middle School Students covered a range of topics this month in Physical Science, After previously studying the relationship between potential and kinetic energy, we began to look at elastic potential energy, especially as demonstrated in springs. We learned about English mathematician Robert Hooke and his namesake equation, Hooke’s Law, which correlates displacement of a spring with force. We conducted several explorations of springs that studied the correlation between restoring force and displacement of the spring using various masses and springs of different spring constants. One of these included the construction of a miniature projectile launcher made with the springs from a pen. In our second unit this month, we again looked at the transfer of forms of energy, this time using a marble track. Students began with the challenge of designing a track that required the longest time for a marble to complete. This led them to focus on conserving potential energy and looking for ways to minimize kinetic energy. We returned to simple machines in our third unit and examined the many manifestations of the 6 simple machines - the screw, wedge, inclined plane, pulley, lever, and the wheel. Students used levers and inclined planes in the design challenge of launching an object as high as possible with materials found in the MakerSpace. We performed a similar challenge in our unit on catapults. Overall, students are enjoying seeing the fundamental concept of energy transfer in various guises and physical phenomena.
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# 5.4 Integration formulas and the net change theorem (Page 4/8)
Page 4 / 8
## Integrating an odd function
Evaluate the definite integral of the odd function $-5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x$ over the interval $\left[\text{−}\pi ,\pi \right].$
The graph is shown in [link] . We can see the symmetry about the origin by the positive area above the x -axis over $\left[\text{−}\pi ,0\right],$ and the negative area below the x -axis over $\left[0,\pi \right].$ We have
$\begin{array}{ll}{\int }_{\text{−}\pi }^{\pi }-5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}xdx\hfill & =-5\left(\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x\right){|}_{\text{−}\pi }^{\pi }\hfill \\ \\ \\ & =5\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x{|}_{\text{−}\pi }^{\pi }\hfill \\ & =\left[5\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\pi \right]-\left[5\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{−}\pi \right)\right]\hfill \\ & =-5-\left(-5\right)\hfill \\ & =0.\hfill \end{array}$
Integrate the function ${\int }_{-2}^{2}{x}^{4}dx.$
$\frac{64}{5}$
## Key concepts
• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positive x -axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.
## Key equations
• Net Change Theorem
$F\left(b\right)=F\left(a\right)+{\int }_{a}^{b}F\text{'}\left(x\right)dx$ or ${\int }_{a}^{b}F\text{'}\left(x\right)dx=F\left(b\right)-F\left(a\right)$
Use basic integration formulas to compute the following antiderivatives.
$\int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)dx$
$\int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)dx=\int {x}^{1\text{/}2}dx-\int {x}^{-1\text{/}2}dx=\frac{2}{3}{x}^{3\text{/}2}+{C}_{1}-2{x}^{1\text{/}2}+{C}_{2}=\frac{2}{3}{x}^{3\text{/}2}-2{x}^{1\text{/}2}+C$
$\int \left({e}^{2x}-\frac{1}{2}{e}^{x\text{/}2}\right)dx$
$\int \frac{dx}{2x}$
$\int \frac{dx}{2x}=\frac{1}{2}\text{ln}|x|+C$
$\int \frac{x-1}{{x}^{2}}dx$
${\int }_{0}^{\pi }\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x-\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)dx$
${\int }_{0}^{\pi }\text{sin}\phantom{\rule{0.1em}{0ex}}xdx-{\int }_{0}^{\pi }\text{cos}\phantom{\rule{0.1em}{0ex}}xdx=\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x{|}_{0}^{\pi }-\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right){|}_{0}^{\pi }=\left(\text{−}\left(-1\right)+1\right)-\left(0-0\right)=2$
${\int }_{0}^{\pi \text{/}2}\left(x-\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)dx$
Write an integral that expresses the increase in the perimeter $P\left(s\right)$ of a square when its side length s increases from 2 units to 4 units and evaluate the integral.
$P\left(s\right)=4s,$ so $\frac{dP}{ds}=4$ and ${\int }_{2}^{4}4ds=8.$
Write an integral that quantifies the change in the area $A\left(s\right)={s}^{2}$ of a square when the side length doubles from S units to 2 S units and evaluate the integral.
A regular N -gon (an N -sided polygon with sides that have equal length s , such as a pentagon or hexagon) has perimeter Ns . Write an integral that expresses the increase in perimeter of a regular N -gon when the length of each side increases from 1 unit to 2 units and evaluate the integral.
${\int }_{1}^{2}Nds=N$
The area of a regular pentagon with side length $a>0$ is pa 2 with $p=\frac{1}{4}\sqrt{5+\sqrt{5+2\sqrt{5}}}.$ The Pentagon in Washington, DC, has inner sides of length 360 ft and outer sides of length 920 ft. Write an integral to express the area of the roof of the Pentagon according to these dimensions and evaluate this area.
A dodecahedron is a Platonic solid with a surface that consists of 12 pentagons, each of equal area. By how much does the surface area of a dodecahedron increase as the side length of each pentagon doubles from 1 unit to 2 units?
With p as in the previous exercise, each of the 12 pentagons increases in area from 2 p to 4 p units so the net increase in the area of the dodecahedron is 36 p units.
An icosahedron is a Platonic solid with a surface that consists of 20 equilateral triangles. By how much does the surface area of an icosahedron increase as the side length of each triangle doubles from a unit to 2 a units?
Write an integral that quantifies the change in the area of the surface of a cube when its side length doubles from s unit to 2 s units and evaluate the integral.
$18{s}^{2}=6{\int }_{s}^{2s}2xdx$
Write an integral that quantifies the increase in the volume of a cube when the side length doubles from s unit to 2 s units and evaluate the integral.
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
find derivative f(x)=1/x
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
What is the first fundermental theory of Calculus?
I want simple integral
for MSc chemistry... simple formulas of integration
aparna
hello?
funny
how are you
funny
I don't understand integration
aparna
r u insane
aparna
integration is so simple not typical..
funny
tell me any questions about integration then i will solve.
funny
we use integration for whole values or for sum of values any there are some basic rule for integration..
funny
I just formulas
aparna
I just want formulas of integration
aparna
value of log ax cot-x cos-x
aparna
there are many formulas about integration
funny
more then one formula are exist about integration..
funny
so I want simple formulas Because I'm studying MSc chem...Nd have done bsc from bio...
aparna
I am M.sc physics now i am studying in m.phil
funny
so what can i do
aparna
I will send you basic formula for integration after two mint first of all i write then i will send you.
funny
send me your messenger id where i can send you formulas about integration because there is no option for image sending..
funny
integration f(X) dx this is basic formula of integration sign is not there you can look integration sign in methematics form... and f(X) my be any function any values
funny
you send me your any ID where i can send you information about integration
funny
funny
Hi
RIZWAN
I don't understand the formula
who's formula
funny
What is a independent variable
a variable that does not depend on another.
Andrew
solve number one step by step
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel
how to prove tan^2 x=csc^2 x tan^2 x-1?
divide by tan^2 x giving 1=csc^2 x -1/tan^2 x, rewrite as: 1=1/sin^2 x -cos^2 x/sin^2 x, multiply by sin^2 x giving: sin^2 x=1-cos^2x. rewrite as the familiar sin^2 x + cos^2x=1 QED
Barnabas
how to prove sin x - sin x cos^2 x=sin^3x?
sin x - sin x cos^2 x sin x (1-cos^2 x) note the identity:sin^2 x + cos^2 x = 1 thus, sin^2 x = 1 - cos^2 x now substitute this into the above: sin x (sin^2 x), now multiply, yielding: sin^3 x Q.E.D.
Andrew
take sin x common. you are left with 1-cos^2x which is sin^2x. multiply back sinx and you get sin^3x.
navin
Left side=sinx-sinx cos^2x =sinx-sinx(1+sin^2x) =sinx-sinx+sin^3x =sin^3x thats proved.
Alif
how to prove tan^2 x/tan^2 x+1= sin^2 x
Rochel
Salim
what is function.
what is polynomial
Nawaz
an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Alif
a term/algebraic expression raised to a non-negative integer power and a multiple of co-efficient,,,,,, T^n where n is a non-negative,,,,, 4x^2
joe
An expression in which power of all the variables are whole number . such as 2x+3 5 is also a polynomial of degree 0 and can be written as 5x^0
Nawaz
what is hyperbolic function
find volume of solid about y axis and y=x^3, x=0,y=1
3 pi/5
vector
what is the power rule
Is a rule used to find a derivative. For example the derivative of y(x)= a(x)^n is y'(x)= a*n*x^n-1.
Timothy
|
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$x=-5$
Multiplying both sides by $2,$ the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{2}x^2+5x+\dfrac{25}{2}=0 \\\\ 2\left( \dfrac{1}{2}x^2+5x+\dfrac{25}{2} \right)=2(0) \\\\ x^2+10x+25=0 .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ equation }$ \begin{array}{l}\require{cancel} x^2+10x+25=0 \end{array} has $c= 25$ and $b= 10 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 5,5 \right\}.$ Using these two numbers, the $\text{ equation }$ above is equivalent to \begin{array}{l}\require{cancel} (x+5)(x+5)=0 .\end{array} Equating each factor to zero (Zero Product Property), and then isolating the variable, the solution is $x=-5 .$
|
# Learning Objectives
### Learning Objectives
By the end of this section, you will be able to do the following:
• Describe simple and complex machines
• Calculate mechanical advantage and efficiency of simple and complex machines
complex machine efficiency output ideal mechanical advantage inclined plane input work lever mechanical advantage output work pulley screw simple machine wedge wheel and axle
# Simple Machines
### Simple Machines
Simple machines make work easier, but they do not decrease the amount of work you have to do. Why can’t simple machines change the amount of work that you do? Recall that in closed systems the total amount of energy is conserved. A machine cannot increase the amount of energy you put into it. So, why is a simple machine useful? Although it cannot change the amount of work you do, a simple machine can change the amount of force you must apply to an object, and the distance over which you apply the force. In most cases, a simple machine is used to reduce the amount of force you must exert to do work. The down side is that you must exert the force over a greater distance, because the product of force and distance, fd, (which equals work) does not change.
Let’s examine how this works in practice. In Figure 9.8(a), the worker uses a type of lever to exert a small force over a large distance, while the pry bar pulls up on the nail with a large force over a small distance. Figure 9.8(b) shows the how a lever works mathematically. The effort force, applied at Fe, lifts the load (the resistance force) which is pushing down at Fr. The triangular pivot is called the fulcrum; the part of the lever between the fulcrum and Fe is the effort arm, Le; and the part to the left is the resistance arm, Lr. The mechanical advantage is a number that tells us how many times a simple machine multiplies the effort force. The ideal mechanical advantage, IMA, is the mechanical advantage of a perfect machine with no loss of useful work caused by friction between moving parts. The equation for IMA is shown in Figure 9.8(b).
Figure 9.8 (a) A pry bar is a type of lever. (b) The ideal mechanical advantage equals the length of the effort arm divided by the length of the resistance arm of a lever.
In general, the IMA = the resistance force, Fr, divided by the effort force, Fe. IMA also equals the distance over which the effort is applied, de, divided by the distance the load travels, dr.
$IMA=FrFe=dedrIMA=FrFe=dedr$
Getting back to conservation of energy, for any simple machine, the work put into the machine, Wi, equals the work the machine puts out, Wo. Combining this with the information in the paragraphs above, we can write
The equations show how a simple machine can output the same amount of work while reducing the amount of effort force by increasing the distance over which the effort force is applied.
### Watch Physics
This video shows how to calculate the IMA of a lever by three different methods: (1) from effort force and resistance force; (2) from the lengths of the lever arms, and; (3) from the distance over which the force is applied and the distance the load moves.
Grasp Check
Two children of different weights are riding a seesaw. How do they position themselves with respect to the pivot point (the fulcrum) so that they are balanced?
1. The heavier child sits closer to the fulcrum.
2. The heavier child sits farther from the fulcrum.
3. Both children sit at equal distance from the fulcrum.
4. Since both have different weights, they will never be in balance.
Some levers exert a large force to a short effort arm. This results in a smaller force acting over a greater distance at the end of the resistance arm. Examples of this type of lever are baseball bats, hammers, and golf clubs. In another type of lever, the fulcrum is at the end of the lever and the load is in the middle, as in the design of a wheelbarrow.
The simple machine shown in Figure 9.9 is called a wheel and axle. It is actually a form of lever. The difference is that the effort arm can rotate in a complete circle around the fulcrum, which is the center of the axle. Force applied to the outside of the wheel causes a greater force to be applied to the rope that is wrapped around the axle. As shown in the figure, the ideal mechanical advantage is calculated by dividing the radius of the wheel by the radius of the axle. Any crank-operated device is an example of a wheel and axle.
Figure 9.9 Force applied to a wheel exerts a force on its axle.
An inclined plane and a wedge are two forms of the same simple machine. A wedge is simply two inclined planes back to back. Figure 9.10 shows the simple formulas for calculating the IMAs of these machines. All sloping, paved surfaces for walking or driving are inclined planes. Knives and axe heads are examples of wedges.
Figure 9.10 An inclined plane is shown on the left, and a wedge is shown on the right.
The screw shown in Figure 9.11 is actually a lever attached to a circular inclined plane. Wood screws (of course) are also examples of screws. The lever part of these screws is a screw driver. In the formula for IMA, the distance between screw threads is called pitch and has the symbol P.
Figure 9.11 The screw shown here is used to lift very heavy objects, like the corner of a car or a house a short distance.
Figure 9.12 shows three different pulley systems. Of all simple machines, mechanical advantage is easiest to calculate for pulleys. Simply count the number of ropes supporting the load. That is the IMA. Once again we have to exert force over a longer distance to multiply force. To raise a load 1 meter with a pulley system you have to pull N meters of rope. Pulley systems are often used to raise flags and window blinds and are part of the mechanism of construction cranes.
Figure 9.12 Three pulley systems are shown here.
### Watch Physics
#### Mechanical Advantage of Inclined Planes and Pulleys
The first part of this video shows how to calculate the IMA of pulley systems. The last part shows how to calculate the IMA of an inclined plane.
Grasp Check
How could you use a pulley system to lift a light load to great height?
1. Reduce the radius of the pulley.
2. Increase the number of pulleys.
3. Decrease the number of ropes supporting the load.
4. Increase the number of ropes supporting the load.
A complex machine is a combination of two or more simple machines. The wire cutters in Figure 9.13 combine two levers and two wedges. Bicycles include wheel and axles, levers, screws, and pulleys. Cars and other vehicles are combinations of many machines.
Figure 9.13 Wire cutters are a common complex machine.
# Calculating Mechanical Advantage and Efficiency of Simple Machines
### Calculating Mechanical Advantage and Efficiency of Simple Machines
In general, the IMA = the resistance force, Fr, divided by the effort force, Fe. IMA also equals the distance over which the effort is applied, de, divided by the distance the load travels, dr.
$IMA=FrFe=dedrIMA=FrFe=dedr$
Refer back to the discussions of each simple machine for the specific equations for the IMA for each type of machine.
No simple or complex machines have the actual mechanical advantages calculated by the IMA equations. In real life, some of the applied work always ends up as wasted heat due to friction between moving parts. Both the input work (Wi) and output work (Wo) are the result of a force, F, acting over a distance, d.
$Wi=Fidi and Wo=FodoWi=Fidi and Wo=Fodo$
The efficiency output of a machine is simply the output work divided by the input work, and is usually multiplied by 100 so that it is expressed as a percent.
Look back at the pictures of the simple machines and think about which would have the highest efficiency. Efficiency is related to friction, and friction depends on the smoothness of surfaces and on the area of the surfaces in contact. How would lubrication affect the efficiency of a simple machine?
### Worked Example
#### Efficiency of a Lever
The input force of 11 N acting on the effort arm of a lever moves 0.4 m, which lifts a 40 N weight resting on the resistance arm a distance of 0.1 m. What is the efficiency of the machine?
### Strategy
State the equation for efficiency of a simple machine, and calculate Wo and Wi. Both work values are the product Fd.
Solution
$Wi=FidiWi=Fidi$ = (11)(0.4) = 4.4 J and $Wo=FodoWo=Fodo$ = (40)(0.1) = 4.0 J, then
Discussion
Efficiency in real machines will always be less than 100 percent because of work that is converted to unavailable heat by friction and air resistance. Wo and Wi can always be calculated as a force multiplied by a distance, although these quantities are not always as obvious as they are in the case of a lever.
# Practice Problems
### Practice Problems
What is the IMA of an inclined plane that is $5m$ long and $2m$ high?
1. $0.4$
2. $2.5$
3. $0.4m$
4. $2.5m$
If a pulley system can lift a 200N load with an effort force of 52 N and has an efficiency of almost 100 percent, how many ropes are supporting the load?
1. 1 rope is required because the actual mechanical advantage is 0.26.
2. 1 rope is required because the actual mechanical advantage is 3.80.
3. 4 ropes are required because the actual mechanical advantage is 0.26.
4. 4 ropes are required because the actual mechanical advantage is 3.80.
Exercise 6
True or false—The efficiency of a simple machine is always less than 100 percent because some small fraction of the input work is always converted to heat energy due to friction.
1. True
2. False
Exercise 7
The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1 cm and the IMA of the machine is 6 , what is the radius of the handle?
1. 0.08 cm
2. 0.17 cm
3. 3.0 cm
4. 6.0 cm
|
# Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions
Students can Download Maths Chapter 1 Number System Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions
Exercise 1.1
(Try These Text book Page No. 2)
7th Maths Term 2 Exercise 1.1 Question 1.
Observe the following and write the fraction of the shaded portion and mention in decimal form also.
Solution:
(i) Total parts = 8
(ii)
(iii)
7th Std Samacheer Kalvi Maths Solutions Term 2 Question 2.
Represent the following fractions in decimal form by converting denominator into ten or powers of 10.
Solution:
7th Standard Maths Number System Exercise 1.1 Question 3.
Give any two life situations where we use decimal numbers.
Solution:
(i) Measuring weight of gold.
(ii) Weighing our height
(Try These Text book Page No. 3)
Samacheer Kalvi 7th Maths Book Answers Term 2 Question 1.
Represent the following decimal numbers pictorially.
(i) 5 ones and 3 tenths
(ii) 6 tenths
(iii) 7 ones and 9 tenths
(iv) 6 ones and 4 tenths
(v) Seven tenths
Solution:
(i) 5 ones and 3 tenths
(ii) 6 tenths
(iii) 7 ones and 9 tenths
(iv) 6 ones and 4 tenths
(v) Seven tenths
(Try These Text book Page No. 5 & 6)
Samacheer Kalvi 7th Maths Term 2 Question 1.
Express the following decimal numbers in an expanded form and place value grid form.
(i) 56.78
(ii) 123.32
(iii) 354.56
Solution:
(i) 56.78
(a) Expanded form
56.78 = 5 × 101 + 6 × 100 + 7 × 10-1 + 8 × 10-2
(b) Place value grid
(ii) 123.32
(a) Expanded form
123.32 = 1 × 102 + 2 × 101 + 3 × 100 + 3 × 10-1 +2 × 10-2
(b) Place value grid
(iii) 354.56
(a) Expande form
354.56 = 3 × 102 + 5 × 101 + 4 × 100+ 5 × 10-1 + 6 × 10-2
(b) Place value grid
Samacheer Kalvi 7th Maths Book Solutions Term 2 Question 2.
Express the following measurements in terms of metre and in decimal form. One is done for you.
Solution:
Samacheer Kalvi 7th Maths Book Answers Question 3.
Write the following numbers in the place value grid and find the place value of the underlined digits.
(i) 36.37
(ii) 267.06
(iii) 0.23
(iv) 27.69
(v) 53.27
Solution:
(i) Place value of 3 in 36.37 is Tenths.
(ii) Place value of 6 in 267.06 is Hundredths.
(iii) Place value of 2 in 0.23 is Tenths.
(iv) Place value of 9 in 27.69 is Hundredths.
(v) Place value of 2 in 53.27 is Tenths.
Exercise 1.2
(Try These Text book Page No. 10)
7th Standard Samacheer Kalvi Maths Solutions Question 1.
Convert the following fractions into the decimal numbers.
(i) $$\frac { 16 }{ 1000 }$$
(ii) $$\frac { 638 }{ 10 }$$
(iii) $$\frac { 1 }{ 20 }$$
(iv) $$\frac { 3 }{ 50 }$$
Solution:
(i) $$\frac { 16 }{ 1000 }$$ = 0.016
(ii) $$\frac { 638 }{ 10 }$$ = 63.8
(iii) $$\frac { 1 }{ 20 }$$ = $$\frac{1 \times 5}{20 \times 5}$$ = $$\frac { 5 }{ 100 }$$ = 0.05
(iv) $$\frac { 3 }{ 50 }$$ = $$\frac{3 \times 2}{50 \times 2}$$ = $$\frac { 6 }{ 100 }$$ = 0.06
Samacheer Kalvi 7th Books Maths Term 2 Question 2.
Write the fraction for each of the following:
(i) 6 hundreds + 3 tens + 3 ones + 6 hundredths + 3 thousandths.
(ii) 3 thousands + 3 hundreds + 4 tens + 9 ones + 6 tenths.
Solution:
(i) 6 hundreds + 3 tens + 3 ones + 6 hundreds + 3 thousandths.
= 6 × 100 + 3 × 10 + 3 × 1 + 0 × $$\frac { 1 }{ 10 }$$ + 6 × $$\frac { 1 }{ 100 }$$ + 3 × $$\frac { 1 }{ 1000 }$$
= 600 + 30 + 3 + 0 + $$\frac { 6 }{ 100 }$$ + $$\frac { 3 }{ 1000 }$$
= 633 + 0.06 + 0.003
= 633.063
(ii) 3 thousands + 3 hundreds + 4 tens + 9 ones + 6 tenths.
= 3 × 1000 + 3 × 100 + 4 × 10 + 9 × 1 + 6 × $$\frac { 1 }{ 10 }$$
= 3000 + 300 + 40 + 9 + $$\frac { 6 }{ 10 }$$
= 3349 + 0.6
= 3349.6
Samacheer Kalvi 7th Maths Question 3.
Convert the following decimals into fractions.
(i) 0.0005
(ii) 6.24
Solution:
(i) 0.0005 = $$\frac { 5 }{ 10000 }$$ = $$\frac{5 \div 5}{10000 \div 5}$$ = $$\frac { 1 }{ 2000 }$$
(ii) 6.24 = $$\frac { 624 }{ 100 }$$ = $$\frac{624 \div 4}{100 \div 4}$$ = $$\frac { 156 }{ 25 }$$
Exercise 1.4
(Try These Text book Page No. 22)
Samacheer Kalvi 7th Maths Book Solutions Question 1.
Mark the following decimal numbers on the number line.
(i) 0.3
(ii) 1.7
(iii) 2.3
Solution:
(i) 0.3
We know that 0.3 is more than 0, but less than 1.
There are 3 tenths in it. Divide the unit lenght between O and 1 on the number line
into 10 equal parts and take 3 parts, which represent 0.3.
(ii) 1.7
We know that 1.7 is more than 1, but less than 2.
There are one ones and 7 tenths in it. Divide the unit length between 1 and 2 on the number line into 10 equal parts and take 7 parts which represents 1.7 = 1 + 0.7
(iii) We know that 2.3 is more than 2 and less than 3.
There are 2 ones and 3 tenths in it. Divide the unit length between 2 and 3 into 10 equal parts and take 3 parts, which represents 2.3 = 2 + 0.3
Samacheer Kalvi 7th Maths Book Term 2 Question 2.
Identify any two decimal numbers between 2 and 3.
Solution:
2.5 and 2.9
Samacheer Kalvi 7th Maths Solutions Question 3.
Write any decimal number which is greater than 1 and less than 2.
Solution:
1.7, 1.9, 1.6, ………………..
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# Republic of the Philippines
Department of Education
Las Navas National High School
School ID: 303560
## Name:_________________________________ Grade Level:____GRADE 9_______________
Section:________________________________ Date: ___________________________________
Score:___________________________________
QUADRATIC EQUATION
## BACKGROUND INFORMATION FOR LEARNERS (BRIEF DISCUSSION OF THE LESSON,
IF POSSIBLE CITE EXAMPLES)
## SOLVING EQUATIONS TRANSFORMABLE TO QUADRATIC EQUATIONS (INCLUDING RATIONAL ALGEBRAIC
EQUATIONS) PROBLEMS INVOLVING QUADRATIC EQUATIONS AND RATIONAL ALGEBRAIC EQUATIONS.
There are equations that are transformable to quadratic equations. These equations may be
given in different forms. Hence, the procedures in transforming these equations to quadratic
equations may also be different. Once the equations are transformed to quadratic
equations, then they can be solved using the different methods of solving quadratic
equations, such as extracting square roots, factoring, completing the square and using
the quadratic formula. An extraneous root of an equation can be derived from an original
equation. However, it is not a solution of the original equation.
## Example 1: Solve x ( x−5 )=36
This is a quadratic equation that is not written in standard form.
To write the quadratic equation in standard form, simplify the expression x ( x−5).
x ( x−5 )=36 ⟶ x2 −5 x −36=0
Write the resulting quadratic equation in standard form.
x −5 x=36 ⟶ x2 −5 x−36=0
2
Use any of the four methods of solving quadratic equations in finding the solutions of
the equation x 2−5 x−36=0.
Try factoring in finding the roots of the equation.
x 2−5 x−36=0 ⟶ ( x−9 ) ( x + 4 ) =0
x−9=0 or x+ 4=0
x=9 or x=−4
Check whether the obtained values of x make the equation x ( x−5 )=36 true.
If the obtained values of x make the equation x ( x−5 )=36 true, then the solutions of
the equation are: x=9 or x=−4.
Example 2: Find the roots of the equation ( x +5)2+(x−2)2=37
The given equation is a quadratic equation but it is not written in standard form.
Transform this equation to standard form, then solve it using any of the methods of
solving quadratic equations.
( x +5)2+(x−2)2=37 ⟶ x 2 +10 x+ 25+ x 2−4 x + 4=37
x 2+ x2 +10 x−4 x+ 25+ 4=37
2 x2 +6 x +29=37⟶ 2 x2 +6 x−8=0
2 x2 +6 x−8=0 ⟶ ( 2 x −2 )( x +4 )=0
2 x−2=0 or x +4=0
References: Grade 9 LM pp. 77- 95
Grade 9 ADM Module pp. 1-19, LCTG 32-42
Solving Rational Algebraic Equations Transformable into Quadratic Equations.
6 x −3
Example 3: Solve the rational algebraic equation + =2
x 4
The given rational algebraic equation can be transformed into a quadratic equation. To solve
the equation, the following procedure can be followed.
Multiply both sides of the equation by the Least Common Multiple (LCM) of all denominators.
In the given equation, the LCM is 4x.
6 x −3 6 x−3
x
+
4
−2⟶ 4 x +
x( 4 )
=4 x (2)
24+ x2 −3 x =8 x
Write the resulting quadratic equation in standard form.
24+ x2 −3 x =8 x ⟶ x2 −11 x +24=0
Find the roots of the resulting equation using any of the methods of solving quadratic
equations. Try factoring inx+
x 2−11 finding
24=0 ⟶the( x−3
roots)( x−8
of the
)=0equation.
x−3=0 or x −8=0
x−3 or x=8
6 x −3
Check whether the obtained values of x make the equation + =2 true.
x 4
If the obtained values of x make the equation true, then the solutions of the equation are:
x=3 or x=8 .
Problems involving Quadratic Equations and Rational Algebraic Equations. The concept of
quadratic equations is illustrated in many real-life situations. Problems that arise from these
situations, such as those involving area, work, profits, and many others, can be solved by applying
the different mathematics concepts and principles previously studied including quadratic equations
and the different ways in solving them. There are conditions in a given problem which when
translated to the equation form in one variable lead to a quadratic equation and two answers are
obtained. In some problems, a set of answers can be discarded. For instance, negative dimensions of
rectangles and the negative time are disregarded.
Example 1: The top of a rectangular table has an area of 27 ft.2 and a perimeter of 24 ft.What is the
length of the top of the table? What is its width?
Solution 1:
a. The sum of twice the length and twice the width of the rectangular
table gives the perimeter. Hence, 2l + 2w = 24.
## b. If we divide both sides of the equation 2l + 2w = 24 by 2, then l + w =12.
c. The product of the length and the width of the rectangular table gives the area. Hence,
l ● w = 27.
We can think of l ● w = 27 and l + w =12 as the equations representing the product and
Remember that if the sum and the product of the roots of a quadratic equation are given, the
sum of roots, respectively, of a quadratic equation.
roots can be determined. This can be done by inspection or by using the equation
b c c b
x2 +
Remember
x +that=if0where
the sum is
andthethe product
sum of theof the roots
roots and ofisathe
quadratic equation are given, the roots
product.
a a a a 2 b c
can be determined. This can be done by inspection or by using the equation x + x + = 0where
a a
c b
is the sum of the roots and is the product.
a a
d. By inspection, the numbers whose product is 27 and whose sum is 12 are 3 and 9.
References: Grade 9 LM pp. 77- 95
Grade 9 ADM Module pp. 1-19, LCTG 32-42
Solution 2:
2 b c
Another method of finding the roots is to use the equation x + x + = 0.
a a
−b c
Let = 12 and = 27. Then substitute these values in the equation.
a a
b c
X2 + x+ =0 x2 + (-12)x + 27 = 0
a a
x2 – 12x + 27 = 0
Solve the resulting equation x2 – 12x + 27 = 0 using any of the methods of solving
quadratic equation. Try factoring.
x2 – 12x + 27 = 0
(x – 3) (x – 9) = 0
x–3=0 x–9=0
x=3 x=9
With the obtained roots of the quadratic equation, the dimensions of the table then
are 3 ft. and 9 ft., respectively.
Example 2. The Local Government of Iligan City wants to place a new rectangular billboard
to inform and give awareness to the Iliganons on how to protect themselves from the spread
of COVID19. Suppose the length of the billboard to be placed is 4 m longer than its width
and the area is 96 m2. What will be the length and the width of the billboard?
Solution:
If we represent the width, in meters, of the billboard by x, then its length is x + 4.
Since the area of the billboard is 96 m2, then (x)(x+4) = 96.
The equation (x)(x+4) = 96 is a quadratic equation that can be written in the form
ax2 + bx + c = 0.
(x)(x+4) = 96 x2 + 4x = 96
x2 + 4x – 96 = 0
## Solve the resulting equation (by factoring)
x2 + 4x – 96 = 0
(x – 8) (x + 12) = 0
x–8=0 x + 12 = 0
x=8 x = -12
The equation x2 + 4x – 96 = 0
has two solutions: x = 8 or x = -12. However, we only consider the positive value of x
since the situation involves measure of length. Hence, the width of the billboard is
8m and its length is 12m.
Example 3: Mrs. Cesa can finish checking 40 item test ahead of Mrs. Saburao by 1 min.
Together they can finish 27 papers in 1 hour. How long does it take each to finish checking
one paper.
Solution:
Representation: Let x be the time it takes for Mrs. Cesa to finish checking a 40-item
test paper. x + 1 be the time it takes for Mrs. Saburao to finish checking a 40-item test
paper.
1 1 27
Equation: + =
x x+1 60
60 x ( x+1) ¿ ⟶ 60 ( x+1 ) +60 x=27 x ( x – 1 )
60(x + 1) + 60x 27x(x + 1)
= ⟶ 20( x +1)+20 x=9 x ( x +1)
3 3
9x2 – 31x – 20 = 0 ⟶ x = -5/9 or x = 4
We reject -5/9 since time should be considered positive. Thus, x = 4 minutes.
## References: Grade 9 LM pp. 77- 95
Grade 9 ADM Module pp. 1-19, LCTG 32-42
Final Answer: Mrs, Cesa finish checking a 40-item test paper in 4 minutes while
Mrs. Saburao takes 5 minutes to finish the same task.
## In this lesson, the students will learn to:
Solves equations transformable to quadratic equations (including rational algebraic
equations). (M9AL-Ic-d-1)
Solves problems involving quadratic equations and rational algebraic equations. (M9AL-Ie-
1)
General Reminders: Use this activity sheet with care. Do not put unnecessary mark/s on
any part of the activity sheet. Use a separate sheet of paper in answering the exercises. Read
the directions carefully before doing each task. Return this activity sheet to your
teacher/facilitator once you are through with it.
EXCERCISES/ACTIVITIES
EXPLORE:
Directions: Answer the puzzle below by simplifying the following expressions. Then shade the box
containing the corresponding answer. The unshaded boxes will reveal the answer to this puzzle.
## Which great mathematician and scientist said:
“Do not worry about difficulties in Mathematics. I can assure you that mine are still greater.”
x 2−16 4 x 2 +4 x +1
1. 7.
x2 +8 x +16 4 x 2−1
x 2−27 1−2 x+ x 2
2. 8.
x2 +6 x +7 1−x3
4 x 2−12 x+ 9 x 4−16
3. 9.
4 x 2−9 x −2
3 x 2−27 25−16 x 2
4. 10.
x 2+3 16 x 2 + 40 x +25
3 x 2−12 x+ 12 9−12 x+ 4 x 2
5. 11.
6 x 2−24 27−8 x 2
x 2−1 9−x 2
6. 12.
x2 +2 x+ 1 C E 27−x 2 I A
2 x +1 ( x−3 )2 1−x
3−x
2 x−1 x +3 1+ x+ x 2
N H S H
x−1 2 x−3 x+ 4
2 3 x−9
References: + x +1 9 LM pp. 77- 95 2 x +3
xGrade x−4
Grade 9 ADM Module pp. 1-19, LCTG 32-42
A T R E
A T R E
2 x−1 x−2 4 x−5
x−3
2 x +1 2 x +4 4 x +5
W L I E
5−4 x x−4 3 x−9 x−1
4 x +5 x+ 4 ( x−3 )2 x+ 1
T N P O
3−x 3 x+ 6 3−2 x
2
2 x−4 x 3+ 2 x 2 +4 x +8
x −3 x +9 4 x 2 +6 x+ 9
LEARN:
Directions: Transform each of the following equations into a quadratic equation in the form
ax 2+ bx + c = 0 to find the solution set.
1. x ( x +3 ) =28
2. 3 s ( s−2 )=12
3. (t +1)2 +( t−8)2=45
4. (3 r +1)2+(r +2)2=65
(x +2) (x−2)
5. + =15
5 3
6. x(x + 5) = 2
7. (s + 6)2 = 15
8. (m – 4) 2 + (m – 7) 2 = 15
2 x2 5 x
9. + =10
5 4
ENGAGE:
## Directions: Solve the following problems.
1. To signal the start of a 10-km race, an official fired a blank bullet vertically up into
the air. The initial velocity of the bullet is 120 ft. per second and the distance(s)
covered above the ground after t seconds is represented by the equation s = 120t –
16t2.
a. How many seconds would it take for the bullet to reach 216 feet?
b. Would the bullet reach a stationary kite 900 feet above? Why or why not?
2. Pedro lives in Iligan City where it is known as the City of Waterfalls. With its
abundance
References: of 9
Grade water, Pedro
LM pp. wants to build a swimming pool for his family. His
77- 95
2
planned rectangular
Grade 9 ADM swimming pool1-19,
Module pp. has aLCTG
perimeter
32-42of 86 m and its area is 450 m .
a. How would you represent the length and the width of the swimming pool?
b. What equation represents the perimeter of the swimming pool? How about the
3. Mindanao Island is the second largest island in the Philippines. To easily access the
provinces and tourist spots of Mindanao, the Department of Transportation plans to design
and construct train stations and railways on selected provinces of Mindanao. They decided to
place trains that travels at a certain average speed for a distance of 63 km and then travels a
distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3
hours to complete the total journey of the train, what is its original average speed?
APPLY:
A. Directions: Transform each of the following equations into a quadratic equation in the
form
ax 2+ bx + c = 0 to find the solution set.
1. x(x + 3) = 28
2. 3s(s – 2) = 12s
3. (3r + 1) 2 + (r + 2) 2 = 65
(x +2)2 (x−2)2 16
4. + =
5 3 3
5 x+2
5. − =x−1
4x 3
## B. Directions: Solve the following problems.
1. Juan has a rectangular garden in their backyard. One day, obeying the “stay at home” policy of
the government, he plans to replant the vegetables in his garden since it was not replanted after
the fruitful harvest last year because of his new work. The length of the rectangular garden is 6
more than the width. The area is 27 sq. units. Find the dimensions of the rectangular garden.
2. The Rural Bus Inc. travels around the Island of Mindanao. Last December, Jose visited Siargao
Island. He observes that the speed of the bus decreased by 10 kmph, it took 2 hours more than
Prepared by:
LIOMER M. CREDO FLORAJANE D. ALARAS
Subject Teacher Subject Teacher
## Checked by: Approved:
LUZETTE N.YU, MT – II MARCO T. TEPACE
Head, Mathematics Department Principal - I
## References: Grade 9 LM pp. 77- 95
Grade 9 ADM Module pp. 1-19, LCTG 32-42
## Fußzeilenmenü
### Holen Sie sich unsere kostenlosen Apps
Copyright © 2021 Scribd Inc.
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# Water's Home
Just another Life Style
0%
## Matrices and Vectors
The dimension of the matrix is going to be written as the number of row times the number of columns in the matrix. A vector turns out to be a special case of a matrix. Matrix with just one column is what we call a vector.
$\begin{bmatrix} 1 & 0 \\ 2 & 5 \\ 3 & 1 \end{bmatrix} + \begin{bmatrix} 4 & 0.5 \\ 2 & 5 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0.5 \\ 4 & 10 \\ 3 & 2 \end{bmatrix}$
#### Scalar Multiplication
$3 * \begin{bmatrix} 1 & 0 \\ 2 & 5 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 6 & 15 \\ 9 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 5 \\ 3 & 1 \end{bmatrix} * 3$
## Matrix Vector Multiplication
$\begin{bmatrix} 1 & 3 \\ 4 & 0 \\ 2 & 1 \end{bmatrix} * \begin{bmatrix} 1 \\ 5 \end{bmatrix} = \begin{bmatrix} 16 = 1 * 1 + 3 * 5 \\ 4 = 4 * 1 + 0 * 5 \\ 7 = 2 * 1 + 1 * 5 \end{bmatrix}$
## Matrix Matrix Multiplication
$\begin{bmatrix} C0 & C1 \\ C2 & C3 \end{bmatrix} = \begin{bmatrix} A0 & A1 \\ A2 & A3 \end{bmatrix} * \begin{bmatrix} B0 & B1 \\ B2 & B3 \end{bmatrix}$
$C0 = A0 * B0 + A1 * B2$ $C1 = A0 * B1 + A1 * B3$ $C2 = A2 * B0 + A3 * B2$ $C3 = A2 * B1 + A3 * B3$
## Matrix Multiplication Properties
$A * B \neq B * A$ $A * (B * C) = (A * B) * C$ Identity matrix, has the property that it has ones along the diagonals, right, and so on and is zero everywhere else. $AA^{-1} = A^{-1}A = I$ $AI = IA = A$
## Inverse and Transpose
#### Inverse
$A^{-1}$
#### Transpose
$A^{T}$
$\begin{vmatrix} a & b\\ c & d\\ e & f \end{vmatrix} ^{T} = \begin{vmatrix} a & c & e\\ b & d & f \end{vmatrix}$
|
You are on page 1of 13
# Unit 1: Real Numbers. Mathematics 4th E.S.O.
AlbertoTeacher:
Lorenzo Miguel
Prieto Angel Hernández Lorenzo.
Sets of Numbers:
## Natural Numbers: ℕ={0,1,2 ,3 , 4,5, 6,...}
In English books of Maths, the set of natural numbers is {1,2 ,3 ,4 ,...} , while the set of whole
numbers is {0,1,2 ,3 ,4 , ...} , so there is a tiny difference between both sets, including or not the
number zero.
## Integers: ℤ={... ,−3,−2,−1,0 ,1 ,2,3 ,...}
The set of integers is formed by the natural numbers including zero and the negatives, the opposites
of the natural numbers.
## Rational Numbers: ℚ= {ab , where a , b∈ℤ , b≠0}
Rational numbers are the fractions. All number that can be written as a fraction is a rational number.
Every fraction, that is, every rational number, has a decimal expression ( we can get it dividing the
numerator by the denominator). The different types of the decimal expression of the rational
numbers are:
– An exact or terminating number is one which does not go on forever, so you can write
2
down all its digits. For example: =0,4
5
– A recurring, periodic or repeating decimal is a decimal number which does go on forever,
but where some of the digits are repeated over and over again. For example:
0,125252525...=0,1 25
We can distinguish:
- Decimals that the repeating part or period starts just after the decimal point (pure periodic
2
or recurring decimal). For example: =0,6666...=0, 6
3
- Decimals that the repeating part or period does not start just after the decimal point (mixed
1
periodic or recurring decimal). For example: =0,16666...=0,1 6
6
Irrational Numbers: I . Irrational numbers have decimal expressions that neither terminate nor
become periodic.
Examples: 2 , 3 , 5 …
π=3,141592 …
1,030030003...
2,010011000111...
## Alberto Lorenzo Prieto
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
## The real numbers include both rational numbers,
such as 24 or −43/127 , and irrational numbers
such as π or 7 .
Your
Turn
Classify the following numbers into the corresponding set:
## 7 -6 21 3,7373... 20 13 0,04343... 1,131331333... − 3 -8765
5 5 4
Natural
Numbers
Integers
Rational
Numbers
Irrational
Numbers
I
Real
Numbers
2
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
Rational Numbers:
As we know, the set of rational numbers is formed by all the numbers that can be written as a
a
fraction , where a and b are integers and b is not 0.
b
## To calculate the decimal expression of a fraction, we divide numerator by denominator. We can
obtain an integer (if the numerator is a multiple of the denominator) or a decimal number of the
following types:
An exact or terminating decimal, if after the simplification of the fraction the numerator only has
as prime factors either 2 or 5.
A pure periodic decimal, if after the simplification of the fraction, 2 and 5 are not factors of the
denominator.
A mixed periodic decimal, if after the simplification of the fraction, 2 and/or 5 are prime factors of
the denominator, and it has other prime factors.
Your
Turn
Without doing the division, try to say what kind of decimal the following fractions generate:
3 7 9
a) b) c)
20 3 2
5 3 16
d) e) f)
12 15 30
4 3 5
g) h) i)
25 75 21
3
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
## Converting decimal into a fraction:
• Exact or terminating decimals: Write on the numerator the number without decimal point
and on the denominator the unit followed by as many “zeroes” as decimal digits the number
has.
17 246 123
Examples: 0,017= 2,46= =
1000 100 50
• Pure periodic decimals: Write on the numerator the number without decimal point and
subtract the whole part of the number. Then, write on the denominator as many “nines” as
decimal digits the “repeating part” of the numbers has.
12−1 11 32−0 32
Examples: 1,2222...=1, 2 = = 0,323232...=0, 32= =
9 9 99 99
• Mixed periodic decimals: Write on the numerator the number without decimal point and
subtract the whole part of the number followed by the “non repeating part”. Then, write on
the denominator as many “nines” as decimal digits the “repeating part” of the number has,
followed by as many zeroes as decimal digits the “non repeating part” has.
1023−10 1013
Examples: 1,0232323...=1,0
23= =
990 990
123−12 111 37
0,123333...=0,12 3= = =
900 900 300
Your
Turn
Convert into fractions the following decimal numbers:
a) 1,2 b) 0, 7 c) 0,66
d) 1,1 6 e) 1,
72 f) 0,45
g) 0,1
56 h) 2,
795 i) 0,00 3
## Be careful with the numbers whose repeating part is 9:
2, 9 10, 9
4
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
## Irrational numbers: A number is irrational if it cannot be expressed as a fraction. Its decimal
expression has an infinite number of digits that are not regularly repeated.
d 1 cm
1 cm
## d) 1,232233... e) 1,11213213... f) 0,122333...
Real Numbers: The set of real numbers ℝ is formed by the set of rational numbers ℚ and
the set of irrational numbers I.
ℝ=ℚ∪I
Real numbers ℝ
{Racional numbers ℚ
{ {
Integers ℤ Natural Numbersℕ
Negative Integers
Terminating and Periodic decimals
Irrational Numbers I
Order in ℝ :
## • a is less than b, and we write ab , when b−a is positive.
• a is greater than b, and we write ab , when b−a is negative.
5
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
## Properties of real numbers:
Associative abc=abc a · b· c=a ·b · c
Identity element a0=0a=a a ·1=1 · a=a
Opposite/Inverse element a−a=0
a· 1
a
=1
## Commutative ab=ba a ·b=b · a
Distributive a ·bc=a · ba · c
## Extracting common factor:
Using the distributivity of real numbers: a ·ba · c=a · bc , we can extract common factor to
do easily some operations:
a) 63+27
2 3 2 1 3 2
b) · − · ·
5 4 5 5 10 5
1 3 1 1 5
c) − · − ·
2 5 2 2 7
Your
Turn
b) 0,03 1
0,0 31 0,
031 0,031
## c) 7,33 5 7,3 5 7, 5 7,5 3 7, 3 7,3 ̂
35
6
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
## a) 3,475 3,447 3,744 3,444 3,457
b) 0,92 0,9 2
0, 92 0, 2 0,
922
c) 5,9 ̂
17 5,9
71 5,97 1 5,91 7 5,
917
## 3. Extract common factor and operate these expressions:
a) 102030405060708090
2 5 2
b) · 3− · 25·
7 7 7
1 3
c) · 7−7 · 27 ·
4 2
1 1 5 1 1 1 4
d) · · − ·
3 2 6 3 3 3 3
Real Line:
The real numbers may be thought of as points on an infinitely long number line. Each point of the
real line corresponds to a real number, and each real number corresponds to a point of the real line.
