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# Is the point 3/4 on the unit circle?
## Is the point 3/4 on the unit circle?
The point (3,4) is on the circle of radius 5 at some angle θ. Find cos(θ) and sin(θ). There are a few cosine and sine values which we can determine fairly easily because the corresponding point on the circle falls on the x or y axis.
## Is each point on the unit circle?
In the case of a unit circle, the equation is x2 + y2 = 1. This equation shows that the points lying on the unit circle have to have coordinates (x- and y-values) that, when you square each of them and then add those values together, equal 1. When you square each coordinate and add those values together, you get 1.
How do you find the coordinates of trigonometric points?
We can find the coordinates of any point on the unit circle. Given any angle t , we can find the x – or y -coordinate at that point using x=cos t x = cos t and y=sin t y = sin t .
### Is the point 1 1 on the unit circle?
Explanation: The unit circle is by definition a circle with radius equal to 1 and center (0,0), so the distance of the points (x,y) in that circle to the center is equal to 1, hence by the distance formula : √a2+b2=1⟺a2+b2=1. If it equals 1, it is on the unit circle.
### What is the unit circle equation?
The unit circle is a circle centered at the origin, with a radius of one. The equation of the unit circle is u2 + v2 = 1.
How do you find the missing point on a unit circle?
How to Find a Missing Coordinate on a Unit Circle
1. Substitute the x-coordinate value into the unit-circle equation.
2. Square the x-coordinate and subtract that value from each side.
3. Take the square root of each side.
#### How do you determine if points are inside or outside a circle?
If the distance is greater than the radius, the point lies outside. If it’s equal to the radius, the point lies on the circle. And if it’s less than the radius, you guessed it right, the point will lie inside the circle.
#### What is the formula for an unit circle?
The unit circle formula is: Unit Circle Formula (Equation) x 2 +y 2 =1 . Where x and y are the coordinate values.
How do you use the unit circle?
The unit circle is a circle, centered at the origin, with a radius of 1. Recall from conics that the equation is x 2 +y 2 =1. This circle can be used to find certain “special” trigonometric ratios as well as aid in graphing. There is also a real number line wrapped around the circle that serves as the input value when evaluating trig functions.
## What is the reference number for unit circle?
The reference number is identified as the shortest distance along the unit 360 degrees circle between the x-axis and the terminal or ending point of the circle angle. Finding the reference number requires you to understand circle angles and the radians of a circle in terms of pi.
## Is sin x or y unit circle?
The Unit Circle. For instance, in the unit circle, for any angle θ, the trig values for sine and cosine are clearly nothing more than sin (θ) = y and cos (θ) = x. Working from this, you can take the fact that the tangent is defined as being tan (θ) = y/x, and then substitute for x and y to easily prove that the value of tan (θ)… |
S k i l l
i n
A L G E B R A
27
Simplest form
2nd level
Simplifying the square roots of powers
WE SAY THAT A SQUARE ROOT RADICAL is simplified, or in its simplest form, when the radicand has no square factors.
A radical is also in simplest form when the radicand is not a fraction.
Example 1. 33, for example, has no square factors. Its factors are 3 · 11, neither of which is a square number. Therefore, is in its simplest form.
Example 2. Extracting the square root. 18 has the square factor 9.
18 = 9 · 2.
Therefore, is not in its simplest form. We have,
=
We may now extract, or take out, the square root of 9:
= = 3.
is now simplified. The radicand no longer has any square factors.
The justification for taking out the square root of 9, is this theorem:
The square root of a product
is equal to the product of the square roots
of each factor.
We will prove that when we come to rational exponents, Lesson 29. Here is a simple illustration:
As for , then, it is equal to the square root of 9 times the square root of 2, which is irrational. 3.
Example 3 Simplify .
Solution. = = 5.
75 has the square factor 25. And the square root of (25 times 3)
is equal to the square root of 25 times the square root of 3.
is now simplified.
Example 4. Simplify .
Solution. We have to factor 42 and see if it has any square factors. We can begin the factoring in any way. For example,
42 = 6 · 7
We can continue to factor 6 as 2 · 3, but we cannot continue to factor 7 because 7 is a prime number. Therefore,
42 = 2 · 3 · 7
We now see that, because no factor is repeated, 42 has no square factors. therefore is in its simplest form.
Compare Example 1 and Problem 2 of the previous Lesson.
· 3 · 7 is the prime factorization of 42.
2, 3, and 7 are prime numbers.
Example 5. Simplify .
Solution. We must look for square factors, which will be factors that are repeated.
180 = 2 · 90 = 2 · · 45 = 2 · 2 · · 5 = 2 · 2 · · 3 · 5
Therefore,
= 2 · 3 = 6.
Problem 1. To simplify a radical, why do we look for square factors?
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
In order to take its square root out of the radical.
Problem 2. Which is correct?
Problem 3. Simplify the following. Do that by inspecting each radicand for a square factor: 4, 9, 16, 25, and so on.
a) =
b) = = = 5
c) = = = 3
d) = = 7
e) = = 4
f) = = 10
g) = = 5
h) = = 4
Reduce to lowest terms.
a) 2 = 2 = 2 =
b) 3 = 3 = 3 = 2
c) 2 = The radical is in its simplest form. The fraction cannot be reduced.
Similar radicals have the same radicand. We add them as like terms.
7 + 2 + 5 + 6 − = 7 + 2 + 6 + 5 − = 7 + 8 + 4.
2 and 6 are similar, as are 5 and −. We combine them by adding their coefficients.
In practice, it is not necessary to change the order of the terms. The student should simply see which radicals have the same radicand.
As for 7, it does not "belong" to any radical.
Problem 5. Simplify each radical, then add the similar radicals.
a) + = 3 + 2 = 5
b) 4 − 2 + = 4 − 2 + = 4· 5 − 2· 7 + = 20 − 14 + = 7
c) 3 + − 2 = 3 + − 2 = 3· 2 + 2 − 2· 4 = 6 + 2 − 8 = 2 − 2
d) 3 + + = 3 + + = 3 + 2 + 3 = 3 + 5
e) 1 − + = 1 − + = 1 − 8 + 3 = 1 − 5
Problem 6. Simplify the following.
a) 2 = 2 = 2 − , on dividing every term in the numerator by 2.
Compare Example 4 here.
To see that 2 was a factor of the radical, you first have to simplify the radical. Compare Problem 4.
b) 5 = 5 = 2 +
c) 6 = 6 = 3 on dividing every term by 2.
2nd Level
First Lesson on Radicals
Next Lesson: Multiplying and dividing radicals
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# Lesson 2
Representations of Equal Groups of Fractions
## Warm-up: Number Talk: Three, Six, Nine, Twelve (10 minutes)
### Narrative
This Number Talk encourages students to use their knowledge of multiplication facts, properties of operations, and the structure of the given expressions to mentally solve problems. The reasoning elicited here will be helpful in upcoming lessons as students find products of whole numbers and non-unit fractions (such as $$3 \times \frac{6}{10}$$ or $$6 \times \frac{9}{4}$$).
### Launch
• Display one expression.
• “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Find the value of each expression mentally.
• $$3 \times 6$$
• $$3 \times 9$$
• $$6 \times 9$$
• $$12 \times 9$$
### Activity Synthesis
• “What did you notice about the factors in all of the expressions?” (They are all multiples of 3.)
• “Did noticing that all the factors are multiples of 3 help you find the values?” (Sample responses:
• Yes, I was able to think of “3 more groups of something.”
• Yes, it helped me see how the factors were related, which helped me reason about the products.
• No, it didn't, but it made me think that the values would be multiples of 3 as well.)
## Activity 1: Card Sort: Expressions and Diagrams (25 minutes)
### Narrative
In this activity, students interpret multiplication expressions and diagrams as the number of groups and amount in each group and match representations of the same quantity. They then use their insight from the matching activity to generate diagrams for expressions without a match and to find their values (MP2).
MLR8 Discussion Supports. Students should take turns finding a match and explaining their reasoning to their partner. Display the following sentence frames for all to see: “I noticed _____ , so I matched . . .” Encourage students to challenge each other when they disagree.
Engagement: Develop Effort and Persistence. Chunk this task into more manageable parts. Give students a subset of the cards to start with and introduce the remaining cards once students have completed their initial set of matches.
Supports accessibility for: Organization, Conceptual Processing
### Required Materials
Materials to Copy
• Expressions and Diagrams
### Required Preparation
• Create a set of cards from the blackline master for each group of 2.
### Launch
• Groups of 2
• Give each group a set of cards from the blackline master.
### Activity
• “Work with your partner to match each expression to a diagram that represents the same equal-group situation and the same amount.”
• “Be prepared to explain how you know the two representations belong together.”
• 5 minutes: partner work time
• Monitor for students who reason about the number of groups and amount in each group as they match.
• Pause for a discussion. Select students to share their matches and reasoning.
• Highlight reasoning that clearly connects one factor in the expression to the number of groups and the other factor to the size of each group.
• “Now you will complete an unfinished diagram for $$7 \times \frac{1}{8}$$ and then draw a new diagram for an expression without a match.”
• 5 minutes: independent work time
### Student Facing
Your teacher will give you a set of cards with expressions and diagrams.
1. Match each expression with a diagram that represents the same quantity.
2. Record each expression without a match.
3. Han started drawing a diagram to represent $$7 \times \frac{1}{8}$$ and did not finish. Complete his diagram. Be prepared to explain your reasoning.
4. Choose one expression that you recorded earlier that didn't have a match.
Draw a diagram that can be represented by the expression. What value do the shaded parts of your diagram represent?
### Student Response
If students are not yet matching expressions to appropriate diagrams, consider asking them to compare the diagrams for $$5 \times 3$$ and $$5 \times \frac{1}{3}$$ and reason about the number of groups and the size of each group. Consider asking: “How are these alike? How are they different?”
### Activity Synthesis
• “What was missing from Han’s diagram? How do you know?” (4 more groups of $$\frac{1}{8}$$ were missing, because $$7 \times \frac{1}{8}$$ means 7 groups of $$\frac{1}{8}$$and there are only 3 in Han’s diagram.)
• “If the expression was for 7 groups of $$\frac{1}{3}$$ instead of $$\frac{1}{8}$$ how would Han’s diagram change?” (Each rectangle representing 1 would have 3 equal parts with 1 shaded.)
• Select students to share the diagrams they drew for the expressions without a match. Ask them to point out the number of groups and size of each group in each diagram.
## Activity 2: Different Representations (10 minutes)
### Narrative
This activity prompts students to use their earlier observations to generate a diagram or expression that represents equal groups of unit fractions when one or the other is given. In one of the problems, only the total quantity ($$\frac{7}{2}$$) is given, so students need to reason in about the number of groups and the size of each group that could lead to this value. Finally, they analyze two different ways of representing $$4 \times\frac{1}{3}$$ with a diagram, which further illustrates that the value of the expression is $$\frac{4}{3}$$.
### Launch
• Groups of 2
• “Turn to a partner and explain what needs to be done to complete the first problem.”
### Activity
• Complete the first problem independently. Afterwards, pause for a class discussion.
• 5 minutes: independent work time
• Pause to discuss the fraction $$\frac{7}{2}$$ in the first problem.
• “How did you know what diagram and expression would have the value $$\frac{7}{2}$$?” (Sample response:
• For the diagram, the numerator, 7, is the number of groups, and the denominator, 2, shows how many parts are in 1 whole.
• For the expression, I multiplied a whole number and a fraction. The whole number was the same as the number in the numerator of $$\frac{7}{2}$$ and the fraction has the same number as the denominator of $$\frac{7}{2}$$.)
• Work on the last problem with your partner.
• 5 minutes: partner work time
### Student Facing
1. Write a multiplication expression that represents the shaded parts of the diagram. Then, find the value of the expression.
Diagram:
Expression:
Value:
2. Draw a diagram that the expression $$6 \times \frac{1}{3}$$ could represent. Then, find the value of the expression.
Diagram:
Expression: $$6 \times \frac{1}{3}$$
Value:
3. Draw a diagram and write an expression that gives the value $$\frac{7}{2}$$.
Diagram:
Expression:
Value: $$\frac{7}{2}$$
1. To represent $$4 \times \frac{1}{3}$$, Diego drew this diagram:
Elena drew this diagram:
Are they representing the same expression and value? Explain or show how you know.
### Student Response
Students may be unsure about how to begin writing expressions for fractions. Remind students that the fraction will be written as a whole number times a unit fraction. Consider asking: “How might this help to write the expression?”
### Activity Synthesis
• See lesson synthesis.
## Lesson Synthesis
### Lesson Synthesis
“Today we analyzed expressions and diagrams that represent equal groups and created some of these representations.”
Display or sketch these diagrams:
“How do we know which diagram represents $$3 \times \frac{1}{5}$$? Where do we see each number in the diagram?” (B represents $$3 \times \frac{1}{5}$$ because it shows 3 groups of $$\frac{1}{5}$$.)
“What expression does the other diagram represent?” (A represents $$5 \times \frac{1}{3}$$, because it shows 5 groups with $$\frac{1}{3}$$ in each group.)
“What is the value of $$3 \times \frac{1}{5}$$? How do we know?” ($$\frac{3}{5}$$. We can count the number of shaded fifths and see that there are 3.)
“What is the value of $$5 \times \frac{1}{3}$$?” ($$\frac{5}{3}$$) |
ACT Math : How to find excluded values
Example Questions
Example Question #1 : How To Find Excluded Values
Which of the following provides the complete solution set for ?
No solutions
Explanation:
The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as . Thus all values except for 2 will work.
Example Question #2 : How To Find Excluded Values
Find the excluded values in the following algebraic fraction.
Explanation:
In order to find excluded values, find the values that make the denominator equal zero.
To do this, you must factor the denominator:
Now, set each part equal to zero and solve for .
Example Question #3 : How To Find Excluded Values
Find the excluded values for for the following algebraic fraction.
Explanation:
In order to find the excluded values for x, find the values that make the denominator zero.
In order to find the zeroes, factor the denominator:
Now, set each part equal to zero.
Example Question #4 : How To Find Excluded Values
Find the excluded values for for the following algebraic fraction.
Explanation:
To find the excluded values, find the values that make the denominator zero.
Example Question #5 : How To Find Excluded Values
Find the excluded values for for the following algebraic fraction.
Explanation:
To find the excluded values, find the values that make the denominator equal zero.
To do that, set the denominator equal to zero and solve for
Begin by taking each multiple in the denominator and setting it equal to zero.
Example Question #6 : How To Find Excluded Values
Find the excluded values for in the following algebraic fraction.
Explanation:
To find the excluded values, find the values that make the denominator equal zero.
To do that, begin by factoring the denominator.
Now, set each part equal to zero.
Now, solve for .
Example Question #7 : How To Find Excluded Values
Find the excluded values for in the following algebraic fraction.
Explanation:
To find the excluded values, find the -values that make the denominator equal zero.
To do this, set the denominator equal to zero:
Next, solve for x.
Example Question #8 : How To Find Excluded Values
Find the excluded values for in the following algebraic fraction.
Explanation:
To find the excluded values, find the -values that make the denominator equal zero.
1) set the denominator equal to zero.
2) solve for .
Example Question #1 : How To Find Excluded Values
Find the excluded values of the following algebraic fraction
The numerator cancels all the binomials in the denomniator so ther are no excluded values.
Explanation:
To find the excluded values of a algebraic fraction you need to find when the denominator is zero. To find when the denominator is zero you need to factor it. This denominator factors into
so this is zero when x=4,7 so our answer is
Example Question #10 : How To Find Excluded Values
Which of the following are answers to the equation below?
I. -3
II. -2
III. 2
III only
II only
I, II, and III
I only
II and III
III only
Explanation:
Given a fractional algebraic equation with variables in the numerator and denominator of one side and the other side equal to zero, we rely on a simple concept. Zero divided by anything equals zero. That means we can focus in on what values make the numerator (the top part of the fraction) zero, or in other words,
The expression is a difference of squares that can be factored as
Solving this for gives either or . That means either of these values will make our numerator equal zero. We might be tempted to conclude that both are valid answers. However, our statement earlier that zero divided by anything is zero has one caveat. We can never divide by zero itself. That means that any values that make our denominator zero must be rejected. Therefore we must also look at the denominator.
The left side factors as follows
This means that if is or , we end up dividing by zero. That means that cannot be a valid solution, leaving as the only valid answer. Therefore only #3 is correct. |
# Ratios and Rates Ratios and Rates ratio is
• Slides: 26
Ratios and Rates
Ratios and Rates ratio – is a comparison of two numbers or more values. Example: 1: 3
EXAMPLE: IF WE COMPARE THE NUMBER OF GIRLS WITH THE NUMBER OF BOYS WE CAN SAY THAT: THERE IS 1 BOY AND 3 GIRLS , SO: 1: 3 (FOR EVERY ONE BOY THERE ARE 3 GIRLS) OR (THERE ARE 3 GIRLS FOR EVERY ONE BOY) 3: 1 )
Ratios and Rates The order of the numbers in a ratio should match the order of the words in a ratio. apples to oranges 2 to 3
Ratios and Rates Ratios can be written in 3 different ways. red to black 3 to 1 3: 1 3/1
Ratios and Rates Ratios can show comparisons between: a) part to part red to green 1 to 2 green to red 2 to 1
Ratios and Rates Ratios can show comparisons between: b) part to whole red to whole 1 to 3 green to whole 2 to 3
Ratios and Rates Ratios can show comparisons between: c) whole to part whole to red 3 to 1 whole to green 3 to 2
Ratios and Rates Compare octagon to all shapes 1 to 8 1: 8 1/8
Ratios and Rates Compare all shapes to trapezoid 8 to 1 8: 1 8/1
Ratios and Rates Compare pentagons to hexagons 2 to 3 2: 3 2/3
NOW, SOME EXERCISES
Ratios and Rates Give each ratio in three forms. 8 roses out of 24 flowers 3 dogs out of 12 animals 3 : 12
SOLVE 1) In the following diagram What is the ratio of orange squares to white squares?
2. WHAT IS THE RATIO OF GREEN BALLS TO RED BALLS? 3. A CLASS OF 32 STUDENTS HAS 12 GIRLS. WHAT IS THE RATIO OF GIRLS TO BOYS?
GIVE EACH RATIO BY SHOWING COMPARISONS BETWEEN : Part to part Part to whole Whole to part
Equivalent Ratios You can have equivalent ratios by dividing or multiplying the numerator and denominator by by the same value ÷ 8 = 4 : 5 is the same as 4× 2 : 5× 2 = 8 : 10
EXAMPLE: A Recipe for pancakes uses 3 cups of flour and 2 cups of milk. So the ratio of flour to milk is 3 : 2 To make pancakes for a LOT of people we might need 4 times the quantity, so we multiply the numbers by 4: 3× 4 : 2× 4 = 12 : 8 In other words, 12 cups of flour and 8 cups of milk. The ratio is still the same, so the pancakes should be just as yummy.
Rate – It is a ratio that compares different kinds of units by division to obtain a unit rate or rate per unit.
To express a ratio as a rate per unit (1), write the ratio as a fraction and then divide the numerator by the denominator. 395 miles in 5 hours miles = hours miles hours 79 miles per hour
Ratios and Rates Express each ratio as a rate. 5 inches of rain in 30 days ÷ 5 = inch of rain per day ÷ 5 120 words in 3 minutes = words per minute 40 words per minute |
Let us learn some important definitions before we get into the concept of real numbers Class 10:
• A natural number is a counting number. Hence a set of natural number can be shown as N = {1, 2, 3, . .}
• A whole number is all the natural numbers including 0. Therefore a set of whole numbers is W = {0, 1, 2, ..}
• An integer is a set of all positive and negative whole numbers. The set of integers can be written as Z = { -4, -3, -2, -1, 0, 1, 2, . . }. The natural numbers (excluding zero) are known as positive integers. When a number gives zero on being added to its corresponding positive value is known as a negative integer. When we consider zero along with the natural numbers, we call it non-negative integers.
• A rational number is any number that can be expressed in p / q form where both the numerator and denominator are integers and the value of q is positive. This includes all integers, natural numbers and rational number. Any two rational numbers has infinite rational numbers between them. The rational number can either be terminating decimal or non-terminating decimal, which can again be recurring or non-recurring in nature. There are certain operations of rational numbers that need to be accounted for. The sum of two rational numbers is a rational number. The same hold for difference and product too. This may or may not be true for division.
• An irrational number is a number that cannot be expressed in p / q format. We can represent the set of irrational numbers on a number line with the help of Pythagoras theorem.
## What are real numbers?
A collection of rational numbers and irrational numbers make up the set of real number. A real number can be expressed on the number line and has some specific properties. They satisfy:
• The commutative law of addition. That is, when a and b are two real numbers then a + b = b + a. For example 1 + 3 = 3 + 1 = 4
• The commutative law of multiplication. That is, when a and b are two real numbers then a x b = b x a. For example 1 x 3 = 3 x 1 = 3
• The associative law of addition. That is, when a, b and c are three real numbers then a + (b + c) = (a + b) + c. For example 1 + (3 + 4) = (1 + 3) + 4 = 8
• The associative law of multiplication. That is, when a, b and c are three real numbers then a x (b x c) = (a x b) x c. For example, 1 x (3 x 4) = (1 x 3) x 4 = 12
• The law of distribution. That is, when a, b, c are three real numbers then
a x (b +c) = (a x b) + (a x c). For example 1 x (3 + 4) = (1 x 3) + (1 x 4)
= 7
There are some laws of exponents as well that is demonstrated by real numbers. They are:
1. ap x aq = a (p+q)
2. ap / aq = a(p-q)
3. (ap)q = apq
4. ap x bp = (ab)p
Here a and b, both are real numbers.
## What is Euclid’s division algorithm?
There is a value q and r for every set of positive integer such that
a = bq +r, where 0 ≤ r < b.
There are specific names for each term, a is dividend, b is divisor, q is quotient and r is remainder. It is a method used to find the highest common factor.
For example, when we need to use the Euclid’s algorithm to find the HCF of 135 and 225.
Since 225 > 135, we can write 225 = 135 x 1 +90
Since 90>0, the remainder is 45. We apply the division rule again to get
135 = 90 x 1 +45
We then consider 90 to be the new divisor with the remainder as 45 and on applying the algorithm again we get, 90 = 2 x 45 + 0. Here, since the remainder is 0, the HCF = 45, which is the divisor at this particular stage.
Some solved questions
1. Prove that √2 is irrational number
Answer: Let us assume that it is rational number
Then, √2=p/q
Where, p and q are co-primes.
Or, q√2=p
Squaring both sides, we get
2q2=p2
Hence, 2 divide p2.
From theorem we know that,
2 will divide p also. p = 2c, where c is any constant
2q2=4c2
or q2=2c2
So q is divisible by 2 also
So 2 divide both p and q, which contradicts our assumption.
So √2 is an irrational number.
1. Show that a positive odd integer is in either of the following forms: 4q+1 or 4q+3 where q is an integer.
Answer: let a be any positive off integer. On applying the division algorithm with a, b = 4, we get that the remainder can be 0, 1, 2, or 3.
So, a can be 4q, 4q+1, 4q+2 or 4q+3 where the quotient is q.
Since a is odd, it cannot be 4q or 4q+2 (since both are divisible by 2)
hence, an odd integer is always of form 4q+1 or 4q+3.
1. Express 140 and 5005 as a product of its prime factor:
140 = 2 x 2 x 5 x 7 = 22 x 5 x 7
5005 = 5 x 7 x 11 x 13
1. Find the LCM and HCF of 26 and 91
Answer: 26 = 2 x 13
91 = 7 x 13
HCF = 13
LCM = 2 x 7 x 13 = 182
Product of 2 numbers = 2366
HCF X LCM = 2366
Hence, proved
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## Section1.5Reflectors
When we decompose matrices into products of other matrices, we often wish to create matrices with many zero entries. A Householder matrix is a unitary matrix which transforms a vector into a vector of the same size that is a nonzero multiple of the first column of an identity matrix, thus creating a vector with just a single nonzero entry. A typical application is to “zero out” entries of a matrix below the diagonal, column-by-column, in order to achieve a triangular matrix.
###### Definition1.5.1.
Given a nonzero vector $\vect{v}\in\complex{n}\text{,}$ the Householder matrix for $\vect{v}$ is
The vector $\vect{v}$ is called the Householder vector.
A Householder matrix is both Hermitian and unitary.
Our aim with a Householder matrix is to convert a vector $\vect{x}$ into a scalar multiple of the first column of the identity matrix, $\vect{e}_1\text{.}$ Which Householder vector should we choose for constructing such a matrix, and which multiple will we get? It is an instructive exercise to reverse-engineer the choice by setting $P\vect{x}=\alpha\vect{e}_1$ (Exercise Checkpoint 1.5.9). Instead, we give the answer and prove that it does the desired job.
We first establish an unmotivated identity.
Then
Consider the vector $\vect{x}$ and construct the Householder vector $\vect{v}\text{.}$
\begin{align*} \vect{x}&=\colvector{4\\4\\7}&\vect{v}&=\vect{x}-9\vect{e}_1=\colvector{-5\\4\\7} \end{align*}
Then the Householder matrix for $\vect{v}$ is
\begin{equation*} P = \begin{bmatrix} \frac{4}{9} & \frac{4}{9} & \frac{7}{9} \\ \frac{4}{9} & \frac{29}{45} & -\frac{28}{45} \\ \frac{7}{9} & -\frac{28}{45} & -\frac{4}{45} \end{bmatrix} \end{equation*}
We can check that the matrix behaves as we expect.
\begin{equation*} P\vect{x}=\colvector{9\\0\\0} \end{equation*}
A Householder matrix is often called a Householder reflection. Why? Any Householder matrix, when thought of as a mapping of points in a physical space, will fix the elements of a hyperplane and reflect any other points about that hyperplane. To see this, consider any vector $\vect{w}$ and compare it with its image, $P\vect{w}$
\begin{align*} P\vect{w}-\vect{w} &=I_n\vect{w} -\left(\frac{2}{\innerproduct{\vect{v}}{\vect{v}}}\right)\vect{v}\adjoint{\vect{v}}\vect{w} -\vect{w}\\ &=-\frac{2}{\innerproduct{\vect{v}}{\vect{v}}}\vect{v}\innerproduct{\vect{v}}{\vect{w}}\\ &=-\frac{2\innerproduct{\vect{v}}{\vect{w}}}{\innerproduct{\vect{v}}{\vect{v}}}\vect{v} \end{align*}
So this difference is always a scalar multiple of the Householder vector $\vect{v}\text{,}$ and thus every point gets moved in the same direction, the direction described by $\vect{v}\text{.}$ Which points are fixed by $P\text{?}$ The difference above is the zero vector precisely when the scalar multiple is zero, which will happen when $\innerproduct{\vect{v}}{\vect{w}}=0\text{.}$ So the set of points/vectors which are orthogonal to $\vect{v}$ will be unmoved. This is a subspace of dimension one less than the full space, which are typically described by the term hyperplanes.
To be more specific, consider the specific situation of Example Example 1.5.5, viewed in $\real{3}\text{.}$ The hyperplane is the subspace orthogonal to $\vect{v}\text{,}$ or the two-dimensional plane with $\vect{v}$ as its normal vector, and equation $-5x+4y+7z=0\text{.}$ The points $(4,4,7)$ and $(9,0,0)$ lie on either side of the plane and are a reflection of each other in the plane, by which we mean the vector $(4,4,7)-(9,0,0)=(-5,4,7)$ is perpendicular (orthogonal, normal) to the plane.
Our choice of $\vect{v}$ can be replaced by $\vect{v}=\vect{x}+\norm{\vect{x}}\vect{e}_1\text{,}$ so in the previous example we would have $\vect{v}=\colvector{13\\4\\7}\text{,}$ and then $P$ would take $\vect{x}$ to $\colvector{-9\\0\\0}\text{.}$ This would be a reflection across the plane with equation $13x+4y+7z=0\text{.}$ Notice that the normal vector of this plane is orthogonal to the normal vector of the previous plane, which is not an accident (Exercise Checkpoint 1.5.6).
As a practical matter, we would choose the Householder vector which moves $\vect{x}$ the furthest, to get better numerical behavior. So in our example above, the second choice would be better, since $\vect{x}$ will be moved a distance $2\norm{\vect{v}}$ and the second $\vect{v}$ has a larger norm.
In the real case, we have two choices for a Householder vector which will “zero out” most of a vector. Show that these two vectors, $\vect{x}+\norm{\vect{x}}\vect{e}_1$ and $\vect{x}-\norm{\vect{x}}\vect{e}_1\text{,}$ are orthogonal to each other.
Prove the following generalization of Theorem Theorem 1.5.4. Given a vector $\vect{x}\in\complex{n}\text{,}$ define $\rho=\vectorentry{\vect{x}}{1}/\modulus{\vectorentry{\vect{x}}{1}}$ and $\vect{v}=\vect{x}\pm\rho\norm{\vect{x}}\vect{e}_1$ and let $P$ be the Householder matrix for the Householder vector $\vect{v}\text{.}$ Then $P\vect{x}=\mp\rho\norm{\vect{x}}\vect{e}_1\text{.}$
Hint
You can establish the same identity as in the first part of the proof of Theorem Theorem 1.5.4.
Suppose that $P$ is a Householder matrix of size $n$ and $\vect{b}\in\complex{n}$ is any vector. Find an expression for the matrix-vector product $P\vect{b}$ which will suggest a way to compute this vector with fewer than the $\orderof{2n^2}$ operations required for a general matrix-vector product.
Begin with the condition that a Householder matrix will accomplish $P\vect{x}=\alpha\vect{e}_1$ and “discover” the choice for the Householder vector described in Theorem Theorem 1.5.4. Hint: The condition implies that the Householder vector $\vect{v}$ is in the span of $\set{\vect{x}, \vect{e}_1}\text{.}$ |
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Fractions, Decimals, Percents, & Ratios
# Fractions, Decimals, Percents, & Ratios
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In This Chapter…
Four Ways to Express Parts of a Whole
Convert 0.25 to 25%: Move The Decimal Point Two Places Right
Convert 0.25 or 25% to 1/4: Put 25 over 100 and Simplify
Convert 1/4 to 0.25 or 25%: Long-Divide 1 by 4
Multiply a Decimal by a Power of Ten: Shift the Decimal Point
Add or Subtract Decimals: Line Up the Decimal Points
Multiply Two Decimals: Ignore Decimal Points At First
Multiply a Decimal and a Big Number: Trade Decimal Places
Divide Two Decimals: Move Points in the Same Direction To Kill Decimals
“20% Of \$55” = 0.2 × \$55
Percent Change: Divide Change in Value by Original Value
Percent Of a Percent Of: Multiply Twice
Ratio: One Quantity Divided By Another
Part : Part : Whole Ratios—Write Part + Part: Whole and Use the Unknown Multiplier
Chapter 5:
Fractions, Decimals,
Percents, & Ratios
In This Chapter:
• Relationships among fractions, decimals, percents, & ratios
Four Ways to Express Parts of a Whole
Say you have the shaded part of this orange. You can express how much you have in four ways.
A. You have of the orange.
B. You have 0.25 of the orange.
C. You have 25% of the orange.
D. The ratio of your piece to the whole orange is 1 to 4, or 1:4.
Fraction
Decimal
Percent
Ratio
Any of these four forms can express a part-to-whole relationship. The main difference between the
forms is how you think about the whole.
= 1 out of 4 pieces of the whole.
0.25 = 0.25 of the whole itself.
25% = 25 out of 100 pieces of the whole.
“1 to 4” = 1 out of 4 pieces of the whole.
In other words, what is each form “out of”? What is the whole that you are dividing by?
Fractions are out of the denominator (4 in this case).
Decimals are out of 1 (the whole). You've already done the division.
Percents are out of 100. Percent literally means “per hundred,” or divided by 100.
Ratios are out of the second term in the ratio (4 in this case). So ratios are very similar to
fractions, and you can quickly rewrite any ratio as a fraction. For instance, a ratio of 3 to 7 is
. Another word for ratio is proportion.
Which form is most useful depends on the problem at hand. You might say any of the following:
The container is full.
The container is filled to 0.5 of its capacity.
The container is 50% full.
The ratio of the contents of the container to its capacity is 1 to 2.
By the way, the “part” can be greater than the whole.
I ate of a box of cereal. (I ate more than one box.)
I ate 1.25 boxes of cereal.
I ate 125% of a box of cereal.
The ratio of what I ate to a whole box of cereal was 5 to 4.
Convert 0.25 to 25%: Move The Decimal Point Two Places Right
Decimals are out of 1. Percents are out of 100. So, to convert a decimal to a percent, move the
decimal point two places to the right. Add zeroes if necessary.
0.53 = 53%
0.4 = 40%
0.03 = 3%
1.7 = 170%
A percent might still contain a visible decimal point when you're done.
0.4057 = 40.57%
0.002 = 0.2%
0.0005 = 0.05%
Just keep track of which decimal is part of the percent and which one is the “pure” decimal.
To convert a percent to a decimal, go in reverse. That is, move the decimal point two places to the
left. If the decimal point isn't visible, it's actually just before the % sign. Add zeroes if necessary as
you move left.
39% = 39.% = 0.39
60% = 0.60 = 0.6
8% = 0.08
225% = 2.25
13.4% = 0.134
0.7% = 0.007
0.001% = 0.00001
If you…
Want to convert a decimal to a
percent
Want to convert a percent to a
decimal
Then you…
Move the decimal point two places to the
right
Move the decimal point two places to the left
Like this
0.036 = 3.6
41.2% =
0.412
1. Convert 0.035 to a percent.
Answer can be found on page 225.
Convert 0.25 or 25% to 1/4: Put 25 over 100 and Simplify
The decimal 0.25 is twenty-five one-hundredths. So rewrite that as 25 over 100:
Now simplify by cancelling common factors from top and bottom.
When you convert a decimal to a fraction, put a power of 10 (10, 100, 1,000, etc.) in the
denominator of the fraction. Which power of 10? It depends on how far the decimal goes to the
right.
Put as many zeroes in your power of 10 as you have digits to the right of the decimal point.
0.3
=
Zero point
three
is
0.23
three tenths, or three
over ten
=
Zero point two
is
three
0.007
twenty-three onehundredths
=
Zero point zero zero
is
seven
seven onethousandths
Don't forget to cancel.
In the second case, you cancel 125 from top and bottom, leaving 3 and 8.
When you put the digits on top, keep any zeroes in the middle, such as the 0 between the 1 and the 2 in
0.0102. Otherwise, drop any zeroes (such as the 0's to the left of the 1).
To convert a percent to a fraction, write the percent “over one hundred.” Remember that
percent literally means “per hundred.”
Alternatively, you can first convert the percent to a decimal by moving the decimal place. Then
follow the process given earlier.
If you don't convert to a decimal first, be sure to write the fraction over 100.
That fraction ultimately reduces to
further on.
If you…
, but we'll look at the process of dividing decimals a little
Then you…
Like this
Want to convert a
decimal to a fraction
Put the digits to the right of the decimal point over the
appropriate power of 10, then simplify
Want to convert a
percent to a fraction
Write the percent “over 100,” then simplify
OR
Convert first to a decimal, then follow the process for
converting decimals to fractions
3.6% 0.0
2. Convert 0.375 to a fraction.
3. Convert 24% to a fraction.
Answers can be found on page 225.
Convert 1/4 to 0.25 or 25%: Long-Divide 1 by 4
A fraction represents division. The decimal equivalent is the result of that division.
To convert a fraction to a decimal, long-divide the numerator by the denominator.
Divide 1 by 4.
Divide 5 by 8.
In some cases, the decimal never ends because the long division never ends. You get a repeating
decimal.
If the denominator contains only 2's and/or 5's as factors, the decimal will end. In that case, you can
take a shortcut to find the decimal equivalent: make the denominator a power of 10 by
multiplication.
Since 4 = 22, you multiply by 25 (= 52) to get 100 (= 102). Likewise, you multiply 8 (= 23) by 125 (=
53) to get 1,000 (= 103).
To convert a fraction to a percent, first convert it to a decimal, then convert the decimal to a
percent.
If you…
Want to convert a
fraction to a decimal
Then you…
Do long division
OR
Convert the denominator to a power of 10, if the denominator
only contains 2's and 5's as factors
Know the following conversions.
Fraction
1/100
1/20
1/10
1/8
1/5
1/4
3/10
1/3
3/8
Decimal
0.01
0.05
0.1
0.125
0.2
0.25
0.3
0.3333…
0.375
Percent
1%
5%
10%
12.5%
20%
25%
30%
33.33…%
37.5%
Like th
2/5
1/2
3/5
5/8
2/3
7/10
3/4
4/5
7/8
9/10
1
6/5
5/4
3/2
0.4
0.5
0.6
0.625
0.6666…
0.7
0.75
0.8
0.875
0.9
1
1.2
1.25
1.5
40%
50%
60%
62.5%
66.66…%
70%
75%
80%
87.5%
90%
100%
120%
125%
150%
4. Change 3/5 to a decimal.
5. Convert 3/8 to a percent.
Answers can be found on page 225.
Multiply a Decimal by a Power of Ten: Shift the Decimal Point
Decimals are tenths, hundredths, thousandths, and so on. One tenth is a power of 10—namely, 10 –1.
One hundredth is also a power of 10—namely, 10–2.
You can write any decimal as a fraction with a power of 10 in the denominator, or as a product
involving a power of 10. The power of 10 determines where the decimal point is.
So if you multiply or divide a decimal by a power of 10, you move the decimal point.
If you multiply by 10 itself, you shift the decimal point one place to the right.
0.004 × 10 = 0.04
The 10 cancels with one power of 10 in the denominator.
You can also see it in terms of exponents. The additional 10 increases the overall exponent from –3 to
–2.
4 × 10–3 × 10 = 4 × 10–2
If you multiply by 100, or 102, you shift the decimal point two places to the right.
0.004 × 100 = 0.4
4 × 10–3 × 102 = 4 × 10–1
That is,
When you multiply by a power of 10, the exponent of that power is the number of places you move the
decimal.
43.8723 × 103 = 43,872.3
Move the decimal 3 places to the right.
If you divide by a power of 10, you just move to the left instead.
782.95 ÷ 10 = 78.295
57,234 ÷ 104 = 5.7234
Move the decimal 1 place to the left.
Move the decimal 4 places to the left.
If you encounter negative powers of 10, flip them to positive powers of 10 and change from
multiplication to division or vice versa.
Multiplying by a negative power of 10 is the same as dividing by a positive power.
0.004 ì 103 = 0.004 ữ 103 = 0.000004
Move 3 places to the left.
Likewise, dividing by a negative power of 10 is the same as multiplying by a positive power.
62 ữ 102 = 62 ì 102 = 6,200
Move 2 places to the right.
All of these procedures work the same for repeating decimals.
× 10 = 0.333…× 10 = 3.33…
Move 1 place to the right.
If you…
Multiply a decimal by a
power of 10
Then you…
Move the decimal point right a number of places,
corresponding to the exponent of 10
0.007 × 1
= 0.7
Divide a decimal by a
power of 10
Move the decimal point left a number of places,
corresponding to the exponent of 10
0.6 ÷ 10
0.0006
6. 32.753 × 102 =
7. 43,681 × 10–4 =
Like thi
Answers can be found on page 225.
Add or Subtract Decimals: Line Up the Decimal Points
When you add or subtract decimals, write the decimals vertically, with the decimal points lined up.
0.3 + 0.65 =
0.65 – 0.5 =
You can add zeroes on the right to help you line up. For instance, turn 0.5 into 0.50 before you
subtract it from 0.65.
If you…
Then you…
Add or subtract decimals
Line up the decimal
points vertically
Like this:
8. 3.128 + 0.045 =
9. 1.8746 – 0.313 =
Answers can be found on page 225.
Multiply Two Decimals: Ignore Decimal Points At First
Consider this example:
0.25 × 0.5 =
First, multiply the numbers together as if they were integers. In other words, ignore the decimal
points.
25 × 5 = 125
Now count all the digits to the right of the original decimal points.
0.25 has 2 digits to the right.
0.5 has 1 digit to the right.
There were a total of 3 digits originally to the right. So we move the decimal point of our answer 3
places to the left, in order to compensate.
125 becomes 0.125
Therefore, 0.25 × 0.5 = 0.125
You can see why this process works using powers of 10.
0.25 = 25 × 10–2
0.5 = 5 × 10–1
0.25 × 0.5 = (25 × 10–2) × (5 × 10–1) = 125 × 10–3 = 0.125
The powers of 10 just tell you where to put the decimal point. Here is another example:
3.5 × 20 =
We originally had one digit to the right of a decimal point.
Move the final decimal point one place to the left.
35 × 20 = 700
3.5 × 20 = 70.0 = 70
Count the zeroes to the right of the decimal point as well.
0.001 × 0.005 =
We originally had six digits to the right, including zeroes. Move the final
decimal point six places to the left.
1×5=5
0.001 × 0.005 =
0.000005
If you…
Multiply two
decimals
Then you…
Like this:
0.2 × 0.5 = ?
Ignore the decimal points, multiply integers, then place the decimal
2 × 5 = 10
point by counting the original digits on the right
10 → 0.10
0.2 × 0.5 = 0.1
10. 0.6 × 1.4 =
11. 0.0004 × 0.032 =
Answers can be found on pages 225–226.
Multiply a Decimal and a Big Number: Trade Decimal Places
Now consider this example:
4,000,000 × 0.0003 =
When one number is very big and the other one is very small, you can trade powers of 10 from the big
one (4,000,000) to the small one (0.0003). In other words, move one decimal point left and the other
one right. Just make sure you move the same number of places.
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# If $\int {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k$, where $k$ is constant of integration, then find the value of$A + B + C$.(1) $\dfrac{{21}}{5}$ (2) $\dfrac{{21}}{{10}}$ (3) $\dfrac{{16}}{5}$ (4) $\dfrac{7}{{10}}$
Hint:Using the trigonometric identity of double angles, transform the integral in form of secant and tangent function. Now use the identity ${\sec ^2}x = {\tan ^2}x + 1$to transform the secant squared function into a tangent function. Take a proper substitution as $\tan x = m \Rightarrow {\sec ^2}xdx = dm$ and further solve the integral to obtain an expression as required in the question.
Here we have to integrate the given integral and compare our result with ${\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k$ to find the value of A, B and C
Let’s get started with the solution of the given integral. We can start with simplifying the radical using the double-angle formula, i.e. $\sin 2x = 2\sin x\cos x$
$\Rightarrow \int {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} = \int {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2 \times 2\sin x\cos x} }}} = \int {\dfrac{{dx}}{{2{{\cos }^3}x\sqrt {\sin x\cos x} }}}$
Now in the denominator, multiply and divide with $\cos x$, which will give:
$\Rightarrow \int {\dfrac{{dx}}{{2{{\cos }^3}x\sqrt {\sin x\cos x} }}} = \int {\dfrac{{dx}}{{\dfrac{{\cos x \times 2{{\cos }^3}x\sqrt {\sin x\cos x} }}{{\cos x}}}}}$
We can take this $\cos x$ inside the radical and then transform it further as:
$\Rightarrow \int {\dfrac{{dx}}{{\dfrac{{\cos x \times 2{{\cos }^3}x\sqrt {\sin x\cos x} }}{{\cos x}}}}} = \int {\dfrac{{dx}}{{2{{\cos }^4}x\sqrt {\dfrac{{\sin x\cos x}}{{{{\cos }^2}x}}} }}} = \int {\dfrac{{dx}}{{2{{\cos }^4}x\sqrt {\dfrac{{\sin x}}{{\cos x}}} }}}$
Since we know that $\dfrac{{\sin x}}{{\cos x}} = \tan x$ and the reciprocal of $\cos x$ is $\sec x$ which can be used in the above integral transforming it further: $\dfrac{1}{{\cos x}} = \sec x \Rightarrow \dfrac{1}{{{{\cos }^4}x}} = {\sec ^2}x \times {\sec ^2}x$
$\Rightarrow \int {\dfrac{{dx}}{{2{{\cos }^4}x\sqrt {\dfrac{{\sin x}}{{\cos x}}} }}} = \int {\dfrac{{{{\sec }^2}x \times {{\sec }^2}xdx}}{{2\sqrt {\tan x} }}}$
Now, let’s use the trigonometric identity ${\sec ^2}x = {\tan ^2}x + 1$ in the numerator and utilize the method of substitution, i.e. assume that $\tan x = m \Rightarrow d\left( {\tan x} \right) = dm \Rightarrow {\sec ^2}xdx = dm$
Therefore, we get the integral as:$\int {\dfrac{{{{\sec }^2}x \times {{\sec }^2}xdx}}{{2\sqrt {\tan x} }}} = \int {\dfrac{{\left( {{{\tan }^2}x + 1} \right){{\sec }^2}xdx}}{{2\sqrt {\tan x} }}} = \int {\dfrac{{\left( {{m^2} + 1} \right)dm}}{{2\sqrt m }}}$
So, now we get an integral with the variable as $'m'$and it can be further split into two parts:
$\Rightarrow \int {\dfrac{{\left( {{m^2} + 1} \right)dm}}{{2\sqrt m }}} = \int {\dfrac{{{m^2}}}{{2\sqrt m }}dm + \int {\dfrac{1}{{2\sqrt m }}dm = } } \dfrac{1}{2}\int {{m^{2 - \dfrac{1}{2}}}dm + \dfrac{1}{2}\int {{m^{\dfrac{{ - 1}}{2}}}dm} }$
Now, using some basic integrals $\int {{a^b}da = \dfrac{{{a^{b + 1}}}}{{b + 1}}} + C$ in the above two integrals, we can further evaluate them as:
$\Rightarrow \dfrac{1}{2}\int {{m^{2 - \dfrac{1}{2}}}dm + \dfrac{1}{2}\int {{m^{\dfrac{{ - 1}}{2}}}dm} } = \dfrac{1}{2}\int {{m^{\dfrac{3}{2}}}dm + \dfrac{1}{2}\int {{m^{\dfrac{{ - 1}}{2}}}dm} } = \dfrac{1}{2}\dfrac{{{m^{\dfrac{3}{2} + 1}}}}{{\left( {\dfrac{3}{2} + 1} \right)}} + \dfrac{1}{2}\dfrac{{{m^{\dfrac{{ - 1}}{2} + 1}}}}{{\left( {\dfrac{{ - 1}}{2} + 1} \right)}} + C$
$\Rightarrow \dfrac{1}{2}\dfrac{{{m^{\dfrac{3}{2} + 1}}}}{{\left( {\dfrac{3}{2} + 1} \right)}} + \dfrac{1}{2}\dfrac{{{m^{\dfrac{{ - 1}}{2} + 1}}}}{{\left( {\dfrac{{ - 1}}{2} + 1} \right)}} + C = \dfrac{1}{2} \times \dfrac{2}{5}{m^{\dfrac{5}{2}}} + \dfrac{1}{2} \times 2{m^{\dfrac{1}{2}}} + C = \dfrac{1}{5}{m^{\dfrac{5}{2}}} + {m^{\dfrac{1}{2}}} + C$
Hence, we got the integral solved in a variable $m$; let’s substitute the assumed value $\tan x = m$ in the result.
$\Rightarrow \dfrac{1}{5}{m^{\dfrac{5}{2}}} + {m^{\dfrac{1}{2}}} + C = \dfrac{1}{5}{\left( {\tan x} \right)^{\dfrac{5}{2}}} + {\left( {\tan x} \right)^{\dfrac{1}{2}}} + k…………………...(1)$
But we need to compare the result of the integration with ${\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k$
$\Rightarrow$ If ${\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k$ is equal to $\dfrac{1}{5}{\left( {\tan x} \right)^{\dfrac{5}{2}}} + {\left( {\tan x} \right)^{\dfrac{1}{2}}} + k$, then the coefficient of the terms must also be the same in each of the expressions.
Therefore, we can see that $A = \dfrac{1}{2},B = \dfrac{5}{2}$ and $C = \dfrac{1}{5}$
Hence, the required value for the expression $A + B + C = \dfrac{1}{2} + \dfrac{5}{2} + \dfrac{1}{5} = \dfrac{6}{2} + \dfrac{1}{5} = 3 + \dfrac{1}{5} = \dfrac{{16}}{5}$
So, the correct answer is “Option C”.
Note:Always solve the integrals step by step. Make the use of trigonometric identities wisely. Be careful while using the substitution method in the integrals, as it requires you to differentiate both sides of the assumed value and then change the variable in the whole integral. The substitution was an important part of this solution. |
# RS Aggarwal Solutions for Class 9 Chapter 8 -Triangles
A triangle is a shape having three angles and three sides. It is mainly divided on the basis of number of sides and angles it possesses. Different types of triangles based on the number of sides are Scalene triangles, Isosceles triangles, Equilateral triangles and the triangles based on the angles are Obtuse angled triangle, Acute angled triangle and Right angled triangle. Scoring good marks in the exams has become much simpler with the help of RS Aggarwal Solutions for Class 9. The exercise wise solutions with steps based on the evaluation in the Class 9 exams improves confidence in students to solve difficult questions. RS Aggarwal Solutions for Class 9 Chapter 8 Triangles are provided here.
## Access RS Aggarwal Solutions for Class 9 Chapter 8: Triangles
Exercise 8 PAGE: 252
1. In △ ABC, if ∠B = 76o and ∠C = 48o, find ∠A.
Solution:
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
By substituting the values in the above equation we get
∠A + 76o + 48o = 180o
On further calculation
∠A = 180o – 76o – 48o
By subtraction we get
∠A = 180o – 124o
∠A = 56o
Therefore, the value of ∠A is 56o.
2. The angles of a triangle are in the ratio 2: 3: 4. Find the angles.
Solution:
Let us consider the measure of the angles in a triangle as 2xo, 3xo and 4xo
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
2x + 3x + 4x = 180o
9x = 180o
By division
x = 180/9
x = 20
By substituting the values of x
2xo = 2 (20) = 40o
3xo = 3 (20) = 60o
4xo = 4 (20) = 80o
Therefore, the angles are 40o, 60o and 80o
3. In △ ABC, if 3 ∠A = 4 ∠B = 6 ∠C, calculate ∠A, ∠B and ∠C.
Solution:
Consider 3 ∠A = 4 ∠B = 6 ∠C = x
So we can write it as
3 ∠A = x
∠A = x/3
4 ∠B = x
∠B = x/4
6 ∠C = x
∠C = x/6
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
By substituting the values in the above equation we get
(x/3) + (x/4) + (x/6) = 180o
LCM of 3, 4 and 6 is 12
So we get
(4x + 3x + 2x)/12 = 180o
9x/12 = 180o
By cross multiplication
9x = 180 × 12
9x = 2160
By division
x = 2160/9
x = 240
By substituting the values of x
∠A = x/3 = 240/3 = 80o
∠B = x/4 = 240/4 = 60o
∠C = x/6 = 240/6 = 40o
Therefore, the value of ∠A, ∠B and ∠C is 80o, 60o and 40o.
4. In △ ABC, if ∠A + ∠B = 108o and ∠B + ∠C = 130o, find ∠A, ∠B and ∠C.
Solution:
It is given that ∠A + ∠B = 108o …… (1)
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
By substituting ∠A + ∠B = 108o in the above equation
108o + ∠C = 180o
On further calculation
∠C = 180o – 108o
By subtraction
∠C = 72o
It is given that ∠B + ∠C = 130o
By substituting the value of ∠C
∠B + 72o = 130o
On further calculation
∠B = 130o – 72o
By subtraction
∠B = 58o
By substituting ∠B = 58o in equation (1)
So we get
∠A + ∠B = 108o
∠A + 58o = 108o
On further calculation
∠A = 108o – 58o
By subtraction
∠A = 50o
Therefore, ∠A = 50o, ∠B = 58o and ∠C = 72o
5. In △ ABC, if ∠A + ∠B = 125o and ∠A + ∠C = 113o, find ∠A, ∠B and ∠C.
Solution:
It is given that ∠A + ∠B = 125o …. (1)
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
By substituting ∠A + ∠B = 125o in the above equation
125o + ∠C = 180o
On further calculation
∠C = 180o – 125o
By subtraction
∠C = 55o
It is given that ∠A + ∠C = 113o
By substituting the value of ∠C
∠A + 55o = 113o
On further calculation
∠A = 113o – 55o
By subtraction
∠A = 58o
By substituting ∠A = 58o in equation (1)
So we get
∠A + ∠B = 125o
58o + ∠B= 125o
On further calculation
∠B = 125o – 58o
By subtraction
∠B = 67o
Therefore, ∠A = 58o, ∠B = 67o and ∠C = 55o
6. In â–³ PQR, if ∠P – ∠Q = 42o and ∠Q – ∠R = 21o, find ∠P, ∠Q and ∠R.
Solution:
It is given that ∠P – ∠Q = 42o
It can be written as
∠P = 42o + ∠Q
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠P + ∠Q + ∠R = 180o
By substituting ∠P = 42o + ∠Q in the above equation
42o + ∠Q +∠Q + ∠R = 180o
On further calculation
42o + 2 ∠Q + ∠R = 180o
2 ∠Q + ∠R = 180o – 42o
By subtraction we get
2 ∠Q + ∠R = 138o …. (i)
It is given that ∠Q – ∠R = 21o
It can be written as
∠R = ∠Q – 21o
By substituting the value of ∠R in equation (i)
2 ∠Q + ∠Q – 21o = 138o
On further calculation
3 ∠Q – 21o = 138o
3 ∠Q = 138o + 21o
3 ∠Q = 159o
By division
∠Q = 159/3
∠Q = 53o
By substituting ∠Q = 53o in ∠P = 42o + ∠Q
So we get
∠P = 42o + 53o
∠P = 95o
By substituting ∠Q in ∠Q – ∠R = 21o
53o – ∠R = 21o
On further calculation
∠R = 53o – 21o
By subtraction
∠R = 32o
Therefore, ∠P = 95o, ∠Q = 53o and ∠R= 32o
7. The sum of two angles of a triangle is 116o and their difference is 24o. Find the measure of each angle of the triangle.
Solution:
Let us consider the sum of two angles as ∠A + ∠B = 116o and the difference can be written as ∠A – ∠B = 24o
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
By substituting ∠A + ∠B = 116o in the above equation
116o + ∠C = 180o
On further calculation
∠C = 180o – 116o
By subtraction
∠C = 64o
It is given that ∠A – ∠B = 24o
It can be written as ∠A = 24o + ∠B
Now by substituting ∠A = 24o + ∠B in ∠A + ∠B = 116o
∠A + ∠B = 116o
24o + ∠B + ∠B = 116o
On further calculation
24o + 2∠B = 116o
By subtraction
2∠B = 116o – 24o
2∠B = 92o
By division
∠B = 92/2
∠B = 46o
By substituting ∠B = 46o in ∠A = 24o + ∠B
We get
∠A = 24o + 46o
∠A = 70o
Therefore, ∠A = 70o, ∠B = 46o and ∠C = 64o
8. Two angles of a triangle are equal and the third angle is greater than each one of them by 18o. Find the angles.
Solution:
Consider ∠A and ∠B in a triangle is xo
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
By substituting the values
xo + xo + ∠C = 180o
2xo + ∠C = 180o ….. (1)
According to the question we get
∠C = xo + 18o ……. (2)
By substituting (2) in (1) we get
2xo + xo + 18o = 180o
On further calculation
3xo + 18o = 180o
By subtraction
3xo = 180o – 18o
3xo = 162o
By division
xo = 162/3
xo = 54o
By substituting the values of x
∠A = ∠B = 54o
∠C = 54o + 18o = 72o
Therefore, ∠A = 54o, ∠B = 54o and ∠C = 72o
9. Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.
Solution:
Consider ∠C is the smallest angle among ∠ABC
According to the question
We can write it as
∠A = 2 ∠C and ∠B = 3 ∠C
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
By substituting the values
2 ∠C + 3 ∠C + ∠C = 180o
6∠C = 180o
By division
∠C = 180/6
∠C = 30o
Now by substituting the value of ∠C we get
∠A = 2 ∠C = 2 (30o) = 60o
∠B = 3 ∠C = 3 (30o) = 90o
Therefore, ∠A = 60o, ∠B = 90o and ∠C = 30o.
10. In a right-angled triangle, one of the acute angles measures 53o. Find the measure of each angle of the triangle.
Solution:
Consider ABC as a right-angled triangle with ∠C = 90o
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
We can write it as
∠A + ∠B = 180o – ∠C
By substituting the values
∠A + ∠B = 180o – 90o
By subtraction
∠A + ∠B = 90o
Let us consider ∠A = 53o
By substituting the value of ∠A in ∠A + ∠B = 90o
We get
53o + ∠B = 90o
On further calculation
∠B = 90o – 53o
By subtraction
∠B = 37o
Therefore, ∠A = 53o, ∠B = 37o and ∠C = 90o.
11. If one angle of a triangle is equal to the sum of the other two, show that the triangle is right angled.
Solution:
Consider ABC as a triangle
According to the question it can be written as
∠A = ∠B + ∠C ….. (1)
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
By substituting ∠A in the above equation
∠B + ∠C + ∠B + ∠C = 180o
2 (∠B + ∠C) = 180o
By division
∠B + ∠C = 180/2
∠B + ∠C = 90o
According to equation (1) we can write it as
∠A = 90o
Therefore, it is proved that the triangle is right angled.
12. If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
Consider ABC as a triangle
According to the question it can be written as
∠A < ∠B + ∠C
Add ∠A to both the sides of the equation
So we get
∠A + ∠A < ∠A + ∠B + ∠C
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
So we get
2 ∠A < 180o
By division we get
∠A < 180/2
∠A < 90o
In the same way we can also write
∠B < ∠A + ∠C
Add ∠B to both the sides of the equation
So we get
∠B + ∠B < ∠A + ∠B + ∠C
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
So we get
2 ∠B < 180o
By division we get
∠B < 180/2
∠B < 90o
So we know that
∠C < ∠A + ∠B
Add ∠C to both the sides of the equation
So we get
∠C + ∠C < ∠A + ∠B + ∠C
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
So we get
2 ∠C < 180o
By division we get
∠C < 180/2
∠C < 90o
Therefore, it is proved that the triangle ABC is acute angled.
13. If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse angled.
Solution:
Consider ABC as a triangle
According to the question it can be written as
∠B > ∠A + ∠C …. (1)
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠A + ∠B + ∠C = 180o
So we get
∠A + ∠C = 180o – ∠B
Substituting ∠A + ∠C in equation (1) we get
∠B > 180o – ∠B
Add ∠B to both the sides of the equation
So we get
∠B + ∠B > 180o – ∠B + ∠B
2 ∠B > 180o
By division we get
∠B > 180/2
∠B > 90o
So we know that ∠B > 90o which means that ∠B is an obtuse angle
Therefore, it is proved that the triangle ABC is obtuse angled.
14. In the given figure, side BC of △ ABC is produced to D. If ∠ACD = 128o, and ∠ABC = 43o, find ∠BAC and ∠ACB.
Solution:
From the figure we know that ∠ACB and ∠ACD form a linear pair of angles
So we get
∠ACB + ∠ACD = 180o
By substituting the values
∠ACB + 128o = 180o
On further calculation
∠ACB = 180o – 128o
By subtraction
∠ACB = 52o
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠ABC + ∠ACB + ∠BAC = 180o
By substituting the values
43o + 52o + ∠BAC = 180o
On further calculation
∠BAC = 180o – 43o – 52o
By subtraction
∠BAC = 180o – 95o
∠BAC = 85o
Therefore, ∠BAC = 85o and ∠ACB = 52o.
15. In the given figure, the side BC of △ ABC has been produced on the left-hand side from B to D and on the right-hand side from C to E. If ∠ABD = 106o and ∠ACE = 118o, find the measure of each angle of the triangle.
Solution:
From the figure we know that ∠DBA and ∠ABC form a linear pair of angles
So we get
∠DBA + ∠ABC = 180o
By substituting the values
106o + ∠ABC = 180o
On further calculation
∠ABC = 180o – 106o
By subtraction
∠ABC = 74o
From the figure we know that ∠ACB and ∠ACE form a linear pair of angles
So we get
∠ACB + ∠ACE = 180o
By substituting the values
∠ACB + 118o = 180o
On further calculation
∠ACB = 180o – 118o
By subtraction
∠ACB = 62o
We know that the sum of all the angles in a triangle is 180o.
So we can write it as
∠ABC + ∠ACB + ∠BAC = 180o
By substituting the values
74o + 62o + ∠BAC = 180o
On further calculation
∠BAC = 180o – 74o – 62o
By subtraction
∠BAC = 180o – 136o
∠BAC = 44o
Therefore, the measure of each angle of the triangle is ∠A = 44o, ∠B = 74o and ∠C = 62o.
16. Calculate the value of x in each of the following figures.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Solution:
(i) From the figure we know that ∠EAB and ∠BAC form a linear pair of angles
So we get
∠EAB + ∠BAC = 180o
By substituting the values
110o + ∠BAC = 180o
On further calculation
∠BAC = 180o – 110o
By subtraction
∠BAC = 70o
From the figure we know that ∠BCA and ∠ACD form a linear pair of angles
So we get
∠BCA + ∠ACD = 180o
By substituting the values
∠BCA + 120o = 180o
On further calculation
∠BAC = 180o – 120o
By subtraction
∠BAC = 60o
We know that the sum of all the angles in a triangle is 180o.
∠ABC + ∠ACB + ∠BAC = 180o
By substituting the values
xo + 70o + 60o = 180o
On further calculation
xo = 180o – 70o – 60o
By subtraction
xo = 180o – 130o
xo = 50o
(ii) We know that the sum of all the angles in triangle ABC is 180o.
∠A + ∠B + ∠C = 180o
By substituting the values
30o + 40o + ∠C = 180o
On further calculation
∠C = 180o – 30o – 40o
By subtraction
∠C = 180o – 70o
∠C = 110o
From the figure we know that ∠BCA and ∠ACD form a linear pair of angles
So we get
∠BCA + ∠ACD = 180o
By substituting the values
110o + ∠ACD = 180o
On further calculation
∠ACD = 180o – 110o
By subtraction
∠ACD = 70o
We know that the sum of all the angles in triangle ECD is 180o.
∠ECD + ∠CDE + ∠CED = 180o
By substituting the values
70o + 50o + ∠CED = 180o
On further calculation
∠CED = 180o – 70o – 50o
By subtraction
∠CED = 180o – 120o
∠CED = 60o
From the figure we know that ∠AED and ∠CED form a linear pair of angles
So we get
∠AED + ∠CED = 180o
By substituting the values
xo + 60o = 180o
On further calculation
xo = 180o – 60o
By subtraction
xo = 120o
(iii) From the figure we know that ∠EAF and ∠BAC are vertically opposite angles
So we get
∠EAF = ∠BAC = 60o
We know that in the triangle ABC exterior angle is equal to the sum of two opposite interior angles
So we can write it as
∠ACD = ∠BAC + ∠ABC
By substituting the values
115o = 60o + xo
On further calculation
xo = 115o – 60o
By subtraction
xo = 55o
(iv) We know that AB || CD and AD is a transversal
We know that the sum of all the angles in triangle ECD is 180o.
∠E + ∠C + ∠D = 180o
By substituting the values
xo + 45o + 60o = 180o
On further calculation
xo = 180o – 45o – 60o
By subtraction
xo = 180o – 105o
xo = 75o
(v) We know that in the triangle AEF exterior angle is equal to the sum of two opposite interior angles
So we can write it as
∠BED = ∠EAF + ∠EFA
By substituting the values
100o = 40o + ∠EFA
On further calculation
∠EFA = 100o – 40o
By subtraction
∠EFA = 60o
From the figure we know that ∠CFD and ∠EFA are vertically opposite angles
So we get
∠CFD = ∠EFA = 60o
We know that in the triangle FCD exterior angle is equal to the sum of two opposite interior angles
So we can write it as
∠BCD = ∠CFD + ∠CDF
By substituting the values
90o = 60o + xo
On further calculation
xo = 90o – 60o
By subtraction
xo = 30o
(vi) We know that the sum of all the angles in triangle ABE is 180o.
∠A + ∠B + ∠E = 180o
By substituting the values in the above equation
75o + 65o + ∠E = 180o
On further calculation
∠E = 180o – 75o – 65o
By subtraction
∠E = 180o – 140o
∠E = 40o
From the figure we know that ∠CED and ∠AEB are vertically opposite angles
So we get
∠CED = ∠AEB = 40o
We know that the sum of all the angles in triangle CED is 180o.
∠C + ∠E + ∠D = 180o
By substituting the values
110o + 40o + xo = 180o
On further calculation
xo = 180o – 110o + 40o
By subtraction
xo = 180o – 150o
xo = 30o
17. In the figure given alongside, AB || CD, EF || BC, ∠BAC = 60o and ∠DHF = 50o. Find ∠GCH and ∠AGH.
Solution:
We know that AB || CD and AC is a transversal
From the figure we know that ∠BAC and ∠ACD are alternate angles
So we get
∠BAC = ∠ACD = 60o
So we also get
∠BAC = ∠GCH = 60o
From the figure we also know that ∠DHF and ∠CHG are vertically opposite angles
So we get
∠DHF = ∠CHG = 50o
We know that the sum of all the angles in triangle GCH is 180o.
