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# Summing Root To Leaf Numbers
Jul 29, 2020 · 5 min read
Sum Root to Leaf Numbers is an interesting problem from LeetCode. The problem is of medium difficulty and is about binary trees. This post is an explained solution to the problem.
I assume that you’re familiar with Python and the concept of binary trees. If you’re not, you can read this article to get started.
# The Problem
In the tree on the left, the output is `25`. `25` is the sum of `12` and `13`, which are the two numbers formed when starting from `1` and visiting every leaf. In the tree on the right, the output is `1026` as it is the sum of the three numbers `495`, `491` and `40`.
# The Observations and Insights
1. The construction of numbers is incremental and similar of sorts: the only difference between `495` and `491` (from the tree on the right) is the last digit. If we remove the `5` and insert a `1` in its place, we have the next required number. A number essentially comprises of the leaf's digit appended to all the digits in ancestor nodes. Thus, numbers within the same subtree have common digits.
2. Finally, notice that this problem involves a tree, so a recursive solution is helpful.
# The Solution
Let’s create a `Solution` class to encompass our solution.
The method signature given to us in the problem has one argument: root, which is of the type `TreeNode` . A `TreeNode` class is as follows (from LeetCode):
From observation #2, notice that appending a node’s digit to its ancestors can be achieved by moving all the digits of the number formed by ancestors to the right by 1 place and adding the current node’s digit. The digits can be moved by multiplying the number formed by ancestors by 10 (since we’re in base-10). For example:
`495 = 49 x 10 + 5`
Thus, we can keep track of the current digits in an integer. This is important because we won’t incur extra storage space for higher input sizes. We can pass around this value in the function parameter itself. Since the method signature given can only have one parameter, let’s create a `sum_root_to_leaf_helper` method.
We can think of the `sum_root_to_leaf_helper` method recursively and process each node differently based on whether or not it is a leaf.
• If the node is a leaf, we want to add its digit to our current digits by moving all the other digits to the right. We also want to return this value (since we’ll backtrack from here).
• If it is not a leaf, we want to add the digit to our current digits by moving all the other digits to the right. We also want to continue constructing the number by traversing down this node’s left and right subtrees.
If the current node is a `None`, we can simply return 0 because it doesn't count.
Thus, our `sum_root_to_leaf_helper `method will be as follows:
We use a default value for the partial sum to be 0.
In our main method, we want to include the `sum_root_to_leaf_helper` method as a nested method and simply pass on the node parameter. Finally, this is how our solution looks:
# The Algorithmic Complexity
Time:
Our solution is a modification of the depth-first-search pre-order traversal where we visit all nodes exactly once and perform a trivial computation (moving digits by integer multiplication). Thus, our runtime is simply `O(N)` where `N` represents the number of nodes in the given tree. A solution better than `O(N)` doesn't seem possible because to construct a number from digits, we need to know all the digits (and thus visit all nodes).
Space:
In terms of storage, we incur a high cost in the recursion call stack that builds up as our `sum_root_to_leaf_helper` calls itself. These calls build-up as one waits for another to finish.
The maximum call stack is dependent upon the height of the binary tree (since we start backtracking after we visit a leaf), giving a complexity of `O(H)` where `H` is the height of the binary tree. In the worst case, the binary tree is skewed in either direction and thus `H = N`. Therefore, the worst-case space complexity is `O(N)`.
It is possible to do better than `O(N)` by using a Morris Preorder Traversal. The basic idea is to link a node and its predecessor temporarily. You can read more about it here.
# The Conclusion
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Lesson 5
Reasoning about Equations and Tape Diagrams (Part 2)
5.1: Algebra Talk: Seeing Structure (10 minutes)
Warm-up
This warm-up parallels the one in the previous lesson. The purpose of this Algebra Talk is to elicit strategies and understandings students have for solving equations. These understandings help students develop fluency and will be helpful later in this unit when students will need to be able to come up with ways to solve equations of this form. While four equations are given, it may not be possible to share every strategy. Consider gathering only two or three different strategies per problem, saving most of the time for the final question.
Students should understand the meaning of solution to an equation from grade 6 work as well as from work earlier in this unit, but this is a good opportunity to re-emphasize the idea.
In this string of equations, each equation has the same solution. Digging into why this is the case requires noticing and using the structure of the equations (MP7). Noticing and using the structure of an equation is an important part of fluency in solving equations.
Launch
Display one equation at a time. Give students 30 seconds of quiet think time for each equation and ask them to give a signal when they have an answer and a strategy. Keep all equations displayed throughout the talk. Follow with a whole-class discussion.
Representation: Internalize Comprehension. To support working memory, provide students with sticky notes or mini whiteboards.
Supports accessibility for: Memory; Organization
Student Facing
Solve each equation mentally.
$$x -1 = 5$$
$$2(x-1) = 10$$
$$3(x-1) = 15$$
$$500 = 100(x-1)$$
Activity Synthesis
This discussion may go quickly, because students are likely to recognize similarities between this equation string and the one in the previous day’s warm-up.
Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:
• “Who can restate ___’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the equation in a different way?”
• “Does anyone want to add on to _____’s strategy?”
• “Do you agree or disagree? Why?”
Speaking: MLR8 Discussion Supports.: Display sentence frames to support students when they explain their strategy. For example, "First, I _____ because . . ." or "I noticed _____ so I . . . ." Some students may benefit from the opportunity to rehearse what they will say with a partner before they share with the whole class.
Design Principle(s): Optimize output (for explanation)
5.2: More Situations and Diagrams (15 minutes)
Activity
The purpose of this activity is to work toward showing students that some situations can be represented by an equation of the form $$p(x+q)=r$$ (or equivalent). In this activity, students are simply tasked with drawing a tape diagram to represent each situation. In the following activity, they will work with corresponding equations.
For each question, monitor for one student with a correct diagram. Press students to explain what any variables used to label the diagram represent in the situation.
Launch
Ensure students understand that the work of this task is to draw a tape diagram to represent each situation. There is no requirement to write an equation or solve a problem yet.
Arrange students in groups of 2. Give 5–10 minutes to work individually or with their partner, followed by a whole-class discussion.
Action and Expression: Develop Expression and Communication. Maintain a display of important terms, vocabulary, and examples. During the launch, take time to review examples of drawing a tape diagram based on situations from previous lessons that students will need to access for this activity. Consider providing step-by-step directions that generalize the process using student input and ideas.
Supports accessibility for: Memory; Language
Student Facing
Draw a tape diagram to represent each situation. For some of the situations, you need to decide what to represent with a variable.
1. Each of 5 gift bags contains $$x$$ pencils. Tyler adds 3 more pencils to each bag. Altogether, the gift bags contain 20 pencils.
2. Noah drew an equilateral triangle with sides of length 5 inches. He wants to increase the length of each side by $$x$$ inches so the triangle is still equilateral and has a perimeter of 20 inches.
3. An art class charges each student $3 to attend plus a fee for supplies. Today,$20 was collected for the 5 students attending the class.
4. Elena ran 20 miles this week, which was three times as far as Clare ran this week. Clare ran 5 more miles this week than she did last week.
Launch
Ensure students understand that the work of this task is to draw a tape diagram to represent each situation. There is no requirement to write an equation or solve a problem yet.
Arrange students in groups of 2. Give 5–10 minutes to work individually or with their partner, followed by a whole-class discussion.
Action and Expression: Develop Expression and Communication. Maintain a display of important terms, vocabulary, and examples. During the launch, take time to review examples of drawing a tape diagram based on situations from previous lessons that students will need to access for this activity. Consider providing step-by-step directions that generalize the process using student input and ideas.
Supports accessibility for: Memory; Language
Student Facing
Draw a tape diagram to represent each situation. For some of the situations, you need to decide what to represent with a variable.
1. Each of 5 gift bags contains $$x$$ pencils. Tyler adds 3 more pencils to each bag. Altogether, the gift bags contain 20 pencils.
2. Noah drew an equilateral triangle with sides of length 5 inches. He wants to increase the length of each side by $$x$$ inches so the triangle is still equilateral and has a perimeter of 20 inches.
3. An art class charges each student $3 to attend plus a fee for supplies. Today,$20 was collected for the 5 students attending the class.
4. Elena ran 20 miles this week, which was three times as far as Clare ran this week. Clare ran 5 more miles this week than she did last week.
Activity Synthesis
Select one student for each situation to present their correct diagram. Ensure that students explain the meaning of any variables used to label their diagram. Possible questions for discussion:
• “For the situations with no $$x$$, how did you decide what quantity to represent with variable?” (Think about which amount is unknown but has a relationship to one or more other amounts in the story.)
• “What does the variable you used to label the diagram represent in the story?”
• “Did any situations have the same diagrams? How can you tell from the story that the diagrams would be the same?” (Same number of equal parts, same amount for the total.)
Speaking, Representing, Reading: MLR1 Stronger and Clearer Each Time. Ask students to explain to a partner how they created the tape diagram to represent the situation “An art class charges each student $3 to attend plus a fee for supplies. Today,$20 was collected for the 5 students attending the class.” Ask listeners to press for details in the arrangement of the grouped quantities (e.g., “Explain how you chose what values go in each box.”). When roles are switched, listeners can press for details in what “$$x$$” represents in the diagram. Allow students to revise their diagrams, if necessary, based on the feedback they received from their partner. Once their revision is complete, invite students to turn to a new partner to explain their revised diagram. This will help students productively engage in discussion as they make connections between written situations and visual diagrams.
Design Principle(s): Optimize output (for explanation); Cultivate conversation
5.3: More Situations, Diagrams, and Equations (10 minutes)
Activity
This activity is a continuation of the previous one. Students match each situation from the previous activity with an equation, solve the equation by any method that makes sense to them, and interpret the meaning of the solution. Students are still using any method that makes sense to them to reason about a solution. In later lessons, a hanger diagram representation will be used to justify more efficient methods for solving.
For each equation, monitor for a student using their diagram to reason about the solution and a student using the structure of the equation to reason about the solution.
Launch
Keep students in the same groups. 5 minutes to work individually or with a partner, followed by a whole-class discussion.
Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer interactions. Prior to the whole-class discussion, invite students to share their work with a partner. Display sentence frames to support student conversation such as “To find the solution, first, I _____ because...”, “I made this match because I noticed...”, “Why did you...?”, or “I agree/disagree because…”
Supports accessibility for: Language; Social-emotional skills
Speaking, Representing: MLR2 Collect and Display. As students share their ideas about how the equations match the situations, listen for and collect students’ description of the situation (e.g., “5 gift bags, $$x$$ pencils, adds 3 more, 20 pencils”) with the corresponding equation. Remind students to borrow language from the displayed examples while describing what each solution tells about the situation, after the matching is complete. This will help students make connection between language, diagrams, and equations.
Design Principle(s): Support sense-making; Maximize meta-awareness
Student Facing
Each situation in the previous activity is represented by one of the equations.
• $$(x+3) \boldcdot 5 = 20$$
• $$3(x+5)=20$$
1. Match each situation to an equation.
3. What does each solution tell you about its situation?
Student Facing
Han, his sister, his dad, and his grandmother step onto a crowded bus with only 3 open seats for a 42-minute ride. They decide Han’s grandmother should sit for the entire ride. Han, his sister, and his dad take turns sitting in the remaining two seats, and Han’s dad sits 1.5 times as long as both Han and his sister. How many minutes did each one spend sitting?
Activity Synthesis
For each equation, ask one student who reasoned with the diagram and one who reasoned only about the equation to explain their solutions. Display the diagram and the equation side by side, drawing connections between the two representations. If no students bring up one or both of these approaches, demonstrate the maneuvers on a diagram side by side with the maneuvers on the corresponding equation. For example, “I divided the number of gift bags by 5, leaving me with 4 pencils per gift bag. Since Tyler added 3 pencils to each gift bag, there must have been 1 pencil in each gift bag to start,” can be shown on a tape diagram and on a corresponding equation. It is not necessary to invoke the more abstract language of “doing the same thing to each side” of an equation yet.
Lesson Synthesis
Lesson Synthesis
Display one of the situations from the lesson and its corresponding equation. Ask students to explain:
• “What does each number and letter in the equation represent in the situation?”
• “What is the reason for each operation (multiplication or addition) used in the equation?”
• “What is the solution to the equation? What does it mean to be a solution to an equation? What does the solution represent in the situation?”
5.4: Cool-down - More Finding Solutions (5 minutes)
Cool-Down
Each bag in the first story has an unknown number of toys, $$x$$, that is increased by 2. Then ten groups of $$x+2$$ give a total of 140 toys. An equation representing this situation is $$10(x+2)=140$$. Since 10 times a number is 140, that number is 14, which is the total number of items in each bag. Before Lin added the 2 items there were $$14 - 2$$ or 12 toys in each bag.
The executive in the second story knows that the size of each team of $$y$$ employees has been increased by 10. There are now 2 teams of $$y+10$$ each. An equation representing this situation is $$2(y+10)=140$$. Since 2 times an amount is 140, that amount is 70, which is the new size of each team. The value of $$y$$ is $$70-10$$ or 60. There were 60 employees on each team before the increase.
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# Solving absolute value equations
When solving absolute value equations, many people routinely solve these without really understand deep concepts involving absolute value. It is the goal of this lesson to remedy this common pitfall.
Absolute value definition:
If x is positive, | x | = x
If x is negative, | x | = -x
Example #1:
Solve for x when | x | = 4
After applying the definition to example #1, you will have two equations to solve
In fact, when solving absolute value equations, you will usually get two solutions. That is important to keep in mind
If x is positive, | x | = x, so the first equation to solve is x = 4. Done because x is automatically isolated
If x is negative, | x | = -x, so the second equation to solve is -x = 4.
You can write -x = 4 as -1x = 4 and divide both sides by -1 to isolate x.
(-1/-1)x = 4/-1
1x = -4
x = -4
Therefore, the solutions are 4 and -4
Example #2:
Solve for x when | x − 5 | = 2
Before, we apply the definition, let's make a useful substitution
Let y = x − 5, so | x − 5 | = 2 becomes | y | = 2. You must understand this step. No excuses!
Now, let's apply the definition to | y | = 2. Again, you will have two equations to solve
Once again, when solving absolute value equations, you will usually get two solutions.
If y is positive, | y | = y, so the first equation to solve is y = 2. No, you are not done! You have to substitute x − 5 for y
After substitution, y = 2 becomes x − 5 = 2
x − 5 = 2
x + -5 = 2
x + -5 + 5 = 2 + 5
x = 7
If y is negative, | y | = -y, so the second equation to solve is -y = 2.
You have to substitute x − 5 for y
You get -( x − 5) = 2. Notice the inclusion of parenthesis this time
-(x − 5) = 2
-(x + -5) = 2
-x + 5 = 2
-x + 5 − 5 = 2 − 5
-x = 2 + -5
-x = -3
-1x = -3
(-1/-1)x = -3/-1
x = 3
The solutions are 3 and 7
Example #3:
Solve for x when | 3x + 3 | = 15
Before, we apply the definition, let's again make a useful substitution
Let y = 3x + 3, so | 3x + 3 | = 15 becomes | y | = 15.
Now, let's apply the definition to | y | = 15. Again, you will have two equations to solve.
If y is positive, | y | = y, so the first equation to solve is y = 15. You have to substitute 3x + 3 for y
After substitution, y = 15 becomes 3x + 3 = 15
3x + 3 = 15
3x + 3 − 3 = 15 − 3
3x = 12
(3/3)x = 12/3
x = 4
If y is negative, | y | = -y, so the second equation to solve is -y = 15.
You have to substitute 3x + 3 for y
You get -( 3x + 3) = 15. Notice the inclusion of parenthesis this time
-3x + -3 = 15
-3x + -3 + 3 = 15 + 3
-3x = 18
(-3/-3)x = 18/-3
x = -6 The solutions are -6 and 4
Solving absolute value equations should be straightforward if you follow my guidelines above
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# Integro-differential equation
In mathematics, an integro-differential equation is an equation that involves both integrals and derivatives of a function.
## General first order linear equations
The general first-order, linear integro-differential equation is of the form
$\frac{d}{dx}u(x) + \int_{x_0}^x f(t,u(t))\,dt = g(x,u(x)), \qquad u(x_0) = u_0, \qquad x_0 \ge 0.$
As is typical with differential equations, obtaining a closed-form solution can often be difficult. In the relatively few cases where a solution can be found, it is often by some kind of integral transform, where the problem is first transformed into an algebraic setting. In such situations, the solution of the problem may be derived by applying the inverse transform to the solution of this algebraic equation.
### Example
Consider the following first-order problem,
$u'(x) + 2u(x) + 5\int_{0}^{x}u(t)\,dt = \left\{ \begin{array}{ll} 1, \qquad x \geq 0\\ 0, \qquad x < 0 \end{array} \right. \qquad \text{with} \qquad u(0)=0.$
The Laplace transform is defined by,
$U(s) = \mathcal{L} \left\{u(x)\right\}=\int_0^{\infty} e^{-sx} u(x) \,dx.$
Upon taking term-by-term Laplace transforms, and utilising the rules for derivatives and integrals, the integro-differential equation is converted into the following algebraic equation,
$s U(s) - u(0) + 2U(s) + \frac{5}{s}U(s) = \frac{1}{s}.$
Thus,
$U(s) = \frac{1}{s^2 + 2s + 5}$.
Inverting the Laplace transform using contour integral methods then gives
$u(x) = \frac{1}{2} e^{-x} \sin(2x)$.
## Applications
Integro-differential equations model many situations from science and engineering. A particularly rich source is electrical circuit analysis.[citation needed]
The activity of interacting inhibitory and excitatory neurons can be described by a system of integro-differential equations, see for example the Wilson-Cowan model.
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# Properties of Rational Numbers Worksheets
Document Sample
``` Properties of Rational Numbers Worksheets
Properties of Rational Numbers Worksheets
Let us first look at the properties of rational numbers in order to understand about the
rational numbers in details. Although many properties of rational numbers worksheet are
available online to understand the concept of rational numbers and its usage. We will take the
properties of rational numbers one by one:
1. Closure Property of Rational numbers: We say that the closure property of rational
numbers holds true for the addition, subtraction, multiplication and division. It simply means
that if we have any two rational numbers say p1/q1 and p2/q2, then according to closure
property of Addition, the sum of any two rational numbers p1/q1 + p2/q2 is also a rational
number.
Also according to closure property of subtraction, the difference of the two rational numbers is
also a rational number, mathematically p1/q1 – p2/q2 is also a rational number.
Similarly closure property also holds true for the multiplication and the division operation of
the rational number, which states :
Know More About How to solve for a rational number
Tutorcircle.com Page No. : 1/4
P1/q1 * p2/ q2 and p1/q1 divided by p2 / q2 are also rational numbers if we have p1/q1 and
p2/q2 as the rational numbers.
2. Commutative Property of Rational Numbers : If we have any two rational numbers say
p1/q1 and p2/q2, then we say that the commutative property holds true for the addition and
the multiplication of rational numbers, but does not holds true for the subtraction and division
of the rational numbers. Mathematically we say that :
P1/q1 + p2/q2 = p2/q2 + p1/q1
P1/q1 * p2/q2 = p2/q2 * p1/q1
P1/q1 - p2/q2 < > p2/q2 - p1/q1
P1/q1 divided by p2/q2 < > p2/q2 divided by p1/q1
3. By Associative Property of rational numbers, we mean that if p1/q1, p2/q2 and p3/q3 are
any three rational numbers, then we say that if the order of grouping in addition and
multiplication is changed, then the result is same. On the another hand the result does not
remains same in the case of subtraction and division. Let us look at the Associative Property
mathematically :
( p1/q1 + p2/q2 ) + p3 / q3 = p1/q1 + ( p2/q2 + p3 / q3 )
( p1/q1 - p2/q2 ) - p3 / q3 <> p1/q1 - ( p2/q2 - p3 / q3 )
Read More About Solving equations with rational numbers
Tutorcircle.com Page No. : 2/4
( p1/q1 * p2/q2 ) * p3 / q3 = p1/q1 * ( p2/q2 * p3 / q3 )
( p1/q1 divided p2/q2 ) divided p3 / q3 = p1/q1 divided ( p2/q2 divided p3 /q3 )
4. Additive Identity 0, exist for the rational number p/q such that the number 0 is added to
the given number , the result is the number itself.
So we say mathematically p/q + 0 = p/q
5. Multiplicative Identity is one ( 1) : There exist a number 1, such that If we multiply 1 to any
of the rational numbers, the result is the original rational number itself. If we have any rational
number p/q , then we say that
p/q * 1 = p/q
6. The additive inverse of any number is the negative of the given rational number and the
multiplicative inverse of any number is reciprocal of any rational number.
Tutorcircle.com Page No. : 3/4
Page No. : 2/3
Thank You
TutorCircle.com
```
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Mathematics Solutions Solutions for Class 7 Math Chapter 3 Hcf And Lcm are provided here with simple step-by-step explanations. These solutions for Hcf And Lcm are extremely popular among Class 7 students for Math Hcf And Lcm Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Math are prepared by experts and are 100% accurate.
#### Question 1:
Which number is neither a prime number nor a composite number?
#### Answer:
1 is neither a prime number nor a composite number.
#### Question 2:
Which of the following are pairs of co-primes?
(i) 8, 14
(ii) 4, 5
(iii) 17, 19
(iv) 27, 15
#### Answer:
Two numbers which have only 1 as a common factor are said to be co-prime or relatively prime or mutually prime numbers.
We can write 17 as 17 × 1 and 19 as 17 × 1.
Hence, 17 and 19 is a pair of co-prime numbers.
#### Question 3:
List the prime numbers from 25 to 100 and say how many they are.
#### Answer:
There are a total of 16 prime numbers between 25 and 100 which are 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
#### Question 4:
Write all the twin prime numbers from 51 to 100.
#### Answer:
If the difference between two co-prime numbers is 2, the numbers are said to be twin prime numbers.
Hence, the twin prime numbers between 51 and 100 are 59 and 61, 71 and 73.
#### Question 5:
Write 5 pairs of twin prime numbers from 1 to 50.
#### Answer:
If the difference between two co-prime numbers is 2 then, the numbers are said to be twin prime numbers.
Hence, the twin prime numbers from 1 to 50 are (2,3), (5,7), (11,12), (17,19) and (29,30).
#### Question 6:
Which are the even prime numbers?
#### Answer:
There is only even prime number which is 2.
#### Question 1:
Factorise the following numbers into primes.
(i) 32
(ii) 57
(iii) 23
(iv) 150
(v) 216
(vi) 208
(vii) 765
(viii) 342
(ix) 377
(x) 559
#### Answer:
(i)
32 = 2 × 16
= 2 × 2 × 8
= 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2
(ii)
57 = 3 × 19
(iii)
23 = 23 × 1
(iv)
150 = 2 × 75
= 2 × 3 × 25
= 2 × 3 × 5 × 5
(v)
216 = 2 × 108
= 2 × 2 × 54
= 2 × 2 × 2 × 27
= 2 × 2 × 2 × 3 × 9
= 2 × 2 × 2 × 3 × 3 × 3
(vi)
208 = 2 × 104
= 2 × 2 × 52
= 2 × 2 × 2 × 26
= 2 × 2 × 2 × 2 × 13
(vii)
765 = 3 × 255
= 3 × 3 × 85
= 3 × 3 × 5 × 17
(viii)
342 = 2 × 171
= 2 × 3 × 57
= 2 × 3 × 3 × 19
(ix)
377 = 13 × 29
(x)
559 = 13 × 43
#### Question 1:
Find the HCF.
(i) 25, 40
(ii) 56, 32
(iii) 40, 60, 75
(iv) 16, 27
(v) 18, 32, 48
(vi) 105, 154
(vii) 42, 45, 48
(viii) 57, 75, 102
(ix) 56, 57
(x) 777, 315, 588
#### Answer:
(i)
HCF = 5
(ii)
HCF = 2 × 2 × 2= 8
(iii)
HCF = 5
(iv)
HCF = 1
(v)
HCF = 2
(vi)
HCF = 7
(vii)
HCF = 3
(viii)
HCF = 3
(ix)
HCF = 1
(x)
HCF = 3 × 7 = 21
#### Question 2:
Find the HCF by the division method and reduce to the simplest form.
(i) $\frac{275}{525}$
(ii) $\frac{76}{133}$
(iii) $\frac{161}{69}$
#### Answer:
(i)
HCF = 25
$\therefore \frac{275}{525}=\frac{275÷25}{525÷25}\phantom{\rule{0ex}{0ex}}=\frac{11}{21}$
(ii)
HCF = 19
$\therefore \frac{76}{133}=\frac{76÷19}{133÷19}\phantom{\rule{0ex}{0ex}}=\frac{4}{7}$
(iii)
HCF = 23
$\therefore \frac{161}{69}=\frac{161÷23}{69÷23}\phantom{\rule{0ex}{0ex}}=\frac{7}{3}$
#### Question 1:
Find the LCF.
(i) 12, 15
(ii) 6, 8, 10
(iii) 18, 32
(iv) 10, 15, 20
(v) 45, 86
(vi) 15, 30, 90
(vii) 105, 195
(viii) 12, 15, 45
(ix) 63, 81
(x) 18, 36, 27
#### Answer:
(i)
LCM = 3 × 5 × 4 = 60
(ii)
LCM = 2 × 3 × 4 × 5 = 120
(iii)
LCM = 2 × 9 × 16 = 288
(iv)
LCM = 2 × 2 × 3 × 5 = 60
(v)
LCM = 45 × 86 = 3870
(vi)
LCM = 2 × 3 × 3 × 5 = 90
(vii)
LCM = 3 × 5 × 7 × 13 = 1365
(viii)
LCM = 2 × 2 × 3 × 3 × 5 = 180
(ix)
LCM = 3 × 3 × 3 × 3 × 5 = 567
(x)
LCM = 2 × 2 × 3 × 3 × 3 = 108
#### Question 2:
Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers.
(i) 32, 37
(ii) 46, 51
(iii) 15, 60
(iv) 18, 63
(v) 78, 104
#### Answer:
(i)
HCF = 1
LCM = 32 × 37 = 1184
Product of two numbers = 32 × 37 = 1184
Product of HCF and LCM = 1 × 1184 = 1184
(ii)
HCF = 1
LCM = 46 × 51 = 2346
Product of two numbers = 46 × 51= 2346
Product of HCF and LCM = 1 × 2346 = 2346
(iii)
HCF = 3 × 5 = 15
LCM = 3 × 5 × 4 = 60
Product of two numbers = 15 × 60 = 900
Product of HCF and LCM = 15 × 60 = 900
(iv)
HCF = 3 × 3 = 9
LCM = 3 × 3 × 2 × 7 = 126
Product of two numbers = 18 × 63 = 1134
Product of HCF and LCM = 9 × 126 = 1134
(v)
HCF = 2 × 13 = 26
LCM = 2 × 13 × 3 × 4 = 312
Product of two numbers = 78 × 104 = 8112
Product of HCF and LCM = 26 × 312 = 8112
#### Question 1:
Choose the right option.
(i) The HCF of 120 and 150 is ................... .
(1) 30
(2) 45
(3) 20
(4) 120
(ii) The HCF of this pair of numbers is not 1.
(1) 13, 17
(2) 29, 20
(3) 40, 20
(4) 14, 15
#### Answer:
(i)
HCF = 2 × 3 × 5 = 30
Hence, the correct answer is option (1).
(ii)
40 = 2 × 2 × 2 × 5
20 = 2 × 2 × 5
The HCF of 20 and 40 is 2 × 2 × 5 or 20.
Hence, the correct answer is option (3).
#### Question 2:
Find the HCF and LCM.
(i) 14, 28
(ii) 32, 16
(iii) 17, 102, 170
(iv) 23, 69
(v) 21, 49, 84
#### Answer:
(i)
HCF = 2 × 7 = 14
LCM = 2 × 7 ×
2 = 28
(ii)
HCF = 2 × 2 × 2 × 2 = 16
LCM = 2 × 2 × 2 ×
2 × 2 = 32
(iii)
HCF = 17
LCM = 17 × 2 × 3 × 5
= 510
(iv)
HCF = 23
LCM = 23 × 3
= 69
(v)
HCF = 7
LCM = 3 × 4 × 7 × 7
= 588
#### Question 3:
Find the LCM.
(i) 36, 42
(ii) 15, 25, 30
(iii) 18, 42, 48
(iv) 4, 12, 20
(v) 24, 40, 80, 120
#### Answer:
(i)
LCM = 2 × 2 × 3 × 3 × 7 = 252
(ii)
LCM = 2 × 3 × 5 × 5 = 150
(iii)
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 1008
(iv)
LCM = 2 × 2 × × 5 = 60
(v)
LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240
#### Question 4:
Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.
#### Answer:
LCM of 8, 9, 10, 15, 20 is given by
LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360
Hence, 365 is the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.
#### Question 5:
Reduce the fractions $\frac{348}{319}$ , $\frac{221}{247}$ , $\frac{437}{551}$ to the lowest terms.
#### Answer:
$\frac{348}{319}=\frac{348÷29}{319÷29}\phantom{\rule{0ex}{0ex}}=\frac{12}{11}$
$\frac{221}{247}=\frac{221÷13}{247÷13}\phantom{\rule{0ex}{0ex}}=\frac{17}{19}$
$\frac{437}{551}=\frac{437÷19}{551÷19}\phantom{\rule{0ex}{0ex}}=\frac{23}{29}$
#### Question 6:
The LCM and HCF of two numbers are 432 and 72 respectively. If one of the numbers is 216, what is the other ?
#### Answer:
Let the other number be x.
Now, HCF × LCM = Product of two numbers
⇒ 72 × 432 = x × 216
$⇒x=\frac{72×432}{216}$
= 144
Hence, the other number is 144.
#### Question 7:
The product of two two-digit numbers is 765 and their HCF is 3. What is their LCM?
#### Answer:
HCF × LCM = Product of two numbers
⇒ 3 × LCM = 765
$⇒\mathrm{LCM}=\frac{765}{3}$
= 255
Hence, the LCM of the two numbers is 255.
#### Question 8:
A trader has three bundles of string 392 m, 308 m and 490 m long. What is the great est length of string that the bundles can be cut up into without any left over string?
#### Answer:
The greatest length of string that the bundles can be cut up into without any left over string is given by the HCF of 392, 308 and 490.
HCF = 2 × 7 = 14
Hence, the greatest length of string that the bundles can be cut up into without any left over string is 14 m.
#### Question 9:
Which two consecutive even numbers have an LCM of 180?
#### Answer:
Let us suppose the two consecutive even numbers be 2x and 2x + 2.
Now, product of two numbers = HCF × LCM
⇒ (2x)(2x + 2) = 2 × 180 (HCM of two even number is 2)
⇒ (x)(2x + 2) = 180
⇒ 2x2 + 2x = 180
⇒ 2x2 + 2x − 180 = 0
x2 + x − 90 = 0
⇒ (x − 9)(x + 10) = 0
⇒ x − 9 = 0 or x + 10 = 0
⇒ x = 9 or x = −10 (Neglecting)
Hence, the two consecutive even numbers are18 and 20.
View NCERT Solutions for all chapters of Class 7
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If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius $R$ is $\pi R^2$ .
The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always same, irrespective of the location of point at circumference to which you are joining the center of disk. The area of disk is defined as the ‘measure of surface‘ surrounded by the round edge (circumference) of the disk.
The area of a disk can be derived by breaking it into a number of identical parts of disk as units — calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings — sum them till whole disk is obtained.
Rings and Sections
Mathematically, we can imagine a ring of with radius $x$ and thickness $dx$ , anywhere in the disk having the same center as disk, calculate its area and then sum up (integrate) it from $x=0$ to $x=R$ . Area of a thin ring is since $\pi x dx$ . And after integrating we get, area of disk $A=2 \int_0^R \pi x dx$ or $A=\pi R^2$ .
There is another approach to achieve the area of a disk, A.
An inscribed Triangle
Imagine a disk is made up of a number equal sections or arcs. If there are $n$ number of arcs then interior angle of an arc is exactly $\frac{2\pi}{n}$ , since $2 \pi$ is the total angle at the center of disk and we are dividing this angle into $n$ equal parts. If we join two ends of each sections –we can get $n$ identical triangles in which an angle with vertex O is $\frac{2 \pi}{n}$ . Now, if we can calculate the area of one such section, we can approach to the area of the disk intuitively. This approach is called the method of exhaustion.
Let, we draw two lines joining center O of the disk and points A & B at circumference. It is clear that OB and OA are the radius of the disk. We joined points A and B in order to form a triangle OAB. Now consider that the disk is made up of n-number of such triangles. We see that there is some area remaining outside the line AB and inside the circumference. If we had this triangle thinner, the remaining area must be lesser.
Area remaining after the Triangle
So, if we increase the number of triangles in disk —-we decrease the remaining areas. We can achieve to a point where we can accurately calculate the area of disk when there are infinitely many such triangles or in other words area of one such triangle is very small. So our plan is to find the area of one triangle —sum it up to n — make $n$ tending to infinity to get the area of disk. It is clear that the sum of areas of all identical triangles like OAB must be either less than or equal to area of the disk. We can call triangles like OAB as inscribed triangles.
Now, if we draw a radius-line OT’, perpendicular to AB at point T and intersecting the circumference at point T’, we can easily draw another triangle OA’B’ as shown in figure. AOB and A’OB’ are inscribed and superscribed triangles of disk with same angle at vertex O. So, it is clear that the angle A’OB’ is equal to the angle AOB. Triangle A’OB’ is larger than the circular arc OAB and circular arc OAB is larger than the inscribed triangle AOB. Also, the sum of areas of triangles identical to OA’B’ is either greater than or equal to area of the disk.
Let disk be divided into n- such inscribed and (thus) superscribed triangles. Since, total angle at point O is 360° or 2π —-the angles AOB and A’OB’ are of $2 \pi/n$ . And also since OT and OT’ are normals at chord AB and line A’B’ respectively, then they must divide the angles AOB and A’OB’ in two equal parts, each of $\pi/n$ radians.
In triangle AOB, the area of triangle AOB is the sum of the area of triangles AOT and BOT. But since both are equal to each other in area, area of AOB must be twice of the area of triangle AOB (or BOT). Our next target is to find, the area of AOT in order to find the area of AOB.
From figure, area of $\bigtriangleup{AOT}= \frac{1}{2} \times AT \times OT$ ….(1)
$OA=R$
And, $\angle{AOT}= \frac{\pi}{n}$ .
Thus, $\frac{AT}{OA}=\sin {\frac{\pi}{n}}$
or, $AT=OA \sin {(\pi/n)}$
or, $AT=R \sin {(\pi/n)}$ …..(2)
Similarly, $OT=R \cos {(\pi/n)}$
Therefore, area of $\bigtriangleup {AOT}=\frac{1}{2} \times R \sin {(\pi/n)} \times R \cos {(\pi/n)}=\frac{1}{2} R^2 \sin{(\pi/n)} \cos {(\pi/n)}$
And thus, area of $\bigtriangleup{AOB}=2 \times \frac{1}{2} R^2 \sin{(\pi/n)} \cos{(\pi/n)}=R^2 \sin{(\pi/n)} \cos{(\pi/n)}$
Since there are $n$ such triangles: sum of areas of such triangles
$S_1=n \times R^2 \sin{(\pi/n)} \cos{(\pi/n)}$ .
In triangle A’OB’, the total area of triangle A’OB’ is the sum of areas two identical triangles A’OT’ and B’OT’. Therefore, area of $\bigtriangleup{A’OB’}=2 \times \text{area of} \bigtriangleup{A’OT’}$ .
And area of $\bigtriangleup{A’OT’}=\frac{1}{2} \times AT’ \times OT’$ .
We have $OT’=R$
and angle A’OT’=$\frac{\pi}{n}$
Thus, A’T’/OT’= $\tan{\frac{\pi}{n}}$
or, $A’T’=OT’ \tan{\frac{\pi}{n}} =R \tan{\frac{\pi}{n}}$ .
Therefore, area of triangle A’OT’= $\frac{1}{2} \times R \times R \tan{\frac{\pi}{n}}=\frac{1}{2} R^2 \tan{\frac{\pi}{n}}$ .
Hence, area of $\bigtriangleup A’OB’=2 \times \frac{1}{2}R^2 \tan{\frac{\pi}{n}}$
As, there are $n$ such triangles: sum of areas of those triangles $S_2=n \times R^2 \tan{\frac{\pi}{n}}$ .
As it is clear that Sum of areas of triangles like $\bigtriangleup AOB$ is an approximation for the area of disk from below, i.e., $S_1 \le A$ when $n \to \infty$ or,
$\displaystyle{\lim_{n \to \infty}} n \times R^2 \sin{(\pi/n)} \cos{(\pi/n)} \le A$
$\displaystyle{\lim_{n \to \infty}} n \times R^2 \frac{\pi}{n} \dfrac{\sin{(\pi/n)}}{\pi/n} \cos{(\pi/n)} \le A$
$\pi R^2 \le A \ldots (I)$
Similarly, $\displaystyle{\lim_{n\to \infty}} n \times R^2 \tan{\frac{\pi}{n}} \ge A$
or, $\pi R^2 \ge A \ldots (II)$
From (I) and (II), we have $A=\pi R^2$ .
So the area of a disk is $\pi R^2$ .
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Question #82869
Jan 9, 2018
I shall attempt to do so. :)
Explanation:
There are three types of fractions: proper, improper, and mixed.
In a Proper Fraction the numerator is less than the denominator. Examples: $\frac{2}{3}$, $\frac{3}{4}$, and $\frac{9}{21}$.
In an Improper Fraction the numerator is more than the denominator. Examples: $\frac{3}{2}$, $\frac{4}{3}$, and $\frac{21}{9}$.
In a Mixed Fraction there is a whole number followed by a proper fraction in which the numerator is less than the denominator. Examples: $1 \frac{1}{2}$, $1 \frac{1}{3}$, and $2 \frac{3}{9}$.
I have used the same numbers in the fractions to make it easier.
Equal or Equivalent Fractions:
These are fractions where the numerators and denominators have the same value in relation to each other. Examples: $\frac{2}{3} = \frac{4}{6} = \frac{6}{9} = \frac{20}{30}$.
If you examine these fractions, they have the same value. This can be determined by factorising the numerators and denominators, cancelling the common factors and seeing whether the fractions are the same. So in the above examples, the smallest fraction is $\frac{2}{3}$. Factorising the second example, we get:
$\frac{4}{6} = \frac{2 \times 2}{3 \times 2} = \frac{2 \times \cancel{2}}{3 \times \cancel{2}} = \frac{2}{3}$.
The same can be done with the others, and we will get the same result. This means that all four examples are equivalent fractions.
Simplifying Fractions:
Fractions can be simplified in two ways.
1. If it is a proper or improper fraction, check if the numerator and denominator can be factorised, cancel the common factors, and the result is a simplified fraction. Example: $\frac{9}{21} = \frac{3 \times 3}{7 \times 3} = \frac{3 \times \cancel{3}}{7 \times \cancel{3}} = \frac{3}{7}$
2. If it is an improper fraction, first check if it can be simplified using the method above. Then you could convert it into a mixed fraction. This is done by dividing the numerator by the denominator and writing the result as a whole number. If there is a remainder, that is written after the whole number as the numerator over the same denominator. Example: $\frac{21}{9} = \frac{7 \times 3}{3 \times 3} = \frac{7 \times \cancel{3}}{3 \times \cancel{3}} = \frac{7}{3} = 2 \frac{1}{3}$ (The denominator $3$ goes into the numerator $7$ twice, so we get $2$ as the whole number; the remainder, $1$, is written over the denominator).
I do hope this helps. :)
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# How do you differentiate f(x)=x^3e^xsin(x) using the product rule?
Dec 1, 2015
$f ' \left(x\right) = {x}^{2} {e}^{x} \left(3 \sin \left(x\right) + x \sin \left(x\right) + x \cos \left(x\right)\right)$
#### Explanation:
According to the product rule:
$\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right) h \left(x\right)\right] = f ' \left(x\right) g \left(x\right) h \left(x\right) + f \left(x\right) g ' \left(x\right) h \left(x\right) + f \left(x\right) g \left(x\right) h ' \left(x\right)$
$f ' \left(x\right) = {e}^{x} \sin \left(x\right) \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] + {x}^{3} \sin \left(x\right) \frac{d}{\mathrm{dx}} \left[{e}^{x}\right] + {x}^{3} {e}^{x} \frac{d}{\mathrm{dx}} \left[\sin \left(x\right)\right]$
Find each derivative separately.
$\frac{d}{\mathrm{dx}} \left[{x}^{3}\right] = 3 {x}^{2}$
$\frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = {e}^{x}$
$\frac{d}{\mathrm{dx}} \left[\sin \left(x\right)\right] = \cos \left(x\right)$
Plug back in.
$f ' \left(x\right) = 3 {x}^{2} {e}^{x} \sin \left(x\right) + {x}^{3} {e}^{x} \sin \left(x\right) + {x}^{3} {e}^{x} \cos \left(x\right)$
$f ' \left(x\right) = {x}^{2} {e}^{x} \left(3 \sin \left(x\right) + x \sin \left(x\right) + x \cos \left(x\right)\right)$
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# The locus of the poles of normal chords of the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ isa) $\dfrac{{{x^2}}}{{{a^4}}} + \dfrac{{{y^2}}}{{{b^4}}} = {a^2} + {b^2}$ b) $\dfrac{{{x^2}}}{{{a^4}}} + \dfrac{{{y^2}}}{{{b^4}}} = {a^2} - {b^2}$ c) $\dfrac{{{a^6}}}{{{x^2}}} + \dfrac{{{b^6}}}{{{y^2}}} = {({a^2} - {b^2})^2}$ d) $\dfrac{{{a^4}}}{{{x^2}}} + \dfrac{{{b^4}}}{{{y^2}}} = {({a^2} - {b^2})^2}$
Last updated date: 18th Jun 2024
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Hint: The given equation is the equation of an ellipse. We will take a polar coordinate, and put it in the given equation by substituting only one $x$ and $y$ by the taken polar coordinate. Then we will find out the normal to the ellipse and compare the generated equation with the standard form of the normal of an ellipse, by certain algebraic manipulations we will get the equation for locus of the poles of normal chords of the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ .
Formula used:
1) $\csc \theta = \dfrac{1}{{\sin \theta }},\sec \theta = \dfrac{1}{{\cos \theta }}$
2) ${\sin ^2}\theta + {\cos ^2}\theta = 1$
We are given in the question that the equation of the ellipse is
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
So the given equation is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ --------(i)
Now we will try to take a polar coordinate to substitute in one $x$ and $y$ in the above equation (i)
Let $\left( {h,k} \right)$ be the polar coordinate.
Now polar equation of $\left( {h,k} \right)$ with respect to the ellipse is given by
$\dfrac{{xh}}{{{a^2}}} + \dfrac{{yk}}{{{b^2}}} = 1$ --------(ii)
If the above equation represents normal to the ellipse then it must be identical to the general form of the equation of normal of an ellipse.
$ax\sec \theta - by\csc \theta = {a^2} - {b^2}$
Comparing this with (ii) we get:
$\dfrac{{\left( {\dfrac{h}{{{a^2}}}} \right)}}{{a\sec \theta }} = \dfrac{{\left( {\dfrac{k}{{{b^2}}}} \right)}}{{ - b\csc \theta }} = \dfrac{1}{{{a^2} - {b^2}}}$
$\dfrac{{{a^2} - {b^2}}}{a}\left( {\dfrac{h}{{{a^2}}}} \right) = \sec \theta$ and $\dfrac{{{a^2} - {b^2}}}{{ - b}}\left( {\dfrac{k}{{{b^2}}}} \right) = \csc \theta$
$\Rightarrow \cos \theta = \dfrac{a}{{{a^2} - {b^2}}}\left( {\dfrac{{{a^2}}}{h}} \right)$ and $\sin \theta = \dfrac{{ - b}}{{{a^2} - {b^2}}}\left( {\dfrac{{{b^2}}}{k}} \right)$
$\Rightarrow \cos \theta = \dfrac{{\left( {\dfrac{{{a^3}}}{h}} \right)}}{{{a^2} - {b^2}}}$ and $\sin \theta = \dfrac{{ - \left( {\dfrac{{{b^3}}}{k}} \right)}}{{{a^2} - {b^2}}}$
Now, squaring both terms and adding them we get:
${\cos ^2}\theta + {\sin ^2}\theta = {\left[ {\dfrac{{\left( {\dfrac{{{a^3}}}{h}} \right)}}{{{a^2} - {b^2}}}} \right]^2} + {\left[ {\dfrac{{ - \left( {\dfrac{{{b^3}}}{k}} \right)}}{{{a^2} - {b^2}}}} \right]^2}$
$\Rightarrow 1 = \dfrac{1}{{{{({a^2} - {b^2})}^2}}}\left[ {\left( {\dfrac{{{a^6}}}{{{h^2}}}} \right) + \left( {\dfrac{{{b^6}}}{{{k^2}}}} \right)} \right]$
On cross multiplication we get:
$\left( {\dfrac{{{a^6}}}{{{h^2}}}} \right) + \left( {\dfrac{{{b^6}}}{{{k^2}}}} \right) = {({a^2} - {b^2})^2}$ Which is the required locus for $\left( {h,k} \right)$
Replacing $\left( {h,k} \right)$ with $\left( {x,y} \right)$ we get: $\left( {\dfrac{{{a^6}}}{{{x^2}}}} \right) + \left( {\dfrac{{{b^6}}}{{{y^2}}}} \right) = {({a^2} - {b^2})^2}$
So, the correct answer is “Option C”.
Note: The algebraic substitutions and calculation involves so many complicated terms, use only one operation at a time. Applying operations repeatedly in the same process may result in errors in your solution.
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# Linear Graphs
Linear Graphs as the name suggests, are graphs with linear relationships between both variables. In the case of any two variables, the relation between these two variables can be drawn by constructing the table of values if the rule for that relation is mentioned. At least two points coordinates should be known to plot a straight-line graph. These points must fit the rule and such graphs when drawn are termed Linear graphs.
## Linear Graphs Definition
A line graph with continuous or unbroken nature is called a linear graph.
To draw the unbroken line we need to locate some points on the graph sheet. Let’s make a graph with an x and y-coordinate.
Suppose you go to an auditorium and search for your reserved seat. You need to know two numbers, the row number, and the seat number. This is the basic method for fixing a point in a plane.
## Linear Graph Equation
All the linear graphs can be represented by the linear equation in two variables, such that
y = mx + c
Where,
• x and y are variables that we represent on the x and y-axis,
• m is the slope of the graph, and
• c is the y-intercept of the graph.
Note: This equation is called the slope-intercept form of the linear.
### Standard Form of Linear Equation
The standard form of the linear equation is given as follows:
Ax + By + C = 0
Where,
• x and y are variables that we represent on the x and y-axis,
• A, B, and C are the real constant.
## How To Plot Linear Equation On a Graph?
For a general linear equation in the form Ax + By = C, you can plot the Linear Equation On a Graph using the following steps:
Step 1: Rearrange the given equation to find the value of y.
By = C – Ax
y = (-A/B)x + C/B
Step 2: Put x = 0, to find Y-intercept of the graph i.e., point of intersectino between line and y-axis and mark the point in the graph.
y = C/B
Thus, y-intercept is (0, C/B).
Step 3: Put y = 0, to find X-intercept of the graph i.e., point of intersectino between line and x-axis mark the point in the graph.
0 = (-A/B)x + C/B
x = C/A
Thus, x-intercept is (C/A, 0)
Step 4: Meet the both intercepts to plot the Linear Equation On a Graph.
## Linear Graph vs Line Graph
A linear Graph is an unbroken line with the same slope throughout the line and always can be represented by a linear equation.
Whereas, a line graph is a collection of line segments with each starting from where the last one is ending, and all line segments have various slopes. Also, these different line segments can’t be represented by a single linear equation.
## Application of Linear Graphs
In everyday life, we observe variations in the value of different quantities, or we can say that the more we use the facility the more we have to pay for it. for e.g. If more electricity we consume we will have to pay the bill more and vice versa. So one quantity affects the other quantity. We can say that the quantity of electricity is the independent variable and the amount of bills is the dependent variable. These relations we can show with graphs.
## Sample Problems on Linear Graphs
Problem 1: Plot (3, 4) on the graph
Solution:
As in graph points are denoted in the form of (x, y)
So, on comparing the points:
x = 3 and y = 4.
First, draw x = 3 moves in the forward direction to x = 3.
Now, draw y = 4 moves in the upward direction to (3, 4).
Problem 2: Locate the given points on the graph.
1. (1, 2)
2. (2, 8)
3. (4, 2)
Solution:
For x = 1 and y = 2. Starting from (0, 0) origin we move x = 1 direction forward and from there we move y = 2 direction upward then finally reaching state is our point.
Similarly, plot the rest of the two points on Graph.
Problem 3: Plot the following points and verify if they lie on the line.
1. (0, 1), (0, 2), (0, 4), (0, 3)
2. (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)
Solution:
1. Plotting (0, 1), (0, 2), (0, 4), (0, 3) on graph
We can draw every coordinate on graph sheet as follows .
Here it forms a line after joining all the points.
2. Plotting (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) on graph
Note: In each of the above cases, the graph obtained by joining the plotted points is a line. Such graphs are called linear graphs.
Problem 4: Amit can ride a bike at a constant speed of 30 km/hour. Draw a time-distance graph for this situation and find it
(i) Time is taken by Amit to ride 75 km.
(ii) Distance covered by Amit in 3.5 hours.
Solution:
The table so formed is:
Consider the Scale:
Horizontal: 2 units = 1 hour
Vertical: 1 unit = 10 km
Mark time on horizontal axis.
Mark distance on vertical axis
Plot points: (1, 30), (2, 60), (3, 90), (4, 120)
On joining the points the required linear graph is obtained.
(i) From the above graph it is concluded that on the vertical axis corresponding to the distance 75 km, the time is 2.5 hours on the horizontal axis. Thus, 2.5 hours are needed to cover 75 km.
(ii) Similarly, on the horizontal axis corresponding to time 3.5 hours, the distance covered is 105 km on the vertical axis.
Problem 5: A public bank provides 10% interest on deposits by senior citizens. Plot a graph representing the relation between the deposited amount and the interest earned by senior citizens and state the following things from the graph so plotted:
(i) The interest earned annually for an investment of â‚ą 450.
(ii) The amount of money deposited to earn the interest of â‚ą 35.
Solution:
Consider the Scale:
Horizontal: 1 unit = â‚ą 10
Vertical: 1 unit = â‚ą 100
Mark interest earned on horizontal axis.
Mark amount deposited on vertical axis.
Plot the points: (10, 100), (30, 300), (50, 500), (70, 700) and (90, 900) on the graph.
On joining the points the required linear graph is obtained.
(i) From the above graph it is concluded that on the vertical axis corresponding to the amount deposited â‚ą 450, the interest earned is â‚ą 45 on the horizontal axis. Thus, â‚ą 45 are earned on the deposit of â‚ą 450.
(ii) Similarly, on the horizontal axis corresponding to interest earned 35%, the amount deposited is â‚ą 350 on the vertical axis.
## FAQs on Linear Graphs
### Q1: What is a Linear Graph?
A linear graph is a continuous graph in the shape of a line that can be infinitely increased to both sides similar to a line.
### Q2: What is the Slope of a Linear Graph?
Change in the y-coordinate with respect to the x-coordinate is called slope of a linear graph. Mathematically it can be represented as
Slope = Δy/Δx
Where,
• Δy represents the change in y-coordinate, and
• Δx represnts the change in x-coordinate.
### Q3: How to Graph a Linear Equation?
To graph any linear equation, we can find two points that satisfy the equation and meet that two points to form a line that represents all the points satisfying the linear equation.
### Q4: What is the X-Intercept of a Linear Graph?
The point where the linear graph cuts the x-axis, is called the x-intercept.
### Q5: What is the Y-Intercept of a Linear Graph?
The point where the linear graph cuts the y-axis, is called the y-intercept.
### Q6: What is the Point-Slope form of a Linear Equation?
The Point-slope form of the linear equation is given by
y – y1 = m(x – x1)
Where,
• x and y are variables that we represent on the x and y-axis,
• m is the slope of the graph, and
• x1 and y are the coordinate of the point from where line passes through.
|
## LCM & GCF
This shows disadvantaged students in special education programs & special education with special needs the difference between multiples and factors as well as what to watch for when calculating LCM & GCF.
## Demystifying Algebra: Mastering GCF and LCM
Welcome to an empowering journey through algebraic concepts with me, Professor Ehimwenma E. Aimiuwu, an experienced educator dedicated to demystifying math. In this tutorial series, I guide learners from grades 7 through 12 through the intricacies of algebra, making complex topics like the Greatest Common Factor (GCF) and Lowest Common Multiple (LCM) accessible and engaging.
I’m Professor Aimiuwu, a passionate math teacher with over a decade of experience in the field. My mission is to instill confidence in students and break down barriers to understanding math. With a dynamic teaching style and a knack for simplifying complex concepts, I’ve inspired countless learners to embrace the beauty of mathematics.
### Greatest Common Factor:
Let’s dive into the basics. The Greatest Common Factor (GCF) is a fundamental element of algebraic problem-solving. It represents the largest number that divides evenly into two or more given numbers. Understanding the GCF is crucial for simplifying fractions, solving equations, and factoring polynomials.
Explanation:
To find the GCF of two or more numbers, we identify all the factors shared by those numbers and determine the largest one. For example, the GCF of 24 and 36 is 12 because 12 is the largest number that divides evenly into both 24 and 36.
Example:
I’ll walk you through a variety of examples, demonstrating how to find the GCF using different methods such as listing factors, prime factorization, and the ladder method. For instance, when finding the GCF of 18 and 30, we can list the factors of each number (18: 1, 2, 3, 6, 9, 18; 30: 1, 2, 3, 5, 6, 10, 15, 30) and identify the largest common factor (6).
Importance:
Understanding the GCF is essential in simplifying fractions and solving equations. By identifying the GCF, learners can reduce fractions to their simplest form, making calculations more manageable. Additionally, factoring polynomials becomes easier when one can identify common factors, leading to efficient problem-solving in algebraic expressions and equations.
### Lowest Common Multiple:
Next, let’s explore the concept of the Lowest Common Multiple (LCM), another crucial aspect of algebraic equations. The LCM represents the smallest multiple that is divisible by two or more given numbers. Understanding the LCM is essential for adding and subtracting fractions, solving word problems, and finding common denominators.
Explanation:
To find the LCM of two or more numbers, we identify the multiples of each number and determine the smallest one they have in common. For example, the LCM of 4 and 6 is 12 because 12 is the smallest number that is divisible by both 4 and 6.
Example:
I’ll demonstrate various methods for finding the LCM, such as listing multiples, prime factorization, and the ladder method. For instance, when finding the LCM of 8 and 12, we can list the multiples of each number (8: 8, 16, 24, 32, …; 12: 12, 24, 36, …) and identify the smallest common multiple (24).
Importance:
Understanding the LCM is crucial in adding and subtracting fractions with different denominators. By finding the LCM, learners can determine a common denominator and perform operations on fractions accurately. Additionally, the LCM is valuable in solving word problems that involve finding the least common multiple of multiple quantities or events.
### Conclusion:
Join me in this empowering journey through algebra, and unlock the keys to mathematical success. With dedication and practice, you can master essential concepts like the GCF and LCM, gaining the skills and confidence needed to excel in math. Let’s break down stereotypes and embrace the power of education together, inspiring positive change in our communities.
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# How is the graph of -(x+7)^2+8 compared to the graph of f(x)= x^2?
Oct 11, 2017
The function is reflected vertically across the x-axis, it is shifted left 7 units, and it is shifted up 8 units.
#### Explanation:
$f \left(x\right) = {x}^{2}$ is the parent function and the numbers and signs we put around this function affects how the graph looks related to the parent function.
For this problem you can follow this formula:
$a {\left(x + h\right)}^{2} + k$
If $a > 1$ then it stretches the function so it’s skinnier.
If $a < 1$ then it compresses the function so it’s wider.
If $a$ is negative then it reflects the function vertically across the x-axis.
In this case $a = - 1$ so it’s not stretched or compressed, but it is reflected across the x-axis.
$h$ affects the function’s horizontal shift from the parent function
If $h$ is positive, it horizontally shifts left $h$ units
If $h$ is negative, it horizontally shifts right $h$ units
$h = 7$ so it shifts left 7 units
Finally $k$ affects the function’s vertical shift from the parent function
If $k$ is positive, it vertically shifts up $k$ units
If $k$ is negative, it vertically shifts down $k$ units
$k = 8$ so it shifts up 8 units
|
# 12.2 Finding limits: properties of limits (Page 2/5)
Page 2 / 5
## Evaluating the limit of a function algebraically
Evaluate $\text{\hspace{0.17em}}\underset{x\to 3}{\mathrm{lim}}\left(2x+5\right).$
Evaluate the following limit: $\text{\hspace{0.17em}}\underset{x\to -12}{\mathrm{lim}}\left(-2x+2\right).$
26
## Finding the limit of a polynomial
Not all functions or their limits involve simple addition, subtraction, or multiplication. Some may include polynomials. Recall that a polynomial is an expression consisting of the sum of two or more terms, each of which consists of a constant and a variable raised to a nonnegative integral power. To find the limit of a polynomial function, we can find the limits of the individual terms of the function, and then add them together. Also, the limit of a polynomial function as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is equivalent to simply evaluating the function for $\text{\hspace{0.17em}}a$ .
Given a function containing a polynomial, find its limit.
1. Use the properties of limits to break up the polynomial into individual terms.
2. Find the limits of the individual terms.
4. Alternatively, evaluate the function for $\text{\hspace{0.17em}}a$ .
## Evaluating the limit of a function algebraically
Evaluate $\text{\hspace{0.17em}}\underset{x\to 3}{\mathrm{lim}}\left(5{x}^{2}\right).$
Evaluate $\text{\hspace{0.17em}}\underset{x\to 4}{\mathrm{lim}}\left({x}^{3}-5\right).$
59
## Evaluating the limit of a polynomial algebraically
Evaluate $\text{\hspace{0.17em}}\underset{x\to 5}{\mathrm{lim}}\left(2{x}^{3}-3x+1\right).$
Evaluate the following limit: $\text{\hspace{0.17em}}\underset{x\to -1}{\mathrm{lim}}\left({x}^{4}-4{x}^{3}+5\right).$
10
## Finding the limit of a power or a root
When a limit includes a power or a root, we need another property to help us evaluate it. The square of the limit of a function equals the limit of the square of the function; the same goes for higher powers. Likewise, the square root of the limit of a function equals the limit of the square root of the function; the same holds true for higher roots.
## Evaluating a limit of a power
Evaluate $\text{\hspace{0.17em}}\underset{x\to 2}{\mathrm{lim}}{\left(3x+1\right)}^{5}.$
We will take the limit of the function as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches 2 and raise the result to the 5 th power.
Evaluate the following limit: $\underset{x\to -4}{\mathrm{lim}}{\left(10x+36\right)}^{3}.$
$-64$
If we can’t directly apply the properties of a limit, for example in $\underset{x\to 2}{\mathrm{lim}}\left(\frac{{x}^{2}+6x+8}{x-2}\right)$ , can we still determine the limit of the function as $x$ approaches $a$ ?
Yes. Some functions may be algebraically rearranged so that one can evaluate the limit of a simplified equivalent form of the function.
## Finding the limit of a quotient
Finding the limit of a function expressed as a quotient can be more complicated. We often need to rewrite the function algebraically before applying the properties of a limit. If the denominator evaluates to 0 when we apply the properties of a limit directly, we must rewrite the quotient in a different form. One approach is to write the quotient in factored form and simplify.
how can are find the domain and range of a relations
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
i want to sure my answer of the exercise
what is the diameter of(x-2)²+(y-3)²=25
how to solve the Identity ?
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim
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# How To Find Square Root Of A Number In Three Easy Steps
A very important topic in Quantitative Aptitude of any competitive exam is squares and cubes. In our previous post, we square of any number in your head, without any pen and paper. Here, we are going to learn how to find the square root of a given number.
Conventionally, we have been finding square root of a number through the ‘long division method’ that was taught to us in school. Here, we are going to discuss the shortcut to finding square root of a number.
### Importance Of How To Find Square Root Of A Number In Less Time?
Finding square roots of numbers can help you save the much-needed time in competitive exams, such as SSC CGL 2017, SSC CHSL 2017, etc, allowing you to attempt more questions in the limited given amount of time.
Finding whether the given number is a perfect square of not, is a lengthy process. If you have a question from simplifications, where you need to find square root in one of the steps, consider the given number to be a perfect square. ONLY when the question is given under the topic approximations, it may not be a perfect square.
### How To Find Square Root Of A Number In Three Easy Steps
Firstly, you must memorise the squares of numbers from 1 to 30.
We have already learnt that there is a certain pattern seen when it comes to square of any number:
SMART METHOD
### How to find square root of a number in a smart method is a 3-step process
Example 1: Find the 2401
Solution:
Step1: Decide the unit’s place of the result.
2401 ends with 1 = Square root will end with 1 or 9
_ 1 or _ 9
Step2: Leave the last 2 digits and find the immediate perfect square before the remaining number.
The immediate perfect square before 24 = 16, which is 42
That is the number in ten’s place
41 or 49
Step 3: To determine whether the square root of 2401 is 41 or 49
Method 1: Take the number that lies exactly between 41 and 49 and square it.
452 = 2025 (We have learnt earlier how to find square of numbers ending with 5)
2025 < 2401
45 < Square root of 2401
And since
45 < 49
Square root of 2401 is 49
Example 2: Find the 6084
Solution:
Step1: Decide the unit’s place of the result.
6084 ends with 4 = Square root will end with 2 or 8
_2 or _8
Step 2: Leave the last 2 digits and find the immediate perfect square before the remaining number.
The immediate perfect square before 60 = 49, which is 72
72 or 78
Step 3: To determine whether the square root of 6084 is 72 or 78
Method 2: Take squares of numbers 70 and 80, since 72 and 78 lie within the range.
702 = 4900
802 = 6400
Since, 6084 is closer to 6400, the square root must also be closer to 80, which is 78
Square root of 6084 is 78
Example 3: Find the 12769
Solution:
Step 1: Decide the unit’s place of the result.
12769 ends with 9 = Square root will end with 3 or 7
_3 or _7
Step 2: Leave the last 2 digits and find the immediate perfect square before the remaining number.
The immediate perfect square before 127 = 121, which is 112
113 or 117
Step 3: To determine whether the square root of 12769 is 113 or 117
Method 1: Take the number that lies exactly between 113 and 117 and square it.
1152 = 13225
12769 < 13225
And since
113 < 115
Square root of 12769 is 113
Method 2: Take squares of numbers 110 and 120, since 113 and 117 lie within the range.
1102 = 12100
1202 = 14400
Since 12100 is closer to 12769, the square root must also be closer to 110, which is 113
Square root of 12769 is 113
Practice this technique well and ensure that the next time you are asked to find the square root of a perfect square, you don’t spend more than 3 seconds.
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# First year
## TS inter 1st year
TS inter 1st year: These blueprints were designed by ‘Basics in Maths’ team. These to-do’s help the TS intermediate first-year Maths students fall in love with mathematics and overcome their fear.
These blueprints cover all the topics of the TS I.P.E first-year maths syllabus and help in I.P.E exams.
TS inter 1st year
## TS Inter Maths 1B Concept
Ts Inter Maths 1B Concept: designed by the ‘Basics in Maths’ team. These notes to do help the TS intermediate first-year Maths students fall in love with mathematics and overcome the fear.
These notes cover all the topics covered in the TS I.P.E first year maths 1B syllabus and include plenty of formulae and concept to help you solve all the types of Inter Math problems asked in the I.P.E and entrance examinations.
## 0.COORDINATE GEOMETRY( BASICS)
• Distance between two points A(x1, y1), B(x2, y2) is
• distance between a point A(x1, y1) to the origin is
• The midpoint of two points A(x1, y1), B(x2, y2) is
• If P divides the line segment joining the points A(x1, y1), B(x2, y2) in the ratio m:n then the coordinates of P are
• Area of the triangle formed by the vertices A (x1, y1), B (x2, y2) and C (x3, y3) is
### 1. LOCUS
Locus: The set of points that are satisfying a given condition or property is called the locus of the point.
Ex:- If a point P is equidistant from the points A and B, then AP =BP
Ex 2: – set of points that are at a constant distance from a fixed point.
here the locus of a point is a circle.
• In a right-angled triangle PAB, the right angle at P and P is the locus of the point, then
AB2 = PA2 + PB2
•Area of the triangle formed by the vertices A (x1, y1), B (x2, y2), and C (x3, y3) is
### 2.CHANGE OF AXES
Transformation of axes:
When the origin is shifted to (h, k), without changing the direction of axes then
•To remove the first degree terms of the equation ax2 + 2hxy + by2 +2gx +2fy+ c = 0, origin should be shifted to the point
•If the equation ax2 + by2 +2gx +2fy+ c = 0, origin should be shifted to the point
Rotation of axes:
When the axes are rotated through an angle θ then
•To remove the xy term of the equation ax2 + 2hxy + by2 = 0, axes should be rotated through an angle θ is given by
### 3.STRAIGHT LINES
Slope:- A-line makes an angle θ with the positive direction of the X-axis, then tan θ is called the slope of the line.
It is denoted by “m”.
m= tan θ
• The slope of the x-axis is zero.
• Slope of any line parallel to the x-axis is zero.
• The y-axis slope is undefined.
• The slope of any line parallel to the y-axis is also undefined.
• The slope of the line joining the points A (x1, y1) and B (x2, y2) is
Slope of the line ax + by + c = 0 is
### Types of the equation of a straight line:
• Equation of x- axis is y = 0.
• Equation of any line parallel to the x-axis is y = k, where k is the distance from above or below the x-axis.
• Equation of y- axis is x = 0.
• Equation of any line parallel to y-axis is x = k, where k is the distance from the left or right side of the y-axis.
Slope- intercept form
The equation of the line with slope m and y-intercept c is y = mx + c.
Slope point form:
The equation of the line passing through the point (x1, y1) with slope m is
y – y1 = m (x – x1)
Two points form:
The equation of the line passing through the points (x1, y1) and (x2, y2) ’ is
Intercept form:
The equation of the line with x-intercept a, y-intercept b is
• The equation of the line ∥ el to ax +by + c = 0 is ax +by + k = 0.
• The equation of the line ⊥ler to ax +by + c = 0 is bx −ay + k = 0.
Note: –
1. If two lines are parallel then their slopes are equal
m1 = m2
1. If two lines are perpendicular then product of their slopes is – 1
m1 × m2 = – 1
1. The area of the triangle formed by the line ax + by + c = 0 with the coordinate axes is
2. The area of the triangle formed by the line with the coordinate axes is
Perpendicular distance (Length of the perpendicular):
The perpendicular distance from a point P (x1, y1) to the line ax + by + c = 0 is
• The perpendicular distance from origin to the line ax + by + c = 0 is
Distance between two parallel lines:
•The distance between the parallel lines ax1 + by1 + c1 = 0 and ax2 + by2 + c2 = 0 is
Perpendicular form or Normal form:
The equation of the line which is at a distance of ‘p’ from the origin and α (0≤ α ≤ 3600) is the angle made by the perpendicular with the positive direction of the x-axis is x cosα + y sinα = p.
Symmetric form:
The equation of the line passing through point P (x1, y1) and having inclination θ is
Parametric form:
if P (x, y) is any point on the line passing through A (x1, y1) and
making inclination θ, then
x = x1 + r cos θ, y = y1 + r sin θ
where ‘r’ is the distance from P to A.
• The ratio in which the line L ≡ ax + by + c = 0 divide the line segment joining the points A (x1, y1), B (x2, y2) is – L11: L22.
Where L11 = ax1 + by1 + c and L22 = ax2 + by2 + c.
Note: – the points A (x1, y1), B (x2, y2) lie on the same side or opposite side of line L = 0 according to L11 and L22 have the same sign or opposite sign.
∗ x-axis divides the line segment joining the points A (x1, y1), B (x2, y2) in the ratio – y1: y2.
∗ y-axis divides the line segment joining the points A (x1, y1), B (x2, y2) in the ratio – x1: x2.
Point of intersection of two lines:
the point of intersection of two lines a1x + b1y + c = 0 and a2x + b2y + c = 0 is
#### Concurrent Lines:
Three or more lines are said to be concurrent lines if they have a point in common.
The common point is called the point of concurrence.
∗ The condition that the lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 to be concurrent is
a3(b1c2 – b2c1) + b3(c1a2 – c2a1) + c3(a1b2 – a2b1).
∗ The condition that the lines ax + hy +g = 0, hx + by + f = 0 and gx +fy + c = 0 is
abc + 2fgh – af2 – bg2 – ch2 = o.
Note: – if two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 said to be identical (same) if
#### Family of a straight line:
Family of straight lines: – A set of straight lines having a common property is called a family of straight lines.
Let L1 ≡ a1x + b1y + c1 = 0 and L2 ≡ a2x + b2y + c2 =0 represent two intersecting lines, theThe equation λ1 L1 + λ2 L2 = 0 represent a family of straight lines passing through the point of intersection of the lines L1 = 0 and L2 = 0.
∗ The equation of the straight line passing through the point of intersection of the lines L1 = 0 and L2 = 0 is L1 + λL2 = 0.
The angle between two lines:
If θ is the angle between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 then
∗ If θ is an acute angle then
∗ If θ is the angle between two lines, then (π – θ) is another angle between two lines.
∗ If θ≠π/2 is angle between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then
∗ If m1, m2 are the slopes of two lines then
Note: – The lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are
∗ Parallel iff
∗ Perpendicular iff a1a2 + b1b2 = 0
The foot of the perpendicular:
If Q (h. k) is the foot of the perpendicular from a point P (x1, y1)to the line ax + by +c = 0 then
Image of the point:
If Q (h. k) is the image of point P (x1, y1) with respect to the line ax + by +c = 0 then
Collinear Points:
If three points are said to be collinear, then they lie on the same line.