## Representing rational numbers on the real line:
Examples:
3 5
7 3
−3 14
8 3
−5 −8
2 5
7
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
## Representing irrational numbers on the real line:
Examples:
2 5
10 3
13 6
Intervals:
A real interval is a subset of real numbers that corresponds to the points of a segment or a half-line
on the real line.
## INTERVAL NOTATION SET NOTATION GEOMETRIC CLASSIFICATION
PICTURE
a , b {x∈ℝ :a xb} Finite; open
a b
[a , b] {x∈ℝ : axb} Finite; closed
a b
a ,b] {x∈ℝ : axb} Finite; half-open
a b
[ a , b {x∈ℝ : axb} Finite; half-open
a b
a ,∞ {x∈ℝ : ax } Infinite; open
a
[ a ,∞ {x∈ℝ : ax } Infinite; closed
a
−∞ , b {x∈ℝ : x b} Infinite, open
b
−∞ ,b ] {x∈ℝ : x b} Infinite; closed
b
−∞ ,∞ ℝ Infinite; open and
closed
8
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
Examples: Write as intervals and represent on the real line the following set of numbers:
c) Less than 0.
## f) Greater or equal than -2 and less than 3.
Your
Turn
1. Represent on the real line and write as intervals the following set:
## a) ∣x∣2 b) ∣x∣2 c) ∣x∣2
9
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
Approximations:
An approximation of a number is a representation of this number that is not exact, but still close
enough to be useful.
Approximation Methods:
## Examples: Approximate by truncation to the hundredths the following numbers:
1,2345 0,03968 1,
382
0,0 53
Rounding: The figures are deleted from a considered order, and the last figure is increased by one
unit if the following digit is greater than or equal to 5.
## Examples: Rounding off to the nearest thousandths the following numbers:
1,2345 0,03968 1,
382
0,0 53
## Absolute Error and Relative Error:
The approximation error in some data is the discrepancy between an exact value and some
approximation to it; an approximation error can occur because
## 1. The measurement of the data is not precise (due to the instruments), or
2. approximations are used instead of the real data (3,14 instead of π).
One commonly distinguishes between the absolute error and the relative error. The absolute error
is the magnitude of the difference between the exact value and the approximation. The relative error
is the absolute error divided by the magnitude of the exact value.
## Absolute Error: E a =∣V Exact −V Approx.∣ Relative Error: Er=
∣ ∣
Ea
V Exact
Example:
a) The height of a house is 4,7 m. If we say the height of the house is 5 m, calculate the absolute
error and the relative error of this approximation.
b) The height of a skyscraper is 115,3 m. If we say the height of the skyscraper is 115 m, calculate
the absolute error and the relative error of this approximation.
10
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
Your
Turn
1. Skeeter, the dog, weighs exactly 36,5 pounds. When weighed on a defective scale, he
weighed 38 pounds.
a) What is the absolute error and the relative error in measurement of the defective scale?
b) If Millie, the cat , weighs 14 pounds on the same defective scale, what is Millie's actual
weight?
2. The actual length of this field is 500 feet. A measurement instrument shows the length to be
508 feet. Find:
a) The absolute error in the measured length of the field.
b) The relative error in the measured length of the field.
250 feet
c) The percentage error on the measured length of the field.
500 feet
3. Find the absolute and relative error of the approximation 3,14 to the value π.
11
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
Keywords:
subtraction / difference = resta, diferencia
multiplication / product = multiplicación, producto
division / quotient = división, cociente
set of numbers = conjunto numérico
Natural numbers = Números Naturales
Integers = Números Enteros
Rational Numbers = Números Racionales
fraction = fración
exact or terminating decimal = decimal exacto
pure periodic decimal= decimal periódico puro
mixed periodic decimal= decimal periódico mixto
Irrational numbers = Números Irracionales
Real numbers = Números Reales
to be included = estar incluido
factor / divisor = factor, divisor
multiple = múltiplo
Highest Common Factor (UK) / Greatest Common Factor (USA) = Máximo Común
Divisor
Lowest Common Multiple (UK) / Least Common Factor (USA) = Mínimo Común
Múltiplo
diagonal = diagonal
height = altura
weight = peso
to weigh = pesar
measurement = medida
to measure = medir
rectangle = rectángulo
triangle = triángulo
a is less than b ab = a es menor que b
a is less or equal than b ab = a es menor o igual que b
a is greatest than b ab = a es mayor que b
a is less or equal than b ab = a es mayor o igual que b
to order from lowest to highest (UK) = ordenar de menor a mayor
to order from least to greatest(USA) = ordenar de menor a mayor
12
Unit 1: Real Numbers. Mathematics 4th E.S.O. Teacher: Miguel Angel Hernández Lorenzo.
## Associative property = Propiedad asociativa
Identity element = elemento neutro
Opposite element = elemento opuesto
Inverse element = elemento inverso
common factor = factor común
Real line = Recta real
Pythagoras' Theorem = Teorema de Pitágoras
Interval = intervalo
segment = segmento
half-line = semirrecta
Open interval = intervalo abierto
Half-open interval = intervalo semiabierto
Infinite = infinito
Absolute value = valor absoluto
approximation = aproximación
Truncation = truncamiento
to round off = redondear
absolute error = error absoluto
relative error = error relativo
percentage = porcentaje
place value of a digit / figure = valor posicional de una cifra
|
# LCM of 38
The LCM of 38 and 25 is 950 . To find the LCM (least common multiple) of 38 and 25, we need to find the multiples of 38 and 25 (multiples of 38 = 38, 76, 114, 152).
Similarly, what are the multiples of 38? Multiples of 38: 38, 76, 114, 152, 190, 228, 266, 304, 342, 380 and so on.
What is the GCF of 38 and 19? The GCF of 19 and 38 is 19 .
What is the least common factor of 38 and 19? Answer: LCM of 19 and 38 is 38 .
What is the second prime factor of 3825? So 3, 5, 17 are prime factors.
## Which number can add up to 38?
2+1+4=38. seriously? As you’ve probably noticed, there are an infinite number of different combinations that add up to 38.
So what are the factors of 3? The factors of 3 are only 1 and 3 . Note that -1 × -3 = 3.
How many factors does 39 have? The factors of 39 are 1, 3, 13 and 39. Since 39 has more than two factors , it is a composite number.
#### What are the common factors of 8?
Solution: The factors of 8 are 1, 2, 4 and 8 . The factors of 7 are 1 and 5. Since 5 is a prime number, the common factor of 8 and 5 is 1.
What is GCF 6? The greatest common factor (GCF or GCD or HCF) of the set of whole numbers is the greatest positive integer that divides evenly into all numbers with zero remainder. For example, GCF = 6 for the set of numbers 18, 30 and 42.
How many factors does 57 have?
The factors of 57 are 1, 3, 19, and 57 .
What is the LCM of 38 and 57? The least common factor of 38 and 57 is 114 .
#### What is the GCF of 38 and 48?
As you can see when you list the factors of each number, 2 is the largest number that can divide 38 and 48.
What is the least common multiple of 3?
How do you find the LCM of 3825? Steps to find LCM
1. 1. Find the prime factorization of 1 = 1.
2. Find the prime factorization of 3825. 3825 = 3 × 3 × 5 × 5 × 17.
3. LCM = 3×3×5×5×17.
4. MCL = 3825.
How many factors are there in 3825? Prime factor of 3,825 3 . is 2 × 5 2 × 17. Since its total is 45915 prime factors, 3,825 is a composite number. ,
#### How do you find prime factors?
Find the prime factors of a composite number using ladder method
1. Divide the number by the smallest prime.
2. Keep dividing by that prime until it is evenly divisible.
3. Divide by the next prime until it is evenly divisible.
4. Continue until the quotient becomes a prime.
What are two similar numbers whose sum is 38? However, two consecutive decimal numbers can give a sum of 38 and they are 18.5 and 19.5 .
Which 4 numbers add up to 38?
So the four numbers whose sum is 38 are 8, 9, 10, and 11 .
What are the integers of 90? The integers are 29 30, and 31 . Hope this helps!
#### What is a factor of 4?
The factors of 4 are 1, 2, and 4 .
What is a factor of 5? 5 is a prime number. Therefore, it can have only two factors, i.e., 1 and the number itself. The factors of 5 are 1 and 5 .
What are the factors of 30?
factors of 30
• 30: Factors 1, 2, 3, 5, 6, 10, 15 and 30.
• Negative factors of 30: -1, -2, -3, -5, -6, -10, -15 and -30.
• 30: prime factors of 2, 3, 5
• Prime factorization of 30: 2 × 3 × 5 = 2 × 3 × 5.
• The sum of the factors of 30:72.
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# 35800 in Words
The number 35800 in words is represented by Thirty-Five Thousand Eight Hundred. For instance, Siya purchased a golden bracelet worth Rs. 35800, then you can say, “Siya purchased a golden bracelet worth Rupees Thirty-Five Thousand Eight Hundred”. 35800 is a cardinal number as it depicts a certain amount. Let’s learn how to transform the number 35800 into words using a place value method in this article.
35800 in Words Thirty-Five Thousand Eight Hundred Thirty-Five Thousand Eight Hundred in numerical form 35800
## 35800 in English Words
In Maths, we generally write the numbers in words using the letters of the English alphabet. Therefore, the number 35800 in words is written as Thirty-Five Thousand Eight Hundred.
## How to Write 35800 in Words?
We can convert the number 35800 into words by finding the position of each digit using the place value system. The place value chart for the number 35800 is shown below.
Ten-Thousands Thousands Hundreds Tens Ones 3 5 8 0 0
Hence, we can write the expanded form as:
3 x Ten Thousand + 5 x Thousand + 8 x Hundred + 0 x Ten + 0 x One
= 3 x 10000 + 5 x 1000 + 8 x 100 + 0 x 10 + 0 x 1
= 30000 + 5000 + 800 + 0 + 0
= 30000 + 5000 + 800
= 35800
= Thirty-Five Thousand Eight Hundred
Therefore, 35800 in words is written as Thirty-Five Thousand Eight Hundred
Interesting way of writing 35800 in words
3 = Three
35 = Thirty-Five
358 = Three Hundred and Fifty-Eight
3580 = Three Thousand Five Hundred Eighty
35800 = Thirty-Five Thousand Eight Hundred
Thus, the word form of the number 35800 is Thirty-Five Thousand Eight Hundred
35800 is a natural number that is one less than 35801 and one greater than 35799
• 35800 in words – Thirty-Five Thousand Eight Hundred
• Is 35800 an odd number? – No
• Is 35800 an even number? – Yes
• Is 35800 a perfect square number? – No
• Is 35800 a perfect cube number? – No
• Is 35800 a prime number? – No
• Is 35800 a composite number? – Yes
## Frequently Asked Questions on 35800 in Words
Q1
### What is 35800 in words?
35800 in words is Thirty-Five Thousand Eight Hundred.
Q2
### Simplify 30000 + 5800, and express in words.
Simplifying 30000 + 5800, we get 35800. Therefore, the number 35800 in words is Thirty-Five Thousand Eight Hundred.
Q3
### Write Thirty-Five Thousand Eight Hundred in numbers.
Thirty-Five Thousand Eight Hundred in numbers is 35800.
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# Surface Area of a Cylinder
The surface area of a cylinder is the total space occupied by the flat circular bases of the cylinder and the curved surface. Simply, it is the sum of an area of a circle, since the base of the cylinder is a circle and the area of the curved surface which is a rectangle.
It is expressed in square units such as m2, cm2, mm2, and in2
A cylinder has 2 types of surface areas:
• Curved (Lateral) Surface Area
• Total Surface Area
## Formulas
### Lateral (Curved) Surface Area
The lateral or the curved surface area (LSA) of a cylinder is the area formed by the curved surface of the cylinder. It is thus the space occupied between the two parallel circular bases. The formula to calculate LSA is given below:
Let us solve some examples to understand the concept better.
Find the curved surface area of a cylinder whose diameter is 44 cm and height is 16 cm.
Solution:
As we know,
Radius (r) = d/2, here d = diameter
= 44/2 cm
= 22 cm
Now, as we know
Curved Surface Area (CSA) = 2πrh, here π = 3.141, r = 22 cm, h = 16
= 2 × 3.141 × 22 × 16
= 2211.26 cm2
The lateral surface area of a cylinder is 122 ft2. If the radius of the cylinder is 5 ft, calculate the height of the cylinder.
Solution:
As we know the curved surface area is the lateral surface area of the cylinder,
Curved Surface Area (CSA) = 2πrh, here CSA =122 ft2, π = 3.141, r = 5 cm
=> 122 = 2 × 3.141 × 5 × h
=> h = 122/2 × 3.141 × 5
=> h = 122/31.41
=> h = 3.8 ft
### Total Surface Area
The total surface area (TSA) of a cylinder is the sum of the curved surface area and the area of two circular bases. The formula to calculate TSA is given below:
Derivation
As we know, a cylinder has 2 flat surfaces which are usually circles, and a curved surface which is in the form of a rectangle.
Let the height of the cylinder be ‘h’, and the circular base has a radius ‘r’. See the diagram below.
Now,
Total Surface Area (TSA) = Curved Surface Area (CSA) + (2 × Area of 1 circular base)
As we know,
Curved Surface Area (CSA) = 2πrh, and
Area of 1 circular base = πr2
∴ The area of 2 circular bases = 2πr2
Thus,
TSA = 2πrh + 2πr2
TSA = 2πr(r + h)
Thus, the equation to calculate TSA of a cylinder is TSA = 2πr(r + h)
Let us solve some examples to understand the concept better.
Find the total surface area of a cylinder whose radius is 8 cm and height is 12 cm.
Solution:
As we know,
Total Surface Area (TSA) = 2πr(r + h), here π = 3.141, r = 8 cm, h = 12 cm
= 2× 3.141 × 8(8 + 12)
= 2× 3.141 × 8 × 20
= 1005.12 cm2
Find the height of a cylinder if its total surface area is 2442 inand the radius is 12 in.
Solution:
As we know,
Total Surface Area (TSA) = 2πr(r + h), here TSA = 1224 in2, π = 3.141, r = 12 in
=> 2442 = 2 × 3.141 × 12(12 + h)
=> 2442 = 75.384(12 + h)
Divide both sides by 75.384, we get
=> 32.39 = 12 + h
Subtracting 12 from both sides, we get
h = 32.39 in
Calculate the radius of a cylinder whose total surface area is 222.46 square feet, and the height is 4 feet.
Solution:
As we know,
Total Surface Area (TSA) = 2πr(r + h), here TSA = 222.46ft2, π = 3.141, h = 4 ft
=> 222.46 = 2 × 3.141 × r(r + 4)
=> 222.46 = 6.282r(r + 4)
By distributive property of multiplication on the R.H.S, we get
=> 222.46 = 6.282r2 + 25.128
Dividing each term by 6.282, we get
=> 35.412 = r2 + 4
=> r2 = 31.412
=> r = 5.6 ft
## Surface Area of an Open Cylinder
An open cylinder is a cylinder without a top. Thus the surface area of an open cylinder can be calculated by finding the area of one base and the curved surface.
The formula to calculate the surface area of an open-top cylinder is given below:
Surface Area (A) = πr(rh + r), here π = 22/7 = 3.141, r = radius, h = height
Let us solve an example to understand the concept better.
Find the surface area of an open cylinder with a height of 10 m and a radius of 18 m.
Solution:
As we know,
Surface Area (A) = πr(rh + r), here π = 3.141, r = 18 m, h = 10 m
= 3.141 × 18(18 × 10 + 18)
= 3.141 × 18 × 198
= 11194.52 m2
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# Question #95e73
Mar 9, 2017
$\frac{\sqrt{2}}{2}$
#### Explanation:
$\sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2}\right)}^{2}}$
First find out what ${\left(\frac{1}{2}\right)}^{2}$ is:
${\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$
So, we now know that the expression is now $\sqrt{\frac{2}{4}}$ which is the same as $\sqrt{\frac{1}{2}}$.
Then, we rationalize the denominator as we cannot have a root as
the denominator:
$\sqrt{1} = 1$, so the expression is now $\frac{1}{\sqrt{2}}$
Multiply both the numerator and the denominator by $\sqrt{2}$
Denominator $= \sqrt{2} \cdot \sqrt{2} = 2$
Numerator $= 1 \cdot \sqrt{2} = \sqrt{2}$
Hence,
$= \frac{\sqrt{2}}{2}$
Mar 9, 2017
The expression is equivalent to $\frac{\sqrt{2}}{2}$.
#### Explanation:
The square of $\frac{1}{2}$ is $\frac{1}{4}$ because $\left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = \frac{1}{2 \cdot 2} = \frac{1}{4}$.
Therefore,
$\sqrt{\frac{1}{4} + \frac{1}{4}}$
$\sqrt{\frac{2}{4}}$
$\sqrt{\frac{1}{2}}$
$\frac{\sqrt{1}}{\sqrt{2}}$
$\frac{1}{\sqrt{2}}$
I would recommend you rationalize the denominators.
$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$
$\frac{\sqrt{2}}{2}$
Hopefully this helps!
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# A triangle has corners at (1 ,3 ), (9 ,2 ), and (6 ,7 ). How far is the triangle's centroid from the origin?
Mar 2, 2017
$\text{The answer is l=6.67 units}$
#### Explanation:
$\text{the triangle A(1,3),B(9,2),C(6,7) is shown in figure below}$
$\text{The centroid of any triangle can be calculated using the formula below.}$
$c e n t r o i d \left(x , y\right)$
$\text{Centroid of any triangle represents by two values (x,y)}$
$x = \frac{{x}_{A} + {x}_{B} + {x}_{C}}{3}$
$y = \frac{{y}_{A} + {y}_{B} + {y}_{C}}{3}$
$A \left(1 , 3\right) \text{ "rArr " "x_A=1" , } {y}_{A} = 3$
$B \left(9 , 2\right) \text{ "rArr" "x_B=9" , } {y}_{B} = 2$
$C \left(6 , 7\right) \text{ "rArr" "x_C=6" , } {y}_{C} = 7$
$x = \frac{1 + 9 + 6}{3} = \frac{16}{3} = 5.33$
$y = \frac{3 + 2 + 7}{3} = \frac{12}{3} = 4$
$c e n t r o i d \left(5.33 , 4\right)$
$\text{let distance between origin and centroid be l;}$
$l = \sqrt{{\left({y}_{\text{centroid")-y_("origin"))^2+(x_("centroid")-x_("origin}}\right)}^{2}}$
$l = \sqrt{{\left(5.33 - 0\right)}^{2} + {\left(4 - 0\right)}^{2}}$
$l = \sqrt{{\left(5.33\right)}^{2} + {4}^{2}}$
$l = \sqrt{28.41 + 16}$
$l = \sqrt{44.41}$
$l = 6.67$
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Lesson Video: Adding One-Digit Numbers by Making Ten | Nagwa Lesson Video: Adding One-Digit Numbers by Making Ten | Nagwa
# Lesson Video: Adding One-Digit Numbers by Making Ten Mathematics • 1st Grade
In this video, we will learn how to add two one-digit numbers by first making ten.
09:12
### Video Transcript
Adding One-Digit Numbers by Making 10
In this lesson, we’re going to learn how to add two one-digit numbers together by first making 10. Now, before we start, let’s remind ourselves about something. And this is the fact that adding a number on to 10 is quick and is also easy. We know that 10 plus two equals 12. Five more than 10 is 15. As we’ve just said, adding a number on to 10 is quick and easy. Perhaps you can see the patterns in the numbers that we’re adding and the totals that we get. Now, in this video, we’re going to be adding two one-digit numbers together. But because adding a number on to 10 is quick and easy, we’re going to make one of the numbers into 10 first. Let’s see how we do this.
Now, let’s imagine that we’ve got a packet of cookies and we open it. And we tip out two piles of cookies. We have a pile of seven cookies and six more cookies. And to find the total, we need to add seven and six together. Now, one way we could model this addition is using ten frames. Seven plus six equals what? Now, we could find the answer by starting with the number seven and counting on six more. But in maths, it’s always good to find quick and easy ways to find the answer.
Now, we know that adding numbers to 10 is quick and easy. What if we split up the number six and move some of our cookies into the first ten frame so that we make 10? Then we’d be adding 10 and another number together. That would be quick and easy to do. What number do we know goes with seven to make 10? Well, we know that seven and three make 10. And so we’re going to have to split up our six cookies into the three cookies that we’re going to move and then three more cookies that are left behind because we know three and three makes six.
Find an expression that has the same answer as six plus eight.
Underneath the numbers six and eight, we can see two ten frames, and these are going to help us to answer the question. There are six pink squares and eight blue squares. We can see that the eight blue squares have been split up. Four of them have been moved across to the first ten frame. To do this, we’ve had to split up the number eight into four. These are the four that we’ve moved. And we have four left. We know that four and four go together to make eight. By putting some of our eight squares into this ten frame, we’ve managed to fill it. Six and four go together to make 10.
All that we have to do to find the same answer is to add on what’s left of the number eight, which is four. So we can say that an expression that has the same answer as six plus eight is 10 plus four. And you know, by splitting up the number eight to make 10 like this, we can make the calculation much easier to work out. The expression that has the same answer as six plus eight is 10 plus four.
Ethan is adding by making 10. Nine plus six equals what. Which calculation should he do first? Nine plus one equals 10 or nine plus five equals 14. And then which calculation should he do next? 10 plus six equals 16 or 10 plus five equals 15.
In this question, we can see that Ethan is adding two one-digit numbers together, nine plus six. And in the first sentence, we’re told how Ethan does this. He’s adding by making 10. Now, if Ethan wants to add nine and six together, why might he want to make 10? Or perhaps he knows that adding any number to 10 is a quick and an easy way to find an answer. And if we look at the diagram, we can see what Ethan’s done. We can see that he started off by writing the addition nine plus six. But he’s done something interesting to the number six. He split up the number six using a part-whole model. And he split it into the number one and five because one and five are two parts that go together to make six.
Now, why might Ethan have split the number six into one and five? Well, he wants to make 10. That’s why. And Ethan knows that if he puts one with nine, he can make the number 10. That’s how we know the calculation that Ethan should do first is nine plus one equals 10. Now, adding a number on to 10 is a quick and easy thing to do. So all Ethan needs to do to find the total is to add what’s left of the number six, the other part, which is five. This is how we know that the calculation Ethan needs to do next is 10 plus five equals 15. Ethan found the answer to nine plus six by splitting up the number six and using it to make 10. His first calculation was nine plus one equals 10. And then he added what was left of the number six. 10 plus five equals 15.
Scarlett added nine and seven by making 10. What number is missing from her solution? Nine plus seven equals 16. 10 plus what equals 16.
To understand what Scarlett’s done here, we could try to answer the question ourselves. And we’re going to try to make the addition, nine plus seven, a little easier to work out by making 10. Here’s what nine plus seven might look like using two ten frames. And one way to find the answer, we could start with the number nine and then count on seven more. But you know, adding a number to the number 10 is always a quick way to find the total. So what we can do to help ourselves is to turn the number nine into 10.
What goes with nine to make 10? Well, we know nine plus one more makes 10. To find the extra one that we need, we’re going to split our number seven up into one and whatever is left. And we know that one and six make seven. We can show this using our ten frames by moving across one counter. Now, we’ve made 10 in the first ten frame. We just need to add the six counters in the second ten frame. We’ve turned our calculation nine plus seven into 10 plus six. 10 plus six is 16. And so we know that nine plus seven must be 16 too. Scarlett added nine and seven by making 10. The number that’s missing from her solution is the number six. 10 plus six equals 16.
Now, what have we learned in this video? Well, firstly, we’ve reminded ourselves that we can quickly add numbers on to 10. And so we’ve learned how to add two one-digit numbers together by first making 10.
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# 5 Best Examples of Basic Multiplication
This video shows five examples of basic multiplication. When looking at a simple multiplication problem, you should be able to immediately be able to know what the answer, or product, is going to be. If it makes it easier, you can rewrite a multiplication problem as an addition problem. For example, you can rewrite the problem 2×6 as 6+6 to get the product 12.
Basic Multiplication
## Basic Multiplication
We’re going to take a look at some basic multiplication. So we just have five problems up here on the board that we’re going to work together to find the product of. Because that’s what the directions say: find the following products.
So the product is what you get after you do the multiplication. It’s what goes after the equals sign. And so it’s important that you can just look at these numbers and be able to immediately tell what the product is going to be. Because for any multiplication that’s going on, as long as there’s no number higher than 12, you want to be able to just look at those numbers and be able to tell what the answer’s going to be because that’s going to save you a lot of time in future math problems. And so notice here throughout these five problems we never see a number higher than 12.
So your goal should be to be able to eventually just look at all of these and know immediately what the product is going to be. But if you don’t, that’s fine because we’re going to go ahead and work through these problems. So we start out with 2 times 6. So this one’s pretty easy because we’re multiplying 2 times 6 and so this is basically 6 plus 6 because we have this number right here is telling us we have it’s the number we’re dealing with and then 2 right here is telling us how many of that number we have. So this is just 6 plus 6. But this is multiplication, so if it makes more sense to you, you can write it like this, so we could also look at it like this: 2 plus 2 plus 2 plus 2 plus 2 plus 2.
I hope I’m not confusing you; I’m just showing you some different ways to approach this problem. So what we’re doing here is we’re just adding two 6s together so the answer is 12. Or we’re adding six 2s together, depending on how you want to look at it so 2 plus 2 is 4, plus 2 is 6, plus 2 is 8, plus 2 is 10, plus 2 is, again 12.
Now when we come to this one, 5 times 3, here we know this has to be a multiple of 5. And so multiples of 5 always end up in 0 or 5. And so here, it’s really easy, because multiples of 5 are really easy to deal with in your head. So 5 plus 5 is 10, plus another 5 is 15.
Now we’re going to go ahead and skip this one because this one may be the hardest one, so we’re going to come back to it. So now we have 4 times 11. And so whenever you are multiplying by 11, it’s really easy because you’re going to have the same digit twice. So if you do 2 times 11, the answer’s going to be 22. If you do 3 times 11, the answer’s 33. So 4 times 11, the answer’s going to be 44. So you’re just going to write this number twice.
Now we come to 12 times 10. Now this seems like a really hard problem, and the easiest thing here may be just to work out this problem. 12 times 10, and work out the multiplication problem. So you get 120 that way. But an easier way to do it, is any time you’re multiplying something by 10, you can just take a 0 and add it to this number right here, so we just add a 0 to the end of 12 and get 120. Now this final problem is going to take a little bit longer to work out: it’s 7 times 9.
So what we’re dealing here is we’re dealing with 9 plus 9 plus 9 plus 9 plus 9 plus 9 plus 9. We’re dealing with seven 9s all just added together. And so one way to approach it is, if you don’t know what 7 times 9 is, think about another multiple of 9. So maybe you know what 5 times 9 is, so you know that’s 45. So we multiply 5 times 9 and we get 45. But notice here that 5 is 2 less than 7, so we still need to add two 9s to 45. So here we can just add 9 plus 9. So 45 plus 9 is going to be 54, plus another 9 is, you guessed it, 63. So these are the products of these five multiplication problems.
So again, you want, as long as there’s not a number higher than 12 in the multiplication problem you’re dealing with, your goal is to eventually just look at all of these and know immediately, 2 times 6, that’s 12. 7 times 9, that’s 63. 12 times 10, that’s 120. Because you’re going to need to know the answers to these multiplication problems many many times in life, probably on a daily basis, and so it’s very important to build this strong foundation in math.
And so I hope my explanations helped you, and remember a couple of the tricks I taught you, especially with 4 times 11: any number times 11, you’re just doubling this number right here. So 4 times 11: 4-4, 44. 5 times 11, answer’s going to be 55. And then any time you’re multiplying by 10, just take the 0 from the 10, and put it at the end of this number, so that’s how we got 120.
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by Mometrix Test Preparation | Last Updated: February 13, 2019
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Friday, February 23, 2024
HomeMath4th Grade Numbers Worksheets | Place Worth Chart
4th Grade Numbers Worksheets | Place Worth Chart
In 4th grade numbers worksheets we are going to resolve learn and
write massive numbers, signify 5 and 6-digit numbers on abacus, use of place worth chart to put in writing a quantity in expanded
kind, signify the massive quantity on the abacus, write the quantity in customary kind, write a big quantity in phrases and figures, place worth of a quantity, customary kind and prolonged type of a quantity, evaluate with one other quantity and organize numbers in ascending and descending
order, successor and predecessor of a quantity and rounding off the
numbers.
We now have already learnt to learn and write the 4-digit numbers. Additionally, we’ve got learnt number-names, comparability of numbers and ordering of numbers. Let’s revise our studying earlier than we go additional:
1. Who am I?
(i) I’m the smallest 5-digit quantity: _______________
(ii) I’m the predecessor of 8,25,481: _______________
(iii) I’m yet one more than 99,999: _______________
(iv) My place worth in 9,83,351 is ten thousand: _______________
(v) I’m the biggest 6 – digit quantity: _______________
2. Fill within the blanks:
(i) 16 is a a number of of 8 and ………….
(ii) …………. is a a number of of 4 and seven
(iii) …………. is an element of each quantity.
(iv) The best issue of 364 is ………….
(v) …………. is the
smallest prime quantity.
(vi) …………. is neither a primary or a composite quantity.
(vii) A quantity which is divisible by each 2 and three is
divisible by ………….
(viii) The least frequent a number of of and 12 is ………….
(ix) 10(^{5}) = ………….
(x) 60(^{0}) = ………….
A. Write the number-names for the next:
(i) 6,309
(ii) 4,209
(iii) 1,777
(iv) 3,408
(v) 4,011
(vi) 8,888
(vii) 9,306
(viii) 5,679
3. Write the quantity names for the next numerals.
(i) 89,53,828
(ii) 25,87,463
(iii) 341,906,235
(iv) 54,928,329
(v) 8,842,935
B. Write the numerals for the given number-names:
(i) Three thousand one hundred-fifty.
(ii) One thousand 4 hundred-four.
(iii) Two thousand 2 hundred fifty-three.
(iv) Three thousand eight hundred twenty-five.
(v) Six thousand 4 hundred seventy-two.
(vi) 4 thousand seven hundred forty-eight.
4. Write the numerals for the next quantity names.
(i) Seventy 9 lakhs ninety six thousand, 2 hundred and
thirty eight.
(ii) 4 hundred and thirty a million, 9 hundred and
seven thousand, 5 hundred and sixty three.
(iii) Two crores, seventy three lakhs, fifty six thousand,
eight hundred and seventeen.
(iv) Seventy million 4 hundred and eight.
5. Full the given desk.
S.No. Numeral Quantity identify (i) 24,998 ________________________________ (ii) __________ Three lakh twenty three thousand 4 hundred two (iii) 1,84,305 ________________________________ (iv) __________ Six lakh ten thousand 300 seven (v) 3,10,706 ________________________________
J. Type the best and the smallest 4-digit numerals utilizing the next digits:
S.No. Digits Best Numeral Smallest Numeral (i) 2, 4, 6, 1 6421 1246 (ii) 2, 9, 0, 8 _____ _____ (iii) 4, 8, 6, 7 _____ _____ (iv) 7, 6, 5, 4 _____ _____
6. Type the biggest and smallest 6–digit numbers utilizing the
given digits.
S.No. Digits Best Quantity Smallest Quantity (i) 3, 9, 4, 1, 7 __________ __________ (ii) 9, 6, 1, 5, 2 __________ __________ (iii) 3, 0, 8, 2, 7, 1 __________ __________
G. Prepare the next numbers in ascending order:
(i) 2916, 4531, 9814, 6005
(ii) 7308, 7038, 7348, 7304
I. Write the next numerals in descending order:
(i) 4,444; 6.666; 7,777; 8,888
(ii) 9,624; 8,400; 2,046; 3,098
(iii) 6,438; 4,846; 6,396; 9,666
(iv) 6,666; 7,077; 8,097; 9,999
7. Write the place worth of underlined digits within the given
blanks.
(i) 2103
(ii) 4,00,493
(iii) 9,14,825
(iv) 2,13,921
8. Rewrite the next numbers utilizing the Indian
place-value chart.
(i) 563,789
(ii) 53,670,245
(iii) 8,634,186
9. Rewrite the next numbers utilizing the Worldwide
place-value chart.
(i) 7,28,70,246
(ii) 85,67,432
(iii) 2,68,961
10. Signify the given numbers on the abacus:
(i) 67,751
(ii) 5,44,301
(iii) 8,54,206
C. Write every of the given numerals within the expanded kind:
(i) 1296
(ii) 2492
(iii) 3004
(iv) 5020
(v) 6999
(vi) 8248
11. Write the given numbers in expanded kind.
(i) 63,498 _______________
(ii) 7,70,052 _______________
(iii) 9,72,855 _______________
(iv) 3,88,941 _______________
(v) 8,00,527 _______________
D. Write the brief type of every of the next:
(i) 3000 + 500 + 80 + 4 = __________
(ii) 6000 + 90 + 8 = __________
(iii) 4000 + 70 + 1 = __________
(iv) 9000 + 400 + 7 = __________
12. Write the given quantity in customary kind.
(i) 80,000 + 5000 + 200 + 90 + 6 _______________
(ii) 60,000 + 7000 + 200 + 8 _______________
(iii) 4,00,000 + 70,000 + 5000 + 50 + 1 _______________
H. Evaluate the numerals utilizing >, < or =
1. 1093 __ 1903
2. 4285 __ 4825
3. 6092 __ 9206
4. 4001 __ 4001
5. 1204 __ 1205
6. 9909 __ 9099
7. 8902 __ 2908
8. 4260 __ 4260
9. 4242 __ 2424
10. 3600 __ b3600
13. Evaluate the given numbers. Put > , < or = signal.
(i) 35,493 _____ 34,395
(ii) 2,90,018 _____ 8,10,092
(iii) 7,44,582 _____ 7,12,298
(iv) 8,17,736 _____ 8,30,559
14. If within the quantity 2,39,987; 3 is changed by 8 and the
hundred place is changed by 1, then what quantity is shaped? Additionally give its
quantity identify.
_______________; _______________
15. Fill within the
blanks:
(i) One lakh
is a ……………………………… digit quantity.
(ii) The
distinction between the best 4-digit quantity and the best 3-digit quantity
is ………………………………
(iii) A 5
digit quantity begins with ……………………………… place.
(iv) The sum
of the place worth of two 5s in 3,58,654 is ………………………………
(v) The
successor of 5 lakh, fifty thousand, 5 hundred fifty is ………………………………
(vi) 1 lakh =
……………………………… thousand
(vii) 7 lakh =
……………………………… hundred
(viii) Type the
largest 6-digit quantity with 9 on the tens place and with out repeating any
digits. ………………………………
(ix) Type the
smallest 5-digit quantity with 0 on the hundreds place and with out repeating any
digits. ………………………………
(x) The
predecessor of seven,85,000 in phrases might be ………………………………
(xi) The
subsequent quantity within the sequence: 7,46,952, 7,57,952, 7,68,952, ………………………………
(xii) The
quantity 100 greater than 6,66,666 is ………………………………
(xiii) 1000
lower than 5,98,456 is ………………………………
16. Given
under is the world of 5 states of South India.
S.NO. Metropolis Space (in Sq. km) (i) Telangana 1,12,077 (ii) Andhra Pradesh 1,62,969 (iii) Kerala 38,763 (iv) Karnataka 1,91,791 (v) Tamil Nadu 1,30,058
(i) Prepare
the states in growing order to their space.
(ii) Which
state has digit 9 on the place worth of hundred and ones.
(iii) Give
the quantity identify for the world of Andhra Pradesh and Karnataka.
(iv)
Signify the world of Telangana on the babacus.
(v) Write
the world of Tamil Nadu in expanded kind.
(vi) Give
the place worth of digit 8 in space of kerala.
E. Discover the successor of every of the next:
(i) 8248 = __________
(ii) 3710 = __________
(iii) 9924 = __________
(iv) 2197 = __________
F Discover the predecessor of every of the next:
(i) 5000 = __________
(ii) 6595 = __________
(iii) 2012 = __________
(iv) 9892 = __________
17. Write
the predecessor and successor of the given numbers.
S.NO. Predecessor Quantity Successor (i) __________ 7,56,993 __________ (ii) __________ 8,99,999 __________ (iii) __________ 6,30,050 __________ (iv) __________ 9,82,311 __________
18. Write the corresponding Roman numerals.
(i) 77
(ii) 58
(iii) 29
(iv) 67
19. Write the elements of the next numbers.
(i) 26
(ii) 60
20. Prime factorize the next numbers utilizing division
methodology.
(i) 256
(ii) 96
21. Discover the L.C.M of the next numbers.
(i) 26, 28, 24
22. Discover the HCF of the next numbers
(i) 96, 80
264532
+ 121684
+ 385086
__________
24. Subtract the
following numbers.
8642710
– 5862412
___________
25. Multiply the next numbers.
56549
× 329
________
26. Divide the next numbers.
56458 ÷ 15
27. Work out the next issues.
(i) The feminine inhabitants of a village is 29,646. The male
inhabitants is 276 lower than the feminine inhabitants. What’s the male inhabitants
of the village? What’s the complete inhabitants?
(ii) A e-book accommodates 468 pages. What number of pages are there in
26 books?
(iii) The entire prepare fare for 12 individuals in \$4200. What’s
the fare for one individual.
Solutions on 4th Grade Numbers Worksheets are given under to verify the precise solutions of the questions.
Solutions:
1. (i) 10,000
(ii) 8,25,480
(iii) 1,00,000
(iv) 8
(v) 9,99,999
2. (i) 2
(ii) 28
(iii) 1
(iv) 364
(v) 2
(vi) 1
(vii) 6
(viii) 12
(ix) 100000
(x) 1
3. (i) Eighty 9 lakhs fifty-three thousand eight hundred twenty eight.
(ii) Twenty 5 lakhs eighty seven thousand 4 hundred sixty three.
(iii) 300 and forty a million, 9 hundred and 6 thousand, 2 hundred and thirty 5.
(iv) Fifty 4 million, 9 hundred twenty eight thousand, 300 twenty 9.
(v) Eight million, eight hundred and forty two thousand, 9 hundred thirty 5.
4. (i) 79,96,238
(ii) 431,907,563
(iii) 2,73,56,817
(iv) 70,000,408
5. (i) Twenty-four thousand, 9 hundred ninety eight
(ii) 3,23,402
(iii) One lakh, eighty-four thousand, 300 5.
(iv) 6,10,307
(v) Three lakh, ten thousand, seven hundred six.
6. (i) 97,431; 13,479
(ii) 96,521; 12,569
(iii) 8,73,210; 1,02,378
7. (i) 3
(ii) 4,00,000
(iii) 20
(iv) 2,00,000
8. Rewrite the next numbers utilizing the Indian
place-value chart.
(i) 563,789
(ii) 53,670,245
(iii) 8,634,186
9. Rewrite the next numbers utilizing the Worldwide
place-value chart.
(i) 7,28,70,246
(ii) 85,67,432
(iii) 2,68,961
10.
11. (i) 60,000 3,000 + 400 + 90 8
(ii) 7,00,000 + 70,000 + 0 + 50 + 2
(iii) 9,00,000 + 70,000 + 2,000 + 800 + 50 + 5
(iv) 3,00,000 + 80,000 + 8,000 + 900 + 40 + 1
(v) 8,00,000 + 0 + 0 + 500 + 20 + 7
12. (i) 85,296
(ii) 67,208
(iii) 4,75,051
13. (i) 35,493 > 34,395
(ii) 2,90,018 < 8,10,092
(iii) 7,44,582 > 7,12,298
(iv) 8,17,736 < 8,30,559
14. 2,89,187; Two lakh, eighty-nine thousand, 100
eighty-seven.
15. (i) 6
(ii) 9000
(iii) ten hundreds
(iv) 50,050
(v) 5,50,551
(vi) 100
(vii) 7,000
(viii) 8,76,590
(ix) 10,234
(x) Seven lakh, eighty-four thousand, 9 hundred ninety-nine
(xi) 7,79,952
(xii) 6,66,766
(xiii) 5,97,456
16. (i) Kerala,
(ii) Andhra
(iii)
Andhra Pradesh – One lakh, sixty-two thousand, 9 hundred sixty-nine,
Karnataka –
One lakh, ninety-one thousand, seven hundred ninety-one,
(iv)
(v) 1,00,000
+ 30,000 + 0 + 0 + 50 + 8
(vi) 8000
17.
S.NO. Predecessor Quantity Successor (i) 7,56,992 7,56,993 7,56,994 (ii) 8,99,998 8,99,999 9,00,000 (iii) 6,30,049 6,30,050 6,30,051 (iv) 9,82,310 9,82,311 9,82,312
18. (i) LXXVII
(ii) LVIII
(iii) XXIX
(iv) LXVII
19. (i) 1, 2, 13 and 26.
(ii) 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
20. (i) 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 or, 28
(ii) 2 × 2 × 2 × 2 × 2 × 3 or, 25 × 3.
21. (i) 2184
22. (i) 16
23. 771302
24. 2780298
25. 18604621
26. Quotient = 3763; The rest = 13
27. (i) Male inhabitants: 29370; Whole inhabitants: 59016
(ii) 12168
(iii) \$350
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# Introduction to Set Theory
## Motivation
Take a look at these objects; what are their similarities and differences? Can you classify them by some criterion?
You can notice that objects denoted by numbers $1, 3$ and $5$ are plane figures, i.e. two – dimensional (2D) shapes, while objects denoted by numbers $2, 4$ and $6$ are solid figures, i.e. three – dimensional (3D) shapes. Therefore, we have two collections: collection of plane figures and collection of solid figures. Obviously, figures in each of the two collections have the same property. We can tell that collection of objects with some property in common is called a set.
Some more examples of sets are: the set of people younger than $25$ years, the set of flowers, the set of four seasons, the set of string instruments and so on.
A set which has no elements is called an empty set or null set. We denote it by $\emptyset$.
## Set notation
Members of set are called elements. Notations for mentioned sets are: A = {$1, 3, 5$} and B = {$2, 4, 6$} or A = {triangle, quadrilateral, circle} and B = {parallelepiped, pentagonal prism, triangular pyramid}. Therefore, we denote sets by a single capital letter, simply list the elements, separate them by comma and put curly brackets.
The notation for set of plants is: {tulip, sunflower, rose, … }; we use the three dots when the set has a large number of elements or when it is infinite. Sets which aren’t infinite are called finite sets.
Size of the set $A$ is called cardinality number and it is denoted by $|A|$. Cardinalities of mentioned sets $A$ and $B$ are: $|A|=3$ and $|B|=3$. Cardinality of empty set is $0$.
Other way of representing a set is describing a property that its element must satisfy. For example, we denote set of all numbers which are divisible by $7$ as:
$\{$ x: x is divisible by 7 $\}$ or $\{$ x|x is divisible by 7 $\}$.
If $x$ is an element of a set $A$, we write: $x \in A$ and say ”$x$ belongs to $A$”.
If $x$ isn’t the element of a set $A$, we write: $x \notin A$ and say ”$x$ doesn’t belong to $A$”.
Example 1: Write a collection of vowels in English alphabet. How many elements does it have? Are letters $b, d$ and $o$ elements of that collection?
Solution:
$V= \{a, e, i, o, u\}$, it has $5$ elements.
$b \notin V$, $d \notin V$, $o \in V$.
## Properties of sets
The order in which elements of a set are listed is not important. For example, $\{a, e, i, o, u\} = \{u, o, a ,i, e\}$.
If some elements are repeated, set is still the same. For example, $\{3, 3, 7, 8\} = \{3, 7, 8\}$.
Two sets are equal if and only if they have the same elements. For example, $A$ = set of primary colors and $B = \{red, blue, yellow\}$ are equal.
## Subsets
A set $A$ is said to be a subset of the set $B$ if and only if every element of set $A$ is also the element of set $B$. We write:
$A \subseteq B$.
In that case, $B$ is the superset of set $A$.
Empty set is a subset of every set.
If $A$ is not the subset of $B$, we write: $A \not \subseteq B$.
Universal set U is a set that is superset of all sets.
Example 2: Is $A$ subset of $B$, where $A = \{$x: x is a whole number greater than $4$ $\}$ and $B = \{$x: x is even number $4<x<15$ $\}$?
Solution:
$A= \{$$5, 6, 7, 8, …$$\}$ and $B = \{$$6, 8, 10, 12, 14$$\}$
$A \not \subseteq B$, $B \subseteq A$
A set $A$ is said to be a proper subset of the set $B$ if and only if every element of set $A$ is also the element of set $B$, but there exists at least one element which is in $B$ and not in $A$. We write:
$A \subset B$.