So we can write it as
∠GCH + ∠CHG + ∠CGH = 180o
By substituting the values
60o + 50o + ∠CGH = 180o
On further calculation
∠CGH = 180o – 60o – 50o
By subtraction
∠CGH = 180o – 110o
∠CGH = 70o
From the figure we know that ∠CGH and ∠AGH form a linear pair of angles
So we get
∠CGH + ∠AGH = 180o
By substituting the values
70o + ∠AGH = 180o
On further calculation
∠AGH = 180o – 70o
By subtraction
∠AGH = 110o
Therefore, ∠GCH = 60o and ∠AGH = 110o.
18. Calculate the value of x in the given figure.
Solution:
Construct a line CD to cut the line AB at point E.
We know that in the triangle BDE exterior angle is equal to the sum of two opposite interior angles
So we can write it as
∠CDB = ∠CEB + ∠DBE
By substituting the values
xo = ∠CEB + 45o …. (1)
We know that in the triangle AEC exterior angle is equal to the sum of two opposite interior angles
So we can write it as
∠CEB = ∠CAB + ∠ACE
By substituting the values
∠CEB = 55o + 30o
∠CEB = 85o
By substituting ∠CEB in equation (1) we get
xo = 85o + 45o
xo = 130o
19. In the given figure, AD divides ∠BAC in the ratio 1: 3 and AD = DB. Determine the value of x.
Solution:
It is given that AD divides ∠BAC in the ratio 1: 3
So let us consider ∠BAD and ∠DAC as y and 3y
According to the figure we know that BAE is a straight line
From the figure we know that ∠BAC and ∠CAE form a linear pair of angles
So we get
∠BAC + ∠CAE = 180o
We know that
∠BAC = ∠BAD + ∠DAC
So it can be written as
∠BAD + ∠DAC + ∠CAE = 180o
By substituting the values we get
y + 3y + 108o = 180o
On further calculation
4y = 180o – 108o
By subtraction
4y = 72o
By division
y = 72/4
y = 18o
We know that the sum of all the angles in triangle ABC is 180o.
So we can write it as
∠ABC + ∠BCA + ∠BAC = 180o
It is given that AD = DB so we can write it as ∠ABC = ∠BAD
From the figure we know that ∠BAC = y + 3y = 4y
By substituting the values
y + x + 4y = 180o
On further calculation
5y + x = 180o
By substituting the value of y
5 (18o) + x = 180o
By multiplication
90o + x = 180o
x = 180o – 90o
By subtraction we get
x = 90o
Therefore, the value of x is 90.
20. If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.
Solution:
Given: â–³ ABC in which AB, BC and CA are produced to points D, E and F.
To prove: ∠DCA + ∠FAE + ∠FBD = 180o
Proof:
From the figure we know that
∠DCA = ∠A + ∠B …. (1)
∠FAE = ∠B + ∠C …. (2)
∠FBD = ∠A + ∠C …. (3)
By adding equation (1), (2) and (3) we get
∠DCA + ∠FAE + ∠FBD = ∠A + ∠B + ∠B + ∠C + ∠A + ∠C
So we get
∠DCA + ∠FAE + ∠FBD = 2 ∠A + 2 ∠B + 2 ∠C
Now by taking out 2 as common
∠DCA + ∠FAE + ∠FBD = 2 (∠A + ∠B + ∠C)
We know that the sum of all the angles in a triangle is 180o.
So we get
∠DCA + ∠FAE + ∠FBD = 2 (180o)
∠DCA + ∠FAE + ∠FBD = 360o
Therefore, it is proved.
21. In the adjoining figure, show that ∠A + ∠B +∠C + ∠D + ∠E + ∠F = 360o.
Solution:
We know that the sum of all the angles in triangle ACE is 180o.
∠A + ∠C + ∠E = 180o ….. (1)
We know that the sum of all the angles in triangle BDF is 180o.
∠B + ∠D + ∠F = 180o ….. (2)
Now by adding both equations (1) and (2) we get
∠A + ∠C + ∠E + ∠B + ∠D + ∠F = 180o + 180o
On further calculation
∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360o
Therefore, it is proved that ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360o.
22. In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70o and ∠ACB = 20o, find ∠MAN.
Solution:
We know that the sum of all the angles in triangle ABC is 180o.
∠A + ∠B + ∠C = 180o
By substituting the values
∠A + 70o + 20o = 180o
On further calculation
∠A = 180o – 70o – 20o
By subtraction
∠A = 180o – 90o
∠A = 90o
We know that the sum of all the angles in triangle ABM is 180o.
∠BAM + ∠ABM + ∠AMB = 180o
By substituting the values
∠BAM + 70o + 90o = 180o
On further calculation
∠BAM = 180o – 70o – 90o
By subtraction
∠BAM = 180o – 160o
∠BAM = 20o
It is given that AN is the bisector of ∠A
So it can be written as
∠BAN = (1/2) ∠A
By substituting the values
∠BAN = (1/2) (90o)
By division
∠BAN = 45o
From the figure we know that
∠MAN + ∠BAM = ∠BAN
By substituting the values we get
∠MAN + 20o = 45o
On further calculation
∠MAN = 45o – 20o
By subtraction
∠MAN = 25o
Therefore, ∠MAN = 25o.
23. In the given figure, BAD || EF, ∠AEF = 55o and ∠ACB = 25o, find ∠ABC.
Solution:
We know that BAD || EF and EC is the transversal
From the figure we know that ∠AEF and ∠CAD are corresponding angles
So we get
∠AEF = ∠CAD = 55o
From the figure we know that ∠CAD and ∠CAB form a linear pair of angles
So we get
∠CAD + ∠CAB = 180o
By substituting the values
55o + ∠CAB = 180o
On further calculation
∠CAB = 180o – 55o
By subtraction
∠CAB = 125o
We know that the sum of all the angles in triangle ABC is 180o.
∠ABC + ∠CAB + ∠ACB = 180o
By substituting the values in the above equation we get
∠ABC + 125o + 25o = 180o
On further calculation
∠ABC = 180o – 125o – 25o
By subtraction
∠ABC = 180o – 150o
∠ABC = 30o
24. In a △ ABC, it is given that ∠A: ∠B: ∠C = 3: 2: 1 and CD ⊥ AC. Find ∠ECD.
Solution:
In a â–³ ABC, it is given that
∠A: ∠B: ∠C = 3: 2: 1
It can also be written as
∠A = 3x, ∠B = 2x and ∠C = x
We know that the sum of all the angles in triangle ABC is 180o.
∠A + ∠B + ∠C = 180o
By substituting the values we get
3x + 2x + x = 180o
6x = 180o
By division
x = 180/6
x = 30o
Now by substituting the value of x we get
∠A = 3x = 3 (30o) = 90o
∠B = 2x = 2 (30o) = 60o
∠C = x = 30o
We know that in the triangle ABC exterior angle is equal to the sum of two opposite interior angles
So we can write it as
∠ACE = ∠A + ∠B
By substituting the values we get
∠ACE = 90o + 60o
∠ACE = 150o
We know that ∠ACE can be written as ∠ACD + ∠ECD
So we can write it as
∠ACE = ∠ACD + ∠ECD
By substituting the values we get
150o = 90o + ∠ECD
It is given that CD ⊥ AC so ∠ACD = 90o
On further calculation
∠ECD = 150o – 90o
By subtraction
∠ECD = 60o
Therefore, ∠ECD = 60o.
25. In the given figure, AB || DE and BD || FG such that ∠ABC = 50o and ∠FGH = 120o. Find the values of x and y.
Solution:
From the figure we know that ∠FGH and ∠FGE form a linear pair of angles
So we get
∠FGH + ∠FGE = 180o
By substituting the values
120o + y = 180o
On further calculation
y = 180o – 120o
By subtraction
y = 60o
We know that AB || DF and BD is a transversal
From the figure we know that ∠ABC and ∠CDE are alternate angles
So we get
∠ABC = ∠CDE = 50o
We know that BD || FG and DF is the transversal
From the figure we know that ∠EFG and ∠CDE are alternate angles
So we get
∠EFG = ∠CDE = 50o
We know that the sum of all the angles in triangle EFG is 180o.
∠FEG + ∠FGE + ∠EFG = 180o
By substituting the values we get
x + y + 50o = 180o
x + 60o + 50o = 180o
On further calculation
x = 180o – 60o – 50o
By subtraction
x = 180o – 110o
x = 70o
Therefore, the values of x = 70o and y = 60o.
26. In the given figure, AB || CD and EF is a transversal. If ∠AEF = 65o, ∠DFG = 30o, ∠EGF = 90o and ∠GEF = xo, find the value of x.
Solution:
We know that AB || CD and EF is a transversal
From the figure we know that ∠AEF and ∠EFD are alternate angles
So we get
∠AEF = ∠EFG + ∠DFG
By substituting the values
65o = ∠EFG + 30o
On further calculation
∠EFG = 65o – 30o
By subtraction
∠EFG = 35o
We know that the sum of all the angles in triangle GEF is 180o.
∠GEF + ∠EGF + ∠EFG = 180o
By substituting the values we get
x + 90o + 35o = 180o
On further calculation
x = 180o – 90o – 35o
By subtraction
x = 55o
Therefore, the value of x is 55o
27. In the given figure, AB || CD, ∠BAE = 65o and ∠OEC = 20o. Find ∠ECO.
Solution:
We know that AB || CD and AE is a transversal
From the figure we know that ∠BAE and ∠DOE are corresponding angles
So we get
∠BAE = ∠DOE = 65o
From the figure we know that ∠DOE and ∠COE form a linear pair of angles
So we get
∠DOE + ∠COE = 180o
By substituting the values
65o + ∠COE = 180o
On further calculation
∠COE = 180o – 65o
By subtraction
∠COE = 115o
We know that the sum of all the angles in triangle OCE is 180o.
∠OEC + ∠ECO + ∠COE = 180o
By substituting the values we get
20o + ∠ECO + 115o = 180o
On further calculation
∠ECO = 180o – 20o – 115o
By subtraction
∠ECO = 45o
Therefore, ∠ECO = 45o
28. In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB = 35o and QP ⊥ EF, find the measure of ∠PQH.
Solution:
We know that AB || CD and EF is a transversal
From the figure we know that ∠EGB and ∠GHD are corresponding angles
So we get
∠EGB = ∠GHD = 35o
From the figure we know that ∠GHD and ∠QHP are vertically opposite angles
So we get
∠GHD = ∠QHP = 35o
We know that the sum of all the angles in triangle DQHP is 180o.
∠PQH + ∠QHP + ∠QPH = 180o
By substituting the values we get
∠PQH + 35o + 90o = 180o
On further calculation
∠PQH = 180o – 35o – 90o
By subtraction
∠PQH = 180o – 125o
∠PQH = 55o
Therefore, ∠PQH = 55o
29. In the given figure, AB || CD and EF ⊥ AB. If EG is the transversal such that ∠GED = 130o, find ∠EGF.
Solution:
We know that AB || CD and GE is the transversal
From the figure we know that ∠EGF and ∠GED are interior angles
So we get
∠EGF + ∠GED = 180o
By substituting the values
∠EGF + 130o = 180o
On further calculation
∠EGF = 180o – 130o
By subtraction
∠EGF = 50o
Therefore, ∠EGF = 50o
### RS Aggarwal Solutions for Class 9 Maths Chapter 8: Triangles
Chapter 8, Triangles, has 1 exercise with solutions prepared by our expert team of faculties having much knowledge about the concepts in Mathematics. Some of the topics in RS Aggarwal Solutions which are explained in brief under this chapter are:
• Triangles
• Types of triangles on the basis of sides
• Types of triangles on the basis of angles
• Some terms related to triangle
• Exterior and interior opposite angles of a triangle
• Some results on triangles
### RS Aggarwal Solutions Class 9 Maths Chapter 8 – Triangles
Students can make use of the RS Aggarwal Solutions which covers the entire syllabus as per the textbook with exercise wise answers in each chapter. The brief solutions prepared by our experts mainly help students to solve the problems of higher difficulty with ease. RS Aggarwal Solutions for Class 9 can be used by the students as a vital resource to boost their exam preparation. Some of the applications of triangles are architecture, design of buildings, height of bridges and finding the area of geometric structures. Here, we provide PDF containing chapter wise solutions of the entire chapter based on the RS Aggarwal textbook to help students study well before appearing for the exams. |
# Question Video: Finding the Indefinite Integral of a Function Containing Exponential Functions by Distributing the Division Mathematics
Determine β« ((8π^(3π₯) β π^(2π₯) + 9)/7π^π₯) dπ₯.
03:32
### Video Transcript
Determine the indefinite integral of eight times π to the power of three π₯ minus π to the power of two π₯ plus nine divided by seven π to the power of π₯ with respect to π₯.
In this question, weβre asked to evaluate the indefinite integral of an exponential expression. And at first, it might look very difficult to evaluate this integral. For example, we might be considering a π’-substitution. However, we should always check if we can simplify our integrand. And we can do this by dividing every term in the numerator by the denominator, where we remember for each term in the numerator, we just need to subtract the exponents of π. This gives us the indefinite integral of eight over seven times π to the power of two π₯ minus one-seventh π to the power of π₯ plus nine over seven times π to the power of negative π₯ with respect to π₯.
Now, we can see we know how to integrate each term in the integrand separately. So, weβre going to split this into three separate integrals. And we can do this because we recall the integral of the sum of functions is equal to the sum of their individual integrals. This gives us the following. We can then simplify each term separately. Weβll take the constant factor outside of each integral. This then gives us eight-sevenths times the indefinite integral of π to the power of two π₯ with respect to π₯ minus one-seventh the indefinite integral of π to the power of π₯ with respect to π₯ plus nine-sevenths the integral of π to the power of negative π₯ with respect to π₯.
Weβre now ready to evaluate each of these indefinite integral separately. We can do this by recalling for any real constant π not equal to zero, the integral of π to the power of ππ₯ with respect to π₯ is one over π times π to the power of ππ₯ plus the constant of integration πΆ. Weβll use this to evaluate each integral separately. Letβs start with the first integral, the integral of π to the power of two π₯ with respect to π₯. Our value of π is two. So, we substitute π is two into our integral rule. We get one-half times π to the power of two π₯. And we need to multiply this by eight over seven. And itβs worth noting since weβre evaluating the sum of multiple indefinite integrals, weβll get a constant of integration for each of these integrals. We can combine all of these constants into one constant at the end of our expression. So, we donβt need to worry about adding πΆ until the end.
Letβs now evaluate our second integral. We know π to the power of π₯ is the same as π to the power of one times π₯. So, our value of π will be one. Of course, this is somewhat not necessary since we know the integral of π to the power of π₯ with respect to π₯ is just π to the power of π₯ with respect to π₯. However, if we did substitute π is equal to one, we would get one over one times π to the power of one π₯, which is just π to the power of π₯. So, evaluating our second integral and multiplying by negative one-seventh, we get negative one-seventh π to the power of π₯.
Finally, letβs evaluate the third integral. Our value of π is negative one. Substituting π is negative one into our integral result gives us one over negative one times π to the power of negative π₯. We need to multiply this by nine over seven. And remember, we also need to add a constant of integration at the end of this expression. We can then simplify this slightly. One over negative one is just equal to negative one. So, instead of adding this term, we can subtract this term. And we can also notice in the first term, we have a shared factor of two in the numerator and denominator. We can cancel this to get a factor of four over seven.
This then gives us our final answer. The indefinite integral of π to the power of three π₯ minus π to the power two π₯ plus nine over seven π to the π₯ with respect to π₯ is equal to four over seven times π to the power of two π₯ minus π to the power of π₯ over seven minus nine-sevenths π to the power of negative π₯ plus the constant of integration πΆ. |
Graphs of Sine and Cosine Functions
Introduction: In this lessons, the basic graphs of sine and cosine will be discussed and illustrated.
The Lesson:
The graphs of have some basic features which makes them easy to identify and accurately diagram. The domain, the possible values for x, is all real numbers. We will use radian measure so that any real number can be used for x. This is because any real number can be associated with an arc length on a circle of radius 1, the definition of a radian measure. Therefore, when using a calculator, set the mode for radians, not degrees.
We first construct a table of some common values for sin(x) and plot those points.
x 0 sin(x) 0 0.5 0.707 0.866 1 0.866 0.707
These points are plotted below. The WINDOW is X: and Y: (– 2, 2, 1).
Since the domain of is all real numbers, we can continue plotting points both to the left and right of the ones shown. Connecting those points gives the graph shown below, which contains the original points plotted.
All sine and cosine graphs will have this general shape which is called a sinusoid. If we focus on the important points which are the maximum, minimum and x-intercepts we can sketch a sinusoid containing those points quite accurately. The graph above of shows a maximum value of 1 at , a minimum of – 1 at and x-intercepts at .
The graph of , shown below, is also a sinusoid. The WINDOW is X: and Y: (– 2, 2, 1). This graph shows a maximum value of 1 at 0 and , a minimum of – 1 at and x-intercepts at .
Facts:
1. The sinusoid graph of or is a continuous repetition of one wave. For cosine, one complete wave can be seen from maximum point (0, 1) to the next maximum point at . For sine, one complete wave can be seen from an x-intercept at (0, 0) to the x-intercept at .
2. The length (distance along the x-axis) of one complete wave is called the period of the wave. In the case of of and both and the period is . Functions with periods other than are discussed in the lesson on period and frequency.
3. The height or amplitude of the wave is the vertical distance from the middle of the wave to a maximum point or minimum point. For both and the amplitude is 1.
Let's Practice:
1. What is the graph of ?
The graph is shown below using a WINDOW X: and Y: (– 2, 2, 1). The only difference between this graph and is that the 2 has made the graph go twice as high and twice as low. Therefore the amplitude is 2.
1. What is the graph of ?
The graph is shown below using a WINDOW X: and Y: (– 3, 3, 1). One difference between this graph and is that the 3 has made the graph go three times as high and three times as low. The amplitude is 3. The other difference is caused by the negative sign. This causes the graph to be reflected or “flipped” over the x-axis. Where had a maximum point at (0, 1), when x = 0, has a minimum point at (0, – 3).
Examples
Describe the graph of ? What is your answer?
Describe the graph of ? What is your answer?
M Ransom
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# Chapter 4 - Test: 6
$2\log_7 x+\dfrac{1}{4}\log_7y-3\log_7z$
#### Work Step by Step
$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to change the form of the given expression, $\log_7\dfrac{x^2\sqrt[4]{y}}{z^3} .$ $\bf{\text{Solution Details:}}$ U sing the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log_7(x^2\sqrt[4]{y})-\log_7z^3 .\end{array} Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log_7x^2+\log_7\sqrt[4]{y}-\log_7z^3 .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_7x^2+\log_7y^{1/4}-\log_7z^3 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} 2\log_7 x+\dfrac{1}{4}\log_7y-3\log_7z .\end{array}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
# Class 7 Maths Chapter 2 Exercise 2.6 Pdf Notes NCERT Solutions
Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.6 pdf notes:-
Exercise 2.6 Class 7 maths Chapter 2 Pdf Notes:-
## Ncert Solution for Class 6 Maths Chapter 2 Fractions and decimals Exercise 2.6 Tips:-
DIVISION OF DECIMAL NUMBERS
Savita was preparing a design to decorate her classroom. She needed a few coloured
strips of paper of length 1.9 cm each. She had a strip of coloured paper of length 9.5 cm.
How many pieces of the required length will she get out of this strip
Take 31.5 Ć· 10 = 3.15. In 31.5 and 3.15, the digits are
same i.e., 3, 1, and 5 but the decimal point has shifted in the
quotient. To which side and by how many digits? The decimal
point has shifted to the left by one place. Note that 10 has one
zero over 1.
Consider now 31.5 Ć· 100 = 0.315. In 31.5 and 0.315 the
digits are same, but what about the decimal point in the quotient?
It has shifted to the left by two places. Note that 100 has two zeros over1.
So we can say that, while dividing a number by 10, 100 or 1000, the digits of the
number and the quotient are same but the decimal point in the quotient shifts to the
left by as many places as there are zeros over 1. Using this observation let us now
quickly find: 2.38 Ć· 10 = 0.238, 2.38 Ć· 100 = 0.0238, 2.38 Ć· 1000 = 0.00238
You get 324. There are two digits to the right of the decimal in
12.96. Making similar placement of the decimal in 324, you will get 3.24.
Note that here and in the next section, we have considered only those
divisions in which, ignoring the decimal, the number would be completely
divisible by another number to give remainder zero. Like, in 19.5 Ć· 5, the
number 195 when divided by 5, leaves remainder zero.
However, there are situations in which the number may not be completely
divisible by another number, i.e., we may not get remainder zero. For example, 195 Ć· 7.
We deal with such situations in later classes. |
# Proving an inequality by induction
I'm trying to prove that $5^n-3^n>5^{n-1}$
I tried using mathematical induction and got stuck at the induction step.
First, I started by rearranging the inequality as: $4 \times5^n>5\times3^n$
• Try $n=1$: $$20>15$$ Therefore true for $n=1$
• Assume true for $n=k$: $$4 \times5^k>5\times3^k$$
• Examine case $n=k+1$: $$4\times5^{k+1}>5\times3^{k+1}$$
I'm not really sure where to go from here. Any help would be appreciated.
• The lhs grows by a factor of $5$, the rhs only by a factor of $3$. Use $4\cdot 5^{k+1}=5\cdot 4\cdot 5^k$. – Hagen von Eitzen Oct 16 '16 at 19:50
Your inequality, after dividing by $5^n$ is equivalent to
$1-(\frac{3}{5})^n>\frac{1}{5}$
or
$(\frac{3}{5})^n<\frac{4}{5}$
for $n=1$, it is true.
Now, let $n\geq 1$ such that
$(\frac{3}{5})^n<\frac{4}{5}$ (induction hypothesis).
as $\frac{3}{5}<1$
if we multiply by $(\frac{3}{5})^n$, we will get
$(\frac{3}{5})^{n+1}<(\frac{3}{5})^n<\frac{4}{5}$
which is the desired inequality.
we conclude that the inequality is satsfied for all integer $n\geq 1$.
First, show that this is true for $n=1$:
$5^1-3^1>5^{1-1}$
Second, assume that this is true for $n$:
$5^n-3^n>5^{n-1}$
Third, prove that this is true for $n+1$:
$5^{n+1}-3^{n+1}=$
$5\cdot5^n-3\cdot3^n>$
$5\cdot5^n-5\cdot3^n=$
$5\cdot(\color\red{5^n-3^n})>$
$5\cdot\color\red{5^{n-1}}=$
$5^n$
Please note that the assumption is used only in the part marked red. |
# Chapter 4 Work and Energy
## Presentation on theme: "Chapter 4 Work and Energy"— Presentation transcript:
Chapter 4 Work and Energy
Section 1: Work and Machines Section 2: Describing Energy Section 3: Conservation of Energy
Section 1: Work and Machines
Work – the transfer of energy that occurs when a force makes an object move no movement, no work direction of the net force indicates where or on what work is being done calculating work: equation for work: work = force x distance, or: 𝑾=𝑭𝒅 Example: How much work is done if Reggie lifts a box, m = 50-kg, 1.75 meters? Solution: Where: W = Work F = force (N) d = distance (m) Units for Work: W =Fd W = (N)(m) = Nm 1Nm = 1 Joule (J) The unit for work is Joules (J) m = 50.0kg d = 1.75m Because Reggie is lifting the box he must exert a force greater than the weight of the box 𝑊=𝑚𝑔 𝑊=50.0 𝑘𝑔 𝑚 𝑠 2 𝑊=490 𝑘𝑔𝑚 𝑠 2 =490.0 𝑁 Solve for work: 𝑊=𝐹𝑑 𝑊=490.0 𝑁 1.75𝑚 𝑊=857.5 𝑁𝑚 𝑾=𝟖𝟓𝟕.𝟓 𝑱
Section 1: Work and Machines
Machine – a device that makes doing work easier Machines make doing work easier in three ways: Increasing the force applied to the object example: a car jack to lift a car to change a flat tire Increasing the distance over which the force is applied example: using a ramp to raise objects to a height Changing the direction of the applied force example: a wedge – the vertical force is changed to a horizontal force Work done by machines Two forces are involved when a machine is used to do work: Effort force – the force applied to the machine Resistance force – the force applied by the machine to overcome resistance Conservation of Energy You transfer energy to a machine, the machine transfers that energy to the object Energy is neither created nor destroyed, so the work done by the machine is never greater than the work done to the machine Because of energy losses due to friction, the work done by the machine is always less than the work done to the machine
Section 1: Work and Machines
Mechanical Advantage – the number of times a machine multiplies the effort force Equation for Mechanical Advantage : Example: A claw hammer is used to pull a nail from a board. If the claw exerts a resistance force of 2500-N to the applied force of 125-N, what is the mechanical advantage of the hammer? Solution: Notice that the force units (N) cancel; mechanical advantage has no units, it is just a number. 𝑴𝑨= 𝒇 𝒐 𝒇 𝒊 Where: MA = mechanical advantage fo = force out (force applied by the machine) fi = force in (force applied to the machine) fo = 2,500.0N fi = 125.0N MA =? 𝑀𝐴= 𝑓 𝑜 𝑓 𝑖 𝑀𝐴= 2,500𝑁 125𝑁 𝑴𝑨=𝟐𝟎
Section 1: Work and Machines
Simple machine – a machine that does work with only one movement There are six (6) simple machines divided into two types: Compound machine – a machine that consists of two or more simple machines used together The lever type The inclined plane type Includes: Lever Pulley Wheel and axle Ramp Wedge Screw
Section 1: Work and Machines
Lever – a bar that is free to pivot, or turn, about a fixed point. There are three classes of levers: 1st class lever – the fulcrum is between the effort and the resistance Multiplies effort force and changes its direction Examples: crow bars, teeter-totters 2nd class lever – the resistance force is between the effort force and the fulcrum Multiplies force without changing direction Examples: wheel barrows, doors 3rd class lever – the effort force is between the fulcrum and the resistance force The effort force is always greater than the resistance force. MA < 1 Examples: the fore-arm, fishing poles If the 3rd class lever has no mechanical advantage, why use one?
Section 1: Work and Machines
Calculating the mechanical advantage of levers Equation: 𝑴𝑨= 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒐𝒇 𝒆𝒇𝒇𝒐𝒓𝒕 𝒂𝒓𝒎 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒐𝒇 𝒓𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒂𝒓𝒎 , or: 𝑴𝑨= 𝒅 𝒆 𝒅 𝒓 the distances are measured from the fulcrum to the point where the forces are acting Example: If the distance of the effort force is 3-m, and the distance of the resistance arm is 1-m, what is the mechanical advantage of the lever? Solution: Notice the distance units cancel. Remember, mechanical advantage is just a number. de = 3.0m dr = 1.0m MA = ? 𝑀𝐴= 𝑑 𝑒 𝑑 𝑟 𝑀𝐴= 3.0𝑚 1.0𝑚 𝑴𝑨=𝟑.𝟎
Section 1: Work and Machines
Pulleys The two sides of the pulley are the effort arm and the resistance arm. A fixed pulley changes the direction of the force only, it does not increase force A moveable pulley will increase the effort Block-and-tackle – a system of pulleys consisting of fixed and moveable pulleys. The block-and-tackle will multiply the effort force Wheel-and-axle – a machine consisting of two wheels of different sizes that rotate together Inclined plane (ramp) – a sloping surface that reduces the amount of force required to do work The same amount of work is done by lifting a box straight up or by sliding it up a ramp. However, the ramp reduces the amount of force required by increasing the distance Mechanical advantage of a ramp: 𝑴𝑨= 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇𝒊𝒏𝒄𝒍𝒊𝒏𝒆 𝒉𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒓𝒂𝒎𝒑 , or: 𝑴𝑨= 𝒍 𝒉 Example: Jessica uses a ramp 5-m long to raise a box to a height of 1-m. What is the mechanical advantage of the ramp? Solution Length =5.0m Height = 1.0m MA = ? 𝑀𝐴= 𝑙 ℎ 𝑀𝐴= 5.0𝑚 1.0𝑚 𝑴𝑨=𝟓.𝟎
Section 1: Work and Machines
Screw – an inclined plane wrapped around a cylinder Wedge – an inclined plane with one or two sloping sides Mechanical Efficiency (ME) Recall that the amount of work done by the machine (work output) is always less than the work done on the machine (work input) Mechanical Efficiency is the measure of how much of the work put into a machine is changed into useful output work by the machine Because of friction no machine is 100% efficient. ME will always be less than 100% Equation: 𝑴𝑬= 𝒘𝒐𝒓𝒌 𝒐𝒖𝒕𝒑𝒖𝒕 𝒘𝒐𝒓𝒌 𝒊𝒏𝒑𝒖𝒕 𝐱𝟏𝟎𝟎%, or: 𝑴𝑬= 𝒘 𝒐 𝒘 𝒊 𝒙𝟏𝟎𝟎% Example: John is changing a flat tire on his truck. He does 2,500J of work on the jack, while the jack does 2,100J of work on the car. How efficient is the jack? Solution wi = 2,500J wo = 2,100J ME = ? 𝑀𝐸= 𝑤 𝑜 𝑤 𝑖 𝑥100% 𝑀𝐸= 2,100𝐽 2,500𝐽 𝑥100% 𝑀𝐸=0.84𝑥100% 𝑴𝑬=𝟖𝟒%
Section 2: Describing Energy
Energy – the ability to cause change Energy comes in different forms chemical, electrical, thermal, etc. We will be looking at three (3) types of energy: kinetic, potential, and mechanical. Kinetic Energy (KE) KE is energy in a moving object Anything that moves has kinetic energy Kinetic energy depends of two things: 1. the mass of the moving object 2. the velocity of at which the object is moving Equation for kinetic energy: Unit for energy: 𝐾𝐸=𝑘𝑔( 𝑚 𝑠 ) 2 𝐾𝐸=𝑘𝑔 𝑚 2 𝑠 2 𝐊𝐄= kg m 2 s 2 =Nm=𝐉 𝑲𝑬= 𝟏 𝟐 𝒎 𝒗 𝟐 Where: KE = kinetic energy M = mass (kg) V = velocity (m/s)
Section 2: Describing Energy
Example: A ball, m = 1.5-kg, is rolling across the floor towards the door at 2 m/s. What is the KE of the rolling ball? Solution Important: Always square the velocity before you do any multiplication Potential Energy Potential energy – energy stored due to an object’s position Three types of potential energy: Elastic – PE stored by things that stretch or compress Ex.: rubber bands, springs, pole vault poles Chemical – PE stored in chemicals bonds Ex.: nuclear weapons and fuels Gravitational – PE stored by things that are elevated Ex.: fruit on trees, bouncing balls m = 1.5-kg v = 2.0-m/s KE = ? 𝐾𝐸= 1 2 𝑚 𝑣 2 𝐾𝐸=( 1 2 )(1.5𝑘𝑔)( 2.0 𝑚 𝑠 ) 2 𝐾𝐸= 𝑘𝑔 𝑚 2 𝑠 2 𝑲𝑬=𝟑.𝟎 𝒌𝒈 𝒎 𝟐 𝒔 𝟐 =𝟑.𝟎 𝑱
Section 2: Describing Energy
The amount of potential energy can be determined mathematically. We will focus on gravitational PE Equation for gravitational PE: Example: An apple, mass = 0.5-kg, is hanging from a branch 4.0-m above the ground. What is its gravitational PE? Solution 𝑷𝑬=𝒎𝒈𝒉 PE = Potential Energy (J) M = mass (kg) g = 9.8 m/s2 H = height (m) m = 0.5 kg h = 4.0 m PE = ? 𝑃𝐸=𝑚𝑔ℎ 𝑃𝐸=0.5𝑘𝑔 9.8 𝑚 𝑠 𝑚 𝑃𝐸=19.6 𝑘𝑔 𝑚 2 𝑠 2 𝑷𝑬=𝟏𝟗,𝟔 𝑱
Section 3: Conservation of Energy
Mechanical Energy – the total amount of potential and kinetic energy in a system Equation: mechanical energy = potential energy + kinetic energy, or: Example: An object held in the air has a gravitational PE of 480.0J. What is its kinetic energy if it has fallen two-thirds of the way to the ground? Law of Conservation of Energy: Energy is neither created nor destroyed On a large scale: total energy in the universe is constant Consequence: energy can change form: potential kinetic kinetic thermal chemical mechanical Solution Before the object started falling ME = PE, so ME = 480.0J. As the object is falling PE is being converted to KE. At anytime during the fall ME = PE + KE. When the object is two-thirds of the way down ME = 1/3PE + 2/3KE. So: KE = 2/3(480.0 J), or KE = 320.0J
Section 3: Conservation of Energy
Power – the amount of work done in a certain amount of time Power is a rate Equation for calculating power: Units for Power: the Watt (W) 1 watt is about equal to the power required to lift a glass of water from a table to your mouth Example 1: It took 20 seconds to move a refrigerator, You did 3,150 J of work in the process. How much power was required to move the refrigerator? Solution Example 2: It took you 1.5 s to lift a 10-kg box of the floor to a height of 1.0-m. How much work did you do on the box, and how much power was required to do this? Solution |
# Commutative Property of Multiplication
Commutative property of multiplication:
$\color{purple}{a\times b=b\times a}$
What does it mean?
It means, that order of numbers doesn't matter.
Indeed, as can be seen from illustration, we can count there are 3 circles in a row, and there are 2 rows, so total number of squares is ${3}\times{2}={6}$.
From another side, we can rotate picture and count circles in another way: 3 rows, and in each row 2 circles: ${2}\times{3}={6}$.
Warning: it doesn't work with division, i.e. $\frac{{a}}{{b}}\ne\frac{{b}}{{a}}$.
For example, $\frac{{3}}{{2}}\ne\frac{{2}}{{3}}$.
However, commutative property of multiplication works for negative numbers (in fact, for real numbers) as well.
Example 1. ${3}\times{\left(-{4}\right)}={\left(-{4}\right)}\times{3}=-{12}$.
Example 2. ${\left(-{2.51}\right)}\times{\left(-{3.4}\right)}={\left(-{3.4}\right)}\times{\left(-{2.51}\right)}$.
Example 3. $-\frac{{5}}{{8}}\times\frac{{2}}{{3}}={\left(-\frac{{5}}{{8}}\right)}\times\frac{{2}}{{3}}=-\frac{{5}}{{12}}$.
Conclusion. So, the basic rule here is following: whenever you see multiplication, you can interchange factors. |
In the previous chapter, we developed an algorithm to draw a triangle filled with a solid color. Our goal for this chapter is to draw a shaded triangle—that is, a triangle filled with a color gradient.
Defining Our Problem
We want to fill the triangle with different shades of a single color. It will look like Figure 8-1.
We need a more formal definition of what we’re trying to draw. To do this, we’ll assign a real value $$h$$ to each vertex, denoting the intensity of the color at the vertex. $$h$$ is in the $$[0.0, 1.0]$$ range, where $$0.0$$ represents the darkest possible shade (that is, black) and $$1.0$$ represents the brightest possible shade (that is, the original color—not white!).
To compute the exact color shade of a pixel given the base color of the triangle $$C$$ and the intensity at that pixel $$h$$, we’ll multiply channel-wise: $$C_h = (R_C \cdot h, G_C \cdot h, B_C \cdot h)$$. Therefore $$h = 0.0$$ yields pure black, $$h = 1.0$$ yields the original color $$C$$, and $$h = 0.5$$ yields a color half as bright as the original one.
In order to draw a shaded triangle, all we need to do is compute a value of $$h$$ for each pixel of the triangle, compute the corresponding shade of the color, and paint the pixel. Easy!
At this point, however, we only know the values of $$h$$ for the triangle vertices, because we chose them. How do we compute values of $$h$$ for the rest of the triangle?
Let’s start with the edges of the triangle. Consider the edge $$AB$$. We know $$h_A$$ and $$h_B$$. What happens at $$M$$, the midpoint of $$AB$$? Since we want the intensity to vary smoothly from $$A$$ to $$B$$, the value of $$h_M$$ must be between $$h_A$$ and $$h_B$$. Since $$M$$ is in the middle of $$AB$$, why not choose $$h_M$$ to be in the middle of $$h_A$$ and $$h_B$$—that is, their average?
More formally, we have a function $$h = f(P)$$ that gives each point $$P$$ an intensity value $$h$$; we know its values at $$A$$ and $$B$$, $$h(A) = h_A$$ and $$h(B) = h_B$$, respectively. We want this function to be smooth. Since we know nothing else about $$h = f(P)$$, we can choose any function that is compatible with what we do know, such as a linear function (Figure 8-2).
This is suspiciously similar to the situation in the previous chapter: we had a linear function $$x = f(y)$$, we knew the values of this function at the vertices of the triangle, and we wanted to compute values of $$x$$ along its sides. We can compute values of $$h$$ along the sides of the triangle in a very similar way, using Interpolate with y as the independent variable (the values we know) and h as the dependent variable (the values we want):
x01 = Interpolate(y0, x0, y1, x1)
h01 = Interpolate(y0, h0, y1, h1)
x12 = Interpolate(y1, x1, y2, x2)
h12 = Interpolate(y1, h1, y2, h2)
x02 = Interpolate(y0, x0, y2, x2)
h02 = Interpolate(y0, h0, y2, h2)
Next, we concatenated the $$x$$ arrays for the “short” sides and then determined which of x02 and x012 was x_left and which was x_right. Again, we can do something very similar here for the $$h$$ vectors.
However, we will always use the $$x$$ values to determine which side is left and which side is right, and the $$h$$ values will just “follow along.” $$x$$ and $$h$$ are properties of actual points on the screen, so we can’t freely mix-and-match left- and right-side values.
We can code this as follows:
// Concatenate the short sides
remove_last(x01)
x012 = x01 + x12
remove_last(h01)
h012 = h01 + h12
// Determine which is left and which is right
m = floor(x012.length / 2)
if x02[m] < x012[m] {
x_left = x02
h_left = h02
x_right = x012
h_right = h012
} else {
x_left = x012
h_left = h012
x_right = x02
h_right = h02
}
This is very similar to the relevant section of the code in the previous chapter (Listing 7-1), except that every time we do something with an x vector, we do the same with the corresponding h vector.
The last step is drawing the actual horizontal segments. For each segment, we know x_left and x_right, as in the previous chapter; now we also know h_left and h_right. But this time we can’t just iterate from left to right and draw every pixel with the base color: we need to compute a value of $$h$$ for each pixel of the segment.
Again, we can assume $$h$$ varies linearly with $$x$$, and use Interpolate to compute these values. In this case, the independent variable is $$x$$, and it goes from the x_left value to the x_right value of the specific horizontal segment we’re shading; the dependent variable is $$h$$, and its corresponding values for x_left and x_right are h_left and h_right for that segment:
x_left_this_y = x_left[y - y0]
h_left_this_y = h_left[y - y0]
x_right_this_y = x_right[y - y0]
h_right_this_y = h_right[y - y0]
h_segment = Interpolate(x_left_this_y, h_left_this_y,
x_right_this_y, h_right_this_y)
Or, expressed in a more compact way:
h_segment = Interpolate(x_left[y - y0], h_left[y - y0],
x_right[y - y0], h_right[y - y0])
Now it’s just a matter of computing the color for each pixel and painting it! Listing 8-1 shows the complete pseudocode for DrawShadedTriangle.
DrawShadedTriangle (P0, P1, P2, color) {
❶// Sort the points so that y0 <= y1 <= y2
if y1 < y0 { swap(P1, P0) }
if y2 < y0 { swap(P2, P0) }
if y2 < y1 { swap(P2, P1) }
// Compute the x coordinates and h values of the triangle edges
x01 = Interpolate(y0, x0, y1, x1)
h01 = Interpolate(y0, h0, y1, h1)
x12 = Interpolate(y1, x1, y2, x2)
h12 = Interpolate(y1, h1, y2, h2)
x02 = Interpolate(y0, x0, y2, x2)
h02 = Interpolate(y0, h0, y2, h2)
// Concatenate the short sides
remove_last(x01)
x012 = x01 + x12
remove_last(h01)
h012 = h01 + h12
// Determine which is left and which is right
m = floor(x012.length / 2)
if x02[m] < x012[m] {
x_left = x02
h_left = h02
x_right = x012
h_right = h012
} else {
x_left = x012
h_left = h012
x_right = x02
h_right = h02
}
// Draw the horizontal segments
❷for y = y0 to y2 {
x_l = x_left[y - y0]
x_r = x_right[y - y0]
❸h_segment = Interpolate(x_l, h_left[y - y0], x_r, h_right[y - y0])
for x = x_l to x_r {
❹shaded_color = color * h_segment[x - x_l]
}
}
}
The pseudocode for this function is very similar to that for the function developed in the previous chapter (Listing 7-1). Before the horizontal segment loop ❷, we manipulate the $$x$$ vectors and the $$h$$ vectors in similar ways, as explained above. Inside the loop, we have an extra call to Interpolate ❸ to compute the $$h$$ values for every pixel in the current horizontal segment. Finally, in the inner loop we use the interpolated values of $$h$$ to compute a color for each pixel ❹.
Note that we’re sorting the triangle vertices as before ❶. However, we now consider these vertices and their attributes, such as the intensity value $$h$$, to be an indivisible whole; that is, swapping the coordinates of two vertices must also swap their attributes.
Source code and live demo >>
Summary
In this chapter, we’ve extended the triangle-drawing code developed in the previous chapter to support smoothly shaded triangles. Note that we can still use it to draw single color triangles by using 1.0 as the value of $$h$$ for all three vertices.
The idea behind this algorithm is actually more general than it seems. The fact that $$h$$ is an intensity value has no impact on the “shape” of the algorithm; we assign meaning to this value only at the very end, when we’re about to call PutPixel. This means we could use this algorithm to compute the value of any attribute of the vertices of the triangle, for every pixel of the triangle, as long as we assume this value varies linearly on the screen.
We will indeed use this algorithm to improve the visual appearance of our triangles in the upcoming chapters. For this reason, it’s a good idea to make sure you really understand this algorithm before proceeding further.
In the next chapter, however, we take a small detour. Having mastered the drawing of triangles on a 2D canvas, we will turn our attention to the third dimension.
Found this interesting? Buy the author a coffee ☕
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Computer Graphics From Scratch · Dedication | Acknowledgements | Table of Contents | Introduction | Introductory Concepts
Part I: Raytracing · Basic Raytracing | Light | Shadows and Reflections | Extending the Raytracer
Part II: Rasterization · Lines | Filled Triangles | Shaded Triangles | Perspective Projection | Describing and Rendering a Scene | Clipping | Hidden Surface Removal | Shading | Textures | Extending the Rasterizer
Appendixes · Linear Algebra | Afterword |
# The Proof of the Polygon Angle Sum Theorem
Introduction
We have learned that the angle sum of a triangle is $180^\circ$. We have also learned that the angle sum of a quadrilateral is $360^\circ$. In getting the angle sum of quadrilaterals, we divided the quadrilateral into two triangles by drawing a diagonal. In this post, we use this method to find the angle sum of the pentagon and other polygons.
Let us extend the method stated above to pentagon (5-sided polygon). Clearly, we can divide the pentagon into three non-overlapping triangles by drawing two diagonals. Since each triangle has an angle sum of $180^\circ$, the angle sum of a pentagon, which is composed of three triangles, is $540^\circ$.
Using the method above, we can see the pattern on the table below. The sum of a polygon with $n$ sides is $180(n-2)$ degrees. Next, we summarize the polygon angle sum theorem and prove it.
Theorem
The angle sum of a polygon with $n$ sides is 180(n-2) degrees.
Proof
A polygon with $n$ sides can be divided into $n - 2$ triangles. Since the angle sum of a triangle is $180^\circ$, the angle sum of a polygon is $180^\circ (n-2)$.
Exercise
In the proof above, we have only considered the angle sum of a convex polygons. Does this theorem also hold on non-convex polygons? Explain your answer. |
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### Course: 4th grade>Unit 1
Lesson 1: Intro to place value
# Place value: FAQ
## What is place value?
Place value refers to the value that each digit in a number has, based on its position. For example, in the number $523$, the $5$ is in the hundreds place, the $2$ is in the tens place, and the $3$ is in the ones place.
Here are a couple of the exercises that build off of place value:
## What are the different ways we can write numbers?
We can write numbers in standard form, expanded form, or written form. Expanded form breaks a number down to show the value of each digit. For example, $724$ can be written in expanded form as $700+20+4$. Written form uses words to write out a number. For example, $724$ can be written in written form as "seven hundred twenty-four." Standard form is the way we usually write numbers, using just the digits. For example, $724$ is in standard form.
Here are a couple of the exercises that build off of writing numbers in different forms:
## Why is $10$ important in place value?
$10$ is important in place value because our number system is based on the number $10$. This means that when we count past $9$ in one place value column, we regroup to the next place value (in this case, the tens place).
Here are a couple of the exercises that build off of using $10$ in place value:
## How do I compare multi-digit numbers?
To compare multi-digit numbers, we start by looking at the largest digit, or the leftmost digit. If one number has a larger digit in that place, we know that number is larger. If the digits are the same, we move to the next digit to the right and continue comparing until we find a difference.
Want to learn more about comparing numbers? Check out this exercise: Compare multi-digit numbers
## Why do I need to know about place value?
Understanding place value is important for many reasons. It can help us add, subtract, multiply, and divide multi-digit numbers. It also helps us understand the value of numbers and compare them. Place value is used in all kinds of real-world situations, from counting money to measuring quantities.
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# 1 Topic 6.2.1 Rules of Exponents. 2 Lesson 1.1.1 California Standards: 2.0 Students understand and use such operations as taking the opposite, finding.
## Presentation on theme: "1 Topic 6.2.1 Rules of Exponents. 2 Lesson 1.1.1 California Standards: 2.0 Students understand and use such operations as taking the opposite, finding."— Presentation transcript:
1 Topic 6.2.1 Rules of Exponents
2 Lesson 1.1.1 California Standards: 2.0 Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0 Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll multiply and divide algebraic expressions using the rules of exponents. Rules of Exponents Topic 6.2.1 Key words: exponent
3 Lesson 1.1.1 You learned about the rules of exponents in Topic 1.3.1. Rules of Exponents Topic 6.2.1 In this Topic, you’ll apply those same rules to monomials and polynomials. We’ll start with a quick recap of the rules of exponents to make sure you remember them all.
4 1) x a · x b = x a + b 2) x a ÷ x b = x a – b (if x 0) 3) ( x a ) b = x ab 4) ( cx ) b = c b x b 5) x 0 = 1 6) x – a = (if x 0) 7) xaxa 1 Lesson 1.1.1 Use the Rules of Exponents to Simplify Expressions Rules of Exponents Topic 6.2.1 These are the same rules you learned in Chapter 1, but this time you’ll use them to simplify algebraic expressions: Rules of Exponents
5 (–2 x 2 m )(–3 x 3 m 3 ) = (–2)(–3)( x 2 )( x 3 )( m )( m 3 ) = 6 x 2+3 · m 1+3 = 6 x 5 m 4 Rules of Exponents Example 1 Topic 6.2.1 Simplify the expression (–2 x 2 m )(–3 x 3 m 3 ). Solution Solution follows… Put all like variables together Use Rule 1 and add the powers Rule 1) x a · x b = x a + b
6 (3 a 2 xb 3 ) 2 = 3 2 · a 2·2 · x 2 · b 3·2 = 9 a 4 x 2 b 6 Rules of Exponents Example 2 Topic 6.2.1 Simplify the expression (3 a 2 xb 3 ) 2. Solution Solution follows… Use Rules 3 and 4 Rule 3) ( x a ) b = x ab Rule 4) ( cx ) b = c b x b
7 = 2 xm 2 Rules of Exponents Example 3 Topic 6.2.1 Solution Solution follows… From Rule 5, anything to the power 0 is 1 Simplify the expression. Separate the expression into parts that have only one variable = =Use Rule 2 and subtract the powers Rule 2) x a ÷ x b = x a – b (if x 0) Rule 5) x 0 = 1
8 1. –3 at (4 a 2 t 3 ) 2. (–5 x 3 yt 2 )(–2 x 2 y 3 t ) 3. (–2 x 2 y 3 ) 3 4. –2 mx (3 m 2 x – 4 m 2 x + m 3 x 3 ) 5. (–3 x 2 t ) 3 (–2 x 3 t 2 ) 2 6. –2 mc (–3 m 2 c 3 + 5 mc ) Simplify each expression. Lesson 1.1.1 Guided Practice Rules of Exponents Topic 6.2.1 Solution follows… (–3 4)( a a 2 )( t t 3 ) = –12 a (1 + 2) t (1 + 3) (Rule 1) = –12 a 3 t 4 (–5 –2)( x 3 x 2 )( y y 3 )( t 2 t ) = 10 x (3 + 2) y (1 + 3) t (2 + 1) (Rule 1) = 10 x 5 y 4 t 3 (–2) (1 3) x (2 3) y (3 3) (Rule 3) = (–2) 3 x 6 y 9 = –8 x 6 y 9 –2 mx (– m 2 x + m 3 x 3 ) = 2 m (1 + 2) x (1 + 1) – 2 m (1 + 3) x (1 + 3) (Rule 1) = 2 m 3 x 2 – 2 m 4 x 4 ((–3) 3 x (2 3) t 3 )((–2) 2 x (3 2) t (2 2) ) (Rule 3) = (–27 x 6 t 3 )(4 x 6 t 4 ) = –108 x (6 + 6) t (3 + 4) (Rule 1) = –108 x 12 t 7 6 m (1 + 2) c (1 + 3) – 10 m (1 + 1) c (1 + 1) (Rule 1) = 6 m 3 c 4 – 10 m 2 c 2
9 Simplify each expression. Lesson 1.1.1 Guided Practice Rules of Exponents Topic 6.2.1 Solution follows… 7. 8. 9. 10. = 5 m (3 – 2) n (8 – 3) z (6 – 1) (Rule 2) = 5 mn 5 z 5 = (14 ÷ 4) a (2 – 7) b (4 – 4) c (8 – 0) (Rule 2) = a –5 c 8 (Rule 5) = (Rule 6) = (12 ÷ 8) j (8 – 2) k (–8 – –10) m (–1 – 4) (Rule 2) = j 6 k 2 m –5 = (Rule 6) = (16 ÷ 32) b (9 – 5 2) a (4 – 3 2) c (–1 2) j 4 (Rule 2) = b –1 a –2 c –2 j 4 = (Rule 6)
10 1. 2. 3. 4 a 2 ( a 2 – b 2 )4. 4 m 2 x 2 ( x 2 + x + 1) 5. a ( a + 4) + 4( a + 4)6. 2 a ( a – 4) – 3( a – 4) 7. m 2 n 3 ( mx 2 + 3 nx + 2) – 4 m 2 n 3 8. 4 m 2 n 2 ( m 3 n 8 + 4) – 3 m 3 n 10 ( m 2 + 2 n 3 ) Simplify. Rules of Exponents Independent Practice Solution follows… Topic 6.2.1 4 a 4 – 4 a 2 b 2 a 2 + 8 a + 16 m 3 n 3 x 2 + 3 m 2 n 4 x – 2 m 2 n 3 2 a 2 – 11 a + 12 m 5 n 10 + 16 m 2 n 2 – 6 m 3 n 13 4 m 2 x 4 + 4 m 2 x 3 + 4 m 2 x 2 1
11 9. 10. 11. 12. Simplify. Rules of Exponents Independent Practice Solution follows… Topic 6.2.1
12 13. m ? ( m 4 + 2 m 3 ) = m 6 + 2 m 5 14. m 4 a 6 (3 m ? a 8 + 4 m 2 a ? ) = 3 m 7 a 14 + 4 m 6 a 9 Find the value of ? that makes these statements true. Rules of Exponents Independent Practice Solution follows… Topic 6.2.1 ? = 2 ? = 3 ? = 4 ? = 7 15. 16.
13 Topic 6.2.1 Round Up Rules of Exponents You can apply the rules of exponents to any algebraic values. In this Topic you just dealt with monomials, but the rules work with expressions with more than one term too.
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# There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from $A$ to $B$. But if 20 students are sent from $\mathrm{B}$ to $\mathrm{A}$, the number of students in $\mathrm{A}$ becomes double the number of students in B. Find the number of students in the two halls.
Given:
There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from $A$ to $B$. But if 20 students are sent from $\mathrm{B}$ to $\mathrm{A}$, the number of students in $\mathrm{A}$ becomes double the number of students in B.
To do:
We have to find the number of students in the two halls.
Solution:
Let the number of students in hall A and the number of students in hall B be $x$ and $y$ respectively.
If 10 candidates are sent from A to B, the number of students in each hall is same.
This implies,
$x-10=y+10$
$x=y+10+10$
$x=y+20$.....(i)
If 20 candidates are sent from B to A, the number of students in A is double the number of students in B.
$x + 20 = 2(y-20)$
$x + 20 = 2y-40$
$y+20+20=2y-40$ (From (i))
$2y-y=40+40$
$y=80$
Substituting $y=80$ in equation (i), we get,
$x=80+20$
$x=100$
Therefore, the number of students in hall A is 100 and the number of students in hall B is 80.
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# Lesson Notes By Weeks and Term - Senior Secondary 1
Data collection and presentation
Term: 1st Term
Week: 7
Class: Senior Secondary School 1
Age: 15 years
Duration: 40 minutes of 2 periods each
Date:
Subject: Economics
Topic:- Data collection and presentation
SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to
1. Define the various measures of central tendency
2. Perform calculations involving the various measures of central tendency
INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source
INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures
INSTRUCTIONAL PROCEDURES
PERIOD 1-2
PRESENTATION TEACHER’S ACTIVITY STUDENT’S ACTIVITY STEP 1 INTRODUCTION The teacher reviews the previous lesson on the basic tools for economic analysis Students pay attention STEP 2 EXPLANATION She defines the various measures of central tendency Students pay attention and participates STEP 3 DEMONSTRATION She performs calculations on each Students pay attention and participate STEP 4 NOTE TAKING The teacher writes a summarized note on the board The students copy the note in their books
NOTE
MEASURES OF CENTRAL TENDENCY
Measures of central tendency means are values which show the degree to
which a given data or any given set of values will converge toward the
central point of the data. It is also called measure of location and is the
statistical information that gives the middle or centre or average of a set of
data. It includes mean, median and mode.
THE MEAN
Mean or arithmetic mean is defined as the sum of series of figures divided
by the number of observations. It is the commonest and the most widely
used among the other types of averages or measures of central tendency.
TYPES OF MEAN
1. The Arithmetic Mean
2. The Geometric Mean
Example
Calculate the arithmetic mean of the following scores of eight students in
an economics test. The scores are: 14, 18, 24, 16, 30, 12, 20, and 10.
Solution
14+18+24+16+30+12+20+10 = 144
Number of observation (students) = 8
Arithmetic Mean =Sum of observations divided by Number of observations
= 144/8 =18
1. It is easy to derive or calculate.
2. It is easy to interpret.
3. It is the best-known average.
4. It has determinate exact value.
5. It provides a good measure of comparison.
1. It is difficult to determine without calculation.
2. Some facts may be concealed.
3. It cannot be obtained graphically.
4. If one or more value is incorrect or missing, the calculation becomes difficult.
5. It may lead to distorted results.
THE MEDIAN
The median is an average which is the middle value when figures are
arranged in their order of magnitude either in ascending or descending
order, especially from ungrouped data.
1. It is easy to determine with little or no calculations
2. It is easy to understand and compute
3. It does not use all values in the distribution
4. It gives a clean idea of the distribution
1. It is not useful in further statistical calculation
2. It ignores very large or small values
3. It does not represent a true average of the set of data.
THE MODE
This is the most frequently recurring number in a set of numbers or data,
that is to say, it is the number or value with the highest frequency. It tells us
the observation which is most popular. The best and easiest way of
calculating the mode of any distribution is to form a frequency table for it.
1. It can be easily understood.
2. It is not affected by extreme values.
3. Easy to calculate from the graph.
4. It is easy to determine.
1. It can be a poor average.
2. It can be difficult to compute if more than one mode exists.
3. It is not useful in further statistical calculations.
4. All the values used in the distribution are not considered.
FORMULATION OF FREQUENCY TABLE FOR UNGROUPED DATA
UNGROUPED DATA:
Ungrouped data is one in which the raw data has occurrences or
frequencies more than and are without class intervals. In the formulation of
a frequency table for ungrouped data, two basic steps are taken.
1. Prepare a tally sheet.
2. Prepare a frequency table.
PREPARATION OF A TALLY SHEET: This is when the variables are taken one after the other with a stroke called tally. The tally of five makes a bundle.
PREPARATION OF A FREQUENCY TABLE: The frequency table is simply obtained by adding the tallies together in a separate column referred to as frequency.
Example: The following are scores of thirty (30) students of SS 1 in an
economics test.
2, 4, 8, 8, 2, 6, 6, 8, 2, 4
8, 0, 8, 6, 0, 10, 2, 2, 0, 10
4, 6, 0, 10, 2, 2, 6, 6, 4, 2
1. Prepare a frequency table inclusive of a tally
2. Find the mean
3. Find the mode
4. Find the median
SOLUTION
NUMBERS FREQUENCY TALLY 0 4 //// 2 8 //// /// 4 4 //// 6 6 //// / 8 5 //// 10 3 ///
1. Arithmetic mean = 0 x 4 + 2 x 8 + 4 x 4 + 6 x 6 + 8 x 5 + 10 x 3
30
= 0 + 16 + 16 + 36 + 40 + 30
30
= 138
30
= 4.6
1. Median = 4 + 4
2
= 8
2
= 4
1. Mode = 2
EVALUATION: 1. Define the following measures of central tendency
a. Mean
b. Median
c. Mode
a. Mean
b. Median
c. Mode
3. The following are scores of 20 students in an Economics test.
5 10 2 9 5 3 4
6 1 3 2 3 6 1
3 3 2 3 4 3
a. Prepare a frequency table with tally
Find the
b. Mean
c. Median
d. Mode
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively |
<meta http-equiv="refresh" content="1; url=/nojavascript/">
Absolute Value Equations
%
Progress
Practice Absolute Value Equations
Progress
%
Absolute Value Equations
What if you were asked to solve an absolute value equation like $|3x-4|=5$ ? How could you interpret the solution? After completing this Concept, you'll be able to interpret the solutions to absolute value equations like this one by graphing them on a number line.
Guidance
In the previous concept, we saw how to solve simple absolute value equations. In this concept, you will see how to solve more complicated absolute value equations.
Example A
Solve the equation $|x-4|=5$ and interpret the answers.
Solution
We consider two possibilities: the expression inside the absolute value sign is non-negative or is negative. Then we solve each equation separately.
$& x-4 = 5 \quad \text{and} \quad x-4=-5\\& \quad \ \ x=9 \qquad \qquad \quad \ x=-1$
$x = 9$ and $x = -1$ are the solutions.
The equation $|x-4|=5$ can be interpreted as “what numbers on the number line are 5 units away from the number 4?” If we draw the number line we see that there are two possibilities: 9 and -1.
Example B
Solve the equation $|x+3|=2$ and interpret the answers.
Solution
Solve the two equations:
$& x+3 = 2 \quad \text{and} \quad \ \ x+3=-2\\& \quad \ \ x=-1 \qquad \qquad \quad \ x=-5$
$x = -5$ and $x = -1$ are the answers.
The equation $|x+3|=2$ can be re-written as: $|x-(-3)|=2$ . We can interpret this as “what numbers on the number line are 2 units away from -3?” There are two possibilities: -5 and -1.
Solve Real-World Problems Using Absolute Value Equations
Example C
A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces, but it is hard to fill each bag to the exact weight. After being filled, each bag is weighed; if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags?
Solution
The weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the difference between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if $x$ is the weight of a bag in ounces, then the equation that describes this problem is $|x-16|=0.25$ .
Now we must consider the positive and negative options and solve each equation separately:
$& x-16 = 0.25 \qquad \text{and} \quad x-16 =-0.25\\& \qquad x=16.25 \qquad \qquad \qquad \ \ x=15.75$
The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces.
We see that $16.25 - 16 = 0.25 \ ounces$ and $16 - 15.75 = 0.25 \ ounces$ . The answers are 0.25 ounces bigger and smaller than 16 ounces respectively.
The answer you just found describes the lightest and heaviest acceptable bags of coffee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again.
Watch this video for help with the Examples above.
Vocabulary
• The absolute value of a number is its distance from zero on a number line.
• $|x|=x$ if $x$ is not negative, and $|x|=-x$ if $x$ is negative.
• An equation or inequality with an absolute value in it splits into two equations, one where the expression inside the absolute value sign is positive and one where it is negative. When the expression within the absolute value is positive , then the absolute value signs do nothing and can be omitted. When the expression within the absolute value is negative, then the expression within the absolute value signs must be negated before removing the signs.
• Inequalities of the type $|x| can be rewritten as “ $-a < x < a$ .”
• Inequalities of the type $|x|>b$ can be rewritten as “ $x < -b$ or $x > b$ .”
Guided Practice
Solve the equation $|2x-7|=6$ and interpret the answers.
Solution
Solve the two equations:
$& 2x-7 = 6 \qquad \qquad \quad 2x-7=-6\\& \quad \ \ 2x=13 \qquad \text{and} \qquad \ \ 2x=1\\& \quad \ \ \ x=\frac{13}{2} \qquad \qquad \qquad \ \ x=\frac{1}{2}$
Answer: $x=\frac{13}{2}$ and $x=\frac{1}{2}$ .
The interpretation of this problem is clearer if the equation $|2x-7|=6$ is divided by 2 on both sides to get $\frac{1}{2}|2x-7|=3$ . Because $\frac{1}{2}$ is nonnegative, we can distribute it over the absolute value sign to get $\left | x-\frac{7}{2} \right |=3$ . The question then becomes “What numbers on the number line are 3 units away from $\frac{7}{2}$ ?” There are two answers: $\frac{13}{2}$ and $\frac{1}{2}$ .
Explore More
Solve the absolute value equations and interpret the results by graphing the solutions on the number line.
1. $|x-5|=10$
2. $|x+2|=6$
3. $|5x-2|=3$
4. $|x-4|=-3$
5. $\left|2x-\frac{1}{2}\right|=10$
6. $|-x+5|=\frac{1}{5}$
7. $\left|\frac{1}{2}x-5\right|=100$
8. $|10x-5|=15$
9. $|0.1x+3|=0.015$
10. $|27-2x|=3x+2$
Vocabulary Language: English
Absolute Value
Absolute Value
The absolute value of a number is the distance the number is from zero. Absolute values are never negative.
linear equation
linear equation
A linear equation is an equation between two variables that produces a straight line when graphed. |
# How Do You Write An Equation From A Graph
To write an equation in slope-intercept form, given a graph of that equation, pick two points on the line and use them to find the slope. This is the value of m in the . You can use this equation to write an equation if you know the slope and the y- intercept. We can find the b-value, the y-intercept, by looking at the graph.
## WRITE THE EQUATION THAT REPRESENTED BY THE GRAPH BASED ON THE Y INTERCEPT AND THE SLOPE
Where m is the slope of the line and b is the y-intercept. You can use this equation to write an equation if you know the slope and the y-intercept. Every straight line can be represented by an equation: y = mx + b. The coordinates of every point on the line will solve the equation if you substitute them in the.
### HOW TO FIND Y INTERCEPT
The y-intercept of an equation is a point where the graph of the equation intersects the Y-axis. Finding the Y-Intercept from the Slope and Point. Learn the slope-intercept form of an equation. When you have a linear equation, the y-intercept is the point where the graph of the line crosses the y-axis. In this tutorial, learn about the y-intercept. Check it.
## SLOPE EQUATION
Learn how to write the slope formula from scratch and how to apply it to find the slope of a line from two points. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. If you a point that a line passes through, and its slope, this page.
## WRITING EQUATIONS FROM GRAPHS WORKSHEET
Elementary Algebra Skill. Writing Equations of Lines Given the Graph. Write the slope-intercept form of the equation of each line. 1). −5 −4 −3 −2 −1 0. 1 2 3 4 5. Worksheet by Kuta Software LLC. Kuta Software - Infinite Pre-Algebra Write the slope-intercept form of the equation of each line. 1). −5 −4 −3 −2 −1 0. 1 2 3 4 5.
### Y=MX+B
Every straight line can be represented by an equation: y = mx + b. The coordinates of every point on the line will solve the equation if you substitute them in the. Equation of a Straight Line. The equation of a straight line is usually written this way: y = mx + b. (or "y = mx + c" in the UK see below).
centrebadalona.com 2019. writing a website report |
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# Ackerman, Toe and a Few Other Random Thoughts
The diagram below outlines the important geometry in determining the motions of
the steer wheels in a vehicle that uses Ackerman steering geometry. Ackerman is an
interesting problem because it is dynamic. That is to say that we have two components
moving together the left and right steering knuckles, but the relationship between their
motions changes as we move them. This is a real head scratcher. Its a bit like having a
bowling ball in a dark room and throwing other bowling balls in an attempt to locate it by
listening for an impact. Every time you find the ball by crashing into it, it moves and
again dont know where it is.
