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# What is the formula of conditional probability?
What is the formula of conditional probability?
## What is the formula of conditional probability?
The formula for conditional probability is derived from the probability multiplication rule, P(A and B) = P(A)*P(B|A). You may also see this rule as P(A∪B). The Union symbol (∪) means “and”, as in event A happening and event B happening.
## What is conditional probability answer?
Conditional Probability Definition The probability of occurrence of any event A when another event B in relation to A has already occurred is known as conditional probability. It is depicted by P(A|B).
What is conditional probability PDF?
Conditional Probability. Definition. The conditional probability of an event given another is the probability of the event given that the other event has occurred. If P(B) > 0, P(A|B) = P(A and B) P(B) With more formal notation, P(A|B) = P(A ∩ B) P(B) , if P(B) > 0.
### What is conditional problem solving in computer?
CONDITIONAL: In a conditional statement you make a test. The result of the test is a Boolean – either True or False. If the result of the test is True you take a certain course of action and if the result of the test is False you take another course of action. IF condition THEN sequence 1 ELSE sequence 2 ENDIF.
### How do you calculate conditional mean?
The conditional expectation (also called the conditional mean or conditional expected value) is simply the mean, calculated after a set of prior conditions has happened….Step 2: Divide each value in the X = 1 column by the total from Step 1:
1. 0.03 / 0.49 = 0.061.
2. 0.15 / 0.49 = 0.306.
3. 0.15 / 0.49 = 0.306.
4. 0.16 / 0.49 = 0.327.
What best defines a conditional probability Mcq?
In probability theory, the probability measure of an event is made if another event has already occurred is referred to as conditional probability.
## What is conditional probability Slideshare?
Conditional Probability: • the probability of an event ( A ), given that another ( B ) has already occurred. • Where two events, A and B, are dependent.
## What is conditional statement with example?
Example. Conditional Statement: “If today is Wednesday, then yesterday was Tuesday.” Hypothesis: “If today is Wednesday” so our conclusion must follow “Then yesterday was Tuesday.” So the converse is found by rearranging the hypothesis and conclusion, as Math Planet accurately states.
How to solve this conditional probability problem?
conditional probability equation from the multiplication rule: step 1: type the multiplication rule: P(A and B) =P(A)*P(B| A) Step 2: Split the two sides of the equation by P(A): P(A and B) / P(A) = P(A) * P(B| A) / / P(A) Step 3: Cancel P(A) on the right side of the equation: P(A and B) / P(A) = P(B| A) Step 4: Rewrite the equation: P(B| A) =P(A
### How to calculate simple conditional probabilities?
Reading comprehension – ensure that you draw the most important information from the related conditional probabilities lesson
• Making connections – use understanding of the concept simple conditional probabilities
• Problem solving – use acquired knowledge to solve simple conditional probabilities practice problems
• ### How do you find conditional probability?
– The conditional probability that event A occurs, given that event B has occurred, is calculated as follows: – P (A|B) = P (A∩B) / P (B) – where: – P (A∩B) = the probability that event A and event B both occur. – P (B) = the probability that event B occurs. |
Question Video: Find the Displacement Vector | Nagwa Question Video: Find the Displacement Vector | Nagwa
# Question Video: Find the Displacement Vector Mathematics • Third Year of Secondary School
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Given π = <6, 1, 4> and π = <3, 1, 2>, find ππ.
02:20
### Video Transcript
Given vector π equals six, one, four and vector π equals three, one, two, find ππ.
We recall that to find ππ, we subtract vector π from vector π. The reason for this can be shown on a two-dimensional grid. This can then be transferred to three-dimensional vectors.
Letβs consider two points on the grid, A and B. The vector π is given from its displacement from the origin O. The same is true of vector π. We wish to travel from point A to point B. This is the same as traveling from point A to point O and then point O to point B. Going from point A to point O would be equivalent to negative vector π. And going from point O to point B is equivalent to vector π. ππ is equal to negative vector π plus vector π. This can be rewritten as vector π minus vector π.
Weβre told in this question that vector π is equal to six, one, four. This could also be written as a column vector as shown. It is also sometimes written in terms of vectors π’, π£, and π€. In this case, six π’ plus π£ plus four π€, where six, one, and four are the coefficients of π’, π£, and π€.
For the purposes of this question, we will stick with the notation given. And weβre also told that vector π is equal to three, one, two. In order to calculate π minus π, we simply subtract the individual components. Three minus six is equal to negative three. One minus one is equal to zero. Finally, two minus four is equal to negative two. ππ is therefore equal to negative three, zero, negative two.
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# Simplification in Decimals
Simplification in decimals can be done with the help of Pemdas Rule.
Pemdas Rule Table
From the above chart we can observe that first we have to work on "P or Parentheses" and then on "E or Exponents", then from left to right doing either "Multiplication" or "Division" as we find them in the question.
Then from left to right doing either "Addition" or "Subtraction" as we find them we need to solve them accordingly.
Let us consider some of the examples on simplification in decimals:
1. 4.5 + 3.7 – 3.6 ÷ 1.2
Solution:
4.5 + 3.7 – 3.6 ÷ 1.2
= 4.5 + 3.7 – 3.0
= 8.2 – 3.0
= 5.2
2. 3 ÷ 16 + 1.2 × 1/4 - {1/5+ (1-0.8) }
Solution:
3 ÷ 16 + 1.2 × 1/4 - {1/5+ 1-0.8}
= 3 ÷ 16 + 1.2 × 1/4 - {(1+5-4.0)/5}
= 3 ÷ 16 + 1.2 × ¼ - {(6-4)/5}
= 3 ÷ 16 + 1.2 × ¼ - {2/5}
= 3 ÷ 16 + 1.2 × ¼ - 2/5
= 3/16 + 1.2/4 – 2/5
= (15+24-32)/80
= 7/80
= 0.0875
Expanded form of Decimal Fractions
Like Decimal Fractions
Unlike Decimal Fraction
Equivalent Decimal Fractions
Changing Unlike to Like Decimal Fractions
Comparison of Decimal Fractions
Conversion of a Decimal Fraction into a Fractional Number
Conversion of Fractions to Decimals Numbers
Subtraction of Decimal Fractions
Multiplication of a Decimal Numbers
Multiplication of a Decimal by a Decimal
Properties of Multiplication of Decimal Numbers
Division of a Decimal by a Whole Number
Division of Decimal Fractions
Division of Decimal Fractions by Multiples
Division of a Decimal by a Decimal
Division of a whole number by a Decimal
Conversion of fraction to Decimal Fraction
Simplification in Decimals
Word Problems on Decimal |
# RS Aggarwal Solutions for Class 8 Maths Chapter 5 - Playing with Numbers Exercise 5A
Students can refer and download RS Aggarwal Solutions for Class 8 Maths Chapter 5- Exercise 5A, Playing with Numbers from the links provided below. Our experts have solved the RS Aggarwal Solutions to ensure that the students have in-depth knowledge of the basic concepts by practicing the solutions.
In Exercise 5A of RS Aggarwal Class 8 Maths, we shall study Numbers in the generalised form- A number is said to be in a generalised form if it is expressed as the sum of the products of its digits with their respective place values.
## Download PDF of RS Aggarwal Solutions for Class 8 Maths Chapter 5- Playing with Numbers Exercise 5A
### Access answers to RS Aggarwal Solutions for Class 8 Maths Chapter 5- Playing with Numbers Exercise 5A
Q1. The units digit of a two-digit number is 3 and seven times the sum of the digits is the number itself. Find the number.
Solution:
We know that the unit place is 3
So let’s consider the tens place as y
So the equation is (10y + 3)…………….. Equation (1)
From the question, seven times the sum of the digits is the number itself
∴from the above condition, 7(y + 3)……….. Equation (2)
Combining equation 1 and 2 we get,
7(y + 3) = (10y + 3)
7y + 21 = 10y + 3
7y -10y = 3 – 21
-3y = -18
∴ y = 6
Substituting the value of y in equation 1 we get,
10y + 3
10(6) + 3 = 60 + 3 = 63
∴ the required number is 63
Q2. In a two-digit number, the digit at the units place is double the digit in the tens place. The number exceeds the sum of its digits by 18. Find the number.
Solution: we know that the digit at the units place is double the digit in the tens place.
So let’s consider the tens place as y
The digit at the unit place is 2y
So the equation is (10y + 2y) = 12y…………….. Equation (1)
From the question, the number exceeds the sum of its digits by 18
∴from the above condition, (y + 2y) + 18……….. Equation (2)
Combining equation 1 and 2 we get,
(y + 2y) + 18 = (10y + 2y)
3y + 18 =12y
18 = 12y -3y
18 = 9y
∴ y = 2
Hence, the digit tens place is 2
The digit unit place is 2y = 2(2) = 4
Substituting the value of y in equation 1 we get,
12y
12(2) = 24
∴ the required number is 24
Q3. A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.
Solution: let us consider the unit place digit as x and tens place digit as y.
The equations becomes 10y + x……..equation (1)
From the question, a two-digit number is 3 more than 4 times the sum of its digits
∴from the above condition, 4(y + x) + 3……… equation (2)
Combining equation 1 and 2
4(y + x) + 3 = 10y + x
4y + 4x + 3 = 10y + x
4x – x + 4y – 10y = -3
3x – 6y = -3
3(x – 2y) = -3
x -2y = -1 …………..equation (3)
From the second condition, If 18 is added to the number, its digits are reversed
∴the reversed number is 10x + y……..equation (4)
∴by the given condition
(10y + x) + 18 = 10x + y
10y – y =10x –x -18
9y – 9x = -18
9(y – x) = -18
y – x = -2 ……….equation (5)
Solving equation 3 and 5 simultaneously we get,
y=3 and x = 5
∴the required number is (10y + x) = (10(3) + 5) = 30 + 5 = 35
### Access other Exercises of RS Aggarwal Solutions for Class 8 Maths Chapter 5- Playing with Numbers
Exercise 5B Solutions 15 Questions
Exercise 5C Solutions 14 Questions
Exercise 5D Solutions 9 Questions
## RS Aggarwal Solutions for Class 8 Maths Chapter 5- Playing with Numbers Exercise 5A
Exercise 5A of RS Aggarwal Class 8, Playing with Numbers contains the basic concepts related to Numbers. This exercise mainly deals with the basic properties of numbers. Some of them include
• Two-digit Numbers.
• Three-digit Numbers.
The RS Aggarwal Solutions can help the students be rigorous in their practice while learning the fundamentals as it provides all the answers to the questions from the RS Aggarwal textbook. Students are suggested to practice the problems on a regular basis which will help them excel in their exams and increase their overall percentage. Practicing more number of times helps in achieving high marks, in turn, helps in time management skills and also improves the confidence level. |
# Equation of the Medians of a Triangle
To find the equation of the median of a triangle we examine the following example: Consider the triangle having vertices $A\left( { - 3,2} \right)$, $B\left( {5,4} \right)$ and $C\left( {3, - 8} \right)$.
If $G$ is the midpoint of side $AB$ of the given triangle, then its coordinates are given as $\left( {\frac{{ - 3 + 5}}{2},\frac{{2 + 4}}{2}} \right) = \left( {\frac{2}{2},\frac{6}{2}} \right) = \left( {1,3} \right)$.
Since the median $CG$ passes through points $C$ and $G$, using the two-point form of the equation of a straight line, the equation of median $CG$ can be found as
If $H$ is the midpoint of side$BC$ of the given triangle, then its coordinates are given as $\left( {\frac{{3 + 5}}{2},\frac{{ - 8 + 4}}{2}} \right) = \left( {\frac{8}{2},\frac{{ - 4}}{2}} \right) = \left( {4, - 2} \right)$.
Since the median $AH$ passes through points $A$ and $H$, using the two-point form of the equation of a straight line, the equation of median $AH$ can be found as
If $I$ is the midpoint of side$AC$ of the given triangle, then its coordinates are given as $\left( {\frac{{ - 3 + 3}}{2},\frac{{2 - 8}}{2}} \right) = \left( {0,\frac{{ - 6}}{2}} \right) = \left( {0, - 3} \right)$.
Since the median $BI$ passes through points $B$ and $I$, using the two-point form of the equation of a straight line, the equation of median $BI$ can be found as |
# Class 6 Maths Playing With Numbers | Model Questions
CLASS 6 MATHEMATICS | Playing With Numbers | Practice Problems
1. Write all the factors of 27.
2. Write first five multiples of 8.
3. Express the following as the sum of two odd primes.
a) 28
b) 18
4. Using divisibility test, determine 2150 is divisible by 4 and by 8.
5. Using divisibility tests, determine which of the following numbers are divisible by 11.
a) 1783695
b) 80124
6. Using divisibility tests, determine 87512 is divisible by 6.
7. Find the common multiples of 3, 4 and 5 and find its LCM?
8. Find the common factors of 4, 8 and 10 and find its HCF?
9. Which of the following numbers are co prime?
a) 23, 42
b) 30, 45
10. Determine the greatest 2 digit number exactly divisible by 4, 8 and 12?
1. Factor pairs of 27 are 1 x 27, 3 x 9.
So factors of 27 are 1, 3, 9 and 27.
2. First five multiples of 8 are 8, 16, 24, 32, and 40.
3. a) 28 = 23 + 5
b) 18 = 11 + 7
4. A number with 3 or more digits is divisible by 4 if the number formed by its last two digits is divisible by 4.
Here the number is 2150.
Its last two digits are 50.
50 is not divisible by 4, therefore the number 2150 is not divisible by 4.
A number with 4 or more digits is divisible by 8, if the number formed by the last three digits is divisible by 8.
Here the number is 2150.
Its last three digits are 150.
150 is not divisible by 8, therefore the number 2150 is not divisible by 8.
5. a) 1783695
Sum at odd place = 5 + 6 +8 +1 = 20
Sum at even place = 9 + 3+7 = 19
Difference = 20- 19 = 1
Therefore 1783695 is not divisible by 11.
b) 80124
Sum at odd place = 4 + 1 + 8 = 13
Sum at even place = 2 + 0 = 2
Difference = 13 – 2 = 11
Therefore, 80124 is divisible by 11.
6. If a number is divisible by 2 and 3 both then it is divisible by 6 also.
Here the number is 87512.
Since it is an even number, it is divisible by 2.
Sum of the digits = 8 + 7 + 5+ 1+2 = 23, which is not a multiple of 3.
So it is not divisible by 3.
Therefore, 87512 is not divisible by 6.
7. Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60 ……..
Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60……………………….
Multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 ……………..
Common multiples are 60, 120, 180 ……………………
LCM = 60
8. Factors of 4 are 1, 2, and 4.
Factors of 8 are 1, 2, 4, and 8.
Factors of 10 are 1, 2, 5, and 10.
Common factors are 1 and 2.
HCF = 2
9. Two numbers having only 1 as a common factor are called co – prime numbers.
a) Factors of 23 are 1 and 23.
Factors of 42 are 1, 2, 3, 6, 7, 14, 21and 42.
But the common factor is 1.
Therefore 23 and 42 are co – prime.
b) Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
Factors of 45 are 1, 3, 5, 9, 15 and 45.
Common factors are 1, 3, 5 and 15.
Therefore 30 and 45 are not co – prime.
10. Here we need to find the LCM of 4, 8 and 12.
Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40………………..
Multiples of 8 are 8, 16, 24, 32, 40, 48……………..
Multiples o 12 are 12, 24, 36, 48, 60 ………………
Common Multiples are 24, 48, 72, 96, 120, 144 …………………………..
Therefore, the greatest two digit number which is exactly divisible by 4, 8 and 12 are 96. |
Question Video: Finding the Width of a Parallelogram given the Perimeter and a Relation between the Parallelogram’s Dimensions | Nagwa Question Video: Finding the Width of a Parallelogram given the Perimeter and a Relation between the Parallelogram’s Dimensions | Nagwa
Question Video: Finding the Width of a Parallelogram given the Perimeter and a Relation between the Parallelogramβs Dimensions
π΄π΅πΆπ· is a parallelogram. Given that the perimeter of π΄π΅πΆπ· is 54, find the length of π΄π΅.
04:53
Video Transcript
π΄π΅πΆπ· is a parallelogram. Given that the perimeter of π΄π΅πΆπ· is 54, find the length of π΄π΅.
Letβs look at the diagram carefully. The lengths of the four sides of the parallelogram have been given in terms of three variables: π₯, π¦, and π§.
In order to work out the length of π΄π΅, weβre going to need to know the values of these variables or at least the value of π¦. Weβre told that the perimeter of the parallelogram π΄π΅πΆπ· is 54, so letβs start with that. The perimeter of a parallelogram is the sum of all four of its sides. So this means if we add together the expressions for π΄π΅, π΅πΆ, πΆπ·, and π·π΄, in terms of π₯, π¦, and π§, we must get 54.
This gives the equation on screen. Now this equation can be simplified by grouping like terms together. Thereβs only one π¦-term, so thatβs still seven π¦. The three π₯ and the five π₯ make eight π₯, and the seven π§ and six π§ make 13π§. The constant on the left-hand side also simplifies to negative 53. We can then add 53 to both sides of this equation, giving us seven π¦ plus eight π₯ plus 13π§ equals 107.
Now this is one equation with three unknown letters: π₯, π¦, and π§. So there isnβt anything else we can do with this equation at this stage. Letβs look back to the diagram to see what else we can do. A key fact about parallelograms is that opposite sides are of equal length. This means that π΄π΅ is equal to πΆπ·, and also π΄π· is equal to π΅πΆ.
Letβs substitute the expressions for each of these sides in terms of the letters π₯, π¦, and π§. For the first pair of sides, we have that seven π¦ minus 38 is equal to six π§ minus one. Adding 38 to both sides simplifies this to seven π¦ equals six π§ plus 37. Now this is one equation with two letters so thereβs nothing further we can do with this. Letβs look at the equations for π΄π· and π΅πΆ. Substituting the expressions for each side gives five π₯ minus four is equal to three π₯ plus seven π§ minus 10.
Subtracting three π₯ from both sides and also adding four simplifies this equation to two π₯ is equal to seven π§ minus six. Now we canβt solve either these equations, but what theyβve done is give us both π¦ and π₯ in terms of π§. Each of these expressions for π₯ and π¦ can be substituted into our equation for the perimeter so that instead of it involving π₯, π¦, and π§, it now just involves π§.
The expression six π§ plus 37 can be substituted for the seven π¦ in the equation. If we think of eight π₯ as four multiplied by two π₯, then our expression for two π₯ of seven π§ minus six can also be substituted into the equation for the perimeter. Making both of these substitutions gives six π§ plus 37 plus four lots of seven π§ minus six plus 13π§ is equal to 107.
Expanding brackets and grouping like terms gives 47π§ plus 13 equals 107. Subtracting 13 from both sides of this equation gives 47π§ equals 94. Finally, dividing both sides of the equation by 47 tells us that the value of π§ is two. Now we could go on to use this value of π§ to calculate the values of π₯ and π¦, but letβs look back at the question.