If $A$ is not a proper subset of $B$, we write: $A \not \subset B$.
Example 3: Find proper subsets of set $A=\{$$a, b$$\}$.
Solution:
Proper subsets are: $\emptyset$, $\{$a$\}$, $\{$b$\}$.
Note: $A=B \Leftrightarrow (A \subseteq B \wedge B \subseteq A)$
## Power set
A power set of a set $S$, denoted by $\mathcal{P}(S)$, is a set of all the subsets of a set. If a set has $n$ elements, power set has $2^{n}$ elements.
Example 4: Find $\mathcal{P}(S)$, where $S=\{a, b\}$.
Solution:
$\mathcal{P}(S)= \{\emptyset, \{a\}, \{b\}, \{a, b\}\}$
|
## Mean: Measure of Central Tendency
The measure of Central Tendency Mean (also know as average or arithmetic mean) is used to describe the data set as a single number (value) which represents the middle (center) of the data, that is average measure (performance, behaviour, etc) of data. This measure of central tendency is also known as measure of central location or measure of center.
Mathematically mean can be defined as the sum of the all values in a given dataset divided by the number of observations in that data set under consideration. The mean is also called arithmetic mean or simply average.
Example: Consider the following data set consists of marks of 15 student in certain examination.
50, 55, 65, 43, 78, 20, 100, 5, 90, 23, 40, 56, 70, 88, 30
The mean of above data values is computed by adding all these values (50 + 55 + 65 + 43 + 78 + 20 + 100 + 5 + 90 + 23 + 40 + 56 + 70 + 88 + 30 = 813) and then dividing by the number of observations added (15) which equals 54.2 marks, that is
$\frac{50 + 55 + 65 + 43 + 78 + 20 + 100 + 5 + 90 + 23 + 40 + 56 + 70 + 88 + 30 }{15}=\frac{813}{15}=54.2$
The above procedure of calculating the mean can be represented mathematically
$\mu= \frac{\sum_{i=1}^n X_i}{N}$
The Greek symbol $\mu$ (pronounced “mu”) is the representation of population mean in statistics and $N$ is the number of observations in the population data set.
The above formula is known as population mean as it is computed for whole population. The sample mean can also be computed in same manner as population mean is computed. Only the difference is in representation of the formula, that is,
$\overline{X}= \frac{\sum_{i=1}^n X_i}{n}$.
The $\overline{X}$ is representation of sample mean and $n$ shows number of observations in the sample.
The mean is used for numeric data only. Statistically the data type for calculating mean should be Quantitative (variables should be measured on either ratio or interval scale), therefore, the numbers in data set can be continuous and/ or discrete in nature.
Note that mean should not be computed for alphabetic or categorical data (data should not belong to nominal or ordinal scale). Mean is influenced by very extreme values in data, i.e. very large or very small values in data changes the mean drastically.
For other measures of central tendencies visit: Measures of Central Tencencies
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A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years
Question:
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.
Solution:
There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is $\frac{1}{15}$.
The given information can be compiled in the frequency table as follows.
$P(X=14)=\frac{2}{15}, P(X=15)=\frac{1}{15}, P(X=16)=\frac{2}{15}, P(X=16)=\frac{3}{15}$
$P(X=18)=\frac{1}{15}, P(X=19)=\frac{2}{15}, P(X=20)=\frac{3}{15}, P(X=21)=\frac{1}{15}$
Therefore, the probability distribution of random variable X is as follows.
Then, mean of X = E(X)
$=\sum X_{i} P\left(X_{i}\right)$
$=14 \times \frac{2}{15}+15 \times \frac{1}{15}+16 \times \frac{2}{15}+17 \times \frac{3}{15}+18 \times \frac{1}{15}+19 \times \frac{2}{15}+20 \times \frac{3}{15}+21 \times \frac{1}{15}$
$=\frac{1}{15}(28+15+32+51+18+38+60+21)$
$=\frac{263}{15}$
$=17.53$
$\mathrm{E}\left(\mathrm{X}^{2}\right)=\sum \mathrm{X}_{i}^{2} \mathrm{P}\left(\mathrm{X}_{i}\right)$
$=(14)^{2} \cdot \frac{2}{15}+(15)^{2} \cdot \frac{1}{15}+(16)^{2} \cdot \frac{2}{15}+(17)^{2} \cdot \frac{3}{15}+$
$(18)^{2} \cdot \frac{1}{15}+(19)^{2} \cdot \frac{2}{15}+(20)^{2} \cdot \frac{3}{15}+(21)^{2} \cdot \frac{1}{15}$
$=\frac{1}{15} \cdot(392+225+512+867+324+722+1200+441)$
$=\frac{4683}{15}$
$=312.2$
$\therefore$ Variance $(X)=E\left(X^{2}\right)-[E(X)]^{2}$
$=312.2-\left(\frac{263}{15}\right)^{2}$
$=312.2-307.4177$
$=4.7823$
$\approx 4.78$
Standard derivation $=\sqrt{\operatorname{Variance}(\mathrm{X})}$
$=\sqrt{4.78}$
$=2.186 \approx 2.19$
|
# Properties of Binary Tree
In this post, properties of binary are discussed.
1) The maximum number of nodes at level ‘l’ of a binary tree is 2l-1.
Here level is number of nodes on path from root to the node (including root and node). Level of root is 1.
This can be proved by induction.
For root, l = 1, number of nodes = 21-1 = 1
Assume that maximum number of nodes on level l is 2l-1
Since in Binary tree every node has at most 2 children, next level would have twice nodes, i.e. 2 * 2l-1
2) Maximum number of nodes in a binary tree of height ‘h’ is 2h – 1.
Here height of a tree is maximum number of nodes on root to leaf path. Height of a tree with single node is considered as 1.
This result can be derived from point 2 above. A tree has maximum nodes if all levels have maximum nodes. So maximum number of nodes in a binary tree of height h is 1 + 2 + 4 + .. + 2h-1. This is a simple geometric series with h terms and sum of this series is 2h – 1.
In some books, height of the root is considered as 0. In this convention, the above formula becomes 2h+1 – 1
3) In a Binary Tree with N nodes, minimum possible height or minimum number of levels is ? Log2(N+1) ?
This can be directly derived from point 2 above. If we consider the convention where height of a leaf node is considered as 0, then above formula for minimum possible height becomes ? Log2(N+1) ? – 1
4) A Binary Tree with L leaves has at least ? Log2L ? + 1 levels
A Binary tree has maximum number of leaves (and minimum number of levels) when all levels are fully filled. Let all leaves be at level l, then below is true for number of leaves L.
``` L <= 2l-1 [From Point 1]
l = ? Log2L ? + 1
where l is the minimum number of levels.```
5) In Binary tree where every node has 0 or 2 children, number of leaf nodes is always one more than nodes with two children.
``` L = T + 1
Where L = Number of leaf nodes
T = Number of internal nodes with two children```
See handshaking Lemma and tree for proof.
In the next article on tree series, we will be discussing different typas of Binary Trees and their properties.
|
# 20 Number Series Questions with Tips and Tricks (Latest Pattern)
## For All Banking Exams, IBPS PO, SBI PO, SSC CGL
How to Solve Number Series Questions ?
First of All for solving these types of questions, You need to practice daily to understand the logic behind it. In these types of questions there is no certain Tricks, You need to judge from the pattern of the questions. But today we will tell you some basic patterns, It may help to understand the Logic – Before solving questions try to understand the pattern from the below questions. Take your time, don't see answers or explanations. First Try to solve it by own.
Generally in these types, Questions may be asked in 2 ways -
• To find the missing Number in the series.
• To find the wrong one in the series.
Today we will discuss both types of questions. Try to solve it by own. If anyone need solution just give your comment. our team members will tell you the logic.
Types of Pattern:
Type 1. Prime Number Series
In these Types of Questions you will get a series of Prime Number or -1, +1, etc. Here, We will explained the types of series in Prime Number. It may be +2, -2, +3, -3, ...... You need to understand the pattern from the Questions........
Series 1 - 2, 3, 5, 7, 11, 13, 17, 19, 23 .......... (Means its based on Prime Numbers)
Series 2 - 1, 2, 4, 6, 10, 12, 16, 18, 22, ..........(Means its based on Prime Numbers -1)
Series 3 - 3, 4, 6, 8, 12, 14, 18, 20, 24, ..........(Means its based on Prime Numbers +1)
Type 2. Difference between the numbers
In these Types of Questions you will get same difference, either in + or - ..... The Series will be either increasing or decreasing. Its very very easy to understand. Now These days you won't get these types of questions directly.
Series 1 - 2, 4, 6, 8, 10, 12, 14, 16, 18 .............. (Means its increasing with +2)
Series 2 - 23, 21. 19, 17, 15, 13, 11, 9, 7, ..........(Means its decreasing with -2)
So you need to be smart wile solving these types of questions- Try to understand the below pattern (Its also based on Difference between numbers)
Type 3. Difference between the numbers - ${{x}^{2}},{{x}^{3}}$
Series 1 - 1, 2, 6, 15, 31, 56 .......
Series 2 - 56, 31, 15, 6, 2, 1 ........ (Just Reverse of Series 1)
Type 4. Difference between the number is - ${{x}^{2}}+1,{{x}^{3}}+1,{{x}^{2}}+x,{{x}^{2}}-x,{{x}^{2}}+2,{{x}^{2}}-2,.....$
In these types you will get the difference like this.... It may be + or - but the order is same. You before solving try to find out the difference then apply all these to find out the actual tricks.
Now Start Solving the below Questions
Directions (Q.1 –Q.3): In the following number series one number is wrong. Find out the wrong number.
Question 1.
• 2, 3, 10, 38, 172, 885
1) 3
2) 172
3) 38
4) 10
5) 885
Question 2.
• 9, 25, 90, 488, 2926, 20480
1) 90
2) 488
3) 20480
4) 25
5) 2926
Question 3.
• 134, 141, 152, 167, 185, 209
1) 167
2) 185
3) 152
4) 141
5) 209
Directions (Q.4-Q.6): What value should come in the place of question mark (?) in the following number series?
Question 4.
• 21, 25, 59, 193, ?
1) 797
2) 787
3) 794
4) 782
5) 789
Question 5.
• 27, 44, 65, 90, ?
1) 109
2) 121
3) 115
4) 119
5) 117
Question 6.
• 19, 19, 133, 1729, ?
1) 32851
2) 34851
3) 32861
4) 32751
5) 33149
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# Rational Numbers
Use our extensive free resources below to learn about Rational Numbers.
This material is an extract from our National 5 Mathematics: Curriculum Breakdown course led by instructor Andrew Eadie. Enrol in the full course now and gain access to over 100 detailed topic breakdowns, 48 video tutorials (20 hours) and 39 quizzes spanning the entire curriculum.
# What are Rational Numbers?
Number Theory is the branch of mathematics that deals with the properties of and relationships between numbers. It is a fundamental area of study which underpins our whole understanding of mathematics. It can be a complex topic, so we will not go into great depth here, but there are some elementary aspects we can use to our advantage.
The very basics of Number Theory involve defining key number sets. In mathematics, a set is a collection of distinct objects which share specific characteristics, with the set then considered as an object in its own right. Different number sets take on different characteristics, and if we know that a number lies within a specific set, then we know that it possesses the characteristics which define the set. This allows us to study numbers and how they work together more easily.
The most common sets of numbers, along with the symbols we use to represent each set, are outlined below:
#### \boldsymbol{ℕ} – Natural Numbers
The Natural Numbers, symbol , can be described in common language as “counting numbers” – the numbers that occur commonly and obviously in nature. They are the set of whole, non-negative numbers beginning with 1 and increasing by 1 at each interval going on to infinity:
#### \boldsymbol{𝕎} – Whole Numbers
The Whole Numbers, symbol 𝕎, are basically just the set of Natural Numbers with the inclusion of the concept of zero:
#### 𝕎: \boldsymbol{{0, 1, 2, 3, 4 …}}
Notice that the Whole Numbers include everything that was in the Natural Numbers (the set that came before it), plus zero. This will be a common theme with each subsequent number set we discuss here:
Each new number set in this list contains all the numbers from the previous set, plus additional terms that belong in the new set.
#### \boldsymbol{ℤ} – Integers
Integers, symbol , include all of the Whole Numbers plus their negative equivalents:
#### \boldsymbol{ℚ} – Rational Numbers
The Real Numbers, symbol , include all of the Rational Numbers, plus Irrational Numbers:
Irrational Numbers are numbers which cannot be written as fractions of Integers and which have decimal expansions which go on forever and do not start to repeat.
Basically, you cannot write Irrational Numbers as exact decimals. Sound familiar? Think surds! In the Surds topic we learned that surds are the exact way to represent the roots of numbers whose roots cannot be expressed exactly using decimals. Since surds cannot be expressed exactly as decimals, surds are irrational and therefore fit within the Real Number set:
#### ℝ: \boldsymbol{\{… -10, -5.5, -\sqrt{5}, -1, 0, \frac{1}{3}, 1, \sqrt{3}, \frac{9}{4}, \pi, 6 …\}}
In the number set above I have just given various random examples of Real Numbers as it is impossible to write them all out in chronological order (the same problem we had with writing out the Rational Numbers). To discuss a few examples:
If we try to figure out \sqrt3 as a decimal, we find \sqrt3=1.73205080757\ldots We see that we cannot find \sqrt3 as an exact decimal – it goes on forever and never starts to repeat itself. Therefore, \sqrt3 is an Irrational Number and it fits within the Real Number set.
If we try to figure out \pi as a decimal, we find \pi=3.14159265359\ldots We see that we cannot find \pi as an exact decimal – it goes on forever and never starts to repeat itself. Mathematicians and computer scientists have actually calculated \pi to trillions of decimal places and have thus far found that indeed it does seem to go on forever and never starts to repeat itself. Let’s be clear that unlike \sqrt{3}, \pi is not a surd, but it is another example of an Irrational Number (one of a special subset of the Irrational Numbers called Transcendental Numbers), and therefore it fits within the Real Number set.
The key principle here – that each new number set we defined contains all the numbers from the previous set, plus additional terms that belong in the new set – can be summarised in the diagram below which shows examples from each number set:
In the diagram above:
• The Natural Numbers () stand alone.
• The Whole Numbers (𝕎) include all of the Natural Numbers, plus zero.
• The Integers () include all of the Whole Numbers, plus their negatives.
• The Rational Numbers () include all of the Integers, plus all fractions of Integers.
• The Real Numbers () include all of the Rational Numbers, plus the Irrational Numbers (i.e. all surds and other special numbers like \pi).
You do not need to worry about memorising the symbols or exact definitions detailed in this section, but it is useful to know the difference between Integers, Rational Numbers and Real Numbers. Have you ever been asked to rationalise a denominator for example? You may know the process to do this, but what does that even mean? We will use the definitions we have learned here to understand this in more detail in the following topics of this module.
### Key Outcomes
In mathematics, a set is a collection of distinct objects which share specific characteristics, with the set then considered as an object in its own right.
Each new number set in this list:
– Natural Numbers
𝕎 – Whole Numbers
– Integers
– Rational Numbers
– Real Numbers
Contains all the numbers from the previous set, plus additional terms that belong in the new set.
A Rational Number () is any number that can be written as a fraction of two integers, i.e. \frac{a}{b} where both a and b belong to the Integers.
Irrational Numbers are numbers which cannot be written as fractions of Integers and which have decimal expansions which go on forever and do not start to repeat.
All surds are Irrational Numbers.
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# Giải vở bài tập Toán lớp 5 tập 2 trang 62, 63: Luyện tập Vận tốc
In this article, we will provide detailed solutions, including answers and step-by-step instructions, for solving the math exercises in Class 5 Part 2, pages 62 and 63, which focus on speed. Whether you need to review the concepts covered in the exercise or you simply want to improve your math skills, we’ve got you covered. Let’s dive into the detailed solutions together.
## Solving Exercise 1 on Page 62, Class 5 Part 2 Math Workbook
Problem: An car travels at a speed of 21.6 km/h. Calculate the car’s speed in the following units:
a. m/minute
b. m/second.
Solution:
a) To calculate the speed, we need to convert the distance and time units.
• Conversion: 21.6 km = 21600 m; 1 hour = 60 minutes.
• Speed is calculated by dividing the distance by the time.
b) To calculate the speed, we need to convert the distance and time units.
• Conversion: 21.6 km = 21600 m; 1 hour = 3600 seconds.
• Speed is calculated by dividing the distance by the time.
a. The car’s speed in m/minute is:
21600/60 = 360 (m/minute).
b. The car’s speed in m/second is:
21600/3600 = 6 (m/second).
## Solving Exercise 2, Class 5 Part 2 Math Workbook, Page 62
Problem: Fill in the blanks (following the given pattern).
Solution:
• To find the speed, we need to convert the time units into decimal numbers with hours as the unit.
• Speed is calculated by dividing the distance by the time.
The speed in the first blank is:
63/1.5 = 42 km/h.
3 hours 30 minutes is equal to 3.5 hours.
The speed in the second blank is:
14.7/3.5 = 4.2 km/h.
1 hour 15 minutes is equal to 1.25 hours.
The speed in the third blank is:
1025/1.25 = 820 km/h.
3 hours 15 minutes is equal to 3.25 hours.
The speed in the fourth blank is:
79.95/3.25 = 24.6 km/h.
## Solving Exercise 3, Class 5 Math Workbook, Page 62, Part 2
Problem: In a running competition, an athlete runs 1500m in 4 minutes. Calculate the athlete’s running speed in m/second.
Solution:
• To calculate the running speed, we need to convert the time units into seconds, keeping in mind that 1 minute is equal to 60 seconds.
• Running speed is calculated by dividing the distance by the time.
4 minutes is equal to 240 seconds.
The athlete’s running speed is:
1500/240 = 6.25 (m/second).
## Solving Exercise 4, Page 63, Class 5 Math Workbook, Part 2
Problem: Two cities, A and B, are 160km apart. A car travels from A at 6:30 am and arrives at B at 11:15 am. The car stops for 45 minutes along the way. Calculate the car’s speed, considering the time spent on the break.
Solution:
• To find the time taken by the car from A to B, including the break time, we subtract the departure time from the arrival time.
• To find the time taken by the car from A to B, excluding the break time, we subtract the total break time from the total travel time.
• To find the car’s speed, we divide the distance by the time.
The time taken by the car from A to B, including the break time, is:
11:15 am – 6:30 am = 4 hours 45 minutes.
The actual travel time from A to B is:
4 hours 45 minutes – 45 minutes = 4 hours.
The car’s speed is:
160/4 = 40 (km/h).
## Click here to visit Kienthucykhoa.com for more educational resources and to download the Math Class 5 Part 2 Exercise Solutions on Pages 62, 63: Practicing Speed in Word and PDF formats, absolutely free of charge.
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# Into Math Grade 4 Module 21 Answer Key Solve Problems with Time and Measurement
We included HMH Into Math Grade 4 Answer Key PDF Module 21 Solve Problems with Time and Measurement to make students experts in learning maths.
## HMH Into Math Grade 4 Module 21 Answer Key Solve Problems with Time and Measurement
How much time do you spend on the bus?
Imagine you ride a bus to school every day. The clocks below show the times in the morning that you are picked up at home and dropped off at school.
Morning Bus Ride to School
Pick up: ________
Drop off : __________
Write the time below each clock. What is the total amount of time you spend on the bus going to school?
Pick up: 8:24 a.m.
Drop off : 8:53 a.m.
Explanation:
The pick up time is 8:24 a.m. and the drop off time is 8:53 a.m. The time difference between 8:24 a.m. and 8:30 a.m. is 6 minutes and 8:30a.m. to 8:53 a.m. is 23 minutes. Add 6 minutes with 23 minutes the sum is equal to 29 minutes. So, the total amount of time spend on the bus going to school is 29 minutes.
Turn and Talk
If it takes you 2 minutes less to ride home from school than to ride to school, how many minutes would you spend on the bus riding to and from school during a 5-day school week?
The pick up time is 8:24 a.m. and the drop off time is 8:51 a.m. The time difference between 8:24 a.m. and 8:30 a.m. is 6 minutes and 8:30a.m. to 8:51 a.m. is 21 minutes. Add 6 minutes with 21 minutes the sum is equal to 27 minutes. So, the total amount of time spend on the bus going to school is 27 minutes.
Complete these problems to review prior concepts and skills you will need for this module.
Time to the Half Hour
Read the clock. Write the time.
Question 1.
The time is 11:30.
Explanation:
The hours hand is in between 11 and 12. The minute hand is at 6 which implies 30 minutes(6 x 5). So, the time shown in the above clock is 11:30.
Question 2.
The time is 5:30.
Explanation:
The hours hand is in between 5 and 6. The minute hand is at 6 which implies 30 minutes(6 x 5). So, the time shown in the above clock is 5:30.
Question 3.
The time is 2:30.
Explanation:
The hours hand is in between 2 and 3. The minute hand is at 6 which implies 30 minutes(6 x 5). So, the time shown in the above clock is 2:30.
Count by Twos and Fives
Count to find the unknown numbers.
Question 4.
Count by fives.
20, _____, 30, 35, _______, 45, 50
Given series is 20, _____, 30, 35, _______, 45, 50.
As we are counting by 5 we have to add 5 to the given first number which is 20.
So, the unknown number in first blank is 20 + 5 = 25.
Likewise, after adding 5 to 35 gives 40(35 + 5).
So, the unknown number in the second blank is 40.
The unknown numbers in the given series is 20, 25, 30, 35, 40, 45, 50.
Question 5.
Count by twos.
30, 32, _______, ______, 38, _______
Given series is 30, 32, _______, ______, 38, _______.
As we are counting by 2 we have to add 2 to the given number which is 32.
So, the unknown number in first blank is 32 + 2 = 34.
Likewise, after adding 2 to 34 gives 36(34 + 2).
So, the unknown number in the second blank is 36.
After adding 2 to the number 38 gives 40(38 + 2).
So, the unknown number in the third blank is 40.
The unknown numbers in the given series is 30, 32, 34, 36, 38, 40.
Question 6.
Count by fives.
9, 14, 19, _____, _____, _____, _____, 44
Given series is 9, 14, 19, _____, _____, _____, _____, 44.
As we are counting by 5 we have to add 5 to the given number which is 19.
So, the unknown number in first blank is 19 + 5 = 24.
Likewise, after adding 5 to 24 gives 29(24 + 5).
So, the unknown number in the second blank is 29.
After adding 5 to the number 29 gives 34(29 + 5).
So, the unknown number in the third blank is 34.
After adding 5 to the number 34 gives 39(34 + 5).
So, the unknown number in the fourth blank is 39.
The unknown numbers in the given series is 9,14, 19, 24, 29, 34, 39, 44.
Elapsed Time
Find the elapsed time.
Question 7.
Start: 6:30 a.m.
End: 7:15 a.m.
The elapsed time is 45 minutes.
Explanation:
The amount of time that passes between two given time stamps(Start time and End time) is called as elapsed time.
The given start time is 6:30 a.m. and the the end time is 7:15 a.m. The time difference between 6:30 a.m. and 7:00 a.m. is 30 minutes and 7:00 a.m. to 7:15 a.m. is 15 minutes. Add 30 minutes with 15 minutes the sum is equal to 45 minutes. So, the elapsed time is 45 minutes.
Question 8.
Start: 10:00 p.m.
End: 10:35 p.m.
The elapsed time is 35 minutes.
Explanation:
The amount of time that passes between two given time stamps(Start time and End time) is called as elapsed time.
The given start time is 10:00 p.m. and the the end time is 10:35 p.m. The time difference between 10:00 p.m. and 10:30 p.m. is 30 minutes and 10:00 p.m. to 10:35 p.m. is 5 minutes. Add 30 minutes with 5 minutes the sum is equal to 35 minutes. So, the elapsed time is 35 minutes.
Question 9.
Start: 12:20 a.m.
End: 1:10 a.m.
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Home | | Maths 8th Std | Summary
# Summary
8th Maths : Chapter 4 : Life Mathematics : Summary
SUMMARY
When the S.P is more than the C.P, then there is a gain or profit. Profit/Gain =S.P – C.P.
When the S.P is less than the C.P, then there is a loss. Loss = C.P – S.P.
The profit or loss is always calculated on the cost price.
Selling price = Marked price – Discount.
Formulae
When the interest is compounded annually but rate of interest differs year by year, where a, b and c are interest rates for the I, II and III years respectively.
When interest is compounded annually but time period is in fraction say a b/c years,
C.I = A−P (Amount − Principal).
The simple interest and the compound interest remains the same for the first year or the first conversion period.
For 2 years, the difference in C.I and S.I is C.I − S.I =P (r/100)2.
For 3 years, the difference in C.I and S.I is C.I − S.I =
x and y are said to vary directly if y = kx always, where k is called the proportionality constant and k > 0 assuming that y depends on x and so k = y/x.
x and y are said to vary inversely, if xy = k always, where k is called the proportionality constant and k > 0.
There will be problems which involve a chain of two or more variations in them. This is called as compound variation.
By finding the proportion, we can use the fact that the product of the extremes is equal to the product of the means to find the unknown (x) in the problem.
By using the formula = [ P1×D1×H1 ] / W1 = [ P2×D2×H2 ] / W2, we can find the unknown (x).
We can also find the unknown (x) by Multiplicative Factor Method.
If two persons X and Y can complete some work individually in a and b days, their one day’s work will be 1/a and 1/b respectively and X and Y together can complete the work in ab / a+b days.
ICT CORNER
Step-1 Open the Browser type the URL Link given below (or) Scan the QR Code. GeoGebra work book named “LIFE MATHEMATICS” will open. Click on the worksheet named “Percentage”.
Step-2 In the given worksheet you can drag the red points E and F to change the Blue rectangle. Find the ratio of Blue with the whole by counting the squares and check the ratio and percentage.
Go through the remaining worksheets given for this chapter
Life Mathematics:
https://www.geogebra.org/m/fqxbd7rz#chapter/409575 or Scan the QR Code.
*Pictures are indicatives only.
*If browser requires, allow Flash Player or Java Script to load the page.
ICT CORNER
Step-1 Open the Browser type the URL Link given below (or) Scan the QR Code. GeoGebra work sheet named “8th Standard III term” will open. Select the work sheet named “Work Day Problem”
Step-2 Click on “ NEW PROBLEM”. Check the calculation and work out yourself.
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# Arithmetic Mean in Statistics
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Last updated date: 19th Jul 2024
Total views: 390k
Views today: 8.90k
## What is Arithmetic Mean in Statistics?
The arithmetic mean is an easiest and most commonly used measure of a mean, or also referred to as an average. People with the slightest knowledge of math and finance skills can calculate it. Computation through arithmetic simply involves taking the sum of a group of numbers, then dividing that sum by the count of the numbers taken in the series.
One of the widely used measures of central tendency, arithmetic mean is denoted by the symbol x. What makes arithmetic mean a useful measure of central tendency is its tendency to render useful results, even with a huge grouping of numbers.
### Arithmetic Mean Formula in Statistics
Arithmetic mean is the sum of all observations divided by a number of observations.
The formula to calculate the arithmetic mean is as follows:
Mean = sum of all observations / number of observations
Arithmetic means the formula = X=Σ Xi/in, where i varies from 1 to n.
### Example of Arithmetic Mean
For example, consider the numbers 14, 26, 70, 33, and 47. The result of 14 + 26 + 70 + 33 + 47 is 190. The Arithmetic Mean is 190 divided by 5 or 38.
### How Does Arithmetic Mean Works In The Field of Finance?
Though not an ideal application in the world of finance, the arithmetic mean maintains its place in some aspects of finance, as well. For example, mean earnings approximate essentially are an arithmetic mean. Suppose that you want to find out the average earnings expectation of the 12 analysts covering a specific stock. Simply, you need to add up all the approximates and divide by 12 to get the arithmetic mean.
Similarly, you can also find out a stock’s average closing price during a particular month. Suppose that there are 22 trading days in the month. Simply, consider all the prices, add them, divide by 22 and you will get the arithmetic mean.
### Limitations of Arithmetic Mean
1. The Arithmetic mean is not an ideal application when computing the performance of investment portfolios, particularly when it involves compounding, or reinvesting dividends and earnings.
2. Not ideal to calculate present and future cash flows, which economic analysts use in making their estimates. Using arithmetic means in the situation is sure to mislead numbers.
3. The arithmetic mean can be misleading when looking at historical returns. In such cases, the geometric mean is most suitable for series that reveal serial correlation, particularly true for investment portfolios.
For example, the arithmetic mean is not feasible, particularly when a single outlier can skew the mean by a huge amount. Suppose you want to estimate the wages of a group of 20 laborers. Fifteen of them get wages between 2000 and 2500 a week. The 20th laborer gets the wages of 4500. That one outlier is not very representative of the group.
### Arithmetic Mean Solved Examples:
Example 1: Find the arithmetic mean of the first five prime numbers.
Solution: First five prime numbers are 2, 3, 5, 7 and 11. The mean or the average formula is = (2 + 3 + 5 + 7 +11)/5 = 5.6.
Example 2 : The Marks Scored by Maria in 5 Subjects are 40, 73, 68, 50, and 54 respectively. Calculate the Mean.
Solution: Marks obtained by Maria in 5 Test Subjects are:
40, 73, 68, 50 and 54
Thus, Mean = Total Marks / Number of subjects
Total Marks = 40 + 73 + 68 + 50 + 54 = 285
Number of Subjects = 5
Mean = 285/5 = 57
Example 3: Find out the Arithmetic Mean of the Squares of the First n Natural Numbers.
Solution:
Sum of Squares of the 1st ‘n’ Natural Numbers = n (n+1) (2n+1)/4
Their Arithmetic Mean = sum/n
= n (n+1) (2n+1)/4n
= (n+1) (2n+1)/4
It is so because the mean of a set of observations is the value that is representative of the whole number as well as occurs most frequently. It considers all the values present in the group and averages them dividing observation into two equal parts.
### Key Takeaways
1. Arithmetic mean (average) of any given data set is the sum of a series of numbers divided by the count of that series of numbers.
2. In the field of finance, the arithmetic mean is generally an inappropriate method to use for calculating an average.
3. Arithmetic mean is not always ideal, particularly when a single outlier can skew the mean by a big amount
## FAQs on Arithmetic Mean in Statistics
1. What is the Best Use of the Arithmetic Mean?
The Arithmetic means enables us to critically categorize the center of the frequency distribution of a quantitative variable by taking into account all of the observations with the same weight allotted to each (as opposed to the weighted arithmetic mean).
2. What are the Methods to Compute the Arithmetic Mean?
We use methods to compute the Arithmetic Mean for 3 types of series which are as follows:
1. Discrete Data Series
2. Individual Data Series
3. Continuous Data Series
3. What is the Best Use of the Arithmetic Mean?
The Arithmetic mean enables us to critically categorize the centre of the frequency distribution of a quantitative variable by taking into account all of the observations with the same weight allotted to each (as opposed to the weighted arithmetic mean).
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Notes On Divisibility of Numbers - CBSE Class 6 Maths
Tests of divisibility: There are certain tests of divisibility that can help us to decide whether a given number is divisible by another number without actual division. Divisibility of numbers by 10: A number that has 0 in its ones place is divisible by 10. Divisibility of numbers by 5: A number that has either 0 or 5 in its ones place is divisible by 5. Divisibility of numbers by 2: A number that has 0, 2, 4, 6 or 8 in its ones place is divisible by 2. Divisibility of numbers by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. Divisibility of numbers by 6: A number is divisible by 6 if that number is divisible by both 2 and 3. Divisibility of numbers by 4: A number is divisible by 4 if the number formed by its last two digits (i.e. ones and tens) is divisible by 4. Divisibility of numbers by 8: A number is divisible by 8 if the number formed by its last three digits (i.e. ones, tens and hundreds) is divisible by 8. Divisibility of numbers by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. Divisibility of numbers by 11: If the difference between the sum of the digits at the odd and even places in a given number is either 0 or a multiple of 11, then the given number is divisible by 11. Co–prime numbers: If the only common factor of two numbers is 1, then the two numbers are called co-prime numbers. General rules of divisibility for all numbers If a number is divisible by another number, then it is also divisible by all the factors of the other number. If two numbers are divisible by another number, then their sum and difference are also divisible by the other number. If a number is divisible by two co-prime numbers, then it is also divisible by the product of the two co-prime numbers.
#### Summary
Tests of divisibility: There are certain tests of divisibility that can help us to decide whether a given number is divisible by another number without actual division. Divisibility of numbers by 10: A number that has 0 in its ones place is divisible by 10. Divisibility of numbers by 5: A number that has either 0 or 5 in its ones place is divisible by 5. Divisibility of numbers by 2: A number that has 0, 2, 4, 6 or 8 in its ones place is divisible by 2. Divisibility of numbers by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. Divisibility of numbers by 6: A number is divisible by 6 if that number is divisible by both 2 and 3. Divisibility of numbers by 4: A number is divisible by 4 if the number formed by its last two digits (i.e. ones and tens) is divisible by 4. Divisibility of numbers by 8: A number is divisible by 8 if the number formed by its last three digits (i.e. ones, tens and hundreds) is divisible by 8. Divisibility of numbers by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. Divisibility of numbers by 11: If the difference between the sum of the digits at the odd and even places in a given number is either 0 or a multiple of 11, then the given number is divisible by 11. Co–prime numbers: If the only common factor of two numbers is 1, then the two numbers are called co-prime numbers. General rules of divisibility for all numbers If a number is divisible by another number, then it is also divisible by all the factors of the other number. If two numbers are divisible by another number, then their sum and difference are also divisible by the other number. If a number is divisible by two co-prime numbers, then it is also divisible by the product of the two co-prime numbers.
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Home / Geometry / ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. Prove that AD>AB.
# ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. Prove that AD>AB.
### Solution: General enunciation: ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. Prove that AD>AB.
Particular enunciation: Given that, ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. It is required to prove that AD>AB.
Proof: In ABC AB = AC
∴∠ABC = ∠ACB
Now ∠ACD be an exterior angle of ABC
∴∠ACD >interior opposite ∠ABC
⟹∠ACD >∠ACB [∵∠ABC = ∠ACB]
Again ∠ACB is the external angle of ACD.
Now In ACD
## Problem: In the ∆ABC, AB = AC and D is any point on BC. Prove that AB>AD.
Solution: General equation: In the ABC, AB = AC and D is any point on BC. Prove that AB>AD.
Particular Enunciation: Given that, in the ABC, AB = AC and D is any point on BC. It is required to prove that AB>AD.
Proof: Given that in ABC, AB = AC
∴∠ABC = ∠ACB [∵ the angles opposite to equal sides are equal]
Again, ∠ADB is an exterior angle of ACD
Now in triangle ABD,
∠ADB>∠ABD [∵ the angles opposite to equal sides are equal]
#### Problem: In the ∆ABC, AB ⊥ AC and D is any point on AC. Prove that BC>BD.
Solution: General enunciation: In the ABC, AB ⊥ AC and D is any point on AC. Prove that BC>BD.
Particular enunciation: Given that, in the ABC, AB ⊥ AC and D is any point on AC. It is required to prove that BC>BD.
Proof: Given that, in ABC, AB ⊥ AC
∠A = 900
∴ BC be the hypotenuse.
∴ ∠BAC > ∠BAC
Again ∠BDC is an exterior angle of ABD.
⟹∠BDC>∠BAC
⟹∠BDC>∠BCA [∵∠BAC > ∠BAC]
⟹∠BDC>∠BCD
Now, in BCD
∠BDC>∠BCD
⟹BC>BD [∵ the angles opposite to equal sides are equal]
∴ BC>BD (Proved)
## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.
If two triangles have the three sides of the one equal to the three sides ...
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# GRE Math : How to divide fractions
## Example Questions
### Example Question #51 : Fractions
Car A traveled 120 miles with 5 gallons of fuel.
Car B can travel 25 miles per gallon of fuel.
Quantity A: The fuel efficiency of car A
Quantity B: The fuel efficiency of car B
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined.
Quantity B is greater.
Explanation:
Let's make the two quantities look the same.
Quantity A: 120 miles / 5 gallons = 24 miles / gallon
Quantity B: 25 miles / gallon
Quantity B is greater.
### Example Question #52 : Fractions
Quantity A:
The -value of the equation when
Quantity B:
The relationship cannot be determined from the information given.
Both quantities are equal
Quantity A is greater.
Quantity B is greater.
Quantity A is greater.
Explanation:
In order to solve quantitative comparison problems, you must first deduce whether or not the problem is actually solvable. Since this consists of finding the solution to an -coordinate on a line where nothing too complicated occurs, it will be possible.
Thus, your next step is to solve the problem.
Since and , you can plug in the -value and solve for :
Plug in y:
Divide by 3/4. To divide, first take the reciprocal of 3/4 (aka, flip it) to get 4/3, then multiply that by 5/3:
Make the improper fraction a mixed number:
Now that you have what x equals, you can compare it to Quantity B.
Since is bigger than 2, the answer is that Quantity A is greater
### Example Question #53 : Fractions
What is equivalent to ?
Explanation:
Remember that when you divide by a fraction, you multiply by the reciprocal of that fraction. Therefore, this division really is:
At this point, it is merely a matter of simplification and finishing the multiplication:
### Example Question #54 : Fractions
Which of the following is equivalent to ?
Explanation:
To begin with, most students find it easy to remember that...
From this, you can apply the rule of division of fractions. That is, multiply by the reciprocal:
Therefore,
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# 2nd PUC Maths Question Bank Chapter 12 Linear Programming Miscellaneous Exercise
Students can Download Maths Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka 2nd PUC Maths Question Bank Chapter 12 Linear Programming Miscellaneous Exercise
Question 1.
Refer to Example 9. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Amount of P(x) Q (y) Calcium 12 3 Iron 4 20 Chlosterol 6 4 Vitamin A 6 3
Z = 6x + 3y subject to the constraints
(i) 12x + 3y ≥240 ⇒ 4x + y ≥ 80
(ii) 4x + 20y ≥ 460 ⇒x + 5y ≥115
(iii) 6x + 4y ≤ 300 ⇒ 3x + 2y≤150
(iv) x ≥0 ; y≥0
ABC is the solution region
A (15,20) Z = 6 x 15+ 3 x 20= 150
B(40,15) Z = 6 x 40+ 3 x 15 = 285
C (2,72) Z = 6 x 2 + 3 x 72 = 228
a
Vitamin A is maximised at B (40,15) to 285 if P contains 40 units and Q containts 15 units.
Question 2.
A farmer mixes two brands P and Q of cattle feed. Brand P, costing ? 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ?200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
ABCD is the unbounded solution region
A (3,6) Z = ₹ 1950
B (9,2) Z = ₹ 2650
C (0,12) Z = ₹ 2400
D (18,0) Z = ₹ 4500
cost is minimum at A (3,6)
Z = ₹ 1950 with 3 packets of P and 6 packets of Q.
Question 3.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
Food Vitamin A Vitamin B Vitamin C X 1 2 3 Y 2 2 1
One kg of food X costs ₹ 16 and one kg of food Y costs ₹ 20. Find the least cost of the mixture which will produce the required diet?
Mininize
Z = 16x +20y Subject to the constraints
(i) x + 2y ≥ 10
(ii) 2x + 2y ≥ 12 ⇒ x + y ≥ 6
(iii) 3x + y ≥ 8
(iv) x ≥ 0; y ≥ 0
shaded region is the un bounded solution region
A (0,8) Z = ₹ 160
B (1,5) Z = ₹ 116
C (2,4) Z = ₹ 112
D (10,0) Z = ₹ 160
The cost is minimised at C (2,4) Z = ₹ 112 2 kg of X and 4 kg of Y.
Question 4.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
Types of Toys Machines I II III A 12 18 6 B 6 0 9
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹ 7.50 and that on each toy of type B is ₹ 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Machines Toys I II III Profit A (x) 12 18 6 ₹ 7.50 B(y) 6 0 9 ₹ 5
$$Z=\frac{15}{2} x+5 y$$ subject to the constraints
(i) 12x + 6y ≤ 360 ⇒ 2x+y ≤ 60
(ii) 18x ≤ 360 ⇒ x ≤ 20
(iii) 6x + 9y ≤ 360 ⇒ 2x +3y ≤ 120
(iv) x ≥ 0 , y ≥ 0
ABCDE is the solution region
A (0, 0) Z = 0
B (20, 0) Z = ₹ 150
C (20, 20) Z = ₹ 250
D (15, 30) Z = ₹ 262.5 ,
E (0, 40) Z = ₹ 200 ‘
At D (15, 30) Z is maximised to ₹ 262. 5
Hence proved.
Question 5.
An aeroplane can carry a maximum of 200 passengers. A profit of t 1000 is made on-each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Z = 1000 x + 600 y
subject to the constreint
(i) x + y ≤ 200
(ii) x ≥ 20
(iii) y ≥ 80
(iv) x ≥ 0; y ≥ 0
A (20,180) Z= ₹ 128000
B(120,80) Z= ₹ 168000
C (20,80) Z = ₹ 68000
Z is maximised at B (120, 80) ₹ 168000
120 executive at B 80 economic class tickets
Question 6.
Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table: How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Transportation Cost Per quintal (in ₹) From/To A B D 6 4 E 3 2 F 2.50 3
(i) 60 – x ≥ 0 ⇒ x ≤ 60
(ii) 50 – y ≥0 ⇒ y ≤ 50
(iii) 100 – (x + y) ≥ 0 ⇒ x + y ≤ 100
(iv) x + y – 60 ≥ 0 ⇒ x + y ≥ 60
ABCD is the solution region
A (10, 50) Z = ₹ 510
B(50, 50) Z = ₹ 610
C (60, 40) Z = ₹ 620
D (60, 0) Z = ₹ 560
Cost is minimised to ₹ 510 at A (10,50)
Question 7.
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
Distance in (Km) From/To A B D 7 3 E 6 4 F 3 2
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
(ii) 3000 – y ≥ 0 = y ≤ 3000
(iii) x + y – 3500 ≥ 0 = x+y ≥ 3500 x ≥ 0
(iv) 7000 – (x + y) ≥ 0= x + y ≥ 3500 y ≥ 0
ABCD is the the solution region
A (500, 3000) Z = ₹ 4400
B (3500,0) Z = ₹ 5000
C (4500,0) Z = ₹ 5300
D (4500, 2500) Z = ₹ 5550
E (4000,3000) Z = ₹ 5450
Z is minimised at (500,3000)
i.e. ₹ 4400
From A to D 500 Ltrs
From A to E 3000 Ltrs
From A to F 3500 Ltrs
From B to D 4000 Ltrs.
Question 8.
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine. If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
kg per bag Brand P Brand Q Nitrogen 3 3.5 Phosphoric acid 1 2 Potash 3 1.5 Chlorine 1.5 2
P(x) Q (y) N2 3 3.5 Phosphoric 1 2 Potash 3 1.5 Chlorine 1.5 2
$$\text { To minimise } \mathrm{N}_{2} \mathrm{Z}=3 \mathrm{x}+\frac{7}{2} \mathrm{y}$$
∴ Amount of nitrogen is minimised equal to 470 kg when 40 bags of P and 100 bags of Q are mixed.
Question 9.
Refer to Question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
The amount of N2 is maximum equal to 595kg when 140 bags of P and 50 bags of Q are mixed.
Question 10.
A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of ₹ 12 and ₹ 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
x be the no. of dolls A
y be the no. of dolls B
Maximize Z = 12x + 16 y subject to the constraints
(i) x + y < 1200
(ii) $$y \leq \frac{x}{2} \Rightarrow x-2 y \geq 0$$
(iii) x ≤ 3y + 600 ⇒ x -3y < 600
(iv) x ≥ 0; y ≥0
ABCD is the solution region
A (0,0) Z = 0
B (600,0) Z = ₹ 7200
C(1050,150) Z = ₹15000
D (800,400) Z = ₹ 16000
∴ Profit is maximum equal to ₹ 16000 when 800 balls A and 400 balls B are manufactured if sold.
### 2nd PUC Maths Linear Programming Miscellaneous Exercise Additional Question and Answers
Question 1.
One kind of cake required 300g of flour and 15g of fat, another kind requires 150gm of flour and 30gm of fat. Find the maximum number of cakes which can be made from 7.5kg of flour and 600gm of fat. (CBSE 2010)
Let the 1st type of cake x and
2nd type of cake be y
∴ maximize z = x + y subject to
300x + 150 y ≤ 7500, 2x + y ≤ 50
15 x + 30 y ≤ 600 x + 2y ≤ 40
x ≤ 0, y ≤0
comer point are (0,0), (25,0) (20, 10), (0, 20)
At 0, Z = 0
A, Z = 25
B, Z = 30 – Maximum
C, C = 20
Maximum at B (20, 10)
∴ 20 pieces of 1 st type and
10 pieces of 2nd type cake
Question 2.
A merchant plans to sell two type of computer and desktop model and a portable model that will cost ₹ 25,000/ and ₹ 40,000. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computer should stock to get maximum profit if he does not want to invest more than 70 lakhs and his profit on the desktop models is ₹ 4500/ and on the portable model is ?5000/ (CBSE 2011)
No. of desktop computer be x and no. of portable be y
Maximize
∴ Z = 4500 x + 5000 y
subject to x + y ≤ 250
25000 x + 4000 y ≤ 70,00,000
or 5x + 8y ≤ 1400, x ≥ 0, y
The corner points are.