D
C
King Pin Center to Center Distance
## Tie Rod Length
Ackerman Angle
B
A
RAA
Wheel Base
Luckily we have some mathematical voodoo that can help us figure it out. Lets
look at the important distances and angles. The two most fundamental distances are the
wheel base of the car and the kingpin center to center distance. If we draw two lines
representing the wheelbase and the distance from the cars center line to one of the king
pins, we can make a triangle. By design, the line that goes through the centers of the
Ackerman arm forms the hypotenuse of this triangle. See below.
Note that the
## King Pin Center to Center Distance
angle with its
vertex at A is
90 degrees by
design, unless
the vehicle has
2
been crashed.
If this angle
experiences an
unplanned C
to an impact, Angle
the car will dog
track. This can
be checked
using a tape
measure and B
A
comparing
distances from
side to side. Also note that the line that forms the Ackerman angle with the hypotenuse is
parallel with the thrust line, again by definition. Because of this, we can say that angle B
and the Ackerman angle are similar, so if we know one, we know the other. But angle B
isnt too hard to come up with. Recall that the tangent function gives the ratio between
the opposite side and the adjacent side of the triangle. So
## TAN Angle B = king pin center to center distance / 2
Wheelbase
The problem is that we know the distances and are trying to find angle B. We need the
inverse function ARCTAN. Rearranging, we get:
## ARCTAN king pin center to center distance / 2 = Angle B
Wheelbase
We can pick distances, turn the crank and find Angle B and by extension, the Ackerman
Angle. For example, lets choose a wheel base of 72 and a king pin to king pin distance
of 36. The formula would look as follows:
ARCTAN 36 / 2 = Angle B
72
## Plugging the number into a calculator or Excel, or looking up in a table,
ARCTAN (.25) = 14.036
So, the Ackerman Angle is 14.036 degrees. We can use this to find the length of the tie
rod.
D
C
King Pin Center to Center Distance
## Tie Rod Length
14.036
B
A
To find the length of the tie rod, we can decompose the trapezoid ABCD into a rectangle
and two triangles.
RAA
Wheel Base
D
C
B
Y
14.036
B
A
If you think logically about the diagram above, the length of the tie rod (segment BC in
the drawing) is equal to the king pin to king pin center distance minus distance Y on each
side. So, what is distance Y? To find out, you have to pick an Ackerman Arm Radius.
You may choose this by purchasing a standard Ackerman arm out of a catalog, or you
may design your own. Either way, it is what we might refer to as a drawing board
problem, meaning that basically this is a parameter that the engineer chooses by his gut.
Lets pick 6 to make life easy. So, how long is distance Y? Well, recall that the SIN of
an angle is the ratio between the side opposite the angle and the hypotenuse. In
shorthand it looks as follows.
## SIN 14.036 = Y/6
As you know, the name of the game in Algebra is getting the variable by itself, so
6 * SIN 14.036 = Y
You can look SIN 14.036 up in a table, or punch it up on a calculator, giving you:
6 * .243 = Y
## A few Dazzling Algebraic Contortions, And:
1.445 = Y
So, the tie rod is 1.445 inches shorter on the bottom and 1.445 inches shorter on the top
than the kingpin center to center distance. Expressed mathematically:
Where:
## LT is the length of the tie rod
DKC is the distance between king pins center to center
RAA is the radius of the Ackerman Arm
## Plugging in our numbers
LT = 36 2*6*SIN 14.036
Looking up the SIN value
LT = 36 2*6*.243
## Turning the crank
LT = 33.084 D
C
So, for a car configured as this one is, the tie rod needs to be 33.084 from Something
the center of one rod end to the center of the other. less than 20
Weve figured out all of the static values. Now the real fun begins.
Lets contemplate a turn as diagrammed in red to the right. Suppose that
the Ackerman arm labeled AB steers 20 degrees to the left as shown. What
angle does the other Ackerman arm transect? You might think 20 degrees,
but this would result in the steering wheels being parallel in a turn, which
B
A
20
would be unsatisfactory as we discussed in the previous packet. In reality, because the
car pictured is turning to the left, the right Ackerman arm (CD) needs to steer something
less than 20 degrees. But how much less?
This becomes a moving target. I tried a lot of high flutin mathematical tricks
until I discovered a rather straightforward way to attack this. Let us consider a line
drawn diagonally from point D to B. This creates three
angles that add together to give the angle of the wheel that
pivots at point D. Well call the first angle K, the second
D angle (pronounced gamma), and the third angle is of course,
C AA
the Ackerman angle.
Now we can set to work on determining each. If you
think about angle k, we can determine it because for any steer
k angle, we know the positions of the ends of the diagonal line.
If we assigned point A the coordinate of 0,0 then point D
would have the coordinates Kingpin Center to Center
Distance,0. In the our case specifically point Ds coordinates
would be 36,0. Point Bs coordinates take a little bit more
elbow grease to find. We can calculate its locations with the
following formulae:
B
A
Point Bs X coordinate = RAA * COS(AA + SAL)
## Point Bs Y coordinate = RAA * SIN(AA + SAL)
Where:
RAA is the Ackerman Arm Radius
AA is the Ackerman Angle
SAL is the steering angle of the left wheel. Zero degrees is straight ahead.
Positive values are a left turn, negative values are a right turn.
D
Plugging in our numbers for a 20 left turn
## Jumping through a few hoops
AA
D Point Bs X coordinate = 4.972
C
## Point Bs Y coordinate = 3.358
k So, the coordinates of Point B at a 20 left turn are
4.972,3.358. We can project straight to the left of point B
and straight up from point A to create a new point called
point E. Because we projected straight left and straight up,
4.972
E
3.358
B
A
the angle at E is by definition 90. Also, because point E falls on segment AD, we can
calculate distance DE with the formula:
DE = 36 3.358
## Crunching the numbers
DE = 32.642
Now that we know EB and ED, we can find the length of BD because it is a hypotenuse
of the triangle formed. Using Pythagorean Theorem:
BD = (EB2 + (DE)2
## Plugging our numbers in
BD = (32.642)2 + (4.972)2
BD = 33.019
Furthermore, because we know the sides of the triangle we can determine angle k in the
following manner:
TAN k = EB/ED
Of course, were trying to find k, so lets get that by itself by taking the ARCTAN
ARCTAN (EB/ED) = k
## Plugging in our numbers
AA
ARCTAN (4.972/32.642) = k C
D
Some mathematical acrobatics and
8.661 = k
k
So now that we know angle k and the Ackerman angle, the problem is two
thirds licked. All we have left is to find angle (pronounced gamma). Note
that is included in triangle BDC. Lets think about what we know about
4.972
E B
3.358
A
this triangle. We know that side DC is the length of the Ackerman arm, which we chose
to be 6. We know that side CB is the length of the tie rod, which we calculated earlier to
be 33.084. Finally, we know the distance BD, which we determined using Pathagorean
Theorem to be 33.019. So we have a triangle and we know the lengths of each of the
three sides.
Luckily, there is a somewhat abstract relationship between the sides of non-right
triangles called law of cosines. It can be expressed a number of ways, but we will use
the permutation shown below.
COS = A2 + B2 C2
2AB
Per usual, we are trying to get the thing we dont know by itself, so well need to beat this
up a little bit to make it useful. Rearranging gives:
ARCCOS A2 + B2 C2 =
2AB
2(33.019)(6)
## Crunching the numbers
85.411 =
Now if we add up angle k, and the Ackerman angle, well have the tires steer angle
from the line that connects the two kingpins. To get the steer angle, we have to subtract
90. The formula is:
## Steer Angle = 18.108
As with any engineering math, we must ask if this is a reasonable number. Lets think
about it. The car is executing a left turn. The left front wheel is steered 20 to the left.
The right wheel is tracing a larger arc, and therefore should have a lesser steer angle. In
short the left side steering angle is 20 and the right side should be something less than
20. Weve passed that test. Additionally, experience on real cars on the alignment rack
indicates that these numbers are reasonable. Attached you will find an Excel file that will
do the heavy lifting for you. |
Question Video: Solving Problems Using Theoretical Probability Mathematics • 7th Grade
A bag contains white, red, and black balls. The probability of drawing a white ball at random is 11/20 and a red ball at random is 3/10. What is the smallest number of red balls and black balls that could be in the bag?
02:59
Video Transcript
A bag contains white, red, and black balls. The probability of drawing a white ball at random is 11 out of 20 and a red ball at random is three out of 10. What is the smallest number of red balls and black balls that could be in the bag?
We are told in the question that there are three different color balls in the bag. The probability of selecting a white ball is 11 out of 20 or eleven twentieths. The probability of selecting a red ball is three out of 10 or three-tenths. We are not given the probability of selecting a black ball.
In order to compare fractions, we need to ensure that the denominators are the same. The lowest common multiple of 10 and 20 is 20, so we need to multiply the denominator of the second fraction by two. Whatever we do to the denominator, we must do to the numerator. Three multiplied by two is equal to six, and 10 multiplied by two is 20. Therefore, the fractions three-tenths and six twentieths are equivalent.
We know that the sum of all probabilities is one. In this question, as our denominator is 20, this is equal to twenty twentieths. Eleven twentieths plus six twentieths is equal to seventeen twentieths. When the denominators are the same, we just add the numerators. Subtracting this from one or twenty twentieths gives us three twentieths. This means that the probability of selecting a black ball is three twentieths.
We now have three probabilities that we can compare as the denominators are all the same. The ratio of white to red to black balls is 11 to 6 to three. This means that the smallest number of balls in total is 20, where 11 would be white, six red, and three black. The smallest number of red balls and black balls that could be in the bag are six and three, respectively.
As we’re only given the probabilities, the total number of balls could be any multiple of 20. For example, we could have 40 balls in total where 22 are white, 12 are red, and six are black. However, as we were looking for the smallest number of red and black balls, the correct answer is six red and three black. |
# 18-Dec-14 Pruning. 2 Exponential growth How many leaves are there in a complete binary tree of depth N? This is easy to demonstrate: Count “going left”
## Presentation on theme: "18-Dec-14 Pruning. 2 Exponential growth How many leaves are there in a complete binary tree of depth N? This is easy to demonstrate: Count “going left”"— Presentation transcript:
18-Dec-14 Pruning
2 Exponential growth How many leaves are there in a complete binary tree of depth N? This is easy to demonstrate: Count “going left” as a 0 Count “going right” as a 1 Each leaf represents one of the 2 N possible N-bit binary numbers This observation turns out to be very useful in certain kinds of problems depth = 0, count = 1 depth = 1, count = 2 depth = 3, count = 4 depth = 4, count = 8 depth = 5, count = 16 depth = N, count = 2 N
3 Pruning Suppose the binary tree represents a problem that we have to explore to find a solution (or goal node) If we can prune (decide we can ignore) a part of the tree, we save effort The higher up in the tree we can prune, the more effort we can save The advantage is exponential saves 3 saves 7 saves 15
4 Sum of subsets Problem: There are n positive integers, and a positive integer W Find a subset of the integers that sum to exactly W Example: The numbers are 2, 5, 7, 8, 13 Find a subset of numbers that sum to exactly 25 We can multiply each number by 1 if it is in the sum, 0 if it is not 2 5 7 8 13 0 0 0 0 0 0 0 0 0 0 1 13 0 0 0 1 0 8 0 0 0 1 1 21 0 0 1 0 1 20 0 0 1 1 0 15 0 0 1 1 1 28 0 1 0 0 0 5 0 1 0 0 1 18 0 1 0 1 0 13 0 1 0 1 1 26 0 1 1 0 0 12 0 1 1 0 1 25
5 Brute force We have a brute-force method for solving the sum of subsets problem For N numbers, count in binary from 0 to 2 N For each 1, include the corresponding number; for each 0, exclude the corresponding number Stop if we get lucky This is clearly an exponential-time algorithm It seems like, with a little cleverness, we could do better It turns out that we can use pruning to do somewhat better But we are still left with an exponential-time algorithm
6 Binary tree representation Suppose our numbers are 3, 8, 9, 17, 26, 39, 43, 56 and our goal is 100 We can describe this as a binary tree search As we search the binary tree, A node is promising if we might be able to get to a solution from it A node is nonpromising if we know we can’t get to a solution When we detect a nonpromising node, we can prune (ignore) the entire subtree rooted at that node How do we detect nonpromising nodes? 3 (yes or no) 8 (yes or no) 9 (yes or no) 17 (yes or no) etc.
7 Detecting nonpromising nodes Suppose we work from left to right in the sequence 3 8 9 17 26 39 43 56 That is, we try things in the order 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0... When we get to 0 0 0 0 0 0 1 0 43 we notice that even if we include all the remaining numbers (in this case, there is only one), we can’t get to 100 There is no need to try the 0 0 0 0 0 0 1 x numbers When we get to 1 1 1 1 1 1 0 0 101 we notice that we have overshot, hence no solution is possible with what we have so far We don’t need to try any of the 1 1 1 1 1 1 x x numbers
8 Still exponential Even with pruning, the sum of subsets typically requires exponential time However, in some cases, pruning can save significant amounts of time Consider trying to find a subset of {23, 29, 35, 41, 43, 46, 48, 51} that sums to 100 Here, pruning can save substantial effort Sometimes, common sense can be a big help Consider trying to find a subset of {16, 20, 28, 34, 44, 48} that sums to 75
9 Boolean satisfaction Suppose you have n boolean variables, a, b, c,..., that occur in a logical expression such as (a or c or not f) and (not b or not d or a) and... The problem is to assign true / false values to each of the boolean variables in such a way as to satisfy (make true ) the logical expression The brute-force algorithm is the same as before ( 0 is false, 1 is true, try all n binary numbers) Again, you can do significant pruning, if you think about the problem Anything you do, you will still end up with a solution that is exponential in n
10 Intractable problems The technical term for a problem that takes exponential time is intractable Intractable problems can only be solved for small input sizes Faster computer speeds will not help much— exponential growth is fast Bottom line: Avoid these problems if at all possible!
11 The End
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# What is the derivative of y = (sin x)^(cos x)?
##### 1 Answer
Jan 4, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\sin x\right)}^{\cos} x \left(- \sin x \ln \left(\sin x\right) + \cos x \cot x\right)$
#### Explanation:
Take the natural logarithm of both sides.
$\ln y = \ln {\left(\sin x\right)}^{\cos} x$
Use the rule $\log {a}^{n} = n \log a$ to simplify:
$\ln y = \cos x \ln \left(\sin x\right)$
Use the implicit differentiation as well as the product and chain rules to differentiate.
$\frac{d}{\mathrm{dx}} \left(\ln \sin x\right) = \frac{1}{\sin} x \cdot \cos x = \cos \frac{x}{\sin} x = \cot x$
Now to the relation as a whole:
$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - \sin x \left(\ln \left(\sin x\right)\right) + \cos x \cot x$
Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(- \sin x \ln \left(\sin x\right) + \cos x \cot x\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\sin x\right)}^{\cos} x \left(- \sin x \ln \left(\sin x\right) + \cos x \cot x\right)$
Hopefully this helps! |
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# Relations and Functions-CBSE CLASS 11
Relation and function-CBSE CLASS 11
CONTENT LIST
## Introduction
In our daily life we come accorss many relations between any two entities like
relation between a father and a son,
relation between a Mother and a son,
relation between a wife and a husband,
relation between a brother and sister,
relation between a student and class,
relation between a student and bag, etc.
In mathematics also, many relations are found between numbers such as
a number x is less than y,
line l is parallel to line m,
x is one more than y,
x is square of y etc.
Relation and function map elements of one set (domain) to the elements of another set (codomain).Functions is a special types of relations that define the precise correspondence between one quantity with the other. By this artilce Relation and function We shall learn ,how to link pairs of elements from two sets and then define a relation between them, different types of relations and functions, and the difference between relation and function.
## Relation
A relation R from a non-empty set A to a non empty set B is a subset of the Cartesian product set A × B.The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
The set of all first elements in a relation R, is called the domain of the relation R, and the set of all second elements called images, is called the range of R.
If A and B are two non-empty sets, then the relation R from A to B is a subset of A x B, i.e., R ⊆ A x B.
Ex: Let, A = {1,2,3} and B = {2,3,4}
A X B = {(1,2),(1,3),(1,4);(2,2),(2,3),(2,4);(3,2),(3,3),(3,4)}
R = {(1,2),(2,3),(3,4)} (R:y = x+1)
## Representation of Relation
A relation may be represented either by the Roster form or by the set builder form, or by an arrow diagram which is a visual representation of a relation.
Consider an example of two sets A = {9, 16, 25} and B = {5, 4, 3, -3, -4, -5}. The relation is that the elements of A are the square of the elements of B.
In set-builder form, R = {(x, y): x is the square of y, x ∈ A and y ∈ B}.
In roster form, R = {(9, 3), (9, -3), (16, 4), (16, -4), (25, 5), (25, -5)}.
In arrow diagram form
## Number of Relations
If n (A) = p, n (B) = q; then the n (A × B) = pq and the total number of possible relations from the set A to set B = (2)^pq
## Functions
Functions - A relation f from a set A to a set B is said to be a function, if every element of set A has one and only one image in set B.
f:A→B show a function from A to B.
For function:: Element of set A has an image in set B.
For function::Element of set A has only image in set B.
b∈B is called image of a∈A under function f .
a∈A is called preimage of b∈B under function f .
Let us take an example Set A={a,b,c,d} Set B={Aarya, Aadhya,Beena,Cema,Eklavya}
The above daigram shows relation between set A and B but it is not a function due to following reasons:
1. a ∈ A, has 2 images in set B.
2. There is no image of d ∈ A in set B.
All functions are relation, but all relation are not function
Example:
Let A={1,2,3} & B={2,3,4} and f1,f2 & f3 are three subsets of AXB such that f1={(1,2),(2,3),(3,4)} f2={(1,2),(1,3),(2,3),(3,4)} & f3={(1,3),(2,4)} . Identify which relation is showing function.
In relation f1: Each element of set A has uniuqe image in set B. so relation f1 is a function.
In relation f2: An element 1 of set A has two image 2 & 3 in set B. so relation f1 is not a function.
In relation f3: An element 3 of set A has two no image in set B. so relation f3 is not a function.
## Domain Co-domain and Range of function
If there are two non empty set A and B then a function f is expressed as the set of ordered pairs of AXB in such a way that Each element of set A has uniuqe image in set B & no two pre image of any element of set B in set A .
Domain of f = Set of all first elements of memeber of f OR Domain of f={a:(a,b)∈f}
Range of f = Set of all second elements of memeber of f OR Range of f={b:(a,b)∈f}
Co-domain = all element of set B
Example:
Let A={-1,0,1,2,} & B={0,1,2,3,4} Consider rule f(x)=x^2
Under this rule when f(-1)=1,f(0)=0,f(1)=1,f(2)=4 Here we see all element od set A associated with unique element of setB.
Therefore f(x)=x^2 is a function.
Domain of f(x)=x^2={-1,0,1,2,}
Range of f(x)=x^2={0,1,4}
Co-domain of f(x)=x^2={0,1,2,3,4}
### Types of functions
1. Identity function
2. Constant function
3. Polynomial function
4. Rational functions
5. Modulus function
6. Signum function
7. Greatest integer function
### Identity functions
A function is said to be an identity function when it returns the same value as the output Which was used as its input(argument).
Let us an example of a function of set A = {1, 2, 3, 4, 5} to itself. f: A → A such that, g = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}.
### Identity functions
A function is said to be an identity function when it returns the same value as the output Which was used as its input(argument).
Let us an example of a function of set A = {1, 2, 3, 4, 5} to itself. f: A → A such that, g = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}.
If we plot a graph of identity function, then it will be a straight line.
Let us draw an identity function f:R→R, f(x)=x for x∈R
When x=1 than y=f(1)=1
When x=0 than y=f(0)=0
When x=-2 than y=f(-2)=-2
### Constant function
A function is said to be a constant function when it returns constant value as the output for all input(argument).
Constant function f:R→R, f(x)=k for x∈R
If we plot a graph of constant function, then it will be a straight line parallel to x axis .
Let us draw an identity function f:R→R, f(x)=5 for x∈R
When x=1 than y=f(1)=5
When x=0 than y=f(0)=5
When x=-2 than y=f(-2)=5
### Polynomial function
Polynomial Function is defined as the real valued function f:R→R for x∈R
### Rational functions
Rational function is the ratio of two polynomial functions where the denominator polynomial is not equal to zero.
where g(x) and t(x) are polynomial functions and t(x) is a non-zero polynomial.
### Modulus function
The modulus function only gives a positive value as ouput of any variable or input. It is also known as the absolute value function because it gives a non-negative value as out put against any independent variable or input, no matter if it is positive or negative. The function f:R→R,Modulus is defined by
Let us draw graph of f(x)=|x|
When x=-5,then y= f(-5)=|-5|=5
When x=-4,then y= f(-4)=|-4|=4
When x=-3,then y= f(-3)=|-3|=3
When x=-2,then y= f(-2)=|-2|=2
When x=-1,then y= f(-1)=|-1|=2
When x=0,then y= f(0)=|0|=0
When x=1,then y= f(1)=|1|=1
When x=1,then y= f(1)=|1|=1
When x=2,then y= f(2)=|2|=2
When x=3,then y= f(3)=|3|=3
When x=4,then y= f(4)=|4|=4
When x=5,then y= f(5)=|5|=5
### Signum function
Domain of f = R, Range of f = {1, 0, – 1}
### Greatest integer function
• Greatest integer function gives the rounds up value of the number to the most neighboring integer that is less than or equal to the provided number. Graph of this function has a step curve and is also known as the step function.
• Any number which is less than or equal to a number x is shown by the notation ⌊x⌋.
• Examples: ⌊3.01⌋ = 3 and ⌊4.57⌋ = 4.
• Greatest integer function is a function Which returns a constant value for each specific interval.
• Greatest integer function: The real function f : R → R defined by
f (x) = [x], x ∈R assumes the value of the greatest integer less than or equal to x, is called the greatest integer function.
Thus f (x) = [x] = – 1 for – 1 ≤ x < 0
f (x) = [x] = 0 for 0 ≤ x < 1
[x] = 1 for 1 ≤ x < 2
[x] = 2 for 2 ≤ x < 3 and so on
Related Topics
• Sets: Chapter Note
• Sets MCQ Test 1
• Sets MCQ Test 2
• Sets MCQ Test 3 |
# How to Prove It - Solutions
## Chapter - 3, Proofs
### Summary
• A theorem is a statement that has been proven on the basis of previously established statements, such as other theorems—and generally accepted statements, such as axioms.
• A theorem that says that if certain assumptions called the hypotheses of the theorem are true, then some conclusion must also be true.
• The hypothesis and conclusion of a theorem often contains free variables. When values are substituted/assigned for these variables, it is called an instance of the theorem.
• A proof of a theorem is simply a deductive argument whose premises are the hypotheses of the theorem and whose conclusion is the conclusion of the theorem.
• To prove a conclusion of the form of $P \to Q$, there can be two strategies:
• Assuming $P$ is true, Prove $Q$. In this way this strategy transformed the problem of proving $P \to Q$ to proving $Q$. After this transformation, $P$ is now the part of the hypothesis.
• Prove contra-positive: assuming $\lnot Q$ is true, prove $\lnot P$. Similarly here also problem is transformed from proving $P \to Q$ to proving $\lnot P$.
• Thus multiple transformations can be applied to convert the problem to simpler one.
• At any point of the proof:
• givens are the set of statements that are known or assumed to be true till that point.
• goal is the final statement that remains to be proven.
• Thus at the start of a proof, givens will be the initial hypothesis and goal will be the final conclusion.
• In the course of the proof, the list of hypothesis will increase(during transformation).
• Similarly, conclusion will change during the course of the proof.
• In writing final proof:
• Reasoning used for figuring the the proof are not used/written.
• Steps to justify the conclusions are used/written but no explanations on how the steps were thought of.
• Similarly Goals and Givens are also not part of the final proof.
• Explanations are avoided in proofs to maintains the distinction between:
• Explanation of thought process.
• Justifying the conclusions.
• Thus proof of the goal $P \to Q$ will look as follows:
• Given: ,
Goal $P \to Q$
• Using strategy-1:
• Scratch work will look as:
Given: $P$ ,
Goal $Q$.
• Final proof will look as:
Suppose $P$ is true.
[Proof of $Q$ ]
Therefore $P \to Q$.
• Using strategy-2:
• Scratch work will look as:
Given: $\lnot Q$ ,
Goal $\lnot P$.
• Final proof will look as: Suppose $Q$ is false.
[Proof of $\lnot P$ ]
Therefore $P \to Q$.
Soln1
(a)
Hypothesis:
• $n$ is an integer and $n > 1$.
• $n$ is not prime.
Conclusion:
$2^n - 1$ is not prime.
• For $n = 6$, hypothesis is true.
• Conclusion is $2^6 - 1 = 63$ is not prime. This is also true.
(b)
• For $n = 15$, hypothesis is true.
• Conclusion is $2^{15} - 1 = 32767$ is not prime. This is also true: $32767 = 7 * 4681$.
(c)
• For $n = 11$, hypothesis is not true.
• Thus theorem does not say anything.
Soln2
(a)
Hypothesis: $b^2 > 4ac$.
Conclusion: $ax^2 + bx + c = 0$ has exactly 2 solutions.
(b) $a,\,b,\, c$ are free variables. $x$ is not a free variable.
(c) Substituting the values: $(-5)^2 > 4 \times 2 \times 3$ gives $25 > 24$. This is true. Thus hypothesis is true.
Conclusion is $2x^2 -5x + 3 = 0$
$\quad = 2x^2 -3x -2x + 3 = 0$
$\quad = x(2x - 3) -3(x - 1) = 0$
$\quad = (2x - 3)(x - 1) = 0$
Thus $x = 1$and$ x = 3/2$\$ are two solutions. That means conclusion is also correct.
(d) Substituting the values: $(4)^2 > 4 \times 2 \times 3$ gives $16 > 24$. This is false. Thus hypothesis is false.
Theorem does not say anything as hypothesis is not true.
Soln3
Hypothesis:
• $n$ is a natural number and $n > 2$.
• $n$ is not prime.
Conclusion:
$2n + 13$ is not prime.
For $n = 9$, Hypothesis is true. Conclusion: $2 \times 9 + 13 = 31$, which is a prime number. Thus conclusion is not true.
Soln4
Suppose $% $. Then $b−a > 0$.
Also $a + b > 0$. Multiplying $b−a > 0$ by $b + a$
We get $(b-a)(b+a) > 0$. Which gives:
$b^2 - a^2 > 0$.
Since $b^2 − a^2 > 0$, it follows that $% $.
Soln5
if $% $ then $b - a > 0$ and $% $
Thus, the product $% $ because product of a positive and negative number is negative.
Expanding the product, we get $% $.
Now since $% $, It follows that $% $.
Soln6
if $% $, Multiplying both sides by positive number $\frac 1 {ab}$.
$% $
$% $
Soln7
Suppose $a^3 > a$. Subtracting $a$ from both sides:
$\quad a^3 - a > 0$.
Multiplying by $a^2 + 1$ on both sides:
$\quad = (a^3 - a)(a^2 + 1) > 0$
Simplifying:
$\quad = a^5 - a^3 + a^3 - a > 0$
$\quad = a^5 - a > 0$
Thus we have: $a^5 > a$.
Soln8
We are given that $A \setminus B \subseteq C \cap D$ and $x \in A$.
Thus we have $\forall y (y \in A \land y \notin B) \to (y in C \land y \in D)$
For $y = x$, we have:
$x \in A \land x \notin B \to (x \in C \land x \in D)$
As $x \in A$ and suppose $x \notin D$, then we get:
$true \land x \notin B \to (x \in C \land false)$
Simplifying:
$x \notin B \to false$
It follows that $x \notin B = false$, or $x \in B$.
Soln9
Given that $% $, then adding $b$ to both sides:
$% $
$% $
Dividing both sides by 2:
$% $
Soln10
We shall prove it by proving contra-positive.
Suppose $x = 8$, we get
$\frac {\sqrt[3]{8} + 5} {8^2 + 6} = \frac 1 8$
Simplifying:
$\quad \frac {7} {70} = \frac 1 8$
$\quad \frac {1} {10} = \frac 1 8$
Clearly which is not correct. Thus if $x = 8$, then $\frac {\sqrt[3]{x} + 5} {8^2 + 6} \neq \frac 1 x$
Thus if $\frac {\sqrt[3]{x} + 5} {8^2 + 6} = \frac 1 x$, then $x \neq 8$
Soln11
We shall prove this using contra-positive. Given that $% $, and $d > 0$
Multiplying $% $, by $d$ on both sides: $% $
Suppose that $% $, Multiplying both sides by $a$ gives: $% $
Thus we have $% $ and $% $, which gives $% $
Or we can say that: $% $.
Thus we proved that if $% $, then $% $.
It follows that if $ac >= bd$, then $c > d$.
Soln12
Suppose $x > 1$, then $3x > 3$
$3x + 2 > 5$, or $2 > 5 - 3x$, which is same as $% $
Also, given that $3x + 2y \le 5$
Subtracting 3x from both sides:
$2y \le 5 - 3x$
Thus we have $% $
Simplifying:
$% $, or $% $
Soln13
Given that $x^2 + y = −3$ and $2x − y = 2$
$x^2 + y + 2x - y = -3 + 2$
Simplifying:
$x^2 + 2x + 1 = 0$
Or ${(x+1)}^2 = 0$
which means $x = -1$.
Soln14
Given that $x > 3$
As $x > 0$ and $3 > 0$, we can apply the theorem: If $% $, then $% $
Thus we have $x^2 > 9$
Also, it is givem that $% $.
Multiplying both sides by -2, and reverting the sign of inequality:
$-2y > -4$
Adding both inequalities: $x^2 - 2y > 9 - 4$
Or $x^2 - 2y > 5$.
Soln15
(a)
The proof is for: if $x = 7$, then $\frac {2x-5} {x -4} = 3$
while the required proof was for: if $\frac {2x-5} {x -4} = 3$, then $x = 7$.
(b)
Given that $\frac {2x-5} {x -4} = 3$
Cross multiplying:
$2x - 5 = 3x - 12$
Simplifying:
$x = 7$.
Soln16
(a)
The issue is that value of $x = -3$ is not taken into consideration. As for this value $x^2 = 9$.
(b)
If x = -3, then the equation gives $y = 1$, thus theorem is not correct as it says if $x \ne 3$, then $y = 0$. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Measures of Central Tendency and Dispersion
## Mean, median, mode, range
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Practice Measures of Central Tendency and Dispersion
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Mean, Median and Mode
The three measures of central tendency are mean, median, and mode. When would it make sense to use one of these measures and not the others?
#### Watch This
http://www.youtube.com/watch?v=h8EYEJ32oQ8 Khan Academy: Statistics Intro: Mean, Median, and Mode
#### Guidance
With descriptive statistics, your goal is to describe the data that you find in a sample or is given in a problem. Because it would not make sense to present your findings as long lists of numbers, you summarize important aspects of the data. One important aspect of the data is the measure of central tendency, which is a measure of the “middle” value of a set of data. There are three ways to measure central tendency:
1. Use the mean, which is the arithmetic average of the data.
2. Use the median, which is the number exactly in the middle of the data. When the data has an odd number of counts, the median is the middle number after the data has been ordered. When the data has an even number of counts, the median is the arithmetic average of the two most central numbers.
3. Use the mode, which is the most often occurring number in the data. If there are two or more numbers that occur equally frequently, then the data is said to be bimodal or multimodal.
Calculating the mean, median and mode is straightforward. What is challenging is determining when to use each measure and knowing how to interpret the data using the relationships between the three measures.
Example A
Five people were called on a phone survey to respond to some political opinion questions. Two people were from the zip code 94061, one person was from the zip code 94305 and two people were from 94062.
Which measure of central tendency makes the most sense to use if you want to state where the average person was from?
Solution: None of the measures of central tendency make sense to apply to this situation. Zip codes are categorical data rather than quantitative data even though they happen to be numbers. Other examples of categorical data are hair color or gender. You could argue that mode is applicable in a broad sense, but in general remember that mean, median, and mode can only be applied to quantitative data.
Example B
Compute the mean, median and mode for the following numbers.
3, 5, 1, 6, 8, 4, 5, 2, 7, 8, 4, 2, 1, 3, 4, 6, 7, 9, 4, 3, 2
Solution:
Mean: The sum of all these numbers is 94 and there are 21 numbers total so the mean is .
Median: When you order the numbers from least to greatest you get:
1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9
The number has ten numbers to the right and ten numbers to the left so it is the median. The median is the number 4.
Mode: the most frequently occurring number is the number 4.
Note: it is common practice to round to 4 decimals in AP Statistics.
Example C
You write a computer code to produce a random number between 0 and 10 with equal probability. Unfortunately, you suspect your code doesn’t work perfectly because in your first few attempts at running the code, it produces the following numbers:
1, 9, 1, 1, 9, 2, 9, 1, 9, 9, 9, 2, 2
How would you argue using mean, median, or mode that this code is probably not producing a random number between 0 and 10 with equal probability?
Solution: This question is very similar to questions you will see when you study statistical inference.
First you would note that the mean of the data is 4.9231. If the data was truly random then the mean would probably be right around the number 5 which it is. This is not strong evidence to suggest that the random number generating code is broken.
Next you would note that the median of the data is 2. This should make you suspect that something is wrong. You would expect that the median is of random numbers between 0 and 10 to be somewhere around 5.
Lastly, you would note that the mode of the data is 9. By itself this is not strong data to suggest anything. Every sample will have to have at least one mode. What should make you suspicious, however, is the fact that only two other numbers were produced and were almost as frequent as the number 9. You would expect a greater variety of numbers to be produced.
Concept Problem Revisited
In order to decide which measure of central tendency to use, it is a good idea to calculate and interpret all three of the numbers.
For example, if someone asked you how many people can sit in the typical car, it would make more sense to use mode than to use mean. With mode, you could find out that a five person car is the most frequent car driven and determine that the answer to the question is 5. If you calculate the mean for the number of seats in all cars, you will end up with a decimal like 5.4, which makes less sense in this context.
On the other hand, if you were finding the central heights of NBA players, using mean might make a lot more sense than mode.
#### Vocabulary
The mean is the arithmetic average of the data.
The median is the number in the middle of a data set. When the data has an odd number of counts, the median is the middle number after the data has been ordered. When the data has an even number of counts, the median is the average of the two most central numbers.
The mode is the most often occurring number in the data. If there are two or more numbers which occur equally frequently, then the data is said to be bimodal or multimodal.
With descriptive statistics, your goal is to describe the data that you find in a sample or is given in a problem.
With inference statistics, your goal is use the data in a sample to draw conclusions about a larger population.
#### Guided Practice
1. Ross is with his friends and they want to play basketball. They decide to choose teams based on the number of cousins everyone has. One team will be the team with fewer cousins and the other team will be the team with more cousins. Should they use the mean, median or mode to compute the cutoff number that will separate the two teams?
2. Compute the mean, median, and mode for the following numbers.
1, 4, 5, 7, 6, 8, 0, 3, 2, 2, 3, 4, 6, 5, 7, 8, 9, 0, 6, 5, 3, 1, 2, 4, 5, 6, 7, 8, 8, 8, 4, 3, 2
3. The cost of fresh blueberries at different times of the year are:
$2.50,$2.99, $3.20,$3.99, $4.99 If you bought blueberries regularly what would you typically pay? Answers: 1. Ross and his friends should use the median number of cousins as the cutoff number because this will allow each team to have the same number of players. If there are an odd number of people playing, then the extra person will just join either team or switch in later. 2. The mean is 4.6061. The median is 5. The mode is 8. 3. The word “typically” is used instead of average to allow you to make your own choice as to whether mean, median, or mode would make the most sense. In this case, mean does make the most sense. The average cost is$3.53.
#### Practice
You surveyed the students in your English class to find out how many siblings each student had. Here are your results:
0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 10, 12
1. Find the mean, median, and mode of this data.
2. Why does it make sense that the mean number of siblings is greater than the median number of siblings?
3. Which measure of central tendency do you think is best for describing the typical number of siblings?
4. So far in math you have taken 10 quizzes this semester. The mean of the scores is 88.5. What is the sum of the scores?
5. Find if 5, 9, 11, 12, 13, 14, 16, and have a mean of 12.
6. Meera drove an average of 22 miles a day last week. How many miles did she drive last week?
7. Find if 2, 6, 9, 8, 4, 5, 8, 1, 4, and have a median of 5.
Calculate the mean, median, and mode for each set of numbers:
8. 11, 15, 19, 12, 21, 34, 15, 28, 24, 15, 27, 19, 20, 13, 15
9. 3, 5, 7, 5, 5, 17, 8, 9, 11, 5, 3, 7
10. -3, 0, 5, 8, 12, 4, 2, 1, 6
Calculate the mean and median for each set of numbers:
11. 12, 88, 89, 90
12. 16, 17, 19, 20, 20, 98
13. For which of the previous two questions was the median less than the mean? What in the set of numbers caused this?
14. For which of the previous two questions was the median greater than the mean? What in the set of numbers caused this?
15. In each of the sets of numbers for problems 11 and 12, there is one number that could be considered an outlier. Which numbers do you think are the outliers and why? What would happen to the mean and median if you removed the outliers?
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 15.1.
### Vocabulary Language: English
arithmetic mean
arithmetic mean
The arithmetic mean is also called the average.
descriptive statistics
descriptive statistics
In descriptive statistics, the goal is to describe the data that found in a sample or given in a problem.
inferential statistics
inferential statistics
With inferential statistics, your goal is use the data in a sample to draw conclusions about a larger population.
measure of central tendency
measure of central tendency
In statistics, a measure of central tendency of a data set is a central or typical value of the data set.
Median
Median
The median of a data set is the middle value of an organized data set.
Mode
Mode
The mode of a data set is the value or values with greatest frequency in the data set.
multimodal
multimodal
When a set of data has more than 2 values that occur with the same greatest frequency, the set is called multimodal .
Outlier
Outlier
In statistics, an outlier is a data value that is far from other data values.
Population Mean
Population Mean
The population mean is the mean of all of the members of an entire population.
resistant
resistant
A statistic that is not affected by outliers is called resistant.
Sample Mean
Sample Mean
A sample mean is the mean only of the members of a sample or subset of a population. |
# Complex Numbers
Formally, a complex number is an expression $$a + bi$$, with $$a$$ and $$b$$ real numbers and $$i$$ an entity satisfying $$i^{2} = -1$$. The square of an arbitrary real number is non-negative, so whatever $$i$$ may be, it is not a real number.
Suppose $$z = 3 + 4i$$ and $$w = -5 + 2i$$ are complex numbers. How might we add them? How might we multiply them? In each case, the goal is represent an expression, $$z + w$$ or $$zw$$, in the form $$a + bi$$ for suitable real numbers $$a$$ and $$b$$. In an exploratory sense, it's reasonable to assume complex numbers satisfy the commutative, associative, and distributive properties, and to gather like terms: $z + w = (3 + 4i) + (-5 + 2i) = (3 - 5) + (4i + 2i) = -2 + 6i.$ Similarly, multiplying out (the distributive property), rearranging (associative and commutative properties), and collecting like terms gives \begin{align*} zw &= (3 + 4i)(-5 + 2i) \\ &= (3)(-5) + 3(2i) + (4i)(-5) + (4i)(2i) \\ &= (-15) + 6i + (-20i) + 8i^{2} \\ &= (-15) + (6 - 20)i - 8 \\ &= -23 - 14i. \end{align*}
Generally, we might write $$z = a + bi$$ and $$w = c + di$$, and calculate \begin{align*} z + w &= (a + bi) + (c + di) \\ &= (a + c) + (bi + di) \\ &= (a + c) + (b + d)i. \end{align*} Since $$a + c$$ and $$b + d$$ are real numbers (being sums of real numbers), the preceding formula defines the sum of complex numbers on the left-most side.
Similarly, \begin{align*} zw &= (a + bi)(c + di) \\ &= (ac) + a(di) + (bi)c + (bi)(di) \\ &= (ac) + (ad + bc)i + (bd)i^{2} \\ &= (ac - bd) + (ad + bc)i. \end{align*}
To emphasize, we have defined \begin{align*} (a + bi) + (c + di) &= (a + c) + (b + d)i, \\ (a + bi)(c + di) &= (ac - bd) + (cb + ad)i. \end{align*} These formulas obviate logical concerns about the formal manipulations that led us to them. But are these definitions useful?
It turns out these definitions obey the associative and commutative properties, and the distributive property, on the set of complex numbers. This justifies our definitions: We have defined a new, logically-consistent set of complex numbers, including the operations of addition and multiplication.
The complex number $$a + bi$$ is said to have real part $$a$$ and imaginary part $$b$$. (Note carefully: The “imaginary part” is real!) A complex number of the form $$a = a + 0i$$ “behaves just like” an ordinary real number. Practically, we may view real numbers as special complex numbers, those whose imaginary part is $$0$$.
## Complex Division
A complex number can be divided by a non-zero complex number. This fact is striking if we try naively to make sense of an expression such as $$(3 + 4i)/(-5 + 2i)$$. On the other hand, there is a sneaky algebra trick that lets us find the reciprocal of a non-zero complex number. Division by a non-zero complex number is multiplication by the reciprocal.
Note first that if we multiply $$-5 + 2i$$ by $$-5 - 2i$$, the number obtained by “reversing” the imaginary part, we get $(-5 + 2i)(-5 - 2i) = (-5)^{2} - (2i)^{2} = 25 - (-4) = 25 + 4 = 29,$ a positive real number! (We used the “difference of squares” identity. The definition of multiplication gives the same result.)
Now, although we do not yet know how to divide arbitrary complex numbers, we can perhaps agree that dividing a non-zero complex number by itself gives a quotient of $$1$$. Thus, \begin{align*} \frac{1}{-5 + 2i} &= \frac{1}{-5 + 2i} \cdot 1 &&\text{multiplying by $$1$$ does not change a number} \\ &= \frac{1}{-5 + 2i} \cdot \frac{-5 - 2i}{-5 - 2i} &&\text{rewriting $$1$$ using the conjugate} \\ &= \frac{-5 - 2i}{(-5 + 2i)(-5 - 2i)} &&\text{multiplying numerators and denominators} \\ &= \frac{-5 - 2i}{29} &&\text{prior calculation} \\ &= -\frac{5}{29} - \frac{2}{29}i &&\text{a complex number!} \end{align*}
Generally, if $$z = a + bi$$ is complex, we define its complex conjugate to be $$\bar{z} = a - bi$$. By the definition of multiplication, the product of a complex number and its conjugate is \begin{align*} z \bar{z} &= (a + bi)(a - bi) \\ &= \bigl(a^{2} - b(-b)\bigr) + \bigl(a(-b) + ba\bigr)i \\ &= a^{2} + b^{2} + 0i. \end{align*} This is a sum of real squares, and is positive unless $$a = b = 0$$, i.e., unless $$z = 0 + 0i$$.
By the reasoning above (with a few steps elided), if $$z = a + bi$$ is non-zero as a complex number (at least one of $$a$$ and $$b$$ is a non-zero real number), then $\frac{1}{z} = \frac{1}{a + bi} = \frac{1}{a + bi} \cdot \frac{a - bi}{a - bi} = \frac{a - bi}{a^{2} + b^{2}}$ as complex numbers.
Now assume $$z = a + bi$$ and $$w = c + di$$ are complex, and $$w$$ is non-zero. As just noted we have $\frac{1}{w} = \frac{1}{c + di} = \frac{c - di}{c^{2} + d^{2}}.$ Consequently, $\frac{z}{w} = z \cdot \frac{1}{w} = (a + bi) \cdot \frac{c - di}{c^{2} + d^{2}},$ which we can multiply out to obtain a formula if desired. In practice the process is faster than the formula. For example, $\frac{3 + 4i}{-5 + 2i} = \frac{(3 + 4i)(-5 - 2i)}{29} = \frac{-15 + 8 - (20 + 6)i}{29} = \frac{-7 - 26i}{29}.$
One philosophical take-away is that complex numbers may be viewed abstractly as a “number system”: They are a well-defined set of mathematical entities that can be added, multiplied, and divided just like real numbers. The next paragraph can be skipped on first reading if this imprecision doesn't bother you.
More carefully, addition and multiplication are associative and commutative, and multiplication distributes over addition. Further, there is a complex number $$0 = 0 + 0i$$ satisfying $$0 + z = z$$ for all complex $$z$$, and every complex number $$z = a + bi$$ has a negative $$-z = -a - bi$$ satisfying $$z + (-z) = 0$$. Similarly, there is a complex number $$1 = 1 + 0i$$ satisfying $$1 \cdot z = z$$ for all complex $$z$$, and every non-zero complex number $$z = a + bi$$ has a reciprocal $$1/z = (a - bi)/(a^{2} + b^{2})$$ satisfying $$z \cdot (1/z) = 1$$.
## Geometry of Complex Operations
A complex number $$a + bi$$ amounts to an ordered pair $$(a, b)$$ of real numbers. We may view the set of complex numbers as the Cartesian plane: The Euclidean plane with a distinguished origin and coordinate axes. The real axis is horizontal, and comprises complex numbers $$a + 0i$$ with imaginary part $$0$$. The imaginary axis is vertical, and comprises complex numbers $$0 + bi$$ with real part $$0$$.
Addition and multiplication have pleasant interpretations in Euclidean geometry. We won't prove these geometric assertions, but will note some useful, intuitive consequences.
Addition translates the plane. Specifically, if $$\alpha$$ (alpha, the first letter of the Greek alphabet) is a fixed complex number, there is a unique translation of the plane carrying $$0$$ to $$\alpha$$. This is the mapping that sends each complex number $$z$$ to $$z + \alpha$$. Particularly, adding $$1$$ shifts the plane one unit to the right, adding $$i$$ shifts the plane one unit upward.
Multiplication rotates and scales the plane about the origin $$0$$. Specifically, if $$\alpha$$ is a fixed complex number, there is a unique way to rotate-and-scale the plane keeping $$0$$ fixed and carrying $$1$$ to $$\alpha$$. This is the mapping that sends each complex number $$z$$ to $$\alpha z$$. Particularly, multiplying by $$1$$ is the identity mapping, while multiplying by $$-1$$ is a half-turn about the origin.
Multiplying by $$i$$ has a concrete meaning as a one-quarter turn counterclockwise about the origin. Performing two quarter turns, namely squaring $$i$$, gives a half-turn, multiplication by $$-1$$! To say “imaginary numbers don't exist” amounts to denying existence of rotations.
Although complex conjugation is not an arithmetic operation, it too has a useful geometric interpretation: The mapping that sends each complex number $$\alpha = a + bi$$ to $$\bar{\alpha} = a - bi$$ is reflection across the real axis. Although reflection is a Euclidean motion (preserves distances), it does not preserve the sense of angles, but instead swaps clockwise and counterclockwise.
## Magnitude and the Unit Circle
Every non-zero complex number lies on a unique circle centered at $$0$$. If we write $$\alpha = a + bi$$ with $$a$$ and $$b$$ real, then the distance from $$0$$ to $$\alpha$$ is $$\sqrt{a^{2} + b^{2}}$$ by the Pythagorean theorem applied to the right triangle with corners $$0$$, $$a$$, and $$\alpha$$. This radius is useful enough to earn a name: We call $\sqrt{a^{2} + b^{2}} = \sqrt{\alpha\bar{\alpha}}$ the magnitude of $$\alpha$$, and denote it with absolute value symbols, $$|\alpha| = \sqrt{a^{2} + b^{2}}$$. It can be checked (by somewhat laborious algebra) that if $$\alpha$$ and $$\beta$$ (beta) are complex, then $|\alpha\beta| = |\alpha| \cdot |\beta|.$
In words, the magnitude of a product is the product of the magnitudes. Particularly, if $$\alpha$$ is non-zero, we may multiply and divide by $$|\alpha|$$, obtaining $\alpha = |\alpha| \cdot \frac{\alpha}{|\alpha|}.$ The first factor $$|\alpha|$$ is a positive real number. Multiplication by $$|\alpha|$$ geometrically effects scaling (zooming in or out) about the origin. The second factor $$\alpha/|\alpha|$$ is a point of the unit circle, a complex number of magnitude $$1$$. Multiplying by a point of the unit circle is rotation about the origin.
Complex multiplication plays a starring role at the Math Art Shop, and the unit circle is a particular favorite. An earlier post on circles and trigonometry noted that every point of the unit circle in the Cartesian plane has coordinates $$(\cos\theta, \sin\theta)$$ for some real number $$\theta$$. If we identify each real ordered pair $$(a, b)$$ with the complex number $$a + bi$$, we are led to describe points of the unit circle in the form $\cos\theta + i\sin\theta,\qquad\text{$$\theta$$ real.}$ A remarkable identity, often called de Moivre's identity or Euler's formula, says $e^{i\theta} = \cos\theta + i\sin\theta,\qquad\text{$$\theta$$ real.}$ Even to define the terms involved, especially the left-hand side, requires calculus. One approach, often seen in the second semester of a careful study of real numbers and functions, uses “power series” to place this formula on a solid footing.
The particular case $$\theta = \pi$$ reads $e^{i\pi} = \cos\pi + i\sin\pi = -1,$ or $$e^{i\pi} + 1 = 0$$. Perhaps more interestingly, the law of exponents holds for complex exponents. Multiplying the equations \begin{align*} e^{i\theta} = \cos\theta + i\sin\theta, \\ e^{i\phi} = \cos\phi + i\sin\phi, \end{align*} gives a formula for $$e^{i(\theta + \phi)} = \cos(\theta + \phi) + i\sin(\theta + \phi)$$ that contains two hard-to-remember trig identities, the sum formulas for cosine and sine. |
# 5.2 Right triangle trigonometry (Page 6/12)
Page 6 / 12
## Algebraic
For the following exercises, use cofunctions of complementary angles.
$\mathrm{cos}\left(\text{34°}\right)=\mathrm{sin}\left(\text{__°}\right)$
$\mathrm{cos}\left(\frac{\pi }{3}\right)=\mathrm{sin}\text{(___)}$
$\frac{\pi }{6}$
$\mathrm{csc}\left(\text{21°}\right)=\mathrm{sec}\left(\text{___°}\right)$
$\mathrm{tan}\left(\frac{\pi }{4}\right)=\mathrm{cot}\left(\text{__}\right)$
$\frac{\pi }{4}$
For the following exercises, find the lengths of the missing sides if side $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is opposite angle $\text{\hspace{0.17em}}A,$ side $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is opposite angle $\text{\hspace{0.17em}}B,$ and side $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is the hypotenuse.
$\mathrm{cos}\text{\hspace{0.17em}}B=\frac{4}{5},a=10$
$b=\frac{20\sqrt{3}}{3},c=\frac{40\sqrt{3}}{3}$
$\mathrm{tan}\text{\hspace{0.17em}}A=\frac{5}{12},b=6$
$\mathrm{tan}\text{\hspace{0.17em}}A=100,b=100$
$a=10,000,c=10,000.5$
$a=5,\text{\hspace{0.17em}}\measuredangle \text{\hspace{0.17em}}A={60}^{\circ }$
$b=\frac{5\sqrt{3}}{3},c=\frac{10\sqrt{3}}{3}$
$c=12,\text{\hspace{0.17em}}\measuredangle \text{\hspace{0.17em}}A={45}^{\circ }$
## Graphical
For the following exercises, use [link] to evaluate each trigonometric function of angle $\text{\hspace{0.17em}}A.$
$\mathrm{sin}\text{\hspace{0.17em}}A$
$\frac{5\sqrt{29}}{29}$
$\mathrm{cos}\text{\hspace{0.17em}}A$
$\mathrm{tan}\text{\hspace{0.17em}}A$
$\frac{5}{2}$
$\mathrm{csc}\text{\hspace{0.17em}}A$
$\mathrm{sec}\text{\hspace{0.17em}}A$
$\frac{\sqrt{29}}{2}$
$\mathrm{cot}\text{\hspace{0.17em}}A$
For the following exercises, use [link] to evaluate each trigonometric function of angle $\text{\hspace{0.17em}}A.$
$\mathrm{sin}\text{\hspace{0.17em}}A$
$\frac{5\sqrt{41}}{41}$
$\mathrm{cos}\text{\hspace{0.17em}}A$
$\mathrm{tan}\text{\hspace{0.17em}}A$
$\frac{5}{4}$
$\mathrm{csc}\text{\hspace{0.17em}}A$
$\mathrm{sec}\text{\hspace{0.17em}}A$
$\frac{\sqrt{41}}{4}$
$\mathrm{cot}\text{\hspace{0.17em}}A$
For the following exercises, solve for the unknown sides of the given triangle.
## Technology
For the following exercises, use a calculator to find the length of each side to four decimal places.
$b=15,\text{\hspace{0.17em}}\measuredangle B={15}^{\circ }$
$a=55.9808,c=57.9555$
$c=200,\text{\hspace{0.17em}}\measuredangle B={5}^{\circ }$
$c=50,\text{\hspace{0.17em}}\measuredangle B={21}^{\circ }$
$a=46.6790,b=17.9184$
$a=30,\text{\hspace{0.17em}}\measuredangle A={27}^{\circ }$
$b=3.5,\text{\hspace{0.17em}}\measuredangle A={78}^{\circ }$
$a=16.4662,c=16.8341$
## Extensions
Find $\text{\hspace{0.17em}}x.$
Find $\text{\hspace{0.17em}}x.$
188.3159
Find $\text{\hspace{0.17em}}x.$
Find $\text{\hspace{0.17em}}x.$
200.6737
A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is $\text{\hspace{0.17em}}36°,$ and that the angle of depression to the bottom of the tower is $\text{\hspace{0.17em}}23°.\text{\hspace{0.17em}}$ How tall is the tower?
A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is $\text{\hspace{0.17em}}43°,$ and that the angle of depression to the bottom of the tower is $\text{\hspace{0.17em}}31°.\text{\hspace{0.17em}}$ How tall is the tower?
498.3471 ft
A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is $\text{\hspace{0.17em}}15°,$ and that the angle of depression to the bottom of the tower is $\text{\hspace{0.17em}}2°.\text{\hspace{0.17em}}$ How far is the person from the monument?
A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is $\text{\hspace{0.17em}}18°,$ and that the angle of depression to the bottom of the tower is $\text{\hspace{0.17em}}3°.\text{\hspace{0.17em}}$ How far is the person from the monument?
1060.09 ft
There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be $\text{\hspace{0.17em}}40°.\text{\hspace{0.17em}}$ From the same location, the angle of elevation to the top of the antenna is measured to be $\text{\hspace{0.17em}}43°.\text{\hspace{0.17em}}$ Find the height of the antenna.
There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be $\text{\hspace{0.17em}}36°.\text{\hspace{0.17em}}$ From the same location, the angle of elevation to the top of the lightning rod is measured to be $\text{\hspace{0.17em}}38°.\text{\hspace{0.17em}}$ Find the height of the lightning rod.
27.372 ft
## Real-world applications
A 33-ft ladder leans against a building so that the angle between the ground and the ladder is $\text{\hspace{0.17em}}80°.\text{\hspace{0.17em}}$ How high does the ladder reach up the side of the building?
how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Got questions? Join the online conversation and get instant answers! |
Tuesday, 7 March 2017
As a follow up to the article on Napier’s Bones, I thought it would be useful to look at other historical (and interesting) methods for multiplication. There were two in particular that I used in my own teaching - the Gelosia and Russian Peasant methods. They were both great for exploring different ways of handling numbers when calculating and practising key skills, but I wouldn’t necessarily use either as a chosen standard method.
## Gelosia Method
This was brought over as a method from India to Europe in the 14th century, with Gelosia meaning ‘lattice’ from the trellis work on windows in Italy. John Napier based his set of bones on this Gelosia arrangement on a grid.
• You are only dealing with multiplication facts up to 9x9 and addition
• It looks quite tidy and efficient - it is surprisingly quick.
• It looks like a trick – difficult to see the reason why it works
• Easy to forget the ‘carried’ number
• You need lattice paper for the layout
Here is how to calculate 214 x 38.
1. Using the lattice squares with diagonals drawn on them, write 214 along the top and 38 down the right of the grid, with one digit for each square.
2. Start with the top right square and multiply together pairs of digits that meet. 4x3 is 12, so write the answer with the tens digit above the diagonal line and the ones below.
3. Continue across for each square and then do the same for the second row. When it is a single digit, eg 1x3, write 0 above the diagonal line.
4. Once the lattice is complete, start in the bottom right corner and look at the numbers in line separated by the diagonals. Working across and reading up, this is (2), (8, 3, 2), (6, 0, 3, 1), (1, 6, 0) and (0).
5. Add these diagonals up, writing the totals (in red) in line with each square. When the total goes beyond 9, carry the 1 and write it in the next diagonal.
6. Now read the number (in red) from top left down and across to bottom right.
It shows that 214 x 38 = 8132
## Russian Peasant Method
This method for multiplying large numbers only involves halving, doubling and adding.
Follow this example for 45 x 26.
STEP 1
• Write each number at the top of a column.
• Keep halving the number in the left column until you reach 1. If it is an odd number, take away 1 and halve that number instead.
• Double the numbers in the right column for the same number of steps as in the left column.
STEP 2
• Cross out any even numbers in the left column and the corresponding numbers in the right column.
STEP 3
• Add the numbers in the right column that are not crossed out. This gives the answer to the multiplication.
While visiting a Ministry school in Abu Dhabi, the girls in a Y6 class were using what they called the ‘lattice method’, which I recognised as the Gelosia method for multiplying. The teacher was from the US and had this as her preferred standard method for long multiplication.
The girls could explain the method very clearly – even in broken English as Arabic speakers – but did not question why it worked. This is a problem with Napier’s Bones and the Gelosia method – a basic understanding of the method is not clear, unlike the grid method for multiplication which partitions and uses place value with transparency.
Although looking at various methods for long multiplication is valulabe to inspire interest and encourage some great mathematical thinking, the grid method is still my preferred standard method for multiplication.
The advantage of the grid method is that the place value position is clear for children to see and this encourages real understanding of numbers and multiplication, rather than having a method that works but does not support children's understanding.
Some teachers interpret the 2014 NC PoS as only allowing the written method of mulitplication to be taught. This is the ultimate aim, to take children on to the formal vertical written method, however I believe that the grid method provides an important step in a child's understanding, which then leads on to the fformal vertical written method.
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# Question: How Do You Write 1 Crore 30 Lakhs In Numbers?
## How do you write 5 lakhs?
Let’s take a 6-digit number 5,46,783.
According to the Indian place value system, it is written as Five lakh, forty-six thousand, seven hundred and eighty-three..
## How many lakhs make a billion?
In Indian numeral system, the place values of the digits are as follows: Ones, Tens, Hundreds, Thousands, Ten Thousand, Lakhs, Ten Lakhs, Crores and so on. So we see that 10000 Lakh is there in one billion.
## How do you write 30 lakhs?
Note: We found that some people call it 30 lakhs or 30 lac, but the correct way to say it is 30 lakh (lakh without the trailing “s”).
## How do you write 4 lakhs in digits?
As you can see, 4 lakh is the same as 0.4 million.
## How many zeros are there in 45 lakhs?
As you can see, 45 lakh is the same as 4.5 million. How many zeros in 45 lakh? When we count the trailing zeros in 45 lakh above, we see that the answer is 5.
## What is next to crore?
In the Indian system, the next powers of ten are called one lakh, ten lakh, one crore, ten crore, one arab (or one hundred crore), and so on; there are new words for every second power of ten (105 + 2n): lakh (105), crore (107), arab (109), etc.
## How many zeros are there in 3 crores?
7How many zeros in 3 crore? When we count the trailing zeros in 3 crore above, we see that the answer is 7.
## How do you write 1 crore 50 lakhs in numbers?
Similarly,10 million (as per the international digit system) or the one crore have 7 zeros (0) in it’s numerical value. To write the 1 crore 50 lakhs in numerical form we just have to replace the first zero with a five, because 50 lakhs is the exact half value of 1 crore. So the numerical value will be; 15000000.
## How do you write 1 crore 75 lakhs in numbers?
1,75,00,000 As you can see, 1 crore and 75 lakh is the same as 17.5 million.
## How do you write 1 crore 45 lakhs in numbers?
1,45,00,000 As you can see, 1 crore and 45 lakh is the same as 14.5 million.
## How many dollars is 5 lakhs?
Convert Indian Rupee to US DollarINRUSD1000 INR13.4261 USD5000 INR67.1305 USD10000 INR134.261 USD50000 INR671.305 USD7 more rows•Nov 2, 2020
## How do you say 2 lakh in English?
Both lakh and lac are correct. In India,we generally write lakh. Lakh is more commonly used here. While lac is mostly used by western nations.
## How do you write 20 lakhs in words?
Note: We found that some people call it 20 lakhs or 20 lac, but the correct way to say it is 20 lakh (lakh without the trailing “s”).
## How can I write 1 lakh in English?
A lakh (/læk, lɑːk/; abbreviated L; sometimes written Lac or Lacs) is a unit in the Indian numbering system equal to one hundred thousand (100,000; scientific notation: 105). In the Indian 2,2,3 convention of digit grouping, it is written as 1,00,000.
## How many lakhs make a crore?
100 lakhA crore (/krɔːr/; abbreviated cr), karor or koti denotes ten million (10,000,000 or 107 in scientific notation) and is equal to 100 lakh in the Indian numbering system. |
# What is 1/1788 as a percentage?
It's very common when learning about fractions to want to know how convert a fraction like 1/1788 into a percentage. In this step-by-step guide, we'll show you how to turn any fraction into a percentage really easily. Let's take a look!
Want to quickly learn or show students how to convert 1/1788 to a percentage? Play this very quick and fun video now!
Before we get started in the fraction to percentage conversion, let's go over some very quick fraction basics. Remember that a numerator is the number above the fraction line, and the denominator is the number below the fraction line. We'll use this later in the tutorial.
When we are using percentages, what we are really saying is that the percentage is a fraction of 100. "Percent" means per hundred, and so 50% is the same as saying 50/100 or 5/10 in fraction form.
So, since our denominator in 1/1788 is 1788, we could adjust the fraction to make the denominator 100. To do that, we divide 100 by the denominator:
100 รท 1788 = 0.05592841163311
Once we have that, we can multiple both the numerator and denominator by this multiple:
1 x 0.05592841163311 / 1788 x 0.05592841163311 = 0.05592841163311 / 100
Now we can see that our fraction is 0.05592841163311/100, which means that 1/1788 as a percentage is 0.0559%.
We can also work this out in a simpler way by first converting the fraction 1/1788 to a decimal. To do that, we simply divide the numerator by the denominator:
1/1788 = 0.0005592841163311
Once we have the answer to that division, we can multiply the answer by 100 to make it a percentage:
0.0005592841163311 x 100 = 0.0559%
And there you have it! Two different ways to convert 1/1788 to a percentage. Both are pretty straightforward and easy to do, but I personally prefer the convert to decimal method as it takes less steps.
I've seen a lot of students get confused whenever a question comes up about converting a fraction to a percentage, but if you follow the steps laid out here it should be simple. That said, you may still need a calculator for more complicated fractions (and you can always use our calculator in the form below).
If you want to practice, grab yourself a pen, a pad, and a calculator and try to convert a few fractions to a percentage yourself.
Hopefully this tutorial has helped you to understand how to convert a fraction to a percentage. You can now go forth and convert fractions to percentages as much as your little heart desires!
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 1/1788 as a percentage?". VisualFractions.com. Accessed on September 22, 2023. http://visualfractions.com/calculator/fraction-as-percentage/what-is-1-1788-as-a-percentage/.
• "What is 1/1788 as a percentage?". VisualFractions.com, http://visualfractions.com/calculator/fraction-as-percentage/what-is-1-1788-as-a-percentage/. Accessed 22 September, 2023. |
# In the Given Figure, If Bp || Cq and Ac = Bc, Then the Measure of X is - Mathematics
MCQ
In the given figure, if BP || CQ and AC = BC, then the measure of x is
• 20°
• 25°
• 30°
• 35°
#### Solution
In the given figure, BP || CQ and AC || BC
We need to find the measure of x
Here, we draw a line RS parallel to BP, i.e BP || RS
Also, using the property, “two lines parallel to the same line are parallel to each other”
As,
BP || RS
BP || CQ
Thus, RS || CQ
Now, BP || RS and BA is the transversal, so using the property, “alternate interior angles are equal”
∠PBA = ∠BAS
∠BAS = 20° .......... (1)
Similarly, CQ|| RS and AC is the transversal
∠QCA = ∠SAC
∠SAC = x ........(2)
Adding (1) and (2), we get
∠SAC + ∠BAS = 20° + x
∠A = 20° + x
Also, as AC = BC
Using the property,”angles opposite to equal sides are equal”, we get
∠ABC = ∠CAB
∠ABC = 20° + x
Further, using the property, “an exterior angle is equal to the sum of the two opposite interior angles”
In ΔABC
ext. ∠C = ∠CAB + ∠ABC
70° + x = 20° + x + 20° + x
70° + x = 40° + 2x
70° - 40° = 2x - x
x = 30°
Thus, x = 30°
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Mathematics for Class 9
Chapter 11 Triangle and its Angles
Q 22 | Page 28 |
Definitions
# Division by zero
In mathematics, a division is called a division by zero if the divisor is zero. Such a division can be formally expressed as $textstylefrac\left\{a\right\}\left\{0\right\}$ where a is the dividend. Whether this expression can be assigned a well-defined value depends upon the mathematical setting. In ordinary (real number) arithmetic, the expression has no meaning.