The question asked us to find the length of π΄π΅. Weβve already stated that the length of π΄π΅ is the same as the length of πΆπ·. We have an expression for the length of πΆπ· in terms of π§, and therefore we can use this to work out the length of π΄π΅. So substituting the known value of π§, which is two, into this expression tells us that the length of π΄π΅ is six multiplied by two minus one, which is 11, and therefore we have our answer to the problem: the length of π΄π΅ is 11.
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We use ratios to make comparisons between two things. When we express ratios in words, we use the word "to"--we say "the ratio of something to something else." Ratios can be written in several different ways: as a fraction, using the word "to", or with a colon.
Let's use this illustration of shapes to learn more about ratios. How can we write the ratio of squares to circles, or 3 to 6? The most common way to write a ratio is as a fraction, 3/6. We could also write it using the word "to," as "3 to 6." Finally, we could write this ratio using a colon between the two numbers, 3:6. Be sure you understand that these are all ways to write the same number.
Which way you choose will depend on the problem or the situation.
• ratio of squares to circles is 3/6
• ratio of squares to circles is 3 to 6
• ratio of squares to circles is 3:6
There are still other ways to make the same comparison, by using equal ratios. To find an equal ratio, you can either multiply or divide each term in the ratio by the same number (but not zero). For example, if we divide both terms in the ratio 3:6 by the number three, then we get the equal ratio, 1:2. Do you see that these ratios both represent the same comparison? Some other equal ratios are listed below. To find out if two ratios are equal, you can divide the first number by the second for each ratio. If the quotients are equal, then the ratios are equal. Is the ratio 3:12 equal to the ratio 36:72? Divide both, and you discover that the quotients are not equal. Therefore, these two ratios are not equal.
Some other equal ratios:
3:6 = 12:24 = 6:12 = 15:30
Are 3:12 and 36:72 equal ratios?
Find 3÷12 = 0.25 and 36÷72 = 0.5
The quotients are not equal —> the ratios are not equal.
You can also use decimals and percents to compare two quantities. In our example of squares to circles, we could say that the number of squares is "five-tenths" of the number of circles, or 50%.
Here is a chart showing the number of goals made by five basketball players from the free-throw line, out of 100 shots taken. Each comparison of goals made to shots taken is expressed as a ratio, a decimal, and a percent. They are all equivalent, which means they are all different ways of saying the same thing. Which do you prefer to use?
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# Lesson video
In progress...
Hallo, I'm Mr.Langton and today we're going to be looking at frequency trees, all we're going to need is something to write with and something to write on.
Try finding a quiet space where you won't be disturbed, when you're ready, we'll begin.
There are 32 students in a class.
17 of these students did their maths homework, and of these 17 students, 15 passed their maths test that week.
Of the students that didn't do their homework, only 3 of them passed.
How could you represent this information? Could you make it more visual? Could you make it easier to follow? If you chose a student at random, what is the probability that they didn't do their homework and they failed their test? Pause the video and see what you can do.
When you're ready, unpause it, we can go through it together.
You can pause it in three, two, one.
How did you get on? I'm going to show you how to put the information into a frequency tree.
We're going to go through the question one line at a time.
The first line says there are 32 students in a class, we're going to make a note to the number 32 there.
The next line says that 17 of these students, did their maths homework.
I'm going to draw two branches coming out of there, showing those that did, the 17 and those that didn't.
Now we know there are 32 students altogether, so 15 students didn't do their homework.
Of the 17 students that did their homework, 15 passed their Maths test that week, I'm just going to take that 17 and split it up into two parts, 15 of them passed, that means that two of them failed because those two have got to add up to make 17 eminently.
Of the students that didn't do their homework, only three of them passed.
So we've got two new branches, three of them passed.
Now we've got 15 students who didn't do their homework, that means that 12 of them must have failed.
So we've now got a nice and visual method, that shows us all the possible outcomes, we've got no missing numbers, which is going to make the question much more straight forward.
If you chose a student at random, what is the probability that they didn't do their homework and they failed their test? Well the students who didn't do their homework, that's 15 of them, 12 of them failed so all together, there were 12 students who didn't do their homework and failed.
And in total, we have 32 students.
So if I pick students at random from the 32, the chance of me picking someone who didn't do their homework and failed their test is 12 out of 32.
Or you could simplify that to down to 6 out of 16 or down to three eighths.
Now it's your turn to have a go, you can pause the video and to the worksheet, there are two questions I want you to get on with, one of them, I've drawn the frequency tree, you just have to fill it in for the second one, you'll have to draw it yourself.
Pause the video and have a go, when you're ready, we can go through it together.
Good luck! How did you get on? Let's go through the questions together.
A 100 students had some homework.
So we're going to label that straight away, that we've got 100 students.
I'm going to take that, tell me that you've already done it.
42 of these students are boys, 42 students are boys.
Now before I go any further, 100 subtract 42, equals 58.
So add on the 58 girls.
We're done with that information I'm going to give that a tick, we don't need to pay attention to that anymore.
8 of the 100 students did not do their homework, now we've got a problem here because that says eight of the 100 students so that means that this one here and this one here adds up to make 8.
I'm just going to leave that for now, I'm not going to tick here, there's nothing I can fill in coz I don't know whether that's four and four, or if it's five and three, if it's six and two, let's move on and see what else we can work out.
53 of the girls did do their homework so the girls who did do their homework that's 53 girls.
So that means if there are 58 girls all together, that means five girls didn't do their homework.
I can tick that coz I've done it and now I can come back to this one.
If I know all together, eight students didn't do their homework, if five girls didn't do their homework, then three boys didn't do their homework, and that means 42 take away three is 39.
39 boys did do their homework.
Now for the second question, we have got to come up with our own frequency tree.
50 students went to visit either Bath or Manchester University.
We start off, we have 50 students and they either went to Bath, or they went to Manchester.
We're going to put two bubbles there, I can tick that line just to let me know that I'm done.
12 girls and 11 boys went to Bath so, 11 boys and 12 girls.
Now that makes in total 23.
So that tells me that 27 people went to Manchester so I've got awful lot of information just from that statement.
13 girls went to Manchester.
So that means 14 boys went to Manchester.
So I've drawn our frequency tree.
Calculate the probability that the students picked at random went to Manchester.
We know that 27 students went to Manchester, and there were 50 students altogether.
The probability that the student picked at random was a girl, we're aware that there were 13 girls in Manchester and 12 girls in Bath so in total that means it's going to be 25 girls.
And as there were 50 students, there's also 25 boys.
Part D, given that they were a boy, that they went to Bath? That's a really tricky question to try and answer.
Altogether, we've got 25 boys so we're looking at the 25 boys, and of those 25 boys, 11 went to Bath and 14 went to Manchester.
So that's a really sneaky question to try and answer.
How did you get on? We're going to finish with this explore activity.
You are to complete the frequency tree, matching these criteria; 3 students passed despite not doing their homework, half of the group did their homework, 8 of those that did their homework failed the test, and twice that number passed the test altogether.
Pause the video and have a go, when you're ready, unpause it and we'll go through it together.
If you're struggling, I'm going to go through it step by step really slowly so maybe leave the video running for a little bit longer, just to see me just start filing in the frequency tree, when you're ready, you can pause it and finish.
You want to start now, pause in three, two, one.
So the first step is to fill in this first one.
Three students passed despite not doing their homework so they didn't do their homework and they passed, that's three students.
That one's rather easy to fill in.
Start the after that.
Half of the group did their homework, we don't know how many people are in the group, we got nearly enough information on there yet, now to just leave that one to one side at the moment, and let's go on to the next bit.
8 of those who did their homework failed the test.
So they did their homework and they failed, that's eight.
That's going to go there.
Okay and tick that, there's nothing more I can use it for yet.
Twice that number passed the test all together, so we're referring to that eight there so that means it's 16 people passed all together.
Now we know already that three people, didn't do their homework and passed.
If 16 people passed all together, that means that 13 people must have done their homework, and passed the test.
Now we can start to work backwards here.
If 13 people did it and passed, eight people did it and failed, that means that 21 people did their homework.
Now we know, we go back here, that half of the group did their homework.
That means half of them didn't.
So that means we've also got 21 students there, so in total, there must be 42 students.
Now the last thing to fill in, is how many students didn't do their homework and then failed the test.
And if 21 students didn't do their homework, three of them passed that means that 18 failed.
Now we can check this, because at each stage, we should be getting a total of 42, we know there are 42 students altogether.
That they did their homework or they didn't.
Now again, we've got 42 students, that they passed or they failed, well 13 passed, add another three, 16 of them passed, eight of them failed, and another 18 failed, that makes 26 that failed, now we can add those together and total once again, we've got 42.
So we can be very confident that we got this one right.
That's it for today, I'll see you later.
Goodbye!. |
# How to Write Ratios
How to Write Ratios. Guidelines on how to write ratios with examples and step by step explanations. What is a ratio? A ratio is two numbers that are in a specified relationship with each other. The simplest form of a ratio is in the format . Commonly used ratios are given names. These names makes them easier to remember. These ratios can be written in simple terms or expanded out into fractions
Ratios are a way to compare 2 different quantities, which makes them incredibly useful for doing everything from graphing to calculating volume. In this introduction to ratios you will learn ways to understand ratios and how to use ratios in everyday life.
A ratio is a mathematical means of comparing one value to another. You hear the term “ratio” used in relation to studies or analyses of various sets of data, such as in demographics or the performance ratings of products. Proportions and fractions are intertwined with ratios. Both proportions and fractions deal with the comparison of multiple values. You can use proportions and fractions as alternative means of writing ratios. You may have to perform this type of task in a middle school, high school or college math course.
# Characteristics of a Ratio
A ratio is a sort of mathematical metaphor, an analogy used to compare different amounts of the same measure. You could almost consider any type of measurement a ratio, since every measurement in the world has to have some sort of reference point. This fact alone makes measurement by ratio one of the most basic of all forms of quantification.
## Units of Measure
A ratio compares two things in the same unit of measure. It doesn’t matter what that unit of measure is — pounds, cubic centimeters, gallons, newton-meters — it matters only that the two are measured in the same units. For instance, you can’t compare 1 part fuel to 14 parts of air if you’re measuring fuel in pounds and air in cubic feet.
## Modes of Expression
You can express a ratio either in narrative form or in symbolic mathematical notation. You can express ratio as “the ratio of A to B,” “A is to B,” “A : B” or the quotient of A divided by B. For example, you can express a ratio of 1 to 4 as 1:4 or 0.25 (1 divided by 4).
## Equality of Ratios
You can use ratios as direct analogies to compare one thing to another, notating it either with an “=” sign or verbally. For instance, you can say “A is to B as C is to D,” or you can say, “A:B = C:D.” In this instance, A and D are the “extremes” and B and C are called the “means.” For example, you can say, “1 is to 4 as 3 is to 12,” or you can say “1:4 = 3:12.”
## Ratios as Fractions
In practice, ratios act something like fractions. You can replace the colon with a division sign and still arrive at the same result. As in the previous example, 1/4 (1 divided by 4) and 3/12 (3 divided by 12) both come out to 0.25. This is consistent with the last mode of expression. So any ratio may be expressed as A divided by B.
## Continued Proportions
Any series of three or more ratios can string together to create a continued or serial proportion. As an example, “1 is to 4 as 3 is to 12 as 4 is to 16” and “1:4 = 3:12 = 4:16” are both continued proportions. Expressing them as decimal figures (dividing the first number by the second in each proportion), you indeed find that 0.25 = 0.25 = 0.25.
## How to Make a Ratio
### Method 1 Make a Ratio
1. Use a symbol to denote the ratio. To indicate that you’re using a ratio, you can use the division sign ( / ), a colon ( : ), or the word to. For example, if you wanted to say, “For every five men at the party, there are three women,” then you could use any of the three symbols to state this.[1] Here’s how you would do it:
• 5 men / 3 women
• 5 men : 3 women
• 5 men to 3 women
2. Write the first given quantity to the left of the symbol. Write down the quantity of the first item before the symbol of your choice. You should also remember to state the units, or the number you’re working with, whether it’s men or women, chickens or goats, or miles or inches.
• Example: 20 g of flour
3. Write the second given quantity to the right of the symbol. After you’ve written the first given quantity followed by the symbol, you should write the second given quantity, along with its units.
• Example: 20 g of flour/ 8g of sugar
4. Simplify your ratio (optional). You may want to simplify your ratio to do something like scale down a recipe. If you’re using 20 g of flour for a recipe, then you know you’ll need 8 g of sugar, and you’re done. But if you’d like to scale down the ratio as much as possible, then you’ll need to simplify it by writing the ratio in its lowest possible terms. You should use the same process as you would use to simplify a fraction. To do this, you have to find the GCF, or the greatest common factor, of both quantities, and then see how many times that number fits into each given quantity.
• To find the GCF of 20 and 8, write down all of the factors of both numbers (the numbers that can multiply to make those numbers and thus can be evenly divided into those numbers) and find the largest number that is evenly divisible into both. Here’s how you do it:
• 20: 1, 2, 4, 5, 10, 20
• 8: 1, 2, 4, 8
• 4 is the GCF of 20 and 8 — it’s the largest number that evenly divides into both numbers. To get your simplified ratio, simply divide both numbers by 4:
• 20/4 = 5
• 8/4 = 2
• Your new ratio is 5 g flour/ 2 g sugar.
5. Turn the ratio into a percentage (optional). If you’d like to turn the ratio into a percentage, you just have to complete the following steps:
• Divide the first number by the second number. Ex: 5/2 = 2.5.
• Multiply the result by 100. Ex: 2.5 * 100 = 250.
• Add a percentage sign. 250 + % = 250%.
• This indicates that for every 1 unit of sugar, there is 2.5 units of flour; it also means that there is 250% as much flour as there is sugar.
1. The order of the quantities doesn’t matter. The ratio simply represents the relationship between two quantities. “5 apples to 3 pears” is the same as “3 pears to 5 apples.” Therefore, 5 apples/ 3 pears = 3 pears/ 5 apples.
2. A ratio can also be used to describe probability. For example, the probability of rolling a 2 on a die is 1/6, or one out of six. Note: if you’re using a ratio to denote probability, then the order of quantities does matter.
3. You can scale a ratio up as well as down. Though you may be used to simplifying numbers whenever you can, it can benefit you to scale a ratio up. For example, if you know that you’ll need 2 cups of water for every 1 cup of pasta you boil (2 cups water/1 cup pasta), but you want to boil 2 cups of pasta, then you’ll need to scale up the ratio to know how much water to use. To scale up a ratio, simply multiply the top and bottom by the same number.
• 2 cups water/ 1 cup pasta * 2/ 2 = 4 cups water/ 2 cups pasta. You’ll need 4 cups of water to boil 2 cups of pasta.
## Conclusion
Ratio is defined as a relationship of one quantity to another quantity. It shows the relationship between two quantities, e.g., the number of students in a class is to the number of teachers in a school. Ratios compare two quantities with each other. When neither of the quantities is zero then it is called a proper ratio. |
# Solving absolute value equations and inequalities
The absolute number of a number a is written as
$\\ \left | a \right | \\$
And represents the distance between a and 0 on a number line.
An absolute value equation is an equation that contains an absolute value expression. The equation
$\\ \left | x \right |=a \\$
Has two solutions x = a and x = -a because both numbers are at the distance a from 0.
To solve an absolute value equation as
$\\ \left | x+7 \right |=14 \\$
You begin by making it into two separate equations and then solving them separately.
$\\\begin{matrix} x+7 =14: &x+7\, {\color{green} -\, 7}\, =14\, {\color{green} -\, 7} \\ & \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x=7 \end{matrix} \\or\\\\ \begin{matrix} x+7 =-14: &x+7\, {\color{green} -\, 7}\, =-14\, {\color{green} -\, 7} \\ & \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x=-21 \end{matrix}$
An absolute value equation has no solution if the absolute value expression equals a negative number since an absolute value can never be negative.
The inequality
$\\ \left | x \right |<2 \\$
Represents the distance between x and 0 that is less than 2
Whereas the inequality
$\\ \left | x \right |>2 \\$
Represents the distance between x and 0 that is greater than 2
You can write an absolute value inequality as a compound inequality.
$\\ \left | x \right |<2\: or \\-2
This holds true for all absolute value inequalities.
$\\ \left | ax+b \right |0 \\=-cc,\: where\: c>0 \\=ax+b<-c\: or\: ax+b>c \\$
You can replace > above with ≥ and < with ≤.
When solving an absolute value inequality it's necessary to first isolate the absolute value expression on one side of the inequality before solving the inequality.
Example:
Solve the absolute value inequality
$\\ 2\left |3x+9 \right |<36 \\\\\\\frac{2\left |3x+9 \right |}{2}<\frac{36}{2} \\\\\left | 3x+9 \right |<18 \\-18<3x+9<18 \\-18\, {\color{green} -\, 9}<3x+9\, {\color{green} -\, 9}<18\, {\color{green} -\, 9} \\-27<3x<9 \\\\\frac{-27}{{\color{green} 3}}<\frac{3x}{{\color{green} 3}}<\frac{9}{{\color{green} 3}} \\\\-9
Videolesson: Solve the absolute value equation
$\\4 \left |2x -1 \right | -2 = 10\\$ |
Maths Codes Practice 3
In this practice exercise, students will replace letters with numbers and solve problems using BIDMAS under timed conditions.
Key stage: KS 2
Curriculum topic: Verbal Reasoning Exam-Style Practice
Curriculum subtopic: Maths Codes
Difficulty level:
QUESTION 1 of 10
If you are unsure of an answer, move on to the next question.
You should come back to questions you found difficult at the end if you have time.
Let’s try one as a reminder and write the answer as a letter:
If a = 4, b = 125, c = 3, d = 11 and e = 7, what is d (c x a) - e?
Where there isn’t a -, +, x or ÷ between the first letter and the bracket, it means we have to multiply the outside number by what is in the brackets.
So the steps we follow are:
c x a = 12
d x 12 = 132
132 - e = 125
We must turn the number 125 back into a letter. This number corresponds to the letter b.
Now it's time to begin this practice exercise.
Good luck!
You have been given the following information: A = 3, B = 5, C = 6, D = 9, E = 11.
Which letter completes the equation: √D + E - A = __?
A
B
C
D
E
You have been given the following information: a = 2, b = 6, c = 4, d = 8, e = 12.
Which letter completes the equation: c² - (b - a) = __?
a
b
c
d
e
You have been given the following information: A = 20, B = 10, C = 4, D = 5, E = 2.
Which letter completes the equation: B² ÷ A = __?
A
B
C
D
E
You have been given the following information: a = 3, b = 6, c = 7, d = 9, e = 11.