0 (0,0), B (250,0), C (200,50), D (0,175)
At 0, Z = 0
At C Z = 1150000
B Z= 1125000
D Z = 875000
Maximum of C (200, 50)
∴ No. of desktop model = 200
No. of portable = 50
a
|
AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.1 Textbook Questions and Answers.
## AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.1
Question 1.
Classify the fractions as proper, improper and mixed.
$$\frac{3}{4}$$, $$\frac{6}{5}$$, $$\frac{3}{2}$$,$$\frac{4}{1}$$, $$\frac{2}{3}$$, $$\frac{1}{4}$$, $$\frac{18}{13}$$, 1$$\frac{5}{7}$$,$$\frac{1}{3}$$, 11$$\frac{1}{2}$$
If in a fraction numerator is less than the denominator, then it is called a proper fraction.
Proper fractions are $$\frac{3}{4}$$, $$\frac{2}{3}$$, $$\frac{1}{4}$$, $$\frac{1}{3}$$
If in a fraction numerator is greater than the denominator, then it is called an improper fraction.
Improper fractions are $$\frac{6}{5}$$, $$\frac{3}{2}$$, $$\frac{4}{1}$$, $$\frac{18}{13}$$
A combination of a whole number and a proper fraction is called a mixed fraction.
Mixed fractions are 1$$\frac{5}{7}$$, 11$$\frac{1}{2}$$
Question 2.
Write the following fractions in an ascending order.
i) $$\frac{3}{4}$$, $$\frac{3}{2}$$, $$\frac{2}{3}$$, $$\frac{1}{5}$$, $$\frac{18}{7}$$
ii) $$\frac{2}{7}$$, $$\frac{3}{8}$$, $$\frac{3}{4}$$, $$\frac{5}{7}$$, $$\frac{4}{9}$$
Given fractions are $$\frac{3}{4}$$, $$\frac{3}{2}$$, $$\frac{2}{3}$$, $$\frac{1}{5}$$, $$\frac{18}{7}$$
These are unlike fractions.
To arrange the unlike fractions in ascending order / descending order first we have to convert them into equivalent fractions with LCM of their denominators, then compare the like fractions (i.e.,) we convert them to like fractions.
LCM of Denominators = 2 × 2 × 3 × 5 × 7 = 420
$$\frac{315}{420}$$, $$\frac{630}{420}$$, $$\frac{280}{420}$$, $$\frac{84}{420}$$, $$\frac{1080}{420}$$
These are like fractions. Now, we can compare them.
$$\frac{84}{420}$$< $$\frac{280}{420}$$ < $$\frac{315}{420}$$ < $$\frac{630}{420}$$ < $$\frac{1080}{420}$$
i.e, $$\frac{1}{5}$$ < $$\frac{2}{3}$$ < $$\frac{3}{4}$$ < $$\frac{3}{2}$$ < $$\frac{18}{7}$$
∴ Ascending order: $$\frac{1}{5}$$, $$\frac{2}{3}$$, $$\frac{3}{4}$$, $$\frac{3}{2}$$, $$\frac{18}{7}$$
Descending order: $$\frac{18}{7}$$, $$\frac{3}{2}$$, $$\frac{3}{4}$$, $$\frac{2}{3}$$, $$\frac{1}{5}$$
ii) $$\frac{2}{7}$$, $$\frac{3}{8}$$, $$\frac{3}{4}$$, $$\frac{5}{7}$$, $$\frac{4}{9}$$
These are unlike fractions.
To arrange the unlike fractions in ascending order / descending order first we have to convert them into equivalent fractions with LCM of their denominators. Then compare the like fractions (i.e.,) we convert them to like fractions.
LCM of denominators = 2 × 2 × 2 × 3 × 3 × 7 = 504
$$\frac{144}{504}$$, $$\frac{189}{504}$$, $$\frac{378}{504}$$, $$\frac{360}{504}$$, $$\frac{224}{504}$$
These are like fractions. Now, we can compare them.
$$\frac{144}{504}$$ < $$\frac{189}{504}$$ < $$\frac{224}{504}$$ < $$\frac{360}{504}$$ < $$\frac{1080}{420}$$
i.e, $$\frac{2}{7}$$ < $$\frac{3}{8}$$ < $$\frac{4}{9}$$ < $$\frac{5}{7}$$ < $$\frac{3}{4}$$
∴ Ascending order: $$\frac{2}{7}$$, $$\frac{3}{8}$$, $$\frac{4}{9}$$, $$\frac{5}{7}$$, $$\frac{3}{4}$$
Descending order: $$\frac{3}{4}$$, $$\frac{5}{7}$$, $$\frac{4}{9}$$, $$\frac{3}{8}$$, $$\frac{2}{7}$$
Question 3.
Without doing calculation, find the result $$\frac{2}{3}$$ + 1$$\frac{3}{4}$$ + $$\frac{1}{3}$$ – $$\frac{1}{4}$$
Given $$\frac{2}{3}$$ + 1$$\frac{3}{4}$$ + $$\frac{1}{3}$$ – $$\frac{1}{4}$$
Question 4.
Neha bought a cake. She ate $$\frac{7}{15}$$ th of the cake immediately add in the afternoon she ate the remaining part. How much part die ate in the afternoon ?
Whole cake = 1 = $$\frac{15}{15}$$
Neha divided the cake into 15 parts.
Part of a cake eaten by Neha = $$\frac{7}{15}$$
Remaining part of cake = Whole – eaten part immediately
= $$\frac{1}{1}$$ – $$\frac{7}{15}$$
= $$\frac{15}{15}$$ – $$\frac{7}{15}$$
= $$\frac{15-7}{15}$$
= $$\frac{8}{15}$$
∴ Part of a cake eaten by Neha in the afternoon = $$\frac{8}{15}$$
Question 5.
Simplify:
i) $$\frac{2}{5}$$ + $$\frac{1}{3}$$
ii) $$\frac{5}{7}$$ + $$\frac{2}{3}$$
iii) $$\frac{3}{5}$$ – $$\frac{7}{20}$$
iv) $$\frac{17}{20}$$ – $$\frac{13}{25}$$
i) $$\frac{2}{5}$$ + $$\frac{1}{3}$$
LCM of denominators = 3 × 5 = 15
ii) $$\frac{5}{7}$$ + $$\frac{2}{3}$$
LCM of denominators = 7 × 3 = 21
iii) $$\frac{3}{5}$$ – $$\frac{7}{20}$$
LCM of denominators = 2 × 2 × 5 = 20
iv) $$\frac{17}{20}$$ – $$\frac{13}{25}$$
LCM of denominators = 2 × 2 × 5 × 5 = 100
Question 6.
Represent $$\frac{16}{5}$$ pictorially
Given fraction is $$\frac{16}{5}$$ (Improper fraction)
$$\frac{16}{5}$$ = Mixed fraction is 3$$\frac{1}{5}$$
3 + $$\frac{1}{5}$$ = 3$$\frac{1}{5}$$
|
What Are Factors?
## What Are Factors?
Factors are numbers that we multiply together to get a number, which we call a product.
Let's find the factors of 6.
What numbers can we multiply to make 6?
1 × 6 = 6
So, 1 and 6 are factors of 6.
Does 6 have other factors?
Let's try to multiply 2 and 3.
2 × 3 = 6
So, 2 and 3 are also factors of 6. 👍
Are there other numbers that we can multiply to make 6?
Nope.
So the factors of 6 are 1, 2, 3, and 6. 👏
### Finding Factors
Let's find the factors of 24.
The possible factors for 24 are the numbers 1 through 24.
We need to go through each possible factor, starting from 1, and check if it really is a factor.
Tip: A factor can divide its product evenly, without remainder.
So we check if 1 can divide 24 evenly. It can. So we know two factors:
1 × 24 = 24
1 and the number itself are always factors of a number.
Now we check the next possible factor, 2.
2 × 12 = 24
2 divides 24 into 12, so we know 2 and 12 are factors too.
Let's keep checking possible factors. The next is 3.
3 × 8 = 24
3 and 8 are another pair of factors of 24.
Factors always come in pairs, or groups of 2.
Let's check the next possible factor, 4.
4 × 6 = 24
4 and 6 are factors as well.
Is 5 a factor of 24?
5 × ? = 24
5 doesn't divide 24 evenly. It's not a factor. ❎
We actually don't have any more possible factors to check, because we already know 6 is a factor.
So we have our answer. 🎉
24 has eight factors:
1, 2, 3, 4, 6, 8, 12, and 24.
Some numbers have more factors than other numbers.
20, for example, has six factors:
1, 2, 4, 5, 10, 20
7 has only two factors, 1 and itself, 7.
Numbers with only two factors, 1 and themselves, are called prime numbers.
Great job learning about factors. 👏
|
# Transposition of
formulae
In mathematics, engineering and science, formulae are used to relate physical quantities to each
other. They provide rules so that if we know the values of certain quantities, we can calculate
the values of others. In this unit we discuss how formulae can be transposed, or transformed,
or rearranged.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• transpose formulae in order to make other variables the subject of the formula
Contents
1. Introduction 2
2. Solving a simple linear equation 2
3. Transposition of simple formulae 3
4. The formula for the simple pendulum 5
5. Further examples of useful formulae 6
1 c mathcentre June 11, 2004
1. Introduction
Consider the formula for the period, T, of a simple pendulum of length l:
T = 2π
l
g
, where l is the length of the pendulum
Now, on Earth, we tend to regard g, the acceleration due to gravity, as being fixed. It varies a
little with altitude but for most purposes we can regard it as a constant.
Just suppose we had a pendulum of a fixed length l, and we took it somewhere else, say the
moon, or Mars. The gravity there would not be the same and so the value of g would be different
from its value on Earth.
Suppose we wanted to measure g. One way would be to take the pendulum, set it swinging and
measure the time, T, for one complete cycle. Once we have measured the time, we could then
use that to calculate g. But before we could do this, we would need to know what g was, in
terms of the rest of the symbols in the formula. So we need to re-arrange the formula so that it
states “g =?”. We will show you how to do this rearrangement later in the video.
To rearrange, transform, or tranpose the formula, we need many of the techniques used to solve
equations. So, the video ‘Solving Linear Equations in One Variable’ might be very useful to
have a look at.
2. Solving a simple linear equation
Before we look at rearranging more complicated formulae we recap by having a look at a simple
linear equation. Suppose we wanted to solve
3x + 5 = 6 −3(5 −2x)
Our aim is to end up with an expression for x, that is ‘x =?’. We start by expanding the
brackets on the right.
3x + 5 = 6 −15 + 6x
3x + 5 = 6x −9
We then proceed to manipulate this to try to get all the terms involving x on to one side. We
must preserve the balance in the original equation by doing exactly the same operations to both
sides.
Subtracting 3x from both sides:
3x −3x + 5 = 6x −3x −9
5 = 3x −9
5 + 9 = 3x −9 + 9
14 = 3x
and so
x =
14
3
We have obtained an expression for x as required.
c mathcentre June 11, 2004 2
With practice the amount of working written down will reduce because you will be able to carry
out many of the stages at the same time. Having gone through the steps of solving an equation,
the same technique, particularly the idea of keeping the balance by doing the same thing to
both sides, is what we are going to do when we look at the transformation of formulae.
3. Transposition of simple formulae
Example
Consider the formula v = u+at. Suppose we wish to transpose this formula to obtain one for t.
Because we want to obtain t on its own we start by subtracting u from each side:
v = u + at
v −u = at
We now divide everything on both sides by a.
v −u
a
=
at
a
= t
and so finally t =
v −u
a
. We have transposed the formula to find an expression for t.
Example
Consider the formula v
2
= u
2
+ 2as and suppose we wish to transpose it to find u.
We want to obtain u on its own and so we begin by subtracting 2as from each side.
v
2
= u
2
+ 2as
v
2
−2as = u
2
Finally, taking the square root of both sides:
u =
v
2
−2as
Notice we need to take the square root of the whole term (
v
2
−2as) in order to find u.
Example
Consider the formula s = ut +
1
2
at
2
. Suppose we want to transpose it to find a.
Because we want a on its own, we begin by subtracting ut from both sides.
s = ut +
1
2
at
2
s −ut =
1
2
at
2
Multiplying both sides by 2:
2(s −ut) = at
2
Dividing both sides by t
2
:
2(s −ut)
t
2
= a
and so
a =
2(s −ut)
t
2
3 c mathcentre June 11, 2004
Example
Suppose we wish to rearrange y(2x + 1) = x + 1 in order to find x.
Notice that x occurs both on the left and on the right. We need to try to get all the terms
involving x together. We begin by expanding the brackets on the left:
y(2x + 1) = x + 1
2xy + y = x + 1
Subtracting x from both sides:
2xy −x + y = 1
The left-hand side now has two terms involving x. We can factorise these as follows:
x(2y −1) + y = 1
Then subtracting y from both sides:
x(2y −1) = 1 −y
and finally, dividing both sides by (2y −1)
x =
1 −y
2y −1
Example
Suppose we wish to rearrange
y
y + x
+ 5 = x to find an expression for y.
We begin by multiplying every term on both sides by (y + x) in order to remove the fractions:
y + 5(y + x) = x(y + x)
Next we multiply out the brackets:
y + 5y + 5x = xy + x
2
We try to get all the terms involving y onto the left-hand side. Subtracting xy from both sides:
6y −xy + 5x = x
2
Subtracting 5x from both sides, and taking out the common factor y we have
y(6 −x) = x
2
−5x
Finally, dividing both sides by 6 −x we obtain
y =
x
2
−5x
(6 −x)
c mathcentre June 11, 2004 4
Exercise 1
Rearrange each of the following formulae to make the quantity shown the subject.
1. v = u + at, u
2. v
2
= u
2
+ 2as, s
3. s = vt −
1
2
at
2
, a
4. p = 2(w + h), h
5. A = 2πr
2
+ 2πrh, h
6. E =
1
2
mv
2
+ mgh, v
7. E =
1
2
mv
2
+ mgh, m
8. a(3b −1) = 2b + 2, b
9.
t
2t −s
= 3s, t
10.
s
2t −s
+ 5 = 3t, s
4. The formula for the simple pendulum
We began with the formula T = 2π
l
g
. Let us now try to rearrange this to find an expression
for g.
We begin by squaring both sides of the equation in order to remove the square root.
T
2
= (2π)
2
l
g
To remove the fraction we multiply both sides by g:
T
2
g = (2π)
2
l
Dividing both sides by T
2
gives
g =
(2π)
2
l
T
2
By observing the two square terms on the right, we note that this formula could be written, if
we wish, in the equivalent form
g =
T
2
l
5 c mathcentre June 11, 2004
5. Further examples of useful formulae
Example - the lens formula
The so-called lens formula, which is used in optics, is given by
1
f
=
1
u
+
1
v
Suppose we want to rearrange this formula to find u.
Because we want to isolate u we begin by subtracting
1
v
from both sides.
1
f
1
v
=
1
u
The left-hand side fractions can be combined by expressing them over a common denominator
v −f
fv
=
1
u
Inverting both sides
fv
v −f
= u
and so u =
fv
v −f
as required.
Example
The formula T =
T
0
1 −
v
2
c
2
1/2
arises in the study of relativity. Suppose we want to rearrange it
to find an expression for
v
c
.
We begin by noticing that if we square both sides this will remove the square root term (i.e. the
power
1
2
) on the right-hand side. So squaring:
T
2
=
T
2
0
1 −
v
2
c
2
We remove the fraction by multiplying both sides by
1 −
v
2
c
2
:
T
2
1 −
v
2
c
2
= T
2
0
Dividing both sides by T
2
:
1 −
v
2
c
2
=
T
2
0
T
2
v
2
c
2
to both sides gives
1 =
T
2
0
T
2
+
v
2
c
2
c mathcentre June 11, 2004 6
Subtracting
T
2
0
T
2
from both sides:
1 −
T
2
0
T
2
=
v
2
c
2
Finally, taking the square root of both sides
v
c
=
1 −
T
2
0
T
2
as required.
Exercise 2
Rearrange each of the following formulae to make the quantity shown the subject.
1. y = a +
1
x
, x
2. y = a +
1
1 −x
, x
3. P =
P
0
1 −r
2
, r
4. m = k
a(1 −x), x
5. V =
V
0
r
2
−1
, r
Exercise 1
1. u = v −at 2. s =
v
2
−u
2
2a
3. a =
2(vt −s)
t
2
4. h =
1
2
(p −2w)
5. h =
A −2πr
2
2πr
6. v =
2(E −mgh)
m
7. m =
2E
v
2
+ 2gh
8. b =
2 + a
3a −2
9. t =
3s
2
6s −1
10. s =
6t
2
−10t
3t −4
.
Exercise 2
1. x =
1
y −a
2. x = 1 −
1
y −a
3. r =
1 −
P
0
P
4. x = 1 −
1
a
m
k
2
5. r =
1 +
V
0
V
2
7 c mathcentre June 11, 2004
So. Subtracting 3x from both sides: 3x − 3x + 5 = 6x − 3x − 9 5 Adding 9 to both sides: = 3x − 9 5 + 9 = 3x − 9 + 9 14 = 3x x= 14 3 We have obtained an expression for x as required. So we need to re-arrange the formula so that it states “g =?”. Introduction Consider the formula for the period. We start by expanding the brackets on the right. on Earth. Suppose we wanted to solve 3x + 5 = 6 − 3(5 − 2x) Our aim is to end up with an expression for x. the video ‘Solving Linear Equations in One Variable’ might be very useful to have a look at. T . that is ‘x =?’. But before we could do this. Just suppose we had a pendulum of a fixed length l. say the moon.1. or Mars. set it swinging and measure the time. It varies a little with altitude but for most purposes we can regard it as a constant. or tranpose the formula. 3x + 5 = 6 − 15 + 6x 3x + 5 = 6x − 9 We then proceed to manipulate this to try to get all the terms involving x on to one side. we could then use that to calculate g. we tend to regard g. The gravity there would not be the same and so the value of g would be different from its value on Earth. transform. we would need to know what g was. T . for one complete cycle. One way would be to take the pendulum. as being fixed. in terms of the rest of the symbols in the formula. Suppose we wanted to measure g. To rearrange. Once we have measured the time. 2. We must preserve the balance in the original equation by doing exactly the same operations to both sides. of a simple pendulum of length l: T = 2π l . g where l is the length of the pendulum Now. Solving a simple linear equation Before we look at rearranging more complicated formulae we recap by having a look at a simple linear equation. 2004 and so 2 . and we took it somewhere else. we need many of the techniques used to solve equations. the acceleration due to gravity. We will show you how to do this rearrangement later in the video. c mathcentre June 11.
2004 1 2 at 2 and so 3 . 1 s = ut + at2 2 s − ut = Multiplying both sides by 2: 2(s − ut) = at2 Dividing both sides by t2 : 2(s − ut) =a t2 a= 2(s − ut) t2 c mathcentre June 11. the same technique. Suppose we wish to transpose this formula to obtain one for t.With practice the amount of working written down will reduce because you will be able to carry out many of the stages at the same time. taking the square root of both sides: √ u = v 2 − 2as √ Notice we need to take the square root of the whole term ( v 2 − 2as) in order to find u. is what we are going to do when we look at the transformation of formulae. 2 Because we want a on its own. Having gone through the steps of solving an equation. Because we want to obtain t on its own we start by subtracting u from each side: v = u + at v − u = at We now divide everything on both sides by a. we begin by subtracting ut from both sides. at v−u = =t a a v−u and so finally t = . particularly the idea of keeping the balance by doing the same thing to both sides. Transposition of simple formulae Example Consider the formula v = u + at. Suppose we want to transpose it to find a. a Example Consider the formula v 2 = u2 + 2as and suppose we wish to transpose it to find u. 3. We have transposed the formula to find an expression for t. Example 1 Consider the formula s = ut + at2 . v 2 = u2 + 2as v 2 − 2as = u2 Finally. We want to obtain u on its own and so we begin by subtracting 2as from each side.
We can factorise these as follows: x(2y − 1) + y = 1 Then subtracting y from both sides: x(2y − 1) = 1 − y and finally. y+x We begin by multiplying every term on both sides by (y + x) in order to remove the fractions: Suppose we wish to rearrange y + 5(y + x) = x(y + x) Next we multiply out the brackets: y + 5y + 5x = xy + x2 We try to get all the terms involving y onto the left-hand side. We need to try to get all the terms involving x together. We begin by expanding the brackets on the left: y(2x + 1) = x + 1 2xy + y = x + 1 Subtracting x from both sides: 2xy − x + y = 1 The left-hand side now has two terms involving x. dividing both sides by (2y − 1) x= Example y + 5 = x to find an expression for y. Notice that x occurs both on the left and on the right. 2004 4 . dividing both sides by 6 − x we obtain y= x2 − 5x (6 − x) 1−y 2y − 1 c mathcentre June 11. Subtracting xy from both sides: 6y − xy + 5x = x2 Subtracting 5x from both sides.Example Suppose we wish to rearrange y(2x + 1) = x + 1 in order to find x. and taking out the common factor y we have y(6 − x) = x2 − 5x Finally.
2. in the equivalent form 2 2π g= l T g= l g l . + 5 = 3t. t = 3s. T 2 = (2π)2 To remove the fraction we multiply both sides by g: T 2 g = (2π)2 l Dividing both sides by T 2 gives (2π)2 l T2 By observing the two square terms on the right. We begin by squaring both sides of the equation in order to remove the square root. 1 3. p = 2(w + h). 2 4. v 2 = u2 + 2as. v = u + at. u s a h h v m b 5. 2t − s 9. s = vt − at2 . E = mv 2 + mgh. The formula for the simple pendulum We began with the formula T = 2π for g. a(3b − 1) = 2b + 2. t 2t − s s 10. 2 1 7. 2004 . 1. E = mv 2 + mgh. 2 8. 1 6. Let us now try to rearrange this to find an expression g 5 c mathcentre June 11. if we wish. we note that this formula could be written. A = 2πr2 + 2πrh. s 4.Exercise 1 Rearrange each of the following formulae to make the quantity shown the subject.
Suppose we want to rearrange it to find an expression for v . So squaring: 2 T2 = 2 T0 2 1 − v2 c We remove the fraction by multiplying both sides by 2 1− v2 : c2 T Dividing both sides by T 2 : v2 1− 2 c v2 c2 2 = T0 1− Adding v2 to both sides gives c2 = 2 T0 T2 2 v2 T0 1= 2 + 2 T c c mathcentre June 11. the power 1 ) on the right-hand side.the lens formula The so-called lens formula. v−f arises in the study of relativity. Because we want to isolate u we begin by subtracting 1 1 1 − = f v u The left-hand side fractions can be combined by expressing them over a common denominator v−f 1 = fv u Inverting both sides fv =u v−f and so u = Example The formula T = T0 1− v 2 1/2 c2 1 from both sides. c We begin by noticing that if we square both sides this will remove the square root term (i. Further examples of useful formulae Example . 2004 6 . which is used in optics.e. is given by 1 1 1 = + f u v Suppose we want to rearrange this formula to find u.5. v fv as required.
1 − r2 4. r2 − 1 r 2. 1−x P0 . s = . s = v 2 − u2 2a 3. 9. y = a + . v = 2πr m 6t2 − 10t 3s2 10. y = a + 3. x 2. h = a(1 − x). V = √ Answers Exercise 1 1. x = 1 − a k 5. m = 1 4. Exercise 2 2 T0 v2 = 2 T2 c 1− 2 T0 T2 Rearrange each of the following formulae to make the quantity shown the subject. m = k 5. x = 1 − 2 1 y−a 1+ 3. h = (p − 2w) 2 v2 2E + 2gh 8. a = 2(vt − s) t2 7.Subtracting 2 T0 from both sides: T2 1− Finally. u = v − at 5. 2004 . taking the square root of both sides v = c as required. x = 1 y−a 2. r = V0 V 2 1− P0 P 1 m 4. P = x x r x 1 . r = 7 c mathcentre June 11. b = 2+a 3a − 2 2(E − mgh) A − 2πr2 6. t = 6s − 1 3t − 4 Exercise 2 1. V0 . 1 1.
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# All But Seven
All But 7 is a game which is designed to help students begin thinking about the composition and decomposition of numbers. Students create their own part- whole relationships using the numbers they roll. This requires students to think on a deeper level about numbers, and how they work together. Initially students may tend to use the operation with which they are most comfortable- however, in order to be successful, students must use both addition and subtraction.
Children also have the opportunity to think strategically about the many ways some numbers can be formed. For example, 1 can be formed 5 ways (6-5, 5-4, 4-3, 3-2, or 2-1). In contrast, some numbers can be formed in only one or two ways. For example, 11 can only be formed in 2 ways (6+5 and 5+6). They can use this information in deciding which operation to use and which number to cover up. This is also an opportunity for students to think about probability. Students will experience probability as they are more likely to roll numbers that will allow them to cover a 1 than an 11.
Recommended # of Players: 2
### Common Core Standards
*Click on the category title (ex: "Counting and Cardinality") to view the entire standards of the Common Core.
Kindergarten
Counting and Cardinality
Know number names and the count sequence.
• K.CC.2. Count forward beginning from a given number within the known sequence (instead of having to begin at 1).
Operations and Algebraic Thinking
Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from.
• K.OA.1. Represent addition and subtraction with objects, fingers, mental images, drawings, sounds (e.g.., claps), acting out situations, verbal explanations, expressions, or equations.
• K.OA.2. Solve addition and subtraction word problems, and add and subtract within 10, e.g.., by using objects or drawings to represent the problem.
• K.OA.3. Decompose numbers less than or equal to 10 into pairs in more than one way, e.g.., by using objects or drawings, and record each decomposition by a drawing or equation (e.g.., 5 = 2 + 3 and 5 = 4 + 1).
• K.OA.4. For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g.., by using objects or drawings, and record the answer with a drawing or equation.
• K.OA.5. Fluently add and subtract within 5.
Number and Operations in Base Ten
Work with numbers 11-19 to gain foundations for place value.
• K.NBT.1. Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g.., by using objects or drawings, and record each composition or decomposition by a drawing or equation (such as 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones.
Operations and Algebraic Thinking
Represent and solve problems involving addition and subtraction.
• 1.OA.1. Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g.., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
• 1.OA.2. Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g.., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Understand and apply properties of operations and the relationship between addition and subtraction.
• 1.OA.3. Apply properties of operations as strategies to add and subtract. Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
• 1.OA.4.Understand subtraction as an unknown-addend problem. For example, subtract 10 - 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
Add and subtract within 20.
• 1.OA.5. Relate counting to addition and subtraction (e.g.., by counting on 2 to add 2).
• 1.OA.6. Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g.., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g.., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g.., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g.., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Work with addition and subtraction equations.
• 1.OA.7. Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 - 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
• 1.OA.8. Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ - 3, 6 + 6 = _.
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### Prompt Cards
These two group activities use mathematical reasoning - one is numerical, one geometric.
### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
##### Stage: 2 Challenge Level:
If you write plus signs between each of the digits $1$ to $9$, this is what you get:
$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$
However, if you alter where the plus signs go, you could also get:
$12 + 3 + 45 + 6 + 7 + 8 + 9 = 90$
Can you put plus signs in so this is true?
$1\;$ $2\;$ $3\;$ $4\;$ $5\;$ $6\;$ $7\;$ $8\;$ $9\;$ $= 99$
How many ways can you do it?
### Why do this problem?
This problem is one on which learners will need to practise much addition and subtraction! It will require systematic work using an approach of trial and improvement.
### Key questions
What do the numbers from $1$ to $9$ add to? What do you still need to make $99$?
How about trying to make $90$ using the digits $1$ to $8$ and leaving the $9$ for adding on at the end?
Why not start by pushing the $1$ and $2$ together to make $12$?
Why not try pushing the $2$ and $3$ together to make $23$? What do you need now to make $90$?
### Possible extension
Learners who find this straightforward could work out all the different numbers with less than four figures that can be made in the same way while still keeping the digits from $1$ to $9$ in order.
### Possible support
Children could use a calculator to work out what the numbers from $1$ to $9$ add to and the differences made by putting two adjacent numbers together to make a two-digit number.
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# Subtracting negative numbers number line
##### 2019-12-16 08:40
Cut, glue, and build this giant, 9foot number line, which includes all integers from negative fifty through positive fifty.Negative numbers on the number line Adding negative numbers Video transcript [Voiceover So we've already spent some time introducing ourselves to the idea of adding or subtracting positive and negative numbers. subtracting negative numbers number line
Addition using number line. Use the number line to find the sum of two integers. The integers may either be positive or negative. Follow the example given in the worksheet. Sheet 1 Sheet 2 Sheet 3. Download All; Writing addition sentence. The addition of integers is represented on a number line. Form an addition equation for each number line.
Mar 20, 2017 So we could draw the number line I could draw a straighter number line than that so draw the number line again. And let's say that this is negative 2, negative 1, 0, 1, and 2 again. We're starting at negative 2, we're This should give an overview of the instructional video, including vocabulary and any special materials needed for the instructional video. We recommend keeping it to 12 paragraphs.subtracting negative numbers number line Aug 03, 2013 This video attempts to explain subtracting negative numbers using a number line while emphasizing that taking away any quantity from its identical self must result in zero.
## Subtracting negative numbers number line free
0 through 15 subtract numbers up to 14 0 through 20 subtract numbers less than 10 0 through 20 subtract numbers up to 1910 through 10 subtract numbers from 10 to 10. Problem Type. Student creates the number line from the problem. Student creates the problem from the number line. Language for the Number Line Worksheet subtracting negative numbers number line How to Add and Subtract Positive and Negative Numbers Numbers Can be Positive or Negative. This is the Number Line: Negative Numbers () Positive Numbers () is the negative sign. Start at 6 on the number line, move forward 3, and you end up at 3 Now Play With It!
Rating: 4.95 / Views: 514
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# Integral of Cos(t^2)
Integral of cos(t^2) along with its formula and proof with examples. Also learn how to calculate integration of cos(t^2) with step by step examples.
Alan Walker-
Published on 2023-04-14
## Introduction to integral of cos(t^2)
In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.
Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function sine. You will also understand how to compute the cos(t^2) integral by using different integration techniques.
## What is the integral of cos(t2)?
The integral of cos(t)^2 is an antiderivative of the cosine function which is done by using Taylor’s series expansion. It is also known as the reverse derivative of the cos(t2) function which is a trigonometric identity. The sine function is the ratio of the opposite side to the hypotenuse of a triangle which is written as:
The integral of cos (t2) is a common integral in calculus. It contains a trigonometric function cos with an angle t^2. It is helpful in solving many integration problems involving such functions.
### Integral of cos(t^2) formula
The formula of the integral of sin contains the integral sign, coefficient of integration, and the function as sine. It is denoted by ∫(cos t2)dx. In mathematical form, the integral of cos t^2 is:
$\int (\cos t^2)dx = t - \frac{t^5}{5×2!} + \frac{t^9}{9×4!} + \frac{t^{13}}{13×6!} + ....+C$
Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. Replacing the cos(t^2) by cos(x^2) will give the integral of cos(x^2).
## How to calculate the integral of cos(t2)?
The integral of cos(t2) is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of cosine by using:
1. Taylor’s series expansion
2. Definite integral
## Integral of cos t2 by using Taylor’s Series
Taylor’s series is an infinite sum of terms that are expressed in terms of a function’s derivative. It can be used to calculate the derivative of a function that is complex to solve. Since cos(t2) is impossible to integrate by using formal integration. Therefore, we will use Taylor’s series to find the integral of cos(t2).
### Proof of integral of cos t2 by using Taylor’s Series
Since we know that the integration is the reverse of the derivative. Therefore, we can calculate the integral of cos t 2 by using Taylor’s series. For this, we have to first assume the sine series that is,
$\cos t=t – \frac{t^2}{2!} + \frac{t^4}{4!} – \frac{t^6}{6!} + …$
We can use the above series in the integral of sin x to calculate the integral of cos t^2. Then,
$I = \int (\cos t^2)dx$
Substituting the series of cosx, we get,
$I = \int [1 – \frac{(t^2)^4}{2!} + \frac{(t^2)^4}{4!} – \frac{(t^2)^6}{6!} + … ] dx$
Now we can easily integrate these terms to get the integral cos(t2). Therefore,
$\int \cos t^2dx = t - \frac{t^5}{5×2!} + \frac{t^9}{9×4!} + \frac{t^{13}}{13×6!} + ....+ C$
Hence the above equation is the integration of cos(t)^2 by using Taylor’s series. Similarly, the integral of cos(x^3) can also be obtained by using Taylor's series expansion.
## Integral of cos t2 by using definite integral
The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:
$\int^b_a f(x) dx = F(b) – F(a)$
Let’s understand the verification of the cos(t^2) integral by using the definite integral.
### Proof of integral of cos(t^2) by using definite integral
To compute the integral of cos t 2 by using a definite integral, we can use the interval from 0 to π or 0 to π/2. Let’s compute the integral of cos t^2 from 0 to π. For this we can write the integral as:
$\int^\pi_0 \cos t^2 dx = \left|t - \frac{t^5}{5×2!} + \frac{t^9}{9×4!} + \frac{t^{13}}{13×6!} + ...\right|^\pi_0$
Now, substituting the limit in the given function.
$\int^\pi_0 \cos t^2 dx = \left[π - \frac{π^5}{5×2!} + \frac{π^9}{9×4!} + \frac{π^{13}}{13×6!} + ...\right]–\left[0 - \frac{0^5}{5×2!} + \frac{0^9}{9×4!} + \frac{0^{13}}{13×6!} + ..\right]$
The remaining terms are:
$\int^\pi_0 \cos t^2 dx= π - \frac{π^5}{20} + \frac{π^9}{216} + \frac{π^{13}}{9360} + …$
Which is the calculation of the definite integral of cos t2. Now to calculate the integral of cos t2 between the interval 0 to π/2, we just have to replace π by π/2. Therefore,
$\int^{\frac{\pi}{2}}_0 \cos t^2 dx = \left|t - \frac{t^5}{5×2!} + \frac{t^9}{9×4!} + \frac{t^{13}}{13×6!} + ...\right|^{\frac{\pi}{2}}_0$
$\int^{\frac{\pi}{2}}_0 \cos t^2 dx= π - \frac{(π/2)^5}{5×2!} + \frac{(π/2)^9}{9×4!} + \frac{(π/2)^{13}}{13×6!} +… -{ 0}$
The remaining terms are:
$\int^{\frac{\pi}{2}}_0 \cos t^2 dx = π - \frac{π^5}{160×2!} + \frac{π^9}{4608×4!} + \frac{π^{13}}{106496×6!} + …$
Hence it is the calculation of the integral of cos t2 by using a definite integral. Since these calculations are tricky and long-term, we offer you to use our definite integral calculator to evaluate this integral.
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# Combined Equation of Pair of Lines – 02
Science >You are Here
### Type – IC: When Lines are Forming Equilateral Triangle With x = k or y = k are Given
#### ALGORITHM:
1. Find angle made by the lines with the positive direction of x-axis and calculate slopes m1 and m2 of the two lines.
2. Use y = mx form to find equations of the two lines.
3. Write equations of lines in the form u = 0 and v = 0.
4. Find u.v = 0.
5. Simplify the L.H.S. of the joint equation.
Example – 17:
• Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line x = 1.
• Solution:
Let OA and OB be two lines such that triangle OAB is an equilateral triangle.
Equation of AB is x = 1 and AB, is parallel to the y-axis.
Now triangle OAB is an equilateral triangle,
by symmetry, m ∠ MOA = m ∠MOB = 30°
Slope of OA = m1 = tan(-30°) = – tan 30° = – 1/3 and
Slope of OB = m2 = tan( 30°) = 1/3
Equation of OA is passing through the origin, is
y = m1x. i.e. y = – 1/3 x
∴ √3 y = – x
∴ x + √3 y = 0 …….. (1)
As OB is also passing through the origin, the form of the equation of OB is
y = m2x. i.e. y = 1/3 x
∴ √3 y = x
∴ x – √3 y = 0 …….. (2)
Hence, the combined equation of the pair OA and OB is
(x + √3 y)(x – √3 y ) = 0
∴ x 2 – 3y 2 = 0.
This is the required combined equation.
#### Note:
• The joint equation of the straight lines through the origin which forms the equilateral triangle with any straight line x = k is always x 2 – 3y 2 = 0.
#### Example 18 :
• Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line y = 2.
• Solution:
Let OA and OB be two lines such that the triangle OAB is an equilateral triangle.
The equation of AB is y = 2 and AB is parallel to the x-axis.
Now since triangle OAB is an equilateral triangle,
m ∠ XOA = 60° and m ∠ XOB = 120°
∴ Slope of OA = tan 60° = 3 and
∴ slope of OB = tan 120° – 3
As OA is passing through the origin, the form of equation of OA is
y = m1x. i.e. y = 3 x
∴ √3 x – y = 0 ………. (1)
As OB is also passing through the origin, the form of equation of OB is y = mx.
y = m2x. i.e. y = – 3 x
∴ √3 x + y = 0 ………. (2)
Hence their combined equation is
( 3 x – y)( 3 x + y) = 0
∴ 3x2 – y2 = 0.
#### Note:
• The joint equation of the straight lines through the origin which forms the equilateral triangle with any straight line y = k is always 3x2 – y2 = 0.
### Type – ID: When Slopes of Lines are Given
#### ALGORITHM :
• Locate slopes m1 and m2 of the two lines.
• Use y = mx form to find equations of the two lines.
• Write equations of lines in the form u = 0 and v = 0. Find u.v = 0.
• Simplify the L.H.S. of the joint equation.
#### Example 19:
• Find the joint equation of a pair of lines through the origin having slopes 1 and 3.
• Solution:
Let l1 and l2 be the two lines. Slope of l1 is 1 and that of l2 is 3
Therefore the equation of line (l1) passing through origin and having slope 1 is
y = 1 (x)
∴ x – y = 0 ……. (1)
Similarly the equation of line (l2) passing through origin and having slope 3 is
y = 3 (x)
∴ 3 x – y = 0 …. (2)
From (1) and (2) the required combined equation is
(x – y) (3x – y) = 0
∴ x(3x – y) – y(3x – y) = 0
∴ 3x2 – xy – 3xy + y2 = 0
∴ 3x2 – 4xy + y2 = 0
This is the required cobined equation.
#### Example 20:
• Find the joint equation of a pair of lines through the origin having slopes 4 and -1.
Solution:
Let l1 and l2 be the two lines. Slope of l1 is 4 and that of l2 is – 1
Therefore the equation of line (l1) passing through origin and having slope 4 is
y = 4 (x)
∴ 4x – y = 0 ……. (1)
Similarly the equation of line (l2) passing through origin and having slope -1 is
y = -1 (x)
∴ x + y = 0 …. (2)
From (1) and (2) the required combined equation is
(4x – y) (x + y) = 0
∴ 4x(x + y) – y(x + y) = 0
∴ 4x2 + 4xy – xy + y2 = 0
∴ 4x2 + 3xy + y2 = 0
This is the required cobined equation.
#### Example 21:
• Find the joint equation of a pair of lines through the origin having slopes 2 and – 1/2.
• Solution:
Let l1 and l2 be the two lines. Slope of l1 is 2 and that of l2 is – 1/2
Therefore the equation of line ( l1) passing through origin and having slope 2 is
y = 2 (x)
∴ 2x – y = 0 ……. (1)
Similarly the equation of line (l2) passing through origin and having slope is
y = -1/2 (x)
∴ 2y = – x
∴ x + 2y = 0 ………. (2)
From (1) and (2) the required combined equation is
(2x – y) (x + 2y) = 0
∴ 2x(x + 2y) – y(x + 2y) = 0
∴ 2x2 + 4xy – xy – 2y2 = 0
∴ 2x2 + 3xy – 2y2 = 0
This is the required cobined equation.
#### Example 22:
• Find the joint equation of a pair of lines through the origin having slopes 1/3 and -1/2.
Solution:
Let l1 and l2 be the two lines. Slope of l1 is 1/3 and that of l2 is – 1/2
Therefore the equation of line (l1) passing through origin and having slope is
y = 1/3 (x)
∴ 3y = x
∴ x – 3y = 0 ……. (1)
Similarly the equation of line (l2) passing through origin and having slope is
y = (x)
∴ 2y = -x
∴ x + 2y = 0 ……. (2)
From (1) and (2) the required combined equation is
(x – 3y) (x + 2y) = 0
∴ x(x + 2y) – 3y(x + 2y) = 0
∴ x2 + 2xy – 3xy – 6y2 = 0
∴ x2 – xy = 6y2 = 0
This is the required cobined equation.
#### Example 23:
• Find the joint equation of a pair of lines through the origin having slopes 1 + 3 and 1 – 3.
• Solution:
Let l1 and l2 be the two lines. Slope of l1 is 1 + 3 and that of l2 is 1 – 3
Therefore the equation of line (l1) passing through origin and having slope is
y = (1 + 3)x
∴ (1 + 3)x – y = 0 …. (1)
Similarly, the equation of the line (l2) passing through the origin and having slope is
y = ( 1 – 3)x
∴ ( 1 – 3)x – y = 0 …. (2)
From (1) and (2) the required combined equation is
This is the required combined equation.
### Type – IE: When Point and Lines Parallel to Co-ordinate Axes are Given
#### ALGORITHM :
• When line passes through (h, k) and is parallel to co-ordinate axes are given. Equations of lines are x = h and y = k
• Write equations of lines in the form u = 0 and v = 0.
• Find u.v = 0.
• Simplify the L.H.S. of the joint equation.
#### Example 24:
• Find the joint equation of a pair of lines which passes through (1, 2) and parallel to co-ordinate axes.
• Solution:
Let l1 and l2 be the two lines.
Equation of l1 passing through (1, 2) and parallel to y-axis is x = 1
∴ x – 1 = 0 …………. (1)
Equation of l2 passing through (1, 2) and parallel to x-axis is y = 2
∴ y – 2 = 0 …………. (2)
The required joint equation is
(x – 1)(y – 2) = 0
∴ xy – 2x – y + 2 = 0
This is the required cobined equation.
#### Example 25:
• Find the joint equation of a pair of lines which passes through (-3, 4) and parallel to co-ordinate axes.
• Solution:
Let l1 and l2 be the two lines.
Equation of l1 passing through (-3, 4) and parallel to y – axis is x = -3
∴ x + 3 = 0 …………. (1)
Equation of l2 passing through (-3, 4) and parallel to x – axis is y = 4
∴ y – 4 = 0 …………. (2)
The required joint equation is
(x + 3)(y – 4) = 0
∴ xy – 4x + 3y – 12 = 0
This is the required cobined equation.
#### Example 27:
• Find the joint equation of a pair of lines which passes through (3, 2) and parallel to the lines x = 2 and y = 3.
• Solution:
Let l1 and l2 be the two lines.
Equation of l1 passing through (3, 2) and parallel to x = 2 is x = 3
∴ x – 3 = 0 …………. (1)
Equation of l2 passing through (3, 2) and parallel to y = 3 is y = 2
∴ y – 2 = 0 …………. (2)
The required joint equation is
(x – 3)(y – 2) = 0
∴ xy – 2x – 3y + 6 = 0
This is the required cobined equation.
#### Example 27:
• Find the joint equation of a pair of lines which passes through (2, 3) and parallel to co-ordinate axes.
• Solution:
Let l1 and l2 be the two lines.
Equation of l1 passing through (1, 2) and parallel to y – axis is x = 2
∴ x – 2 = 0 …………. (1)
Equation of l2 passing through (1, 2) and parallel to x – axis is y = 3
∴ y – 3 = 0 …………. (2)
The required joint equation is
(x – 2)(y – 3) = 0
∴ xy – 3x – 2y + 6 = 0
This is the required cobined equation.
#### Example 28:
• Find the joint equation of a pair of lines which are at a distance of 9 units from the y-axis and parallel to it.
• Solution:
Let l1 and l2 be the two lines.
Equation of l1 which is at a distance of 9 from y – axis and parallel to it is x = – 9 i.e. x + 9 = 0 …. (1)
Equation of l2 which is at a distance of 9 from y – axis and parallel to it is x = 9 i.e. x – 9 = 0 …… (1)
The required joint equation is
(x + 9)(x – 9) = 0
∴ x2 – 81 = 0
This is the required cobined equation.
#### Example – 29:
• Find the joint equation of a pair of lines which are at a distance of 5 units from the x-axis and parallel to it.
• Solution:
Let l1 and l2 be the two lines.
Equation of l1 which is at a distance of 5 from x – axis and parallel to it is y = 5i.e. y – 5 = 0 ……… (1)
Equation of l2 which is at a distance of 5 from x – axis and parallel to it is y = -5 i.e. y + 5 = 0 …… (2)
The required joint equation is
(y – 5)(y + 5) = 0
∴ y2 – 25 = 0
This is the required cobined equation.
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Order of Operations. Section 1.3. 1.3. Order of Operations. 1. GOAL. Use the order of operations to evaluate algebraic expressions. To solve real-life problems, such as calculating the cost of admission for a family to a state fair. What you should learn. Why you should learn it. 1.3.