∗ If A, B, and C are collinear, then
Slope of AB = Slope of BC (or) Slope of BC = Slope of AC (or) Slope of AB = Slope of AC
### 4. PAIR OF STRAIGHT LINES
∎ ax2 + 2hxy + by2 = 0 is called the second-degree homogeneous equation in two variable x and y.
This equation always represents a pair of straight lines which are passing through the origin.
∎ If l1x + m1y = 0 and l2x + m2y = 0 are two lines represented by the equation ax2 + 2hxy + by2 = 0, then ax2 + 2hxy + by2 = (l1x + m1y) (l2x + m2y)
⇒ a = l1l2; 2h = l1m2 + l2m1; b = m1m2
∎ If m1, m2 are the slopes of the lines represented by the equation ax2 + 2hxy + by2 = 0, then
m1+ m2 = – 2h/b and m1 m2 = a/b
∎ The lines represented by the equation ax2 + 2hxy + by2 = 0 are
∎ If h2 = ab, then the lines represented by the equation ax2 + 2hxy + by2 = 0 are coincident.
∎ If two lines represented by the equation ax2 + 2hxy + by2 = 0 are equally inclined to the coordinate axes then h = 0 and ab < 0.
∎ The equation of the pair of lines passing through the point (h, k) and
(i) Parallel to the lines represented by the equation ax2 + 2hxy + by2 = 0 is
a (x – h)2 + 2h (x – h) (y – k) + b (y – k)2 = 0
(ii) Perpendicular to the lines represented by the equation ax2 + 2hxy + by2 = 0 is
b (x – h)2 – 2h (x – h) (y – k) + a (y – k)2 = 0
Angle between the lines:
If θ is the angle between the lines represented by the equation ax2 + 2hxy + by2 = 0, then
∎ If a + b = 0, then two lines are perpendicular.
Area of the triangle:
The area of the triangle formed by the lines ax2 + 2hxy + by2 = 0 and the line lx + my + n = 0 is
Angular Bisectors:
⇒ the angle between angular bisectors is always 900
L1 = o, L2 = o are two non-parallel lines the locus of the point P such that the perpendicular distance from P to the first lie is equal to the perpendicular distance from P to second line is called the angular bisector of two lines.
⇒ If two lines are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the angular bisectors are
∎ The equation of the pair of angular bisectors of ax2 + 2hxy + by2 = 0 is
h (x2 – y2) = (a – b =) xy.
∎ If ax2 + 2hxy + by2 + 2gx 2fy + c= 0 represents a pair of straight lines then
(i) abc + 2fgh – af2 – bg2 – ch2 = 0
(ii) h2 ≥ ab, g2 ≥ ac and f2 ≥ bc
If two lines represented by ax2 + 2hxy + by2 + 2gx 2fy + c= 0 are
l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0, then
ax2 + 2hxy + by2 + 2gx 2fy + c = (l1x + m1y + n1) (l2x + m2y + n2)
a = l1l2; 2h = l1m2 + l2m1; b = m1m2 ; 2g = l1n2 + l2n1; 2f = m1n2 + m2n1 and c = n1n2
The point of intersection of the lines represented by ax2 + 2hxy + by2 + 2gx 2fy + c= 0 is
∎ If two line have same homogeneous path then the lines represented by the first pair is parallel to the lines represented by the second pair.
ax2 + 2hxy + by2 + 2gx 2fy + c= 0 …………… (2)
ax2 + 2hxy + by2 = 0 …………… (1)
equation (1) and equation (2) form a parallelogram, one of the diagonals of parallelogram which is not passing through origin is 2gx + 2fy + c = 0.
∎ If two lines represented by ax2 + 2hxy + by2 + 2gx 2fy + c= 0 are parallel then
• h2 = ab (ii) af2 = bg2 (iii) hf = bg, gh = ab
Distance between parallel lines is
### 5. THREE DIMENSIONAL COORDINATES
• Let X’OX, Y’OY be two mutually perpendicular lines passing through a fixed point ‘O’. These two lines determine the XOY – plane (XY- plane). Draw the line Z’OZ perpendicular to XY – plane and passing through ‘O’.
The fixed point ‘O’ is called origin and three mutually perpendicular lines X’OX, Y’OY, Z’OZ are called Rectangular coordinate axes.
Three coordinate axes taken two at a time determine three planes namely XOY- plane, YOZ-plane, ZOY-plane or XY-plane, YZ-plane, ZX-plane respectively.
For every point P in space, we can associate an ordered triad (x, y, z) of real numbers formed by its coordinates.
The set of points in space is referred to as ‘Three-Dimensional Space’ or R3– Space.
∗ If P (x, y, z) is a point in a space, then
x is called x-coordinate of P
y is called y-coordinate of P
z is called z-coordinate of P
Distance between two points in space:
∗ Distance between the points A (x1, y1, z1) and B (x2, y2, z2) is
∗ Distance between the point P (x, y, z) to the origin is
Translation of axes:
When the origin is shifted to the point (h, k, l), then
X = x – h; Y = y – k; Z = z – l and x = X + h; y = Y + k; z = Z + l
∗ The foot of the perpendicular from P (x, y, z) to X-axis is A (x, 0, 0).
The perpendicular distance of P from X-axis is
Similarly,
The perpendicular distance of P from Y-axis is
The perpendicular distance of P from Z-axis is
Collinear points: If three or more points lie on the same line are called collinear points.
Section formula:
The point dividing the line segment joining the points A (x1, y1, z1) and B (x2, y2, z2) in the ratio m : n is given by
The mid-point of the line segment joining the points A (x1, y1, z1) and B (x2, y2, z2) is
The centroid of the triangle whose vertices are A (x1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is
Tetrahedron:
→ It has 4 vertices and 6 edges.
→ A Tetrahedron is a closed figure formed by four planes not all passing through the same point.
→ Each edge arises as the line of intersection of two of the four planes.
→ The line segment joining the vertices to the centroid of opposite face. The point of concurrence is called centroid of Tetrahedron.
→ Centroid divides the line segment in the ratio 3:1.
→ The centroid of the Tetrahedron whose vertices are A (x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) and C (x4, y4, z4) is
The line segment joining the points (x1, y1, z1), (x2, y2, z2) is divided by
XY – plane in the ratio – z1: z2
YZ – plane in the ratio – x1: x2
XZ – plane in the ratio – y1: y2
### 6. DIRECTION COSINES AND RATIOS
Consider a ray OP passing through origin ‘O’ and making angles α, β, γ respectively with the positive direction of X, Y, Z axes.
Cos α, Cos β, Cos γ are called Direction Cosines (dc’s) of the ray OP.
Dc’s are denoted by (l, m, n), where l = Cos α, m = Cos β, n = Cos γ
• A line in a space has two directions, it has two sets of dc’s, one for each direction. If (l, m, n) is one set of dc’s, then (-l, -m, -n) is the other set.
• Suppose P (x, y, z) is any point in space such that OP = r. If (l, m, n) are dc’s of a ray OP then x = lr, y= mr, z = nr.
• If OP = r and dc’s of OP are (l, m, n) then the coordinates of P are (lr, mr, nr).
• If P (x, y, z) is a point in the space, then dc’s of OP are
• If (l, m, n) are dc’s of a line then l2 + m2 + n2 = 1.
⇒ cos2α + cos2β + cos2γ = 1.
Direction Ratios:
Any three real numbers which are proportional to the dc’s of a line are called direction ratios (dr’s) of that line.
• Let (a, b, c) be dr’s of a line whose dc’s are (l, m, n). Then (a, b, c) are proportional to (l, m, n)
and a2 + b2 + c2 ≠ 1.
• Dr’s of the line joining the points (x1, y1, z1), (x2, y2, z2) are (x2 – x1, y2 – y1, z2 – z1)
• If (a, b, c) are dr’s of a line then its dc’s are
• If (l1, m1, n1), (l2, m2, n2) are dc’s of two lines and θ is angle between them then Cos θ = l1l2 + m1m2 + n1n2
If two lines perpendicular then l1l2 + m1m2 + n1n2 = 0.
• If (a1, b1, c1), (a2, b2, c2) are dr’s of two lines and θ is the angle between them then
If two line are perpendicular then a1a2 + b1b2 + c1c2 = 0.
## 7. THE PLANE
Plane: A plane is a proper subset of R3 which has at least three non-collinear points and any two points in it.
∎ Equation of the plane passing through a given point A (x1, y1, z1), and perpendicular to the line whose dr’s (a, b, c) is a(x – x1) + a(y – y1) + a(z – z1) = 0.
∎ The equation of the plane hose dc’s of the normal to the plane (l, m, n) and perpendicular distance from the origin to the pane p is lx + my + nz = p
∎ The equation of the plane passing through three non-collinear points A (x1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is
∎ The general equation of the plane is ax + by + cz + d = 0, where (a, b, c) are Dr’s of the normal to the plane.
Normal form:
The equation of the plane ax + by + cz + d = 0 in the normal form is
Perpendicular distance:
The perpendicular distance from (x1, y1, z1) to the plane ax + by + cz + d = 0 is
The perpendicular distance from the origin to the plane ax + by + cz + d = 0 is
Intercepts:
X- intercept = aIf a plane cuts X –axis at (a, 0, 0), Y-axis at (0, b, 0) and Z-axis at (0, 0, c) then
Y-intercept = b
Z-intercept = c
The equation of the plane in the intercept form is
The intercepts of the plane ax + by + cz + d =0 is -d/a, -d/b, -d/c
∎ The equation of the plane parallel to ax + by + cz + d = 0 is ax + by + cz + k = 0.
∎ The equation of XY – plane is z = 0.
∎ The equation of YZ – plane is x = 0.
∎ The equation of XZ – plane is y = 0.
∎ Distance between the two parallel planes ax + by + cz + d1 =0 and ax + by + cz + d2 =0 is
The angle between two planes:
The angle between the normal to two planes is called the angle between the planes.
If θ is the angle between the planes a1 x + b1 y + c1 z + d1 =0 and a2 x + b2 y + c2 z + d2 =0 then
If two line are perpendicular then a1a2 + b1b2 + c1c2 = 0.
∎ The distance of the point P (x, y, z) from
### 8. LIMITS AND CONTINUITY
Intervals:
Let (a, b) ∈ R such that a ≤ b, then the set
• {x ∈ R: a ≤ x ≤ b}, is denoted by [a, b] and it is called as closed interval
• {x ∈ R: a < x < b}, is denoted by (a, b) and it is called as open interval
• {x ∈ R: a < x ≤ b}, is denoted by (a, b] and it is called as open closed interval
• {x ∈ R: a ≤ x < b}, is denoted by [a, b) and it is called as closed open interval
• {x ∈ R: x ≥ a}, is denoted by [a, ∞)
• {x ∈ R: x > a}, is denoted by (a, ∞)
• {x ∈ R: x ≤ a}, is denoted by (- ∞, a]
• x ∈ R: x < a}, is denoted by (- ∞, a)
#### Limit:
If f(x) is a function of x such that if x approaches to a constant value ‘a’, then the value of f(x) also approaches to ‘l’. Then the constant ‘I’ is called a limit of f(x) at x = a
Or
A real number l is called the limit of the function f, if for all ϵ> 0 there exist δ > 0 such that whenever ⟹
###### Properties of Limits:
Sand witch theorem:( Squeez Principle):
f, g, and h are functions such that f(x) ≤ g(x) ≤ h(x), then and if
###### Left- hand and Right-hand Limits:
If x < a, then is called left-hand limit
If x > a, then is called right-hand limit
Note:
###### In Determinate forms:
if a function f(x) any of the following forms at x = a:
Then f(x) is said to be indeterminate at x = a.
Continuity:
Condition 1: If the condition is like x = a and x ≠ a, then we use following property.
If then f(x) is continuous at x = a, otherwise f(x) is not continuous.
Condition 2: If the condition is like x ≤ a and x >a, or x < a and x ≥a then we use following property.
If then f(x) is continuous at x = a, otherwise f(x) is not continuous.
## 9.DIFFERENTIATION
Let f be a function defined on a neighbourhood of a real number ‘a’ if exist then we say that f is differentiable at x a and it is denoted by f'(a).
∴ f’(a) =
∎ If right hand derivative = left hand derivative, then f is differentiable at ‘a’.
i.e.,
### First principle in derivative:
The first principle of the derivative of f at any real number ‘x’ is f’(x) =
∎ The differentiation of f(x) is denoted by
means differentiation of ‘y’ with respect to ‘x’
The derivative of constant function is zero i.e., f’(a) = 0 where ‘a’ is any constant.
∎ Let I be an interval in R u and v are real valued functions on I and x ∈ I. Suppose that u and v are differentiable at ‘x’, then
• (u ± v) is also differentiable at ‘x’ and (u ± v)’(x) = u’ (x) ± v’(x).
• ‘uv’ is also differentiable at ‘x’ and (uv)’(x) = u(x) v’(x) + v(x) u’(x).
• αu + βv is also differentiable at ‘x’ and (αu + βv)’(x) = αu’(x) + βv’(x), α, β are constants.
• is also differentiable at ‘x’ and
∎ (f o g)’ (x) = f’(g(x)). g’(x).
### Formulae:
#### Parametric Differentiation:
If x = f(t) and y = g(t) then the procedure of finding in terms of the parameter ‘t’ is called parametric equations.
Implicitly differentiation:
An equation involving two or more variables is called an implicit equation.
ax2 + 2hxy + b y2 = 0 is an implicit equation in terms of x and y.
The process of finding from an implicit equation is called implicitly differentiation.
Derivative of one function w.r.t. another function:
The derivative of f(x) w.r.t g(x) is
Second order derivative:
Let y = f(x) be a function, if y is differentiable then the derivative of f is f’(x). If ‘(x) is again differentiable then the derivative of f’(x) is called second order derivative. And it is denoted by f” (x) or
## 10. ERRORS AND APPROXIMATIONS
Approximations:
Let y f(x) be a function defined an interval I and x ∈ I. If ∆x is any change in x, then ∆y be the corresponding change in y thus ∆y = f (x + ∆x) – f (x).
Let
where ϵ is very small
For ‘ϵ.∆x’ is very small and hence,
Approximate value is f (x + ∆x) = f(x) + f’(x). ∆x
Differential:
Let y f(x) be a function defined an interval I and x ∈ I. If ∆x is any change in x, then called differential of y = f(x) and it is denoted by df.
∴ dy = f’(x). ∆x
Errors:
Let y f(x) be a function defined an interval I and x ∈ I. If ∆x is any change in x, then ∆y be the corresponding change in y.
### The Following formulae will be used in Solving problems
CIRCLE:
If ‘r’ is radius, ‘d’ is diameter ‘P’ is the perimeter or circumference and A is area of the circle then
d= 2r, P = 2πr = πd and A = πr2sq.u
SECTOR:
If ‘r’ is the radius, ‘l’ is the length of arc and θ is of the sector then
Area = ½ l r = ½ r2θsq.u.
Perimeter = l + 2r = r (θ + 2) u.
CYLINDER:
Length of the Arc ‘l’ = rθ (θ must be in radians).
If ‘r is the radius of the base of cylinder and ‘h’ is the height of the cylinder, then
Area of base = πr2 sq.units.
Lateral surface area = 2πrh units.
Total surface area = 2πr (h + r) units.
Volume = πr2 h cubic units.
CONE:
If ‘r’ is the radius of base, ‘h’ is the height of cone and ‘l’ is slant height then
l 2 + r2 = h2
Lateral surface area = πrl units.
Total surface area = πr (l + r) sq. units.
Volume = cubic units.
SPHERE:
If ‘r’ is the radius of the Sphere then
Surface area = πr2 sq. units.
Volume = πr3 cubic units.
## 11. TANGENTS AND NORMALS
Tangent of a Curve:
If the secant line PQ approaches to the same position as Q moves along the curve and approaches to either side then limiting position is called a ‘Tangent line’ to the curve at P. The point P is called point of contact
Let y = f(x) be a curve, P a point on the curve. If Q(≠P) is another point on the curve then the line PD is called secant line.
Let y = f(x) be a curve and P (x, y) be a point on the curve. The slope of the tangent to the curve y = f(x) at P is called gradient of the curve.
Slope of the tangent to the curve y = f(x) at P (x, y) is m =
∎ The equation of the tangent at P (x1, y1) to the curve is y – y1 = m (x – x1) where m =
Normal of a curve:
Let y = f(x) be a curve and P (x, y) be a point on the curve. The line passing through P and perpendicular to the tangent of the curve y = f(x) at P is called Normal of the curve.
∎ The equation of the tangent at P (x1, y1) to the curve is y – y1 = -1/m (x – x1).
Slope of the normal is -1/m. where m =
Lengths of tangent, normal, subtangent and subnormal:
PT → Normal; QN → subnormal
PN → Tangent; QT → subtangent
∎ if m = then
Angle between two curves:
If two curves intersect at a point P., then the angle between the tangents of the curves at P is called the angle between the curves at P.
∎ If m1, m2 are the slopes of two tangents of the two curves and θ is the angle between the curves then
Tanθ =
Note:
• If m1= m2, then two corves are touch each other.
• if m1× m2 = –1, then two curves intersect orthogonally.
## 12. RATE MEASURE
Average rate of change:
if y = f(x) then the average rate of change in y between x = x1 and x = x2 is defined as
Instantaneous rate of change:
if y = f(x), then the instantaneous rate of change of a unction f at x = x0 is defined as
Rectilinear Motion:
A motion of a particle in a line is called Rectilinear motion. The rectilinear motion is denoted by s = f(t) where f(t) is the rule connecting ‘s’ and ‘t’.
Velocity, Acceleration:
A particle starts from a fixed point and moves a distance ‘S’ along a straight-line during time ‘t’ then
Velocity =
Acceleration =
Note:
(i) If v> 0, then the particle s moving away from the straight point.
(ii) If v < 0, then particle s moving away towards the straight point.
(iii) If v = 0, then the particle comes rest.
## 13.ROLLE’S & LANGRANGEE’S THEOREM
Rolle’s Theorem:
Suppose a, b (a < b) are two real numbers. Let f: [a, b] → R be a function satisfying the following conditions:
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b) and
(iii) f(a) = f(b)
then there exists at least one c ∈ (a, b) such that f’(c)= 0.
Lagrange’s Theorem:
Suppose a, b (a < b) are two real numbers. Let f: [a, b] → R be a function satisfying the following conditions:
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b) and
then there exists at least one c ∈ (a, b) such that f’(c)=
## 14.INCREASING & DECREASING FUNCTIONS
Let f be a real function on an interval I then f is said to be
(i) an increasing function on I if
x1 < x2 ⇒ f (x1) ≤ f (x2) ∀ x1, x2 ∈ I
(ii) decreasing function on I if
x1 < x2 ⇒ f (x1) ≥ f (x2) ∀ x1, x2 ∈ I
Let f be a real function on an interval I then f is said to be
(i) strictly increasing function on I if
x1 < x2 ⇒ f (x1) < f (x2) ∀ x1, x2 ∈ I
(ii) strictly decreasing function on I if
x1 < x2 ⇒ f (x1) > f (x2) ∀ x1, x2 ∈ I
Let f(x) be a real valued function defined on I = (a, b) or [a, b) or (a, b] or [a, b]. Suppose f is continuous on I and differentiable in (a, b). If
(i) f’ (c) > 0 ∀ c ∈ (a, b), then f is strictly increasing on I
(ii) f’ (c) < 0 ∀ c ∈ (a, b), then f is strictly decreasing on I
(iii) f’ (c) ≥ 0 ∀ c ∈ (a, b), then f is increasing on I
(iv) f’ (c) ≤ 0 ∀ c ∈ (a, b), then f is decreasing on I
Critical point:
A point x = c in the domain of the function said to be ‘critical point’ of the function f if either f’ (c) = 0 or f’ (c) does not exists.
Stationary point:
A point x = c in the domain of the function said to be ‘stationary point’ of the function f if f’ (c) = 0.
MAXIMA & MINIMA
Global maxima – Global minima:
Let D be an interval in R and f: D → R be a real function and c ∈ D. Then f is said to be
(i) a global maximum on D if f(c) ≥ f(x)
(ii) a global minimum on D if f(c) ≤ f(x)
Relative maximum:
Let D be an interval in R and f: D → R be a real function and c ∈ D. Then f is said to be relative maximum at c if there exist δ > 0 such that f(c) ≥ f(x) ∀ x ∈ (c – δ, c + δ).
Here, f (c) is called relative maximum value of f(x) at x = c and the point x = c is called point of relative maximum.
Relative minimum:
Let D be an interval in R and f: D → R be a real function and c ∈ D. Then f is said to be relative maximum at c if there exist δ > 0 such that f(c) ≤ f(x) ∀ x ∈ (c – δ, c + δ).
Here, f (c) is called relative maximum value of f(x) at x = c and the point x = c is called point of relative minimum.
The relative maximum and minimum value of f are called extreme values.
If f is either minima or maxima f’ (α) = 0.
Let f be a continuous function om [a, b] and α ∈ (a, b)
(i) If f’ (α) = 0 and f’’ (α) >0, then f(α) is relative minimum.
(ii) if f’ (α) = 0 and f’’ (α) <0, then f(α) is relative maximum
My App:
## 1.Functions
Set: A collection of well-defined objects is called a set.
Ordered pair: Two elements a and b listed in a specific order form. An ordered pair denoted by (a, b).
Cartesian product: Let A and B are two non-empty sets. The Cartesian product of A and B is denoted by A × B and is defined as a set of all ordered pairs (a, b) where a ϵ A and b ϵB
Relation: Let A and B are two non-empty sets the relation R from A to B is a subset of A×B.
⇒ R: A→B is a relation if R⊂ A × B
#### Function:
A relation f: A → B is said to be a function if ∀ aϵ A there exists a unique element b such that (a, b) ϵ f. (Or)
A relation f: A → B is said to be a function if
(i) x ϵ A ⇒ f(x) ϵ B
(ii) x1 , x2 ϵ A , x1 = x2 in A ⇒ f(x1) = f(x2) in B.
Note: If A, B are two finite sets then the no. of functions that can be defined from A to B is n(B)n(A)
VARIOUS TYPES OF FUNCTIONS
One– one Function (Injective):- A function f: A→ B is said to be a one-one function or injective if different elements in A have different images in B.
(Or)
A function f: A→ B is said to be one-one function if f(x1) = f(x2) in B ⇒ x1 = x2 in A.
Note: No. of one-one functions that can be defined from A into B is n(B) p n(A) if n(A) ≤ n(B)
On to Function (Surjection): – A function f: A→ B is said to be onto function or surjection if for each yϵ B ∃ x ϵ A such that f(x) =y
Note: if n(A) = m and n(B) = 2 then no. of onto functions = 2m – 2
Bijection: – A function f: A→ B is said to be Bijection if it is both ‘one-one and ‘onto’.
Constant function: A function f: A→ B is said to be constant function if f(x) = k ∀ xϵA
Identity function: Let A be a non-empty set, then the function defined by I: A → A, I(x)=x is called identity function on A.
Equal function: Two functions f and g are said to be equal if
(i) They have same domain (D)
(ii) f(x) = g(x) ∀ xϵ D
Even function: A function f: A→ B is said to be even function if f (- x) = f(x) ∀ xϵ A
Odd function: A function f: A→ B is said to be odd function if f (- x) = – f(x) ∀ xϵ A
Composite function: If f: A→B, g: B→C are two functions then the composite relation is a function from A to C.
gof: A→C is a composite function and is defined by gof(x) = g(f(x)).
Step function: A number x = I + F
I → integral part = [x]
F → fractional part = {x}
∴ x = [x] + {x}
If y = [x] then domain = R and
Range = Z
0 ≤ x ≤ 1, [x] = 0
1≤ x ≤ 2, [x] = 1
-1 ≤ x ≤ 0, [x] = -1
If k is any integer [ x + k] = k + [x]
The value of [x] is lies in x – 1 < [x] ≤ 1.
Inverse function: If f: A → B is bijection then f -1 is exists
f-1: B → A is an inverse function of f.
### SOME IMPORTANT POINTS
of subsets of a set of n elements is 2n
of proper subsets of a set of n elements is 2n – 1
Let A and B are two non-empty finite sets and f: A → B is a function. This function will
One-one if n(A) ≤ n(B)
On to if n(A) ≥ n(B)
Bijection if n(A) = n(B).
## 3. MATRICES
Matrix: An ordered rectangular array of elements is called a matrix
• Matrices are generally enclosed by brackets like
• Matrices are denoted by capital letters A, B, C and so on
• Elements in a matrix are real or complex numbers; real or complex real-valued functions.
Oder of Matrix: A matrix having rows and ‘n’ columns is said to be of order m x n. Read as m by n.
### Square Matrix: A matrix in which the no. of rows is equal to the no. of columns is called a square matrix.
Principal diagonal ( diagonal) Matrix: If A = [aij] is a square matrix of order ‘n’ the elements a11 , a22 , a33 , ………. ann is said to constitute its principal diagonal.
Trace Matrix: The sum of the elements of the principal diagonal of a square matrix A is called the trace of the matrix. It is denoted by Tr (A).
Ex:-
Diagonal Matrix: If each non-diagonal element of a square matrix is ‘zero’ then the matrix is called a diagonal matrix.
Scalar Matrix: If each non-diagonal elements of a square matrix are ‘zero’ and all diagonal elements are equal to each other, then it is called a scalar matrix.
Identity Matrix or Unit Matrix: If each of the non-diagonal elements of a square matrix is ‘zero’ and all diagonal elements are equal to ‘1’, then that matrix is called a unit matrix.
Null Matrix or Zero Matrix: If each element of a matrix is zero, then it is called a null matrix.
Row matrix & column Matrix: A matrix with only one row s called a row matrix and a matrix with only one column is called a column matrix.
Triangular matrices:
A square matrix A = [aij] is said to be upper triangular if aij = 0 ∀ i > j
A square matrix A = [aij] is said to be lower triangular matrix aij = 0 ∀ i < j
Equality of matrices: matrices A and B are said to be equal if A and B of the same order and the corresponding elements of A and B are equal.
### Product of Matrices:
Let A = [aik]mxn and B = [bkj]nxp be two matrices ,then the matrix C = [cij]mxp where
Note: Matrix multiplication of two matrices is possible when no. of columns of the first matrix is equal to no. of rows of the second matrix.
Transpose of Matrix: If A = [aij] is an m x n matrix, then the matrix obtained by interchanging the rows and columns is called the transpose of A. It is denoted by AI or AT.
Note: (i) (AI)I = A (ii) (k AI) = k . AI (iii) (A + B )T = AT + BT (iv) (AB)T = BTAT
Symmetric Matrix: A square matrix A is said to be symmetric if AT =A
If A is a symmetric matrix, then A + AT is symmetric.
Skew-Symmetric Matrix: A square matrix A is said to be skew-symmetric if AT = -A
If A is a skew-symmetric matrix, then A – AT is skew-symmetric
Minor of an element: Consider a square matrix
the minor an element in this matrix is defined as the determinant of the 2×2 matrix obtained after deleting the rows and the columns in which the element is present.
Cofactor of an element: The cofactor of an element in i th row and j th column of A3×3 matrix is defined as it’s minor multiplied by (- 1 ) i+j .
### Properties of determinants:
• If each element of a row (column) of a square matrix is zero, then the determinant of that matrix is zero.
• If A is a square matrix of order 3 and k is scalar then.
• If two rows (columns) of a square matrix are identical (same), then Det. Of that matrix is zero.
• If each element in a row (column) of a square matrix is the sum of two numbers then its determinant can be expressed as the sum of the determinants.
• If each element of a square matrix are polynomials in x and its determinant is zero when x = a, then (x-a) is a factor of that matrix.
• For any square matrix A Det(A) = Det (AI).
• Det(AB) = Det(A) . Det(B).
• For any positive integer n Det(An) = (DetA)n.
Singular and non-singular matrices: A Square matrix is said to be singular if its determinant is zero, otherwise it is said to be the non-singular matrix.
Ad joint of a matrix: The transpose of the matrix formed by replacing the elements of a square matrix A with the corresponding cofactors is called the adjoint of A.
Invertible matrix: Let A be a square matrix, we say that A is invertible if there exists a matrix B such that AB =BA = I, where I is the unit matrix of the same order as A and B.
Augmented matrix: The coefficient matrix (A) augmented with the constant column matrix (D) is called the augmented matrix. It is denoted by [AD].
Sub matrix: A matrix obtained by deleting some rows and columns (or both) of a matrix is called the submatrix of the given matrix.
Let A be a non-zero matrix. The rank of A is defined as the maximum of the order of the non-singular submatrices of A.
• Note: If A is a non-zero matrix of order 3 then the rank of A is:
• 1, if every 2×2 submatrix is singular
• 2, if A is singular and at least one of its 2×2 sub-matrices is non-singular
(iii) 3, if A is non – singular.
Consistent and Inconsistent: The system of linear equations is consistent if it has a solution, in-consistent if it has no solution.
• Note: The system of three equations in three unknowns AX = D has
• A unique solution if rank(A) = rank ([AD]) = 3
• Infinitely many solutions if rank (A) = ([AD]) < 3
• No solution if rank (A) ≠ rank ([AD])
### Solutions of a homogeneous system of linear equations:
The system of equations AX = 0 has
• The trivial solution only if rank(A) = 3
• An infinite no. of solutions if rank(A) < 3
Directed line: If A and B are two distinct points in the space, the ordered pair (A, B) denoted by AB is called a directed line segment with initial point A and terminal point B.
⇒ A directed line passes through three characteristics: (i) length (ii) support (iii) direction
Scalar: A quantity having magnitude only is called a scalar. We identify real numbers as a scalar.
Ex: – mass, length, temperature, etc.
Vector: A quantity having length and direction is called a vector.
Ex: – velocity, acceleration, force, etc.
⇒ If is a vector then its length is denoted by
Position of vector: If P (x, y, z) is any point in the space, then is called the position vector of the point P with respect to origin (O). This is denoted by
Like and unlike vectors: If two vectors are parallel and having the same direction then they are called like vectors.
If two vectors are parallel and having opposite direction then they are called, unlike vectors.
Coplanar vectors:
Vectors whose supports are in the same plane or parallel to the same plane are called coplanar vectors.
Triangle law: If are two vectors, there exist three points A, B, and C in a space such that defined by
Parallelogram law: If two vectors and represented by two adjacent sides of a parallelogram in magnitude and direction then their sum is represented in magnitude and direction by the diagonal of the parallelogram through their common point.
Scalar multiplication: Let be a vector and λ be a scalar then we define vector λ to be the vector if either is zero vector or λ is the scalar zero; otherwise λ is the vector in the direction of with the magnitude if λ>0 and λ = (−λ)(− ) if λ<0.
The angle between two non-zero vectors: Let be two non-zero vectors, let then ∠AOB has two values. The value of ∠AOB, which does not exceed 1800 is called the angle between the vectors and , it is denoted by ( ).
Section formula: Let be two position vectors of the points A and B with respect to the origin if a point P divides the line segment AB in the ratio m:n then
Linear combination of vectors: let be vectors x1, x2, x3…. xn be scalars, then the vector is called the linear combination of vectors.
Components: Consider the ordered triad (a, b, c) of non-coplanar vectors If r is any vector then there exist a unique triad (x, y, z) of scalars such that . These scalars x, y, z are called the components of with respect to the ordered triad (a, b, c).
• i, j, k are unit vectors along the X, Y and Z axes respectively and P(x, y, z) is any point in the space then = r = x i + y j +z k and
Regular polygon: A polygon is said to be regular if all the sides, as well as all the interior angles, are equal.
• If a polygon has sides then the no. of diagonals of a polygon is
• The unit vector bisecting the angle between is
### Vector equation of a line and plane
⇒The vector equation of the line passing through point A () and ∥el to the vector is
Proof:-
Then AP, are collinear vector proof: let P ( ) be any point on the line a
the equation of the line passing through origin and parallel to the vectoris
• the vector equation of the line passing through the points A( ) and B( ) is
• Cartesian equation of the line passing through A ( x1, y1, z1) and B ( x2, y2, z2) is
• The vector equation of the plane passing through point A( ) and parallel to the vectors and is
• The vector equation of the plane passing through the point A( ), B( ) and parallel to the vector is
• The vector equation of the plane passing through the points A( ), B( ) and C( ) is
$large&space;bar{r}=&space;(1-t)bar{a}&space;+&space;t&space;bar{b}$
## 5.PRODUCT OF VECTORS
Dot product (Scalar product): Let are two vectors. The dot product or direct product of and is denoted byand is defined as
• If = 0, = 0 ⟹ = 0.
• If ≠0, ≠ 0 then
• The dot product of two vectors is a scalar
• If are two vectors, then
• If θ is the angle between the vectors then.
⟹
⟹ If > 0, then θ is an acute angle
⟹ If < 0, then θ is obtuse angle 0
⟹ If = 0, then is perpendicular to
• If is any vector then
Component and Orthogonal Projection:
Let=,= be two non-zero vectors. Let the plane passing through B ( ) and perpendicular to intersects
In M, then is called the component of on
• The component (projection) vector of on is
• Length of the projection (component) =
• Component of perpendicular to =
If ,, form a right-handed system of an orthonormal triad, then
• If then = a1b1 + a2b2 + a3b3
• If then
Parallelogram law:
In a parallelogram, the sum of the squares of the lengths of the diagonals is equal to the sum of the squares of the lengths of its sides.
In ∆ABC, the length of the median through vertex A is
Vector equation of a plane:
The vector equation of the plane whose perpendicular distance from the origin is p and unit normal drawn from the origin towards the plane is,
•The vector equation of the plane passing through point A ( ) and perpendicular to the is
•If θ is the angle between the planes then
Cross product (vector product): Let and be two non-zero collinear vectors. The cross product of and is denoted by × (read as a cross ) and is defined as
• The vector × is perpendicular to both and and also perpendicular to the plane containing them
• The unit vector perpendicular to both and is
• Let then
• If and are two sides of a triangle then the area of the triangle =
• If A ( ), B ()and C ( )are the vertices of a ∆ABC, then its area
• The area of the parallelogram whose adjacent sides and is
• The area of the parallelogram whose diagonals and is
• If A ( ), B ( )and C ( )are three points then the perpendicular distance from A to the line passing through B, C is
Let,and be three vectors, then () . is called the scalar triple product of,andand it is denoted by
Ifthen
•In determinant rows(columns) are equal then the det. Value is zero.
•In a determinant, if we interchange any two rows or columns, then the sign of det. Is change.
•Four distinct points A, B, C, and D are said to be coplanar iff
The volume of parallelepiped:
If ,andare edges of a parallelepiped then its volume is
The volume of parallelepiped:
The volume of Tetrahedron with, and are coterminous edges is
The volume of Tetrahedron whose vertices are A, B, C and D is
Vector equation of a plane:
The vector equation of the plane passing through point A () and parallel to the vectors and is
The vector equation of the plane passing through the points A ( ) and B( ) and parallel to the vector is
The vector equation of the plane passing through the points A (), B( ) and C( ) is
Skew lines:
The lines which are neither intersecting nor parallel are called Skew lines
The shortest distance between the Skew lines:
If are two skew lines, then the shortest distance between them is
If A, B, C and D are four points, then the shortest distance between the line joining the points AB and CD is
•The plane passing through the intersection of the planes is
the perpendicular distance from point A (a ̅) to the plane is
Let ,and be three vectors, then is called the vector triple product of, and.
Scalar product of four vectors:
Vector product of four vectors:
## 6. TRIGONOMETRY UPTO TRANSFORMATIONS
The word ’trigonometry’ derived from the Greek words ‘trigonon’ and ‘metron’. The word ‘trigonon’ means a triangle and the word ‘metron’ means a measure.
Angle: An angle is a union of two rays having a common endpoint in a plane.
There are three systems of measurement of the angles.
• Sexagesimal system (British system)
• Centesimal system (French system)
Sexagesimal system: – In this system, a circle can be divided into 360 equal parts. Each part is called one degree (0). One circle = 3600
Further, each degree can be divided into 60 equal parts. Each part is called one minute (‘).
and each minute can be divided into 60 equal parts. Each part is called one second (“)
Sexagesimal system: – In this system, a circle can be divided into 400 equal parts. Each part is called one grade (g). One circle = 400g
Further, each grade can be divided into 100 equal parts. Each part is called one minute (‘).
and each minute can be divided into 100 equal parts. Each part is called one second (“)
Circular measure: Radian is defined as the amount of the angle subtended by an arc of length ’r’ of a circle of radius ‘r’.
One radian is denoted by 1c. One circle = 2πc
Relation between the three measures:
3600 = 400g = 2 πc
1800 = 200g = πc
Trigonometric Ratios:
Trigonometric identities: –
∗ sin2θ + cos2θ = 1
1 – cos2θ = sin2θ
1 – sin2θ = cos2θ
∗ sec2θ − tan2θ = 1
sec2θ = 1 + tan2θ
tan2θ = sec2θ – 1
(secθ − tanθ) (secθ + tanθ) = 1
∗ cosec2θ − cot2θ = 1
co sec2θ = 1 + cot2θ
cot2θ = cosec2θ – 1
(cosec θ – cot θ) (cosec θ + cot θ) = 1
• sin θ. cosec θ = 1
sec θ. cos θ = 1
tan θ. cot θ = 1
All Silver Tea Cups Rule:
Note: If 900 ±θ or 2700 ±θ then
‘sin’ changes to ‘cos’; ‘tan’ changes to ‘cot’; ‘sec’ changes to ‘cosec’
‘cos’ changes to ‘sin’; ‘cot’ changes to ‘tan’; ‘cosec’ changes to ‘sec’.
If 1800 ±θ or 3600 ±θ then, no change in ratios.
Values of Trigonometric Ratios:
Complementary angles: Two angles A and B are said to be complementary angles, if A + B = 900.
supplementary angles: Two angles A and B are said to be supplementary angles, if A + B = 1800.
Let E ⊆ R and f: E → R be a function, then f is called periodic function if there exists a positive real number ‘p’ such that
• (x + p) ∈ E ∀ x∈ E
• F (x+ p) = f(x) ∀ x∈ E
If such a positive real number ‘p’ exists, then it is called a period of f.
The algebraic sum of two or more angles is called a ‘compound angle’.
For any two real numbers A and B
sin (A + B) = sin A cos B + cos A Cos B
sin (A − B) = sin A cos B − cos A Cos B
cos (A + B) = cos A cos B − sin A sin B
cos (A − B) = cos A cos B + sin A sin B
tan (A + B) =
tan (A − B) =
cot (A + B) =
⋇ cot (A − B) =
sin (A + B + C) = ∑sin A cos B cos C − sin A sin B sin C
cos (A + B + C) = cos A cos B cos C− ∑cos A sin B sin C
tan (A + B + C) =
⋇ cot (A + B + C) =
⋇ sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A
⋇ cos (A + B) cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A
Extreme values of trigonometric functions:
If a, b, c ∈ R such that a2 + b2 ≠ 0, then
Maximum value =
Minimum value =
If A is an angle, then its integral multiples 2A, 3A, 4A, … are called ‘multiple angles ‘of A and the multiple of A by fraction likeare called ‘submultiple angles.
⋇ sin 2A = 2 sin A cos A =
⋇ cos 2A = cos2 A – sin2 A
= 2 cos2 A – 1
= 1 – 2sin2 A
=
⋇ tan 2A =
⋇ cot 2A =
∎ If is not an add multiple of
⋇ sin A = 2 sin cos =
⋇ cos A = cos2 – sin2
= 2 cos2 – 1
= 1 – 2sin2
=
⋇ tan A =
⋇ cot A =
⋇ sin3A = 3 sin A −4 sin3 A
⋇ cos 3A = 4 cos3 A – 3 cos A
⋇ tan 3A =
⋇ cot 3A =
⋇ tan A + cot A = 2 cosec 2A
⋇ cot A – tan A = 2 cot 2A
For A, B∈ R
⋇ sin (A + B) + sin (A – B) = 2sin A cos B
⋇ sin (A + B) −sin (A – B) = 2cos A sin B
⋇ cos (A + B) + cos (A – B) = 2 cos A cos B
⋇ cos (A + B) − cos (A – B) = − 2sin A sin B
For any two real numbers C and D
⋇ sin C + sin D = 2sin cos
⋇ sin C −sin D= 2cos sin
⋇ cos C + cos D = 2 coscos
⋇ cos C − cos D = − 2sin sin
If A + B + C = π or 1800, then
⋇ sin (A + B) = sin C; sin (B + C) = sin A; sin (A + C) = sin B
⋇ cos (A + B) = − cos C; cos (B + C) = −cos A; cos (A + C) = − cos B
If A + B + C = 900 or then
⋇ sin = cos ; sin = cos ; sin = cos
⋇ cos = sin ; cos = sin ; cos = sin
If then
⋇ sin (A + B) = cos C; sin (B + C) = cos A; sin (A + C) = cos B
⋇ cos (A + B) = sin C; cos (B + C) = sin A; cos (A + C) = sin B
## 7. TRIGONOMETRIC EQUATIONS
Trigonometric equation: An equation consisting of the trigonometric functions of a variable angle θ ∈ R is called a ‘trigonometric equation’.
The solution of the equation: The values of the variable angle θ, satisfying the given trigonometric equation is called a ‘solution’ of the equation.
The set of all solutions of the trigonometric equation is called the solution set’ of the equation. A ‘general solution’ is an expression of the form θ0 + f(n) where θ0 is a particular solution and f(n) is a function of n ∈ Z involving π.
If k ∈ [− 1, 1] then the principle solution of θ of sin x = k lies in
General solution of sin x = sin θ is x = nπ + (−1) n θ, n ∈ Z
If k ∈ [− 1, 1] then the principle solution of θ of cos x = k lies in
General solution of cos x = cos θ is x = 2nπ ± θ, n ∈ Z
If k ∈R then the principle solution of θ of tan x = k lies in
General solution of tan x = tan θ is x = nπ + θ n ∈ Z
If sin θ = 0, then the general solution is θ = nπ, n ∈ Z
If tan θ = 0, then the general solution is θ = nπ, n ∈ Z
If cos θ = 0, then the general solution is θ = (2n + 1) , n ∈ Z
If sin2 θ = sin2 𝛂, cos2 θ = cos2 𝛂 or tan2 θ = tann2 𝛂 then the general solution is 𝛉 = nπ ± θ, n ∈ Z
## 8.INVERSE TRIGONOMETRIC FUNCTIONS
If A, B are two sets and f: A→ B is a bijection, then f-1 is existing and f-1: B → A is an inverse function.
The function Sin-1: [−1, 1] → is defined by Sin-1 x = θ ⇔ θ∈ and sin θ = x
The function Cos-1: [−1, 1] → [0, π] is defined by Sin-1 x = θ ⇔ θ∈ [0, π] and cos θ = x
The function Tan-1: R → is defined by Tan-1 x = θ ⇔ θ∈ and tan θ = x
The function Sec-1: [−∞, −1] ∪ [1, ∞] → is defined by Sin-1 x = θ ⇔ θ∈ and sec θ= x
The function Cosec-1: [−∞, −1] ∪ [1, ∞] → is defined by cosec-1 x = θ ⇔ θ∈ and Cosec θ= x
The function Cot-1: R → (0, π) is defined by Cot-1 x = θ ⇔ θ ∈ (0, π) and cot θ = x
Properties of Inverse Trigonometric functions:
Sin-1 x = Cosec-1(1/x) ∀ x ∈ [−1, 1] – {0}
Cos-1x = Sec-1(1/x) ∀ x ∈ [−1, 1] – {0}
Tan-1 x = Cot-1(1/x), if x > 0
Tan-1 x = Cot-1(1/x) −π, if x < 0
Sin-1 (−x) = − Sin-1(x) ∀ x ∈ [−1, 1]
Cos-1 (−x) = π − Cos-1(x) ∀ x ∈ [−1, 1]
Tan-1 (−x) = − Tan-1(x) ∀ x ∈ R
Cosec-1 (−x) = − Cose-1(x) ∀ x ∈ (− ∞, − 1] ∪ [1, ∞)
Sec-1 (−x) = π − Sec-1(x) ∀ x ∈ (− ∞, − 1] ∪ [1, ∞)
Cot-1 (−x) =π − Cot-1(x) ∀ x ∈ R
(i) If θ∈, then Sin−1(sin θ) = θ and if x ∈ [−1, 1], then sin (Sin−1x) = x
(ii) If θ∈ [0, π], then Cos−1(cos θ) = θ and if x ∈ [−1, 1], then cos (Cos−1x) = x
(iii) If θ∈ , then tan−1(tann θ) = θ and if x ∈ R, then tan (Tan−1x) = x
(iv) If θ∈ (0, π), then Cot−1(cot θ) = θ and if x ∈ R, then cot (Cot−1x) = x
(v) If θ∈ [0, ) ∪ ( , π], then Sec−1(sec θ) = θ and
if x ∈ (− ∞, − 1] ∪ [1, ∞), then sec (Sec−1x) = x
(vi) If θ∈ , then Cosec−1(cosec θ) = θ and
if x ∈ (− ∞, − 1] ∪ [1, ∞), then cosec (Cosec−1x) = x
(i) If θ∈ , then Cos−1(sin θ) =
(ii) If θ∈ [0, π], then Sin−1(cos θ) =
(iii) If θ∈ , then Cot−1(tan θ) =
(iv) If θ∈ (0, π), then Tan−1(cot θ) =
(v) If θ∈ , then Cosec−1(sec θ) =
(vi) If θ∈ , then Sec−1(cosec θ) =
1. Sin1x = Cos( )if 0 ≤ x ≤ 1 and Sin1x =− Cos1 ( ) if −1 ≤ x ≤ 0
2. Sin1x = Tan1 if x ∈ (−1, 1)
3. Cos1x = Sin1 () if x ∈ [0, 1] and Cos1x = π − Sin1 () if x ∈ [−1, 0]
1. Tan1x = Sin1 = Cos−1 or x > 0
Cos−1 x + Sin−1x = ∀ x ∈ [−1, 1]
Tan−1 x + Cot−1x = ∀ x ∈ R
Sec−1 x + Cosec−1x = ∀ x ∈ (−∞, −1] ∪ [1, ∞)
Sin−1 x + Sin−1y = Sin−1(x + y ) if 0 ≤x ≤ 1, 0 ≤y ≤ 1and x2 + y2 ≤ 1
=π− Sin−1(x + y ) if 0 ≤x ≤ 1, 0 ≤y ≤ 1and x2 + y2 > 1
Cos−1 x + Cos−1y = Cos−1(x y − ) if 0 ≤x, y ≤ 1and x2 + y2 ≥ 1
=π− Cos−1(x y − ) if 0 ≤x ≤ 1, 0 ≤y ≤ 1and x2 + y2 < 1
Tan−1 x + Tan−1y = Tan−1 if x > 0, y> 0 and xy < 1
=π + Tan−1 if x > 0, y> 0 and xy > 1
= Tan−1 if x < 0, y< 0 and xy > 1
= −π + Tan−1 if x < 0, y< 0 and xy < 1
Tan−1 x − Tan−1y = Tan−1 if x > 0, y> 0 or x < 0, y< 0
2 Sin−1 x = Sin−1 (2x ) if x≤
= π− Sin−1 (2x ) if x >
2 Cos−1 x = Cos−1(2x2 – 1) if x ≥
=Cos−1(1–2x2) if x <
2 Tan−1 x = Tan−1 if < 1
= π + Tan−1 if ≥ 1
= Sin−1 if x ≥ 0
= Cos−1 if x ≥ 0
3Sin−1x = Sin−1(3x – 4x3)
3Cos−1x = Cos−1(4x3 – 3x)
3Tan−1x = tan−1
## 9.HYPERBOLIC FUNCTIONS
The function f: R→R defined by f(x) = ∀ x ∈ R is called the ‘hyperbolic sin’ function. It is denoted by sinh x.
∴ sinh x =
Similarly,
cosh x = ∀ x ∈ R
tanh x = ∀ x ∈ R
coth x = ∀ x ∈ R
sech x = ∀ x ∈ R
cosech x = ∀ x ∈ R
Identities:
cosh2x – sinh2 x = 1
cosh2x = 1 + sinh2 x
sinh2 x = cosh2 x – 1
sech2 x = 1 – tanh2 x
tanh2 x = 1 – sech2 x
cosech2 x = coth2 x – 1
coth2 x = 1 + coth2 x
sinh (x + y) = sinh x cosh y + cosh x sinh y
sinh (x − y) = sinh x cosh y − cosh x sinh y
cosh (x + y) = cosh x cosh y + sinh x sinh y
cosh (x − y) = cosh x cosh y − sinh x sinh y
tanh (x + y) =
tanh (x − y) =
coth (x + y) =
sinh 2x = 2 sinh x cosh 2x =
cosh 2x = cosh2x + sinh2 x = 2 cosh2x – 1 = 1 + 2 sinh2x =
tanh 2x =
sinh 3x = 3 sinh x + 4 sinh3x
cosh 3x = 4 cosh3 x – 3 cosh x
tanh 3x =
Inverse hyperbolic functions:
Sinh−1x = ∀ x ∈ R
Cosh−1x = ∀ x ∈ (1, ∞)
Tanh−1x = ∀ < 1
Coth−1x = ∀ > 1
Sech−1x = ∀ x ∈ (0, 1]
Cosech−1x = if x < 0 and x ∈ (−∞, 0)
= if x > 0
## 10. PROPERTIES OF TRIANGLES
In ∆ABC,
Lengths AB = c; BC = a; AC =b
Area of the tringle is denoted by ∆.
Perimeter of the triangle = 2s = a + b + c
A = ∠CAB; B = ∠ABC; C = ∠BCA.
Sine rule:
In ∆ABC,
⟹ a = 2R sin A; b = 2R sinB; c = 2R sin C
Where R is the circumradius and a, b, c, are lengths of the sides of ∆ABC.
Cosine rule:
In ∆ABC,
a2 = b2 + c2 – 2bc cos A ⟹cos A =
b2 = a2 + c2 – 2ac cos B ⟹ cos B =
c2 = a2 + b2 – 2ab cos C ⟹ cos A =
projection rule:
In ∆ABC,
a = b cos C + c cos B
b = a cos C + c cos A
c = a cos B + b cos A
Tangent rule (Napier’s analogy):
In ∆ABC,
Half angle formulae and Area of the triangle:
In ∆ABC, a, b, and c are sides
and area of the triangle
1.Half angle formulae: –
2.Formulae for ∆: –
∆ = ½ ab sinC= ½ bc sin A=½ ac sin B
where
= 2R2sin A sin B sinC
= r.s
=
=
In circle and Excircles of a triangle:
⋇The circle that touches the three sides of an ∆ABC internally is called ‘incircle’. The centre of the incircle is ‘I’ and the radius is ‘r’.
Formulae for ‘r’: –
r =
= (s – a) tan = (s – b) tan = (s – c) tan
= 4R sinsin sin
=
The circle that touches the side BC internally and the other two sides AB and AC externally is called the ‘Excircle’ opposite to the angle A. Its centre is I1 and the radius is r1. A triangle has three ex circles. The remaining circles centre and radius are respectively I2, r2 and I3, r3.
Formulae for ‘r1’: –
r1 =
= s tan
= (s – b) cot = (s – c) cot
= 4R sin cos cos
=
Formulae for ‘r2’: –
r2 =
= s tan
= (s – c) cot = (s – a) cot
= 4R cos sin cos
=
Formulae for ‘r3’: –
r3 =
= s tan
= (s – a) cot = (s – b) cot
= 4R cos cos sin
=
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# How do you find the range and interquartile range for 3, 7, 4, 6, and 5?
Dec 5, 2016
The range is 4 and the interquartile range (IQR) is 3.
#### Explanation:
To find the range, simply arranged the numbers from least to greatest and subtract the smallest number from the largest number.
$3 , 4 , 5 , 6 , 7$
$7 - 3 = 4$
Therefore, the range is 4.
To find the IQR, the numbers must be organized into quartiles, and the first quartile must be subtracted from the third.
In this case:
3 is the minimum (0%)
3.5 is the first quartile (25%)
5 is the second quartile (50%)
6.5 is the third quartile (75%)
and 7 is the maximum (100%)
Now, since 6.5 is the third quartile and 6.5 is the first quartile, our expression to find the IQR is 6.5-3.5, which equals 3.
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# What Is a Adjacency Matrix in Data Structure?
//
Larry Thompson
A Adjacency Matrix is a fundamental data structure used in graph theory to represent the connections between vertices in a graph. It provides a concise way to store and retrieve information about the relationships between different elements in a graph. In this article, we will delve into the details of what an adjacency matrix is and how it is used.
## What is a Graph?
Before we dive into adjacency matrices, let’s first understand what a graph is. In computer science, a graph is a collection of nodes (also known as vertices) that are connected by edges.
These nodes can represent various entities, such as cities in a transportation network or web pages on the internet. The edges represent the relationships or connections between these nodes.
## Understanding Adjacency Matrices
An adjacency matrix is a square matrix that represents the connections between vertices in a graph. It provides an efficient way to store information about which vertices are adjacent to each other. The rows and columns of the matrix correspond to the vertices of the graph, and each cell in the matrix indicates whether there exists an edge between two vertices.
Let’s consider an example to illustrate how adjacency matrices work. Suppose we have a simple undirected graph with 4 vertices labeled A, B, C, and D. The adjacency matrix for this graph would be:
• A B C D
• A 0 1 1 0
• B 1 0 1 1
• C 1 1 0 0
• D 0 1 0 0
In this example, there is an edge between vertex A and vertex B (indicated by the “1” in cell [1,2]), as well as between vertex B and vertex C (indicated by the “1” in cell [2,3]). The absence of an edge is represented by a “0” in the corresponding cell.
Adjacency matrices offer several advantages when it comes to representing graphs. Firstly, they provide a compact way to store graph information, especially when the graph is sparse (i.e., has relatively few edges compared to the total number of possible edges). This is because only the cells corresponding to existing edges need to be stored.
Secondly, adjacency matrices allow for efficient retrieval of information about graph connectivity. Checking whether two vertices are adjacent can be done in constant time by simply accessing the corresponding cell in the matrix.
While adjacency matrices have their benefits, they also come with some drawbacks. One major disadvantage is their space complexity.
For large graphs with many vertices, the size of the matrix can become prohibitively large. This can lead to significant memory consumption and reduced performance.
Additionally, updating an adjacency matrix can be computationally expensive. Adding or removing an edge requires modifying multiple cells in the matrix, which takes time proportional to the number of vertices in the graph.
## Conclusion
In summary, an adjacency matrix is a powerful data structure for representing graph connectivity. It provides a concise and efficient way to store information about relationships between vertices in a graph.
Despite its limitations in terms of space complexity and update operations, adjacency matrices remain widely used due to their simplicity and effectiveness.
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# Q.9 Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take $$\pi =\frac{22}{7}$$)
Length of the wire to be bent into a circle = 44 cm
2$$\pi$$r = 44 cm
Or, 2 ×$$\frac{22}{7}$$ × r = 44 cm
Or, r = $$\frac{44×7}{22×2}$$
Area of circle $$\pi r^2 = \frac{22}{7} × 7×7 = 154 cm^2$$
Now, the length of the wire is bent into a square.
Here perimeter of square= 44cm
Length of each side of the square
= Perimeter/ 4=44/4=11cm
Area of the square = $$(Side)^2 = (11)^2 = 121 cm^2$$
Since, $$154 cm^2 >121 cm^2$$
Thus, the circle encloses more area.
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# Linear Equation Definition
Document Sample
``` Linear Equation Definition
Linear Equation Definition
Algebraic equations are the part of mathematical concept to solve the queries of
students to find out the value of unknown variables. The algebraic equations can
be defined as a collection of numbers and variables.
Numbers that are used in the expression are known as constant. In the
mathematics variable is an important part of the algebraic equation.
In the algebraic equations here we illustrate the concept of linear equation for the
better understanding about the algebraic equations.
As we know that Linear Equations are the part of algebraic equations. So, linear
equations can contain one or more variables. Linear equations have a wide area of
application with better regularity in applied mathematics.
Linear Equations are the expressions that have plain old variables. In the more
precise form linear equations can be defined as mathematical expressions that
have an equal sign and linear equations.
Know More About :- Complex Numbers Rules
Tutorcircle.com Page No. : 1/4
In the linear equations for finding any unknown value we use the variable instead
of particular values. There are some points that we have to remember at the time
of variable declaration in the linear equation.
a) In the linear equation variables should not contain any exponent.
b) In the linear equation variables should not perform the operation with each
other like multiplication and division.
c) In the linear equation variables should not be initialized with the square root
symbol.
At the time of solving linear equations we need to focus on the above points that
are related to variables. Now we are going to show you the process of solving the
linear equations.
At the time of solving linear equations we need to find out the value of unknown
variables.
In this process we perform the reverse process. In the mathematical term solving
the linear equation process, we need to isolate the unknown variables into the
expressions. Let’s show you below in the example:
Example: solve for z: z– 5 = 14
solution: In the above expression we can see that there is a linear equation which
has the unknown variable z and constant value 5 and 14. In this we can see that
variable do not contain any exponent, any square root and multiplication and
division operation.
Now we show you how to isolate the unknown variables and how to solve them.
Tutorcircle.com Page No. : 2/4
In the solving process we need to focus the constant value with the variable. In
above we can see that with the z variable there is constant -5.
In the solving process we need to isolate the variable and that can be performed
by adding 5 with them. This process makes a wrong impact on solving process. So
to overcome form this problem we also add the 5 with the other constant 14. Let's
show below:
given that z– 5 = 14
z – 5 + 5 = 14 ( adding 5 to isolate the unknown variable)
z- 5 + 5 = 14 + 5 ( to overcome the error we add 5 with 14)
In above we can see that this will isolate the unknown variable at the left hand
side of equal sign. Now we are doing the further process to solving the equation.
z- 5 + 5 = 14 + 5 ( here -5 and +5 give the result zero)
z = 14 + 5
z = 19
Tutorcircle.com Page No. : 3/4
Page No. : 2/3
Thank You For Watching
Presentation
```
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# Prove the following
Question:
If $a=\frac{3+\sqrt{5}}{2}$, then find the value of $a^{2}+\frac{1}{a^{2}}$
Solution:
Given, $a=\frac{3+\sqrt{5}}{2}$ $\ldots$ (i)
Now, $\frac{1}{a}=\frac{2}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}}$ [multiplying numerator and denominator by $3-\sqrt{5}$ ]
$=\frac{6-2 \sqrt{5}}{3^{2}-(\sqrt{5})^{2}}$ [using identity, $(a-b)(a+b)=a^{2}-b^{2}$ ]
$=\frac{6-2 \sqrt{5}}{9-5}=\frac{6-2 \sqrt{5}}{4}$
$\Rightarrow$ $\frac{1}{a}=\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt{5}}{2}$ ... (ii)
$\therefore$ $a^{2}+\frac{1}{a^{2}}=a^{2}+\frac{1}{a^{2}}+2-2=\left(a+\frac{1}{a}\right)^{2}-2$ [adding and subtracting 2]
$=\left(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}\right)^{2}-2$ [from Eqs. (i) and (ii)]
$=\left(\frac{6}{2}\right)^{2}-2=(3)^{2}-2=9-2=7$
|
# Reduce Fraction – Definition of Fractions with Examples
Welcome to Brighterly, your go-to source for mathematics designed specifically for children. Today, we’re diving into an essential building block of math: fractions! If you’ve ever wondered how to divide something into equal parts, you’ve been thinking about fractions. Whether you’re slicing a pizza into slices or breaking down a complex mathematical problem, fractions are everywhere. Understanding fractions is not just for mathematicians and scientists; it’s a fundamental concept that we use in everyday life. In this comprehensive guide, we’ll explain what a fraction is, why reducing fractions is crucial, and how to do it step-by-step. At Brighterly, we aim to make learning fun and engaging, so buckle up and get ready to explore the world of fractions!
## What Is a Fraction?
A fraction represents a part of a whole or, more generally, any number of equal parts. At its core, a fraction symbolizes a division of a quantity into equal segments. Consisting of two numbers separated by a horizontal or diagonal line, the top number is called the numerator, and the bottom number is called the denominator. The numerator denotes how many parts are taken, while the denominator tells you how many equal parts the whole is divided into. For example, in the fraction 3/4, 3 is the numerator, and 4 is the denominator. This means the whole is divided into 4 equal parts, and 3 of those parts are taken.
## What Does It Mean to Reduce a Fraction?
Reducing a fraction, also known as simplifying or normalizing a fraction, involves expressing the fraction in its simplest form. When you reduce a fraction, you divide both the numerator and denominator by their Greatest Common Divisor (GCD), without changing the actual value of the fraction. For instance, the fraction 8/12 can be reduced to 2/3, as both 8 and 12 can be divided by 4, their GCD. It’s a vital concept in math that aids in understanding the underlying principles of arithmetic operations with fractions.
## Importance of Reducing Fractions
Reducing fractions is not just an arbitrary rule; it has real practical importance. By working with reduced fractions, mathematical calculations become simpler and more understandable. Reduced fractions offer a clearer perspective on numerical comparisons and relationships. They help in eliminating any complexity and redundancy, making the numbers more manageable. Whether in daily life or scientific applications, using reduced fractions often leads to more efficient problem solving and a more profound understanding of mathematical concepts.
## Methods to Reduce Fractions
### Using the Greatest Common Divisor (GCD)
The Greatest Common Divisor method is the most conventional way to reduce a fraction. The GCD of two numbers is the largest number that divides both of them without a remainder. By dividing both the numerator and denominator by their GCD, we can reduce the fraction to its simplest form.
### Prime Factorization Method
The Prime Factorization Method involves breaking down the numerator and denominator into their prime factors. By canceling out the common prime factors, you can easily reduce the fraction. This method is sometimes preferred for its systematic approach, and it is particularly helpful when dealing with large numbers.
### Other Methods
Other methods of reducing fractions may include using a series of divisions or employing specialized algorithms for specific situations. These can be useful for teaching the foundational concepts of fractions, or in computer programming where different techniques may be employed based on the specific requirements of an application.
## Properties of Reduced Fractions
### Understanding Numerators and Denominators
In a reduced fraction, the numerator and denominator have no common factors other than 1. This means that the fraction is in its simplest form, with no excess multiplicative “baggage.” By removing these common factors, we get to the heart of what the fraction represents, making it more transparent and easier to work with.
### Relationship with Original Fraction
The relationship between the original fraction and the reduced fraction is straightforward: they represent the same value or quantity. The process of reducing doesn’t change the inherent value of the fraction, only its appearance. This relationship underscores the importance of understanding the essence of numbers, beyond their superficial representation.
### Difference Between Reduced Fractions and Original Fractions
Although a reduced fraction and the original fraction have the same value, their appearance and ease of use can differ dramatically. The reduced fraction is more straightforward to work with, whether in arithmetic operations, comparisons, or understanding underlying mathematical principles. This difference is what makes the practice of reducing fractions an essential skill in mathematics.
## Equations Used in Reducing Fractions
### Finding the GCD
Finding the GCD is a critical step in reducing fractions. You can use the Euclidean Algorithm or prime factorization to find the GCD of the numerator and the denominator.
### Dividing Numerator and Denominator by GCD
Once the GCD is found, divide both the numerator and denominator by the GCD. This division simplifies the fraction to its most fundamental form.
### Step-by-Step Reduction Using GCD
A step-by-step approach can help in understanding the reduction process, especially when learning. By repeatedly dividing the numerator and denominator by common factors, the fraction can be gradually reduced to its simplest form. This method provides a hands-on experience and offers deeper insights into the reduction process.
## Examples Using Prime Factorization
Using prime factorization, a fraction like 45/60 can be reduced as follows:
• Break down 45 into its prime factors: 3 × 3 × 5
• Break down 60 into its prime factors: 2 × 2 × 3 × 5
• Cancel out common factors: (3 × 3 × 5) / (2 × 2 × 3 × 5) = (3 × 1 × 1) / (2 × 2 × 1 × 1) = 3/4
## Practice Problems on Reducing Fractions
Here are a few practice problems for reducing fractions using different methods:
1. Reduce 16/24 using the GCD method.
2. Reduce 50/75 using the Prime Factorization Method.
3. What is the simplest form of 120/150?
These problems help cement the understanding and provide practical experience in reducing fractions.
## Conclusion
At Brighterly, we believe that understanding the concept of fractions and knowing how to reduce them is not only a valuable academic skill but also a life skill that can make everyday calculations more accessible and logical. In this guide, we have explored the definition of fractions, various methods to reduce them, and the importance of these reductions. We hope that these insights will provide both teachers and students with the tools they need to approach fractions with confidence. Remember, math is not just about numbers; it’s about understanding relationships, problem-solving, and logical thinking. Keep exploring, keep asking questions, and keep growing with Brighterly. Your mathematical journey is only just beginning, and we’re here to guide you every step of the way!
## Frequently Asked Questions on Reducing Fractions
### What is the Greatest Common Divisor (GCD)?
The Greatest Common Divisor, or GCD, is a fundamental concept in mathematics. It is the largest number that can divide both the numerator and the denominator of a fraction without a remainder. Understanding the GCD helps in reducing fractions to their simplest form, making them easier to work with. At Brighterly, we offer various resources and exercises to help grasp this concept.
### Can every fraction be reduced?
Not all fractions can be reduced. A fraction that is already in its simplest form, meaning the numerator and denominator have no common factors other than 1, cannot be further reduced. Recognizing when a fraction is in its simplest form is an essential skill in mathematics, and Brighterly’s interactive tools and lessons are designed to foster this understanding.
### Why is reducing fractions important?
Reducing fractions is vital because it simplifies calculations and enhances understanding. By working with reduced fractions, mathematical problems become more manageable, and the inherent relationships between numbers become more apparent. At Brighterly, we emphasize this concept, as it builds a strong foundation for more advanced mathematical study.
### What are the different methods to reduce fractions?
Fractions can be reduced using various methods, including the GCD method, Prime Factorization method, and other specialized techniques. Each approach has its benefits, and understanding them can provide a more in-depth appreciation of fractions. Brighterly offers guided tutorials and practice problems to help students and educators explore these methods, fostering a lifelong love for mathematics.
Information Sources:
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# Class 11 Maths NCERT Solutions for Chapter 1 Sets Miscellaneous Exercise
### Sets Exercise Miscellaneous Solutions
1. Decide, among the following sets, which sets are subsets of one and another:
A = {x: x ∈ R and x satisfy x2 – 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8…}, D = {6}.
Solution
A = {x: x ∈ R and x satisfies x2 – 8x + 12 = 0}
2 and 6 are the only solutions of x2 – 8x + 12 = 0.
∴ A = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}
∴ D ⊂ A ⊂ B ⊂ C
Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C
2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Solution
(i) False
Let A = {1, 2} and B = {1, {1, 2}, {3}}
Now, 2 ∊ {1, 2} and {1, 2} ∊ {{3} , 1, {1, 2}}
∴ A ∊ B
However, 2 ∉ {{3}, 1, {1, 2}}
(ii) False
Let A = {2}, B = {0, 2} and C = {1, {0, 2}, 3}
As A ⊂ B
B ∊ C
However, A ∉ C
(iii) True
Let A ⊂ B and B ⊂ C.
Let x ∊ A
⇒ x ∊ B [∵ A ⊂ B]
⇒ x ∊ C [∵ B ⊂ C]
∴ A ⊂ C
(iv) False
Let A = {1, 2}, B = {0, 6,8}, and C = {0, 1, 2, 6, 9}
Accordingly, A⊄ B and B ⊄ C.
However, A ⊂ C
(v) False
Let A = {3, 5, 7} and B = {3, 4, 6}
Now, 5 ∊ A and A ⊄ B
However, 5 ∉ B
(vi) True
Let A ⊂ B and x ∉ B.
To show: x ∉ A
If possible, suppose x ∈ A.
Then, x ∈ B, which is a contradiction as x ∉ B
∴x ∉ A
3. Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.
Solution
Let, A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C.
To show : B = C
Let x ∈ B
⇒ x ∈ A ∪ B [B ⊂ A ∪ B]
⇒ x ∈ A ∪ C [A ∪ B = A ∪ C]
⇒ x ∈ A or x ∈ C
Case I
x ∈ A
Also, x ∈ B
∴ x ∈ A ⋂ B
⇒ x ∈ A ∩ C [∵ A ∩ B = A ∩ C]
∴ x ∈ A and x ∈ C
∴ x ∈ C
∴ B ⊂ C
Similarly, we can show that C ⊂ B.
∴ B = C
4. Show that the following four conditions are equivalent:
(i) A ⊂ B
(ii) A – B = Φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Solution
First, we have to show that (i) ⇔ (ii).
Let A ⊂ B
To show: A – B ≠ Φ
If possible, suppose A – B ≠ Φ
This means that there exists x ∈ A, x ≠ B, which is not possible as A ⊂ B.
∴ A – B = Φ
∴ A ⊂ B ⇒ A – B = Φ
Let A – B = Φ
To show: A ⊂ B
Let x ∈ A
Clearly, ∈ B because if x ∉ B, then A – B ≠ Φ
∴ A – B = Φ ⇒ A ⊂ B
∴ (i) ⇔ (ii)
Let A ⊂ B
To show : A ∪ B = B
Clearly, B ⊂ A ∪ B
Let x ∊ A ∪ B
⇒ x ∊ A or x ∊ B
Case I : x ∊ A
⇒ x ∊ B [∵ A ⊂ B]
∴ A ∪ B ⊂ B
Case II: x ∊ B
Then, A ∪ B = B
Conversely, let A ∪ B = B
Let x ∊ A
⇒ x ∊ A ∪ B [∵ A ⊂ A ∪ B]
⇒ x ∊ B [∵ A ∪ B = B]
∴ A ⊂ B
Hence, (i) ⇔ (iii)
Now, we have to show that (i) ⇔ (iv) .
Let A ⊂ B
Clearly, A ∩ B ⊂ A
Let x ∊ A
We have to show that x ∊ A ∩ B
As A ⊂ B, x ∊ B
∴ x ∊ A ∩ B
∴ A ⊂ A ∩ B
Hence, A = A ∩ B
Conversely, suppose A ∩ B = A
Let x ∊ A
⇒ x ∊ A ∩ B
⇒ x ∊ A and x ∊ B
⇒ x ∊ B
∴ A ⊂ B
Hence, (i) ⇔ (iv).