Which letter completes the equation: (d x e) ÷ (b x c - d) = __?
a
b
c
d
e
You have been given the following information: A = 24, B = 9, C = 8, D = 2, E = 1.
Which letter completes the equation: (B x C) ÷ (D + E) = __?
A
B
C
D
E
You have been given the following information: a = 16, b = 7, c = 6, d = 8, e = 12.
Which letter completes the equation: e x d ÷ c = __?
a
b
c
d
e
You have been given the following information: A = 32, B = 4, C = 8, D = 24, E = 14.
Which letter completes the equation: (D + A) ÷ B = __?
A
B
C
D
E
You have been given the following information: a = 4, b = 8, c = 9, d = 2, e = 32.
Which letter completes the equation: d (b² ÷ a) = __?
a
b
c
d
e
You have been given the following information: A = 81, B = 9, C = 17, D = 4, E = 6.
Which letter completes the equation: √A (D + E) - A = __?
A
B
C
D
E
You have been given the following information: a = 36, b = 6, c = 25, d = 10, e = 110.
Which letter complete the equation: d (√a + √c) = __?
a
b
c
d
e
• Question 1
You have been given the following information: A = 3, B = 5, C = 6, D = 9, E = 11.
Which letter completes the equation: √D + E - A = __?
E
EDDIE SAYS
The question tells us that D = 9, E = 11 and A = 3.
If we substitute these numbers for the letters in the equation then √D + E - A = __ becomes √9 + 11 - 3 = __.
Using the rules of BIDMAS, we can tackle the steps of the calculation in exactly the order in which they have been written.
So 3 + 11 - 3 = 11.
The letter that corresponds to 11 is E, so this is the correct answer.
• Question 2
You have been given the following information: a = 2, b = 6, c = 4, d = 8, e = 12.
Which letter completes the equation: c² - (b - a) = __?
e
EDDIE SAYS
The question tells us that c = 4, b = 6 and a = 2.
If we substitute these numbers for the letters in the equation then c² - (b - a) = __ becomes 4² - (6 - 2) = __.
Using the rules of BIDMAS, we need to work out the sum in the brackets first, then the power (²), and then subtract them from each other.
So (6 - 2) = 4 and 4² = 16. 16 - 4 = 12.
The letter that corresponds to 12 is e, so this is the correct answer.
• Question 3
You have been given the following information: A = 20, B = 10, C = 4, D = 5, E = 2.
Which letter completes the equation: B² ÷ A = __?
D
EDDIE SAYS
The question tells us that B = 10 and A = 20.
If we substitute these numbers for the letters in the equation then B² ÷ A = __ becomes 10² ÷ 20 = __.
Using the rules of BIDMAS, we can tackle the steps of the calculation in exactly the order in which they have been written.
So 10² ÷ 20 = 100 ÷ 20 = 5.
The letter that corresponds to 5 is D, so this is the correct answer.
• Question 4
You have been given the following information: a = 3, b = 6, c = 7, d = 9, e = 11.
Which letter completes the equation: (d x e) ÷ (b x c - d) = __?
a
EDDIE SAYS
The question tells us that d = 9, e = 11, b = 6 and c = 7.
If we substitute these numbers for the letters in the equation then (d x e) ÷ (b x c - d) = __ becomes (9 x 11) ÷ (6 x 7 - 9) = __.
Using the rules of BIDMAS, we must tackle the two sets of brackets separately first, and then divide the answers we find from each.
So 9 x 11 = 99. And 6 x 7 - 9 = 33.
99 ÷ 33 = 3.
The letter that corresponds to 3 is a, so this is the correct answer.
• Question 5
You have been given the following information: A = 24, B = 9, C = 8, D = 2, E = 1.
Which letter completes the equation: (B x C) ÷ (D + E) = __?
A
EDDIE SAYS
The question tells us that B = 9, C = 8, D = 2 and E = 1.
If we substitute these numbers for the letters in the equation then (B x C) ÷ (D + E) = __ becomes (9 x 8) ÷ (2 + 1) = __. Using the rules of BIDMAS, we must tackle the two sets of brackets first, then divide the answer we reach.
So (9 x 8) = 72. And (2 + 1) = 3.
72 ÷ 3 = 24.
The letter that corresponds to 24 is A, so this is the correct answer.
• Question 6
You have been given the following information: a = 16, b = 7, c = 6, d = 8, e = 12.
Which letter completes the equation: e x d ÷ c = __?
a
EDDIE SAYS
The question tells us that e = 12, d = 8 and c = 6.
If we substitute these numbers for the letters in the equation then e x d ÷ c = __ becomes 12 x 8 ÷ 6 = __.
Using the rules of BIDMAS, we can work through the steps of this calculation in exactly the order in which they have been written.
So 12 x 8 ÷ 6 = 96 ÷ 6 = 16.
The letter that corresponds to 16 is a, so this is the correct answer.
• Question 7
You have been given the following information: A = 32, B = 4, C = 8, D = 24, E = 14.
Which letter completes the equation: (D + A) ÷ B = __?
E
EDDIE SAYS
The question tells us that D = 24, A = 32 and B = 4.
If we substitute these numbers for the letters in the equation then (D + A) ÷ B = __ becomes (24 + 32) ÷ 4 = __.
Using the rules of BIDMAS, we can tackle the steps of the calculation in the exact order they have been written.
So (24 + 32) ÷ 4 = 56 ÷ 4 = 14.
The letter that corresponds to 14 is E, so this is the correct answer.
• Question 8
You have been given the following information: a = 4, b = 8, c = 9, d = 2, e = 32.
Which letter completes the equation: d (b² ÷ a) = __?
e
EDDIE SAYS
The question tells us that d = 2, b = 8 and a = 4.
If we substitute these numbers for the letters in the equation then d (b² ÷ a) = __ becomes 2 x (8² ÷ 4) = __.
Using the rules of BIDMAS, we must work out the sum in the brackets first, then multiply this answer by 2.
So 2 x (8² ÷ 4) = 2 x 16 = 32.
The letter that corresponds to 32 is e, so this is the correct answer.
• Question 9
You have been given the following information: A = 81, B = 9, C = 17, D = 4, E = 6.
Which letter completes the equation: √A (D + E) - A = __?
B
EDDIE SAYS
The question tells us that A = 81, D = 4 and E = 6.
If we substitute these numbers for the letters in the equation then √A (D + E) - A = __ becomes √81 x (4 + 6) - 81 = __.
Using the rules of BIDMAS, we must work out the sum in the brackets first, then multiply this by the answer from the square root, before subtracting 81 from our total.
So √81 x (4 + 6) - 81 = 9 x 10 - 81 = 9.
The letter that corresponds to 9 is B, so this is the correct answer.
• Question 10
You have been given the following information: a = 36, b = 6, c = 25, d = 10, e = 110.
Which letter complete the equation: d (√a + √c) = __?
e
EDDIE SAYS
The question tells us that d = 10, a = 36 and c = 25.
If we substitute these numbers for the letters in the equation then d (√a + √c) = __ becomes 10 x (√36 + √25) = __.
Using the rules of BIDMAS, must work out the sum in the brackets first then multiply this answer by 10.
So 10 x (6 + 5) = 10 x 11 = 110.
The letter that corresponds to 110 is e, so this is the correct answer.
---- OR ----
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Contents
## Problem
There are four people A, B, C, D. A box with a red ball inside it is given to A, who can leave things unchanged with probability $$p$$ or replace the red ball with a white one with probability $$q=1-p$$. The box is passed to B who can leave the box unchanged with probability $$p$$, or change the color of the ball from red to white or vice versa. Then it is the turn of C and then of D, with the same modalities and probability.
At the end a check is made and there is a red ball in the box.
What is the probability that A originally left the red ball?
Other equivalent formulations of the problem can be given. For example, this one: there are four people A, B, C, D who tell the truth with probability $$p$$. Initially A makes a statement; so D says that C says that B says that A has told the truth. Compute the probability that A has told the truth.
For the solution of this exercise it’s necessary to know some elementary concepts of Theory of Probability. For an introduction to Probability with numerous exercises performed, see for example [1].
### Solution
We define the following events:
• $$X$$: A left the red ball in the box
• $$Y$$: at the end there is a red ball in the box
• $$X \cap Y$$: intersection event, that is true if $$X$$ and $$Y$$ are both true
To solve the problem we have to compute the following conditional probability:
$P(X|Y)= \frac {P(X \cap Y)}{P(Y)}$
where the symbol $$P(X|Y)$$ indicates the probability that $$X$$ is true, given the knowledge that the event $$Y$$ has occurred.
1) Calculation of $$P(Y)$$
There must have been an even number of color changes, since the ball was initially red. Three cases are possible with the following probabilities:
• 0 changes $$p_{1}= p^4$$
• 2 changes $$p_{2}= {\binom{4}{2}}p^2q^2$$
• 4 changes $$p_{3}= q^4$$
Then we have:
$P(Y) = p_{1} + p_{2} + p_{3}=p^4 + {\binom{4}{2}}p^2q^2 + q^4$
2) Calculation of $$P (X \cap Y)$$
Even in this case there must have been an even number of color changes. However, A has made no changes if it is true the event $$X$$. So we only have these two cases, with the following probabilities:
• 0 changes $$p_{1}= p^4$$
• 2 changes $$p_{2}= {\binom{3}{2}}p^2q^2$$
Then we have
$P(X \cap Y) = p_{1} + p_{2}= p^4 + {\binom{3}{2}}p^2q^2$
Putting together the two results we finally have:
$P(X|Y)= \frac {P(X \cap Y)}{P(Y)} = \frac {p^4 + {\binom{3}{2}}p^2q^2}{p^4 + {\binom{4}{2}}p^2q^2 + q^4}$
For example, if $$p=\frac{1}{3}$$ the probability is $$P(X|Y)= \frac{13}{41}$$.
If $$p=\frac{1}{2}$$ the probability is $$P(X|Y)= \frac{1}{2}$$, as was to be expected for reasons of symmetry.
## Bibliography
[1]Murray Spiegel – Probability and Statistics (McGraw-Hill) |
# Samacheer Kalvi Books: Tamilnadu State Board Text Books Solutions
## Samacheer Kalvi 9th Maths Chapter 7 Mensuration Notes PDF Download: Tamil Nadu STD 9th Maths Chapter 7 Mensuration Notes
Samacheer Kalvi 9th Maths Chapter 7 Mensuration Notes PDF Download: Tamil Nadu STD 9th Maths Chapter 7 Mensuration Notes
We bring to you specially curated Samacheer Kalvi 9th Maths Chapter 7 Mensuration Notes PDF which have been prepared by our subject experts after carefully following the trend of the exam in the last few years. The notes will not only serve for revision purposes, but also will have several cuts and easy methods to go about a difficult problem.
Board Tamilnadu Board Study Material Notes Class Samacheer Kalvi 9th Maths Subject 9th Maths Chapter Chapter 7 Mensuration Format PDF Provider Samacheer Kalvi Books
## How to Download Samacheer Kalvi 9th Maths Chapter 7 Mensuration Notes PDFs?
2. Click on the Samacheer Kalvi 9th Maths Notes PDF.
3. Look for your preferred subject.
4. Now download the Samacheer Kalvi 9th Maths Chapter 7 Mensuration notes PDF.
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Additional Questions
Exercise 7.1
Question 1.
Using Heron’s formula, find the area of a triangle whose sides are 41 m, 15 m, 25 m.
Solution:
Question 2.
Find the area of an equilateral triangle whose perimeter is 150 m.
Solution:
3a = 150
Question 3.
Find the area of a quadrilateral whose sides are PQ = 15 cm, QR = 8 cm, RS = 25 cm, PS = 12 cm and LQ = 90°
Solution:
Exercise 7.2
Question 1.
Find the TSA and LSA of a cuboid whose length, breadth and height are 10 cm, 12 cm and 14 cm respectively.
Solution:
TSA = 2 (lb + bh + Ih)
= 2 (10 × 12 + 12 × 14 + 10 × 14) = 2 (120 + 168 + 140) = 856 cm2
LSA = 2 (bh + Ih) = 2 (12 × 14 + 10 × 14) = 2 (168 + 140) =2 (308) = 616 cm2
Question 2.
A cuboid has total surface area of 40 m2 and its lateral surface area is 26 m2. Find the area of its base.
Solution:
Question 3.
Find the surface area of a cube whose edge is
(i) 27 cm
(ii) 3 cm
(iii) 6 cm
(iv) 2.1 cm
Solution:
(i) TSA = 6a2 = 6 × 272 = 6 × 729 = 4374 cm2
(ii) TSA = 6a2 = 6 × 32 = 54 cm2
(iii) TSA = 6a2 = 6 × 62 = 6 × 36 = 216 cm2
(iv) TSA = 6a2 = 6 × 2.12 = 6 × 4.41 = 26.46 cm2
Exercise 7.3
Question 1.
Find the volume of a cube whose surface area is a 96 cm2.
Solution:
6a2 = 96 cm2
a = 6
Volume = a3 = 63 = 36 × 6 = 216 cm3
Question 2.
The volume of a cuboid is 440 cm3 and the area of its base is 88 cm2, find its height.
Solution:
l × b × h = 440 cm3
Question 3.
How many 3 metre cubes can be cut from a cuboid measuring 18 m × 12 m × 9 m?
Solution:
Question 4.
The outer dimensions of a closed wooden box are 10 cm by 8 cm by 7 cm. Thickness
of the wood is 1 cm3. Find the total cost of wood required to make box if 1 cm3 of wood costs ₹ 2.00?
Solution:
Volume of wood = 12 × 10 × 9 – 10 × 8 × 7 = 1080 – 560 = 520 cm3 × 2.00 = ₹ 1040
## How to Prepare using Samacheer Kalvi 9th Maths Chapter 7 Mensuration Notes PDF?
Students must prepare for the upcoming exams from Samacheer Kalvi 9th Maths Chapter 7 Mensuration Notes PDF by following certain essential steps which are provided below.
• Use Samacheer Kalvi 9th Maths Chapter 7 Mensuration notes by paying attention to facts and ideas.
• Pay attention to the important topics
• Refer TN Board books as well as the books recommended.
• Correctly follow the notes to reduce the number of questions being answered in the exam incorrectly
• Highlight and explain the concepts in details.
## Frequently Asked Questions on Samacheer Kalvi 9th Maths Chapter 7 Mensuration Notes
#### How to use Samacheer Kalvi 9th Maths Chapter 7 Mensuration Notes for preparation??
Read TN Board thoroughly, make separate notes for points you forget, formulae, reactions, diagrams. Highlight important points in the book itself and make use of the space provided in the margin to jot down other important points on the same topic from different sources.
#### How to make notes for Samacheer Kalvi 9th Maths Chapter 7 Mensuration exam?
Read from hand-made notes prepared after understanding concepts, refrain from replicating from the textbook. Use highlighters for important points. Revise from these notes regularly and formulate your own tricks, shortcuts and mnemonics, mappings etc.
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# The area of a circle whose area and circumference
Question:
The area of a circle whose area and circumference are numerically equal, is
(a) $2 \pi$ sq. units
(b) $4 \pi$ sq. units
(c) $6 \pi$ sq. units
(d) $8 \pi$ sq. units
Solution:
We have given that circumference and area of a circle are numerically equal.
Let it be x.
Let r be the radius of the circle, therefore, circumference of the circle is and area of the circle will be.
Therefore, from the given condition we have,
$2 \pi r=x \ldots \ldots \ldots(1)$
$\pi r^{2}=x$.........(2)
Therefore, from equation (1) get $r=\frac{x}{2 \pi}$. Now we will substitute this value in equation (2) we get, $\pi\left(\frac{x}{2 \pi}\right)^{2}=x$
Simplifying further we get,
$\pi \times \frac{x^{2}}{4 \pi^{2}}=x$
Cancelling x we get,
$\pi \times \frac{x}{4 \pi^{2}}=1$
Now we will cancel $\pi$
$\frac{x}{4 \pi}=1$.........(3)
Now we will multiply both sides of the equation (3) by $4 \pi$ we get,
$x=4 \pi$
Therefore, area of the circle is $4 \pi s q$.units.
Hence, option (b) is correct. |
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# RD Sharma solutions for Class 9 Mathematics chapter 13 - Quadrilaterals
## Chapter 13: Quadrilaterals
Ex. 13.10Ex. 13.20Ex. 13.30Ex. 13.40Others
#### Chapter 13: Quadrilaterals Exercise 13.10 solutions [Page 4]
Ex. 13.10 | Q 1 | Page 4
Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle
Ex. 13.10 | Q 2 | Page 4
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angles of the quadrilateral
Ex. 13.10 | Q 3 | Page 4
The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13 Find all the angles of the quadrilateral.
Ex. 13.10 | Q 4 | Page 4
In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that
∠COD = 1/2 (∠A+ ∠B).
#### Chapter 13: Quadrilaterals Exercise 13.20 solutions [Pages 19 - 20]
Ex. 13.20 | Q 1 | Page 19
Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each angle of the parallelogram .
Ex. 13.20 | Q 2 | Page 20
If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram .
Ex. 13.20 | Q 3 | Page 20
Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle
Ex. 13.20 | Q 4 | Page 20
The perimeter of a parallelogram is 22 cm . If the longer side measures 6.5 cm what is the measure of the shorter side?
Ex. 13.20 | Q 5 | Page 20
In a parallelogram ABCD, ∠D = 135°, determine the measures of ∠A and ∠B
Ex. 13.20 | Q 6 | Page 20
ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D .
Ex. 13.20 | Q 7 | Page 20
In Fig. below, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute
Ex. 13.20 | Q 8.1 | Page 20
The following statement are true and false .
In a parallelogram, the diagonals are equal
Ex. 13.20 | Q 8.2 | Page 20
The following statement are true and false.
In a parallelogram, the diagonals bisect each other.
Ex. 13.20 | Q 8.3 | Page 20
The following statement are true and false .
In a parallelogram, the diagonals intersect each other at right angles .
Ex. 13.20 | Q 8.4 | Page 20
The following statement are true and false .
In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.
Ex. 13.20 | Q 8.5 | Page 20
The following statement are true and false .
If all the angles of a quadrilateral are equal, it is a parallelogram .
Ex. 13.20 | Q 8.6 | Page 20
The following statement are true and false .
If three sides of a quadrilateral are equal, it is a parallelogram .
Ex. 13.20 | Q 8.7 | Page 20
The following statement are true and false .
If three angles of a quadrilateral are equal, it is a parallelogram .
Ex. 13.20 | Q 8.8 | Page 20
The following statement are true and false .
If all the sides of a quadrilateral are equal it is a parallelogram.
Ex. 13.20 | Q 9 | Page 20
In Fig., below, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.
Ex. 13.20 | Q 9 | Page 20
In Fig., below, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.
Ex. 13.20 | Q 10 | Page 20
In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.