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Order of Operations
Section 1.3
1.3
Order of Operations
1
GOAL
Use the order of operations to evaluate algebraic expressions.
To solve real-life problems, such as calculating the cost of admission for a family to a state fair.
Whatyou should learn
Why you should learn it
1.3
Order of Operations
USING THE ORDER OF OPERATIONS
1
GOAL
WHAT IS THE ORDER OF OPERATIONS?
It’s the method we use to evaluate an expression involving more than one operation:
1.
2.
3.
4.
Parenthesis (innermost first, work to the outermost)
Powers
Multiplication and division
left to right
left to right
EXAMPLE 2
What happens when operations that have the same priority appear in the same expression, such as 5 – 4 +1?
5 – 4 + 1
1 + 1
2
5 – 4 + 1
5 – 5
0
Which is correct?
Perform the operations from
left to right!
Simplify the expression. Click to see each step.
Example 4
Hint: The fraction bar serves as a grouping symbol. Simplify the numerator and denominator separately before the final step.
Simplify the power.
Simplify the numerator.
Work from left to right.
Subtract.
Simplify.
Simplify the expression.
Checkpoint
• 1. Evaluate the expression when x = 5.
• a. 2x2 + 8 b. 100 ÷ x2 + 6
58
10
• Simplify the expression.
• a. 25 + 10 – 8 b. 24 ÷ 2 • 3
27
36
4
Homework
• Worksheet 1.3 B
#’s 1-24 evens
SHOW YOUR WORK!!
Can use a calculator to CHECK your answers but you MUST show your work for each step!
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## Related Articles
• RD Sharma Class 10 Solutions
# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 1
### Find the average expenditure (in rupees) per household.
Solution:
Let the assumed mean (A) = 275
It’s seen that A = 275 and h = 50
So,
Mean = A + h x (Σfi ui/N)
= 275 + 50 (-35/200)
= 275 – 8.75
= 266.25
### Which method did you use for finding the mean, and why?
Solution:
From the given data,
To find the class interval we know that,
Class marks (xi) = (upper class limit + lower class limit)/2
Now, let’s compute xi and fixi by the following
Here,
Mean = Σ fiui/N
= 162/ 20
= 8.1
Thus, the mean number of plants in a house is 8.1
We have used the direct method as the values of class mark xi and fi is very small.
### Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Let us assume mean (A) = 150
It’s seen that,
A = 150 and h = 20
So,
Mean = A + h x (Σfi ui/N)
= 150 + 20 x (-12/50)
= 150 – 24/5
= 150 = 4.8
= 145.20
### Question 4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Solution:
Using the relation (xi) = (upper class limit + lower class limit)/ 2
And, class size of this data = 3
Let the assumed mean (A) = 75.5
So, let’s calculate di, ui, fiui as following:
From table, it’s seen that
N = 30 and h = 3
So, the mean = A + h x (Σfi ui/N)
= 75.5 + 3 x (4/30
= 75.5 + 2/5
= 75.9
Therefore, the mean heart beats per minute for those women are 75.9 beats per minute.
### Question 5. Find the mean of each of the following frequency distributions:
Solution:
Let’s consider the assumed mean (A) = 15
From the table it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σfi ui/N)
= 15 + 6 x (3/40)
= 15 + 0.45
= 15.45
### Question 6. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 100
From the table it’s seen that,
A = 100 and h = 20
Mean = A + h x (Σfi ui/N)
= 100 + 20 x (61/100)
= 100 + 12.2
= 112.2
### Question 7. Find the mean of the following frequency distribution:
Solution:
From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 8 x (7/40)
= 20 + 1.4
= 21.4
### Question 8. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 15
From the table it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σfi ui/N)
= 15 + 6 x (5/40)
= 15 + 0.75
= 15.75
### Question 9. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 25
From the table it’s seen that,
A = 25 and h = 10
Mean = A + h x (Σfi ui/N)
= 25 + 10 x (8/60)
= 25 + 4/3
= 79/3 = 26.333
### Question 10. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 20
From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 8 x (5/40)
= 20 + 1
= 21
### Question 11. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 20
From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σfi ui/N)
= 20 + 6 x (-9/20)
= 20 – 72/20
= 20 – 3.6
= 16.4
### Question 12. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 60
From the table it’s seen that,
A = 60 and h = 20
Mean = A + h x (Σfi ui/N)
= 60 + 20 x (14/50)
= 60 + 28/5
= 60 + 5.6
= 65.6
### Question 13. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 50
From the table it’s seen that,
A = 50 and h = 10
Mean = A + h x (Σfi ui/N)
= 50 + 10 x (-2/40)
= 50 – 0.5
= 49.5
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# SOLUTION: you order 2 tacos and 2 enchiladas for \$4.80. you friend order 3 tacos and 1 enchiladas for \$4. how much does each cost
Algebra -> Algebra -> Coordinate Systems and Linear Equations -> SOLUTION: you order 2 tacos and 2 enchiladas for \$4.80. you friend order 3 tacos and 1 enchiladas for \$4. how much does each cost Log On
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Question 604444: you order 2 tacos and 2 enchiladas for \$4.80. you friend order 3 tacos and 1 enchiladas for \$4. how much does each costAnswer by nerdybill(6961) (Show Source): You can put this solution on YOUR website! you order 2 tacos and 2 enchiladas for \$4.80. you friend order 3 tacos and 1 enchiladas for \$4. how much does each cost . Let t = cost of a taco and e = cost of an enchilada . 2t + 2e = 4.80 3t + e = 4.00 . multiply bottom equation by -2 and add both equations together: 2t + 2e = 4.80 -6t - 2e = -8.00 ------------------- -4t = -3.20 t = -3.20/(-4) t = \$0.80 (or 80 cents for a taco) . enchilada: 3t + e = 4.00 3(0.80) + e = 4.00 2.40 + e = 4.00 e = 4.00 - 2.40 e = \$1.60
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Chapter 3
Theorems in Geometry
In our study of geometry, we will be deal with many geometric figures such as triangles and circles, and we will be concerned mostly with their properties. A property of a geometric figure is some interesting or important thing that is true about the figure. For example, a property of a triangle is that it has 3 sides; this property comes from the definition of a triangle. But once we define "triangle", we might notice that it has other properties.
Draw 2 different triangles on a piece of paper, and measure each side length with a ruler, and each angle measurement with a proctactor as shown in the example below:
Do you think there is a relationship between the lengths of the sides and the sizes of the angles? If so, this would be an important property of triangles! In the diagram above, notice the measures of the sides and angles. What appears to be true about the relationship between the side and angle measures? (If you have geometry software, you can try this on a computer. When you drag a vertex of the triangle, the lengths of the sides change.You will notice that the measures of the angles change also.)
Did you notice that the sides of the first triangle are all equal and the angles are all equal also, but in the second triangle the sides are unequal and the angles are unequal? This is an important property of triangles! Also, do you notice anything else about the angles and sides? In the triangle on the right, which side is the longest, and which angle is the largest? In that same triangle, which side is the shortest, and which angle is the smallest? If you draw more triangles and measure them, you will find that the largest angle in a triangle is always opposite the longest side, the "medium sized" angle in the triangle is opposite the "medium sized" side, and the smallest angle is opposite the shortest side. This is an important, and interesting property of triangles!
Mathematicians have, for centuries, explored geometric figures in order to discover their properties. Sometimes we have a feeling that we have discovered a property, something that seems to always be true, and we call this feeling a conjecture or hypothesis or theory. A conjecture is a belief that something is true. But even though we may feel very strongly that it is true, we are not absolutely positive that it's true: maybe there is one exception, one type of triangle in which it isn't true!
How do we discover geometric properties?
Where do they come from? Perhaps you have taken a science class, and done lab experiments. If you are trying to find the properties of a piece of rock, you might experiment with the rock, trying to determine its properties. You might see if it floats by putting it in water, or check to see if it is flammable by lighting a match to it. Whatever you found in your experiments you might then consider a property of the rock. If no one had ever discovered these properties for this particular rock, then these would beyour discoveries, and you could call them by your own name. If it turned out to be a very significant scientific discovery, you might even become famous! Much the same is true in mathematics. Mathematicians study geometric figures, make conjectures, experiment and test the relationships, and then try to prove that their conjectures are true.
As a mathematician works on a math problem, he or she may notice something interesting about the problem, something that seems to relate to another problem or seem to be true in more than just this one situation. The mathematician might then explore this idea in a number of ways. One way would be to do some very accurate diagrams, measure them, and try to check the validity of their conjecture. Software such as the Geometer's Sketchpad can make this process much easier, and more visual. If the conjecture seems to be true, then the mathematician could either accept that it is true without proof and hope that he or she is correct, or prove that it is true. A mathematical proof is a written verification that a conjecture is true. Once proved, the conjecture is called a theorem. (Occasionally, a proof is discovered to be incorrect, and the theorem is then in question, and the mathematician may be a bit embarrassed!)
And now, a bit of history!
"Euclid was a Greek mathematician who lived around 300 B.C. He established a mathematical school in Alexandria And created much of the geometry we study today.The name of Euclid is often considered synonymous with geometry. His book The Elements is one of the most important and influential works in the history of mathematics, having served as the basis, if not the actual text, for most geometrical teaching in the West for the past 2000 years. It contributed greatly to the 'geometrization' of mathematics and set the standard for rigor and logical structure for mathematical works.
In the thirteen books of The Elements, Euclid presents, in a very logical way, all of the elementary Greek geometrical knowledge of his day. This includes the theorems and constructions of plane geometry and solid geometry, along with the theory of proportions, number theory, and a type of geometrical algebra. The Elements is a textbook which gathers into one place the concepts and theorems which constitute the foundation of Greek mathematics. Euclid was not the first to write such a work. It is known that Hippocrates of Chios (440 B.C.) and others had composed books of elements before him. However, Euclid's treatise was quickly recognized as being superior to all previous Elements and none of the earlier works have survived.
Euclid's book The Elements contained Definitions, Postulates (sometimes called Axioms), and Propositions.
A Postulate or Axiom is a mathematical property that is assumed to be true without proof. (The Greeks made a distinction between postulates and axioms, but modern mathematicians usually do not.)
Some of the original axioms of Euclid, which he calls "common notions", are the following:
1.Things which are equal to the same thing are also equal to one another.
2.If equals are added to equals, the wholes are equal.
3.If equals are subtracted from equals, the remainders are equal.
4.The whole is greater than the part."
Propositions
A proposition is usually a statement about the properties of a geometric object. We call them theorems, in modern geometry. They are accepted as true after having been proved. Most of what we will be studying in modern geometry are theorems, and the great majority of these theorems are based on Euclid's original Propositions. The Elements contained 353 Propositions. It is really amazing that these Propositions are still used today, after so many centuries!
According to another author, "Euclid's Elements form one of the most beautiful and influential works of science in the history of humankind. Its beauty lies in its logical development of geometry and other branches of mathematics. It has influenced all branches of science but none so much as mathematics and the exact sciences. The Elements have been studied 24 centuries in many languages starting, of course, in the original Greek, then in Arabic, Latin, and many modern languages."
This quote was taken from a very interesting website. If you would like to visit this site, click on the link below:
http://aleph0.clarku.edu/~djoyce/java/elements/elements.html
How do Mathematicians Decide if a Conjecture is True?
In this chapter of Connecting Geometry, you will use The Geometer's Sketchpad to experiment with some geometric figures. You will make discoveries about these figures, and you will be asked to write a brief paragraph proving that your conjecture is true. In future chapters, you will often be asked to experiment with other geometric figures. Your conjectures will be verified using various methods. A description of one of these methods of verification is written below.
Deductive proof is one way to prove things in mathematics. It depends on a kind of reasoning called deductive reasoning. When we use deductive reasoning, we write a series of statements, each of which is either some information that we were given (called the Given), a definition, a postulate, or a previously proved theorem. In the proof of the right angle theorem below, you can probably see which is the Given. A definition used in this proof is the definition of right angle, a postulate used in this proof is Euclid's first postulate. The final result of this proof is the Right Angles Theorem.
The sequential steps in this proof form what is called a chain of reasoning. In a chain of reasoning, each thought is dependent on, and follows from the previous thought, and all lead to the final conclusion. At the end, we have proved that which we set out to prove. This is what we mean by deductive proof. Notice that the theorem is written using the words if . . . and then . . . This is the standard way in which theorems are written. Although the theorem may seem to be obviously true, to be called a Theorem it nevertheless requires proof.The proof may seem short and lacking in real substance, but we do want to start off with a nice, short, easy one. Don't worry, they will get more interesting (and, more complex!).
Theorem: if two angles are right angles then they are equal.
GIVEN: angle A is a right angle, angle B is a right angle
PROVE: angle A = angle B
PROOF: since angles A and B are each right angles, angle A= 90° and angle B=90° by the definition of right angle. Since angle A and B are equal to the same thing (90°), they are equal to each other.
Some deductive proofs are written informally, as in the proof above: this informal, wordy style is called a Paragraph Proof. Sometimes proofs are written in a more formal style: a form called a Two-Column Proof. An example of a Two -Column Proof is shown below. Both forms of proof shown here prove the same theorem.
Theorem: if two angles are right angles then they are equal.
Mathematicians prove theorems in a variety of ways. Sometimes the proofs of theorems can be very brief and informal, such as the Paragraph Proof. Some are more formal, such as the Two Column Proof. Proofs can also be written using algebra and using coordinate geometry. In this course, we will explore each of these different types of proof as they apply to our studies in geometry.
Project
Your project is to do some research of your own on the internet, on the History of Geometry. I would suggest you begin with the following links, to get warmed up. Then do a Net Search using Geometry as the Key Word. Write a 2 to 3 page report on what you discover, typing in a GSP file. Please do not just copy and print pages off the internet. You should search, read what you find, think about it, and write a summary/discussion of what you find. You may include direct quotations (a few sentences or even a paragraph), but be sure to put these portions in quotation marks, and include the web addresses to give credit to your sources as I have done above.
http://forum.swarthmore.edu/
http://www-groups.dcs.st-and.ac.uk/~history/
Go to Chapter 4 Congruent Triangles
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# Ordinal Multiplication is Left Distributive
## Theorem
Let $x$, $y$, and $z$ be ordinals.
Let $\times$ denote ordinal multiplication.
Let $+$ denote ordinal addition.
Then:
$x \times \left({ y + z }\right) = \left({ x \times y }\right) + \left({ x \times z }\right)$
## Proof
The proof shall proceed by Transfinite Induction, as follows:
### Basis for the Induction
Let $0$ denote the ordinal zero.
$\displaystyle x \times \left({ y + 0 }\right)$ $=$ $\displaystyle x \times y$ Definition of Ordinal Addition $\displaystyle$ $=$ $\displaystyle \left({ x \times y }\right) + 0$ Definition of Ordinal Addition $\displaystyle$ $=$ $\displaystyle \left({ x \times y }\right) + \left({ x \times 0 }\right)$ Definition of Ordinal Multiplication
This proves the basis for the induction.
### Induction Step
$\displaystyle x \times \left({ y + z }\right)$ $=$ $\displaystyle \left({ x \times y }\right) + \left({ x \times z }\right)$ Inductive Hypothesis $\displaystyle \implies \ \$ $\displaystyle x \times \left({ y + z^+ }\right)$ $=$ $\displaystyle x \times \left({ y + z }\right)^+$ Definition of Ordinal Addition $\displaystyle$ $=$ $\displaystyle \left({ x \times \left({ y + z }\right) }\right) + x$ Definition of Ordinal Multiplication $\displaystyle$ $=$ $\displaystyle \left({ \left({ x \times y }\right) + \left({ x \times z }\right) }\right) + x$ Inductive Hypothesis $\displaystyle$ $=$ $\displaystyle \left({ x \times y }\right) + \left({ \left({ x \times z }\right) + x }\right)$ Ordinal Addition is Associative $\displaystyle$ $=$ $\displaystyle \left({ x \times y }\right) + \left({ x \times z^+ }\right)$ Definition of Ordinal Multiplication
This proves the induction step.
### Limit Case
The inductive hypothesis for the limit case states that:
$x \times \left({ y + w }\right) = \left({ x \times y }\right) + \left({ x \times w }\right)$ for all $w \in z$ and $z$ is a limit ordinal.
The proof shall proceed by cases:
### Case 1
Suppose $x = 0$.
$\displaystyle x \times \left({ y + z }\right)$ $=$ $\displaystyle 0$ Ordinal Multiplication by Zero $\displaystyle$ $=$ $\displaystyle 0 + 0$ Definition of Ordinal Addition $\displaystyle$ $=$ $\displaystyle \left({ x \times y }\right) + \left({ x \times z }\right)$ Ordinal Multiplication by Zero
### Case 2
Suppose that $x \ne 0$.
Since $w$ is a limit ordinal, $y + w$ and $x \times w$ are limit ordinals by Limit Ordinals Preserved Under Ordinal Addition and Limit Ordinals Preserved Under Ordinal Multiplication.
$\displaystyle x \times \left({ y + z }\right)$ $=$ $\displaystyle \bigcup_{w \mathop \in \left({ y + z }\right)} \left({ x \times w }\right)$ Definition of Ordinal Multiplication $\displaystyle \left({ x \times y }\right) + \left({ x \times z }\right)$ $=$ $\displaystyle \bigcup_{v \mathop \in \left({ x \times z }\right)} \left({ x \times y }\right) + v$ Definition of Ordinal Addition
Take any $w \in \left({ y + z }\right)$.
It follows that $w \in y \lor \left({ y \subseteq w \land w \in \left({ y + z }\right) }\right)$ by Relation between Two Ordinals and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
Thus, $\left({ w \in y \lor w = \left({ y + u }\right) }\right)$ for some $u \in z$ by Ordinal Subtraction when Possible is Unique.
If $w < y$, then:
$\displaystyle \left({ x \times w }\right)$ $\in$ $\displaystyle \left({ x \times y }\right)$ Membership is Left Compatible with Ordinal Multiplication $\displaystyle$ $\subseteq$ $\displaystyle \left({ x \times y }\right) + v$ Ordinal is Less than Sum
If $w = \left({ y + u }\right)$, then:
$\displaystyle x \times w$ $=$ $\displaystyle x \times \left({ y + u }\right)$ definition of $w$ $\displaystyle$ $=$ $\displaystyle \left({ x \times y }\right) + \left({ x \times u }\right)$ Inductive Hypothesis $\displaystyle \left({ x \times u }\right)$ $\in$ $\displaystyle \left({ x \times z }\right)$ Membership is Left Compatible with Ordinal Multiplication $\displaystyle$ $=$ $\displaystyle \left({ x \times y }\right) + v$ setting $v$ to $x \times u$
$x \times \left({ y + z }\right) \subseteq \left({ x \times y }\right) + \left({ x \times z }\right)$
Conversely, if $v \in \left({ x \times z }\right)$, then:
$\displaystyle \exists w \in z: \ \$ $\displaystyle v$ $\in$ $\displaystyle \left({ x \times w }\right)$ Ordinal is Less than Ordinal times Limit $\displaystyle \implies \ \$ $\displaystyle \left({ x \times y }\right) + v$ $=$ $\displaystyle \left({ x \times y }\right) + \left({ x \times w }\right)$ Substitutivity of Class Equality $\displaystyle$ $=$ $\displaystyle x \times \left({ y + w }\right)$ Inductive Hypothesis
$\left({ x \times y }\right) + \left({ x \times z }\right) \subseteq x \times \left({ y + z }\right)$
By definition of set equality:
$x \times \left({ y + z }\right) = \left({ x \times y }\right) + \left({ x \times z }\right)$
This proves the limit case.
$\blacksquare$
|
# Constant Function
🏆Practice increasing and decreasing intervals of a function
We will say that a function is constant when, as the value of the independent variable $X$ increases, the dependent variable $Y$ remains the same.
Let's assume we have two elements $X$, which we will call $X1$ and $X2$, where the following is true: $X1, that is, $X2$ is located to the right of $X1$.
• When $X1$ is placed in the domain, the value $Y1$ is obtained.
• When $X2$ is placed in the domain, the value $Y2$ is obtained.
The function is constant when: $X2>X1$ and also $$Y2=Y1). The function can be constant in intervals or throughout its domain. ## Test yourself on increasing and decreasing intervals of a function! In what interval is the function increasing? Purple line: \( x=0.6$$
If you are interested in this article, you might also be interested in the following articles:
Graphical representation of a function
Algebraic representation of a function
Notation of a function
Domain of a function
Indefinite integral
Assignment of numerical value in a function
Variation of a function
Increasing function
Decreasing function
Intervals of increase and decrease of a function
In the blog of Tutorela you will find a variety of articles with interesting explanations about mathematics
## Constant Function Exercises
### Exercise 1
Assignment
Find the decreasing and increasing area of the function
$f(x)=5x^2-25$
Solution
In the first step, let's consider that $a=5$
Therefore $a>0$ and the parabola is at the minimum
In the second step, we find $x$ of the vertex
according to the data we know:
$a=5,b=0,c=-25$
We replace the data in the formula:
$x=\frac{-b}{2\cdot a}$
$x=\frac{-0}{2\cdot5}$
$x=\frac{-0}{10}$
$x=0$
$x<0$ Decreasing
$0 Increasing
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### Exercise 2
Assignment
Given the linear function of the graph
What is the domain of negativity of the function?
Solution
Keep in mind that the function is always above the axis: $x$
That is, the function is always positive and has no negative domain. Therefore, no $x$
The function is always positive
### Exercise 3
Assignment
Find the increasing area of the function
$f(x)=6x^2-12$
Solution
In the first step, let's consider that $a=6$
Therefore $a>0$ and the parabola is a minimum
In the second step, we find $x$ of the vertex
according to the data we know that:
$a=6,b=0,c=-12$
We replace the data in the formula
$x=\frac{-b}{2\cdot a}$
$x=\frac{-0}{2\cdot6}$
$x=\frac{0}{12}$
$x=0$
Therefore
$0 Increasing
$x<0$ Decreasing
$0
Do you know what the answer is?
### Exercise 4
Assignment
To find the increasing and decreasing area of the function, you need to find the intersection point of the vertex
True
### Exercise 5
Assignment
Given the function in the diagram, what is its domain of positivity?
Solution
Note that the entire function is always above the axis: $x$
Therefore, it will always be positive. Its area of positivity will be for all $x$
For all $x$
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hw9soln
# hw9soln - Math 113 HW#9 Solutions Fraleigh 23.8 We rst try...
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Math 113 HW #9 Solutions Fraleigh 23.8. We first try to find a single generator. We have 2 11 = 1, so 2 is not a generator. The same problem occurs with 3 and 4. Let’s try - 2 = 21 instead. The successive powers of - 2 are - 2, 4, - 8, 16, - 9, 18, - 13, 3, - 6, 12, - 1, 2, - 4, 8, - 16, 9, - 18, 13, - 3, 6, - 12, 1. So - 2 has order 22 and is a generator. By Corollary 6.16, the set of all generators consists of the elements ( - 2) n where n is relatively prime to 22, i.e. n = 1, 3, 5, 7, 9, 13, 15, 17, 19, 21. Reading these off from the above list, the generators are - 2, - 8, - 9, - 13, - 6, - 4, - 16, - 18, - 3, - 12. Adding 23 to these to make them positive and sorting, we get 5, 7, 10, 11, 14, 15, 17, 19, 20, 21. Fraleigh 23.10. To do this we look for a root α in Z 7 by trying all 7 possibilities, then divide by x - α , then repeat this process with the quotient. 0 and 1 are not roots, but 2 is, so we divide to get x 3 + 2 x 2 + 2 x + 1 = ( x - 2)( x 2 + 4 x + 3). Now we see that the quotient x 2 + 4 x + 3 factors over Z , and hence over Z 7 , as ( x + 1)( x + 3). Thus the final answer is ( x - 2)( x + 1)( x + 3). Fraleigh 23.18. Yes, use p = 3. Fraleigh 23.19. Yes, use p = 3 again. Fraleigh 23.25. (a) True because it has degree 1 and we are over a field. (b) True as in (a). (c) True by Eisenstein with p = 3. (d) False, because 5 is a root, so in Z 7 [ x ] we have x 2 + 3 = ( x + 2)( x - 2). (e) True by a previous problem since the units in a field are the nonzero elements, by the definition of a field. (f) True; this seems to be exactly the same as (e). (?!?) (g) True; this follows from the factor theorem as we showed in class. (h) True. This follows from (g) since a polynomial in F [ x ] can be regarded as a polynomial in E [ x ] also. (i) True. A polynomial of degree 1 in F [ x ] has the form ax + b with a 6 = 0, and then - b/a is a zero. (j) True; this follows from (g).
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Section 6.1 Factoring Expressions
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1 Section 6.1 Factoring Expressions The first method we will discuss, in solving polynomial equations, is the method of FACTORING. Before we jump into this process, you need to have some concept of what it means to FACTOR using numbers that are more familiar. Factoring Whole Numbers To FACTOR the number 60, you could write down a variety of responses some of which are below:! 60 = 1 \$ 60 (not very interesting but true)! 60 = 2 \$ 30! 60 = 3 \$ 20! 60 = 4 \$ 3 \$ 5 All of these are called FACTORIZATIONS of 60, meaning to write 60 as a product of some of the numbers that divide it evenly. The most basic factorization of 60 is as a product of its prime factors (remember that prime numbers are only divisible by themselves and 1). The PRIME FACTORIZATION of 60 is: 60 = 2 \$ 2 \$ 3 \$ 5 There is only one PRIME FACTORIZATION of 60 so we can now say that 60 is COMPLETELY FACTORED when we write it as 60 = 2 \$ 2 \$ 3 \$ 5. When we factor polynomial expressions, we use a similar process. For example, to factor the expression 24x 2, we would first find the prime factorization of 24 and then factor x = 2 \$ 2 \$ 2 \$ 3 and x 2 = x \$ x Putting these factorizations together, we obtain the following: 24x 2 = 2 \$ 2 \$ 2 \$ 3 x x Let s see how the information above helps us to factor more complicated polynomial expressions. 191
2 Problem 1 WORKED EXAMPLE Factoring Using GCF Method Factor 3x 2 + 6x. Write your answer in completely factored form. The building blocks of 3x 2 + 6x are the terms 3x 2 and 6x. Each is written in FACTORED FORM below. 3x 2 = 3 \$ x \$ x and 6x = 3 \$ 2 \$ x Let s rearrange these factorizations just slightly as follows: 3x 2 = (3 \$ x) \$ x and 6x = (3 \$ x) \$ 2 We can see that (3 \$ x) = 3x is a common FACTOR to both terms. In fact, 3x is the GREATEST COMMON FACTOR (GCF) to both terms. Let s rewrite the full expression with the terms in factored form and see how that helps us factor the expression: 3x 2 + 6x = (3 \$ x) \$ x + (3 \$ x) \$ 2 = (3x) \$ x + (3x) \$ 2 = (3x) (x + 2) = 3x (x + 2) Always CHECK your factorization by multiplying the final result. 3x(x + 2) = 3x 2 + 6x CHECKS! Problem 2 MEDIA/CLASS EXAMPLE Factoring Using GCF Method Factor the following quadratic expressions. Write your answers in completely factored form. a) 11a 2 4a b) 55w 2 + 5w Problem 3 YOU TRY Factoring Using GCF Method Factor the following quadratic expression. Write your answers in completely factored form. a) 64b 2 16b b) 8c 2 12c
3 Always check to see if there is a greatest common factor before proceeding to use other factoring methods. Another method used to factor polynomial expressions is shown below. Factoring trinomials of the form x 2 + bx + c by TRIAL AND ERROR x 2 + bx + c = (x + p)(x + q), where b = p + q and c = p q Factoring trinomials of the form ax 2 + bx + c by TRIAL AND ERROR ax 2 + bx + c = (mx + p) (nx + q), where a = m n, c = p q and b = p n + q m Factoring Differences of Squares a 2 b 2 = (a b)(a + b) Note: The sum of squares a 2 + b 2 is NOT factorable. Problem 4 WORKED EXAMPLE Factoring Using Trial and Error Factor the quadratic expression x 2 + 5x 6. Write your answer in completely factored form. Step 1: Look to see if there is a common factor in this expression. If there is, then you can use the GCF method to factor out the common factor. The expression x 2 + 5x 6 has no common factors. Step 2: For this problem, b = 5 and c = 6. We need to identify p and q. In this case, these will be two numbers whose product is 6 and sum is 5. One way to do this is to list different numbers whose product is 6, then see which pair has a sum of 5. Product = 6 Sum = 5 3 \$ 2 No 3 \$ 2 No 1 \$ 6 YES 1 \$ 6 No 193
4 Step 3: Write in factored form. x 2 + 5x 6 = (x + ( 1))(x + 6) Step 4: Check by multiplying the factors. x 2 + 5x 6 = (x 1)(x + 6) (x 1)(x + 6) = x 2 + 6x x 6 = x 2 + 5x 6 CHECKS! Problem 5 MEDIA/CLASS EXAMPLE Factor the Expression Factor each of the following expressions completely. Check your answers. a) a 2 + 7a + 12 b) w 2 + w 20 c) x
5 Problem 6 YOU TRY Factor the Expression Factor each of the following expressions completely. Check your answers. a) 4x 2 + 2x 6 b) m 3 m 2 30m c) 6r r 5 d) 16 y 4 e) n 4 + 5n 2 6 f) m 3 + 3m 2 4m
6 Factoring Sum and Difference of Cubes a 3 + b 3 = (a + b) (a 2 ab + b 2 ) Sum of Cubes a 3 b 3 = (a b) (a 2 + ab + b 2 ) Difference of Cubes NOTE: The trinomials (a 2 ab + b 2 ) and (a 2 + ab + b 2 ) cannot be factored further. NOTE: For cubes, their sum and difference are factorable. However, for squares, the difference is factorable but not the sum. That is, a 2 b 2 = (a + b) (a b) but a 2 + b 2 is NOT factorable. Problem 7 WORKED EXAMPLE Factor the Expression a) x Rewrite the expression as a sum of cubes: x Let a = x and b = 2. Then use the above formula to factor: x = (x + 2) (x 2 2x ) = (x + 2) (x 2 2x + 4) b) 27x 3 1 Rewrite the expression as a difference of cubes: (3x) Let a = 3x and b = 1. Then use the above formula to factor: 27x 3 1 = (3x 1) [(3x) 2 + (3x)(1) ] = (3x 1) (9x 2 + 3x + 1) Problem 8 WORKED EXAMPLE Factor the Expression a) x 3 64 b) m c) 1 + 8n 3 d) 27a
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# $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ Proof
The limit of $1/x$-th power of $1+x$ as $x$ approaches $0$ is a standard result in limits and it is used as a rule to evaluate the limits of algebraic functions which are exponential form. So, let’s us first prove this in calculus to use it as a formula in mathematics.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$
### Expand the Binomial function
The exponential function in algebraic form is in the form of Binomial Theorem. So, it can be expanded infinitely on the basis of this theorem.
${(1+x)}^{\displaystyle n}$ $\,=\,$ $1$ $+$ $\dfrac{n}{1!} x$ $+$ $\dfrac{n(n-1)}{2!} x^2$ $+$ $\dfrac{n(n-1)(n-3)}{3!} x^3$ $+$ $\cdots$
There is no much difference between them but replace $n$ by $\dfrac{1}{x}$.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{\Bigg(\dfrac{1}{x}\Bigg)}{1!}{x}$ $+$ $\dfrac{\Bigg(\dfrac{1}{x}\Bigg)\Bigg(\dfrac{1}{x}-1\Bigg)}{2!}{x^2}$ $+$ $\dfrac{\Bigg(\dfrac{1}{x}\Bigg)\Bigg(\dfrac{1}{x}-1\Bigg)\Bigg(\dfrac{1}{x}-2\Bigg)}{3!}{x^3} + \cdots \Bigg]$
Now, simplify each term in this series and it helps us to evaluate the limit of this exponential function in the next few steps.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{\Bigg(\dfrac{1}{x}\Bigg)}{1!}{x}$ $+$ $\dfrac{\Bigg(\dfrac{1}{x}\Bigg)\Bigg(\dfrac{1-x}{x}\Bigg)}{2!}{x^2}$ $+$ $\dfrac{\Bigg(\dfrac{1}{x}\Bigg)\Bigg(\dfrac{1-x}{x}\Bigg)\Bigg(\dfrac{1-2x}{x}\Bigg)}{3!}{x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{\Bigg(\dfrac{1}{x}\Bigg)}{1!}{x}$ $+$ $\dfrac{\Bigg(\dfrac{1 \times (1-x)}{x^2}\Bigg)}{2!}{x^2}$ $+$ $\dfrac{\Bigg(\dfrac{1 \times (1-x) \times (1-2x)}{x^3}\Bigg)}{3!}{x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{\Bigg(\dfrac{1}{x}\Bigg)}{1!}{x}$ $+$ $\dfrac{\Bigg(\dfrac{1-x}{x^2}\Bigg)}{2!}{x^2}$ $+$ $\dfrac{\Bigg(\dfrac{(1-x)(1-2x)}{x^3}\Bigg)}{3!}{x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{1}{1! \times x}{x}$ $+$ $\dfrac{1-x}{2! \times x^2}{x^2}$ $+$ $\dfrac{(1-x)(1-2x)}{3! \times x^3}{x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{x}{1! \times x}$ $+$ $\dfrac{(1-x)(x^2)}{2! \times x^2}$ $+$ $\dfrac{(1-x)(1-2x)(x^3) }{3! \times x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\require{cancel} \Bigg[1 + \dfrac{\cancel{x}}{1! \times \cancel{x}}$ $+$ $\require{cancel} \dfrac{(1-x)(\cancel{x^2})}{2! \times \cancel{x^2}}$ $+$ $\require{cancel} \dfrac{(1-x)(1-2x)(\cancel{x^3}) }{3! \times \cancel{x^3}}$ $+$ $\cdots$ $\Bigg]$
$\therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{1}{1!}$ $+$ $\dfrac{(1-x)}{2!}$ $+$ $\dfrac{(1-x)(1-2x)}{3!}$ $+$ $\cdots$ $\Bigg]$
### Evaluate the Limit of exponential function
Evaluate the limit of the infinite series as $x$ approaches $0$ and it is equal to the limit of exponential function in algebraic form as $x$ tends to $0$.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $1$ $+$ $\dfrac{1}{1!}$ $+$ $\dfrac{(1-(0))}{2!}$ $+$ $\dfrac{(1-(0))(1-2(0))}{3!}$ $+$ $\cdots$
$\,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1 \times 1}{3!} + \cdots$
$\therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!} + \cdots$
### Evaluate the series
The infinite series represents the expansion of natural exponential function when its exponent is equal to one.
$e^{\displaystyle x} \,=\,$ $1 + \dfrac{x}{1!}$ $+$ $\dfrac{x^2}{2!}$ $+$ $\dfrac{x^3}{3!} + \cdots$
Put $x = 1$
$e^1 \,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1^2}{2!}$ $+$ $\dfrac{1^3}{3!} + \cdots$
$e \,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!} + \cdots$
The value of $e$ in infinite series and the limit of exponential function as $x$ approaches $0$ are same.
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $1$ $+$ $\dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!}$ $+$ $\cdots$ $\,=\,$ $e$
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $e$
Therefore, the limit of the exponential function ${(1+x)}^{\frac{1}{x}}$ as $x$ approaches zero is equal to $e$.
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# 7th Class Mathematics The Triangle and its Properties Triangles
Triangles
Category : 7th Class
Triangles
• A triangle is a simple closed figure bounded by three line segments. It has three vertices three sides and three angles. The three sides and three angles of a triangle are called its six elements. It is denoted by the symbol A.
In$\Delta$ABC. Sides: $\overline{AB}\,,\overline{BC}\,and\,\overline{CA}$; Angles:
$\angle$BAC,$\angle$ABC and $\angle$BCA ; Vertices: A, Band C
• A triangle is said to be
(a) an acute angled triangle, if each one of its
(b) a right angled triangle, if any one of its angles measures ${{90}^{o}}$.
(c) an obtuse angled triangle, if any one of its angles measures more than ${{90}^{o}}$
Note: A triangles cannot have more than one right angle.
A triangles cannot have more than one abuts angle.
In a right triangle, the sum of the acute angles is $\mathbf{9}{{\mathbf{0}}^{\mathbf{o}}}$
• Angle sum property: The sum of the angles of a triangle is ${{180}^{o}}$.
• Properties of sides:
(i) The sum of any two sides of a triangle is greater than the third side.
(ii) The difference of any two sides is less than the third side.
(iii) Property of exterior angles: If a side of a triangle is produced, the exterior angle so formed
is equal to the sum of interior opposite angles.
e.g., Exterior angle,
${{x}^{o}}=\angle A+\angle B={{70}^{o}}+{{40}^{o}}=\,{{110}^{o}}$
• A triangle is said to be
(a) an equilateral triangle, if all of its sides are equal.
(b) an isosceles triangle, if any two of its sides are equal.
(c) a scalene triangle, if all of its sides are of different lengths.
• The medians of a triangle are the line segments joining the vertices of the triangle to the midpoints of the opposite sides.
Here AD, BE and CF are medians of $\Delta$ABC.
• The medians of a triangle are concurrent.
• The centroid of a triangle is the point of concurrence of its medians. The centroid is denoted by G.
• Triangle divides the medians in the ratio 2:1.
• The medians of an equilateral triangle are equal.
• The medians to the equal sides of an isosceles triangle are equal.
• The centroid of a triangle always lies in the interior of the triangle.
• Altitudes of triangle are the perpendiculars drawn from the vertices of a triangle to the opposite sides.
Here AL, BM and CN are the altitudes of $\Delta$ABC.
The altitudes of a triangle are concurrent,
• The orthocenter is the point of concurrence of the altitudes of triangle. Orthocenter is denoted by H.
• The orthocenter of an acute angled triangle lies in the interior B of the triangle.
• The orthocentre'of a right angled triangle is the vertex containing the right angle.
• The orthocenter of an obtuse angled triangle lies in the exterior of the triangle.
• Properties:
(i) The altitudes drawn on equal sides of an isosceles triangle are equal.
(ii) The altitude bisects the base of an isosceles triangle.
(iii) The altitudes of an equilateral triangle are equal.
(iv) The centroid of an equilateral triangle coincides with its orthocenter.
• In a right angled triangle, the side opposite to the right angle is called the hypotenuse and the other two sides are known as its legs.
• Pythagoras ‘Theorem:
In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the remaining two sides.
• In the right angled triangle
$\operatorname{ABC}, A{{C}^{2}} = A{{B}^{2}}+B{{C}^{2}}.$
• In a right angled triangle, the hypotenuse is the longest side.
#### Other Topics
##### Notes - Triangles
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# REVIEW: Write each statement as an inequality and then graph the inequality.
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1 LESSON 1-5 NOTES (Part A): SOLVING INEQUALITIES Words like "at most" and "at least" suggest a relationship in which two quantities may not be equal. These relationships can be represented by a mathematical inequality. Inequality symbols include less than (<), less than or equal to ( < ), great than ( >), greater than or equal to ( > ), or not equal to ( ). REVIEW: Write each statement as an inequality and then graph the inequality. Statement Inequality Graph x is greater than 4 x is greater than or equal to x is less than 4 x is less than or equal to 4 Open dot Closed dot Value is not a solution to the inequality. Value is one of the solutions to the inequality. EXAMPLE 1: What inequality represents the sentence "5 fewer than a number is at least 12."? PRACTICE: Write the inequality that represents the sentence. 1) Five less than a number is at least 28. 2) The product of a number and four is at most 10. 3) Six more than a quotient of a number and three is greater than 14. As with an equation, the solutions of an inequality are numbers that make it true. The procedure for solving a linear inequality is much like the one for solving linear equations. To isolate the variable on one side of the inequality, perform the same algebraic operation on each side of the inequality symbol. Below are some properties of inequalities to keep in mind when solving inequalities. The Addition and Subtraction Properties of Inequality state that adding or subtracting the same number from both sides of the inequality does not change the inequality. If a < b, then a + c < b + c. If a < b, then a c < b c. The Multiplication and Division Properties of Inequality state that multiplying or dividing both sides of the inequality by the same positive number does not change the inequality. If a < b and c > 0, then ac < bc. If a < b and c > 0, then a c < b c. * The procedure for solving an inequality is similar to the procedure for solving an equation but with one important exception. The Multiplication and Division Properties of Inequality also state that, when you multiply or divide each side of an inequality by a negative number, you must reverse the inequality symbol. If a < b and c < 0, then ac > bc. If a < b and c < 0, then a b c c.
2 * Remember, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol. Example: 5 > 3 multiply both sides by (-2) < 3. (-2) -10 < - 6 EXAMPLE 2: What is the solution of 3(x + 2) 5 21 x? Graph the solution. Justify each line in the solution by naming one of the properties of inequalities. 3x x Distributive Property 3x x Simplify. 4x Addition Property of Inequality 4x 20 Subtraction Property of Inequality x 5 Division Property of Inequality To graph the solution, locate the boundary point. Plot a point at x = 5. Because the inequality is "less than or equal to, the boundary point is part of the solution set. Therefore, use a closed dot to graph the boundary point. Shade the number line to the left of the boundary point because the inequality is less than. Graph the solution on a number line. EXAMPLE 3: What is the solution of 2x 3(x 1) < x + 5? Graph the solution. Justify each line in the solution by naming one of the properties of inequalities. 2x 3(x 1) < x + 5 2x 3x + 3 < x + 5 Distributive Property x + 3 < x + 5 Simplify. 2x < 2 Subtraction Property of Inequality x > 1 Division Property of Inequality The direction of the inequality changed in the last step because we divided both sides of the inequality by a negative number. Graph the solution on a number line. PRACTICE: Solve each inequality. Graph the solution. 4) -3x ) x < -23 6) 2x + 4(x 2) > 4 7) 4 (2x 4) 5 (4x + 3)
3 LESSON 1-5 NOTES: SOLVING INEQUALITIES EXAMPLE: Using inequalities to solve a problem (Problem 3, Page 35 of textbook) A movie rental company offers two subscription plans. You can pay \$36 a month and rent as many movies as desired, or you can pay \$15 a month and \$1.50 to rent each movie. How many movies must you rent in a month for the first plan to cost less than the second plan? Define the variable. Write an expression for the cost of each plan. Write an inequality for the problem situation. Solve the inequality. Write the answer in a complete sentence. PRACTICE 1: (Got It? #3, Page 35 of textbook) A digital music service offers two subscription plans. The first has a \$9 membership fee and charges \$1 per download. The second has a \$25 membership fee and charges \$0.50 per download. How many songs must you download for the second plan to cost less than the first plan? Define the variable. Write an expression for the cost of each plan. Write an inequality for the problem situation. Solve the inequality. Write the answer in a complete sentence. PRACTICE 2: The width of a rectangle is 4 cm less than the length. The perimeter is at most 48 cm. What are the restrictions on the dimensions of the rectangle? Solve the problem using an inequality.
4 LESSON 1-5 NOTES (Part B): SOLVING INEQUALITIES INEQUALITIES WITH NO SOLUTION -OR- ALL REAL NUMBERS AS SOLUTIONS EXAMPLES: Is the inequality always, sometimes or never true? (Problem 4, Pg. 36 of text) A) -2(3x + 1) > -6x + 7 B) 5(2x - 3) - 7x 3x + 8 PRACTICE: Is the inequality always, sometimes, or never true? 3) 5(x 2) 2x + 1 4) 2x + 8 2(x + 1) 5) 6x + 1 3(2x 4) 6) 2(3x + 3) 2(3x + 1) COMPOUND INEQUALITY: Two inequalities can be combined with the word and or the word or to form a compound inequality. Compound and inequalities can be combined into a simpler form. For example, the and inequality 5 < x + 1 and x + 1 < 13 can be written as 5 < x + 1 < 13. It is read as " x + 1 is greater than 5 and less than 13." Graphs of compound and inequalities are "segments." The graphs of compound or inequalities are "rays" with different endpoints that go in opposite directions. EXAMPLES: SOLVING AN "AND" INEQUALITY - To solve a compound inequality containing and, find all the values that make both inequalities true. Find and graph the solution to each compound and inequality. A) 7 < 2x + 1 and 3x 16 B) -3 < 2x + 5 7
5 PRACTICE: Find and graph the solution to each compound and inequality. 7) 3x and 2x < 12 8) -5 x - 3 < 2 EXAMPLE: SOLVING AN "OR" INEQUALITY - To solve a compound inequality containing or, find all the values of the variable that make at least one of the inequalities true. Find and graph the solution to: 7 + k 6 or 8 + k < 3 PRACTICE: Find and graph the solution to each compound or inequality. 9) 7w + 3 > 11 or 4w - 1 < ) 16 < 5x + 1 or 3x + 9 < 6 PRACTICE: Write a compound inequality to represent each sentence. 10) The average of Shondra s test scores in Physics is between 88 and ) The Morgans are buying a new house. They want to buy either a house more the 75 years old or a house less than 10 years old.
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# Cube root
(Redirected from Cube roots)
Plot of y = 3x. The plot is symmetric with respect to origin, as it is an odd function. At x = 0 this graph has a vertical tangent.