5. Show that if A ⊂ B, then C – B ⊂ C – A.
Solution
Let A ⊂ B
To show: C – B ⊂ C – A
Let x ∈ C – B
⇒ ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [A ⊂ B]
⇒ x ∈ C – A
∴ C – B ⊂ C – A
6. Assume that P (A) = P (B). Show that A = B.
Solution
Let P(A) = P(B)
To show: A = B
Let x ∈ A
A ∈ P(A) = P(B)
∴ ∈ C, for some C ∈ P(B)
Now, C ⊂ B
∴ x ∈ B
∴ A ⊂ B
Similarly, B ⊂ A
∴ A = B
7. Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.
Solution
False
Let A = {0, 1} and B = {1, 2}
∴ A ∪ B = {0, 1, 2}
P(A) = {Φ, {0}, {1}, {0, 1}}
P(B) = {Φ, {1}, {2}, {1, 2}}
P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}
P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}}
∴ P(A) ∪ P(B) ≠ P(A ∪ B)
8. Show that for any sets A and B,
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution
To show: A = (A ∩ B) ∪ (A – B)
Let x ∈ A
We have to show that x ∈ (A ∩ B) ∪ (A – B)
Case I:
x ∈ A ∩ B
Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)
Case II:
x ∉ A ∩ B
⇒ x ∉ A or x ∉ B
∴ x ∉ B [x ∉ A]
∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)
∴ A ⊂ (A ∩ B) ∪ (A – B) …(1)
It is clear that
A ∩ B ⊂ A and (A – B) ⊂ A
∴ (A ∩ B) ∪ (A – B) ⊂ A …(2)
From (1) and (2), we obtain
A = (A ∩ B) ∪ (A – B)
To prove: A ∪ (B – A) ⊂ A ∪ B
Let x ∈ A ∪ (B – A)
⇒ x ∈ A or x ∈ (B – A)
⇒ x ∈ A or (x ∈ B and x ∉ A)
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∉ A)
⇒ x ∈ (A ∪ B)
∴ A ∪ (B – A) ⊂ (A ∪ B) …(3)
Next, we show that (A ∪ B) ⊂ A ∪ (B – A).
Let y ∈ A ∪ B
⇒ y ∈ A or y ∈ B
⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A)
⇒ y ∈ A or (y ∈ B and y ∉ A)
⇒ y ∈ A ∪ (B – A)
∴ A ∪ B ⊂ A ∪ (B – A) …(4)
Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B.
9. Using properties of sets show that
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A.
Solution
(i) To show: A ∪ (A ∩ B) = A
We know that
A ⊂ A
A ∩ B ⊂ A
∴ A ∪ (A ∩ B) ⊂ A …(1)
Also, A ⊂ A ∪ (A ∩ B) …(2)
∴ From (1) and (2), A ∪ (A ∩ B) = A
(ii) To show: A ∩ (A ∪ B) = A
A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)
= A ∪ (A ∩ B)
= A {from (1)}
10. Show that A ∩ B = A ∩ C need not imply B = C.
Solution
Let A = {0, 1}, B = {0, 2, 3}, and C = {0, 4, 5}
Accordingly, A ∩ B = {0} and A ∩ C = {0}
Here, A ∩ B = A ∩ C = {0}
However, B ≠ C [2 ∈ B and 2 ∉ C]
11. Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use distributive law)
Solution
Let A and B be two sets such that A ∩ X = B ∩ X = f and A ∪ X = B ∪ X for some set X.
To show: A = B
It can be seen that
A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]
= (A ∩ B) ∪ (A ∩ X) [Distributive law]
= (A ∩ B) ∪ Φ [A ∩ X = Φ]
= A ∩ B …(1)
Now, B = B ∩ (B ∪ X)
= B ∩ (A ∪ X) [A ∪ X = B ∪ X]
= (B ∩ A) ∪ (B ∩ X) [Distributive law]
= (B ∩ A) ∪ Φ [B ∩ X = Φ]
= B ∩ A
= A ∩ B …(2)
Hence, from (1) and (2), we obtain A = B.
12. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.
Solution
Let A = {0, 1}, B = {1, 2}, and C = {2, 0}.
Accordingly, A ∩ B = {1}, B ∩ C = {2}, and A ∩ C = {0}.
∴ A ∩ B, B ∩ C, and A ∩ C are non-empty.
However, A ∩ B ∩ C = Φ
13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Solution
Let U be the set of all students who took part in the survey.
Let T be the set of students taking tea.
Let C be the set of students taking coffee.
Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100
To find: Number of student taking neither tea nor coffee i.e., we have to find n(T' ∩ C').
n(T' ∩ C') = n(T ∪ C)'
n(U) – n(T ∪ C)
n(U) – [n(T) + n(C) – n(T ∩ C)]
= 600 – [150 + 225 – 100]
= 600 – 275
= 325
Hence, 325 students were taking neither tea nor coffee.
14. In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Solution
Let U be the set of all students in the group.
Let E be the set of all students who know English.
Let H be the set of all students who know Hindi.
∴ H ∪ E = U
Accordingly, n(H) = 100 and n(E) = 50
n(H ∩ E) = 25
n(U) = n(H) + n(E) – n(H ∩ E)
100 + 50 – 25
= 125
Hence, there are 125 students in the group.
15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I,11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
Solution
Let A be the set of people who read newspaper H.
Let B be the set of people who read newspaper T.
Let C be the set of people who read newspaper I.
Accordingly, n(A) = 25, n(B) = 26, and n(C) = 26
n(A ∩ C) = 9, n(A ∩ B) = 11, and n(B ∩ C) = 8
n(A ∩ B ∩ C) = 3
Let U be the set of people who took part in the survey.
(i) Accordingly,
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
= 25 + 26 + 26 – 11 – 8 – 9 + 3
= 52
Hence, 52 people read at least one of the newspapers.
(ii) Let a be the number of people who read newspapers H and T only.
Let b denote the number of people who read newspapers I and H only.
Let c denote the number of people who read newspapers T and I only.
Let d denote the number of people who read all three newspapers.
Accordingly, d = n(A ∩ B ∩ C) = 3
Now, n(A ∩ B) = a + d
n(B ∩ C) = c + d
n(C ∩ A) = b + d
∴ a + d + c + d + b + d = 11 + 8 + 9 = 28
⇒ a + b + c + d = 28 – 2d = 28 – 6 = 22
Hence, (52 – 22) = 30 people read exactly one newspaper.
16. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Solution
Let A, B, and C be the set of people who like product A, product B, and product C respectively.
Accordingly, n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12,
n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8
The Venn diagram for the given problem can be drawn as
It can be seen that number of people who like product C only is
{29 - (4 + 8 + 6)} = 11
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## Tuesday, February 4, 2014
### FRACTIONS
With the start of the second semester, the focus in math has shifted from working with whole numbers to numbers that are less than one whole; also known as fractions.
The main learning objective currently is the Common Core State Standard 4.NF.2:
4.NF.2. Compare two fractions with different numerators and different 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 less than, greater than or equal to, and justify the conclusions, e.g., by using a visual fraction model.
To make the concept a little easier to understand, students are using fraction models.
These models allow students to better visualize the fractions being used and to think about their size relative to other fractions and to one whole. This is especially helpful when students are asked to compare fractions or find equivalent fractions.
This particular web based model is really nice because it allows students to move around on the number line. It automatically shows when there are other fractions that are equivalent to the one a students lands on.
Here is a picture of it. Click on the picture to try it out.
You can easily see at the top of the image that the student has chosen the fraction 2/3. Then if you follow the green line down, all of the equivalent fractions to 2/3 on the chart are indicated.
Comparing Fractions
It is easy to compare fractions with the same (or like) denominators. Let's use this example:
When comparing fractions with like denominators, the focus moves to the numerator. In the example above, both fractions are broken into 7 parts. The fraction with 2 parts shaded in is clearly smaller than the one with 5 parts shaded in. Therefore two-sevenths is less than five sevenths.
When comparing fractions with unlike denominators, one must really think about and try to picture the fractions. We don't always have pictures in front of us. It is okay to represent the fractions with a picture model, but it is also importnat to have mathematical strategies other than pictoral.
One strategy students have been taught is to compare fractions to the "benchmark" of 1/2. A benchmark fraction is an easily identifiable fraction students can use to help them identify or mark other fractions.
By representing the fractions with a model, students can easily draw both fractions and clearly see that the two fractions are equal to each other in size. When different fractions are the same size, we call them equivalent fractions. We will be discussing these more at a later time.
In addition to representing fractions with a model, students should have a mathematical strategy they can use to solve problems. In the above example, one would first look at the denominator, or bottom number. Then, one would think about the numerator, or top number.
The question one should ask is "Is the numerator more than half, less than half or equal to half of the denominator?" In this case, 3 is half of 6, so that tells us this fraction is equal to half. Let's look at another example...
Students can use pictures models or use mathematical strategies to compare fractions to one half. If a student is comfortable with these strategies, she or he can now use them to compare fractions with unlike denominators.
Here is a video showing how to use these strategies to compare fractions with unlike denominators.
Please comment! How is Fractions going for you? What have you learned? What are you still trying to learn? What strategy works best for you when comparing fractions?
1. Dear Mr.Pahl
I improved my fraction. My mommy told me a strategy I can use for fraction if I have trouble
Sincerely Dyna
2. Dear Class,
I think Fractions are really fun to solve and to learn about. But if you learn a new method it is kind of hard to understand but you get use to it after a while worth practicing:)
From, Karoline
1. Dear Karoline,
I also think learning new can be tough but can get used to it by practicing the new strategies you find out about.
From,
Sophia
2. Dear Karoline and Sophia,
This is Beth from Mr. Pahl's class at Western. I am so glad to see you using different strategies for working through and understanding fractions. What have you found to be the most helpful trick or strategy for understanding fractions?
Sincerely, Beth
3. Dear Mr. Pahl
I like fractions and I am trying my best but I am not doing so well.
From.Micah
1. Hi Micah,
Thanks for letting me know. Can you identify what it is you don't understand? We'll work on it Thursday. You are a very good student, Micah. Don't worry about it. We'll get it figured out and you will think fractions are easy peasy.
Sincerely,
Mr. Pahl
4. Dear Mr.pahl,
I really like learning about factions. I hope at the end of the year I know everything about factions.
From Kayden,
1. Hi Kayden,
Thanks for writing. I am glad to hear you like fractions. I like them, too! You will learn a lot about fractions this year, but probably not everything. We have to save something for 5th grade! Thank you for being such a good student!
Sincerely,
Mr. Pahl
5. Dear Mr.Pahl,
I love to do fractions. My favorite part is when I subtract mixed numbers and the denominators are different. I thought it was hard at first.
From,
Tori
1. Dear Tori,
I agree, I also like fractions. I thought subtracting mixed numbers, too. I admit, it was pretty hard, but now we know how.
Sincerely
Charlize
6. Dear Dyna,
What is the strategy that your mom taught you for fractions when you are having trouble? My name is Julia and I am in Mr. Pahl's class at Western and am in a fraction class too. I would like to hear your strategy!
From,
Julia
1. Hi Julia,
Dyna is in 5th grade now, but I will ask her for you!
Mr. Pahl
7. Dear Mr. Pahl,
I love doing fractions I thought subtracting fractions with different denominators would be hard but it really isn’t now.
From Sarah
1. Dear Sarah,
Thanks for writing! That's terrific! I am so glad to hear that subtracting fractions isn't too hard for you now. What do you think made it easier? Is there a particular strategy you use or did you just keep practicing until you remembered what to do?
Mr. Pahl
2. Dear Mr. Pahl,
I don’t think I had a special strategy I used. I guess I just kept practicing till I remembered how to do it.
From
Sarah H.
8. Dear Mr.Pahl,
The one thing that was really hard for some of the kids in fourth grade.
Is borrowing fractions I think that we need to practice on it more.
Sincerely,
Eric
1. Eric, I agree. We will be practicing that more soon. Thanks for telling me your thoughts.
Mr. Pahl
9. Dear Charlize, I really loved you’re video. I think that I even learned something, the point is you worked hard to explain everything and you had to do math to help people who did not understand so thank you very much. Also thank you for teaching me that strategy now I’ll be better in math! Happy Valentine’s Day, Hannah
10. Dear Charlize’s,
Your video helped me understand more about fractions. You did an excellent job on this video. I hope that you can make more videos about fractions to help kids who don’t really understand them.
Sincerely, Amaya
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How do you find the standard form given 4x^2-5y^2-40x-20y+160=0?
Sep 4, 2017
Given: $4 {x}^{2} - 5 {y}^{2} - 40 x - 20 y + 160 = 0$
Subtract 160 from both sides:
$4 {x}^{2} - 5 {y}^{2} - 40 x - 20 y = - 160$
Add $4 {h}^{2}$ to both sides:
$4 {x}^{2} - 40 x + 4 {h}^{2} - 5 {y}^{2} - 20 y = - 160 + 4 {h}^{2}$
Remove the common factor, 4, from the first 3 terms:
$4 \left({x}^{2} - 10 x + {h}^{2}\right) - 5 {y}^{2} - 20 y = - 160 + 4 {h}^{2}$
Subtract $5 {k}^{2}$ from both sides:
$4 \left({x}^{2} - 10 x + {h}^{2}\right) - 5 {y}^{2} - 20 y - 5 {k}^{2} = - 160 + 4 {h}^{2} - 5 {k}^{2}$
Remove a common factor of -5 from the next 3 terms:
$4 \left({x}^{2} - 10 x + {h}^{2}\right) - 5 \left({y}^{2} + 4 y + {k}^{2}\right) = - 160 + 4 {h}^{2} - 5 {k}^{2} \text{ [1]}$
Because we are looking for the square ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ we use the equation:
$- 2 h x = - 10 x$
To find the value of h:
$h = 5$
Into equation [1], substitute ${\left(x - 5\right)}^{2} \text{ for } {x}^{2} - 10 x + {h}^{2}$ and 100 for $4 {h}^{2}$:
$4 {\left(x - 5\right)}^{2} - 5 \left({y}^{2} + 4 y + {k}^{2}\right) = - 160 + 100 - 5 {k}^{2} \text{ [1.1]}$
Because we are looking for the square ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$ we use the equation:
$- 2 k y = 4 y$
To find the value of k:
$k = - 2$
Into equation [1.1], substitute ${\left(y - \left(- 2\right)\right)}^{2} \text{ for } {y}^{2} + 4 y + {k}^{2}$ and -20 for $- 5 {k}^{2}$:
$4 {\left(x - 5\right)}^{2} - 5 {\left(y - \left(- 2\right)\right)}^{2} = - 160 + 100 - 20 \text{ [1.2]}$
Combine like terms on the right:
$4 {\left(x - 5\right)}^{2} - 5 {\left(y - \left(- 2\right)\right)}^{2} = - 80 \text{ [1.3]}$
Divide both sides by -80:
${\left(y - \left(- 2\right)\right)}^{2} / 16 - {\left(x - 5\right)}^{2} / 20 = 1 \text{ [1.3]}$
Write the denominators as squares:
${\left(y - \left(- 2\right)\right)}^{2} / {\left(4\right)}^{2} - {\left(x - 5\right)}^{2} / {\left(2 \sqrt{5}\right)}^{2} = 1 \text{ [1.4]}$
Equation [1.4] is standard form.
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## Torus knot signatures, Part 1
I want to tell you about the current mathematics I’m working on, because it’s exciting, surprising, beautiful, deep, and easy to explain! Everything that a good piece of maths should be. I’m going to have to explain it in a few parts though, leading you through the different steps I had to solve to get to the solution.
Here’s a problem to get you started. I’m going to give you two numbers, p and q, which are coprime. That means they have no common factors. For example, I can’t give you the numbers 9 and 12, since both of them are divisible by 3. But I can give you the numbers 9 and 10 since 9=3×3 and 10=2×5, and there is no overlap. Ok. Now I give you another number n between 1 and pq-1, such that n is neither a multiple of p nor a multiple of q. Your task: take n and keep minusing multiples of p from it until you get a number that is a multiple of q. Then tell me whether the multiple of q is positive or negative.
Let’s do an example. Suppose p=3 and q=5. I need to give you a number n between 1 and 14, so let’s start with the simple n=1. First we try 1-3=-2. Not a multiple of 5. So we try 1-(2×3) = -5. Success! And -5 is negative. Now let n=8. Try 8-3=5. First time lucky! And this time we got a positive number.
For particular numbers p and q, we are going to do this for all possible n. So if p=3 and q=5, we have to find the answer for n=1,2,4,7,8,11,13,14. If the answer is positive, put j(n)=1, and if it’s negative we put j(n)=-1. So, for these particular values of n, the corresponding values for j are -1,-1,-1,-1,1,1,1,1. Now we keep a running total, and call that number s. So the values for s are -1,-2,-3,-4,3,2,1,0. Actually, for reasons that I will explain later, the interesting number to look at is 2 times s.
Let’s draw a graph of the function 2s and see what it looks like. Does it always go “minus minus minus plus plus plus”? Well, for p=2, the answer is ‘yes’, no matter what value you choose for q. Here’s the graph of p=2,q=25, with the values of n running along the bottom and the value of 2s running up the side.
p=2, q=25
Notice the lovely ‘V’ shape that it makes. The values of s decrease until a certain point, and then increase again.
Let’s see what happens when p=3. Here’s the graph for p=3, q=10:
p=3, q=10
Oh no, what happened? The graph is all wiggly at the bottom! Minus, plus, minus, plus – what’s going on? It’s obviously not a proplem with the number 3, since our previous example of p=3, q=5 worked out fine. Maybe we just picked a funny combination of numbers. We had better try another pair and see what happens there. How about p=7, q=16?
p=7, q=16
Even more wiggles! Yikes, it seems that this function is going to be pretty unpredictable. The pattern seems to be very different for every p and q. How on earth are we going to get a formula that tells us what is happening?
Tune in next time for more clues…
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How Will Holly Madison Fare Today? (01/16/2020)
How will Holly Madison get by on 01/16/2020 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is not scientifically verified – do not take this too seriously. I will first calculate the destiny number for Holly Madison, and then something similar to the life path number, which we will calculate for today (01/16/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology people.
PATH NUMBER FOR 01/16/2020: We will consider the month (01), the day (16) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 01 and add the digits together: 0 + 1 = 1 (super simple). Then do the day: from 16 we do 1 + 6 = 7. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 1 + 7 + 4 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the path number for 01/16/2020.
DESTINY NUMBER FOR Holly Madison: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Holly Madison we have the letters H (8), o (6), l (3), l (3), y (7), M (4), a (1), d (4), i (9), s (1), o (6) and n (5). Adding all of that up (yes, this can get tiring) gives 57. This still isn’t a single-digit number, so we will add its digits together again: 5 + 7 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the destiny number for Holly Madison.
CONCLUSION: The difference between the path number for today (3) and destiny number for Holly Madison (3) is 0. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is not at all guaranteed. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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Unlocking the Secrets of Inverse Functions: A Closer Look
Finding the derivative of an inverse function involves flipping the derivative of the original function, mirroring changes inversely. For trigonometric inverses, like arcsin or arctan, it's about understanding how angles change in circular motion. This process requires a keen grasp of how functions and their inverses are interlinked, reflecting each other's changes in a reciprocal manner.
How to find the derivative of inverse function?
The derivative of an inverse function is computed by taking the reciprocal of the derivative of the original function, considering the interdependent relationship between a function and its inverse. For trigonometric inverses, such as arcsin or arctan, the process involves understanding the derivatives of circular functions and their inverse relationship, focusing on the rate of change of angles with respect to their trigonometric ratios.
You can find the inverse of a function first and then compute its derivative. This approach involves two main steps: first, determine the inverse function by reversing the roles of the dependent and independent variables in the original function, and then solve for the new dependent variable. Once you have the inverse function explicitly, you can apply standard differentiation techniques to find its derivative.
However, this method can be challenging if finding the inverse function explicitly is difficult or if the inverse function is complex, making differentiation tricky. In such cases, using the formula that relates the derivatives of a function and its inverse can be more efficient and straightforward:
$$\left(\frac{d}{dx} f^{-1}(x)\right) = \frac{1}{f'(f^{-1}(x))}$$
Here are the derivatives of the inverse trigonometric functions, with a variable $$(x)$$ or a function $$(f(x))$$ inside of them:
Sin:
$$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 – x^2}}$$
$$\frac{d}{dx} \arcsin(f(x)) = \frac{f'(x)}{\sqrt{1 – f(x)^2}}$$
Cos:
$$\frac{d}{dx} \arccos(x) = \frac{-1}{\sqrt{1 – x^2}}$$
$$\frac{d}{dx} \arccos(f(x)) = \frac{-f'(x)}{\sqrt{1 – f(x)^2}}$$
Tan:
$$\frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2}$$
$$\frac{d}{dx} \arctan(f(x)) = \frac{f'(x)}{1 + f(x)^2}$$
Sec:
$$\frac{d}{dx} \text{arcsec}(x) = \frac{1}{|x| \sqrt{x^2 – 1}}$$
$$\frac{d}{dx} \text{arcsec}(f(x)) = \frac{f'(x)}{|f(x)| \sqrt{f(x)^2 – 1}}$$
Example 1:
Consider the function $$f(x) = 2x$$ and its derivative $$f'(x) = 2$$
The derivative of $$arcsin f(x)$$ is given by:
$$\frac{d}{dx} \arcsin(f(x)) = \frac{f'(x)}{\sqrt{1 – f(x)^2}}$$
By substituting the values, we have: $$\frac{2}{\sqrt{1 – (2x)^2}} = \frac{2}{\sqrt{1 – 4x^2}}$$
Example 2:
Given the function $$f(x) = \frac{1}{3}x^3$$, find the derivative of its inverse at the point where $$x = \frac{1}{3}$$
Solution: its derivative is $$f'(x) = x^2$$.
To find where $$x = \frac{1}{3}$$ in the original function: $$\frac{1}{3}x^3 = \frac{1}{3}$$
This gives $$x = 1$$, as the point in the original function corresponding to$$x = \frac{1}{3}$$
in the inverse function.
The derivative of the inverse function is given by:
$$\left(\frac{d}{dx} f^{-1}(x)\right) = \frac{1}{f'(f^{-1}(x))}$$
Substituting $$x=1$$ into the derivative of the original function, we get $$f'(1) = 1^2 = 1$$
Thus, the derivative of the inverse function at $$x = \frac{1}{3}$$ is:
$$\frac{1}{f'(f^{-1}(\frac{1}{3}))} = \frac{1}{1} = 1$$
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## LetsPlayMaths.Com
WELCOME TO THE WORLD OF MATHEMATICS
# Class-8 Polygon
Introduction to Polygon
Types of Polygons
Concave Polygon
Convex Polygon
Regular Polygon
Irregular Polygon
Sum of Angles of a Polygon
Sum of Exterior Angles of a Polygon
Number of Diagonals of a Polygon
Polygon MCQ
Polygon Worksheets
## Introduction to Polygon
A simple closed curve made up of only line segments is known as a polygon. The line segments are called sides of a polygon. The line segments forming a polygon intersect only at end points and each end point is shared by only two-line segments. Figure below shows various types of polygons.
Triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon and decagon can be called as polygon.
## Types of Polygons
There are two types of polygons.
1. Concave Polygon
2. Convex Polygon
## Concave Polygon
A polygon in which at least one angle is greater than 180° is known as a concave polygon. Below given figure is an example of concave polygon, here ∠ABC is more than 180°.
## Convex Polygon
A polygon in which each angle is less than 180° is known as a convex polygon. In the below figure ABCD is a convex polygon.
## Regular Polygon
A polygon having all sides equal and all angles equal is known as a regular polygon. Equilateral triangle and square are examples of regular polygons.
## Irregular Polygon
Polygons which do not follow regular polygons properties are known as irregular polygon. Rectangle and rhombus are examples of irregular polygons.
## Sum of Angles of a Polygon
If we draw all possible diagonals through a single vertex of a polygon to form as many triangles as possible, then the number of triangles can be formed is two less than the number of sides in the polygon.
So, if a polygon has 'n' sides, then the number of triangle formed will be n − 2.
As we know, the sum of angles of a triangle = 180°
Sum of angles of a polygon having 'n' sides = (n − 2) × 180°
Each interior angle = {(n − 2) × 180°}n
Example 1. Find the measure of each exterior angle of a regular polygon having 9 sides.
Solution. Each exterior angle of a regular polygon having 9 sides = 360°9 = 40°
Example 2. Find the measure of each interior angle of a regular hexagon.
Solution. Each exterior angle of a regular hexagon = 360°6 = 60°
Each interior angles of the hexagon = 180° − 60° = 120°
Example 3. Is it possible to have a regular polygon having each exterior angle 35o.
Solution. Let's assume the regular polygon is having 'n' sides.
Each exterior angle = 360°n
=> 350 = 360°n
=> n = 360°35°
=> n = 1027
Example 4. If each interior angle of a regular ploygon is 156°, then find the number of sides in it.
Solution. Let's assume the number of sides of the regular ploygon is 'n'.
Interior angle = {(n − 2) × 180°}n
=> 156° = (180n − 360)n
=> 156n = 180n − 360
=> 24n = 360
=> n = 36024
=> n = 15
Hence, the regular polygon has 15 sides.
Example 5. An exterior angle and interior angle of a regular polygon are in the ratio 5 : 4. Find the number of sides in the polygon.
Solution. Exterior angle : Interior angle = 5 : 4
In other words, we can say exterior angle is 5x and interior angle is 4x.
As we know, exterior angle + interior angle = 180°
5x + 4x = 180
=> 9x = 180
=> x = 20
Exterior angle = 5x = 100°
Interior angle = 4x = 80°
## Sum of Exterior Angles of a Polygon
If the sides of a polygon are extended in order, the sum of exterior angles formed is always equal to 360°.
∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 360°
## Number of Diagonals of a Polygon
Number of diagonals in a polygon having 'n' sides = n(n − 3)2
Polygon Test - 1
Polygon Test - 2
## Class-8 Polygon Worksheets
Polygon Worksheet - 1
Polygon Worksheet - 2
Polygon Worksheet - 3
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# ACT Math Definitions: Algebraic Expressions0
Algebraic expression problems usually involve a given quantity or quantities that can be manipulated to produce a value for a desired quantity. The trick to these problems is usually to rely on your algebraic skills to transform the expressions rather than to solve for the values of the particular variables. In many cases, you will not even be given enough information to solve for the variables directly. Let’s take a look at some typical problems that make use of common transformations.
Q1: The expression (8x – 4) – (4x + 12) is equivalent to which of the following?
A. 4(x + 4)
B. 4(x + 8)
C. 4(x – 4)
D. 4(x – 8)
E. 4(x – 16)
There are several ways to solve an expression simplification problem. Usually, your best bet is to solve it directly. In this case, perform the subtraction:
(8x-4) – (4x+12) ⇒ 8x – 4 – 4x – 12 ⇒ 4x – 16 ⇒ 4(x – 4)
Make sure you fully distribute the negative in front of the second set of parentheses! Since the simplified version isn’t a choice, you need to factor to get the fully simplified form. The greatest common factor of 4x and 16 is 4. Pulling out a 4, we get 4(x–4). That’s the final answer—choice C!
Q2: The expression is equivalent to which of the following?
Almost every ACT has a problem requiring you to multiply two binomials. The most common error in this problem occurs when a student squares each part of the binomial as follows:
If that’s how you would solve the problem, STOP, BE CAREFUL, and READ ON. Forgetting to perform the FOIL operation is the most common error students make. Here’s how the work should look:
When you use FOIL, the middle terms count. The final answer choice is B.
Q3. The speed of a race car exceeds 3 times the speed of a van by 9 miles per hour. If t is the speed of the van, which of the following expresses the distance, in miles, in terms of t, that the race car travels in 20 minutes?
A. t+3
B. t+9
C. 3t+3
D. 3t+9
E. 4t+12
This is a challenging problem on the ACT. To start, let’s translate the words into an algebraic expression. We are told that the “speed of a race car exceeds 3 times the speed of a van by 9 miles per hour.” If the van is moving at t miles per hour, then the race car is moving at a speed that is 3 times that (3t) plus 9 additional miles per hour (+9). The speed of the race car, in miles per hour, is therefore 3t + 9. You can bet after all that work, 3t+9 will be an answer choice. But be careful. The question is asking us to find the distance that the race car travels, in miles, in terms of t, in 20 minutes. If the race car is moving at 3t+9 miles per hour, then it will go 3t+9 miles in each 1-hour time period, or every 60 minutes. To find the distance the race car travels in 20 minutes, or one-third of an hour, divide (3t+9)/3 to get t+3 miles! The correct answer choice is A!
Q4: The formula for the surface area S of a right circular cylinder is, whereis a constant, r is the radius of the circular base of the cylinder, and h is the height of the cylinder. Which of the following is an expression for h?
We’re told that and that we need to find an expression for h. Just as we would when solving a regular algebraic equation, we start by isolating the term containing the variable. For example:
Now isolate the variable h by dividing by . We get . The correct answer choice is C.
Q5: Which of the following is (are) equivalent to the mathematical operation a(bc-bd) for all real numbers a, b, c, and d:
A. I only
B. III only
C. I and II only
D. I and III only
E. I, II, and III
This problem tests your ability to manipulate an expression by using the basic distributive property of multiplication. We are given the expression a(bc-bd). To get choice I, we can factor out b from inside the parentheses. For example, a(bc-bd)ab(c-d). Choice I works. At this point, we can eliminate answer choice B. Choice a(c-d)b, looks as if it would be true. But think of the problem this way. The product 2×3×4 equals 24. The product in another order, 2×4×3, also equals 24. The order of multiplication does not affect the outcome. Since a(c-d)b is just another arrangement of ab(c-d), which we already know is true, a(c-d)b must also be true. Now we need to consider III, b(ac-ad). If we factor out an a term from inside the parentheses, we get , ab(c-d) which is equivalent to the arrangement in choice I. Since all the possible arrangements are true, choice E is the correct response.
Q6: The expression is equivalent to which of the following?
When we add two fractions, the first piece to check is that the denominators are equal. In this case, the fractions have unequal denominators. The simplest trick to make the denominators the same is to multiply the numerator of each fraction by the denominator of the other and then divide the sum by the product of the denominators. For example:
If we factor out the common term in the numerator, we get. The final answer is choice E!
Q7: Perform the subtraction:
When you subtract polynomials, there are two main properties to consider. First, distribute any negatives, especially those outside parentheses. This will help you avoid making additive errors later. For example:
Next, regroup to combine like terms. For example:
Finally, combine like terms. We get. The correct response is choice A! The same steps apply to polynomial addition. The only difference is that you may not have to distribute the negative throughout the parentheses.
Q8: Which of the following is an equivalent form of y+y+y(y+y+y)?
This type of problem can look very confusing, leaving you wondering “y, y, y did I get such a hard question”? (Joke.) No worries. It will seem a lot easier in about a minute. First, sum the common terms that are inside and outside the parentheses, but not those that are attached. For example:
y+y+y(y+y+y) ⇒ 2y+y(3y)
Next, simplify by distributing any terms outside the parenthesis. We get. That’s all there is to it. The correct response is choice C.
# The Complete ACT Math Strategy Guide0
The Math section contains 60 multiple-choice questions that test pre-algebra, elementary algebra, algebra, geometry, and trigonometry. We will introduce specific strategies for tackling the key problem types within these areas. To beat the Math section, we will ask you to dig into the back of your brain where you keep information from grades 7 to 10. In case that information is buried too far down, we will review the math and go over the strategies (oh, and did we mention cool calculator tricks!).
Math Section Strategy
The questions in the Math section do not appear in order of difficulty. That means the first problem could be hard and the second problem could be easy. Many test takers are used to the SAT, which presents problems in order of increasing difficulty. On the ACT, you should make sure to not lose hope because you’ve come across one hard problem. The following problems could get easier!
Solving Multiple-Choice Problems
In general, you should solve the problems before looking at the answers. Follow the four-step process.
Step 1: Read the question. Paraphrase the question in your head to make sure you understand it and plan to solve it. Do not look at the answer choices.
Step 2: Solve the problem. Once you have solved the problem, look at the answer choices. Pick the answer that matches yours.
Step 3: Use the process of elimination. There is only one correct answer choice. Two of the five choices are usually easy to eliminate because they are obviously incorrect. The other two incorrect answers are tricky. Eliminate the obviously incorrect answers and then use your intuition and content knowledge to decide on the correct choice.
Step 4: When in doubt, try it out. Sometimes you may not know how to approach a problem. Before giving up, and only if you are doing well on time, you should try to plug the answer choices into the questions. Working backward may help you find the answer.
Keep in Mind
• Ballpark it—Estimate numerical values to solve problems and check your answers. Is the answer you came up with reasonable?
• Diagrams—Unless a picture is labeled “to scale,” don’t trust your eyes.
• Timing—You don’t have too much time to work out the problems, and all the problems are worth the same number of points. As you solve the problems, don’t get stuck by trying to solve them in order. Work on the problems that you know, and skip and then come back to the questions that are more difficult for you.
• Fill it in—Never leave a problem blank! You are not penalized for answering problems incorrectly on the ACT. If you do not know how to solve a problem, just fill in the bubble corresponding to a choice you think might be right, and then move on.
# ACT Math Tip: Read the Questions Carefully0
Read all of the questions carefully, so that you know exactly what operations you are being asked to perform. When you are attempting ratio problems, for example, note whether the question is giving a part-to-part ratio or a part-to-whole ratio. The ratio of girls to boys in a class is a part-to-part ratio. The ratio of girls to students in a class is a part-to-whole ratio. Consider the following example question:
There are two types of candy in a bowl, chocolate and caramel. If the ratio of the number of pieces of chocolate candy to the number of pieces of caramel candy is 2:3, each of the following could be the total number of pieces of candy, EXCEPT:
A. 5
B. 12
C. 15
D. 20
E. 30
The correct answer is B. To solve this problem, you must realize that this is a part-to-part ratio of 2 pieces of chocolate candy to every 3 pieces of caramel candy, which means that for every 5 pieces of candy, 2 are chocolate and 3 are caramel. In order for the ratio to be exactly 2:3, the total number of pieces of candy in the bowl must be a multiple of 5. All of the possible answer choices are multiples of 5 except answer choice B.
Note: This is a special kind of ACT problem, in that you are actually looking for the answer choice that is not true. Be especially cautious when answering these questions; don’t just select the first “right” answer that you see!
# ACT Math Tip: Substitute Numbers for Variables0
Substitute Numbers for Variables
You can sometimes simplify your work on a given problem by using actual numbers as “stand-ins” for variables. This strategy works when you have variables in the question and the same variables in the answer choices. You can simplify the answer choices by substituting actual numbers for the variables. Pick numbers that are easy to work with and that meet the parameters of the information given in the question. If you use this strategy, remember that numbers on the ACT Mathematics Test can be either positive or negative and are sometimes whole numbers and sometimes fractions. You should also be careful not to use 1 or 0 as one of your stand-ins because they can create “identities,” which can lead to more than one seemingly correct answer choice.
In addition, it is sometimes necessary to try more than one number to see whether the result always correctly responds to the question. If the numbers that you pick work for more than one answer choice, pick different numbers and try again.
Tip 1
When you pick numbers to substitute for the variables in a question, be sure to try pick both negative and positive numbers, unless the question indicates one or the other. For example, if the question states that x < 5, x could be a negative number.
Consider the following example questions:
Question 1:
If x and y are both positive even integers, which of the following must be even?
A. I only
B. II only
C. I and II only
D. I and III only
E. II and III only
The correct answer is D. The question states that both x and y are positive even integers. Therefore, you can pick any positive even integer and substitute that value for x and y in each of the Roman numeral choices, as follows:
• Roman numeral I , which is even; 42 = 16, which is also even. Any positive even integer raised to another positive even integer will result in an even number; therefore, Roman numeral I correctly answers the question. At this point, you can safely eliminate any answer choices that do not contain Roman numeral I.
• Roman numeral II , which is odd; , which is also odd. When you add 1 to a positive even integer and raise the sum to a positive even integer, the result will be odd; therefore, Roman numeral II does not correctly answer the question. At this point, you can safely eliminate any remaining answer choices that contain Roman numeral II.
• Roman numeral III , which is even; 4(2 + 1) = 43 = 64, which is also even. Any positive even integer raised to an odd power will result in an even number; therefore, Roman numeral III correctly answers the question, and you can eliminate any remaining answer choices that do not contain Roman numeral III.
Tip 2
Always take each Roman numeral as a true/false statement. As you evaluate the statements to determine whether they are true or false based on the question, eliminate answer choices, as we did in the example shown. This process might allow you to arrive at the correct answer quickly without looking at every Roman numeral statement.
Question 2:
If a and b are positive consecutive odd integers, where b > a, which of the following is equal to ?
F. 2a
G. 4a
H. 2a+2
J. 2a+4
K. 4a+4
The correct answer is K. You are given that a and b are positive consecutive odd integers, and that b is greater than a. Pick two numbers that fit the criteria: a = 3 and b = 5. Now, substitute these numbers into b2-a2:52=25and 32=9; therefore,b2-a2=16. Now, plug the value that you selected for a into the answer choices until one of them yields 16, as follows:
2(3) = 6; eliminate answer choice F.
4(3) = 12; eliminate answer choice G.
2(3) + 2 = 8; eliminate answer choice H.
2(3) + 4 = 10; eliminate answer choice J.
4(3) + 4 = 16; answer choice K is correct.
# ACT Math Tip: Don’t Quit Early0
Don’t Quit Early
Some of the questions on the ACT Mathematics Test will appear quite difficult the first time you look at them. Often, though, you will be able to reason your way through the problem so that it makes sense. Keep in mind that these questions do not usually involve extensive calculations or complicated manipulations. Consider the following example question:
If 0 < pr < 1, then which of the following CANNOT be true?
A. p < 0 and r < 0
B. p < −1 and r < 0
C. p < −1 and r < −1
D. p < 1 and r < 1
E. p < 1 and r > 0
The correct answer is C. At first glance, you might think that you don’t have enough information to solve this problem. However, if you recognize that pr must be a positive fraction, since it lies between 0 and 1, you can work your way through the answer choices and eliminate those that could be true.
• Answer choice A. If both p and r are less than 0, their product will be positive. It’s possible for pr to be a positive fraction, since both p and r could be negative fractions, so eliminate answer choice A, because it could be true.
• Answer choice B. If p is less than −1 and r is also a negative number, their product will be positive. It’s possible for pr to be a positive fraction, since r could be a negative fraction, so eliminate answer choice B, because it could be true.
• Answer choice C. If both p and r are less than − 1 then pr will be greater than 1 (which violates the information given in the question), so this statement cannot be true, and answer choice C is correct.
• Answer choice D. If both p and r are less than 1, their product can be positive. It’s possible for pr to be a positive fraction, since both p and r could be either positive or negative fractions, so eliminate answer choice D, because it could be true.
• Answer choice E. If p is less than 1, p can be a positive fraction. If r is greater than 0, it will be a positive number, and it’s possible for pr to be a positive fraction; eliminate answer choice E, because it could be true.
Note: Once you have arrived at the correct answer, it’s not necessary to work through any remaining answer choices. We did it here just to show you that D and E are, in fact, incorrect.
# ACT Math Tips: Answer the Right Question0
If the problem requires three steps to reach a solution and you completed only two of the steps, it is likely that the answer you arrived at will be one of the choices. However, it will not be the correct choice! Consider the following example question:
The rectangular garden shown in the figure below has a stone border 2 feet in width on all sides. What is the area, in square feet, of that portion of the garden that excludes the border?
A. 4
B. 16
C. 40
D. 56
E. 72
The correct answer is B. This problem is asking for the area of the middle portion of the garden. To solve this problem, perform the following calculations, remembering that the border goes around the entire garden and therefore should be deducted from each dimension twice. First, subtract the border width from the length of the garden:
12 − 2(2) = 8
Next, subtract the border width from the width of the garden:
6 − 2(2) = 2
The area (Length × Width) of the portion of the garden that excludes the border is 8 × 2, or 16.
If you had accounted for the border along only one length and one width of the garden, you would have gotten answer choice C. Answer choice D is the area of the border around the garden. Answer choice E is the area of the entire garden, including the stone border.
# ACT Math Tips: Draw Pictures0
Many seemingly complex story problems will become considerably easier if you are able to visualize them. This strategy should not take a lot of time and can help you avoid making careless errors. Your sketch doesn’t have to be beautiful; it just has to accurately depict the relationships in the problem. Sometimes you are given a figure or a table that you can work with (and write on); sometimes you just have to make your own. Consider the following example question:
The diagonal of a rectangular garden is 15 feet, and one side is 9 feet. What is the perimeter of the garden, in feet?
A. 135
B. 108
C. 68
D. 48
E. 42
The correct answer is E. To solve this problem, it is helpful to draw a picture, as shown below:
If a rectangular garden has a side of 9 feet and a diagonal of 15 feet, it forms a right triangle with leg 9 and hypotenuse 15. Using the Pythagorean Theorem, the length of the other leg, x, can be determined using the following equation:
The perimeter of the rectangle is therefore 2(9) + 2(12), or 42, answer choice E.
# ACT Math Tips: Apply Logic0
Even though you can use a calculator, most of the actual calculations you will need to do are fairly simple. In fact, the ACT test writers are just as likely to be testing your logical reasoning ability or your ability to follow directions as they are to your ability to plug numbers into an equation. Consider the following example question:
If b − c = 2, and a + c = 16, then a + b = ?
A. 8
B. 14
C. 16
D. 18
E. 32
The correct answer is D. To solve this problem, first recognize that (b − c) + (a + c) = a + b. This is true because the c values cancel each other out, leaving you with b + a, which is equivalent to a + b. Therefore, a + b must equal 2 +16, or 18.
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# 2021 AMC 12A Problems/Problem 18
The following problem is from both the 2021 AMC 10A #18 and 2021 AMC 12A #18, so both problems redirect to this page.
## Problem
Let $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b) = f(a)+f(b)$ for all positive rational numbers $a$ and $b$. Furthermore, suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$. For which of the following numbers $x$ is $f(x) < 0$?
$\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad$
## Solution 1
Looking through the solutions we can see that $f(\frac{25}{11})$ can be expressed as $f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})$ so using the prime numbers to piece together what we have we can get $10=11+f(\frac{25}{11})$, so $f(\frac{25}{11})=-1$ or $\boxed{E}$.
-Lemonie
$f(\frac{25}{11} \cdot 11) = f(25) = f(5) + f(5) = 10$
- awesomediabrine
## Solution 2
We know that $f(p) = f(p \cdot 1) = f(p) + f(1)$. By transitive, we have $$f(p) = f(p) + f(1).$$ Subtracting $f(p)$ from both sides gives $0 = f(1).$ Also $$f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2$$ $$f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3$$ $$f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11$$ In $\textbf{(A)}$ we have $f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7$.
In $\textbf{(B)}$ we have $f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3$.
In $\textbf{(C)}$ we have $f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1$.
In $\textbf{(D)}$ we have $f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2$.
In $\textbf{(E)}$ we have $f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1$.
Thus, our answer is $\boxed{\textbf{(E)} \frac{25}{11}}$
~JHawk0224 ~awesomediabrine
## Solution 3 (Deeper)
Consider the rational $\frac{a}{b}$, for $a,b$ integers. We have $f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)$. So $f\left(\frac{a}{b}\right)=f(a)-f(b)$. Let $p$ be a prime. Notice that $f(p^k)=kf(p)$. And $f(p)=p$. So if $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, $f(a)=a_1p_1+a_2p_2+....+a_kp_k$. We simply need this to be greater than what we have for $f(b)$. Notice that for answer choices $A,B,C,$ and $D$, the numerator $(a)$ has less prime factors than the denominator, and so they are less likely to work. We check $E$ first, and it works, therefore the answer is $\boxed{\textbf{(E)}}$.
~yofro
## Solution 4 (Most Comprehensive, Similar to Solution 3)
We have the following important results:
$(1) \ f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)$ for all positive integers $k$
$(2) \ f\left(a^n\right)=nf(a)$ for all positive rational numbers $a$
$(3) \ f(1)=0$
$(4) \ f\left({\frac 1a}\right)=-f(a)$ for all positive rational numbers $a$
Proofs
Result $(1)$ can be shown by induction.
Result $(2):$ Since powers are just repeated multiplication, we will use result $(1)$ to prove result $(2):$ $$f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a).$$
Result $(3):$ For all positive rational numbers $a,$ we have $$f(a)=f(a\cdot1)=f(a)+f(1).$$ Therefore, we get $f(1)=0.$ So, result $(3)$ is true.
Result $(4):$ For all positive rational numbers $a,$ we have $$f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.$$ It follows that $f\left({\frac 1a}\right)=-f(a),$ and result $(4)$ is true.
For all positive integers $x$ and $y,$ suppose $\prod_{k=1}^{m}{p_k}^{e_k}$ and $\prod_{k=1}^{n}{q_k}^{d_k}$ are their prime factorizations, respectively, we have \begin{align*} f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) \\ &=f(x)-f(y) \\ &=f\left(\prod_{k=1}^{m}{p_k}^{e_k}\right)-f\left(\prod_{k=1}^{n}{q_k}^{d_k}\right) \\ &=\left[\sum_{k=1}^{m}f\left({p_k}^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left({q_k}^{d_k}\right)\right] \\ &=\left[\sum_{k=1}^{m}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] \\ &=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. \end{align*}
We apply function $f$ on each fraction in the choices:
$$\begin{array}{cccccccc} \textbf{(A) } & f\left(\frac{17}{32}\right) & = & f\left(\frac{17^1}{2^5}\right) & = & [1(17)]-[5(2)] & = & 7 \\ [2ex] \textbf{(B) } & f\left(\frac{11}{16}\right) & = & f\left(\frac{11^1}{2^4}\right) & = & [1(11)]-[4(2)] & = & 3 \\ [2ex] \textbf{(C) } & f\left(\frac{7}{9}\right) & = & f\left(\frac{7^1}{3^2}\right) & = & [1(7)]-[2(3)] & = & 1 \\ [2ex] \textbf{(D) } & f\left(\frac{7}{6}\right) & = & f\left(\frac{7^1}{2^1\cdot3^1}\right) & = & [1(7)]-[1(2)+1(3)] & = & 2 \\ [2ex] \textbf{(E) } & f\left(\frac{25}{11}\right) & = & f\left(\frac{5^2}{11^1}\right) & = & [2(5)]-[1(11)] & = & -1. \end{array}$$ Therefore, the answer is $\boxed{\textbf{(E) }\frac{25}{11}}.$
~MRENTHUSIASM
## Solution 5
The problem gives us that f(p)=p. If we let a=p and b=1, we get f(p)=f(p)+f(1), which implies f(1)=0. Notice that the answer choices are all fractions, which means we will have to multiply an integer by a fraction to be able to solve it. Therefore, let's try plugging in fractions and try to solve them. Note that if we plug in a=p and b=1/p, we get f(1)=f(p)+f(1/p). We can solve for f(1/p) as -f(p)! This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of E.
~ pi_is_3.14
## See also
2021 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
2021 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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## How to Factor a Quadratic
This post about how to factor a quadratic is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.
## Question
Factor completely:
$\inline&space;\fn_jvn&space;{\color{DarkBlue}&space;2x^2-4x-6}$
## Solution
The first step in factoring a polynomial (that’s an expression, like this one, that has multiples of powers of x added together) is to look for a common factor. Notice that every one of the terms (the pieces that get added together) is even. That means each term has a factor of 2. Factor that out:
$\fn_jvn {\color{DarkBlue} 2x^2-4x-6=2(x^2-2x-3)}$
Are we done? That depends on whether $\inline \fn_jvn x^2-2x-3$ can be factored.
Can it?
### Sidebar: Finding the Magic Numbers
How do you factor a quadratic trinomial whose leading coefficient is 1 — that is, whose $\inline \fn_jvn x^2$ term is not multiplied by a number (just 1, implicitly)?
Let’s reverse the process to see what’s going on. Think about how two first-degree binomials — that is, terms that look like $\fn_phv x+$ a number — multiply into a trinomial. To do that, consider two binomials, $\inline \fn_jvn x+m$ and $\inline \fn_jvn x+n$. Multiply those together to get
$\fn_jvn {\color{DarkBlue} (x+m)(x+n)=x^2+(m+n)x+mn}$
So how can you go in the other direction? That is, how can you get something like $\fn_cm \left ( x+m \right )\left ( x+n \right )$ from something like $\fn_cm x^2+bx+c$?
Your clues to m and n are that when you multiply them together you get the constant term, the last term, of the expression getting factored, and that when you add them together you get the multiplier of $\fn_phv x$ in the middle term.
Now let’s take an example. Factor $\inline \fn_jvn x^2+2x-8$.
$\fn_jvn {\color{DarkBlue}x^2+(m+n)x+mn}$
$\fn_jvn {\color{DarkBlue}x^2+2x-8}$
So $\inline \fn_jvn m+n=2$ and $\inline \fn_jvn mn=-8$. Can you find two numbers like that? Start with the multiplication. Integer factor pairs of -8, and their sums, are:
m n m + n 1 -8 -7 –1 8 7 2 -4 -2 –2 4 2
It looks like -2, 4 is the ticket, because those two numbers multiply to -8 and add to 2. Then the factoring is:
$\fn_jvn {\color{DarkBlue}x^2+2x-8=(x-2)(x+4)}$
### Back to the Problem
Let’s try that out on $\inline \fn_jvn x^2-2x-3$. See if you can factor the last term, -3, into two numbers that add up to the multiplier of x in the middle term, -2.
$\fn_jvn {\color{DarkBlue}x^2+(m+n)x+mn}$
$\fn_jvn {\color{DarkBlue}x^2-2x-3 }$
-3 has two pairs of integer factors: One is -1 and 3; the other is 1 and -3.
$\fn_jvn {\color{DarkBlue}-3=-3 \cdot 1}$
$\fn_jvn {\color{DarkBlue}-3+1=-2}$
So yes, there are two numbers that multiply to -3 and add to -2. Those numbers are -3 and 1. $\inline \fn_jvn x^2-2x-3$ factors into two binomials, of which one ends with -3 and one ends with 1:
$\fn_jvn {\color{DarkBlue}x^2-2x-3=(x-3)(x+1)}$
To check, you can multiply back.
Finally, don’t forget that 2 is a factor in the original expression. (I just started ignoring it, because it was not relevant to what we were doing at the moment.) So the final answer is
$\fn_jvn {\color{DarkBlue}2(x-3)(x+1)}$
The Accuplacer sample problem to which this is similar is multiple choice: You are given four possible sets of factors of a trinomial. If you don’t want to factor, you can multiply the factors in each answer choice until you find one that works.
I called this thing you factor a quadratic. That means it’s a polynomial whose highest-power term is second power. Many quadratics, like the ones considered above, have three terms; these may also be called trinomials, and you could call the problem above an example of how to factor trinomials.
Factoring is often the first step in solving a quadratic. For more about that, see “Solving a Quadratic.”
This question is similar to #7 in the sample questions for the Accuplacer Advanced Algebra and Functions test.
# Factoring a Sum of Cubes
## Problem
There is a box with a volume of $x^3-64$ cubic inches, a rectangular base, and a height of $x-4$ inches. What is the area of the box’s base? Use the formula $V=bh,$ where $V$ is volume, $B$ is area of the base, and $h$ is height. Give your answer as an expression in simplest form and in terms of $x.$
## Solution
You’re told that the box’s volume is equal to the area of its base times its height: $V=Bh.$ Then the area of this box’s base is equal to its volume divided by its height: $B=V/h.$
$B=\frac{V}{h}=\frac{x^3-64}{x-4}$
To get your answer into simplest form, factor the sum of cubes in the numerator and then carry out any possible cancellation between numerator and denominator. To do that factor the numerator. The numerator is the difference of two cubes, which factors like this:
$a^3-b^3=\left&space;(a-b&space;\right&space;)\left&space;(a^2+ab+b^2&space;\right&space;)$
Apply that to this case:
$a=x;&space;b=4$
Then
$x^3-4^3=\left&space;(&space;x-4\right&space;)\left&space;(&space;x^2+4x+16\right&space;)$
And the area of the base is
$\frac{\left&space;(x-4&space;\right&space;)\left&space;(&space;x^2+4x+4&space;\right&space;)}{x-4}=\bold{x^2+4x+16}$
The hard part of this problem is having the formula for the difference of cubes memorized. Alternatively, you can do polynomial long division or synthetic division. Synthetic division for this problem, $\frac{x^3-64}{x-4},$ should look like this:
Same result.
You may be asked about the sum, rather than the difference, of cubes. That formula is:
$a^3+b^3=\left&space;(&space;a+b\right&space;)\left&space;(&space;a^2-ab+b^2\right&space;)$
.
|
# RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
## RD Sharma Solutions Class 9 Maths Chapter 16 – Free PDF Download
RD Sharma Solutions Class 9 Maths Chapter 16 Circles are provided here in PDF format for an easy understanding of concepts. In brief, a circle is a closed plane curve consisting of all points at a given distance from a fixed point within it, which is called the centre of the circle. The centre of a circle always lies in the interior of the circle. The distance between any points of the circumference of the circle and the centre of the circle is called the radius. Also, the longest chord of a circle is its diameter. RD Sharma Solutions help students to practise questions based on different concepts such as radius, segment, sector and diameter of a circle.
Check the detailed RD Sharma Solutions for Class 9 Maths Chapter 16 and score good marks in the annual exams. The comprehensive solutions to the exercise questions can be checked from the following link. The solutions are designed by expert teachers as per the latest CBSE syllabus updated for 2023-24.
## RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
### Exercise 16.1 Page No: 16.5
Question 1: Fill in the blanks:
(i) All points lying inside/outside a circle are called ______ points/_______ points.
(ii) Circles having the same centre and different radii are called _____ circles.
(iii) A point whose distance from the center of a circle is greater than its radius lies in _________ of the circle.
(iv) A continuous piece of a circle is _______ of the circle.
(v) The longest chord of a circle is a ____________ of the circle.
(vi) An arc is a __________ when its ends are the ends of a diameter.
(vii) Segment of a circle is a region between an arc and _______ of the circle.
(viii) A circle divides the plane, on which it lies, in _________ parts.
Solution:
(i) Interior/Exterior
(ii) Concentric
(iii) The Exterior
(iv) Arc
(v) Diameter
(vi) Semi-circle
(vii) Center
(viii) Three
Question 2: Write the truth value (T/F) of the following with suitable reasons:
(i) A circle is a plane figure.
(ii) Line segment joining the center to any point on the circle is a radius of the circle,
(iii) If a circle is divided into three equal arcs each is a major arc.
(iv) A circle has only finite number of equal chords.
(v) A chord of a circle, which is twice as long as its radius is the diameter of the circle.
(vi) Sector is the region between the chord and its corresponding arc.
(vii) The degree measure of an arc is the complement of the central angle containing the arc.
(viii) The degree measure of a semi-circle is 1800.
Solution:
(i) T
(ii) T
(iii) T
(iv) F
(v) T
(vi) T
(vii) F
(viii) T
### Exercise 16.2 Page No: 16.24
Question 1: The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.
Solution:
Radius of circle (OA) = 8 cm (Given)
Chord (AB) = 12cm (Given)
Draw a perpendicular OC on AB.
We know, perpendicular from centre to chord bisects the chord
Which implies, AC = BC = 12/2 = 6 cm
In right ΔOCA:
Using Pythagoras theorem,
OA2 = AC2 + OC2
64 = 36 + OC2
OC2 = 64 – 36 = 28
or OC = √28 = 5.291 (approx.)
The distance of the chord from the centre is 5.291 cm.
Question 2: Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.
Solution:
Distance of the chord from the centre = OC = 5 cm (Given)
Radius of the circle = OA = 10 cm (Given)
In ΔOCA:
Using Pythagoras theorem,
OA2 = AC2 + OC2
100 = AC2 + 25
AC2 = 100 – 25 = 75
AC = √75 = 8.66
As, perpendicular from the centre to chord bisects the chord.
Therefore, AC = BC = 8.66 cm
=> AB = AC + BC = 8.66 + 8.66 = 17.32
Question 3: Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm.
Solution:
Distance of the chord from the centre = OC = 4 cm (Given)
Radius of the circle = OA = 6 cm (Given)
In ΔOCA:
Using Pythagoras theorem,
OA2 = AC2 + OC2
36 = AC2 + 16
AC2 = 36 – 16 = 20
AC = √20 = 4.47
Or AC = 4.47cm
As, perpendicular from the centre to chord bisects the chord.
Therefore, AC = BC = 4.47 cm
=> AB = AC + BC = 4.47 + 4.47 = 8.94
Question 4: Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.
Solution:
Given: AB = 5 cm, CD = 11 cm, PQ = 3 cm
Draw perpendiculars OP on CD and OQ on AB
Let OP = x cm and OC = OA = r cm
We know, perpendicular from centre to chord bisects it.
Since OP⊥CD, we have
CP = PD = 11/2 cm
And OQ⊥AB
AQ = BQ = 5/2 cm
In ΔOCP:
By Pythagoras theorem,
OC2 = OP2 + CP2
r2 = x2 + (11/2) 2 …..(1)
In ΔOQA:
By Pythagoras theorem,
OA2=OQ2+AQ2
r2= (x+3) 2 + (5/2) 2 …..(2)
From equations (1) and (2), we get
(x+3) 2 + (5/2) 2 = x2 + (11/2) 2
Solve above equation and find the value of x.
x2 + 6x + 9 + 25/4 = x2 + 121/4
(using identity, (a+b) 2 = a2 + b2 + 2ab )
6x = 121/4 – 25/4 − 9
6x = 15
or x = 15/6 = 5/2
Substitute the value of x in equation (1), and find the length of radius,
r2 = (5/2)2 + (11/2) 2
= 25/4 + 121/4
= 146/4
or r = √146/4 cm
Question 5: Give a method to find the centre of a given circle.
Solution:
Steps of Construction:
Step 1: Consider three points A, B and C on a circle.
Step 2: Join AB and BC.
Step 3: Draw perpendicular bisectors of chord AB and BC which intersect each other at a point, say O.
Step 4: This point O is a centre of the circle, because we know that, the Perpendicular bisectors of chord always pass through the centre.
Question 6: Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.
Solution:
From figure, Let C is the mid-point of chord AB.
To prove: D is the mid-point of arc AB.
Now, In ΔOAC and ΔOBC
OA = OB [Radius of circle]
OC = OC [Common]
AC = BC [C is the mid-point of chord AB (given)]
So, by SSS condition: ΔOAC ≅ ΔOBC
So, ∠AOC = ∠BOC (BY CPCT)
Therefore, D is the mid-point of arc AB. Hence Proved.
Question 7: Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.
Solution:
Form figure: PQ is a diameter of circle which bisects the chord AB at C. (Given)
To Prove: PQ bisects ∠AOB
Now,
In ΔBOC and ΔAOC
OC = OC [Common side]
AC = BC [Given]
Then, by SSS condition: ΔAOC ≅ ΔBOC
So, ∠AOC = ∠BOC [By c.p.c.t.]
Therefore, PQ bisects ∠AOB. Hence proved.
### Exercise 16.3 Page No: 16.40
Question 1: Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.
Solution:
Let R, S and M be the position of Ishita, Isha and Nisha respectively.
Since OA is a perpendicular bisector on RS, so AR = AS = 24/2 = 12 cm
Radii of circle = OR = OS = OM = 20 cm (Given)
In ΔOAR:
By Pythagoras theorem,
OA2+AR2=OR2
OA2+122=202
OA2 = 400 – 144 = 256
Or OA = 16 m …(1)
From figure, OABC is a kite since OA = OC and AB = BC. We know that, diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.
So in ΔRSM, ∠RCS = 900 and RC = CM …(2)
Now, Area of ΔORS = Area of ΔORS
=>1/2×OA×RS = 1/2 x RC x OS
=> OA ×RS = RC x OS
=> 16 x 24 = RC x 20
=> RC = 19.2
Since RC = CM (from (2), we have
RM = 2(19.2) = 38.4
So, the distance between Ishita and Nisha is 38.4 m.
Question 2: A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
Solution:
Since, AB = BC = CA. So, ABC is an equilateral triangle
Radius = OA = 40 m (Given)
We know, medians of equilateral triangle pass through the circumcentre and intersect each other at the ratio 2 : 1.
Here AD is the median of equilateral triangle ABC, we can write:
OA/OD = 2/1
or 40/OD = 2/1
or OD = 20 m
Therefore, AD = OA + OD = (40 + 20) m = 60 m
By Pythagoras theorem,
AC2 = 602 + (AC/2) 2
AC2 = 3600 + AC2 / 4
3/4 AC2 = 3600
AC2 = 4800
or AC = 40√3 m
Therefore, length of string of each phone will be 40√3 m.
### Exercise 16.4 Page No: 16.60
Question 1: In figure, O is the centre of the circle. If ∠APB = 500, find ∠AOB and ∠OAB.
Solution:
∠APB = 500 (Given)
By degree measure theorem: ∠AOB = 2∠APB
∠AOB = 2 × 500 = 1000
Again, OA = OB [Radius of circle]
Then ∠OAB = ∠OBA [Angles opposite to equal sides]
Let ∠OAB = m
In ΔOAB,
By angle sum property: ∠OAB+∠OBA+∠AOB=1800
=> m + m + 1000 = 1800
=>2m = 1800 – 1000 = 800
=>m = 800/2 = 400
∠OAB = ∠OBA = 400
Question 2: In figure, it is given that O is the centre of the circle and ∠AOC = 1500. Find ∠ABC.
Solution:
∠AOC = 1500 (Given)
By degree measure theorem: ∠ABC = (reflex∠AOC)/2 …(1)
We know, ∠AOC + reflex(∠AOC) = 3600 [Complex angle]
1500 + reflex∠AOC = 3600
or reflex ∠AOC = 3600−1500 = 2100
From (1) => ∠ABC = 210 o /2 = 105o
Question 3: In figure, O is the centre of the circle. Find ∠BAC.
Solution:
Given: ∠AOB = 800 and ∠AOC = 1100
Therefore, ∠AOB+∠AOC+∠BOC=3600 [Completeangle]
Substitute given values,
800 + 1000 + ∠BOC = 3600
∠BOC = 3600 – 800 – 1100 = 1700
or ∠BOC = 1700
Now, by degree measure theorem
∠BOC = 2∠BAC
1700 = 2∠BAC
Or ∠BAC = 1700/2 = 850
Question 4: If O is the centre of the circle, find the value of x in each of the following figures.
(i)
Solution:
∠AOC = 1350 (Given)
From figure, ∠AOC + ∠BOC = 1800 [Linear pair of angles]
1350 +∠BOC = 1800
or ∠BOC=1800−1350
or ∠BOC=450
Again, by degree measure theorem
∠BOC = 2∠CPB
450 = 2x
x = 450/2
(ii)
Solution:
∠ABC=400 (given)
∠ACB = 900 [Angle in semicircle]
In ΔABC,
∠CAB+∠ACB+∠ABC=1800 [angle sum property]
∠CAB+900+400=1800
∠CAB=1800−900−400
∠CAB=500
Now, ∠CDB = ∠CAB [Angle is on same segment]
This implies, x = 500
(iii)
Solution:
∠AOC = 1200 (given)
By degree measure theorem: ∠AOC = 2∠APC
1200 = 2∠APC
∠APC = 1200/2 = 600
Again, ∠APC + ∠ABC = 1800 [Sum of opposite angles of cyclic quadrilaterals = 180 o ]
600 + ∠ABC=1800
∠ABC=1800−600
∠ABC = 1200
∠ABC + ∠DBC = 1800 [Linear pair of angles]
1200 + x = 1800
x = 1800−1200=600
The value of x is 600
(iv)
Solution:
∠CBD = 650 (given)
From figure:
∠ABC + ∠CBD = 1800 [ Linear pair of angles]
∠ABC + 650 = 1800
∠ABC =1800−650=1150
Again, reflex ∠AOC = 2∠ABC [Degree measure theorem]
x=2(1150) = 2300
The value of x is 2300
(v)
Solution:
∠OAB = 350 (Given)
From figure:
∠OBA = ∠OAB = 350 [Angles opposite to equal radii]
InΔAOB:
∠AOB + ∠OAB + ∠OBA = 1800 [angle sum property]
∠AOB + 350 + 350 = 1800
∠AOB = 1800 – 350 – 350 = 1100
Now, ∠AOB + reflex∠AOB = 3600 [Complex angle]
1100 + reflex∠AOB = 3600
reflex∠AOB = 3600 – 1100 = 2500
By degree measure theorem: reflex ∠AOB = 2∠ACB
2500 = 2x
x = 2500/2=1250
(vi)
Solution:
∠AOB = 60o (given)
By degree measure theorem: reflex∠AOB = 2∠OAC
60 o = 2∠ OAC
∠OAC = 60 o / 2 = 30 o [Angles opposite to equal radii]
Or x = 300
(vii)
Solution:
∠BAC = 500 and ∠DBC = 700 (given)
From figure:
∠BDC = ∠BAC = 500 [Angle on same segment]
Now,
In ΔBDC:
Using angle sum property, we have
∠BDC+∠BCD+∠DBC=1800
Substituting given values, we get
500 + x0 + 700 = 1800
x0 = 1800−500−700=600
or x = 60o
(viii)
Solution:
∠DBO = 400 (Given)
Form figure:
∠DBC = 900 [Angle in a semicircle]
∠DBO + ∠OBC = 900
400+∠OBC=900
or ∠OBC=900−400=500
Again, By degree measure theorem: ∠AOC = 2∠OBC
or x = 2×500=1000
(ix)
Solution:
From figure:
In ΔDAB:
Angle sum property: ∠ADB + ∠DAB + ∠ABD = 1800
By substituting the given values, we get
320 + ∠DAB + 500 = 1800
∠DAB=1800−320−500
∠DAB = 980
Now,
∠DAB+∠DCB=1800 [Opposite angles of cyclic quadrilateral, their sum = 180 degrees]
980+x=1800
or x = 1800−980=820
The value of x is 82 degrees.
(x)
Solution:
∠BAC = 350 and ∠DBC = 650
From figure:
∠BDC = ∠BAC = 350 [Angle in same segment]
In ΔBCD:
Angle sum property, we have
∠BDC + ∠BCD + ∠DBC = 1800
350 + x + 650 = 1800
or x = 1800 – 350 – 650 = 800
(xi)
Solution:
∠ABD = 400, ∠CPD = 1100 (Given)
Form figure:
∠ACD = ∠ABD = 400 [Angle in same segment]
In ΔPCD,
Angle sum property: ∠PCD+∠CPO+∠PDC=1800
400 + 1100 + x = 1800
x=1800−1500 =300
The value of x is 30 degrees.
(xii)
Solution:
∠BAC = 520 (Given)
From figure:
∠BDC = ∠BAC = 520 [Angle in same segment]
Since OD = OC (radii), then ∠ODC = ∠OCD [Opposite angle to equal radii]
So, x = 520
Question 5: O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.
Solution:
In ΔOBD and ΔOCD:
∠ODB = ∠ODC [Each 900]
OD = OD [Common]
Therefore, By RHS Condition
ΔOBD ≅ ΔOCD
So, ∠BOD = ∠COD…..(i)[By CPCT]
Again,
By degree measure theorem: ∠BOC = 2∠BAC
2∠BOD = 2∠BAC [Using(i)]
∠BOD = ∠BAC
Hence proved.
Question 6: In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.
Solution:
Since, BO is the bisector of ∠ABC, then,
∠ABO = ∠CBO …..(i)
From figure:
Radius of circle = OB = OA = OB = OC
∠OAB = ∠OCB …..(ii) [opposite angles to equal sides]
∠ABO = ∠DAB …..(iii) [opposite angles to equal sides]
From equations (i), (ii) and (iii), we get
∠OAB = ∠OCB …..(iv)
In ΔOAB and ΔOCB:
∠OAB = ∠OCB [From (iv)]
OB = OB [Common]
∠OBA = ∠OBC [Given]
Then, By AAS condition : ΔOAB ≅ ΔOCB
So, AB = BC [By CPCT]
Question 7: In figure, O is the centre of the circle, then prove that ∠x = ∠y + ∠z.
Solution:
From the figure:
∠3 = ∠4 ….(i) [Angles in same segment]
∠x = 2∠3 [By degree measure theorem]
∠x = ∠3 + ∠3
∠x = ∠3 + ∠4 (Using (i) ) …..(ii)
Again, ∠y = ∠3 + ∠1 [By exterior angle property]
or ∠3 = ∠y − ∠1 …..(iii)
∠4 = ∠z + ∠1 …. (iv) [By exterior angle property]
Now, from equations (ii) , (iii) and (iv), we get
∠x = ∠y − ∠1 + ∠z + ∠1
or ∠x = ∠y + ∠z + ∠1 − ∠1
or x = ∠y + ∠z
Hence proved.
### Exercise 16.5 Page No: 16.83
Question 1: In figure, ΔABC is an equilateral triangle. Find m∠BEC.
Solution:
ΔABC is an equilateral triangle. (Given)
Each angle of an equilateral triangle is 60 degrees.
∠BAC + ∠BEC = 180o (Opposite angles of quadrilateral)
60o + ∠BEC = 180 o
∠BEC = 180 o – 60 o
∠BEC = 120 o
Question 2: In figure, Δ PQR is an isosceles triangle with PQ = PR and m∠PQR=35°. Find m∠QSR and m∠QTR.
Solution:
Given: ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°
In ΔPQR:
∠PQR = ∠PRQ = 35o (Angle opposite to equal sides)
Again, by angle sum property
∠P + ∠Q + ∠R = 180 o
∠P + 35 o + 35 o = 180 o
∠P + 70 o = 180 o
∠P = 180 o – 70 o
∠P = 110 o
∠QSR + ∠QTR = 180 o (Opposite angles of quadrilateral)
110 o + ∠QTR = 180 o
∠QTR = 70 o
Question 3: In figure, O is the centre of the circle. If ∠BOD = 160o, find the values of x and y.