#### Chapter 13: Quadrilaterals Exercise 13.30 solutions [Pages 42 - 43]
Ex. 13.30 | Q 1 | Page 42
In a parallelogram ABCD, determine the sum of angles ∠C and ∠D .
Ex. 13.30 | Q 1 | Page 42
In a parallelogram ABCD, determine the sum of angles ∠C and ∠D .
Ex. 13.30 | Q 2 | Page 42
In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles .
Ex. 13.30 | Q 2 | Page 42
In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles .
Ex. 13.30 | Q 3 | Page 42
ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.
Ex. 13.30 | Q 3 | Page 42
ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.
Ex. 13.30 | Q 4 | Page 42
ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC .
Ex. 13.30 | Q 5 | Page 42
The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
Ex. 13.30 | Q 5 | Page 42
The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
Ex. 13.30 | Q 6 | Page 42
P and Q are the points of trisection of the diagonal BD of a parallelogram AB Prove that CQ is parallel to AP. Prove also that AC bisects PQ.
Ex. 13.30 | Q 6 | Page 42
P and Q are the points of trisection of the diagonal BD of a parallelogram AB Prove that CQ is parallel to AP. Prove also that AC bisects PQ.
Ex. 13.30 | Q 7 | Page 43
ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
Ex. 13.30 | Q 8 | Page 43
ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles
Ex. 13.30 | Q 9 | Page 43
ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.
#### Chapter 13: Quadrilaterals Exercise 13.40 solutions [Pages 62 - 65]
Ex. 13.40 | Q 1 | Page 62
In a ∆ABC, D, E and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ∆DEF.
Ex. 13.40 | Q 2 | Page 62
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of
the triangle formed by joining the mid-points of the sides of this triangle.
Ex. 13.40 | Q 3 | Page 63
In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC =
21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.
Ex. 13.40 | Q 4 | Page 63
In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a
parallelogram.
Ex. 13.40 | Q 4 | Page 63
In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a
parallelogram.
Ex. 13.40 | Q 5 | Page 63
In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.
Ex. 13.40 | Q 6 | Page 63
In a ΔABC, BM and CN are perpendiculars from B and C respectively on any line passing
through A. If L is the mid-point of BC, prove that ML = NL.
Ex. 13.40 | Q 7 | Page 63
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
Ex. 13.40 | Q 8 | Page 63
In Fig. below, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.
Ex. 13.40 | Q 9 | Page 63
In Fig. below, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC.
Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram
Ex. 13.40 | Q 9 | Page 63
In Fig. below, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC.
Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram
Ex. 13.40 | Q 10 | Page 63
ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the
mid-points of the sides, in order, is a rectangle.
Ex. 13.40 | Q 11 | Page 64
Let Abc Be an Isosceles Triangle in Which Ab = Ac. If D, E, F Be the Mid-points of the Sides Bc, Ca and a B Respectively, Show that the Segment Ad and Ef Bisect Each Other at Right Angles.
Ex. 13.40 | Q 12 | Page 64
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral
bisect each other.
Ex. 13.40 | Q 13.1 | Page 64
Fill in the blank to make the following statement correct
The triangle formed by joining the mid-points of the sides of an isosceles triangle is
Ex. 13.40 | Q 13.2 | Page 64
Fill in the blank to make the following statement correct:
The triangle formed by joining the mid-points of the sides of a right triangle is
Ex. 13.40 | Q 13.3 | Page 64
Fill in the blank to make the following statement correct:
The figure formed by joining the mid-points of consecutive sides of a quadrilateral is
Ex. 13.40 | Q 14 | Page 64
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively
intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of
ΔABC
Ex. 13.40 | Q 15 | Page 64
In Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.
Ex. 13.40 | Q 16 | Page 64
ABC is a triang D is a point on AB such that AD = 1/4 AB and E is a point on AC such that AE = 1/4 AC. Prove that DE = 1/4 BC.
Ex. 13.40 | Q 17 | Page 64
In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = 1/4 AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.
Ex. 13.40 | Q 18 | Page 64
In the below Fig, ABCD and PQRC are rectangles and Q is the mid-point of Prove thaT
i) DP = PC (ii) PR = 1/2 AC
Ex. 13.40 | Q 19 | Page 65
ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH
Ex. 13.40 | Q 20 | Page 65
BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If
L is the mid-point of BC, prove that LM = LN.
#### Chapter 13: Quadrilaterals solutions [Pages 68 - 70]
Q 1 | Page 68
In a parallelogram ABCD, write the sum of angles A and B.
Q 1 | Page 68
In a parallelogram ABCD, write the sum of angles A and B.
Q 2 | Page 68
In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.
Q 2 | Page 68
In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.
Q 3 | Page 68
PQRS is a square such that PR and SQ intersect at O. State the measure of ∠POQ.
Q 4 | Page 68
If PQRS is a square, then write the measure of ∠SRP.
Q 5 | Page 68
If ABCD is a rhombus with ∠ABC = 56°, find the measure of ∠ACD.
Q 6 | Page 68
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of shorter side?
Q 7 | Page 68
If the angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13, then find the measure of the smallest angle.
Q 8 | Page 68
In a parallelogram ABCD, if ∠A = (3x − 20)°, ∠B = (y + 15)°, ∠C = (x + 40)°, then find the values of xand y.
Q 8 | Page 68
In a parallelogram ABCD, if ∠A = (3x − 20)°, ∠B = (y + 15)°, ∠C = (x + 40)°, then find the values of xand y.
Q 9 | Page 69
If measures opposite angles of a parallelogram are (60 − x)° and (3x − 4)°, then find the measures of angles of the parallelogram.
Q 10 | Page 69
In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Find AB : AD.
Q 10 | Page 69
In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Find AB : AD.
Q 11 | Page 69
In the given figure, PQRS is an isosceles trapezium. Find x and y.
Q 12 | Page 69
In the given figure, ABCD is a trapezium. Find the values of x and y.
Q 13 | Page 69
In the given figure, ABCD and AEFG are two parallelograms. If ∠C = 58°, find ∠F.
Q 14.1 | Page 69
Complete the following statement by means of one of those given in brackets against each:
If one pair of opposite sides are equal and parallel, then the figure is ........................
• parallelogram
• rectangle
• trapezium
Q 14.2 | Page 69
Complete the following statement by means of one of those given in brackets against each:
If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is ................
• square
• rectangle
• trapezium
Q 14.3 | Page 69
Complete the following statement by means of one of those given in brackets against each:
A line drawn from the mid-point of one side of a triangle .............. another side intersects the third side at its mid-point.
• perpendicular to parallel to
• to meet
Q 14.4 | Page 69
Complete the following statement by means of one of those given in brackets against each:
If one angle of a parallelogram is a right angle, then it is necessarily a .................
• rectangle
• square
• rhombus
Q 14.5 | Page 69
Complete the following statement by means of one of those given in brackets against each:
Consecutive angles of a parallelogram are ...................
• supplementary
• complementary
Q 14.6 | Page 69
Complete the following statement by means of one of those given in brackets against each:
If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a ...............
• rectangle
• parallelogram
• rhombus
Q 14.7 | Page 69
Complete the following statement by means of one of those given in brackets against each:
If opposite angles of a quadrilateral are equal, then it is necessarily a ....................
• parallelogram
• rhombus
• rectangle
Q 14.8 | Page 69
Complete the following statement by means of one of those given in brackets against each:
f consecutive sides of a parallelogram are equal, then it is necessarily a ..................
• kite
• rhombus
• square
Q 15 | Page 69
In a quadrilateral ABCD, bisectors of angles A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.
Q 16 | Page 69
The diagonals of a rectangle ABCD meet at O, If ∠BOC = 44°, find ∠OAD.
Q 17 | Page 70
If ABCD is a rectangle with ∠BAC = 32°, find the measure of ∠DBC.
Q 18 | Page 70
If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O such that ∠C + ∠D = k ∠AOB, then find the value of k.
Q 19 | Page 70
In the given figure, PQRS is a rhombus in which the diagonal PR is produced to T. If ∠SRT = 152°, find x, y and z.
Q 20 | Page 70
In the given figure, ABCD is a rectangle in which diagonal AC is produced to E. If ∠ECD = 146°, find ∠AOB.
#### Chapter 13: Quadrilaterals solutions [Pages 70 - 73]
Q 1 | Page 70
Mark the correct alternative in each of the following:
The opposite sides of a quadrilateral have
• no common point
• one common point
• two common points
• infinitely many common points
Q 2 | Page 70
The consecutive sides of a quadrilateral have
• no common point
• one common point
• two common points
• infinitely many common points
Q 3.1 | Page 71
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
∠P = 100°, ∠Q = 80°, ∠R = 95°
Q 3.1 | Page 71
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
∠P = 100°, ∠Q = 80°, ∠R = 95°
Q 3.2 | Page 71
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
∠P =85°, ∠Q = 85°, ∠R = 95°
Q 3.3 | Page 71
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
PQ = 7 cm, QR = 7 cm, RS = 8 cm, SP = 8 cm
Q 3.4 | Page 71
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm
Q 4 | Page 71
Which of the following quadrilateral is not a rhombus?
• All four sides are equal
• Diagonals bisect each other
• Diagonals bisect opposite angles
• One angle between the diagonals is 60°
Q 5 | Page 71
Diagonals necessarily bisect opposite angles in a
• rectangle
• parallelogram
• isosceles trapezium
• square
Q 6 | Page 71
The two diagonals are equal in a
• parallelogram
• rhombus
• rectangle
• trapezium
Q 7 | Page 71
We get a rhombus by joining the mid-points of the sides of a
• parallelogram
• rhombus
• rectangle
• triangle
Q 7 | Page 71
We get a rhombus by joining the mid-points of the sides of a
• parallelogram
• rhombus
• rectangle
• triangle
Q 8 | Page 71
The bisectors of any two adjacent angles of a parallelogram intersect at
• 30°
• 45°
• 60°
• 90°
Q 9 | Page 71
The bisectors of the angle of a parallelogram enclose a
• parallelogram
• rhombus
• rectangle
• square
Q 10 | Page 71
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
• parallelogram
• rectangle
• square
• rhombus
Q 11 | Page 71
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
• square
• rhombus
• trapezium
• none of these
Q 12 | Page 71
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
• square
• rectangle
• trapezium
• none of these
Q 12 | Page 71
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
• square
• rectangle
• trapezium
• none of these
Q 13 | Page 71
The figure formed by joining the mid-points of the adjacent sides of a square is a
• rhombus
• square
• rectangle
• parallelogram
Q 14 | Page 71
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
• rectangle
• parallelogram
• rhombus
• square
Q 14 | Page 71
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
• rectangle
• parallelogram
• rhombus
• square
Q 15 | Page 71
If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is
• 176°
• 68°
• 112°
• 102°
Q 16 | Page 71
In a parallelogram ABCD, if ∠DAB = 75° and ∠DBC = 60°, then ∠BDC =
• 75°
• 60°
• 45°
• 55°
Q 17 | Page 71
ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCDrespectively, then EF =
• AE
• BE
• CE
• DE
Q 18 | Page 72
ABCD is a parallelogram, M is the mid-point of BD and BM bisects ∠B. Then ∠AMB =
• 45°
• 60°
• 90°
• 75°
Q 19 | Page 72
If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is
• 108°
• 54°
• 72°
• 81°
Q 20 | Page 72
If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10x, what is the sum of the measures of the smallest angle and largest angle?
• 140°
• 150°
• 168°
• 180°
Q 21 | Page 72
If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to
• 16 cm
• 15 cm
• 20 cm
• 17 cm
Q 22 | Page 72
ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =
• 70°
• 110°
• 90°
• 120°
Q 22 | Page 72
ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =
• 70°
• 110°
• 90°
• 120°
Q 23 | Page 72
In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =
• 70°
• 45°
• 50°
• 60°
Q 24 | Page 72
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are
• 70°, 70°, 40°
• 60°, 40°, 80°
• 30°, 40°, 110°
• 60°, 70°, 50°
Q 25 | Page 72
The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB =
• 40°
• 50°
• 10°
• 90°
Q 26 | Page 72
ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and the respectively. If AB = 12 cm, MN = 14 cm, then CD =
• 10 cm
• 12 cm
• 14 cm
• 16 cm
Q 27 | Page 72
Diagonals of a quadrilateral ABCD bisect each other. If ∠A= 45°, then ∠B =
• 115°
• 120°
• 125°
• 135°
Q 28 | Page 72
P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
• 5 cm
• 6 cm
• 8 cm
• 10 cm
Q 28 | Page 72
P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
• 5 cm
• 6 cm
• 8 cm
• 10 cm
Q 29 | Page 72
In ΔABC, E is the mid-point of median AD such that BE produced meets AC at F. IF AC = 10.5 cm, then AF =
• 3 cm
• 3.5 cm
• 2.5 cm
• 5 cm
Q 30 | Page 72
ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =
• $\frac{3}{2}AB$
• 2 AB
• 3 AB
• $\frac{5}{4}AB$
Q 31 | Page 73
In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B=
• 60°
• 80°
• 120°
• 80°
• None of these
Q 31 | Page 73
In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B=
• 60°
• 80°
• 120°
• 80°
• None of these
Q 32 | Page 73
The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC =
• 70°
• 90°
• 80°
• 100°
## Chapter 13: Quadrilaterals
Ex. 13.10Ex. 13.20Ex. 13.30Ex. 13.40Others
## RD Sharma solutions for Class 9 Mathematics chapter 13 - Quadrilaterals
RD Sharma solutions for Class 9 Maths chapter 13 (Quadrilaterals) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics for Class 9 by R D Sharma (2018-19 Session) solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 9 Mathematics chapter 13 Quadrilaterals are Concept of Quadrilaterals, Angle Sum Property of a Quadrilateral, Types of Quadrilaterals, Another Condition for a Quadrilateral to Be a Parallelogram, The Mid-point Theorem, Theorem : a Diagonal of a Parallelogram Divides It into Two Congruent Triangles, Theorem : a Diagonal of a Parallelogram Divides It into Two Congruent Triangles, In a Parallelogram, Opposite Sides Are Equal. Ab = Cd and Bc = Da, Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram., In a Parallelogram, Opposite Angles Are Equal., Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram., The Diagonals of a Parallelogram Bisect Each Other., Theorem : If the Diagonals of a Quadrilateral Bisect Each Other, Then It is a Parallelogram.
Using RD Sharma Class 9 solutions Quadrilaterals exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 9 prefer RD Sharma Textbook Solutions to score more in exam.
Get the free view of chapter 13 Quadrilaterals Class 9 extra questions for Maths and can use Shaalaa.com to keep it handy for your exam preparation
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# Equations that Describe Patterns
## Describe numerical sequences by finding a rule.
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Cheez Its
Teacher Contributed
## Real World Applications – Algebra I
### Topic
How are algebraic relationships represented in the nutrition facts of snacks?
### Student Exploration
Below is a copy of the nutrition facts for a box of Cheez Its (Four Cheese), one of the most popular snack foods.
There are a lot of different ways that we can represent the information in this picture.
Let’s say that I’m addicted to Cheez Its and I can never really have just one serving size, which is 36 crackers. We can represent this relationship in a table.
# of serving size 0 1 2 3 4
# of crackers total 0 36 72 108 144
According to this table, 144 crackers eaten is equivalent to 4 servings! That’s a lot!
If we were to calculate the number of crackers in 10 servings, we can write this as an expression and figure it out. If x\begin{align*}x\end{align*} represents the number of serving sizes, the expression is 36x\begin{align*}36x\end{align*}. So, in 10 servings, I could eat 360 crackers. (I might also have a stomach ache too.)
We can also represent this relationship as an equation. If y represents the number of crackers, our equation would be y=36x\begin{align*}y = 36x\end{align*}. If I wanted to eat 200 crackers we can figure out how many serving sizes that would equate to. 200=36x\begin{align*}200 = 36x\end{align*}. We would divide both sides by 36, and find that x=5.6\begin{align*}x = 5.6\end{align*}, or a little over 5 and a half serving sizes.
When you get older, you will start to see that a lot of adults care about their calorie intake. According to the nutrition facts, one serving size has about 140 calories. People usually like to base what they eat on a 2000 calorie diet. Taking this fact into account, what is the maximum number of cheez its that I can eat in a day? Let’s create another equation to represent this relationship.
Let x\begin{align*}x\end{align*} represent the number of cheez its I eat, and let y\begin{align*}y\end{align*} represent the number of calories. Our equation is xy=36140\begin{align*}\frac{x}{y} = \frac{36}{140}\end{align*}. This special equation is also called a proportion. If we use multiplication to eliminate the denominators, our new equation is (140x)36=y\begin{align*}\frac{(140x)}{36} = y\end{align*}. Since we can’t exceed our 2000 calorie diet, our inequality is (140x)362000\begin{align*}\frac{(140x)}{36} \le 2000\end{align*}. What value of x\begin{align*}x\end{align*} would make this statement true?
We can use just two steps to figure this out! First, multiply both sides by 36 to eliminate the denominator. We now have 140x72000\begin{align*}140x \le 72000\end{align*}. Next, we divide both sides by 140. So, x514.3\begin{align*}x \le 514.3\end{align*}. This means that I can’t eat more than 514 crackers in one day, assuming that’ll be the ONLY food I eat all day.
Now we have the maximum amount of crackers I can have in one day, how many servings is this?
51436=14\begin{align*}\frac{514}{36} = 14\end{align*} servings.
Note: It is not suggested that a person eat 514 Cheez Its!
### Extension Investigation
Try looking at the nutrition facts label on your favorite food, and figure out what the maximum amount you can eat of that particular food, based on a 2000 calorie diet.
### Notes/Highlights Having trouble? Report an issue.
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Write-up #8
Exploring Altitudes in Relation to Circumcircles
by
Holly Anthony
Problem: Construct any acute triangle ABC and its circumcircle. Construct the three altitudes AD, BE, and CF. Extend each altitude to its intersection with the circumcircle at corresponding points P,Q, and R. Then find:
First, let's look at our construction:
To manipulate this graph for yourself with measures and ratios of
You should discover that = 4
We find that the sum of these ratios always equals four regardless of your manipulation of the graph. However, the measurements in GSP are not enough to prove this.
For a proof, see below.
Proof
Prove: = 4
For the proof, refer to the illustration below:
First, consider the two triangles: AHE and AQE.
Angles AHQ and AQH are congruent since they both subtend the same chord on the circumcircle.
AEH and AEQ are both 90 degrees since BE is an altitude and they share the side AE. This shows that they are congruent, thus HE = EQ and similarly HF = FR and HD = DP.
So we have the following:
Now we have:
Now let's consider:
First look at the triangle again.
We see that AD*BC is twice the area of ABC since AD is the height for the base BC.
CF*AB and BE*AC are also twice the area of ABC. We will call this number X.
Now notice that the area of ABC could be computed by summing the areas of AHC, BHC, and AHB.