In computer programming, integer division by zero may cause a program to terminate or, as in the case of floating point numbers, may result in a special not-a-number value (see below).
## In elementary arithmetic
When division is explained at the elementary arithmetic level, it is often considered as a description of dividing a set of objects into equal parts. As an example, consider having 10 apples, and these apples are to be distributed equally to five people at a table. Each person would receive $textstylefrac\left\{10\right\}\left\{5\right\}$ = 2 apples. Similarly, if there are 10 apples, and only one person at the table, each person would receive $textstylefrac\left\{10\right\}\left\{1\right\}$ = 10 apples.
So for dividing by zero — what if there are 10 apples to be distributed, but no one comes to the table? How many apples does each "person" at the table receive? The question itself is meaningless — each "person" can't receive zero, or 10, or an infinite number of apples for that matter, because there are simply no people to receive anything in the first place. So $textstylefrac\left\{10\right\}\left\{0\right\}$, at least in elementary arithmetic, is said to be meaningless, or undefined.
Another way to understand the nature of division by zero is by considering division as a repeated subtraction. For example, to divide 13 by 5, 5 can be subtracted twice, which leaves a remainder of 3 — the divisor is subtracted until the remainder is less than the divisor. The result is often reported as $textstylefrac\left\{13\right\}\left\{5\right\}$ = 2 remainder 3. But, in the case of zero, repeated subtraction of zero will never yield a remainder less than zero. Dividing by zero by repeated subtraction results in a series of subtractions that never ends. This connection of division by zero to infinity takes us beyond elementary arithmetic (see below).
### Early attempts
The Brahmasphutasiddhanta of Brahmagupta (598–668) is the earliest known text to treat zero as a number in its own right and to define operations involving zero. The author failed, however, in his attempt to explain division by zero: his definition can be easily proven to lead to algebraic absurdities. According to Brahmagupta,
"A positive or negative number when divided by zero is a fraction with the zero as denominator. Zero divided by a negative or positive number is either zero or is expressed as a fraction with zero as numerator and the finite quantity as denominator. Zero divided by zero is zero."
In 830, Mahavira tried unsuccessfully to correct Brahmagupta's mistake in his book in Ganita Sara Samgraha:
"A number remains unchanged when divided by zero."
Bhaskara II tried to solve the problem by defining $textstylefrac\left\{n\right\}\left\{0\right\}=infty$. This definition makes some sense, as discussed below, but can lead to paradoxes if not treated carefully. These paradoxes were not treated until modern times.
## In algebra
It is generally regarded among mathematicians that a natural way to interpret division by zero is to first define division in terms of other arithmetic operations. Under the standard rules for arithmetic on integers, rational numbers, real numbers and complex numbers, division by zero is undefined. Division by zero must be left undefined in any mathematical system that obeys the axioms of a field. The reason is that division is defined to be the inverse operation of multiplication. This means that the value of $textstylefrac\left\{a\right\}\left\{b\right\}$ is the solution x of the equation $bx = a$ whenever such a value exists and is unique. Otherwise the value is left undefined.
For b = 0, the equation bx = a can be rewritten as 0x = a or simply 0 = a. Thus, in this case, the equation bx = a has no solution if a is not equal to 0, and has any x as a solution if a equals 0. In either case, there is no unique value, so $textstylefrac\left\{a\right\}\left\{b\right\}$ is undefined. Conversely, in a field, the expression $textstylefrac\left\{a\right\}\left\{b\right\}$ is always defined if b is not equal to zero.
### Fallacies based on division by zero
It is possible to disguise a special case of division by zero in an algebraic argument, leading to spurious proofs that 2 = 1 such as the following:
With the following assumptions:
begin\left\{align\right\}
0times 1 &= 0 0times 2 &= 0. end{align}
The following must be true:
$0times 1 = 0times 2.,$
Dividing by zero gives:
$textstyle frac\left\{0\right\}\left\{0\right\}times 1 = frac\left\{0\right\}\left\{0\right\}times 2.$
Simplified, yields:
$1 = 2.,$
The fallacy is the implicit assumption that dividing by 0 is a legitimate operation with $0/0=1$.
Although most people would probably recognize the above "proof" as fallacious, the same argument can be presented in a way that makes it harder to spot the error. For example, if 1 is denoted by $x$, $0$ can be hidden behind $x-x$ and $2$ behind $x+x$. The above mentioned proof can then be displayed as follows:
begin\left\{align\right\}
(x-x)x &= x^2-x^2 = 0 (x-x)(x+x) &= x^2-x^2 = 0 end{align}
hence:
$\left(x-x\right)x = \left(x-x\right)\left(x+x\right).,$
Dividing by $x-x,$ gives:
$x = x+x,$
and dividing by $x,$ gives:
$1 = 2.,$
The "proof" above requires the use of the distributive law. However, this requirement introduces an asymmetry between the two operations in that multiplication distributes over addition, but not the other way around. Thus, the multiplicative identity element, 1, has an additive inverse, -1, but the additive identity element, 0, does not have a multiplicative inverse.
## In calculus
### Extended real line
At first glance it seems possible to define $textstylefrac\left\{a\right\}\left\{0\right\}$ by considering the limit of $textstylefrac\left\{a\right\}\left\{b\right\}$ as b approaches 0.
For any positive a, it is known that
$lim_\left\{b to 0^\left\{+\right\}\right\} \left\{a over b\right\} = \left\{+\right\}infty$
and for any negative a,
$lim_\left\{b to 0^\left\{+\right\}\right\} \left\{a over b\right\} = \left\{-\right\}infty.$
Therefore, if $textstylefrac\left\{a\right\}\left\{0\right\}$ as +∞ is defined for positive a, and −∞ for negative a. However, taking the limit from the right is arbitrary. The limits could be taken from the left as well and defined $textstylefrac\left\{a\right\}\left\{0\right\}$ to be −∞ for positive a, and +∞ for negative a. This can be further illustrated using the equation (assuming that several natural properties of reals extend to infinities)
$+infty = frac\left\{1\right\}\left\{0\right\} = frac\left\{1\right\}\left\{-0\right\} = -frac\left\{1\right\}\left\{0\right\} = -infty$
which would lead to the result +∞ = −∞, inconsistent with standard definitions of limit in the extended real line. The only workable extension is introducing an unsigned infinity, discussed below.
Furthermore, there is no obvious definition of $textstylefrac\left\{0\right\}\left\{0\right\}$ that can be derived from considering the limit of a ratio. The limit
$lim_\left\{\left(a,b\right) to \left(0,0\right)\right\} \left\{a over b\right\}$
does not exist. Limits of the form
$lim_\left\{x to 0\right\} \left\{f\left(x\right) over g\left(x\right)\right\}$
in which both f(x) and g(x) approach 0 as x approaches 0, may equal any real or infinite value, or may not exist at all, depending on the particular functions f and g (see l'Hôpital's rule for discussion and examples of limits of ratios). These and other similar facts show that the expression $textstylefrac\left\{0\right\}\left\{0\right\}$ cannot be well-defined as a limit.
#### Formal operations
A formal calculation is one which is carried out using rules of arithmetic, without consideration of whether the result of the calculation is well-defined. Thus, as a rule of thumb, it is sometimes useful to think of $textstylefrac\left\{a\right\}\left\{0\right\}$ as being $infty$, provided a is not zero. This infinity can be either positive, negative or unsigned, depending on context. For example, formally:
$limlimits_\left\{x to 0\right\} \left\{frac\left\{1\right\}\left\{x\right\} =frac\left\{limlimits_\left\{x to 0\right\} \left\{1\right\}\right\}\left\{limlimits_\left\{x to 0\right\} \left\{x\right\}\right\}\right\} = frac\left\{1\right\}\left\{0\right\} = infty.$
As with any formal calculation, invalid results may be obtained. A logically rigorous as opposed to formal computation would say only that
$limlimits_\left\{x to 0^+\right\} \left\{frac\left\{1\right\}\left\{x\right\}\right\} = frac\left\{1\right\}\left\{0^+\right\} = +infty$ and $limlimits_\left\{x to 0^-\right\} \left\{frac\left\{1\right\}\left\{x\right\}\right\} = frac\left\{1\right\}\left\{0^-\right\} = -infty.$
(Since the one-sided limits are different, the two-sided limit does not exist in the standard framework of the real numbers. Also, the fraction $textstylefrac\left\{1\right\}\left\{0\right\}$ is left undefined in the extended real line, therefore it and $textstylefrac\left\{limlimits_\left\{x to 0\right\} \left\{1\right\}\right\}\left\{limlimits_\left\{x to 0\right\} \left\{x\right\}\right\}$ are meaningless expressions that should not rigorously be used in an equation.)
### Real projective line
The set $mathbb\left\{R\right\}cup\left\{infty\right\}$ is the real projective line, which is a one-point compactification of the real line. Here $infty$ means an unsigned infinity, an infinite quantity which is neither positive nor negative. This quantity satisfies $-infty = infty$ which is necessary in this context. In this structure, $textstylefrac\left\{a\right\}\left\{0\right\} = infty$ can be defined for nonzero a, and $textstylefrac\left\{a\right\}\left\{infty\right\} = 0$. It is the natural way to view the range of the tangent and cotangent functions of trigonometry: tan(x) approaches the single point at infinity as x approaches either $textstyle+frac\left\{pi\right\}\left\{2\right\}$ or $textstyle-frac\left\{pi\right\}\left\{2\right\}$ from either direction.
This definition leads to many interesting results. However, the resulting algebraic structure is not a field, and should not be expected to behave like one. For example, $infty + infty$ has no meaning in the projective line.
### Riemann sphere
The set $mathbb\left\{C\right\}cup\left\{infty\right\}$ is the Riemann sphere, of major importance in complex analysis. Here, too, $infty$ is an unsigned infinity, or, as it is often called in this context, the point at infinity. This set is analogous to the real projective line, except that it is based on the field of complex numbers. In the Riemann sphere, $1/0=infty$, but $0/0$ is undefined, as well as $0timesinfty$.
### Extended non-negative real number line
The negative real numbers can be discarded, and infinity introduced, leading to the set $\left[0, infty\right]$, where division by zero can be naturally defined as $textstylefrac\left\{a\right\}\left\{0\right\} = infty$ for positive a. While this makes division defined in more cases than usual, subtraction is instead left undefined in many cases, because there are no negative numbers.
## In higher mathematics
Although division by zero cannot be sensibly defined with real numbers and integers, it is possible to consistently define it, or similar operations, in other mathematical structures.
### Non-standard analysis
In the hyperreal numbers and the surreal numbers, division by zero is still impossible, but division by non-zero infinitesimals is possible.
### Distribution theory
In distribution theory one can extend the function $textstylefrac\left\{1\right\}\left\{x\right\}$ to a distribution on the whole space of real numbers (in effect by using Cauchy principal values). It does not, however, make sense to ask for a 'value' of this distribution at $x = 0$; a sophisticated answer refers to the singular support of the distribution.
### Linear algebra
In matrix algebra (or linear algebra in general), one can define a pseudo-division, by setting $textstylefrac\left\{a\right\}\left\{b\right\}=a b^+$, in which b+ represents the pseudoinverse of b. It can be proven that if b−1 exists, then b+ = b−1. If b equals 0, then 0+ = 0; see Generalized inverse.
### Abstract algebra
Any number system which forms a commutative ring, as do the integers, the real numbers, and the complex numbers, for instance, can be extended to a wheel in which division by zero is always possible, but division has then a slightly different meaning.
The concepts applied to standard arithmetic are similar to those in more general algebraic structures, such as rings and fields. In a field, every nonzero element is invertible under multiplication, so as above, division poses problems only when attempting to divide by zero. This is likewise true in a skew field (which for this reason is called a division ring). However, in other rings, division by nonzero elements may also pose problems. For example, the ring Z/6Z of integers mod 6. The meaning of the expression $textstylefrac\left\{2\right\}\left\{2\right\}$ should be the solution x of the equation $2x = 2$. But in the ring Z/6Z, 2 is not invertible under multiplication. This equation has two distinct solutions, x = 1 and x = 4, so the expression $textstylefrac\left\{2\right\}\left\{2\right\}$ is undefined.
In field theory, the expression $textstylefrac\left\{a\right\}\left\{b\right\}$ is only shorthand for the formal expression $ab^\left\{-1\right\}$, where $b^\left\{-1\right\}$ is the multiplicative inverse of $b$. Since the field axioms only guarantee the existence of such inverses for nonzero elements, this expression has no meaning when $b$ is zero. In modern texts the axiom $textstyle 0neq 1$ is included in order to avoid having to consider the one-element field where the multiplicative identity coincides with the additive identity. In such 'fields' however, $0=1$, and $textstylefrac\left\{0\right\}\left\{0\right\}=frac\left\{0\right\}\left\{1\right\}=0$, and division by zero is actually noncontradictory.
## In computer arithmetic
The IEEE floating-point standard, supported by almost all modern processors, specifies that every floating point arithmetic operation, including division by zero, has a well-defined result. In IEEE 754 arithmetic, a ÷ 0 is positive infinity when a is positive, negative infinity when a is negative, and NaN (not a number) when a = 0. The infinity signs change when dividing by −0 instead. This is possible because in IEEE 754 there are two zero values, plus zero and minus zero, and thus no ambiguity.
Integer division by zero is usually handled differently from floating point since there is no integer representation for the result. Some processors generate an exception when an attempt is made to divide an integer by zero, although others will simply continue and generate an incorrect result for the division. (That result is often zero.)
Because of the improper algebraic results of assigning any value to division by zero, many computer programming languages (including those used by calculators) explicitly forbid the execution of the operation and may prematurely halt a program that attempts it, sometimes reporting a "Divide by zero" error. Some programs (especially those that use fixed-point arithmetic where no dedicated floating-point hardware is available) will use behavior similar to the IEEE standard, using large positive and negative numbers to approximate infinities. In some programming languages, an attempt to divide by zero results in undefined behavior.
In two's complement arithmetic, attempts to divide the smallest signed integer by $-1$ are attended by similar problems, and are handled with the same range of solutions, from explicit error conditions to undefined behavior.
Most calculators will either return an error or state that 1/0 is undefined, however some TI graphing calculators will evaluate 1/02 to ∞.
### Historical accidents
• On September 21, 1997, a divide by zero error in the USS Yorktown (CG-48) Remote Data Base Manager brought down all the machines on the network, causing the ship's propulsion system to fail.
## References
• Patrick Suppes 1957 (1999 Dover edition), Introduction to Logic, Dover Publications, Inc., Mineola, New York. ISBN 0-486-40687-3 (pbk.). This book is in print and readily available. Suppes's §8.5 The Problem of Division by Zero begins this way: "That everything is not for the best in this best of all possible worlds, even in mathematics,is well illustrated by the vexing problem of defining the operation of division in the elementary theory of artihmetic" (p. 163). In his §8.7 Five Approaches to Division by Zero he remarks that "...there is no uniformly satisfactory solution" (p. 166)
• Charles Seife 2000, Zero: The Biography of a Dangerous Idea, Penguin Books, NY, ISBN 0 14 02.9647 6 (pbk.). This award-winning book is very accessible. Along with the fascinating history of (for some) an abhorent notion and others a cultural asset, describes how zero is misapplied with respect to multiplication and division.
• Alfred Tarski 1941 (1995 Dover edition), Introduction to Logic and to the Methodology of Deductive Sciences, Dover Publications, Inc., Mineola, New York. ISBN 0-486-28462-X (pbk.). Tarski's §53 Definitions whose definiendum contains the identity sign discusses how mistakes are made (at least with respect to zero). He ends his chapter "(A discussion of this rather difficult problem [exactly one number satisfying a definiens] will be omitted here.*)" (p. 183). The * points to Exercise #24 (p. 189) wherein he asks for a proof of the following: "In section 53, the definition of the number '0' was stated by way of an example. In order to be certain that this definition does not lead to a contradiction, it should be preceded by the following theorem: There exists exactly one number x such that, for any number y, we have: y + x = y" |
# In a Group of Buffaloes and Ducks, the Number of Legs are 24 More than Twice the Number of Heads. What is the Number of Buffaloes in the Group?
### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs | Set 6
What are you looking for? Let’s dig in quickly
## Explanation
Let suppose “b” represents buffalo and “d” ducks. Each buffalo has 4 legs (4b) and each duck has 2 legs (2d). So in the group of buffaloes and ducks, total legs will be “4b + 2d” and total heads “b + d”. Now condition of the question says;
4b + 2d – 24 = 2(b + d)
Through this equation, we can easily figure out buffaloes.
Buffaloes = ?
## Solution
Let suppose
Buffalo = b
Duck = d
Total legs = 4b + 2d
Total heads = b + d
According to given condition;
4b + 2d – 24 = 2 (b + d)
4b + 2d – 24 = 2b + 2d
4b – 24 = 2b
4b – 2b = 24
2b = 24
b = 24/2
## Conclusion
In a group of buffaloes and ducks, the number of legs are 24 more than twice the number of heads. There will be 12 buffaloes in the group of ducks and buffaloes. |
# CHANGE OF VARIABLES AND THE JACOBIAN. Contents 1. Change of variables: the Jacobian 1
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1 CHANGE OF VARIABLES AND THE JACOBIAN Contents 1. Change of variables: the Jacobian 1 1. Change of variables: the Jacobian So far, we have seen three examples of situations where we change variables to help us evaluate integrals: when we change from rectangular coordinates in R 2 to polar coordinates, when we change from rectangular in R 3 to cylindrical coordinates, and when we change from rectangular to spherical coordinates. These are all special instances of change of variables, where we replace variables in an integral with other variables, according to how they are related to each other. In each case, there was a formula which connected an integral in one coordinate system to an integral in the other coordinate system, and in each case an extra factor appeared in the integrand. We want to give a brief idea of just why these factors appear, although a full explanation requires linear algebra. Let us take a step back and think about just how we change variables before we actually think about integration. In each situation, we re given two or three variables, such as x, y, z, and two or three other variables, such as r, θ, z, or ρ, θ, φ, which we express x, y, z in terms of. Let s look at the case of polar coordinates, where x = r cos θ, y = r sin θ. We can encapsulate the relationship between x, y, r, θ by defining a function T : R 2 R 2 which takes (r, θ) to (r cos θ, r sin θ). In other words, T takes the polar coordinates of some points to the rectangular coordinates of that point. In particular, suppose we want to integrate a function f(x, y) over a region R using polar coordinates. Then we start by finding a region S which describes R in polar coordinates: in other words, we want T (S) = R. Furthermore, we want T to be a one-to-one map on S; ie, we only want one point in S to map to a given point in R. This manifests itself in the requirement that r 0, 0 β α 2π when we integrate over a polar rectangle, for example. Strictly speaking, we can slightly relax the one-to-one condition, and possibly allow T to not be one-to-one for points on the boundary of S. In any case, we end up with a formula which describes the integral of f(x, y) over R as an integral of f(t (r, θ)) = f(r cos θ, r sin θ) over S. This new terminology brings out the essential features when changing variables. In the case where we are dealing with two variables, we have a function T, defined on points (u, v), which sends (u, v) to (x, y), where x = g(u, v), y = h(u, v), are some functions of x, y. If S is a region in the uv plane on which T is a one-to-one function, except possibly at the boundary of S, and R = T (S), then we want an expression which relates an integral using xy coordinates to one using uv coordinates. 1
2 2 CHANGE OF VARIABLES AND THE JACOBIAN For technical reasons, we require that T be a C 1 transformation, which is a fancy way of saying that g(u, v), h(u, v) should have continuous first order partial derivatives. How are integrals using uv coordinates related to integrals using xy coordinates? Recall that when defining an integral, we used Riemann sums, whose individual terms looked like f(x, y ) x y, where x, y was a point inside a box of dimensions x y. If we are integrating over S in the uv plane, a typical term in a Riemann sum will be over a rectangle with side lengths u, v. Suppose the rectangle this term represents has a vertex (u, v) in the lower left hand corner. This is a small rectangle in the uv plane. What is the image of this rectangle under the transformation T? Well, it might be hard to determine exactly what this image is (T might well be a non-linear map), but we can approximate the image with a parallelogram by pretending that T is linear. In particular, the way we pretend that T is linear is by looking at partial derivatives of g, h. For example, suppose we want to approximate the image of (u + u, v) under T. Then we are incrementing the u-coordinate by a small amount u. The x coordinate of T (u, v) is g(u, v). The x coordinate of T (u + u, v) is given by g(u + u, v), but this can be approximated by g(u) + (u, v) u because the partial derivative of g with respect to u, at (u, v), is equal to the rate of change in the u-variable of g at (u, v). Therefore, the xy coordinates of T (u + u, v) are approximately equal to ( g(u) + ) (u, v) u, h(u) + (u, v) u. In a similar fashion, the xy coordinates of T (u, v + v) are approximately equal to ( g(u) + ) (u, v) v, h(u) + (u, v) v. We approximate the image of the small rectangle under T by the parallelogram which has these three vertices as adjacent vertices. The two sides of the parallelogram are represented by the vectors (u, v) u, (u, v) u, (u, v) v, (u, v) v. Recall that we have a method to calculate the area of a parallelogram spanned by two vectors! If we pretend these are two vectors in R 3 by adding a z coordinate of 0, then the absolute value of their cross product is equal to the area of this parallelogram. The cross product of these two vectors (thought of as in R 3 is i j k (u, v) u (u, v) u 0 (u, v) v (u, v) v 0 = ( (u, v) u (u, v) v ) (u, v) u (u, v) v k.
3 CHANGE OF VARIABLES AND THE JACOBIAN 3 The end result of all of these approximations and calculations is that the image of the small u v rectangle has area approximately equal to (u, v) u (u, v) v (u, v) u (u, v) v = (u, v) (u, v) (u, v) Therefore, if we use the images of all the small uv rectangles to form a Riemann sum in the xy plane, we end up with a sum of terms of the form f(g(u, v ), h(u, v ) (u, v)(u, v) (u, v) (u, v) u v. Define the function J(u, v), called the Jacobian of T, to be J(u, v) = (u, v)(u, v) (u, v) (u, v). The sums of these terms approximates the double integral f(x, y) da, R on the one hand, since the sum is composed of terms over small pieces which make up R. On the other hand, this sum is also a Riemann sum for f(g(u, v), h(u, v)) J(u, v) du dv. S Therefore, it is reasonable to expect these two integrals to be equal to each other: f(x, y) da = f(g(u, v), h(u, v)) J(u, v) du dv. R S Notice that this looks like a more general form of the various integration formulas we know for different coordinate systems: the integral of a function over a region R is expressed as an integral over a region S, which describes R using a different coordinate system, with an additional factor J(u, v) inserted into the integrand. We remark here that, at least in the two variable case, the Jacobian J(u, v) is equal to the determinant of a 2 2 matrix: (x, y) J(u, v) = (u, v) = g u(u, v) g v (u, v) h u (u, v) h v (u, v). [ ] a b (The determinant of a 2 2 matrix is equal to ad bc.) c d Example. Let x = r cos θ, y = r sin θ be the change of coordinates from (r, θ) to (x, y). What is the Jacobian of this transformation? We begin by calculating the partial derivatives of x, y with respect to r, θ. We have x r = cos θ, x θ = r sin θ, y r = sin θ, y θ = r cos θ. Therefore, the Jacobian is (u, v) u v.
4 4 CHANGE OF VARIABLES AND THE JACOBIAN J(r, θ) = (x, y) (r, θ) = cos θ sin θ r sin θ r cos θ = r cos2 θ + r sin 2 θ = r. This explains the factor of r which appears in the integrand when switching to polar coordinates. In the case where we use change of variables with three coordinates, from a uvw coordinate system to an xyz coordinate system, such as in cylindrical or spherical coordinates, it turns out that the Jacobian is given by the determinant of a 3 3 matrix whose entries are the various partial derivatives of x, y, z in terms of u, v, w. This is not an accident, because a fundamental property of determinants you learn in linear algebra is that the absolute value of the determinant of a matrix is equal to the volume of the parallelepiped spanned by its rows (or columns). Example. Calculate the Jacobian of the transformation for rectangular coordinates; ie, the Jacobian of x = r cos θ, y = r sin θ, z = z. The relevant partial derivatives are x r = cos θ, x θ = r sin θ, x z = 0, y r = sin θ, y θ = r cos θ, y z = 0, z r = 0, z θ = 0, z z = 1. Therefore, the Jacobian of this transformation is given by the determinant J(r, θ, z) = (x, y, z) (r, θ, z) = cos θ r sin θ 0 sin θ r cos θ One can calculate this determinant using whatever formulas you may know; some knowledge of linear algebra tells us that this determinant is in fact equal to cos θ sin θ r sin θ r cos θ 1 = r. (This is found by performing so-called Laplace expansion on the third row.) Recall that the factor which appears in a change of variable formula when integrating is the Jacobian, which is the determinant of a matrix of first order partial derivatives. Example. Check that the Jacobian of the transformation to spherical coordinates is ρ 2 sin φ. The formulas relating rectangular to spherical coordinates are x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. Therefore, the Jacobian is given by x ρ x θ x φ y ρ y θ y φ z ρ z θ z φ =. sin φ cos θ ρ sin φ sin θ ρ cos φ cos θ sin φ sin θ ρ sin φ cos θ ρ cos φ sin θ cos φ 0 ρ sin φ If one expands this determinant in the usual way, after a lot of gathering terms and using the identity cos 2 + sin 2 = 1, the above expression eventually equals ρ 2 sin φ. Here are a few remarks about the Jacobian:.
5 CHANGE OF VARIABLES AND THE JACOBIAN 5 The Jacobian and change of variable formula is a generalization of u-substitution. If we make a u-substitution u = u(x), then we have a formula f(u) du = f(u(x))u (x) dx. The Jacobian of the transformation u = u(x) is u (x), which is exactly the factor appearing on the right side of this equation. The missing absolute value sign is accounted for by the fact that if u (x) is negative, the bounds of integration are interchanged. If you have been carefully paying attention to the definition of a Jacobian, you might be somewhat bothered by the fact that the Jacobian seems to depend on the ordering of the variables in each of the two variable systems, since the ordering determines the order of the rows and columns of the matrix in the Jacobian. It turns out that interchanging rows and columns of a matrix may change the sign of a determinant, but never the absolute value of the determinant, so the ordering of the variables does not particularly matter when calculating a Jacobian. You may end up with an answer which differs by a minus sign from someone else with a different ordering, but when using the change of variable formula, you take the absolute value of the Jacobian so the sign ambiguity will disappear. Here is a typical example of using a change of variables which is not polar, cylindrical, or spherical. You need to first decide which change of coordinates you should use to bring the problem to a manageable form, and then make the appropriate coordinate change. Deciding which change to use is more of an art than a science, and requires a lot of practice to get consistently right. Nevertheless there are some clues which might help you make the correct coordinate change. Example. Evaluate the double integral e (y x)/(y+x) da. D where D is the triangle with vertices (0, 0), (2, 0), (0, 2). The presence of y x, y + x in a fraction suggests we should make a change of variable like u = y x, v = y + x. (It is still not obvious at this point that this is the correct change to make.) We start by determining what the corresponding domain of integration, say R, in the uv plane is. The triangle D lies in the xy plane, so to determine R we should look at the image of D under the map u = y x, v = y + x. In particular, we can determine the boundary of R by looking at how this change of variables to uv coordinates acts on the boundary of the triangle D. For example, the side with endpoints (0, 0), (2, 0) is given by 0 x 2, y = 0. The corresponding uv coordinates are then u = x, v = x, 0 x 2, so this side maps to the side v = u, 2 u 0. In particular, in the uv plane this is a line segment with endpoints (0, 0), ( 2, 2). We also find that the other two line segments map to line segments, and that R is actually a triangle with vertices (0, 0), ( 2, 2), (2, 2). Therefore we can describe R using inequalities 0 v 2, v u v. We now need to calculate the Jacobian of this transformation. We need to solve for
6 6 CHANGE OF VARIABLES AND THE JACOBIAN x, y in terms of u, v. Fortunately, in this example this is easy, and we see that x = (v u)/2, y = (v + u)/2. The Jacobian is given by the following determinant: x u x v y u y v = 1/2 1/2 1/2 1/2 = 1/2. Therefore, the original integral is equal to the integral e u/v (1/2) da = 1 2 R 2 v 0 v e u/v du dv = ve u/v u=v u= v dv = ve ve 1 = e 1/e.
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# NCERT Solutions for Class 6 Maths Exercise 11.4
NCERT solutions for Maths Algebra
## NCERT Solutions for Class 6 Maths Algebra
###### (Ex. 11.4)
(a) Take Sarita’s present age to be {tex}y{/tex} years.
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
(b)The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is {tex}b{/tex} meters?
(c)A rectangular box has height {tex}h{/tex} cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
(d)Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step {tex}s,{/tex} Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using {tex}s.{/tex}
(e)A bus travels at {tex}v{/tex} km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours. Beespur is still 20 km away. What id the distance from Daspur to Beespur? Express it using {tex}v.{/tex}
(a)(i) {tex}y + 5{/tex}
(ii) {tex}y – 3{/tex}
(iii) {tex}6y{/tex}
(iv) {tex}6y – 2{/tex}
(v) {tex}3y + 5{/tex}
(b) Length = {tex}3b{/tex} and Breadth = {tex}\left( {3b – 4} \right){/tex} meters
(c) Height of the box = {tex}h{/tex} cm
Length of the box = 5 times the height = {tex}5h{/tex} cm
Breadth of the box = 10 cm less than length = {tex}\left( {5h – 10} \right){/tex} cm
(d) Meena’s position = {tex}s{/tex}
Beena’s position = 8 steps ahead = {tex}s + 8{/tex}
Leena’s position = 7 steps behind = {tex}s – 7{/tex}
Total number of steps = {tex}4s – 10{/tex}
(e) Speed of the bus = {tex}v{/tex} km/h
Distance travelled in 5 hours = {tex}5v{/tex} km
Remaining distance = 20 km
Therefore, total distance = {tex}\left( {5v + 20} \right){/tex} km
NCERT Solutions for Class 6 Maths Exercise 11.4
###### Question 2.Change the following statements using expressions into statements in ordinary language.
(For example, given Salim scores {tex}r{/tex} runs in a cricket match, nalin scores {tex}\left( {r + 15} \right){/tex} runs. In ordinary language – Nalin scores 15 runs more than Salim).
(a)A note book costs {tex}p.{/tex} A book costs {tex}3p.{/tex}
(b)Tony puts {tex}q{/tex} marbles on the table. He has {tex}8q{/tex} marbles in his box.
(c)Our class has {tex}n{/tex} students. The school has {tex}20n{/tex} students.
(d)Jaggu is {tex}z{/tex} years old. His uncle is {tex}4z{/tex} years old and his aunt is {tex}\left( {4z – 3} \right){/tex} years old.
(e)In an arrangement of dots there are {tex}r{/tex} rows. Each row contains 5 dots.
(a) A book cost 3 times the cost of a notebook.
(b) The number of marbles in box is 8 times the marble on the table.
(c) Total number of students in the school is 20 times that in our class.
(d) Jaggu’s uncle’s age is 4 times the age of Jaggu. Jaggu’s aunt is 3 years younger than his uncle.
(e) The total number of dots is 5 times the number of rows.
NCERT Solutions for Class 6 Maths Exercise 11.4
###### Question 3.
(a) Given, Munnu’s age to be {tex}x{/tex} years. Can you guess what {tex}\left( {x – 2} \right){/tex} may show? (Hint: Think of
Munnu’s younger brother). Can you guess what {tex}\left( {x + 4} \right){/tex} may show? What {tex}\left( {3x + 7} \right){/tex} may show?
(b) Given Sara’s age today to be {tex}y{/tex} years. Think of her age in the future or in the past. What will the following expression indicate? {tex}y + 7,y – 3,y + 4\frac{1}{2},y – 2\frac{1}{2}{/tex}
(c) Given, {tex}n{/tex} students in the class like football, what may {tex}2n{/tex} show? What may {tex}\frac{n}{2}{/tex} show? (Hint: Think of games other than football).
(a) Munnu’s age = {tex}x{/tex} years
His younger brother is 2 years younger than him = {tex}\left( {x – 2} \right){/tex} years
His elder brother’s age is 4 years more than his age = {tex}\left( {x + 4} \right){/tex} years
His father is 7 year’s more than thrice of his age = {tex}\left( {3x + 7} \right){/tex} years
(b) Her age in past = {tex}\left( {y – 3} \right),\left( {y – 2\frac{1}{2}} \right){/tex}
Her age in future = {tex}\left( {y + 7} \right),\left( {y + 4\frac{1}{2}} \right){/tex}
(c) Number of students like hockey is twice the students liking football, i.e., {tex}2n{/tex}
Number of students like tennis is half the students like football, i.e., {tex}\frac{n}{2}{/tex}
## NCERT Solutions for Class 6 Maths Exercise 11.4
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# Section 6-5 Sample Spaces and Probability
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1 492 6 SEQUENCES, SERIES, AND PROBABILITY 52. How many committees of 4 people are possible from a group of 9 people if (A) There are no restrictions? (B) Both Juan and Mary must be on the committee? (C) Either Juan or Mary, but not both, must be on the committee? 53. A 5-card hand is dealt from a standard 52-card deck. Which is more likely: the hand contains exactly king or the hand contains no hearts? 54. A 0-card hand is dealt from a standard 52-card deck. Which is more likely: all cards in the hand are red or the hand contains all four aces? Section 6-5 Sample Spaces and Probability Experiments Sample Spaces and Events Probability of an Event Equally Likely Assumption Empirical Probability This section provides an introduction to probability, a topic that has whole books and courses devoted to it. Probability studies involve many subtle notions, and care must be taken at the beginning to understand the fundamental concepts on which the studies are based. First, we develop a mathematical model for probability studies. Our development, because of space, must be somewhat informal. More formal and precise treatments can be found in books on probability. Experiments Our first step in constructing a mathematical model for probability studies is to describe the type of experiments on which probability studies are based. Some types of experiments do not yield the same results, no matter how carefully they are repeated under the same conditions. These experiments are called random experiments. Familiar examples of random experiments are flipping coins, rolling dice, observing the frequency of defective items from an assembly line, or observing the frequency of deaths in a certain age group. Probability theory is a branch of mathematics that has been developed to deal with outcomes of random experiments, both real and conceptual. In the work that follows, the word experiment will be used to mean a random experiment. Sample Spaces and Events Associated with outcomes of experiments are sample spaces and events. Our second step in constructing a mathematical model for probability studies is to define these two terms. Set concepts will be useful in this regard. Consider the experiment, A single six-sided die is rolled. What outcomes might we observe? We might be interested in the number of dots facing up, or whether the number of dots facing up is an even number, or whether the number of dots facing up is divisible by 3, and so on. The list of possible outcomes appears endless. In general, there is no unique method of analyzing all possible outcomes of an experiment. Therefore, before conducting an experiment, it is important to decide just what outcomes are of interest. In the die experiment, suppose we limit our interest to the number of dots facing up when the die comes to rest. Having decided what to observe, we make a
2 6-5 Sample Spaces and Probability 493 list of outcomes of the experiment, called simple events, such that in each trial of the experiment, one and only one of the results on the list will occur. The set of simple events for the experiment is called a sample space for the experiment. The sample space S we have chosen for the die-rolling experiment is S {, 2, 3, 4, 5, 6} Now consider the outcome, The number of dots facing up is an even number. This outcome is not a simple event, since it will occur whenever 2, 4, or 6 dots appear, that is, whenever an element in the subset E {2, 4, 6} occurs. Subset E is called a compound event. In general, we have the following definition: DEFINITION EVENT Given a sample space S for an experiment, we define an event E to be any subset of S. If an event E has only one element in it, it is called a simple event. If event E has more than one element, it is called a compound event. We say that an event E occurs if any of the simple events in E occurs. EXAMPLE Choosing a Sample Space A nickel and a dime are tossed. How will we identify a sample space for this experiment? Solutions There are a number of possibilities, depending on our interest. We will consider three. (A) If we are interested in whether each coin falls heads (H) or tails (T), then, using a tree diagram, we can easily determine an appropriate sample space for the experiment: Start Nickel Outcomes H T Dime Outcomes H T H T Combined Outcomes HH HT TH TT Thus, S {HH, HT, TH, TT} and there are four simple events in the sample space.
3 494 6 SEQUENCES, SERIES, AND PROBABILITY (B) If we are interested only in the number of heads that appear on a single toss of the two coins, then we can let S 2 {0,, 2} and there are three simple events in the sample space. (C) If we are interested in whether the coins match (M) or don t match (D), then we can let S 3 {M, D} and there are only two simple events in the sample space. MATCHED PROBLEM An experiment consists of recording the boy girl composition of families with 2 children. (A) What is an appropriate sample space if we are interested in the sex of each child in the order of their births? Draw a tree diagram. (B) What is an appropriate sample space if we are interested only in the number of girls in a family? (C) What is an appropriate sample space if we are interested only in whether the sexes are alike (A) or different (D)? (D) What is an appropriate sample space for all three interests expressed above? In Example, sample space S contains more information than either S 2 or S 3. If we know which outcome has occurred in S, then we know which outcome has occurred in S 2 and S 3. However, the reverse is not true. In this sense, we say that S is a more fundamental sample space than either S 2 or S 3. Important Remark: There is no one correct sample space for a given experiment. When specifying a sample space for an experiment, we include as much detail as necessary to answer all questions of interest regarding the outcomes of the experiment. If in doubt, include more elements in the sample space rather than fewer. Now let s return to the 2-coin problem in Example and the sample space S {HH, HT, TH, TT} Suppose we are interested in the outcome, Exactly head is up. Looking at S, we find that it occurs if either of the two simple events HT or TH occurs.* Thus, to say that the event, Exactly head is up occurs is the same as saying the experiment has an outcome in the set E {HT, TH} This is a subset of the sample space S. The event E is a compound event. *Technically, we should write {HT} and {TH}, since there is a logical distinction between an element of a set and a subset consisting of only that element. But we will just keep this in mind and drop the braces for simple events to simplify the notation.
4 6-5 Sample Spaces and Probability 495 EXAMPLE 2 Rolling Two Dice Consider an experiment of rolling 2 dice. A convenient sample space that will enable us to answer many questions about interesting events is shown in Figure. Let S be the set of all ordered pairs listed in the figure. Note that the simple event (3, 2) is to be distinguished from the simple event (2, 3). The former indicates a 3 turned up on the first die and a 2 on the second, while the latter indicates a 2 turned up on the first die and a 3 on the second. What is the event that corresponds to each of the following outcomes? (A) A sum of 7 turns up. (B) A sum of turns up. (C) A sum less than 4 turns up. (D) A sum of 2 turns up. FIGURE A sample space for rolling two dice. (, ) (, 2) SECOND DIE (, 3) (, 4) (, 5) (, 6) (2, ) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) FIRST DIE (3, ) (4, ) (3, 2) (4, 2) (3, 3) (4, 3) (3, 4) (4, 4) (3, 5) (4, 5) (3, 6) (4, 6) (5, ) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, ) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Solutions (A) By A sum of 7 turns up, we mean that the sum of all dots on both turnedup faces is 7. This outcome corresponds to the event {(6, ), (5, 2), (4, 3), (3, 4), (2, 5), (, 6)} (B) A sum of turns up corresponds to the event {(6, 5), (5, 6)} (C) A sum less than 4 turns up corresponds to the event {(, ), (2, ), (, 2)} (D) A sum of 2 turns up corresponds to the event {(6, 6)} MATCHED PROBLEM 2 Refer to the sample space in Example 2 (Fig. ). What is the event that corresponds to each of the following outcomes? (A) A sum of 5 turns up. (B) A sum that is a prime number greater than 7 turns up.
5 496 6 SEQUENCES, SERIES, AND PROBABILITY Informally, to facilitate discussion, we often use the terms event and outcome of an experiment interchangeably. Thus, in Example 2 we might say the event A sum of turns up in place of the outcome A sum of turns up, or even write E A sum of turns up {(6, 5), (5, 6)} Technically speaking, an event is the mathematical counterpart of an outcome of an experiment. Probability of an Event The next step in developing our mathematical model for probability studies is the introduction of a probability function. This is a function that assigns to an arbitrary event associated with a sample space a real number between 0 and, inclusive. We start by discussing ways in which probabilities are assigned to simple events in S. DEFINITION 2 PROBABILITIES FOR SIMPLE EVENTS Given a sample space S {e, e 2,..., e n } with n simple events, to each simple event e i we assign a real number, denoted by P(e i ), that is called the probability of the event e i. These numbers may be assigned in an arbitrary manner as long as the following two conditions are satisfied:. 0 P(e i ) 2. P(e i ) P(e 2 )... P(e n ) The sum of the probabilities of all simple events in the sample space is. Any probability assignment that meets conditions and 2 is said to be an acceptable probability assignment. Our mathematical theory does not explain how acceptable probabilities are assigned to simple events. These assignments are generally based on the expected or actual percentage of times a simple event occurs when an experiment is repeated a large number of times. Assignments based on this principle are called reasonable. Let an experiment be the flipping of a single coin, and let us choose a sample space S to be S {H, T} If a coin appears to be fair, we are inclined to assign probabilities to the simple events in S as follows: P(H) and P(T) 2 2
6 6-5 Sample Spaces and Probability 497 These assignments are based on reasoning that, since there are two ways a coin can land, in the long run, a head will turn up half the time and a tail will turn up half the time. These probability assignments are acceptable, since both of the conditions for acceptable probability assignments in Definition 2 are satisfied:. 0 P(H), 0 P(T) 2. P(H) P(T) 2 2 But there are other acceptable assignments. Maybe after flipping a coin,000 times we find that the head turns up 376 times and the tail turns up 624 times. With this result, we might suspect that the coin is not fair and assign the simple events in the sample space S the probabilities P(H).376 and P(T).624 This is also an acceptable assignment. But the probability assignment P(H) and P(T) 0 though acceptable, is not reasonable, unless the coin has 2 heads. The assignment P(H).6 and P(T).8 is not acceptable, since.6.8.4, which violates condition 2 in Definition 2. In probability studies, the 0 to the left of the decimal is usually omitted. Thus, we write.8 and not 0.8. It is important to keep in mind that out of the infinitely many possible acceptable probability assignments to simple events in a sample space, we are generally inclined to choose one assignment over another based on reasoning or experimental results. Given an acceptable probability assignment for simple events in a sample space S, how do we define the probability of an arbitrary event E associated with S? DEFINITION 3 PROBABILITY OF AN EVENT E Given an acceptable probability assignment for the simple events in a sample space S, we define the probability of an arbitrary event E, denoted by P(E), as follows:. If E is the empty set, then P(E) If E is a simple event, then P(E) has already been assigned. 3. If E is a compound event, then P(E) is the sum of the probabilities of all the simple events in E. 4. If E is the sample space S, then P(E) P(S). This is a special case of 3. EXAMPLE 3 Finding Probabilities of Events Let s return to Example, the tossing of a nickel and dime, and the sample space S {HH, HT, TH, TT}
7 498 6 SEQUENCES, SERIES, AND PROBABILITY Since there are four simple outcomes and the coins are assumed to be fair, it appears that each outcome should occur in the long run 25% of the time. Let s assign the same probability of to each simple event in S: Simple event, e i HH HT TH TT 4 P(e i ) This is an acceptable assignment according to Definition 2 and a reasonable assignment for ideal coins that are perfectly balanced or coins close to ideal. (A) What is the probability of getting exactly head? (B) What is the probability of getting at least head? (C) What is the probability of getting a head or a tail? (D) What is the probability of getting 3 heads? Solutions (A) E Getting head {HT, TH} Since E is a compound event, we use item 3 in Definition 3 and find P(E ) by adding the probabilities of the simple events in E. Thus, P(E ) P(HT) P(TH) (B) E 2 Getting at least head {HH, HT, TH} P(E 2 ) P(HH) P(HT) P(TH) (C) E 3 {HH, HT, TH, TT} S P(E 3 ) P(S) (D) E 3 Getting 3 heads Empty set P( ) 0 STEPS FOR FINDING PROBABILITIES OF EVENTS Step. Set up an appropriate sample space S for the experiment. Step 2. Assign acceptable probabilities to the simple events in S. Step 3. To obtain the probability of an arbitrary event E, add the probabilities of the simple events in E. The function P defined in steps 2 and 3 is called a probability function. The domain of this function is all possible events in the sample space S, and the range is a set of real numbers between 0 and, inclusive.
8 6-5 Sample Spaces and Probability 499 MATCHED PROBLEM 3 Return to Matched Problem, recording the boy girl composition of families with 2 children and the sample space S {BB, BG, GB, GG} Statistics from the U.S. Census Bureau indicate that an acceptable and reasonable probability for this sample space is Simple event, e i BB BG GB GG P(e i ) Find the probabilities of the following events: (A) E Having at least one girl in the family (B) E 2 Having at most one girl in the family (C) E 3 Having two children of the same sex in the family Equally Likely Assumption In tossing a nickel and dime (Example 3), we assigned the same probability, 4, to each simple event in the sample space S {HH, HT, TH, TT}. By assigning the same probability to each simple event in S, we are actually making the assumption that each simple event is as likely to occur as any other. We refer to this as an equally likely assumption. In general, we have Definition 4. DEFINITION 4 PROBABILITY OF A SIMPLE EVENT UNDER AN EQUALLY LIKELY ASSUMPTION If, in a sample space S {e, e 2,..., e n } with n elements, we assume each simple event e i is as likely to occur as any other, then we assign the probability /n to each. That is, P(e i ) n Under an equally likely assumption, we can develop a very useful formula for finding probabilities of arbitrary events associated with a sample space S. Consider the following example. If a single die is rolled and we assume each face is as likely to come up as any other, then for the sample space S {, 2, 3, 4, 5, 6}
9 500 6 SEQUENCES, SERIES, AND PROBABILITY we assign a probability of 6 to each simple event, since there are 6 simple events. Then the probability of is E Rolling a prime number {2, 3, 5} P(E) P(2) P(3) P(5) Thus, under the assumption that each simple event is as likely to occur as any other, the computation of the probability of the occurrence of any event E in a sample space S is the number of elements in E divided by the number of elements in S. THEOREM PROBABILITY OF AN ARBITRARY EVENT UNDER AN EQUALLY LIKELY ASSUMPTION If we assume each simple event in sample space S is as likely to occur as any other, then the probability of an arbitrary event E in S is given by P(E) Number of elements in E Number of elements in S n(e) n(s) EXAMPLE 4 Finding Probabilities of Events If in rolling 2 dice we assume each simple event in the sample space shown in Figure (page 495) is as likely as any other, find the probabilities of the following events: (A) E A sum of 7 turns up (B) E 2 A sum of turns up (C) E 3 A sum less than 4 turns up (D) E 4 A sum of 2 turns up Solutions Referring to Figure, we see that: (A) P(E ) n(e ) n(s) (C) P(E 3 ) n(e 3) n(s) (B) (D) P(E 2 ) n(e 2) n(s) P(E 4 ) n(e 4) n(s) 36 MATCHED PROBLEM 4 Under the conditions in Example 4, find the probabilities of the following events: (A) E 5 A sum of 5 turns up (B) E 6 A sum that is a prime number greater than 7 turns up
10 6-5 Sample Spaces and Probability 50 Explore/Discuss A box contains 4 red balls and 7 green balls. A ball is drawn at random and then, without replacing the first ball, a second ball is drawn. Discuss whether or not the equally likely assumption would be appropriate for the sample space S {RR, RG, GR, GG}. We now turn to some examples that make use of the counting techniques developed in Section 6-4. EXAMPLE 5 Solution Drawing Cards In drawing 5 cards from a 52-card deck without replacement, what is the probability of getting 5 spades? Let the sample space S be the set of all 5-card hands from a 52-card deck. Since the order in a hand does not matter, n(s) C 52,5. The event we seek is E Set of all 5-card hands from 3 spades Again, the order does not matter and n(e) C 3,5. Thus, assuming each 5-card hand is as likely as any other, P(E) n(e) n(s) C 3,5 3!/5!8! C 52,5 52!/5!47! 3! 5!8! 5!47! ! MATCHED PROBLEM 5 In drawing 7 cards from a 52-card deck without replacement, what is the probability of getting 7 hearts? EXAMPLE 6 Solution Selecting Committees The board of regents of a university is made up of 2 men and 6 women. If a committee of 6 is chosen at random, what is the probability that it will contain 3 men and 3 women? Let S Set of all 6-person committees out of 28 people: n(s) C 28,6 Let E Set of all 6-person committees with 3 men and 3 women. To find n(e), we use the multiplication principle and the following two operations: O : Select 3 men out of the 2 available N : C 2,3 O 2 : Select 3 women out of the 6 available N 2 : C 6,3
11 502 6 SEQUENCES, SERIES, AND PROBABILITY Thus, and n(e) C 2,3 C 6,3 P(E) n(e) n(s) C 2,3 C 6,3 C 28,6.327 MATCHED PROBLEM 6 What is the probability that the committee in Example 6 will have 4 men and 2 women? Empirical Probability In the earlier examples in this section, we made a reasonable assumption about an experiment and used deductive reasoning to assign probabilities. For example, it is reasonable to assume that an ordinary coin will come up heads about as often as it will come up tails. Probabilities determined in this manner are called theoretical probabilities. No experiments are ever conducted. But what if the theoretical probabilities are not obvious? Then we assign probabilities to simple events based on the results of actual experiments. Probabilities determined from the results of actually performing an experiment are called empirical probabilities. As an experiment is repeated over and over, the percentage of times an event occurs may get closer and closer to a single fixed number. If so, this single fixed number is generally called the actual probability of the event. Explore/Discuss 2 Like a coin, a thumbtack tossed into the air will land in one of two positions, point up or point down [see Fig. 2(a)]. Unlike a coin, we would not expect both events to occur with the same frequency. Indeed, the frequencies of landing point up and point down may well vary from one thumbtack to another [see Fig. 2(b)]. Find two thumbtacks of different sizes and guess which one is likely to land point up more frequently. Then toss each tack 00 times and record the number of times each lands point up. Did the experiment confirm your initial guess? FIGURE 2 (a) Point up or point down (b) Two different tacks Suppose when tossing one of the thumbtacks in Explore/Discuss 2, we observe that the tack lands point up 43 times and point down 57 times. Based on this experiment, it seems reasonable to say that for this particular thumbtack
12 6-5 Sample Spaces and Probability 503 P(Point up) P(Point down) Probability assignments based on the results of repeated trials of an experiment are called approximate empirical probabilities. In general, if we conduct an experiment n times and an event E occurs with frequency f(e), then the ratio f(e)/n is called the relative frequency of the occurrence of event E in n trials. We define the empirical probability of E, denoted by P(E), by the number, if it exists, that the relative frequency f(e)/n approaches as n gets larger and larger. Of course, for any particular n, the relative frequency f(e)/n is generally only approximately equal to P(E). However, as n increases in size, we expect the approximation to improve. DEFINITION 5 EMPIRICAL PROBABILITY If f(e) is the frequency of event E in n trials, then P(E) Frequency of occurrence of E Total number of trials f(e) n If we can also deduce theoretical probabilities for an experiment, then we expect the approximate empirical probabilities to approach the theoretical probabilities. If this does not happen, then we should begin to suspect the manner in which the theoretical probabilities were computed. If P(E) is the theoretical probability of an event E and the experiment is performed n times, then the expected frequency of the occurrence of E is n P(E). EXAMPLE 7 Finding Approximate Empirical and Theoretical Probabilities Two coins are tossed 500 times with the following frequencies of outcomes: 2 heads: 2 head: heads: 7 (A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome. (C) Compute the expected frequency for each outcome. Solutions (A) P(2 heads) P( head) P(0 heads)
13 504 6 SEQUENCES, SERIES, AND PROBABILITY (B) A sample space of equally likely simple events is S {HH, HT, TH, TT}. Let Then E 2 heads {HH} E 2 head {HT, TH} E 3 0 heads {TT} P(E ) n(e ) n(s) 4.25 P(E 2 ) n(e 2) n(s) P(E 3 ) n(e 3) n(s) 4.25 (C) The expected frequencies are E : 500(.25) 25 E 2 : 500(.5) 250 E 3 : 500(.25) 25 The actual frequencies obtained from performing the experiment are reasonably close to the expected frequencies. Increasing the number of trials of the experiment would produce even better approximations. MATCHED PROBLEM 7 One die is rolled 500 times with the following frequencies of outcomes: Outcome Frequency (A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome. (C) Compute the expected frequency for each outcome. FIGURE 3 Using a random number generator. Tossing two coins 500 times is certainly a tedious task and we did not do this to generate the data in Example 7. Instead, we used a random number generator on a graphing utility to simulate this experiment. Specifically, we used the command randint(i,k,n) on a Texas Instruments TI-83, which generates a random sequence of n integers between i and k, inclusively. If we let 0 represent tails and represent heads, then a random sequence of 0s and s can be used to represent the outcomes of repeated tosses of one coin (see the first two lines of Fig. 3). Thus, in six tosses, we obtained 2 heads and 4 tails. To simulate tossing two coins, we simply add together two similar statements, as shown in lines three through
14 6-5 Sample Spaces and Probability 505 five of Figure 3. We see that in these six tosses 0 heads occurred once, head occurred four times, and 2 heads occurred once. Of course, to obtain meaningful results, we need to toss the coins many more times. Figure 4(a) shows a command that will simulate 500 tosses of two coins. To determine the frequency of each outcome, we construct a histogram [see Figs. 4(b) and 4(c)] and use the TRACE command to determine the following frequencies [see Figs. 5(a), 5(b), and 5(c)]. FIGURE 4 Simulating 500 tosses of two coins. (a) Generating the (b) Setting up the (c) Selecting the random numbers histogram window variables FIGURE 5 Results of the simulation. (a) 0 heads: 7 (b) head: 262 (c) 2 heads: 2 If you perform the same simulation on your graphing utility, you are not likely to get exactly the same results. But the approximate empirical probabilities you obtain will be close to the theoretical probabilities. Explore/Discuss 3 This discussion assumes that your graphing utility has the ability to generate and manipulate sequences of random integers. (A) As an alternative to using the histogram in Figure 5 to count the outcomes of the sequence of random integers in Figure 4(a), enter the following function and evaluate it for x 0,, and 2: y sum(seq(l (I) X,I,,dim(L ))) (B) Simulate the experiment of rolling a single die and compare your empirical results with the results in Matched Problem 7. EXAMPLE 8 Empirical Probabilities for an Insurance Company An insurance company selected,000 drivers at random in a particular city to determine a relationship between age and accidents. The data obtained are listed in Table. Compute the approximate empirical probabilities of the following events for a driver chosen at random in the city: (A) E : being under 20 years old and having exactly 3 accidents in year (B) E 2 : being years old and having or more accidents in year
15 506 6 SEQUENCES, SERIES, AND PROBABILITY (C) E 3 : having no accidents in year (D) E 4 : being under 20 years old or having exactly 3 accidents in year TABLE Accidents in Year Age Over 3 Under Over Solutions (A) (B) (C) (D) P(E ) 35, P(E 2 ) P(E 3 ) P(E 4 ) , , , Notice that in this type of problem, which is typical of many realistic problems, approximate empirical probabilities are the only type we can compute. MATCHED PROBLEM 8 Referring to Table in Example 8, compute the approximate empirical probabilities of the following events for a driver chosen at random in the city: (A) E : being under 20 years old with no accidents in year (B) E 2 : being years old and having fewer than 2 accidents in year (C) E 3 : not being over 49 years old Approximate empirical probabilities are often used to test theoretical probabilities. Equally likely assumptions may not be justified in reality. In addition to this use, there are many situations in which it is either very difficult or impossible to compute the theoretical probabilities for given events. For example, insurance companies use past experience to establish approximate empirical probabilities to predict future accident rates, baseball teams use batting averages, which are approximate empirical probabilities based on past experience, to predict the future performance of a player, and pollsters use approximate empirical probabilities to predict outcomes of elections.
16 6-5 Sample Spaces and Probability 507 Answers to Matched Problems. (A) S {BB, BG, GB, GG}; Sex of Sex of Combined First Child Second Child Outcomes B BB B G BG B GB G G GG (B) S 2 {0,, 2} (C) S 3 {A, D} (D) The sample space in part A. 2. (A) {(4, ), (3, 2), (2, 3), (, 4)} (B) {(6, 5), (5, 6)} 3. (A).74 (B).76 (C).5 4. (A) P(E 5 ) 9 (B) P(E 6 ) 8 5. C 3,7 /C 52, C 2,4 C 6,2 /C 28, (A) P(E ).78, P(E 2 ).66, P(E 3 ).54, P(E 4 ).82, P(E 5 ).44, P(E 6 ).76 (B) 6.67 for each (C) 83.3 for each 8. (A) P(E ).05 (B) P(E 2 ).57 (C) P(E 3 ).82 EXERCISE 6-5 A. How would you interpret P(E)? 2. How would you interpret P(E) 0? 3. A spinner can land on 4 different colors: red (R), green (G), yellow (Y), and blue (B). If we do not assume each color is as likely to turn up as any other, which of the following probability assignments have to be rejected, and why? (A) P(R).5, P(G).35, P(Y).50, P(B).70 (B) P(R).32, P(G).28, P(Y).24, P(B).30 (C) P(R).26, P(G).4, P(Y).30, P(B) Under the probability assignments in Problem 3, part C, what is the probability that the spinner will not land on blue? 5. Under the probability assignments in Problem 3, part C, what is the probability that the spinner will land on red or yellow? 6. Under the probability assignments in Problem 3, part C, what is the probability that the spinner will not land on red or yellow? 7. A ski jumper has jumped over 300 feet in 25 out of 250 jumps. What is the approximate empirical probability of the next jump being over 300 feet? 8. In a certain city there are 4,000 youths between 6 and 20 years old who drive cars. If 560 of them were involved in accidents last year, what is the approximate empirical probability of a youth in this age group being involved in an accident this year? 9. Out of 420 times at bat, a baseball player gets 89 hits. What is the approximate empirical probability that the player will get a hit next time at bat? 0. In a medical experiment, a new drug is found to help 2,400 out of 3,000 people. If a doctor prescribes the drug for a particular patient, what is the approximate empirical probability that the patient will be helped? B. A small combination lock on a suitcase has 3 wheels, each labeled with the 0 digits from 0 to 9. If an opening combination is a particular sequence of 3 digits with no repeats, what is the probability of a person guessing the right combination? 2. A combination lock has 5 wheels, each labeled with the 0 digits from 0 to 9. If an opening combination is a particular sequence of 5 digits with no repeats, what is the probability of a person guessing the right combination? An experiment consists of dealing 5 cards from a standard 52-card deck. In Problems 3 6, what is the probability of being dealt each of the following hands? 3. 5 black cards 4. 5 hearts 5. 5 face cards if aces are considered to be face cards 6. 5 nonface cards if an ace is considered to be and not a face card 7. If 4-digit numbers less than 5,000 are randomly formed from the digits, 3, 5, 7, and 9, what is the probability of forming a number divisible by 5? Digits may be repeated; for example,,355 is acceptable.
18 6-6 Binomial Formula 509 C An experiment consists of rolling 2 fair dice and adding the dots on the 2 sides facing up. Each die has dot on two opposite faces, 2 dots on two opposite faces, and 3 dots on two opposite faces. Compute the probabilities of obtaining the indicated sums in Problems An odd sum 56. An even sum An experiment consists of dealing 5 cards from a standard 52-card deck. In Problems 57 64, what is the probability of being dealt the following cards? cards, jacks through aces cards, 2 through aces of a kind 6. Straight flush, ace high; that is, 0, jack, queen, king, ace in one suit 62. Straight flush, starting with 2; that is, 2, 3, 4, 5, 6 in one suit aces and 3 queens kings and 3 aces APPLICATIONS 65. Market Analysis. A company selected,000 households at random and surveyed them to determine a relationship between income level and the number of television sets in a home. The information gathered is listed in the table: Televisions per Household Yearly Income Above 3 Less than \$2, \$2,000 9, \$20,000 39, \$40,000 59, \$60,000 or more Compute the approximate empirical probabilities: (A) Of a household earning \$2,000 \$9,999 per year and owning exactly 3 television sets (B) Of a household earning \$20,000 \$39,999 per year and owning more than television set (C) Of a household earning \$60,000 or more per year or owning more than 3 television sets (D) Of a household not owning 0 television sets 66. Market Analysis. Use the sample results in Problem 65 to compute the approximate empirical probabilities: (A) Of a household earning \$40,000 \$59,999 per year and owning 0 television sets (B) Of a household earning \$2,000 \$39,999 per year and owning more than 2 television sets (C) Of a household earning less than \$20,000 per year or owning exactly 2 television sets (D) Of a household not owning more than 3 television sets Section 6-6 Binomial Formula Pascal s Triangle The Binomial Formula Proof of the Binomial Formula The binomial form (a b) n where n is a natural number, appears more frequently than you might expect. It turns out that the coefficients in the expansion are related to probability concepts that we have already discussed.
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Q) In the given figure, PQ is tangent to a circle centred at O and ∠BAQ = 30°; show that BP = BQ.
Ans: Here, We need to prove that BP = BQ,
it means we need to get ∠ BQP = ∠ BPQ
Since AB is a straight line and passing thru the circle’s center O, hence AB is a diameter
∴ ∠ AQB = 90 …….. (i)
since it is given that ∠ QAB = 30,
∴ ∠ ABQ = 60 (sum of all angles in Δ ABQ is 180)
Since OP is a straight line,
∴ ∠ ABQ + ∠ QBP = 180
∴ ∠ QBP = 180 – 60 = 120
Step 2: Next, we look at Δ QBP,
∴ ∠ QBP + ∠ BPQ + ∠ PQB = 180
∴ 120 + ∠ BPQ + ∠ PQB = 180
∴ ∠ BPQ + ∠ PQB = 180 – 120
∴ ∠ BPQ + ∠ PQB = 60 ………. (ii)
Step 3: Next, let’s connect OQ:
Now we look in Δ OQP,
Since OQ is a radius and PQ is tangent to the circle,
∴ ∠ OQP = 90 …………… (iii)
Step 4: Now we look in Δ AOQ,
Since OA and OQ are radii of same circle,
∴ OA = OQ
∴ ∠ AQO = ∠ QAO = 30 (given) …… (iv)
Step 5: From equation (i) we calculated:
∠ AQB = 90
∴ ∠ AQO + ∠ OQB = 90
∴ 30 + ∠ OQB = 90 [from equation (iv)]
∴ ∠ OQB = 60
Step 6: From equation (iii) we calculated:
∠ OQP = 90
∴ ∠ OQB + ∠ PQB = 90
∴ 60 + ∠ PQB = 90 (from above)
∴ ∠ PQB = 30
Step 7: From equation (ii) we calculated:
∴ ∠ BPQ + ∠ PQB = 60
∴ ∠ BPQ + 30 = 60
∴ ∠ BPQ = 60 – 30 = 30
Step 8: Now we get, ∠ BPQ = ∠ PQB
Therefore in Δ BPQ, opposite sides to equal angles will also be equal
∴ BP = BQ
Hence Proved !
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## Translating Algebraic Expressions From Words
### Learning Outcomes
• Translate word phrases into algebraic expressions
• Write an algebraic expression that represents the relationship between two measurements
## Translate Words to Algebraic Expressions
In the previous section, we listed many operation symbols that are used in algebra, and then we translated expressions and equations into word phrases and sentences. Now we’ll reverse the process and translate word phrases into algebraic expressions. The symbols and variables we’ve talked about will help us do that. They are summarized below.
Operation Phrase Expression
Addition $a$ plus $b$
the sum of $a$ and $b$
$a$ increased by $b$
$b$ more than $a$
the total of $a$ and $b$
$b$ added to $a$
$a+b$
Subtraction $a$ minus $b$
the difference of $a$ and $b$
$b$ subtracted from $a$
$a$ decreased by $b$
$b$ less than $a$
$a-b$
Multiplication $a$ times $b$
the product of $a$ and $b$
$a\cdot b$ , $ab$ , $a\left(b\right)$ , $\left(a\right)\left(b\right)$
Division $a$ divided by $b$
the quotient of $a$ and $b$
the ratio of $a$ and $b$
$b$ divided into $a$
$a\div b$ , $a/b$ , $\frac{a}{b}$ , $b\overline{)a}$
Look closely at these phrases using the four operations:
• the sum of $a$ and $b$
• the difference of $a$ and $b$
• the product of $a$ and $b$
• the quotient of $a$ and $b$
Each phrase tells you to operate on two numbers. Look for the words of and and to find the numbers.