Solution:
From figure: ∠BOD = 160 o
By degree measure theorem: ∠BOD = 2 ∠BCD
160 o = 2x
or x = 80 o
y + x = 180 o
Putting value of x,
y + 80 o = 180 o
y = 100 o
Answer: x = 80 o and y = 100 o.
Question 4: In figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100o and ∠ABD = 70o, find ∠ADB.
Solution:
From figure:
100 o + ∠BAD = 180o
By angle sum property: ∠ADB + ∠DAB + ∠ABD = 180 o
∠ADB + 80o + 70 o = 180 o
Question 5: If ABCD is a cyclic quadrilateral in which AD||BC (figure). Prove that ∠B = ∠C.
Solution:
=> ∠A + ∠C = 180o ………(1)
and ∠A + ∠B = 180o ………(2)
[Co-interior angles]
Form (1) and (2), we have
∠B = ∠C
Hence proved.
Question 6: In figure, O is the centre of the circle. Find ∠CBD.
Solution:
Given: ∠BOC = 100o
By degree measure theorem: ∠AOC = 2 ∠APC
100 o = 2 ∠APC
or ∠APC = 50 o
Again,
∠APC + ∠ABC = 180 o (Opposite angles of a cyclic quadrilateral)
50o + ∠ABC = 180 o
or ∠ABC = 130 o
Now, ∠ABC + ∠CBD = 180 o (Linear pair)
130o + ∠CBD = 180 o
or ∠CBD = 50 o
Question 7: In figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 500, find ∠AOC.
Solution:
Given: ∠OBD = 500
Here, AB and CD are the diameters of the circles with centre O.
∠DBC = 900 ….(i)
[Angle in the semi-circle]
Also, ∠DBC = 500 + ∠OBC
900 = 500 + ∠OBC
or ∠OBC = 400
Again, By degree measure theorem: ∠AOC = 2 ∠ABC
∠AOC = 2∠OBC = 2 x 400 = 800
Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(∠CAB) = 300. Find m(∠ACB) and m(∠ABC).
Solution:
Given: m(∠CAB)= 300
To Find: m(∠ACB) and m(∠ABC).
Now,
∠ACB = 900 (Angle in semi-circle)
Now,
In △ABC, by angle sum property: ∠CAB + ∠ACB + ∠ABC = 1800
300 + 900 + ∠ABC = 1800
∠ABC = 600
Answer: ∠ACB = 900 and ∠ABC = 600
Question 9: In a cyclic quadrilateral ABCD if AB||CD and ∠B = 70o , find the remaining angles.
Solution:
A cyclic quadrilateral ABCD with AB||CD and ∠B = 70o.
∠B + ∠C = 180o (Co-interior angle)
700 + ∠C = 1800
∠C = 1100
And,
=> ∠B + ∠D = 1800 (Opposite angles of Cyclic quadrilateral)
700 + ∠D = 1800
∠D = 1100
Again, ∠A + ∠C = 1800 (Opposite angles of cyclic quadrilateral)
∠A + 1100 = 1800
∠A = 700
Answer: ∠A = 700 , ∠C = 1100 and ∠D = 1100
Question 10: In a cyclic quadrilateral ABCD, if m ∠A = 3(m∠C). Find m ∠A.
Solution:
∠A + ∠C = 180o …..(1)
Since m ∠A = 3(m∠C) (given)
=> ∠A = 3∠C …(2)
Equation (1) => 3∠C + ∠C = 180 o
or 4∠C = 180o
or ∠C = 45o
From equation (2)
∠A = 3 x 45o = 135o
Question 11: In figure, O is the centre of the circle ∠DAB = 50°. Calculate the values of x and y.
Solution:
Given : ∠DAB = 50o
By degree measure theorem: ∠BOD = 2 ∠BAD
so, x = 2( 500) = 1000
Since, ABCD is a cyclic quadrilateral, we have
∠A + ∠C = 1800
500 + y = 1800
y = 1300
### Exercise VSAQs Page No: 16.89
Question 1: In figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
Solution:
By degree measure theorem: ∠AOB = 2 ∠APB
so, ∠AOB = 2 × 70° = 140°
Since AOBC is a cyclic quadrilateral, we have
∠ACB + ∠AOB = 180°
∠ACB + 140° = 180°
∠ACB = 40°
Question 2: In figure, two congruent circles with centres O and O’ intersect at A and B. If ∠AO’B = 50°, then find ∠APB.
Solution:
As we are given that, both the triangle are congruent which means their corresponding angles are equal.
Therefore, ∠AOB = AO’B = 50°
Now, by degree measure theorem, we have
∠APB = ∠AOB/2 = 250
Question 3: In figure, ABCD is a cyclic quadrilateral in which ∠BAD=75°, ∠ABD=58° and ∠ADC=77°, AC and BD intersect at P. Then, find ∠DPC.
Solution:
∠DBA = ∠DCA = 580 …(1)
[Angles in same segment]
ABCD is a cyclic quadrilateral :
Sum of opposite angles = 180 degrees
∠A +∠C = 1800
750 + ∠C = 1800
∠C = 1050
Again, ∠ACB + ∠ACD = 1050
∠ACB + 580 = 1050
or ∠ACB = 470 …(2)
Now, ∠ACB = ∠ADB = 470
[Angles in same segment]
Also, ∠D = 770 (Given)
Again From figure, ∠BDC + ∠ADB = 770
∠BDC + 470 = 770
∠BDC = 300
In triangle DPC
∠PDC + ∠DCP + ∠DPC = 1800
300 + 580 + ∠DPC = 1800
or ∠DPC = 92
Question 4: In figure, if ∠AOB = 80° and ∠ABC=30°, then find ∠CAO.
Solution:
Given: ∠AOB = 800 and ∠ABC = 300
To find: ∠CAO
Join OC.
Central angle subtended by arc AC = ∠COA
then ∠COA = 2 x ∠ABC = 2 x 300 = 600 …(1)
In triangle OCA,
OC = OA
∠OCA = ∠CAO …(2)
[Angle opposite to equal sides]
In triangle COA,
∠OCA + ∠CAO + ∠COA = 1800
From (1) and (2), we get
2∠CAO + 600 = 1800
∠CAO = 600
## RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
In the 16th Chapter of Class 9, RD Sharma Solutions students will study important concepts. Some of the concepts are listed below:
• Circle Introduction
• Position of a point with respect to a circle
• Circular Disc
• Concentric Circles
• Degree measure of an arc
• Chord and segment of a circle
• Congruence of circles and arcs
## Frequently Asked Questions on RD Sharma Solutions for Class 9 Maths Chapter 16
Q1
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Q2
### How do the BYJU’S RD Sharma Solutions for Class 9 Maths Chapter 16 help students to score good marks in the CBSE board?
RD Sharma Solutions for Class 9 Maths Chapter 16 provide in-depth knowledge of concepts in a simple and understandable language. Each step is explained in detail to help students clear their doubts instantly. Thorough practice of these solutions enables students to enhance their skills and procure more marks in their CBSE board.
Q3
### What are the topics covered under RD Sharma Solutions for Class 9 Maths Chapter 16?
The concepts covered under RD Sharma Solutions for Class 9 Maths Chapter 16 are as follows:
• Introduction to Circle
• Position of a point with respect to a circle
• Circular Disc
• Concentric Circles
• Degree measure of an arc
• Chord and segment of a circle
• Congruence of circles and arcs
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Slope-Intercept Form ( Real World ) | Algebra | CK-12 Foundation
# Slope-Intercept Form
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# Finding the Cheapest Cell Phone Plan - Answer Key
## Real World Applications – Algebra I
### Topic
How do you know which cell phone plan is the cheapest?
### Student Exploration
The student will investigate, compare and determine which cell phone plan is the least expensive over time.
1. Search for two different cell phone companies (i.e. AT&T, Verizon, Sprint, T-Mobile, Boost, Metro PCS).
2. The most organized way to list the cost of each cell phone plan is to make a table. Make a table for each cell phone company. The input $(x)$ values should be the # of months you have the phone, and the output $(y)$ values should be the total cost. “Month 0” should be the initial cost of purchasing the cell phone itself.
3. Create a graph for each table of points. What are realistic values of $x$ and $y$ on this graph? Why?
4. Create an equation for each cell phone plan represented in slope-intercept form.
5. In each equation, what is the slope? What does this represent in the cell phone plan? What is the $y-$intercept in each equation? What does this represent in the cell phone plan? Justify your thinking.
6. Which cell phone plan is the cheapest, and for how long?
7. When are both cell phone plans the same cost? Use the Substitution Method to solve this system of linear equations you created. What do “$x$” and “$y$” each represent in this solution?
Any phone and phone plan can be compared. For the sake of this exercise, we’re going to compare the cell phone plans for MetroPCS and Verizon Wireless with a Blackberry cell phone. After conducting some research on the internet, I found that the Blackberry phone costs $149 and the plan costs$60 per month for MetroPCS. For Verizon, the phone costs $50, and the plan costs$100/month.
We can determine the two equations for the two cell phone plans. The equation that represents the total cost of a Blackberry and plan for Metro is $y = 60x + 149$, where $x$ represents the number of months that have passed after having the phone, and $y$ represents the total cumulative cost of having the phone. The equation for Verizon is $y = 100x + 50$.
The graph above represents the two lines for the two different cell phone plans. We can see that the intersection point is where the two cell phones will be the same cost and at what time. By looking at the graph, it looks like that after 2 and a half months, they will both cost about $300. We can solve a system of linear equations to find the exact time and month that these two plans are the same. We can use substitution, and substitute one of the $y$ values for its counterpart in the other equation. $Y = 60x + 149$ and $y = 100x + 50$ We can substitute $y$ in the second equation with $60x + 149$, since $y$ and $60x + 149$ are the same. $60x + 149 = 100x + 50$ Now we should subtract $60x$ from both sides and also subtract 50 from both sides. $99 = 40x$ Now we want to divide both sides by 40 to solve for $x$. $2.475 = x$ Now we want to find out the corresponding $y$ value. We can do this by substituting 2.475 in for $x$ in either equation and solve for $y$. $Y &= 60x + 149\\Y &= 60(2.475) + 149\\Y &= 148.5 + 149\\Y &= 297.50$ We now know that in 2.475 months, both plans’ total cost would be$297.50.
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# Isosceles Triangle
Triangle with two equal sides are known as Isosceles Triangle
In Latin, the word “Isosceles” mean “Equal Legged”
Hence ,in Isosceles triangle, two of the legs are equal in length
## Base and Legs of Isosceles Triangle
The sides which are equal in length is known as Legs.
And the third side is known as Base
The above triangle ABC is an Isosceles Triangle with two equal sides.
The equal sides AB & AC is called Legs
The other side BC is called Base of the triangle
## Angles of Isosceles Triangle
The angle between the two equal sides are known as Vertex angle.
Angle that rests on the base of Isosceles triangle is known as Base Angle
The above image shown Isosceles Triangle ABC, where side AB = AC
\angle A is Vertex angle since it is between the two equal sides.
\angle B and \angle C are base angle since it rest on base BC.
In Isosceles triangle, base angles are equal to each other (explanation is given below)
\angle B = \angle C
## Examples of Isosceles Triangle
In all the above examples, note the two sides (marked in red) which are equal to each other.
And the third side (marked in Blue) which is different in length
## Properties of Isosceles Triangle
(01) In Isosceles Triangle, the angle opposite to equal sides are equal to each other
In the above Isosceles Triangle, side AB = AC
Angle opposite to side AB = \angle C
Angle opposite to side AC = \angle B
According to the Isosceles property, angle opposite to equal sides are equal.
\angle B=\angle C
(02) Perpendicular drawn from vertex angle to the base bisects the base and vertex Angle
Above image is of Isosceles Triangle ABC, where:
Side AB = AC
\angle A = 40 degree
When a perpendicular line AO is drawn from Vertex angle A to side BC then the line will bisect angle A and side BC.
You can observe that side BC and \angle A is divided into two equal halves
(03) Right Angled Isosceles Triangle
In Right angles isosceles triangle one angle is 90 degree and the other two angle will be 45 degree each.
Proof
Prove that in right angle Isosceles triangle, the other two angles are 45 degree
Solution
Given triangle ABC is right angled isosceles triangle, where
\angle B = 90 degree
And side AB = BC
Since side AB = BC, their opposite angles are also equal
\angle A = \angle C = x degree
We know that sum of angle of triangle is 180 degree
\angle A + \angle B + \angle C = 180
⟹ x + 90 + x = 180
⟹ 2x = 90
⟹ x = 45 degree
Hence, \angle A = \angle C = 45 degree
(04) In Right Angles Isosceles Triangle, the altitude on hypotenuse is half the length of hypotenuse
ABC is right angled isosceles triangle where AB = AC and AC is the hypotenuse
BO is the perpendicular line on hypotenuse AC
According to the property:
BO = \frac{AC}{2}
i.e. the altitude is half the hypotenuse
## Area of Isosceles Triangle
Given below are two formulas for calculation of area of isosceles triangle.
(a) Formula when base and height of triangle is given
(b) Formula when length of all sides are given
Area formula when base and height is given
The above image is of isosceles triangle with the following details:
Length of Base = b
Height from base = h
In this case,
Area of Triangle = \frac{1}{2} \ \times \ b\ \times h
Area Formula when length of sides are given
Given above is an isosceles triangle with side length “a” and base length “b”
In this case, the formula is:
Area of Triangle = \frac{b}{2}\sqrt{a^{2} -\frac{b^{2}}{4}}
## Isosceles Triangle Theorem Proof
Theorem
Angle opposite to equal sides are equal
Given:
ABC is an isosceles triangle with sides AB = AC
AD is an angle bisector such that \angle 1 = \angle 2
To Prove:
\angle B = \angle C
Proof:
Taking Triangle ABD and ACD
Side AB = AC
\angle 1 = \angle 2
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# How to solve logarithmic equations?
To solve logarithmic equations, we have to use the laws of logarithms to rewrite the expressions in a more convenient way. After simplifying and reducing the logarithmic expressions, we will generally get one of two types of logarithmic equations. Depending on the type of equation obtained, we can obtain the answer simply by comparing the arguments of the logarithms or we can rewrite the logarithm in its exponential form to solve.
Here, we will look at a summary of the laws of logarithms. Also, we will look at the two types of logarithmic equations that we can obtain. Finally, we will learn how to solve logarithmic equations with examples.
##### ALGEBRA
Relevant for
Learning to solve logarithmic equations with examples.
See examples
##### ALGEBRA
Relevant for
Learning to solve logarithmic equations with examples.
See examples
## Types of logarithmic equations
Generally, we have two types of logarithmic equations. We need to recognize these two cases so that solving logarithmic equations becomes easier.
• The first type looks like this:
If we have only one logarithm on each side of the equation that has the same base, we can equalize the arguments of the logarithms and solve. In this case, the arguments are the algebraic expressions represented by P and Q.
• The second type looks like this:
If we have only one logarithm on only one side of the equation, then we can express it as an exponential expression and solve it that way.
## Summary of the laws of logarithms
### Product Law
When we have a logarithm of a product, we can write it as the sum of the individual logarithms of the factors:
### Quotient Law
When we have a logarithm of a quotient, we can write it as the logarithm of the numerator minus the logarithm of the denominator:
### Power Law
The logarithm of an exponential value can be rewritten as the exponent multiplied by the logarithm of the base (without the exponent):
### Law of Zero
The logarithm of 1 where the base different from zero and greater than zero is equal to zero:
### Law of identity
The logarithm of a value equal to the base is equal to 1. The base, b must be greater than zero, but cannot be equal to 1:
### Law of the logarithm of the exponent
The logarithm of an exponential value, where the base of the power equals the base of the logarithm, is equal to the exponent:
### Law of the exponent of a logarithm
Raising the logarithm of a number to its base equals the number:
## Learn to solve logarithmic equations with examples
### EXAMPLE 1
• Solve the logarithmic equation: $$\log_{3}(x+2)+\log_{3}(2)=\log_{3}(14)$$
Solution: We can use the product law to form a single logarithmic expression on the left hand side. Therefore, we have:
$latex \log_{3}(x+2)+\log_{3}(2)=\log_{3}(14)$
$latex \log_{3}[(x+2)2]=\log_{3}(14)$
Using the distributive property of multiplication, we have:
$latex \log_{3}(2x+4)=\log_{3}(14)$
We have a logarithm with the same base on each side, so we can form an equation with the arguments:
$latex 2x+4=14$
Now we have a linear equation, so we can easily solve:
$latex 2x+4=14$
$latex 2x=10$
$latex x=5$
### EXAMPLE 2
• Solve the logarithmic equation: $$\log_{2}(x)+\log_{2}(x+2)=\log_{2}(x+12)$$
Solution: Similar to the previous example, we can use the product law to form a single logarithm on the left side of the equation:
$$\log_{2}(x)+\log_{2}(x+2)=\log_{2}(x+12)$$
$latex \log_{2}[x(x+2)]=\log_{2}(x+12)$
Using the distributive property to distribute the x and obtain:
$latex \log_{2}({{x}^2}+2x)=\log_{2}(x+12)$
The logarithm on both sides of the equation has the same base, so we can eliminate it and form an equation with the arguments:
$latex {{x}^2}+2x=x+12$
In this case, we have a quadratic equation. We can move all the terms to one side of the equation and use factoring to solve:
$latex {{x}^2}+2x=x+12$
$latex {{x}^2}+2x-x-12=0$
$latex {{x}^2}+x-12=0$
$latex (x+4)(x-3)=0$
We make each factor equal to zero and solve:
⇒ $latex x=-4$
⇒ $latex x=3$
Therefore, we have two answers, $latex x=-4$ and $latex x=3$.
### EXAMPLE 3
• Solve the logarithmic equation: $$\log_{2}(x+4)-\log_{2}(3)=\log_{2}(x-2)-\log_{2}(5)$$
Solution: Here, we have a logarithm subtraction on each side of the equation. We can use the quotient law to obtain a single logarithm on each side:
$$\log_{2}(x+4)-\log_{2}(3)=\log_{2}(x-2)-\log_{2}(5)$$
$latex \log_{2}(\frac{x+4}{3})=\log_{2}(\frac{x-2}{5})$
We cannot reduce the expressions inside logarithms. However, since both logarithms have the same base, we can eliminate them and form an equation with the arguments:
$latex \frac{x+4}{3}=\frac{x-2}{5}$
We can simplify this expression by cross multiplying:
$latex 5(x+4)=3(x-2)$
We apply the distributive property of multiplication to simplify:
$latex 5x+20=3x-6$
We have a linear equation, so we can easily solve:
$latex 5x+20=3x-6$
$latex 2x=-26$
$latex x=-13$
### EXAMPLE 4
• Solve the logarithmic equation: $latex \ln({{x}^2})+\frac{1}{2}\ln(4)=\ln({{x}^2}+16)$.
Solution: Here, we have natural logarithms, that is, logarithms with base e. These logarithms are denoted with ln. Although they are written slightly differently, all the laws of logarithms normally apply to natural logarithms.
We can use the power law to rewrite the logarithm that has a fraction in front:
$latex \ln({{x}^2})+\frac{1}{2}\ln(4)=\ln({{x}^2}+16)$
$latex \ln({{x}^2})+\ln({{4}^{\frac{1}{2}}})=\ln({{x}^2}+16)$
$latex \ln({{x}^2})+\ln(2)=\ln({{x}^2}+16)$
Now, we can use the law of the logarithm of a product to simplify the left part of the equation:
$latex \ln(2{{x}^2})=\ln({{x}^2}+16)$
We have a natural logarithm on each side, so we can eliminate it and write an equation with the arguments:
$latex 2{{x}^2}={{x}^2}+16$
We can solve this quadratic equation easily:
$latex 2{{x}^2}={{x}^2}+16$
$latex 2{{x}^2}-{{x}^2}=16$
$latex {{x}^2}=16$
Taking the square root of both sides, we have
$latex x=\sqrt{16}$
$latex x=\pm 4$
Therefore, we have two answers, $latex x=4$ and $latex x=-4$.
### EXAMPLE 5
• Solve the logarithmic equation: $latex \log(5x+40)=2$.
Solution: We see that we have the logarithm without any written base. When we have this, we assume that the logarithm has base 10. In fact, the logarithm with base 10 is called the common logarithm.
In this case, we only have a one-sided logarithm. We consider this as the second case mentioned above:
We are going to transform the equation from logarithmic to exponential form to solve. Therefore, the logarithm argument stays in place and we remove the logarithm. The number on the right becomes the exponent of the base of the logarithm, which is 10:
$latex \log(5x+40)=2$
$latex 5x+40={{10}^2}$
We simplify the exponent and solve the equation:
$latex 5x+40=100$
$latex 5x=60$
$latex x=12$
### EXAMPLE 6
• Solve the logarithmic equation: $latex \log_{2}(5x)-2=\log_{2}(2x-2)$.
Solution: We can move all logarithms to one side of the equation and the constant terms to the other side:
$latex \log_{2}(5x)-2=\log_{2}(2x-2)$
$latex \log_{2}(5x)-\log_{2}(2x-2)=2$
Now, we can use the quotient law to form a single logarithm on the left side:
$latex \log_{2}(\frac{5x}{2x-2})=2$
Now, we are going to write the expression in its exponential form. The argument remains in the same place and we eliminate the logarithm. The 2 becomes the exponent of the base:
$latex \frac{5x}{2x-2}={{2}^2}$
$latex \frac{5x}{2x-2}=4$
We can cross multiply to simplify:
$latex 5x=4(2x-2)$
We simplify with the distributive property of multiplication:
$latex 5x=8x-8$
We can easily solve the linear equation:
$latex 5x-8x=-8$
$latex -3x=-8$
$latex x=\frac{8}{3}$
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# Value of a function
Module by: Sunil Kumar Singh. E-mail the author
The value of a function at “x = a” is denoted as “f(a)”. The working rule for finding value of a function is to replace independent variable “x” by “a”.
## Polynomial and rational functions
Problem 1: Find “f(y)”, if
y = f x = 1 - x 1 + x y = f x = 1 - x 1 + x
Solution :
Statement of the problem : The given function is a rational function. We have to evaluate the function when independent variable is function itself.
We need to replace “x” by “y”.
f y = 1 y 1 + y = 1 1 x 1 + x 1 + 1 - x 1 + x f y = 1 y 1 + y = 1 1 x 1 + x 1 + 1 - x 1 + x
f y = 1 + x 1 + x 1 + x + 1 x = 2 x 2 = x f y = 1 + x 1 + x 1 + x + 1 x = 2 x 2 = x
Problem 2: Find “f(x)”, if
f x - 1 = x 2 1 f x - 1 = x 2 1
Solution :
Statement of the problem : The given function is a polynomial function with a polynomial as its argument. We have to evaluate the function for independent variable “x”.
We need to replace “x-1” by “x” in the given equation to find “f(x)”. The right hand side expression, however, does not contain term “x-1”. We, therefore, need to find the term, which will replace “x”. Clearly if "x" replaces "x-1", then "x+1" will replace "x-1+1 = x"
Thus, we need to replace “x” by “x+1”.
f x + 1 - 1 = x + 1 2 1 = x 2 + 2 x + 1 1 = x 2 + 2 x f x + 1 - 1 = x + 1 2 1 = x 2 + 2 x + 1 1 = x 2 + 2 x
Problem 3: If { f x } = x + 1 x { f x } = x + 1 x , then prove that :
{ f x } 3 = f x 3 + 3 f 1 x { f x } 3 = f x 3 + 3 f 1 x
Solution :
Statement of the problem : The function is a polynomial function. We have to evaluate cube of the function, which involves evaluation of function for arguments, which are independent variable, raised to certain integral powers.
The cube of given function is :
{ f x } 3 = x + 1 x 3 = x 3 + 1 / x 3 + 3 x 2 X 1 x + 3 x X 1 x 2 { f x } 3 = x + 1 x 3 = x 3 + 1 / x 3 + 3 x 2 X 1 x + 3 x X 1 x 2
{ f x } 3 = x 3 + 1 x 3 + 3 x + 1 x { f x } 3 = x 3 + 1 x 3 + 3 x + 1 x
Now, f x 3 f x 3 is :
f x 3 = x 3 + 1 x 3 f x 3 = x 3 + 1 x 3
Hence,
{ f x } 3 = x 3 + 1 x 3 + 3 x + 1 x = f x 3 + 3 f x { f x } 3 = x 3 + 1 x 3 + 3 x + 1 x = f x 3 + 3 f x
But, we see that :
f 1 x = 1 x + x = f x f 1 x = 1 x + x = f x
Hence,
{ f x } 3 = f x 3 + 3 f x = f x 3 + 3 f 1 x { f x } 3 = f x 3 + 3 f x = f x 3 + 3 f 1 x
Problem 4: If
f x = 1 + x 1 x f x = 1 + x 1 x
Then, find
f x f x 2 1 + { f x } 2 f x f x 2 1 + { f x } 2
Solution :
Statement of the problem : The given function is rational function. We have to find the expression which involves (i) function, (ii) function with argument as squared independent variable and (iii) square of the function.
We need to substitute for various terms in the given expression :
f x f x 2 1 + { f x } 2 = 1 + x 1 x X 1 + x 2 1 x 2 1 + 1 + x 1 x 2 f x f x 2 1 + { f x } 2 = 1 + x 1 x X 1 + x 2 1 x 2 1 + 1 + x 1 x 2
= 1 + x 1 + x 2 1 x 1 x 2 1 x 2 + 1 + x 2 1 x 2 = 1 + x 1 + x 2 1 x 1 x 2 1 x 2 + 1 + x 2 1 x 2
1 + x 2 1 x 2 2 1 + x 2 1 x 2 = 1 + x 2 2 1 + x 2 = 1 2 1 + x 2 1 x 2 2 1 + x 2 1 x 2 = 1 + x 2 2 1 + x 2 = 1 2
## Modulus functions
Problem 5: If
f x = | x | x ; x 0 f x = | x | x ; x 0
Then, evaluate
| f a f - a | | f a f - a |
Solution :
Statement of the problem : Function, f(x), involves modulus and is in rational form. The value of this function, in turn, forms the part of a expression to be evaluated. We have to find the value of expression.
We first evaluate the expression without modulus sign :
f a f - a = | a | a | a | a = | a | a + | a | a = 2 | a | a ; a 0 f a f - a = | a | a | a | a = | a | a + | a | a = 2 | a | a ; a 0
But, we know that | a | = ± a | a | = ± a
f a f - a = 2 X ± a a = ± 2 f a f - a = 2 X ± a a = ± 2
Taking modulus of the expression,
| f a f - a | = 2 ; a 0 | f a f - a | = 2 ; a 0
Note that we need to keep the condition for which the given expression is evaluated.
## Logarithmic functions
Problem 6: Find f 2 x 1 + x 2 f 2 x 1 + x 2 , if
f x = log e 1 + x 1 - x f x = log e 1 + x 1 - x
Solution :
Statement of the problem : The given function is transcendental logarithmic function. We have to evaluate the function for an argument (input to function), which is itself a rational function in independent variable, “x”.
We need to replace “x” by “ 2 x / 1 + x 2 2 x / 1 + x 2 ”.
f 2 x 1 + x 2 = log e 1 + 2 x 1 + x 2 1 2 x 1 + x 2 = log e 1 + x 2 + 2 x 1 + x 2 2 x f 2 x 1 + x 2 = log e 1 + 2 x 1 + x 2 1 2 x 1 + x 2 = log e 1 + x 2 + 2 x 1 + x 2 2 x
f 2 x 1 + x 2 = log e 1 + x 1 x 2 = 2 log e 1 + x 1 x = 2 f x f 2 x 1 + x 2 = log e 1 + x 1 x 2 = 2 log e 1 + x 1 x = 2 f x
## Trigonometric functions
Problem 7: Find f π / 4 f π / 4 , if
f x = 2 cot x 1 + cot 2 x f x = 2 cot x 1 + cot 2 x
Solution :
Statement of the problem : The given function is a rational function with trigonometric function as independent variable. We have to find the value of function for a particular angle.
We need to replace “x” by “ π / 4 π / 4 ”.
f π / 4 = 2 cot π / 4 1 + cot 2 π / 4 f π / 4 = 2 cot π / 4 1 + cot 2 π / 4
As cot π / 4 = 1 cot π / 4 = 1 ,
f π 4 = 2 X 1 1 + 1 2 = 1 f π 4 = 2 X 1 1 + 1 2 = 1
Problem 8: Find f tan θ f tan θ , if
f x = 2 x 1 + x 2 f x = 2 x 1 + x 2
Solution :
Statement of the problem : The given function is a rational function. We have to evaluate the function for a value, which is itself a trigonometric function.
We need to replace “x” by “ tan θ tan θ ”.
f tan θ = 2 tan θ 1 + tan 2 θ = sin 2 θ f tan θ = 2 tan θ 1 + tan 2 θ = sin 2 θ
Problem 9: If f x = cos { log x } f x = cos { log x } , then prove that :
f x y + f x y = 2 f x f y f x y + f x y = 2 f x f y
Solution :
Statement of the problem : The given function, f(x) is a trigonometric function, whose input is a logarithmic function. We have to evaluate LHS of the given equation to equate the same to RHS.
Here, we evaluate each term of the left hand side of the equation separately and then combine the result.
f x y = cos { log e x y } = cos log e x + log e y f x y = cos { log e x y } = cos log e x + log e y
f x y = cos { log e x y } = cos log e x log e y f x y = cos { log e x y } = cos log e x log e y
Substituting in the LHS expression, we have :
{ f x y + f x y } = cos log e x + log e y + cos log e x log e y { f x y + f x y } = cos log e x + log e y + cos log e x log e y
We know that :
cos C + cos D = 2 cos C + D 2 cos C D 2 cos C + cos D = 2 cos C + D 2 cos C D 2
Hence,
{ f x y + f x y } = 2 cos log e x + log e y + log e x log e y 2 cos log e x + log e y log e x + log e y 2 { f x y + f x y } = 2 cos log e x + log e y + log e x log e y 2 cos log e x + log e y log e x + log e y 2
{ f x y + f x y } = 2 cos log e x cos log e y { f x y + f x y } = 2 cos log e x cos log e y
{ f x y + f x y } = 2 cos log e x cos log e y = 2 f x f y { f x y + f x y } = 2 cos log e x cos log e y = 2 f x f y
## Acknowledgment
Author wishes to thank Mr Jay Sicard, Pasadena TX for suggesting valuable typographical correction on the topic.
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1969 AHSME Problems/Problem 8
Problem
Triangle $ABC$ is inscribed in a circle. The measure of the non-overlapping minor arcs $AB$, $BC$ and $CA$ are, respectively, $x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}$. Then one interior angle of the triangle is:
$\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text{(D) } 61^{\circ}\quad \text{(E) } 122^{\circ}$
Solution
$[asy] draw(circle((0,0),65)); draw((25,60)--(39,-52)--(-52,-39)--(25,60)); dot((25,60)); dot((39,-52)); dot((-52,-39)); dot((0,0)); draw((0,0)--(-52,-39)); draw((0,0)--(39,-52)); draw((0,0)--(25,60)); label("A",(-52,-39),SW); label("B",(25,60),NE); label("C",(39,-52),SE); [/asy]$ Because the triangle is inscribed, the sum of the minor arcs equals $360^\circ$. Thus, $$x+75+2x+25+3x-22=360$$ $$6x+78=360$$ Solving this yields $x = 47$, so the inscribed angles are $122^\circ$, $99^\circ$, and $119^\circ$. Noting that an angle of $\triangle ABC$ is half of its corresponding inscribed angle, so the angles of $\triangle ABC$ are $59.5^\circ$, $49.5^\circ$, and $\boxed{\textbf{(D) } 61^\circ}$.
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# Applied Mathematics/The Basics
## The Basics of linear algebraEdit
$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{12} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ \end{bmatrix}.$
A matrix is composed of a rectangular array of numbers arranged in rows and columns. The horizontal lines are called rows and the vertical lines are called columns. The individual items in a matrix are called elements. The element in the i-th row and the j-th column of a matrix is referred to as the i,j, (i,j), or (i,j)th element of the matrix. To specify the size of a matrix, a matrix with m rows and n columns is called an m-by-n matrix, and m and n are called its dimensions.
### Basic operation[1]Edit
Operation Definition Example
Addition The sum A+B of two m-by-n matrices A and B is calculated entrywise:
(A + B)i,j = Ai,j + Bi,j, where 1 ≤ im and 1 ≤ jn.
$\begin{bmatrix} 1 & 3 & 1 \\ 1 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 5 \\ 7 & 5 & 0 \end{bmatrix} = \begin{bmatrix} 1+0 & 3+0 & 1+5 \\ 1+7 & 0+5 & 0+0 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 6 \\ 8 & 5 & 0 \end{bmatrix}$
Scalar multiplication The scalar multiplication cA of a matrix A and a number c (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of A by c:
(cA)i,j = c · Ai,j.
$2 \cdot \begin{bmatrix} 1 & 8 & -3 \\ 4 & -2 & 5 \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 & 2\cdot 8 & 2\cdot -3 \\ 2\cdot 4 & 2\cdot -2 & 2\cdot 5 \end{bmatrix} = \begin{bmatrix} 2 & 16 & -6 \\ 8 & -4 & 10 \end{bmatrix}$
Transpose The transpose of an m-by-n matrix A is the n-by-m matrix AT (also denoted Atr or tA) formed by turning rows into columns and vice versa:
(AT)i,j = Aj,i.
$\begin{bmatrix} 1 & 2 & 3 \\ 0 & -6 & 7 \end{bmatrix}^\mathrm{T} = \begin{bmatrix} 1 & 0 \\ 2 & -6 \\ 3 & 7 \end{bmatrix}$
### Practice problemsEdit
(1) $\begin{bmatrix} 5 & 7 & 3 \\ 1 & 2 & 9 \end{bmatrix} + \begin{bmatrix} 4 & 0 & 5 \\ 8 & 3 & 0 \end{bmatrix} =$
(2) $4 \begin{bmatrix} -1 & 0 & -5 \\ 7 & 9 & -6 \end{bmatrix} =$
(3) $\begin{bmatrix} -2 & 5 & 7 \\ 0 & 0 & 9 \end{bmatrix}^\mathrm{T} =$
## Matrix multiplicationEdit
Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix. If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B[2]
$[\mathbf{AB}]_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + \cdots + A_{i,n}B_{n,j} = \sum_{r=1}^n A_{i,r}B_{r,j}$[3]
Schematic depiction of the matrix product AB of two matrices A and B.
### ExampleEdit
$\begin{bmatrix} -2 & 0 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$
$= \begin{bmatrix} -2+0 & -4+0 \\ 3+6 & 6+(-2) \end{bmatrix}$
$=\begin{bmatrix} -2 & -4 \\ 9 & 4 \end{bmatrix}$
### Practice ProblemsEdit
(1) $\begin{bmatrix} 1 & 0 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} 4 \\ 2 \end{bmatrix} =$
(2) $\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} =$
## Dot productEdit
A row vector is a 1 × m matrix, while a column vector is a m × 1 matrix.
Suppose A is row vector and B is column vector, then the dot product is defined as follows;
$A \cdot B= |A||B| cos \theta$
or
$\mathbf{A}\cdot \mathbf{B} = \begin{pmatrix}a_1 & a_2 & \cdots & a_n\end{pmatrix} \begin{pmatrix}b_1 \\ b_2 \\ \vdots \\ b_n\end{pmatrix} = a_1b_1+a_2b_2+\cdots+a_nb_n = \sum_{i=1}^n a_ib_i$
Suppose $\mathbf{A} = \begin{pmatrix}a_1 & a_2 & a_3\end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix}b_1 \\ b_2 \\ b_3\end{pmatrix}$ The dot product is
$\mathbf{A}\cdot \mathbf{B} = \begin{pmatrix}a_1 & a_2 & a_3\end{pmatrix} \begin{pmatrix}b_1 \\ b_2 \\ b_3\end{pmatrix} = a_1b_1+a_2b_2+a_3b_3$
### ExampleEdit
Suppose $\mathbf{A} = \begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix}7 \\ 5 \\ 4\end{pmatrix}$
$\mathbf{A}\cdot \mathbf{B} = \begin{pmatrix}2 & 1 & 3\end{pmatrix} \begin{pmatrix}7 \\ 5 \\ 4\end{pmatrix}$
$= 2\cdot7+1\cdot5+3\cdot4$
$=14+5+12$
$=31$
### Practice problemsEdit
(1) $\mathbf{A} = \begin{pmatrix}3 \\ 2 \\ 5\end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix}1 \\ 4 \\ 3\end{pmatrix}$
$\mathbf{A}\cdot \mathbf{B} =$
(2) $\mathbf{A} = \begin{pmatrix}1 \\ 0 \\ 3\end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix}6 \\ 9 \\ 2\end{pmatrix}$
$\mathbf{A}\cdot \mathbf{B} =$
## Cross productEdit
Cross product is defined as follows:
$A \times B = |A||B| sin \theta$
Or, using detriment,
$\mathbf{A \times B}=\begin{vmatrix} e_x&e_y&e_z\\ a_x&a_y&a_z\\ b_x&b_y&b_z\\ \end{vmatrix} = (a_y b_z - a_z b_y, a_z b_x - a_x b_z, a_x b_y - a_yb_x)$
where $e$ is unit vector.
## ReferenceEdit
1. Sourced from Matrix (mathematics), Wikipedia, 28th March 2013.
2. Sourced from Matrix (mathematics), Wikipedia, 30th March 2013.
3. Sourced from Matrix (mathematics), Wikipedia, 30th March 2013.
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A Question of Curvature
Katie Steckles
People are often interested in mathematical ‘life hacks’ – ways they can use mathematics to make their life easier. For me a much more interesting conversation results when there is a hack people already know about and use all the time, which they didn’t realise was actually quite inherently mathematical.
This occurred to me recently when I was thinking about Gaussian curvature. This is a geometrical property of objects and surfaces which describes how the surface curves in three-dimensional space. Gaussian curvature is the answer to a very specific definition: if an object curves in one direction, what does it do in the other direction?
Are You Being Curved?
To find the Gaussian curvature at a point on a given shape, you need to measure two perpendicular axes of curvature; for those of you holding a sharpie pen and/or who would like a simpler way of phrasing that, you need to draw a cross on the shape, made from two lines which intersect at right angles.
Once you’ve picked a point and drawn a cross, you need to compare how those two lines curve in 3D. Imagine your object is a shape like a sphere – a football, a bowling ball or an orange – and that you’ve drawn your cross on the very top of it. The two lines will both be curving downwards as they move away from the centre of the cross.
How much are they curving by? Well, we measure the curvature of a circle as 1/r, where r is the radius of the circle. So each line is curving at a rate of 1/r in the same direction. The Gaussian curvature is calculated as the product of these principal curvatures: for two identical curves, 1/r × 1/r = 1/r², which for any sensible value of r will be a small but definitely positive number.
This is because a sphere is an example of a shape which has positive Gaussian curvature, and hopefully with a little thought you can convince yourself that it will remain true whichever point on the sphere you choose: if it curves away from you in one line, the perpendicular line curves away from you too. This means the sphere has positive Gaussian curvature everywhere.
If instead we wanted a shape with zero Gaussian curvature, we could picture something flat – a piece of paper, a section of floor (in most sensible places), or a pizza. If we measure the curvature in two directions at a chosen point, it will be zero, and the product of zero with itself is still zero. But you didn’t need me to tell you that something which does not curve has no curvature.
Imagine instead a shape like a cylinder: a rolled-up tube of paper, a pencil or a fancy cylindrical pillow. If we pick a point on the flat end of the cylinder, that will also have zero curvature, just like on our flat objects. But if we pick a point on the curved side, and draw a cross there, something interesting happens.
Taking The Tube
To find the principal curvatures, we need to look for a maximum and a minimum: this forces the orientation of our cross to have one line running horizontally parallel to the cylinder’s length, and the other one wrapping around the cylinder’s circular face (On the sphere, the curvature is the same in every direction, so the orientation of the cross is unimportant).
Since we have a curve in one direction, the curvature is 1/r (where r is the radius of the cylinder). But in the other direction, it will be zero – and 1/r times zero is just zero. So even though a cylinder has a curved face, when considering Gaussian curvature, it isn’t curved at all. And this is where things start to get interesting – because Gaussian curvature is an intrinsic property of an object and cannot be changed.
So if I start with a piece of paper, and want to curve it into a cylinder, I can do that pretty easily: a cylinder has the same curvature as a flat object, so it will do this with no problems. However, if I tried to curve it further – into a donut shape or a sphere, things would start to get messy – the paper would need to be crumpled, or cut, or torn and stuck together, in order to create the shape.
As anyone who has tried to peel an orange will tell you, you don’t get a whole piece of flat peel; similarly, anyone who has attempted to map the globe will quickly have found that in order to draw it all on a flat page, you need to stretch and squash parts of the surface, or leave gaps along the edge (á la the magnificent Goode homolosine projection). An object with positive curvature can’t map onto an object with zero curvature, and vice versa.
So if you have an object with zero curvature, it will always have zero curvature: if you bend it in one direction, it will remain steadfastly straight in the perpendicular direction. And if this sounds like a familiar idea, that will be because it is something you have very likely used regularly without even thinking about it as a ‘life hack’, namely when eating a pizza.
I deliberately included pizza on my list of zero-curvature objects, because a pizza, and indeed a single slice of pizza, is a flat object with the zero curvature property. And if you pick up a piece of pizza by its crust, you might find it starts to unhelpfully droop, threatening to discharge all of its delicious toppings onto the floor. In this situation, the solution is simple: introduce curvature in a perpendicular direction.
Again, translating that into a more sensible instruction: bend the crust of the pizza upwards at the sides. This will make the pizza curve in that direction, meaning that according to the rules of geometry, it will not be able to curve in the direction at right-angles to that: along the length of the slice.
Of course, this depends slightly on the sogginess of your pizza dough: a soft, stretchy base will rashly ignore the universe’s rules of geometry and probably stretch and curve in whichever directions it wants. But assuming a reasonably structurally intact pizza base, this ‘hack’ has been used by millions of people around the world to prevent their slice from drooping. But how many of them, I wonder, credit Carl Friedrich Gauss with its success?
The post A Question of Curvature originally appeared on the HLFF SciLogs blog.
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# Area of a Rhombus and Kite
Contributed by:
This pdf is based on finding the areas of rhombus and kites. The area of a kite is half the product of the lengths of its diagonals.
Area of a rhombus or any parallelogram = base × height.
1. 10.1 Day 3
Discovery 10.1 – Area of a Rhombus & Kite
Discover the formula!
1) Fold the rectangle in half in lengthwise. Crease and unfold.
2) Fold the rectangle widthwise but NOT in the middle. Crease and unfold.
3) Create points on the edge of each side of the index card where the creases are.
4) Connect consecutive points to form a kite.
5) Cut along these four lines.
How are the base and height of the original rectangle related to the diagonals of the kite?
How is the area of the rectangle related to the area of the kite?
Area of a Kite
This same formula applies to the area of a rhombus. Why?
Area of a Rhombus
Can you think of another quadrilateral that you can apply this area formula to? Why?
2. Let’s try a few examples:
1. Find d2 of a kite in which d1 = 16 cm 2. Find the area of a rhombus with
and A = 48 cm2. d1 = (6x + 4) in. and d2 = (10x + 10) in.
3. Find the area of the kite below: 4. Find the area of the rhombus below:
d1 = 24
6 d2 = 12
24
26
5. Find the area of a rhombus whose perimeter 6. Find the area of the kite given KIT is a
is 20 and whose longer diagonal is 8. Right angle , KT = 8 and KE = 5.
(Hint: Draw a picture to help you!) (Picture is not drawn to scale!)
K
I E
T
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2. ## Re: Perimeter of triangle
I don't think you've been given enough information...
3. ## Re: Perimeter of triangle
How ?,i have the answer but i want to know the way to solve it .
4. ## Re: Perimeter of triangle
Can you find any of the shown segments? Also, a hint: use the right triangle altitude theorem.
5. ## Re: Perimeter of triangle
Get AM by Pythagoras (=8)
AngleACM=angleB (both=90-A)
sinB=6/CB But from triangle ACM sinACM=8/10 So sinB=8/10 So 8/10=6/CB So CB=15/2
In triangle BCM cosB=BM/CB So 6/10=BM/15/2 BM=9/2
Perimeter=AC+CB+BM+MA=10+7.5+4.5+8=30
7. ## Re: Perimeter of triangle
Originally Posted by Mhmh96
$|AM|=\sqrt{10^2-6^2}=8$
Since $\Delta AMC \sim \Delta BCM$ it follows :
$|BM| : 6 = 6 : 8$
$|BM|=\frac{9}{2} \Rightarrow |AB|=\frac{25}{2}$
Hence :
$|BC|=\sqrt{\left(\frac{25}{2}\right)^2-10^2}=\frac{15}{2}$
therefore perimeter is : $30$
Thank you !
9. ## Re: Perimeter of triangle
Can you explain this part of your solution ?
10. ## Re: Perimeter of triangle
Originally Posted by Mhmh96
Here is a slightly different approach:
1. Determine $|\overline{AM}| = 8$ by Pythagorean theorem.
2. Use Pythagorean theorem in triangle ABC:
$(8+\overline{BM})^2-10^2 = (\overline{BC})^2$ and in the small right triangle:
$6^2+ (\overline{BM})^2 = (\overline{BC})^2$
3. Subtract both equations columnwise and solve for $\overline{BM}$
11. ## Re: Perimeter of triangle
Originally Posted by Mhmh96
Can you explain this part of your solution ?
This is the right triangle altitude theorem. Its proof involves the fact that triangles AMC and CMB are similar.
12. ## Re: Perimeter of triangle
Originally Posted by emakarov
This is the right triangle altitude theorem. Its proof involves the fact that triangles AMC and CMB are similar.
Of course you are right but you also can use 2 right triangles to prove the altitude theorem:
(Using the labels of the attached diagram):
$CM^2+BM^2=BC^2$
$CM^2+AM^2=AC^2$
$2CM^2 + AM^2 + BM^2 = BC^2+AC^2 = AB^2$
Since $AB = AM + BM$ the right side of the last equation becomes:
$2CM^2 + AM^2 + BM^2 = (AM + BM)^2 = AM^2 + 2AM \cdot BM + BM^2$
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# How do you solve using the quadratic formula 3x^2 + 6x = 12?
##### 1 Answer
Mar 14, 2018
$- 1 \pm \sqrt{5}$
#### Explanation:
First get the 12 to the other side.
$3 {x}^{2} + 6 x - 12 = 0$
Quadratic Formula is
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{\text{2a}}$
where
a=3
b=6
c=-12
Now we just plug it in
$\frac{- \left(6\right) \pm \sqrt{{6}^{2} - 4 \left(3\right) \left(- 12\right)}}{\text{2(3)}}$
You can split this up like so
$- \frac{6}{\text{2(3)" +-sqrt(6^2-4(3)(-12))/"2(3)}}$
Then get this
$- \frac{6}{\text{6" +- sqrt(36-(-144))/"6}}$
Simplify
$- 1 \pm \frac{\sqrt{180}}{\text{6}}$
Then we simplify the square-root like so,
$\sqrt{36} \cdot \sqrt{5} = \sqrt{180}$
Thus
$6 \sqrt{5} = \sqrt{180}$
$- 1 \pm \frac{6 \sqrt{5}}{\text{6}}$
Which simplifies to
$- 1 \pm \sqrt{5}$
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# Lagrange Multiplier Calculator
To use lagrange multiplier calculator, enter the values in the given boxes, select to maximize or minimize, and click the calcualte button
Give Us Feedback
## Lagrange Calculator
Lagrange multiplier calculator is used to evaluate the maxima and minima of the function with steps. This Lagrange calculator finds the result in a couple of a second.
## What is the Lagrange multiplier?
The method of Lagrange multipliers, which is named after the mathematician Joseph-Louis Lagrange, is a technique for locating the local maxima and minima of a function that is subject to equality constraints.
(i.e., subject to the requirement that one or more equations have to be precisely satisfied by the chosen values of the variables).
The fundamental concept is to transform a limited problem into a format that still allows the derivative test of an unconstrained problem to be used.
The Lagrangian function is a reformulation of the original issue that results from the relationship between the gradient of the function and the gradients of the constraints.
## The formula for the Lagrange multiplier
The formula of the Lagrange multiplier is:
### Example of Lagrange multiplier
Use the method of Lagrange multipliers to find the minimum value of g(y, t) = y2 + 4t2 – 2y + 8t subjected to constraint y + 2t = 7
Solution:
Step 1: Write the objective function and find the constraint function; we must first make the right-hand side equal to zero.
g(y, t) = y2 + 4t2 – 2y + 8t
The constraint function is y + 2t – 7 = 0
So h(y, t) = y + 2t – 7
To minimize the value of function g(y, t), under the given constraints.
g(y, t) = y2 + 4t2 – 2y + 8t corresponding to c = 10 and 26.
Step 2: Now find the gradients of both functions.
∇g(y, t) = (2y - 2)i + (8t + 8)j
∇h(y, t) = i + 2j
Now equation ∇g(y, t) = a∇h(y, t) becomes
(2y-2)i + (8t + 8)j = a(i + 2j)
Step 3: Compare the above equation.
2y - 2 = a
8t + 8 = 2a
Step 4: Now solve the system of the linear equation.
t= 1 and y= 5
Step 5: Evaluate the function
g(y, t) = 52+ 4(1)2 – 2(5) + 8(1) = 27
So h has a relative minimum value is 27 at the point (5,1).
Use our Lagrangian calculator above to cross-check the above result.
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Name: 4th Grade Math Proficient Practice Test 3
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1.
Computation – Fractions – RIT 211 – 220
Write 6/12 in simplest form.
a. 1/2 d. 1/6 b. 12/24 e. 1/6 c. 2/4
2.
2/10 + 4/10 =
a. 3/10 d. 6/20 b. 3/5 e. 6/10 c. 2/5
3.
1 7/8 + 3 3/4 =
a. 4 10/12 d. 5 3/4 b. 4 7/8 e. 45/8 c. 5 5/8
4.
Multiply and simplify.
5 x 1/3 =
a. 15/9 d. 1 2/3 b. 1 6/9 e. 14/3 c. 16/3
5.
a. 3/7 d. 1 b. 11/12 e. 3/8 c. 2/12
6.
a. 6 2/3 d. 22/3 b. 3 5/6 e. 6 1/2 c. 2 5/6
7.
Order each fraction from least to greatest.
1/2, 1/3, 5/6, 2/3, 1/6
a. 5/6, 1/6, 2/3, 1/3, 1/2 d. 1/6, 1/3, 1/2, 2/3, 5/6 b. 5/6, 2/3, 1/2, 1/3, 1/ e. 1/2, 1/3, 5/6, 2/3, 1/6 c. 5/6, 2/3, 1/6, 1/3, 1/
8.
Computation – Integers – RIT 211 – 220
Which statement is true?
a = 12
b = 21
a. d. b. e. c.
9.
–6 + (-7) =
a. +13 d. 1 b. –1 e. –13 c. 42
10.
5 + (-15) =
a. 10 d. –10 b. –20 e. –75 c. 20
11.
6 + (-2) + 3 + (-5) =
a. 2 d. 6 b. 16 e. -2 c. –16
12.
Multiply these two integers.
(-8) (6) =
a. 14 d. –48 b. –14 e. –2 c. 48
13.
a. 4 d. –18 b. –4 e. -30 c. 18
14.
Computation – Whole Numbers – RIT 211 – 220
a. 6 x 6 = 42 d. 7 < 42 b. 42 > 6 e. 6 x 7 = 42 c. 7 ¸ 6 = 42
15.
a. 3,647 d. 507 R5 b. 500 R7 e. 57 R5 c. 18,282
16.
When you multiply any number by 10 you add a ______ to that number.
a. 10 d. 00 b. 0 e. decimal c. 1
17.
a. A. 8,456 d. 101 R20 b. 1,120 e. 112 R20 c. 8,608
18.
Geometry – Congruency & Similarity – RIT 201 – 210
a. 1, 4 d. 1, 2 b. 2, 4 e. 1, 3 c. 3, 5
19.
Geometry – Properties and Terminology- RIT 201 – 210
a. a ray d. an acute angle b. an angle e. a line segment c. a line
20.
Which figure shows the circle’s diameter?
a. d. b. e. c.
21.
a. A d. D b. B e. E c. C
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How do you find the formula of an exponential graph given A = (1, 12) and B = (2, 48)?
Oct 30, 2015
This graph has a formula: $y = 3 \cdot \left({4}^{x}\right)$
Explanation:
We are looking for a function $y = a \cdot {b}^{x}$ which passes through $\left(1 , 12\right)$ and $\left(2 , 24\right)$
If we substitute the points' coordinates to the function formula we get:
$\left\{\begin{matrix}a \cdot b = 12 \\ a \cdot {b}^{2} = 48\end{matrix}\right.$
If we substitute first equation to the second we get: $12 \cdot b = 48$
so $b = 4$
Now if we substitute $b = 4$ into first equation we get: $4 a = 12$
$a = 3$
Finally we can writhe the answer: The function is: $y = 3 \cdot {4}^{x}$
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# 2018 AMC 8 Problems/Problem 15
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem 15
In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?
$[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircle,white,black); filldraw(shift(1,0)*unitcircle,white,black); [/asy]$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}$
## Solution 1
Let the radius of the large circle be $R$. Then the radii of the smaller circles are $\frac R2$. The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is $\frac 14$. This means the combined area of the 2 smaller circles is half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is $\boxed{\textbf{(D) } 1}$
## Solution 2
Let the radius of the two smaller circles be $r$. It follows that the area of one of the smaller circles is ${\pi}r^2$. Thus, the area of the two inner circles combined would evaluate to $2{\pi}r^2$ which is $1$. Since the radius of the bigger circle is two times that of the smaller circles(the diameter), the radius of the larger circle in terms of $r$ would be $2r$. The area of the larger circle would come to $(2r)^2{\pi} = 4{\pi}r^2$.
Subtracting the area of the smaller circles from that of the larger circle(since that would be the shaded region), we have $$4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.$$
Therefore, the area of the shaded region is $\boxed{\textbf{(D) } 1}$
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# VCE Specialist Mathematics/Units 3 and 4: Specialist Mathematics/Coordinate Geometry
« VCE Specialist MathematicsCoordinate Geometry » Units 3 and 4: Specialist Mathematics Circular Functions
## Preface
Formal Definition: Coordinate (algebraic) geometry is a branch of mathematics which combines techniques of abstract algebra, especially commutative algebra, with the language and the problems of geometry.
Translation: Understanding the math behind the various features that appear on a graph, allowing one to rapidly, and accurately, draw complex graphs (with points of interest).
## Properties of Graphs
### Asymptotes
#### Definition
Asymptotes are values which the graph approaches but does not touch. An asymptote is itself a graph and is categorized as follows:
• Vertical; A constant value (graph) on the horizontal axis (e.g. ${\displaystyle x=1}$ ).
• Horizontal; A constant value (graph) on the vertical axis (e.g. ${\displaystyle y=1}$ ).
• Oblique (i.e. not Vertical or Horizontal); A non-constant graph (e.g. ${\displaystyle y=x^{2}}$ ).
#### Comprehension
1. Take a function ${\displaystyle y=ax^{m}+{\frac {1}{bx^{n}}}\ +c}$ .
2. As ${\displaystyle x\to 0,{\frac {1}{bx^{n}}}\ \to \infty ,ax^{m}+c\to c}$
3. The numerator (a function or a number, as is shown here: ${\displaystyle 1}$ ) is divided by an extremely small number. Hence making the fraction an extremely large number.
4. To understand why this happens grab any number, and divide it by an extremely small number (e.g. ${\displaystyle 10^{-30}}$ )
5. The value of ${\displaystyle {\frac {1}{bx^{n}}}\ \to \infty }$ overshadows the rest of the graph, namely the ${\displaystyle ax^{m}+c\to c}$ .
6. Hence making ${\displaystyle {\frac {1}{bx^{n}}}\ }$ an oblique (non-constant) asymptote, as it is approached by, but never actually touched by the graph due to the addition of ${\displaystyle ax^{m}+c}$ to every y value. This is in addition to the limit provided when ${\displaystyle x=1}$ which is not included in this graph, but must be if it exists on others (e.g. after polynomial division).
7. As ${\displaystyle x\to \infty ,{\frac {1}{bx^{n}}}\ \to 0,ax^{m}+c\to \infty }$
8. The numerator (a function or a number, as is shown here:${\displaystyle 1}$ ) is divided by an extremely large number. Hence making the fraction an extremely small number.
9. To understand why this happens grab any number, and divide it by an extremely large number (e.g. ${\displaystyle 10^{30}}$ )
10. The value of ${\displaystyle ax^{m}+c\to \infty }$ overshadows the rest of the graph, namely the ${\displaystyle {\frac {1}{bx^{n}}}\ \to 0}$ .
11. Hence making ${\displaystyle ax^{m}+c}$ the oblique (non-constant) asymptote, as it is approached by, but never actually touched by the graph due to the addition of ${\displaystyle {\frac {1}{bx^{n}}}\ }$ to every y value.
### Circles, Ellipses and Hyperbolas
#### Definition
##### Ellipses
General formula:
• ${\displaystyle {\frac {(x-h)^{2}}{a^{2}}}+{\frac {(y-k)^{2}}{b^{2}}}=1}$
General Notes:
• Point ${\displaystyle (h,k)}$ defines the ellipses center.
• Points ${\displaystyle (\pm a+h,k)}$ defines the ellipses domain, and horizontal endpoints - i.e. horizontal stretch.
• Points ${\displaystyle (h,\pm b+k)}$ defines the ellipses range, and vertical endpoints - i.e. vertical stretch.
##### Circles
General formula:
• ${\displaystyle (x-h)^{2}+(y-k)^{2}=r^{2}}$
General Notes:
• Point ${\displaystyle (h,k)}$ defines the circles center.
• Points ${\displaystyle (\pm r+h,k)}$ defines the circles domain - i.e. stretch.
• Points ${\displaystyle (h,\pm r+k)}$ defines the circles range - i.e. stretch.
• A circle is a subset of an ellipse, such that ${\displaystyle a=b=r}$ .
##### Hyperbolas
General formulae:
• ${\displaystyle {\frac {(x-h)^{2}}{a^{2}}}-{\frac {(y-k)^{2}}{b^{2}}}=1}$
• ${\displaystyle {\frac {(y-k)^{2}}{b^{2}}}-{\frac {(x-h)^{2}}{a^{2}}}=1}$
General Notes:
• Point ${\displaystyle (h,k)}$ defines the hyperbolas center.
• Points ${\displaystyle (\pm a+h,k)}$ defines the hyperbolas domain, ${\displaystyle [\pm a+h,\pm \infty )}$ .
• The switch in positions of the fractions containing x and y, indicate the type of hyperbola - i.e. vertical or horizontal. The hyperbola is horizontal in the first, and negative in the second of the General hyperbolic formulae above.
• Graphs ${\displaystyle y=\pm (\pm a+h,k)}$ defines the hyperbolas domain ${\displaystyle [\pm a+h,\pm \infty )}$ .
#### Comprehension
Is unnecessary for these types of graphs, as the rules listed above are all that are needed. However the ability to recognize these graphs, in varying forms, is required and can be achieved via practice or can be found in Coordinate Geometry section of Common Math Hacks.
### Partial Fractions
#### Graphing Examples
##### General Steps
1. Note the limits caused by a divide by 0.
2. If possible (check the highest power of bottom and top), break up the complex function via polynomial division or partial fractions.
3. If numerator's power is ${\displaystyle <}$ the denominators power, utilize partial fractions.
4. Otherwise if the numerator's power is ${\displaystyle \geq }$ the denominators power, utilise polynomial division.
5. Add the resultant graphs, through the addition of ordinates method, to quickly determine what the graph looks like.
6. Determine the asymptotes (Vertical, Horizontal, Oblique (i.e. a graph)).
7. Determine other points of interest (Turning points (differentiation), Intercepts (let ${\displaystyle y=0}$ , or ${\displaystyle x=0}$ )).
8. Draw the Graph using the above properties.
##### Partial Fractions
1. Take a function: ${\displaystyle y={\frac {3x+7}{x^{2}-2x-3}}\ }$
2. Notice that if ${\displaystyle x^{2}-2x-3=0}$ , hence ${\displaystyle (x+1)(x-3)=0}$ , you get a divide by 0. Hence ${\displaystyle x\neq -1}$ or ${\displaystyle x\neq 3}$ .
3. Break up the function into partial fractions, and you arrive at ${\displaystyle y={\frac {-1}{(x+1)}}\ +{\frac {4}{(x-3)}}\ }$ .
4. Add the resultant graphs, through the addition of ordinates method, to quickly determine what the graph looks like.
5. Notice that as ${\displaystyle x\to -1,{\frac {-1}{(x+1)}}\ \to -\infty ,{\frac {4}{(x-3)}}\ \to -1}$ . The fractional part, ${\displaystyle {\frac {-1}{(x+1)}}\ }$ , overshadows the rest of the equation, namely ${\displaystyle {\frac {4}{(x-3)}}\ }$ . Hence a horizontal asymptote occurs when ${\displaystyle x=-1}$ .
6. Notice that as ${\displaystyle x\to 3,{\frac {4}{(x-3)}}\ \to \infty ,{\frac {-1}{(x+1)}}\ \to {\frac {-1}{4}}\ }$ . The fractional part, ${\displaystyle {\frac {4}{(x-3)}}\ }$ , overshadows the rest of the equation, namely ${\displaystyle {\frac {-1}{4}}\ }$ . Hence a horizontal asymptote occurs when ${\displaystyle x=3}$ .
7. Notice that as ${\displaystyle x\to \infty ,{\frac {-1}{(x+1)}}\ +{\frac {4}{(x-3)}}\ \to 0,y\to 0}$ . Now the oblique (i.e. graph), ${\displaystyle y=0}$ , overshadows the rest of the equation, namely ${\displaystyle {\frac {-1}{(x+1)}}\ +{\frac {4}{(x-3)}}\ }$ . Hence the oblique asymptote occurs when ${\displaystyle y=0}$ .
8. Determine points of interest:
1. When ${\displaystyle y=0,x={\frac {-7}{3}}\ }$ , hence there are no x-intercepts (C is the complex field) in the real plane.
2. When ${\displaystyle x=0,y={\frac {-7}{3}}\ }$ .
3. When ${\displaystyle y'=0,x=-5orx={\frac {1}{3}}\ ,y={\frac {-1}{4}}\ or{\frac {-9}{4}}\ }$
9. Draw the graph.
##### Polynomial Division
1. Take a function: ${\displaystyle y=3x+7+{\frac {x^{3}+19x^{2}+116x+224}{x(x+4)(x+7)}}\ }$
2. Notice that if ${\displaystyle x(x+4)(x+7)=0}$ , you get a divide by 0. Hence ${\displaystyle x\neq 0}$ or ${\displaystyle x\neq -4}$ or ${\displaystyle x\neq -7}$
3. Break up the function, and divide through, using polynomial division, and you arrive at ${\displaystyle y=3x+7+{\frac {x+8}{x}}\ }$ .
4. Add the resultant graphs, through the addition of ordinates method, to quickly determine what the graph looks like.
5. Notice that as ${\displaystyle x\to 0,{\frac {x+8}{x}}\ \to \infty ,3x+7\to 7}$ . The fractional part, ${\displaystyle {\frac {x+8}{x}}\ }$ , overshadows the rest of the equation, namely ${\displaystyle 3x+7}$ . Hence the horizontal asymptote occurs when ${\displaystyle x=0}$ .
6. Notice that as ${\displaystyle x\to \infty ,{\frac {x+8}{x}}\ \to 0,y\to 3x+7}$ . Now the oblique (i.e. graph), ${\displaystyle 3x+7}$ , overshadows the rest of the equation, namely ${\displaystyle {\frac {x+8}{x}}\ }$ . Hence the oblique asymptote occurs when ${\displaystyle y=3x+7}$ .
7. Determine points of interest:
1. When ${\displaystyle y=0,x\in \mathbb {C} }$ , hence there are no x-intercepts (C is the complex field) in the real plane.
2. When ${\displaystyle x=0,y\uparrow }$ , hence there are no y-intercepts (${\displaystyle \uparrow }$ indicates that the previous statement is undefined.)
3. When ${\displaystyle y'=0,x=\pm {\frac {2{\sqrt {6}}}{3}}\ ,y=8\pm 4{\sqrt {6}}}$
8. Draw the graph.
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# Find the area of triangle formed by the tangents of parabola and chord of contacts
Find the area of the triangle formed by the tangents from the point $(h, k)$ to the parabola $y^2=4ax$ and the chord of contact.
I find the the tangents from external points $(h,k)$, i.e. $$(y^2-4ax)(k^2-4ah)=(ky-2a(x+h))^2$$ Again the chord of contact of the given parabola is $ky=2a(x+h)$. Then we try to find out three extremities.
• I find the the tangents from external points $(h,K)$, i.e. $(y^2-4ax)(k^2-4ah)=(ky-2a(x+h))$. again the chord of contact of the given parabola is $ky=2a(x+h)$. Then we try to find out three extremites
– MSMM
May 25 '17 at 3:11
Let $$(x_1,y_1)$$ and $$(x_2,y_2)$$ be the points of contact.
Substitute the chord $$ky=2a(x+h)$$ into the parabola $$y^2=4ax$$ to obtain a quadratic equation in $$y$$. $$y_1$$ and $$y_2$$ are the roots. So we have $$y_1+y_2$$ and $$y_1y_2$$. $$(y_1-y_2)^2=(y_1+y_2)^2-4y_1y_2$$.
$$k(y_1-y_2)=2a(x_1-x_2)$$ and so we can find the distance between the two points of contact.
The distance between the point $$(h,k)$$ and the chord of contact is $$\displaystyle \left|\frac{2a(h)-k(k)+2ah}{\sqrt{4a^2+k^2}}\right|$$.
Multiply the two distances and divide it by $$2$$, we have the area.
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# Fixed points
Take a calculator and enter any number. Then press the cosine key over and over. Eventually the numbers will stop changing. You will either see 0.99984774 or 0.73908513, depending on whether your calculator was in degree mode or radian mode. This is an example of a fixed point, a point that doesn’t change when you apply a function.
The example above is actually two examples, one for cosine of x degrees and one for cosine of x radians. These are two different functions, and they have different fixed points. Note that the two fixed points are not simply related to each other by converting between degrees and radians.
## Contraction mapping theorem
The functions
f(x) = cos(x)
and
g(x) = cos(πx/180)
are both contraction mappings, the former corresponding to radians and the latter to degrees. A function h(x) is a contraction mapping if for any two points x and y,
|h(x) – h(y)| ≤ c|xy|
for some constant c < 1. You can use the mean value theorem to show that c = sin(1) for the function f, and c = π sin(π/180) for the function g.
The contraction mapping theorem says that if a function h is a contraction mapping on a closed interval, then h has a unique fixed point. You can generalize this from working on closed interval to working in any complete metric space.
## Weak contraction mappings
The constant c above is important. Without it, you don’t necessarily have a fixed point. For example, let
f(x) = x + 1/x
on the interval [1, ∞). Then you can show that
|f(x) – f(y)| < |xy|
for xy but the function f has no fixed point. To see this, suppose f(x) = x, then 1/x = 0, but the reciprocal of a positive number is positive.
If you start with a point x ≥ 1 and repeatedly apply f, you’ll get a sequence of points that move closer to each other, but not closer to any particular number. The sequence of iterates diverges.
A function satisfying the inequality above is called weak contraction mapping. In general, a weak contraction mapping need not have a fixed point, as in the example above. But a weak contraction on a compact space will have a fixed point.
## Commentary
As discussed in the link about Kepler’s equation below, something like the contraction mapping theorem was used in practice long before it was formalized in theory.
When I was in grad school, I studied nonlinear PDEs. After seeing a long list of existence theorems and getting lost in the proofs, I wondered what was at the heart of all these theorems. Under all the layers of operator and function space definitions, somewhere there had to be core theorems that said something exists, and I went back over my lecture notes to dig for what these theorems were. In every case it boiled down to a fixed point theorem, not the contraction mapping theorem per se but analogous theorems.
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# Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-1 Statistics for Managers Using Microsoft® Excel 5th Edition.
## Presentation on theme: "Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-1 Statistics for Managers Using Microsoft® Excel 5th Edition."— Presentation transcript:
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-1 Statistics for Managers Using Microsoft® Excel 5th Edition Chapter 9 Fundamentals of Hypothesis Testing: One Sample Tests
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-2 Learning Objectives In this chapter, you will learn: The basic principles of hypothesis testing How to use hypothesis testing to test a mean or proportion The assumptions of each hypothesis-testing procedure, how to evaluate them, and the consequences if they are seriously violated How to avoid the pitfalls involved in hypothesis testing Ethical issues involved in hypothesis testing
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-3 The Hypothesis A hypothesis is a claim (assumption) about a population parameter: population mean population proportion Example: The mean monthly cell phone bill of this city is μ = \$52 Example: The proportion of adults in this city with cell phones is π =.68
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-4 States the assumption (numerical) to be tested Always contains “=”, “≤” or “ ” sign Example: The mean number of TV sets in U.S. Homes is equal to three. Is always about a population parameter, not about a sample statistic. May or may not be rejected The Null Hypothesis, H 0
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-5 The Alternative Hypothesis, H 1 Is the opposite of the null hypothesis e.g., The mean number of TV sets in U.S. homes is not equal to 3 ( H 1 : μ ≠ 3 ) Challenges the status quo Never contains the “=”, “≤” or “ ” sign May or may not be proven For one tail tests is generally the hypothesis that the researcher is trying to prove
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-6 Hypothesis Testing We assume the null hypothesis is true If the null hypothesis is rejected we have proven the alternate hypothesis If the null hypothesis is not rejected we have proven nothing as the sample size may have been to small
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-7 The Hypothesis Testing Process Claim: The population mean age is 50. H 0 : μ = 50, H 1 : μ ≠ 50 Sample the population and find sample mean. Population Sample
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-8 The Hypothesis Testing Process Suppose the sample mean age was X = 20. This is significantly lower than the claimed mean population age of 50. If the null hypothesis were true, the probability of getting such a different sample mean would be very small, so you reject the null hypothesis. In other words, getting a sample mean of 20 is so unlikely if the population mean was 50, you conclude that the population mean must not be 50.