In mathematics, a cube root of a number x is a number such that a3 = x. All real numbers (except zero) have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8, denoted 38, is 2, because 23 = 8, while the other cube roots of 8 are −1 + 3i and −1 − 3i. The three cube roots of −27i are
${\displaystyle 3i,\quad {\frac {3{\sqrt {3}}}{2}}-{\frac {3}{2}}i,\quad {\text{and}}\quad -{\frac {3{\sqrt {3}}}{2}}-{\frac {3}{2}}i.}$
The cube root operation is not associative or distributive with addition or subtraction.
In some contexts, particularly when the number whose cube root is to be taken is a real number, one of the cube roots (in this particular case the real one) is referred to as the principal cube root, denoted with the radical sign 3. The cube root operation is associative with exponentiation and distributive with multiplication and division if considering only real numbers, but not always if considering complex numbers: for example, the cube of any cube root of 8 is 8, but the three cube roots of 83 are 8, −4 + 4i3, and −4 − 4i3.
## Formal definition
The cube roots of a number x are the numbers y which satisfy the equation
${\displaystyle y^{3}=x.\ }$
### Real numbers
For any real number y, there is one real number x such that x3 = y. The cube function is increasing, so does not give the same result for two different inputs, plus it covers all real numbers. In other words, it is a bijection, or one-to-one. Then we can define an inverse function that is also one-to-one. For real numbers, we can define a unique cube root of all real numbers. If this definition is used, the cube root of a negative number is a negative number.
The three cube roots of 1
If x and y are allowed to be complex, then there are three solutions (if x is non-zero) and so x has three cube roots. A real number has one real cube root and two further cube roots which form a complex conjugate pair. This can lead to some interesting results.
For instance, the cube roots of the number one are:
${\displaystyle {\sqrt[{3}]{1}}={\begin{cases}\ \ 1\\-{\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i\\-{\frac {1}{2}}-{\frac {\sqrt {3}}{2}}i.\end{cases}}}$
The last two of these roots lead to a relationship between all roots of any real or complex number. If a number is one cube root of any real or complex number, the other two cube roots can be found by multiplying that number by one or the other of the two complex cube roots of one.
### Complex numbers
Plot of the complex cube root together with its two additional leaves. The first picture shows the main branch which is described in the text
Riemann surface of the cube root. One can see how all three leaves fit together
For complex numbers, the principal cube root is usually defined as the cube root that has the largest real part, or, equivalently, the cube root whose argument has the least absolute value. It is related to the principal value of the natural logarithm by the formula
${\displaystyle x^{\tfrac {1}{3}}=\exp({\tfrac {1}{3}}\ln {x}).}$
If we write x as
${\displaystyle x=r\exp(i\theta )\,}$
where r is a non-negative real number and θ lies in the range
${\displaystyle -\pi <\theta \leq \pi }$,
then the principal complex cube root is
${\displaystyle {\sqrt[{3}]{x}}={\sqrt[{3}]{r}}\exp({\tfrac {1}{3}}i\theta ).}$
This means that in polar coordinates, we are taking the cube root of the radius and dividing the polar angle by three in order to define a cube root. With this definition, the principal cube root of a negative number is a complex number, and for instance 3−8 will not be −2, but rather 1 + i3.
This limitation can easily be avoided if we write the original complex number x in three equivalent forms, namely
${\displaystyle x={\begin{cases}r\exp {\bigl (}i(\theta ){\bigr )},\\r\exp {\bigl (}i(\theta +2\pi ){\bigr )},\\r\exp {\bigl (}i(\theta -2\pi ){\bigr )}.\end{cases}}}$
The principal complex cube roots of these three forms are then respectively
${\displaystyle {\sqrt[{3}]{x}}={\begin{cases}{\sqrt[{3}]{r}}\exp {\bigl (}i({\tfrac {1}{3}}\theta ){\bigr )},\\{\sqrt[{3}]{r}}\exp {\bigl (}i({\tfrac {1}{3}}\theta +{\tfrac {2}{3}}\pi ){\bigr )},\\{\sqrt[{3}]{r}}\exp {\bigl (}i({\tfrac {1}{3}}\theta -{\tfrac {2}{3}}\pi ){\bigr )}.\end{cases}}}$
In general, these three complex numbers are distinct, even though the three representations of x were the same. For example, 3−8 may then be calculated to be −2, 1 + i3, or 1 − i3.
In programs that are aware of the imaginary plane, the graph of the cube root of x on the real plane will not display any output for negative values of x. To also include negative roots, these programs must be explicitly instructed to only use real numbers.
## Impossibility of compass-and-straightedge construction
Cube roots arise in the problem of finding an angle whose measure is one third that of a given angle (angle trisection) and in the problem of finding the edge of a cube whose volume is twice that of a cube with a given edge (doubling the cube). In 1837 Pierre Wantzel proved that neither of these can be done with a compass-and-straightedge construction.
## Numerical methods
Newton's method is an iterative method that can be used to calculate the cube root. For real floating-point numbers this method reduces to the following iterative algorithm to produce successively better approximations of the cube root of a:
${\displaystyle x_{n+1}={\frac {1}{3}}\left({\frac {a}{x_{n}^{2}}}+2x_{n}\right).}$
The method is simply averaging three factors chosen such that
${\displaystyle x_{n}\times x_{n}\times {\frac {a}{x_{n}^{2}}}=a}$
at each iteration.
Halley's method improves upon this with an algorithm that converges more quickly with each step, albeit consuming more multiplication operations:
${\displaystyle x_{n+1}=x_{n}\left({\frac {x_{n}^{3}+2a}{2x_{n}^{3}+a}}\right).}$
With either method a poor initial approximation of x0 can give very poor algorithm performance, and coming up with a good initial approximation is somewhat of a black art. Some implementations manipulate the exponent bits of the floating-point number; i.e. they arrive at an initial approximation by dividing the exponent by 3. This has the disadvantage of requiring knowledge of the internal representation of the floating-point number, and therefore a single implementation is not guaranteed to work across all computing platforms.
Also useful is this generalized continued fraction, based on the nth root method:
If x is a good first approximation to the cube root of z and y = zx3, then:
${\displaystyle {\sqrt[{3}]{z}}={\sqrt[{3}]{x^{3}+y}}=x+{\cfrac {y}{3x^{2}+{\cfrac {2y}{2x+{\cfrac {4y}{9x^{2}+{\cfrac {5y}{2x+{\cfrac {7y}{15x^{2}+{\cfrac {8y}{2x+\ddots }}}}}}}}}}}}}$
${\displaystyle =x+{\cfrac {2x\cdot y}{3(2z-y)-y-{\cfrac {2\cdot 4y^{2}}{9(2z-y)-{\cfrac {5\cdot 7y^{2}}{15(2z-y)-{\cfrac {8\cdot 10y^{2}}{21(2z-y)-\ddots }}}}}}}}.}$
The second equation combines each pair of fractions from the first into a single fraction, thus doubling the speed of convergence. The advantage is that x and y are only computed once.
## Appearance in solutions of third and fourth degree equations
Cubic equations, which are polynomial equations of the third degree (meaning the highest power of the unknown is 3) can always be solved for their three solutions in terms of cube roots and square roots (although simpler expressions only in terms of square roots exist for all three solutions if at least one of them is a rational number). If two of the solutions are complex numbers, then all three solution expressions involve the real cube roots of two real numbers, while if all three solutions are real numbers then each solution is expressed in terms of the complex cube roots of two complex numbers.
Quartic equations can also be solved in terms of cube roots and square roots.
## History
The calculation of cube roots can be to traced back to Babylonian mathematicians from as early as 1800 BCE.[1] In the fourth century BCE Plato posed the problem of doubling the cube, which required a compass-and-straightedge construction of the edge of a cube with twice the volume of a given cube; this required the construction, now known to be impossible, of the length 32.
A method for extracting cube roots appears in The Nine Chapters on the Mathematical Art, a Chinese mathematical text compiled around the 2nd century BCE and commented on by Liu Hui in the 3rd century CE.[2] The Greek mathematician Hero of Alexandria devised a method for calculating cube roots in the 1st century CE. His formula is again mentioned by Eutokios in a commentary on Archimedes.[3] In 499 CE Aryabhata, a mathematician-astronomer from the classical age of Indian mathematics and Indian astronomy, gave a method for finding the cube root of numbers having many digits in the Aryabhatiya (section 2.5).[4]
## References
1. ^ Saggs, H. W. F. (1989). Civilization Before Greece and Rome. Yale University Press. p. 227. ISBN 978-0-300-05031-8.
2. ^ Crossley, John; W.-C. Lun, Anthony (1999). The Nine Chapters on the Mathematical Art: Companion and Commentary. Oxford University Press. p. 213. ISBN 978-0-19-853936-0.
3. ^ Smyly, J. Gilbart (1920). "Heron's Formula for Cube Root". Hermathena (Trinity College Dublin) 19 (42): 64–67.
4. ^ Aryabhatiya Marathi: आर्यभटीय, Mohan Apte, Pune, India, Rajhans Publications, 2009, p.62, ISBN 978-81-7434-480-9
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# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## AP Board Class 7 Maths Chapter 9 Construction of Triangles Ex 2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 9 Construction of Triangles Ex 2 Book Answers
AP Board Class 7 Maths Chapter 9 Construction of Triangles Ex 2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 9 Construction of Triangles Ex 2 Book Answers
## Andhra Pradesh State Board Class 7th Maths Chapter 9 Construction of Triangles Ex 2 Books Solutions
Board AP Board Materials Textbook Solutions/Guide Format DOC/PDF Class 7th Subject Maths Chapters Maths Chapter 9 Construction of Triangles Ex 2 Provider Hsslive
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## AP Board Class 7th Maths Chapter 9 Construction of Triangles Ex 2 Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 7th Maths Chapter 9 Construction of Triangles Ex 2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:
Question 1.
Draw ΔCAR in which CA = 8 cm, ∠A = 60° and AR = 8 cm. Measure CR, ∠R and ∠C. What kind of triangle is this?
Solution:
CA = 8 cm, ∠A = 60°, AR = 8 cm
Step. -1: Draw a rough sketch of a triangle and label it with the given measurements.
Step -2: Draw a line segment CA of length 8 cm.
Step -3: Draw a ray AX−→− making an angle 60° with CA.
Step -4: Draw an arc of radius 8 cm fromA which cuts AX−→− at C.
Step -5: Join C, R to get the required
Δ CAR. CR = 8 cm, ∠C = 60° and ∠R = 60°.
∴ This is an equilateral triangle.
Question 2.
Construct ΔABC in which AB = 5cm, ∠B = 45° and BC = 6cm.
Solution:
AB = 5cm, ∠B = 45° and BC = 6cm.
Step -1: Draw a rough sketch of a triangle and label it with the given measurements.
Step -2: Draw a line segment AB of length 5cm.
Step -3: Draw a ray BY−→− making an angle 45° with AB.
Step -4: Draw an arc of radius 6 cm from B, which cuts BY−→− at C.
Step -5: Join A, B to get the required ΔABC.
Question 3.
Construct ΔPQR such that ∠R = 100°, QR = RP = 5.4 cm.
Solution:
∠R= 100°,QR= RP = 5.4cm.
Step -1: Draw a rough sketch of a triangle and label it with the given measUrements.
Step -2: Draw a line segment QR of length 5.4 cm.
Step -3: Draw a ray RX−→− making an angle 100° with QR.
Step -4: Draw an arc of radius 5.4 cm from R, which cuts RX−→− at P.
Step -5: Join P, Q to get the required ΔPQR
Question 4.
Construct ΔTEN such that TE = 3 cm, ∠E = 90° and NE = 4 cm.
Solution:
TE = 3cm, ∠E = 90°, NE = 4cm.
Step -1: Draw a rough sketch of the triangle and label it with the given measurements.
Step -2: Draw a line segment TE of length 3 cm.
Step -3: Draw a ray EX−→− making an angle 90° with TE.
Step -4: Draw an arc of radius 4 cm from E, which cuts EX−→− at N.
Step -5: Join N, T to get the required ΔTEN.
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# Find the Maximum Value of 2x3 − 24x + 107 in the Interval [1,3]. Find the Maximum Value of the Same Function in [ − 3, − 1]. - CBSE (Arts) Class 12 - Mathematics
#### Question
Find the maximum value of 2x3$-$ 24x + 107 in the interval [1,3]. Find the maximum value of the same function in [ $-$ 3, $-$ 1].
#### Solution
$\text { Given:} f\left( x \right) = 2 x^3 - 24x + 107$
$\Rightarrow f'\left( x \right) = 6 x^2 - 24$
$\text { For a local maximum or a local minimum, we must have }$
$f'\left( x \right) = 0$
$\Rightarrow 6 x^2 - 24 = 0$
$\Rightarrow 6 x^2 = 24$
$\Rightarrow x^2 = 4$
$\Rightarrow x = \pm 2$
$\text { Thus, the critical points of f in the interval } \left[ 1, 3 \right] \text { are 1, 2 and 3 } .$
$\text { Now,}$
$f\left( 1 \right) = 2 \left( 1 \right)^3 - 24\left( 1 \right) + 107 = 85$
$f\left( 2 \right) = 2 \left( 2 \right)^3 - 24\left( 2 \right) + 107 = 75$
$f\left( 3 \right) = 2 \left( 3 \right)^3 - 24\left( 3 \right) + 107 = 89$
$\text { Hence, the absolute maximum value when x = 3 in the interval } \left[ 1, 3 \right] is 89 .$
$\text { Again, the critical points of f in the interval } \left[ - 3, - 1 \right] \text {are - 1, - 2 and } - 3 .$
$\text { So },$
$f\left( - 3 \right) = 2 \left( - 3 \right)^3 - 24\left( - 3 \right) + 107 = 125$
$f\left( - 2 \right) = 2 \left( - 2 \right)^3 - 24\left( - 2 \right) + 107 = 139$
$f\left( - 1 \right) = 2 \left( - 1 \right)^3 - 24\left( - 1 \right) + 107 = 129$
$\text { Hence, the absolute maximum value when } x = - 2 \text { is } 139 .$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution Find the Maximum Value of 2x3 − 24x + 107 in the Interval [1,3]. Find the Maximum Value of the Same Function in [ − 3, − 1]. Concept: Graph of Maxima and Minima.
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# What is secant method example?
## What is secant method example?
Example 1. As an example of the secant method, suppose we wish to find a root of the function f(x) = cos(x) + 2 sin(x) + x2. A closed form solution for x does not exist so we must use a numerical technique. We will use x0 = 0 and x1 = -0.1 as our initial approximations.
## What is secant method in numerical analysis?
In numerical analysis, the secant method is a root-finding algorithm that uses a succession of roots of secant lines to better approximate a root of a function f. The secant method can be thought of as a finite-difference approximation of Newton’s method.
How do you solve a secant method?
Secant Method
1. Step 1: Choose two starting points x0 and x1.
2. Step 2: Let.
3. Step 3: if |x2-x1|
4. Step 4: End.
What is the modified secant method?
It is an open numerical method and a modified or improved version of the regular Secant method. It asks for only one initial guesses and a (fractional) constant. It generally converges to the true root faster compared to the regular Secant method.
### How do you use Newton’s method in Matlab?
Newton’s Method in Matlab
1. g(x)=sin(x)+x cos(x). Since.
2. g'(x)=2cos(x)-xsin(x), Newton’s iteration scheme,
3. xn+1=xn-g(xn)/g'(xn) takes the form.
4. xn+1=xn-[sin(xn)+x cos(xn)]/[2cos(xn)-xsin(xn)]. To check out in which range the root is, we first plot g(x) in the range 0£x£2.5 using the command.
### Why do we use secant method?
Advantages of Secant Method: The speed of convergence of secant method is faster than that of Bisection and Regula falsi method. It uses the two most recent approximations of root to find new approximations, instead of using only those approximations which bound the interval to enclose root.
What is another name of secant method?
2-point method
Explanation: Secant Method is also called as 2-point method. In Secant Method we require at least 2 values of approximation to obtain a more accurate value. 3. Secant method converges faster than Bisection method.
What is the other name of secant method?
Explanation: Secant Method is also called as 2-point method. In Secant Method we require at least 2 values of approximation to obtain a more accurate value.
## When secant method is fail?
If f ( a n ) f ( b n ) ≥ 0 at any point in the iteration (caused either by a bad initial interval or rounding error in computations), then print “Secant method fails.” and return None .
## What are the advantages of secant method?
Convergence is guarenteed: Bisection method is bracketing method and it is always convergent.
• Error can be controlled: In Bisection method,increasing number of iteration always yields more accurate root.
• Does not involve complex calculations: Bisection method does not require any complex calculations.
• How to write program for secant method in Mathematica?
bk is the current iterate,i.e.,the current guess for the root of f;
• ak is the “contrapoint,” i.e.,a point such that f(ak) and f(bk) have opposite signs,so the interval[ak,bk]contains the solution.
• bk-1 is the previous iterate (for the first iteration,we set bk-1 = a0 ).
• What is the secant method?
THE SECANT METHOD Newton’s method was based on using the line tangent to the curve of y = f(x), with the point of tangency (x 0;f(x 0)).When x 0 ˇ , the graph of the tangent line is approximately the same as the
### How to use secant?
fun = the function you’re trying to find the root for,
• x0 = first value (initial guess)
• x1 = second value (initial guess)
• eps = tolerance for the algorithm’s convergence
• maxit = maximum number of iterations
• silent = logical statement which decides if the iterations should be printed (or not)
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# Displacement and Distance with Examples
In this article, examples of displacement and distance in physics are presented along with their definitions which is helpful for high school physics.
## Definition of displacement and distance
Displacement is a vector quantity describes a change in position of an object or how far an object is displaced from its initial position and is given by below formula
$\Delta \vec x = x_f - x_i$
Where the starting and ending positions are denoted by $x_i$ and $x_f$, respectively.
Distance is a scalar quantity indicating the total path traveled by a moving object.
Since in both cases, the interval between two points is measured- one as the shortest line and the other as the total traveled path- so the SI unit of displacement and distance is the meter.
The illustration below shows the difference between them clearly.
## 5 Examples of displacement and distance
Now with this brief definition, we can go further and explain those concepts more concisely with numerous examples.
Example (1): A boy is playing around a rectangular path. He starts his game from one corner ($i$) and ends at the same one.
Solution: In this path, since his initial and final points are the same, by definition, his displacement is zero.
On the other hand, the total distance traveled by him is the perimeter of that rectangular shape.
In this way, distance traveled (in physics) is consistent with the concept of displacement in everyday language.
Example: A car travels a circle of the circumference of $100\,\rm m$. In each of the following, find the quantity wanted:
(a) If the car moves once around the track, what is the distance traveled by it? What if it moves twice?
(b) If the car moves once around the track, what is its displacement?
Solution: Another example of displacement that clarifies its difference with the distance traveled.
Distance is defined as the whole path covered by a moving object, but displacement is the difference between the initial and final positions of the moving body.
(a) The whole path taken by the car around the circle is its circumference. Thus, when the car travels once around the circle, its distance traveled is $100\,\rm m$.
Once the car moves twice around the circle, its distance traveled is $200\,\rm m$.
(b) When the car travels a complete circle, its initial and final positions coincide with each other. Thus, according to the definition of displacement, $\Delta x=x_f-x_i$, its displacement becomes zero.
This example explicitly shows us that the distance traveled cannot never get zero.
Example (2): A biker ride into a horizontal loop of a radius of 10 meters and covers three fourth of it as follows. What are the displacement and distance traveled by him?
Solution: As mentioned earlier, displacement is a quantity that depends only on the position of initial and final points.
Thus, the straightest line between those points, which gives the magnitude of the displacement vector, is computed by Pythagorean theorem as
\begin{align*} D^2 &= r^2 + r^2 \\ &= 2 r^2 \\ \Rightarrow D &= r\sqrt{2} \end{align*} where the magnitude of displacement denoted by $D$. Putting the values gives, $D=10\sqrt{2}\,\rm m$.
Distance is simply the circumference of three fourth of the circle, so \begin{align*} \text{distance}&= \frac{3}{4} \, {\text{perimeter}} \\\\ &=\frac{3}{4} (2\pi\,r) \\\\ &= \frac{3 \times 2\times \pi \times 10}{4} \\\\ &=15\,\pi \,\rm m \end{align*}
Example (3): A person starts at position 0 and walks 4 meters to the right, returns, and walks 6 meters to the left.
(a) What is displacement?
(b) What is the distance traveled by him?
Solution: denote the initial position as $x_i=0$ and the final position as $x_f$. As you can see, from the starting position to the turning point, the person walks 4 meters to the right. From that point ($B$) count 6 steps to the left to reach the final point whose position is on the negative side of the axis i.e. $x_f=-2$.
(a) Displacement in physics is a vector and is defined to be the final position minus the initial position. So, $\Delta x=x_f-x_i=-2-0=-2\,{\rm m}$ Hence, the person displaced 2 meters to the left (green arrow). Here, the minus sign indicates the direction of the displacement.
(b) For the first part of the trip, the person walks 4 meters, and next, 6 meters more additional to the left. Thus, the total distance traveled is calculated to be 10 meters.
Example (4): If you walk exactly 4 times around a quarter-meter path, what is your displacement?
Solution: let the initial position be $x_i=0$. By gluing 4 of these quarter tracks, you reach your initial position. So, the initial and final positions are the same and consequently, the displacement becomes zero.
Example (5): Consider a car to be a point particle and move in one dimension. To specify the location of the particle in one dimension we need only one axis that we call $x$ and lies along the straight-line path.
Solution: First, we must define an important quantity that the other kinematics quantities are made from it, displacement.
To describe the motion of the car, we must know its position and how that position changes with time.
The change in the car's position from initial position $x_i$ to final position $x_f$ is called displacement, $\Delta \vec x= x_f-x_i$ (in physics we use the Greek letter $\Delta$ to indicate the change in a quantity).
This quantity is a vector point from $A$ to $B$ and in $1$-D denoted by $\Delta \vec x=x_B-x_A$.
In the figure below, the car moves from point $A$ at $x=2\, {\rm m}$ and after reaching to $x=9\,{\rm m}$ returns and stops at position $x=6\,{\rm m}$ at point $B$.
Therefore, the car's displacement is $\Delta x=6-2=+4\,{\rm m}$.
Another quantity which sometimes confused with displacement is the distance traveled (or simply distance) which is defined as the overall distance covered by the particle.
In the above example, the distance from the initial position is computed as follows:
First, calculate the distance to the return point $d_1=x_C-x_A=9-2=7\,{\rm m}$ then from that point ($x_C$) to the final point $x_B$ i.e. $d_2=x_B-x_C=6-9=-3$.
But we should pick the absolute value of it since the distance is a scalar quantity and for them, a negative value is non-sense.
Therefore, the total distance covered by our car is $d_{tot}=d_1+|d_2|=7+|-3|=7+3=10\,{\rm m}$ .
In case of several turning points along the straight path or once the motion's path is on a plane or even three-dimensional cases, one should divide the overall path (one, two, or three dimensions) into straight lines (without any turning point), calculate the difference of those initial and final points and then add their absolute values of each path to reach to the distance traveled by that particle on that specific path (see examples below).
## Displacement in two and three dimensions
In more than one dimension, the computations are a bit involved and we need to be armed with additional concepts.
In this section, we can learn how by using vectors one can describe the position of an object, and by manipulating them to characterize the displacement and other related kinematical quantities (like ).
In a coordinate system, the position of an object is described by a so-called position vector that extends from reference origin $O$ to the location of the object $P$ and is denoted by $\vec{r}=\overrightarrow{OP}$.
These vectors, in the Cartesian coordinate system (or other related coordinates), can be expressed as a linear combination of unit vectors, $\hat{i},\hat{j},\hat{k}$, (the ones with unit length) as
$\textbf{r}=\sum_1^{n} r_x \hat{i}+r_y \hat{j} +r_z \hat{k}$
where $n$ denotes the dimension of the problem, i.e., in two and three dimensions $n=2,3$, respectively. $r_x , r_y$ and $r_z$ are called the components of the vector $\vec{r}$.
Now the only thing that remains is adding or subtracting these vectors, known as vector algebra, to provide the kinematical quantities.
To do this, simply add or subtract the terms (components) along a specific axis with each other (as below).
Consider adding two vectors $\vec{a}$ and $\vec{b}$ in two dimensions, \begin{array}{cc} \textbf{a}+\textbf{b}&=&\left(a_x \hat{i}+a_y \hat{j} \right)+\left(b_x \hat{i}+b_y \hat{j}\right)\\ &=&\left(a_x+b_x\right)\hat{i}+\left(a_y+b_y\right)\hat{j}\\ &=&c_x\, \hat{i}+c_y\, \hat{j} \end{array}In the last line, the components of the final vector (or resultant vector) are denoted by $c_x$ and $c_y$.
As we learned from vector practice problems, the magnitude and direction of the obtained vector are represented by the following relations \begin{array}{cc} |\textbf{c}| &=& \sqrt{\left(a_x+b_x\right)^2 +\left(a_y+b_y\right)^2 } \qquad \text{magnitude}\\ \theta &=& \tan^{-1} \left(\frac{a_y+b_y}{a_x+b_x}\right) \qquad \text{direction} \end{array} where $\theta$ is the angle with respect to the $x$ axis.
We have two types of problems on the topic of displacement.
In the first case, the initial and final coordinates (position) of an object are given.
Write position vectors for every point. The vector which extends from the tail of the initial point to the tail of the final point is a displacement vector and computed as the difference of those vectors i.e. $\vec{c}=\vec{b}-\vec{a}$.
In the second case, the overall path of an object, between the initial and final points, is given as consecutive vectors as in the figure below.
Here, one should decompose each vector with respect to its origin, then add components along $x$ and $y$ axes separately.
The displacement vector is the one that points from the tip of the first vector to the tail of the last vector and its magnitude is the vector addition of those vectors i.e. $\vec{d}=\vec{a}+\vec{b}+\vec{c}$.
## More examples of displacement:
Example 1: A moving object is displaced from A(2,-1) to B(-5,3) in a two-dimensional plane. What is the displacement vector of this object?
Solution: First, this is the first case mentioned above. So construct the position vectors of point $A$ and $B$ as below \begin{gather*} \overrightarrow{OA}=2\,\hat{i}+(-1)\,\hat{j}\\ \overrightarrow{OB}=-5\,\hat{i}+3\,\hat{j} \end{gather*} Now, by definition, the difference of initial and final points or simply position vectors gets the displacement vector $\vec{d}$ as \begin{align*}\vec{d} &=\overrightarrow{OB}-\overrightarrow{OA}\\\\ &= \left(-5\,\hat{i}+3\,\hat{j}\right)-\left(2\,\hat{i}+(-1)\,\hat{j}\right)\\\\ &= -7\,\hat{i}+4\,\hat{j} \end{align*} its magnitude and direction is also obtained as follows \begin{align*} |\vec{d}|&=\sqrt{(-7)^2 +4^2} \\\\ &=\sqrt{49+16} \\\\ &\approx 8.06\,{\rm m} \end{align*} and $\theta = \tan^{-1} \left(\frac{d_y}{d_x}\right)=\tan^{-1}\left(\frac{4}{-7}\right)$ The angle $\theta$ may be $-29.74^\circ$ or $150.25^\circ$ but since $d_x$ is negative and $d_y$ is positive so the resultant vector lies on the second quarter of coordinate system. Therefore, the desired angle with $x$-axis is $150.25^\circ$.
Example (2): An airplane flies $276.9\,{\rm km}$ $\left[{\rm W}\, 76.70^\circ\, {\rm S}\right]$ from Edmonton to Calgary and then continues $675.1\,{\rm km}$ $\left[{\rm W}\, 11.45^\circ\,{\rm S}\right]$ from Calgary to Vancouver. Using components, calculate the plane's total displacement. (Nelson 12, p. 27).
Solution: In these problems, there is a new thing that appears in many textbooks which is the compact form of direction as stated in brackets. $\left[{\rm W}\, 76.70^\circ\, {\rm S}\right]$ can be read as "point west, and then turn $76.70^\circ$ toward the south".
To solve such practices, first sketch a diagram of all vectors, decompose them, and next using vector algebra, explained above, compute the desired quantity (here $\vec{d}$).
Two successive paths denoted by vectors $\vec{d_1}$ and $\vec{d_2}$ and in terms of components read as \begin{align*} \vec{d_1} &= |\vec{d_1}|\cos \theta (-\hat{i})+|\vec{d_1}|\sin \theta (-\hat{j}) \\ \vec{d_2} &=|\vec{d_2}|\cos \alpha (-\hat{i})+|\vec{d_2}| \sin \alpha (-\hat{j}) \end{align*} Substituting the numerical values into the above expression, we can find \begin{align*} \vec{d_1} &= 276.9\cos 76.70^\circ (-\hat{i})+276.9\sin 76.7^\circ (-\hat{j}) \\ &= 63.700 (-\hat{i})+269.47 (-\hat{j}) \quad [{\rm km}] \\\\ \vec{d_2} &= 675.1\cos 11.45^\circ (-\hat{i})+675.1\sin 11.45^\circ (-\hat{j}) \\ &= 661.664 (-\hat{i})+134.016(-\hat{j}) \quad [{\rm km}] \end{align*} The total displacement is drawn from the tail of $\vec{d_1}$ to the tip of $\vec{d_2}$. In the language of vector addition $\vec{d}=\vec{d_2}+\vec{d_1}$, so \begin{align*} \vec{d} &= \vec{d_2}+\vec{d_1}\\\\ &= 725.364(-\hat{i})+403.486(-\hat{j}) \quad [{\rm km}] \end{align*} Therefore, the length of total path flied by the airplane from Edmonton to Vancouver and its direction with $x$-axis is \begin{align*} |\vec{d}| &=\sqrt{(725.364)^2 +(403.486)^2}\\\\ &= 830.032\quad {\rm km} \end{align*} and \begin{align*} \gamma &=\tan^{-1} \left(\frac{d_y}{d_x}\right) \\\\ &= \tan^{-1} \left(\frac{403.486}{725.364}\right) \\\\ &= 29.09^\circ \end{align*} As one can see, the resultant vector points to the south west or $\left[{\rm W}\,29.09^{\circ}\,{\rm S}\right]$
Example (3): A moving particle moves over the surface of a solid cube in such a way that passes through $A$ to $B$. What is the magnitude of the displacement vector in this change of location of the particle?
Solution: In three-dimensional cases like $2-D$ ones, we should only know the location(coordinates) of the object, then use the following relations to obtain the displacement of a moving particle.Points $A$ and $B$ lies on the $x-z$ plane and $y$ axis, respectively so their coordinates are $(10,0,10)$ and $(0,10,0)$ which parenthesis denote the $(x,y,z)$. This question is of type one, so \begin{align*} \overrightarrow{OA}&= 10\,\hat{i}+0\,\hat{j}+10\,\hat{k} \\ \overrightarrow{OB} &= 0\,\hat{i}+10\,\hat{j}+0\,\hat{k} \end{align*} and displacement vector is \begin{align*} \vec{d} &=\overrightarrow{OB}-\overrightarrow{OA} \\ &= -10\,\hat{i}+10\,\hat{j}-10\,\hat{k} \end{align*} Therefore, the wanted vector in terms of its components was computed as above. Also, its magnitude is the square root of the sum of the squares of each component, i.e., $|\vec{d}|=\sqrt{(-10)^2 +(-10)^2 + (10)^2}=10\sqrt{3}\rm m$
Example (4): A car moves around a circle of radius of $20\,{\rm m}$ and returns to its starting point. What is the distance and displacement of the car? ($\pi = 3$)
Solution: As mentioned above, displacement depends on the initial and final points of the motion. Because the car returns to its initial position so no displacement is made by the car. But the amount of distance traveled is simply the perimeter of the circle (since this scalar quantity depends on the form of the path). So $d=2 \pi r=2 \times 3 \times 20 =120\,{\rm m}$, where $r$ is radius of the circle.
Example (5): A moving object is covering a square path -with one end left open- as shown in the figure below. What is the desired displacement and distance traveled between the specified points? point $p$ lies in the middle of $BC$.
Solution: Displacement is the shortest and straightest line between the initial and final points.
So using the Pythagorean theorem, we get \begin{align*} D^2 &= (iB)^2 +(Bf)^2 \\\\ &= 5^2+(2.5)^2 \\\\ \Rightarrow D &= \sqrt{25 + 6.25} \\\\ &\approx 5.6\quad (\rm m) \end{align*} The direction of the displacement vector is also obtained as \begin{align*} \tan \theta &= \frac{Bp}{iB} \\\\ &= \frac{2.5}{5} \\\\ \Rightarrow \theta &= \arctan \frac{2.5}{5} \\\\ &= 26.5^\circ\, \left[\text{South east}\right] \end{align*} Distance is simply the perimeter of the path traveled, so \begin{align*} \text{distance} &=5 + 2.5 \\ &=7.5\quad \rm m \end{align*}
Example (6): you walk once around an oval track for 6.0 minutes at an average speed of 1.7 m/s.
(a) What is the distance traveled?
(b) What is the displacement?
Solution:(a) here, the distance covered by the person is the perimeter of the ellipse. Since the time elapsed and average speed are given, we can use the definition of average speed and solve for the distance traveled to find it as below \begin{align*} \text{average speed}&=\frac{\text{distance}}{\text{time interval}}\\1.7&=\frac{d}{6\times 60\,{\rm s}}\\ &\\ \Rightarrow d&=\big(1.7\,{\rm \frac ms}\big)(360\,{\rm s})\\&=612\quad {\rm m} \end{align*}
(b) Displacement is the difference between initial and final positions. In this problem, since the person had returned to his original position so his displacement is zero
Example (7): The velocity versus time graph of a car is shown below. Find the displacement in 4 seconds.
Solution: The area under a velocity-time graph in a given time interval indicates the displacement in that interval.
In this case, the bounded area is a triangle whose base is 5 units and height is 2 units. So, its area is \begin{align*} \text{area=displacement}&=\frac 12 \times base\times height\\\\&=\frac 12 \times 5\times 2\\\\&=5\quad {\rm m}\end{align*} Thus, this car displaces 5 meters in 5 s.
## Conclusion
As you can see in the examples of displacement above, displacement is a vector that depends only on the initial and final positions of the object, and not on the details of the motion and path.
These vector quantities require both a length and a direction to be identified.
It is also possible to find the displacement vector using a position vs. time graph.
On the contrary, distance is a quantity that is characterized only by a simple value, which is called scalar, and is path dependence.
In general, the distance traveled and the magnitude of the displacement vector between the two points is not the same.
If the moving object changes its direction in the course of travel, then the total distance traveled is greater than the magnitude of the displacement between those points.
The SI units of both quantities are meters.
Distance and displacement are frequently used in solving velocity and acceleration problems.
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Difference between revisions of "1987 AJHSME Problems/Problem 18"
Problem
Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. The original number of people in the room was
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$
Solution
Let the original number of people in the room be $x$. Half of them left, so $\frac{x}{2}$ of them are left in the room.
After that, one third of this group is dancing, so $\frac{x}{2}-\frac{1}{3}\left( \frac{x}{2}\right) =\frac{x}{3}$ people are not dancing.
This is given to be $12$, so $$\frac{x}{3}=12\Rightarrow x=36$$
$\boxed{\text{C}}$
Solution #2
First note that of the $\frac{1}{2}$ people remaining in the room, $\frac{2}{3}$ are not dancing. Therefore $\frac{1}{2}\cdot\frac{2}{3}= \frac{1}{3}$ of the original amount of people in the room is $12$. The answer is $\boxed{C}$.
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# Introduction to volume
Volume is simply that numerical value that you see in your bottle of coke indicating how much liter is contained in that bottle. It’s simply the amount of water stored in your water tank, and how much of it used up the moment you hit the showers, wash your hands and do the dishes It’s all around us, which is why we have a huge need to understand it more.
This chapter will be all about volume. In the first part we will be looking at the different 3D shapes that are around us like the prisms, cube, cylinder, sphere, pyramids, and more. Each of these shapes have formulas that we can use to solve for their volume. Generally you will need to multiply length (L), width(W) and height (H). Other 3D shapes have more specific equation of volume because of their shapes.
In the second part of the chapter we will get to look at how to solve for the volume of a prism. There are two kinds of prisms namely, the triangular prism and the rectangular prism. In solving for the volume of a triangular prism you need the formula ½ ( base x height x length) since a triangular prism is comprised of triangles, where you need to apply the Pythagorean theorem 1/2Base x Height, and rectangles which has an area of LW. For the volume of the rectangular prism, which is comprised of rectangles, the formula is length x width x height.
The third part of this chapter will teach us how to solve for the volume of cylinders. A cylinder is comprised of two circular bases and one rectangle, which is why the formula to use would be ?r2h.
The last part would be about the solving word problems involving prisms and cylinders. This part would require you to have a good understanding of the previous parts of the chapter. At the end of the day, when you’re all great in solving volume problems you might want to test yourself with some free volume games online.
### Introduction to volume
In this lesson, we will make use of what we learn from the surface area chapter to calculate the volume of 3D shapes like triangular prisms and cylinders.
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## Trigonometry
### Course: Trigonometry>Unit 1
Lesson 6: Modeling with right triangles
# Right triangle trigonometry review
Review right triangle trigonometry and how to use it to solve problems.
## What are the basic trigonometric ratios?
$\mathrm{sin}\left(\mathrm{\angle }A\right)=$$\frac{\text{opposite}}{\text{hypotenuse}}$
$\mathrm{cos}\left(\mathrm{\angle }A\right)=$$\frac{\text{adjacent}}{\text{hypotenuse}}$
$\mathrm{tan}\left(\mathrm{\angle }A\right)=$$\frac{\text{opposite}}{\text{adjacent}}$
## Practice set 1: Solving for a side
Trigonometry can be used to find a missing side length in a right triangle. Let's find, for example, the measure of $AC$ in this triangle:
We are given the measure of angle $\mathrm{\angle }B$ and the length of the $\text{hypotenuse}$, and we are asked to find the side $\text{opposite}$ to $\mathrm{\angle }B$. The trigonometric ratio that contains both of those sides is the sine:
$\begin{array}{rl}\mathrm{sin}\left(\mathrm{\angle }B\right)& =\frac{AC}{AB}\\ \\ \mathrm{sin}\left({40}^{\circ }\right)& =\frac{AC}{7}\phantom{\rule{1em}{0ex}}\mathrm{\angle }B={40}^{\circ },AB=7\\ \\ 7\cdot \mathrm{sin}\left({40}^{\circ }\right)& =AC\end{array}$
Now we evaluate using the calculator and round:
$AC=7\cdot \mathrm{sin}\left({40}^{\circ }\right)\approx 4.5$
Problem 1.1
$BC=$
Want to try more problems like this? Check out this exercise.
## Practice set 2: Solving for an angle
Trigonometry can also be used to find missing angle measures. Let's find, for example, the measure of $\mathrm{\angle }A$ in this triangle:
We are given the length of the side $\text{adjacent}$ to the missing angle, and the length of the $\text{hypotenuse}$. The trigonometric ratio that contains both of those sides is the cosine:
$\begin{array}{rl}\mathrm{cos}\left(\mathrm{\angle }A\right)& =\frac{AC}{AB}\\ \\ \mathrm{cos}\left(\mathrm{\angle }A\right)& =\frac{6}{8}\phantom{\rule{1em}{0ex}}AC=6,AB=8\\ \\ \mathrm{\angle }A& ={\mathrm{cos}}^{-1}\left(\frac{6}{8}\right)\end{array}$
Now we evaluate using the calculator and round:
$\mathrm{\angle }A={\mathrm{cos}}^{-1}\left(\frac{6}{8}\right)\approx {41.41}^{\circ }$
Problem 2.1
$\mathrm{\angle }A=$
${}^{\circ }$
Want to try more problems like this? Check out this exercise.
## Practice set 3: Right triangle word problems
Problem 3.1
Howard is designing a chair swing ride. The swing ropes are $5$ meters long, and in full swing they tilt in an angle of ${29}^{\circ }$. Howard wants the chairs to be $2.75$ meters above the ground in full swing.
How tall should the pole of the swing ride be?
meters
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• This is not correct. The path of the swing is an arc so at the point where it is parallel to the support pole it would closer to the ground than at the point of full swing which is 2.75 meters. To make this example correct the 2,75 meters needs to be applied to the point where the swing is parallel to the supporting pole.
• You are correct that it is an arc. However, the key to the question is the phrase "in full swing". The swing will be closer than 2.75 meters at the bottom of the arc. That is an interesting point that I hadn't considered, but not what the question is asking.
• Shouldn't we take in account the height at which the MIB shoots its laser. I'm guessing it would be somewhere from his shoulder. Maybe the answer wouldn't differ that much but it might make it a little more challenging to figure out. You would even be able to calculate the height the agent is holding his gun at with stretched arms when you know the angle he's keeping his arms at, his arm length and the length from his shoulders to the ground.
• Good point, let's estimate :D.
Men In Black are generally rather tall so it is fair to estimate the man is about two meters tall. The average arm length of an adult human is ~25 inches which equates to about 0.6 meters. If we assume that the man holds his arms directly above his head (not technically realistic but a fair assumption) then we can estimate the height of the LASER to be about 2.5 meters. If we subtract 2.5 from 324 we get 321.5. Arctan(321.5/54) = 80.465.
That deviates from the ground angle by only 0.09%, so it probably wouldn't affect how he aimed the laser.
If we assume he shot from shoulder height with his arms straight out, then that would be arctan(322/53.5) = ~80.567 which deviates from the ground angle by only 0.04%.
• This is really fun to do
• What is the value of sine, cosine, and tangent?
• The Sine, Cosine, and Tangent are three different functions. They do not have a value outright, it would be like trying to ask what the value of f(x) = x + 1 is. The trig functions give outputs in terms of the ratios of two sides of a triangle when we feed them the input of an angle measure.
Sine outputs the ratio of opposite to hypotenuse
Cosine outputs the ratio of adjacent to hypotenuse
Tangent outputs the ratio of opposite to adjacent
• I am so confused...I try my best but I still don't get it . I need someone to Break it down further for me? I never not understand math but this one really has me stuck.Thank you.
• You might not be taking trig anymore, but there is SOHCAHTOA - Sine = Opposite * Hypotenuse. sine is when you are given the opposite side from the angle and the hypotenuse of the triangle. Cosine = Adjacent * Hypotenuse. cosine is when you are given the adjacent side of the angle and the hypotenuse of the triangle. Finally, Tangent = Opposite * Adjacent. This is when you are given the opposite and adjacent sides of the angle. A side could also have x as its value and you solve for x.
• in question 1.1 the given answer is approx 5.44 my calculator is giving 0.91 as an answer even in degrees mode
• when solving for an angle why does cos have a -1 on top?
• Trig functions like cos^-1(x) are called inverse trig functions. THey are the inverse functions of the normal trig functions. Where cos(x) would take in an angle and output a ratio of side lengths, cos^-1(x) takes in the ratio of adjacent/hypotenuse and gives you an angle, which is why we use it when solving for unknown angles.
Note that cos^-1(x), (cos(x))^-1, and cos(x^(-1)) give three completely separate results.
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These worksheets will give your students practice deriving the midpoints of given segments and coordinates.
#### What is a Midpoint of a Line Segment? The line segment is defined as the portion of a line with two endpoints. It suggests that a line segment has a definite starting point, and it ends at a defined point. We use letters to denote the two endpoints of the line segment. The midpoint of a line segment is defined as the average of its two endpoints. Just like we divide the number by two to find their average, we will divide the measurements of line segments by two. In some cases, we need to locate or find out the point present midway between two endpoints. For example, you might have to bisect the given line segment and divide it into two equal parts. The formula for the midpoint of the line segment works like any other formula used for finding the average. It is typically expressed as: M = ((x1 + x2)/2 , (y1 + y2)/2). Where (x1, y1) and (x2, y2) are the two endpoints.
The exact middle of a line segment or side of a geometric figure is found in the dead center. Because of this both sides of the line segment are therefore equal. While many students discount this skill of calculating the middle, it is an imperative task for engineers, construction workers, and architects. Every time you travel over a bridge or tunnel you had better hope the engineers behind the construction knew where the anchor mid-points should be set. These worksheets explain how to find the midpoint of a segment. Please note that some answers may contain variables rather than integers.
# Print MidPoint of a Line Segment Worksheets
## Midpoint of the Segment Lesson
This worksheet explains how to find the midpoint of a line segment. A sample problem is solved, and two practice problems are provided.
## Midpoint of the Segment Worksheet
Students will find the midpoint and learn a slightly new technique. Ten problems are provided.
## Midpoint of the Segment Practice
Students find the midpoint of a segment, the slope, and write the equation. Ten problems are provided.
## Review and Practice
We will review this skill and see if we can add a few nuggets of info for you. Six practice problems are provided.
## Midpoint of the Segment Quiz
Students will demonstrate their proficiency in finding the midpoint of segments. Ten problems are provided.
## Line Segment Check
Students will determine the midpoint of the segments. Three problems are provided, and space is included for students to copy the correct answer when given.
## Finding the Midpoint Lesson
This worksheet explains how to find the midpoint of a segment connecting two points. A sample problem is solved, and two practice problems are provided.