### example
Translate each word phrase into an algebraic expression:
1. The difference of $20$ and $4$
2. The quotient of $10x$ and $3$
Solution
1. The key word is difference, which tells us the operation is subtraction. Look for the words of and and to find the numbers to subtract.
$\begin{array}{}\\ \text{the difference of }20\text{ and }4\hfill \\ 20\text{ minus }4\hfill \\ 20 - 4\hfill \end{array}$
2. The key word is quotient, which tells us the operation is division.
$\begin{array}{}\\ \text{the quotient of }10x\text{ and }3\hfill \\ \text{divide }10x\text{ by }3\hfill \\ 10x\div 3\hfill \end{array}$
This can also be written as $\begin{array}{l}10x/3\text{ or}\frac{10x}{3}\hfill \end{array}$
### example
Translate each word phrase into an algebraic expression:
1. How old will you be in eight years? What age is eight more years than your age now? Did you add $8$ to your present age? Eight more than means eight added to your present age.
2. How old were you seven years ago? This is seven years less than your age now. You subtract $7$ from your present age,a. Seven less than means seven subtracted from your present age,a.
### example
Translate each word phrase into an algebraic expression:
1. five times the sum of $m$ and $n$
2. the sum of five times $m$ and $n$
### try it
Watch the video below to better understand how to write algebraic expressions from statements.
Later in this course, we’ll apply our skills in algebra to solving equations. We’ll usually start by translating a word phrase to an algebraic expression. We’ll need to be clear about what the expression will represent. We’ll see how to do this in the next two examples.
### example
The height of a rectangular window is $6$ inches less than the width. Let $w$ represent the width of the window. Write an expression for the height of the window.
### example
Blanca has dimes and quarters in her purse. The number of dimes is $2$ less than $5$ times the number of quarters. Let $q$ represent the number of quarters. Write an expression for the number of dimes.
### try it
In the following video we show more examples of how to write basic algebraic expressions from words, and simplify. |
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Home » News » Financial Aid » GMAT Tip: Inequalities…and then some…(Part 1)
# GMAT Tip: Inequalities…and then some…(Part 1)
When it comes to tricky quantitative questions, inequalities take the prize for serving as one of the biggest deceivingly easy questions. Test takers should just treat inequalities as algebraic “equal to” equations with just a < or > symbol in place of the = sign, right?
If only it was so simple. Inequalities are designed to assess your critical thinking skills, working beyond solving for a value of a variable to considering the range of variable possibilities. Furthermore, inequalities questions are set up as data sufficiency questions the majority of the time. Smart test takers know to be on their toes.
Is n between 0 and 1?
Statement 1: n^2 < n
Statement 2: n^3 > 0
Start by evaluating the question. What does it mean when n is between 0 and 1? It means n is not an integer. It is also a positive fraction or decimal, such as ½ or 0.25.
Now that we know what we are working with, so let’s move on to Statement 1. What would make n^2 less than n?
The variable n cannot possibly be a negative number, because by squaring it n becomes a positive number.
The only way the n^2 could be less than n is if n was a fraction, like ½. When ½ is squared (½ * ½) it becomes ¼, which is smaller than ½.
Indeed, Statement 1 is sufficient to prove that n lies between 0 and 1.
Okay, so next up is Statement 2, where n^3 > 0. If we have the same mindset as Statement 1, we work to assess the options.
If n^3 is greater than 0, then it cannot possibility be negative. But, then, does n definitely lie between 0 and 1?
Not necessarily. The expression n^3 means that, sure, if n = ½ then (½)^3 is ⅛ and therefore greater than 0. The variable n could also be 1, or 2, which would make n^3 greater than 0, but not a fraction or decimal.
Statement 2 is not sufficient. The correct answer to this question is (A).
When looking at inequality questions, the key to getting the right answer is being flexible in your thinking.
But inequalities are rarely this simple. Often, these trickster questions are complicated by absolute value, multiple variables, and square roots.
How do we tackle those questions? Check out Part 2 in next week’s post.
The above GMAT Tip comes from Veritas Prep. Since its founding in 2002, Veritas Prep has helped more than 100,000 students prepare for the GMAT and offers the most highly rated GMAT Prep course in the industry.
Posted in: Financial Aid, GMAT, GMAT - Quantitative, GMAT Tips
## Lena Maratea
Lena Maratea is the Digital Marketing Manager at Clear Admit. She's a South Philadelphia native who graduated from Temple University’s Fox School of Business with a BBA in Marketing. She creates and curates essential digital content for the Clear Admit community. |
Question 22
# Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to
Solution
According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle.
Hence, $$\angle\ ABC=60^{\circ\ }$$, $$\angle\ CAB=30^{\circ\ }$$, which implies $$\angle\ ACB=180^{\circ\ }-\left(30^{\circ\ }+60^{\circ\ }\right)=90^{\circ\ }$$
Therefore, The triangle is a right-angled triangle with base = BC, height = AC, and hypotenuse = AB.
We know that $$\cos\ \angle\ ABC\ =\ \frac{\ BC}{AB}=>\ \cos60^{\circ\ }=\frac{BC}{500}\ =>\ 500\times\ \frac{1}{2}=250\ km$$
Similarly, $$\sin\ \angle\ ABC\ =\ \frac{\ AC}{AB}=>\ \sin60^{\circ\ }=\frac{AC}{500}\ =>\ 500\times\ \frac{\sqrt{\ 3}}{2}=250\sqrt{\ 3}\ km$$
So, total time taken by train to travel B to C is is (250/50) = 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.
Time taken by Rahim to travel by car is around $$\ \frac{\ 250\sqrt{\ 3}}{70}=6.19$$ hrs (Around 6.2 hrs = 6 hrs 12 minutes). So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.
### Video Solution
• All Quant CAT complete Formulas and shortcuts PDF
• 35+ CAT previous year papers with video solutions PDF |
New Zealand
Level 8 - NCEA Level 3
# Integration of sin/cos (non-linear functions)
Lesson
The integrations considered in this chapter make use of a pattern that arises when a function of a function is differentiated.
According to the function-of-a-function rule, the derivative of a function $h(x)=f\left(g(x)\right)$h(x)=f(g(x)) is
$h'(x)=f'(g(x).g'(x)$h(x)=f(g(x).g(x)
This is also called the chain rule because of the way it looks when expressed using Leibniz's notation. If $w=g(x)$w=g(x) and $y=f(w)$y=f(w), we have
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}w}\cdot\frac{\mathrm{d}w}{\mathrm{d}x}$dydx=dydw·dwdx.
By reversing this process, we can obtain antiderivatives of the form $\int\ f'(g(x).g'(x)\ \mathrm{d}x$ f(g(x).g(x) dx or, equivalently, $\int\ \frac{\mathrm{d}y}{\mathrm{d}w}\cdot\frac{\mathrm{d}w}{\mathrm{d}x}\ \mathrm{d}x$ dydw·dwdx dx
#### Example 1
Let $y=2x.\cos(x^2)$y=2x.cos(x2). The antiderivative of this function can be found because $2x$2x is the derivative of $x^2$x2.
The $\cos$cos function is the derivative of the $\sin$sin function and we observe that differentiating $\sin(x^2)$sin(x2) gives $2x\cos(x^2)$2xcos(x2). So,
$\int\ 2x.\cos(x^2)\ \mathrm{d}x=\sin(x^2)+C$ 2x.cos(x2) dx=sin(x2)+C
We can use Leibniz notation to find $\int\ 2x.\cos(x^2)\ \mathrm{d}x$ 2x.cos(x2) dx by first putting $u=x^2$u=x2 and then $\frac{\mathrm{d}u}{\mathrm{d}x}=2x$dudx=2x. On making the substitutions, the integral becomes $\int\ \cos u.\frac{\mathrm{d}u}{\mathrm{d}x}\ \mathrm{d}x$ cosu.dudx dx.
It can be shown rigorously that this is the same as $\int\ \cos u\ \mathrm{d}u$ cosu du. We recognise this as $\sin u+C$sinu+C which is just $\sin(x^2)+C$sin(x2)+C.
#### Example 2
Find the definite integral $I=\int_0^{\pi}\ \left(3x^2+1\right)\sin(x^3+x)\ \mathrm{d}x$I=π0 (3x2+1)sin(x3+x) dx.
Since $3x^2+1$3x2+1 is the derivative of $x^3+x$x3+x we can write $I=\left[-\cos x^3+x\right]_0^{\pi}$I=[cosx3+x]π0
This is $I=-\cos\left(\pi^3+\pi\right)+\cos0\approx1.917$I=cos(π3+π)+cos01.917.
#### Example 3
Find the antiderivative $\int\ \tan u\ \mathrm{d}u$ tanu du.
Since $\tan u\equiv\frac{\sin u}{\cos u}$tanusinucosu and $-\sin u$sinu is the derivative of $\cos u$cosu, we try putting $w=\cos u$w=cosu and then $\frac{\mathrm{d}w}{\mathrm{d}u}=-\sin u$dwdu=sinu.
This means $\int\ \tan u\ \mathrm{d}u=\int\ \frac{1}{w}\cdot-\frac{\mathrm{d}w}{\mathrm{d}u}\mathrm{d}u=-\int\ \frac{\mathrm{d}w}{w}$ tanu du= 1w·dwdudu= dww.
So, $\int\ \tan u\ \mathrm{d}u=-\ln w+C=-\ln\cos u+C$ tanu du=lnw+C=lncosu+C.
#### Worked Examples
##### Question 1
Consider the function $y=\sin\left(x^5\right)$y=sin(x5).
1. Find the derivative $\frac{dy}{dx}$dydx.
2. Hence find the value of $\int x^4\cos\left(x^5\right)dx$x4cos(x5)dx.
You may use $C$C to represent the constant of integration.
##### Question 2
Consider the function $y=\cos\left(6x^2-\frac{\pi}{3}\right)$y=cos(6x2π3).
1. Find the derivative $\frac{dy}{dx}$dydx.
2. Hence find the exact value of $\int_{\sqrt{\frac{\pi}{18}}}^{\sqrt{\frac{\pi}{6}}}\left(-12x\sin\left(6x^2-\frac{\pi}{3}\right)\right)dx$π6π18(12xsin(6x2π3))dx.
##### Question 3
Find the exact value of $I$I, where $I=\int_0^{\frac{\pi^2}{9}}\frac{\sin\sqrt{x}}{\sqrt{x}}dx$I=π290sinxxdx.
### Outcomes
#### M8-11
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
#### 91579
Apply integration methods in solving problems |
# Chapter 9 Notes Honors Pre-Calculus.
## Presentation on theme: "Chapter 9 Notes Honors Pre-Calculus."— Presentation transcript:
Chapter 9 Notes Honors Pre-Calculus
9.1 Intro to Conics EQUATION: EQUATIONS:
A Circle is the set of all points, in a plane, such that each point is equidistant from a given point, the center. EQUATION: A Parabola is the set of all points, in a plane, that are equidistant from a given point, the focus, and a given line, the directrix. EQUATIONS:
9.1 cont’d. PICTURES Opens up/down Opens left/right Vertex : (h,k)
Axis of Symmetry : x = h y = k Focus: (h, y+p) (h+p, k) Directrix: y = k – p x = h – p p is the distance from the vertex to the focus or vertex to the directrix. (p is directional) PICTURES
9.1 Examples Find the Standard Form Equation for each Parabola:
1. V (2, 3) Focus (2, 5) 2. V (-1, 4) D: x = 1 Find Everything: Equation: Vertex: Focus: Directrix:
9.2 Ellipses An Ellipse is the set of all points, in a plane, such that the sum of each points’ distances from two fixed points, the foci, is constant. Equations: Center : (h, k) Horizontal -Major Axis- Vertical Vertical -Minor Axis- Horizontal C is the distance from the Center to the Foci. Length of Major Axis – 2a Length of Minor Axis – 2b
9.3 Hyperbolas Equations:
A Hyperbola is the set of all points, in a plane, such that the difference of each points’ two distances from two fixed points, the foci, is constant. Equations: Center: (h,k) Right/Left OPENS Up/Down a is the distance from Center to vertices. “box” method for graphing, asymptotes. 2 “parabolas” Slopes of asymptotes:
9.5 Parametric Equations Equations:
If f and g are continuous functions of t on an interval, I. The set of ordered pairs (f(t), g(t)) is a plane curve, C. Parametric Equations t is the parameter Equations: Plot points in order of increasing values of t, you trace the curve in a specific direction. Orientation of the curve. Examples: Eliminating the Parameter 1.Solve for t in one eq. 2.Substitute into 2nd eq. 3.Rectangular Equation
Finding Parametric Equations:
9.5 cont’d. Finding Parametric Equations: Answers:
9.6/9.7 Polar Coordinates & Polar Graphs
Directed distance Directed angle r Plot points and give three equivalent points. Polar Axis Coordinate Conversions Rect. To Polar r y x Polar to Rect, P to R R to P Convert:
9.6/9.7 cont’d. GRAPHS: Rose Curves Circle Line n is odd, n petals
n is even, 2n petals a is the length of petals Circles Equation Conversion Need to know when sin and cos are (1 or -1). For sin, # petals = # of values. For cos, # petals need 1 more value. |
## A duality of pictures
Duality relates objects, which seem different at first but turn out to be similar. The concept of duality occurs almost everywhere in maths. If two objects seem different but are actually the same, we can view each object in a “usual” way, and in a “dual” way – the new vantage point is helpful for new understanding of the object. In this blog post we’ll see a pictorial example of a mathematical duality.
How are these two graphs related?
In the first graph, we have five vertices, the five black dots, and six green edges which connect them. For example, the five vertices could represent cities (San Francisco, Oakland, Sausalito etc. ) and the edges could be bridges between them.
In the second graph, the role of the cities and the bridges has swapped. Now the bridges are the vertices, and the edges (or hyperedges) are the cities. For example, we can imagine that the cities are large metropolises and the green vertices are the bridge tolls between one city and the next.
Apart from swapping the role of the vertices and the edges, the information in the two graphs is the same. If we shrink each city down to a dot in the second graph, and grow each bridge toll into a full bridge, we get the first graph. We will see that the graphs are dual to each other.
We represent each graph by a labeled matrix: we label the rows by the vertices and the columns by the edges, and we put a $1$ in the matrix whenever the vertex is in the edge. For example, the entry for vertex $1$ and edge $a$ is $1$, because edge $a$ contains vertex $1$. The matrix on the left is for the first graph, and the one on the right is for the second graph.
We can see that the information in the two graphs is the same from looking at the two matrices – they are the same matrix, transposed (or flipped). The matrix of a hypergraph is the transpose of the matrix of the dual hypergraph.
Mathematicians are always on the look-out for hidden dualities between seemingly different objects, and we are happy when we find them. For example, in a recent project we studied the connection between graphical models, from statistics, and tensor networks, from physics. We showed that the two constructions are the duals of each other, using the hypergraph duality we saw in this example.
## Understanding the brain using topology: the Blue Brain project
ALERT ALERT! Applied topology has taken the world has by storm once more. This time techniques from algebraic topology are being applied to model networks of neurons in the brain, in particular with respect to the brain processing information when exposed to a stimulus. Ran Levi, one of the ‘co-senior authors’ of the recent paper published in Frontiers in Computational Neuroscience is based in Aberdeen and he was kind enough to let me show off their pictures in this post. The paper can be found here.
So what are they studying?
When a brain is exposed to a stimulus, neurons fire seemingly at random. We can detect this firing and create a ‘movie’ to study. The firing rate increases towards peak activity, after which it rapidly decreases. In the case of chemical synapses, synaptic communication flows from one neuron to another and you can view this information by drawing a picture with neurons as dots and possible flows between neurons as lines, as shown below. In this image more recent flows show up as brighter.
Numerous studies have been conducted to better understand the pattern of this build up and rapid decrease in neuron spikes and this study contains significant new findings as to how neural networks are built up and decay throughout the process, both at a local and global scale. This new approach could provide substantial insights into how the brain processes and transfers information. The brain is one of the main mysteries of medical science so this is huge! For me the most exciting part of this is that the researchers build their theory through the lens of Algebraic Topology and I will try to explain the main players in their game here.
Topological players: cliques and cavities
The study used a digitally constructed model of a rats brain, which reproduced neuron activity from experiments in which the rats were exposed to stimuli. From this model ‘movies’ of neural activity could be extracted and analysed. The study then compared their findings to real data and found that the same phenomenon occurred.
Neural networks have been previously studied using graphs, in which the neurons are represented by vertices and possible synaptic connections between neurons by edges. This throws away quite a lot of information since during chemical synapses the synaptic communication flows, over a miniscule time period, from one neuron to another. The study takes this into account and uses directed graphs, in which an edge has a direction emulating the synaptic flow. This is the structural graph of the network that they study. They also study functional graphs, which are subgraphs of the structural graph. These contain only the connections that fire within a certain ‘time bin’. You can think of these as synaptic connections that occur in a ‘scene’ of the whole ‘movie’. There is one graph for each scene and this research studies how these graphs change throughout the movie.
The main structural objects discovered and consequentially studied in these movies are subgraphs called directed cliques. These are graphs for which every vertex is connected to every other vertex. There is a source neuron from which all edges are directed away, and a sink neuron for which all edges are directed towards. In this sense the flow of information has a natural direction. Directed cliques consisting of n neurons are called simplices of dimension (n-1). Certain sub-simplices of a directed clique for their own directed cliques, when the vertices in the sub-simplices contain their own source and sink neuron, called sub-cliques. Below are some examples of the directed clique simplices.
And the images below show these simplices occurring naturally in the neural network.
The researchers found that over time, simplices of higher and higher dimension were born in abundance, as synaptic communication increased and information flowed between neurons. Then suddenly all cliques vanished, the brain had finished processing the new information. This relates the neural activity to an underlying structure which we can now study in more detail. It is a very local structure, simplices of up to 7 dimensions were detected, a clique of 8 neurons in a microcircuit containing tens of thousands. It was the pure abundance of this local structure that made it significant, where in this setting local means concerning a small number of vertices in the structural graph.
As well as considering this local structure, the researchers also identified a global structure in the form of cavities. Cavities are formed when cliques share neurons, but not enough neurons to form a larger clique. An example of this sharing is shown below, though please note that this is not yet an example of a cavity. When many cliques together bound a hollow space, this forms a cavity. Cavities represent homology classes, and you can read my post on introducing homology here. An example of a 2 dimensional cavity is also shown below.
The graph below shows the formation of cavities over time. The x-axis corresponds to the first Betti number, which gives an indication of the number of 1 dimensional cavities, and the y-axis similarly gives an indication of the number of 3 dimensional cavities, via the third Betti number. The spiral is drawn out over time as indicated by the text specifying milliseconds on the curve. We see that at the beginning there is an increase in the first Betti number, before an increase in the third alongside a decrease in the first, and finally a sharp decrease to no cavities at all. Considering the neural movie, we view this as an initial appearance of many 1 dimensional simplices, creating 1 dimensional cavities. Over time, the number of 2 and 3 dimensional simplices increases, by filling in extra connections between 1 dimensional simplices, so the lower dimensional cavities are replaced with higher dimensional ones. When the number of higher dimensional cavities is maximal, the whole thing collapses. The brain has finished processing the information!
The time dependent formation of the cliques and cavities in this model was interpreted to try and measure both local information flow, influenced by the cliques, and global flow across the whole network, influenced by cavities.
So why is topology important?
These topological players provide a strong mathematical framework for measuring the activity of a neural network, and the process a brain undergoes when exposed to stimuli. The framework works without parameters (for example there is no measurement of distance between neurons in the model) and one can study the local structure by considering cliques, or how they bind together to form a global structure with cavities. By continuing to study the topological properties of these emerging and disappearing structures alongside neuroscientists we could come closer to understanding our own brains! I will leave you with a beautiful artistic impression of what is happening.
There is a great video of Kathryn Hess (EPFL) speaking about the project, watch it here.
For those of you who want to read more, check out the following blog and news articles (I’m sure there will be more to come and I will try to update the list)
Frontiers blog
Wired article
Newsweek article
## Combing braids
I’m going to a conference next week, and it’s all about braids! So I thought I would write a wee post on combing, a technique which dates back to Artin in the 1940s. In fact the paper where he introduces the concept of combing finishes with the following amusing warning:
“Although it has been proved that every braid can be deformed into a similar normal form the writer is convinced that any attempt to carry this out on a living person would only lead to violent protests and discrimination against mathematics. He would therefore discourage such an experiment.” – Artin 1946
but I really don’t see it as so bad!
Combing is a technique for starting with any braid (see my introductory post on braids here) and ending up with a braid in which first the leftmost strand moves and the others stay put, then the next strand moves while the rest stay put etc etc. It’s much nicer to show this in pictures.
We want to start with any old braid, say this one:
and transform it into a braid where the strands move one at a time, like the following one. I’ve coloured the strands here so you can see that, reading the braid from top to bottom, first the red strand moves (i.e. all crossing involve the red strand, until it is finished), and then the green, and then the blue.
For convenience I’ll only look at braids called pure braids, where each strand starts and ends at the same position. You can easily comb non-pure braids, you just need to add an appropriate twist right at the end to make them finish in the correct positions.
So how do we do this? Consider the first strand, I’ve coloured it red to make it clear. We want all the crossings between red and black strands to happen before (higher up than) a crossing of two black strands. So in this case the crossing circled in yellow are okay, because they happen lower down than any crossing involving the red strand. The crossings circled in blue and green need to be changed.
We can slide some crossings of black strands down past the red and black crossings, as they don’t interfere. Here we can do it with the crossing circled in blue, as shown:
We can start to do it with the crossing circled in green, but we encounter a problem as it wont simply slide past the red strand crossing below it. Moving this crossing down requires using some of the braid relations (see braid post) to replace a few crossings with an equivalent section in which the red strand moves first, as follows:
Even though this braid looks different than the previous one they are in fact the same (you can always test this with string!). Now we have a braid in which the first strand moves before any others. Since all the first stand action is now at the top of the braid, we can now ignore the first strand all together, and consider the rest of the braid, as show below:
we only need to consider the following section now, and again we can put this into a form where only the first strand moves.
In this case using braid relations gives us the following:
And we can now ignore the green strand!
Colouring the first strand in this final section gives us no crossing that don’t involve the first strand:
and we colour the last strand yellow for fun!
Remembering all the pieces we have ignored gives us the full combed braid, where we focus on the leftmost strand until it ‘runs out of moves’ before looking to the next one.
And this is exactly the same as the original braid, which looks a lot messier when coloured:
Why might we want to do this? In some cases it makes mathematical proofs a lot easier. For me, recently I have been focusing only on what the first strand is doing, and so I want a technique to push the other strands down and away!
## Tea with (Almond) Milk
Making a cup of tea in a hurry is a challenge. I want the tea to be as drinkable (cold) as possible after a short amount of time. Say, 5 minutes. What should I do: should I add milk to the tea at the beginning of the 5 minutes or at the end?
The rule we will use to work this out is Newton’s Law of Cooling. It says “the rate of heat loss of the tea is proportional to the difference in temperature between the tea and its surroundings”.
This means the temperature of the tea follows the differential equation $T' = -k (T - T_s)$, where the constant $k$ is a positive constant of proportionality. The minus sign is there because the tea is warmer than the room – so it is losing heat. Solving this differential equation, we get $T = T_s + (A - T_s) e^{-kt}$, where $A$ is the initial temperature of the tea.
We’ll start by defining some variables, to set the question up mathematically. Most of them we won’t end up needing. Let’s say the tea, straight from the kettle, has temperature $T_0$. The cold milk has temperature $m$. We want to mix tea and milk in the ratio $L:l$. The temperature of the surrounding room is $T_s$.
Option 1: Add the milk at the start
We begin by immediately mixing the tea with the milk. This leaves us with a mixture whose temperature is $\frac{T_0 L + m l }{L + l}$. Now we leave the tea to cool. Its cooling follows the equation $T = T_s +\left( \frac{T_0 L + m l }{L + l} - T_s \right) e^{-kt}$. After five minutes, the temperature is
Option 1 $= T_s +\left( \frac{T_0 L + m l }{L + l}- T_s \right) e^{-5k} .$
Option 2: Add the milk at the end
For this option, we first leave the tea to cool. Its cooling follows the equation $T = T_s + (T_0 - T_s) e^{-kt}$. After five minutes, it has temperature $T = T_s + (T_0 - T_s) e^{-5k}$. Then, we add the milk in the specified ratio. The final concoction has temperature
Option 2 $= \frac{(T_s + (T_0 - T_s) e^{-5k}) L + m l }{L + l}.$
So which temperature is lower: the “Option 1” temperature or the “Option 2” temperature?
It turns out that most of the terms in the two expressions cancel out, and the inequality boils down to a comparison of $e^{-5k} (T_s L - ml)$ (from Option 2) with $(T_s L - ml)$ (from Option 1). The answer depends on whether $T_s L - ml > 0$. For our cup of tea, it will be: there’s more tea than milk ($L > l$) and the milk is colder than the surroundings ($m < T_s$). [What does this quantity represent?] Hence, since $k$ is positive, we have $e^{-5k} < 1$, and option 2 wins: add the milk at the end.
But, does it really make a difference? (What’s the point of calculus?)
Well, we could plug in reasonable values for all the letters ($T_0 = 95^o C$, etc.) and see how different the two expressions are.
So, why tea with Almond milk?
My co-blogger Rachael is vegan. She inspires me to make my tea each morning with Almond milk.
Finally, here’s a picture of an empirical experiment from other people (thenakedscientists) tackling this important question:
## Mapping class groups and curves in surfaces
Firstly, thanks to Rachael for inviting me to write this post after meeting me at the ECSTATIC conference at Imperial College London, and to her and Anna for creating such a great blog!
My research is all about surfaces. One of the simplest examples of a surface is a sphere. We are all familiar with this – think of a globe or a beach ball. Really we should think of this beach ball as having no thickness at all, in other words it is 2-dimensional. We are allowed to stretch and squeeze it so that it doesn’t look round, but we can’t make every surface in this way. The next distinct surface we come to is the torus. Instead of a beach ball, this is like an inflatable ring (see this post by Rachael). We say that the genus of the torus is 1 because it has one “hole” in it. If we have $g$ of these holes then the surface has genus $g$. The sphere doesn’t have any holes so has genus 0. We can also alter a surface by cutting out a disc. This creates an edge called a boundary component. If we were to try to pass the edge on the surface, we would fall off. Here are a few examples of surfaces.
As with the sphere, topology allows us to deform these surfaces in certain ways without them being considered to be different. The classification of surfaces tells us that if two surfaces have the same genus and the same number of boundary components then they are topologically the same, or homeomorphic.
Now that we have a surface, we can start to think about its properties. A recurring theme across mathematics is the idea of symmetries. In topology, the symmetries we have are called self-homeomorphisms. Strictly speaking, all of the self-homeomorphisms we will consider will be orientation-preserving.
Let’s think about some symmetries of the genus 3 surface.
Here is a rotation which has order 2, that is, if we apply it twice, we get back to what we started with.
Here is another order 2 rotation.
And here is a rotation of order 3. Remember that we are allowed to deform the surface so that it looks a bit different to the pictures above but still has genus 3.
However, not all symmetries of a surface have finite order. Let’s look at a Dehn twist. The picture (for the genus 2 surface) shows the three stages – first we cut along a loop in the surface, then we rotate the part of the surface on just one side of this loop by one full turn, then we stick it back together.
A Dehn twist has infinite order, that is, if we keep on applying it again and again, we never get back to what we started with.
If we compose two homeomorphisms (that is, apply one after the other) then we get another homeomorphism. The self-homeomorphisms also satisfy some other properties which mean that they form a group under composition. However, this group is very big and quite nasty to study, so we usually consider two homeomorphisms to be the same if they are isotopic. This is quite a natural relationship between two homeomorphisms and roughly means that there is a nice continuous way of deforming one into the other. Now we have the set of all isotopy classes of orientation-preserving self-homeomorphisms of the surface, which we call mapping classes. These still form a group under composition – the mapping class group. This group is much nicer. It still (usually) has infinitely many elements, but now we can find a finite list of elements which form a generating set for the group. This means that every element of the group can be made by composing elements from this list. Groups with finite generating sets are often easier to study than groups which don’t have one.
An example of a mapping class group appears in Rachael’s post below. The braid group on $n$ strands is the mapping class group of the disc with $n$ punctures (where all homeomorphisms fix the boundary pointwise). Punctures are places where a point is removed from the surface. In some ways punctures are similar to boundary components, where an open disc is removed, but a mapping class can exchange punctures with other punctures.
So how can we study what a mapping class does? Rachael described in her post how we can study the braid group by looking at arcs on the punctured disc. Similarly, in the pictures above of examples of self-homeomorphisms the effect of the homeomorphism is indicated by a few coloured curves. More precisely, these are simple closed curves, which means they are loops which join up without any self-intersections. Suppose we are given a mapping class for a surface but not told which one it is. If we are told that it takes a certain curve to a certain other curve then we can start to narrow it down. If we get information about other curves we can narrow it down even more until eventually we know exactly what the mapping class is.
Now I can tell you a little about what I mainly think about in my research: the curve graph. In topology, a graph consists of a set of points – the vertices – with some pairs of vertices joined by edges.
Each vertex in the curve graph represents an isotopy class of curves. As in the case of homeomorphisms, isotopy is a natural relationship between two curves, which more or less corresponds to pushing and pulling a curve into another curve without cutting it open. For example, the two green curves in the picture are isotopic, as are the two blue curves, but green and blue are not isotopic to each other.
Also, we don’t quite want to use every isotopy class of curves. Curves that can be squashed down to a point (inessential) or into a boundary component (peripheral) don’t tell us very much, so we will ignore them. Here are a few examples of inessential and peripheral curves.
We now have infinitely many vertices, one for every isotopy class of essential, non-peripheral curves, and it is time to add edges. We put an edge between two vertices if they have representative curves which do not intersect. So if two curves from these isotopy classes cross each other we can pull one off the other by an isotopy. Here’s an example of some edges in the curve graph of the genus 2 surface. In the picture, all of the curves are intersecting minimally, so if they intersect here they cannot be isotoped to be disjoint.
I should emphasise that this is only a small subgraph of the curve graph of the genus 2 surface. Not only does the curve graph have infinitely many vertices, but it is also locally infinite – at each vertex, there are infinitely many edges going out! This isn’t too hard to see – if we take any vertex, this represents some curve (up to isotopy). If we cut along this curve we get either one or two smaller surfaces. These contain infinitely many isotopy classes of curves, none of which intersects the original curve.
So why is this graph useful? Well, as we noted above, we can record the effect of a mapping class by what it does to curves. Importantly, the property of whether two curves are disjoint is preserved by a mapping class. So not only does a mapping class take vertices of the curve graph (curves) to vertices, but it preserves whether or not two vertices are connected by an edge. Thus a mapping class gives us a map from the curve graph back to itself, where the vertices may be moved around but, if we ignore the labels, the graph is left looking the same. We say that the mapping class group has an isometric action on the curve graph, so to every element of the group we associate an isometry of the graph, which is a map which preserves distances between elements. The distance between two points in the graph is just the smallest number of edges we need to pass along to get from one to the other. When we have an isometric action of a group on a space, this is really useful for studying the geometry of the group, but that would be another story.
## Defining topology through interviews. Interview seven with Jeremy Mann.
The final interview (*cry*) in the Defining topology through interviews series is with Jeremy Mann, who is a PhD student in mathematics at the University of Notre Dame, studying geometry and topology.
1. What would your own personal description of “topology” be?
Topology studies features we call “qualitative”: ones that don’t change if the system is gently* disturbed. In some sense, we created topology in order to give precise answers to qualitative questions. In my day to day life, I reason qualitatively. I rarely wonder “Will the temperature outside be greater than 23 degrees?” I ask: “Is it warm outside?” I would call the first question quantitative, and my second one topological. In other words, topology is created to give precise answers to the types of questions we, as humans, are naturally interested in.
* What one means by “gently” depends enormously on the context, and one has a lot of freedom in choosing what that means. For these reasons, despite being wonderfully vivid, topology is at times unavoidably abstract.
2. What do you say when trying to explain your work to non-mathematicians?
I fudge the details and I lie. If a careful mathematician were listening, they might interject with a few “well, actually—”s. But the details can obscure content, and people enjoy fiction, so I try not to lose sleep over it. That being said, I might tell a story like this:
By the age of three, I could pick two peaches out of a bag without knowing the first thing about the symbol “2.” A number was something like a bunch of stuff contained within a box. A number could bounce around and bruise. I could hold it in my hands.
If I had a sack of plums and a sack of peaches, I could add them together by pouring them both into a bigger sack.
But these terms didn’t help me add the grains of sand in a bucket, or the stars spread before my eyes. So I dropped this way of adding, in favor of an algebra with lots of symbols like “2,” and “376,” and eventually “x.”
Since then, I’ve made another shift. These days, my conceptualization of arithmetic is a lot closer to a child’s. This approach has many names, but my favorite is Factorization Algebras. I see a number as a collection of objects contained within a region of space.
But now, my numbers can interact. Symbols are no longer rich enough to capture their structure. Sometimes my numbers feel like exotic creatures. They can circle each other suspiciously.
Symbols see this as “2=2=2=2,” but this picture shows us there’s a lot more going on.
“1 + 1 = 2”.
Two numbers can be enemies. When I add them together, they remove each other from existence. “1 + (-1) = 0”. Sometimes, I play this in reverse, watching two enemies spontaneously born from empty tranquility.
I guess I’m interested in more than just writing down the final answer. I want to see their costumes. I want to know how they come together. I want to feel the content in their choreography. My work helps me do this.
3. How does your work relate, if at all, to the Nobel prize work?
The Nobel Prize was awarded for insights into the behavior of certain forms of matter at very low energy, where their behavior becomes “topological.” Strictly speaking, the structures I consider are not “topological” — despite “being a topologist,” my work does in fact know the difference between a coffee cup and a donut. It’s much higher-energy.*
Many physicists are interested in a material’s low energy behavior because these conditions contain a huge amount of information about a material’s possible phases. This even includes more “exotic” phases of matter, some with potential applications to quantum computers.
I’d like to point out the following: often, “exotic” means “outside of one’s comfort zone.” So, when physicists say “exotic topological phases of matter,” I suspect they are expressing how the low energy behaviors of certain materials are outside the comfort of zone of many members of the physics (and mathematics) community. This “exotic” behavior defies common intuition. However, when a material behaves in this manner, to a topologist, it enters very familiar territory. The topological is not exotic to a topologist.
** hotter, but certainly not sexier.
## Defining topology through interviews. Interview six with Cécile Repellin.
The penultimate interview in this Defining topology through interviews series is with Cécile Repellin, who is a postdoc at the Max Planck Institute for the physics of complex systems, Dresden and works on condensed matter theory.
1. What do you say when trying to explain your work to non-mathematicians?
Since I’m not a mathematician myself, let’s rather pretend that I’m trying to explain my work to a non-physicist.
Understanding the different phases of matter is one of the most important goals of my field, condensed matter physics. Sometimes we can intuitively grasp the difference between various phases: you can think of water, which can appear in liquid form, but also as solid ice, or as steam, which is in gas form. Sometimes, it is more subtle, like the difference between a material that can carry an electrical current — a metal — and an insulator. Quantum mechanics leads to phases of matter even less intuitive than this, with electrical properties that are neither those of an electrical conductor like copper, nor those of a simple insulator, or even semiconductor like silicium. In these materials, everything happens as if the electrons carrying the electrical current were split into three or more parts. This phenomenon arises from the collective behavior and interactions (electric and magnetic) between the electrons.
My work consists in finding new phases of (quantum) matter, and more specifically new ways that electrons could split up. Among the many forces competing at the microscopic level, I try to figure out which ones are essential, to help predict in which materials these phases might appear.
2. What would your own personal description of “topology” be?
One often gives the example of a mug and a donut to explain the concept of topology. Imagine that you have a mug made of an extremely elastic material. By stretching it, you can transform your mug into an object with the shape of a donut. Had you started from a bun, it would have been necessary to pierce a hole to achieve the same result. The number of holes is a global property of an object, or topological invariant: if you stand too close, you can’t tell how many holes there are. You need to take a step back and look at the whole object. On the other hand, the details of the mug do not matter to determine the number of holes. In the context of physics, you would not be talking about the number of holes, but about something that you can measure in an experiment, like the conductivity or resistivity. In a quantum Hall experiment, a thin layer of semiconductor is sandwiched between two thicker layers, and subjected to a large magnetic field. If you apply a voltage on either side, you will observe the apparition of a voltage in the opposite direction. Another way of saying this is that the transverse resistance (or Hall resistance) is finite. If the temperature is low enough (around -273C), this resistance evolves step by step by forming plateaus as you tune the magnetic field. There is something very special about the value of the Hall resistance on the plateaus: it does not depend on the sample that you are looking at, nor does it depend on the material. It is in fact related by a simple proportionality rule to physical constants: the Planck constant h and the charge of the electron e, or rather the ratio e^2/h. This property is very unique and is so robust that it is used in metrology to define the ratio e^2/h. The robustness is a consequence of the Hall resistance being a topological invariant, much like the number of holes in an object.
3. How does your work relate, if at all, to the Nobel prize work?
A lot of my research relates to the work of David Thouless and Duncan Haldane, two of this year’s Nobel prize winners in physics. In 1988, Haldane proposed a lattice model where the quantum Hall effect could be realized in the absence of a magnetic field. The first projects I worked on as a PhD student consisted in understanding the physics of this model (and other similar ones) when the electrons hopping on the lattice strongly repel one another. One way or another, my research interests are in large part related to topology in condensed matter physics. I was attending a conference on topological phases of matter when I heard about the Nobel prize, and it was very nice to share this moment with colleagues and see the community react and celebrate the great news. |
Lesson Objectives
• Learn how to solve direct variation problems
• Learn how to solve inverse variation problems
• Learn how to solve joint variation problems
• Learn how to solve variation word problems
## How to Solve Variation Problems
In this lesson, we will discuss the various types of variation problems (direct, inverse, and joint) that we will encounter. Let's begin with the most common type of variation problem, which is known as "direct variation".
### Solving Direct Variation Problems
We will see direct variation whenever y depends on a multiple of x. For example, the circumference of a circle is found by using the formula:
C = 2πr
The circumference "C" is always found by multiplying the radius "r" by the constant 2π (2 pi). Therefore, as the radius "r" increases, the circumference will increase and as the radius "r" decreases, the circumference will decrease. Because of this, we can say that the circumference of a circle varies directly with the radius.
In general, we can say that "y" varies directly with "x" if there is some real number "k" such that:
y = kx
or
k = y/x
y is also said to be proportional to x. The number k is called the constant of variation or the constant of proportionality.
When x > 0, as x increases, y increases and when x decreases, y decreases.
If k > 0, then as x ↑ by 1 unit, y ↑ k units.
If k > 0, then as x ↓ by 1 unit, y ↓ k units.
Let's suppose at a local gas station, the cost per gallon is $7. We can say our total cost "y" is equal to the constant cost per gallon of "7" times the number of gallons purchased "x". This gives us the following equation: y = 7x y = 7x x = 0 y = 0 x = 1 y = 7 x = 2 y = 14 x = 3 y = 21 x = 4 y = 28 x = 5 y = 35 We can see that each time x or the number of gallons purchased increases by 1, y or the total cost increases by 7. When we solve a direct variation problem, we can use the following steps. • Write the variation equation: y = kx or k = y/x • Use substitution to find the value for k • Rewrite the variation equation: y = kx with the known value for k • Find the required answer using substitution Let's look at an example. Example 1: Solve each variation problem If y varies directly with x, and y is 60 when x is 5, find y when x is 20. Step 1) Write the variation equation: $$k = \frac{y}{x}$$ Step 2) Use substitution to find the value for k: $$k = \frac{60}{5} = 12$$ Step 3) Rewrite the variation equation: y = kx with the known value for k: $$y = 12x$$ Step 4) Find the required answer using substitution: $$y = 12(20)$$ $$y = 240$$ y is 240 when x is 20. ### Direct Variation as a Power We will also see direct variation as a power. We can say that "y" varies directly with the nth power of "x" if there is some real number "k" such that: $$y = kx^n$$ To solve a direct variation as a power problem, we use the same procedure. The only difference is we change up the variation equation to include our power. Let's look at an example. Example 2: Solve each variation problem If y varies directly with the square of x, and y is 27 when x is 3, find y when x is 11. Step 1) Write the variation equation: $$y = kx^{2}$$ $$k = \frac{y}{x^{2}}$$ Step 2) Use substitution to find the value for k: $$k = \frac{27}{3^{2}} = \frac{27}{9} = 3$$ Step 3) Rewrite the variation equation: y = kx2 with the known value for k: $$y=3x^2$$ Step 4) Find the required answer using substitution: $$y=3(11)^2$$ $$y = 3(121) = 363$$ y is 363 when x is 11. ### Inverse Variation Similar to direct variation, we have inverse variation. Inverse variation occurs when y depends on some constant number divided by x. For example, we have all used the distance formula to solve a motion word problem. $$d = rt$$ Where d is the distance traveled, r is the rate of speed, and t is the amount of time traveled. Let's first solve this equation for t: $$t = \frac{d}{r}$$ Over a given distance, time varies inversely with the rate of speed. This should be fairly intuitive. If we travel a set distance, as we increase speed, the trip will be shorter. Similarly, as we decrease speed, our trip will take longer. As an example, suppose we take a day trip that is 200 miles. Let's set our first rate of speed as 40 miles per hour. $$t = \frac{200}{40} = 5$$ This trip would take 5 hours. What happens if we increase our speed to 100 miles per hour? $$t =\frac{200}{100} = 2$$ Now we can see our same trip is only going to take 2 hours. We increased our speed and the time of our trip went down. What happens if we decrease our speed to 20 miles per hour? $$t = \frac{200}{20} = 10$$ This trip will take 10 hours. This is longer because we decreased our speed. $$t=\frac{200}{r}$$ r = 10 t = 20 r = 20 t = 10 r = 40 t = 5 r = 80 t = 2.5 r = 160 t = 1.25 r = 320 t = 0.625 If we look at the table above, we can see that the time our trip takes decreases as we increase our speed and increases as we decrease our speed. In general, we can say that "y" varies inversely with "x" if there is some real number "k" such that: $$y = \frac{k}{x}$$ or $$k=xy$$ To solve an inverse variation problem, we can use the following steps. • Write the variation equation: y = k/x or k = yx • Use substitution to find the value for k • Rewrite the variation equation: y = k/x with the known value for k • Find the required answer using substitution Let's look at an example. Example 3: Solve each variation problem If y varies inversely with x, and y is 25 when x is 5, find y when x is 10. Step 1) Write the variation equation: k = yx $$k = yx$$ Step 2) Use substitution to find the value for k: $$k = 25(5) = 125$$ Step 3) Rewrite the variation equation: y = k/x with the known value for k: $$y = \frac{125}{x}$$ Step 4) Find the required answer using substitution: $$y = \frac{125}{10} = \frac{25}{2}$$ y is 25/2 or 12.5 when x is 10. ### Inverse Variation as a Power We will also see inverse variation as a power. We say that "y" varies inversely with the nth power of "x" if there is some real number "k" such that: $$y=\frac{k}{x^n}$$ To solve an inverse variation as a power problem, we use the same procedure. The only difference is we change up the variation equation to include our power. Let's look at an example. Example 4: Solve each variation problem If y varies inversely with the cube root of x, and y is 1/40 when x is 10, find y when x is 15. Step 1) Write the variation equation: y = k/x3 or k = yx3: $$k = yx^3$$ Step 2) Use substitution to find the value for k: $$k = \frac{1}{40} \cdot 10^3$$ $$k = \frac{1}{40} \cdot 1000$$ $$k = \frac{1000}{40} = 25$$ Step 3) Rewrite the variation equation: y = k/x3 with the known value for k: $$y = \frac{25}{x^3}$$ Step 4) Find the required answer using substitution: $$y = \frac{25}{15^3} = \frac{25}{3375}$$ $$y = \frac{1}{135}$$ y is 1/135 when x is 15. ### Joint Variation In some cases, we will have one variable that depends on several others. For example, we can think about the formula which is used to solve simple interest word problems. $$I = prt$$ For a given principal (amount invested) "p", the amount of simple interest earned "I" varies jointly with the interest rate "r" and the time "t". Let's suppose we have 5000 dollars to invest in a savings account that pays annual simple interest. If we invest our money at a rate of 3% for 6 years: $$I=5000(.03)(6) = 900$$ This means we would earn 900 dollars in simple interest over a period of 6 years. What happens if we increase the rate to 7%, using the same time period of 6 years? $$I = 5000(.07)(6) = 2100$$ This means we would earn 2100 dollars in simple interest over a period of 6 years. We can see that as the interest rate increased, the amount of simple interest earned increased. Let's suppose we again use an interest rate of 7%, but decrease our time period to 3 years. $$I = 5000(.07)(3) = 1050$$ We can see as the time period decreased from 6 years to 3 years, the simple interest earned decreased. I = 5000rt r = .01 t = 5 I = 250 r = .02 t = 5 I = 500 r = .02 t = 10 I = 1000 r = .2 t = 2 I = 2000 r = .45 t = 1 I = 2250 r = .9 t = .5 I = 2250 From the table above, we can see that increasing rate or time increases simple interest earned, while a decrease in rate or time results in a decrease in simple interest earned. In general, we can say that "y" varies jointly with "x" and "z" if there is some real number "k" such that: $$y = kxz$$ or $$k=\frac{y}{xz}$$ To solve an inverse variation problem, we can use the following steps. • Write the variation equation: y = kxz or k = y/xz • Use substitution to find the value for k • Rewrite the variation equation: y = kxz with the known value for k • Find the required answer using substitution Let's look at an example. Example 5: Solve each variation problem If y varies jointly with x2 and z, and y is 816 when x is 4 and z is 3, find y when x is 5 and z is -1. Step 1) Write the variation equation: k = y/x2z: $$k = \frac{y}{x^2z}$$ Step 2) Use substitution to find the value for k: $$k = \frac{816}{4^2 \cdot 3}$$ $$k = \frac{816}{16 \cdot 3}$$ $$k = \frac{816}{48} = 17$$ Step 3) Rewrite the variation equation: y = kx2z with the known value for k: $$y = 17x^2z$$ Step 4) Find the required answer using substitution: $$y = 17(5)^2(-1)$$ $$y = 17(25)(-1)$$ $$y = -425$$ y is -425 when x is 5 and z is (-1). ### Solving Variation Word Problems To solve a variation word problem, we first read and interpret our problem. We can then set up and solve the problem using the methods discussed earlier in the lesson. Let's look at an example. Example 6: Solve each variation word problem A local church sells raffle tickets to supplement revenue. The total revenue from raffle tickets sold varies directly with the cube of the anticipated winnings. If the raffle sold$5,000,000 worth of tickets when the anticipated winnings were $100, find the revenue from raffle tickets sold when the anticipated winnings are$25.
$$k = \frac{y}{x^3}$$ Let's plug in for x and y, we are told that the raffle sold $5,000,000 worth of tickets (y) when the anticipated winnings were$100 (x). $$k = \frac{5,\hspace{-.1em}000,\hspace{-.1em}000}{100^3}$$ $$k = \frac{5,\hspace{-.1em}000,\hspace{-.1em}000}{1,\hspace{-.1em}000,\hspace{-.1em}000}$$ $$k = 5$$ Now we can set up our variation equation with the known value for k: $$y = 5x^3$$ We want to find the revenue from raffle tickets sold when the anticipated winnings are $25, let's plug in 25 for x: $$y = 5(25)^3$$ $$y = 5(15,\hspace{-.1em}625)$$ $$y = 78,\hspace{-.1em}125$$ When the anticipated winnings are$25, the church will sell \$78,125 worth of raffle tickets. |
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Numbers - Important Concept and Formulae Finding the number of zeroes at the end of any factorial: How to find the number of zeroes ...
# 05. Aptitude For C.A.T.-- Numbers - Theory - Remainder Theorem
Numbers - Important Concept and Formulae
Finding the number of zeroes at the end of any factorial:
How to find the number of zeroes at the end of the product of numbers
The number of zeroes at the end of the product depends upon the number of 2‘s and 5’s in the expression.
The number of zeroes is the same as the number of pair of (5 x 2) units.
Since the number of two’s will be more than the number of fives
So, we have to count the number of five’s to get the number of zeroes at the end.
Example : Find the number of zeroes in 10!?
Solution : 10!= 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.
Since the number of 2 is more than the number of 5’s, so the number of 5’s will be counted to form the combination of 2 x 5.
We have two 5’s, one digit 4 and another 5 is 10 ( since 10= 5 x2)
Thus, there are only 2 zeroes at the end of 10!
Example: Find the number of zeroes at the of the product of 2111 x 5222
Solution : In the given product, the number of 2’s are less than the number of 5’s, hence we take the number of 2’s into consideration.
Thus there will be only 111 pairs of(2 x5).
Therefore, the number of zeroes at the end of the product of the given expression will be 111.
Finding the largest power of a number contained in a factorial:
How to find the highest power of x that can exactly divide n! : We divide n by x, n by x2 … and so on till we get $\frac{n}{x^{k}}$ equal to 1.
Example : Find the largest power of 5 contained in 120!?
Solution : [$\frac{120}{5}$ ] + [$\frac{120}{5^{2}}$ ] = 24 + 4 =28
We need not further, because 120 is not divisible by 53 (=125)
Example 2: Find the largest power of 7 that can exactly divide 800!?
Solution : $\frac{800}{7}$ + $\frac{800}{7^{2}}$ + $\frac{800}{7^{3}}$
= 114 + 16 +2 = 132
Thus the highest power of 7 is 132 by which 800! can be divided.
Concept of Remainder :
Definition of Remainder: The part of the dividend that is not evenly divided by the divisor, is called the remainder.
Rule 1 :
When x1, x2, x3 … are divided by n individually, the respective remainders obtained are R1, R2, and R3 etc .. when ( x1 + x2 + x3) is divided by n the remainders can be obtained by (R1+ R2+R3+ ..) by n.
Example : Find the remainder when 65+ 75+ 87 is divided by 7?
Solution: When we divided 65, 75 and 87, we obtain the respective remainders as 2, 5 and 3.
Now we can get the required remainder by dividing the sum of the remainders by 7 = $\frac{(2 + 5 + 3)}{7}$ =3
Instead of dividing the sum of original numbers (65 +75+87 =227) by 7, which will also give the same remainder 3.
Example: Find the remainder when 719 x 121 x 237 x 725 x 727 is divided by 14 ?
Solution :
First, we find the remainders when each individual number is divided by 14.
The remainder when 719 is divided by 14 is 5
The remainder when 121 is divided by 14 is 9
The remainder when 237 is divided by 14 is 13
The remainder when 725 is divided by 14 is 11
The remainder when 727 is divided by 14 is 13
Now we find the remainder of the remainders.
Hence, the remainder of $\frac{ 719 \,\times\, 121 \,\times\, 237 \,\times\, 725 \,\times\, 727}{14}$
=$\frac{5\,\times\, 9\,\times\, 13\,\times\, 11\,\times\, 13}{14}$
ð $\frac{45 \,\times\, 13 \,\times\, 11 \,\times\, 13}{14}$
ð $\frac{3 \,\times\, 13 \,\times\, 11 \,\times\, 13}{14}$
ð $\frac{39 \,\times\, 143}{14}$
ð $\frac{11 \,\times\, 3}{4}$
ð $\frac{33}{4}$
ð 5
Example: What is the remainder when 79 is divided by 4?
Solution : $\frac{7^{9}}{4}$ = $\frac{7\,\times\, 7 \,\times\, 7 \,\times\, 7 \,\times\, 7\,\times\, 7 \,\times\, 7 \,\times\, 7 \,\times\, 7}{4}$
= $\frac{3 \,\times 3 \,\times\, 3 \,\times\, 3\,\times\, 3 \,\times\, 3 \,\times\, 3 \,\times\, 3 \,\times\, 3}{4}$
= $\frac{27 \,\times\, 27 \,\times\, 27}{4}$
= $\frac{3 \,\times\, 3 \,\times\, 3}{4}$
=4
Thus the required remainder is 4.
Rule 2 :
A certain number when successively divided by two different numbers leaves some remainder. The same number, when divided by the product of the two divisors, will leave remainder equal to :
(First Divisor x Second Remainder ) + First Remainder.
Example: A certain number when successively divided by 5 and 9 leaves the remainders 2 and 7 respectively. What is the remainder when the same number is divided by 45?
Solution :
Remainder =( First Divisor x Second Remainder ) + First remainder
= (5 x 7) +2 =37
Rule 3 :
If two numbers M and N when separately divided by x, leaving the remainders a and b respectively. Then Mp x Nq, when divided by x, will leave the remainder am x bn
Example : Find the remainder when 802 x 533 is divided by 26.
Solution :80 and 53, when divided by 26, leave remainders 4 and 1 respectively.
Therefore, 802 x 533 when divided by 26 leaves remainder 42 x 13 =16
Rule 4 :
Finding whether (an – k) is divisible by x?
Method :
1. an is divided into some parts in such a way that the value of each part is slightly more or less than the multiple of x
2. Take remainder when each part is divided by x
3. Multiply all the remainders obtained from each part
4. Subtract k from the product of remainders
5. If the result is a multiple of x or zero, then (an – k) is exactly divisible by y.
Example: Is 78 -1 divisible by 24?
Solution : 78 -1 = 72 x 72 x72 x72 -1= 49 x 49 x 49 x 49 – 1
The remainder we get when each part 49 is divided by 24 = 1 x 1 x 1 x 1 -1 =0
Therefore, 78 -1 divisible by 24
Concept of Negative Remainders:
Remainders are always non-negative by their definition.
When 14 is divided by 5, the usual remainder would be 4. But it can also be expressed as -1. It is a negative remainder. The concept of negative remainders sometimes helps the student in reducing calculations.
For numbers like 12,19,26,33 etc., we can represent them as 7k+5 or 7k-5.
Example: Find the remainder when 262 x 68 is divided by 18?
Thus , $\frac{262 \,\times\, 68 }{18}$ will give remainder $\frac{-10 \,\times\, -4 }{18}$ = $\frac{40 }{18}$ = 14
This will give 14 as remainder.
So, using the concept of negative remainders, we can reduce our calculations
Example: What is the remainder when 11215 is divided by 12?
Solution: The remainder we get when 11 divided by 12 is 11 = -1
The negative remainder is -1.
So 11215 =(-1)215 = -1 =11
Using the concept of the negative remainder, we can say the remainder is 11. |
# How do you solve 1/6 = 5/12 + p?
Apr 16, 2016
$- \frac{3}{12} = p$
#### Explanation:
We need to isolate $p$.
1) Find the inverse operation of addition to remove $\frac{5}{12}$ from the right side of the equal sign and make sure that $p$ is alone.
2) The inverse operation of addition is subtraction, so we subtract $\frac{5}{12}$ from both sides of the equal sign.
$\frac{1}{6} - \frac{5}{12} = \frac{5}{12} - \frac{5}{12} + p$
$\frac{2}{12} - \frac{5}{12} = p$
$- \frac{3}{12} = p$ |
### Rearrange:
Rearrange the equation by individually what is to the best of the equal sign from both sides of the equation : y^2-(7*y+18)=0
## Step 1 :
Trying to variable by dividing the middle term
1.1Factoring y2-7y-18 The first term is, y2 that coefficient is 1.The center term is, -7y that coefficient is -7.The critical term, "the constant", is -18Step-1 : main point the coefficient of the very first term by the constant 1•-18=-18Step-2 : find two factors of -18 whose sum equates to the coefficient of the middle term, which is -7.
-18 + 1 = -17 -9 + 2 = -7 That"s it
Step-3 : Rewrite the polynomial splitting the center term making use of the two components found in step2above, -9 and also 2y2 - 9y+2y - 18Step-4 : include up the first 2 terms, pulling out favor factors:y•(y-9) include up the last 2 terms, pulling out usual factors:2•(y-9) Step-5:Add up the four terms that step4:(y+2)•(y-9)Which is the preferred factorization
Equation in ~ the end of action 1 :
(y + 2) • (y - 9) = 0
## Step 2 :
Theory - roots of a product :2.1 A product of number of terms equates to zero.When a product of 2 or more terms equals zero, climate at least one that the terms should be zero.We shall currently solve each term = 0 separatelyIn other words, we room going to fix as numerous equations together there space terms in the productAny systems of ax = 0 solves product = 0 together well.
Solving a single Variable Equation:2.2Solve:y+2 = 0Subtract 2 indigenous both political parties of the equation:y = -2
Solving a single Variable Equation:2.3Solve:y-9 = 0Add 9 come both political parties of the equation:y = 9
### Supplement : resolving Quadratic Equation Directly
Solving y2-7y-18 = 0 directly Earlier us factored this polynomial by separating the center term. Permit us now solve the equation by completing The Square and by making use of the Quadratic Formula
Parabola, detect the Vertex:3.1Find the crest oft = y2-7y-18Parabolas have actually a highest or a lowest allude called the Vertex.Our parabola opens up and appropriately has a lowest allude (AKA absolute minimum).We understand this even prior to plotting "t" since the coefficient that the an initial term,1, is positive (greater 보다 zero).Each parabola has actually a vertical line of symmetry that passes through its vertex. Because of this symmetry, the heat of the opposite would, for example, pass v the midpoint that the two x-intercepts (roots or solutions) the the parabola. That is, if the parabola has actually indeed two genuine solutions.Parabolas deserve to model numerous real life situations, such as the height over ground, of an object thrown upward, after some duration of time. The vertex of the parabola can administer us with information, such together the maximum height that object, thrown upwards, can reach. For this reason we want to have the ability to find the works with of the vertex.For any kind of parabola,Ay2+By+C,the y-coordinate that the peak is given by -B/(2A). In our instance the y coordinate is 3.5000Plugging into the parabola formula 3.5000 because that y we can calculate the t-coordinate:t = 1.0 * 3.50 * 3.50 - 7.0 * 3.50 - 18.0 or t = -30.250
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : t = y2-7y-18 Axis of symmetry (dashed) y= 3.50 Vertex in ~ y,t = 3.50,-30.25 y-Intercepts (Roots) : source 1 at y,t = -2.00, 0.00 source 2 in ~ y,t = 9.00, 0.00
Solve Quadratic Equation by perfect The Square
3.2Solvingy2-7y-18 = 0 by completing The Square.Add 18 come both side of the equation : y2-7y = 18Now the clever bit: take the coefficient the y, which is 7, division by two, giving 7/2, and also finally square it giving 49/4Add 49/4 come both sides of the equation :On the best hand side us have:18+49/4or, (18/1)+(49/4)The common denominator of the two fractions is 4Adding (72/4)+(49/4) gives 121/4So adding to both sides we ultimately get:y2-7y+(49/4) = 121/4Adding 49/4 has completed the left hand side right into a perfect square :y2-7y+(49/4)=(y-(7/2))•(y-(7/2))=(y-(7/2))2 things which room equal to the same thing are likewise equal come one another. Sincey2-7y+(49/4) = 121/4 andy2-7y+(49/4) = (y-(7/2))2 then, according to the legislation of transitivity,(y-(7/2))2 = 121/4We"ll describe this Equation together Eq.
You are watching: Which equation could be used to solve y 2+7y=18 by factoring?
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#3.2.1 The Square source Principle says that once two things are equal, your square roots are equal.Note the the square source of(y-(7/2))2 is(y-(7/2))2/2=(y-(7/2))1=y-(7/2)Now, applying the Square source Principle to Eq.#3.2.1 we get:y-(7/2)= √ 121/4 add 7/2 come both sides to obtain:y = 7/2 + √ 121/4 due to the fact that a square root has actually two values, one positive and also the various other negativey2 - 7y - 18 = 0has two solutions:y = 7/2 + √ 121/4 ory = 7/2 - √ 121/4 note that √ 121/4 have the right to be written as√121 / √4which is 11 / 2 |
# What are the step to solve simultaneous differential EQ by matrix methods?
## What are the step to solve simultaneous differential EQ by matrix methods?
The process of solving the above equations and finding the required functions, of this particular order and form, consists of 3 main steps….Brief descriptions of each of these steps are listed below:
1. Finding the eigenvalues.
2. Finding the eigenvectors.
3. Finding the needed functions.
How do you solve two differential equations?
Second Order Differential Equations
1. Here we learn how to solve equations of this type: d2ydx2 + pdydx + qy = 0.
2. Example: d3ydx3 + xdydx + y = ex
3. We can solve a second order differential equation of the type:
4. Example 1: Solve.
5. Example 2: Solve.
6. Example 3: Solve.
7. Example 4: Solve.
8. Example 5: Solve.
How do you solve a vector differential equation?
In order to make some headway in solving them, however, we must make a simplifying assumption: The coefficient matrix A consists of real constants. Recall that an n × n matrix A may be diagonalized if and only if it is nondefective. When this happens, we can solve the homogeneous vector differential equation x = Ax.
### What is matrix algebra method?
Matrix and computer methods Matrix algebra is a mathematical notation that simplifies the presentation and solution of simultaneous equations. It may be used to obtain a concise statement of a structural problem and to create a mathematical model of the structure.
How do you find the fundamental matrix?
In other words, a fundamental matrix has n linearly independent columns, each of them is a solution of the homogeneous vector equation ˙x(t)=P(t)x(t). Once a fundamental matrix is determined, every solution to the system can be written as x(t)=Ψ(t)c, for some constant vector c (written as a column vector of height n).
What is the Jacobian matrix and why it is needed?
The Jacobian matrix is used to analyze the small signal stability of the system. The equilibrium point Xo is calculated by solving the equation f(Xo,Uo) = 0. This Jacobian matrix is derived from the state matrix and the elements of this Jacobian matrix will be used to perform sensitivity result.
#### What do you need to know about matrix differential equation?
A matrix differential equation contains more than one function stacked into vector form with a matrix relating the functions to their derivatives. {\\displaystyle n imes n} matrix of coefficients. where λ1, λ2., λn are the eigenvalues of A; u1, u2., un are the respective eigenvectors of A ; and c1, c2.., cn are constants.
Which is the derivative notation used in the matrix equation?
The derivative notation x ′ etc. seen in one of the vectors above is known as Lagrange’s notation, (first introduced by Joseph Louis Lagrange. It is equivalent to the derivative notation dx/dt used in the previous equation, known as Leibniz’s notation, honoring the name of Gottfried Leibniz .)
What are the stability conditions of the matrix differential equation?
In the n = 2 case (with two state variables), the stability conditions that the two eigenvalues of the transition matrix A each have a negative real part are equivalent to the conditions that the trace of A be negative and its determinant be positive. evaluated using any of a multitude of techniques. are simple first order inhomogeneous ODEs.
## How to calculate the size of a new matrix?
The new matrix will have size 2 × 4 2 × 4. The entry in row 1 and column 1 of the new matrix will be found by multiplying row 1 of A A by column 1 of B B. This means that we multiply corresponding entries from the row of A A and the column of B B and then add the results up. Here are a couple of the entries computed all the way out. |
# Adapted from Walch Education The standard form of a quadratic function is f ( x ) = ax 2 + bx + c, where a is the coefficient of the quadratic term,
## Presentation on theme: "Adapted from Walch Education The standard form of a quadratic function is f ( x ) = ax 2 + bx + c, where a is the coefficient of the quadratic term,"— Presentation transcript:
The standard form of a quadratic function is f ( x ) = ax 2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term. The x -intercepts are the points at which the graph crosses the x -axis, and are written as ( x, 0). The y -intercept is the point at which the graph crosses the y -axis and is written as (0, y ). 5.3.1: Creating and Graphing Equations Using Standard Form 2 Key Concepts
The axis of symmetry of a parabola is the line through the vertex of a parabola about which the parabola is symmetric. The axis of symmetry extends through the vertex of the graph. The vertex of a parabola is the point on a parabola where the graph changes direction, ( h, k ), where h is the x -coordinate and k is the y -coordinate. The equation of the axis of symmetry is or 5.3.1: Creating and Graphing Equations Using Standard Form 3 Axis of Symmetry
To find the y -coordinate, substitute the value of x into the original function, The maximum is the largest y -value of a quadratic and the minimum is the smallest y -value. If a > 0, the parabola opens up and therefore has a minimum value. If a < 0, the parabola opens down and therefore has a maximum value. 5.3.1: Creating and Graphing Equations Using Standard Form 4 Key Concepts
x -intercepts, y -intercept, where the function is increasing and decreasing, where the function is positive and negative, relative minimums and maximums, symmetries, and end behavior of the function. 5.3.1: Creating and Graphing Equations Using Standard Form 5 Key Features of a Quadratic
h ( x ) = 2 x 2 – 11 x + 5 is a quadratic function. Determine the direction in which the function opens, the vertex, the axis of symmetry, the x -intercept(s), and the y - intercept. Use this information to sketch the graph. 5.3.1: Creating and Graphing Equations Using Standard Form 6 Practice # 1
1.Determine whether the graph opens up or down. h ( x ) = 2 x 2 – 11 x + 5 is in standard form; therefore, a = 2. Since a > 0, the parabola opens up. 5.3.1: Creating and Graphing Equations Using Standard Form 7 Solution
2.Find the vertex and the equation of the axis of symmetry. h ( x ) = 2 x 2 – 11 x + 5 is in standard form; therefore, a = 2 and b = –11. The vertex has an x -value of 2.75. 5.3.1: Creating and Graphing Equations Using Standard Form 8 Solution, continued Equation to determine the vertex Substitute 2 for a and –11 for b. x = 2.75Simplify.