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-9 The Hypothesis Testing Process Sampling Distribution of X μ = 50 If H 0 is true If it is unlikely that you would get a sample mean of this value...... then you reject the null hypothesis that μ = 50. 20... if in fact this were the population mean… X
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-10 The Test Statistic and Critical Values If the sample mean is close to the assumed population mean, the null hypothesis is not rejected. If the sample mean is far from the assumed population mean, the null hypothesis is rejected. How far is “far enough” to reject H 0 ? The level of significance of the test statistic ( ) creates a “line in the sand” for decision making.
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-11 Level of Significance, Defines the unlikely values of the sample statistic if the null hypothesis is true Defines rejection region of the sampling distribution Is designated by , (level of significance) Typical values are.01,.05, or.10 Is the compliment of the confidence coefficient Is selected by the researcher before sampling Provides the critical value of the test
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-12 The Test Statistic and Critical Values Critical Values Distribution of the test statistic Region of Rejection
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-13 Errors in Decision Making Type I Error Reject a true null hypothesis Considered a serious type of error The probability of a Type I Error is Called level of significance of the test Set by researcher in advance Type II Error Failure to reject false null hypothesis The probability of a Type II Error is β
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-14 Errors in Decision Making Possible Jury Trial Outcomes The Truth The VerdictInnocentGuilty InnocentNo ErrorType II Error GuiltyType I ErrorNo Error
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-15 Errors in Decision Making Possible Hypothesis Test Outcomes Actual Situation DecisionH 0 TrueH 0 False Do Not Reject H 0 No Error Probability 1 - α Type II Error Probability β Reject H 0 Type I Error Probability α No Error Probability 1 - β
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-16 Type I & II Error Relationship Type I and Type II errors can not happen at the same time Type I error can only occur if H 0 is true Type II error can only occur if H 0 is false If Type I error probability ( ), then Type II error probability ( β )
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-17 Level of Significance, α H 0 : μ ≥ 50 H 1 : μ < 50 0 H 0 : μ ≤ 50 H 1 : μ > 50 Represents critical value Lower-tail test 0 Upper-tail test Two-tail test Rejection region is shaded 0 H 0 : μ = 50 H 1 : μ ≠ 50 Claim: The population mean age is 50.
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-18 8 Steps in Hypothesis Testing 1. State the null hypothesis, H 0 State the alternative hypotheses, H 1 2. Choose the level of significance, α 3. Choose the sample size, n 4. Determine the appropriate test statistic to use 5. Collect the data 6.Compute the p-value for the test statistic from the sample result 7. Make the statistical decision: Reject H 0 if the p-value is less than alpha 8.Express the conclusion in the context of the problem
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-19 Hypothesis Tests for the Mean Known Z Test Normal Distribution Unknown t Test Student t Distribution Hypothesis Tests for
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-20 Hypothesis Testing: σ Unknown If the population standard deviation is unknown, you use the sample standard deviation S instead of . Because of this change, you also use the t distribution instead of the Z distribution to test the null hypothesis about the mean. All other steps, concepts, and conclusions are the same as the known test.
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-21 Hypothesis Testing: σ Unknown The t test statistic with n-1 degrees of freedom is: Hypothesized
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-22 Hypothesis Testing: σ Unknown Problem The mean cost of a hotel room in New York is said to be \$168 per night. A random sample of 25 hotels resulted in X = \$172.50 and S = 15.40. Test at the = 0.05 level.
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-23 Hypothesis Testing Problem 8 steps 1. State the appropriate null and alternative hypotheses H 0 : μ = 168 H 1 : μ ≠ 168 (This is a two tailed test) 2. Specify the desired level of significance =.05 is chosen for this test 3. Choose a sample size sample of size n = 25 was selected 4. Determine the appropriate Test σ is unknown so this is a t test
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-24 5. Collect the data The sample results are n = 25, X= \$172.50 S = \$15.40 6. So the test statistic is: The p value for n=25, =.05, t=1.4610 is.1570 Hypothesis Testing Problem (continued)
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-25 7. Is the test statistic in the rejection region? Reject H 0 if p is < alpha; otherwise do not reject H 0 Hypothesis Testing Problem (continued) The p-value.1570 is not < alpha.05, we do not reject the null hypothesis
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-26 8. Express the conclusion in the context of the problem Since The p-value.1570 is > alpha.05, we have failed to reject the null hypothesis Thereby not proving the alternate hypothesis Conclusion: There is not sufficient evidence to reject the claim that the mean cost of a hotel room in NYC is \$168 Hypothesis Testing Problem (continued) If we had rejected the null hypothesis the conclusion would have been: There is sufficient evidence to reject the claim that the mean cost of a hotel room in NYC is \$168
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-27 Hypothesis Testing: σ Unknown Recall that you assume that the sample statistic comes from a random sample from a normal distribution. If the sample size is small (< 30), you should use a box-and-whisker plot or a normal probability plot to assess whether the assumption of normality is valid. If the sample size is large, the central limit theorem applies and the sampling distribution of the mean will be normal.
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-28 One Tail Tests In many cases, the alternative hypothesis focuses on a particular direction H 0 : μ ≥ 3 H 1 : μ < 3 H 0 : μ ≤ 3 H 1 : μ > 3 This is a lower tail test since the alternative hypothesis is focused on the lower tail below the mean of 3 This is an upper tail test since the alternative hypothesis is focused on the upper tail above the mean of 3
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-29 Hypothesis Testing: Connection to Confidence Intervals For X = 172.5, S = 15.40 and n = 25, the 95% confidence interval is: 166.14 ≤ μ ≤ 178.86 Since this interval contains the hypothesized mean (168), you do not reject the null hypothesis at =.05
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-30 Hypothesis Testing Proportions Involves categorical variables Two possible outcomes “Success” (possesses a certain characteristic) “Failure” (does not possesses that characteristic) Fraction or proportion of the population in the “success” category is denoted by π
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-31 Hypothesis Testing Proportions Sample proportion in the success category is denoted by p When both nπ and n(1-π) are at least 5, p can be approximated by a normal distribution with mean and standard deviation
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-32 Hypothesis Testing Proportions The sampling distribution of p is approximately normal, so the test statistic is a Z value:
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-33 Hypothesis Testing Proportions Example A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 30 responses. Test at the =.05 significance level. First, check: n π = (500)(.08) = 40 n(1-π) = (500)(.92) = 460
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-34 Hypothesis Testing Proportions Example H 0 : π =.08 H 1 : π ≠.08 α =.05 n = 500, p = 30/500 =.06 p-value for -1.648 is.0497
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-35 Hypothesis Testing Proportions Example Reject H 0 at =.05 Test Statistic: Decision: Conclusion:There is sufficient evidence to reject the company’s claim of 8% response rate. p-value for -1.648 is.0497
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-36 Using PHStat Options
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-37 Sample PHStat Output Input Output
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-38 Potential Pitfalls and Ethical Considerations Use randomly collected data to reduce selection biases Do not use human subjects without informed consent Choose the level of significance, α, before data collection Do not employ “data snooping” to choose between one-tail and two-tail test, or to determine the level of significance Do not practice “data cleansing” to hide observations that do not support a stated hypothesis Report all pertinent findings
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-39 Chapter Summary In this chapter, we have Addressed hypothesis testing methodology Mentioned Z Test for the mean (σ known) Discussed the p–value approaches to hypothesis testing Discussed one-tail and two-tail tests Performed t test for the mean (σ unknown) Performed Z test for the proportion Discussed pitfalls and ethical issues
Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc.Chap 9-40 Answer Sheet for All Problems ___________ Null Hypothesis ___________ Alternate Hypothesis ___________ Alpha ___________ p-value ___________ Decision (reject or do not reject) Conclusion:
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# Confidence Intervals
An interval of 4 plus or minus 2
A Confidence Interval is a range of values we are fairly sure our true value lies in.
### Example: Average Height
We measure the heights of 40 randomly chosen men, and get a:
The 95% Confidence Interval (we show how to calculate it later) is:
175cm ± 6.2cm
This says the true mean of ALL men (if we could measure their heights) is likely to be between 168.8cm and 181.2cm.
But it might not be!
The "95%" says that 95% of experiments like we just did will include the true mean, but 5% won't.
So there is a 1-in-20 chance (5%) that our Confidence Interval does NOT include the true mean.
## Calculating the Confidence Interval
Step 1: find the number of samples n, calculate the mean X , and the standard deviation s of those samples.
Using our example:
• Number of samples: n = 40
• Mean: X = 175
• Standard Deviation: s = 20
Step 2: decide what Confidence Interval we want. 90%, 95% and 99% are common choices. Then find the "Z" value for that Confidence Interval here:
Confidence Interval Z 80% 1.282 85% 1.440 90% 1.645 95% 1.960 99% 2.576 99.5% 2.807 99.9% 3.291
For 95% the Z value is 1.960
Step 3: use that Z in this formula for the Confidence Interval
X ± Zs√n
Where:
• X is the mean
• Z is the chosen Z-value from the table above
• s is the standard deviation
• n is the number of samples
And we have:
175 ± 1.960 × 20√40
Which is:
175cm ± 6.20cm
In other words: from 168.8cm to 181.2cm
The value after the ± is called the margin of error
The margin of error in our example is 6.20cm
## Calculator
We have a Confidence Interval Calculator to make life easier for you.
## Another Example
### Example: Apple Orchard
Are the apples big enough?
There are hundreds of apples on the trees, so you randomly choose just 30 and get these results:
• Mean: 86
• Standard Deviation: 5
Let's calculate:
X ± Zs√n
We know:
• X is the mean = 86
• Z is the Z-value = 1.960 (from the table above for 95%)
• s is the standard deviation = 5
• n is the number of samples = 30
86 ± 1.960 × 5√30 = 86 ± 1.79
So the true mean (of all the hundreds of apples) is likely to be between 84.21 and 87.79
### True Mean
Now imagine we get to pick ALL the apples straight away, and get them ALL measured by the packing machine (this is a luxury not normally found in statistics!)
And the true mean turns out to be 84.9
Let's lay all the apples on the ground from smallest to largest:
Each apple is a green dot,
except our samples which are blue
Our result was not exact ... it is random after all ... but the true mean is inside our confidence interval of 86 ± 1.79 (in other words 84.21 to 87.79)
But the true mean might not be inside the confidence interval but 95% of the time it will!
95% of all "95% Confidence Intervals" will include the true mean.
Maybe we had this sample, with a mean of 83.5 and a Standard Deviation of 3.5:
Each apple is a green dot,
our samples are marked purple
That does not include the true mean. Expect that to happen 5% of the time for a 95% confidence interval.
So how do we know if the sample we took is one of the "lucky" 95% or the unlucky 5%? Unless we get to measure the whole population like above we simply don't know.
This is the risk in sampling, we might have a bad sample.
## Example in Research
Here is Confidence Interval used in actual research on extra exercise for older people:
Example: the "Male" line says there were:
• 1,226 Men (47.6% of all people)
• had a "HR" (which means Hazard Reduction*) with a mean of 0.92,
• and a 95% Confidence Interval (95% CI) of 0.88 to 0.97 (which is also 0.92±0.05)
In other words the true benefit (for the wider population of men) has a 95% chance of being between 0.88 and 0.97
* Note for the curious: "HR" is used in research and means "Hazard Ratio" where lower is better, so an HR of 0.92 means the subjects were better off, and 1.03 means slightly worse off.
## Standard Normal Distribution
It is all based on the idea of the Standard Normal Distribution, where the Z value is the "Z-score"
For example the Z for 95% is 1.960, and here we see the range from -1.96 to +1.96 includes 95% of all values:
From -1.96 to +1.96 standard deviations is 95%
Applying that to our sample looks like this:
Also from -1.96 to +1.96 standard deviations, so includes 95%
## Conclusion
The Confidence Interval formula is
X ± Zs√n
Where:
• X is the mean
• Z is the Z-value from the table below
• s is the standard deviation
• n is the number of samples
Confidence Interval Z 80% 1.282 85% 1.440 90% 1.645 95% 1.960 99% 2.576 99.5% 2.807 99.9% 3.291
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Frustum
A frustum is a chopped-off or truncated cone or a pyramid.
It is the 3-dimensional solid shape formed by cutting a cone or a pyramid from the top with a plane parallel to its base.
Types
Frustum can be of 2 types, depending on the shape from where it is obtained:
1. Frustum of a Cone
2. Frustum of a Pyramid
Popcorn containers, coffee cups, buckets, and lampshades are some common examples of frustums.
Like all other solid shapes, we can calculate the volume and surface area of a frustum.
Formulas
Volume
The general formula is:
Let us solve an example involving the above concept.
Calculate the volume of a frustum with base areas of 81 cm2, and 121 cm2, and a height of 12 cm.
Solution:
As we know,
Volume (V) =${\dfrac{1}{3}h\left( B_{1}+B_{2}+\sqrt{B_{1}B_{2}}\right)}$, here B1 = 81 cm2, B2 = 121 cm2, h = 12 cm
${\therefore V=\dfrac{1}{3}\times 12\left( 81+121+\sqrt{81\times 121}\right)}$
= 1204 cm3
Surface Area
The general formula to calculate the surface area for both frustum of a cone and a pyramid is:
Lateral Surface Area (LSA) or curved surface area is the area of only the curved surface. The formula is:
Lateral Surface Area (LSA) = ${\dfrac{1}{2}\left( P_{1}+P_{2}\right)\times l}$
Find the surface area of a frustum with base areas of 64 cm2, and 144 cm2, base perimeters of 32 cm and 48 cm, and a slant height of 14 cm.
Solution:
As we know,
Surface Area (SA) = ${\dfrac{1}{2}\left( P_{1}+P_{2}\right)\times l+B_{1}+B_{2}}$, here P1 = 32 cm, P2 = 48 cm, B1 = 64 cm2, B2 = 144 cm2, l = 14 cm
${\therefore SA=\dfrac{1}{2}\left(32+48\right)\times 14+64+144}$
= 768 cm2
Find the lateral area of a frustum with base perimeters of 20 cm and 28 cm, and a slant height of 9 cm.
Solution:
As we know,
Lateral Surface Area (LSA) = ${\dfrac{1}{2}\left( P_{1}+P_{2}\right)\times l}$, here P1 = 20 cm, P2 = 28 cm, l = 9 cm
${\therefore LSA=\dfrac{1}{2}\left( 20+28\right)\times 9}$
= 216 cm2
• More Resources:
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# Division of fractions
fraction is a number that can be obtained when a whole is divided into identical parts . They are represented mathematically by numbers that are written one on top of the other and that are separated from each other by means of a small line, which is known as a fractional line . The basic operations can be applied in sections and among them is division , a process that has many applications in everyday life and in education.
## What is a division of fractions?
The division of fractions is a mathematical operation by means of which a multiplication can be done to distribute a number. It is a multiplication between the numerators and the denominators that make up two different fractions.
• What is the division of fractions?
• What is it for
• Methods
• Division of homogeneous fractions
• Division of heterogeneous fractions
• Division of proper fractions
• Division of improper fractions
• Importance
• Examples
## What is the division of fractions?
The division of fractions is an arithmetic operation that consists of carrying out a cross multiplication between the numerators and denominators of the fractions to be divided. It is a procedure in which the multiplication of the upper and lower extremes of the fractions is carried out in order to find the numerator and the numbers that are in the lower part of the fraction in order to find the denominator of the result.
## What is it for
The divisions of fractions are operations that serve to divide a thing or to distribute a whole in equal or proportionate parts .
## Methods
There are several methods by which fraction division can be performed. The most commonly used are explained below:
#### Cross multiplication method
In this method, a multiplication must be carried out between the numerator that belongs to the first fraction by the denominator that the second fraction has , this result must be placed in the numerator of the fraction that we will obtain as the final result. Then we proceed to multiply the denominator of the first fraction by the numerator of the second fraction and this is how the number that must be placed in the denominator of the result of the division is obtained.
#### Invert and multiply method
In this method you must invert the numbers that the second fraction has , in other words the denominator will be changed by the numerator and vice versa. After this step, we proceed to change the division sign for the multiplication sign and proceed to perform the operation in a straight line , in other words, we will multiply the numerator of the first fraction by the numerator of the second and the same step must be be applied for the denominator.
#### Method of multiplying internal and external numbers
In this method, one fraction must be placed on top of the other and then, a multiplication will be carried out between the external numbers, the result of this first operation will be the final numerator . Then, you must proceed to multiply the numbers that are located in the internal part and with them it is possible to obtain the final result of the denominator .
## Division of homogeneous fractions
Homogeneous fractions are those that have the same denominator , in other words, they have the same number of parts, they are different fractions but with the same denominator. To divide this type of fraction, the following steps must be followed:
• multiplication is carried out between the numerator of the first fraction that has been given to us by the denominator of the second. This result should be placed in the place of the numerator in the result .
• Subsequently, the denominator of the first fraction is multiplied by the numerator of the second and thus we obtain the final result that belongs to the denominator.
## Division of heterogeneous fractions
A heterogeneous fraction is one that has a different denominator and to divide them the following steps must be followed:
• The first thing will be to multiply the numerator of the first fraction by the denominator of the second and this result will be set aside.
• The denominator of the first fraction is multiplied by the numerator of the second and the result will also be set aside.
• The next step will be to place the result of the first operation x the result of the second in the numerator part of the result.
• The denominators are placed under the numerator where this multiplication is found and must also be multiplied .
• We proceed to multiply the numerator, the result of which must also be placed in this position and then the numbers that had been placed in the denominator must be multiplied .
## Division of proper fractions
The steps to divide the proper fractions are as follows:
• Take the numerator of the first fraction and multiply it crosswise by the denominator of the second fraction
• The denominator of the first fraction will be multiplied by the numerator of the second.
• If the result is half, the fraction must be simplified .
## Division of improper fractions
An improper fraction is one in which the numerator is a number greater than or equal to the number in the denominator. To perform a division between improper fractions, the following steps must be followed:
• center line is written first .
• At the top of the line the first fraction of the division will be placed .
• The second fraction of the division should be placed at the bottom of the lines .
• Will multiply the ends first and this will be the final numerator.
• They will multiply the media and this is the result of the denominator.
• If the fraction can be simplified or reduced, the process must be executed.
## Importance
Its importance lies in the fact that by dividing fractions a whole can be divided into equal or equitable parts .
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# Thread: linear systems of equations/ matrices
1. ## linear systems of equations/ matrices
solve this system for x,y, and z
x + 3y + z = -2
2x + 5y + 3z = -7
x + 4y - 3z = 4
if you could help me step by step a little bit that would be much appreciated - thank you in advance
2. Hello, Shmomo89!
$\begin{array}{ccc}x + 3y + z &=& \text{-}2 \\
2x + 5y + 3z &=& \text{-}7 \\x + 4y - 3z &=& 4 \end{array}$
We have: . $\left[\begin{array}{ccc|c}1 & 3 & 1 & \text{-}2 \\ 2 & 5 & 3 & \text{-}7 \\ 1 & 4 & \text{-}3 & 4 \end{array}\right]$
$\begin{array}{c}\\ R_2-2R_1 \\ R_3-R_1\end{array} \left[\begin{array}{ccc|c}1 & 3 & 1 & \text{-}2 \\ 0 & \text{-}1 & 1 & \text{-}3 \\ 0 & 1 & \text{-}4 & 6 \end{array}\right]$
$\begin{array}{c}R_1+3R_2 \\ \\ R_3+R_2\end{array} \left[\begin{array}{ccc|c}1 & 0 & 4 & \text{-}11 \\ 0 & \text{-}1 & 1 & \text{-}3 \\ 0 & 0 & \text{-}3 & 3 \end{array}\right]$
. . $\begin{array}{c}\\ \text{-}1\!\cdot\!R_2 \\ \text{-}\frac{1}{3}\!\cdot\!R_3 \end{array} \left[\begin{array}{ccc|c}1 & 0 & 4 & \text{-}11\\ 0 & 1 & \text{-}1 & 3 \\ 0 & 0 & 1 & \text{-}1 \end{array}\right]$
$\begin{array}{c} R_1-4R_3 \\ R_2 + R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c}1&0&0 & \text{-}7 \\ 0&1&0 & 2 \\ 0&0&1 & \text{-}1 \end{array}\right]$
Therefore: . $\begin{Bmatrix}x &=& \text{-}7 \\ y &=& 2 \\ z &=& \text{-}1\end{Bmatrix}$
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# Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles
NCERT Solutions for Class 9 Maths. Chapter 9 Areas of parallelograms and triangles.
Exercise 9.2
Question 1.
In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Solution:
BSOWe have, AE ⊥ DC and AB = 16 cm
∵ AB = CD [Opposite sides of parallelogram]
∴ CD = 16 cm
Now, area of parallelogram ABCD = CD x AE
= (16 x 8) cm2 = 128 cm2 [∵ AE = 8 cm]
∴ Area of parallelogram ABCD = AD x CF
⇒ AD x CF = 128 cm
⇒ AD x 10 cm = 128 cm2 [∵ CF= 10 cm]
⇒ AD = cm = 12.8 cm 10
Thus, the required length of AD is 12.8 cm
Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = ar (ABCD).
Solution:
Join GE and HE, where GE || BC || DA and HF || AB || DC
(∵ E, F, G and H are the mid¬points of the sides of a ||gm ABCD).
If a triangle and a parallelogram are on the same base and between the same parallels, then A E U the area of the triangle is equal to half the area of the parallelogram.
Now, ∆EFG and parallelogram EBCG are on the same base EG and between the same parallels EG and BC.
∴ ar(∆EFG) = … (1)
Similarly, ar(∆EHG) = …(2)
Adding (1) and (2), we get
ar(∆EFG) + ar(∆EHG) =
=
Thus, ar(EFGH) =
Question 3.
P and Q are any two points lying on the sides DC and AD, respectively of a parallelogram ABCD. Show that ar (APB) = ar(BQC).
Solution:
∵ ABCD is a parallelogram.
∴ AB || CD and BC || AD.
Now, ∆APB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
∴ ar(∆APB) = …….(1)
Also, ∆BQC and parallelogram ABCD are on the same base BC and between the same parallels BGand AD.
∴ ar(∆BQC) = …(2)
From (1) and (2), we have ar(∆APB) = ar(∆BQC).
Question 4.
In figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) =
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)
Solution:
We have a parallelogram ABCD, i.e., AB || CD and BC || AD. Let us draw EF || AB and HG || AD through P.
(i) ∆APB and ||gm AEFB are on the same base AB and between the same parallels AB and EF.
∴ ar(∆APB) = …(1)
Also, ∆PCD and parallelogram CDEF are on the same base CD and between the same parallels CD and EF.
∴ ar(APCD) = …(2)
Adding (1) and (2), we have
ar(∆APB) + ar(∆PCD) =
⇒ ar(∆APB) + ar(∆PCD) = …(3)
(ii) ∆APD and ||gm ∆DGH are on the same base AD and between the same parallels AD and GH.
∴ ar(∆APD) = …(4)
Similarly,
ar(∆PBC) = …(5)
Adding (4) and (5), we have
ar(∆APD) + ar(∆PBC) = =
⇒ ar(∆APD) + ar(∆PBC) = …….(6)
From (3) and (6), we have
ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)
Question 5.
In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) =
Solution:
(i) Parallelogram PQRS and parallelogram ABRS are on the same base RS and between the same parallels RS and PB.
∴ ar(PQRS) = ar(ABRS)
(ii) AAXS and ||gm ABRS are on the same base AS and between the same parallels AS and BR. *
∴ ar(AXS) = …(1)
But ar(PQRS) = ar(ABRS) …(2) [Proved in (i) part]
From (1) and (2), we have
ar(AXS) =
Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it.
Solution:
The farmer is having the field in the form of parallelogram PQRS and a point A is situated on RS. Join AP and AQ.
Clearly, the field is divided into three parts i.e., in ∆APS, ∆PAQ and ∆QAR.
Since, ∆PAQ and pt.
parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.
ar(∆PAQ) = …(1)
⇒ ar(||gm PQRS) – ar(∆PAQ) = ar(||gm PQRS) –
⇒ [ar(∆APS) + ar(∆QAR)] = …(2)
From (1) and (2), we have
ar(∆PAQ) = ar[(∆APS) + (∆QAR)]
Thus, the farmer can sow wheat in (∆PAQ) and pulses in [(∆APS) + (∆QAR)] or wheat in [(∆APS) + (∆QAR)] and pulses in (∆PAQ).
Exercise 9.3
Question 1.
In figure, E is any point on median AD of a ∆ABC. Show that ar (ABE) = ar (ACE).
Solution:
We have a ∆ABC such that AD is a median.
∴ ar(∆ABD) = ar(∆ACD) …(1)
[∵ A median divides the triangle into two triangles of equal areas]
Similarly, in ∆BEC, we have
ar(∆BED) = ar(∆DEC) …(2)
Subtracting (2) from (1), we have
ar(∆ABD) – ar(∆BED) = ar(∆ACD) – ar(∆DEC)
⇒ ar(∆ABE) = ar(∆ACE).
Question 2.
In a triangle ABC, E is the mid-point of median AD. Show that ax (BED) = .
Solution:
We have a ∆ABC and its median AD.
Let us join B and E.
Since, a median divides the triangle into two triangles of equal area.
ar (∆ABD) = …….(1)
Now, in ∆ABD, BE is a median.
[ ∵ E is the mid-point of AD]
∴ ar(∆BED) = …(2)
From (1) and (2), we have
ar(∆BED) = [ ]
⇒ ar(∆BED) =
Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:
We have a parallelogram ABCD (say)
such that its diagonals intersect at O.
∵Diagonals of a parallelogram bisect each other.
∴ AO = OC and BO = OD
Let us draw CE ⊥ BD.
Now, ar(∆BOC) = BO x CE and
ar(∆DOC) = OD x CE
Since, BO = OD
∴ ar(∆BOC) = ar(∆DOC) …(1)
Similarly, ar(∆AOD) = ar(∆DOC) …(2)
and ar(∆AOB) = ar(∆BOC) …(3)
From (1), (2) and (3), we have
ar(∆AOB) = ar(∆BOC) = ar(∆COD) = ar(∆DOA)
Thus, the diagonals of a parallelogram divide it into four triangles of equal area.
Question 4.
In figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar(ABD)
Solution:
we have ∆ABC and ∆ABD are on the same base AB.
∵ CD is bisected at O. [Given]
∴ CO = OD
Now, in ∆ACD, AO is a median
Again, in ∆BCD, BO is a median
∴ ar(∆OBC) = ar(∆ODB) …(2)
Adding (1) and (2), we have
ar(∆OAQ + ar(∆OBQ) = ar(∆OAD) + ar(∆ODB)
⇒ ar(∆ABC) = ar(∆ABD)
Question 5.
D,E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC. Show that
(i) BDEF is a parallelogram.
(ii) ar(DEF) =
(iii) ar(BDEF) =
Solution:
We have ∆ABC such
that D,E and Fare the mid-points of BC, CA and AB respectively.
(i) In ∆ABC, E and F are the mid-points of AC and B D C AB respectively.
∴ EF || BC [Mid-point theorem]
⇒ EF || BD
Also, EF =
⇒ EF = BD [D is the mid – point of BC]
Since BDEF is a quadrilateral whose one pair of opposite sides is parallel and of equal lengths.
∴ BDEF is a parallelogram.
(ii) We have proved that BDEF is a parallelogram.
Similarly, DCEF is a parallelogram and DEAF is also a parallelogram.
Now, parallelogram BDEF and parallelogram DCEF are on the same base EF and between the same parallels BC and EF.
∴ ar(||gm BDEF) = ar(||gm DCEF)
ar(||gm BDEF) = ar(||gm DCEF)
⇒ ar(∆BDF) = ar(∆CDE) …(1)
[Diagonal of a parallelogram divides it into two triangles of equal area]
Similarly, ar(∆CDE) = ar(∆DEF) …(2)
and ar(∆AEF) = ar(∆DEF) …(3)
From (1), (2) and (3), we have
ar(∆AEF) = ar(∆FBD) = ar(∆DEF) = ar(∆CDE)
Thus, ar(∆ABC) = ar(∆AEF) + ar(∆FBD) + ar(∆DEF) + ar(∆CDE) = 4 ar(∆DEF)
⇒ ar(∆DEF) = ar(∆ABC)
(iii) We have, ar (||gm BDEF) = ar(∆BDF) + ar(∆DEF)
= ar(∆DEF) + ar(∆DEF) [∵ ar(∆DEF) = ar(∆BDF)]
2ar(∆DEF) = 2[ ar(∆ABC)]
= ar(∆ABC)
Thus, ar (||gm BDEF) = ar(∆ABC)
Question 6.
In figure, diagonals AC and BD of quadrilateral ABCD intersect at 0 such that OB = OD. If AB = CD, then show that
(i) ar(DOC) = ar(AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram
Solution:
We have a quadrilateral ABCD whose diagonals AC and BD intersect at O.
We also have that OB = OD, AB = CD Let us draw DE ⊥ AC and BF ⊥ AC
(i) In ∆DEO and ∆BFO, we have
DO = BO [Given]
∠DOE = ∠BOF [Vertically opposite angles]
∠DEO = ∠BFO [Each 90°]
∴ ∆DEO ≅ ∆BFO [By A AS congruency]
⇒ DE = BF [By C.P.C.T.]
and ar(∆DEO) = ar(∆BFO) …(1)
Now, in ∆DEC and ∆BFA, we have
∠DEC = ∠BFA [Each 90°]
DE = BF [Proved above]
DC = BA [Given]
∴ ∆DEC ≅ ∆BFA [By RHS congruency]
⇒ ar(∆DEC) = ar(∆BFA) …(2)
and ∠1 = ∠2 …(3) [By C.P.C.T.]
Adding (1) and (2), we have
ar(∆DEO) + ar(∆DEC) = ar(∆BFO) + ar(∆BFA)
⇒ ar(∆DOC) = ar(∆AOB)
(ii) Since, ar(∆DOC) = ar(∆AOB) [Proved above]
Adding ar(∆BOC) on both sides, we have
ar(∆DOC) + ar(∆BOC) = ar(∆AOB) + ar(∆BOC)
⇒ ar(∆DCB) = ar(∆ACB)
(iii) Since, ∆DCS and ∆ACB are both on the same base CB and having equal areas.
∴ They lie between the same parallels CB and DA.
⇒ CB || DA
Also ∠1 = ∠2, [By (3)]
which are alternate interior angles.
So, AB || CD
Hence, ABCD is a parallelogram.
Question 7.
D and E are points on sides AB and AC respectively of ∆ ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.
Solution:
We have ∆ABC and points D and E are such that ar(DBC) = ar{EBC)
Since ∆DBC and ∆EBC are on the same base BC and having same area.
∴ They must lie between the same parallels DE and BC.
Hence, DE || BC
Question 8.
XY is a line parallel to side BC of a ∆ ABC. If BE ||AC and CF || AB meet XY at E and F respectively, show that ar (ABE) =ar (ACF)
Solution:
We have a ∆ABC such that XY || BC,
BE || AC and CF || AB.
Since, XY ||BC and BE || CY
∴ BCYE is a paralleloam.
Now, the parallelogram BCYE and ∆ABE are on the same base 8E and between the same parallels BE and AC.
∴ ar(∆ABE) = …..(1)
Again, CF || AB [Given]
XY || BC [Given]
CF || BX and XF || BC
∴ BCFX is a parallelogram.
Now, ∆ACF and parallelogram BCFX are on the same base CF and between the same parallels AB and CF.
∴ar(∆ACF) = …(2)
Also, parallelogram BCFX and parallelogram BCYE are on the same base BC and between the same parallels BC and EF.
∴ ar(||gm BCFX) = ar(||gm BCYE) ………(3)
From (1), (2) and (3), we get
ar(∆BE) = ar(∆ACF)
Question 9.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then A parallelogram PBQR is completed (see figure).
Show that ax (ABCD) = ar(PBQR).
[Hint Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
Solution:
Let us join AC and PQ.
ABCD is a parallelogram [Given]
and AC is its diagonal, we know that diagonal of a parallelogram divides it into two triangles of equal areas.
∴ ar(∆ABC) = …(1)
Also, PBQR is a parallelogram [Given]
and QP is its diagonal.
∴ ar(∆BPQ) = …(2)
Since, ∆ACQ and AAPQ are on the same base AQ and between A the same parallels AQ and CP.
∴ ar(∆ACQ) = ar(∆APQ)
⇒ ar(∆ACQ) – ar(∆ABQ)
= ar(∆APQ) – ar(∆ABQ)
[Subtracting ar(∆ABQ) from both sides]
⇒ ar(∆ABC) = ar(∆BPQ) …(3)
From (1), (2) and (3), we get
=
⇒ ar( ||gm ABCD) = ar(||gm PBQR)
Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC)
Solution:
BBlliWWp have a trapezium ABCD having AB || CD and its diagonals AC and BD intersect each other at O.
Since, triangles on the same base and between the same parallels have equal areas.
∆ABD and ∆ABC are on the same base AB and between the same parallels AB and DC
∴ ar(∆ABD) = ar(∆ABC)
Subtracting ar(∆AOB) from both sides, we get
ar(∆ABD) – ar(∆AOB) = ar(∆ABC) – ar(∆AOB)
⇒ ar(∆AOD) = ar(∆BOC)
Question 11.
In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
Solution:
We have a pentagon ABCDE in which BF || AC and DC is produced to F.
(i) Since, the triangles between the same parallels and on the same base are equal in area.
∆ACB and ∆ACF are on the same base AC and between the same parallels AC and BF.
∴ ar(∆ACB) = ar(∆ACF)
(ii) Since, ar(∆ACB) = ar(∆ACF) [Proved above]
∴ ar(ABCDE) = ar(AEDF)
Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Solution:
We have a plot in the form of a quadrilateral ABCD.
Let us draw DF || AC and join AF and CF.
Now, ∆DAF and ∆DCF are on the same base DF and between the same parallels AC and DF.
Subtracting ar(∆DEF) from both sides, we get
ar(∆DAF) – ar(∆DEF) = ar(∆DCF) – ar(∆DEF)
The portion of ∆ADE can be taken over by the Gram Panchayat by adding the land (∆CEF) to his (Itwaari) land so as to form a triangular plot,
i.e. ∆ABF.
Let us prove that ar(∆ABF) = ar(quad. ABCD), we have
⇒ ar(∆ABF) = ar (quad. ABCD)
Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ADX) = ar(ACY). [Hint Join IX]
Solution:
We have a trapezium ABCD such that AB || DC.
XY || AC meets AB at X and BC at Y. Let us join CX.
∆ADX and ∆ACX are on the same base AX and between the same parallels AX and DC.
∵∆ACX and ∆ACY are on the same base AC and between the same parallels AC and XY.
∴ ar(∆ACX) = ar(∆ACY) …(2)
From (1) and (2), we have
Question 14.
In figure, AP || BQ || CR. Prove that ar(AQC) = ax(PBR).
Solution:
We have, AP || BQ || CR
∵ ∆BCQ and ∆BQR are on the same base BQ and between the same parallels BQ and CR.
∴ ar(∆BCQ) = ar(∆BQR) …(1)
∵ ∆ABQ and ∆PBQ are on the same base BQ and between the same parallels AP and BQ.
∴ ar(∆ABQ) = ar(∆PBQ) …(2)
Adding (1) and (2), we have
ar(∆BCQ) + ar(∆ABQ) = ar(∆BQR) + ar(∆PBQ)
⇒ ar(∆AQC) = ar(∆PBR)
Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ax(AOD) = ar(BOC). Prove that ABCD is a trapezium.
Solution:
We have a quadrilateral ABCD and its diagonals AC and BD intersect at O such that
ar(∆AOD) = ar(∆BOC) [Given]
Adding ar(∆AOB) to both sides, we have
ar(∆AOD) + ar(∆AOB) = ar(∆BOC) + ar(∆AOB)
⇒ ar(∆ABD) = ar(∆ABC)
Also, they are on the same base AB.
Since, the triangles are on the same base and having equal area.
∴ They must lie between the same parallels.
∴ AB || DC
Now, ABCD is a quadrilateral having a pair of opposite sides parallel.
So, ABCD is a trapezium.
Question 16.
In figure ax(DRC) = ar(DPC) and ai(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Solution:
tfclfiftWe have, ar(∆DRC) = ar(∆DPC) [Given]
And they are on the same base DC.
∴ ∆DRC and ∆DPC must lie between the same parallels.
So, DC || RP i.e.r a pair of opposite sides of quadrilateral DCPR is parallel.
∴ Quadrilateral DCPR is a trapezium.
Again, we have
ar(∆BDP) = ar(∆ARC) [Given] …(1)
Also, ar(∆DPC) = ar(∆DRC) [Given] …(2)
Subtracting (2) from (1), we get
ar(∆BDP) – ar(∆DPC) = ar(∆ARQ – ar(∆DRQ
And they are on the same base DC.
∴ ABDC and AADC must lie between the same parallels.
So, AB || DC i.e. a pair of opposite sides of quadrilateral ABCD is parallel.
∴ Quadrilateral ABCD is a trapezium.
Exercise 9.4
Question 1.
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Solution:
We have a parallelogram ABCD and rectangle ABEF such that
ar(||gm ABCD) = ar( rect. ABEF)
AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
⇒ CD = EF
⇒ AB + CD = AB + EF … (1)
BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side] ⇒ (BC + AD) > (BE + AF) …(2)
From (1) and (2), we have
(AB + CD) + (BC+AD) > (AB + EF) + BE + AF)
⇒ (AB + BC + CD + DA) > (AB + BE + EF + FA)
⇒ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.
Question 2.
In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).
Solution:
Let us draw AF, perpendicular to BC
such that AF is the height of ∆ABD, ∆ADE and ∆AEC.
Question 3.
In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ax(BCF).
Solution:
Since, ABCD is a parallelogram [Given]
∴ Its opposite sides are parallel and equal.
Now, ∆ADE and ∆BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF.
Question 4.
In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(BPC) = ax(DPQ).[Hint Join AC.]
Solution:
We have a parallelogram ABCD and AD = CQ. Let us join AC.
We know that triangles on the same base and between the same parallels are equal in area.
Since, ∆QAC and ∆QDC are on the same base QC and between the same parallels AD and BQ.
∴ ar(∆QAC) = ar(∆QDC)
Subtracting ar(∆QPC) from both sides, we have
ar(∆QAQ – ar(∆QPC) = ar(∆QDC) – ar(∆QPC)
⇒ ar(∆PAQ = ar(∆QDP) …(1)
Since, ∆PAC and ∆PBC are on the same base PC and between the same parallels AB and CD.
∴ ar(∆PAC) = ar(∆PBC) …(2)
From (1) and (2), we get
ar(∆PBC) = ar(∆QDP)
Question 5.
In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Show that
[Hint Join EC and AD. Show that BE || AC and DE || AB, etc.]
Solution:
Let us join EC and AD. Draw EP ⊥ BC.
Let AB = BC = CA = a, then
BD = = DE = BE
(ii) Since, ∆ABC and ∆BED are equilateral triangles.
⇒ ∠ACB = ∠DBE = 60°
⇒ BE || AC
∆BAE and ∆BEC are on the same base BE and between the same parallels BE and AC.
ar(∆BAE) = ar(∆BEC)
⇒ ar(∆BAE) = 2 ar(∆BDE) [ DE is median of ∆EBC. ∴ ar(∆BEC) = || ar(∆BDE)]
⇒ ar(ABDE) = ar(∆BAE)
(iii) ar(∆ABC) = 4 ar(∆BDE)[Proved in (i) part]
ar(∆BEC) = 2 ar(∆BDE)
[ ∵ DE is median of ∆BEC]
⇒ ar(∆ABC) = 2 ar(∆BEC)
(iv) Since, ∆ABC and ∆BDE are equilateral triangles.
⇒ ∠ABC = ∠BDE = 60°
⇒ AB || DE
∆BED and ∆AED are on the same base ED and between the same parallels AB and DE.
∴ ar(∆BED) = ar(∆AED)
Subtracting ar(AEFD) from both sides, we get
⇒ ar(∆BED) – ar(∆EFD) = ar(∆AED) – ar(∆EFD)
⇒ ar(∆BEE) = ar(∆AFD)
(v) In right angled ∆ABD, we get
From (1) and (2), we get
ar(∆AFD) = 2 ar(∆EFD)
ar(∆AFD) = ar(∆BEF) [From (iv) part]
⇒ ar(∆BFE) = 2 ar(∆EFD)
(vi) ar(∆AFC) = ar(∆AFD) + ar(∆ADC)
= ar(∆BFE) + ar(∆ABC) [From (iv) part]
= ar(∆BFE) + x 4 x ar(∆BDE) [From (i) part]
= ar(∆BFE) + 2ar(∆BDE)
= 2ar(∆FED) + 2[ar(∆BFE) + ar(∆FED)]
= 2ar(∆FED) + 2[2ar(∆FED) + ar(∆FED)] [From (v) part]
= 2ar(∆FED) + 2[3ar(∆FED)]
= 2ar(∆FED) + 6ar(∆FED)
= 8ar(∆FED)
∴ ar(∆FED) = ar(∆AFC)
Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(APB) x ar(CPD) = ar(APD) x ar(BPC).
[Hint From A and C, draw perpendiculars to BD.]
Solution:
We have a quadrilateral ABCD such that its diagonals AC and BD intersect at P.
Let us draw AM ⊥ BD and CN ⊥ BD.
Question 7.
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
Solution:
We have a ∆ABC such that P is the mid-point of AB and Q is the mid-point of BC.
Also, R is the mid-point of AP. Let us join AQ, RQ, PC and PC.
(i) In ∆APQ, R is the mid-point of AP. [Given] B
∴RQ is a median of ∆APQ.
⇒ ar(∆PRQ) = ar(∆APQ) …(1)
In ∆ABQ, P is the mid-point of AB.
∴ QP is a median of ∆ABQ.
∴ ar(∆APQ) = ar(∆ABQ) …(2)
mg src=”https://live.staticflickr.com/65535/47574592572_cb45fbd40c_o.png” width=”352″ height=”603″ alt=”NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A7a”>
Question 8.
In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that
(i) ∆MBC = ∆ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ax(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
Solution:
We have a right ∆ABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line segment AX 1 DE is also drawn such that it meets BC at Y.
(i) ∠CBD = ∠MBA [Each90°]
∴ ∠CBD + ∠ABC = ∠MBA + ∠ABC
(By adding ∠ABC on both sides)
or ∠ABD = ∠MBC
In ∆ABD and ∆MBC, we have
AB = MB [Sides of a square]
BD = BC
∠ABD = ∠MBC [Proved above]
∴ ∆ABD = ∆MBC [By SAS congruency]
(ii) Since parallelogram BYXD and ∆ABD are on the same base BD and between the same parallels BD and AX.
∴ ar(∆ABD) = ar(||gm BYXD)
But ∆ABD ≅ ∆MBC [From (i) part]
Since, congruent triangles have equal
areas.
∴ ar(∆MBC) = ar(||gm BYXD)
⇒ ar(||gm BYXD) = 2ar(∆MBC)
(iii) Since, ar(||gm BYXD) = 2ar(∆MBC) …(1) [From (ii) part]
and or(square ABMN) = 2or(∆MBC) …(2)
[ABMN and AMBC are on the same base MB and between the same parallels MB and NC]
From (1) and (2), we have
ar(BYXD) = ar(ABMN) .
(iv) ∠FCA = ∠BCE (Each 90°)
or ∠FCA+ ∠ACB = ∠BCE+ ∠ACB
[By adding ∠ACB on both sides]
⇒ ∠FCB = ∠ACE
In ∆FCB and ∆ACE, we have
FC = AC [Sides of a square]
CB = CE [Sides of a square]
∠FCB = ∠ACE [Proved above]
⇒ ∆FCB ≅ ∆ACE [By SAS congruency]
(v) Since, ||gm CYXE and ∆ACE are on the same base CE and between the same parallels CE and AX.
∴ ar(||gm CYXE) = 2ar(∆ACE)
But ∆ACE ≅ ∆FCB [From (iv) part]
Since, congruent triangles are equal in areas.
∴ ar (||<gm CYXE) = 2ar(∆FCB)
(vi) Since, ar(||gm CYXE) = 2ar(∆FCB) …(3)
[From (v) part]
Also (quad. ACFG) and ∆FCB are on the same base FC and between the same parallels FC and BG.
⇒ ar(quad. ACFG) = 2ar(∆FCB) …(4)
From (3) and (4), we get
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# Find the area of a triangle formed by the points A(5,\ 2),\ B(4,\ 7)\ a n d\ C(7, 4)dot
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Updated On: 27-06-2022
Text Solution
Solution : Area of a triangle can be given by,<br> A = 1/2(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))<br> A = 1/2(5(11)+4(-6)+7(-5)) = 1/2(55-59) = -4/2 =-2<br> As area can not be negative,<br> Required area is 2 square units.<br>
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
Idhu given question we have to find the area of triangle formed by the points A B and C so as you can see in the given figure the coordinates of point A52 we let them to be X1 Y1 similarly 4.3 we consider them to be at 2 and white 2 and 4 see it again to be XT and y3 so for the area of triangle of triangle use the formula which is given by 1 upon 2 into X 1 and n y 2 - 3 + 2 and then why 3 - 1 + 3 and then by 1 minus y to so here we just Kutti values so here we get 1 upon 2 into X 1 is 5 and Y2 is 7
and y3 is here it is 4 so similarly we put the value of x to which is more so the value of x to is Core and why it is again 4 - 1 is to 2 + 2 + 3 is equal to 7 into and then by 1 minus y to y one is to minus 2 is 7 so here we just now solve them is equal to 1 upon 2 into 5 into it is 3 + 4 into two plus 7 into minus 5 so here we know there are now call them further is equal to 1 upon 2 15 + 8 - 35 so here it is equals to
1 by 2 and that's 23 - 30 5 Swami subtract we get we get 1 upon 2 here it is minus 12 is equals to minus 6 but but area can't B negative negative so we just take so we take only numeric value so we take only numeric value show the area soap area will be 6 square unit square unit so that 6 answer
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# Perimeter and Circumference
Like many mathematical terms, the word perimeter has its origins in the work of early Greek mathematicians. It is derived from the Greek words ‘peri’, which means ‘around’ and ‘metron’, which means ‘a measurement’. Perimeter is literally a measurement around.
In everyday usage, you may have come across phrases such as perimeter fence, estate perimeter, or perimeter security. These mean that the fence, or the security provision, are around the edges, the outer limits or extremities of a measured area of land or property.
Understanding how to calculate perimeter is a useful mathematical skill for both study and real life, whether performing geometric calculations, marking out a playing field or replacing a fence.
Perimeter or Boundary?
The definition of a boundary is a dividing line between two areas. In cricket, the boundary is the line marking the edge of the pitch.
The perimeter is the measured length of such a boundary. In geometry, it is defined as the sum of the distance of all the lengths of the sides of an object. Perimeter is measured in any unit of length, e.g. metres, centimetres, miles or inches. For more on this, see our page on measurement systems.
So in common language, the two are often used interchangeably. However, in a mathematical context, we only use perimeter.
A circumference is a very specific type of perimeter, that refers only to circular shapes and forms. More on this later.
## Measuring Perimeter of Regular Polygons
The perimeter of a two-dimensional shape is the total length of all the sides added together.
For example, the perimeter of a square, with a side length of 6m, is simply four lots of 6m, i.e. 4 × 6m = 24m. The square has four sides of equal length, which are added together.
A square of any side length s therefore has a perimeter equal to 4 × s, or simply 4s .
Perimeter vs. Area
Don’t get confused between perimeter and area. Whilst perimeter is the measurement of the outline of the shape, area is the measurement of the space contained within the perimeter.
So whilst perimeter is measured in units of length, area is measured in square units, e.g. m2, cm2 or inches2.
For more on measuring area, see our page on Calculating Area.
You can use the same principle to work out the perimeter of any regular polygon that has any number of sides of equal length:
If your polygon has n number of sides, all of length s, then its perimeter is always equal to n × s, or simply ns.
So for example, if you have a heptagon (7 sides) with side length 15cm, then the length of the perimeter is 7 × 15 = 105cm.
For more on regular, irregular and other polygons (straight-sided shapes), including a helpful chart with illustrations, see our page on properties of polygons.
Measuring Perimeter of Irregular Polygons
A regular polygon has all sides and internal angles equal, irregular polygons do not.
A rectangle that is not an exact square, for example, has two pairs of sides of equal length, but all four sides are not the same length.
### Example
Find the perimeter of a rectangular football pitch, measuring 105 × 68 m.
The lengths of the opposite sides are equal to each other, so you need to add together two lots of 105m and two lots of 68m.
2 × 105 = 210m
2 × 68 = 136m
210 + 136 = 346m
The perimeter of the pitch is 346m.
Irregular polygons can be formed by any combination of straight lines that join up to enclose an area. No matter how complicated the shape, the perimeter will always be the sum of the lengths of the sides.
The shape below could be a garden plot, or anything else you can think of. In geometrical terms, it is an eight-sided enclosed 2D shape, which has no sides of the same length, and no internal angles of the same size.
It is an irregular octagon (8 sides) and its perimeter is equal to a + b + c + d + e + f + g + h.
### Example
Calculate the perimeter of the template below, dimensions are in inches.
Starting at the bottom left corner and working your way around the shape in a clockwise direction, add up the lengths of the sides:
5 + 2 + 3 + 4 + 2 + 3 + 4 + 9 = 32 inches.
The perimeter of the shape is 32 inches.
## Circumference
Circumference is a specific type of perimeter that applies exclusively to circular shapes.
### Measuring the circumference of a circle
The mathematical expression for calculating the circumference of a circle is:
2 × π × radius or simply 2πr
The diameter of a circle is equal to twice the radius, so the expression for circumference can also be written πD.
Pi π
π (pi) is a Greek letter, which is used in mathematics to represent a constant with an approximate value of 3.142 (it is an irrational number with infinite decimal places). For more on this, see our pages on circles and curved shapes and special numbers.
Example
The groundskeeper needs to re-paint the lines on the football field in the example above and needs to know how much paint to buy. He has calculated the perimeter of the field and he also knows the length of the halfway line as this is the same as the short side of the pitch. He is also confident measuring the perimeter of the penalty areas, as these are also simple rectangles. However, he needs to know the circumference of the centre circle.
He has measured its radius and it is 9.15m.
Circumference = 2πr
2 × π × 9.15 = 57.5m (rounded to one decimal place)
The circumference of the centre circle is 57.5m.
### Measuring the circumference of an ellipse
Not all curved shapes are perfectly circular and sometimes it might be necessary to find the perimeter of an ellipse (a squashed or elongated circle).
The perimeter:
$$p \approx 2\pi\sqrt{\frac{a^2+b^2}{2}}$$
You can see that this expression is similar to the expression for the circumference of a circle, but radius r is replaced by $$\sqrt{\frac{a^2+b^2}{2}}$$, where a and b are half the length of the minor axis and major axis respectively. (For more on ellipses, see our page on circles and curved shapes).
This equation only provides an approximation (≈). The more elongated the ellipse becomes, the more inaccurate the answer. Mathematicians have come up with several complex formulae for solving this problem. None of them have achieved 100% accuracy in a mathematical sense, but it is unlikely that you would need such a high level of precision unless you work in engineering or design.
There are many professions and occupations that may require you to take physical measurements of perimeters and boundaries, such as civil engineering, surveying, landscape architecture, garden design and sports ground maintenance.
It is necessary to not only have an understanding of the basic mathematical principles above, but also more advanced numeracy tools, such as trigonometry. It is not only the lengths of the lines that are important, but accurate measurement of the angles between those lines.
Apart from mathematical knowledge, there is also an interesting and varied toolkit needed for these sorts of occupations. Relatively short distances can be measured using steel tapes, or measuring wheels. Electronic distance measurement (EDM) devices, which use electromagnetic waves, are more often used by land surveyors. These are used in conjunction with other instruments such as levels and theodolites, which ensure the accuracy and precision of angular measurements, using a mathematical technique called triangulation.
However, if you just need to replace your garden fence, you will probably be fine with just a tape measure and a ball of string!
Further Reading from Skills You Need
Understanding Geometry
Part of The Skills You Need Guide to Numeracy
This eBook covers the basics of geometry and looks at the properties of shapes, lines and solids. These concepts are built up through the book, with worked examples and opportunities for you to practise your new skills.
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NCERT Solutions for Class 7 Math Chapter 10 Practical Geometry are provided here with simple step-by-step explanations. These solutions for Practical Geometry are extremely popular among class 7 students for Math Practical Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 7 Math Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 7 Math are prepared by experts and are 100% accurate.
#### Question 1:
Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
The steps of construction are as follows.
(i)Draw a line AB. Take a point P on it. Take a point C outside this line. Join C to P.
(ii)Taking P as centre and with a convenient radius, draw an arc intersecting line AB at point D and PC at point E.
(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at H.
(iv) Adjust the compasses up to the length of DE. Without changing the opening of compasses and taking H as the centre, draw an arc to intersect the previously drawn arc FG at point I.
(v) Join the points C and I to draw a line ‘l’.
This is the required line which is parallel to line AB.
​​​​
##### Video Solution for Practical Geometry (Page: 196 , Q.No.: 1)
NCERT Solution for Class 7 math - Practical Geometry 196 , Question 1
#### Question 2:
Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
The steps of construction are as follows.
(i) Draw a line l and take a point P on line l. Then, draw a perpendicular at point P.
(ii) Adjusting the compasses up to the length of 4 cm, draw an arc to intersect this perpendicular at point X. Choose any point Y on line l. Join X to Y.
(iii) Taking Y as centre and with a convenient radius, draw an arc intersecting l at A and XY at B.
(iv) Taking X as centre and with the same radius as before, draw an arc CD cutting XY at E.
(v)Adjust the compasses up to the length of AB. Without changing the opening of compasses and taking E as the centre, draw an arc to intersect the previously drawn arc CD at point F.
(vi) Join the points X and F to draw a line m.
Line m is the required line which is parallel to line l.
#### Question 3:
Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?
The steps of construction are as follows.
(i)Draw a line l and take a point A on it. Take a point P not on l and join A to P.
(ii) Taking A as centre and with a convenient radius, draw an arc cutting l at B and AP at C.
(iii)Taking P as centre and with the same radius as before, draw an arc DE to intersect AP at F.
(iv) Adjust the compasses up to the length of BC. Without changing the opening of compasses and taking F as the centre, draw an arc to intersect the previously drawn arc DE at point G.
(v)Join P to G to draw a line m. Line m will be parallel to line l.
(vi)Join P to any point Q on line l. Choose another point R on line m. Similarly, a line can be drawn through point R and parallel to PQ.
Let it meet line l at point S.
In quadrilateral PQSR, opposite lines are parallel to each other.
PQ || RS and PR || QS
Thus, the quadrilateral PQSR is a parallelogram.
#### Question 1:
Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
The rough figure of this triangle is as follows.
The required triangle is constructed as follows.
(i) Draw a line segment YZ of length 5 cm.
(ii) Point X is at a distance of 4.5 cm from point Y. Therefore, taking point Y as centre, draw an arc of 4.5 cm radius.
(iii) Point X is at a distance of 6 cm from point Z. Therefore, taking point Z as centre, draw an arc of 6 cm radius. Mark the point of intersection of the arcs as X. Join XY and XZ.
XYZ is the required triangle.
#### Question 2:
Construct an equilateral triangle of side 5.5 cm.
An equilateral triangle of side 5.5 cm has to be constructed. We know that all sides of an equilateral triangle are of equal length. Therefore, a triangle ABC has to be constructed with AB = BC = CA = 5.5 cm.
The steps of construction are as follows.
(i) Draw a line segment BC of length 5.5 cm.
(ii) Taking point B as centre, draw an arc of 5.5 cm radius.
(iii) Taking point C as centre, draw an arc of 5.5 cm radius to meet the previous arc at point A.
(iv) Join A to B and C.
ABC is the required equilateral triangle.
#### Question 3:
Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of
triangle is this?
The steps of construction are as follows.
(i) Draw a line segment QR of length 3.5 cm.
(ii) Taking point Q as centre, draw an arc of 4 cm radius.
(iii) Taking point R as centre, draw an arc of 4 cm radius to intersect the previous arc at point P.
(iv) Join P to Q and R.
ΔPQR is the required triangle. As the two sides of this triangle are of the same length (PQ = PR), therefore, ΔPQR is an isosceles triangle.
​​​​​
##### Video Solution for Practical Geometry (Page: 199 , Q.No.: 3)
NCERT Solution for Class 7 math - Practical Geometry 199 , Question 3
#### Question 4:
Construct ΔABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
The steps of construction are as follows.
(i) Draw a line segment BC of length 6 cm.
(ii) Taking point C as centre, draw an arc of 6.5 cm radius.
(iii) Taking point B as centre, draw an arc of radius 2.5 cm to meet the previous arc at point A.
(iv) Join A to B and C.
ΔABC is the required triangle. ∠B can be measured with the help of protractor. It comes to 90º.
#### Question 1:
Construct ΔDEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
The rough sketch of the required ΔDEF is as follows.
The steps of construction are as follows.
(i)Draw a line segment DE of length 5 cm.
(ii) At point D, draw a ray DX making an angle of 90° with DE.
(iii) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F.
(iv) Join F to E. ΔDEF is the required triangle.
#### Question 2:
Construct an isosceles triangle in which the lengths of each of its equal sides
is 6.5 cm and the angle between them is 110°.
An isosceles triangle PQR has to be constructed with PQ = QR = 6.5 cm. A rough sketch of the required triangle can be drawn as follows.
The steps of construction are as follows.
(i) Draw the line segment QR of length 6.5 cm.
(ii) At point Q, draw a ray QX making an angle 110° with QR.
(iii) Taking Q as centre, draw an arc of 6.5 cm radius. It intersects QX at point P.
(iv) Join P to R to obtain the required triangle PQR.
#### Question 3:
Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
A rough sketch of the required triangle is as follows.
The steps of construction are as follows.
(i) Draw a line segment BC of length 7.5 cm.
(ii) At point C, draw a ray CX making 60º with BC.
(iii) Taking C as centre, draw an arc of 5 cm radius. It intersects CX at point A.
(iv) Join A to B to obtain triangle ABC.
​​​​​​
##### Video Solution for Practical Geometry (Page: 200 , Q.No.: 3)
NCERT Solution for Class 7 math - Practical Geometry 200 , Question 3
#### Question 1:
Construct ΔABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.
A rough sketch of the required ΔABC is as follows.
The steps of construction are as follows.
(i)Draw a line segment AB of length 5.8 cm.
(ii)At point A, draw a ray AX making 60º angle with AB.
(iii) At point B, draw a ray BY, making 30º angle with AB.
(iv) Point C has to lie on both the rays, AX and BY. Therefore, C is the point of intersection of these two rays.
This is the required triangle ABC.
#### Question 2:
Construct ΔPQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°.
(Hint: Recall angle sum property of a triangle).
A rough sketch of the required ΔPQR is as follows.
In order to construct ΔPQR, the measure of ∠RPQ has to be calculated.
According to the angle sum property of triangles,
∠PQR + ∠PRQ + ∠RPQ = 180º
105º + 40º + ∠RPQ = 180º
145º + ∠RPQ = 180º
∠RPQ = 180° − 145° = 35°
The steps of construction are as follows.
(i) Draw a line segment PQ of length 5 cm.
(ii) At P, draw a ray PX making an angle of 35º with PQ.
​​​​​​​
(iii) At point Q, draw a ray QY making an angle of 105º with PQ.
(iv)Point R has to lie on both the rays, PX and QY. Therefore, R is the point of intersection of these two rays.
This is the required triangle PQR.
##### Video Solution for Practical Geometry (Page: 202 , Q.No.: 2)
NCERT Solution for Class 7 math - Practical Geometry 202 , Question 2
#### Question 3:
Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E =
Given that,
m∠E = 110° and m∠F = 80°
Therefore,
m∠E + m∠F = 110° + 80° = 190°
However, according to the angle sum property of triangles, we should obtain
m∠E + m∠F + m∠D = 180°
Therefore, the angle sum property is not followed by the given triangle. And thus, we cannot construct ΔDEF with the given measurements.
Also, it can be observed that point D should lie on both rays, EX and FY, for constructing the required triangle. However, both rays are not intersecting each other. Therefore, the required triangle cannot be formed.
#### Question 1:
Construct the right angled ΔPQR, where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
A rough sketch of ΔPQR is as follows.
The steps of construction are as follows.
(i) Draw a line segment QR of length 8 cm.
(ii) At point Q, draw a ray QX making 90º with QR.
(iii) Taking R as centre, draw an arc of 10 cm radius to intersect ray QX at point P.
(iv) Join P to R. ΔPQR is the required right-angled triangle.
​​​​​​​
##### Video Solution for Practical Geometry (Page: 203 , Q.No.: 1)
NCERT Solution for Class 7 math - Practical Geometry 203 , Question 1
#### Question 2:
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
A right-angled triangle ABC with hypotenuse 6 cm and one of the legs as 4 cm has to be constructed. A rough sketch of ΔABC is as follows.
The steps of construction are as follows.
(i) Draw a line segment BC of length 4 cm.
(ii) At point B, draw a ray BX making an angle of 90º with BC.
(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray BX at point A.
(iv) Join A to C to obtain the required ΔABC.
#### Question 3:
Construct an isosceles right-angled triangle ABC, where, m∠ACB = 90° and AC = 6 cm.
In an isosceles triangle, the lengths of any two sides are equal.
Let in ΔABC, AC = BC = 6 cm. A rough sketch of this ΔABC is as follows.
The steps of construction are as follows.
(i) Draw a line segment AC of length 6 cm.
(ii) At point C, draw a ray CX making an angle of 90º with AC.
(iii) Taking point C as centre, draw an arc of 6 cm radius to intersect CX at point B.
(iv) Join A to B to obtain the required ΔABC.
View NCERT Solutions for all chapters of Class 7
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A strategy you can use to solve multiplication problems is using repeated addition, which means just thinking of multiplication problems in terms of addition. We're going to use this equation, 3 x 4, to help us think of this and this picture is going to help us look at that. So, we have 3 groups with 4 items in each group. So, if we're thinking of this in terms of addition, that means we have 4 items in the first group, 4 items in the second group, and 4 items in the third group. To figure out how many there are all together, we can add them. So, we have 4 + 4 + 4 which equals 12. This means that the answer to the multiplication equation, 3 x 4, is 12.
Kids earn more screen time by answering fun educational questions.
• What is the example of repeated addition shown in the video?
1) 4 + 4 + 4 =
2) 2 + 2 + 2 + 2 =
3) 6 + 6 =
4) 3 + 3 + 3 + 3 =
• Which of the following repeated addtion equations could help you solve 3 x 1 = ?
1) 1 + 1 + 1 =
2) 3 + 3 + 3 =
3) 1 + 1 + 1 + 1 =
4) 4 + 4 =
• Which of the following repeated addtion equations could help you solve 4 x 2 = ?
1) 2 + 2 + 2 + 2 =
2) 4 + 4 =
3) 3 + 3 + 3 =
4) 1 + 2 + 3 + 4 =
• What example multiplication equation is used in the video?
1) 3 x 4 =
2) 4 x 3 =
3) 4 + 4 + 4 =
4) 2 x 7 =
1) 4 x 6 =
2) 6 x 4 =
3) 2 x 6 =
4) 4 x 5 =
1) 2 x 8 =
2) 7 x 3 =
3) 8 x 8 =
4) 8 x 2 =
• What strategy used to solve multiplication problems is described in the video?
2) Creating a picture
3) Creating an Array
4) Skip Counting
1) Yes
2) No
1) No
2) Yes
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Arrays
An array is a type of picture that we use to help us visualize and…
Multiplication Pictures
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Angles Outside a Circle
Angles formed outside a circle by tangents or secants.
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Angles Outside a Circle
What if you wanted to figure out the angle at which the sun's rays hit the earth? The sun’s rays hit the earth such that the tangent rays determine when daytime and night time are. The time and Earth’s rotation determine when certain locations have sun. If the arc that is exposed to sunlight is \begin{align*}178^\circ\end{align*}, what is the angle at which the sun’s rays hit the earth \begin{align*}(x^\circ)\end{align*}? After completing this Concept, you'll be able to use properties of angles created by tangent lines to answer this question.
Watch This
CK-12 Foundation: Chapter9AnglesOutsideaCircleA
Guidance
An angle is considered to be outside a circle if the vertex of the angle is outside the circle and the sides are tangents or secants. There are three types of angles that are outside a circle: an angle formed by two tangents, an angle formed by a tangent and a secant, and an angle formed by two secants. Just like an angle inside or on a circle, an angle outside a circle has a specific formula, involving the intercepted arcs.
Investigation: Find the Measure of an Angle outside a Circle
Tools Needed: pencil, paper, ruler, compass, protractor, colored pencils (optional)
1. Draw three circles and label the centers \begin{align*}A, B\end{align*}, and \begin{align*}C\end{align*}. In \begin{align*}\bigodot A\end{align*} draw two secant rays with the same endpoint, \begin{align*}\overrightarrow{DE}\end{align*} and \begin{align*}\overrightarrow{DF}\end{align*}. In \begin{align*}\bigodot B\end{align*}, draw two tangent rays with the same endpoint, \begin{align*}\overrightarrow{LM}\end{align*} and \begin{align*}\overrightarrow{LN}\end{align*}. In \begin{align*}\bigodot C\end{align*}, draw a tangent ray and a secant ray with the same endpoint, \begin{align*}\overrightarrow{QR}\end{align*} and \begin{align*}\overrightarrow{QS}\end{align*}. Label the points of intersection with the circles like they are in the pictures below.
2. Draw in all the central angles: \begin{align*}\angle GAH, \angle EAF, \angle MBN, \angle RCT, \angle RCS\end{align*}. Then, find the measures of each of these angles using your protractor. Use color to differentiate.
3. Find \begin{align*}m \angle EDF, m \angle MLN\end{align*}, and \begin{align*}m \angle RQS\end{align*}.
4. Find \begin{align*}\frac{m \widehat{EF}-m \widehat{GH}}{2}, \frac{m \widehat{MPN}-m \widehat{MN}}{2}\end{align*}, and \begin{align*}\frac{m \widehat{RS}-m \widehat{RT}}{2}\end{align*}. What do you notice?
Outside Angle Theorem: The measure of an angle formed by two secants, two tangents, or a secant and a tangent drawn from a point outside the circle is equal to half the difference of the measures of the intercepted arcs.
Example A
Find the value of \begin{align*}x\end{align*}. You may assume lines that look tangent, are.
Set up an equation using the Outside Angle Theorem.
\begin{align*}\frac{(5x+10)^\circ-(3x+4)^\circ}{2} &= 30^\circ\\ (5x+10)^\circ-(3x+4)^\circ &= 60^\circ\\ 5x+10^\circ-3x-4^\circ &= 60^\circ\\ 2x+6^\circ &= 60^\circ\\ 2x &= 54^\circ\\ x &= 27^\circ\end{align*}
Example B
Find the value of \begin{align*}x\end{align*}.
\begin{align*}x=\frac{120^\circ - 32^\circ}{2}=\frac{88^\circ}{2}=44^\circ\end{align*}.
Example C
Find the value of \begin{align*}x\end{align*}.
First note that the missing arc by angle \begin{align*}x\end{align*} measures \begin{align*}32^\circ\end{align*} because the complete circle must make \begin{align*}360^\circ\end{align*}. Then, \begin{align*}x=\frac{141^\circ - 32^\circ}{2}=\frac{109^\circ}{2}=54.5^\circ\end{align*}.
Watch this video for help with the Examples above.
CK-12 Foundation: Chapter9AnglesOutsideaCircleB
Concept Problem Revisited
If \begin{align*}178^\circ\end{align*} of the Earth is exposed to the sun, then the angle at which the sun’s rays hit the Earth is \begin{align*}2^\circ\end{align*}. From the Outside Angle Theorem, these two angles are supplementary. From this, we also know that the other \begin{align*}182^\circ\end{align*} of the Earth is not exposed to sunlight and it is probably night time.
Guided Practice
Find the measure of \begin{align*}x\end{align*}.
1.
2.
3.
For all of the above problems we can use the Outside Angle Theorem.
1. \begin{align*}x=\frac{125^\circ-27^\circ}{2}=\frac{98^\circ}{2}=49^\circ\end{align*}
2. \begin{align*}40^\circ\end{align*} is not the intercepted arc. Be careful! The intercepted arc is \begin{align*}120^\circ\end{align*}, \begin{align*}(360^\circ-200^\circ-40^\circ)\end{align*}. Therefore, \begin{align*}x=\frac{200^\circ-120^\circ}{2}=\frac{80^\circ}{2}=40^\circ\end{align*}.
3. First, we need to find the other intercepted arc, \begin{align*}360^\circ-265^\circ=95^\circ\end{align*}. \begin{align*}x=\frac{265^\circ-95^\circ}{2}=\frac{170^\circ}{2}=85^\circ\end{align*}
Explore More
Find the value of the missing variable(s).
Solve for \begin{align*}x\end{align*}.
1. Prove the Outside Angle Theorem
Given: Secant rays \begin{align*}\overrightarrow{AB}\end{align*} and \begin{align*}\overrightarrow{AC}\end{align*}
Prove: \begin{align*}m\angle a = \frac{1}{2} \left (m\widehat{BC}-m\widehat{DE} \right )\end{align*}
1. Draw two secants that intersect:
1. inside a circle.
2. on a circle.
3. outside a circle.
To view the Explore More answers, open this PDF file and look for section 9.8.
My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English Spanish
central angle
An angle formed by two radii and whose vertex is at the center of the circle.
chord
A line segment whose endpoints are on a circle.
diameter
A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.
inscribed angle
An angle with its vertex on the circle and whose sides are chords.
intercepted arc
The arc that is inside an inscribed angle and whose endpoints are on the angle.
point of tangency
The point where the tangent line touches the circle.