Then twice the area of AHC is HE*AC, the area of BHC is HD*BC, and the area of BHA is HF*AB.
So now consider again:
Notice the values of the denominators on the right-hand side of the above equation. The are all two times the area of ABC or X. We can factor this out:
Remember that:
so we have:
Therefore,
End of Proof. |
#### Ordered Pair
Simply, an ordered pair is a pair of numbers which are used to locate the point on a coordinate plane. Ordered pairs are also used to show the position on a graph where the horizontal value (x') is the first and vertical value (y') is second. If the ordered pairs have two elements, it is written in the form (x, y) in which x is fixed as the second component. For example,
We can pair off the elements and member diagrammatically as follows: -
Elements Numbers x 20 y 30 z 40
We have drawn an arrow from the elements to the numbers. Such figure is called a balloon diagram or arrow diagram. The arrows are used to show the relationship between the ordered pairs.
### Equity of Ordered Pairs
When the first component and the second component of an ordered pair are correspondingly equal then it is called equality of ordered pairs. For examples,
(x, y) = (x, y)
(-2, -5) = (-2, -5)
(1, 2) = (1, 2)
(4, 5) = ($$\frac{8}{2}$$, $$\frac{10}{2}$$)
### Cartesian Product
Cartesian product is simply defined as the set of all possible ordered pairs with first element x and second element y.
Mathematically, the cartesian product of two sets X and Y is written as,
X × Y = {(x, y) ; x ∈ X and y ∈ Y}
### Tree diagram representation of a Cartesian Product
The cartesian product can be represented in 3 ways. They are as follows:
• Tree Diagram
• Mapping Diagram
• Graphical Representation
Tree Diagram
The tree diagram can be represented as follows:
Let, suppose two sets X = (x, y, z) and Y = (3, 4).
Now, Taking x-component from set x and y-component from set y, then we can do all possible pairs as given below:
$$\therefore$$ X × Y = {(x, 3), (x, 4), (y, 3), (y, 4), (z, 3), (z, 4)}
Similarly, we can find the cartesian product B × A as follows:
B × A = {(3, x) (3, y) (3, z) (4, x) (4, y) (4, z)}
Mapping Diagram
Mapping diagram can be represented as given below:
X= {3, 4} and Y = {5, 6}
X×Y = {3, 4} × {5,6} = {(3, 5) (3, 6) (4, 5) (4, 6)}
Here, Each arrow represented an ordered pairs of A × B.
Graphical Representation
Graphical representation can be represented as below:
Suppose, X = {1, 2, 3} and Y = {2, 4, 6}
Then {X×Y} = {1, 2, 3} × {2, 4, 6} = {(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 2), (3, 4), (3, 6)}
• An ordered pair is a pair of numbers which are used to locate the point on a coordinate plane.
• When the first component and the second component of an ordered pair are correspondingly equal then it is called equality of ordered pairs.
• The set of all possible ordered pairs with first element and second element y is called Cartesian Product.
• The arrows are used to show the relationship between the ordered pairs.
0%
a=2, b=5
a=4, b=2
a=5, b=3
a=3, b=3
a= -3, b=9
a= -2, b= -3
a=3, b=6
a=2, b= -2
• ### Find the values of a and b.(2a-3, 4) = (4a, b+5)
a= -2, b=4
a= -3, b= (frac{2}{3})
a=2, b=4
a= -(frac{3}{2}), b= -1
x=6, y=6
x=5, y=5
x=5, y=6
x=6, y=5
2,2,
3,2
2,3
3,3
3,3
2,3
2,2
3,2
2,1
1,1
2,2
1,2
6,6
3,3
3,6
6,3
7,2
2,7
2,2
7,2
1,0
3,4
3,3
4,4
2,2
2,4
3,3
4,4
2,2
3,3
3,4
4,4
2,1
1,1
1,2
2,2 |
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# What is a mathematical proof why is it important?
## What is a mathematical proof why is it important?
All mathematicians in the study considered proofs valuable for students because they offer students new methods, important concepts and exercise in logical reasoning needed in problem solving. The study shows that some mathematicians consider proving and problem solving almost as the same kind of activities.
## How do you write a proof in math?
Write out the beginning very carefully. Write down the definitions very explicitly, write down the things you are allowed to assume, and write it all down in careful mathematical language. Write out the end very carefully. That is, write down the thing you’re trying to prove, in careful mathematical language.
## What is flowchart proof?
A flow chart proof is a concept map that shows the statements and reasons needed for a proof in a structure that helps to indicate the logical order. Statements, written in the logical order, are placed in the boxes. The reason for each statement is placed under that box.
## What is the first step in a proof?
Writing a proof consists of a few different steps. Draw the figure that illustrates what is to be proved. The figure may already be drawn for you, or you may have to draw it yourself. List the given statements, and then list the conclusion to be proved.
## How do you prove a theorem?
Summary — how to prove a theorem Identify the assumptions and goals of the theorem. Understand the implications of each of the assumptions made. Translate them into mathematical definitions if you can. Make an assumption about what you are trying to prove and show that it leads to a proof or a contradiction.
## What is used to prove theorems?
Postulates may be used to prove theorems true. The term “axiom” may also be used to refer to a “background assumption”. Example of a postulate: Through any two points in a plane there is exactly one straight line.
## What is difference between postulate and theorem?
A postulate is a statement that is assumed true without proof. A theorem is a true statement that can be proven. Postulate 1: A line contains at least two points.
## Do axioms Need proof?
Unfortunately you can’t prove something using nothing. You need at least a few building blocks to start with, and these are called Axioms. Mathematicians assume that axioms are true without being able to prove them. If there are too few axioms, you can prove very little and mathematics would not be very interesting.
## What is difference between theorem and Axiom?
The axiom is a statement which is self evident. But,a theorem is a statement which is not self evident. An axiom cannot be proven by any kind of mathematical representation. A theorem can be proved or derived from the axioms.
## What are the basic axioms of mathematics?
An Axiom is a mathematical statement that is assumed to be true. There are five basic axioms of algebra. The axioms are the reflexive axiom, symmetric axiom, transitive axiom, additive axiom and multiplicative axiom.
## What are the 5 axioms?
AXIOMS
• Things which are equal to the same thing are also equal to one another.
• If equals be added to equals, the wholes are equal.
• If equals be subtracted from equals, the remainders are equal.
• Things which coincide with one another are equal to one another.
• The whole is greater than the part.
## What is an axiom example?
In mathematics or logic, an axiom is an unprovable rule or first principle accepted as true because it is self-evident or particularly useful. “Nothing can both be and not be at the same time and in the same respect” is an example of an axiom.
five axioms
## Can axioms be wrong?
Unfortunately there is no set of axioms that will let you prove or disprove every statement. True and false aren’t really meaningful when applied to axioms.
## What are Euclid axioms?
Some of Euclid’s axioms were : (1) Things which are equal to the same thing are equal to one another. (2) If equals are added to equals, the wholes are equal. (3) If equals are subtracted from equals, the remainders are equal. (4) Things which coincide with one another are equal to one another.
## Is our world Euclidean?
In the small, the world is Euclidean. Curved space does not become obvious until it is extended. That is why so many people in ancient time believed the earth was flat.
## What did Euclid say about circles?
Euclid typically names a circle by three points on its circumference. Perhaps a better translation than “circumference” would be “periphery” since that is the Greek word while “circumference” derives from the Latin.
Aristotle
## What are axioms postulates?
Axioms and postulates are essentially the same thing: mathematical truths that are accepted without proof. Their role is very similar to that of undefined terms: they lay a foundation for the study of more complicated geometry. Axioms are generally statements made about real numbers.
## What is postulate example?
A postulate is a statement that is accepted without proof. Axiom is another name for a postulate. For example, if you know that Pam is five feet tall and all her siblings are taller than her, you would believe her if she said that all of her siblings are at least five foot one.
Category: Uncategorized
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# Working with Expressions and Equations Part 2
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## Objective
SWBAT: ⢠Create algebraic expressions and equations based on word problems. ⢠Use substitute to evaluate algebraic expressions. ⢠Identify dependent and independent variables.
#### Big Idea
Students use substitution to find the values of equations and expressions they wrote using appropriate math symbols.
## Do Now
7 minutes
Part of my class routine is a do now at the beginning of every class. Students walk into class and pick up the packet for the day. They get to work quickly on the problems. Often, I create do nows that have problems that connect to the task that students will be working on that day. For this lesson I want students to practice identifying the independent and dependent variables.
I ask for volunteers to define the independent and dependent variables in their own words. We refine the definitions. I call on students to identify the variables in #1 and #2. I want students to be able to articulate that the number of miles Dan drives depends on the number of hours he drives. The longer he drives the further he’ll go. I want students to be able to articulate that the amount of money you spend on a data plan depends on the number of months you use the plan. The longer you use the plan, the more money you will spend on the data plan. If students struggle to articulate these patterns I will set up a table and have students generate values as the hours/months increase.
## Income, Cost, and Profit
13 minutes
After the Do Now, I have a student read the objectives for the day. I tell students that they will be creating expressions and equations to model situations. These expressions and equations will be more complicated than the ones that we created previously.
On page 2 we review the equation for the amount of money collected and identify the independent variables. I introduce the concepts of income, costs, and profit. Students share out ideas for costs. Students will likely identify cost of paying employees, cost of tools and materials (tractor, gas, clippers, signs ), advertising (tv/radio ad, website, etc), and utilities (water, electricity). These are all things that cost money! Profit is the amount of money a business still has after paying its costs.
I have students brainstorm for a minute to create an expression for #1 on page 3. Students share out their ideas. Some students may struggle to connect 8n (the amount of money collected) and 75 (the cost of one day of running the maze). Other students may struggle with whether the expression should be 8n – 75 or 75 – 8n. I will revisit the idea of profit being the difference between the amount of money the maze takes in and the costs of running the maze.
Once we have agreed on the expression for the profit, students will work in partners to answer question 2. I am looking to see that students are correctly using substitution. A common mistake is that a student will plug in a value for n, but forget to multiply that value by 8. For example, for 2a a student may substitute 80 for n and get 880 – 75, instead of multiplying 8 times 80 and getting 640 – 75.
We move on to part C on page 4. I ask students what they think each line graph represents. I am looking for students to connect the top graph with the graph that they created in the last lesson. Students should notice that the bottom graph represents the profit that the business makes. I am looking for students to observe that (0,-75) means that if there are no visitors, the profit is -\$75, or the maze loses \$75. How could that be? That means that the business spent more money than it brought it.
Before moving on to the next section, I will ask students to use the graph and their equation to predict what the smallest number of visitors it would take for the maze to make a profit. Some students may look at the graph and see that when there are 10 visitors, the amount of profit is above \$0. Other students may use substitution to show that 8(10) – 75 = 5, so when there are 10 visitors there is a profit of \$5. Other students may create a table or chart showing that with 9 visitors the maze has a loss of \$3 and that with 10 visitors the maze has a profit of \$5.
## Partner Practice
30 minutes
I have students work on the Practice section (pages 5-6) in homogeneous partner pairs. See my video Creating Homogeneous Groups in my Strategy Folder. I walk around and monitor student progress.
If students are struggling, I may ask them the following questions:
• What is the expression that we used to represent the amount of money collected by the maze in one day? How can we use this expression to help us show the amount of money collected on Saturday and Sunday?
• What does the variable r represent? What does the variable u represent?
• What does it mean when two things are equivalent?
• What operation is occurring when the 8 is outside of the parentheses?
• How can you test to see if your expression is equivalent to 8 (r + u)?
• If the cost for running the maze is \$75 for one day, what is the cost of running the maze for 2 days?
• How do we find profit? Look back to our notes on page 2.
If students are correctly working through the examples, they can move on to the extra practice on pages 7 and 8.
## Closure
10 minutes
I have students work on the Practice section (pages 5-6) in partners. I walk around and monitor student progress.
If students are struggling, I may ask them the following questions:
• What is the expression that we used to represent the amount of money collected by the maze in one day? How can we use this expression to help us show the amount of money collected on Saturday and Sunday?
• What does the variable r represent? What does the variable u represent?
• What does it mean when two things are equivalent?
• What operation is occurring when the 8 is outside of the parentheses?
• How can you test to see if your expression is equivalent to 8 (r + u)?
• If the cost for running the maze is \$75 for one day, what is the cost of running the maze for 2 days?
• How do we find profit? Look back to our notes on page 2.
If students are correctly working through the examples, they can move on to the extra practice on pages 7 and 8. |
# Comparison of Ratios in Decimal Form
## Rewrite ratios as decimals.
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Comparison of Ratios in Decimal Form
Erinda went with her dad to the toy store. She was looking for stuffed animals to give to children in need during the holidays. Luckily for Erinda, the store was having a big sale, and all the stuffed animals were marked down one third. Some of the toys that Erinda would like to buy were originally priced at 99 cents. How can she convert \begin{align*}\frac{1}{3}\end{align*} to a decimal to calculate how much money she can save on each stuffed animal?
In this concept, you will learn how to convert and compare ratios in decimal form.
### Comparing Ratios in Decimal Form
A ratio is a relationship between two numbers by division.
Fractions are ratios.
Decimals are ratios.
Remember that decimals are based on multiples of 10, so if possible, convert fractions to one of these multiples.
To convert a ratio to its decimal form:
First, write the ratio as a fraction.
Next, divide the numerator by the denominator. "Denominator" and "down" both start with the letter "d."
Look at this example.
Rewrite the ratio 1 to 4 in decimal form.
First, write the ratio as a fraction.
\begin{align*}\frac{1}{4}\end{align*}
Next, divide the numerator by the denominator.
\begin{align*}\frac{1}{4} = {4 \overline{ ) 1 \;\;\;}}\end{align*}
Since 1 cannot be evenly divided by 4, rewrite 1 as a decimal with a zero after the decimal point. Any number of zeroes can be annexed because \begin{align*}1 = 1.0 = 1.00 = 1.000\end{align*}, etc.
Before you divide, remember to put a decimal point in the quotient directly above the decimal point in the dividend.
Then, divide.
\begin{align*}& \overset{ \ \ 0.2}{4 \overline{ ) {1.0 \;}}}\\ & \quad \underline{-8}\\ & \quad \ \ 2 \end{align*}
Continue adding zeroes after the decimal point and dividing until the quotient has no remainder.
\begin{align*}& \overset{ \ \ 0.25}{4 \overline{ ) {1.00 \;}}}\\ & \quad \underline{-8 \;\;}\\ & \quad \ \ 20\\ & \ \ \ \underline{-20}\\ & \qquad \ 0 \end{align*}
The answer is 0.25. The decimal form of the ratio \begin{align*}\frac{1}{4}\end{align*} is 0.25.
### Examples
#### Example 1
Earlier, you were given a problem about Erinda and the one-third-off sale at the toy store.
She found some stuffed animals that were originally 99 cents.How much will she save on each toy?
First, write the ratio as a fraction.
\begin{align*}\frac{1}{3}\end{align*}
Next, convert the fraction to a decimal by dividing the numerator by the denominator.
\begin{align*}& \overset{ \ \ 0.333}{3 \overline{ ) { 1.000 \;}}}\\ & \quad \ \underline{-9\;}\\ & \qquad 10\\ & \quad \ \underline{-9}\\ & \qquad \quad 1 \end{align*}
Then, recognize that .333 is a repeating decimal and round to two places.
\begin{align*}.333\approx .33\end{align*}
The answer is \begin{align*}\frac{1}{3}\end{align*}off the price of each toy is 33 cents. Erinda will save 33 cents on each toy.
#### Example 2
Rewrite the ratio 9:5 in decimal form.
First, rewrite the ratio as a fraction.
\begin{align*}\frac{9}{5}\end{align*}
Next, divide the numerator by the denominator, remembering to annex zeroes and continue until there is no remainder.
\begin{align*}& \overset{ \ \ 1.8}{5 \overline{ ) {9.0 \;}}}\\ & \quad \underline{-5\;}\\ & \quad \ 40\\ & \ \ \underline{-40}\\ & \qquad 0 \end{align*}
The answer is 1.8 The decimal form of 9:5 is 1.8
#### Example 3
Compare the two ratios by converting them to decimals. Are they equivalent?
\begin{align*}\frac{7}{14}\end{align*}and \begin{align*}\frac{11}{20}\end{align*}
First, rewrite \begin{align*}\frac{7}{14}\end{align*} in decimal form by dividing the numerator by the denominator.
\begin{align*}& \overset{ \ \ 0.5}{14 \overline{ ) {\ 7.0 \;}}}\\ & \ \ \underline{-70\;}\\ & \qquad 0\\ & \quad \ \underline{-0}\\ & \qquad 0 \end{align*}
Next, rewrite \begin{align*}\frac{11}{20}\end{align*} in decimal form.
\begin{align*}& \overset{ \ \ 0.55}{20 \overline{ ) { 11.00 \;}}}\\ & \quad \ \underline{-100\;}\\ & \qquad 100\\ & \quad \ \underline{-100}\\ & \qquad \quad 0 \end{align*}
Then, compare the two values by writing each one to the same number of decimal places.
\begin{align*}\frac{7}{14}=0.50\end{align*}
\begin{align*}\frac{11}{20}=0.55\end{align*}
\begin{align*}0.50<0.55\end{align*}
The answer is that the ratios are not equivalent. \begin{align*}\frac{7}{14}<\frac{11}{20}\end{align*}
#### Example 4
Compare 5 to 10 and \begin{align*}\frac{2}{5}\end{align*} and tell if they are equivalent.
First, write 5 to 10 as a fraction.
\begin{align*}\frac{5}{10}\end{align*}
Next, write each fraction as a decimal.
\begin{align*}\frac{5}{10}=0.5\end{align*}
Then, remember that decimals are based on multiples of 10, and convert \begin{align*}\frac{2}{5}\end{align*} to tenths by multiplying both the numerator and denominator times 2.
\begin{align*}\frac{2\times 2}{5\times 2}=\frac{4}{10}\end{align*}= 0.4
Next, compare the decimals.
\begin{align*}0.5>0.4\end{align*}
The answer is that the ratios are not equivalent.
#### Example 5
Compare 6 to 10 and 1:4
First, write the ratios as fractions.
\begin{align*}\frac{6}{10}\end{align*}and \begin{align*}\frac{1}{4}\end{align*}
Next, convert the fractions to decimals.
\begin{align*}\frac{6}{10}=0.60\end{align*}
\begin{align*}\frac{1}{4}=0.25\end{align*}
Then, compare the decimals.
\begin{align*}0.60>0.25\end{align*}
The answer is that the ratios are not equivalent.
### Review
Write each ratio as a decimal. Round to the nearest hundredth when necessary.
1. 1 to 4
2. 3 to 6
3. 3:4
4. 8 to 5
5. 7 to 28
6. 8 to 10
7. 9 to 100
8. 15:20
9. 18:50
10. 3 to 10
11. 6 to 8
12. 15 to 35
Compare the following ratios using <, >, or =.