The y -value of the vertex is –10.125. The vertex is the point (2.75, –10.125). Since the axis of symmetry is the vertical line through the vertex, the equation of the axis of symmetry is x = 2.75. 9 Solution, continued h(x) = 2x 2 – 11x + 5 Original equation h(2.75) = 2(2.75) 2 – 11(2.75) + 5 Substitute 2.75 for x. h(2.75) = –10.125Simplify.
3.Find the y -intercept. h ( x ) = 2 x 2 – 11 x + 5 is in standard form, so the y - intercept is the constant c, which is 5. The y -intercept is (0, 5). 5.3.1: Creating and Graphing Equations Using Standard Form 10 Solution, continued
4.Find the x -intercepts, if any exist. The x -intercepts occur when y = 0. Substitute 0 for the output, h ( x ), and solve. Solved by factoring: h ( x ) = 2 x 2 – 11 x + 5 0 = 2 x 2 – 11 x + 5 0 = (2 x – 1)( x – 5) 0 = 2 x – 1 or 0 = x – 5 x = 0.5 or x = 5 The x -intercepts are (0.5, 0) and (5, 0). 5.3.1: Creating and Graphing Equations Using Standard Form 11 Solution, continued
5.3.1: Creating and Graphing Equations Using Standard Form 12 Plot the points
Thanks for Watching! Ms. Dambreville
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# Solve Algebra Word Problems with Ease Using a Calculator
Algebra word problems can be challenging for students, but calculators can be a valuable tool for solving them. Calculators can help students to:
• Solve problems more quickly and accurately.
• Learn how to identify key information in a problem and translate it into mathematical expressions.
• Develop problem-solving skills.
To use a calculator to solve an algebra word problem, students should follow these steps:
1. Identify the key information in the problem. This includes the known quantities, the unknown quantity, and any key words that indicate what operation needs to be performed.
2. Translate the problem into a mathematical expression or equation.
3. Use the calculator to solve the equation.
4. Check your answer by substituting the known quantities back into the original problem.
Here are some examples of algebra word problems that can be solved using a calculator:
## Rate problem:
A train travels 600 kilometers in 5 hours. What is the train’s speed?
Solution:
Speed = Distance / Time
Speed = 600 kilometers / 5 hours
Speed = 120 kilometers per hour
## Proportion problem:
• A recipe calls for 2 cups of flour and 3 cups of sugar. If I have 4 cups of flour, how many cups of sugar do I need?
Solution:
Flour : Sugar = 2 : 3
Flour : Sugar = 4 : x
x = (4 * 3) / 2
x = 6 cups
## Mixture problem:
• A chemist mixes 2 liters of a 10% acid solution with 3 liters of a 20% acid solution. What is the concentration of the resulting acid solution?
Solution:
Total acid = (0.1 * 2 liters) + (0.2 * 3 liters) = 0.9 liters
Total solution = 2 liters + 3 liters = 5 liters
Concentration = Acid / Solution
Concentration = 0.9 liters / 5 liters
Concentration = 18%
## Work problem:
• Two people can complete a task in 6 hours. If one person works for 3 hours, how long will it take the other person to complete the task?
Solution:
Work = Rate * Time
Rate = 1 task / 6 hours
Work = 1/2 task
Time = Work / Rate
Time = (1/2 task) / (1 task / 6 hours)
Time = 3 hours
## Interest problem:
• A loan of \$1000 is borrowed at an interest rate of 5% per year. How much interest is owed after 2 years?
Solution:
Interest = Principal * Rate * Time
Interest = \$1000 * 0.05 * 2 years
Interest = \$100
## Conclusion
Calculators can be a valuable tool for solving algebra word problems. By following the tips and examples in this article, students can learn to use calculators to solve these problems efficiently and effectively.
## FAQs
### Q.How do I solve algebra word problems?
There are a few steps you can follow to solve algebra word problems:
1. Identify the unknown quantity. What are you trying to find?
2. Write down the given information. What do you know about the situation?
3. Translate the problem into an equation. Use variables to represent the unknown and known quantities.
4. Solve the equation. Use algebraic methods to solve the equation for the unknown quantity.
5. Check your answer. Make sure your answer makes sense in the context of the problem.
Here is an example:
Problem:
A train travels 200 miles in 4 hours. What is the train’s speed?
Solution:
Let x be the train’s speed in miles per hour.
Equation:
distance = speed * time
Solving the equation:
200 miles = x miles/hour * 4 hours 200 miles = 4x miles/hour x miles/hour = 200 miles / 4 hours x miles/hour = 50 miles/hour
Answer:
The train’s speed is 50 miles per hour.
### Q.What are some common mistakes to avoid when solving algebra word problems?
Here are some common mistakes to avoid when solving algebra word problems:
• Not identifying the unknown quantity. Make sure you know what you are trying to find before you start solving the problem.
• Not translating the problem into an equation. This is the most important step in solving algebra word problems. Make sure you use variables to represent the unknown and known quantities, and that your equation accurately reflects the information in the problem.
• Making careless algebraic errors. Be careful when solving your equation to avoid making mistakes.
• Not checking your answer. Make sure your answer makes sense in the context of the problem.
### Q.How can I get better at solving algebra word problems?
The best way to get better at solving algebra word problems is to practice. Try to solve as many word problems as you can, and don’t be afraid to ask for help if you get stuck.
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# 8.8: Transformations in the Coordinate Plane
Difficulty Level: At Grade Created by: CK-12
## Introduction
The King’s Chamber
In one room of the museum was a King’s bedroom. The furniture in the room was large and wooden and old with great golden cloths. On the walls was a beautiful red, blue and gold pattern.
Jessica thought that the pattern was the most beautiful one that she had ever seen.
“I love this,” she said to Mrs. Gilman. “I want to draw it, but I’m not sure how.”
“Well, you could break it up into a coordinate grid, since the pattern repeats itself, and use what we have learned about transformations to draw it in.”
“How could I get started?” Jessica asked.
“Well, start by drawing the coordinate grid, then use these coordinates for one of the diamonds. See if you can figure it out from there.”
In Jessica’s notebook, Mrs. Gilman wrote down the following coordinates.
\begin{align*}(4, 1)\!\\ (5, 2)\!\\ (5, 0)\!\\ (6, 1)\end{align*}
Jessica began to draw it in. Then she got stuck.
This is where you come in. This lesson will teach you all about drawing transformations. Follow along closely and you can help Jessica draw in the diamonds in each quadrant at the end of the lesson .
What You Will Learn
In this lesson you will learn how to demonstrate the following activities and skills:
• Identify and describe transformations in the coordinate plane.
• Translate a figure in the coordinate plane using coordinate notation, and graph the resulting image.
• Reflect a figure in the coordinate plane using coordinate notation, and graph the resulting image.
• Rotate a figure in the coordinate plane using coordinate notation, and graph the resulting image.
Teaching Time
I. Identify and Describe Transformations in the Coordinate Plane
In the last lesson you learned how to identify and perform different transformations. Remember that a transformation is when we move a figure in some way, even though we don’t change the figure at all. This lesson will teach you how to identify and perform transformations in the coordinate plane.
The coordinate plane is a representation of two-dimensional space. It has a horizontal axis, called the \begin{align*}x-\end{align*}axis, and a vertical axis, called the \begin{align*}y-\end{align*}axis. We can graph and move geometric figures on the coordinate plane.
Do you remember the three types of transformations?
The first is a translation or slide. A translation moves a figure up, down, to the right, to the left or diagonal without altering the figure.
The second is a reflection or flip. A reflection makes a mirror image of the figure over a line of symmetry. The line of symmetry can be vertical or horizontal.
The third is a rotation or turn. A rotation moves a figure in a circle either clockwise or counterclockwise.
Now let’s look at performing each type of transformation in the coordinate plane.
II. Translate a Figure in the Coordinate Plane Using Coordinate Notation, and Graph the Resulting Image
As we have said, when we perform translations, we slide a figure left or right, up or down. This means that on the coordinate plane, the coordinates for the vertices of the figure will change. Take a look at the example below.
Now let’s look at performing a translation or slide of this figure.
We can choose the number of places that we want to move the triangle and the direction that we wish to move it in. If we slide this triangle 3 places down, all of its vertices will shift 3 places down the \begin{align*}y-\end{align*}axis. That means that the ordered pairs for the new vertices will change. Specifically, the \begin{align*}y-\end{align*}coordinate in each pair will decrease by 3.
Let’s see why this happens.
We can see the change in all of the \begin{align*}y-\end{align*}coordinates. Compare the top points. The \begin{align*}y-\end{align*}coordinate on the right is 2. The \begin{align*}y-\end{align*}coordinate for the corresponding point in the triangle after it moves is -1. The \begin{align*}y-\end{align*}coordinate decreased by 3. Now compare the left-hand point of each triangle. The \begin{align*}y-\end{align*}coordinate originally is -2, and the \begin{align*}y-\end{align*}coordinate after the translation is -5. Again, the difference shows a change of -3 in the \begin{align*}y-\end{align*}coordinate. For the last point, the \begin{align*}y-\end{align*}coordinate starts out as -6, and shifts to -9 after the downward slide. For each point, then, the \begin{align*}y-\end{align*}coordinate decreases by 3 while the \begin{align*}x-\end{align*}coordinates stay the same. This means that we slid the triangle down 3 places.
We can translate figures in other ways, too. As you might guess, we move figures right or left on the coordinate grid by their \begin{align*}x-\end{align*}coordinates. We can also move figures diagonally by changing both their \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates. One way to recognize translations, then, is to compare their points. The \begin{align*}x-\end{align*}coordinates will all change the same way, and the \begin{align*}y-\end{align*}coordinates will all change the same way.
To graph a translation, we perform the same change for each point. Let’s try graphing a translation.
Example
Slide the following figure 5 places to the right.
In this translation, we will move the figure to the right. That means the \begin{align*}x-\end{align*}coordinates for each point will change but the \begin{align*}y-\end{align*}coordinates will not. We simply count 5 places to the right from each point and make a new point.
Once we relocate each point 5 places to the right, we can connect them to make the new figure that shows the translation.
We can check to see if we performed the translation correctly by adding 5 to each \begin{align*}x-\end{align*}coordinate (because we moved to the right) and then checking these against the ordered pairs of the figure you drew. This is called coordinate notation. Notice that each point is represented by coordinates.
\begin{align*}& (-4, 3) \qquad (-6, -2) \qquad (-1, -6) \qquad (2, -1)\\ & +5 \qquad \qquad +5 \qquad \qquad +5 \qquad \qquad +5\\ & (1, 3) \qquad \quad (-1, -2) \qquad (4, -6) \qquad \ \ (7, -1)\end{align*}
These are the points we graphed, so we have performed the translation correctly.
Let’s try another.
Example
Slide the following figure 4 places to the left and 2 places up.
This time we need to perform two movements, both left and up. That means we will change both the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates of the ordered pairs. We graph each point by counting 4 places to the left first, and from there 2 places up (2 places up from where you started, not 2 places up from the \begin{align*}y-\end{align*}axis!). Make a mark and repeat this process for each point. Then connect the new points.
Again, we can check that we performed the translation correctly by changing the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates in the ordered pairs and then comparing these to the points we graphed. This time we subtract 4 from each \begin{align*}x-\end{align*}coordinate (because we moved left; imagine a number line) and add 2 to each \begin{align*}y-\end{align*}coordinate. Let’s see what happens.
\begin{align*}& \ \ (3, \ 2) \qquad \quad (4, -2) \quad \qquad (1, -4)\\ & -4 +2 \qquad -4 +2 \qquad \ \ -4 +2\\ & (-1, \ 4) \qquad \ \ \ (0, \ 0) \qquad \quad \ (-3, -2)\end{align*}
These are the points we graphed, so we performed the translation correctly.
8O. Lesson Exercises
Use coordinate notation to write the coordinates of each translated triangle. The vertices of the original figure have been given to you.
1. Triangle \begin{align*}ABC\end{align*} (0, 1)(1, 3)(4, 0) translate this figure up 4.
2. Triangle \begin{align*}DEF\end{align*} (-3, 2)(1, 6)(2, 1) translate this figure right 8.
Take a few minutes to check your work with a neighbor. Be sure that your answers are accurate. Correct any errors before continuing.
III. Reflect a Figure in the Coordinate Plane Using Coordinate Notation, and Graph the Resulting Image
We can also identify a reflection by the changes in its coordinates. Recall that in a reflection, the figure flips across a line to make a mirror image of itself. Take a look at the reflection below.
We usually reflect a figure across either the \begin{align*}x-\end{align*} or the \begin{align*}y-\end{align*}axis. In this case, we reflected the figure across the \begin{align*}x-\end{align*}axis. If we compare the figures in the first example vertex by vertex, we see that the \begin{align*}x-\end{align*}coordinates change but the \begin{align*}y-\end{align*}coordinates stay the same. This is because the reflection happens from left to right across the \begin{align*}x-\end{align*}axis. When we reflect across the \begin{align*}y-\end{align*}axis, the \begin{align*}y-\end{align*}coordinates change and the \begin{align*}x-\end{align*}coordinates stay the same. Take a look at this example.
Now let’s compare some of the vertices. In the figure above the coordinates for the upper-left vertex of the original figure are (-5, 5). After we reflect it across the \begin{align*}y-\end{align*}axis, the coordinates for the corresponding vertex are (-5, -5). How about the lower-right vertex? It starts out at (-1, 1), and after the flip it is at (-1, -1). As you can see, the \begin{align*}x-\end{align*}coordinates stay the same while the \begin{align*}y-\end{align*}coordinates change. In fact, the \begin{align*}y-\end{align*}coordinates all become the opposite integers of the original \begin{align*}y-\end{align*}coordinates. This indicates that this is a vertical (up/down) reflection or we could say a reflection over the \begin{align*}x-\end{align*}axis.
In a horizontal (left/right) reflection or a reflection over the \begin{align*}y-\end{align*}axis, the \begin{align*}x-\end{align*}coordinates would become integer opposites. Let’s see how.
This is a reflection across the \begin{align*}x-\end{align*}axis. Compare the points. Notice that the \begin{align*}y-\end{align*}coordinates stay the same. The \begin{align*}x-\end{align*}coordinates become the integer opposites of the original \begin{align*}x-\end{align*}coordinates. Look at the top point of the triangle, for example. The coordinates of the original point are (-4, 6), and the coordinates of the new point are (4, 6). The \begin{align*}x-\end{align*}coordinate has switched from -4 to 4.
We can recognize reflections by these changes to the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates. If we reflect across the \begin{align*}x-\end{align*}axis, the \begin{align*}x-\end{align*}coordinates will become opposite. If we reflect across the \begin{align*}y-\end{align*}axis, the \begin{align*}y-\end{align*}coordinates will become opposite.
We can also use this information to graph reflections. To graph a reflection, we need to decide whether the reflection will be across the \begin{align*}x-\end{align*}axis or the \begin{align*}y-\end{align*}axis, and then change either the \begin{align*}x-\end{align*} or \begin{align*}y-\end{align*}coordinates. Let’s give it a try.
Example
Draw a reflection of the figure below across the \begin{align*}x-\end{align*}axis.
We need to reflect the rectangle across the \begin{align*}x-\end{align*}axis, so the “flip” will move the rectangle down. Because the reflection is across the \begin{align*}x-\end{align*}axis, we’ll need to change the "y" coordinates (which determine where points are up and down). Specifically, we need to change them to their integer opposites. An integer is the same number with the opposite sign. This gives us the new points.
\begin{align*}& (3, 6) \qquad \quad (5, 6) \qquad \quad (3, 1) \qquad \ \ (5, 1)\\ & (3, -6) \qquad (5, -6) \qquad (3, -1) \qquad (5, -1)\end{align*}
Now we graph the new points. Remember to move right or left according to the \begin{align*}x-\end{align*}coordinate and up or down according to the \begin{align*}y-\end{align*}coordinate.
Here is the completed reflection. Let’s practice with coordinate notation.
8P. Lesson Exercises
Write each set of coordinates to show a reflection in the \begin{align*}y-\end{align*}axis.
1. (-3, 1) (0, 3) (1, 2)
2. (-3, 6) (-2, 3) (2, 3) (3, 6)
Take a few minutes to check your work.
IV. Rotate a Figure in the Coordinate Plane Using Coordinate Notation, and Graph the Resulting Image
Now let’s look at the third kind of transformation: rotations. A rotation is a transformation that turns the figure in either a clockwise or counterclockwise direction. The figure below has been rotated. What are its new coordinates?
The new coordinates of the rectangle’s vertices are (1, -3), (1, 2), (3, 2), and (3, -3). As you can see, both the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates changed. Unlike a translation or reflection, a rotation can change both of the coordinates in an ordered pair. Now look closely. One of the points remained exactly the same! We say that we rotated the figure about this point. Imagine you put your finger on this corner of the rectangle and spun it. That’s what happened in the rotation. The rectangle has been rotated \begin{align*}90^{\circ}\end{align*} clockwise.
How do we graph a rotation?
When we graph a rotation, we first need to know how much the figure will be rotated. Rotating the above rectangle \begin{align*}90^{\circ}\end{align*} stands it up on end. Rotating it \begin{align*}180^{\circ}\end{align*} would make it flat again. We also need to know which point we will rotate it around. This is the point that stays the same.
Next, we need to count how many units long and wide the figure is. The figure above stretches from 1 on the \begin{align*}x-\end{align*}axis to -4 on the \begin{align*}x-\end{align*}axis. This is a total of 5 units along the \begin{align*}x-\end{align*}axis. When we rotate a figure \begin{align*}90^{\circ}\end{align*}, the distance on the \begin{align*}x-\end{align*}axis becomes the distance on the \begin{align*}y-\end{align*}axis. Look at the rectangle. The long sides are horizontal at first, but after we rotate it, they become the vertical sides. This means that the \begin{align*}x-\end{align*}distance of 5 will become a \begin{align*}y-\end{align*}distance of 5.
Now, remember the point (1, -3) stays the same, so it is one corner of the rotated figure. We add 5 to the \begin{align*}y-\end{align*}coordinate to find the next vertex of the rectangle. \begin{align*}-3 + 5 = 2\end{align*}. This puts a vertex at (1, 2).
To find the other points of the rotated rectangle, we need to think about its width. Find the width, or short side, of the original rectangle by counting the units between vertices along the \begin{align*}y-\end{align*}axis. The rectangle covers 2 units on the \begin{align*}y-\end{align*}axis. As you might guess, this becomes the \begin{align*}x-\end{align*}distance in the rotated figure. In other words, we add 2 to the \begin{align*}x-\end{align*}coordinate of the point that stays the same. \begin{align*}1 + 2 = 3\end{align*}, so another vertex of the rectangle will be (3, -3). To find the fourth and final vertex, add 2 to the \begin{align*}x-\end{align*}coordinate of the other ordered pair we know, (1, 2). This puts the last vertex at (3, 2).
Let’s go back to the problem in the introduction and use what we have learned to figure it out!
## Real – Life Example Completed
The King’s Chamber
Here is the original problem once again. Reread it before working on the drawing.
In one room of the museum was a King’s bedroom. The furniture in the room was large and wooden and old with great golden cloths. On the walls was a beautiful red, blue and gold pattern.
Jessica thought that the pattern was the most beautiful one that she had ever seen.
“I love this,” she said to Mrs. Gilman. “I want to draw it, but I’m not sure how.”
“Well, you could break it up into a coordinate grid, since the pattern repeats itself, and use what we have learned about transformations to draw it in.”
“How could I get started?” Jessica asked.
“Well, start by drawing the coordinate grid, then use these coordinates for one of the diamonds. See if you can figure it out from there.”
In Jessica’s notebook, Mrs. Gilman wrote down the following coordinates.
\begin{align*}(4, 1)\!\\ (5, 0)\!\\ (5, 2)\!\\ (6, 1)\end{align*}
Jessica began to draw it in. Then she got stuck.
Now we can draw this diamond in on a coordinate grid. It belongs in Quadrant one. Now we want to draw a diamond into each of the other three quadrants. We can draw this, but we can also use mathematics to figure out the coordinates for each of the other diamonds first.
The diamond in the second quadrant is reflected over the \begin{align*}y-\end{align*}axis. Therefore, the \begin{align*}x-\end{align*}coordinate is going to change and become negative in each of the four vertices of the diamond. Here are the coordinates.
\begin{align*}(-4,1)\!\\ (-5,0)\!\\ (-5, 2)\!\\ (-6, 1)\end{align*}
Next, we can reflect the original diamond in the first quadrant over the \begin{align*}x-\end{align*}axis into the fourth quadrant. Here the \begin{align*}y-\end{align*}coordinates will be negative.
\begin{align*}(4, -1)\!\\ (5, 0)\!\\ (5, -2)\!\\ (6, -1)\end{align*}
Finally we can reflect this diamond over the \begin{align*}y-\end{align*}axis into the third quadrant. Notice that here the \begin{align*}x\end{align*} and \begin{align*}y -\end{align*}coordinates will both be negative.
\begin{align*} (-4, -1)\!\\ (-5, 0)\!\\ (-5, -2)\!\\ (-6, -1)\end{align*}
Did you notice any patterns? Take a minute and create this pattern of diamonds in a coordinate grid. Then you will have an even deeper understanding of how a pattern like this one is created.
If you wanted to add in the gold \begin{align*}X\end{align*} that crosses through the original pattern could you do it? Explain your thinking with a friend and then add in the \begin{align*}X\end{align*} to the coordinate grid with the diamonds.
## Vocabulary
Here are the vocabulary words found in this lesson.
Transformation
a figure that is moved in the coordinate grid is called a transformation.
Coordinate Plane
a representation of a two-dimensional plane using an \begin{align*}x -\end{align*} axis and a \begin{align*}y -\end{align*} axis.
\begin{align*}x-\end{align*}axis
the horizontal line in a coordinate plane.
\begin{align*}y-\end{align*}axis
the vertical line in a coordinate plane.
Translation
a slide. A figure is moved up, down, left or right.
Reflection
a flip. A figure can be flipped over the \begin{align*}x-\end{align*}axis or the \begin{align*}y-\end{align*}axis.
Rotation
a turn. A figure can be turned clockwise or counterclockwise.
Coordinate Notation
notation that shows where the figure is located in the coordinate plane. The vertices of the figure are represented using ordered pairs.
## Time to Practice
Directions: Identify the transformations shown below as a translation, reflection, or rotation.
9. True or false. This figure has been translated 5 places to the right.
10. True or false. This is a picture of a reflection.
11. True or false. The figure below is an image of a reflection.
12. True or false. This figure has been rotated \begin{align*}180^{\circ}\end{align*}.
Directions: Independent work. Have the students work to draw five different figures in the coordinate plane. Then with a partner have them switch papers and ask them to create a rotation, translation and a reflection of each.
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Subsections
# Graphs in Polar Coordinates
## Purpose
The purpose of this lab is to help you become familiar with graphs in polar coordinates.
## Getting Started
To assist you, there is a worksheet associated with this lab that contains examples and even solutions to some of the exercises. You can copy that worksheet to your home directory with the following command, which must be run in a terminal window, not in Maple.
cp /math/calclab/MA1023/Polar_start.mws My_Documents
You can copy the worksheet now, but you should read through the lab before you load it into Maple. Once you have read to the exercises, start up Maple, load the worksheet Polar_start.mws, and go through it carefully. Then you can start working on the exercises.
## Background
There are many times in math, science, and engineering that coordinate systems other than the familiar one of Cartesian coordinates are convenient. In this lab, we consider one of the most common and useful such systems, that of polar coordinates.
The main reason for using polar coordinates is that they can be used to simply describe regions in the plane that would be very difficult to describe using Cartesian coordinates. For example, graphing the circle in Cartesian coordinates requires two functions - one for the upper half and one for the lower half. In polar coordinates, the same circle has the very simple representation .
### Cardioids, Limaçons, and Roses
These are three types of well-known graphs in polar coordinates. The table below will allow you to identify the graphs in the exercises.
Name Equation cardioid or limaçon or rose or
### Intersections of Curves in Polar Coordinates
Finding where two graphs in Cartesian coordinates intersect is straightforward. You just set the two functions equal and solve for the values of . In polar coordinates, the situation is more difficult. Most of the difficulties are due to the following considerations.
1. A point in the plane can have more than one representation in polar coordinates. For example, , is the same point as , . In general a point in the plane can have an infinite number of representations in polar coordinates, just by adding multiples of to . Even if you restrict a point in the plane can have several different representations.
2. The origin is determined by . The angle can have any value.
These considerations can make finding the intersections of two graphs in polar coordinates a difficult task. As the exercises demonstrate, it usually requires a combination of plots and solving equations to find all of the intersections.
## Exercises
1. For each of the following polar equations, plot the graph in polar coordinates using the plot command and identify the graph as a cardioid, limaçon, or rose.
2. Find all points of intersection for each pair of curves in polar coordinates.
1. and for .
2. and for .
3. and for .
3. Find the domain of values that are necessary to trace the inner loop of the limacon . Using these values, plot the inner loop in polar coordinates. Then find the area of the inner loop. |
Succeed with maths – Part 1
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# 2 Equivalent fractions
If you multiply the numerator (the number on top) and the denominator (the number underneath) of any fraction by the same number (except zero), you will get a fraction that is equivalent to the original one, as Figure 5 shows:
Figure 5 Equivalent fractions by multiplying the numerator and denominator
Note that you must multiply the numerator and denominator by the same number. This is actually the same as multiplying by one, so it doesn’t change the value of the fraction. You can also generate equivalent fractions by dividing the numerator (top number) and the denominator (bottom number) of the fraction by the same number (again, not zero). This is the method that you will be employing frequently since fractions should usually be shown in their simplest form.
A fraction in its simplest form is one where you can no longer find a number (except one) to divide into both the numerator and the denominator to give you whole number answers.
This process of dividing the numerator (the number above the line) and the denominator (the number below the line) by the same number is known as cancelling or simplifying the fraction – hence ultimately a fraction in its simplest form.
Figure 6 shows an example of simplifying a fraction, using division. Before you look at the method used, think about how you would go about simplifying . Look for numbers that you can divide exactly into the top and bottom of the fraction.
Figure 6 Equivalent fractions by dividing the numerator and the denominator
It is important to realise that there is often more than one way to simplify a fraction, so it is fine if you did use different steps here. For example, you could have started by dividing by 20, 5 or 2. The most efficient way in terms of the number of steps would be to start with 20 because you would need to use only one step.
Try these different starting steps for yourself now to see how else this could have been approached; just remember to divide the top and the bottom of the fraction by the same number. If there is no whole number apart from 1 that can be divided into both the numerator and the denominator, the fraction is said to be in its simplest form.
The next section shows a way of visualing this process. |
# Find the value of x, if $\begin {bmatrix} 3x+y & -y \\ 2y-x & 3 \end{bmatrix} = \begin {bmatrix} 1 & 2 \\ -5 & 3 \end{bmatrix}$
$\begin{array}{1 1} x = 1\; y = -1 \\ x = 1\; y = -2 \\ x = 1\; y = 2 \\ x = 0\; y = -2 \end{array}$
Toolbox:
• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
• We can then match the corresponding elements and solve the resulting equations to find the values of the unknown variables.
Step1:
Given
$\begin {bmatrix} 3x+y & -y \\ 2y-x & 3 \end{bmatrix} = \begin {bmatrix} 1 & 2 \\ -5 & 3 \end{bmatrix}$
The given two matrices are equal,hence their corresponding elements should be equal.
$\Rightarrow -y=2$
y=-2.
Step2:
3x+y=1-----(1)
Substitute the value of y in equation (1),we get
3x-2=1
3x=3
x=1 |
Intuition For The Golden Ratio
Summary: The Golden Ratio is special because it perfectly balances addition and multiplication.
The Golden Ratio (1.618...) is often presented with an air of mysticism as "the perfect proportion". Setting aside whether we can find the Golden Ratio in the leaves of a nearby houseplant, what makes it special from a math perspective?
Well, let's try to make a pattern that's as balanced (symmetric) as possible.
A quick guess is something like 1, 1, 1, 1, 1. Every item is identical, but it's not very interesting -- it's a song where every word is the same. Does it rhyme? Yeah. Do I care? Not really.
Ok, let's try 1, 2, 3, 4, 5. There's a symmetry in the relationship that every element is one more than the previous. But if we skip around to random elements, there's no real connection: what do 3, 8, and 17 have in common?
Ok, fine. What about 1, 2, 4, 8, 16? Each element is twice the previous, and all the numbers are clean powers of two. But... what about addition? Can we create new elements in our pattern from the previous ones?
• 1 + 2 = 3 [not quite 4...]
• 1 + 2 + 4 = 7 [not quite 8...]
• 1 + 2 + 4 + 8 = 15 [not quite 16...]
Argh. We can concoct a rule for addition ("Every element is the sum of all previous elements... plus an extra 1"), but that's not clean and you know it. Let's figure this out.
Our goal is a pattern that's connected with both addition and multiplication. No varying number of additions, no cute "+1" on the end. Just a clean, simple relationship.
First off, we need to make every item connected by multiplication. We need powers:
1 x x^2 x^3 x^4
Whatever number the pattern uses (x), everything will be a power of it (just like 1, 2, 4, 8, although that sequence wasn't symmetric enough).
Next, we need an "addition symmetry" that connects the items. Every element should be buildable from the previous ones without extra rules. Throwing a few ideas against the wall:
• Can we have 1 + x = x? If we subtract x from both sides, that leads to 1 = 0. Uh oh, we're breaking math.
• How about 1 + 1 = x? Maybe we can allow multiple copies of everything. But that's just another way of getting x=2, you sneak.
• How about 1 + x = 1? Ack, that just implies x = 0. 0, 0, 0, 0 is symmetric but not interesting.
Hrm. The next addition pattern that might make sense is 1 + x = x^2. If we follow this idea, here's what happens:
A few notes:
• Starting from a single connection 1 + x = x^2, we can multiply through by x and get x + x^2 = x^3. Every element is made from the previous two.
• $x^0$ is another way to write 1. Makes the pattern easier to see, I think.
• Everything in the pattern grid is just our original pattern, shifted a bit. Neat.
Ok. That's the goal for the pattern, let's try to solve it.
$\displaystyle{(1 + x) = x^2 }$
$\displaystyle{x^2 - x - 1 = 0}$
We can find the "x" that makes this relationship true using the quadratic formula:
$\displaystyle{ x=\frac{-b + \sqrt {b^2-4ac}}{2a}\quad\text{and}\quad x=\frac{-b - \sqrt {b^2-4ac}}{2a} }$
Plugging in a=1, b=-1, and c=-1, we get:
$\displaystyle{ x=\frac{1 + \sqrt {1+4}}{2}\quad\text{and}\quad x=\frac{1 - \sqrt {1+4}}{2} }$
We only want a positive solution (the new part can't be negative), so we have
$\displaystyle{ x=\frac{1 + \sqrt {1+4}}{2}\quad\text = \frac{1 + \sqrt {5}}{2} = 1.618... = \phi }$
We label the solution Phi ($\phi$).
You can see it happening below. There are slight differences as the decimals go on -- computers have fixed precision.
Visualize the Intuition
Many descriptions of the Golden Ratio describe splitting a whole into parts, each of which is in the golden ratio:
I prefer the "growth factor" scenario, where we start with a single item (1.0) and evolve it, while keeping it linked to its ancestors with both arithmetic and multiplication. Just describing a ratio doesn't call out the symmetry we're able to achieve.
Visually, I see a growing blob, like this:
As we scale by Phi each time (1.618) we get:
• Addition symmetry: Each element is the sum of the previous two
• Multiplication symmetry: Each element is a scaled version of the previous
• Growth symmetry: Addition and multiplication change the pattern identically
Aha! That's a nice combo if I ever saw one. The "growing blob" can represent the length of a line, a 2d shape, an angle -- which can lead to interesting patterns:
The key relationship is we move from one blob to the next, such that multiplication and addition have the same effect:
We've described the Golden Ratio with different phrasing: the scaling factor (f) must equal the original (1) plus the previous item (1/f). We divide by our growth factor to find the previous element given the current one.
Solving $f = 1 + \frac{1}{f}$ yields the Golden Ratio as before.
Is the Golden Ratio Everywhere?
The Golden Ratio tends to be oversold in its occurrences. While it may appear occasionally in nature, buildings, and portraits, if you draw lines thick enough many things have a ratio of about 1.5 to 1.
I think the deeper intuition comes from realizing we've made addition and multiplication symmetric.
The Fibonacci sequence is built from having every piece built from the two before:
Using a certain formula, we can jump to a Fibonacci number by repeated scaling (exponents) instead of laboriously adding the parts. And maybe we'd come to expect the Golden Ratio here, since it's the scaling factor that allows two parts to add to the next item in the sequence. (The specifics of dividing by $\sqrt 5$ are because we need to adjust based on the starting pieces.)
Rather than hunting for examples of the Golden Ratio in the produce aisle, let's soak in the beauty of balancing the forces of multiplication and addition.
Happy math. |
Question
1. # If Sine Of X Equals 1 Over 2, What Is Cos(X) And Tan(X)? Explain Your Steps In Complete Sentences.
The Trigonometric Functions of Sine, Cosine, and Tangent (also known as Sin, Cos, and Tan) are essential tools for any math student’s toolbox. In this article, we will explore a step-by-step process of solving a trigonometry problem involving the Sine function. We will be specifically looking at the equation “sin(x) = 1/2” and determining what the values of cos(x) and tan(x) would be. By working through this example, you can get a better understanding of how to use trigonometric functions in your own work.
## The Sine of X
The sine of x is defined as the ratio of the side opposite to the angle x in a right angled triangle to the hypotenuse of the triangle. Sine of x is represented by sin(x).
If we take a look at the unit circle, we can see that the sine of x is equal to the y-coordinate of the point on the unit circle where x intersects it. This means that if we were to draw a line from (0,0) to (x,y), where y is equal to sin(x), then the angle between this line and the positive x-axis would be equal to x.
We can also see from the unit circle that cos(x) is equal to the x-coordinate of the point on the unit circle where x intersects it. This means that if we draw a line from (0,0) to (x,y), where y is equal to cos(x), then the angle between this line and the positive x-axis would be equal to x.
Similarly, we can see that tan(x) is equal to sin(x)/cos(x). This means that if we draw a line from (0,0) to (x,y), where y is equal to tan(x), then the angle between this line and the positive x-axis would be equal to x.
## Cos(X) and Tan(X)
The cosine of x is equal to the sine of x divided by the cosine of x. The tangent of x is equal to the sine of x divided by the cosine of x.
## How to find Cos(X) and Tan(X)
To find cos(x), we can use the Pythagorean theorem. First, we draw a right triangle. Then, we label the sides of the triangle. The hypotenuse is the side opposite the 90 degree angle, and is always labeled “c.” The side adjacent to angle x is labeled “a,” and the remaining side is labeled “b.” Now, we can use the Pythagorean theorem to solve for “a.” We know that a^2 + b^2 = c^2, so we can rearrange this equation to get a = sqrt(c^2 – b^2). Now that we know the value of “a,” we can use it to find cos(x). Cos(x) = a/c. To find tan(x), we use a similar method. First, we draw a right triangle and label the sides. Then, we use the Pythagorean theorem to solve for “b.” We know that b^2 = c^2 – a^2, so b = sqrt(c^2 – a^2). Now that we know the value of “b,” we can use it to find tan(x). Tan(x) = b/a.
## Conclusion
In conclusion, when given a sine value of 1/2, one can calculate the cosine and tangent value by using the Pythagorean identity along with some basic trigonometric functions. By beginning with the definition of sine as “opposite side over hypotenuse”, we can then use the reciprocal identities to solve for cos and tan in terms of sin. This method provides an efficient way to attain both values quickly and easily.
2. In mathematics, trigonometry is the study of relationships between the sides and angles of triangles. Specifically, the sine, cosine, and tangent functions are used to relate a given angle in a triangle to its sides. If we know one side or angle of a triangle, it’s possible to calculate the other two using these three functions.
For example, if we know that sin(x) = 1/2 then by using basic trigonometric identities we can find out that cos(x) =√3/2 and tan(x) = √3/3. To find this solution, we first use Pythagoras’ theorem to calculate the length of one side: x^2 + (1/2)^2 = 1.
3. Are you ready to explore the world of trigonometry and find out what happens when the sine of an angle is equal to one-half? Let’s dive in!
When the sine of an angle x is equal to one-half, the cosine and tangent of x can be calculated. In order to find out cos(x) and tan(x), we first need to understand the relationship between sine and cosine.
The sine and cosine of an angle x are related by the equation:
sin(x) = cos(90° – x)
This means that if we know the sine of an angle, we can use this equation to calculate the cosine of the angle.
In this case, the sine of x is equal to one-half, so the cosine of x can be calculated by plugging this value into the equation above:
cos(x) = sin(90° – x)
cos(x) = sin(90° – (1/2))
cos(x) = sin(89.5°)
Using a scientific calculator, we can calculate the sine of 89.5° which gives us a result of 0.987688. Therefore, if the sine of an angle x is equal to one-half, the cosine of x is equal to 0.987688.
Now that we know the cosine of x, we can calculate the tangent of x. The tangent of an angle x is related to the sine and cosine of x by the equation:
tan(x) = sin(x) / cos(x)
Using the values we just calculated, we can plug these into the equation above:
tan(x) = (1/2) / 0.987688
tan(x) = 0.506446
Therefore, if the sine of an angle x is equal to one-half, the cosine of x is equal to 0.987688 and the tangent of x is equal to 0.506446.
We hope this explanation was helpful and that you now understand how to calculate the cosine and tangent of an angle given the sine of the angle.
4. Have you ever been confused by a trigonometry question involving sine, cosine and tangent? If so, you’ve come to the right place!
In this blog post, we’ll answer the question, “If sine of x equals 1/2, what is cos(x) and tan(x)?” and explain our steps in complete sentences.
First, let’s begin by reviewing the definitions of sine, cosine and tangent. Sine is the ratio of the opposite side to the hypotenuse of a right triangle. Cosine is the ratio of the adjacent side to the hypotenuse of a right triangle. Tangent is the ratio of the opposite side to the adjacent side of a right triangle.
Now we can answer our original question. If the sine of x is 1/2, we can use the Pythagorean theorem to find the hypotenuse of the right triangle. The hypotenuse is equal to the square root of 2. Therefore, since the sine of x is 1/2, the opposite side is equal to 1.
Since we now know the opposite side and the hypotenuse, we can calculate the cosine of x: cos(x) = adjacent side/hypotenuse. Therefore, cos(x) = 1/√2.
We can also calculate the tangent of x: tan(x) = opposite side/adjacent side. Therefore, tan(x) = 1/1 = 1.
Therefore, if the sine of x equals 1/2, then the cosine of x equals 1/√2 and the tangent of x equals 1.
We hope this blog post has helped you understand the answer to the question, “If sine of x equals 1/2, what is cos(x) and tan(x)?” |
Events and the sample space
Target: On completion of this worksheet you should be able to find the probability of simple events and mutually exclusive events. You should also be able to find the probability of independent events and the probability of a combination of events using ‘and’ and ‘or’.
# 1 Introduction
If we toss an ordinary coin the chance of getting a head is 1 out of 2. We say that the probability of getting a head is . This is because there is only one way of getting a head but two possible outcomes. Each possible outcome of an experiment (in this case either a head or a tail) is called an event. The set of all possible outcomes (in this case the set including heads and tails) is called the sample space.
Similarly the probability of getting a tail is also .
If we consider a six sided dice. Each number has an equal chance of appearing when the dice is thrown.
So the probability of throwing a six is .
We write P( throwing a 6 ) = .
In general, P(event)
So, P(even number) .
If we toss a biased coin we do not know what the probability of getting a head is. The only way we can find out is to experiment by tossing the coin many times. The more tosses we do the more accurate will be the estimate of the probability.
If an event cannot occur then the probability is 0.
If an event is certain to occur then the probability is 1.
## 1.1 Examples
A card is picked from a well shuffled pack of 52 playing cards. Find the probability that the card is:
1. A king
2. A spade
Solutions:
1. There are 4 kings in the pack of cards. The probability of a king is therefore .
2. There are 13 spades in the pack of cards. The probability of a spade is therefore
## 1.2 Examples
1. A biased dice is thrown 100 times and a 6 appears 40 times.
a) What is the probability of getting a 6
b) What is the probability of not getting a 6?
Solutions:
a) P(6), the probability of getting a 6 is .
b) P(not a 6), the probability of not getting a 6 is .
2. A manufacturer produces bolts of which 2 out of 100 are defective. What is the probability of a bolt not being defective?
P(defective bolt) = .
P(bolt not defective) = 1 – 0.02 = 0.98.
Mutually exclusive events
# 2 Mutually exclusive events
If a dice is thrown then two possible events are ‘getting a 5’ and ‘getting a 6’. If the dice is thrown once then we cannot have both of these events occurring . We say that these events are mutually exclusive. If we consider two different events ‘getting a 3’ and ‘getting an odd number’ then these are not mutually exclusive because if we throw a 3 then both events have occurred.
## 2.1 Examples
1. A dice is thrown. Find the probability of getting a 1 or 6.
Solution:
Since a 1 and a 6 cannot be thrown simultaneously, they are mutually exclusive events. Consequently
P(1 or 6) = P(1) + P(6) .
2. Two students, Ali and Ben, take part in a race. The probability that Ali wins the race is and the probability that Ben wins is . What is the probability that either Ali or Ben wins the race? ( Assume a deadheat is not possible ).
The two events ‘Ali wins the race’ and ‘Ben wins the race’ are mutually exclusive. So P(Ali or Ben wins) = P(Ali wins) + P(Ben wins) = .
# 3 Independent events
Suppose a coin is tossed and a dice is thrown. The events ‘get a head’ and ‘get a 4’ are independent events as one event does not affect the other. In this case
P(head and 4) P(head) P(4) .
In general if A and B are two independent events then P(A and B) P(A) P(B).
## 3.1 Examples
1. If a red dice and a blue dice are thrown what is the probability of getting a 2 on the red dice and a 3 on the blue dice?
Solution: These events are independent so P(2 on red and 3 on blue)
P(2 on red) P(3 on blue) .
2. A coin is tossed twice. Find the probability of getting 2 heads.
Solution: Let the probability of getting 2 heads be denoted by P(2 H)
P(H on 1st throw) P(H on 2nd throw) .
3. A manufacturer produces components and packs them in boxes of 20. In one particular box it is known that there are 3 defective items. If two components are picked from this box what is the probability that both are defective?
Solution:
P(1st defective) .
Now, having picked one item, there are only 19 left in the box and 2 defectives. So
P(2nd defective).
P(both defective) .
4. A bag contains 5 red sweets and 7 blue sweets. Find the probability that if two sweets are picked they are not both red.
P(2 red).
P(not both red) .
5. A company manufactures CDs of which 3% are faulty. What is the probability of getting exactly one faulty CD in a box of 3?
P(faulty CD) = 0.03 P( good CD)
P(faulty, good, good)
But the order could be good, faulty, good or good, good, faulty. In each of these the probability is the same as above. These are mutually exclusive events so we can add the probabilities ie multiply by 3
P(1 faulty CD in box of 3) to 3 sig fig.
6. A bag contains 5 red balls and 7 green balls. Four balls are picked out of the bag. What is the probability that two are red and two are green?
P(R,R,G,G) .
The denominator decreases by 1 each time a ball is taken out. We must find the probability for all the different ways of getting 2 red and 2 green and then add these together.
P(R,G,R,G) .
Notice that we have the same numerators but in a different order (the denominators are exactly the same). This will be the same for all cases so we must find how many ways there are of picking 2 red and 2 green balls:
RRGG, RGRG, RGGR, GGRR, GRGR, GRRG. There are 6 different ways so
P(2R &2G).
Notation: In the above we take account of what has already happened. We say ‘probability of a red given that a red has already been picked’ and write P(red | red already picked).
In general if A and B are two events then P(B|A) means probability of B occurring given that A has already occurred. Also
P(A and B) = P(A) P(B|A).
Suppose we draw 1 card from a well shuffled pack of 52 playing cards. Consider the 2 events ‘an ace is drawn’ and ‘a club is drawn’. We want to find
P(an ace or a club). We can list the possible cards for each event:
Event 1 Event 2 Ace of ♦, Ace of ♠,Ace of ♦, Ace of ♠ Ace of ♣, 2 ♣,3 ♣, 4 ♣, …,queen ♣, king ♣
The Ace of ♣ is in both columns so the events are not mutually exclusive. Now
P(ace) , P(♣) , P(ace ♣) .
P(an ace or a club)P(ace) P(♣) P(ace ♣)
Includes ace ♣ace ♣ Takes one of the ace ♣ away
P(ace or club).
In general given 2 events A and B
P(A or B) = P(A) + P(B) – P(A and B) |
Beginner
# 218. Find the right most set bit of a number
Objective: Given a number, write and algorithm to find the right most set bit in it (In binary representation).
Example:
```Number : 1
Binary representation: 1
Position of right most set bit: 1
Number : 6
Binary representation: 1 1 0
Position of right most set bit: 2
Number : 11
Binary representation: 1 0 1 1
Position of right most set bit: 1
Number : 24
Binary representation: 1 1 0 0 0
Position of right most set bit: 4
```
Approach:
If N is a number then the expression below will give the right most set bit.
`N & ~ (N -1)`
• Let’s dig little deep to see how this expression will work.
• We know that N & ~N = 0
• If we subtract 1 from the number, it will be subtracted from the right most set bit and that bit will be become 0.
• So if we negate the remaining number from step above then that bit will become 1.
• Now N & ~(N-1) will make all the bits 0 but the right most set bit of a number.
Example:
```Say N =10, so N = 1 0 1 0, then ~N = 0 1 1 0 => N & ~ N =0
N – 1 = 1 0 1 0 – 0 0 0 1 = 1 0 0 1
~(N-1) = 0 1 1 0
N & ~(N-1) = 0 0 1 0 => 2nd bit```
Output:
```Right most set bit for 1 is : 1
Position: 1.0
``` |
offers hundreds of practice questions and video explanations. Go there now.
# GMAT Math: Factors
“How many odd factors does 210 have?”
“If n is the smallest integer such that 432 times n is the square of an integer, what is the value of y?”
“How many prime numbers are factors of 33150?”
If questions like these make you cringe, I’d like to convince you that only a few easy-to-understand concepts stand between you and doing these flawlessly.
## Idea #1: Prime Numbers
This is probably review, but just for a refresher: a prime number is any positive integer that is divisible by only 1 and itself. In other words, a prime number has only two factors: itself and 1. Numbers that have more than two factors are called composite. (By mathematical convention, 1 is the only positive integer considered neither prime nor composite.) The first few prime numbers are:
2 3 5 7 11 13 17 19 23 29
In preparation for the GMAT, it would be good to be familiar with this list. If you verify for yourself why each number from 2 to 30 is prime or composite, it will help you remember this list.
Occasionally, the GMAT will expect you know whether a larger two-digit number, like 67, is prime. Of course, if the number is even, it’s not prime. If it ends in a digit of 5, it’s divisible by five. For divisibility by 3, a good trick to know: if the sum of the digits is divisible by three, then the number is divisible by three. Here 6 + 7 = 13, not divisible by three, so 67 is not divisible by three.
To see whether a number less than 100 is prime, all we have to do is see whether it is divisible by one of the single digit prime numbers: 2, 3, 5, or 7. We’ve already checked 2, 3, and 5. The number 67 is not divisible by 7: 7 goes evenly into 63 and 70, not 67. That’s enough checking to establish irrevocably that 67 is prime.
## Idea #2: Prime Factorization
Every positive integer greater than 1 can be written in a unique way as a product of prime numbers; this is called its prime factorization. The prime factorization is analogous to the DNA of the number, the unique blueprint by which to construct the number. In other words, when you calculate the prime factorization of a number, you have some powerful information at your disposal.
How does one calculate the prime factorization of a number? In grade school, you may remember making “factor trees”: that’s the idea. To find the prime factorization of, for example, 48, we simply choose any two factors — say 6 and 8 — and then choose factors of those number, and then of those numbers, until we are left with nothing but primes.
Typically, once we are done, we sort the prime factors in numerical order:
Once we have the prime factorization, what can we do with it? See the next two items.
## Idea #3: The Number of Factors
Suppose the GMAT asks: how many factors does 1440 have? It would be quite tedious to count them all, but there’s a fast trick once you have the prime factorization. First of all, the prime factorization of 1440 is
Each prime factor has an exponent (the exponent of 5 is 1).
To find the total number of factors:
a) Find the list of exponents in the prime factorization — here {5, 2, 1}
b) Add one to each number on the list — here {6, 3, 2}
c) Multiply those together —
The number 1440 has thirty-six factors, including 1 and itself.
Suppose the GMAT asked the number of odd factors of 1440. We know that odd factors cannot contain any factor of 2 at all, so basically we repeat that procedure with all the factors except the factors of 2. Here: {2, 1} –> {3, 2} –> . The number 1440 has 6 odd factors, including 1. Just for verification, the odd factors of 1440 are
{1, 3, 5, 9, 15, and 45}
This also means it has 36 – 6 = 30 even factors.
## Idea #4: GCF and LCM
GCF = greatest common factor
LCM = least common multiple.
(Note: LCM and LCD are the same thing: a least common denominator, LCD, of two number is always their LCM.)
Suppose a GMAT Math question involves finding, say, the LCM (or LCD) of 30 and 48. There’s a very straightforward procedure to find the LCM.
• Find the prime factorizations of the two numbers: 30 = 2*3*5 and 48 = 2*2*2*2*3
• Find the factors they have in common – the product of these is the GCF. Here, the GCF = 2*3 = 6
• Express each number as the GCF*(other stuff): 30 = 6*5 and 48 = 6*8
• The LCM = GCF*(other stuff from first number)*(other stuff from the second number): LCM = 6*5*8 = 240
Let’s do one more, just for practice. Suppose, on a GMAT math problem, we need to find the LCM/LCD of 28 and 180
Step (a): 28 = 2*2*7, 180 = 2*2*3*3*5
Step (b): 28 = 2*2*7, 180 = 2*2*3*3*5; GCF = 2*2 = 4
Step (c) 28 = 4*7, 180 = 4*45
Step (d) LCM = 4*7*45 = 1260
## Practice Questions:
1) The number of boxes in a warehouse can be divided evenly into 6 equal shipments by boat or 27 equal shipments by truck. What is the smallest number of boxes that could be in the warehouse?
(A) 27
(B) 33
(C) 54
(D) 81
(E) 162
2) How many odd factors does 210 have?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
3) If n is the smallest integer such that 432 times n is the square of an integer, what is the value of n?
(A) 2
(B) 3
(C) 6
(D) 12
(E) 24
4) How many distinct prime numbers are factors of 33150?
(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight
5) If n is a positive integer, then n(n + 1)(n – 1) is
(A) even only when n is even
(B) odd only when n is even
(C) odd only when n is odd
(D) always divisible by 3
(E) always one less than a prime number
1) C
2) E
3) B
4) B
5) D
## Explanations
1) This tells us that the number of boxes is evenly divisible by both 6 and 27; in other words, it’s a common multiple of 6 and 27. The question says: what’s the smallest value it could have? In other words, what’s the LCM of 6 and 27? (This question is one example of a real-world set-up where the question is actually asking for the LCM.)
Step (a): 6 = 2*3 27 = 3*3*3
Step (b): 6 = 2*3 27 = 3*3*3 GCF = 3
Step (c): 6 = 3*2 27 = 3*9
Step (d) LCM = 3*2*9 = 54
Thus, 54 is the LCM of 6 and 27.
For odd factors, we put aside the factor of two, and look at the other prime factors.
set of exponents = {1, 1, 1}
plus 1 to each = {2, 2, 2}
product = 2*2*2 = 8
Therefore, there are 8 odd factors of 210. In case you are curious, they are {1, 3, 5, 7, 15, 21, 35, and 105}
3) The prime factorization of a square has to have even powers of all its prime factors. If the original number has a factor, say of 7, then when it’s squared, the square will have a factor of 7^2. Another way to say that is: any positive integer all of whose prime factors have even powers must be a perfect square of some other integer. Look at the prime factorization of 432
432 = (2^4)*(3^3)
The factor of 2 already has an even power —- that’s all set. The factor of 3 currently has an odd power. If n = 3, then 432*n would have an even power of 2 and an even power of 3; therefore, it would be a perfect square. Thus, n = 3 is a choice that makes 432*n a perfect square.
33150 = 50*663 = (2*5*5)*3*221 = (2)*(3)*(5^2)*(13)*(17)
There are five distinct prime factors, {2, 3, 5, 13, and 17}
5) Notice that (n – 1) and n and (n + 1) are three consecutive integers. This question is about the product of three consecutive integers.
If n is even, then this product will be (odd)*(even)*(odd) = even
If n is odd, this this product will be (even)*(odd)*(even) = even
No matter what, the product is even. Therefore, answers (A) & (B) & (C) are all out.
Let’s look at a couple examples, to get a feel for this
3*4*5 = 60
4*5*6 = 120
5*6*7 = 210
6*7*8 = 336
7*8*9 = 504
Notice that one of the three numbers always has to be a multiple of 3: when you take any three consecutive integers, one of them is always a multiple of 3. Therefore the product will always be divisible by 3.
BTW, for answer choice E of that question, you will notice that for some trios of positives integers, adding one to the product does result in a prime, but for others, it doesn’t.
3*4*5 + 1 = 61 = prime
4*5*6 + 1 = 121 = 11^2 (not prime)
5*6*7 + 1 = 211 = prime
6*7*8 + 1 = 337 = prime
7*8*9 + 1 = 505 = 5*101 (not prime)
This is a mathematical idea far far more advanced than anything on the GMAT, but it is mathematically impossible to create an easy rule or formula that will always result in prime numbers. The prime numbers follow an astonishingly complicated pattern, which is the subject of the single hardest unanswered question in modern mathematics: the Riemann Hypothesis. Fascinating stuff for leisure reading, but absolutely 100% not needed for the GMAT :).
By the way, sign up for our 1 Week Free Trial to try out Magoosh GMAT Prep!
### 53 Responses to GMAT Math: Factors
1. Carmelite April 18, 2016 at 7:51 am #
What am I?
I am a square number that is less than 100. I am also a multiple of 4 and can be evenly divided by 3.
• Magoosh Test Prep Expert April 18, 2016 at 9:06 am #
Hi Carmelite,
The number fitting that description would be 36. 🙂
2. Jenny July 2, 2015 at 8:04 pm #
Hi Mike, this post was highly informative. Thanks! I just had a question about the first question. In your trick above for finding the LCM, you multiply the common factors to get the GCF. Although it is obvious that 3 is the GCF for 6 and 27, what happened to multiplying the common factors. If the common factors were multiplied, the GCF would have been 9. Is there an exception to this when the numbers have the same factors, in this case 3, where you don’t need to multiply? Thanks!
3. Karin April 8, 2015 at 8:47 am #
Hi Mike,
I have a question regarding the solution for the second question mentioned above. I understand the approach how we get to these 8 odd factors of the number 210, which is the main task of the question. However, looking at these 8 factors {1, 3, 5, 7, 15, 21, 35, and 105} I can’t properly follow why we have e.g. 35 as a factor of 210 but not 70 since 210 is divisible by 70 as well. May you explain me this situation?
Karin
• Mike April 9, 2015 at 11:52 am #
Karin,
I’m happy to respond! 🙂 My friend, remember that math is having an eye for details. That question did NOT ask the number of factors of 210. That question asked the number of ODD factors of 210. The eight factors listed there are all odd numbers. Certainly 70 and 2 and 10 and 210 are all factors of 210, but these are even numbers, and so don’t count in the list. Altogether, 210 has 16 factors, 8 odd and 8 even.
Does all this make sense?
Mike 🙂
• Vish May 12, 2016 at 5:06 am #
Hi Karin
As author asked for only odd factors hence we cannot consider 70 as one of the factor whereas 35 is an odd factor.
4. asu January 5, 2015 at 8:38 pm #
Hi Mike,
To find the number of the odd factors you said you can’t have a factor of two. For example we have 2^3, 5^3, 6^4. To find the number of odd factors, do we just exclude 2 or 6 too? Since 2 is a factor of 6. If that’s the case we basically eliminate all the even numbers right? Thanks!
• Mike January 6, 2015 at 1:14 pm #
Dear Asu,
What I said is that if you have a prime factorization, then you would eliminate only two, because two is the only even prime number. A factor of 6 would not appear in a prime factorization, because 6 is not a prime number.
Does this make sense?
Mike 🙂
5. Tanya November 24, 2014 at 5:24 pm #
Can you please give a definition of a distinct prime number? more specifically why is 1 not one of them?
Thank you!
• Mike November 25, 2014 at 10:33 am #
Dear Tanya,
I’m happy to respond! 🙂 The word “distinct” simply means “different.” For example, 12 = 2*2*3 — it has three prime factors but two of them are the same number, so it has just two distinct prime factors. Similarly, 900 = 2*2*3*3*5*5 has six prime factors in total, but many repeats among them: it has only three distinct prime factors. Does this make sense?
As for why 1 is not prime number, see this post:
https://magoosh.com/gmat/2012/gmat-math-one-is-not-a-prime-number/
Mike 🙂
6. Devyani November 20, 2014 at 9:18 am #
What will be the difference in the number of factors of N^2013 and N^2015 ?
• Mike November 20, 2014 at 10:19 am #
Dear Devyani,
Normally, we don’t answer outside questions in blog comments, but I’ll address this, because it’s quick. My friend, the question as you pose it is not well-defined. The number of factors would depend very much on N. If N is a prime number, then N^2015 has only two more factors — N^2014 and N^2015. By contrast, if N were a number loaded with factors, then the number of new factors in N^2015 would increase considerably. Even if N has just two prime factors (e.g. N = 6 or N = 15), the number of new factors would be in the 1000s. The question really doesn’t make sense without specifying N.
Does all this make sense?
Mike 🙂
7. Rachael August 15, 2014 at 7:34 am #
Hello,
Thanks a lot for your very helpful explanations. This is pertaining to a question on the GMAT review number 13 on the diagnostic test. Although I have read the topic on remainder and tried the practice questions therein I still was unable to apply the format used to solve this particular question: if s and t are positive integers such that s/t = 64.12, which of the following could be the remainder when s is divided by t?
Please can you help me with this.
Regards,
Rachael
• Mike August 15, 2014 at 10:36 am #
Dear Rachael,
Normally, we do not answer outside questions on the blog, even official questions, but your question is so directly pertinent to this blog article that I decided to answer it here. 🙂
Let R be the remainder. R/t = 0.12 = 12/100 = 3/25. So, it could be that R = 3 and t = 25, or they both could be multiples of those numbers. (3k)/(25k). R would absolutely have to be a multiple of 3. The only multiple of 3 is (E) 45.
Does this make sense?
Mike 🙂
8. Karan August 6, 2014 at 3:12 pm #
Hey Mike,
From Practice Question No. 5’s explanation, is it safe to assume that for a set of N consecutive integers, their product will ALWAYS be divisible by N?
Thanks! 🙂
• Mike August 6, 2014 at 3:21 pm #
Dear Karan,
Yes, you are perfectly correct. Even if the set of N consecutive integers includes 0, the product will be zero, and zero is divisible by every integer. You are 100% correct.
Mike 🙂
9. Leszek July 13, 2014 at 11:11 am #
There’s a typo in Idea #3: The Number of Factors
the number should be 1440 and not 1400
Best
• Mike July 13, 2014 at 11:23 am #
Dear Leszek,
Thanks. I just fixed that! Best of luck to you, my friend.
Mike 🙂
10. Will July 9, 2014 at 1:05 pm #
Hey Mike!
Thanks for the post! It’s really helpful!
I have an issue with Question #5.
If n=1, the result would be 1*2*0 = 0, which is not divisible by 3.
Then there would ne no correct answer.
Maybe shoud mention n is positive integer >1?
Am I wrong?
Thanks a lot!
William
• Mike July 9, 2014 at 1:28 pm #
Dear William,
Zero is divisible by every positive integer, precisely because zero is a multiple of every positive integer.
Mike 🙂
11. sanjoy June 8, 2014 at 11:53 pm #
Hi. My problem with LCM and GCF (word problems) is that I don’t know when to use each.how can I tell if they are asking me to calculate the LCM or the GCF? any language tricks????!!!! plz clarify.Thanks.
• Mike June 9, 2014 at 11:54 am #
Dear Sanjoy,
I’m happy to respond. 🙂 It’s true, these two ideas can show up in many different ways. I guess the only think I would say is: if the problem is collecting smaller groups to make a larger group, and asks what’s the smallest larger group that would do X — chances are good that’s a LCM problem. If we are adjusting the size of the smaller group, trying to make it as large as possible to make up some larger group, that’s a GCF. Does this make sense?
Mike 🙂
12. vivian March 17, 2014 at 9:29 am #
Hi,
I have a question regarding idea #3, I am having hard time understanding the concept here when you said that ” each prime factor has an exponent”. I am not seeing where these numbers coming from. can you help me out here? thanks,
• Mike March 17, 2014 at 10:29 am #
Vivian,
I’m happy to help. 🙂
The number 432 is even, so let’s start by dividing it by 2.
432/2 = 216
Well, here, it would be helpful to have memorized that 6^3 = 216. I think it’s good to have memorized the perfect cubes of the first ten positive integers. But suppose you don’t have that memorized. Well, 216 is still even, so keep dividing by 2
216/2 = 108
108/2 = 54
54/2 = 27
That was four divisions by two, so the original number has four factors of two:
432 = (2^4)*27
Now, 27 is just 3^3 — that’s one you definitely have to have memorized! That gives us
432 = (2^4)(3^3)
All of those calculations you should be able to do without a calculator. Once we have it in this form, we can see the exponents, and see which ones are even or odd.
Does all this make sense?
Mike 🙂
13. Misty January 14, 2014 at 4:10 pm #
Mike – great blog post as always. Quick question – how come there is a good trick to finding the number of odd factors in an equation, but not even factors? In the example at the beginning of the post, there were 3×2=6 odd factors but not 6 even factors (5+1). Instead you had to find the total number of factors and subtract the number of odd factors to find even factors. Thank you!!
• Mike January 14, 2014 at 5:41 pm #
Misty,
It’s easy to count all the factors. It’s also easy to count all the odd factors, because we simply exclude the 2’s and count everything else. For even factors, we would have to include 2 (or powers of 2) and count everything that includes it: it just a more complicated procedure, especially if there are several powers of two involves. The subtraction trick is easy to explain and works 100% of the time.
Technically, if there are n powers of 2 in the prime factorization of the number, then 1/(n + 1) of the total factors will be odd, and n/(n + 1) of the total factors will be even. Now, would you rather remember that, or just the simple subtraction trick?
Does all this make sense?
Mike 🙂
• Misty January 15, 2014 at 4:43 pm #
Makes sense to me! I wouldn’t have realized it unless you pointed out the subtraction method at the beginning of the post, which I am glad you did. Perhaps it’s something worth highlighting in future posts/discussions on this topic. Thanks again Mike and keep up the great work!
• Mike January 15, 2014 at 4:47 pm #
Misty,
You are quite welcome. Best of luck to you, my friend.
Mike 🙂
14. Kumar January 9, 2014 at 12:13 am #
Is problem 5, what if value of n is 1. Then the whole equation turns out to be zero. The answer option D does not seem to be correct in all situations.
• Mike January 9, 2014 at 9:43 am #
Dear Kumar,
This is a little appreciated mathematical fact, but zero is a multiple of every positive integer, and hence, zero is divisible by every positive integer. When we divide 0 by 3, we get a quotient of 0 with no remainder — if we can divide one number into another with no remainder, that’s the definition of divisibility. Does all this make sense?
Mike 🙂
15. Allan July 22, 2013 at 6:06 pm #
Hello Mike. You state that 1 is not prime. You also state that all positive integers can be written as a product of prime numbers. You did not intend to include 1, in the latter statement, did you?