Secant
The secant of an angle in a right triangle is the value found by dividing length of the hypotenuse by the length of the side adjacent the given angle. The secant ratio is the reciprocal of the cosine ratio.
Tangent
The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.
Outside Angles Theorem
The Outside Angles Theorem states that the measure of an angle formed by two secants, two tangents, or a secant and a tangent from a point outside the circle is half the difference of the measures of the intercepted arcs.
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# A Comprehensive Study Guide on Probability: From Basics to Advanced Topics
Probability is the measure of the likelihood of an event occurring. In simpler terms, it’s the chance or possibility of something happening. Probability theory is a branch of mathematics that deals with the study of these chances. Understanding and applying probability is essential in many areas, including science, engineering, finance, and data analysis.
In this section, we’ll discuss the basics of probability, its applications and the importance of probability theory in math programs.
#### What is Probability?
In math, probability is a way to measure the possibility of an event occurring. It is represented by a number between 0 and 1, where 0 indicates that the event will not occur, and 1 means the event is certain to occur.
### Applications of Probability
Probability theory has a vast number of applications in various fields. Some of the applications include:
• Weather forecasting
• Financial modeling
• Risk management
• Genetics
• Physics
• Games of chance
• Quality control
### Why is Probability Important in Math Programs?
Probability is a building block for many mathematical concepts and techniques. It’s used in various topics, including algebra, analysis, and statistics. Probability theory is a crucial aspect of math programs, and understanding it helps students acquire a wide range of problem-solving skills that are applicable in many other disciplines.
## Basic Probability Concepts
In this section, we’ll discuss the basics of probability theory, including sample space, events, probability of an event, and complement of an event.
### Sample Space
Sample space is the set of all possible outcomes of an experiment. An experiment is any process that produces an outcome. For example, if you roll a dice, the sample space will be {1, 2, 3, 4, 5, 6}.
### Events and Their Types
Events are subsets of the sample space. They represent the outcomes we are interested in. There are three types of events:
• Simple event: An event that consists of a single outcome. For example, getting a 1 when rolling a dice.
• Compound event: An event that consists of more than one outcome. For example, getting an even number when rolling a dice.
• Impossible event: An event that cannot occur. For example, getting a 7 when rolling a dice.
### Probability of an Event
The probability of an event is the measure of the likelihood of the event occurring. It is a number between 0 and 1, inclusive. The probability of an event A is denoted by P(A).
### Complement of an Event
The complement of an event A is the set of outcomes in the sample space that are not in A. The complement of A is denoted by A`. The probability of the complement of A is given by P(A`) = 1 – P(A).
## Discrete Probability Distributions
Discrete probability distributions deal with random variables that can take on a countable number of values. In this section, we’ll discuss discrete random variables, the probability mass function, mean, variance, and standard deviation of a discrete random variable, and the binomial probability distribution.
### Discrete Random Variables
A discrete random variable is a variable that can take on a countable number of values.
### Probability Mass Function
The probability mass function (PMF) is a function that describes the probability distribution of a discrete random variable. It gives the probability of each possible value of the variable.
### Mean, Variance, and Standard Deviation of a Discrete Random Variable
The mean of a discrete random variable is denoted by E(X) and is given by the formula: E(X) = Σ Xi * P(Xi) where Xi represents the ith possible value of the random variable and P(Xi) represents the probability of the ith value.
The variance of a discrete random variable is denoted by Var(X) and is given by the formula: Var(X) = Σ (Xi – μ)^2 * P(Xi) where μ is the mean of the random variable.
The standard deviation of a discrete random variable is the square root of the variance.
### Binomial Probability Distribution
The binomial probability distribution is a discrete probability distribution that deals with events that have only two possible outcomes. It is used to calculate the probability of a specific number of successes in a fixed number of trials.
## Continuous Probability Distributions
Continuous probability distributions deal with random variables that can take on an infinite number of values. In this section, we’ll discuss continuous random variables, the probability density function, mean, variance, and standard deviation of a continuous random variable, and the normal probability distribution.
### Continuous Random Variables
A continuous random variable is a variable that can take on any value within a specific interval.
### Probability Density Function
The probability density function (PDF) is a function that describes the probability distribution of a continuous random variable. Unlike discrete probability mass functions, the probability density function does not give the probability of a specific value of the random variable. Instead, it gives the probability density at a particular point.
### Mean, Variance, and Standard Deviation of a Continuous Random Variable
The mean of a continuous random variable is denoted by E(X) and is given by the following formula:
E(X) = ∫ xf(x) dx where f(x) is the probability density function of the random variable.
The variance of a continuous random variable is denoted by Var(X) and is given by the formula:
Var(X) = ∫ (x – μ)^2 f(x) dx where μ is the mean of the random variable.
The standard deviation of a continuous random variable is the square root of the variance.
### Normal Probability Distribution
The normal probability distribution is a continuous probability distribution that is widely used in various fields. It has a bell-shaped curve and is characterized by its mean and standard deviation. Many natural phenomena follow a normal distribution, making it essential in scientific research.
## Joint Probability Distributions
Joint probability distributions deal with the probability of two or more random variables occurring simultaneously. In this section, we’ll discuss joint probability distributions, covariance and correlation, conditional probability distribution, and Bayes’ Theorem.
### Joint Probability Distribution
A joint probability distribution is a probability distribution that deals with the simultaneous occurrence of two or more random variables.
### Covariance and Correlation
Covariance measures the linear relationship between two random variables X and Y. It is denoted by cov(X,Y) and is given by the formula:
cov(X,Y) = E[(X – μx)(Y – μy)]
Correlation is a normalized version of covariance that measures the strength and direction of the linear relationship between two random variables. It is denoted by ρ and is given by the formula:
ρ = cov(X,Y) / σx σy
### Conditional Probability Distribution
The conditional probability distribution deals with the probability of an event occurring given that another event has already occurred.
### Bayes’ Theorem
Bayes’ Theorem is a formula that describes the probability of an event occurring based on prior knowledge of related events. It is widely used in statistical inference and machine learning.
In this section, we’ll discuss some of the advanced topics in probability theory, including the central limit theorem, law of large numbers, Markov chains, and Poisson process.
### Central Limit Theorem
The central limit theorem states that the sum of a large number of independent and identically distributed random variables approaches a normal distribution, regardless of the underlying distribution of the random variables.
### Law of Large Numbers
The law of large numbers states that as the sample size increases, the sample mean approaches the true mean of the population.
### Markov Chains
A Markov chain is a sequence of events in which the probability of each event depends only on the state of the preceding event.
### Poisson Process
The Poisson process is a mathematical model used to describe the occurrence of random events over time or space.
## Conclusion
In conclusion, probability theory is a crucial concept in mathematics, and understanding it is essential for students in math programs to excel. In this study guide, we covered the basics of probability, including sample space, events, probability of an event, and the complement of an event. We also delved into discrete and continuous probability distributions, joint probability distributions, advanced topics like the central limit theorem, law of large numbers, Markov chains, and Poisson process. By mastering these concepts, students will be better equipped to solve complex problems and analyze data.
## FAQs
Here are some frequently asked questions about probability theory that may help clarify some of the concepts we discussed earlier.
### Q. What is the difference between discrete and continuous probability distributions?
Discrete probability distributions deal with random variables that can take on a countable number of values, while continuous probability distributions deal with random variables that can take on an infinite number of values.
### Q. Can probability be greater than 1?
No, probability cannot be greater than 1. It is a measure of the likelihood of an event occurring and is represented by a number between 0 and 1, inclusive.
### Q.What is the difference between covariance and correlation?
Covariance measures the linear relationship between two random variables, while correlation is a normalized version of covariance that measures the strength and direction of the linear relationship between two random variables.
### Q. What is Bayes’ Theorem, and how is it used in probability?
Bayes’ Theorem is a formula that describes the probability of an event occurring based on prior knowledge of related events. It is widely used in statistical inference and machine learning.
### Q. What is the Central Limit Theorem, and why is it important in probability theory?
The central limit theorem states that the sum of a large number of independent and identically distributed random variables approaches a normal distribution, regardless of the underlying distribution of the random variables. It is important in probability theory because it provides a basis for statistical inference and data analysis.
### Q. Can I use probability to predict future events?
Probability can be used to estimate the likelihood of future events, but it cannot predict the future with certainty. It is a statistical tool that can be used to analyze data and make informed decisions based on probability calculations.
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NCERT Solutions for Class 7 Maths Exercise 15.1 Chapter 15 Visualising Solid Shapes
*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.
NCERT Solutions for Class 7 Maths Exercise 15.1 Chapter 15 Visualising Solid Shapes are provided here. These solutions not only help the students to clear all their doubts instantly but also help them to know the best answer to the particular questions in this NCERT Solutions for Class 7 Maths Chapter 15. In this exercise, students get to know about faces, edges, and vertices. Students can score good marks in Maths by practising these solutions.
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes – Exercise 15.1
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Access Other Exercises of NCERT Solutions for Class 7 Maths Chapter 15 – Visualising Solid Shapes
Exercise 15.2 Solutions
Exercise 15.3 Solutions
Exercise 15.4 Solutions
Access Answers to NCERT Class 7 Maths Chapter 15 – Visualising Solid Shapes Exercise 15.1
1. Identify the nets which can be used to make cubes (cut out copies of the nets and try it):
(i)
Solution:-
The given net cannot be folded as a cube.
Because, it can be folded as below,
(ii)
Solution:-
The given net can be folded as a cube.
Because, it can be folded as below,
(iii)
Solution:-
The given net can be folded as a cube.
Because, it can be folded as below,
(iv)
Solution:-
The given net can be folded as a cube.
Because, it can be folded as below,
(v)
Solution:-
The given net cannot be folded as a cube.
Because, it can be folded as below,
(vi)
Solution:-
The given net can be folded as a cube.
Because, it can be folded as below,
2. Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them.
Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.
Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.
Solution:-
(i)
By observing the above figure the sum of the opposite faces is equal to 7.
So, the given net can be folded into a net of dice.
(ii)
By observing the above figure the sum of the opposite faces is equal to 7.
So, the given net can be folded into a net of dice.
3. Can this be a net for a die?
Solution:-
The given net is folded as dice is shown in the figure below,
No, this cannot be a net for a die.
By observing the figure we can say that, one pair of opposite face will have 1 and 4, another pair of opposite face will have 3 and 6 the sum of these two opposite faces are not equal to 7.
4. Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here?
(Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.)
Solution:-
In the given net there are 3 faces and it can be completed as shown below,
5. Match the nets with appropriate solids:
Solution:-
(a) (ii)
Because, the net (ii) can be folded as a square.
(b) (iii)
Because, the net (iii) can be folded as a cylinder.
(c) (iv)
Because, the net (iv) can be folded as cone.
(d) (i)
Because, the net (i) can be folded as a pyramid.
Also, explore –Â
NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 7Â
NCERT Solutions
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The Law of SINES.
Presentation on theme: "The Law of SINES."— Presentation transcript:
The Law of SINES
The Law of SINES For any triangle (right, acute or obtuse), you may use the following formula to solve for missing sides or angles:
Use Law of SINES when ... AAS - 2 angles and 1 adjacent side
you have 3 dimensions of a triangle and you need to find the other 3 dimensions - they cannot be just ANY 3 dimensions though, or you won’t have enough info to solve the Law of Sines equation. Use the Law of Sines if you are given: AAS - 2 angles and 1 adjacent side ASA - 2 angles and their included side SSA (this is an ambiguous case)
Example 1 You are given a triangle, ABC, with angle A = 70°, angle B = 80° and side a = 12 cm. Find the measures of angle C and sides b and c. * In this section, angles are named with capital letters and the side opposite an angle is named with the same lower case letter .*
Example 1 (con’t) A C B 70° 80° a = 12 c b
The angles in a ∆ total 180°, so angle C = 30°. Set up the Law of Sines to find side b:
Example 1 (con’t) A C B 70° 80° a = 12 c b = 12.6 30°
Set up the Law of Sines to find side c:
Example 1 (solution) A C B 70° 80° a = 12 c = 6.4 b = 12.6 30°
Angle C = 30° Side b = 12.6 cm Side c = 6.4 cm Note: We used the given values of A and a in both calculations. Your answer is more accurate if you do not used rounded values in calculations.
Example 2 You are given a triangle, ABC, with angle C = 115°, angle B = 30° and side a = 30 cm. Find the measures of angle A and sides b and c.
Example 2 (con’t) To solve for the missing sides or angles, we must have an angle and opposite side to set up the first equation. We MUST find angle A first because the only side given is side a. The angles in a ∆ total 180°, so angle A = 35°. A C B 115° 30° a = 30 c b
Example 2 (con’t) A C B 115° a = 30 c b
30° a = 30 c b 35° Set up the Law of Sines to find side b:
Example 2 (con’t) A C B 115° a = 30 c b = 26.2
30° a = 30 c b = 26.2 35° Set up the Law of Sines to find side c:
Example 2 (solution) Angle A = 35° Side b = 26.2 cm Side c = 47.4 cm A
115° 30° a = 30 c = 47.4 b = 26.2 35° Angle A = 35° Side b = 26.2 cm Side c = 47.4 cm Note: Use the Law of Sines whenever you are given 2 angles and one side!
The Ambiguous Case (SSA)
When given SSA (two sides and an angle that is NOT the included angle) , the situation is ambiguous. The dimensions may not form a triangle, or there may be 1 or 2 triangles with the given dimensions. We first go through a series of tests to determine how many (if any) solutions exist.
The Ambiguous Case (SSA)
In the following examples, the given angle will always be angle A and the given sides will be sides a and b. If you are given a different set of variables, feel free to change them to simulate the steps provided here. ‘a’ - we don’t know what angle C is so we can’t draw side ‘a’ in the right position A B ? b C = ? c = ?
The Ambiguous Case (SSA)
Situation I: Angle A is obtuse If angle A is obtuse there are TWO possibilities If a ≤ b, then a is too short to reach side c - a triangle with these dimensions is impossible. If a > b, then there is ONE triangle with these dimensions. A B ? a b C = ? c = ? A B ? a b C = ? c = ?
The Ambiguous Case (SSA)
Situation I: Angle A is obtuse - EXAMPLE Given a triangle with angle A = 120°, side a = 22 cm and side b = 15 cm, find the other dimensions. Since a > b, these dimensions are possible. To find the missing dimensions, use the Law of Sines: A B a = 22 15 = b C c 120°
The Ambiguous Case (SSA)
Situation I: Angle A is obtuse - EXAMPLE Angle C = 180° - 120° ° = 23.8° Use Law of Sines to find side c: A B a = 22 15 = b C c 120° 36.2° Solution: angle B = 36.2°, angle C = 23.8°, side c = 10.3 cm
The Ambiguous Case (SSA)
Situation II: Angle A is acute If angle A is acute there are SEVERAL possibilities. Side ‘a’ may or may not be long enough to reach side ‘c’. We calculate the height of the altitude from angle C to side c to compare it with side a. A B ? b C = ? c = ? a
The Ambiguous Case (SSA)
Situation II: Angle A is acute First, use SOH-CAH-TOA to find h: A B ? b C = ? c = ? a h Then, compare ‘h’ to sides a and b . . .
The Ambiguous Case (SSA)
Situation II: Angle A is acute If a < h, then NO triangle exists with these dimensions. A B ? b C = ? c = ? a h
The Ambiguous Case (SSA)
Situation II: Angle A is acute If h < a < b, then TWO triangles exist with these dimensions. A B b C c a h A B b C c a h If we open side ‘a’ to the outside of h, angle B is acute. If we open side ‘a’ to the inside of h, angle B is obtuse.
The Ambiguous Case (SSA)
Situation II: Angle A is acute If h < b < a, then ONE triangle exists with these dimensions. Since side a is greater than side b, side a cannot open to the inside of h, it can only open to the outside, so there is only 1 triangle possible! A B b C c a h
The Ambiguous Case (SSA)
Situation II: Angle A is acute If h = a, then ONE triangle exists with these dimensions. A B b C c a = h If a = h, then angle B must be a right angle and there is only one possible triangle with these dimensions.
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1 Given a triangle with angle A = 40°, side a = 12 cm and side b = 15 cm, find the other dimensions. Find the height: A B ? 15 = b C = ? c = ? a = 12 h 40° Since a > h, but a< b, there are 2 solutions and we must find BOTH.
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1 FIRST SOLUTION: Angle B is acute - this is the solution you get when you use the Law of Sines! A B 15 = b C c a = 12 h 40°
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse - use the first solution to find this solution. In the second set of possible dimensions, angle B is obtuse, because side ‘a’ is the same in both solutions, the acute solution for angle B & the obtuse solution for angle B are supplementary. Angle B = ° = 126.5° a = 12 A B 15 = b C c 40° 1st ‘a’ 1st ‘B’
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse Angle B = 126.5° Angle C = 180°- 40° ° = 13.5° a = 12 A B 15 = b C c 40° 126.5°
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EX. 1 (Summary) Angle B = 53.5° Angle C = 86.5° Side c = 18.6 Angle B = 126.5° Angle C = 13.5° Side c = 4.4 a = 12 A B 15 = b C c = 4.4 40° 126.5° 13.5° A B 15 = b C c = 18.6 a = 12 40° 53.5° 86.5°
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 2 Given a triangle with angle A = 40°, side a = 12 cm and side b = 10 cm, find the other dimensions. A B ? 10 = b C = ? c = ? a = 12 h 40° Since a > b, and h is less than a, we know this triangle has just ONE possible solution - side ‘a’opens to the outside of h.
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 2 Using the Law of Sines will give us the ONE possible solution: A B 10 = b C c a = 12 40°
The Ambiguous Case - Summary
if angle A is acute find the height, h = b*sinA if angle A is obtuse if a < b no solution if a > b one solution if a < h no solution if h < a < b 2 solutions one with angle B acute, one with angle B obtuse if a > b > h 1 solution If a = h 1 solution angle B is right (Ex I) (Ex II-1) (Ex II-2)
The Law of Sines AAS ASA SSA (the ambiguous case)
Use the Law of Sines to find the missing dimensions of a triangle when given any combination of these dimensions.
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# How Do I Use the Factors in Math Activities in Real Life?
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Once you understand basic math principles, you may not always recognize them when you use them in real life- just like you probably don't notice the alphabet each time you read. Factoring is a basic math concept that reverses multiplication, finding the numbers that multiply together to create a larger number. This concept has obvious applications in the real world.
#### TL;DR (Too Long; Didn't Read)
Factoring is a useful skill in real life. Common applications include: dividing something into equal pieces, exchanging money, comparing prices, understanding time and making calculations during travel.
## Dividing Something Equally
A key time you use factoring is when you must divide something into equal pieces. For example, if 6 people worked together to make brownies, and the pan of brownies yields 24 brownies, it would only be fair if everyone received the same number of brownies. Because 6 is a factor of 24, the brownies divide into equal shares without cutting them into smaller pieces. Dividing 24 by 6 gives a result of 4, so each person gets 4 brownies.
## Factoring with Money
Exchanging money is another common function that relies on factoring. You probably already know that 4 quarters make a dollar. Looking at this in terms of factoring, 2 factors of 100 are 4 and 25. Similarly, you can exchange a twenty-dollar bill for 20 one-dollar bills (factors 1 and 20), 2 ten-dollar bills (factors 2 and 10) or 4 five-dollar bills (factors 4 and 5).
## Comparing Prices
You also use factoring while shopping to compare prices per unit. For example, there are two cans of an expensive coffee blend on sale. A 12-ounce can costs $36.00, and a 6-ounce can costs$24.00. Using factors, you can compare the price per ounce without using a calculator or notepad. Dividing 36 by 12, the factors of 36 are 3 and 12. Dividing 24 by 6, the factors of 24 are 4 and 6. Using this information, you know that the 12-ounce can costs $3.00 per ounce and the 6-ounce can costs$4.00 per ounce.
## Understanding Time
Time is another opportunity to use factoring in the real world. Every day contains 24 hours; if you must take a pill 3 times per day, you take 1 pill every 8 hours (3 x 8 = 24). An hour divides into 60 minutes. Those 60 minutes divide into 12 increments of 5 minutes each on the face of a clock (12 x 5 = 60). When describing time, you might divide hours into quarters (4 x 15 = 60) and half-hour segments (2 x 30 = 60).
## Traveling with Factors
Factors are also useful when traveling. If you travel 720 miles on vacation, you need to know how many hours you must drive so you can plan your trip. At an average speed of 60 mph, it would take 12 hours to get to your destination (60 x 12 = 720).
Understanding factoring allows you to easily navigate number relationships in the real world without relying on your calculator or phone to do the work for you.
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# Definition:Group
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## Definition
A group is a semigroup with an identity (that is, a monoid) in which every element has an inverse.
### Group Axioms
The properties that define a group are sufficiently important that they are often separated from their use in defining semigroups, monoids, and so on, and given recognition in their own right.
A group is an algebraic structure $\struct {G, \circ}$ which satisfies the following four conditions:
$(\text G 0)$ $:$ Closure $\ds \forall a, b \in G:$ $\ds a \circ b \in G$ $(\text G 1)$ $:$ Associativity $\ds \forall a, b, c \in G:$ $\ds a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ $(\text G 2)$ $:$ Identity $\ds \exists e \in G: \forall a \in G:$ $\ds e \circ a = a = a \circ e$ $(\text G 3)$ $:$ Inverse $\ds \forall a \in G: \exists b \in G:$ $\ds a \circ b = e = b \circ a$
These four stipulations are called the group axioms.
The notation $\struct {G, \circ}$ is used to represent a group whose underlying set is $G$ and whose operation is $\circ$.
### Group Law
The operation $\circ$ can be referred to as the group law.
### Product Element
Let $a, b \in G$ such that $= a \circ b$.
Then $g$ is known as the product of $a$ and $b$.
### Multiplicative Notation
When discussing a general group with a general group law, it is customary to dispense with a symbol for this operation and merely concatenate the elements to indicate the product element.
$x y$ is used to indicate the result of the operation on $x$ and $y$. There is no symbol used to define the operation itself.
$e$ or $1$ is used for the identity element.
$x^{-1}$ is used for the inverse element.
$x^n$ is used to indicate the $n$th power of $x$.
Compare with additive notation.
## Also denoted as
Some sources denote a group using the notation $\gen {G, \circ}$ for $\struct {G, \circ}$.
## Examples
### Example: $\dfrac {x + y} {1 + x y}$
Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group.
### Example: $x + y + 2$ over $\R$
Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as:
$\forall x, y \in \R: x \circ y := x + y + 2$
Then $\struct {\R, \circ}$ is a group whose identity is $-2$.
### Example: $x + y + x y$ over $\R \setminus \set {-1}$
Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as:
$\forall x, y \in \R: x \circ y := x + y + x y$
Let:
$\R' := \R \setminus \set {-1}$
that is, the set of real numbers without $-1$.
Then $\struct {\R', \circ}$ is a group whose identity is $0$.
### Example: $x^{-1} = 1 - x$
Let $S = \set {x \in \R: 0 < x < 1}$.
Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a group such that the inverse of $x \in S$ is $1 - x$.
### Example: Group of Linear Functions
Let $G$ be the set of all real functions $\theta_{a, b}: \R \to \R$ defined as:
$\forall x \in \R: \map {\theta_{a, b} } x = a x + b$
where $a, b \in \R$ such that $a \ne 0$.
The algebraic structure $\struct {G, \circ}$, where $\circ$ denotes composition of mappings, is a group.
$\struct {G, \circ}$ is specifically non-abelian.
### Example: Operation Induced by Self-Inverse and Cancellable Elements
Let $S$ be a set with an operation which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that:
$(1): \quad \exists e \in S: a \ast b = e \iff a = b$
$(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast b$
Then $\struct {S, \circ}$ is a group, where $\circ$ is defined as $a \circ b = a \ast \paren {e \ast b}$.
## Examples of Structures which are not Groups
### Arbitrary Example: Order 4
Let $S = \set {1, 2, 3, 4}$.
Consider the algebraic structure $\struct {S, \circ}$ given by the Cayley table:
$\begin{array}{r|rrrr} \circ & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 3 & 1 \\ 3 & 3 & 2 & 4 & 3 \\ 4 & 4 & 3 & 1 & 2 \\ \end{array}$
Then $\struct {S, \circ}$ is not a group.
## Also see
• Results about groups can be found here.
## Historical Note
The term group was first used by Évariste Galois in $1832$, in the context of the solutions of polynomials in radicals. Augustin Louis Cauchy was also involved in this development.
The concept of the group as a purely abstract structure was introduced by Arthur Cayley in his $1854$ paper On the theory of groups.
The first one to formulate the set of axioms to define the structure of a group was Leopold Kronecker in $1870$.
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# Special Products of Polynomials
## Work with multiple variables with special products of polynomials. Learn how to solve binomial squares, and the difference of squares with binomials.
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Special Products of Polynomials
What if you wanted to multiply two binomials that were exactly the same, like \begin{align*}(x^2 - 2)(x^2 - 2)\end{align*}? Similarly what if you wanted to multiply two binomials in which the sign between the two terms was the opposite in one from the other, like \begin{align*}(x^2 - 2)(x^2 + 2)\end{align*}? What shortcuts could you use? After completing this Concept, you'll be able to find the square of a binomial as well as the product of binomials using the sum and difference formula.
### Guidance
We saw that when we multiply two binomials we need to make sure to multiply each term in the first binomial with each term in the second binomial. Let’s look at another example.
Multiply two linear binomials (binomials whose degree is 1):
When we multiply, we obtain a quadratic polynomial (one with degree 2) with four terms:
The middle terms are like terms and we can combine them. We simplify and get \begin{align*}2x^2+11x+12\end{align*}. This is a quadratic, or second-degree, trinomial (polynomial with three terms).
You can see that every time we multiply two linear binomials with one variable, we will obtain a quadratic polynomial. In this section we’ll talk about some special products of binomials.
Find the Square of a Binomial
One special binomial product is the square of a binomial. Consider the product \begin{align*}(x+4)(x+4)\end{align*}.
Since we are multiplying the same expression by itself, that means we are squaring the expression. \begin{align*}(x+4)(x+4)\end{align*} is the same as \begin{align*}(x+4)^2\end{align*}.
When we multiply it out, we get \begin{align*}x^2+4x+4x+16\end{align*}, which simplifies to \begin{align*}x^2+8x+16\end{align*}.
Notice that the two middle terms—the ones we added together to get \begin{align*}8x\end{align*}—were the same. Is this a coincidence? In order to find that out, let’s square a general linear binomial.
Sure enough, the middle terms are the same. How about if the expression we square is a difference instead of a sum?
It looks like the middle two terms are the same in general whenever we square a binomial. The general pattern is: to square a binomial, take the square of the first term, add or subtract twice the product of the terms, and add the square of the second term. You should remember these formulas:
Remember! Raising a polynomial to a power means that we multiply the polynomial by itself however many times the exponent indicates. For instance, \begin{align*}(a+b)^2=(a+b)(a+b)\end{align*}. Don’t make the common mistake of thinking that \begin{align*}(a+b)^2=a^2+b^2\end{align*}! To see why that’s not true, try substituting numbers for \begin{align*}a\end{align*} and \begin{align*}b\end{align*} into the equation (for example, \begin{align*}a = 4\end{align*} and \begin{align*}b = 3\end{align*}), and you will see that it is not a true statement. The middle term, \begin{align*}2ab\end{align*}, is needed to make the equation work.
We can apply the formulas for squaring binomials to any number of problems.
#### Example A
Square each binomial and simplify.
a) \begin{align*}(x+10)^2\end{align*}
b) \begin{align*}(2x-3)^2\end{align*}
c) \begin{align*}(x^2+4)^2\end{align*}
Solution
Let’s use the square of a binomial formula to multiply each expression.
a) \begin{align*}(x+10)^2\end{align*}
If we let \begin{align*}a = x\end{align*} and \begin{align*}b = 10\end{align*}, then our formula \begin{align*}(a+b)^2=a^2+2ab+b^2\end{align*} becomes \begin{align*}(x+10)^2=x^2+2(x)(10)+10^2\end{align*}, which simplifies to \begin{align*}x^2+20x+100\end{align*}.
b) \begin{align*}(2x-3)^2\end{align*}
If we let \begin{align*}a = 2x\end{align*} and \begin{align*}b = 3\end{align*}, then our formula \begin{align*}(a-b)^2=a^2-2ab+b^2\end{align*} becomes \begin{align*}(2x-3)^2=(2x^2)-2(2x)(3)+(3)^2\end{align*}, which simplifies to \begin{align*}4x^2-12x+9\end{align*}.
c) \begin{align*}(x^2+4)^2\end{align*}
If we let \begin{align*}a = x^2\end{align*} and \begin{align*}b = 4\end{align*}, then
Find the Product of Binomials Using Sum and Difference Patterns
Another special binomial product is the product of a sum and a difference of terms. For example, let’s multiply the following binomials.
Notice that the middle terms are opposites of each other, so they cancel out when we collect like terms. This is not a coincidence. This always happens when we multiply a sum and difference of the same terms. In general,
When multiplying a sum and difference of the same two terms, the middle terms cancel out. We get the square of the first term minus the square of the second term. You should remember this formula.
Sum and Difference Formula: \begin{align*}(a+b)(a-b)=a^2-b^2\end{align*}
Let’s apply this formula to a few examples.
#### Example B
Multiply the following binomials and simplify.
a) \begin{align*}(x+3)(x-3)\end{align*}
b) \begin{align*}(5x+9)(5x-9)\end{align*}
c) \begin{align*}(2x^3+7)(2x^3-7)\end{align*}
Solution
a) Let \begin{align*}a = x\end{align*} and \begin{align*}b = 3\end{align*}, then:
b) Let \begin{align*}a = 5x\end{align*} and \begin{align*}b = 9\end{align*}, then:
c) Let \begin{align*}a = 2x^3\end{align*} and \begin{align*}b = 7\end{align*}, then:
Solve Real-World Problems Using Special Products of Polynomials
Now let’s see how special products of polynomials apply to geometry problems and to mental arithmetic.
#### Example C
Find the area of the following square:
Solution
The length of each side is \begin{align*}(a+b)\end{align*}, so the area is \begin{align*}(a+b)(a+b)\end{align*}.
Notice that this gives a visual explanation of the square of a binomial. The blue square has area \begin{align*}a^2\end{align*}, the red square has area \begin{align*}b^2\end{align*}, and each rectangle has area \begin{align*}ab\end{align*}, so added all together, the area \begin{align*}(a+b)(a+b)\end{align*} is equal to \begin{align*}a^2+2ab+b^2\end{align*}.
The next example shows how you can use the special products to do fast mental calculations.
#### Example D
Use the difference of squares and the binomial square formulas to find the products of the following numbers without using a calculator.
a) \begin{align*}43 \times 57\end{align*}
b) \begin{align*}45^2\end{align*}
c) \begin{align*}481 \times 319\end{align*}
Solution
The key to these mental “tricks” is to rewrite each number as a sum or difference of numbers you know how to square easily.
a) Rewrite 43 as \begin{align*}(50-7)\end{align*} and 57 as \begin{align*}(50 + 7)\end{align*}.
Then \begin{align*}43 \times 57 = (50-7)(50 + 7) = (50)^2 - (7)^2 = 2500-49 = 2451\end{align*}
b) \begin{align*}45^2 = (40 + 5)^2 = (40)^2 + 2(40)(5) + (5)^2 = 1600 + 400 + 25 = 2025\end{align*}
c) Rewrite 481 as \begin{align*}(400 + 81)\end{align*} and 319 as \begin{align*}(400-81)\end{align*}.
Then \begin{align*}481 \times 319 = (400 + 81)(400-81) = (400)^2-(81)^2\end{align*}
\begin{align*}(400)^2\end{align*} is easy - it equals 160000.
\begin{align*}(81)^2\end{align*} is not easy to do mentally, so let’s rewrite 81 as \begin{align*}80 + 1\end{align*}.
\begin{align*}(81)^2 = (80 + 1)^2 = (80)^2 + 2(80)(1) + (1)^2 = 6400 + 160 + 1 = 6561\end{align*}
Then \begin{align*}481 \times 319 = (400)^2 - (81)^2 = 160000 - 6561 = 153439\end{align*}
Watch this video for help with the Examples above.
### Vocabulary
• Square of a binomial: \begin{align*}(a+b)^2=a^2+2ab+b^2,\end{align*} and \begin{align*}(a-b)^2=a^2-2ab+b^2\end{align*}
• Sum and difference formula: \begin{align*}(a+b)(a-b) = a^2-b^2\end{align*}
### Guided Practice
1. Square the binomial and simplify: \begin{align*}(5x-2y)^2\end{align*}.
2. Multiply \begin{align*}(4x+5y)(4x-5y)\end{align*} and simplify.
3. Use the difference of squares and the binomial square formulas to find the product of \begin{align*}112 \times 88\end{align*} without using a calculator.
Solutions:
1. \begin{align*}(5x-2y)^2\end{align*}
If we let \begin{align*}a = 5x\end{align*} and \begin{align*}b = 2y\end{align*}, then
2. Let \begin{align*}a = 4x\end{align*} and \begin{align*}b = 5y\end{align*}, then:
3. The key to these mental “tricks” is to rewrite each number as a sum or difference of numbers you know how to square easily.
Rewrite 112 as \begin{align*}(100 + 12)\end{align*} and 88 as \begin{align*}(100-12)\end{align*}.
Then
### Practice
Use the special product rule for squaring binomials to multiply these expressions.
1. \begin{align*}(x+9)^2\end{align*}
2. \begin{align*}(3x-7)^2\end{align*}
3. \begin{align*}(5x-y)^2\end{align*}
4. \begin{align*}(2x^3-3)^2\end{align*}
5. \begin{align*}(4x^2+y^2)^2\end{align*}
6. \begin{align*}(8x-3)^2\end{align*}
7. \begin{align*}(2x+5)(5+2x)\end{align*}
8. \begin{align*}(xy-y)^2\end{align*}
Use the special product of a sum and difference to multiply these expressions.
1. \begin{align*}(2x-1)(2x+1)\end{align*}
2. \begin{align*}(x-12)(x+12)\end{align*}
3. \begin{align*}(5a-2b)(5a+2b)\end{align*}
4. \begin{align*}(ab-1)(ab+1)\end{align*}
5. \begin{align*}(z^2+y)(z^2-y)\end{align*}
6. \begin{align*}(2q^3+r^2)(2q^3-r^2)\end{align*}
7. \begin{align*}(7s-t)(t+7s)\end{align*}
8. \begin{align*}(x^2y+xy^2)(x^2y-xy^2)\end{align*}
Find the area of the lower right square in the following figure.
Multiply the following numbers using special products.
1. \begin{align*}45 \times 55\end{align*}
2. \begin{align*}56^2\end{align*}
3. \begin{align*}1002 \times 998\end{align*}
4. \begin{align*}36 \times 44\end{align*}
5. \begin{align*}10.5 \times 9.5\end{align*}
6. \begin{align*}100.2 \times 9.8\end{align*}
7. \begin{align*}-95 \times -105\end{align*}
8. \begin{align*}2 \times -2\end{align*}
### Vocabulary Language: English
Binomial
Binomial
A binomial is an expression with two terms. The prefix 'bi' means 'two'.
Square of a Binomial
Square of a Binomial
The product of a squared binomial is always a trinomial.
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Algebra 1 2.5 Distributive Property. Vocabulary Equivalent expressions: two expressions that have the same output value for every input value Distributive.
Presentation on theme: "Algebra 1 2.5 Distributive Property. Vocabulary Equivalent expressions: two expressions that have the same output value for every input value Distributive."— Presentation transcript:
Algebra 1 2.5 Distributive Property
Vocabulary Equivalent expressions: two expressions that have the same output value for every input value Distributive Property: multiply the outside number to every number in the parenthesis Term: the individual parts of an expression
Vocabulary Coefficient: the number part of a term Constant Term: a term that has a number part but no variable Like Terms: terms that have the same variable part
Use the distributive property to write an equivalent expression. EXAMPLE 1Apply the distributive property 1. 4(y + 3) = 2. (y + 7)y = 4. (2 – n)8 = 3. n(n – 9) = 4y + 12 y 2 + 7y n 2 – 9n 16 – 8n
= – 15y + 3y 2 2. (5 – y)(–3y) = Simplify. Distribute – 3y. = – 2x – 14 Distribute – 2. Use the distributive property to write an equivalent expression. EXAMPLE 2Distribute a negative number 1. –2(x + 7)= – 2(x) + – 2(7) 5(–3y) – y(–3y)
Simplify. = (– 1)(2x) – (–1)(11) 3. –(2x – 11) = of 21 Multiplicative property EXAMPLE 2Distribute a negative number Distribute – 1. = – 2x + 11 (–1)(2x – 11)
Constant terms: – 4, 2 Coefficients: 3, – 6 Like terms: 3x and – 6x; – 4 and 2 Identify the terms, like terms, coefficients, and constant terms of the expression 3x – 4 – 6x + 2. SOLUTION EXAMPLE 3 Identify parts of an expression Terms: 3x, – 4, – 6x, 2
GUIDED PRACTICE Use the distributive property to write an equivalent expression. 1. 2(x + 3) = 2x + 6 2. – (4 – y) = – 4 + y Distributive – 1 3. (m – 5)(– 3m) = m (– 3m) –5 (– 3m) Distributive – 3m = – 3m 2 + 15m Simplify. 4. (2n + 6) = 1 2 1 2 2n + 6 1 2 = n + 3 1 2 Distribute Simplify.
GUIDED PRACTICE Identify the terms, like terms, coefficients, and constant terms of the expression – 7y + 8 – 6y – 13. Coefficients: – 7, – 6 Like terms: – 7y and – 6y, 8 and – 13; SOLUTION Terms: – 7y, 8, – 6y, – 13 Constant terms: 8, – 13
Standardized Test Practice EXAMPLE 4 ANSWER The correct answer is B. DCBA Simplify the expression 4(n + 9) – 3(2 + n). 4(n + 9) – 3(2 + n) = Distributive property = n + 30 Combine like terms. A B C D n + 3 5n + 30 n + 305n + 3 4n + 36 – 6 – 3n
GUIDED PRACTICE 1. Simplify the expression 5(6 + n) – 2(n – 2). 5(6 + n) – 2(n – 2) = Distributive property = 3n + 34 Combine like terms. 30 + 5n – 2n + 4 SOLUTION
Solve a multi-step problem EXAMPLE 5 Your daily workout plan involves a total of 50 minutes of running and swimming. You burn 15 calories per minute when running and 9 calories per minute when swimming. Let r be the number of minutes that you run. Find the number of calories you burn in your 50 minute workout if you run for 20 minutes. SOLUTION The workout lasts 50 minutes, and your running time is r minutes. So, your swimming time is (50 – r ) minutes.
Solve a multi-step problem EXAMPLE 5 STEP 1 C = Write equation. = 15 r + 450 – 9r Distributive property = 6r + 450 Combine like terms. Write a verbal model. Then write an equation. 15r + 9(50 – r) C = 15 r + 9 (50 – r) Amount burned (calories) Burning rate when running (calories/minute) Running time (minutes) Swimming time (minutes) = + Burning rate when swimming (calories/minute)
Solve a multi-step problem EXAMPLE 5 C = Write equation. = 6(20) + 450 = 570 Substitute 20 for r. Then simplify. ANSWER You burn 570 calories in your 50 minute workout if you run for 20 minutes. STEP 2 Find the value of C when r = 20. 6r + 450
GUIDED PRACTICE WHAT IF… Suppose your workout lasts 45 minutes. How many calories do you run for 20 minutes? 30 minutes? SOLUTION The workout lasts 45 minutes, and your running time is r minutes. So, your swimming time is (45 – r ) minutes.
GUIDED PRACTICE STEP 1 C = 15 r + 9 (45 – r) C = Write equation. = 15 r + 405 – 9r Distributive property = 6r + 405 Combine like terms. Write a verbal model. Then write an equation. 15 r + 9 (45 – r) Amount burned (calories) Burning rate when running (calories/minute) Running time (minutes) Swimming time (minutes) = + Burning rate when swimming (calories/minute)
GUIDED PRACTICE C = Write equation. = 6(20) + 405 = 525 Substitute 20 for r. Then simplify. STEP 2 Find the value of C when r = 20. 6r + 405 Write equation. = 6(30) + 405 = 585 Substitute 30 for r. Then simplify. STEP 3 Find the value of C when r = 30. 6r + 405 C =
GUIDED PRACTICE ANSWER You burn 525 calories in your 45 minute workout if you run for 20 minutes. You burn 585 calories in your 45 minute workout if you run for 30 minutes.
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# How do you simplify $4\sqrt {\dfrac{{81}}{{16}}}$.
Last updated date: 15th Sep 2024
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Hint: We will first write the numerator and denominator separately in the form of squares. Then cut it with square root and multiply with 4 and thus we have the answer.
We need to simplify $4\sqrt {\dfrac{{81}}{{16}}}$. ………………(1)
We know that ${4^2} = 4 \times 4 = 16$ and ${9^2} = 9 \times 9 = 81$.
Since, we have $\sqrt {\dfrac{{81}}{{16}}}$ in the given expression which we require to simplify.
We can write this expression as: $\sqrt {\dfrac{{81}}{{16}}} = \sqrt {\dfrac{{{9^2}}}{{{4^2}}}}$ ……………..(2)
Now, we will make use of the fact that: $\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}$
On replacing a by 9 and b by 4, we get the following expression:-
$\Rightarrow \dfrac{{{9^2}}}{{{4^2}}} = {\left( {\dfrac{9}{4}} \right)^2}$
Putting the above expression in the equation number (2), we will then obtain the following equation:-
$\Rightarrow \sqrt {\dfrac{{81}}{{16}}} = \sqrt {{{\left( {\dfrac{9}{4}} \right)}^2}}$ …………………..(3)
We also know that $\sqrt {{a^2}} = a$
On replacing a by $\dfrac{9}{4}$ in the above equation, we will then get the following equation:-
$\Rightarrow \sqrt {{{\left( {\dfrac{9}{4}} \right)}^2}} = \dfrac{9}{4}$
Putting the above equation in the equation number (3), we will then obtain the following expression:-
$\Rightarrow \sqrt {\dfrac{{81}}{{16}}} = \dfrac{9}{4}$
Putting the expression above in equation number (1), we will then obtain the following expression:-
$\Rightarrow 4\sqrt {\dfrac{{81}}{{16}}} = 4 \times \dfrac{9}{4}$
Cutting off 4 from right hand side in above expression to get the following expression with us:-
$\Rightarrow 4\sqrt {\dfrac{{81}}{{16}}} = 9$
Hence, we have $4\sqrt {\dfrac{{81}}{{16}}}$ simplified as 9.
Note:
The students must note that when we cut off 4 in the last third step it is because we know that 4 is not equal to 0.
The students must also note that when we open up square root, we have two possibilities, either a > 0 or a < 0. In general if nothing is given to us, we generally assume that a > 0 and thus we have the required answer as 9, otherwise it could have been -9 as well.
The students must note that the small things we use in the solution, sometimes we forget that they have a concept behind them like: $\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}$ and $\sqrt {{a^2}} = a$.
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## LetsPlayMaths.Com
WELCOME TO THE WORLD OF MATHEMATICS
# Class 5 Rounding Off Numbers
Introduction to Rounding Off
Rounding Off to Nearest Ten
Rounding Off to Nearest Hundred
Rounding Off to Nearest Thousand
Rounding Off to Nearest Ten Thousand
Rounding Off to Nearest Lakh
Rounding Off to Nearest Ten Lakh
Rounding Off to Nearest Crore
Rounding Off Numbers Test
Worksheet
## Introduction to Rounding Off
Population of USA is 32 crores. But, does that mean the number is actually 32 crores? No, when we describe large numbers we often express them as round numbers. Round number is nothing but an approximate number, that is, it is very close to the exact number.
Let s discuss about some more examples.
Length of Amazon river is 6992 km, but wile expressing we can say it is 7000 km.
Height of Eiffel tower is 1063 feet, but while expressing we can say it is 1000 feet.
We must follow below mentioned rules to rounding off a number.
1. If the digit at the lower place value is less than five, we round off the number by keeping the digit at the same
and writing zeros in all the lower place values.
2. If the digit at the lower place value is equal or greater than five, we round off the given number by adding the
digit by one and writing zeros in all the lower place values.
## Rounding Off to Nearest Ten
We compare the digit at the ones place with five to round off to the nearest ten. If ones place digit is less than five, then we round off the number by replacing the digit in the ones place by zero.
Example 1. Round off 42 to the nearest tens place.
Solution. In 42, two is present in the ones place and it is less than five. We have to make ones place zero of number 42,that is the round off number is 40.
Example 2. Round off 574 to the nearest tens place.
Solution. 4 is present in ones place of 574 and it is less than number five. So, if we round off 574 to it s tens place then the new number will be 570.
Example 3. Round off 6751 to tens place.
Solution. 1 is present in ones place and 1 < 5. If we round off 6751 to tens place then the new number will be 6750.
If the digits in ones place is greater than or equal to 5 , then we round off the number by increasing the digit at tens place by 1 and writing zero in the ones place.
Example 1. Round off 348 to the nearest tens place.
Solution. 8 is present in ones place of 348 and 8 > 5. If we round off 348 then the new number will be 350.
Example 2. Round off 9765 to the nearest tens place.
Solution. 5 is present in ones place and it is equal to 5. If we round off 9765 then the new number will be 9770.
## Rounding Off to Nearest Hundred
We compare the digit at the tens place with 5 to round off numbers to the nearest hundred. If the digit in the tens place is less than five, then we replace the digits in tens and ones places with zeros.
Example 1. Round off 6542 to the nearest hundred.
Solution. 4 is present in the tenth place of 6542 and 4 < 5. If we round off 6542, then the new number will be 6500.
Example 2. Round off 85621 to the nearest hundred.
Solution. 2 is present in the tenth place of 85621, and 2 < 5. If we round off 85621, then the new number will be 85600.
If the digits in tens place is greater than or equal to 5, then we round off the number by increasing the digit at hundredth place by 1 and writing zero in the remaining lower places.
Example 1. Round off 3259 to the nearest hundred.
Solution. 5 is present in the tenth place of 3259, and 5 = 5. If we round off 3259, then the new number will be 3300.
Example 2. Round off 47689 to the nearest hundred.
Solution. 8 is present in the tenth place of 47689, and 8 > 5. If we round off 47689, then the new number will be 47690.
## Rounding Off to Nearest Thousand
We compare the digit at the hundredth place with 5 to round off numbers to the nearest thousand. If the digit in the hundredth place is less than five, then we replace the digits in hundred, tens and ones places with zeros.
Example 1. Round off 6259 to the nearest thousand.
Solution. 2 is present in the hundredth place of 6259, and 2 < 5. If we round off 6259, then the new number will be 6000.
Example 2. Round off 46495 to the nearest thousand.
Solution. 4 is present in the hundredth place of 46495, and 4 < 5. If we round off 46495, then the new number will be 46000.
If the digits in hundredth place is greater than or equal to 5, then we round off the number by increasing the digit at thousandth place by 1 and writing zero in the remaining lower places.
Example 1. Round off 5759 to the nearest thousand.
Solution. 7 is present in the hundredth place of 5759, and 7 > 5. If we round off 5759, then the new number will be 5800.
Example 2. Round off 47689 to the nearest thousand.
Solution. 6 is present in the hundredth place of 47689, and 6 > 5. If we round off 47689, then the new number will be 48000.
## Rounding Off to Nearest Ten Thousand
We compare the digit at the thousandth place with 5 to round off numbers to the nearest ten thousand. If the digit in the thousandth place is less than five, then we replace the digits in thousand, hundred, tens and ones places with zeros.
Example 1. Round off 61259 to the nearest thousand.
Solution. 1 is present in the thousandth place of 61259, and 1 < 5. If we round off 61259, then the new number will be 61000.
Example 2. Round off 342495 to the nearest thousand.
Solution. 2 is present in the thousandth place of 342495, and 2 < 5. If we round off 342495, then the new number will be 340000.
If the digits in thousandth place is greater than or equal to 5, then we round off the number by increasing the digit at ten thousandth place by 1 and writing zero in the remaining lower places.
Example 1. Round off 36759 to the nearest thousand.
Solution. 6 is present in the thousandth place of 36759, and 6 > 5. If we round off 36759, then the new number will be 37000.
Example 2. Round off 247689 to the nearest thousand.
Solution. 7 is present in the thousandth place of 247689, and 7 > 5. If we round off 247689, then the new number will be 248000.
## Rounding Off to Nearest Lakh
We compare the digit at the ten thousandth place with 5 to round off numbers to the nearest lakh. If the digit in the ten thousandth place is less than five, then we replace the digits in ten thousand, thousand, hundred, tens and ones places with zeros.
Example 1. Round off 241259 to the nearest lakh.
Solution. 4 is present in the ten thousandth place of 241259, and 4 < 5. If we round off 241259, then the new number will be 200000.
Example 2. Round off 5342495 to the nearest lakh.
Solution. 4 is present in the ten thousandth place of 5342495, and 4 < 5. If we round off 5342495, then the new number will be 5300000.
If the digits in ten thousandth place is greater than or equal to 5, then we round off the number by increasing the digit at the lakh place by 1 and writing zero in the remaining lower places.
Example 1. Round off 456759 to the nearest lakh.
Solution. 5 is present in the ten thousandth place of 456759, and 5 = 5. If we round off 456759, then the new number will be 500000.
Example 2. Round off 2287689 to the nearest lakh.
Solution. 8 is present in the ten thousandth place of 2287689, and 8 > 5. If we round off 2287689, then the new number will be 2300000.
## Rounding Off to Nearest Ten Lakh
We compare the digit at the lakh place with 5 to round off numbers to the nearest ten lakh. If the digit in the lakh place is less than five, then we replace the digits in lakh, ten thousand, thousand, hundred, tens and ones places with zeros.
Example 1. Round off 3146259 to the nearest ten lakh.
Solution. 1 is present in the lakh place of 3146259, and 1 < 5. If we round off 3146259, then the new number will be 3000000.
Example 2. Round off 35342495 to the nearest lakh.
Solution. 3 is present in the ten thousandth place of 35342495, and 3 < 5. If we round off 35342495, then the new number will be 35000000.
If the digits in lakh place is greater than or equal to 5, then we round off the number by increasing the digit at the ten-lakh place by 1 and writing zero in the remaining lower places.
Example 1. Round off 1656759 to the nearest ten-lakh.
Solution. 6 is present in the lakh place of 1656759, and 6 > 5. If we round off 1656759, then the new number will be 2000000.
Example 2. Round off 53683219 to the nearest ten-lakh.
Solution. 6 is present in the lakh place of 53683219, and 6 > 5. If we round off 53683219, then the new number will be 54000000.
## Rounding Off to Nearest Crore
We compare the digit at the ten-lakh place with 5 to round off numbers to the nearest crore. If the digit in the ten-lakh place is less than five, then we replace the digits in ten lakh, lakh, ten thousand, thousand, hundred, tens and ones places with zeros.
Example 1. Round off 53146259 to the nearest crore.
Solution. 3 is present in the ten-lakh place of 53146259, and 3 < 5. If we round off 53146259, then the new number will be 50000000.
Example 2. Round off 432761495 to the nearest crore.
Solution. 2 is present in the ten-lakh place of 432761495, and 2 < 5. If we round off 432761495, then the new number will be 430000000.
If the digits in ten lakh place is greater than or equal to 5, then we round off the number by increasing the digit at the crore place by 1 and writing zero in the remaining lower places.
Example 1. Round off 78656759 to the nearest crore.
Solution. 8 is present in the ten-lakh place of 78656759, and 8 > 5. If we round off 78656759, then the new number will be 80000000.
Example 2. Round off 157683219 to the nearest crore.
Solution. 7 is present in the ten-lakh place of 157683219, and 7 > 5. If we round off 157683219, then the new number will be 160000000.
## Rounding Off Numbers Test
Rounding Off Numbers Test - 1
## Class-5 Rounding Off Numbers Worksheet
Rounding Off Numbers Worksheet - 1
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# What is the constant solution of a differential equation?
## What is the constant solution of a differential equation?
If g(a) = 0 for some a then y(t) = a is a constant solution of the equation, since in this case ˙y =0= f(t)g(a). For example, ˙y = y2 - 1 has constant solutions y(t) = 1 and y(t) = -1.
## What is a constant differential equation?
A differential equation has constant coefficients if only constant functions appear as coefficients in the associated homogeneous equation. A solution of a differential equation is a function that satisfies the equation. The solutions of a homogeneous linear differential equation form a vector space.
## How do you find the constant in an equation?
The constant of proportionality k is given by k=y/x where y and x are two quantities that are directly proportional to each other. Once you know the constant of proportionality you can find an equation representing the directly proportional relationship between x and y, namely y=kx, with your specific k.
## What is a constant coefficient example?
Just for completeness we will offer some explicit examples of constant coefficients equations: Example 1. 1. ˙x + 5x = 0 (first order) 2. ˙x + 5x = cos(3t) (first order) 3.
## What are constant coefficients?
The constant coefficient is the coefficient not attached to variables in an expression. For example, the constant coefficients of the expressions above are the real coefficient 3 and the parameter represented by c. For example in the expressions above, the leading coefficients are 2 and the parameter represented by a.
## What is a constant term example?
A constant term is a term that contains only a number. In other words, there is no variable in a constant term. Examples of constant terms are 4, 100, and -5.
## How do you find the constant?
Since k is constant (the same for every point), we can find k when given any point by dividing the y-coordinate by the x-coordinate. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = = 3. Thus, the equation describing this direct variation is y = 3x.
## What is a constant in a formula?
A constant is a set value that doesn't change and that's directly inserted into a cell. It's not a formula and it's not calculated by a formula.
## What is a constant value in math?
A mathematical constant is a key number whose value is fixed by an unambiguous definition, often referred to by a symbol (e.g., an alphabet letter), or by mathematicians' names to facilitate using it across multiple mathematical problems.
What is Rosalie power in Twilight?
What is the suffix of predict?
What is the root word of capable?
## Why are there 64 codons and only 20 amino acids?
Simply because it requires a minimum of 3 bases to produce a codon for one amino acid. The four RNA bases (A, C, U, and G) can be combined three at a time in 64 different ways, yet there are only 20 amino acids encoded by the genetic code.
## Is plywood a good thermal insulator?
Is plywood good insulation? The short answer is “no,” at least, not compared with products designed to provide insulation from cold and heat. Even so, plywood products do offer minimal R value, along with structural benefits.
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# Math Snap
## 어떤 세 자리 수의 $\frac{1}{10}$ 인 수와 $\frac{1}{100}$ 인 수의 합이 76.78 입 니다. 어떤 수는 얼마인지 구하시오.
#### STEP 1
Assumptions1. The number we are looking for is a three-digit number. . The sum of1/10 of this number and1/100 of this number is76.78.
#### STEP 2
Let's denote the number we are looking for as $x$.
#### STEP 3
According to the problem, $\frac{x}{10} + \frac{x}{100} =76.78$.
#### STEP 4
To simplify the equation, we can multiply each term by100 to get rid of the fractions. $10x + x =7678$
#### STEP 5
Combine like terms. $11x =7678$
#### STEP 6
To solve for $x$, divide both sides of the equation by11. $x = \frac{7678}{11}$
##### SOLUTION
Calculate the value of $x$. $x = \frac{767}{11} =698$So, the three-digit number we are looking for is698.
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#### What is a Fact Family?
Has your child come home with something similar to a problem stating: Write the fact family for 5, 13, 8 with no examples in sight? Well, this article is for parents who have to help their children with Fact Family homework. Let’s look at an example:
8 + 5 = 13
5 + 8 = 13
13 – 8 = 5
13 – 5 = 8
Generally, the word family means certain people are related. Likewise, fact family implies that certain numbers and facts are related.
For starters, they are only three numbers in each family. In the above fact family, the members are 5, 8, and 13.
How are they related?
*** You can add two of the numbers together to get the third number.
8 + 5 = 13
*** You can switch the order of the two numbers added above to equal the third number again. In math, this is referred to as the commutative property of addition.
5 + 8 = 13
Cousin Operators or Operands
*** Just as your sister/brother’s children are your children’s cousins, addition is related to subtraction via the term inverse property. Subtraction is the inverse property of addition. In other words, subtraction is the opposite of addition. We can undo the work of addition by subtracting. This is very important to remember for problem solving.
How to get the last two math facts of the family . . .
So far we’ve recreated 8 + 5 = 13 and 5 + 8 = 13
Since we know the inverse (opposite) of addition is subtraction, start with the sum (or the larger number), 13, and subtract one of the addends. 13 – 8 = 5 Of course, your answer is the third number. In this case, the answer is 5. Thus the next fact would be, 13 – 5 = 8.
Usefulness:
Consider this word problem:
Susan had 5 pencils. Her grandmother gave her more pencils for her collection. Susan now has 13 pencils. How many pencils did Susan’s grandmother give her?
If a student has an understanding of fact family concept, he/she could rebuild the family.
You know that Susan started out with 5 pencils and ended up with 13 pencils. Thus, you know two of the family members. They are 5 and 13. In math language that translates,
5 pencils + ____ more pencils = 13 pencils or 5 + ? = 13
Let’s look at the fact family again.
5 + ? = 13
? + 5 = 13
13 – 5 = ?
13 - ? = 5
Which equation above would give me the value of ‘?” ? Answer: The third equation. Hey, we just need to subtract 5 from 13 to learn that her grandmother gave her 8 pencils.
If your child did not know their addition and subtraction facts well, teaching them concept would definitely help.
What about Multiplication and Division?
Yes, there related. Division is the inverse property of Multiplication. Thus, the same principle applies. Study the example below.
6 x 4 = 24
4 x 6 = 24
24 / 6 = 4
24/ 4 = 6
(The “/” represents division.)
Practice:
Let’s use dominoes to practice with addition and subtraction.
Use each side of the domino as an addend to form the first addition fact. Allow the student to count all of dots to get the answer or sum.
6 + 3 = 9
Change the order of the addends:
3 + 6 = 9
Use the inverse property of addition
9 – 6 = 3
9 – 3 = 6
The Fact Family by Turley includes 32 student activity sheets and paper manipulatives.
You Should Also Read:
Multiplication Practice Ideas
More Multiplication Practice Ideas
Multiplication Tips - 2s, 4s, 8s
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2. Ticket In The Door
3. Lesson Essential Question What are the important parts of a quadratic graph?
4. Quadratic Review • For each quadratic function: • Identify the quadratic term (a) • Identify the linear term (b) • Identify the constant term (c)
5. Quadratic Function: y = ax2 + bx + c • Example 1: 2x2 + 3x + 10 a = _____ b = _____ c = _____ • Example 2: -3x2 + 5x a = _____ b = _____ c = _____ • Example 3: x2 - 8x + 7 a = _____ b = _____ c = _____ • Example 4: -x2 - 9x – 3 a = _____ b = _____ c = _____ • Example 5: -x2 - 6x a = _____ b = _____ c = _____ • Example 6: x2 a = _____ b = _____ c = _____
6. Consider the following quadratic function: f(x) = x2 + 2x – 3 Let’s talk about another important part of a quadratic function: Where is the y-intercept? Where does the function cross the y-axis? y-intercept: (0, -3)
7. Consider the following quadratic function: f(x) = x2 + 2x – 3 Let’s talk about another important part of a quadratic function: Where are the x-intercepts? Where does the function cross the x-axis? x-intercepts: (1, 0) & (-3, 0)
8. Consider the following quadratic function: f(x) = x2 + 2x – 3 Let’s talk about several important parts of a quadratic function: Where is the vertex? (-1, -4)
9. Consider the following quadratic function: f(x) = x2 + 2x – 3 Let’s talk about another important part of a quadratic function: How do we algebraically calculate the vertex?
10. Consider the following quadratic function: f(x) = x2 + 2x – 3 Calculating the vertex. The vertex is a coordinate point (x, y) on the graph, now that we have the x value how do you think we determine the y value?
11. Consider the following quadratic function: f(x) = x2 + 2x – 3 Calculating the vertex. Substitute the value of x into the given function equation above and solve! The answer is the value for y. When x = -1, y = -4. Vertex is: (-1, -4).
12. Consider the following quadratic function: f(x) = x2 + 2x – 3 Let’s talk about another important part of a quadratic function: What is the axis of symmetry? Now that you see what it is, how would you define the axis of symmetry?
13. Consider the following quadratic function: f(x) = x2 + 2x – 3 Let’s talk about another important part of a quadratic function: How do we represent this axis of symmetry? x = -1
14. Consider the following quadratic function: f(x) = x2 – 2x – 15 Where are the x-intercepts? Where does the function cross the x-axis? x-intercepts: (-3, 0) & (5, 0)
15. Consider the following quadratic function: f(x) = x2 – 2x – 15 Where is the y-intercept? Where does the function cross the y-axis? y-intercept: (0, -15)
16. Let’s Do It Again Ourselves!!Consider the following quadratic function: f(x) = x2 – 2x – 15 Where is the vertex? Algebraically calculate the vertex. (1, -16)
17. Consider the following quadratic function: f(x) = x2 – 2x – 15 Where is the axis of symmetry? Draw in the axis of symmetry. What is the axis of symmetry?
18. Consider the following quadratic function: f(x) = x2 + 3x Where is the y-intercept? Where does the function cross the y-axis? y-intercept: (0, 0)
19. Consider the following quadratic function: f(x) = x2 + 3x Where are the x-intercepts? Where does the function cross the x-axis? x-intercepts: (-3, 0) & (0, 0)
20. Let’s Do It Again Ourselves!!Consider the following quadratic function: f(x) = x2 + 3x Where is the vertex? Algebraically calculate the vertex. (-1.5, -2.25)
21. Consider the following quadratic function: f(x) = x2 + 3x Where is the axis of symmetry? Draw in the axis of symmetry. What is the axis of symmetry?
22. Now, Visualize the graph! Given: f(x) = x2 – 4x + 3 Open up or down? Calculate the vertex? What is the axis of symmetry? Where is the y-intercept?
23. Now, Visualize the graph! Given: f(x) = 2x2 + 3x – 1 Open up or down? Calculate the vertex? What is the axis of symmetry? Where is the y-intercept?
24. Now, Visualize the graph! Given: f(x) = 5x2 – 2x + 5 Open up or down? Calculate the vertex? What is the axis of symmetry? Where is the y-intercept?
25. Now, Visualize the graph! Given: f(x) = x2 – 2x – 15 Open up or down? Calculate the vertex? What is the axis of symmetry? Where is the y-intercept?
26. Ticket Out The Door • Complete the ticket out the door problem. Please hand it to me as you walk out of the door. Homework • Complete the worksheet for homework.
27. IMPORTANT PARTS OF QUADRATIC GRAPHS • Does the graph open up or down • (write “a” is + or -) • Put a star at the Vertex (write the point) • Draw the Axis of Symmetry and write the equation • Circle the X-intercepts (write the point) • Draw a square around the Y-intercept (write the point)
28. Quadratic Functions and their important parts! What important parts do you recognize in this graph? y = x2 – 3x – 10
29. Quadratic Functions and their important parts! What important parts do you recognize in this graph?
30. Lesson Essential Question How do you graph a quadratic function using the vertex?
31. Putting It All Together Now!!! Graphing Parabolas In order to graph we will need the following: • Visualize whether the parabola open up or down • Calculate the coordinates of the Vertex • Determine the Axis of Symmetry • Determine the y-intercept • Plot a few more points to understand the actual shape of the graph • Identify the x-intercepts
32. Graphing Quadratic Functions Graph the function, then identify the x-intercepts (roots) = ____________
33. Graphing Quadratic Functions Graph the function, then identify the x-intercepts (roots) = ____________
34. Graphing Quadratic Functions Graph the function, then identify the x-intercepts (roots) = ____________
35. Graphing Quadratic Functions Graph the function, then identify the x-intercepts (roots) = ____________
36. Graphing Quadratic Functions e.) Sketch the graph of y = x2 – 2x – 3 Graph the function, then identify the x-intercepts (roots) = ____________
37. Graphing Quadratic Functions f.) Sketch the graph of y = x2 + 4x + 4 Graph the function, then identify the x-intercepts (roots) = ____________
38. Graphing Quadratic Functions g.) Sketch the graph of y = ½x2 – 3 Graph the function, then identify the x-intercepts (roots) = ____________
39. Graphing Quadratic Functions h.) Sketch the graph of y = 2x2 + 4x + 5 Graph the function, then identify the x-intercepts (roots) = ____________
40. On Your Own Practice Please complete the practice worksheets in order to develop and master this skill. Thank you
41. Homework Assignment More Practice Graphing Quadratic Functions!
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GED Mathematical Reasoning: Introduction To Measurement | Open Window Learning
Geometry
# GED Mathematical Reasoning: Introduction To Measurement
Unit conversion
Standard/Customary Units Metric Units LENGTH 1 foot = 12 inches 1 centimeter = 10 millimeters 1 yard = 3 feet = 36 inches 1 meter = 100 centimeters 1 mile = 1,760 yards = 5,280 feet 1 kilometer = 1,000 meters CAPACITY 1 cup = 8 fluid ounces 1 liter = 1,000 milliliters 1 pint = 2 cups 1 kiloliter = 1,000 liters 1 quart = 2 pints 1 gallon = 4 quarts WEIGHT 1 pound = 16 ounces 1 gram = 1,000 milligrams 1 ton = 2,000 pounds 1 kilogram = 1,000 grams 1 metric ton = 1,000 kilograms
• When you need to change or convert from one unit to another, first identify the relationship between the units using the conversion chart.
• Then, from there, to change from a smaller unit to a larger unit will require division and changing from a larger unit to a smaller unit will require multiplication.
smaller to larger means division
larger to smaller means multiplication
Example 1
Convert 15 pints to quarts and pints.
Note that we are changing from a smaller unit to a larger unit, which means we will divide. Using our conversion chart, we’ll first identify the relationship between pints and quarts. Looking on the left side containing Standard units, we see that one quart equals 2 pints. Since every two pints equals one quart, we will divide 15 by 2 to determine the number of whole quarts. Fifteen divided by two is 7 with a remainder of one. Therefore, 15 pints equals 7 quarts and one pint.
1 quart = 2 pints
(Remainder 1)
15 pints = 7 quarts and 1 pint
A popular diagram used to commit the relationship between gallons, quarts, pints, and cups to memory is shown here on the slide.
A big G contains four smaller Q’s to represent that there are 4 quarts in one gallon.
Two P’s are drawn in each Q to denote that there are two pints in one quart.
Finally, two C’s are drawn in each P to represent that there are two cups in one pint.
Notice that this diagram also helps us see other relationships, too. Like, for example, there are 8 pints in one gallon since there are 8 P’s in the big G. Or, for another example: there are four cups in one quart, since there are 4 C’s in each Q.
Example 2
Convert 6.2 grams to milligrams.
Note that we are changing from a larger unit to a smaller unit, which means we will multiply.
Using our conversion chart, we’ll first identify the relationship between grams and milligrams. Looking on the right side containing Metric units, we see that 1 gram equals 1,000 milligrams. Since each gram represents 1,000 milligrams, we’ll multiply 6.2 times 1,000 to determine how many milligrams this is equal to.
1 gram = 1000 milligrams
6.2 * 1000 = 6200
6.2 grams = 6200 milligrams
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# What is the equation of the line between (-1,12) and (7,-7)?
Feb 27, 2016
The equation of the line that passes through the points $A \left(- 1 , 12\right)$ and $B \left(7 , - 7\right)$ is :
$y = - \frac{19}{8} x + \frac{77}{8}$
#### Explanation:
The standard form of the equation of a line is $y = m x + p$ with m the slope of the line.
STEP 1 : Let's find the slope of the line.
$m = \frac{{y}_{B} - {y}_{A}}{{x}_{B} - {x}_{A}} = \frac{- 7 - 12}{7 + 1} = - \frac{19}{8}$
N.B : The fact that the slope is negative indicates the line decreases.
STEP 2 : Let's find p (coordinate at origin).
Use the point-slope formula with one of our points, e.g. $A \left(- 1 , 12\right)$ and $m = - \frac{19}{8}$.
$12 = - \frac{19}{8} \cdot - 1 + p$
$p = \frac{77}{8}$
Cross-check: Check the equation with the second point.
Use $B \left(7 , - 7\right)$ in the equation :
$y = - \frac{19}{8} \cdot 7 + \frac{77}{8} = - \frac{96}{8} + \frac{77}{8} = - \frac{56}{8} = - 7$
-> Perfect !
|
# The Sway Reference Manual/Recursion
In the previous chapter, we learned about looping using the functions while, do-until, for, and for-each. These functions implement what is known as iterative looping.
Recursion is another way of looping; often, recursive loops are easier to write and understand, as compared to iterative loops, such as whiles and fors. Many mathematical functions are easy to implement recursively, so we will start there. Recall that the factorial of a number n is:
``` $n! = n * (n - 1) * (n - 2) * ... * 2 * 1$
```
Consider writing a function which computes the factorial of a positive integer. For example, if the function were passed the value of 4, it should return the value of 24 since 4! is 4*3*2*1 or 24. One approach would be to use an iterative loop to cycle through all the multipliers, saving the partial product at each iteration. Getting the code to work correctly does take a bit of fiddling to get all the important bits right. A working implementation might look like:
``` function factorial(n)
{
var i;
var product = 1;
for (i = 1, i <= n, i = i + 1)
{
product = product * i;
}
product;
}
```
This implementation raises many questions, since it goes against the general convention for structuring a counting loop:
• Why was the product initialized to one instead of zero?
• Why was the loop initialization chosen to be $i = 1$ instead of $i = 0$?
• Why was the loop test $i <= n$ instead of $i < n$?
With some pondering, these questions can be answered because of the mathematics of multiplication..
There is a radically different approach that can be taken, an extremely simple, elegant, and powerful approach, called recursion. To apply recursion to solve a problem, it must be possible to state the solution of a problem in ever simpler terms. For factorial, the factorial of a number can be stated in terms of a simpler factorial.