1. .55 ____1 to 2
2. 3:8 _____ 4 to 9
3. 1 to 2 _____ 4:8
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Decimal In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths). |
# Video: Plot Linear Equations Using a Table of Values
Lucy Murray
Learn how to evaluate 𝑥- and 𝑦-coordinates for a linear function, record them in a table of values, and then plot them on a straight line graph. We also look at rearranging equations into a suitable form to help us calculate 𝑦-coordinates more easily.
07:46
### Video Transcript
Plot Linear Equations
We know that every single linear equation is in the form ๐ฆ equals ๐๐ฅ plus ๐, where ๐ is the gradient and ๐ is the ๐ฆ-intercept or where it cuts the ๐ฆ-axis โ cept meaning to cut. Now in this video weโre just going to be able to plot graphs and get them into the form ๐ฆ equals ๐๐ฅ plus ๐ to be able to plot them. So if we want to plot the graph ๐ฆ equals three ๐ฅ minus one or any graph for that matter, the first thing we need is a table of values. Well you have to choose at least three ๐ฅ-values. So we are choosing at least three ๐ฅ-coordinates. And the reason you choose three is basically just in case we get something wrong. So if you choose two and they give you those two points, we could say okay Iโll draw a line in between them. And here we are, hereโs my graph. But actually what happens if we needed the third one. It was here and that tells you either the first, the second, or the third point is wrong. So by doing at least three, it gives us a way of checking what points actually correct into our calculations. Now I like to choose negative one, zero, and one because it involves the least calculation, but you can choose whichever points you like.
So if we do choose negative one, zero, and one, what we need to do is substitute each of those ๐ฅ-coordinates into our function to find the ๐ฆ. So in the first case, weโre going to do ๐ฆ equals three multiplied by negative one minus one. Well, three multiplied by negative one is negative three. And then subtracting one, we get negative four. So our first ๐ฆ-coordinate is negative four. And substituting our next ๐ฅ-coordinate of zero, weโve got three multiplied by zero, which we know is zero. And subtracting one, so zero minus one is minus one. So our next ๐ฆ-coordinate is negative one. And then finally three multiplied by one minus one, so three multiplied by one is three and minus one is equal to two.
So now we need a set of axes to be able to plot our points. Now we can see that our table of value leads us to coordinates to be able to plot. For example, the first one, weโve got negative one as the ๐ฅ-coordinate and then negative four as the ๐ฆ. And the next one is zero as the ๐ฅ and negative one as the ๐ฆ. And then finally, weโve got one in the ๐ฅ-coordinate and two in the ๐ฆ. Now plotting each of these, negative one, negative four, then zero, negative one, and then one, two. And then joining them up, we should get something like this that I prepared earlier.
Now in our next example, we havenโt got it in the form ๐ฆ equals ๐๐ฅ plus ๐. So weโre going to have to do that first. Plot the graph four ๐ฅ plus two ๐ฆ equals ten. So as we can see that as I just said itโs not in the form ๐ฆ equals ๐๐ฅ plus ๐, and it must do- first do a table of values. So first we want it to be in that form. So basically what we need to do is rearrange this to get ๐ฆ as the subject. So what weโre gonna do first is subtract four ๐ฅ from both sides. And that will give us just two ๐ฆ on the left and then ten minus four ๐ฅ on the right. And then this is two multiplied by ๐ฆ or the opposite of multiplied or times by is divide by, so you must now divide both sides by two. Be careful as we need to divide every single term by two. On the left-hand side, we will just have ๐ฆ now. Now weโve got five minus two ๐ฅ.
Now although this isnโt in the form exactly โ๐ฆ equals ๐๐ฅ plus ๐,โ weโve got the ๐ฆ by itself and weโve got the ๐๐ฅ and the ๐ on the same side. So as long as itโs roughly in the same form like that, then youโre okay. So now weโve got this; we can do a table of values. And again Iโm gonna choose the same three coordinates as we did last time, so negative one, zero, and one. And substituting them each individually into our function, five minus two multiplied by negative one. So the two negatives will cancel out giving us a positive two. So the five plus two, and we know the answer to five plus two is seven.
And then for the next case where ๐ฅ is equal to zero, so five minus two times zero, well two times zero is zero. So itโs five minus zero, which we all know is just five. And then for the last ๐ฅ-coordinate of one, weโll substitute that into the function. So weโve got five minus two times one. Two times one is two; so itโs five minus two and five minus two is three. So there we have our table of values and we know that the table of values leads us to coordinates: the first one being negative one, seven, then zero, five, and one, three.
Finally, we just need to plot the points and join them up with a line. So looking for negative one, seven will be minus [minus one] in the ๐ฅ and seven in the ๐ฆ. Then zero, five, zero in the ๐ฅ, five in the ๐ฆ, and then one, three, so across one and up three. And then joining them up, it gives us our straight line here. So we can see it goes through five on the ๐ฆ-axis and it goes down. And this down is related to whatโs in front of the two, being a negative. So this goes on to gradients. So there we have it. To plot linear graphs, what we need is a table of values. But before itโs in a table of values, we must rearrange it to the form ๐ฆ equals ๐๐ฅ plus ๐. Now once we have our table of values, we have some coordinates. Obviously weโve chosen the ๐ฅ-coordinates, but the ๐ฆ-coordinates weโve got from substituting the ๐ฅ-coordinates into the functions individually. And then these give us the coordinates to plot the points on the graph. And once weโve plotted the points, we get a straight edge or a ruler which I havenโt done here and draw a nice line through all these points, giving us the function. |
# Common Numerator
Home » Math Vocabulary » Common Numerator
## Common Numerator – Introduction
When numbers are written as a fraction, they can be represented as $\frac{a}{b}$. Now the fraction has two main parts: the numerator and the denominator.
When the numerators of two or more fractions are the same or like, the fractions are said to have a like numerator, irrespective of whether the denominators are the same or not.
In the above example, the first figure is divided into 3 parts, and 1 part is shaded, whereas in the second figure, there are 4 equal parts, and only 1 is shaded. Thus, the fractions ⅓ and ¼ have like numerators.
When two or more fractions have different numerators, we can convert them into fractions with the same numerators by finding the common multiple of the numerators. Today, we will learn about the definition of Common Numerators and its facts. Let us get right into it.
## What Is a Common Numerator?
A common multiple of the numerators of two or more fractions is called a common numerator. Let’s study more by considering the following example.
Take the fractions $\frac{4}{5}$ and $\frac{6}{7}$. Both fractions have different numerators. To find their common numerator, we find the common multiples of the numerators 4 and 6.
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, . . .
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, . . .
Common multiples of 4 and 6: 12, 24, 36, . . .
So, the least common numerator for 4⁄5 and 6⁄7 can be 12, 24, 36, . . .
If we want the common numerator to be 12, we must find the equivalent fraction of $\frac{4}{5}$ and $\frac{6}{7}$, which has 12 as the numerator.
Now, we have two fractions with the same numerator.
So, $\frac{12}{15}$ and $\frac{12}{14}$ have a common numerator. It is 12.
So, now that we know how to find common numerators, how do we put them to use? How do they help us solve problems in fractions? Let’s take a look at some of the applications of common numerators. The most common use of it is to compare fractions. We can easily compare two or more fractions when their numerators are common or like. Let’s now figure out how we can do this.
## Comparing Fractions with Like Numerators
We can use multiple strategies for comparing fractions. One of these is to compare the fraction by finding the common numerator strategy. Comparing fractions with like or same numerators is an easy task. For example, let us compare $\frac{2}{5}$ and $\frac{4}{7}$.
Look at the two numerators (4 and 2) and the two denominators (5 and 7). It is much easier to find a common multiple of the numerators (4) than to find a common denominator. So, find a fraction equivalent to $\frac{2}{5}$ with 4 as a numerator $(\frac{2}{5} \times \frac{2}{2} = \frac{4}{10} )$.
You can now compare $\frac{4}{10}$ and $\frac{4}{7}$ by comparing the size of each piece. Since sevenths are larger than tenths, $\frac{4}{7}$ is greater than $\frac{4}{10}$ or $\frac{2}{5}$ .
To compare two or more fractions with the same or like numerators, we just have to compare their denominators. The fraction with the smallest denominator is the greatest.
If we compare $\frac{8}{11}$ and $\frac{8}{17}$, we take a look at the denominators. Since 17 is more than 11. So, $\frac{8}{11}\gt\frac{8}{17}$.
## Solved Examples
1. Identify which fractions have the same numerator.
$\frac{3}{8},\frac{7}{8}, \frac{3}{7}$ and $\frac{2}{7}$
Solution:
The fractions given are: $\frac{3}{8},\frac{7}{8}, \frac{3}{7}$ and $\frac{2}{7}$
Of these, the fractions $\frac{3}{8}$, and $\frac{3}{7}$ have the same numerator. Their numerators are like as they have the number 3 as the common numerator.
2. Find out the common numerator of $\frac{5}{8}$ and $\frac{7}{9}$.
Solution:
To find their common numerator, we find the common multiples of the numerators 5 and 7.
Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, . . .
Multiples of 7: 7, 14, 21, 28, 35, 42, . . .
As we can seen, a common multiple of 5 and 7 is 35
So, a common numerator for $\frac{5}{8}$ and $\frac{7}{9}$ can be 35.
If we want the common numerator to be 35, we need to multiply both the numerator and denominator as shown:
$\frac{5}{8}\times \frac{7}{7} = \frac{35}{56}$
$\frac{7}{9} \times \frac{5}{5} = \frac{35}{45}$
Now, we have two fractions with the same numerator.
$\frac{35}{56} \gets$ common numerator $\to \frac{35}{45}$
3. Which fraction is greater: $\frac{5}{11}$ or $\frac{5}{13}$?
Solution:
The fractions $\frac{5}{11}$ and $\frac{5}{13}$ have the same numerator: 5. We know that if two or more fractions have like numerator, the fraction with the smallest denominator is the greatest. So, elevenths are greater than thirteenths.
Therefore, $\frac{5}{11} \gt \frac{5}{13}$.
4. Arrange the following fractions: $\frac{7}{11} ; \frac{7}{17}; \frac{7}{10}; \frac{7}{12}$
a) in the ascending order b) in the descending order
Solution:
The fractions $\frac{7}{11} ; \frac{7}{17}; \frac{7}{10}$ and $\frac{7}{12}$ have the same numerator: 7. We know that if two or more fractions have like numerators, the fraction with the smallest denominator is the greatest.
a) in the ascending order (smallest to greatest): $\frac{7}{17} \lt \frac{7}{12} \lt \frac{7}{11} \lt \frac{7}{10}$
b) in the descending order (greatest to smallest): $\frac{7}{10} \gt \frac{7}{11} \gt \frac{7}{12} \gt \frac{7}{17}$
## Practice Problems
1
### What is the common numerator of these fractions: $\frac{7}{9} ; \frac{7}{11}; \frac{7}{12}$?
7
9
11
12
CorrectIncorrect
We know that the numerator is the part of a fraction that is above the fraction bar. So, the numerator of these fractions is 7.
2
### Which of these fractions is the greatest?
$\frac{5}{7}$
$\frac{5}{6}$
$\frac{5}{11}$
$\frac{5}{9}$
CorrectIncorrect
Correct answer is: $\frac{5}{6}$
We know that if two or more fractions have like numerators, the fraction with the smallest denominator is the greatest. Since 6 is the smallest, $\frac{5}{6}$ is the greatest of these fractions.
3
### Choose the correct symbol to make the statement true.$\frac{2}{7}$ ____ $\frac{4}{13}$
$\gt$
$\lt$
$=$
None of these
CorrectIncorrect
Correct answer is: $\lt$
Look at the relationship between the two numerators (4 and 2). Let’s make the numerator the same or like by finding the common multiple of the numerators, which is 4.
$\frac{2}{7}=\frac{4}{14}$. Now compare the fractions $\frac{4}{14}$ and $\frac{4}{13} . \frac{4}{14} \lt \frac{4}{13}$
4
### Which of the following fractions is greater than $\frac{3}{5}$?
$\frac{3}{4}$
$\frac{3}{7}$
$\frac{3}{8}$
$\frac{3}{9}$
CorrectIncorrect
Correct answer is: $\frac{3}{4}$
If two or more fractions have a common numerator, then the fraction with the smallest denominator is the greatest. Here 4 is smaller than 5, making $\frac{3}{4} \gt \frac{3}{5}$.
The numerator is the number that sits on the top of the fraction bar and represents the number of parts out of the whole. For example, in the fraction $\frac{18}{11}$, the number 18 is the numerator.
In the context of division, the numerator is the dividend. We know that the number to be divided or distributed into a certain number of equal parts is called the dividend. Let us understand this with an example: If we were to divide the number 24 by 7 or $\frac{24}{7}$, the number 24 in this division is the numerator or dividend.
In a fraction, a denominator represents the number of equal parts a whole is divided into. A common denominator is a common multiple of the denominators of two or more fractions. For example, consider the fractions $\frac{1}{8}$ and $\frac{1}{6}$. 24 is a common multiple of denominators 8 and 6. Thus, 24 is a common denominator of fractions $\frac{1}{8}$ and $\frac{1}{6}$. |
# Math: Elementary School: 1st and 2nd Grade Quiz - Which Symbol Should We Use? - Addition, Subtraction and Equals Signs (Questions)
This Math quiz is called 'Which Symbol Should We Use? - Addition, Subtraction and Equals Signs' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is a fun way to learn if you are in the 1st or 2nd grade - aged 6 to 8.
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Recognising, selecting and using the right symbol in a number sentence is vital. A good understanding of what the addition, subtraction and equals signs look like, what they mean in math, and what they do to numbers is a really important area of development. Understanding the language associated with the addition, subtraction and equals signs will help children calculate accurately.
How much do you know about the symbols used in math?
1. What does this symbol mean: +
[ ] Adding or addition [ ] Subtracting or subtraction [ ] Is the answer [ ] Is the same as
2. What does this symbol mean: -
[ ] Multiplication [ ] Addition or adding [ ] Subtracting or subtraction [ ] Division
3. What does this symbol mean: =
[ ] My answer is [ ] Is equal to or the same as [ ] Dividing [ ] Multiplying
4. Which symbol is missing from this calculation:
3 ? 6 = 9
[ ] + [ ] - [ ] = [ ] x
5. Which symbol is missing from this calculation:
6 - 2 ? 4
[ ] x [ ] + [ ] = [ ] -
6. Which symbol is missing from this calculation:
4 ? 1 = 3
[ ] ÷ [ ] = [ ] + [ ] -
7. What is wrong with this number sentence:
7 + 3 = 4
[ ] The addition sign should be a subtraction sign [ ] The addition sign is in the wrong place [ ] The equal sign is in the wrong place [ ] The numbers are too small
8. Which symbol could be used instead of the words in brackets?
5 (plus) 5 = 10
[ ] - [ ] = [ ] + [ ] ÷
9. Which symbol could be used instead of the words in brackets?
5 (minus) 5 = 0
[ ] x [ ] = [ ] + [ ] -
10. Which symbol could be used instead of the words in brackets?
5 + 4 (is equal to) 9
[ ] + [ ] = [ ] - [ ] x
Math: Elementary School: 1st and 2nd Grade Quiz - Which Symbol Should We Use? - Addition, Subtraction and Equals Signs (Answers)
1. What does this symbol mean: +
[x] Adding or addition [ ] Subtracting or subtraction [ ] Is the answer [ ] Is the same as
It is used when numbers are added together, for example 4 + 5 = 9
2. What does this symbol mean: -
[ ] Multiplication [ ] Addition or adding [x] Subtracting or subtraction [ ] Division
This symbol is used in subtraction calculations, such as 5 - 3 = 2
3. What does this symbol mean: =
[ ] My answer is [x] Is equal to or the same as [ ] Dividing [ ] Multiplying
Many people think this symbol means 'is the answer' but it really means 'is equal to'
4. Which symbol is missing from this calculation:
3 ? 6 = 9
[x] + [ ] - [ ] = [ ] x
5. Which symbol is missing from this calculation:
6 - 2 ? 4
[ ] x [ ] + [x] = [ ] -
6 - 2 is equal to, or the same as 4
6. Which symbol is missing from this calculation:
4 ? 1 = 3
[ ] ÷ [ ] = [ ] + [x] -
Subtracting 1 from 4 gives an answer of 3
7. What is wrong with this number sentence:
7 + 3 = 4
[x] The addition sign should be a subtraction sign [ ] The addition sign is in the wrong place [ ] The equal sign is in the wrong place [ ] The numbers are too small
If you change - for +, the answer becomes correct
8. Which symbol could be used instead of the words in brackets?
5 (plus) 5 = 10
[ ] - [ ] = [x] + [ ] ÷
Plus is another word for add
9. Which symbol could be used instead of the words in brackets?
5 (minus) 5 = 0
[ ] x [ ] = [ ] + [x] -
Minus is another way of saying subtract or take away
10. Which symbol could be used instead of the words in brackets?
5 + 4 (is equal to) 9
[ ] + [x] = [ ] - [ ] x
This symbol means 'is equal to' or 'is the same as' |
# Difference Between Equations and Functions (With Table)
When a student is learning the subject of álgebra, the difference between function and an equation is always unclear. Equations and Functions are two different topics of the subject of algebra. There often arises a blur between a function and an equation as both use variables to solve their equation. Although they both use variables, they do have their differences.
## Equations vs Functions
The difference between an equation and a function is that in an equation, a person solving an equation can have either one or two values based on the number of variables they have put into use for solving that equation, and on the other hand, is a function a person will always have solutions based on the Input they have chosen to solve their question.
Equations are a topic used in algebra to solve problems through variables. These equations act as a statement to denote the equality of a variable on both the left and right sides of a statement. When a person wants to show equality between two variables, the sign = is used to denote equality.
Functions are a topic in algebra used by a person to solve problems using variables. When it comes to explaining what functions are in algebra, it seems to be a pretty wide topic to be understood. A function is said to be formed when a value of one variable of a set is paired with a value of another variable from another set.
## What are Equations?
Equations are a topic used in algebra to solve problems through variables. These equations act as a statement to denote the equality of a variable on both the left and right sides of a statement. When a person wants to show equality between two variables, the sign = is used to denote equality.
In a function, the right and the left side are always told to be equal. They are always said to have an inverse relationship being unitary by nature when solved. An equation always contains more than one variable. They are most commonly used for Algebraic calculations to make it easier to find the solutions. There are also some types of equations such as Quadratic equations, linear equations, etc.
An equation, in short, means finding the value of a specific variable given in the problem. Following are some examples of the equation.
1. 2a + 3a = 15, what are the values of (a)?
2. 4a + 6a = 24, what are the values of (a)?