• Mike July 23, 2013 at 9:58 am #
Allan,
Good catch. That’s a niggling little exception that I forget to mention. I edited the text above to include this. The idea of prime factorizations and the Fundamental Theorem of Arithmetic apply to every positive integer *greater than 1*.
Mike 🙂
16. Allison June 13, 2013 at 2:50 pm #
Hi Mike. Could you please explain how you arrived at the answer for #3 in further detail? I do not understand what you mean when you say that you “have to add additional factors for anything with odd exponents.”
• Mike June 13, 2013 at 3:20 pm #
Allison,
I’m happy to explain this. 🙂
As I say in the solution to #3 above, every prime factor in a perfect square must have an *even* exponent. If *any* of the prime factors has an odd exponents, then automatically the number cannot possibly be a perfect square. That’s the really BIG idea.
Given that big idea, suppose we have a problem such as #3, — given some big number, say, N = 150, by what factor do we have to multiply this number, so that the product is a perfect square?
Well, first we find the prime factorization of 150 –> (2^1)*(3^1)*(5^2). That has some odd exponents and some even exponents. The even exponents — they’re fine. The odd exponents are the problem — that’s precisely what prevents this number from being a perfect square right now. If we want to change this number, 150, into a perfect square, we have to multiply it by something that changes those odd exponents into even exponents. That’s precisely what I mean when I say that we “have to include additional factors for anything with odd exponents.” Here, the factor of 5, with an even exponent, is already all set. It’s those factors of 2 & 3, each with an odd exponent, that pose a problem and need reconjiggering. To turn this into a perfect square, we would have to bump each one of those exponents of 1 up to 2, the next biggest even number. This means, we need to multiply the original number, 150, by one factor of 2 and by one factor of 3, that is, by a factor of 6 altogether. Sure enough, if we multiply 150 by 6, we get 900, which is a perfect square, 30^2.
Any number in which some of the exponents of prime factors are odd and some are even is thereby not a perfect square. If we want to multiply it by something to turn it into a perfect square, then for each factor with an odd exponent, we need to multiply once by that factor, to bump that exponent up from odd to even. In other words, we “have to include additional factors for anything with odd exponents.”
Does all this make sense?
Mike 🙂
• Allison June 13, 2013 at 4:05 pm #
Thanks so much Mike, that clears it up 🙂
• Mike June 13, 2013 at 4:31 pm #
Allison,
You are quite welcome, my friend. Best of luck to you!
Mike 🙂
• Aniket June 26, 2015 at 6:28 am #
Gr8 explanation!!
17. Nafiz May 30, 2013 at 2:48 am #
Hi Mike,
I wanted to ask a silly question. At question 4, when prime factoring 33,150, I got stuck when coming to 2 * 5 * 5 * 3 * 221.
221 can be factored into 13 * 17 which takes some work. Is there any quick way to go about these kind of numbers ???
• Mike May 30, 2013 at 10:13 am #
Naifz,
My friend, that is not at all a silly question. I will admit, I kinda cheated here — I picked a number for which I happen to know the factors, but which would probably be particularly difficult for most folks to factor on the fly without a calculator. In other words, I was being a little mean here — the GMAT will not be that mean to you.
BTW, in case you’re curious, here’s a slick factoring trick, if you happen to notice it. If you happen to know that 15^2 = 225, and notice that 221 is exactly 4 less than 225 — of course, 4 = 2^2 — then we can express 221 as a Difference of Two Squares.
a^2 – b^2 = (a + b)(a – b)
221 = 225 – 4 = 15^2 – 2^2 = (15 + 2)(15 – 2) = 17*13
I discuss these factoring techniques more in this post:
Again, all this is at the very outer edge of what you might have to understand if you get absolutely everything else correct on the quant section and the CAT is throwing the hardest possible questions at you. Most folks will not have to worry about this at all.
Does all this make sense?
Mike 🙂
• Nafiz May 30, 2013 at 11:15 pm #
Hi Mike,
Many thanx for your detailed explanation :). Do you think these kind of questions can come at GRE ? I was actually watching the magoosh video for counting divisors of large numbers, from there I came to your related post here.
And what is the difference between GRE and GMAT quant questions qualitatively ? I have heard GMAT math is more hard, hence, do you think practicing GMAT type questions will improve my quant score at GRE ?
Thanx again.
• Mike May 31, 2013 at 2:45 pm #
Dear Nafiz,
This is a notch harder than anything the GRE is likely to throw at you, except perhaps in a very very hard challenge problem if one happens to be included in your Quant section.
Yes, GMAT math is a tad harder than GRE math, and on the GMAT, you are not allowed to use a calculator, so it forces you to use all the mental math tricks and shortcuts which, if you master them, can save you a ton of time on the GRE. Many GRE students think the calculator to which they have access will help them on most problems, whereas actually, ETS designs many many problems to *punish* all who reflexively reach for the calculator rather than doing critical thinking first. Practicing GMAT math is a good way of avoiding all the “calculator traps.”
Does all this make sense?
Mike 🙂
• Subh August 25, 2013 at 8:56 am #
I took more than 15 secs to realize that 221 is composite. But the derivation you provided using a^2-b^2 was wonderful. I’ll try to keep it in my mind next time I reach such a stage. Thanks
• Mike August 25, 2013 at 11:55 am #
Dear Subh,
Mike 🙂
• D January 14, 2016 at 6:28 am #
Hi, Mike!
#4 was confusing to me too, and your method of using the difference of squares was awesomely cool!
So, quick follow-up question here for a situation where this method doesn’t work: in your post you mention that to check if a number is divisible by {2,3,5,7} for primes below 100. Is there a similar rule of thumb which I could use for numbers below 1000 etc.?
Thanks a lot!
• Magoosh Test Prep Expert February 1, 2016 at 11:56 pm #
Hi D,
Sorry about the late reply! The basic rule of thumb is that you only need to check for primes up to the square root of the number. So for numbers which are less than 100, you only need to check divisibility by prime numbers under √100 which is 10. This means that it is sufficient to check 2, 3, 5, and 7 to determine whether something is prime under 100.
Over 100, you just need to figure out how far to check, really. If I need to check if 201 is prime, I should check up to the square root of 201 which is somewhere between 14 and 15, so I should check until 14. That means 2, 3, 5, 7, 11, and 13 are all possible divisors. For your question about 1000, I should check up to 32, so 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. I hope that helps! 🙂
18. test website March 17, 2013 at 6:05 am #
Hiya! I just would like to give a huge thumbs up for the great information you have right here on this post. I shall be coming again to your blog for more soon.
• Mike March 17, 2013 at 11:35 pm #
Mike 🙂
19. taru February 26, 2013 at 6:05 am #
Hi Mike
In question #3, u checked straight away for 432×3 because it was odd or should we check with 2 first n then proceed?am i missing out something?because checking with 2 will kill time?is it a shortcut or something?
Taru
• Mike February 26, 2013 at 10:14 am #
Dear Taru,
What I checked first were the EXPONENTS of the factors. Anything with an even exponent could be ignored, and we had to add additional factors for anything with an odd exponents. That question is all about the exponents of the prime factors. It has nothing to do with whether the factors, 2 or 3, are odd. It has everything to do with whether the exponents of the prime factors are odd.
Does that make sense?
Mike 🙂
20. jp27 October 24, 2012 at 9:56 pm #
If n is the smallest integer such that 432 times n is the square of an integer, what is the value of y?
(A) 2
(B) 3
(C) 6
(D) 12
(E) 24
I think there is a typo in this question, shouldn’t it read what is the value of N? (instead of Y)
• Mike October 25, 2012 at 10:57 am #
Ooops! That’s a silly mistake I made there. I just corrected it. Thank you very much for the heads up.
Mike 🙂
21. Ravi Sankar Vemuri April 28, 2012 at 1:19 am #
“The prime factorization is analogous to the DNA of the number, the unique blueprint by which to construct the number” – Its a great analogy mike.
The examples on number of prime factors and odd factors are good.
• Mike April 30, 2012 at 12:06 pm #
Thank you. I’m glad you found it helpful. These are powerful strategies if you master them. Best of luck to you!
Mike 🙂
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# What is topology?
Updated on August 4, 2014
## Meet the topology
Topology is branch of geometry that deals with the properties of surfaces and general shapes but is unconcerned with the measurements of angles and lengths. High on the agenda are qualities that do not change when shapes are transformed into other shapes. We are allowed to push and pull the shape in any direction and for this reason topology is sometimes described as 'rubber sheet geometry'. Topologists are the people that can not tell the difference between a donut and a coffee cup!
A donut is a surface with a single hole in it. A coffee cup is the same where the hole takes place of the handle.
## Classifying polyhedra
The most basic shapes studied by topologists are polyhedra ‘poly means many and hedra means faces). An example of a poly hedron is a cube, with 6 square faces ,8vertices (points at the junction of the faces) and 12edges (the lines joining the vertices). The cube is a regular polyhedron because:
• all the faces are the same regular shape,
• all the angles between edges meeting at a vertex are equal.
Topology is a relatively new subject, but it can still be traced back to the Greeks, and indeed the culminating result of Euclid’s Elements is to show that there are exactly five regular polyhedra. These are the Platonic solids:
• tetrahedron (with 4 triangular faces),
• cube (with 6 square faces),
• octahedron (with 8 triangular faces),
• dodecahedron (with 12 pentagonal faces),
• icosahedron (with 20 triangular faces).
## Euler's formula
Euler’s formula is that the number of vertices V, edges E and faces F, of a polyhedron are connected by the formula: V – E + F = 2
For example, for a cube, V=8, E=12 and F=6 so V – E + F = 8 – 12 + 6 = 2 and, for buckminsterfullerene, V – E + F = 60 – 90 + 32 = 2. This theorem actually challenges the very notion of a polyhedron.
## Classification of surfaces
A topologist might regard the donut and the coffee cup as identical but what sort of surface is different from the donut? A candidate here is the rubber ball. There is no way of transforming the donut into a ball since the donut has a hole but the ball does not. This is a fundamental difference between the two surfaces. So a way of classifying surfaces is by the number of holes they contain. Let’s take a surface with r holes and divide it into regions bounded by edges joining vertices planted on the surface. Once this is done, we can count the number of vertices, edges, and faces. For any division, the Euler expression V – E + F always has the same value, called the Euler characteristic of the surface:
V – E + F = 2 – 2r.
If the surface has no holes (r=0) as was the case with ordinary polyhedra, the formula reduces to Euler’s V–E+F=2. In the case of one hole (r=1), as was the case with the cube with a tunnel, V–E+F=0.
## One-sided surfaces
Ordinarily a surface will have two sides. The outside of a ball is different from the inside and the only way to cross from one side to the other is to drill a hole in the ball– a cutting operation which is not allowed in topology (you can stretch but you can not cut). A piece of paper is another example of a surface with two sides. The only place where one side meets the other side is along the bounding curve formed by the edges of the paper.
The idea of a one-sided surface seems far-fetched. Nevertheless, a famous one was discovered by the German mathematician and astronomer August Möbius in the 19th century. The way to construct such a surface is to take a strip of paper, give it one twist and then stick the ends together. The result is a ‘Möbius strip’, a one-sided surface with a boundary curve. You can take your pencil and start drawing a line along its middle. Before long you are back where you started! It is even possible to have a one-sided surface that does not have a boundary curve. This is the ‘Klein bottle’ named after the German mathematician Felix Klein. What’s particularly impressive about this bottle is that it does not intersect itself. However, it is not possible to make a model of the Klein bottle in three-dimensional space without a physical intersection, for it properly lives in four dimensions where it would have no intersections. Both these surfaces are examples of what topologists call ‘manifolds’– geometrical surfaces that look like pieces of two-dimensional paper when small portions are viewed by themselves. Since the Klein bottle has no boundary it is called a closed 2-manifold.
## The Poincare conjecture
For more than a century, an outstanding problem in topology was the celebrated Poincaré conjecture, named after Henri Poincaré. The conjecture centres on the connection between algebra and topology. The part of the conjecture that remained unsolved until recently applied to closed 3-manifolds. These can be complicated– imagine a Klein bottle with an extra dimension .Poincaré conjectured that certain closed3-manifolds which had all the algebraic hallmarks of being three-dimensional spheres actually had to be spheres. It was as if you walked around a giant ball and all the clues you received indicated it was a sphere but because you could not see the big picture you wondered if it really was a sphere. No one could prove the Poincaré conjecture for 3-manifolds. Was it true or was it false? It had been proven for all other dimensions but the 3-manifold case was obstinate.There were many false proofs, until in 2002 when it was recognized that Grigori Perelman of the Steklov Institute in St Petersburg had finally proved it. Like the solution to other great problems in mathematics, the solution techniques for the Poincaré conjecture lay out side its immediate area, in a technique related to heat diffusion.
7
15
150
4
31
10
25 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Geometry Concepts Go to the latest version.
# Chapter 9: Circles
Difficulty Level: At Grade Created by: CK-12
## Introduction
Finally, we dive into a different shape, circles. First, we will define all the parts of circles and then explore the properties of tangent lines, arcs, inscribed angles, and chords. Next, we will learn about the properties of angles within circles that are formed by chords, tangents and secants. Lastly, we will place circles in the coordinate plane, find the equations of, and graph circles.
## Summary
This chapter begins with vocabulary associated with the parts of circles. It then branches into theorems about tangent lines; properties of arcs and central angles; and theorems about chords and how to apply them. Inscribed angles and inscribed quadrilaterals and their properties are explored. Angles on, inside, and outside a circle are presented in detail and the subsequent relationships are used in problem solving. Relationships among chords, secants, and tangents are discovered and applied. The chapter ends with the connection between algebra and geometry as the equations of circles are discussed.
### Chapter Keywords
• Circle
• Center
• Radius
• Chord
• Diameter
• Secant
• Tangent
• Point of Tangency
• Congruent Circles
• Concentric Circles
• Tangent to a Circle Theorem
• Central Angle
• Arc
• Semicircle
• Minor Arc
• Major Arc
• Congruent Arcs
• Arc Addition Postulate
• Inscribed Angle
• Intercepted Arc
• Inscribed Angle Theorem
• Inscribed Polygon
• Standard Equation of a Circle
### Chapter Review
Match the description with the correct label.
1. minor arc - A. CD¯¯¯¯¯\begin{align*}\overline{CD}\end{align*}
2. chord - B. AD¯¯¯¯¯¯\begin{align*}\overline{AD}\end{align*}
3. tangent line - C. CB\begin{align*}\overleftrightarrow{CB}\end{align*}
4. central angle - D. EF\begin{align*}\overleftrightarrow{EF}\end{align*}
5. secant - E. A\begin{align*}A\end{align*}
6. radius - F. D\begin{align*}D\end{align*}
7. inscribed angle - G. BAD\begin{align*}\angle BAD\end{align*}
8. center - H. BCD\begin{align*}\angle BCD\end{align*}
9. major arc - I. BDˆ\begin{align*}\widehat{BD}\end{align*}
10. point of tangency - J. BCDˆ\begin{align*}\widehat{BCD}\end{align*}
### Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9694.
At Grade
Aug 09, 2013
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Dec 10, 2015
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# Geometric Figures Practice Problems
based on 1 rating
By McGraw-Hill Professional
Updated on Sep 26, 2011
## Geometric Figures Practice Problems
### Set 1: Introduction to Geometric Figures in Word Problems - Solving for Three Unknowns
To review algebra problems involving geometric figures, go to Geometric Figures Help
#### Practice
1. A box’s width is its length. The perimeter of the box is 40 inches. What are the box’s length and width?
2. A rectangular yard is twice as long as it is wide. The perimeter is 120 feet. What are the yard’s dimensions?
#### Solutions
1. The perimeter of the box is 40 inches, so P = 2 l + 2 w becomes 40 = 2 l + 2 w . The width is its length, and w = (2 l /3), so 40 = 2 l + 2 w becomes 40 = 2 l + 2(2 l /3) = 2 l + (4 l /3).
The length of the box is 12 inches and its width is inches.
2. The perimeter of the yard is 120 feet, so P = 2 l + 2 w becomes 120 = 2 l + 2 w . The length is twice the width, so l = 2 w , and 120 = 2 l + 2 w becomes 120 = 2(2 w ) + 2 w .
The yard’s width is 20 feet and its length is 2 l = 2(20) = 40 feet.
### Set 2: Increasing and Decreasing Dimensions - Rectangles and Squares
To review geometric problems where one or more dimensions changed, go to Geometric Figures Help.
#### Practice
1. A rectangular piece of cardboard starts out with its width being three-fourths its length. Four inches are cut off its length and two inches from its width. The area of the cardboard is 72 square inches smaller than before it was trimmed What was its original length and width?
2. A rectangle’s length is one-and-a-half times its width. The length is increased by 4 inches and its width by 3 inches. The resulting area is 97 square inches more than the original rectangle. What were the original dimensions?
#### Solutions
1. Let l represent the original length and w , the original width. The original area is A = lw . The new length is l – 4 and the new width is w – 2. The new area is then ( l – 4)( w – 2). But the new area is 72 square inches smaller than the original area, so ( l – 4)( w – 2) = A – 72 = lw – 72. So far, we have ( l – 4)( w – 2) = lw – 72.
The original width is three-fourths its length, so l . We will now replace w with
The original length was 16 inches and the original width was inches.
2. Let l represent the original length and w , the original width. The original area is then given by A = lw . The new length is l + 4 and the new width is w + 3. The new area is now ( l + 4)( w + 3). But the new area is also the old area plus 97 square inches, so A + 97 = ( l + 4)( w + 3). But A = lw , so A + 97 becomes lw + 97. We now have
lw + 97 = ( l + 4)( w + 3).
Since the original length is of the original width, . Replace each l by .
The original width is 10 inches and the original length is inches.
### Set 3: Increasing and Decreasing Dimensions - Circles
To review geometric problems where the radius of a circle changed, go to Geometric Figures Help.
#### Practice
A circle’s radius is increased by 5 inches and as a result, its area is increased by 155π square inches. What is the original radius?
#### Solution
Let r represent the original radius. Then r + 5 represents the new radius, and A = πr 2 represents the original area. The new area is 155π square inches more than the original area, so 155π + A = π( r + 5) 2 = 155π + πr 2 .
The original radius is 13 inches.
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#### Revision Notes for Ch 5 Arithmetic Progression Class 10th Maths
Arithmetic Progression
Arithmetic Progression: An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term.
Term: The fixed number d is called the common difference of the A.P.
Common Difference: Each number in the list of arithmetic progression is called term. It can be positive, negative or zero.
General form of an AP: If a is first term and d is common difference of an A.P. The general form of an AP is: a, a+d, a+2d, a+3d
(iii) ax3 + bx+ cx + d is a polynomial of degree 3 cubic polynomial.
Finding General Term of an AP: We can find the nth term of an A.P by using Tn = a + (n-1)d
We can find the required term by putting the numeric value of n(no. of term) which we have to find from starting.
nth term from the end: l - (n-1)d where, l is the last term
The number occurring at nth place is denoted by Tn . The first term is T, second term is Tand so on.
Common difference d = T n-1 - T n
Examples:-
3, 7, 11, 15,19………….
This is an arithmetic progression which starts from 3. The common difference between two consecutive terms is 4.
That is
7-3=4, 11-7=4, 15-11=4
So, this is an AP with first term as 3, and common difference 4.
Or, we can say the AP is
3, 3+4, 3+4+4=3+2×4, 3+4+4+4=3+3×4
That is 4 is added to each term to get the next term.
Arithmetic Mean
If a,A, b are in AP we say that A is arithmetic mean between a and b. It is also written as AM
Let A is the AM between two numbers a and b
Let A be the AM between a and b, then
As we know that common difference is difference between two consecutive term of an AP, and it is same between two consecutive terms of the number.
So,
⇒A-a=b-A
⇒2A=b+a
A=(a+b)/2
So, arithmetic mean of a and b is (a + b)/2.
Sum of n-terms of an AP
Sum of n term of an AP is
Here n is number of terms of AP, a is first term and l is last term of the AP. Here, d is the common difference.
Properties of Arithmetic Progression
If a constant is added or subtracted or multiplied to each term of an A.P. then the resulting sequence is also an A.P.
If each term of an A.P. is divided by a non-zero constant then the resulting sequence is also an A.P. |
# Graphing Linear Equations in Two-Variables
Document Sample
``` 3.3 Graphing Linear Equations in Two-Variables page 14
Build a table of ordered pairs where the x-values are in a sequence and the y-values are in a
sequence. When you plot these points (ordered pairs) you will see that they form a straight line.
Always choose only an integer as the replacement for the x-value so that the resulting y-value
will be an integer, and visa versa. If we put a “sequence of integers" in the table for x , complete
the computation for each x-value, and put each corresponding y-value in the table, then the
y-values in the table will be a “sequence of integers". Write the common differences.
1. Given 2x − 3y = 6 build the table for using an arithmetic sequence as replacement values.
Note: In this equation if you choose consecutive integers for x you will find some y-values that
are fractions. By noting that the constant 6 is a multiple of 3 and the coefficient of y is 3, you
can choose your sequence for the x-values to be multiples of 3 . Now, when you are solving
for the y-values and divide by the coefficient-3 you will obtain a sequence of integers for the
y-values.
x y
2x − 3y = 6
-3 -4
dx = 3 0 -2 dy= 2
Let y = - 4: 2x − 3(- 4) = 6, x= -3
3 0
6 2
repeat for the other numbers 9 4
Check to see that both columns
of values are arithmetic sequences.
y
Start at the leftmost point and count the •
blocks up to the line of the next point.
•
Ratio: dy 2 • x
= •
dx 3
Now count the blocks right to the next •
point.
Compare these to dy and dx.
See the next page for another approach to graphing this equation.
page 15 Graphing Linear Equations 3.3
2. Build the table for 2x − 3y = 6 using an arithmetic sequence as replacement values.
For another approach:
If both the x-intercept and the x-intercept are integers, we can place them in the middle of the
table. Then we can find the difference in the x-values for dx and the difference in the y-values for
dy and use these differences to build arithmetic sequences for the table.
Let x = 0, then 2( 0 ) − 3y = 6 or y = -2 x y
-3 -4
which gives the point: ( 0 , -2 ) -3 -2
0 -2
dx= 3 3 0 dy= 2
Let y = 0, then 2x − 3( 0 ) = 6 or x = 3
6 2
+3 +2
which gives the point: ( 3 , 0 ) 9 4
Check the dy and dx on your table. dy 2
Ratio: = 3
dx
Now you must CHECK the top and bottom points in the table to be sure that they are points on
the line:
For the point (- 3, - 4) replace x = - 3 and y = - 4 in the given equation 2x − 3y = 6
?
CHECK: 2(- 3 ) − 3(- 4 ) = 6
- 6 + 12 = 6 !
For the point (9, 4) replace x = 9 and y = 4 in the given equation 2x − 3y = 6
?
2( 9 ) − 3( 4 ) = 6
CHECK: 18 − 12 = 6 !
Students should always form the practice of checking all of their work whenever they can.
page 16 Graphing Linear Equations 3.3
3. Graph 2x + 3y = 7.
Look at the coefficients and note that if we choose a sequence of odd integers as replacement
for y-values , such as: { -3, -1, 1, 3, 5}, the resulting x-sequence will be integers
Build a table for 2x + 3y = 7.
Put these values in the
middle of the table, then
2x + 3y = 7 Let y = 1 write the “common x y
differences” for both x -3
2x + 3( 1) = 7 or x=2 and y . -1
dx= - 3 2 1 dy= 2
Use the value for dx to -1 3
complete the table 5
repeat for y = 3 values for x.
2x + 3( 3) = 7, or x =-1 Note the value for dy in
the table values for y. For complete table see below.
Note adding ± 3 to given x-values x y
forms an arithmetic sequence. 8 -3
+3
5 -1 -2
Note adding m 2 to given y-values
dx= - 3 2 1 dy= 2
forms an arithmetic sequence.
-1 3
-3 +2
-4 5
To be sure you should always check the “outer points”.
?
Check: (- 4, 5) " 2( - 4) + 3( 5 ) = 7 or - 8 + 15 = 7 !
?
Check: (8, - 3) " 2( 8 ) + 3( - 3 ) = 7 or + 16 – 9 = 7 !
Plot the points, draw line:
y
Check the dy and dx on your graph. •
Ratio:
dy
=
-2 -2 •
dx 3 +3 • x
- 2•
•
+3
page 17 Graphing Linear Equations 3.3
For other examples study the coefficients for combinations that will yield integer points (x1, y1).
3. 3x + 5y = 7
Since 10 − 3 = 7, choose x = -1
and y = 2 to satisfy this difference.
3( -1) + 5( 2) = 7 (-1, 2)
Repeat for the another pair of numbers
Since 5 + 7 = 12, choose y = -1
and solve for x. Put these values in the x y
middle of the table,
3x + 5( -1) = 7 then write their
3x − 5 = 7 (4, -1) “common differences”
+5 +5 dx= + 5 -1 2 dy=- 3
for as dx and dy .
3x = 12 or x = 4 4 -1
Use the value for dx to
complete the table
dy - 3 3 values for x.
= = dy -3
dx 5 -5 and Ratio: =
dx 5
dy to complete the
table values for y.
Note adding m 5 to given x-values
x y
NOTE: Opposite signs -11 8
and ± 3 to given y-values forms above given values. -5 +3
-6 5
an arithmetic sequence for each. dx= + 5 - 1 2 dy=- 3
4 -1
+5 -3
9 -4
To be sure you should always check the “outer points”.
Check: (- 11, 8) " 3( - 11) + 5( 8 ) ? 7 or
= - 33 + 40 = 7 ! Plot the points, draw line:
? y
Check: (9, - 4) " 3( 9 ) + 5( - 4 ) = 7 or + 27 – 20 = 7 !
•
-3
+5
•
•
-3 x
•
dy
Check the dy and dx on your graph. Ratio: =
dx 5 -3
+5
•
page 18 Graphing Linear Equations 3.3
Three special types of lines:
1. Horizontal lines, y = b 2. Vertical lines, x = a 3. Lines through the origin, y = mx
1. Horizontal lines, y = b (any real number) (- 6, 2), (- 3, 2), (0, 2), (3, 2), (6, 2)
Equation: y = 2
Horizontal (y = b) (y is the same for every x-value)
Every point on a horizontal line has the same second number, for all x-values.
Example 1: Graph the line for the Example 2: Given two points with the same
equation: y = - 3 second number: (- 7, 5), (2, 5)
Equation: y = 5
y y
(- 7, 5) (2,
• •
5)
x x
(0, - 3)
•
2. Vertical lines, x = a (any real number)
(3, - 4), (3, - 2), (3, 0), (3, 2), (3, 4)
Equation: x = 3
Vertical (x = a) For every y-value.
Every point on a vertical line has the same first number, for all y-values.
Example 1: Graph the line for the Example 2: Given two points with the same
Equation x = 4. second number: (- 7, 5), (- 7, - 3)
Equation: x = - 7 -8
y y
(- 7, 5), (- 7, - 3)
0
(- 7, 5) •
• x x
(4, 0)
(- 7, - 3) •
page 19 Graphing Linear Equations 3.3
3. Lines through the origin: y = m x ⇒ m is the coefficient, any real number , times x.
If a line goes through the origin then (0, 0) is one point on the line, ⇒ y = m x
The Line is: The coefficient is: Example:
Rising as x moves from left to right m > 0, (Positive)
Falling as x moves from left to right m < 0, (Negative)
⇒ Given an equation in the form y = mx where there is no constant, it is equal to zero, the line
will always go through the origin (0, 0).
Example 1: Equation: y = 5/2 x, m>0 Example 2: Equation: y = - 2/3 x, m<0
Find and plot the points (- 2, - 5), (0, 0), (2, 5), Find and plot the two points (- 3, 2), (0, 0), (3, - 2)
Draw the line through the points. Draw the line through the points.
y y
•(2, (- 3, 2)
5)
•
x x
• ( 3, - 2)
•
(- 2, - 5)
```
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# HCF Tips and Tricks and Shortcuts
## Tips and Tricks of HCF Questions
This page is all about Tips, Tricks and shortcuts of HCF Questions. You will also get to know about its definition and methods to solve it.
Suppose there are two integers a and b. There is another Suppose there are two integers a and b. There is another integer c which divides each of the integers i.e. a and b. Then, integer c is the Highest Common Factor.
## HCF – Tips, Tricks and Shortcuts
• Here, are some easy tips and tricks for you on HCF problems Learn tricks on HCF that are asked in most in most competitive exams and other recruitment exams.
• The H.C.F of two or more numbers is smaller than or equal to the smallest number of given numbers
• The greatest number which divides a, b and c to leave the remainder R is H.C.F of (a – R), (b – R) and (c – R)
• The greatest number which divide x, y, z to leave remainders a, b, c is H.C.F of (x – a), (y – b) and (z – c)
• HCF of a given number always divides its LCM.
### Methods to find HCF
• Prime factorization method :
One of the simplest methods to calculate the HCF of a number. The following steps help to calculate HCF of a number.
a) Expand the given number as the product of their prime factors.
b) Check for common prime
• Division Method :
Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.
### How to calculate HCF for Fractions :
$HCF of fraction = \frac{HCF of Numerators}{LCM of Denominators}$
Example :
Find out HCF of 3/5, 9/15, 18/25
HCF = HCF of Numerators / LCM of Denominators
HCF of 3, 9 and 18 = 3
LCM of 5, 15 and 25 = 75
HCF of fraction = 3/75
### How to calculate HCF for Decimals
Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed.
Step 2 : Now find the LCM/HCF of these numbers without decimal.
Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers
### Question and Answer with solution
Question 1 .The H.C.F. of two numbers is 40 and the other two factors of their L.C.M. are 15 and 20. Find the larger number?
Options
A. 600
B. 460
C. 800
D. 1400
Solution: The numbers are 40 * 15 and 40 * 20
40 * 15 = 600
40 * 20 = 800
Correct option: C
Question 2. Find the HCF of $\frac{3}{10}, \frac{4}{7}, \frac{6}{5}, and \frac{2}{9}$
Options
A. $\frac{1}{630}$
B. $\frac{3}{630}$
C. $\frac{2}{543}$
D. 630
Solution We know that
HCF = HCF of Numerator/LCM of Denominators
HCF = $\frac{HCF(3,4,6,2)}{LCM(10,7,6,9)}$
HCF = $\frac{1}{630}$
Correct Option : A
Question 3 : Find the HCF of .63, 1.05, 2.1
Options:
A. .32
B. .21
C. .12
D. .65
Solution : the numbers can be written as .63, 1.05, 2.10
Now find the HCF of these number without decimal.
HCF of 63, 105 and 210 = 21
we need to put decimal point in the result obtained in step 2 leaving two digits on its right.
HCF (.63, 1.05, 2.1) = .21
Question 4 : For any integer n, what is HCF (22n + 7, 33n + 10) equal to?
A. n
B. 1
C. 11
D. None of these
Solution :
HCF of (22n + 7, 33n + 10) is always 1.
On placing different values in the given equation; n = 1, 2, 3, ……
We get,
For n = 1, HCF (22n + 7, 33n + 10) = (29, 43) ⇒ HCF = 1
For n = 2, HCF (22n + 7, 33n + 10) = (51, 76) ⇒ HCF = 1
For n = 3, HCF (22n + 7, 33n + 10) = (73, 109) ⇒ HCF = 1.
Question 5 : The greatest number which can divide 1356, 1868 and 2764 leaving the same remainder 12 in each case is
A. 64
B. 124
C.128
D. 132
Solution :
Required number = HCF of (1356 – 12) , (1868 – 12 ), (2764 – 12)
HCF of 1344, 1856 and 2752 = 64.
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# How to Calculate a .25% Increase
Most calculators have a percentage button, but percentage can easily be worked out with no mechanical help.
Calculating percentages can be a constant source of confusion if you are not comfortable with applying mathematics in real life. Calculating increases in percentage is useful for many situations, especially financial ones, and it is very easy to do. Many calculators have a percentage function on them, but the calculations are so simple, it is worth learning them anyway. If you learn the principles of making the calculation, you will be able to do it yourself if you need to in the future. There will almost certainly come a time when you need a percentage but don't have a calculator.
## Step 1
Find the figure you want to calculate the increase of. For example, you may earn \$30,000 per year and be getting a 0.25 percent pay raise. This calculation is very simple if you break the idea down into its constituent parts. A percentage is a fraction out of 100. So if there are 50 men out of 100 people in a room, 50 percent of people in the room are men. We could also express it as a fraction, because we know that exactly 1/2 of the people present are men. In decimals, 1/2 is displayed as 0.5.
Video of the Day
## Step 2
Think about your figure. If a percentage is a fraction out of 100, then you can always work out 1 percent of a figure really easily. All you have to do is divide the number by 100, and you have 1 percent. So in the example, 30,000 divided by 100 is 300. Dividing by 100 is always easy because you just move the decimal point two places to the left. To divide by 10, move the decimal one place over.
## Step 3
Apply your knowledge. Now that you have identified what 1 percent is, you can work out any percentage of that figure. For example, if you wanted to know what 52 percent of 30,000 is, you'd simply calculate 300 x 52. This brings the 52 percent calculation out at 15,600. If you wanted to work out what a 52 percent increase on 30,000 was, you would add 15,600 to 30,000, and get the answer, 45,600.
## Step 4
Work out the rate for 0.25 percent. This is a decimal expression of a quarter, so we can take it to mean that the \$30,000 salary would be raised by a quarter of a percent. So to work out the value of a .25 percent increase, we have to work out what a quarter of a single percent is. We divide 300 by four, and get 75. This tells us that a 0.25 percent increase on a \$30,000 salary would bring the figure up to \$30,075.
Video of the Day |
# Category: Numbers
### Recurring Decimals to Fractions
Writing numbers containing recurring decimals into fractions involves a few steps. These are described here using a few examples. First, we start with… Key to the above process is to. read more…
### Multiplication – Lattice Method
There are few ways to multiply numbers. The method shown in the video below is one of them
### Rationalise the Denominator
Rationalise the denominator means eliminating the SURD from the denominator. Here are a few examples showing how to do this operation. A full video lesson on working with SURDS is. read more…
### Finding the Upper Bound and Lower Bound
With rounding, you often get asked to find two values. These are the Upper Bound (UB) and the Lower Bound (LB) and when these 2 values are written in the. read more…
### Rounding to – the nearest… / a significant figure
There are two ways to round a number. You can round a number to a specific place value or to a specific significant figure. Rounding to a specific place value. read more…
### Negative Numbers – Multiplying & Dividing
The rules for multiplying and dividing involving negative numbers are relatively simple compared to adding and subtracting involving negative numbers. Here they are: Dividing works the EXACT SAME WAY as. read more…
### Negative Numbers – Adding & Subtracting
Adding and subtracting involving negative numbers can be challenging to some students if the rules on negatives numbers are not well understood. This article aims to provide clarity on this. read more…
### Numbers – Factors, Multiples & Prime Factors
Getting confused with Factors, Multiples and Prime factors ??? Well! here are some examples that will hopefully shed some light on the meaning of these terms. Let’s starts with Factors:. read more…
### Numbers – Highest Common Factor (HCF) & Lowest Common Multiple (LCM)
The best way to work out the HCF and LCM of two numbers is to use the prime factor decomposition and Venn Diagram method as it is suited for both. read more…
### Numbers – Multipliers
The concept of ‘Multipliers‘ is a useful one to know when dealing with percentage increase (or decrease) as it allows you to work out the ‘final’ value or the answer. read more… |
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# Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18A
### Solution 3
We form the table as below:
Class Frequency fi Class mark xi (fi × xi) 10-30 15 20 300 30-50 18 40 720 50-70 25 60 1500 70-90 10 80 800 90-110 2 100 200 𝛴fi = 70 𝛴(fi × xi) = 3520
Therefore, mean
### Solution 4
We form the table as below:
Class Frequency fi Class mark xi (fi × xi) 0-20 6 10 60 20-40 8 30 240 40-60 10 50 500 60-80 12 70 840 80-100 6 90 540 100-120 5 110 550 120-140 3 130 390 𝛴fi = 50 𝛴(fi × xi) = 3120
Therefore, mean
### Solution 5
We form the table as below:
Class Frequency fi Class mark xi (fi × xi) 0-6 10 3 30 6-12 11 9 99 12-18 7 15 105 18-24 4 21 84 24-30 4 27 108 30-36 3 33 99 36-42 1 39 39 𝛴fi = 40 𝛴(fi × xi) = 564
Therefore, mean number of days
### Solution 6
We form the table as below:
Class Frequency fi Class mark xi (fi × xi) 500-700 6 600 3600 700-900 8 800 6400 900-1100 10 1000 10000 1100-1300 9 1200 10800 1300-1500 7 1400 9800 𝛴fi = 40 𝛴(fi × xi) = 40600
Therefore, mean daily expenses
### Solution 7
We form the table as below:
Class Frequency fi Class mark xi (fi × xi) 64-68 6 66 396 68-72 8 70 560 72-76 10 74 740 76-80 12 78 936 80-84 3 82 246 84-88 1 86 86 𝛴fi = 40 𝛴(fi × xi) = 2964
Therefore, mean heartbeats per minute
### Solution 8
Here, class size, h = 10.
Let the assumed mean be A = 35.
For calculating the mean age, we prepare the following table:
Class Frequency fi Class mark xi (fi × ui) 0-10 105 5 -3 -315 10-20 222 15 -2 -444 20-30 220 25 -1 -220 30-40 138 35 0 0 40-50 102 45 1 102 50-60 113 55 2 226 60-70 100 65 3 300 𝛴fi = 1000 𝛴(fi × ui) = -351
Therefore,
Thus, the mean age of persons visiting the marketing centre on that day is 31.5 years.
### Solution 9
Class Frequency fi Class mark xi (fi × xi) 0-20 12 10 120 20-40 15 30 450 40-60 32 50 1600 60-80 x 70 70x 80-100 13 90 1170 𝛴fi = 72 + x 𝛴(fi × xi) = 3340 + 70x
Arithmetic mean
### Solution 10
Class Frequency fi Class mark xi (fi × xi) 11-13 3 12 36 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 f 20 20f 21-23 5 22 110 23-25 4 24 96 𝛴fi = 40 + f 𝛴(fi × xi) = 704 + 20f
Arithmetic mean
### Solution 14
We have
Class Frequency Mid Value 0 - 20 7 10 70 20 - 40 30 30 40 - 60 12 50 600 60 - 80 =18 - 70 1260 - 70 80 - 100 8 90 720 100 - 120 5 110 550 = 50
### Solution 15
Class Frequency fi Class mark xi (fi × xi) 0-80 20 40 800 80-160 25 120 3000 160-240 x 200 200x 240-320 y 280 280y 320-400 10 360 3600 𝛴fi = 55 + x + y 𝛴(fi × xi) = 7400 + 200x + 280y
Now,
Also, arithmetic mean = 188
### Solution 17
We have, Let A = 25 be the assumed mean.
Marks Frequency Mid value Deviation 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 12 18 27 20 17 6 5 15 25 = A 35 45 55 -20 -10 0 10 20 30 -240 -180 0 200 340 180 = 100 = 300
Hence, mean marks per student = 28
### Solution 18
Let the assumed mean be 150, h = 20
Marks Frequency Mid value Deviation di = - 150 di 100 - 120 120 - 140 140 - 160 160 - 180 180 - 200 10 20 30 15 5 110 130 150=A 170 190 -40 -20 0 20 40 -400 -400 0 300 200 = 80 di=-300
Hence, Mean = 146.25
### Solution 19
Let A = 50 be the assumed mean, we have
Marks Frequency Mid value Deviation 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 20 35 52 44 38 31 10 30 50 = A 70 90 110 -40 -20 0 20 40 60 -800 -700 0 880 1520 1860 = 220
### Solution 25
Let h = 20 and assume mean = 550, we prepare the table given below:
Age Frequency Mid value 500 - 520 520 - 540 540 - 560 560 - 580 580 - 600 600 - 620 14 9 5 4 3 5 510 530 550 = A 570 590 610 -2 -1 0 1 2 3 -27 -9 0 4 6 15 = 40
Thus, A = 550, h = 20, and = 40,
Hence, the mean of the frequency distribution is 544.
### Solution 26
The given series is an inclusive series, making it an exclusive series, we have
Class Frequency Mid value 24.5 - 29.5 29.5 - 34.5 34.5 - 39.5 39.5 - 44.5 44.5 - 49.5 49.5 - 54.5 54.5 - 59.5 4 14 22 16 6 5 3 27 32 37 42 = A 47 52 57 -3 -2 -1 0 1 2 3 -12 -28 -22 0 6 10 9 = 70
Thus, A = 42, h = 5, = 70 and
Hence, Mean = 39.36 years
### Solution 27
The given series is an inclusive series making it an exclusive series,we get
class Frequency Mid value 4.5 - 14.5 14.5 - 24.5 24.5 - 34.5 34.5 - 44.5 44.5 - 54.5 54.5 - 64.5 6 11 21 23 14 5 9.5 19.5 29.5=A 39.5 49.5 59.5 -2 -1 0 1 2 3 -12 -11 0 23 28 15 = 80
Thus, A = 29.5, h = 10, = 80 and
Hence, Mean = 34.87 years
## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18B
### Solution 2
We prepare the frequency table, given below
Marks No. of students C.F. 0 - 7 7 - 14 14 - 21 21 - 28 28 - 35 35 - 42 42 - 49 3 4 7 11 0 16 9 3 7 14 25 25 41 50 N = = 50
Now,
The cumulative frequency is 25 and corresponding class is 21 - 28.
Thus, the median class is 21 - 28
l = 21, h = 7, f = 11, c = C.F. preceding class 21 - 28 is 14 and = 25
Hence the median is 28.
### Solution 3
We prepare the frequency table given below:
Daily wages Frequency C.F. 0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 40 32 48 22 8 40 72 120 142 150 N = = 150
Now, N = 150, therefore
The cumulative frequency just greater than 75 is 120 and corresponding class is 200 - 300.
Thus, the median class is 200 - 300
l = 200, h = 100, f = 48
c = C.F. preceding median class = 72 and
Hence the median of daily wages is Rs. 206.25.
Hence the median is 28.
### Solution 4
We prepare the frequency table, given below:
Class Frequency C.F 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35-40 40 - 45 5 6 15 10 5 4 2 2 5 11 26 36 41 45 47 49 = 49
Now, N = 49
The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 - 20.
Thus, the median class is 15 - 20
l = 15, h = 5, f = 15
c = CF preceding median class = 11 and
Median of frequency distribution is 19.5
### Solution 5
We prepare the cumulative frequency table as given below:
Consumption Frequency C.F 65 - 85 85 - 105 105 - 125 125 - 145 145 - 165 165 - 185 185 - 205 4 5 13 20 14 7 4 4 9 22 42 56 63 67 N = = 67
Now, N = 67
The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 - 145.
Thus, the median class is 125 - 145
l = 125, h = 20, and c = CF preceding the median class = 22, = 33.5
Hence median of electricity consumed is 136.5
### Solution 6
Frequency table is given below:
Height Frequency C.F 135 - 140 140 - 145 145 - 150 150 - 155 155 - 160 160 - 165 165 - 170 170 - 175 6 10 18 22 20 15 6 3 6 16 34 56 76 91 97 100 N = =100
N = 100,
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 - 155
Thus, the median class is 150 - 155
l = 150, h = 5, f = 22, c = C.F.preceding median class = 34
Hence, Median = 153.64
### Solution 7
The frequency table is given below. Let the missing frequency be x.
Class Frequency C.F 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 5 25 x 18 7 5 30 30 + x 48 + x 55 + x
Median = 24 Median class is 20 - 30
l = 20, h = 10, f = x, c = C.F. preceding median class = 30
Hence, the missing frequency is 25.
### Solution 10
Let be the frequencies of class intervals 0 - 10 and 40 - 50
Median is 32.5 which lies in 30 - 40, so the median class is 30 - 40
l = 30, h = 10, f = 12, N = 40 and
### Solution 11
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
Class Frequency C.F 18.5 - 25.5 25.5 - 32.5 32.5 - 39.5 39.5 - 46.5 46.5 - 53.5 53.5 - 60.5 35 96 68 102 35 4 35 131 199 301 336 340 fi = N = 340
The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 - 39.5.
Median class is 32.5 - 39.5
l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131
Hence median is 36.5 years
### Solution 12
Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get
Wages per day (in Rs) Frequency C.F 60.5 - 70.5 70.5 - 80.5 80.5 - 90.5 90.5 - 100.5 100.5 - 110.5 110.5 - 120.5 5 15 20 30 20 8 5 20 40 70 90 98 fi = N =98
The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 - 100.5.
median class is 90.5 - 100.5
l = 90.5, h = 10, f = 30, c = CF preceding median class = 40
Hence, Median = Rs 93.50
### Solution 13
The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get
Marks Frequency C.F 0.5 - 5.5 5.5 - 10.5 10.5 - 15.5 15.5 - 20.5 20.5 - 25.5 25.5 - 30.5 30.5 - 35.5 35.5 - 40.5 40.5 - 45.5 7 10 16 32 24 16 11 5 2 7 17 33 65 89 105 116 121 123 fi = N =123
The cumulative frequency just greater than 61.5 is 65.
The corresponding median class is 15.5 - 20.5.
Then the median class is 15.5 - 20.5
l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33
Hence, Median = 19.95
### Solution 14
Marks Frequency C.F 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 12 20 25 23 12 24 48 36 12 32 57 80 92 116 164 200 N =
The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 - 60.
Thus the median class is 50 - 60
l = 50, h = 10, f = 24, c = C.F. preceding median class = 92, = 100
Hence, Median = 53.33
## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18C
### Solution 4
As the class 26 - 30 has maximum frequency so it is modal class
Hence, mode = 28.5
### Solution 5
As the class 1500 - 2000 has maximum frequency, so it os modal class
Hence the average expenditure done by maximum number of workers = Rs. 1820
### Solution 6
As the class 5000 - 10000 has maximum frequency, so it is modal class
Hence, mode = Rs. 7727.27
### Solution 7
As the class 15 - 20 has maximum frequency so it is modal class.
Hence mode = 17.3 years
### Solution 8
As the class 85 - 95 has the maximum frequency it is modal class
Hence, mode = 85.71
### Solution 9
The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get
Class Frequency 0.5 - 5.5 5.5- 10.5 10.5 - 15. 5 15.5 - 20.5 20.5 - 25. 5 25.5 - 30.5 30.5 - 35.5 35.5 - 40.5 40.5 - 45.5 45.5 - 50.5 3 8 13 18 28 20 13 8 6 3
As the class 20.5 - 25.5 has maximum frequency, so it is modal class
Hence, mode = 23.28
## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18D
### Solution 1
Mean:
Class Frequency fi Cumulative frequency Class mark xi (fi × xi) 0-10 4 4 5 20 10-20 4 8 15 60 20-30 7 15 25 175 30-40 10 25 35 350 40-50 12 37 45 540 50-60 8 45 55 440 60-70 5 50 65 325 𝛴fi = 50 𝛴(fi × xi) = 1910
Therefore, mean
Median:
Cumulative frequency just greater than 25 is 37 and the corresponding class is 40-50.
Thus, the median class = 30-40
Mode:
Mode = 3(Median) - 2(Mean)
= 3(40) - 2(38.2)
= 120 - 76.4
= 43.6
### Solution 5
Let the assumed mean A be 145.Class interval h = 10.
Class Frequency Mid-Value C.F. 120-130 130-140 140-150 150-160 160-170 2 8 12 20 8 125 135 145=A 155 165 -2 -1 0 1 2 -4 -8 0 20 16 2 10 22 42 50 N = 50
(i)Mean
(ii)N = 50,
Cumulative frequency just after 25 is 42
Corresponding median class is 150 - 160
Cumulative frequency before median class, c = 22
Median class frequency f = 20
(iii)Mode = 3 median - 2 mean
= 3 151.5 - 2 149.8 = 454.5 - 299.6
= 154.9
Thus, Mean = 149.8, Median = 151.5, Mode = 154.9
### Solution 6
Class Frequency Mid-value C.F. 100-120 120-140 140-160 160-180 180-200 12 14 8 6 10 110 130 150= A 170 190 -2 -1 0 1 2 -24 -14 0 6 20 12 26 34 40 50 N = 50
Let assumed mean A = 150 and h = 20
(i)Mean
(ii)
Cumulative frequency just after 25 is 26
Corresponding frequency median class is 120 - 140
So, l = 120, f = 14, h = 20, c = 12
(iii)Mode = 3Median - 2Mode
= 3(138.6) - 2(145.2)
= 415.8 - 190.4
= 125.4
Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4
### Solution 7
Class Frequency Mid-value C.F. 100-150 150-200 200-250 250-300 300-350 6 7 12 3 2 125 175 225 275 325 -2 -1 0 1 2 -12 -7 0 3 4 6 13 25 28 30 N = 30
Let assumed mean = 225 and h = 50
(i)Mean =
(ii)
Cumulative frequency just after 15 is 25
corresponding class interval is 200 - 250
Median class is 200 - 250
Cumulative frequency c just before this class = 13
Hence, Mean = 205 and Median = 208.33
## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18E
### Solution 1
We plot the points (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the 'less than type' ogive as follows:
At y = 26.5, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 68 units
Hence, median marks = 68
### Solution 2
Number of wickets Less than 15 Less than 30 Less than 45 Less than 60 Less than 75 Less than 90 Less than 105 Less than 120 Number of bowlers 2 5 9 17 39 54 70 80
We plot the points (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70) and (120, 80) to get the 'less than type' ogive as follows:
At y = 40, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 78 units
Hence, median number of wickets = 78
### Solution 3
'More than type' distribution is as follows:
We plot the points (0, 100), (10, 96), (20, 90), (30, 80), (40, 70), (50, 45), (60, 23) and (70, 5) to get the 'more than type' ogive as follows:
### Solution 4
'More than type' distribution is as follows:
We plot the points (135, 50), (140, 45), (145, 37), (150, 28), (155, 16) and (160, 2) to get the 'more than type' ogive as follows:
### Solution 5
'More than type' distribution is as follows:
We plot the points (140, 156), (160, 153), (180, 145), (200, 130), (220, 90), (240, 40) and (260, 10) to get the 'more than type' ogive as follows:
### Solution 6
We plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) to get the 'more than type' ogive as follows:
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 70.5 units
Hence, median production yield = 70.5 kg/ha
### Solution 7
'More than type' distribution is as follows:
Production yield Number of farms More than 65 24 More than 60 54 More than 55 74 More than 50 90 More than 45 96 More than 40 100
On a graph paper, we plot the points A(40, 100), B(45, 96), C(50, 90), D(55, 74), E(60, 54) and F(65, 24).
Join these points to get a 'More than Ogive'.
### Solution 8
More than series
We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)
Hence,
Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM x-axis intersecting x-axis at M
Then,OM = 590
Hence median = 590
### Solution 9
(i) Less than series:
Marks No. of students Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 Less than 45 Less than 50 2 7 13 21 31 56 76 94 98 100
Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)
Join these points free hand to get the curve representing "less than" cumulative curve.
(ii)From the given table we may prepare the 'more than' series as shown below
Marks No. of students More than 45 More than 40 More than 35 More than 30 More than 25 More than 20 More than 15 More than 10 More than 5 More than 0 2 6 24 44 69 79 87 93 98 100
Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)
Join these points free hand to get the required curve.
Here
Two curves intersect at point P(28, 50)
Hence, the median = 28
### Solution 10
We may prepare less than series and more than series
(i)Less than series
Height in (cm) Frequency Less than 140 Less than 144 Less than 148 Less than 152 Less than 156 Less than 160 Less than 164 Less than 168 Less than 172 Less than 176 Less than 180 0 3 12 36 67 109 173 248 330 416 450
Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)
(ii)More than series
Height in cm C.F. More than 140 More than 144 More than 148 More than 152 More than 156 More than 160 More than 164 More than 168 More than 172 More than 176 More than 180 450 447 438 414 383 341 277 202 120 34 0
Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)
The curves intersect at (167, 225).
Hence, 167 is the median.
## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18F
### Solution 2
Class having maximum frequency is the modal class.
Here, maximum frequency = 27
Hence, the modal class is 40 - 50.
Thus, the lower limit of the modal class is 40.
## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise MCQ
### Solution 1
Correct option: (d)
Range is not a measure of central tendency.
### Solution 2
Correct option: (a)
Mean cannot be determined graphically.
### Solution 3
Correct option: (a)
Since mean is the average of all observations, it is influenced by extreme values.
### Solution 4
Correct option: (c)
Mode can be obtained graphically from a histogram.
### Solution 5
Correct option: (d)
Ogives are used to determine the median of a frequency distribution.
### Solution 6
Correct option: (b)
The cumulative frequency table is useful in determining the median.
### Solution 7
Correct option: (b)
Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.
### Solution 10
Correct option: (c)
di's are the deviations from A of midpoints of the classes.
### Solution 11
Correct option: (b)
While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.
### Solution 12
Correct option: (b)
Mode = (3 x median) - (2 x mean)
### Solution 13
Correct option: (c)
Since the abscissa of the point of intersection of both the ogives gives the median, we have median = 20.5
### Solution 15
Correct option: (c)
Class having maximum frequency is the modal class.
Here, maximum frequency = 30
Hence, the modal class is 30 - 40.
### Solution 18
Correct option: (c)
Mean = 8.9
Median = 9
Mode = 3Median - 2Mean
= 3 x 9 - 2 x 8.9
= 27 - 17.8
= 9.2
### Solution 23
Correct option: (c)
For a symmetrical distribution, we have
Mean = mode = median
### Solution 24
Correct option: (c)
Number of families having income more than Rs. 20000 = 50
Number of families having income more than Rs. 25000 = 37
Hence, number of families having income range 20000 to 25000 = 50 - 37 = 13
### Solution 26
Correct option: (d)
Mean of 20 numbers = 0
Hence, sum of 20 numbers = 0 x 20 = 0
Now, the mean can be zero if
sum of 10 numbers is (S) and the sum of remaining 10 numbers is (-S),
sum of 11 numbers is (S) and the sum of remaining 9 numbers is (-S),
…….
sum of 19 numbers is (S) and the 20th number is (-S), then their sum is zero.
So, at the most, 19 numbers can be greater than zero.
### Solution 29
(a) - (s)
The most frequent value in a data is known as mode.
(b) - (r)
Mean cannot be determined graphically.
(c) - (q)
An ogive is used to determine median.
(d) - (p)
Standard deviation is not a measure of central tendency.
## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Test Yourself
### Solution 1
Correct option: (b)
The cumulative frequency table is useful in determining the median.
### Solution 2
Correct option: (c)
Mean = 27
Median = 33
Mode = 3Median - 2Mean
= 3 x 33 - 2 x 27
= 99 - 54
= 45
### Solution 6
Number of athletes who completed the race in less than 14.6 seconds
= 2 + 4 + 15 + 54
= 75
### Solution 8
The frequency table is as follows:
Classes Profit (in lakhs Rs.) Frequency Number of shops 5 - 10 2 10 - 15 12 15 - 20 2 20 - 25 4 25 - 30 3 30 - 35 4 35 - 40 3
The frequency corresponding to the class 20 - 25 is 4.
### Solution 13
Marks Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Number of students 3 11 28 48 70
We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve as follows:
### Solution 15
Marks obtained Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 Less than 100 Number of students 2 7 17 40 60 82 85 90 100
We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90) and (100, 100) to get the cumulative frequency curve as follows:
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 56.
Hence, median = 56
Mode :
### Solution 20
Less Than Series:
Class interval Frequency Less than 10 2 Less than 15 14 Less than 20 16 Less than 25 20 Less than 30 23 Less than 35 27 Less than 40 30
We plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get less than ogive.
More Than Series:
Class interval Frequency More than 5 30 More than 10 28 More than 15 16 More than 20 14 More than 25 10 More than 30 7 More than 35 3
We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) to get more than ogive.
The two curves intersect at L. Draw LM OX.
Thus, median = OM = 16
### Solution 21
Less Than Series:
Class interval Frequency Less than 45 1 Less than 50 10 Less than 55 25 Less than 60 43 Less than 65 83 Less than 70 109 Less than 75 125 Less than 80 139 Less than 85 149
We plot the points (45, 1), (50, 10), (55, 25), (60, 43), (65, 83), (35, 27), (70, 109), (75, 125), (80, 139) and (85, 149) to get less than ogive.
More Than Series:
Class interval Frequency More than 40 149 More than 45 148 More than 50 139 More than 55 124 More than 60 106 More than 65 66 More than 70 40 More than 75 24 More than 80 10
We plot the points (40, 149), (45, 148), (50, 139), (55, 124), (60, 106), (65, 66), (70, 40), (75, 24) and (80, 10) to get more than ogive.
Mode : |
# AP Statistics Curriculum 2007 Distrib Multinomial
(Difference between revisions)
Revision as of 00:28, 3 May 2008 (view source)IvoDinov (Talk | contribs) (→Examples of Multinomial experiments: Added another example to demonstrate complementation in multinomial probabilities.)← Older edit Revision as of 00:29, 3 May 2008 (view source)IvoDinov (Talk | contribs) m (→Synergies between Binomial and Multinomial processes/probabilities/coefficients: typo)Newer edit → Line 40: Line 40: * The Binomial vs. Multinomial '''Formulas''' * The Binomial vs. Multinomial '''Formulas''' - : $(a+b)^n = \sum_{i=1}^n{{n\choose i}a^1 \times b^{n-i}}$ + : $(a+b)^n = \sum_{i=1}^n{{n\choose i}a^i \times b^{n-i}}$ : $(a_1+a_2+\cdots +a_k)^n = \sum_{i_1+i_2\cdots +i_k=n}^n{ {n\choose i_1,i_2,\cdots, i_k} : [itex](a_1+a_2+\cdots +a_k)^n = \sum_{i_1+i_2\cdots +i_k=n}^n{ {n\choose i_1,i_2,\cdots, i_k} a_1^{i_1} \times a_2^{i_2} \times \cdots \times a_k^{i_k}}$ a_1^{i_1} \times a_2^{i_2} \times \cdots \times a_k^{i_k}}[/itex]
## General Advance-Placement (AP) Statistics Curriculum - Multinomial Random Variables and Experiments
The multinomial experiments (and multinomial distributions) directly extend the their bi-nomial counterparts.
### Multinomial experiments
A multinomial experiment is an experiment that has the following properties:
• The experiment consists of k repeated trials.
• Each trial has a discrete number of possible outcomes.
• On any given trial, the probability that a particular outcome will occur is constant.
• The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.
#### Examples of Multinomial experiments
• Suppose we have an urn containing 9 marbles. Two are red, three are green, and four are blue (2+3+4=9). We randomly select 5 marbles from the urn, with replacement. What is the probability (P(A)) of the event A={selecting 2 green marbles and 3 blue marbles}?
• To solve this problem, we apply the multinomial formula. We know the following:
• The experiment consists of 5 trials, so k = 5.
• The 5 trials produce 0 red, 2 green marbles, and 3 blue marbles; so r1 = rred = 0, r2 = rgreen = 2, and r3 = rblue = 3.
• For any particular trial, the probability of drawing a red, green, or blue marble is 2/9, 3/9, and 4/9, respectively. Hence, p1 = pred = 2 / 9, p2 = pgreen = 1 / 3, and p3 = pblue = 4 / 9.
Plugging these values into the multinomial formula we get the probability of the event of interest to be:
$P(A) = {5\choose 0, 2, 3}p_1^{r_1}p_2^{r_2}p_3^{r_3}$
$P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (4/9)^3=0.0975461.$
Thus, if we draw 5 marbles with replacement from the urn, the probability of drawing no red, 2 green, and 3 blue marbles is 0.0975461.
• Let's again use the urn containing 9 marbles, where the number of red, green and blue marbles are 2, 3 and 4, respectively. This time we select 5 marbles from the urn, but are interested in the probability (P(B)) of the event B={selecting 2 green marbles}! (Note that 2 < 5)
• To solve this problem, we classify balls into green and others! Thus the multinomial experiment consists of 5 trials (k = 5), r1 = rgreen = 2, r2 = rother = 3. In this case, the probabilities of drawing a green or other marble are 3/9, and 6/9, respectively. Notice now the P(other) is the sum of the probabilities of the other colors (complement of green)! Hence,
$P(B) = {5\choose 2, 3}p_1^{r_1}p_2^{r_2} = {5! \over 2! \times 3! }\times (3/9)^2 \times (6/9)^3=0.329218.$
This probability is equivalent to the binomial probability (success=green; failure=other color), B(n=5, p=1/3).
### Synergies between Binomial and Multinomial processes/probabilities/coefficients
${n\choose i}=\frac{n!}{k!(n-k)!}$
${n\choose i_1,i_2,\cdots, i_k}= \frac{n!}{i_1! i_2! \cdots i_k!}$
• The Binomial vs. Multinomial Formulas
$(a+b)^n = \sum_{i=1}^n{{n\choose i}a^i \times b^{n-i}}$
$(a_1+a_2+\cdots +a_k)^n = \sum_{i_1+i_2\cdots +i_k=n}^n{ {n\choose i_1,i_2,\cdots, i_k} a_1^{i_1} \times a_2^{i_2} \times \cdots \times a_k^{i_k}}$
$p=P(X=r)={n\choose r}p^r(1-p)^{n-r}, \forall 0\leq r \leq n$
$p=P(X_1=r_1 \cap X_2=r_2 \cap \cdots \cap X_k=r_k | r_1+r_2+\cdots+r_k=n)={n\choose i_1,i_2,\cdots, i_k}p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}, \forall r_1+r_2+\cdots+r_k=n$
### Example
Suppose we study N independent trials with results falling in one of k possible categories labeled $1,2, \cdots, k$. Let pi be the probability of a trial resulting in the ith category, where $p_1+p_2+ \cdots +p_k = 1$. Let Ni be the number of trials resulting in the ith category, where $N_1+N_2+ \cdots +N_k = N$.
For instance, suppose we have 9 people arriving at a meeting according to the following information:
P(by Air) = 0.4, P(by Bus) = 0.2, P(by Automobile) = 0.3, P(by Train) = 0.1
• Compute the following probabilities
P(3 by Air, 3 by Bus, 1 by Auto, 2 by Train) = ?
P(2 by air) = ?
### SOCR Multinomial Examples
Suppose we row 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={3 ones, 3 twos, 2 threes, and 2 fours}. Assume the dice are loaded to the small outcomes according to the following probabilities of the 6 outcomes (one is the most likely and six is the least likely outcome).
x 1 2 3 4 5 6 P(X=x) 0.286 0.238 0.19 0.143 0.095 0.048
P(A)=?
Of course, we can compute this number exactly as:
$P(A) = {10! \over 3!\times 3! \times 2! \times 2! } \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 = 0.00586690138260962656816896.$
However, we can also find a pretty close empirically-driven estimate using the SOCR Dice Experiment.
For instance, running the SOCR Dice Experiment 1,000 times with number of dice n=10, and the loading probabilities listed above, we get an output like the one shown below.
Now, we can actually count how many of these 1,000 trials generated the event A as an outcome. In one such experiment of 1,000 trials, there were 8 outcomes of the type {3 ones, 3 twos, 2 threes and 2 fours}. Therefore, the relative proportion of these outcomes to 1,000 will give us a fairly accurate estimate of the exact probability we computed above
$P(A) \approx {8 \over 1,000}=0.008$.
Note that that this approximation is close to the exact answer above. By the Law of Large Numbers (LLN), we know that this SOCR empirical approximation to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000. |
Factors the 65 are the number which once multiplied in pairs provide the product as 65. These factors can be an unfavorable as well. 65 is a one-of-a-kind two-digit number due to the fact that apart from 1 and 65, it has actually only one other pair factor, i.e. 5 and also 13. In this lesson, we will certainly learn the determinants of 65, the prime factors of 65, and the determinants of 65 in pairs in addition to solved examples.
You are watching: What are the factor pairs of 65
Factors of 65: 1, 5, 13, and 65Prime factorization of 65: 65 = 5 × 13
1 What space the factors of 65? 2 How to calculation the determinants of 65? 3 Factors the 65 by prime Factorization 4 Factors of 65 in Pairs 5 FAQs on determinants of 65
What space the determinants of 65?
Let us first understand the an interpretation of factors. A aspect is the number i m sorry divides any kind of given number without leaving a remainder at the end. The number 65 is claimed to it is in an strange composite number. A composite number is a number that is composed of an ext than 2 factors. For instance, consider the number 65. The components of 65 room 1, 5, 13, and 65.
How to calculation the factors of 65?
Let us begin calculating the determinants of 65.We start with the number 1.Divide 65 with 1. Is the remainder zero?Yes! By variable definition, the number 65 is divided equally by 1 without leaving a remainder.
65 ÷ 1 = 6565 × 1 = 65
Next, let"s shot the number 2. Since 65 is one odd number, it cannot be separated by 2 or multiples of 2. Hence, we require to check odd numbers only. Let"s try v number 3.65 ÷ 3 = 21.66Therefore, 3 is no a aspect of 65.Now, let"s try with number 5.