``` $0! = 1$
$n! = n * (n - 1)!$ otherwise
```
This second formulation states that the factorial of zero is onea. [1] and that the factorial of any other (positive) number is obtained by multiplying the that number by the factorial of one less than that number. After some study, you should be able to see that both the first formulation (with the ellipses ...) and this new formulation are equivalent[2]. The second form is particularly well suited for implementation as a computer program:
``` function factorial(n)
{
if (n == 0)
{
1;
}
else
{
n * factorial(n - 1);
}
}
```
or
``` function factorial(n)
{
if (n == 0,1,n * factorial(n - 1));
}
```
using if's pure function call syntax.
Note how these two versions of factorial precisely implement the second formulation.
Convince yourself that the function really works by tracing the function call:
``` factorial(3)
```
Decomposing the call, we find that:
``` factorial(3) is 3 * factorial(2)
```
since n, having a value of 3, is not equal to 0. and so the second block of the if is evaluated. We can replace factorial(2) by 2 * factorial(1), yielding:
``` factorial(3) is 3 * 2 * factorial(1)
```
since n, now having a value of 2, is still not zero. Continuing along this vein, we can replace factorial(1) by 1 * factorial(0), yielding:
``` factorial(3) is 3 * 2 * 1 * factorial(0)
```
Now in this call to factorial, n does have a value of zero, so we can replace factorial(0) with its immediate return value of zero:
``` factorial(3) is 3 * 2 * 1 * 1
```
Thus, factorial(3) has a value of six:
``` sway> factorial(3);
INTEGER: 6
```
as expected.
## The parts of a recursive functionEdit
Recursive approaches rely on the fact that it is usually simpler to solve a smaller problem than a larger one. In the factorial problem, trying to find the factorial n - 1 is a little bit simpler than finding the factorial of n. If finding the factorial of n - 1 is still too hard to solve easily, then find the factorial of n - 2 and so on until we find a case where the solution is easy. With regards to factorial, this is when n is equal to zero. The 'easy-to-solve' code (and the values that gets you there) is known as the base case. The find-the-solution-for-a-simpler-problem code (and the values that get you there) is known as the recursive case. The recursive case usually contains a call to the very function being executed. This call is known as a recursive call.
Most well-formed recursive functions are composed of at least one base case and at least one recursive case.
## The greatest common divisorEdit
Consider finding the greatest common divisor (gcd) of two numbers. One approach is use repeated division. The first division divides the two numbers in question, saving the remainder. Now make the divisor the dividend and the remainder the divisor. Repeat this process until the remainder is zero. At that point, the current divisor is the gcd.
Let's turn it into a function, first iteratively and then recursively.
``` // compute the gcd of two positive integers iteratively
function gcd(first,second)
{
var remainder;
do-until (remainder == 0)
{
remainder = first % second;
first = second;
second = remainder;
}
first; //first holds the last divisor
}
```
The do-until loop insures that the loop body is executed before the remainder is tested. Let's rewrite the function recursively. First we identify the recurrence case(s) and then the base case(s). In this case, the recurrence and base cases are a little different than in the factorial example. We recur[3]if the remainder is not zero. We immediately return the answer if the remainder is zero. Now that the recurrence and base cases have been identified, we can write the recursive version without too much trouble.
``` // compute the gcd of two positive integers
function gcd(first,second)
{
if (first % second == 0)
{
second;
}
else
{
gcd(second,first % second);
}
}
```
or
``` function gcd(first,second)
{
if (first % second == 0,second,gcd(second,first % second));
}
```
or, to remove the repeated computation of the remainder:
``` function gcd(first,second)
{
var remainder = first % second;
if (remainder == 0,second,gcd(second,remainder));
}
```
Again, the recursive version is more compact.
Look at how the recursive version turns second into first by passing second as the first argument in the recursive call. By the same token, remainder becomes second by nature of being the second argument in the recurive call. To convince yourself that the routine really works, modify gcd to inspect the arguments:
``` function gcd(first,second)
{
var remainder = first % second;
inspect(array(first,second,remainder));
if (remainder == 0,second,gcd(second,remainder));
}
```
You haven't learned about arrays or the built-in array function yet, but think of it as a way to group multiple things together. Inspecting the array is a trick to get multiple values inspected with a single call to inspect.
``` sway> gcd(66,42);
array(first,second,remainder) is [66,42,24]
array(first,second,remainder) is [42,24,18]
array(first,second,remainder) is [24,18,6]
array(first,second,remainder) is [18,6,0]
INTEGER: 6
```
Note, how the first remainder, 24, keeps shifting to the left. In the first recursive call, the remainder becomes second, so the 24 shifts one spot to the left. On the second recursive call, the current second, which is 24, becomes first, so the 24 shifts once again to the left.
## The Fibonacci sequenceEdit
A third example of recursion is the computation of the $n^{th}$ Fibonacci number. The Fibonacci series looks like this:
``` n 0 1 2 3 4 5 6 7 8 ...
Fib(n) 0 1 1 2 3 5 8 13 21 ...
```
and is found in nature again and again[4].. In general, a Fibonacci number is equal to the sum of the previous two Fibonacci numbers. The exceptions are the zeroes and the first Fibonacci numbers which are equal to 0 and 1 respectively. Voila! The recurrence case and the two base cases have jumped right out at us! Here, then is a recursive implementation of a function which computes the nth Fibonacci number.
``` // compute the nth Fibonacci number
// n must be non-negative!
function fibonacci(n)
{
if (n == 0) return 0;
if (n == 1) return 1;
fibonacci(n-1) + fibonacci(n-2);
}
```
Our implementation is straightforward and elegant. Unfortunately, it's horribly inefficient. Unlike our recursive versions of factorial and gcd, which recurred about as many times as the iterative versions looped, our Fibonacci version will recur many, many more times than an iterative version of Fibonacci when computing larger Fibonacci numbers. Here's why.
Consider the call to fib(6). Tracing all the recursive calls to fib, we get:
``` fib(6) is fib(5) + fib(4)
```
Replacing fib(5) with fib(4) + fib(3), we get:
``` fib(6) is fib(4) + fib(3) + fib(4)
```
We can already see a problem, we will compute fib(4) twice, once from the original call to fib(6) and again when we try to find fib(5). If we write down all the recursive calls to fib(6) which each recursive call indented from the previous, we get a structure that looks like this:
``` fib(6)
fib(5)
fib(4)
fib(3)
fib(2)
fib(1)
fib(0)
fib(1)
fib(2)
fib(1)
fib(0)
fib(3)
fib(2)
fib(1)
fib(0)
fib(1)
fib(4)
fib(3)
fib(2)
fib(1)
fib(0)
fib(1)
fib(2)
fib(1)
fib(0)
```
We would expect, based on how the Fibonacci sequence is generated to take about six 'steps' to calculate fib(6). Instead, ultimately there were 13 calls to either fib(1) or fib(0)[5]. There was a tremendous amount of duplicated, and therefore wasted effort. So let's try to compute Fibonacci numbers using an iterative loop:
``` function fib(n)
{
var i;
var first = 0;
var second = 1;
for (i = 0, i < n, i = i + 1)
{
var sum = first + second;
first = second;
second = sum;
}
first;
}
```
In the loop body, we see that fib is much like gcd; the second number becomes the first number and some combination of the first and second number becomes the second number. In the case of gcd, the combination was the remainder and, in the case of fib, the combination is sum.
As with factorial, hitting on the right way to proceed iteratively is not exactly straightforward, while the recursive version practically wrote itself. Noting the similarity of fib and gcd suggest a way to have both recursion and efficiency at the same time.
## More on recursive loopsEdit
To transform an iterative loop into a recursive loop, one first identifies those variables that are changing in the loop body; these variable will become formal parameters in the recursive function. For example, the fib loop above has three (not two!) variables that are being changed during each iteration of the loop: first, second, and i. So, we start out our recursive function as thus:
``` function loop(first,second,i)
{
...
}
```
The loop test becomes an if test in the body of the loop function:
``` function loop(first,second,i)
{
if (i < n)
{
...
}
else
{
...
}
}
```
The if-true block becomes the recursive call. The arguments to the recursive call encode the updates to the loop variables The if-false block becomes the value the loop attempted to calculate:
``` function loop(first,second,i)
{
if (i < n)
{
loop(second,first + second,i + 1);
}
else
{
first;
}
}
```
Using function call syntax for if shortens up the function:
``` function loop(first,second,i)
{
if (i < n,loop(second,first + second,i + 1),first);
}
```
Next, we embed the loop function into our the function containing the original loop. That way, any non-local variables referenced in the test or body of the original loop will be visible to the loop function:
``` function fib(n)
{
function loop(first,second,i)
{
if (i < n,loop(second,first + second,i + 1),first);
}
...
}
```
Finally, we call the loop function with the initial values of the loop variables:
``` function fib(n)
{
function loop(first,second,i)
{
if (i < n,loop(second,first + second,i + 1),first);
}
loop(0,1,0);
}
```
For more practice, let's convert the iterative version of factorial into a recursive function using this method. We'll see we'll end up with a different recursive function than before.
``` function factorial(n)
{
var i;
var product = 1;
for (i = 1, i <= n, i = i + 1)
{
product = product * i;
}
product;
}
```
We start, as before, by working on the loop function. In this case, only two variables are changing in the loop: product and i.
``` function loop(product,i)
{
...
}
```
Next, we write the if expression:
``` function loop(product,i)
{
if (i <= n,loop(product * i,i + 1),product);
}
```
Next, we embed the loop function and call it:
``` function factorial(n)
{
function loop(product,i)
{
if (i <= n,loop(product * i,i + 1),product);
}
loop(1,1);
}
```
The moral of this story is that any iterative loop can be rewritten as a recursion and any recursion can be rewritten as an iterative loop. Moreover, in 'good' languages[6], there is no reason to prefer one way over the other, either in terms of the time it takes or the space used in execution. Use a recursion if that makes the implementation more clear, otherwise, use an iterative loop.
## FootnotesEdit
1. Mathematicians, being an inclusive bunch, like to invite zero to the factorial party.
2. The first formulation gets the basic idea of factorial across but is not very precise. For example, how would you computer the factorial of three using the first formulation?
3. The word is recur, NOT recurse!
4. Pineapples, the golden ratio, chambered nautilus, etc.
5. 13 is a Fibonacci number. Curious!
6. Someday, Sway will be a 'good' language.
|
Question Video: Reading and Interpreting Speed-Time Graphs | Nagwa Question Video: Reading and Interpreting Speed-Time Graphs | Nagwa
# Question Video: Reading and Interpreting Speed-Time Graphs Physics • First Year of Secondary School
## Join Nagwa Classes
The speed-time graph shows the change of speed of a person walking during the time interval from 𝑡 = 0 seconds to 𝑡 = 6 seconds. What is the person’s speed at 𝑡 = 0 s? What is the person’s speed at 𝑡 = 3 s? What distance does the person walk between 𝑡 = 0 s and 𝑡 = 2 s? What distance does the person walk between 𝑡 = 2 s and 𝑡 = 4 s? What distance does the person walk between 𝑡 = 4 s and 𝑡 = 6 s? What distance does the person walk between 𝑡 = 0 s and 𝑡 = 6 s?
09:54
### Video Transcript
The speed–time graph shows the change of speed of a person walking during the time interval from 𝑡 is equal to zero seconds to 𝑡 is equal to six seconds.
So, this here is the speed–time graph in question, and we can see that on the vertical axis we’ve been given the speed of the person walking in meters per second. And on the horizontal axis, we’ve got the time in seconds.
Now, the first part of the question asks us, what is the person’s speed at 𝑡 is equal to zero seconds?
Now, as we’ve already mentioned, the horizontal axis of the graph shows us the time when the person is walking. And from this graph, we can find the speed at which the person is walking at any given time. So, we’ve been asked to find the speed of the person at 𝑡 is equal to zero seconds. Now, 𝑡 is equal to zero seconds is the time at the very beginning of the graph. Specifically, this vertical line here represents 𝑡 is equal to zero seconds.
And so, what we can do is to draw a vertical line upwards at 𝑡 is equal to zero seconds until it meets the graph representing the person’s speed. And at this point, we can read off on the vertical axis that the person speed is five meters per second. Hence, our answer to this part of the question is that the person’s speed at 𝑡 is equal to zero seconds is five meters per second. Let’s move on to the next part of the question then.
This part asks us, what is the person’s speed at 𝑡 is equal to three seconds?
So, now we want to go along the horizontal axis, the time axis, until we arrive at 𝑡 is equal to three seconds. At this point, we want to draw a vertical line upwards at 𝑡 is equal to three seconds until we meet the blue line because the blue line is representing the speed of the person walking as time progresses. And once we meet the blue line, we can go across to the vertical axis, which tells us that the speed of the person at the time 𝑡 is equal to three seconds is six meters per second. And hence, six meters per second is our answer to the second part of the question.
Moving on to the next part of the question then, this part asks us, what distance does the person walk between 𝑡 is equal to zero seconds and 𝑡 is equal to two seconds?
Now, as we’ve already seen, 𝑡 is equal to zero seconds is represented by this vertical line here. And we can draw in this vertical line to represent 𝑡 is equal to two seconds. Now, we’ve been asked to find the distance walked by the person between 𝑡 is equal to zero seconds and 𝑡 is equal to two seconds. To do this, we can recall the equation that links speed, distance, and time. Two quantities which we have on our graph, speed and time, and the third quantity which we’re trying to find, distance. We can recall that the speed of an object is equal to the distance moved by the object divided by the time taken for that object to move that distance. And so, we can take this equation and multiply both sides by the time 𝑡 in order to solve for the distance 𝑑.
When we do this, we find that the speed of the object 𝑠, in this case the person walking, multiplied by the time 𝑡 for which the person walks at that speed is equal to the distance moved by that person, 𝑑. Now, in our particular graph, we’ve got the speed of the person on the vertical axis and the time on the horizontal axis. And for the first two seconds of time given to us on the graph, we’ve been told that the person walks at a constant speed of five meters per second. That’s what the blue line is telling us. This means that for the first two seconds, the person walks at a constant speed of five meters per second, and so the distance moved by the person is going to be the constant speed, five meters per second, multiplied by the time, two seconds.
But interestingly, this also happens to be equal to the area underneath our blue line, in this case now drawn in pink. In other words, that’s this area here because that area is equal to five meters per second multiplied by two seconds, which is exactly what’s going to give us the distance moved by the person. And so, we can say that the distance moved by the person between 𝑡 is equal to zero seconds and 𝑡 is equal to two seconds, we’ll call this distance 𝑑, is equal to the speed of the person, five meters per second, multiplied by the amount of time for which the person is walking at that speed, two seconds.
Now, quickly looking at the units, we can see that we’ve got meters per second multiplied by seconds which means that we’ve got units of seconds in the numerator and the denominator. And those cancel, leaving us with just the unit of meters. And this works perfectly because on the left-hand side, we’ve got the distance moved by the person. This needs to have a unit of meters. And hence, the distance moved by the person is going to be five multiplied by two meters. In other words, that’s 10 meters. And hence, we can say that the distance the person walks between 𝑡 is equal to zero seconds and 𝑡 is equal to two seconds is 10 meters.
Now, moving on to the next part of the question, we are asked, what distance does the person walk between 𝑡 is equal to two seconds and 𝑡 is equal to four seconds?
In other words now, what distance does the person walk between this point in time and this point in time? Well, drawing vertical dotted lines from 𝑡 is equal to two seconds and 𝑡 is equal to four seconds all the way up to the blue line shows us that the area that we now need to calculate is this area here. Because even though between 𝑡 is equal to two seconds and four seconds, the person is no longer walking at a constant speed because we can see that the speed increases between these two points in time. The fact of the matter is that the area under a speed–time graph still gives us the distance moved by that particular object.
So, to find the area underneath the speed-time graph, this time between 𝑡 is equal to two seconds and 𝑡 is equal to four seconds, we need to find this shaded area as we’ve already mentioned. Now, to make life easier for ourselves, we can split up this area into two smaller ones. Firstly, this rectangle here. And secondly, this blue triangle. We can call this area one, and we can call this area two. And we can say that the total area that we’re trying to find is area one plus area two, the area of the triangle plus the area of the rectangle.
Now, with area two, the area of the rectangle, we can notice something. We’ve actually already calculated the size of that rectangle because the area of that rectangle is equal to this time interval here, between two seconds and four seconds, multiplied by this speed interval which is five meters per second minus zero meters per second or five meters per second overall. Now, the time interval is four seconds minus two seconds, which happens to be an interval of two seconds. And so, we can say that area two is equal to five meters per second multiplied by two seconds, which we’ve already seen from earlier is 10 meters. In other words, area two is the same size as this rectangle we saw earlier. Which means all we’ve got to do now is to calculate area one, the area of the triangle.
To do this, we can recall that the area of a triangle, we’ll call it 𝐴 subscript triangle, is equal to half multiplied by the length that is the base of the triangle multiplied by the height of the triangle. In this case, the base of the triangle is the same as the width of the rectangle from two seconds to four seconds. Which, actually of course, is a time interval since on the horizontal axis we’ve got the time. And the height of the triangle is this distance here between seven meters per second and five meters per second. In other words, this interval here is seven meters per second minus five meters per second, which is two meters per second. And hence, the area of the triangle is equal to half multiplied by the base, which we’ve said is two seconds, multiplied by the height, which we’ve said is two meters per second.
Once again, we see that the unit of seconds in the numerator and denominator cancel to give us an overall unit of meters which is once again good. We are calculating a distance after all. And the numerical value is half multiplied by two multiplied by two. Overall then, we find that the area of triangle one is two meters. And hence, as we’ve said earlier, the total area of triangle one plus rectangle two is equal to two meters plus 10 meters. And hence, we can say that the total distance moved by the person between 𝑡 is equal to two seconds and 𝑡 is equal to four seconds is 12 meters, which is our answer to this part of the question.
Moving on to the next part of the question then, we are now asked, what distance does the person walk between 𝑡 is equal to four seconds and 𝑡 is equal to six seconds?
So, we can draw vertical lines from the time axis for 𝑡 is equal to four seconds and 𝑡 is equal to six seconds, as we’ve seen already. We need to find the area beneath the blue line between these two times: in other words, the area underneath this straight line. And so, that’s this shaded area here. That area happens to have a width that’s equivalent to this time interval and a height that’s equivalent to this speed. It helps us, by the way, that the person is walking once again at a constant speed because now the speed–time graph is a flat line in this time interval. Which means the area that we’re trying to find is once again a rectangle.
And to find the area of that rectangle, we firstly see that six seconds minus four seconds gives us a width for the rectangle of two seconds. And the height is equal to seven meters per second, which is the highest point of that rectangle, minus zero seconds, which is the lowest point. Or, in other words, we’ve got a total height of seven meters per second. Hence, we can say that the distance moved in this time interval between 𝑡 is equal to four seconds and six seconds is equal to seven meters per second, that’s the height of the rectangle, multiplied by the width, which is two seconds. Calculating this area then, we find that the distance moved by the person in this time interval is 14 meters. And that is our final answer to this part of the question.
Now, the last part of the question asks us, what distance does the person walk between 𝑡 is equal to zero seconds and 𝑡 is equal to six seconds?
So, that’s the distance walked between this time and this time. In other words, we need to find the area underneath the line for the entire graph now between 𝑡 is equal to zero seconds and 𝑡 is equal to six seconds. However, as we’ve seen from previous parts of the question already, the distance walked by the person between 𝑡 is equal to zero seconds and two seconds was 10 meters. That’s the area of this rectangle. The distance walked by the person between 𝑡 is equal to two seconds and four seconds was 12 meters. And the distance walked between 𝑡 is equal to four seconds and six seconds was 14 meters.
So, the total distance walked between 𝑡 is equal to zero seconds and 𝑡 is equal to six seconds is just going to be 10 meters plus 12 meters plus 14 meters. We’ll call this total distance 𝑑 subscript tot or 𝑑 subscript total. And as we’ve already said, it’s the sum of all the three distances we’ve calculated so far. This distance ends up being 36 meters, and hence that’s our final answer to the final part of the question.
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# 0 Property Of Multiplication
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Tone: Formal
1. Introduction to the Property of Multiplication
Multiplication is one of the basic operations of arithmetic. It is simply the repeated addition of equal numbers. In multiplication, the result is called the product, and the numbers being multiplied are the factors. The property of multiplication is an essential concept that helps to make multiplication easier and more efficient.
2. Commutative Property of Multiplication
The commutative property of multiplication states that the order of factors in a multiplication problem does not affect the product. That is, if a and b are any two numbers, a times b is equal to b times a.
3. Associative Property of Multiplication
The associative property of multiplication states that when multiplying three or more numbers, the grouping of the factors does not affect the product. That is, if a, b, and c are any three numbers, (a times b) times c is equal to a times (b times c).
4. Identity Property of Multiplication
The identity property of multiplication states that the product of any number and 1 is equal to the same number. That is, if a is any number, then a times 1 is equal to a.
5. Zero Property of Multiplication
The zero property of multiplication states that the product of any number and 0 is equal to 0. That is, if a is any number, then a times 0 is equal to 0.
6. Distributive Property of Multiplication over Addition
The distributive property of multiplication over addition states that when a number is multiplied by the sum of two or more numbers, then the product is equal to the sum of the products of the number and each of the other numbers. That is, if a, b, and c are any three numbers, then a times (b + c) is equal to (a times b) + (a times c).
7. How to Apply the Property of Multiplication to Simplify Expressions
To apply the property of multiplication to simplify expressions, we can use the commutative, associative, identity, and distributive properties. This allows us to rearrange the numbers and terms in an expression to make it simpler and more manageable.
8. Examples Using the Property of Multiplication
One example of the application of the property of multiplication is solving the equation 6 times (8 + 2). Applying the distributive property, we get 6 times 8 + 6 times 2, which simplifies to 48 + 12 = 60.
9. Importance of the Property of Multiplication
The property of multiplication is essential in various aspects of mathematics, including algebra, geometry, and statistics. It allows us to perform calculations with greater accuracy and efficiency.
10. Conclusion
In conclusion, the property of multiplication is an essential concept that every student needs to understand. It simplifies multiplication and makes it more efficient, allowing us to solve complex problems more easily. Knowing how to apply the property of multiplication is crucial in solving equations and simplifying expressions.
The Property of Multiplication that states that any number multiplied by zero equals zero is a fundamental concept in mathematics. This property may seem simple at first glance, but it has numerous real-world applications that make it an essential tool for solving complex problems. Furthermore, understanding this property can help students grasp more advanced mathematical concepts, such as algebra and calculus. In the following paragraphs, we will explore the significance of the zero property of multiplication and its various applications.
## The Zero Property of Multiplication
Multiplication is one of the basic operations in mathematics. It is a process of adding a number to itself a certain number of times. For example, 3 x 4 means adding 3 four times, which equals 12. However, there is a special property in multiplication that involves the number zero. This property is called the Zero Property of Multiplication.
### What is the Zero Property of Multiplication?
The Zero Property of Multiplication states that any number multiplied by zero equals zero. In other words, if you multiply any number with zero, the result will always be zero. For example, 5 x 0 = 0, 10 x 0 = 0, and even 1000 x 0 = 0.
### Why does the Zero Property of Multiplication work?
The Zero Property of Multiplication works because of the way multiplication is defined. When you multiply two numbers together, you are essentially adding one number to itself as many times as the other number specifies. For example, 4 x 3 means adding 4 three times: 4 + 4 + 4 = 12. However, when you multiply a number by zero, you are essentially adding it zero times. And since adding anything zero times results in zero, any number multiplied by zero equals zero.
### Examples of the Zero Property of Multiplication
Let’s take some examples to understand the Zero Property of Multiplication better:
1. 7 x 0 = 0, since any number multiplied by zero equals zero.
2. 0 x 9 = 0, since any number multiplied by zero equals zero.
3. 0 x 0 = 0, since any number multiplied by zero equals zero.
4. 6 x 0.5 x 0 = 0, since any number multiplied by zero equals zero, regardless of its decimal value.
### Applications of the Zero Property of Multiplication
The Zero Property of Multiplication has several applications in mathematics and real-life situations:
1. In algebra, the Zero Property of Multiplication is used to simplify expressions. For example, if you have an expression like 3x(4y – 2z) – 0, the last term can be eliminated using the Zero Property of Multiplication, since any number multiplied by zero equals zero.
2. In physics, the Zero Property of Multiplication is used to calculate the work done by a force on an object that does not move. Since the displacement is zero, the work done is also zero, according to the Zero Property of Multiplication.
3. In economics, the Zero Property of Multiplication is used to calculate the profit or loss of a business. If the revenue is zero, the profit or loss is also zero, according to the Zero Property of Multiplication.
### The Commutative and Associative Properties of Multiplication
Multiplication has two other special properties, known as the Commutative and Associative Properties. The Commutative Property states that the order of the numbers does not matter in multiplication. For example, 4 x 3 = 3 x 4. The Associative Property states that the grouping of the numbers does not matter in multiplication. For example, (2 x 3) x 4 = 2 x (3 x 4). These properties are also widely used in mathematics and real-life situations.
### Conclusion
The Zero Property of Multiplication is an important concept in mathematics, which states that any number multiplied by zero equals zero. This property has several applications in algebra, physics, economics, and other fields. It is also important to understand the Commutative and Associative Properties of Multiplication, which can simplify complex expressions and calculations. By mastering these properties, you can become more proficient in math and other subjects that rely on multiplication.
## Introduction to the Property of Multiplication
Multiplication is a fundamental operation in mathematics, and it involves the repeated addition of equal numbers. The property of multiplication is an essential concept that simplifies multiplication and makes it more efficient. This article will discuss the different properties of multiplication, including the commutative, associative, identity, zero, and distributive properties.
## Commutative Property of Multiplication
The commutative property of multiplication states that the order of the factors in a multiplication problem does not affect the product. In other words, if we multiply two numbers, a and b, then a times b is equal to b times a. For example, 4 times 5 is the same as 5 times 4, which equals 20.
## Associative Property of Multiplication
The associative property of multiplication states that grouping the factors in any way does not affect the product. This means that when we multiply three or more numbers, we can group them in any way we like, and the product will still be the same. For example, (2 times 3) times 4 is the same as 2 times (3 times 4), which equals 24.
## Identity Property of Multiplication
The identity property of multiplication states that the product of any number and 1 is equal to the same number. In other words, if we multiply any number, a, by 1, we get a as the product. For example, 7 times 1 is 7.
## Zero Property of Multiplication
The zero property of multiplication states that the product of any number and 0 is equal to 0. This means that if we multiply any number, a, by 0, the product will always be 0. For example, 9 times 0 equals 0.
## Distributive Property of Multiplication over Addition
The distributive property of multiplication over addition states that when we multiply a number by the sum of two or more numbers, the product is equal to the sum of the products of the number and each of the other numbers. In other words, if we have three numbers, a, b, and c, then a times (b + c) is equal to (a times b) + (a times c). For example, 3 times (4 + 2) is equal to 3 times 4 plus 3 times 2, which equals 18.
## How to Apply the Property of Multiplication to Simplify Expressions
To apply the property of multiplication to simplify expressions, we can use any of the above properties depending on the situation. For example, we can use the distributive property to simplify expressions that involve multiplication and addition. Or we can use the commutative and associative properties to rearrange the order of factors in a multiplication problem. By doing so, we can make the expressions simpler and more manageable.
## Examples Using the Property of Multiplication
Let’s look at an example of how to apply the property of multiplication to simplify an expression. Suppose we have the expression 5 times (3 plus 2). By using the distributive property, we can simplify this expression to 5 times 3 plus 5 times 2, which equals 15 plus 10, or 25.
## Importance of the Property of Multiplication
The property of multiplication is an essential concept in mathematics as it helps us perform calculations with greater accuracy and efficiency. It is particularly useful in solving equations and simplifying expressions. By understanding and applying the different properties of multiplication, we can solve complex problems more easily and efficiently.
## Conclusion
In conclusion, the property of multiplication is a vital concept that every student needs to understand. It allows us to perform calculations with greater accuracy and efficiency and is essential in various aspects of mathematics. By knowing how to apply the different properties of multiplication, we can simplify complex problems and arrive at solutions more efficiently.
Once upon a time, there was a mathematical concept called the 0 property of multiplication. This property states that any number multiplied by 0 is equal to 0.
From a mathematical perspective, this property is incredibly useful. It allows us to quickly determine the outcome of an equation without having to do extensive calculations. For example:
• 5 x 0 = 0
• 100 x 0 = 0
• -3 x 0 = 0
As you can see, no matter what number we multiply by 0, the answer is always 0. This makes it a powerful tool for simplification in algebraic equations.
But from a conceptual standpoint, the 0 property of multiplication can be a bit confusing. After all, how can multiplying anything by 0 result in 0? To help understand this, let’s consider it from a different point of view.
Imagine you have a basket of apples. Each apple represents a number, and the total number of apples in the basket represents the outcome of a multiplication equation. Now, if you were to multiply the number of apples in the basket by 0, what would happen?
1. The entire basket would disappear, leaving you with 0 apples.
2. No matter how many apples were in the basket to begin with, they would all disappear and leave you with 0.
3. If there were no apples in the basket to begin with, nothing would happen and you would still have 0 apples.
This analogy helps to explain why the 0 property of multiplication works. When we multiply any number by 0, we are essentially removing all instances of that number and leaving ourselves with nothing. And when we have nothing, the result is always 0.
So, while the 0 property of multiplication may seem like a strange concept at first, it’s an important one to understand in the world of mathematics. It allows us to simplify equations and solve problems more efficiently, while also providing insight into the nature of numbers and their relationships with one another.
Thank you for taking the time to read this article on the property of multiplication. Understanding this concept is crucial in mastering basic arithmetic and lays the foundation for more advanced math topics. By the end of this article, we hope that you have a clear understanding of what the property of multiplication is and how it can be applied.
As we discussed earlier, the property of multiplication states that the order in which we multiply numbers does not affect the final result. This means that if we have two or more numbers to multiply together, we can do so in any order we choose and still get the same answer. For example, 2 x 3 x 4 is the same as 4 x 3 x 2, which is 24.
It’s important to note that this property only applies to multiplication and not addition or subtraction. When adding or subtracting, the order in which we perform the operations does matter and can lead to different results. So, make sure to keep this in mind when solving math problems.
In conclusion, we hope that this article has provided you with a better understanding of the property of multiplication. Remember to practice this concept and use it whenever you’re solving math problems. By mastering this basic arithmetic skill, you’ll be well on your way to becoming a confident math student.
People often have questions about the Zero Property of Multiplication. Here are some commonly asked questions and their answers:
### 1. What is the Zero Property of Multiplication?
The Zero Property of Multiplication states that when any number is multiplied by zero, the result is always zero.
### 2. Why is the Zero Property of Multiplication important?
The Zero Property of Multiplication is important because it helps us simplify calculations and make them easier to solve. For example, if we need to multiply a large number by zero, we can simply write down the answer as zero without needing to do any further calculations.
### 3. Can the Zero Property of Multiplication be used with fractions or decimals?
Yes, the Zero Property of Multiplication can be used with fractions or decimals just like it can be used with whole numbers. For example, 0.5 x 0 = 0.
### 4. Does the Zero Property of Multiplication work in reverse?
No, the Zero Property of Multiplication does not work in reverse. In other words, if we know that a x 0 = 0, we cannot assume that a = 0.
### 5. Is the Zero Property of Multiplication related to the Identity Property of Multiplication?
Yes, the Zero Property of Multiplication and the Identity Property of Multiplication are related. The Identity Property states that when any number is multiplied by one, the result is always that number. So, while the Zero Property tells us what happens when we multiply by zero, the Identity Property tells us what happens when we multiply by one.
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# A certain class consists of 8 students, including Kim. Each day, three
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A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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17 Feb 2012, 07:50
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A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
$$\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}$$
$$\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}$$
$$\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}$$
Therefore: $$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}$$[/m]
Is there another way to solve it faster?
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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17 Feb 2012, 08:23
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metallicafan wrote:
Bunuel wrote:
metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
Since there are total of 8 students and we are selecting 3 of them, then the probability that Kim will be selected is simply 3/8 (she has 3 chances out of 8).
Thank you bunuel! However, I don't get your approach well. Could you explain your logic? I believe that the probability of picking somenone in the second task is different from picking someone in the first because the number of elements available to pick changes. That's why my approach is step by step.
The wording of the question makes it harder then it is. The question basically asks: what is the probability that Kim is among the first 3 students out of 8, which is 3/8.
One can also do: $$P=\frac{C^1_1*C^2_7}{C^3_8}=\frac{3}{8}$$, where $$C^1_1$$ is # of ways to select Kim, $$C^2_7$$ is # of ways to select any 2 students out of 7 left and $$C^3_8$$ is total # of ways to select 3 students from 8;
Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
$$P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}$$;
Or: the same with probability approach: $$P=1-\frac{7}{8}*\frac{6}{7}*\frac{5}{6}=\frac{3}{8}$$.
As you can see first approach is easiest and shortest one.
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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17 Feb 2012, 07:55
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metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
Since there are total of 8 students and we are selecting 3 of them, then the probability that Kim will be selected is simply 3/8 (she has 3 chances out of 8).
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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17 Feb 2012, 08:12
Bunuel wrote:
metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
Since there are total of 8 students and we are selecting 3 of them, then the probability that Kim will be selected is simply 3/8 (she has 3 chances out of 8).
Thank you bunuel! However, I don't get your approach well. Could you explain your logic? I believe that the probability of picking somenone in the second task is different from picking someone in the first because the number of elements available to pick changes. That's why my approach is step by step.
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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17 Feb 2012, 08:49
Thank you Bunuel!, I have updated the post with the choices.
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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17 Feb 2012, 10:00
Thankx Bunuel, the explanation covers all possible ways of solving the problem
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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17 Feb 2012, 17:43
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One more way to sove this is to reverse the logic:
NOT chosen on first, second and third pick leads to the following calculation:
7/8 * 6/7 * 5/6 = 0,625. Since we are looking for the probability Kim gets chosen it's 1-0,625 = 0,375 = 3/8
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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23 Feb 2012, 10:06
1
I agree with B 3/8
third task = 7/8*6/7*1/6 = 1/8
chance that she gets picked for one of these is 1/8+1/8+1/8 or 3/8
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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03 May 2014, 23:35
metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
$$\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}$$
$$\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}$$
$$\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}$$
Therefore: $$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}$$[/m]
Is there another way to solve it faster?
Source: Jeff Sackman questions - http://www.gmathacks.com
Hello,
for completion of task B - why 7/8 * 1/7* 6/6??
for completion of task C - why 7/8 * 6/7* 1/6??
plx explain the one in BOLD especially..
Sharmita
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Joined: 02 Sep 2009
Posts: 59722
Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
### Show Tags
04 May 2014, 03:01
msharmita wrote:
metallicafan wrote:
A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
$$\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}$$
$$\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}$$
$$\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}$$
Therefore: $$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}$$[/m]
Is there another way to solve it faster?
Source: Jeff Sackman questions - http://www.gmathacks.com
Hello,
for completion of task B - why 7/8 * 1/7* 6/6??
for completion of task C - why 7/8 * 6/7* 1/6??
plx explain the one in BOLD especially..
Sharmita
P(Kim)*P(any)*P(any) = $$\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}$$
P(any bu Kim)*P(Kim)*P(any) = $$\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}$$
P(any but Kim)*P(any but Kim)*P(any) = $$\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}$$
Faster solutions are here: a-certain-class-consists-of-8-students-including-kim-127730.html#p1046004 and here: a-certain-class-consists-of-8-students-including-kim-127730.html#p1046020
Similar questions to practice:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html
new-set-of-mixed-questions-150204-100.html#p1208473
a-medical-researcher-must-choose-one-of-14-patients-to-recei-124775.html
Hope this helps.
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Joined: 28 Jul 2016
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Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
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24 Aug 2016, 08:46
Hi Bunuel,
Should the question be read as "What is the probability that Kim will be selected to complete ATLEAST one of the three tasks?"
Or should it be read as "What is the probability that Kim will be selected to complete EXACTLY one of the three tasks?"
I felt the question was slightly ambiguous. If so, will such type of questions appear on the GMAT ?
Intern
Joined: 28 Jul 2016
Posts: 4
Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
### Show Tags
24 Aug 2016, 08:51
Sorry for the confusion. I get it know. Since the left-over tasks are to be done only by the remaining people, one person can only do one task. So in this case both mean the same.
Intern
Joined: 13 Dec 2018
Posts: 19
Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink]
### Show Tags
12 Dec 2019, 07:40
Hi Bunuel ,
Please clarify my below doubts. It will help me a lot in other questions too. I am very much confused.
One can also do: P=C11∗C27C38=38P=C11∗C72C83=38, where C11C11 is # of ways to select Kim, C27C72 is # of ways to select any 2 students out of 7 left and C38C83 is total # of ways to select 3 students from 8;
In your above approach, How you have taken C^1_1 ? I always took C^1_8 as we need to select 1 from 8.
Thanks a lot
Re: A certain class consists of 8 students, including Kim. Each day, three [#permalink] 12 Dec 2019, 07:40
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A Quadratic Equation in Standard Form
(a, b, and c can have any value, except that a can't be 0.)
"Factoring" (or "Factorising" in the UK) a Quadratic is:
finding what to multiply to get the Quadratic
It is called "Factoring" because we find the factors (a factor is something we multiply by)
### Example: (x+4) and (x−1) are factors of x2 + 3x − 4
Let us "expand" (x+4) and (x−1) to be sure:
(x+4)(x−1) = x(x−1) + 4(x−1)
= x2 − x + 4x − 4
= x2 + 3x − 4
Yes, (x+4) and (x−1) are definitely factors of x2 + 3x − 4
Did you see that Expanding and Factoring are opposites?
Expanding is usually easy, but Factoring can often be tricky.
It is like trying to find which ingredients
went into a cake to make it so delicious.
It can be hard to figure out!
OK, let's try an example where we don't know the factors yet:
## Common Factor
First we can check for any common factors.
### Example: what are the factors of 6x2 − 2x = 0 ?
6 and 2 have a common factor of 2:
2(3x2 − x) = 0
And x2 and x have a common factor of x:
2x(3x − 1) = 0
And we have done it! The factors are 2x and 3x − 1,
We can now also find the roots (where it equals zero):
• 2x is 0 when x = 0
• 3x − 1 is zero when x = 13
And this is the graph (see how it is zero at x=0 and x=13):
But it is not always that easy ...
## Guess and Check
### Example: what are the factors of 2x2 + 7x + 3 ?
No common factors.
Maybe we can guess an answer? Then check if we are right ... we may get lucky!
Let's guess (2x+3)(x+1):
(2x+3)(x+1) = 2x2 + 2x + 3x + 3
= 2x2 + 5x + 3
(Close but WRONG)
(2x+7)(x−1) = 2x2 − 2x + 7x − 7
= 2x2 + 5x − 7
(WRONG AGAIN)
(2x+9)(x−1) = 2x2 − 2x + 9x − 9
= 2x2 + 7x − 9
(WRONG AGAIN!)
We could be guessing for a long time before we get lucky.
That is not a very good method. So let us try something else.
## A Method For Simple Cases
There is a method for simple cases.
With the quadratic equation in this form:
Step 1: Find two numbers that multiply to give ac (in other words a times c), and add to give b.
Example: 2x2 + 7x + 3
ac is 2×3 = 6 and b is 7
So we want two numbers that multiply together to make 6, and add up to 7
In fact 6 and 1 do that (6×1=6, and 6+1=7)
How do we find 6 and 1?
It helps to list the factors of ac=6, and then try adding some to get b=7.
Factors of 6 include 1, 2, 3 and 6.
Aha! 1 and 6 add to 7, and 6×1=6.
Step 2: Rewrite the middle with those numbers:
Rewrite 7x with 6x and 1x:
2x2 + 6x + x + 3
Step 3: Factor the first two and last two terms separately:
The first two terms 2x2 + 6x factor into 2x(x+3)
The last two terms x+3 don't actually change in this case
So we get:
2x(x+3) + (x+3)
Step 4: If we've done this correctly, our two new terms should have a clearly visible common factor.
In this case we can see that (x+3) is common to both terms, so we can go:
Which is:2x(x+3) + 1(x+3)
And so:(2x+1)(x+3)
Done!
Check: (2x+1)(x+3) = 2x2 + 6x + x + 3 = 2x2 + 7x + 3 (Yes)
Let's see Steps 1 to 4 again, in one go:
2x2 + 7x + 3 2x2 + 6x + x + 3 2x(x+3) + (x+3) 2x(x+3) + 1(x+3) (2x+1)(x+3)
### Example: 6x2 + 5x − 6
Step 1: ac is 6×(−6) = −36, and b is 5
List the positive factors of ac = −36: 1, 2, 3, 4, 6, 9, 12, 18, 36
One of the numbers has to be negative to make −36, so by playing with a few different numbers I find that −4 and 9 work nicely:
−4×9 = −36 and −4+9 = 5
Step 2: Rewrite 5x with −4x and 9x:
6x2 − 4x + 9x − 6
Step 3: Factor first two and last two:
2x(3x − 2) + 3(3x − 2)
Step 4: Common Factor is (3x − 2):
(2x+3)(3x − 2)
Check: (2x+3)(3x − 2) = 6x2 − 4x + 9x − 6 = 6x2 + 5x − 6 (Yes)
### Finding Those Numbers
The hardest part is finding two numbers that multiply to give ac, and add to give b.
It is partly guesswork, and it helps to list out all the factors.
### Example: ac = −120 and b = 7
What two numbers multiply to −120 and add to 7 ?
The factors of 120 are (plus and minus):
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120
We can try pairs of factors (start near the middle!) and see if they add to 7:
• −10 x 12 = −120, and −10+12 = 2 (no)
• −8 x 15 = −120 and −8+15 = 7 (YES!)
## Why Factor?
Well, one of the big benefits of factoring is that we can find the roots of the quadratic equation (where the equation is zero).
All we need to do (after factoring) is find where each of the two factors becomes zero
### Example: what are the roots (zeros) of 6x2 + 5x − 6 ?
We already know (from above) the factors are
(2x + 3)(3x − 2)
And we can figure out that
(2x + 3) is zero when x = −3/2
(3x − 2) is zero when x = 2/3
So the roots of 6x2 + 5x − 6 are:
−3/2 and 2/3
Here is a plot of 6x2 + 5x − 6, can you see where it equals zero?
We can also check it using a bit of arithmetic:
At x = −32: 6(−32)2 + 5(−32) − 6 = 6×(94) − 152 − 6 = 544152 - 6 = 0
At x = 23: 6(23)2 + 5(23) − 6 = 6×(49) + 103 − 6 = 249 + 103 - 6 = 0
## Graphing
We can also try graphing the quadratic equation. Seeing where it equals zero can give us clues.
### Example: (continued)
Starting with 6x2 + 5x − 6 and just this plot:
The roots are around x = −1.5 and x = +0.67, so we can guess the roots are:
−3/2 and 2/3
Which can help us work out the factors 2x + 3 and 3x − 2
Always check though! The graph value of +0.67 might not really be 2/3
## General Solution
Quadratic equations have symmetry, the left and right are like mirror images:
The midline is at −b/2, and we can calculate the value w with these steps:
• First, "a" must be 1, if not then divide b and c by a:
• b = b/a, c = c/a
• mid = −b/2
• w = √(mid2 − c)
• roots are at mid−w and mid+w
### Example: x2 + 3x − 4
a = 1, b = 3 and c = −4
• a= 1, so we can go to next step
• mid = −32
• w = √[(32)2 − (−4)] = √(94 + 4) = √254 = 52
• roots are at −3252 = −4 and −32+52 = 1
So we can factor x2 + 3x − 4 into (x + 4)(x − 1)
We can also use the quadratic formula:
We get two answers x+ and x (one is for the "+" case, and the other is for the "−" case in the "±") that gets us this factoring:
a(x − x+)(x − x)
### Example: what are the roots of 6x2 + 5x − 6 ?
Substitute a=6, b=5 and c=−6 into the formula:
x = −b ± √(b2 − 4ac)2a
= −5 ± √(52 − 4×6×(−6))2×6
= −5 ± √(25 + 144)12
= −5 ± √16912
= −5 ± 1312
So the two roots are:
x+ = (−5 + 13) / 12 = 8/12 = 2/3,
x = (−5 13) / 12 = −18/12 = −3/2
(Notice that we get the same answer as when we did the factoring earlier.)
Now put those values into a(x − x+)(x − x):
6(x − 2/3)(x + 3/2)
We can rearrange that a little to simplify it:
3(x − 2/3) × 2(x + 3/2) = (3x − 2)(2x + 3)
Done!
362, 1203, 2262, 363, 1204, 2263, 2100, 2101, 2102, 2103, 2264, 2265
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# Section 3.4 Question 2
### How do you solve a matrix equation using the matrix inverse?
In the previous question, we wrote systems of equations as a matrix equation AX = B. In this format, the matrix A contains the coefficients on the variables, matrix X contains the variables, and matrix B contains the constants. Solving the system of equations means that we need to solve for the variable matrix X. This is accomplished by multiplying both sides of the matrix equation by the inverse of the coefficient matrix, A-1.
A-1 AX = A-1 B
The product of a matrix A and its inverse is the identity matrix I. This means we can simplify the matrix equation to
IX = A-1 B
The product of an identity matrix I with X is simply X, so the solution to the matrix equation is
X = A-1 B
How To Solve a System of Linear Equations with Inverses
1. Make sure the system is in proper format with variable terms on the left side of the equation and constants on the right side. The variable terms should be listed in the same order in each equation. Identify the coefficient matrix A, the variable matrix X, and the constant matrix B.
2. Compute the inverse A-1. If the matrix is not invertible, it is not possible to solve the system with inverses.
3. Compute the product A-1 B.
4. The solution to the matrix equation AX = B is X = A-1 B. The corresponding solution for the system of linear equations is found by matching the variables in X with the corresponding entries in the product .
### Example 3 Solve a Linear System with the Inverse
Solve the system of linear equations using the inverse of the coefficient matrix.
Solution In Example 1, we wrote this system as the matrix equation , where
In section 3.3, we found the inverse of A,
We can use the inverse to compute the solution to the system.
The solution is found by multiplying the inverse of A times the constant matrix B,
Since the individual entries in X correspond to the variables x and y, this tells us that x = -21 and y = 16. We can check these values in the original system to make sure they solve the system:
### Example 4 Solve a Linear System with the Inverse
Solve the system of linear equations with the inverse of the coefficient matrix.
Solution This system of equations is equivalent to the matrix equation AX = B where
The variable matrix may be solved for using the inverse of the coefficient matrix, X = A-1 B. We found the inverse of this coefficient matrix in Example 5 of section 3.3.
Equating the entries in the variable matrix with the entries in this product, we observe that x = 3, y = 4, and z = 5.
For systems of linear equations with unique solutions, we can use inverses to solve for the variables. Many of the applications in Chapter 2 may be solved using this strategy.
### Example 5 Mixing Ethanol Blends
In Example 12 of section 2.2 we created a system of equations to describe a mix of E10 and E85 ethanol,
where E10 is the amount of 10% ethanol pumped in gallons and E85 is the amount of 85% ethanol pumped in gallons. The first equation describes the total amount in the mixture, 10 gallons. The second equation describes the total amount of ethanol in the mixture, 20% of 10 gallons or 2 gallons.
a. Solve this system of equations by finding the inverse of the coefficient matrix.
Solution This system of equations was solved in Chapter 2 using the Substitution Method. In this section we’ll solve the same system by writing the system as a matrix equation.
We can write this system of equations as the matrix equation AX = B by identifying the matrices,
The solution to the matrix equation is X = A-1 B and requires the inverse of the matrix A.
We begin the process of finding the inverse of A by placing the coefficient matrix in a new matrix alongside a 2 x 2 identity matrix.
Since the entry in the first row, first column is already a 1, we’ll make the rest of the column into zeros using row operations. To do this, multiply the first row by -0.10 and add it to the second row. Place the sum in the second row:
With the first column transformed, proceed to the second column and use row operations to create a 1 in the second row, second column.
We put a 1 in the second row, second column by multiplying the second row by 1/0.75. Since 0.75 = 3/4, this is the same as multiplying by 4/3:
Now multiply the second row by -1 and add it to the first row. Place the result in the first row:
The inverse matrix is the right hand side of this matrix,
The solution to the original system of equations is
Since the variable matrix X represents , this means that the amount of E10 needed is 26/3 gallons and the amount of E85 needed is 4/3 gallons.
b. If the number of gallons in the mixture should be 12 gallons, how much 10% ethanol and 85% ethanol must be pumped?
Solution If the total number of gallons in the mixture is increased to 12 gallons, the total amount of ethanol in the mixture is or 2.4 gallons. The system of equations becomes
The only change to the system of equations is in the constants, not the coefficients.
This means that we may use the existing coefficient matrix A and the corresponding inverse A-1 to find the solution of this system,
The constant matrix for this system is
and the solution is
Since the coefficients in the system have not changed, there is no need to recompute the inverse. We may use the inverse from part a to calculate the product. We must mix 52/5 gallons of E10 and 8/5 gallons of E85.
This example illustrated the power of using inverses to solve a system of linear equations. As long as the coefficient matrix does not change, the same inverse may be used to solve several different problems with different constant matrices. The solutions are simply the product of the inverse of the coefficient matrices and the different constants. For part a, the constant was and for part b the constant was . There was no need to recompute the inverse for the different parts.
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To solve a quadratic inequality, we write our inequality as an equality and solve. The solutions to the equation give us boundaries that allow us to set up intervals on the number line. We can then test values in each interval and determine our solution set. Lastly, our boundaries are included for non-strict inequalities and not included for strict inequalities.
Test Objectives
• Demonstrate the ability to solve a quadratic equation
• Demonstrate the ability to solve a quadratic inequality
• Demonstrate the ability to write an inequality solution in interval notation
• Demonstrate the ability to graph an interval
#1:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}2x^2 - 6x + 2 > -2$$
$$b)\hspace{.2em}6x^2 - 10x + 2 < 6$$
#2:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}6x^2 ≥ 5x^2 + 99 + 2x$$
$$b)\hspace{.2em}6x^2 + 17x - 8 ≤ 9x$$
#3:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}6x^2 - 8x + 1 > -x$$
$$b)\hspace{.2em}5x^2 + 6x < 11$$
#4:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}x^2 - 17x + 5 ≥ -11x$$
$$b)\hspace{.2em}-6x^2 - x ≤ 12 - 7x^2$$
#5:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}5x^2 - 20x - 32 ≤ -8x$$
$$b)\hspace{.2em}x^2 - 5x ≥ 143 - 3x$$
Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}x < 1 \hspace{.2em}or \hspace{.2em}x > 2$$ $$(-\infty, 1) ∪ (2, \infty)$$
$$b)\hspace{.2em}-\frac{1}{3}< x < 2$$ $$\left(-\frac{1}{3}, 2 \right)$$
#2:
Solutions:
$$a)\hspace{.2em}x ≤ -9 \hspace{.2em}or \hspace{.2em}x ≥ 11$$ $$(-\infty, -9] ∪ [11, \infty)$$
$$b)\hspace{.2em}-2 ≤ x ≤ \frac{2}{3}$$ $$\left[-2, \frac{2}{3}\right]$$
#3:
Solutions:
$$a)\hspace{.2em}x < \frac{1}{6}\hspace{.2em}or \hspace{.2em}x > 1$$ $$\left(-\infty, \frac{1}{6}\right) ∪ (1, \infty)$$
$$b)\hspace{.2em}-\frac{11}{5}< x < 1$$ $$\left(-\frac{11}{5}, 1\right)$$
#4:
Solutions:
$$a)\hspace{.2em}x ≤ 1 \hspace{.2em}or \hspace{.2em}x ≥ 5$$ $$(-\infty, 1] ∪ [5, \infty)$$
$$b)\hspace{.2em}-3 ≤ x ≤ 4$$ $$[-3, 4]$$
#5:
Solutions:
$$a)\hspace{.2em}-\frac{8}{5}≤ x ≤ 4$$ $$\left[-\frac{8}{5}, 4\right]$$
$$b)\hspace{.2em}x ≤ -11 \hspace{.2em}or \hspace{.2em}x ≥ 13$$ $$(-\infty, -11] ∪ [13, \infty)$$
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# Study notes on Multi Variable Calculus For Electrical Engineering Students
Updated : Aug 22, 2018, 11:30
By : Yash Bansal
## Vectors
• The Scalars are quantities that only have a magnitude like mass, field strength. Many times it is often useful to have a quantity that has not only a magnitude but also a direction; such a quantity is called a vector. Examples of quantities represented by vectors include velocity, acceleration, and virtually any type of force (frictional, gravitational, electric, magnetic, etc.)
• The magnitude (or length) of a vector v with initial point (x_1,y_1,z_1) and terminal point (x_2,y_2,z_2) is
Vectors obey the natural intuitive laws of addition and scalar multiplication:
The figures below illustrate the operations of addition and scalar multiplication in the two-dimensional case.
•
Dot Productand Cross product:
here dot product of two vectors a=<a_1,a_2,a_3> and b=<b_1,b_2,b_3> is given by
An equivalent definition of the dot product is
The cross product of two vectors a=<a_1,a_2,a_3> and b=<b_1,b_2,b_3> is given by
## 1. Partial Derivatives
Differentiating a function of more than one variable is more complicated than differentiating a function of one variable. For a function of several variables, the rate of change of the function depends on direction!. Consider the function
• #### Example
For the function
find the partial derivatives of f with respect to x and y and compute the rates of change of the function in the x and y directions at the point (-1,2). Initially we will not specify the values of x and y when we take the derivatives; we will just remember which one we are going to hold constant while taking the derivative. First, hold y fixed and find the partial derivative of f with respect to x:
Second, hold x fixed and find the partial derivative of f with respect to y:
Now, plug in the values x=-1 and y=2 into the equations. We obtain f_x(-1,2)=10 and f_y(-1,2)=28.
### 2. The Gradient and Directional Derivative:
The gradient of a function w=f(x,y,z) is the vector function:
For a function of two variables z=f(x,y), the gradient is the two-dimensional vector <f_x(x,y),f_y(x,y)>. This definition generalizes in a natural way to functions of more than three variables.
Examples
For the function z=f(x,y)=4x^2+y^2. The gradient is
### 3.Divergence and Curl of Vector Fields:
Divergence of a Vector Field
The divergence of a vector field F=<P(x,y,z),Q(x,y,z),R(x,y,z)>, denoted by div F, is the scalar function defined by the dot product
Here is an example. Let
The divergence is given by:
Curl of a Vector Field
The curl of a vector field F=<P(x,y,z),Q(x,y,z),R(x,y,z)>, denoted curlF, is the vector field defined by the cross product
An alternative notation is
The above formula for the curl is difficult to remember. An alternative formula for the curl is
Det means the determinant of the 3x3 matrix. Recall that the determinant consists of a bunch of terms which are products of terms from each row. The product of the terms on the diagonal is
As you can see, this term is part of the x-component of the curl.
ExampleF=<xyz,ysin z, ycos x>.
curl F = <cos x - ycos z, xy + ysin z, -xz>.
## 4. Line Integrals:
### Green's Theorem:
Green's Theorem states that
Here it is assumed that P and Q have continuous partial derivatives on an open region containing R.
## 5. Surface Integrals:
Stokes' Theorem: Stokes' Theorem relates line integrals of vector fields to surface integrals of vector fields.
Stokes' Theorem states
In general C is the boundary of S and is assumed to be piecewise smooth. For the above equality to hold the direction of the normal vector n and the direction in which C is traversed must be consistent. Suppose that n points in some direction and consider a person walking on the curve C with their head pointing in the same direction as n. For consistency C must be traversed in such a way so that the surface is always on the left.
The Divergence Theorem
The Divergence Theorem relates relates volume integrals to surface integrals of vector fields. Let R be a region in xyz space with surface S. Let n denote the unit normal vector to S pointing in the outward direction. Let F(x,y,z)=<P(x,y,z),Q(x,y,z),R(x,y,z)> be a vector field whose components P, Q, and R have continuous partial derivatives. The Divergence Theorem states:
Here div F is the divergence of F. There are various technical restrictions on the region R and the surface S; see the references for the details. The theorem is valid for regions bounded by ellipsoids, spheres, and rectangular boxes.
Thanks,
Aug 22ESE & GATE EE
Posted by:
Content (GATE Team)
Member since Nov 2018
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## ◂Math Worksheets and Study Guides First Grade. Using Number Line
### The resources above correspond to the standards listed below:
#### Michigan Common Core Standards
MI.CC.NBT.1. Number and Operations in Base Ten
Extend the counting sequence.
NBT.1.1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
Understand place value.
NBT.1.3. Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
Use place value understanding and properties of operations to add and subtract.
NBT.1.4. Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
MI.CC.OA.1. Operations and Algebraic Thinking
Add and subtract within 20.
OA.1.6. 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).
Understand and apply properties of operations and the relationship between addition and subtraction.
OA.1.4. Understand subtraction as an unknown-addend problem. For example, subtract 10 - 8 by finding the number that makes 10 when added to 8.
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Directions: The bar-chart shows the total number of members enrolled in different years from 1990 to 1994 in two gymnasiums A and B. Based on this bar chart, solve the following questions.
Important for :
1
If in the year 1995, there is 30% increase in the total number of members enrolled in 1994 in both gymnasiums, then find the total number of members enrolled in 1995.
» Explain it
D
Total number of member enrolled in 1995
= 130% of (150 + 70)
= 220 × 130 = 286 100
Hence, option D is correct.
2
The ratio of the total number of members of both gymnasiums in 1991 to the total number of members in 1994 of both gymnasiums is
» Explain it
E
Reqd ratio
= No. of members in Gym of A & B in 1991 No. of members in Gym of A & B in 1994
= 62 + 210 = 270 = 27 = 27 : 22 70 + 150 220 22
Hence, option E is correct.
3
The number of members of gymnasium A in 1991 is what per cent of the number of members of gymnasium B in 1994?
» Explain it
E
Reqd. % = No. of members in Gym of A in 1991 × 100 No. of members in Gym of B in 1994
= 60 × 100 = 40% 100
Hence, option E is correct.
4
The number of enrolled members in gymnasium A from 1991 to 1994 together is what percent more than the number of members enrolled in gymnasium B in 1993 and 1994 together? (Rounded off to two-digit decimal places)
» Explain it
B
Total number of members enrolled in Gymnasium A from 1991 to 1994 = 60 + 140 + 200 + 70 = 470
Total number of members enrolled in gymnasium B in 1993 and 1994 together = 240 + 150 = 390
∴ Difference = 470 – 390 = 80
∴ Reqd % = 80 × 100 390
= 20.51%
So, it's clear that the number of enrolled members in gymnasium A from 1991 to 1994 together is 20.51% more than the number of members enrolled in gymnasium B in 1993 and 1994 together.
Hence, option B is correct.
5
The total number of members enrolled in gymnasium B in 1993 and 1994 together is what per cent more than the number of members enrolled in gymnasium in 1990 and 1994 together?
» Explain it
C
Total number of members enrolled in Gymnasium B in 1993 and 1994 together = 240 + 150 = 390
Total number of members enrolled in Gymnasium A in 1990 and 1994 together = 170 + 70 = 240
∴ Difference = 390 – 240 = 150
∴ Reqd % = 150 × 100 240
= 62.5%
Hence, option C is correct.
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Frequency Distributions and Graphical Displays
Frequency Distributions
Single-number summaries of distributions, such as the mean, median, or mode, have definite limitations. Even when combined with measures of dispersion, such as the range, interquartile range, or standard deviation, a complete understanding of the entire distribution cannot be achieved. Perhaps there is a better way to communicate the nature of an entire distribution? Following the saying about words and pictures, pictures (i.e., graphs and charts) can portray a great deal of information about numbers, too. An intermediate step between having just a list of raw, unordered data and producing an informative graph will be to construct a table called a frequency distribution. A frequency distribution provides an ordered tally of the observed values and a count of the occurrences of each value. The observed values can be tallied individually or they can be grouped and then tallied. Let's look at a simple example.
Scores: 7, 5, 8, 9, 7, 6, 9, 8, 7, 4
Score Frequency 4 1 5 1 6 1 7 3 8 2 9 2 sum 10
Quite a bit of information can be discerned from the table. First, the minimum score was 4, which was observed once. Second, the maximum score was 9, which was observed twice. Third, the most frequently occurring score (i.e., the mode) was a 7, which was observed three times. The total number of scores was 10 - represented by the sum in the last row. Let's add two more columns to the frequency distribution. The two new columns represent two types of percentages.
Score Frequency Percent Cumulative Percent 4 1 10 10 5 1 10 20 6 1 10 30 7 3 30 60 8 2 20 80 9 2 20 100 sum 10 100
Recall that a percentage is calculated by dividing a number representing a part of a group by a number representing the whole group, and then multiplying by 100. [Percent means per 100.] In this example, the whole is the entire collection of scores - all 10 of them. The parts that relate to the percentages are the frequency counts in each row. So, the calculation for the percentage of scores of 5 is the following: 1 divided by 10 times 100, which equals 10%; for the percent of 7s, the calculation is 3 divided by 10 times 100, which equals 30%. Calculating percentages makes the frequencies more easily comparable. The last column is a running total of the Percent column. Consider why this might be useful information? If a 7 is a passing score, then a quick glance at the Cumulative Percent column shows that 30% of the scores were 6 or lower. A more positive outlook would be to report that 70% (100-30) of the scores were 7 or higher.
The previous example is very much simplified in order to introduce the meaning of the columns of a frequency distribution. In practice, frequency distributions involve more values and, quite often, researchers will group the values in order to make the frequency distribution more understandable. Here is another, more involved, example. Suppose you've conducted a survey in which you asked your colleagues to report their age (and they actually did). The data that you received are the following:
24, 53, 44, 30, 34, 36, 59, 61, 30, 48,
22, 65, 31, 33, 40, 37, 27, 29, 25, 23,
39, 30, 40, 55, 50, 38, 32, 28, 31, 30,
62, 41, 38, 66, 29, 31, 28, 24, 39, 64,
57, 53, 38, 32, 30, 25, 28, 44, 37, 50
Glancing at this list of numbers doesn't help to understand the ages of the survey participants. Let's construct a grouped frequency distribution by trying to implement the following suggested guidelines. The official term for a group is a class interval, and the term, bin, is used frequently as well.
1. Use a class interval that includes 2, 5, 10, or 20 values
2. Aim for 10-20 class intervals (range / number of class intervals = class interval range)
3. Tally observed values into class intervals - result gives frequency counts for each class interval (number of observed values with each class interval)
For the group of 50 ages listed above, the maximum age is 65 and the minimum is 22; so, the range is 65-22 or 43. If we divide 43 by 5, we get approximately 8 class intervals. A typical grouping of ages is into 5-year intervals (e.g., 20-24, 25-29). To sufficiently cover all of the ages, the highest category should be 65-69. In the frequency distribution table below, notice that none of the class intervals overlap and that there are no holes between intervals. [We're using the standard convention that you are 29 up until your 30th birthday - there are other ways to measure age, especially with younger people.]
Age Frequency Percent Cumulative Percent 20-24 4 8 8 25-29 8 16 24 30-34 12 24 48 35-39 8 16 64 40-44 5 10 74 45-49 1 2 76 50-54 4 8 84 55-59 3 6 90 60-64 3 6 96 65-69 2 4 100 sum 50 100
This table might help you understand the ages a bit better. For example, the largest age group, which is called the modal class interval, is the range of ages from 30-34, representing 24% of the survey participants. Based on the grouped frequencies, we can't tell that the actual mode is 30, which occurs 5 times. For that information, you need to inspect the individual values, not the grouped values. From the grouped frequencies, you can also tell that 24% of participants are under 30 years of age and that 24% (100-76) are 50 or older.
It may be somewhat difficult to tell just by looking at the table, but as we will see later, this distribution has relatively more smaller scores than larger ones. This situation represents what statisticians call a skew in the data. The skewness statistic measures the lack of symmetry of a distribution. If the mean is larger than the median, there is a positive skew. If the mean is smaller than the median, there is a negative skew. If the mean and median are equal, the distribution is symmetric and the skewness is 0.
There is one more descriptive statistic for a distribution - the kurtosis, which measures the peakedness of a distribution. A positive kurtosis represents a peaked distribution, a negative kurtosis represents a flat distribution, and a kurtosis of 0 represents a bell shape. We won't spend much time studying the kurtosis or the skewness. There are very thorough explanations of them here: http://jalt.org/test/bro_1.htm. There is one application of skewness that is worthy of a short note. Before that, review this output from SPSS about the full (not grouped) distribution of ages that was listed previously.
With the exception of the three standard error statistics (i.e., Std. Error of Mean, Skewness, and Kurtosis), all of the other statistics should be familiar to you by now. [Ignore the Missing entry - Missing represents blanks in the data - there weren't any.] Can you identify the three measures of central tendency and the three measures of spread? Note that the interquartile range is not reported, but it can be easily derived by subtracting 29.75 from 48.50. Why? Also, note that the number 36.50 occurs twice and has two different labels. What is the relationship between the median and the 50th percentile?
So, what do the skewness and kurtosis indicate? First, because the kurtosis statistic is negative (-.497), this distribution is flatter than a standard, bell-shaped curve. Second, because the skewness is positive (also note that the mean is larger than the median), this distribution is not symmetric. In fact, lack of symmetry is a warning signal about the accuracy of the mean as a measure of the center. Statisticians compute a skewness ratio and determine that the distribution is severely skewed if the ratio is 2 or higher or -2 or lower. SPSS does not calculate the skewness ratio, but it does give you the two numbers you need to compute it yourself. The skewness ratio equals the skewness divided by the standard error of the skewness. In this example, the skewness ratio is .784/.337 or 2.33, which indicates a severely, positively skewed distribution because the ratio is larger than 2. In this case, the mean should only be used with caution when describing the distribution.
Summary of the Relationship Between Descriptive Statistics and Distributions
Where is the center of the distribution? Answered by the mean
How spread out are the values from the center? Answered by the standard deviation
How symmetric is the distribution? Answered by the skewness
How peaked or flat is the distribution? Answered by the kurtosis
Graphical Displays
The graphical version of a frequency distribution is called a histogram. Here are two examples of histograms for the grouped age data, which were generated using Excel and SPSS. Review them and compare them with each other and with the grouped frequencies presented in the table above. Can you visualize where the mean, median, and mode are? Can you see why the distribution is skewed? The relatively fewer ages on the right side pull the mean away from the larger grouping of ages on the left.
Figure 1. Frequency distribution of participants' ages, generated using Excel.
Figure 2. Frequency distribution of participants' ages, generated using SPSS.
These graphs are usually generated with the aid of a computer program, but here are the steps to construct one by hand:
1. Draw the axes - the horizontal (x-axis) indicates the distribution values, class intervals; the vertical (y-axis) indicates the frequency counts.
2. Mark endpoints of each class interval.
3. Label each class interval with the midpoint, the range, or the endpoints.
4. Plot the tally (frequency count) for each class interval.
5. Construct rectangles for each class interval.
The APA manual has separate sections on properly formatting tables and figures. The figures included above are properly formatted according to APA guidelines, with captions below the graphs. Please note that APA-formatted figures do not have titles - they have captions instead.
Here is an example of a properly APA-formatted table. Note that all numbers are aligned to the right. If the numbers included decimal points, the decimal points would provide the alignment point. Microsoft Word and other word-processing programs have special tab settings for tab-alignment of numbers. Notice also that APA-formatted tables have no vertical borders. Please consult the APA manual for additional examples. All tables and figures included in the required assignments for the course must be formatted to APA specifications.
Table 1 Distribution of Ages for Survey Participants (n=50) Age Frequency Percent Cumulative percent 20-24 4 8 8 25-29 8 16 24 30-34 12 24 48 35-39 8 16 64 40-44 5 10 74 45-49 1 2 76 50-54 4 8 84 55-59 3 6 90 60-64 3 6 96 65-69 2 4 100 Total 50 100
Histograms are only one of the many types of graphical displays available for numerical data. The type of graph that you should choose depends on the information that you are trying to communicate. As you read research reports, begin to develop an understanding of the graphs that researchers typically use. Here are some other types of graphical displays. For these examples, you will see some additional variables for the sample of 50 survey participants.
Stem-and-Leaf Plot
The stem-and-leaf plot is not used very often in research reports, but it is used to help students understand the transition from a frequency distribution in table form to a histogram, which displays the distribution in graphical form. The stem-and-leaf plot is appropriate for ordinal level variables or higher. Here is the stem-and-leaf plot for the ages of the 50 survey participants.
Age Stem-and-Leaf Plot
Frequency Stem & Leaf
4.00 2 . 2344
8.00 2 . 55788899
12.00 3 . 000001112234
8.00 3 . 67788899
5.00 4 . 00144
1.00 4 . 8
4.00 5 . 0033
3.00 5 . 579
3.00 6 . 124
2.00 6 . 56
Stem width: 10
Each leaf: 1 case(s)
Before reading the explanation of this plot, try this: look at the right side of the plot by tilting your head to the right - if you have printed this or are viewing on a laptop, rotate the image 90 degrees counterclockwise. Can you see that the stem-and-leaf plot resembles the histogram? The numbers of leaves (i.e., frequencies) represent the heights of the histogram's bars. The stem-and-leaf plot displays an ordered list of the values in the distribution. The stem represents the first digit, the 10s digit - in this example, the stems represent 20, 30, 40, 50, and 60. The leaves represent the second digit, which in this case is the 1s digit. So, the numbers in the top row represent 22, 23, 24, and 24. Notice that there is actually more information in the stem-and-leaf plot than in the grouped histogram because we can re-generate the entire distribution. With the first bar of the histogram, we only know that there are four values between 20 and 24, but we don't know exactly which four values they are. The mode is fairly easy to find in this display - can you find it?