## What estão Functions?
Functions are a topic in algebra used by a person to solve problems using variables. When it comes to explaining what functions are in algebra, it seems to be a pretty wide topic to be understood. A function is said to be formed when a value of one variable of a set is paired with a value of another variable from another set.
In most schools, a function is always taught to a child as a rule which is taken as each set of a member x and mapped to the same value of y on the page. When a person wants to denote a Function between two variables, he must express it as F → function → y.
A letter such as F, a, or g is used to denote the word function in any kind of algebraic expression. Following are some examples where the problems are solved by using functions.
1. F(x) = 3x + 5
2. F(g) (x) = 6y+9
## Main Differences Between Equations and Functions
1. In an equation, the sums are solved based on the value that they are equated with, and on the other hand, it is a function. The sums are solved on the basis of values that are assigned for the variables.
2. Equations are supersets of functions, and on the other hand, functions are subsets of equations.
3. All equations can be represented on a graph, and on the other hand, not all functions can be represented on a graph.
4. An equation is capable of having more than one value for its variable, and on the other hand, a function cannot have two values for its variable.
5. In a vertical test for equations in a graph, a line can intersect at one or two points, and on the other hand, in a vertical test for functions, a line can intersect at multiple points in a graph.
## Conclusão
Equations and Function are both statements in mathematics used For solving problems by a person. When a student is learning the subject of algebra, the difference between both a function and an equation is always unclear. Equations and Functions are two different topics of the subject of algebra.
There often arises a blur between a function and an equation as both use variables to solve their equation. Although they both use variables, they do have their differences. One should know the differences between them to be able to solve problems related to those concepts. They both are very essential and most commonly used for solving algebraic sums. |
# Rectangle Formula
View Notes
## Area of Rectangle Formula
The area of a rectangle is the region encompassed by the rectangle in a two-dimensional closed plane figure. In simple words, the area of a rectangle is the total space covered by the figure. On the other hand, the space within the bounds of the perimeter of the rectangle is also the area of a rectangle. Here, you will also get to know why the area of a rectangle is the product of its two sides and also the units of measurement.
### What is a Rectangle?
A rectangle is a 2 dimensional geometric shape consisting of four sides, four angles and four vertices. Opposite sides of the rectangle are equivalent and parallel to each other in a rectangle. Remember that all the interior four angles of the rectangle are right angles. Pointing out, a parallelogram also has its opposite sides equal and parallel to each other but the angles do not make 90 degrees of measurement.
### Important Rectangle Formulas
The formula of the area and perimeter of the rectangle is given below.
Given that ‘l’ represents the length and ‘b’ represents the breadth of the rectangle, then the
Element Formula Perimeter of rectangle formula 2(l + b) units Length of the rectangle formula P/2 - b units Breadth of the rectangle P/2 - l units Area of rectangle formula l × b sq. units Length of the rectangle A/b units Breadth of the rectangle A/l units Diagonal of the rectangle √(l²+b²) units
### Derive and Calculate the Area of a Rectangle
To derive the area of a rectangle, we use the unit squares. Divide the rectangle MNOP into unit squares. The area of a rectangle MNOP is the total number of unit squares in it.
Now, finding the area of a rectangle
Rectangle MNOP in unit squares each with 1 sq. inch
Therefore, the total area of the rectangle MNOP is equal to 48 sq. inch.
Also, following the perspective, we discovered that the area of a rectangle is mandatorily the product of its two sides. Here, the length of MN is 6 inches and the breadth of NO is 8 inches. The area of MNOP is the product of 6 and 8, which is 48.
Instead, formulas to find the area of a rectangle is derived by dividing the figure into two equal-sized right triangles. For example, in the given rectangle MNOP, a diagonal from the vertex M is drawn to O.
The diagonal MO divides the rectangle into 2 equivalent right-angle triangles.
Therefore, the area of MNOP will be:
  ⇒ Area (MNOP) = Area (MNO) + Area (MPO)
  ⇒ Area (MNOP) = 2 × Area (MNO)
  ⇒ Area (MNO) = ½ × base × height
  ⇒ Area (MNOP) = 2 × (½ × b × h)
   ⇒ Area (MNOP) = b × h
### Application of Rectangle Formula
The early documentation script of Babylonian culture represents the use of geometric objects with lengths, width, angles, and areas for construction and astronomy. The skill and understanding of stone cutting in standard shapes such as squares, triangles and rectangles along with principles having reference to area and perimeter helped ancient Egyptians in constructing giant structures like pyramids. In modern times, these rectangle volume formula concepts are greatly helpful in object modelling, land surveying and others.
### Examples of Rectangle
Some examples of rectangular figures are:
• Parks and agricultural fields,
• Painting Canvas
• Tiles and walkway with rectangular tiles
• Daily life objects such as a table, serving tray, glass, etc.
### Solved Examples
Here, we will perform practical problems using step-by-step solutions in order to find the area of a rectangle as well as other measurements.
Example:
Calculate the area and the perimeter of the rectangle box that measures 15 cm in length and 8cm in breadth.
Solution:
Given:
Length = 15 cm,
Now applying the area of rectangle formula i.e. = length × breadth
= (15 × 8) cm²
= 120 cm²
Using the formula for Perimeter of rectangle = 2 (length + breadth)
 = 2 (15 + 8) cm
= 2 × 23 cmÂ
= 46 cm
Example:
Evaluate the breadth of the Television set rectangular in shape whose area is 320 m2 and whose length is 40 m. Find its perimeter.
Solution:
We know that the breadth of the rectangular television set = Area/length
= 320m/40m
= 8 m
Thus, the perimeter of the rectangular television set = 2 (length + breadth)
= 2(40 + 8) m
 = 2 × 48 m
= 96 m
### Fun Facts
• In a rectangle, both the diagonals are equal in length.
• We also have a rectangular figure that we call a cyclic rectangle. A circle that contains a rectangle with its entire vertex touching the circumference is a cyclic rectangle.
• This concept of area calculation is also very useful in land surveying, map designing, modelling, and others. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 10.3: Solving Quadratic Equations Using Square Roots
Difficulty Level: At Grade Created by: CK-12
Suppose you needed to find the value of \begin{align*}x\end{align*} such that \begin{align*}x^2=81\end{align*}. How could you solve this equation?
• Make a table of values.
• Graph this equation as a system.
• Cancel the square using its inverse operation.
The inverse of a square is a square root.
By applying the square root to each side of the equation, you get:
\begin{align*}x &= \pm \sqrt{81}\\ x &= 9 \ or \ x=-9\end{align*}
In general, the solution to a quadratic equation of the form \begin{align*}0=ax^2-c\end{align*} is:
\begin{align*}x=\sqrt{\frac{c}{a}} \ \text{or} \ x=- \sqrt{\frac{c}{a}}\end{align*}
Example 1: Solve \begin{align*}(x-4)^2-9=0\end{align*}.
Solution: Begin by adding 9 to each side of the equation.
\begin{align*}(x-4)^2=9\end{align*}
Cancel square by applying square root.
\begin{align*}x-4=3 \ or \ x-4=-3\end{align*}
Solve both equations for \begin{align*}x: x=7 \ or \ x=1\end{align*}
In the previous lesson, you learned Newton’s formula for projectile motion. Let’s examine a situation in which there is no initial velocity. When a ball is dropped, there is no outward force placed on its path; therefore, there is no initial velocity.
A ball is dropped from a 40-foot building. When does the ball reach the ground?
Using the equation from the previous lesson, \begin{align*}h(t)=-\frac{1}{2} (g) t^2+v_0 t+h_0\end{align*}, and substituting the appropriate information, you get:
\begin{align*}&& 0 &=-\frac{1}{2} (32)t^2+(0)t+40\\ \text{Simplify} && 0 &= -16t^2+40\\ \text{Solve for} \ x: && -40 &= -16t^2\\ && 2.5 &= t^2\\ && t & \approx 1.58 \ and \ t \approx -1.58\end{align*}
Because \begin{align*}t\end{align*} is in seconds, it does not make much sense for the answer to be negative. So the ball will reach the ground at approximately 1.58 seconds.
Example: A rock is dropped from the top of a cliff and strikes the ground 7.2 seconds later. How high is the cliff in meters?
Solution: Using Newton’s formula, substitute the appropriate information.
\begin{align*}&& 0 &= -\frac{1}{2} (9.8)(7.2)^2+(0)(7.2)+ h_0\\ \text{Simplify:} && 0 &= -254.016+h_0\\ \text{Solve for} \ h_0: && h_0 &= 254.016\end{align*}
The cliff is approximately 254 meters tall.
## Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
Solve each quadratic equation.
1. \begin{align*}x^2=196\end{align*}
2. \begin{align*}x^2-1=0\end{align*}
3. \begin{align*}x^2-100=0\end{align*}
4. \begin{align*}x^2+16=0\end{align*}
5. \begin{align*}9x^2-1=0\end{align*}
6. \begin{align*}4x^2-49=0\end{align*}
7. \begin{align*}64x^2-9=0\end{align*}
8. \begin{align*}x^2-81=0\end{align*}
9. \begin{align*}25x^2-36=0\end{align*}
10. \begin{align*}x^2+9=0\end{align*}
11. \begin{align*}x^2-16=0\end{align*}
12. \begin{align*}x^2-36=0\end{align*}
13. \begin{align*}16x^2-49=0\end{align*}
14. \begin{align*}(x-2)^2=1\end{align*}
15. \begin{align*}(x+5)^2=16\end{align*}
16. \begin{align*}(2x-1)^2-4=0\end{align*}
17. \begin{align*}(3x+4)^2=9\end{align*}
18. \begin{align*}(x-3)^2+25=0\end{align*}
19. \begin{align*}x^2-6=0\end{align*}
20. \begin{align*}x^2-20=0\end{align*}
21. \begin{align*}3x^2+14=0\end{align*}
22. \begin{align*}(x-6)^2=5\end{align*}
23. \begin{align*}(4x+1)^2-8=0\end{align*}
24. \begin{align*}(x+10)^2=2\end{align*}
25. \begin{align*}2(x+3)^2=8\end{align*}
26. How long does it take a ball to fall from a roof to the ground 25 feet below?
27. Susan drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the splash?
28. It takes a rock 5.3 seconds to splash in the water when it is dropped from the top of a cliff. How high is the cliff in meters?
29. Nisha drops a rock from the roof of a building 50 feet high. Ashaan drops a quarter from the top-story window, which is 40 feet high, exactly half a second after Nisha drops the rock. Which hits the ground first?
30. Victor drops an apple out of a window on the \begin{align*}10^{th}\end{align*} floor, which is 120 feet above ground. One second later, Juan drops an orange out of a \begin{align*}6^{th}\end{align*}-floor window, which is 72 feet above the ground. Which fruit reaches the ground first? What is the time difference between the fruits’ arrival to the ground?
Mixed Review
1. Graph \begin{align*}y=2x^2+6x+4\end{align*}. Identify the following:
1. Vertex
2. \begin{align*}x-\end{align*}intercepts
3. \begin{align*}y-\end{align*}intercepts
4. axis of symmetry
2. What is the difference between \begin{align*}y=x+3\end{align*} and \begin{align*}y=x^2+3\end{align*}?
3. Determine the domain and range of \begin{align*}y=-(x-2)^2+7\end{align*}.
4. The Glee Club is selling hot dogs and sodas for a fundraiser. On Friday the club sold 112 hot dogs and 70 sodas and made $154.00. On Saturday the club sold 240 hot dogs and 120 sodas and made$300.00. How much is each soda? Each hot dog?
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Solved (Free): The American Cancer Society wants to estimate the proportion of high school seniors who smokes regularly. In a random sample of
ByDr. Raju Chaudhari
Apr 4, 2021
The American Cancer Society wants to estimate the proportion of high school seniors who smokes regularly. In a random sample of 1,000 seniors, 200 smoke regularly. Construct 90% confidence interval. Interpret your confidence interval.
Solution
Given that sample size $n = 1000$, observed $X = 200$.
Thus the sample proportion is
$\hat{p}=\frac{X}{n}=\frac{200}{1000}=0.2$.
The procedure for $0.9$ % confidence interval for the proportion of high school seniors who smoke regularly is as follows:
Step 1 Specify the confidence level $(1-\alpha)$
Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.
Step 2 Given information
Given that sample size $n =1000$, observed data $X=200$.
The estimate of the proportion of is $\hat{p} =\frac{X}{n} =\frac{200}{1000}=0.2$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval for population proportion is
\begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}
where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.
Step 4 Determine the critical value
The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.
Thus $Z_{\alpha/2} = Z_{0.05} = 1.64$.
Step 5 Compute the margin of error
The margin of error for proportions is
\begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.64 \sqrt{\frac{0.2*(1-0.2)}{1000}}\\ & =0.021. \end{aligned}
Step 6 Determine the confidence interval
$90$% confidence interval estimate for population proportion is
\begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.2 - 0.021 & \leq p \leq 0.2 + 0.021\\ 0.1793 & \leq p \leq 0.2207. \end{aligned}
Thus, $90$% confidence interval estimate for population proportion $p$ is $(0.1793,0.2207)$. |
# Binary to decimal calculation process in detail
### Binary to decimal, how to calculate ？？？？？？？？？？？
Binary to decimal conversion: the base is multiplied by the power, and then added, simplifying the operation can be left out of the term where the number of digits is 0, (because 0 multiplied by any other number that is not 0 is 0). The decimal part is the same, but with less precision.
Method: “Expanding the sum by power”
Example: 10001111
1×2⁷+1×2³+1×2²+1×2¹+1×2⁰=143, so the decimal representation of 10001111 is 143.
The rule: the number of digits in the single digit is 0, the number of digits in the tens digit is 1, …… , in increasing order, while the number of times the number in the tenths place is -1 and the number of times the number in the hundredths place is -2, …… , in decreasing order.
### Binary to decimal conversion steps?
Method 1: from right to left with each number of binary to multiply by the appropriate power of 2, after the decimal point is from left to right
Example: binary number 1101.01 converted to decimal
1101.01 (2)
=1*2 [0] +0*2 [1] +1*2 [2] +1*2 [3] +1*2 [-1] + 0*2 [-2]
=1+0+4+8+0+0.25=13.25 (10)
This is the first time that a binary number is converted to decimal. +1*2 [-2]
=1+0+4+8+0+0+0.25=13.25 (10)
Method 2: The binary number is first written as a weighted coefficient expansion, and then summed according to the rules of decimal addition. This is called the “weighted sum” method.
For binary numbers with n integers and m decimals, the weighted coefficients are expressed as follows: N(2) = an-1×2n-1+an-2×2n-2+……+a1×21+a0×20+a-1×2-1+a-2×2-2+……+a-m×2-2+a-m×2-2 ……+a-m×2-m (10) where aj denotes the coefficient in the jth place, which is one of 0 and 1.
Example: binary number 1101 converted to decimal
1101 (2)
=1*2 [3] + 1*2 [2] + 0*2 [1] + 1*2 [0]
=13 (10)
Note:
1, [] the number in parentheses represent the sub-square, such as [2] for the second square, [-1] is negative primary.
2. The number in the () bracket represents the decimal number, (2) is binary, (10) is decimal.
### Binary to decimal method
Binary to decimal method is as follows:
1, the unsigned integer binary into decimal numbers, from the first bit of the right side of the binary number, from right to left, the first binary position multiplied by the number of the corresponding digit of the power of 2, and then add the product of each can be obtained by the binary number of the corresponding decimal number.
2, with the sign of the binary number into decimal numbers, first observe the binary number of the highest bit is what number, if it is 1, that is, the negative number, if it is 0, that is, the positive number, to determine the sign of the decimal number and then converted to decimal numbers.
3, the decimal number of binary numbers into decimal numbers, from left to right, with the number of binary digits multiplied by 2 to the power of the negative number of digits, and then add all the products can be obtained.
Binary to decimal:
Binary to decimal is the use of certain mathematical means to convert the binary digits into decimal digits, widely used in programming and other fields.
Binary is a number system widely used in computing technology. Binary data is a number represented by two digits, 0 and 1. Its base is 2, the rule of rounding is “two into one”, and the rule of borrowing is “borrowing one as two”.
The decimal system is a method of counting. The use of decimal system in human arithmetic may be related to the fact that humans have ten fingers. Aristotle claimed that the universal use of the decimal system was simply a result of the anatomical fact that the vast majority of humans are born with 10 fingers.
### How to convert binary to decimal
First of all, let’s take an example: decimal 123 = 1 * 100 + 2 * 10 + 3 * 1, where 100 is 10 squared, 10 is 10 to the first power, 1 is 10 to the zero power, respectively, in the hundredths, tenths, and digits, which is also called the bit weights of the corresponding positions, then binary is also the same reason.
Binary to decimal conversion is: the binary number in each position multiplied by the bit weights in the position, and then add. Such as 101101 converted to decimal is 1 * 2 of the 5th + 0 * 2 of the 4th + 1 * 2 of the 3rd + 1 * 2 of the 2nd + 0 * 2 of the 1st + 1 * 2 of the 0th = 32 + 8 + 4 + 1 = 45; this 45 is the corresponding decimal number.
### Binary how to convert decimal
Binary is divided into two categories of integer binary and decimal binary, the two binary into decimal is converted to the practice of summing by weighted expansion. The next step is to explain in detail.
Integer binary to decimal conversion example: the binary bit 1010 into decimal.
The first step: the first 1010 complement to 8 bits, that is, 00001010.
The second step: as the first digit is 0 for positive, then the algorithm is shown below.
Step 3: Calculate the result, the result is 10, that is, 1010 is converted to decimal as 10.
Decimal binary conversion to decimal
Example: convert 1011.01 to decimal.
Step 1: The algorithm is shown below.
Step 2: The result of the calculation is 11.25, i.e. the result of converting 1011.01 to decimal is 11.25.
Decimal to binary
Example: convert decimal to 42 to binary. The method of converting decimal to binary is “divide by two and take the remainder”.
Step 1: 42/2=21……0
21/2=10……0
10/2=5……0
5/2=2……0
2/2=1……0
1/2=0……1
Step 2: So 42 corresponds to the to the binary bit 101010.
### How do you count binary to decimal conversion?
The decimal numbers, the individual bits, are… Thousands, Hundreds, Tens, Individuals….
Binary numbers, individual bits, are:… Eight, four, two, one…..
The other bits, it’s up to you, just to figure it out.
The decimal number, 8031, is: 8 thousands, 0 hundreds, 3 tens, 1 one.
The binary number, 1101, is: 1 eight, 1 four, 0 two, 1 one, or 13 in decimal. |
# What is 1/351 as a decimal?
## Solution and how to convert 1 / 351 into a decimal
1 / 351 = 0.003
Fraction conversions explained:
• 1 divided by 351
• Numerator: 1
• Denominator: 351
• Decimal: 0.003
• Percentage: 0.003%
The basis of converting 1/351 to a decimal begins understanding why the fraction should be handled as a decimal. Both represent numbers between integers, in some cases defining portions of whole numbers Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). So let’s dive into how and why you can convert 1/351 into a decimal.