65 ÷ 5 = 135 × 13 = 65
Factors that 65 By prime Factorization
Prime factorization is the process that breaking down a composite number right into its prime factors.To get the element factorization that 65, we divide it through its smallest prime element which is 5.
65 ÷ 5 = 13The procedure of prime factorization walk on till we acquire the quotient together 1.The element factorization of 65 is presented below:
Prime administer of 65 can additionally be represented as follows:
65 = 5 × 13 × 1
Now the we have done the element factorization of our number, we can multiply them and get the various other factors. Can you shot and discover out if all the factors are covered or not?And as you can have currently guessed, because that prime numbers, there room no various other factors.
Explore determinants using illustrations and interactive examples
Challenging Question:
Mike needs to divide 65 student in his course into different groups with the same number of students in every the groups. Each team must have more than one student and not all students can be in one group. In how plenty of ways deserve to Mike type these groups?
Factors of 65 in Pairs
The pair of numbers that give 65 as soon as multiplied with each various other is referred to as the pair determinants of 65.Therefore, the pair factors of 65 are (1,65) and also (5,13).Since the product of two an unfavorable numbers is positive, i.e. (-) × (-) = (+), (-1,-65), (-5,-13) are additionally factor bag of 65.
See more: How Long Does It Take To Get To Hawaii By Boat From California To Hawaii
Important Notes:
The numbers which we multiply to get 65 are the determinants of 65.The determinants of 65 are 1, 5, 13, and also 65.As the number 65 is an odd composite number, every one of its determinants will also be odd.The number 65 is neither a perfect square nor a perfect cube. |
# How To Explain To A Child What An Area Is
## Video: How To Explain To A Child What An Area Is
Adults sometimes do not even think that they are using mathematical concepts in speech. They calmly talk about the area of an apartment or land plot, without even thinking that the child may not understand this. Meanwhile, a child will need the concept of area when studying geometry, physics, geography and a number of other sciences.
## It is necessary
• - White paper;
• - colored paper;
• - pencil;
• - ruler;
• - the cloth:
• - furniture;
• - country cottage area;
• - household items.
• - garden tools.
## Instructions
### Step 1
Teach your child to measure different objects. If the process does not appeal to him by itself, come up with practical tasks or create play situations. For example, ask him to find out if the table that you have been planning to take out there will pass through the country gate. To do this, you need to know the dimensions of the table and wicket. Explain that the zero mark should match the corner of the table and the end of the post that defines the wicket. Invite your child to write down the results and compare them. He will do it with ease if he already knows how to count.
### Step 2
Ask your assistant if there is enough space for a table in the corner where you decide to put it. Say that for this you need to know the area of the table itself and the space allotted for it in the country. You already know one size, but is it enough? Most likely, the child himself will understand that you need to know not only the length, but also the width of the table. Invite him to measure it and write down the result.
### Step 3
Invite your child to draw an area on the floor in the room where the table will fit. Let him do it with regular chalk. As a result, you will have a rectangle that occupies the same area on the floor as the table. Explain that area is what is inside the drawn line. It can be counted.
### Step 4
Show how the area is calculated. To do this, each side must be divided into equal segments - for example, 1 cm each. This can be imagined, or you can cut the same exact square from graph paper. Show your student the easiest way to calculate the area of a square or rectangle. To do this, you need to multiply the lengths of its sides. If the figure is not too large, the child can check the result by counting the small squares.
### Step 5
Explain that you measured the length and width with a regular ruler or tape measure. Pay the child's attention to the fact that the divisions are there every 1 cm. With a ruler with exactly the same divisions, you measured the width of the object. You divided the resulting shape into small 1cm squares. Such a square is called a square centimeter. There are also square decimeters, meters and kilometers. For now, these names will be enough for the child.
### Step 6
It is possible to explain what an area is on other subjects as well. For example, tell your child that you want to sew a scarf of a certain size, but are not sure if the remaining piece of fabric from the dress is enough. Invite your child to draw a strip on paper that matches the size of the scarf. What is inside the lines is called the area. Have your assistant cut a strip and place it on the fabric. Explain how to calculate the area without a pattern.
### Step 7
Use didactic games. Cut out a rectangle from cardboard, and from colored paper - several smaller geometric shapes. Ask the child to answer whether it is possible to place all of them on the large card or only part of them. The condition for such a game exercise should be that the figures cannot be tried on, but everything must be determined in advance. Invite your child to measure them, and then determine the area that each figure will occupy on the card. |
## How do you find the equation of a linear function?
Substitute the slope m and the y-value of the y-intercept b into the equation y=mx+b y = m x + b . Finding a linear equation is very straightforward if the slope and y-intercept are given.
## How do you write an equation for a function?
When you are given an equation and a specific value for x, there should only be one corresponding y-value for that x-value. For example, y = x + 1 is a function because y will always be one greater than x. Equations with exponents can also be functions.
## How do you write the equation of a linear equation?
Write the equation y = mx + b, substituting the values for m and b you calculated or determined. The m will be your slope, and the b will be your y-intercept. Leave the y and x variables in the equation as letter variables. Include the sign of the numbers you plug in.
## How do you find a linear equation in two variables?
If a, b, and r are real numbers (and if a and b are not both equal to 0) then ax+by = r is called a linear equation in two variables. (The “two variables” are the x and the y.) The numbers a and b are called the coefficients of the equation ax+by = r. The number r is called the constant of the equation ax + by = r.
## What is an equation in math?
An equation says that two things are equal. It will have an equals sign “=” like this: 7 + 2 = 10 − 1. That equation says: what is on the left (7 + 2) is equal to what is on the right (10 − 1) So an equation is like a statement “this equals that”
## How do I write an equation in standard form?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
## How do you graph a linear equation?
To graph a linear equation, we can use the slope and y-intercept.Locate the y-intercept on the graph and plot the point.From this point, use the slope to find a second point and plot it.Draw the line that connects the two points.
You might be interested: Oscillation frequency equation
## What is linear equation in algebra?
Linear equations are equations of the first order. These equations are defined for lines in the coordinate system. An equation for a straight line is called a linear equation. The general representation of the straight-line equation is y=mx+b, where m is the slope of the line and b is the y-intercept.
## What is standard form in math?
Standard form is a way of writing down very large or very small numbers easily. So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form.
## What is the Y intercept formula?
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.
## What are the 3 ways to graph a linear equation?
There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. The third is applying transformations to the identity function f(x)=x f ( x ) = x .
### Releated
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How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […] |
# Linear Algebra - Finding row space and column space
My question is as follows:
Find a basis for the row space and a basis for the column space by first reducing the matrix to row echelon form:
$$A =\left[\begin{array}{rrr} 5 & 9 & 3 \\ 3 & -5 & -6 \\ 1 & 5 & 3 \\ \end{array}\right]$$
Row operations:
$$(1/5) R_1 \rightarrow R_1$$ $$(-3) R_1 + R_2 \rightarrow R_2$$ $$(-1) R_1 + R_3 \rightarrow R_3$$ $$(-5/52) R_2 \rightarrow R_2$$ $$(-16/5) R_2 + R_3\rightarrow R_3$$
Gives us the matrix $A$ in row echelon form:
$$A =\left[\begin{array}{rrr} 1 & 9/5 & 3/5 \\ 0 & 1 & 3/4 \\ 0 & 0 & 0 \\ \end{array}\right]$$
Now, we find the column space:
$$C_1 =\left[\begin{array}{rrr} 1 \\ 0 \\ \end{array}\right] \in \mathbb{R^2}$$
$$C_2 =\left[\begin{array}{rrr} 9/5 \\ 1 \\ \end{array}\right] \in \mathbb{R^2}$$
$$C_3 =\left[\begin{array}{rrr} 3/5 \\ 3/4 \\ \end{array}\right] \in \mathbb{R^2}$$
Now, we find the row space:
$$r_1 =\left[\begin{array}{rrr} 1 & 9/5 & 3/5 \\ \end{array}\right] \in \mathbb{R^3}$$
$$r_2 =\left[\begin{array}{rrr} 0 & 1 & 3/4 \\ \end{array}\right] \in \mathbb{R^3}$$
Column spaces $$C_1, C_2, C_3$$ and my row spaces $$r_1, r_2$$
Is this all wrong? How is this done correctly?
• Doing the row operations showed you that you can take linear combinations and delete the last row which means the first two rows in the original matrix span the rows. And now looking at A transpose and putting that in row echelon will show you that the first two columns span. – Mr.Fry May 12 '15 at 22:28
Note that $$\DeclareMathOperator{rref}{rref}\rref \begin{bmatrix} 5 & 9 & 3 \\ 3 & -5 & -6 \\ 1 & 5 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -3/4\\ 0 & 1 & 3/4\\ 0 & 0 & 0 \end{bmatrix}$$ This tells us that $\DeclareMathOperator{rank}{rank}\rank(A)=2$ so any basis for $\DeclareMathOperator{Row}{Row}\Row(A)$ and $\DeclareMathOperator{Col}{Col}\Col(A)$ consists of two vectors.
For $\Col(A)$ we need only select two linearly independent columns of $A$. A nice trick is to select the two columns of $A$ that correspond to the pivot columns of $\rref(A)$. This gives \begin{align*} \begin{bmatrix} 5\\3\\1 \end{bmatrix}&& \begin{bmatrix} 9\\-5\\5 \end{bmatrix} \end{align*} as a basis for $\Col(A)$.
To find a basis for $\Row(A)$ we can choose the first two columns. This gives \begin{align*} \begin{bmatrix} 5&9&3 \end{bmatrix}&& \begin{bmatrix} 3&-5&-6 \end{bmatrix} \end{align*} as a basis for $\Row(A)$.
To find a basis for $\Row(A)$ in general we can compute $\rref(A^\top)$ and select the rows of $A$ that correspond to the pivot columns of $\rref(A^\top)$. |
# How Does Pythagorean Theorem Help Us Today?
## How Pythagorean theorem changed the world?
For the past 2500 years, the Pythagoras’ theorem, arguably the most well-known theorem in the world, has greatly helped mankind to evolve.
Its useful right angles are everywhere, whether it is a building, a table, a graph with axes, or the atomic structure of a crystal..
## Who is the father of mathematics?
ArchimedesThe present scientists can follow Archimedes’ footprints, who is the father of mathematics, to contribute to society and bring laurels to the nation. What is the Mathematics behind Covid-19?
## What is the formula a2 b2 c2 used for?
The last equation, a2 + b2 = c2, is called the Pythagorean Theorem. We say “The sum of the squares of the legs of a right triangle equals the square of its hypotenuse.” Good hint. Note that the hypotenuse sits by itself on one side of the equation a2 + b2 = c2.
## What is the formula of A² B²?
(A²-B²) = (A-B)² + 2AB.
## Is Pythagorean theorem only for right triangles?
Pythagoras’ theorem only works for right-angled triangles, so you can use it to test whether a triangle has a right angle or not. In the triangle above, if a 2 < b 2 + c 2 the angle is acute.
## Why is the Pythagorean theorem so important?
Being able to find the length of a side, given the lengths of the two other sides makes the Pythagorean Theorem a useful technique for construction and navigation.
## Why is Pythagoras important today?
He is best known in the modern day for the Pythagorean Theorem, a mathematical formula which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.
## How does Pythagoras theorem work?
Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. … The sides of a right triangle (say a, b and c) which have positive integer values, when squared, are put into an equation, also called a Pythagorean triple.
## How is the Pythagorean theorem used in construction?
The Pythagorean theorem is used extensively in carpentry and construction. Almost every carpentry project involves some combination of squares and triangles. The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. … It is the best way to square two walls.
## How do you solve a2 b2 c2?
Introduction: Pythagorean Theorem The formula is A2 + B2 = C2, this is as simple as one leg of a triangle squared plus another leg of a triangle squared equals the hypotenuse squared.
## Who invented math?
Beginning in the 6th century BC with the Pythagoreans, the Ancient Greeks began a systematic study of mathematics as a subject in its own right with Greek mathematics. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof.
## How do we use the Pythagorean theorem today?
The Pythagorean theorem is used any time we have a right triangle, we know the length of two sides, and we want to find the third side. For example: I was in the furniture store the other day and saw a nice entertainment center on sale at a good price. |
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# Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, – b)
Distance formula =$$\sqrt { (x_2 - x_1)^2 + (y_2 - y_1)^2}$$
(i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = $$\sqrt { (4 - 2)^2 + (1 - 3)^2}$$
$$= \sqrt{ 2^2 + (-2) ^2} = \sqrt{4 + 4}$$
$$= \sqrt{8}$$
$$= 2\sqrt{2}$$
(ii) Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get
d = $$\sqrt { (-5 + 1)^2 + (7 - 3)^2}$$
$$= \sqrt{ ((-4)^2 + (4) ^2} = \sqrt{16 + 16}$$
$$= \sqrt{32}$$
$$= 4\sqrt{2}$$
(iii) Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get
d = $$\sqrt { (-a - a)^2 + (-b - b)^2}$$
$$= \sqrt{ ((-2a)^2 + (-2b) ^2}$$
$$= \sqrt{4a^2 + 4b^2}$$
$$= 2\sqrt{a^2 + b^2}$$ |
# Polar Coordinates
In this section we'll learn about polar coordinates: what they are and how to create them in Doodle. We're interested in polar coordinates because they allow us to easily create regular polygons. We'll construct regular polygons in the final part of this section.
## Points in Two Dimensions
A point in two dimensions is most commonly specified using x and y coordinates. These are known as Cartesian coordinates after their inventor, René Descartes. We can visualize a Cartesian coordinate as projecting a vertical line from the x-axis and a horizontal line from the y-axis. The intersection of these two lines is the chosen point. The animation below shows this.
An alternate representation is polar coordinates. In polar coordinates, a point is specified by a distance from the origin and an angle. The animation below illustrates this.
In Doodle we can construct points using either Cartesian or polar coordinates. To create a Cartesian point we pass the x and y values to the Point constructor.
val cartesian = Point(3, 4)
To create a polar point we need a length and an angle. The length is just a number, but in Doodle angles are their own type called Angle. We usually construct angles in terms of degrees. Here's an example.
val ninety = 90.degrees
As well as degrees we can also use radians or turns. 2π radians make a full circle. A turn is a proportion of a full circle: 1.0 is a full circle (360 degrees), 0.5 is half a circle, and so on. Here are some examples.
val radians = 2.radians
val fullTurn = 1.turns
Now we know how to create an Angle we can create a Point using polar coordinates.
val polar = Point(5, 45.degrees)
## From Points to Polygons
Drawing polygons is our ultimate goal, and polar coordinates allow us to easily specify the corners, or vertices, of regular polygons. Look at the hexagon below. To specify the vertices in Cartesian coordinates we'd have to do some involved geometry. With polar coordinates, however, it's very simple. Each vertex is the same distance from the center but differs in angle, as the animation shows. In the case of the hexagon, each vertex is a 60 degree rotation from the one before. (This is because a full turn of 360 degrees must visit all vertices, and there are 6 evenly spaced vertices, so each vertex is a turn of 360/60 = 60 degrees.)
We can use this idea to draw circles, or another Image, at the vertices of a regular polygon. Here's an example demonstrating what I mean.
val dot = Image.circle(5).fillColor(Color.darkViolet)
val vertices =
dot.at(Point(100, 0.degrees))
.on(dot.at(Point(100, 60.degrees)))
.on(dot.at(Point(100, 120.degrees)))
.on(dot.at(Point(100, 180.degrees)))
.on(dot.at(Point(100, 240.degrees)))
.on(dot.at(Point(100, 300.degrees)))
Drawing this gives the output shown below.
### Flexible Layout with at
The example above uses a method we haven't seen before: at. This is another tool for laying out images, like on, beside, and above. at changes the position of an image relative to its origin. The understand this, and why we have to place the dots on each other, we need to understand how layout works in Doodle.
Every Image in Doodle has a point called its origin, and a bounding box which determines the extent of the image. By convention the origin is in the center of the bounding box but this is not required. We can see the origin and bounding box of an Image by calling the debug method.
The example code below uses debug to draw the origin and bounding box for:
• the entire image, consisting of three circles;
• the individual circles after they have been moved with at; and
• the individual circles before shifting them with at.
val c = Image.circle(40)
val c1 = c.beside(c.at(10, 10)).beside(c.at(10, -10)).debug
val c2 = c.debug.beside(c.at(10, 10).debug).beside(c.at(10, -10).debug)
val c3 = c.debug.beside(c.debug.at(10, 10)).beside(c.debug.at(10, -10))
val example = c1.above(c2).above(c3)
This produces the output shown below.
When we layout Images using above, beside, or on it is the bounding boxes and origins that determine how the individual components are positioned relative to one another. For on the rule is that the origins are placed on top of one another. For beside the rule is that origins are horizontally aligned and placed so that the bounding boxes just touch. The origin of the compound image is placed equidistant from the left and right edges of the compound bounding box on the horizontal line that connects the origins of the component images. The rule for above is the same as beside, but we use vertical alignment instead of horizontal alignment.
Using at moves an Image relative to its origin. So c.at(10, 10) moves the circle 10 units up and to the right relative to its origin.
Taken together these rules explain the results we've seen. When drawing the polygon vertices, we want all the elements to share the same origin, so we use on to combine Images that we have moved using at.
#### Exercise: Get To The Point
Implement a method polygonPoints that produces an Image with circles (or something else of your choice) at the vertices of a regular polygon as shown above. The method should accept two parameters:
• the number of sides of the polygon; and
• the radius (distance of the vertices from the center)
Below shows the output of
polygonPoints(3, 50)
.fillColor(Color.crimson)
.beside(polygonPoints(5, 50).fillColor(Color.lawngreen))
.beside(polygonPoints(7, 50).fillColor(Color.dodgerBlue))
(I've used color to make it clearer which points belong to which polygon.)
This is a structural recursion over the natural numbers, but we need a helper method to actually do the counting. Here's my implementation. I used turns to specify the angle turn, because I felt was the most natural way to express it.
def polygonPoints(sides: Int, radius: Double): Image = {
val turn = (1.0 / sides).turns
def loop(count: Int): Image =
count match {
case 0 => Image.empty
case n =>
Image
.circle(5)
.at(Point(radius, turn * n))
.on(loop(n - 1))
}
loop(sides)
} |
Problems on Divisibility Rules
Here are few problems on the divisibility rules of 2, 3, 4, 5, 6, 7, 8, 9, and 10 which will help the learners in revising their concepts on the divisibility rules.
1. Check whether 3456 is divisible by 2?
Solution:
The last digit is an even number (i.e. 6) hence 3456 is divisible by 2
2. Check whether 8577 is divisible by both 3 and 9?
Solution:
Um of the digits = 8 + 5 + 7 + 7 = 27
27 is divisible by 3 hence, the number is also divisible by 3.
Now, 27 is also divisible by 9, soothe number is divisible by 9
Hence, 8577 is divisible by both 3 and 9
3. Check whether 40800 is divisible by:
i) 5 ii) 10 iii) 25
Solution:
i) We know that if the last digit is 0 or 5 then the number is divisible by 5
Hence 40800 is divisible by 5
ii) As the last digit is 0 hence the number is divisible by 10
iii) According to the divisibility rule if a number has last two digits 0 or the number formed by last two digits is a multiple of 25 then the number is divisible by 25. Here in 40800 the last two digits is 0 hence the number in divisible by 25.
iii) If a number has last two digits 00, 25, 50 or 75 then the number is divisible by 25. Here, in 40800 the number ends with 00 hence, it is divisible by 25.
4. Is 2584 is divisible by both 2 and 4?
Solution:
The last digit is an even number hence the number is divisible by 2.
The number formed by the last two digits of 2584 is 84
Now, 84 is divisible by 4 and it gives 21
Hence, 2584 is divisible by both 2 and 4
5. Check which of the following numbers are divisible by 5 or 10 or both of them?
(i) 545
(ii) 8795
(iii) 3400
(iv) 6490
(v) 45220
Solution:
(i) 545
The last digit is 5 hence it is divisible by 5 but not divisible by 10.
(ii) 8795
The last digit is 5 hence it is divisible by 5 but not divisible by 10.
(iii) 3400
The last digit is 0 hence divisible by both 5 and 10.
(iv) 6490
The last digit is 0 hence divisible by both 5 and 10
(v) 45220
45220 has the last digit 0 hence, divisible by both 5 and 10
6. Is 9486 is divisible by 9? State whether the number is also divisible by 18 also?
Solution:
Sum of the digits = 9 + 4 + 8 + 6 = 27
As, 27 is divisible by 9 hence, 9486 is also divisible by 9
Moreover, the last digit is a even number that is 6 hence it is divisible by 2 as well.
Now, if a number is divisible by both 9 and 2 then the number is also divisible by 18
Yes, 9486 is divisible by 18
7. Check whether 1302 is divisible by 6?
Solution:
Sum of the digits = 1 + 3 + 0 + 2 = 6
As, 6 is divisible by 3, hence 1302 is also divisible by 3
Moreover, the last digit of 1302 is an even number (i.e. 2)
Hence, it is divisible by 2
As, 1302 is divisible by both 3 and 2 hence, it is also divisible by 6
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## How do you write an equation of a line that passes through the given point and is parallel to the given line?
Step 1: Find the slope of the line. To find the slope of the given line we need to get the line into slope-intercept form (y = mx + b), which means we need to solve for y: The slope of the line 3x – 5y = 9 is m = 3/5. Therefore, the slope of the line parallel to this line would have to be m = 3/5.
## How do you find the equation of a perpendicular line that passes through a given point?
First, put the equation of the line given into slope-intercept form by solving for y. You get y = -2x +5, so the slope is –2. Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6.
## How do you write an equation given two points?
Equation from 2 points using Slope Intercept FormCalculate the slope from 2 points.Substitute either point into the equation. You can use either (3,7) or (5,11)Solve for b, which is the y-intercept of the line.Substitute b, -1, into the equation from step 2.
## What is the equation of the line?
The equation of a straight line is usually written this way: y = mx + b.
## What is a line that passes through parallel lines?
Parallel Lines Conjectures A line passing through two or more other lines in a plane is called a transversal. A transversal intersecting two parallel lines creates three different types of angle pairs.
## How do you put something into standard form?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
## How do you find the equation of a line with one point and no slope?
If no slope is given we can let the slope = m .Given the point we can use the point-slope formula to write the equation for this problem. The point-slope formula states: (y−y1)=m(x−x1)(y−34)=m(x−−3)(y−34)=m(x+3)
You might be interested: Froude number equation
## How do you determine if two lines are parallel?
We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. Unlike parallel lines, perpendicular lines do intersect.
## Do parallel lines have the same slope?
In other words, the slopes of parallel lines are equal. Note that two lines are parallel if their slopes are equal and they have different y-intercepts. In other words, perpendicular slopes are negative reciprocals of each other.
## How do u find the midpoint of a line?
Midpoint of a Line SegmentAdd both “x” coordinates, divide by 2.Add both “y” coordinates, divide by 2.
## What is the Y intercept in an equation?
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis. © 2000-2005 Math.com. All rights reserved.
## How do you write an equation in slope intercept form with two points?
If you know two points on a line, you can use them to write the equation of the line in slope-intercept form. The first step will be to use the points to find the slope of the line. This will give you the value of m that you can plug into y = mx + b. The second step will be to find the y-intercept.
### Releated
#### Equation of vertical line
How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]
#### Bernoulli’s equation example
What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […] |
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# Factoring Special Products in Difference of Squares
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### Factoring Special Products in Difference of Squares
1. 1. Problem 1 9x2-4
2. 2. Problem 1 9x2 - 4 3x(3x) 2(2)Both of these can be squared so I will show you whatthey are squared by under them. To make our problemwe will try to fit the formula a2 + b2 = (a + b)(a-b). Sincethe first one is a2 (9x2), a is 3x, using this way ofthinking, I would say that b is 2. Our answer is on thenext page.
3. 3. Problem 1 9x2 – 4 = (3x + 2)(3x – 2)
4. 4. Example Problems a2 – 81 36m2 – 25 4x2 – y2 a2 + 64 Remember that both must be PERFECT squares.
5. 5. Example Problems a2 – 81 (a + 9)(a – 9) 36m2 – 25 (6m + 5)(6m – 9) 4x2 – y2 (2x + y)(2x – y) a2 + 64 This cannot be factored since this method doesn’t work with addition problems, only subtraction.
6. 6. Mini Lesson If you feel you are just doing the problems blindly, check them with F.O.I.L. and you will find that two of the numbers cancel out together.
7. 7. Problem 2 2a2 – 200
8. 8. Problem 2 2a2 – 200To make this problem work so that we have squareswe will have to divide it by 2. a2 – 100Now we can solve that to get (a + 10)(a – 10). We addthe 2 back by placing it next to the problem formultiplication, making our final answer look like theslide on the next page.
9. 9. Problem 2 2a2 – 200 = 2(a + 10)(a – 10)
10. 10. Problem 3 -4c2 + 36
11. 11. Problem 3 -4c2 + 36To make this work we will remember what we did inthe last problem and divide the problem by -4, makingit c2 - 9. Solving this the normal way we will get(c + 3)(c-3) which will change to be -4(c + 3)(c-3).
12. 12. Problem 3 -4c2 + 36 = -4(c + 3)(c-3)
13. 13. Formula a2 + 2ab + b2 = (a + b)(a + b) or (a + b)2
14. 14. Problem 4 25x2+ 10x + 1
15. 15. Problem 4 25x2+ 10x + 1We will answer this using the formula on slide 13.a2 + 2ab + b2 = (a + b)(a + b) or (a + b)225x2+ 10x + 1We will first get the square root the 25x2 (5x) and placeit in the ‘a’ place of the formula. Then we will get thesquare root of 1 (1) and place it in the ‘b’ place of theformula. The answer will be on the next slide.
16. 16. Problem 4 25x2+ 10x + 1 = (5x + 1)2
17. 17. Problem 5 144y2 - 120y + 25
18. 18. Problem 5 144y2 - 120y + 25We need to find a way to accommodate the negativesign in the middle so just blindly using our formula toget (12y + 5)(12y + 5) won’t work. We can however,make it (12y - 5)(12y – 5), which will achieve our goalsperfectly. |
# DAV Class 4 Maths Chapter 1 Worksheet 1 Solutions
The DAV Class 4 Maths Book Solutions and DAV Class 4 Maths Chapter 1 Worksheet 1 Solutions of Numbers up to 999999 offer comprehensive answers to textbook questions.
## DAV Class 4 Maths Ch 1 WS 1 Solutions
Question 1.
What number does the abacus show?
35132
61035
40000
45232
64046
44100
Question 2.
Represent the following numbers on the abacus
### DAV Class 4 Maths Chapter 1 Worksheet 1 Notes
• When we add 1 to any greatest number of 1, 2, 3… digit number it becomes the smallest number of 2, 3, 4… digit number respectively.
For example, when we add 1 to the greatest 3-digit number 999 it becomes smallest 4-digit number 1000.
• A 6-digit number have the place value of lakh. The smallest 6-digit number is 100000 and the largest 6-digit number is 999999.
• There are six places of a 6-digit number i.e., Ones, Tens, Hundreds, Thousands and Lakhs.
• The place value of a digit in a numeral is the product of the digit and its place in the numeral.
• When we read numerals of a number, all the digits in same period are read together, and the name of period (except ones) is read along with them.
• When we write, we separate periods by leaving space or inserting commas between the periods.For example, 682934 is written as 6 82 934 or 6,82,934.
• The expanded form of a numeral is the sum of the place values of each digit of the numeral.
• While comparing digits as greater or smaller, we first compare digits in lakhs place, then in thousands place followed by hundreds, tens and ones place.
Number One More Than 9999:
9999 + 1 = 10000
9999 is the greatest 4 digit number.
10000 is the smallest 5 digit number.
The 5 places of a 5-digit number are
Ones, Tens, Hundreds, Thousands and Ten thousands.
On abacus, 10000 is shown
10000 – Ten thousand
Some 5 – digit numbers |
# How do you solve 2(x=1)=3x-3?
Mar 25, 2015
Solve: $2 \left(x - 1\right) = 3 x - 3$
Remove the parentheses (by distributing the multiplication). Then Collect all terms involving the unknown ($x$) on one side and all other terms on the other side. Finish by dividing both sides by the coefficient of $x$ (the number in front of $x$.
It looks like this:
$2 \left(x - 1\right) = 3 x - 3$
$2 x - 2 = 3 x - 3$
Method 1, collect $x$-terms on the left.
$2 x - 2 = 3 x - 3$
(add $2$ and subtract $3 x$ from both sides. Then divide)
$2 x - 3 x = - 3 + 2$
$- x = - 1$
$\frac{- x}{- 1} = \frac{- 1}{- 1}$
$x = 1$
(There are other ways of convincing yourself that ix $- x = - 1$ is to be true, then we'll need to have $x = 1$)
Method 2, avoid negatives in front of $x$
$2 x - 2 = 3 x - 3$
(add $3$ and subtract $2 x$ on both sides)
$- 2 + 3 = 3 x - 2 x$
$1 = x$
$x = 1$ |
# How do you factor x^4 + 4?
May 15, 2015
${x}^{4} + 4$ has no linear factors since ${x}^{4} + 4 > 0$ for all real values of $x$.
${x}^{4} + 4 = \left({x}^{2} + a x + b\right) \left({x}^{2} + c x + d\right)$
$= {x}^{4} + \left(a + c\right) {x}^{3} + \left(b + d + a c\right) {x}^{2} + \left(a d + b c\right) x + b d$
Comparing coefficients of ${x}^{3}$ we must have $a + c = 0$, so $c = - a$
... $= {x}^{4} + \left(b + d - {a}^{2}\right) {x}^{2} + a \left(d - b\right) x + b d$
Looking at the coefficients of $x$, we either have $a = 0$ or $b = d$.
If $a = 0$ then $b + d = 0$ so $d = - b$ and $b d = - {b}^{2}$, which would require ${b}^{2} = - 4$ - not possible for real values of $b$.
If $b = d$, then since $b d = 4$, $b = d = 2$, which would make ${a}^{2} = b + d = 4$, so $a = \pm 2$.
Indeed ${x}^{4} + 4 = \left({x}^{2} + 2 x + 2\right) \left({x}^{2} - 2 x + 2\right)$ |
## GMAT Geometry: Central Angles
In poetry, a rose is a rose is a rose. On GMAT problems with central angle “slices” in circles, a fraction is a fraction is a fraction.
This may seem like common sense. Cut a pizza into six slices. If you cut it evenly, each slice now has one-sixth the cheese, one sixth the crust, and an angle of one sixth the way around a circle—that is, 60 degrees. However, though this may seem obvious, it’s actually a very useful technique for resolving certain geometry problems.
Consider the following Data Sufficiency question:
In a radius 6 circle, two points A and B are connected to the center, point O. What is angle AOB?
1) The length of the minor arc defined by sector O is 1.5π
2) The area of the sector defined by angle AOB is 4.5π
It looks all GMAT-like and formal, but if you actually think about it…, it’s just the pizza I described above. We’re taking a slice, and want to know what’s the angle of that slice. The “cheese” of the pizza is the area of a radius 6 circle, which is the radius squared times pi, or 36π. The “crust” is it’s circumference, which in this case would be the diameter times pi, 12π.
Statement 1) tells us that the length of the arc/crust is 1.5π, which is 1.5/12 = 1/8 of the circumference. And an eighth is an eighth is an eighth. Our slice, which goes one eighth of the way around the outside of the circle, and it goes an eighth of the way around the inside of our circle as well—its central angle is 1/8 * 360 = 45 degrees. Sufficient!
And statement 2) says that the area of our slice/sector is 4.5π, or 4.5/36 = 1/8 the total area of the whole pie. Once again, an eighth is an eighth is an eighth, so by the exact same reasoning we get 45 degrees. Sufficient!
This is a pretty basic rule, but it’s widely applicable to many circle problems. It can be especially useful with central triangles, which have central angles by definition. Take a look at today’s question of the day to test your skills, and when you get it right, treat yourself to a slice of pepperoni as a reward!
Question:
What is the area of the circle above with center O?
(1) The area of D AOC is 18.
(2) The length of arc ABC is 3π.
Step 1: Analyze the Question Stem
This is a Value question. For any Circle question, we
only need one defining parameter of a circle (area,
circumference, diameter, or radius) in order to calculate
any of the other parameters. Also, all radii of the same
circle will have the same length. So AO = CO. That makes
the triangle an isosceles right triangle (or 45-45-90) for
which we know the ratio of the sides. Not only would the
circle’s circumference, diameter, or radius be sufficient,
but information that gave us any side length of triangle
AOC would also be sufficient, as it would give us the length
Step 2: Evaluate the Statements
Statement (1): We are given the area, and we already know
that the base and the height are equal. So if we call the
radius of the circle r, then the area of the triangle is equal to
1/2 (Base)(Height) = 1/2 (r)(r) = 1/2 r ^2 = 18
Remember that we are not asked to calculate the actual
value of r. Because we have set up an equation with
one variable, and we know in this case that r can only
be positive since it is part of a geometry figure, we have
enough information to determine the area of the circle (the
solution would have been r ^2 = 36, r = 6).
Therefore, Statement (1) is sufficient. We can eliminate
choices (B), (C), and (E).
Statement (2): Because O is the center of the circle and
angle AOC measures 90 degrees, we know that the length
of arc ABC is one-fourth of the circumference. Because this
statement allows us to solve for the circumference, it is
sufficient.
(D) is correct. |
# Fraction notation activities and cards: 3 activities to illustrate fraction notation and concepts
Why are fractions difficult?
There are several reasons fractions are extremely difficult for students with dyscalculia. Both the notation, using two numbers close to each other, and the different ways we show fractions can cause confusion.
• When using the fraction notation, students need to work with two numbers at the same time and think of them in relation to each other. Not only is seeing the two numbers so close to each other confusing, the words used to ‘read out’ a fraction leave the impression it is ‘one entity’, ‘one number’ and does not bring in mind the ‘whole’ that is being divided.
• Not only is a fraction is written with two numbers, the numerator and the denominator, those two numbers actually have an opposite working on the size of the fraction: A larger numerator, the number written above the division line, makes the quantity of the fraction larger, which is in sync with children’s experience with positive numbers. In contrast to that, a larger denominator, the number written below the division line, makes the quantity of the fraction smaller. This is counter-intuitive to students who have learned that bigger numbers mean larger quantities! Students with dyscalculia usually do not understand and remember that these numbers have a totally different meaning depending on the place they are written.
• Another reason students with dyscalculia often get confused with fractions is because they do not automatically see the similarity between different models (such as folded squares, fraction strips, fraction circles, or pizza pies) that are used to illustrate fractions in their textbook or are used in class presentations.
• Last but not least, most students with dyscalculia are slower in copying from the board and prone to making reversals in the numbers as well as in the numerator and denominator of the fractions making for very confusing written notes.
The fraction notation cards and the symbol and number tiles in this activity are designed to show the different meanings of the top and bottom number, focusing on one number at a time, instead of both. The words numerator and denominator are unfamiliar vocabulary and do not add to understanding, so they are not used in this lesson, we call it top and bottom number. The top number shows how many equal parts you count. A larger top number makes the amount of the fraction larger. The bottom number shows what type of parts you count. A larger bottom number makes the amount of the fraction smaller.
# Graph paper is beneficial for all math students!
When you are shopping for school supplies, think about an extra journal or package of three ring binder paper: quad lined paper is the first and most economical help for your struggling Math student. A small minority of students gets visually confused by the squares on the page. So it is always best to search for paper that is very lightly printed with ‘unobtrusive’ squares: the emphasis should be on what your student is writing and drawing not on the grid.
Neatly lining up calculations prevents errors. Quad paper can also help illustrate many concepts that are more complicated to explain in words but are easily seen on paper, such as multiplication, area and perimeter, for making graphs, and drawing congruent shapes, when showing transformations in size, slides, flips, and rotations and many more.
# Shapin’up: first, second, and third place
This new free download is a game to practice the use of cardinal numbers.
It uses shapes in different colors and sizes, and other manipulatives like lego’s and is suitable for the youngest group of Math learners.
Go to Free Activities and Downloads and print the Shapin’up game board, rule cards and shapes. Laminate or glue on card-stock and cut out the cards and shapes.
Before you start, let your child play with the shapes and manipulatives, ask him/her to tell you about them and to sort them according to color and shape.
The actual activity:
Tell your child that you are going to play a game with the shapes. The first contest is the biggest one wins.
Show your child a row with three triangles in order from large to small and announce the winner, second and third place.
Play the game with other shapes. Your child announces first, second, and third place.
Alternate with your child to choose a rule (smallest – largest, longest – shortest, most corners – least corners, highest – lowest) and to put together the row of three and announce places.
Make variations with small toys or manipulatives you have available.
Have fun!
Children learn by doing
Research in education tells us over and over that hands on activities have more impact than verbal or written explanations alone. Simple manipulatives and pictures are used as key elements to illustrate math concepts.
The connections with your child’s everyday life experiences make it all sink in. Safely anchored in his memory he can now start to apply it in other situations.
Download our free guidelines and templates for games, outright playful and more serious activities to develop the math centers in his brain. There are sample activities to develop number sense, visuo-spatial activities and templates. Have fun!
We never share your information with any other organization, see our Terms and Conditions – Privacy Policy. |
## Math Story : Writing Multiplication Tables
#### Cirho’s Numbers Phobia
Cirho is a naughty and jolly child. He loves playing with his friends.
But he has a weird phobia. A phobia? Yes! A phobia of numbers. He is afraid of numbers. But what does this phobia look like?
Once Squarho and Cirho were playing a game of arranging the cups in equal groups on a table. “How many cups did you use in total?” asked Squarho. “1,2,3,4…Ummm! What is there in counting? Let me make a big tower using the cups”, says Cirho ignoring his question.
Another day Cirha and Cirho were counting their coins of value 2. “I have eight coins of value 2 making it 16 in total. How many do you have in total?” she asks. “1,2,3,4…Ummm! What is there in counting? Let us quickly go and buy some ice creams”, says Cirho ignoring her question again.
The next day he visits Triho. “Hey, Triho! What are you doing?” he says. “Nothing much! I wanted to give a new look to my room. So I am just completing the tile design on this wall by placing the tiles in equal groups as you can see”, says Triho. “Wow! This looks beautiful. Can I join too?” he asks.
Triho arranges the tiles in groups of 2s. Cirho arranges them in groups of 3s. “How many tiles have you used in total?” asks Triho. “1,2,3,4… Umm! What is there in counting? Let me make a beautiful design out of it”, says Cirho ignoring Triho’s question.
“No no! This is going to be fun. Find the total number of tiles quickly”, encourages Triho. “Haha! What is there in counting? Making a new design is more fun than counting”, he says and continues designing his tiles.
“Why do you run away from counting? Do you have a fear of numbers?” asks Triho curiously. “Yes! Whenever I start counting or find the total number of objects, it is like all the numbers start dancing on and I get confused”, says Cirho disappointedly.
“Cirho you are so cute! It takes a lot of time for you to find the total objects because you count them one by one. What if I show you a magic trick to count faster?” says Triho. “A magic trick? Really? Wow! Yes please tell”, says excited Cirho. What do you think is the magic trick to count faster?
“Look here, there are 2 tiles in each row and there are 4 such rows. So I have used 8 tiles in total”, says Triho confidently. “How did you count so fast? Did you use the magic trick? I also want to learn this. Tell me”, says Cirho.
“Haha! Yes definitely. All I did was use repeated addition and i.e. multiplication. In the first row, there are 2 tiles. In the second row, there are 2 more tiles. Now the total number of tiles is 2+2 which can be written as 2✕2 as we are adding two twice. The answer here is 4.
Similarly, the third row has 2 more tiles. So the total number of tiles is 2+2+2 which can be written as 2âś•3 as two is added thrice. The answer here is 6. Now can you try finding the total tiles till the fourth row?” says Triho. “Oh yes! It will be 2âś•4 as 2+2+2+2 will give the total tiles here. The answer is 8″, says Cirho confidently.
Triho continues to arrange the tiles on his wall. “7th row now! Adding 2 seven times will give me the total tiles here. So 2+2 +2.. Oh no! This is so time taking?” says Cirho. Can you help him find the total tiles using the multiplication method?
“Yes, to reduce this time and avoid adding 2 seven times, again and again, all you can do is remember 2×7 is 14. So there are 14 tiles in total now”, explains Triho. Cirho finally understands that math is all about shortcuts. He can use a multiplication table to avoid repeated addition every time. So simple.
“Hurrah! This is so easy. The numbers in my head have stopped dancing too. Thanks a ton, Triho”, says Cirho. He wants to find the total number of tiles he has arranged in groups of 3. Can you help Cirho again?
“See, you got it right this time. Isn’t this the best magic trick?” says Triho. Cirho is elated. “Yes, it is the best trick! To find the total objects, all I have to do is remember these multiplication tables. Hurrah!” he says. Triho is glad to help him.
From that day onwards, Cirho never ran away from finding the total number of objects. He overcame his fear. The numbers in his head stopped dancing. He confidently finds the total number of objects now using the multiplication table. Facing fears is always fun. Do you agree?
We Learnt That…
• The multiplication table shows the product of two numbers.
• To write the multiplication table of any number, we can use repeated addition.
• We should not run away from our fears, instead, we should face them bravely.
Let’s Discuss
• What kind of phobia do I have?
• How does Triho help me overcome my fear?
• What are multiplication tables? How can you write them?
• How we wrote multiplication tables of 2 and 3, similarly, can you write the tables of 4,5,6,7,8,9 and 10 using repeated addition?
• Facing fears is always fun. Do you agree? Why or why not? |
# 8.1: Integration by Substitution
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Calculus, Chapter 7, Lesson 1.
In this activity, you will explore:
• Integration of standard forms
• Substitution methods of integration
## Problem 1 – Introduction
1. Consider the integral \begin{align*}\int\limits \sqrt{2x+3}dx\end{align*}. Let \begin{align*}u = 2x + 3\end{align*}. Evaluate the integral using substitution.
Use the table below to guide you.
\begin{align*}f(x) =\end{align*} \begin{align*}\sqrt{2x+3}\end{align*}
\begin{align*}u = \end{align*} \begin{align*}2x+3\end{align*}
\begin{align*}du =\end{align*}
\begin{align*}g (u) =\end{align*}
\begin{align*}\int\limits g(u)du = \end{align*}
\begin{align*}\int\limits f(x)dx = \end{align*}
2. Try using substitution to integrate \begin{align*}\int\limits \sin (x) \cos (x) dx\end{align*}. Let \begin{align*}u = \sin(x)\end{align*}.
3. Now integrate the same integral, but let \begin{align*}u = \cos(x)\end{align*}. How does the result compare to the one above?
4. \begin{align*}\sin (x) \cos (x) dx\end{align*} can be rewritten as \begin{align*}\frac{1}{2} \ \sin(2x)\end{align*} using the Double Angle formula.
What is the result when you integrate \begin{align*}\int\limits \frac{1}{2} \ \sin(2x)\end{align*} using substitution?
## Problem 2 – Common Feature
Find the result of the following integrals using substitution.
5. \begin{align*}\int\limits \frac{x+1}{x^2+2x+3} dx\end{align*}
6. \begin{align*}\int\limits \sin(x) \ e^{\cos(x)} dx\end{align*}
7. \begin{align*}\int\limits \frac{x}{4x^2+1}dx\end{align*}
8. What do these integrals have in common that makes them suitable for the substitution method?
## Extension
Use trigonometric identities to rearrange the following integrals and then use the substitution method to integrate.
9. \begin{align*}\int\limits \tan(x) dx\end{align*}
10. \begin{align*}\int\limits \cos^3 (x)\end{align*}
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Fractions on a Number Line
```Fractions on a Number Line
Focus: Locate and name fractions on a number line.
Students will understand where a fraction falls on the number line between 0 and
1. Students will determine how to divide a number line into equal parts to
represent given fractions.
Materials:
sentence strips, (4-6 per student)
Or adding machine tape (4-6 per student cut in the same
length)
Activity: Activate student’s prior knowledge about numbers on the number line.
Discuss what a fraction is; a part of a whole or a part of a set. Students should
understand that a whole is equal to 1, (one whole) or that a set is (1 whole set).
Review what students know about fractions in relationship to the denominator
and the numerator. Ask students to visualize
3
. Students should be able to
4
describe the given fraction. For example, if you ask students to explain what
3
4
is, they should be able to explain that it is a whole divided into four equal parts
with three parts shaded. Or that it is a set of 4 fruit, where 3 are apples and one
is a banana. Repeat this activity with
2 6
3
,
and in order for students to keep a
3 8
6
visual of the fraction in mind.
Students have made fraction sets by folding sheets of paper and labeling equal
parts. Today students will use a number line to name fractions. Students will use
sentence strips to represent a number line. Students will fold the strips into equal
parts. Students will then label the fractions on the number line.
Give each student 4-6 sentence strips. Ask students where a fraction would land
on a number line. Will the fractions be greater than 1? Less than 1? Greater than
0? Once it is determined that fractions are greater than 0 and less than 1, have
students label the number lines on the sentence strips with a 0 at one end and a
1 at the other end.
Ask students to explain how they would show
1
on the sentence strip? Students
2
should be able to share that it is directly in the middle and that folding the whole
sentence strip in half will give them 2 equal parts. They should understand that
the fold is exactly half. Have students label
1
on the fold.
2
2 equal
parts
Ask students how would we divide our sentence strip into fourths? Once students
understand that folding for equal parts will give them the points on the number
line have students begin labeling their fraction number lines. Have students make
number lines that show thirds, sixths and eighths.
Once students have completed their fractions on the number lines ask them the
following questions.
1
fall on the number line? Closer to 0, ½ or 1?
8
3
3
Where does
fall on the number line? What is another name for ?
6
6
7
1
Is
closer to
or 1?
8
2
Where does
If my number line is divided into 4 equal parts and represents fourths, how
many points are on the number line between 0 and 1? (there are three, the
points are at
1 2
3
,
and ).
4 4
4
If my number line is divided into 3 equal parts and represents thirds, how
many points are on the number line between 0 and 1? (there are two, the
points are at
1
2
and ).
3
3
Additional Practice: Texas Student Activity Book pg. 39 and Fractions on a
Name: __________________________________ Date: _______________
Fractions on a Number Line Additional Practice
1. Look at the number line.
1
3
4
5
Circle the number
0 on the number line that is NOT shown correctly.
1
8
8
8
2. Look at the number
line. 8
0 equal parts the number line is divided into between
1
Determine how many
0 and
1 and label the fractions on the number line.
3. Look at the number line.
G
0
1
2
3
4
Label the points on the number line that are not labeled. What fraction is at
point G?
4. Look at the number line.
B
5
Label point B on the number line.
6
6
1
4
6
1
2
6
3
4
7
``` |
# How to Solve Integers and Absolute Value Problems? (+FREE Worksheet!)
Two vertical lines around a number or expression are used to indicate the absolute value of that number or expression. Here, you can learn how to find the absolute value of a number and how to solve math problems containing absolute values and integers.
The absolute value of the real number $$a$$ Is written in the form of $$| a |$$ and is a positive number. Two vertical lines around a number or expression are used to indicate the absolute value of that number or expression. The output value of the absolute value is always greater than or equal to zero. Absolute value is used to indicate the distance of a number from zero on the line of real numbers.
## Related Topics
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## Step-by-step guide to solving integers and absolute value problems
1. The absolute value of a positive number is equal to the same positive number.
2. The absolute value of zero is equal to zero.
3. The absolute value of a negative number is the positive value of that number.
• Note: To find the absolute value of a number, just find its distance from $$0$$ on a number line! For example, the distance of $$12$$ and $$- \ 12$$ from zero on number line is $$12$$!
### Integers and Absolute Value – Example 1:
Solve. $$|8 \ – \ 2| \ × \ \frac{ |- \ 4 \ × \ 6|}{3}=$$
Solution:
First solve $$|8 \ – \ 2|, →|8 \ – \ 2|=|6|$$, the absolute value of $$6$$ is $$6$$, $$|6|=6$$
$$6 \ × \ \frac{ |- \ 4 \ × \ 6|}{3}=$$
Now solve $$|- \ 4 \ × \ 6|, → |- \ 4 \ × \ 6|=|- \ 24|$$, the absolute value of $$- \ 24$$ is $$24$$, $$|- \ 24|=24$$
Then: $$6 \ × \ \frac{ 24}{3}= 6 \ × \ 8=48$$
### Integers and Absolute Value – Example 2:
Solve. $$\frac{ |- \ 12|}{3} \ × \ |9 \ – \ 4|=$$
Solution:
First find $$|- \ 12| , →$$ the absolute value of $$- \ 12$$ is $$12$$, then: $$|- \ 12|=12$$
$$\frac{12}{3} \ × \ |9 \ – \ 4|=$$
Next, solve $$|9 \ – \ 4|, → |9 \ – \ 4|=| \ 5|$$, the absolute value of $$\ 5$$ is $$5$$, $$| \ 5|=5$$
Then: $$\frac{12}{3} \ × \ 5=4 \ × \ 5=20$$
### Integers and Absolute Value – Example 3:
Solve. $$\frac{ |-18|}{9}×|5-8|=$$
Solution:
First find $$|-18| , →$$ the absolute value of $$-18$$ is $$18$$, then: $$|-18|=18$$
$$\frac{18}{9}×|5-8|=$$
Next, solve $$|5-8|, → |5-8|=|-3|$$, the absolute value of $$-3$$ is $$3$$, $$|-3|=3$$
Then: $$\frac{18}{9}×3=2×3=6$$
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### Integers and Absolute Value – Example 4:
Solve. $$|10-5|×\frac{ |-2×6|}{3}=$$
Solution:
First solve $$|10-5|, →|10-5|=|5|$$, the absolute value of $$5$$ is $$5, |5|=5$$
$$5×\frac{ |-2×6|}{3}=$$
Now solve $$|-2×6|, → |-2×6|=|-12|$$, the absolute value of $$-12$$ is $$12, |-12|=12$$
Then: $$5×\frac{ 12}{3}= 5×4=20$$
## Exercises for Solving Integers and Absolute Value Problems
### Evaluate.
• $$\color{blue}{|-43| – |12| + 10}$$
• $$\color{blue}{76 + |-15-45| – |3|}$$
• $$\color{blue}{30 + |-62| – 46}$$
• $$\color{blue}{|32| – |-78| + 90}$$
• $$\color{blue}{|-35+4| + 6 – 4}$$
• $$\color{blue}{|-4| + |-11|}$$
• $$\color{blue}{41}$$
• $$\color{blue}{133}$$
• $$\color{blue}{46}$$
• $$\color{blue}{44}$$
• $$\color{blue}{33}$$
• $$\color{blue}{15}$$
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# Question a04c1
Sep 4, 2017
$4.56 \cdot {10}^{- 3}$
#### Explanation:
Your goal here is to write your number in normalized scientific notation, which has
$\textcolor{w h i t e}{a a} \textcolor{b l u e}{m} \times {10}^{\textcolor{p u r p \le}{n} \textcolor{w h i t e}{a} \stackrel{\textcolor{w h i t e}{a a a a a a}}{\leftarrow}} \textcolor{w h i t e}{a \textcolor{b l a c k}{\text{the")acolor(purple)("exponent}} a a}$
$\textcolor{w h i t e}{\frac{a}{a} \textcolor{b l a c k}{\uparrow} a a a a}$
$\textcolor{w h i t e}{\textcolor{b l a c k}{\text{the")acolor(blue)("mantissa}} a}$
and
1 <= |color(blue)(m)| < 10" " " "color(darkorange)("(*)")#
As you know, you have
${10}^{0} = 1$
This means that you can write your initial number as
$\textcolor{b l u e}{0.00456} \cdot {10}^{\textcolor{p u r p \le}{0}}$
Now, to start converting the number to scientific notation, multiply it by $1 = \frac{\textcolor{b l u e}{10}}{\textcolor{p u r p \le}{10}}$ $\to$ keep in mind that we can multiply the number by $1$ because that leaves its value unchanged!
$\textcolor{b l u e}{0.00456} \cdot {10}^{\textcolor{p u r p \le}{0}} \cdot \frac{\textcolor{b l u e}{10}}{\textcolor{p u r p \le}{10}}$
You can rewrite this as
$\textcolor{b l u e}{0.00456} \cdot \textcolor{b l u e}{10} \cdot {10}^{\textcolor{p u r p \le}{0}} / \textcolor{p u r p \le}{10} = \textcolor{b l u e}{0.0456} \cdot {10}^{\textcolor{p u r p \le}{- 1}}$
At this point, you must check to see if the new value of the mantissa satisfies condition $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$.
Since
$1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\le}}} \textcolor{b l u e}{0.0456} \textcolor{red}{\cancel{\textcolor{b l a c k}{<}}} 10$
you must repeat the process again. This time, you have
$\textcolor{b l u e}{0.0456} \cdot {10}^{\textcolor{p u r p \le}{- 1}} \cdot \frac{\textcolor{b l u e}{10}}{\textcolor{p u r p \le}{10}}$
which is equivalent to
$\textcolor{b l u e}{0.0456} \cdot \textcolor{b l u e}{10} \cdot {10}^{\textcolor{p u r p \le}{- 1}} / \textcolor{p u r p \le}{10} = \textcolor{b l u e}{0.456} \cdot {10}^{\textcolor{p u r p \le}{- 2}}$
Condition $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ is still not satisfied, so you must repeat the process again. This time, you have
$\textcolor{b l u e}{0.456} \cdot {10}^{\textcolor{p u r p \le}{- 2}} \cdot \frac{\textcolor{b l u e}{10}}{\textcolor{p u r p \le}{10}}$
which is equivalent to
$\textcolor{b l u e}{0.456} \cdot \textcolor{b l u e}{10} \cdot {10}^{\textcolor{p u r p \le}{- 2}} / \textcolor{p u r p \le}{10} = \textcolor{b l u e}{4.56} \cdot {10}^{\textcolor{p u r p \le}{- 3}}$
Finally, you have
$1 \le \textcolor{b l u e}{4.56} < 10 \text{ " " } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$
so you can say that
$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.00456 = 4.56 \cdot {10}^{- 3}}}}$
Notice that the number written in scientific notation has $3$ sig figs, just like the number written in standard form. |
# Graphing arccos(x) functions
A step by step tutorial on graphing and sketching arccos(x) functions and also the domain and range of these functions and other properties are discussed.
In what follows, arccos(x) is the inverse function of f(x) = cos(x) for 0 ≤ x ≤ π.
The domain of y = arccos(x) is the range of f(x) = cos(x) for 0 ≤ x ≤ π and is given by the interval [-1 , 1]. The range of arccos(x) is the domain of f which is given by the interval [0 , π].
The graph, domain and range of both f(x) = cos(x) for 0 ≤ x ≤ π and arccos(x) are shown below.
A table of values of arccos(x) can made as follows:
x -1 0 1 y = arccos(x) π π/2 0
Note that there are 3 key points that may be used to graph arccos(x). These points are: (-1,π) , (0,π/2) and (1,0).
Example 1: Find the domain and range of y = arccos(x - 1) and graph it.
Solution to Example 1:
The graph of y = arccos(x - 1) will be that of arccos(x) shifted 1 unit to the right. The domain is found by stating that -1 ≤ x - 1 ≤ 1. Solve the double inequality to find the domain:
0 ≤ x ≤ 2
The 3 key points of arccos(x) can also be used in this situation as follows:
x - 1 -1 0 1 y = arccos(x-1) π π/2 0 x 0 1 2
The value of x is calculated from the value of x - 1. For example when x - 1 = -1, solve for x to find x = 0 and so on.
The domain is given by the interval [0,2] and the range is given by the interval [0,π]
The three points will now be used to graph y = arccos(x - 1).
Example 2: Find the domain and range of y = 2 arccos(x + 1) and graph it.
Solution to Example 2: We use the 3 key points in the table as follows, then find the value 2 arccos(x + 1) and x.
x + 1 -1 0 1 arccos(x+1) π π/2 0 y = 2 arccos(x+1) 2π π 0 x -2 -1 0
domain = [-2,0] , range = [0 , 2 π]
The graph is that of arccos(x) shifted one unit to the left and stretched vertically by a factor of 2.
Example 3: Find the domain and range of y = - arccos(x - 1) and graph it.
Solution to Example 3: We use the 3 key points in the table as follows, then find the value - arccos(x - 1) and x.
x - 1 -1 0 1 arccos(x-1) π π/2 0 y = - arccos(x-1) -π -π/2 0 x 0 1 2
domain = [0 , 2] , range = [- π , 0]
The graph is that of arccos(x) shifted one unit to the right and reflected on the x axis.
More references and links on Graphing Functions |
# FACTOR THEOREM EXAMPLES AND SOLUTIONS
## About "Factor Theorem Examples and Solutions"
Factor Theorem Examples and Solutions :
Here we are going to see some example problems to understand factor theorem.
To know the steps in factor theorem, please visit the page "Solving determinants using factor theorem".
## Factor Theorem Examples and Solutions - Questions
Question 1 :
Solve the following problems by using Factor Theorem :
(1) Solve
Solution :
By applying x = 0, we get identical rows and columns.
Hence the determinant will become 0.
So, x2 is a factor.
By adding row 1, row 2 and row 3, we get
4 - x + 4 + x + 4 + x = 0
12 + x = 0
x = -12
Hence -12 is the value which make the determinant zero. So the answers are 0, 0 and -12.
Question 2 :
Show that
Solution :
let us apply, x = y
Column 1 and 2 are identical. So the determinant will become zero.
Hence (x - y) is a factor. In the same way, we may show that (y - z) and (z - x) are factors.
Sum of exponents of leading diagonal = 3
A number of factors that we get so far = 3
Hence the required factor is a constant (k).
1(18 - 12) - 1(9 - 3) + 1(4 - 2) = k(-1)(-1)(2)
6 - 6 + 2 = 2k
k = 1
By applying the value of k, we get the given proof.
Question 3 :
In a triangle ABC, if
prove that triangle ABC is an isosceles triangle.
Solution :
By putting sin A = sin B, we get
That is, by putting sin A = sin B we see that, the given equation is satisfied.
Similarly by putting sin B = sin C and sin C = sin A, the given equation is satisfied.
Thus, we have A = B or B = C or C = A.
In all cases atleast two angles are equal. Thus the triangle is isosceles.
After having gone through the stuff given above, we hope that the students would have understood, "Factor Theorem Examples and Solutions".
Apart from the stuff given in "Factor Theorem Examples and Solutions" if you need any other stuff in math, please use our google custom search here.
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OTHER TOPICS
Profit and loss shortcuts
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Domain and range of rational functions with holes
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Graphing rational functions with holes
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Decimal representation of rational numbers
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Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
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Sum of all three four digit numbers formed with non zero digits
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Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Lecture 17: Chebychev's inequality and the weak law of large numbers
• Reading: MCS 20.2 20.4
• Chebychev's inequality
• statement, proof, example
• Weak law of large numbers
• statement, proof, setting up sample space / random variables
## Chebychev's inequality
Claim (Chebychev's inequality): For any random variable $$X$$, $Pr(|X - E(X)| \geq a) \leq \frac{Var(X)}{a^2}$
Proof: Note that $$|X - E(X)| \geq a$$ if and only if $$(X - E(X))^2 \geq a^2$$. Therefore $$Pr(|X - E(X)| \geq a) = Pr((X - E(X))^2 \geq a^2)$$. Applying Markov's inequality to the variable $$(X - E(X))^2$$ gives
\begin{aligned} Pr(|X - E(X)| \geq a) &= Pr((X - E(X))^2 \geq a^2) \\ &\leq \frac{E((X - E(X))^2)}{a^2} \\ &= \frac{Var(X)}{a^2} \end{aligned}
by definition.
Example: Last time we used Markov's inequality and the fact that the average height is 5.5 feet to show that if a door is 55 feet high, then we are guaranteed that at least 90% of people can fit through it.
If we also know that the standard deviation of height is $$σ = 0.2$$ feet, we can use Chebychev's inequality to build a smaller door. Let $$X$$ be the height random variable. $$Var(X) = σ^2 = 0.04$$.
If $$x - E(X) \geq a$$ then $$|x - E(X)| \geq a$$. Therefore, the event $$(X - E(X))$$ is a subset of the event $$(|X - E(X)| \geq a)$$, and thus $$Pr(X - E(X) \geq a) \leq Pr(|X - E(X)| \geq a)$$. This lets us apply Chebychev's inequality to conclude $$Pr(X - E(X) \geq a) \leq \frac{Var(X)}{a^2}$$.
Solving for $$a$$, we see that if $$a \geq .6$$, then $$Pr(X -E(X) \geq a) \leq 0.10$$. This in turn gives us $$Pr(X \lt a + E(X)) = Pr(X - E(X) \lt a) \geq 0.9$$. Thus, if the door is at least $$6.1$$ feet tall, then 90% of the people can fit through.
## Weak law of large numbers
Suppose we wish to estimate the average value of the height of a population by sampling $$n$$ people from the population and averaging their height. The weak law of large numbers says that this will give us a good estimate of the "real" average.
Formally, we can model this experiment by letting our outcomes be sequences of $$n$$ people. We can define several random variables: $$X_1$$ is the height of the first person sampled; $$X_2$$ is the height of the second person sampled, $$X_3$$ is the height of the third and so forth.
Since these are all measures of height, $$E(X_1) = E(X_2) = \cdots = E(X_n)$$ (let's call this value $$\mu$$) and $$Var(X_1) = \cdots = Var(X_n)$$ (let's call this value $$\sigma^2$$). The result of our sampled average is given by the random variable $$(X_1 + X_2 + \cdots + X_n)/n$$. The weak law of large numbers says that this variable is likely to be close to the real expected value:
Claim (weak law of large numbers): If $$X_1, X_2, \dots, X_n$$ are independent random variables with the same expected value $$\mu$$ and the same variance $$σ^2$$, then $Pr\left(\left|\frac{X_1 + X_2 + \cdots + X_n}{n} - μ\right| \geq a\right) \leq \frac{σ^2}{na^2}$
Proof: By Chebychev's inequality, we have $Pr\left(\left|\sum X_i/n - E(\sum X_i/n)\right| \geq a\right) \leq \frac{Var(\sum X_i/n)}{a^2}$
Now, by linearity of the expectation, we have $E(\sum X_i/n) = \sum E(X_i)/n = nμ/n = μ$
As was shown in homework 5, $$Var(cX) = c^2Var(X)$$, and we also know that if $$X$$ and $$Y$$ are independent, that $$Var(X + Y) = Var(X) + Var(Y)$$. Therefore, we have $Var(\sum X_i/n) = \sum Var(X_i)/n^2 = nσ^2/n^2 = σ^2/n$
Plugging these into the result from Chebychev's, we have $Pr\left(\left|\sum X_i/n - μ\right| \geq a\right) \leq \frac{σ}{na^2}$
which is what we were trying to show. |
Question Video: Finding the Relation between Two Given Sets | Nagwa Question Video: Finding the Relation between Two Given Sets | Nagwa
# Question Video: Finding the Relation between Two Given Sets
Given that π = {2, 4}, π = {4, 5}, and the universal set π’ = {5, 4, 9, 2}, find the complement of the intersection of π and the complement of π.
03:02
### Video Transcript
Given the set π is two, four, the set π is four, five, and the universal set π’ is five, four, nine, two, find the complement of the intersection of π and the complement of π.
Before we begin figuring out exactly what this means, letβs draw a Venn diagram to help organize all of our elements. So here weβve begun to draw the Venn diagram. π has elements two and four. And π has elements four and five. So π and π share the element four. And this tiny place where they overlap is where we should put the elements that they have in common. So four will go here. And then π also has the element two. And π also has the element five.
And the universal set is essentially the entire set. The universal set holds all of the elements. And if thereβs another set, it must be a subset of the universal set. So π and π must be subsets of the universal set. Therefore, they must be within the universal set.
So the universal set has five, four, nine, and two. Five, four, and two are already in π and π. So itβs nine that we need to introduce. And nine is not in π. And itβs not in π. So we need to write it outside of these, as we have.
So letβs begin dissecting this. So we have the intersection of π and the complement of π. And we want the complement of the entire thing. So first of all, this symbol means intersection. And the intersection of sets are the elements that are in both sets. So we have the set π. And then we have the set of the complement of π.
So the complement of π, which we know by the line on top of π, means we want all of the elements that are not in π. So if we look at the set π, π has elements four and five. So the only other elements that are not in π would be nine and two. So the complement of π holds the elements nine and two.
And we want the intersection of the complement of π and π. Well, π has elements two and four. So if we want the intersection of these two sets, what numbers do they have in common? They have the two in common. So to represent π and the intersection of the complement of π, so just whatβs inside the parentheses we have here, itβs just two. Two is the only element thatβs in π and isnβt in π.
But now it says that we need to take the complement of that. So we want everything thatβs not in that. So all of the other elements that are inside that pink would be nine, four, and five. So this will be our final answer. So the complement of the intersection of π and the complement of π would be nine, four, and five. And the order that we write these elements does not matter. So once again, this will be our final answer.
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