## Practice Worksheet
Students will find the midpoint of a segment connecting two points. Ten problems are provided.
## Practice Worksheet
Given two points, students will find the midpoint of the segment connecting them. Ten problems are provided.
## Review and Practice
Students Review how to find the midpoint of a segment connecting two points. Six practice problems are provided.
## Two Points Quiz
Students will demonstrate their proficiency in finding the midpoint of a segment connecting two points. Ten problems are provided.
## Skills Check
Students will find the midpoint of a segment connecting two points. Three problems are provided, and space is included for students to copy the correct answer when given.
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You are viewing an older version of this Concept. Go to the latest version.
# Fractions
## Identity, inverse, and zero product properties
0%
Progress
Practice Fractions
Progress
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Identify and Apply Number Properties in Fraction Operations
Do you know how to simplify an expression with fractions by using properties? Take a look at this dilemma.
Simplify:
To simplify this expression, you will need to know how to work with number properties and fractions. This Concept will show you how to accomplish this task successfully.
### Guidance
Now that we are working with fractions, you will have a chance to investigate how the different properties of addition and subtraction can help us when we work with fractions in expressions.
Here are a few properties.
Additive Identity Property
The sum of any number and zero is that number:
Additive Inverse Property
The sum of any number and its inverse is zero:
Which of the following shows the Additive Inverse Property?
a.
b.
c.
Consider choice a.
This equation states that a number added to zero is equal to zero. This is not necessarily correct, unless is also equal to zero.
Consider choice b.
This equation states that the sum of a number and zero is equal to that number. This is correct, but illustrates the additive identity, not the additive inverse property.
Consider choice c.
This equation states that the sum of a number and its inverse is equal to zero. This illustrates the additive inverse property, so this is the correct equation.
We can also use these properties to help us when we simplify a numerical expression. Remember that a numerical expression is a group of numbers and operations. Because we are working with fractions, the numerical expressions in this section will be made up of fractions.
Simplify:
You can use addition properties to reorganize this expression to make it easier to simplify.
First apply the commutative property:
Then apply the associate property:
Now you can easily simplify to find the sum.
The answer is .
Using properties to reorganize fractions can help us to work with these fractions. Notice that we reorganized the common denominators together and this simplified our work.
Now let’s look at how the properties of multiplication and division can help us when working with fractions. You have already learned the Commutative Property, the Associative Property, and the Distributive Property.
Multiplicative Identity
The product of any number and one is that number:
Zero Property
The product of any number and zero is zero:
Multiplicative Inverse
The product of any number and its reciprocal is one:
Which of the following shows the Multiplicative Inverse Property?
a.
b.
c.
Consider choice a.
This equation states that the product of a number and zero is equal to that number. This is not correct.
Consider choice b.
This equation states that the product of a number and its reciprocal is equal to zero. This is not correct.
Consider choice c.
This equation states that the product of a number and its reciprocal is equal to one. This illustrates the multiplicative inverse property, so this is the correct equation.
Take a few minutes to write these properties down in your notebooks. Be sure to include an example of each.
You can also use properties to help you simplify numerical expressions.
We could also use properties when working with a variable. Take a look at this one.
First, we can apply the commutative property:
Now we apply the multiplication inverse property:
Our simplified expression is .
If we had a value substituted in for , then that would be our answer.
#### Example A
Name the property:
Solution: Zero Property
#### Example B
Name the property:
Solution: Multiplicative Inverse
#### Example C
Simplify:
Solution:
Now let's go back to the dilemma from the beginning of the Concept.
Simplify:
You can use multiplication properties to reorganize this expression to make it easier to simplify.
First apply the commutative property:
Then apply the associate property:
Then apply the multiplicative inverse property:
Finally, apply the multiplicative identity property:
This is our answer.
### Vocabulary
Additive Identity Property
any number plus zero is still that number.
Additive Inverse Property
any number plus it’s opposite or inverse is equal to 0.
Multiplicative Identity
any number times 1 is the same number.
Zero Property
any number times 0 is zero.
Multiplicative Inverse
any number times it’s reciprocal is 1.
### Guided Practice
Here is one for you to try on your own.
Simplify:
Solution
First, we find a common denominator so that we can add the fractions. The lowest common denominator for 5 and 2 is 10. Let's rename both fractions in terms of tenths.
Now we can add the fractions.
Change the improper fraction to a mixed number.
This is our simplified expression.
### Practice
Directions: Identify each property shown below.
1.
2.
3.
4.
5.
6.
7.
8.
Directions: Simplify each expression.
9.
10.
11.
12.
13.
14.
15.
### Vocabulary Language: English
Additive Identity Property
Additive Identity Property
The sum of any number and zero is the number itself.
Additive inverse
Additive inverse
The additive inverse or opposite of a number x is -1(x). A number and its additive inverse always sum to zero.
Multiplicative Identity
Multiplicative Identity
The multiplicative identity for multiplication of real numbers is one.
Multiplicative Inverse
Multiplicative Inverse
The multiplicative inverse of a number is the reciprocal of the number. The product of a real number and its multiplicative inverse will always be equal to 1 (which is the multiplicative identity for real numbers).
Zero Property
Zero Property
The zero property of multiplication says that the product of any number and zero is zero. The zero property of addition states that the sum of any number and zero is the number.
Please wait...
Please wait...
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$$\DeclareMathOperator{cosec}{cosec}$$
# Number and Algebra
## Furthermore
Official Guidance, clarification and syllabus links:
Calculator or computer notation is not acceptable. For example, 5.2E30 is not acceptable and should be written as $$5.2 \times 10^{30}$$.
If you use the TI-Nspire calculator you can find instructions for entering numbers in standard form on the GDC Essentials page.
Numbers in standard form, also known as scientific notation, are expressed as $$a \times 10^n$$, where $$1 \leq |a| < 10$$ and $$n$$ is an integer. This notation is particularly useful when dealing with very large or very small numbers, as it allows for a concise representation.
When performing operations with numbers in standard form, it is crucial to follow the rules of arithmetic carefully. Here are some examples to illustrate the operations:
When adding numbers in standard form, it is essential to have the same exponent. If the exponents are different, adjust them appropriately before performing the addition.
$$(3 \times 10^4) + (5 \times 10^3) = (3 \times 10^4) + (0.5 \times 10^4) \\ = 3.5 \times 10^4$$
Subtraction:
Similar to addition, when subtracting numbers in standard form, ensure that the exponents are the same before performing the subtraction.
$$(7 \times 10^6) - (2 \times 10^5) = (7 \times 10^6) - (0.02 \times 10^6) \\ = 6.98 \times 10^6$$
Multiplication:
When multiplying numbers in standard form, multiply the coefficients (the numbers in front of the power of 10) and then add the exponents of the powers of 10.
$$(12 \times 10^3) \times (4 \times 10^5) = 48 \times 10^8$$
Remember to always express your final answer in standard form, ensuring that the coefficient is a number between 1 and 10 (including 1 but excluding 10). Adjust the power of 10 to compensate any changes you have made to the coefficient.
$$= 4.8 \times 10^9$$
Division:
For division, divide the coefficients and then subtract the exponent in the denominator from the exponent in the numerator.
$$\frac{{3 \times 10^9}}{{6 \times 10^2}} = 0.5 \times 10^7$$
Again remember to always express your final answer in standard form, ensuring that the coefficient is a number between 1 and 10 (including 1 but excluding 10). Adjust the power of 10 to compensate any changes you have made to the coefficient.
$$= 5 \times 10^6$$
This video on Scientific Notation is from Revision Village and is aimed at students taking the IB Maths 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.
For Students:
For All:
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# NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1
Get Free NCERT Solutions for Class 10 Maths Chapter 5 Ex 5.1 PDF. Arithmetic Progressions Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 5.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Maths Class 10 Arithmetic Progressions Exercise 5.1 provided in NCERT TextBook.
Topics and Sub Topics in Class 10 Maths Chapter 5 Arithmetic Progressions:
Section Name Topic Name 5 Arithmetic Progressions 5.1 Introduction 5.2 Arithmetic Progressions 5.3 Nth Term Of An AP 5.4 Sum Of First N Terms Of An AP 5.5 Summary
## NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1
Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter 5 Chapter Name Arithmetic Progressions Exercise Ex 5.1 Number of Questions Solved 4 Category NCERT Solutions
We also solved 106 questions from Chapter 9 – Arithmetic Progressions of RD Sharma Class 10 Maths Textbook.
Page No: 99
Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes $\frac { 1 }{ 4 }$of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every meter of digging, when it costs ₹ 150 for the first meter and rises by ₹ 50 for each subsequent meter.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.
Solution:
Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = $\frac { 1 }{ 2 }$
(v) a = -1.25, d = -0.25
Solution:
You can also download the free PDF of Ex 5.1 Class 10 Arithmetic Progressions NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.
Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, ……
(ii) -5, -1, 3, 7, ……
(iii) $\frac { 1 }{ 3 }$, $\frac { 5 }{ 3 }$, $\frac { 9 }{ 3 }$, $\frac { 13 }{ 3 }$, ……..
(iv) 0.6, 1.7, 2.8, 3.9, …….
Solution:
Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, …….
(ii) 2, $\frac { 5 }{ 2 }$, 3, $\frac { 7 }{ 2 }$, …….
(iii) -1.2, -3.2, -5.2, -7.2, ……
(iv) -10, -6, -2,2, …..
(v) 3, 3 + $\sqrt { 2 }$, 3 + 2$\sqrt { 2 }$, 3 + 3$\sqrt { 2 }$, …..
(vi) 0.2, 0.22, 0.222, 0.2222, ……
(vii) 0, -4, -8, -12, …..
(viii) $\frac { -1 }{ 2 }$, $\frac { -1 }{ 2 }$, $\frac { -1 }{ 2 }$, $\frac { -1 }{ 2 }$, …….
(ix) 1, 3, 9, 27, …….
(x) a, 2a, 3a, 4a, …….
(xi) a, a2, a3, a4, …….
(xii) $\sqrt { 2 }$, $\sqrt { 8 }$, $\sqrt { 18 }$, $\sqrt { 32 }$, …..
(xiii) $\sqrt { 3 }$, $\sqrt { 6 }$, $\sqrt { 9 }$, $\sqrt { 12 }$, …..
(xiv) 12, 32, 52, 72, ……
(xv) 12, 52, 72, 73, ……
Solution:
+
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Combining Functions to get New Functions
Combining Functions to get New Functions
Numbers can be ‘connected’ to get new numbers: for example, $\,1+2=3\,.$ Indeed, addition ‘+’ is a connective for numbers.
Sets can be ‘connected’ to get new sets: for example, $\,\{1,2,3\}\cap \{2\} = \{2\}\,.$ Indeed, the intersection operator ‘$\,\cap\,$’ is a connective for sets.
Functions can also be ‘connected’ to get new functions: the most important way to do this, called function composition, is the subject of the next lesson. In this lesson, we look at some other ways that functions can be combined to get new functions.
The Sum Function
Consider the box below:
It shows how two functions $\,f\,$ and $\,g\,$ are ‘combined’ to get a new function, named $\,f+g\,.$
(Don't be intimidated by the multi-symbol function name ‘$\,f+g\,$’! It's a very natural name, as you'll see.)
Here's how this new function $\,f+g\,$ works:
• Drop an input $\,x\,$ in the top.
• This input gets ‘copied’: one copy gets dropped in the $\,\color{green}{f}\,$ box, the other in the $\,\color{green}{g}\,$ box.
• The two outputs $\,\color{purple}{f(x)}\,$ and $\,\color{purple}{g(x)}\,$ get dropped in the ‘+’ box, which adds them.
• The number $\,\color{red}{f(x) + g(x)}\,$ comes out the bottom.
Thus, $\,(f+g)(x) := f(x) + g(x)\,.$
$$\cssId{s22}{(} \overbrace{\strut \cssId{s23}{f+g}}^{\cssId{s25}{\text{this function}}}\cssId{s24}{)} \cssId{s26}{(}\overbrace{\strut \ \ \cssId{s27}{x}\ \ }^{\cssId{s29}{\text{acts on } x}}\cssId{s28}{)} \overbrace{\strut \ \ \cssId{s30}{:=}\ \ }^{\cssId{s31}{\text{and gives}}} \overbrace{\strut \cssId{s32}{f(x) + g(x)}}^{\cssId{s33}{\text{this output}}}$$
When using function notation, parentheses should always be put around the function name $\,f+g\,$ for clarity. Remember that ‘$\,:=\,$’ means ‘equals, by definition’.
The domain of $\,f+g\,$ is the set of all inputs that both $\,f\,$ and $\,g\,$ know how to act on. This is all we have to worry aboutafter getting the outputs $\,f(x)\,$ and $\,g(x)\,,$ any two real numbers can be added.
\begin{align} &\cssId{s40}{\text{dom}(f+g)}\cr\cr &\quad\cssId{s41}{= \{x\ |\ x\in\text{dom}(f) \text{ and } x\in\text{dom}(g)\}}\cr\cr &\quad\cssId{s42}{= \{x\ |\ x\in \bigl(\text{dom}(f) \cap \text{dom}(g)\bigr)\}}\cr\cr &\quad\cssId{s43}{= \text{dom}(f) \cap \text{dom}(g)} \end{align}
Difference, Product, and Quotient Functions
The idea is the same for the difference, product, and quotient functions:
Function: $f - g$ Definition: $(f-g)(x) := f(x) - g(x)$ Domain: $\text{dom}(f-g) = \text{dom}(f) \cap \text{dom}(g)$ Function: $fg$ Definition: $(fg)(x) := f(x)g(x)$ Domain: $\text{dom}(fg) = \text{dom}(f) \cap \text{dom}(g)$ Function: $\displaystyle\frac{f}{g}$ Definition: $\displaystyle\bigl(\frac{f}{g}\bigr)(x) := \frac{f(x)}{g(x)}$ Domain: $$\text{dom}\bigl(\frac{f}{g}\bigr) = \text{dom}(f) \cap \text{dom}(g) \cap \{x\ |\ g(x)\ne 0\}$$ Since division by zero is not allowed, the domain of the quotient function is a bit more complicated.
Example
Let $\,f(x) = x^2\,$ and $\,g(x) = x+3\,.$
Find $\,f+g\,,$ $\,f-g\,,$ $\,fg\,,$ and $\,\frac{f}{g}\,.$
Solution:
\begin{align} \cssId{s63}{(f+g)(x)}\ &\cssId{s64}{:= f(x) + g(x)}\cr &\cssId{s65}{ = x^2 + x + 3}\cr \cr \cssId{s66}{(f-g)(x)}\ &\cssId{s67}{:= f(x) - g(x)}\cr &\cssId{s68}{ = x^2 - (x + 3)}\cr &\cssId{s69}{ = x^2 - x - 3}\cr \cr \cssId{s70}{(fg)(x)}\ &\cssId{s71}{:= f(x)g(x)}\cr &\cssId{s72}{ = x^2(x+3)}\cr &\cssId{s73}{ = x^3 + 3x^2}\cr \cr \cssId{s74}{\bigl(\frac{f}{g}\bigr)(x)}\ &\cssId{s75}{:= \frac{f(x)}{g(x)}}\cr &\cssId{s76}{ = \frac{x^2}{x+3}}\cssId{s77}{\ \ \text{for}\ \ x\ne -3} \end{align}
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### Math Texts for TX-TEKS 4th Grade
Subject
Category
Standards
Try Now
text
Unicorn Knows Number Values
Place Value and Division
4.2.A and 1 more
TX-TEKS
Try Now
text
Panda Breakout: Subtract Large Numbers
Add and Subtract Multi-Digit Whole Numbers
4.4.A
TX-TEKS
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text
Numberock Prime, Composite & Square Song
Prime and Composite Numbers
5.4.A
TX-TEKS
## Math Activities and Teaching Resources for 4th Grade
Whole numbers are easy, fractions are what separate the true mathematicians from the beginners. (This is not true, but hopefully your fourth graders feel that way).
In fourth grade, students will build on the multiplication and division skills mastered in the previous year to solve more complex problems using a variety of strategies. Pattern recognition is emphasized and place value is revisited. Students can now fluently add and subtract large numbers, compare fractions and decimals, and begin applying operations to those numbers. Various units of measurement are explored and angles and lines serve as the primary themes for most geometry work.
Enjoy this sampling of instructional videos, games, and activities for your 4th grade classroom!
Some of the skills students will master in eSpark include:
Operations and Algebraic Thinking
• Interpret a multiplication equation as a comparison; represent verbal statements of multiplicative comparisons as multiplication statements.
• Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number, distinguishing multiplicative comparison from additive comparison.
• Solve multi-step word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted.
• Find all factor pairs for a whole number in the range of 1-100; recognize that a whole number is a multiple of each of its factors; determine whether a given whole number 1-100 is a multiple of a given one-digit number.
• Generate a number of shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself.
Number and Operations in Base Ten
• Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right.
• Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results.
• Use place value understanding to round multi-digit whole numbers to any place.
• Fluently add and subtract multi-digit whole numbers using the standard algorithm.
• Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
Number and Operations Fractions
• Explain why a fraction a/b is equivalent to a fraction (n x a) / (n x b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions are the same size; use this principle to recognize and generate equivalent fractions.
• Compare two fractions with different numerators and denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or < and justify the conclusions.
• Understand a fraction a/b with a > 1 as a sum of fractions 1/b.
• Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.
• Use decimal notation for fractions with denominators 10 or 100.
Measurement and Data
• Know relative sizes of measurement units within one system of units including km, m, cm; kg, g; lb, oz; l, ml; hr, min, sec. Express measurements in a larger unit in terms of a smaller unit.
• Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit.
• Apply the area and perimeter formulas for rectangles in real world and mathematical problems.
• Make a line plot to display a data set of measurements in fractions of a unit; solve problems involving addition and subtraction of fractions by using information presented in line plots.
• Recognize angles as geometric shapes formed when two rays share a common endpoint, and understand angle measurement.
• Measure angles in whole-number degrees using a protractor; sketch angles of specified measure.
• Recognize angle measure as additive; solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems.
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State its domain and range. Introduction We plan to introduce the calculus on Rn, namely the concept of total derivatives of multivalued functions f: Rn!Rm in more than one variable. Exercises13 Chapter 2. Inverse Functions. (b). Example 1: List the domain and range of the following function. Find tangent line at point (4, 2) of the graph of f -1 if f(x) = x3 + 2x … Example: Differentiate . If we know the derivative of f, then we can nd the derivative of f 1 as follows: Derivative of inverse function. (2). The line y = x is shown to so you can clearly see that the graphs are symmetric with respect to that line. ()= 1 +2 As stated above, the denominator of fraction can never equal zero, so in this case +2≠0. How to get the Inverse of a Function step-by-step, algebra videos, examples and solutions, What is a one-to-one function, What is the Inverse of a Function, Find the Inverse of a Square Root Function with Domain and Range, show algebraically or graphically that a function does not have an inverse, Find the Inverse Function of an Exponential Function y= arcsinxif and only if x= sinyand ˇ 2 y ˇ 2. y= arccosxif and only if x= cosyand 0 y ˇ. Figure 2.1: Plot of Gaussian Function and Cumulative Distribution Function When the mean is set to zero ( = 0) and the standard deviation or variance is set to unity (˙= 1), we get the familiar normal distribution G(x) = 1 p 2ˇ e x2=2dx (1.2) which is shown in the curve below. Scroll down the page for more examples and solutions on how to use the formulas. We are indeed familiar with the notion of partial derivatives @ if … Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . Derivative of the inverse function at a point is the reciprocal of the derivative of the function at the corresponding point . Inverse Functions 1. Example 2 Use inverse functions to find range of functions. Example 1: Integration with Inverse Trigonometric Functions (a). Here is a figure showing the function, f(x) (the solid curve) and its inverse function f−1(x) (the dashed curve). The Derivative of an Inverse Function. Exam Questions – Inverse functions. Example: Compute the inverse Laplace transform q(t) of Q(s) = 3s (s2 +1)2 You could compute q(t) by partial fractions, but there’s a less tedious way. Examples of rates of change18 6. 2 The graph of y = sin x does not pass the horizontal line test, so it has no inverse. art’s Calculus Early Transcendentals, and many of the examples included were taken from these sources. Start with . Complete any partial fractions leaving the e asout front of the term. Example 6.24 illustrates that inverse Laplace transforms are not unique. 1. and invert it using the inverse Laplace transform and the same tables again and obtain t2 + 3t+ y(0) With the initial conditions incorporated we obtain a solution in the form t2 + 3t Without the Laplace transform we can obtain this general solution y(t) = t2 + 3t+ C1 Info. Inverse functions and Implicit functions10 5. Answer 1. Informal de nition of limits21 2. 1) View Solution Helpful Tutorials Example \( \PageIndex{4}\): Finding an Antiderivative Involving the Inverse Tangent Function. The function ˜(x) must also obey the homogeneous boundary conditions we require of y(x). Therefore, the inverse is not a function based on it fails the Horizontal Line that intersect the graph more than once. For if not, the two di erentiations applied to a jump function would give us the derivative of a delta function, and we want only … If we restrict the domain (to half a period), then we can talk about an inverse function. Let us first show that function f given above is a one to one function. Here is a set of practice problems to accompany the Inverse Functions section of the Graphing and Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University. Solution. Inverse-Implicit Function Theorems1 A. K. Nandakumaran2 1. Definition 6.25. Using function machine metaphor, forming an inverse function means running the function machine backwards.The backwards function machine will work only if the original function machine produces a unique output for each unique input. Limits and Continuous Functions21 1. 4. =? Now (5.10) tells us that ˜(x) must be continuous at x= ˘. Inverse Laplace With Step Functions - Examples 1 - 4 Tips for Inverse Laplace With Step/Piecewise Functions Separate/group all terms by their e asfactor. the ones which pass the horizontal ... inverse function of f(x) = jxjrestricted to (1 ;0] is the inverse function of Exercises18 Chapter 3. In Chapter 1, you have studied that the inverse of a function f, denoted by f –1, exists if f is one-one and onto.There are many functions which are not one-one, onto or both and hence we can not talk of their inverses. Example … They are also termed as arcus functions, antitrigonometric functions or cyclometric functions. If \(f(x)\) is both invertible and differentiable, it seems reasonable that the inverse of \(f(x)\) is … Solutions of all exercise questions, examples are given, with detailed explanation.In this chapter, first we learnWhat areinverse trigonometry functions, and what is theirdomain and rangeHow are trigonometry and inverse t Derivatives of Inverse Trigonometric Functions. INVERSE FUNCTION Example 1: Find the inverse function of 푓(?) The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined. Inverse Laplace Transform Table 2 + 2 if it exists. 1 Inverse Trigonometric Functions De nition 1.1. The concepts of inverse trigonometric functions is also used in science and engineering. If you are not sure what an inverse function is or how to find one then this video should hopefully show you.Example:In this tutorial you will be shown how to find the inverse of the following:If f(x) = (3x - 2) / 8, find f- 1(x) Inverse Example on Handling more than Get NCERT Solutions of Chapter 2 Class 12 Inverse Trigonometry free atteachoo. 2.2 Basic Concepts In Class XI, we have studied trigonometric functions, which are defined as follows: sine function, i.e., sine : R → [– 1, 1] elementary 2 Slope of the line tangent to at = is the reciprocal of the slope of at = . Math 135Functions: The Inverse Solutions 1.In the ”Functions: Examples” worksheet from Week 5 do the following: (a)Determine whether each function is one-to-one. However, it can be shown that, if several functions have the same Laplace transform, then at most one of them is continuous. original function is to find its inverse function, and the find the domain of its inverse. Find the range of function f give by f(x) = 2 x / (x - 3) Solution to example 2: We know that the range of a one to one function is the domain of its inverse. These inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios. Chapter 1: Relations and Functions – Download NCERT Solutions PDF. 22 Derivative of inverse function 22.1 Statement Any time we have a function f, it makes sense to form is inverse function f 1 (although this often requires a reduction in the domain of fin order to make it injective). {Partial fraction decomposition only works for polynomial nu-merators. Rates of change17 5. An example { tangent to a parabola16 3. Example \( \PageIndex{1}\): Evaluating a Definite Integral Using Inverse Trigonometric Functions ... To close this section, we examine one more formula: the integral resulting in the inverse tangent function. We begin by considering a function and its inverse. If we calculate their derivatives, we see that: ( ) () ( ) ( ) The derivatives are reciprocals of one another, so the slope of one line is the reciprocal of the slope of its inverse line. The inverse of a complex function F(s) to generate a real-valued function f(t) is an inverse Laplace transformation of the function. An inverse function is a function that undoes the action of the another function. Here is a set of practice problems to accompany the Derivatives of Inverse Trig Functions section of the Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. INVERSE TRIGONOMETRIC FUNCTION.pdf - 7001_AWLThomas_ch01p001-057.qxd 2:24 PM Page 46 46 Chapter 1 Functions Solution From Example 1 Section 1.5 with P = {The e asonly a ects nal inverse step. Then find the inverse function and list its domain and range. Inverse functions mc-TY-inverse-2009-1 An inverse function is a second function which undoes the work of the first one. This prompts us to make the following definition. f(a) = f(b) 2 a / (a - 3) = 2 b / (b - 3) The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. This function is therefore an exponentially restricted real function. Table Of Derivatives Of Inverse Trigonometric Functions. For xsatisfying 1 x 1, we de ne the arcsine and arccosine functions as follows. The inverse trigonometric functions play an important role in calculus for they serve to define many integrals. Instantaneous velocity17 4. Finding inverse trig values with a calculator (or trig tables) Example: Find Sin I (-.68) between 90 and 270 Step 1: Check mode I check my calculator: degree mode Step 2: Input value and calculate the inverse function The common reference angle is 30, so our solution is 30 and 330 for the range 0 < < 360 Finding inverse trig values a calculator The normal distribution function … NCERT Solutions For Class 12 Maths Chapter 2 – Inverse Trigonometric Functions . 7.2 Derivatives of Inverse Functions We calculated the inverse of the function ( ) as ( ) in Example 1. Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. 7. The one-to-one functions, i.e. The tangent to a curve15 2. Solution: This quadratic function does not have a restriction on its domain. An inverse function goes the other way! The following table gives the formula for the derivatives of the inverse trigonometric functions. In this unit we describe two methods for finding inverse functions, and we also explain that the domain of a function may need to be restricted before an inverse function can exist. If a unique function is continuous on 0 to ∞ limit and also has the property of Laplace Transform. Derivatives (1)15 1. Inverse Functions
Finding the Inverse
2. polynomial Comment. Inverse Trigonometry Functions and Their Derivatives. An inverse function will always have a graph that looks like a mirror Solution: We can use the above formula and the chain rule. 1st example, begin with your function
f(x) = 3x – 7 replace f(x) with y
y = 3x - 7
Interchange x and y to find the inverse
x = 3y – 7 now solve for y
x + 7 = 3y
= y
f-1(x) = replace y with f-1(x)
Finding the inverse
p388 Section 5.9: Inverse Trigonometric Functions: Integration Theorem 5.19: Integrals Involving Inverse Trigonometric Functions Let u be a differentiable function of x, and let a > 0 (1). 3 Definition notation EX 1 Evaluate these without a calculator. Is also used in science and engineering or cyclometric functions inverse function examples and solutions pdf one-to-one functions and inverse! An exponentially restricted real function following function continuous on 0 to ∞ and! Of fraction can never equal zero, so in this case +2≠0 the (!: List the domain and range of functions in trigonometry are used to get the angle with of. This function is a one to one function for xsatisfying 1 x 1 we! A function that undoes the action of the other trigonometric functions play an important role in Calculus for they to... X 1, we de ne the arcsine and arccosine functions as follows: derivative of,... 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# Example: Center of mass
## Understanding the situation
In our discussion of the center of mass, we did everything with abstract equations. Let's do some examples that show how they play out in explicit cases. These illustrate two important general ideas:
1. In applying general (powerful) equations to specific cases, putting things in terms of a specific well-chosen coordinate system helps a lot in carrying out the calculation.
2. Considering limiting cases can help a lot in making sense of what an equation is telling you.
## Presenting a sample problem
Let's begin by considering the two-mass case that we considered in the introduction to the idea of center of mass (CM), explicitly put in a coordinate system, and carry out the calculation to find the position of the CM.
In the window below, we have created a geogebra window that allows you to change each of two masses connected by a massless rod and to see what happens to the position of the CM as a result.
(a) For an object consisting of two masses, show that the CM of the combined object is along the line connecting them and find an expression showing how far along that line the CM is.
(b) For the three cases mA = mB , m>> mB , and mA << mB , show that the result makes physical sense.
## Solving this problem
(a) The equation for the position of the CM of two objects is
$$\overrightarrow{r}_{CM} = (\frac{m_A}{m_A + m_B}) \overrightarrow{r}_A + (\frac{m_B}{m_A + m_B}) \overrightarrow{r}_B$$
Let's choose our coordinate system so that mass A is at the origin. This makes our calculation a lot easier. Since we will not be changing x and y directions during this problem, we can take i and j as fixed and therefore use (x,y) notation. Mass A is at the origin so its position vector is (0,0). Mass B is on the x axis a distance L from the origin so its position vector is (L,0). Our result for the CM vector is therefore
$$\overrightarrow{r}_{CM} = (\frac{m_A}{m_A + m_B}) (0,0)+ (\frac{m_B}{m_A + m_B}) (L,0)$$
$$\overrightarrow{r}_{CM} = (\frac{m_B}{m_A+m_B}L,0)$$
Since there is no y component, the answer lies along the x-axis. Since $m_B$ is always less than $m_A+m_B$, the fraction in front of the $L$ must be between 0 (if $m_B = 0$) and 1 (if $m_B$ gets very large so $m_A$ can be ignored). Therefore, the CM of the two masses lies along the line connecting them a distance $\frac{m_B }{m_A + m_B}L$ from mass A.
(b) If the two masses are both equal (to, say, m), we get the result is mL/2m = L/2, halfway between the two masses. This agrees with what symmetry demands.
If mass A gets very large compared to B, the denominator of the fraction mB/(mA+mB) gets large while the numerator stays fixed, so the fraction gets small (approaches 0). So if the mass A is much bigger, the CM is near 0 -- almost at the same position as mass A.
If mass B gets very large compared to A, you can ignore the mA in the denominator of the fraction mB/(mA+mB), so the fraction gets approaches 1. So if the mass B is much bigger, the CM is almost at the same position as mass B.
These both make sense and the symmetry of the result is important. We could have chosen either mass to be at the origin. Notice that since we have described where the CM is with respect to the masses rather than in terms of its coordinates, the result does NOT depend on the coordinate system chosen -- which was arbitrary!
This calculation illustrates a number of useful general ideas: (1) the choice of your coordinate system can simplify your calculations; (2) taking limiting cases (a parameter gets large or goes to 0) can help you both check your work and see that it makes sense; and (3) expressing your results in terms of physical things rather than coordinates gives you a result that doesn't depend on an arbitrary choice that you are making -- and that you might make differently the next time!
Joe Redish and Mark Eichenlaub 8/17/15
Article 354
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In mathematics, the additive inverse of a number a is the number that, when added to a, yields zero. This number is also known as the opposite (number),[1] sign change,[2] and negation.[3] For a real number, it reverses its sign: the additive inverse (opposite number) of a positive number is negative, and the additive inverse of a negative number is positive. Zero is the additive inverse of itself.
The additive inverse of a is denoted by unary minus: a (see also § Relation to subtraction below).[4][5] For example, the additive inverse of 7 is −7, because 7 + (−7) = 0, and the additive inverse of −0.3 is 0.3, because −0.3 + 0.3 = 0.
Similarly, the additive inverse of ab is −(ab) which can be simplified to ba. The additive inverse of 2x − 3 is 3 − 2x, because 2x − 3 + 3 − 2x = 0.[6]
The additive inverse is defined as its inverse element under the binary operation of addition (see also § Formal definition below), which allows a broad generalization to mathematical objects other than numbers. As for any inverse operation, double additive inverse has no net effect: −(−x) = x.
These complex numbers, two of eight values of 81, are mutually opposite
Common examples
For a number (and more generally in any ring), the additive inverse can be calculated using multiplication by −1; that is, n = −1 × n. Examples of rings of numbers are integers, rational numbers, real numbers, and complex numbers.
Relation to subtraction
Additive inverse is closely related to subtraction, which can be viewed as an addition of the opposite:
ab = a + (−b).
Conversely, additive inverse can be thought of as subtraction from zero:
a = 0 − a.
Hence, unary minus sign notation can be seen as a shorthand for subtraction (with the "0" symbol omitted), although in a correct typography, there should be no space after unary "−".
Other properties
In addition to the identities listed above, negation has the following algebraic properties:
• −(−a) = a, it is an Involution operation
• −(a + b) = (−a) + (−b)
• −(a - b) = ba
• a − (−b) = a + b
• (−a) × b = a × (−b) = −(a × b)
• (−a) × (−b) = a × b
notably, (−a)2 = a2
Formal definition
The notation + is usually reserved for commutative binary operations (operations where x + y = y + x for all x, y). If such an operation admits an identity element o (such that x + o ( = o + x ) = x for all x), then this element is unique (o′ = o′ + o = o). For a given x, if there exists x′ such that x + x′ ( = x′ + x ) = o, then x′ is called an additive inverse of x.
If + is associative, i.e., (x + y) + z = x + (y + z) for all x, y, z, then an additive inverse is unique. To see this, let x′ and x″ each be additive inverses of x; then
x′ = x′ + o = x′ + (x + x″) = (x′ + x) + x″ = o + x″ = x″.
For example, since addition of real numbers is associative, each real number has a unique additive inverse.
Other examples
All the following examples are in fact abelian groups:
• Complex numbers: −(a + bi) = (−a) + (−b)i. On the complex plane, this operation rotates a complex number 180 degrees around the origin (see the image above).
• Addition of real- and complex-valued functions: here, the additive inverse of a function f is the function f defined by (−f )(x) = − f (x), for all x, such that f + (−f ) = o, the zero function (o(x) = 0 for all x).
• More generally, what precedes applies to all functions with values in an abelian group ('zero' meaning then the identity element of this group):
• Sequences, matrices and nets are also special kinds of functions.
• In a vector space, the additive inverse v is often called the opposite vector of v; it has the same magnitude as the original and opposite direction. Additive inversion corresponds to scalar multiplication by −1. For Euclidean space, it is point reflection in the origin. Vectors in exactly opposite directions (multiplied to negative numbers) are sometimes referred to as antiparallel.
• In modular arithmetic, the modular additive inverse of x is also defined: it is the number a such that a + x ≡ 0 (mod n). This additive inverse always exists. For example, the inverse of 3 modulo 11 is 8 because it is the solution to 3 + x ≡ 0 (mod 11).
Non-examples
Natural numbers, cardinal numbers and ordinal numbers do not have additive inverses within their respective sets. Thus one can say, for example, that natural numbers do have additive inverses, but because these additive inverses are not themselves natural numbers, the set of natural numbers is not closed under taking additive inverses.
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# How do you multiply (m+8)(m-8)?
Jan 28, 2017
$\left(m + 8\right) \left(m - 8\right) = {m}^{2} - 64$ See explanation.
#### Explanation:
If you multiply this you get:
$\left(m + 8\right) \left(m - 8\right) = {m}^{2} \cancel{+ 8 m} \cancel{- 8 m} - 64 = {m}^{2} - 64$
Jan 28, 2017
${m}^{2} - 64$
#### Explanation:
$\left(m + 8\right) \left(m - 8\right)$
$= m \left(m - 8\right) + 8 \left(m - 8\right)$
$= {m}^{2} - 8 m + 8 m - 64$=>simplify:
$= {m}^{2} - 64$
Alternatively the difference 2 squares formula can be used:
${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
hence:
$\left(m + 8\right) \left(m - 8\right) = {m}^{2} - 64$
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# What is the product and sum of 2, 4 and 9?
## What is the product and sum of 2, 4 and 9?
There are several possible answers: What are the product and sum of 2, 4 and 9? Work out the product of 2, 4 and 9. The product means that you need to multiply the three numbers together. The sum means that you need to add the three numbers together. Question: What do you get if you double 15?
When to use sum and difference in Algebra?
We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.
### What do you need to know about sum calculator?
About Sum (Summation) Calculator The Sum (Summation) Calculator is used to calculate the total summation of any set of numbers. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. How to use Sum (Summation) Calculator Video
How to find the product and sum of five numbers?
Answer: Product means multiply, so the 5 whole numbers that multiply together to give 3 are 1,1,1,1 and 3 (that is fours 1’s and one 3). Since “sum” means “add,” then 1 + 1 + 1 + 1 + 3 gives a final answer of 7. Question: Can you find two numbers which have a product of 741 and sum of 70?
#### How to find the product and sum of three numbers?
Answer: Product means multiply, so the 5 whole numbers that multiply together to give 3 are 1,1,1,1 and 3 (that is fours 1’s and one 3). Since “sum” means “add,” then 1 + 1 + 1 + 1 + 3 gives a final answer of 7. Question: Can you find two numbers which have a product of 741 and sum of 70? Answer: The two number you are looking for are 57 and 13.
Do you have to multiply two numbers to find the product?
If you are asked to work out the product of two or more numbers, then you need to multiply the numbers together. If you are asked to find the sum of two or more numbers, then you need to add the numbers together.
## How does the sum calculator work in math?
The Sum (Summation) Calculator is used to calculate the total summation of any set of numbers. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. How does this summation calculator work?
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# Midpoints of the Sides of a Paralellogram
Alignments to Content Standards: G-CO.C.11
Suppose that $ABCD$ is a parallelogram, and that $M$ and $N$ are the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively. Prove that $MN = AD$, and that the line $\overleftrightarrow{MN}$ is parallel to $\overleftrightarrow{AD}$.
## IM Commentary
This is a reasonably direct task aimed at having students use previously-derived results to learn new facts about parallelograms, as opposed to deriving them from first principles. The solution provided (among other possibilities) uses the SAS trial congruence theorem, and the fact that opposite sides of parallelograms are congruent.
## Solution
The diagram above consists of the given information, and one additional line segment, $\overline{MD}$, which we will use to demonstrate the result. We claim that triangles $\triangle AMD$ and $\triangle NDM$ are congruent by SAS:
1. We have $\overline{MD}=\overline{DM}$ by reflexivity.
2. We have $\angle AMD= \angle NDM$ since they are opposite interior angles of the transversal $MD$ through parallel lines $AB$ and $CD$.
3. We have $\overline{MA}=\overline{ND}$, since $M$ and $N$ are midpoints of their respective sides, and opposite sides of parallograms are congruent: $$\overline{MA}=\frac{1}{2}(\overline{AB})=\frac{1}{2}(\overline{CD})=\overline{ND}.$$
Now since corresponding parts of congruent triangles are congruent, we have $DA=NM$, as desired. Similarly, we have congruent opposite interior angles $\angle DMN\cong \angle MDA$, so $\overleftrightarrow{MN}$ is parallel to $\overleftrightarrow{AD}$.
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Here we will talk about the to organize of amount of the interiorangles of an n-sided polygon and some related instance problems.
You are watching: The sum of the interior angles of a polygon with n sides is
The sum of the inner angles the a polygon the n sides isequal to (2n - 4) appropriate angles.
Given: permit PQRS .... Z it is in a polygon of n sides.
To prove: ∠P + ∠Q + ∠R + ∠S + ..... + ∠Z = (2n – 4) 90°.
Statement Reason 1. Together the polygon has n sides, n triangles room formed,namely, ∆OPQ, ∆QR, ...., ∆OZP. 1. On each side of the polygon one triangle has actually been drawn. 2. The amount of all the angles of the n triangle is 2n rightangles. 2. The amount of the angle of every triangle is 2 appropriate angles. 3. ∠P + ∠Q + ∠R + ..... + ∠Z + (sum of every anglesformed at O) = 2n appropriate angles. 3. From statement 2. 4. ∠P + ∠Q + ∠R + ..... + ∠Z + 4 ideal angles = 2n rightangles. 4. Amount of angles roughly the suggest O is 4 ideal angles. 5. ∠P + ∠Q + ∠R + ..... + ∠Z = 2n best angles - 4 best angles = (2n – 4) best angles =(2n – 4) 90°. (Proved) 5. Indigenous statement 4.
Note:
1. In a consistent polygon of n sides, every angles space equal.
Therefore, each interior angle = (frac(2n - 4) × 90°n).
2. A quadrilateral is a polygon because that which n = 4.
Therefore, the amount of internal angles that a quadrilateral =(2 × 4 – 4) × 90° = 360°
Solved instances on finding the sum of the interior angles ofan n-sided polygon:
1. find the amount of the internal angles of a polygon of sevensides.
Solution:
Here, n = 7.
Sum the the internal angles = (2n – 4) × 90°
= (2 × 7 - 4) × 90°
= 900°
Therefore, the sum of the inner angles of a polygon is 900°.
2. amount of the interior angles the a polygon is 540°. Uncover thenumber of political parties of the polygon.
Solution:
Let the number of sides = n.
Therefore, (2n – 4) × 90° = 540°
⟹ 2n - 4 = (frac540°90°)
⟹ 2n - 4 = 6
⟹ 2n = 6 + 4
⟹ 2n = 10
⟹ n = (frac102)
⟹ n = 5
Therefore, the number of sides of the polygon is 5.
3. discover the measure of each interior angle of a regularoctagon.
Solution:
Here, n = 8.
The measure up of each inner angle = (frac(2n– 4) × 90°n)
= (frac(2 × 8 – 4) × 90°8)
= (frac(16 – 4) × 90°8)
= (frac12 × 90°8)
= 135°
Therefore, the measure up of each inner angle of a regularoctagon is 135°.
4.
See more: What Is The Difference Between Ice And Snow And Ice? What'S The Difference Between Snow And Ice
The proportion of the number of sides of two continuous polygonsis 3:4, and also the ratio of the sum of their internal angles is 2:3. Discover thenumber of sides of each polygon.
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## Is ordinary notation the same as scientific notation?
Explanation: Scientific notation (also referred to as scientific form or standard index form, or standard form in the UK) is a way of expressing numbers that are too big or too small to be conveniently written in decimal form. Standard notation is the normal way of writing numbers.
## What is ordinary decimal notation?
A representation of a fraction or other real number using the base ten and consisting of any of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and a decimal point.
What are the 3 standard notations?
In this section, we will introduce the standard notation used to define sets, and give you a chance to practice writing sets in three ways, inequality notation, set-builder notation, and interval notation.
### What is a normal notation?
Normal form (scientific notation) is a way to write very large or very small numbers in a more compact form. It has two parts: A number, usually in the range 0 – 10, called the coefficient. A power of ten to multiply it by called the exponent.
### What does standard notation look like?
An example of scientific notation is 1.3 ×106 which is just a different way of expressing the standard notation of the number 1,300,000. Standard notation is the normal way of writing numbers….Scientific Notation.
Scientific Notation Standard Form
1.23 ×106 1,230,000
What is the correct form of scientific notation?
The proper format for scientific notation is a x 10^b where a is a number or decimal number such that the absolute value of a is greater than or equal to one and less than ten or, 1 ≤ |a| < 10. b is the power of 10 required so that the scientific notation is mathematically equivalent to the original number.
## What is normal decimal notation?
Decimal notation is simply a form of a number using a decimal point. As the decimal expands into the hundredths, thousandths, etc. places, the denominator of the fraction similarly changes. For example, 0.70 uses the hundredths place.
## Which is the best example of a number written in scientific notation?
A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10. For example, 650,000,000 can be written in scientific notation as 6.5 ✕ 10^8.
What is standard set notation?
Common Set Notation |A|, called cardinality of A, denotes the number of elements of A. For example, if A={(1,2),(3,4)}, then |A|=2. A=B if and only if they have precisely the same elements. For example, if A={4,9} and B={n2:n=2 or n=3}, then A=B. A⊆B if and only if every element of A is also an element of B.
### How do you write 0.0021 in scientific notation?
∴ The scientific notation of the number 0.0021 is 2.1×10−3.
### What is notation example?
The definition of a notation is a system of using symbols or signs as a form of communication, or a short written note. An example of a notation is a chemist using AuBr for gold bromide. An example of a notation is a short list of things to do. She made a notation in the margin of the book.
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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2019 AMC 8 Problems/Problem 3"
## Problem 3
Which of the following is the correct order of the fractions $\frac{15}{11},\frac{19}{15},$ and $\frac{17}{13},$ from least to greatest?
$\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}$
## Solution 1
Consider subtracting 1 from each of the fractions. Our new fractions would then be $\frac{4}{11}, \frac{4}{15},$ and $\frac{4}{13}$. Since $\frac{4}{15}<\frac{4}{13}<\frac{4}{11}$, it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$
-will3145
## Solution 2
We take a common denominator: $$\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.$$
Since $2717<2805<2925$ it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
## Solution 3
When $\frac{x}{y}>1$ and $z>0$, $\frac{x+z}{y+z}<\frac{x}{y}$. Hence, the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$. ~ ryjs
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# A 5 digits number divisible by 3 to be formed by the numerals 0, 1, 2, 3, 4 & 5 without repetition. The total number of ways this can be done is:A. 3125B. 600C. 240D. 216
Last updated date: 16th Jun 2024
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Hint: This question is solved with the help of multiplication principle, which states that if an event can occur in m different ways, following another event which can occur in n different ways, then the total number of occurrence of the events in the given order is $m \times n$.