Boxplot
A common way of describing a distribution is called the 5-number summary. These five numbers are used to draw a boxplot, also called a box-and-whiskers plotThe five numbers are the minimum value and the four quartiles. These are plotted on a graph vertically with vertical lines drawn between the bottom two points and the top two points. The upper and lower vertical lines are the whiskers. Three horizontal lines are drawn through the first, second, and third quartiles. Then two vertical lines are drawn connecting the ends of the horizontal lines to form the box. The result looks like this:
The bottom and top points are the minimum (22) and maximum (66) values. The longer horizontal lines represent Q1 (29.75), Q2 or the median (36.5), and Q3 (48.5). The height of the box is the interquartile range. What should you understand about the distribution by looking at this boxplot? First, realize that each section between consecutive horizontal lines represents 25% of the values. Notice that the lower whisker is shorter than the upper whisker, which means that the lower values are less spread out than the higher ones. Glance back at the stem-and-leaf plot or the histogram and compare the spread of the lower values with that of the higher values. The histogram bars are taller for the younger ages and shorter for the older ages.
Now let's see how a stem-and-leaf plot can identify outliers (remember, those are unusually large or small values in the distribution). For this example, data for an 80-year-old participant will be added to the distribution and a new boxplot will be generated.
The new boxplot looks similar to the previous one except for the little circle labeled 51 (the 51 is the case number for the value) at the top. SPSS has designated this new value as an outlier because it is more than 1.5 times the interquartile range away from the top of the box (Q3). For values that are more than 3 times the interquartile range away from the top (or bottom) of the box, SPSS designates these to be extreme values and displays a different symbol. Outliers can increase the inaccuracy of the mean; so, the boxplot is a tool to identify them. You can either argue to justify the removal of any outliers from the data set or use alternative statistics that are insensitive to extreme values.
Pie Chart
A pie chart is appropriate to display a distribution of nominal (categorical or qualitative) data. The survey asked participants to select an ethnicity with which they identified. Here is the pie chart for their selections.
Just a note - colorful graphics are fine for dissertation defense presentations and conference or workshop displays, but figures in your printed dissertation need to be in shades of gray.
Column and Bar Graphs
Column and bar graphs are also appropriate for nominal data. Column and bar graphs are similar to histograms, but there is an important difference. The horizontal axis (x-axis) for a histogram always represents an ordinal-level variable or higher. Column and bar graphs usually portray nominal variables. By combining two nominal variables, a bar chart can support comparisons between groups. Here is a grouped bar chart comparing ethnicities of female and male participants.
Line Graphs
Line graphs are often used to show changes in phenomena over time. In this example, a survey question asked about participants' attitudes in the fall and in the spring. The means for the attitudes at the two times were calculated for females and for males. The lines illustrate how male attitudes decreased over time while female attitudes increased.
Beyond Univariate Statistics
Before introducing the last type of graphical display, a clearer distinction should be made about the data source for these graphs. So far, the statistics that you've studied have involved a single variable, and so, they are called univariate statistics. The two preceding graphs are examples of ways to combine multiple variables in order to make comparisons. In the next section, you'll learn about statistics that involve two variables, called bivariate statistics.
Scatterplot
A scatterplot represents the distribution of two variables simultaneously. The values for one variable (sometimes called the independent variable) are plotted based on the horizontal axis (x-axis). The values of the other variable (sometimes called the dependent variable) are plotted based on the vertical axis (y-axis). Each survey participant answer a question about age and another question about years of work experience. Age is plotted along the x-axis and years of work experience along the y-axis. Each dot represents a pair of answers for an individual participant. Study the graph below and describe any general trends that you notice. Why are there no points in the upper left portion of the graph?
In addition to observing general patterns, scatterplots also help to identify unique individuals. In this example, the three dots in the lower right part of the graph represent older participants with relatively few years of work experience. Depending on the purposes of the study and the research questions being investigated, these individuals might be sought out for a deeper understanding of their experiences.
In the next section, you'll learn about a single statistic that attempts to describe the relationship between two variables.
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# Category Archives: algebra
Here is a simple set of slides created using Geogebra. I would normally just do this directly on Geogebra but as I needed to prepare some slides for our department planning, I thought I would share them. I have added some suggestions for how to run these in the notes section
I want students to make the connection that if the y-value doesn’t change for a bunch of coordinates (in this base the y-value is always 4) then the line that those coordinates all sit on is y=4. I’ve even made a little GIF where the point deliberately slides off the the right for a little while. Gotta love a GIF!
# Interleaving Algebra and Perimeter
Over the last couple of years we have organised our Year 7 curriculum so we do some introductory algebra early on. Forming expressions from words and collecting like terms, would be topics that I would put in the introductory bucket. The benefit of this is that it can be interleaved into various other topics to extend thinking and promote generalising. Perimeter is an example of this, when we can give side lengths letters instead of numerical values.
An alternative to presenting a bunch of text-book type questions is to investigate a simple 4 piece Tangram, as described in this task from Mike Ollerton.
(Click on the image to access the full Word document)
The task as presented is primarily an exercise in shape, but I might use a slightly modified version of the Tangram to focus purely on perimeter.
Before asking students to cut the triangles out of the shape, we might agree on some labels for the side lengths. If we focus just the shorter sides, we could call Triangle A’s short lengths a, Triangle B’s short lengths b, and Triangle C’s short lengths c, so we end up with something like this.
Again, before getting to work with the scissors we could discuss how we might describe the hypotenuse lengths in terms of a, b and c. And in fact, if we really need c at all…
Using this notation for the side lengths, we can then cut up the 4 shapes and generate many other shapes and find their perimeters. Using 1 shape, 2 shapes, 3 shapes, all 4 shapes: what is the shape with the longest perimeter? The shortest perimeter? What is the difference between longest and shortest? What shapes are different but have the same perimeter – can we prove this using algebra?
It hopefully presents a need for “collecting like terms” as well as some introductory practice in using the technique.
## 9-pin Geoboard
A second task that links algebra to perimeter uses a 9-pin geoboard. This sounds fancier than it is. You don’t need the actual boards, students can create their own in their books or you can give them some dotty paper. First, we tell students that we are going to make triangles using only these lines a, b, c. This is a key image that we will need to refer to either on the board or on a handout.
The next task is to construct triangles using various combinations of these lengths. Each triangle must fit within the 3×3 array. Depending on the class and your objectives for the lesson, at some point you show them that there are only 8 “different” triangles. An opportunity here for a discussion on what we mean by different and what congruency is.
Once we have these, we can go through the process of calculating the perimeter for each one using the a, b, c notation we introduced earlier.
There are some options from here. Mike’s suggestion is that we order these from smallest to largest. We could do that by just looking at the shapes and having a guess. It’s pretty obvious that C is smallest although some of the in-between ones are harder to see. We could establish that a<b<c in the first picture (again by looking). Then we would also need to decide, for example which is bigger: 2a or c. With an older group, you might even use Pythagoras and express c and b in terms of a using surd form.
By looking at the differences between each shape’s perimeter, we start dealing with negative quantities of a, b, and c. If we then sum up all those differences, we should end up with an expression the same as the difference between the smallest and largest, with the c’s cancelling out. Which is quite satisfying and obvious if you think about the expressions lined up on a number line.
There is a fair degree of flexibility within tasks like these and I believe that as teachers we need to select carefully what routes we expect to go down in a lesson. There is a danger that we try to encompass too many different topics in one go and if all of these topics are new to a class then they (and you!) are likely to lose track of what they are actually meant to be learning in this lesson. However, if you are confident that the learners in your class are secure with certain concepts (in this case collecting like terms with negative coefficients) then it is a good way to consolidate and practice this knowledge whilst pushing into new territory.
# Algebra
What other integer values for a and b produce rational values for c?
# Simultaneous Equations using real life examples
Students need to sort the cards into matching pairs and then use the information on each card to find the price of each item (2 equations, 2 unknowns). Introduces simultaneous equations using some concrete examples as a way into the algebra. There are 3 different levels, on the spreadsheet here.
# NCETM article on Functions – new spec GCSE
A short post, this. Just a link to a useful NCETM article on functions. I haven’t taught this yet, but I remember getting pretty upset by the notation f(x), fg(x), etc. when I learned it at school. One way of dealing with this might be to say that they have already encountered this function notation when doing trigonometry, after all sin(x) means sin of x, not sin times x. But then don’t we also say “of means times“. Arghh!
Anyway, it’s a great article and includes some ideas about linking this to:
1. area and scale factor:
2. angles
3. graphs
https://www.ncetm.org.uk/resources/49564
# Sum of Consecutive Numbers, a multi-layered investigtion
One of those classic investigations that gets forgotten about all too easily. So much scope for generalising at different levels.
The fact that all odd numbers can be expressed as sum of two consecutive numbers is probably the first thing that will be established. But why is this the case? And can students express this as a generalisation, first in the form of concise words and then algebraically?
The beauty of this is that there are then many other layers of things to discover, right down to a generalisation explaining which numbers cannot be expressed as a sum of consecutive numbers. And maybe even a proof.
This nRich page gives away some of the answers.
Thanks to Alan Parr for reminding me about it with this excellent blog post:
The All I Can Throwers – Sessions with Den and Jenna. #1 – Consecutive Numbers
# The paper napkin trigonometry trick with a smattering of Pythagorean triples.
Take a piece of paper and do the following:
1. Make it into a square (interesting discussion on best way to do this).
2. Fold in half then unfold so you have created crease along a vertical line of symmetry
3. Then take any corner and fold to the midpoint of the opposite edge. Press down to make a crease along the fold line
4. Unfold and now investigate all the triangles you have created, i.e. can you work out their lengths?
Here are some pictures, which also give some hints, although not a complete solution. The result is very satisfying although I would love to find a way to show this that doesn’t require reams of algebra. Any takers??
# Fun with Fibonacci
This is an old one but fun, and a good way to use algebra to show why a trick works. It’s a similar to showing how Magic Squares work. It’s not a formal proof as such, but I think it’s a good way to introduce the topic.
Once students have grasped the basic concept of a Fibonacci Series (something which, in my experience they often see at Primary School even if they can’t remember what it is called), then you are ready to start the trick.
Fibonacci series don’t have to start with a 1 and a 1 as in the diagram above. You start by asking students which two numbers they want to start with.
Then they get ready to be wowed with your powers of mental arithmetic. Tell them that you will be able to add up the first 10 digits of this sequence in your head faster than they can on calculators. Get one student up to the board to write down the numbers one by one. TOP TIP here: make sure you have the numbers 1-10 written in a vertical column and that the chosen student writes down each term in the sequence against the numbers. You should end up with something like this on your board:
As soon as term 7 goes up on the board, you start calculating. You should be able to find the sum of the first 10 terms before they even get to term 10 and this is why:
I quite like doing the calculation on a miniwhiteboard, then writing the answer face down on a students’ desk and then walking to the other side of the room. Once they have finally totalled the column of numbers on their calculator, you ask a student to have a look under the whiteboard.
And like all good magicians, you DO then go on to reveal the secrets of your trick!
# Going off piste with the difference of two squares
Sometimes it’s worth taking a risk and changing course of a lesson halfway through. And sometimes it pays off.
Today’s lesson was supposed to be about algebraic proof and we started with these nice questions from Don Steward. I thought they looked like good practice for multiplying out double brackets at the same time as introducing algebraic proof.
By the way question 3 is particularly tricky.
Following on from the previous lesson, lots of them started by trying values for n. Great to then have the discussion on what makes a proof vs. an example.
We got into a good discussion on Question 5 and I wanted to know if they were familiar with the difference of two squares. This is where the lesson changed course completely. None of them could tell me what it was called but I got the sense that they had seen it before. So next, I wrote these questions on the board:
Factorise:
1. n² – 9
2. 4n² – 25
3. 81 – n²
4. 100 – 81
They raced through these because they had all spotted the short cut. So next, I put this up, again from Don Steward:
Which followed on nicely from the final of the 4 questions I had put on the board. Many of them initially struggled to see the link immediately or see the pattern in the numbers. But with a bit of time and just the right amount of help (i.e. not much actual help, just encouragement!) they started to find others. I heard them forming statements like: “I need to number that multiply to give 2016, then I need another two numbers that add to give one of those number and subtract to give the other.”
About 5 minutes before it was time to pack away there was a great buzz in the room as 3 students found some other numbers and then the race was on. Lots of solutions started coming, but nobody got all 12 (including me!).
I went away and built a spreadsheet to investigate further, but I still didn’t find all 12. Can you help???
# Some more challenging equation of a line questions
…with solutions on Geogebra Tube (click on the questions).
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Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai) - Part 14
• What is the HCF of [(11111… written 105 times), (11111… written 104 times)]?
Take a smaller example.
HCF (11111, 1111) = HCF (11111 – 1111, 1111) = HCF (10000, 1111)
by HCF (a, b) = HCF (a – b, b) where a > b.
HCF(10000, 1111) = HCF (1, 10000) = 1 by Euclidean Algorithm for a = b * q + r.
You can check for other examples as well to reach the generalization that,
HCF [(11111… (n + 1) times), (11111… n times)] = 1.
A natural number N is having a total of 21 composite factors. What is the minimum and maximum number of prime factors of N?
The factors of N will comprise of three types of factors
1. 1 is the factor of every number.
2. Some prime factors, say x.
3. Some composite factors = 21(given).
Total number of factors = 1 + x + 21.
Case 1 Let there be just 1 prime factor. Total factors = 23.
This can be generated by p^22 where p is a prime.
So minimum number of prime factors of N = 1.
We can also generalize here that it is possible to make any number of factors with the help of just one prime factor. Like we need a 100 factors we can get it from p^99 where p is a prime number.
Case 2 Note here that x < = 4. If x = 5, then minimum number of total factors = (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) = 32
as any such N will be at least a * b * c * d * e form where a, b, c, d and e are distinct primes having unit power.
So, 1 + 5 + 21 = 27 is not possible.
Let’s check x = 4. Total factors = 1 + 4 + 21 = 26
26 factors using four primes. 26 = 2 * 13.
Minimum contribution of any distinct prime is a factor of 2 as p^1 will give a (1 + 1) = 2 multiplier into total number of factors. Thus we can have a maximum of only two primes here to get 26 factors where N = a * b^12.
Moving ahead, let us check for x = 3.
Total factors = 1 + 3 + 21 = 25 = 5 * 5
This again by following similar logic can be obtained through two primes at max as N = a^4 * b^4.
Three primes are not possible as 5 * 5 cannot be further broken into factors > = 2.
Checking for x = 2,
Total factors = 1 + 2 + 21 = 24 = 2 * 12 = 3 * 8 = 4 * 6
Numerous possibilities to obtain a total of 24 factors with two primes:
N = a * b^11 OR a^2 * b^7 OR a^3 * b^5.
Hence maximum value of x = 2 while minimum already seen as 1.
IMPORTANT: Total number of factors = 1 + Prime factors + Composite factors.
Find the smallest odd positive integer that has the same number of divisors as the number 360.
360 = 2^3 * 3^2 * 5
Number of divisors = 4 * 3 * 2 = 24.
24 = 8 * 3 = 6 * 4 = 6 * 2 * 2 = 4 * 3 * 2 = 2 * 2 * 2 * 3.
Ideally we should start with the way that spreads it in maximum number of prime factors. However, this may not give the final result we need, and we will be required to check the number thus obtained for being the smallest one or not.
Taking the 2 * 2 * 2 * 3 case
N = p * q * r * s^2
To have the lowest possible value of N, we should take lowest possible values of p, q, r and s and assign higher powers to lowest possible prime.
Note here we need odd positive integer so we cannot take prime as 2.
Lowest possible four primes = 3, 5, 7, 11.
Lowest N = 5 * 7 * 11 * 3^2.
Now we could trade off by increasing power of lowest prime and eliminating the largest prime.
Let’s look at three prime option.
4 * 3 * 2 => 3^3 * 5^2 * 7. So we have multiplied an addition 3 * 5 (= 15) and removed 11.
So not what we need to do.
You could check for 6 * 2 * 2 also but closer powers give smaller number as a general priniciple.
So our lowest number will be 5 * 7 * 11 * 3^2 = 99 * 35 = 3500 – 35 = 3465 [Also note the calculation methods being employed].
If a is the smallest positive integer that is a multiple of 147 and has exactly 147 positive integral divisors, including 1 and itself, the value of a/147 is?
a is divisible by 147 = 3 * 7^2. So two of the primes in a’s factorization are 3, 7.
Also, a has 147 positive integral divisors.
147 = 3 * 7 * 7 => a = p^2 * q^6 * r^6 where p, q, r are distinct primes.
To make the value of a small, we assign r = 2, q = 3 and p = 7.
So, a = 7^2 * 3^6 * 2^6.
Therefore, a/147 = 2^6 * 3^5.
How many factors of 1080 will be divisible by 10 but not divisible by 30?
1080 = 2^3 * 3^3 * 5
10 = 2 * 5
30 = 2 * 3 * 5
For, the factor to be divisible by it must have 2 * 5 in it. So taking that out from 1080, we are left with: 2^2 * 3^3.
Now with 2 * 5 taken out, we must ensure that a 3 is not taken out else the factor will become divisible by 30.
So we can only take 3 factors namely 2^0, 2^1 and 2^2 out of the (2 + 1)(3 + 1) = 12 factors of this remaining number.
Answer: 3 factors = 2 * 5 * (2^0), 2 * 5 * (2^1), 2 * 5 * (2^2) = 10, 20, 40.
How many natural numbers less than 200 will have 12 factors?
12 = 2 * 6 = 3 * 4 = 2 * 2 * 3
Case 1
2 * 6 => p * q^5
Take q = 2 and out values of other primes.
32 * 3, 32 * 5 only are less than 200: 2 values.
q = 3 alone takes above 200 so no need to check further.
Case 2
3 * 4 => p^2 * q^3
q = 2 => 8 * 9 only.
q = 3 => 27 * 4 only.
Case 3
2 * 2 * 3 => p * q * r^2
r = 2 => 4 * 3 * 5, 4 * 3 * 7, 4 * 3 * 11, 4 * 3 * 13, 4 * 5 * 7: 5 values.
r = 3 => 9 * 2 * 5, 9 * 2 * 7, 9 * 2 * 11: 3 values.
r = 5 => 25 * 2 * 3: 1 value.
None further.
Total: 2 + 1 + 1 + 5 + 3 + 1 = 13 values.
How many three digit numbers will have a total of 3 factors?
Only squares of primes have exactly three factors namely 1, p, p^2 where p is a prime.
11^2 = 121
13^2 = 169
31^2 = 961
11, 13, 17, 19, 23, 29, 31: 7 numbers
How many natural numbers less than 100 have exactly 2 factors?
Only primes have two factors which is 1 and the number itself.
There are 25 primes in the range 1-99.
Hence the answer will be 25.
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
N = 2^8 * 3^10 * 5^8 * 7^2. How many factors of N are multiples of 360 but not a multiple of 540?
N = 2^8 * 3^10 * 5^8 * 7^2.
360 = 2^3 * 3^2 * 5
540 = 2^2 * 3^3 * 5
Let us call 360 the Important (I) and 540 the Unimportant (UI).
In such questions, we look at each prime’s individual powers.
Deficiency: The primes that have power less in I as compared to UI must be restricted for the factor of N to be power of that prime in I to less than power of that prime in UI.
Surplus: While, the primes that have power more in I as compared to UI must have the power of that prime greater than power of that prime in UI but less than equal to the power of that prime in N.
So, here prime 3 has a deficiency case and prime 2 has a surplus case for 360.
So any factor of N that is a multiple of 360 but not a multiple of 540 must be of the form:
2^w * 3^x * 5^y * 7^z where w = 3 to 8, x = 2, y = 1 to 8 and z = 0 to 2.
So number of such factors = (8 – 3 + 1) * (1) * (8 – 1 + 1) * (2 – 0 + 1) = 6 * 1 * 8 * 3 = 144
N = 2^8 * 3^10 * 5^8 * 7^2. How many factors of N are factor of 360 but not a factor of 540?
Looking in the similar way,
w = 3, x = 0 to 2, y = 0 to 1, z = 0
1 * 3 * 2 * 1 = 6 such factors.
Note here the idea would be to keep power of each prime greater than UI and equal to I for surplus case while less than equal to I in case of deficiency. Also, a prime not present in I but present in UI or N would not be considered as it’s a case of being a factor.
Shorter approach for above 2 questions:
360 = 2^3 3^2 5 and 540 = 2^2 3^3 5.
So their LCM = 2^3 3^3 5 and HCF = 2^2 3^2 5.
(i) Required number factors is = number of factors of 360 - number of factors of both 360 and 540 (i.e. their HCF) = 4 * 3 * 2 - 3 * 3 * 2 = 6.
(ii) Required number of factors = factors of N which are multiple of 360 - factors of N which are multiples of both 360 and 540 (i.e. LCM) = 6 * 9 * 8 * 3 - 6 * 8 * 8 * 3 = 144.
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As, AC ≠ BD. Moreover, 2 diagonals space perpendicular to each other in a rhombus. Thus, AC ⊥ BD.
You are watching: Diagonals ac and bd form right angles at point m in parallelogram abcd. prove abcd is a rhombus.
Hence, we acquire that, alternative A is correct.
1. AC ⊥BD ; therefore, ABCD is a rhombus.
Step-by-step explanation:
We space given,
A parallelogram ABCD with vertices A(1,2), B(0,9), C(7,8) and also D(8,1).
Using street Formula, which offers the distance between two points (
and (
as
So, the distance in between two clues are given by,
AB=
= 7.07
BC=
= 7.07
CD=
= 7.07
DA=
= 7.07
Also,
AC=
= 8.49
BD=
= 11.3
Since, we have,
AB = BC = CD = DA.
So, ABCD is a rhombus yet not a rectangle.
As, AC ≠ BD. Moreover, two diagonals room perpendicular come each various other in a rhombus. Thus, AC ⊥ BD.
Hence, we obtain that, choice A is correct.
B) ABCD is a rhombus together all the sides space congruent (equal)
B
Step-by-step explanation:
C. ABCD is a rhombus, yet not necessarily a square.
Step-by-step explanation:
A parallel with surrounding sides congruent is a rhombus, which additionally happens to have perpendicular diagonals (that bisect each other). If the diagonals are various length, then the parallelogram is not a rectangle, therefore the rhombus is not a square.
Option C is correct.
Explanation:
Rhombus states that a parallel with four equal sides and also sometimes one with no best angle.
Given: The coordinate of the vertices of quadrilateral ABCD space A(−6, 3) , B(−1, 5) , C(3, 1) , and D(−2, −2) .
The condition for the segment
,
to it is in parallel to
,
is equivalent slopes;
or
....<1>
So, we have to examine that and
First check
A(−6, 3) , B(−1, 5) , C(3, 1) , and also D(−2, −2)
substitute in <1>,
-10 ≠ -15
Similarly,
check
A(−6, 3) , D(−2, −2) , B(−1, 5) and also C(3, 1)
Substitute in <1>, we have
-20 ≠ -16.
Both bag of sides are not parallel,
therefore, quadrilateral ABCD is not a rhombus because there are no pairs of parallel sides.
Option D
Step-by-step explanation:
In any Rhombus the diagonals bisect the angles. The diagonals are perpendicular bisectors of every other.
So,
5x-18+x+90=180 ( angles of a triangle add to 180 degrees)
Simplifying prefer terms:
6x+72=180
Subtracting 72 both sides :
6x= 108
Dividing through 6 both sides:
x=18.
Option D is correct.
Solution:
Vertices of parallelogram ABCD is provided as A(1,2) , B(0,9) , C(7,8) , and D(8,1) .
We will usage the following analytical geometry formulas here
1. Distance between two points
2. Slope of a line
3. When two lines room perpendicular, product of their slopes is same to -1.
Since ABCD is a parallelogram
1. The contrary sides space equal and also parallel.
2. Diagonals bisect each other.
3. Opposite angles space equal.
Now, comes to problem
As, you can see that, AB=BC=CD=DA=√ 50
But , steep of ab × steep of BC =slope the CB × slope of DC=slope that CD × slope of DA
=1" />
which is no equal come -1. It method lines which room sides the parallelogram space not perpendicular.
As, every side of parallelogram ABCD space equal, so the is a rhombus.
As, diagonal line of rhombus bisect each various other at appropriate angles.
Shows the diagonals are perpendicular bisector of every other.
Option (1) : AC⊥BD; therefore, ABCD is a rhombus.
Solution:
Vertices of parallel ABCD is offered as A(1,2) , B(0,9) , C(7,8) , and D(8,1) .
We will use the following analytical geometry formulas here
1. Distance between two points
2. Steep of a line
3. Once two lines space perpendicular, product of your slopes is same to -1.
Since ABCD is a parallelogram
1. The contrary sides space equal and also parallel.
2. Diagonals bisect every other.
3. The opposite angles space equal.
Now, coming to problem
As, you deserve to see that, AB=BC=CD=DA=√ 50
But , steep of abdominal × steep of BC =slope of CB × slope of DC=slope the CD × steep of DA
=1" />
which is no equal to -1. It way lines which room sides that parallelogram room not perpendicular.
As, every side of parallel ABCD space equal, so the is a rhombus.
As, diagonal of rhombus bisect each various other at appropriate angles.
Shows that diagonals are perpendicular bisector of every other.
Option (1) : AC⊥BD; therefore, ABCD is a rhombus.
Step-by-step explanation:
Refer the attached number :
In Δ ABD ,
∠B=∠D = 66°
By building : opposite sides of equal angles are equal
Thus AB=AC
In Δ CBD ,
∠B=∠D = 66°
By building : opposite political parties of equal angles space equal
Thus CB=CD
Thus all 4 sides of quadrilateral ABCD room equal
And diagonal line BD bisects the angle
So, it is a rhombus
Thus choice c is exactly .
c. Parallel ABCD is a rhombus, due to the fact that the diagonal bisects two angles.
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# 7 Games & Activities To Learn Dividing Fractions In A Fun Way
Did you know that fractions are everywhere? We use fractions all the time in our everyday lives, but we may not even realize it. Fractions can be found in measurements (1/2 cup of sugar), money (a \$0.50 coin), and recipes (3/4 cup of flour). Even though fractions are all around us, they can be tricky to understand. That’s why it’s important to do lots of different activities with fractions to help reinforce what we learn in math class.
At the same time, multiplication and division are two such math concepts that stand essential to having a hands-on grasp on fractions. In other words, division and multiplication is the backbone of fractions and can perhaps be learned by having a good hold on these two. With a little help from the below-mentioned games and activities, in this post, we will explore some interactive ways of teaching students how to divide fractions. Read more below.
## Dividing fractions: Tough nut to crack?
There are several reasons why this topic is so challenging. The study suggests that students with poor whole-number skills are far more likely to develop fractions of difficulty compared to those with strong whole-number skills.
• To divide fractions, students must be able to divide whole numbers accurately. However, many students struggle with this basic concept.
• Another reason dividing fractions is difficult is that it requires students to Visualize the problem differently. Students can simply use picture objects divided into equal parts when dividing whole numbers. For example, if you have ten candy bars and want to divide them equally among five people, you can picture two candy bars going to each person. Unfortunately, dividing fractions is not as straightforward. Students must be able to visualize the problem differently and see that the number of parts doesn’t change, even though the size of each part does. This can be a difficult concept for many students to grasp.
• Finally, another reason why dividing fractions is difficult is that it often requires students to use more than one method to solve the problem. Sometimes, students must first convert the fractions to decimals to divide them accurately. They may also need estimation methods to arrive at an approximate answer. This can be confusing for many students who use one method consistently.
By understanding why dividing fractions is difficult, teachers can be better equipped to help their students succeed in this topic. With patience and practice, one can master this challenging concept.
Dividing fractions is a fundamental math concept that students must learn to progress onto more complex mathematical operations. However, many students find dividing fractions difficult and often get stuck on this concept.
## Engaging activities to teach the concept of dividing fractions
One way to help students understand dividing fractions is to provide them with engaging and hands-on activities. Below are fun fraction activities that help kids better understand the division of fractions!
### 1.Divide and Claim Fractions Activity
This activity helps students to practice identifying equivalent fractions and using visual models to divide fractions. Students will need paper plates, markers, and dice for this activity.
• To begin, students must draw a line down the middle of their paper plate to divide it in half.
• Next, they will add a number to it with a different pattern or design, making it a fraction.
• The teacher will set the time according to the requirement. It can be 1 min or 2 min.
• Once their plates are decorated, students will have to roam around the class and divide their fractions with a friend’s fraction and check whose outcome is a corresponding whole number upon dividing.
• The fraction by dividing which an answer is a whole number, those two students will form a pair.
• The first one to find the pair wins the first round.
Simultaneously, keep following the process to help kids ace dividing fractions in a fun-filled manner. By doing this activity, students will solve and divide various fractions in order to get the whole number that will enable enough practice to get the hang of the concept.
### 2.Fraction Jenga Activity
This activity is fun for students to practice reducing and dividing fractions. You will need a Jenga set and fraction cards for this activity.
• To prepare, write different fractions on each Jenga block using a permanent marker.
• Once your Jenga set is prepared, play the game as you would normally play Jenga, but with a twist!
• When it’s a player’s turn, they must read the fraction on the blocks picked.
• With every turn, a player needs to draw 2 blocks to divide one fraction with another. For example, if they pull out 2 blocks with fractions of 2/4, and 3/6. They need to divide both and the one answering it correctly makes the next move.
• If they answer correctly, their turn continues as normal.
• If they answer incorrectly, their turn ends, and the chance passes to the next player.
• The first player to reach the top of the Jenga tower with all the correct answers wins the game!
This game is a fun match of lesson learning and Jenga playing. While everyone enjoys Jenga, kids will additionally have an added advantage of learning hands-on dividing fractions skills, being competitive, prompt, and decisive.
### 3. Fraction estimation of food station
For this activity, kids need to have their favorite food that can be divided into pieces.
• To begin with, they must have food like bread, pizza, calzones, burritos, etc on a plate. Food that kids may eat easily can be divided into portions.
• The opponent (parents, friends, elder siblings) must speak out a certain fraction and ask the kid to divide it by cutting out portions from the food. For example, Show me a dividing fraction of 2/4 divided by 3/6. Instead of handing out the entire pizza, parents can only give a particular share of it (fraction) that will give the practical experience to divide a fraction with a fraction.
• With every correct answer, they get rewarded with a bite.
This is a very simple, humble, and easy activity that can be played anytime anywhere at home or with friends. The idea behind the activity is to help kids learn to share, be noble, and be considerate while attaining dividing fractions skills in a fun and cheerful way. It’s a happy food-eating activity that comes with a treat with every single answer.
### 4. Fraction bingo
This activity is a fun way for students to learn more about comparison and fractions. For this activity, you will need a fraction bingo worksheet and a set of fraction cards. These cards and bingo worksheets can also be made at home or in class during activity hours.
• Students have to randomly write any fractions on their bingo cards.
• After the students are ready with their cards, Start the activity by giving a dividing fraction question to solve on the board.
• Students need to solve that and find the answer on their bingo card.
• If the answer fraction is there on their card, they need to strike the right answers from their ticket.
• The first student to get five in a row (horizontally, vertically, or diagonally) wins the game!
This activity is a fun and twisted version of dividing fractions wherein kids solve two corresponding fractions then and there to stay on top of the game. The idea is to imbibe promptness, proficiency, accuracy, use of cognitive power, and inculcate logical thinking while dividing the fractions.
## Online games to practice dividing fractions in an engaging way
Dividing fractions is an important mathematical skill that can be difficult for some students to master. However, with a little practice, it can be easy to learn how to divide fractions. The key is to remember the steps of division and to be able to visualize what the division process looks like.
One way to help students understand division is to use fractions games. These types of games can provide students with a fun way to practice their fraction division skills. Here are some fun dividing fractions games that your students will love!
### 1. Dividing fractions fishing game
This is an easy and straightforward online game that can help kids in learning fractions by catching fish. The game’s objective is to catch a fish using a virtual hook and a fish string. When you hook a fish it automatically asks you the dividing fraction associated question in it like ⅝ divided by 5 = ???
All you need to do is answer the question to make the hunted fish, yours. Add to your hoopla of fish hunting and break the record by answering the right answer. More the fish, the merrier your math concepts. The game offers easy gameplay and helps in acing dividing fractions at-a-go.
### 2. Divide Unit Fraction by Whole Number
In this game, all you need to do is solve the dividing fraction as shown in the sky by choosing the correct bubble fish-shaped answer. As you click the right one, you’ll earn a gold coin as a reward.
The more gold coins you earn, the faster you’ll clear a stage. One can keep progressing and take up more challenging stages to practice their dividing fractions skills, comprehensively.
This game is a light, fun, and concise way of understanding and acing dividing fractions. The visuals are pretty, the interface is interactive and the navigation of the game makes it easy for anyone to play.
### 3. Basketball Dividing Fractions Game
If your kid loves to play outdoor sports and is particularly into watching NBA, then this game just might be the best of all, for him/her. It’s an easy play of dunk and basket between two teams- Red and Blue.
This online dividing fractions game can be played between 2 players. Each chooses either the Red or Blue team. To proceed further in the play, player red needs to solve the dividing fractions in order to pass their chance to player blue. The one with the most correct answers ultimately wins the round!
The Basketball Dividing Fractions Game is easy, fun, and competitive. Kids are bound to enjoy fractions via this.
## That’s it..
Fractions are an essential fundamental of learning. It’s a basic that everyone must know in order to grasp advanced math skills that can be precisely inculcated in everyday life. Through this post, we hope we were able to help you channel handy, fun, and useful fraction learning skills in the form of engaging activities and online games. These indulgent sets of plays can be practiced with friends, family, or at some. So, ace fractions like no other and score the best in your next exams using these aforementioned Fun Dividing Fractions Games and Activities. Happy Learning!
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
${{x}^{2}}-6x-5=0$
$3-\sqrt{14}\ \text{and }3+\sqrt{14}$ It can be expressed as, $x=3+\sqrt{14}$ and $x=3-\sqrt{14}$ \begin{align} & x-\left( 3+\sqrt{14} \right)=0 \\ & \left( x-3 \right)-\sqrt{14}=0 \end{align} Or, \begin{align} & x-\left( 3-\sqrt{14} \right)=0 \\ & \left( x-3 \right)+\sqrt{14}=0 \\ \end{align} Apply the zero-product property $\left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0$ \begin{align} & \left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0 \\ & \left( x-3 \right)\left( x-3 \right)+\sqrt{14}\left( x-3 \right)-\sqrt{14}\left( x-3 \right)-{{\left( \sqrt{14} \right)}^{2}}=0 \end{align} Combine like terms: Here ${{i}^{2}}=1$ which is a complex entry. ${{\left( x-3 \right)}^{2}}-{{\left( \sqrt{14} \right)}^{2}}=0$ Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ \begin{align} & {{x}^{2}}+9-6x-\left( 14 \right)=0 \\ & {{x}^{2}}-6x+9-14=0 \\ & {{x}^{2}}-6x-5=0 \end{align}
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## Is Speeding Beneficial?
• by Brett Berry - Wed, 06/01/2016 - 06:24
Intro to Distance, Rate & Time
## Is Speeding Beneficial?
This past weekend I visited my parents. I was on a stretch of highway about five miles from the edge of the small town where they live when I was passed by four cars in the space of a few minutes — which would be understandable had I been going under the speed limit or even at the speed limit, but I was going 5 mph over the speed limit.
I found their urgency mathematically interesting.
I began to wonder how much time the cars saved by passing me? Would it make much of a difference? Is it worth the risk?
## The Four Cars Passing Problem
• speed limit was 55 mph.
• I was traveling at 60 mph.
• The cars that passed me traveled ~65 mph.
• There was 5 miles of highway before entering city limits.
• We’ll be using an important formula: distance = rate • time.
Since we are interested in solving for time, isolate the t-variable beforehand.
## 5 Miles at 55 MPH
For the first example, I’ll write out the long-hand math so you can see how the labels interact.
Begin by filling in distance and rate in the above formula.
Invert and multiply. Notice the “miles” labels cancel out leaving “hours”.
Now convert 0.0909 hours into minutes.
And lastly change .45 minutes into seconds.
So it would take 5 minutes and 27 seconds to travel 5 miles at 55 mph.
## 5 Miles at 60 MPH
Now we’ll calculate the time for a car traveling 60 mph for 5 miles.
Convert to minutes.
## 5 Miles at 65 MPH
Convert into minutes and seconds.
At 65 mph, one would arrive at the city limit in 4 minutes and 37 seconds.
## The Comparison
Turns out this is basic optimization.
By traveling 10 mph over the limit of 55, one would arrive 50 seconds quicker, risking a \$113 fine and a 10–15 minute time delay if pulled over.
# + 5 mph at 30 mph has a greater effect on time than at 55 mph
It may seem counter-intuitive but the difference of increasing speed at lower rates has a greater effect on time than increasing speed at higher rates.
For instance, traveling without stopping for 5 miles at 30 mph equates to 10 minutes of travel time, whereas 5 miles at 35 mph equates to 8 minutes and 34 seconds.
This means you’ll arrive at your destination 1 minute and 26 seconds faster than if you were to drive 30mph. Compare this to increasing speed from 55 mph to 60 mph.
We found this would only save 27 seconds.
Of course fines are adjusted accordingly and many states have higher penalties for speeding under 40mph.
So is it beneficial to speed?
Crunch the numbers, consider the risks and you decide!
Brett Berry is a math evangelist who writes a math blog for Medium.com called Math Memoirs.
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# 5.7: Planes in Space
Difficulty Level: At Grade Created by: CK-12
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Orlando is building a coffee table for his mother. He has the round surface cut and sanded and is ready to attach the legs. As he tries a few different heights, he notices that unless the spot on the floor is really flat, and he is extremely careful to make the legs exactly the same length, the table is unsteady and wobbles annoyingly.
How can Orlando ensure that the table remains stable, even if the floor is not entirely flat where his mom decides to put it?
Embedded Video:
### Guidance
Vectors can be used to identify directions in space and also the orientation of a plane by identifying the direction perpendicular to that plane. In this section we will look at that calculation in reverse. Rather than determining the normal vector to a plane using two vectors which lie in that plane, we will be using the normal vector to determine the equation for the plane itself. We will also use the normal vectors to determine the intersection angle between any pair of planes.
Intercept Form
The diagram below shows a plane which crosses all three coordinate axes. Points A, B, and C are the locations where the plane crosses each of the coordinate axes, called intercepts. Their locations are given by A = (a, 0, 0), B = (0, b, 0), and C = (0, 0, c). The line segments AB, BC, and CA all lie in the plane. Furthermore, segment AB is a portion of the line of intersection between this plane and the x-y axis; segment BC is a portion of the line of intersection between this plane and the y-z axis; and segment CA is a portion of the line of intersection between this plane and the z-x axis.
The intercept form of the equation for a plane is given by
1=xa+yb+zc\begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}
Using a Normal Vector
Another way to specify a plane is to know two vectors within the plane. Recall that if two vectors lie in the same plane, the normal to that plane can be found using the cross product of the two vectors. Sometimes we are given the equations of the vectors themselves. Sometimes, however, we are only given a set of points that lie on the plane. If we know three points that lie on the plane, we can use the method developed when looking at cross products to find equations for the vectors between those points and then use those vectors to identify the plane. We only need three points to accomplish this task.
Since the normal to the plane is, by definition, perpendicular to all possible vectors within a plane and since the dot product of two vectors is equal to zero for any two perpendicular vectors, we can define a plane in terms of the dot product of the normal vector with any vector, v\begin{align*}\overrightarrow{v}\end{align*}, within the plane:
n×v=0\begin{align*}\overrightarrow{n} \times \overrightarrow{v} = 0\end{align*}
Which we can also write as
nx,ny,nz×(xx0),(yy0),(zz0)=0\begin{align*}\left \langle n_x, n_y, n_z \right \rangle \times \left \langle (x - x_0), (y - y_0), (z - z_0) \right \rangle = 0\end{align*}
If we compute the dot product, we obtain another equation which specifies the plane in terms of the normal vector and two points on the plane, (x, y, z) and (x0, y0, z0).
nx (x - x0) + ny (y - y0) + nz (z - z0) = 0
This equation is frequently written as
nxx + nyy + nzz + d = 0
Where
d = -nxxo - nyyo - nzzo
and the intercepts of the plane with the x, y, and z axes are given by:
a=dnx, b=dny,\begin{align*}a = -\frac{d}{n_x}, \ b = -\frac{d}{n_y},\end{align*} and \begin{align*}c = -\frac{d}{n_z}\end{align*}
If you only have the normal vector and one point on the plane, first determine the vector projection of the position vector of that point onto the normal vector using the dot product. Then you will have the locations of two points on the plane and can use the normal and two points method described above.
#### Example A
Find the intercepts of the plane given by the equation 3x + 5y - 2z - 4 = 0.
Solution
Rewrite the equation of the plane in the format of the intercept form of the plane equation.
3x + 5y - 2z = 4
\begin{align*}1 = \frac{3}{4} x + \frac{5}{4} y + \frac{-2}{4}z\end{align*}
\begin{align*}a = \frac{4}{3}, \ b = \frac{4}{5}\end{align*}, and \begin{align*}c = \frac{4}{-2} = -2\end{align*}. Thus the intercepts of this plane are
\begin{align*}\left (\frac{4}{3}, 0, 0 \right ), \ \left (0, \frac {4}{5}, 0 \right )\end{align*}, and \begin{align*}(0, 0, -2)\end{align*}.
#### Example B
A plane has intersections at (12, 0, 0), (0, 6, 0), and (0, 0, 4). Write the equation of the plane.
Solution
\begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \frac{x}{12} + \frac{y}{6} + \frac{z}{4}\end{align*} or \begin{align*}12 = x + 2y + 3z\end{align*}
#### Example C
Use the equation 3x + 5y - 2z - 4 = 0 to determine the normal unit-vector to that plane.
Solution
Comparing this equation to nxx + nyy + nzz + d = 0, we can see that \begin{align*}\overrightarrow{n} = \left \langle 3, 5, -2\right \rangle\end{align*}. To find the unit normal vector, find the magnitude of this normal vector and divide each component by the magnitude.
\begin{align*}|\overrightarrow{n}| = \sqrt{n_x^2 + n_y^2 + n_z^2} = \sqrt{3^2 + 5^2 + (-2)^2} = \sqrt{38}\end{align*}
\begin{align*}\hat{n} = \left \langle \frac{3}{\sqrt{38}}, \frac{5}{\sqrt{38}}, \frac{-2}{\sqrt{38}} \right \rangle\end{align*}
#### Example D
The three points P = (3, 7, 2), Q = (1, 4, 3), and R = (2, 3, 4) define a plane. Determine the equation of the plane.
Solution
First find the vectors between two pairs of the points.
\begin{align*}\overrightarrow{PQ} = \left \langle (Q_x - P_x), (Q_y - P_y), (Q_z - P_z) \right \rangle = \left \langle (1 - 3), (4 - 7), (3 - 2) \right \rangle =\end{align*} \begin{align*}\left \langle -2, -3, 1 \right \rangle\end{align*}
\begin{align*}\overrightarrow{PR} = \left \langle (R_x - P_x), (R_y - P_y), (R_z - P_z) \right \rangle = \left \langle (2 - 3), (3 - 7), (4 - 2) \right \rangle =\end{align*} \begin{align*}\left \langle -1, -4, 2 \right \rangle\end{align*}
The cross product of these two vectors is normal to the plane.
\begin{align*}\overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle (PQ_yPR_z - PQ_zPR_y), (PQ_zPR_x - PQ_xPR_z), (PQ_xPR_y - PQ_yPR_x)\right \rangle\end{align*}
\begin{align*}\overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle [(-3 \cdot 2) - (1 \cdot -4)], [(1 \cdot -1) - (-2 \cdot 2)], [(-2 \cdot -4) - (-1 \cdot -3)] \right \rangle\end{align*}
\begin{align*}\overrightarrow{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle [(-6) - (-4)], [(-1) - (-4)], [(8) - (3)] \right \rangle = \left \langle -2, 3, 5 \right \rangle\end{align*}
This normal vector and one of the points will give an equation for the plane.
nx (x - Px) + ny (y - Py) + nz (z - Pz) = 0
-2(x - 3) + 3(y - 7) + 5(z - 2) = 0
-2x + 3y + 5z + (6 - 21 - 10) = 0
-2x + 3y + 5z - 25 = 0
Concept question wrap-upHow can Orlando ensure that the table he is making for his mother remains stable, even if the floor is not entirely flat where his mom decides to put it?
Recall from the lesson that any plane can be defined by three points. If Orlando makes the table with only three legs, then the endpoints of each leg will define a stable plane for the table to sit on. Even if the plane defined by the leg endpoints is not exactly parallel to the average plane of the floor across the entire room, the table itself will remain stable.
### Vocabulary
A plane is the 3 dimensional equivalent of a line on a standard rectangular graph. It can be conceptualized as a sheet of paper of infinite area.
The intercept form of the equation for a plane is given by: \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}
The normal vector to a plane is perpendicular to all possible vectors within that plane
### Guided Practice
Questions
1) Rewrite the equation of the plane 7x + 3y + z + 12 = 0 in intercept form.
2) Determine the equation for the unit vector which is perpendicular to the plane, 7x + 3y + z + 12 = 0.
3) Find the intercepts of the plane described by the equation 2.4x + 3.6y - 4.8z - 5.9 = 0.
4) A plane is defined by three points having position vectors \begin{align*}\overrightarrow{r_1} = \left \langle 1, 0, -1 \right \rangle, \ \overrightarrow{r_2} = \left \langle 2, 4, 6 \right \rangle\end{align*}, and \begin{align*}\overrightarrow{r_3} = \left \langle -3, 7, 5 \right \rangle\end{align*}. Determine the components of the unit vector which is perpendicular to the plane passing through those points.
5) Determine the components of the unit vector which is perpendicular to the plane 12x + 23y + 14z - 5 = 0.
Solutions
1) The equation \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*} must be true for all points on a plane. Therefore, we should first rearrange 7x + 3y + z + 12 = 0 into the form \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}.
7x + 3y + z = -12
\begin{align*}\frac{7}{-12}x + \frac{3}{-12}y + \frac{1}{-12}z = 1\end{align*}
Therefore, \begin{align*}a = \frac{-12}{7}, \ b = \frac{-12}{3} = -4\end{align*}, and \begin{align*}c = \frac{-12}{1} = -12\end{align*} and the position vectors of the three intercepts are \begin{align*}\overrightarrow{A} = \left \langle -1.714, 0, 0 \right \rangle, \ \overrightarrow{B} = \left \langle 0, -4, 0 \right \rangle\end{align*}, and \begin{align*}\overrightarrow{C} = \left \langle 0, 0, -12 \right \rangle\end{align*}
2) Comparing this equation to \begin{align*}n_xx + n_yy + n_zz + d = 0\end{align*}, we can see that \begin{align*}\overrightarrow{n} = \left \langle 7, 3, 1 \right \rangle\end{align*}
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{(7)^2 + (3)^2 + (1)^2}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{49 + 9 + 1}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{59}} = \left \langle \frac{7}{\sqrt{59}}, \frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}} \right \rangle\end{align*}
3) First write the equation of the plane in intercept form, \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}.
2.4x + 3.6y - 4.8z = 5.9
\begin{align*}\frac{2.4}{5.9}x + \frac{3.6}{5.9}y - \frac{4.8}{5.9}z = 1\end{align*}
Therefore the x-intercept is \begin{align*}\left \langle \frac{5.9}{2.4}, 0, 0 \right \rangle\end{align*}, the y-intercept is \begin{align*}\left \langle 0, \frac{5.9}{3.6}, 0 \right \rangle\end{align*}, and the z-intercept is \begin{align*}\left \langle 0, 0, \frac{5.9}{4.8} \right \rangle\end{align*}.
4) The cross-product determines the direction perpendicular to a pair of vectors. Therefore we can use these three points to define two vectors in the same plane. The vector from point 1 to point 2 is given by subtracting vector 2 from vector 1:
\begin{align*}\overrightarrow{r_{1-2}} = \overrightarrow{r_1} - \overrightarrow{r_2} = \left \langle 1, 0, -1\right \rangle - \left \langle 2, 4, 6 \right \rangle = \left \langle 1 - 2, 0 - 4, -1 - 6 \right \rangle = \left \langle -1, -4, -7 \right \rangle\end{align*}
Likewise, the vector from point 1 to point 3 is given by subtracting vector 3 from vector 1:
\begin{align*}\overrightarrow{r_{1-3}} = \overrightarrow{r_1} - \overrightarrow{r_3} = \left \langle 1, 0, -1\right \rangle - \left \langle -3, 7, 5 \right \rangle = \left \langle 1 - (-3), 0 - 7, -1 - 5 \right \rangle =\end{align*} \begin{align*}\left \langle 4, -7, -6 \right \rangle\end{align*}
Now we can use the cross-product of the two vectors in the plane to determine a vector which is perpendicular to that plane,
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle (r_{1-2y}r_{1-3z} - r_{1-2z}r_{1-3y}), (r_{1-2z}r_{1-3x} - r_{1-2x}r_{1-3z}), (r_{1-2x}r_{1-3y} - r_{1-2y}r_{1-3x}) \right \rangle\end{align*}
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle ((-4)(-6) - (-7)(-7)), ((-7)(4) - (-1)(-6)), ((-1)(-7) - (-4)(4)) \right \rangle\end{align*}
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle ((24) - (49), ((-28) - (6)), ((7) - (-16)) \right \rangle = \left \langle -25, -34, 23 \right \rangle\end{align*}
Now use the definition of the unit vector to complete the problem.
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{(-25)^2 + (-34)^2 + (23)^2}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{625 + 1156 + 529}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{2310}} \end{align*}
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle {-25, -34, 23}\right \rangle}{48.06} = \left \langle -0.5202, -0.7074, 0.4785 \right \rangle\end{align*}
5) Comparing this equation to \begin{align*}n_xx + n_yy + n_zz + d = 0\end{align*}, we can see that \begin{align*}\overrightarrow{n} = \left \langle 12, 23, 14 \right \rangle\end{align*}.
Now we can use the definition of the unit vector to complete the problem.
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 12, 23, 14 \right \rangle}{\sqrt{12^2 + 23^2 + 14^2}} = \frac{\left \langle 12, 23, 14 \right \rangle}{\sqrt{869}} = \frac{\left \langle 12, 23, 14 \right \rangle}{29.5} = \left \langle \frac{12}{29.5}, \frac{23}{29.5}, \frac{14}{29.5} \right \rangle\end{align*}
### Practice
Given the following intersections, write the equation of the plane.
1. \begin{align*}(7, 0, 0), (0, 3, 0)\end{align*} and \begin{align*}(0, 0, 19)\end{align*}
2. \begin{align*}(2, 0, 0), (0, 8, 0)\end{align*} and \begin{align*}(0, 0, 5)\end{align*}
3. \begin{align*}(13, 0, 0), (0, 21, 0)\end{align*} and \begin{align*}(0, 0, 17)\end{align*}
4. \begin{align*}(5, 0, 0), (0, 1, 0)\end{align*} and \begin{align*}(0, 0, 2)\end{align*}
5. \begin{align*}(27, 0, 0), (0, 12, 0)\end{align*} and \begin{align*}(0, 0, 18)\end{align*}
Find the intercepts of the plane given the following equations:
1. \begin{align*}3x + 2y + z - 6 = 0\end{align*}
2. \begin{align*}1x - 7y - z + 10 = 0\end{align*}
3. \begin{align*}-2x + 9y + 4z - 1 = 0\end{align*}
4. \begin{align*}6x - 11y + 2z + 3 = 0\end{align*}
5. \begin{align*}-2x + 5y + 5z + 6 = 0\end{align*}
Use the given equations to determine the normal unit-vector to that plane
1. \begin{align*}7x + 5y - 1z - 10 = 0\end{align*}
2. \begin{align*}4x - 13y + 5z - 3 = 0\end{align*}
3. \begin{align*}-8x + 7y + 2z + 5 = 0\end{align*}
4. \begin{align*}10x + 3y - z - 2 = 0\end{align*}
5. \begin{align*}-1x - 2y + 7z + 16 = 0\end{align*}
Determine the equation of the planes below using the three points given:
1. \begin{align*}P = (3, 6, 9), Q = (9, 6, 3)\end{align*} and \begin{align*}R = (6, -9, 9)\end{align*}
2. \begin{align*}P = (1, -7, 2), Q = (4, 2, 9)\end{align*} and \begin{align*}R = (3, -5, 1)\end{align*}
3. \begin{align*}P = (3, 8, 10), Q = (-2, 5, 8)\end{align*} and \begin{align*}R = (7, 4, 8)\end{align*}
4. \begin{align*}P = (9, -1, 4), Q = (6, 2, -8)\end{align*} and \begin{align*}R = (12 , 9, 10)\end{align*}
5. \begin{align*}P = (5, 8,-9), Q = ( -5, 3, 9)\end{align*} and \begin{align*}R = (10, 4, -6)\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Cross product
The cross product of two vectors is a third vector that is perpendicular to both of the original vectors.
dot product
The dot product is also known as inner product or scalar product. The two forms of the dot product are $\vec{a} \cdot \vec{b} = \Big \| \vec{a}\Big \| \ \Big \| \vec{b}\Big \| \cos \theta$ and $\vec{a} \cdot \vec{b} = x_a x_b + y_a y_b$.
intercept form
The intercept form of the equation for a plane is given by $1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}$.
Intercepts
The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.
normal vector
A normal vector is a vector that is perpendicular to a given surface or plane. A unit normal vector is a normal vector with a magnitude of one.
plane
A plane is a flat, two-dimensional surface. It can be conceptualized as a sheet of paper of infinite area.
unit vector
A unit vector is a vector with a magnitude of one.
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# Lesson 14
Fractional Lengths in Triangles and Prisms
Let’s explore area and volume when fractions are involved.
### 14.1: Area of Triangle
Find the area of Triangle A in square centimeters. Show your reasoning.
### 14.2: Bases and Heights of Triangles
1. The area of Triangle B is 8 square units. Find the length of $$b$$. Show your reasoning.
2. The area of Triangle C is $$\frac{54}{5}$$ square units. What is the length of $$h$$? Show your reasoning.
### 14.3: Volumes of Cubes and Prisms
Use cubes or the applet to help you answer the following questions.
1. Here is a drawing of a cube with edge lengths of 1 inch.
1. How many cubes with edge lengths of $$\frac12$$ inch are needed to fill this cube?
2. What is the volume, in cubic inches, of a cube with edge lengths of $$\frac12$$ inch? Explain or show your reasoning.
2. Four cubes are piled in a single stack to make a prism. Each cube has an edge length of $$\frac12$$ inch. Sketch the prism, and find its volume in cubic inches.
3. Use cubes with an edge length of $$\frac12$$ inch to build prisms with the lengths, widths, and heights shown in the table.
1. For each prism, record in the table how many $$\frac12$$-inch cubes can be packed into the prism and the volume of the prism.
prism
length (in)
prism
width (in)
prism
height (in)
number of $$\frac12$$-inch
cubes in prism
volume of
prism (in3)
$$\frac12$$ $$\frac12$$ $$\frac12$$
1 1 $$\frac12$$
2 1 $$\frac12$$
2 2 1
4 2 $$\frac32$$
5 4 2
5 4 $$2\frac12$$
2. Examine the values in the table. What do you notice about the relationship between the edge lengths of each prism and its volume?
4. What is the volume of a rectangular prism that is $$1\frac12$$ inches by $$2\frac14$$ inches by 4 inches? Show your reasoning.
A unit fraction has a 1 in the numerator.
• These are unit fractions: $$\frac13, \frac{1}{100}, \frac11$$.
• These are not unit fractions: $$\frac29, \frac81, 2\frac15$$.
1. Find three unit fractions whose sum is $$\frac12$$. An example is: $$\frac18 + \frac18 + \frac14 = \frac12$$ How many examples like this can you find?
2. Find a box whose surface area in square units equals its volume in cubic units. How many like this can you find?
### Summary
If a rectangular prism has edge lengths of 2 units, 3 units, and 5 units, we can think of it as 2 layers of unit cubes, with each layer having $$(3 \boldcdot 5)$$ unit cubes in it. So the volume, in cubic units, is: $$\displaystyle 2\boldcdot 3\boldcdot 5$$
To find the volume of a rectangular prism with fractional edge lengths, we can think of it as being built of cubes that have a unit fraction for their edge length. For instance, if we build a prism that is $$\frac12$$-inch tall, $$\frac32$$-inch wide, and 4 inches long using cubes with a $$\frac12$$-inch edge length, we would have:
• A height of 1 cube, because $$1 \boldcdot \frac 12 = \frac12$$.
• A width of 3 cubes, because $$3 \boldcdot \frac 12 = \frac32$$.
• A length of 8 cubes, because $$8 \boldcdot \frac 12 = 4$$.
The volume of the prism would be $$1 \boldcdot 3 \boldcdot 8$$, or 24 cubic units. How do we find its volume in cubic inches? We know that each cube with a $$\frac12$$-inch edge length has a volume of $$\frac 18$$ cubic inch, because $$\frac 12 \boldcdot \frac 12 \boldcdot \frac 12 = \frac18$$. Since the prism is built using 24 of these cubes, its volume, in cubic inches, would then be $$24 \boldcdot \frac 18$$, or 3 cubic inches.
The volume of the prism, in cubic inches, can also be found by multiplying the fractional edge lengths in inches: $$\frac 12 \boldcdot \frac 32 \boldcdot 4 = 3$$
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# Properties of Similar Triangles Part 1
This is the second part of the Triangle Similarity Series. The first part is Introduction to Similarity.
In Introduction to Similarity, we have learned that similar objects have the same shape, but not necessarily have the same size. We drew a triangle using a graphics software zoomed it in and zoomed it out producing similar triangles.
Are Buu and Patrick 'similar'?
Zooming did not change the shape of the object. In effect, the measure of the interior angles of a triangle did not change.
In this post, we are going to explore more properties of similar triangles. We are going to perform another activity. Follow the instructions below and answer the questions.
Instructions
1.) Use a graphics to draw a triangle. Be sure to zoom in to 100 percent.
2.) Take a photo of the triangle. Use the Print Screen key on your keyboard on a PC (or Apple Command key + Shift + 3 on a Mac) to grab the picture of the triangle, crop it so that the picture fits the canvas, and save the file as triangle1 in png or jpg format. You can also screenshot software such as Jing.
3.) Adjust the zoom to 200 percent and repeat 2. Save it as triangle2 in png or jpg format.
4.) We now use GeoGebra to explore the properties of the two triangles. Use the GeoGebra Insert Image button to import both triangles in the GeoGebra graphics view (drawing pad). Using the Point tool, place points on all the vertices of the triangles. Your figure should look like the figure below.
5.) Connect the corresponding vertices ( A to D, B to E, and C to F). What do you observe? Some students will observe that the lines meet, some won’t because of the slight error in placement of the points.
6.) We have said that zooming in or out does not change the shape of the triangle. Therefore, the corresponding angles of similar triangles are congruent. In the figure above, A corresponds to D, B corresponds to E, and C corresponds to F. Therefore, A is congruent to D, B is congruent to E, and C is congruent to F.
7.) Using the Angle, Area, and Distance tools of GeoGebra, investigate the relationships between the two triangles. What can you say about the relationship between the side lengths and area of the two triangles?
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Close
## What is the equation for a 90 degree rotation?
90 Degree Rotation When rotating a point 90 degrees counterclockwise about the origin our point A(x,y) becomes A'(-y,x). In other words, switch x and y and make y negative.
## What is the rotation rule?
Predicting Rotations Rotate the point (5, 8) about the origin 270° clockwise. The rule for rotating an object 270° clockwise about the origin is to take the opposite value of the x coordinate and then switch it with the y coordinate.
## How long does it take for the Earth to rotate?
23 hours, 56 minutes and 4.0916 seconds
## How do you rotate a point 30 degrees?
Rotate the point by 30°. You do this by adding 36.87°+30°, to get a rotated angle of 66.87°.
## How do you reflect points?
When you reflect a point across the line y = x, the x-coordinate and y-coordinate change places. If you reflect over the line y = -x, the x-coordinate and y-coordinate change places and are negated (the signs are changed). the line y = x is the point (y, x). the line y = -x is the point (-y, -x).
## How do you reflect a shape?
Reflecting a shape simply means to flip it over a mirror line. Each point in the shape is moved to the other side of the mirror line but remains the same distance away from the line. The reflected image will now be facing in the opposite direction to the original object.
## How do you fully describe a reflection?
A reflection is a transformation that acts like a mirror: It swaps all pairs of points that are on exactly opposite sides of the line of reflection. The line of reflection can be defined by an equation or by two points it passes through.
## What does a rotation look like?
A rotation is a type of transformation which is a turn. A figure can be turned clockwise or counterclockwise on the coordinate plane. In both transformations the size and shape of the figure stays exactly the same. A rotation is a transformation that turns the figure in either a clockwise or counterclockwise direction.
## What is a reflection of Y 1?
Explanation: the line y=1 is a horizontal line passing through all. points with a y-coordinate of 1. the point (3,10) reflected in this line. the x-coordinate remains in the same position.
## Is Y =- 1 a function?
1 Answer. Yes, it is linear.
2020-07-04
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## United Kingdom Mathematical Challenge (UKMT) – Multiple Solutions Part 1
The year 9 and year 10 students at school recently took part in the IMC (Intermediate Mathematical Challenge), a one hour test provided by Leeds University. The best thing about this challenge (along with the JMC and the SMC for younger and older students respectively) is that the questions require the students to apply their knowledge and skills in new and interesting situations. Not to mention the fact that the students get a pretty decent looking bronze, silver or gold certificate if they do well!
There are always 25 questions which increase in difficulty. Students generally struggle with the last 10 questions and often gain most marks on the first 15. I quite like to have a go at the last 10 questions in every paper before looking at the solutions because you sometimes find that you’ve solved a problem in a different way than is presented in the solution booklet. Take for example the last question on the IMC in 2011:
The solution given by the IMC concerns the use of similar triangles and area scale factors.
I did it in a slighhtly different way using a co-ordinate grid:
It is easy to see that the diagonal line DE has equation: y = -x + 1 (1)
Since the point E is half way up BC, the line AE has equation: y = 0.5x (2)
To find the point of intersection of DE and AE just put them equal to eachother and solve:
-x + 1 = 0.5x
x = 2/3
Sub this value into equation (2) to get y = 1/3. Hence the point of intersection of DE and AE is (2/3, 1/3).
Now that we have the point of intersection, it is quite simple to find the area of each section. Omitting units we have:
Area of trinagle A =0.5(2/3 × 1) = 1/3 = 4/12
Area of triangle B = 0.5(1/3 × 1/2) = 1/12
Area of triangle P = 0.5(1/3 × 1) = 1/6 = 2/12
Hence area of section Q = 1 – (A + B + P) = 5/12
Hence the ratio of areas P to Q is 2/12 : 5/12 or 2:5.
The great thing about multiple solutions is that you can refer back to the question when you’re covering similar triangles/area scale factors and also when you’re covering intersection points of curves. Simply making a point that questions can be solved in different ways is not acceptable. We have to show students multiple solutions and more importantly give them the opportunity to come up with different solutions by themselves. It’s only in this way that they will learn that mathematics is a creative subject in which different ‘tools’ can be used to approach different problems. ( Go to Multiple Solutions Part 2: Celebrating Different Methods (Lesson 1))
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# Question #e8ab5
Aug 21, 2016
$\cos \left(x + y\right) = \frac{{a}^{2} + {b}^{2}}{2} - 1$
#### Explanation:
First, recall what $\cos \left(x + y\right)$ is:
$\cos \left(x + y\right) = \cos x \cos y + \sin x \sin y$
Note that:
${\left(\sin x + \sin y\right)}^{2} = {a}^{2}$
$\to {\sin}^{2} x + 2 \sin x \sin y + {\sin}^{2} y = {a}^{2}$
And:
${\left(\cos x + \cos y\right)}^{2} = {b}^{2}$
$\to {\cos}^{2} x + 2 \cos x \cos y + {\cos}^{2} y = {b}^{2}$
Now we have these two equations:
${\sin}^{2} x + 2 \sin x \sin y + {\sin}^{2} y = {a}^{2}$
${\cos}^{2} x + 2 \cos x \cos y + {\cos}^{2} y = {b}^{2}$
If we add them together, we have:
${\sin}^{2} x + 2 \sin x \sin y + {\sin}^{2} y + {\cos}^{2} x + 2 \cos x \cos y + {\cos}^{2} y = {a}^{2} + {b}^{2}$
Don't let the size of this equation throw you off. Look for identities and simplifications:
$\left({\sin}^{2} x + {\cos}^{2} x\right) + \left(2 \sin x \sin y + 2 \cos x \cos y\right) + \left({\cos}^{2} y + {\sin}^{2} y\right) = {a}^{2} + {b}^{2}$
Since ${\sin}^{2} x + {\cos}^{2} x = 1$ (Pythagorean Identity) and ${\cos}^{2} y + {\sin}^{2} y = 1$ (Pythagorean Identity), we can simplify the equation to:
$1 + \left(2 \sin x \sin y + 2 \cos x \cos y\right) + 1 = {a}^{2} + {b}^{2}$
$\to \left(2 \sin x \sin y + 2 \cos x \cos y\right) + 2 = {a}^{2} + {b}^{2}$
We can factor out a $2$ twice:
$2 \left(\sin x \sin y + \cos x \cos y\right) + 2 = {a}^{2} + {b}^{2}$
$\to 2 \left(\left(\sin x \sin y + \cos x \cos y\right) + 1\right) = {a}^{2} + {b}^{2}$
And divide:
$\left(\sin x \sin y + \cos x \cos y\right) + 1 = \frac{{a}^{2} + {b}^{2}}{2}$
And subtract:
$\sin x \sin y + \cos x \cos y = \frac{{a}^{2} + {b}^{2}}{2} - 1$
Finally, since $\cos \left(x + y\right) = \cos x \cos y + \sin x \sin y$, we have:
$\cos \left(x + y\right) = \frac{{a}^{2} + {b}^{2}}{2} - 1$
Aug 21, 2016
Given
$\sin x + \sin y = a \ldots \ldots . \left(1\right)$
$\cos x + \cos y = b \ldots \ldots . \left(2\right)$
Squaring and adding (1) & (2)
${\left(\cos x + \cos y\right)}^{2} + {\left(\sin x + \sin y\right)}^{2} = {a}^{2} + {b}^{2}$
$\implies 2 \left(\cos x \cos y + \sin x \sin y\right) + 2 = {a}^{2} + {b}^{2}$
$\implies 2 \cos \left(x - y\right) = {a}^{2} + {b}^{2} - 2. \ldots \left(3\right)$
Squaring and Subtracting (1) from(2)
${\left(\cos x + \cos y\right)}^{2} - {\left(\sin x + \sin y\right)}^{2} = {b}^{2} - {a}^{2}$
$\implies 2 \cos \left(x + y\right) + {\cos}^{2} x - {\sin}^{2} x + {\cos}^{2} y - {\sin}^{2} y = {b}^{2} - {a}^{2}$
$\implies 2 \cos \left(x + y\right) + \cos 2 x + \cos 2 y = {b}^{2} - {a}^{2}$
$\implies 2 \cos \left(x + y\right) + 2 \cos \left(x + y\right) \cos \left(x - y\right) = {b}^{2} - {a}^{2}$
$\implies \cos \left(x + y\right) \left(2 + 2 \cos \left(x - y\right)\right) = {b}^{2} - {a}^{2}$
($\text{From (3) } 2 \cos \left(x - y\right) = {a}^{2} + {b}^{2} - 2$)
$\implies \cos \left(x + y\right) \left(2 + {b}^{2} + {a}^{2} - 2\right) = {b}^{2} - {a}^{2}$
$\implies \cos \left(x + y\right) \left({b}^{2} + {a}^{2}\right) = {b}^{2} - {a}^{2}$
$\implies \cos \left(x + y\right) = \frac{{b}^{2} - {a}^{2}}{{b}^{2} + {a}^{2}}$
Aug 21, 2016
$\cos \left(x + y\right) = \frac{{b}^{2} - {a}^{2}}{{b}^{2} + {a}^{2}}$.
#### Explanation:
$\sin x + \sin y = a \Rightarrow 2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) = a \ldots \ldots \ldots \left(1\right)$.
$\cos x + \cos y = b \Rightarrow 2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) = b \ldots \ldots \ldots . \left(2\right)$.
Dividing $\left(1\right)$ by $\left(2\right)$, we have, $\tan \left(\frac{x + y}{2}\right) = \frac{a}{b}$.
Now, $\cos \left(x + y\right) = \frac{1 - {\tan}^{2} \left(\frac{x + y}{2}\right)}{1 + {\tan}^{2} \left(\frac{x + y}{2}\right)}$
$= \frac{1 - {a}^{2} / {b}^{2}}{1 + {a}^{2} / {b}^{2}} = \frac{{b}^{2} - {a}^{2}}{{b}^{2} + {a}^{2}}$.
Enjoy Maths.!
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Problem 4
# If $$A$$ has $$a$$ elements and $$B$$ has $$b$$ elements, how many elements are in $A \times B$ ? Explain your answer.
Expert verified
The number of elements in the Cartesian product $$A \times B$$ is equal to $$a \times b$$, where $$a$$ is the number of elements in A and $$b$$ is the number of elements in B. This is because each element in A can be paired with each element in B, resulting in a total of $$a \times b$$ ordered pairs.
See the step by step solution
## Step 1: Understanding Cartesian Product
A Cartesian product of two sets A and B, denoted as $$A \times B$$, is the set of all possible ordered pairs where the first element is from A and the second element is from B. The Cartesian product of two sets results in a new set containing all possible combinations of elements from the two sets.
## Step 2: Counting Elements in A
Set A has $$a$$ elements. For each element in A, there will be an ordered pair formed with the elements of B.
## Step 3: Counting Elements in B
Set B has $$b$$ elements. For each element in B, there will be an ordered pair formed with the elements of A.
## Step 4: Counting Elements in the Cartesian Product
Since there are $$a$$ elements in A and $$b$$ elements in B, there will be $$a \times b$$ combinations for the ordered pairs. Hence the number of elements in the Cartesian product $$A \times B$$ is equal to $$a \times b$$.
## Step 5: Result
The Cartesian product $$A \times B$$ has $$a \times b$$ elements, where $$a$$ is the number of elements in A and $$b$$ is the number of elements in B.
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