The above principle can be generalised for any finite number of events.
Complete step by step solution: A five digit number is formed by using digits 0, 1, 2, 3, 4 and 5 which is divisible by 3.
A number is divisible by 3 only when the sum of digits is a multiple of 3.
There are two ways to form a 5 digit number using the digits 0, 1, 2, 3, 4 and 5 which is divisible by 3
Case 1: Using the digits 0, 1, 2, 4, 5
In this case, the first place cannot be filled with 0 as it will become a 4 digit number therefore there are only 4 ways to fill the first place. On the second place also there are 4 ways to fill it. As repetition is not allowed, we cannot fill the same number again.
Therefore, the number of ways = $4 \times 4 \times 3 \times 2 \times 1 = 96$
Case 2: Using the digits 1, 2, 3, 4, 5
In this case, there are 5 ways to fill the first place, 4 ways to fill the second place and so on.
Therefore, the number of ways = $5 \times 4 \times 3 \times 2 \times 1 = 120$
Now, the total number of ways is the sum of both cases.
Hence, total number formed = 120 + 96 = 216
∴ Option (D) is correct.
Note: A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.
The formula for permutation is ${}^n{P_r} = \dfrac{{\left| \!{\underline {\, n \,}} \right. }}{{\left| \!{\underline {\, {n - r} \,}} \right. }}$
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# MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2
In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Pdf, These solutions are solved subject experts from the latest edition books.
## MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2
polynomial root calculator is a free online too in my website.
Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Solution:
(i) we have p(x) = x2 – 2x – 8
= x2 + 2x + – 4x – 8
= x(x + 2) – 4(x + 2)
= (x – 4)(x + 2)
For p(x) = 0, we must have (x – 4)(x + 2) = 0
Either x – 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = -2
∴ The zeroes of x2 – 2x – 8 are 4 and -2
Now, sum of zeroes
Thus, the relationship between the zeroes and the coefficients in the polynomial x2 – 2x – 8 is verified.
(ii) We have p(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1
= 2s(2s – 1) – l(2s – 1)
= (2s – 1)(2s – 1)
Thus, the relationship between the zeroes and coefficients in the polynomial 4s2 – 4s + 1 is verified.
(iii) We have p(x) = 6x2 – 3 – 7x
= 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
For p(x) = 0, we have,
Thus, the relationship between the zeroes and coefficients in the polynomial 6x2 – 3 – 7x is verified.
(iv) We have, p(u) = 4u2 + 8u = 4u(u + 2)
For p(u) = 0,
we have
Either 4u = 0 ⇒ u = 0
or u + 2 = 0 ⇒ u = -2
∴ The zeroes of 4u2 + 8u are 0 and – 2.
Now, 4u2 + 8u can be written as 4u2 + 8u + 0.
Thus, the relationship between the zeroes and coefficients in the polynomial 4u2 – 4s + 1 is verified.
(v) We have, p(t) = t2 – 15 = (t)2 – $$(\sqrt{15})^{2}$$
Thus, the relationship between zeroes and the coefficients in the polynomial t2 – 15 is verified.
(vi) We have, p(x) = 3x2 – x – 4
= 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1)(3x – 4)
For p(x) = 0 we have,
Either (x + 1) = 0 ⇒ x = -1
Thus, the relationship between the zeroes and coefficients in the polynomial 3x2 – x – 4 is verified
Solve Polynomial problems with our Polynomial calculator and problem solver.
Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $$\frac{1}{4}$$, -1
(ii) $$\sqrt{2}, \frac{1}{3}$$
(iii) 0, $$\sqrt{5}$$
(iv) 1, 1
(v) $$-\frac{1}{4}, \frac{1}{4}$$
(vi) 4,1
Solution:
(i) Since, sum of zeroes, $$(\alpha+\beta)=\frac{1}{4}$$
Product of zeroes, αβ = -1
∴ The required quadratic polynomial is
have samezeroes, therefore (4x2 – x – 4) is the required quadratic polynomial.
(ii) Since, sum of zeroes, $$(\alpha+\beta)=\sqrt{2}$$
product of zeroes, αβ = $$\frac{1}{3}$$
∴ The required quadratic polynomial is
(iii) Since, sum of zeroes, (α + β) = 0
Product of zeroes, αβ = $$\sqrt{5}$$
∴ The required quadratic polynomial is
(iv) Since, sum of zeroes, (α + β) = 1
Product of zeroes, αβ = 1
∴ The required quadratic polynomial is
(v) Since, sum of zeroes, (α + β) = $$-\frac{1}{4}$$
Product of zeroes, αβ = [\frac{1}{4}/latex]
∴ The required quadratic polynomial is
same zeroes, therefore, the required quadratic polynomial is (4x2 + x + 1)
(vi) Since, sum of zeroes, (α + β) = 4 and
product of zeroes, αβ = 1
∴ The required quadratic polynomial is
x2 – (α + β)x + αβ = x2 – 4x + 1
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# How do I find the constant term from a given binomial without expanding the whole expression?
## ${\left(x - \frac{1}{x} ^ \left(\frac{1}{3}\right)\right)}^{12}$
See below for an idea:
#### Explanation:
The constant term in a binomial expansion is the one, if it exists, that has no variable terms within it.
This prior answer on Socratic has a great explanation on how to find the constant term:
Following the instructions in that answer, we first set up the general term:
${T}_{r + 1} = \left(\begin{matrix}12 \\ r\end{matrix}\right) {x}^{12 - r} {\left(- \frac{1}{x} ^ \left(\frac{1}{3}\right)\right)}^{r}$
and now simplifying:
${T}_{r + 1} = \left(\begin{matrix}12 \\ r\end{matrix}\right) {x}^{12 - r} {\left(- 1\right)}^{r} \left({x}^{- \frac{r}{3}}\right)$
${T}_{r + 1} = \left(\begin{matrix}12 \\ r\end{matrix}\right) {x}^{12 - r - \left(\frac{r}{3}\right)} {\left(- 1\right)}^{r}$
${T}_{r + 1} = \left(\begin{matrix}12 \\ r\end{matrix}\right) {x}^{12 - \frac{4 r}{3}} {\left(- 1\right)}^{r}$
The constant term is the one that will have $x = 0$, so we have:
${x}^{12 - \frac{4 r}{3}} = {x}^{0}$
giving
$12 - \frac{4 r}{3} = 0$
$12 = \frac{4 r}{3}$
$4 r = 36 \implies r = 9$
Therefore, the 9th term in the expansion is the one with no $x$ terms within it:
$\left(\begin{matrix}12 \\ 9\end{matrix}\right) {x}^{3} {\left(- \frac{1}{x} ^ \left(\frac{1}{3}\right)\right)}^{9}$
and to show that (i.e. checking our work):
$220 \times {x}^{3} \times \left(- \frac{1}{x} ^ 3\right) = - 220$
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# How to Find Velocity from Position-Time Graph
## How to Find Velocity from Position-Time Graph:
When dealing with the motion of objects, understanding their velocity is very important. Velocity gives us insights into how fast an object is moving and in what direction. A position-time graph is a graphical representation that shows an object’s position at different points in time. In this article, we will explore how to find velocity from a position-time graph.
We will break down the process step by step, providing you with a clear understanding of the concepts and calculations involved. Velocity is the rate of change of an object’s position with respect to time. To find velocity from a position-time graph, follow these steps:
1. Understand the Slope:
The slope of a position-time graph represents the object’s velocity. A steeper slope indicates higher velocity, while a gentle slope indicates slower velocity. The formula for calculating slope is rise over run: (change in position) / (change in time).
2. Identify the Initial and Final Positions:
Determine the positions corresponding to the starting and ending times on the graph. The difference in these positions is the change in position (Δx).
3. Find the Time Interval:
Calculate the time interval (Δt) by subtracting the initial time from the final time.
4. Calculate Velocity:
Use the formula: Velocity (v) = Change in Position (Δx) / Time Interval (Δt). Make sure to include units in your calculations.
5. Assign Direction:
Velocity is a vector quantity, meaning it has both magnitude and direction. Determine the direction by considering whether the slope is positive or negative.
6. Interpret the Results:
Once you’ve calculated the velocity, interpret the value. A positive velocity indicates motion in one direction, while a negative velocity indicates motion in the opposite direction.
## Looking at the Concepts Further:
### The Importance of Slope:
Understanding the concept of slope is essential for interpreting position-time graphs. When the slope is steeper, it signifies faster movement. Conversely, a shallower slope indicates slower movement.
### Positive and Negative Slopes:
A positive slope on a position-time graph indicates forward motion. For instance, when an object’s position increases as time progresses, it has a positive velocity. Conversely, a negative slope represents backward motion or a negative velocity.
### Zero Slope – Stationary Objects:
If the slope of a position-time graph is zero, the object is stationary. In this case, the position does not change over time, leading to a velocity of zero.
### Calculating Average Velocity:
Average velocity can be calculated over a specific time interval. Divide the total change in position by the total time taken to get the average velocity.
### Instantaneous Velocity:
Instantaneous velocity refers to the velocity of an object at a specific moment in time. To find this, consider a smaller time interval and calculate the velocity using the methods mentioned above.
## FAQs
• What if the position-time graph is a horizontal line?
If the graph is a horizontal line, the object is not moving, and its velocity is zero.
• Can velocity be negative even on a positive slope?
Yes, if an object is moving in the opposite direction of the positive slope, its velocity will be negative.
• Is velocity the same as speed?
No, velocity takes into account both the magnitude and direction of motion, while speed only considers magnitude.
• Can a position-time graph have a curved line?
Yes, a curved line indicates changing velocity over time. In such cases, calculate instantaneous velocity at specific points.
• What does a steep negative slope represent?
A steep negative slope indicates rapid backward motion of the object.
• How does constant velocity appear on a graph?
Constant velocity is represented by a straight line on a position-time graph. The slope remains consistent.
You may also like to read:
How to Find Acceleration from Position-Time Graph
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Q. 21
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Found in: Page 29
Precalculus Enhanced with Graphing Utilities
Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465
In Problem 17-36, solve each equation algebraically. Verify your solution using a graphing utility.$\frac{x+1}{3 }+\frac{x+2}{7}=5$
The required solution is $x=\frac{46}{5}$. The graph of the solution is given below:
See the step by step solution
Step 1. Given information.
We have:
$\frac{x+1}{3 }+\frac{x+2}{7}=5$
Step 2. Solve the equation algebraically.
The equation is $\frac{x+1}{3 }+\frac{x+2}{7}=5$.
Find the least common multiplier of $3,7$:$21$
Multiply by LCM (Least common multiple) to both sides of the equation.
$\frac{x+1}{3}· 21+\frac{x+2}{7}· 21=5· 21\phantom{\rule{0ex}{0ex}}7\left(x+1\right)+3\left(x+2\right)=105\phantom{\rule{0ex}{0ex}}7x+7+3x+6=105\phantom{\rule{0ex}{0ex}}10x+13=105$
Subtract $13$ to both sides:
$10x+13-13=105-13\phantom{\rule{0ex}{0ex}}10x=92\phantom{\rule{0ex}{0ex}}\frac{10x}{10}=\frac{92}{10}\phantom{\rule{0ex}{0ex}}x=\frac{46}{5}$
Step 3. Verify the solution using a graphing utility.
The solution of the equation is $x=\frac{46}{5}$.
Draw the graph of the equation $\frac{x+1}{3 }+\frac{x+2}{7}=5$ using the graphing utility:
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# Number Theory and Cryptography
Introduction.
Since the theory of numbers concerns itself with the familiar numbers 1, 2, 3, . . . , it might
seem at first glance to be little more than grade school arithmetic. We shall see that this is
far from true. The first clue that we are dealing with something more than a simple subject
is the three dots in the expression "1, 2, 3, . . . " The three dots, which translate into \and
so on" is the clue that tells us that we are dealing with infinitely many numbers. As such,
since we cannot examine all of the integers one by one, we may well expect to find many
mysteries and unsolved problems regarding these numbers. In fact this is true, as we shall
point out. However, we shall also be able to solve many problems that seem at first sight to
be intractable.
Throughout the text, when we speak of numbers, we understand ordinary whole numbers,
including zero and negative numbers. These are called integers. Much of what many students
know about numbers has been handed down as fact, and these are by now taken for granted.
In what follows, we shall investigate many of these \facts" a little more deeply. In many
cases, we will explain why they are true by giving proofs. Along the way, however, many new
ideas will be introduced. We should mention at the outset, that the topic of number theory
was once considered to be a field of mathematics with no practical applications. Recently,
however, it has proved extremely useful in the study and applications of cryptography. In a
later chapter, we shall explore this further.
We shall take for granted that you are familiar with some simple facts about integers. These
include the following.
Arithmetic operations . For example 43 + 58 = 101, 21 × 65 = 1, 365, 312 = 961. These
are easily found on a calculator, which we assume you have. However, a calculator has
limitations
, since it can usually accept at most nine or ten digits. So if you have to add
or multiply two 40 digit numbers, you would have to revert to the old grade school way
of computation without a calculator or else have a powerful computer program to do this
exactly. In all arithmetic calculations, the answer is always understood to be exact. So if you
use a calculator to find that you do not have the exact answer.
All you have is the first 12 digits of a 16 digit number. When dealing with integers, we will
usually want the exact answer. Calculators only give approximations for very large numbers.
For example, if you are used to writing 1/3 = .333, you are using an approximation. The
calculator which gives 1/3 = 0.333333333 is also giving an approximation.
Algebra. For example, you should know that (x+y)2 = x2+2xy+y2. and n(n+1) = n2+n,
and you should be able to solve the equation 2x+4 = 7x−3, and understand that the solution
need not be an integer.
For the present, we take it for granted that the reader of these notes knows what it means
for one number to divide another . We also take it for granted that the reader knows that
a prime number is a number larger than 1, which is divisible only by itself and by 1. A
few examples of numbers that are not primes (called composite numbers) are 51, 91, and
543,678,967,805. Do you see why? Here are a few questions about numbers. How many can
• What is the 20th prime number?
• Is 571,435,871,001 prime number? What about 34,571?
• How many numbers divide 120?
• Is the fraction 28, 841=33, 043 in lowest terms ?
• The numbers 1− 1 + 41 = 41, 4 −2 + 41 = 43, 9 − 3+41 = 47, 16− 4+41 = 53 are
all primes. Is it true that n2 − n + 41 is always a prime for all positive integers n?
• The numbers 4 = 2+2, 6 = 3+3, 8 = 5+3, and 10 = 7+3 are all sums of two primes.
Is every even number greatere than 2 a sum of two primes?
• The pairs (3,5), (5,7), (11,13), (17,19), (29,31) all consist of twin prime numbers.
(These are primes whose difference is 2.) Are there infinitely many twin primes?
• The odd perfect squares 1, 9, and 25 all leave a remainder of 1 when divided by 8. Is
this true for the square of every odd number?
• The numbers 2 = 1 + 1, 9 = 9+0, 13 = 4 + 9, and 34 = 9 + 25 are all sums of two
squares. However, 3, 14, and 21 are not. Which numbers are the sum of two squares?
Which are not?
• Which numbers are the sum of 4 squares?
The above list consists of a few questions in number theory. Though simple to ask, some are
questions, including two of the ones stated above, that nobody today still knows the answer
to. (These are the twin prime problem as well as the problem of the sum of two primes.) The
problem of factoring very large numbers turns out to be important in cryptography, where
the fastest computers still have to spend many hours deciding if a number is prime. We shall
analyze some of these problems in lecture and in lab to discover some of the methods used
to solve these and similar problems.
## 1 Division
Quotients and Remainders. We start by reviewing something probably learned in grade
school: how to divide two number to get a quotient and remainder. We will want to do this
on a calculator and on a computer. We first start with a simple example.
Example 1.1 Divide 57 by 13 and find the quotient and remainder.
Method:
This is the way I did it in grade school. Since teaching methods change, you might not have
seen this before!
So the quotient is 4 and the remainder is 5.
The method is as follows. To divide 57 by 13, we estimate 4 as the approximate integer
quotient. Multiply 4 by 13 to get 52, subtract from 57 to get the remainder 5. Here, the
relationship of the quotient q and remainder r is
57 = 13 4 + 5 = 13q + r
Dividing this equation by 13, we obtain 57/13 = q + r/13, where r/13 is the fractional part
of the quotient 57/13. To do this with a calculator, we find 57/13 = 4:3846, and we can
read the quotient q = 4. The fractional part is .3846. As above, this is r/13, so we should
get the remainder r if we multiply by 13. If we do this on the calculator we get 4.9998. We
understand that this is only approximate, as most decimals are , and since we must have an
integer for the answer, we make the sensible guess that r = 5. The recommended (and safe)
way is to use whole numbers. Thus 57 = 13q + r and so r = 57 − 13q = 57 − 52 as before.
Let's illustrate with large numbers.
Example 1.2 Divide 68,934 by 5,791 and find the quotient q and remainder r. Express the
relationship between these numbers in a simple formula .
Method: Using a calculator, we find 68, 934/5, 791 = 11.9036. Therefore q = 11 and r =
68, 934 − 11(5, 791) = 5, 233. The relationship is 68, 934 = 5, 791(11) + 5, 233 = 5, 791q + r.
This computation can easily be set up on a spreadsheet. The lab for this course does this
on an Excel spreadsheet called Division. In theory this can done without a calculator or
computer, if you are willing to undergo a process called \long division." Happily we shall
not do this.
Summarizing: If a and b are integers with b > 0, we can always find a quotient q and a
remainder r such that
a = bq + r with 0 ≤ r < b (1)
Equation (1) is called the Division Algorithm. The quotient q is called a div b in most
computer languages. The remainder is called a mod b. The text introduced the \div" and
\mod" notations on page 67. The notation a div b is supposed to remind you that you
are dividing a by b but are conveniently dropping any remainder or fractional part. The
spreadsheet Excel does not have a div function, but it uses INT(a/b) instead. (Think: the
integer part of a/b.) In Excel, the remainder a mod b is written MOD(a, b).
Note that the remainder r is always less than the denominator b , and can be 0 (if the division
\comes out even.")
Definition 1.3 We say that b divides a, or that b is a factor of a, if a/b is an integer, or
equivalently that a = bq for some integer q. The standard way of writing this is b|a. (Read:
b divides a.) We also say that a is a multiple of b
Another way of putting this is that r = 0 in Equation (1).
In high school algebra, it was usually taken for granted that variables such as a, b, x, y des-
ignated real numbers. However, throughout this course we shall assume that they represent
integers, either positive, negative , or zero. This is an important change in usage. The word
number will similarly refer to integers only.
Such basic ideas as "even" and "odd" are de fined using the division algorithm. A number n
is even if 2|n, it is odd if (read as 2 does not divide n). Equivalently, a number is odd
if the remainder when divided by 2 is 1.
Do not confuse the "divides" sign | with the "divided by" sign /. Thus, we have 2|6, but
6/2 = 3.
Examples.
(a) Clearly 1|a since a = 1٠ a. Similarly aja when a > 0, since a = a ٠ 1.
(b) If a, b > 0 then b|ab.
(c) If a > 0, then a|0.
(d) If c|b and b|a then c|a. For we have b = cq1 and a = bq2 for integers q1 and q2. So
a = cq1q2, and therefore c|a.
The statement in (d) is called transitivity of division.
For example, if a number is divisible by 21 then it is divisible by 7. This follows from (d).
For suppose 21|n. Since 7| 21, we get 7|n by transitivity.
The following result is useful and fairly easy to prove.
Prev Next
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Solutions to Simple Differential Equations
# Solutions to Simple Differential Equations
We will now look at two different types of elementary differential equations that we can solve without must hassle.
## Differential Equations with One Isolated Derivative
The simplest type of differential equations to solve are in the form:
(1)
\begin{align} \quad \frac{dy}{dx} = f(x) \end{align}
In fact, you have more than likely solved many of this type of differential equations already. If we integrate both sides of the equation above with respect to $x$, then by the Fundamental Theorem of Calculus Part 1, we have that:
(2)
$$y = F(x)$$
We note that $F(x)$ is any antiderivative of $f$. Thus, the process of antidifferentiation solves differential equations in the form $\frac{dy}{dx} = f(x)$. We summarize all of this in the following theorem.
Theorem 1: If $\frac{dy}{dx} = f(x)$, then the set of solutions for this differential equation is the set of antiderivatives $F(x)$ of $f$ which can be obtained by integrating both sides of this differential equation with respect to $x$.
Let's look at an some examples.
### Example 1
Solve the differential equation $\frac{dy}{dx} = x^2 + \cos (2x)$.
If we integrate both sides with respect to $x$ and apply the Fundamental Theorem of Calculus, we get that:
(3)
\begin{align} \quad \int \frac{dy}{dx} \: dx = \int x^2 + \cos (2x) \: dx \\ \quad y = \frac{x^3}{3} + \frac{1}{2} \sin (2x) + C \end{align}
Therefore all solutions to $\frac{dy}{dx} = x^2 + \cos (2x)$ are given by $y = \frac{x^3}{3} + \frac{1}{2} \sin (2x) + C$ for $C \in \mathbb{R}$.
### Example 2
Solve the differential equation $\frac{d^2y}{dx^2} = 3e^x - \sin x$.
If we integrate both sides with respect to $x$ twice and apply the Fundamental Theorem of Calculus, we get that:
(4)
\begin{align} \quad \int \frac{d^2y}{dx^2} \: dx = \int 3e^x - \sin x \\ \quad \frac{dy}{dx} = 3e^x + \cos x + C \\ \quad \int \frac{dy}{dx} \: dx = \int 3e^x + \cos x + C \: dx \\ \quad y = 3e^x + \sin x + Cx + D \end{align}
Therefore all solutions to $\frac{d^2y}{dx^2} = 3e^x - \sin x$ are given by $y = 3e^x + \sin x + Cx + D$ for $C, D \in \mathbb{R}$.
## Basic Linear Differential Equations
We will now look at another type of differential equations that are relatively easy to solve. Consider a differential equation in the following form (where both $a$ and $b$ are constants):
(5)
\begin{align} \frac{dy}{dx} = ay + b \end{align}
The process for solving a differential equation in the form above is best explained with an example. Consider the differential equation $\frac{dy}{dx} = 3y - 210$. We will now begin to find the solutions to this differential equation. We will start by factoring the righthand side of this equation to get $\frac{dy}{dx} = 3(y - 70)$. We now divide both sides of this equation by $y - 70$, and so for $y \neq 70$ we have rewritten the above differential equation as:
(6)
\begin{align} \quad \frac{\left ( \frac{dy}{dx} \right )}{y - 70} = 3 \end{align}
We ultimately want to get rid of $\frac{dy}{dx}$ in the above equation. To do so, we notice that if $p(x) = \ln \mid y - 70 \mid$ then by applying the differentiation rule for the natural logarithm function, we get that $p'(x) = \frac{\frac{dy}{dx}}{\mid y - 70 \mid}$. Therefore we see that for our particular differential equation example, we have that:
(7)
\begin{align} \quad \frac{d}{dx} \ln \mid y - 70 \mid = \frac{\frac{dy}{dx}}{\mid y - 70 \mid} \\ \quad \frac{d}{dx} \ln \mid y - 70 \mid = 3 \end{align}
We have now reduced our original differential equation to the first type of differential equation we mentioned at the beginning of this page. We will now integrate both sides with respect to $x$ and apply the Fundamental Theorem of Calculus to get:
(8)
\begin{align} \quad \int \frac{d}{dx} \ln \mid y - 70 \mid \: dx = \int 3 \: dx \\ \quad \ln \mid y - 70 \mid = 3x + C \\ \mid y - 70 \mid = e^{3x + C} \\ y - 70 = \pm e^{3x + C} \\ y = 70 \pm e^{3x + C} \\ y = 70 \pm e^{3x}e^C \end{align}
We will now clean up and condense our solution. Note that since $C \in \mathbb{R}$ is a constant that was introduced after integration, then $e^C$ will be some positive constant $e^C = D > 0$, and so we can write our solution as $y = 70 \pm De^{3x}$ where $D > 0$. Now notice that we have a "$\pm$" in our solution. To get rid of this, if we instead let $G \in \mathbb{R} \setminus \{ 0 \}$ then $y = 70 \pm De^{3x}$ is equivalent to $y = 70 + Ge^{3x}$. Lastly, we should note that if $G = 0$, then we get that $y = 70$ which is in fact a solution to our differential equation (verify this), and so the best way to write all of the solutions to $\frac{dy}{dx} = 3y - 210$ is $y = 70 + Ke^{3x}$, $K \in \mathbb{R}$.
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# Question cdf3b
May 3, 2017
${\text{0.6083 mol kg}}^{- 1}$
#### Explanation:
Every time you're looking for a solution's molality, you are looking for the number of moles of solute present for every $\text{1 kg}$ of solvent.
Your first step here will be to convert the number of grams of glucose to moles by using the compound's molar mass
0.9813 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156 color(red)(cancel(color(black)("g")))) = "0.005447 moles glucose"
Next, convert the mass of water to kilograms
8.9547 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.0089547 kg"
Now, use the known composition of the solution--remember, solutions are homogeneous mixtures, which implies that they have the same composition throughout--to calculate the number of moles of glucose present in '1 kg" of solvent.
1 color(red)(cancel(color(black)("kg water"))) * "0.005447 moles glucose"/(0.0089547 color(red)(cancel(color(black)("kg water")))) = "0.6083 moles glucose"
Therefore, you can say that this solution has a molality of
$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 0.6083 mol kg}}^{- 1}}}}$
The answer is rounded to four sig figs, the number of sig figs you have for the mass of glucose.
May 3, 2017
$m = 0.614$
#### Explanation:
Molality is defined as
$\textcolor{w h i t e}{\forall \forall \forall \forall \forall \forall \forall A} \textcolor{m a \ge n t a}{m = \left(\text{moles of solute")/(" kg of solvent}\right)}$
The $\text{glucose}$ is our $\text{solute}$. It is being dissolved in the $\text{solvent}$, $\text{water}$.
We take our mass in grams and convert it to moles, but first, we need to find the molar mass of glucose $\left({C}_{6} {H}_{12} {O}_{6}\right)$ which we will look up or it could be memorized $\left(180 \frac{g}{\text{mole}}\right)$
color(white)(aaaaaaaa)(0.9813 cancel"g")/1 *(1" mole")/(180 cancel"g") = "0.0055 moles of Glucose"
Then we convert our $\text{grams of water}$ into $\text{kg of water}$ (the solvent)
color(white)(aaaaaaaa)(8.957 cancel"g")/(1) * (1" kg")/(1*10^3 cancel"g") = 0.008957" kg of water"
Solve
color(magenta)(m = ("moles of solute")/(" kg of solvent"))->(0.0055" moles of Glucose")/(0.008957" kg of water") = 0.614 m
$\textcolor{b l u e}{\text{Answer: m = 0.614, where m equals molal}}$
May 3, 2017
The molality of the glucose solution is $\text{0.6083 molal}$, or $6.083 \textcolor{w h i t e}{.} m$.
#### Explanation:
Molality$\left(m\right) = \left(\text{moles of solute")/("kg of solvent}\right)$
The solute is glucose, and the solvent is water.
Step 1: Convert the given mass of glucose into moles of glucose by multiplying by the reciprocal of its molar mass.
Molar mass glucose=180.156 g/mol
https://www.ncbi.nlm.nih.gov/pccompound?term=glucose
0.9813color(red)cancel(color(black)("g glucose"))xx(1"mol glucose")/(180.156color(red)cancel(color(black)("g glucose")))="0.0054471 mol glucose"
Step 2: Convert the mass of water from grams to kilogram: $\text{1 kg=1000 g}$.
8.9547color(red)cancel(color(black)("g water"))xx(1"kg")/(1000color(red)cancel(color(black)("g")))="0.0089547 kg water"
Step 3: Calculate molality of the solution using the equation at the top.
m=(0.0054471"mol glucose")/(0.0089547"kg water")="0.6083 molal"=0.6083 m# (rounded to four significant figures)
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What is the least three digit number, which is multiple of $6$?Find the sum of all three digit numbers which are multiple of $6$ ?
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Hint: Here we get the three digit numbers which are multiple of 6 in the form of AP series .To find the sum of digits use the formula
${S_n} = \dfrac{n}{2}({\text{first term}} + {\text{last term}})$
We know that least three digit number is $100$
If we divided $100$ by $6$ we get remainder as $4$
We know that greatest two digit number that is multiple of $6$=$100 - 4 = 96$
Now the least three digit number which is multiple of $6$ =$96 + 6 = 102$
From this we can say that the least three digit number which is multiple of $6$ is $102$
To find the sum of all three digit numbers that are multiple of $6$
Let us add $6$to the first which mean least three digit number $102$ and lets continue the processing adding ‘$6$’ to the resultant number to get next numbers.
Then the next number will be $108,114,120.....$
Then series is $102,108,114,120....$
The above series is of AP where the first term a=$102$, d=$6$
To find the sum of the numbers we have to find the n value
We know that ${n^{th}}$of AP is ${a_n} = a + (n - 1)d$
And again here we need the ${n^{th}}$term value nothing but maximum value that is multiple of $6$
We know that greatest three digit number =$999$
So here if we divide $999$ with $6$ the remainder will be $3$
So to get the maximum three digit number which is multiple of $6$ let us subtract $3$ from $999$ which gives the $3$-digit number that is multiple of$6$.
$\Rightarrow 999 - 3 = 996$
So here the maximum $3$-digit number that is multiple of $6$ is $996$.
Then here ${a_n} = 996$
Now let us find n value
$\Rightarrow {a_n} = a + (n - 1)d \\ \Rightarrow 996 = 102 + (n - 1)6 \\ \Rightarrow (n - 1)6 = 894 \\ \Rightarrow n - 1 = 149 \\ \Rightarrow n = 150 \\ \therefore n = 150 \\$
From this we can say there are total $150$ numbers in the series that are multiple by $6$
Sum of the terms$\Rightarrow$${S_n} = \dfrac{n}{2}({\text{first term}} + {\text{last term}})$
Let us substitute the value
$\Rightarrow {S_n} = \dfrac{{150}}{2}(102 + 996) \\ \Rightarrow {S_n} = 75 \times 1098 \\ \Rightarrow {S_n} = 82350 \\$
Therefore sum of all three digit numbers which are multiple of $6$=$82350$
Note: Make note that to find the sum of all three digit numbers multiple of $6$ it’s important to find the n value. So in this problem we have to find least number that is multiple of 6 and to find sum of number 3 digit numbers that are multiple of 6 we have the how many 3digit numbers are present that are divisible by 6.So we have used nth term of AP to get n value, as the numbers that are multiple of 6 are in AP. And finally we have used the sum of n terms formula to get a sum of numbers that are multiple of 6.
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# Change Units in Algebraic Contexts
In this worksheet, students will convert algebraic amounts between units by applying one (e.g. kg to g, m to mm, etc.) or two (e.g. m/s to km/h, g/m to kg/km, etc.) conversion factors.
Key stage: KS 4
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR
Difficulty level:
### QUESTION 1 of 10
You may already be familiar with converting between metric units.
Let's review this now so you can be sure of what this means.
e.g. What is 17 cm in mm?
We know that there are 10 mm in 1 cm, so we have to multiply by 10.
17 × 10 = 170 mm
How to Convert Algebraic Units
When we are dealing with quantities as algebra, the same rules will apply.
We still need to use conversion factors, but we just have to calculate them in terms of the original algebraic quantity.
Let's explore an example to see how to do this.
e.g. Convert K cm into mm.
We know that to change cm into mm, we need to multiply by 10.
This means that K cm = 10 K mm.
Makes sense, doesn't it?
Let's try another, more complex, example now.
e.g. Convert P m/s into km/h.
This is a two-step problem as we have two units of measurement present: distance (m/km) and time (s/h).
1) We know that there are 3600 seconds in 1 hour.
So P metres = 1 second
3600P metres = 1 hour
2) We know there are 1000 metres in a kilometre.
3600 P metres = 1 hour
3.6P km = 1 hour
Therefore, P m/s = 3.6P km/hr.
In this activity, you will practise converting algebraic amounts between different units by finding and applying one or more conversion factors.
Add a word into the gap to complete the sentence below.
If we convert P metres (m) into cm, how will our new amount be represented?
100P cm
0.01P cm
10P cm
If we convert T cm into km, how will our new amount be represented?
0.001T km
0.00001T km
0.01T km
0.0001T km
Match the algebraic conversions below to their new amounts.
## Column B
T cm into metres
10T
T cm into mm
0.01T
T cm into kilometres
0.000001T
If we convert R cm2 into mm2, how will our new amount be represented?
Just write a number in the space as the unit has already been provided for you.
## Column B
T cm into metres
10T
T cm into mm
0.01T
T cm into kilometres
0.000001T
Which of the following can be used to convert m/s into km/h?
× 3.6
÷ 3.6
× 18/5
× 3600/1000
× 5/18
Which of these options is the correct conversion for T g/cm3 in kg/m3?
1000T kg/m3
0.1T kg/m3
0.001T kg/m3
If we convert 36U km/h into m/s, how will our new amount be represented?
Just write a number in the space as the unit has already been provided for you.
1000T kg/m3
0.1T kg/m3
0.001T kg/m3
If we convert D kg/m3 into g/m3, how will our new amount be represented?
Just write a number in the space as the unit has already been provided for you.
1000T kg/m3
0.1T kg/m3
0.001T kg/m3
If we convert 750N g/cm3 into kg/cm3, how will our new amount be represented?
Just write a number in the space as the unit has already been provided for you.
1000T kg/m3
0.1T kg/m3
0.001T kg/m3
• Question 1
Add a word into the gap to complete the sentence below.
EDDIE SAYS
If you are not sure about this one, why not go back and read over the Introduction again? When we are dealing with quantities as algebra, the same rules apply as with numbers, so we still use conversion factors. However, we need to apply the same process, but in terms of the original algebraic quantity. Let's explore what this means now in the rest of this activity.
• Question 2
If we convert P metres (m) into cm, how will our new amount be represented?
100P cm
EDDIE SAYS
We know that the conversion factor between m and cm is 100. As cm are smaller than metres, there will be more of them used to represent the same length, so we have to multiply P by this conversion factor: P × 100 = 100P cm Does that make sense?
• Question 3
If we convert T cm into km, how will our new amount be represented?
0.00001T km
EDDIE SAYS
We know that the conversion factor between cm and m is 100, then m and km is 1000. If we multiply these together, we find that the conversion factor we need to use here is 100,000. Which means that there are 100,000 cm in 1 km. As km are (much!) larger than cm, there will be much less of them used to represent the same length, so we have to divide T by this conversion factor: T ÷ 100000 = 0.00001T km Hopefully, all these decimals didn't trip you up!
• Question 4
Match the algebraic conversions below to their new amounts.
## Column B
T cm into metres
0.01T
T cm into mm
10T
T cm into kilometres
0.000001T
EDDIE SAYS
This question is all about knowing your conversion factors: Centi > m = 100 Milli > centi = 10 Centi > kilo = 100 x 1000 = 100,000 In each case, we need to decide if our starting unit is larger or smaller than our new unit. If it is larger, then we need to multiply. If it is smaller, then we need to divide. Can you apply these conversion rates and rules to match the pairs accurately?
• Question 5
If we convert R cm2 into mm2, how will our new amount be represented?
Just write a number in the space as the unit has already been provided for you.
EDDIE SAYS
This is a bit more complicated. Changing cm into mm uses a conversion factor of x 10, but this is an area, so we need to multiply by 102: R × 102 = 100R mm2 A little bit of a trickier one, hey?
• Question 6
Which of the following can be used to convert m/s into km/h?
× 3.6
× 18/5
× 3600/1000
EDDIE SAYS
There's a number of ways we can convert m/s into km/h. We need to remember that we have two measurements to convert here - distance (m/km) and time (s/h). Let's start with the distance element - what is the conversion rate between m and km? It's ÷ 1000. And between seconds and hours? There are 60 seconds in a minute and 60 minutes in an hour, so we need to work out: 60 × 60 = 3600 So our conversion factor here is multiply by 3600. 3600 ÷ 1000 = 3.6 This amount can be written in three different ways that are all equivalent - did you find all three in this list?
• Question 7
Which of these options is the correct conversion for T g/cm3 in kg/m3?
1000T kg/m3
EDDIE SAYS
Firstly, we need to know the conversion factors for each. g > kg = ÷ 1000 Changing cm3 into m3 is not as easy as it looks. There are 100 cm in a metre, but we are dealing with cubes too. 1m3 = 100 x 100 x 100 = 10,00,000 cm3 If we tie these two conversions together, we find that: 10000000 ÷ 1000 = 100 So our final answer should be 100T kg/m3
• Question 8
If we convert 36U km/h into m/s, how will our new amount be represented?
Just write a number in the space as the unit has already been provided for you.
EDDIE SAYS
We need to remember that we have two measurements to convert here - distance (km/m) and time (h/s). Let's start with the distance element - what is the conversion rate between km and m? It's × 1000. And between hours and seconds? There are 60 seconds in a minute and 60 minutes in an hour, so we need to work out: 60 × 60 = 3600 So our conversion factor here is divide by 3600. 1000 ÷ 3600 = 3.6 36U ÷ 3.6 = 10U m/s Did you remember to just type the number in, as the units have already been provided for you?
• Question 9
If we convert D kg/m3 into g/m3, how will our new amount be represented?
Just write a number in the space as the unit has already been provided for you.
EDDIE SAYS
We only need to convert the weight (kg/g) element this time, as the distance (m3) stays the same so does not require converting. So what is the conversion rate between kg and g? It's × 1000. D × 1000 = 1000D kg/m3 Did you remember to just type the number in?
• Question 10
If we convert 750N g/cm3 into kg/cm3, how will our new amount be represented?
Just write a number in the space as the unit has already been provided for you.
EDDIE SAYS
We only need to convert the weight (g/kg) element this time, as the distance (cm3) stays the same again. So what is the conversion rate between g and kg? It's ÷ 1000. 750N ÷ 1000 = 0.75D kg/cm3 You can now convert algebraic amounts into different units using one or two conversion factors combined - great job!
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Chapter 7: TE Solving Systems of Equations and Inequalities
Difficulty Level: At Grade Created by: CK-12
## Overview
In this chapter, students discover the methods that determine solutions to systems of linear equations and inequalities. Students begin by solving systems of equations graphically, realizing that the solution for each system is the point of intersection between the two lines represented by the given equations.
Suggested Pacing:
Linear Systems by Graphing - 1hr\begin{align*}1 \;\mathrm{hr}\end{align*}
Solving Linear Systems by Substitution - 1hr\begin{align*}1 \;\mathrm{hr}\end{align*}
Solving Linear Systems by Elimination through Addition or Subtraction - 1hr\begin{align*}1 \;\mathrm{hr}\end{align*}
Solving Systems of Equations by Multiplication - 12hrs\begin{align*}1-2 \;\mathrm{hrs}\end{align*}
Special Types of Linear Systems - 0.5hr\begin{align*}0.5 \;\mathrm{hr}\end{align*}
Systems of Linear Inequalities - 2hrs\begin{align*}2 \;\mathrm{hrs}\end{align*}
## Problem-Solving Strand for Mathematics
In this chapter, Systems of Equations and Inequalities, several different methods of solving equations and inequalities are taught. Throughout, real-world problems are incorporated into the exercises, and practical applications of important concepts such as constraints, optimum solutions, feasibility regions, and maximum and minimum values are outlined.
In this chapter you might try giving students limited choices about which problems to do independently. This allows you to assess which of the problems students feel comfortable with, which they wish to avoid, and where some re-teaching might be helpful. With the section Comparing Methods of Solving Linear Systems, students might be asked to use all three primary methods on a single exercise rather than doing three different exercises with the same method. In addition to giving necessary practice it may help students become more discerning about which method is simpler in a given situation.
### Alignment with the NCTM Process Standards
Using the principle of choice and, occasionally, asking students to write about a problem of their preference can address many of the NCTM Process Standards. Chief among them would be the Communications and Connections standards. Students solving and presenting solutions for equation and inequality problems must organize and consolidate their mathematical thinking (COM.1), communicate their mathematical thinking coherently and clearly to peers, teachers, and others (COM.2), and use the language of mathematics to express mathematical ideas precisely (COM.4). When students share their various insights and approaches to problems as common classroom practice, they learn to analyze and evaluate the mathematical thinking and strategies of others (COM.3).
When students are allowed to select specific problems as part of an assignment rather than to merely mimic the sample problems given in the text, they strive to recognize and use connections among mathematical ideas (CON.1) and to understand how mathematical ideas interconnect and build upon one another to produce a coherent whole (CON.2). They are encouraged to apply and adapt a variety of appropriate strategies to solve problems (PS.3) and use representations to model and interpret physical, social, and mathematical phenomena (R.3).
• COM.1 - Organize and consolidate their mathematical thinking through communication.
• COM.2 - Communicate their mathematical thinking coherently and clearly to peers, teachers, and others.
• COM.3 - Analyze and evaluate the mathematical thinking and strategies of others.
• COM.4 - Use the language of mathematics to express mathematical ideas precisely.
• CON.1 - Recognize and use connections among mathematical ideas.
• CON.2 - Understand how mathematical ideas interconnect and build on one another to produce a coherent whole.
• PS.3 - Apply and adapt a variety of appropriate strategies to solve problems.
• R.3 - Use representations to model and interpret physical, social, and mathematical phenomena.
Chapter Outline
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Feb 22, 2012
Aug 22, 2014
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CK.MAT.ENG.TE.1.Algebra-I.7
Here
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# Chord of a Circle
Sub Topics
A line segment whose end points lie on a circle is called a chord.
In the adjoining figure, each of the line segments PQ, RS, AB and CD is a chord of the circle with centre O. Clearly, an infinite number of chords may be drawn in a circle.
## What is a Chord of a Circle?
Two end points on the circumference of a circle joined together is known as chord of a circle. Here the diameter is the greatest chord of the circle.
## Chord of a Circle Formula
If ‘ r ’ is the radius of the circle and ‘ a ‘ is the length of the arc, then length of the chord made by the arc is 2r * Sin($\frac{\theta}{2}$)
Proof:
If ‘ r ’ is the radius of the circle and ‘ $\theta$ ‘ is the angle made by the chord at the centre of the circle and length of the arc ‘ a ‘
Arc AB subtends an angle ‘ $\theta$ ’ at the centre. Where ‘ $\theta$ ‘ is the measure of angle in radians.
The rule that involves length of the arc ‘ a ‘ and the central angle ‘ $\theta$ ’ and radius ‘ r ‘ is
a = r$\theta$ - - - - - - - - - - - - - - - - - - - (i)
AB is the chord of the circle, ‘ O ‘ is the centre of the circle, draw OM perpendicular to AB.
M is the midpoint of AB and OM bisects $\angle$ AOB.
Therefore, $\angle$ AOM = $\angle$ BOM = $\frac{\theta}{2}$
In triangle AMO, $\angle$ AMO = 900
Sin($\frac{\theta}{2}$) = $\frac{AM}{OA}$
Sin($\frac{\theta}{2}$) = $\frac{AM}{r}$
r * Sin($\frac{\theta}{2}$) = AM
Since the length of the chord AB = 2 * AM = 2r * Sin($\frac{\theta}{2}$)
Therefore, formula to find the length of the chord = 2r * Sin( $\frac{\theta}{2}$ )
## Chord Length of a Circle
Find the length of the chord of a circle if radius of the circle and central angle made by the chord are given.
Proof:
If ‘ r ’ is the radius of the circle and ‘ $\theta$ ‘ is the angle made by the chord at the centre of the circle.
AB is the chord of the circle, ‘ O ‘ is the centre of the circle, draw OM perpendicular to AB.
M is the midpoint of AB and OM bisects $\angle$ AOB.
Therefore, $\angle$ AOM = $\angle$ BOM = $\frac{\theta}{2}$
In triangle AMO, $\angle$ AMO = 900
Sin($\frac{\theta}{2}$) = $\frac{AM}{OA}$
Sin($\frac{\theta}{2}$) = $\frac{AM}{r}$
r * Sin($\frac{\theta}{2}$) = AM
Since the length of the chord AB = 2 * AM = 2r * Sin($\frac{\theta}{2}$)
Therefore, formula to find the length of the chord = 2r * Sin($\frac{\theta}{2}$)
## How to Find the Chord of a Circle
If ‘ r ‘ is the radius of the circle and ‘ p ‘ is the length of the perpendicular drawn to the chord from the centre of the circle then the length of the chord is given by 2
Proof:
Given, a circle of radius ‘ r ‘ and ‘ p ‘ be the length of the perpendicular drawn from centre ‘O’ on the chord.
AB is the chord of the circle, ‘ O ‘ is the centre of the circle, draw OM perpendicular to AB.
In triangle AMO, $\angle$ AMO = 900
Using Pythagoras theorem, we get
OA2 = AM2 + OM2
r2 = AM2 + p2
r2 - p2 = AM2
AM = 2
Since the length of the chord AB = 2 * AM = 2
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