1 / 351 as a percentage 1 / 351 as a fraction 1 / 351 as a decimal
0.003% - Convert percentages 1 / 351 1 / 351 = 0.003
## 1/351 is 1 divided by 351
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. To solve the equation, we must divide the numerator (1) by the denominator (351). Here's how our equation is set up:
### Numerator: 1
• Numerators represent the number of parts being taken from a denominator. Comparatively, 1 is a small number meaning you will have less parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Values like 1 doesn't make it easier because they're small. So how does our denominator stack up?
### Denominator: 351
• Denominators represent the total number of parts, located below the vinculum or fraction bar. 351 is a large number which means you should probably use a calculator. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Overall, two-digit denominators are no problem with long division. So without a calculator, let's convert 1/351 from a fraction to a decimal.
## How to convert 1/351 to 0.003
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 351 \enclose{longdiv}{ 1 }$$
We will be using the left-to-right method of calculation. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 351 \enclose{longdiv}{ 1.0 }$$
Because 351 into 1 will equal less than one, we can’t divide less than a whole number. So we will have to extend our division problem. Add a decimal point to 1, your numerator, and add an additional zero. This doesn't add any issues to our denominator but now we can divide 351 into 10.
### Step 3: Solve for how many whole groups you can divide 351 into 10
$$\require{enclose} 00.0 \\ 351 \enclose{longdiv}{ 1.0 }$$
Now that we've extended the equation, we can divide 351 into 10 and return our first potential solution! Multiply this number by 351, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 351 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 1/351 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. And the same is true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 1/351 into a decimal
Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions.
### When to convert 0.003 to 1/351 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 1/351 = 0.003 what would it be as a percentage?
• What is 1 + 1/351 in decimal form?
• What is 1 - 1/351 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.003 + 1/2?
### Convert more fractions to decimals
From 1 Numerator From 351 Denominator What is 1/352 as a decimal? What is 2/351 as a decimal? What is 1/353 as a decimal? What is 3/351 as a decimal? What is 1/354 as a decimal? What is 4/351 as a decimal? What is 1/355 as a decimal? What is 5/351 as a decimal? What is 1/356 as a decimal? What is 6/351 as a decimal? What is 1/357 as a decimal? What is 7/351 as a decimal? What is 1/358 as a decimal? What is 8/351 as a decimal? What is 1/359 as a decimal? What is 9/351 as a decimal? What is 1/360 as a decimal? What is 10/351 as a decimal? What is 1/361 as a decimal? What is 11/351 as a decimal? What is 1/362 as a decimal? What is 12/351 as a decimal? What is 1/363 as a decimal? What is 13/351 as a decimal? What is 1/364 as a decimal? What is 14/351 as a decimal? What is 1/365 as a decimal? What is 15/351 as a decimal? What is 1/366 as a decimal? What is 16/351 as a decimal? What is 1/367 as a decimal? What is 17/351 as a decimal? What is 1/368 as a decimal? What is 18/351 as a decimal? What is 1/369 as a decimal? What is 19/351 as a decimal? What is 1/370 as a decimal? What is 20/351 as a decimal? What is 1/371 as a decimal? What is 21/351 as a decimal?
### Convert similar fractions to percentages
From 1 Numerator From 351 Denominator 2/351 as a percentage 1/352 as a percentage 3/351 as a percentage 1/353 as a percentage 4/351 as a percentage 1/354 as a percentage 5/351 as a percentage 1/355 as a percentage 6/351 as a percentage 1/356 as a percentage 7/351 as a percentage 1/357 as a percentage 8/351 as a percentage 1/358 as a percentage 9/351 as a percentage 1/359 as a percentage 10/351 as a percentage 1/360 as a percentage 11/351 as a percentage 1/361 as a percentage |
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Solutions of equations reducible to the quadratic equations
In this section we will discuss the equations which are not quadratic but can be reduced to quadratic equations.
Type I
Equations of the form
Example 1: Solve the equation
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Let
Therefore equation (1) becomes,
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Hence solution set is *
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Type II
Equations of the form (
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, a scalar
Where
Example 2:
Solve (
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Solution:
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Rearranging equation (1), we have
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Type III Exponential Equations
Equations in which variable occurs in exponent, are called exponential equations.
Example 3: Solve the equation
Solution:
Now let
, we have
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Example 4: Solve the equation
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Hence solution set is *
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Solution: When Multiplying by Now let , we have On dividing by 2, we get When ( ) ( ) Hence solution set is { } ( )( )
Type IV Reciprocal Equations
An equation which remain unchanged when
is replaced by
.
Example 5: Solve the equation Solution: Dividing by ( )
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Exercise 4.2
Solve the following equations.
Q #1:
When
Solution:
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Hence solution set = {
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Q #2:
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Solution:
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Putting values in (1)
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Q #3:
Solution:
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Q # 4:
Solution:
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Q #5:
Solution:
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Hence solution set is *
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Q# 6:
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Solution:
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Re arranging equation (1), we have
( )( )( )( ) ( )( ) Let ( )( )
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( ) When ( ) When √ ( )( ) √ √ √
Hence solution set = {
Q #7:( )( )( )( ) Solution:
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Hence solution set = {
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Q #8:
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Solution:
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Re arranging equation (1), we have
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Let
Q #9:
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Solution:
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Re arranging equation (1), we have
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When ( ) ( ) ( )( ) ( ) ( ) When ( ) ( ) ( )( ) ( ) ( ) Hence solution set = * +
Q #10: ( )( )( )( ) Solution: ( )( )( )( ) Re arranging equation (1), we have ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) Let ( ) ( )
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When
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Hence solution set = {
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Q #11:
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Solution:
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Hence solution set = {
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Q #12:
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Let
Solution:
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Re arranging equation (2), we have
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Let
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Q #13:
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Solution:
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Re arranging equation (2), we have
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Let
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Hence solution set = {
Q #14:
Solution:
( ) Then equation (2) becomes ( ) ( ) ( )( ) When When Hence solution set = * +
Q #15:
}
(
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Solution:
(
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(
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Then equation (2) becomes
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(
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When
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Hence solution set = *
+
Q #16:
Solution:
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Then equation (2) becomes
When
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Hence solution set = *
+
When
Q #18: .
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.
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Solution:
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When
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Using these values in (1), we have
Hence solution set = *
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Q #17:
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M. Shahid Nadeem, Scholars Academy, Shah Wali Colony, Wah Cantt. (03332823123,) P-13
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Q #19:
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Q #20: .
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M. Shahid Nadeem, Scholars Academy, Shah Wali Colony, Wah Cantt. (03332823123,) P-14
Using these values in (1), we have
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Q #21:
Solution:
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M. Shahid Nadeem, Scholars Academy, Shah Wali Colony, Wah Cantt. (03332823123,) P-15
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Q #22:
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Q #23:
Solution:
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M. Shahid Nadeem, Scholars Academy, Shah Wali Colony, Wah Cantt. (03332823123,) P-16
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Q #24:
Solution:
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M. Shahid Nadeem, Scholars Academy, Shah Wali Colony, Wah Cantt. (03332823123,) P-17
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“I could never have gone far in any science because on the path of every science the lion Mathematics lies in wait for you.”
M. Shahid Nadeem, Scholars Academy, Shah Wali Colony, Wah Cantt. (03332823123,) P-18
} |
反三角函数
主值
（注意：某些數學教科書的作者將${\displaystyle \operatorname {arcsec} }$的值域定為${\displaystyle [0,{\frac {\pi }{2}})\cup [\pi ,{\frac {3\pi }{2}})}$因為當${\displaystyle \tan }$的定義域落在此區間時，${\displaystyle \tan }$的值域${\displaystyle \geq 0}$，如果${\displaystyle \operatorname {arcsec} }$的值域仍定為${\displaystyle [0,{\frac {\pi }{2}})\cup ({\frac {\pi }{2}},\pi ]}$，將會造成${\displaystyle \tan(\operatorname {arcsec} x)=\pm {\sqrt {x^{2}-1}}}$，如果希望${\displaystyle \tan(\operatorname {arcsec} x)={\sqrt {x^{2}-1}}}$，那就必須將${\displaystyle \operatorname {arcsec} }$的值域定為${\displaystyle [0,{\frac {\pi }{2}})\cup [\pi ,{\frac {3\pi }{2}})}$，基於類似的理由${\displaystyle \operatorname {arccsc} }$的值域定為${\displaystyle (-\pi ,-{\frac {\pi }{2}}]\cup (0,{\frac {\pi }{2}}]}$
反三角函数之间的关系
${\displaystyle \arccos x={\frac {\pi }{2}}-\arcsin x}$
${\displaystyle \operatorname {arccot} x={\frac {\pi }{2}}-\arctan x}$
${\displaystyle \operatorname {arccsc} x={\frac {\pi }{2}}-\operatorname {arcsec} x}$
${\displaystyle \arcsin(-x)=-\arcsin x\!}$
${\displaystyle \arccos(-x)=\pi -\arccos x\!}$
${\displaystyle \arctan(-x)=-\arctan x\!}$
${\displaystyle \operatorname {arccot}(-x)=\pi -\operatorname {arccot} x\!}$
${\displaystyle \operatorname {arcsec}(-x)=\pi -\operatorname {arcsec} x\!}$
${\displaystyle \operatorname {arccsc}(-x)=-\operatorname {arccsc} x\!}$
${\displaystyle \arccos {\frac {1}{x}}\,=\operatorname {arcsec} x}$
${\displaystyle \arcsin {\frac {1}{x}}\,=\operatorname {arccsc} x}$
${\displaystyle \arctan {\frac {1}{x}}={\frac {\pi }{2}}-\arctan x=\operatorname {arccot} x,\ }$ ${\displaystyle \ x>0}$
${\displaystyle \arctan {\frac {1}{x}}=-{\frac {\pi }{2}}-\arctan x=-\pi +\operatorname {arccot} x,\ }$ ${\displaystyle \ x<0}$
${\displaystyle \operatorname {arccot} {\frac {1}{x}}={\frac {\pi }{2}}-\operatorname {arccot} x=\arctan x,\ }$ ${\displaystyle \ x>0}$
${\displaystyle \operatorname {arccot} {\frac {1}{x}}={\frac {3\pi }{2}}-\operatorname {arccot} x=\pi +\arctan x,\ }$ ${\displaystyle \ x<0}$
${\displaystyle \operatorname {arcsec} {\frac {1}{x}}=\arccos x}$
${\displaystyle \operatorname {arccsc} {\frac {1}{x}}=\arcsin x}$
${\displaystyle \arccos x=\arcsin {\sqrt {1-x^{2}}},}$ ${\displaystyle \ 0\leq x\leq 1}$
${\displaystyle \arctan x=\arcsin {\frac {x}{\sqrt {x^{2}+1}}}}$
${\displaystyle \arcsin x=2\arctan {\frac {x}{1+{\sqrt {1-x^{2}}}}}}$
${\displaystyle \arccos x=2\arctan {\frac {\sqrt {1-x^{2}}}{1+x}},}$ ${\displaystyle -1
${\displaystyle \arctan x=2\arctan {\frac {x}{1+{\sqrt {1+x^{2}}}}}}$
三角函數與反三角函數的關係
${\displaystyle \theta }$ ${\displaystyle \sin \theta }$ ${\displaystyle \cos \theta }$ ${\displaystyle \tan \theta }$ 圖示
${\displaystyle \arcsin x}$ ${\displaystyle \sin(\arcsin x)=x}$ ${\displaystyle \cos(\arcsin x)={\sqrt {1-x^{2}}}}$ ${\displaystyle \tan(\arcsin x)={\frac {x}{\sqrt {1-x^{2}}}}}$
${\displaystyle \arccos x}$ ${\displaystyle \sin(\arccos x)={\sqrt {1-x^{2}}}}$ ${\displaystyle \cos(\arccos x)=x}$ ${\displaystyle \tan(\arccos x)={\frac {\sqrt {1-x^{2}}}{x}}}$
${\displaystyle \arctan x}$ ${\displaystyle \sin(\arctan x)={\frac {x}{\sqrt {1+x^{2}}}}}$ ${\displaystyle \cos(\arctan x)={\frac {1}{\sqrt {1+x^{2}}}}}$ ${\displaystyle \tan(\arctan x)=x}$
${\displaystyle \operatorname {arccot} x}$ ${\displaystyle \sin(\operatorname {arccot} x)={\frac {1}{\sqrt {1+x^{2}}}}}$ ${\displaystyle \cos(\operatorname {arccot} x)={\frac {x}{\sqrt {1+x^{2}}}}}$ ${\displaystyle \tan(\operatorname {arccot} x)={\frac {1}{x}}}$
${\displaystyle \operatorname {arcsec} x}$ ${\displaystyle \sin(\operatorname {arcsec} x)={\frac {\sqrt {x^{2}-1}}{x}}}$ ${\displaystyle \cos(\operatorname {arcsec} x)={\frac {1}{x}}}$ ${\displaystyle \tan(\operatorname {arcsec} x)={\sqrt {x^{2}-1}}}$
${\displaystyle \operatorname {arccsc} x}$ ${\displaystyle \sin(\operatorname {arccsc} x)={\frac {1}{x}}}$ ${\displaystyle \cos(\operatorname {arccsc} x)={\frac {\sqrt {x^{2}-1}}{x}}}$ ${\displaystyle \tan(\operatorname {arccsc} x)={\frac {1}{\sqrt {x^{2}-1}}}}$
一般解
${\displaystyle \sin y=x\ \Leftrightarrow \ (\ y=\arcsin x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=\pi -\arcsin x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )}$
${\displaystyle \cos y=x\ \Leftrightarrow \ (\ y=\arccos x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=2\pi -\arccos x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )}$
${\displaystyle \tan y=x\ \Leftrightarrow \ \ y=\arctan x+k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} }$
${\displaystyle \cot y=x\ \Leftrightarrow \ \ y=\operatorname {arccot} x+k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} }$
${\displaystyle \sec y=x\ \Leftrightarrow \ (\ y=\operatorname {arcsec} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=2\pi -\operatorname {arcsec} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )}$
${\displaystyle \csc y=x\ \Leftrightarrow \ (\ y=\operatorname {arccsc} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=\pi -\operatorname {arccsc} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )}$
反三角函数的导数
{\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}\arcsin x&{}={\frac {1}{\sqrt {1-x^{2}}}};\qquad |x|<1\\{\frac {\mathrm {d} }{\mathrm {d} x}}\arccos x&{}={\frac {-1}{\sqrt {1-x^{2}}}};\qquad |x|<1\\{\frac {\mathrm {d} }{\mathrm {d} x}}\arctan x&{}={\frac {1}{1+x^{2}}}\\{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arccot} x&{}={\frac {-1}{1+x^{2}}}\\{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arcsec} x&{}={\frac {1}{|x|\,{\sqrt {x^{2}-1}}}};\qquad |x|>1\\{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arccsc} x&{}={\frac {-1}{|x|\,{\sqrt {x^{2}-1}}}};\qquad |x|>1\\\end{aligned}}}
${\displaystyle {\frac {d\arcsin x}{dx}}={\frac {d\theta }{d\sin \theta }}={\frac {1}{\cos \theta }}={\frac {1}{\sqrt {1-\sin ^{2}\theta }}}={\frac {1}{\sqrt {1-x^{2}}}}}$
表达为定积分
{\displaystyle {\begin{aligned}\arcsin x&{}=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\arccos x&{}=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\arctan x&{}=\int _{0}^{x}{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arccot} x&{}=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arcsec} x&{}=\int _{1}^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\\\operatorname {arccsc} x&{}=\int _{x}^{\infty }{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\end{aligned}}}
${\displaystyle x}$等于1时，在有极限的域上的积分是瑕积分，但仍是良好定义的。
无穷级数
{\displaystyle {\begin{aligned}\arcsin z&{}=z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{2n+1}}{(2n+1)}};\qquad |z|\leq 1\end{aligned}}}
{\displaystyle {\begin{aligned}\arccos z&{}={\frac {\pi }{2}}-\arcsin z\\&{}={\frac {\pi }{2}}-\left[z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \right]\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{2n+1}}{(2n+1)}};\qquad |z|\leq 1\end{aligned}}}
{\displaystyle {\begin{aligned}\arctan z&{}=z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}};\qquad |z|\leq 1\qquad z\neq i,-i\end{aligned}}}
{\displaystyle {\begin{aligned}\operatorname {arccot} z&{}={\frac {\pi }{2}}-\arctan z\\&{}={\frac {\pi }{2}}-\left(z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \right)\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}};\qquad |z|\leq 1\qquad z\neq i,-i\end{aligned}}}
{\displaystyle {\begin{aligned}\operatorname {arcsec} z&{}=\arccos \left(z^{-1}\right)\\&{}={\frac {\pi }{2}}-\left[z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{-7}}{7}}+\cdots \right]\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{-(2n+1)}}{(2n+1)}};\qquad \left|z\right|\geq -4\end{aligned}}}
{\displaystyle {\begin{aligned}\operatorname {arccsc} z&{}=\arcsin \left(z^{-1}\right)\\&{}=z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{-7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{-(2n+1)}}{2n+1}};\qquad \left|z\right|\geq 1\end{aligned}}}
${\displaystyle \arctan x=\infty {x}{1+x^{2}}\sum _{n=0}^{\infty }\prod _{k=1}^{n}{\frac {2kx^{2}}{(2k+1)(1+x^{2})}}}$
（注意对x= 0在和中的项是空积1。）
反三角函数的不定积分
{\displaystyle {\begin{aligned}\int \arcsin x\,dx&{}=x\,\arcsin x+{\sqrt {1-x^{2}}}+C,\qquad x\leq 1\\\int \arccos x\,dx&{}=x\,\arccos x-{\sqrt {1-x^{2}}}+C,\qquad x\leq 1\\\int \arctan x\,dx&{}=x\,\arctan x-{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arccot} x\,dx&{}=x\,\operatorname {arccot} x+{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arcsec} x\,dx&{}=x\,\operatorname {arcsec} x-\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C=x\,\operatorname {arcsec} x+\operatorname {sgn}(x)\ln \left|x-{\sqrt {x^{2}-1}}\right|+C\\\int \operatorname {arccsc} x\,dx&{}=x\,\operatorname {arccsc} x+\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C=x\,\operatorname {arccsc} x-\operatorname {sgn}(x)\ln \left|x-{\sqrt {x^{2}-1}}\right|+C\\\end{aligned}}}
舉例
{\displaystyle {\begin{aligned}u&{}=&\arcsin x&\quad \quad \mathrm {d} v=\mathrm {d} x\\\mathrm {d} u&{}=&{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}&\quad \quad {}v=x\end{aligned}}}
${\displaystyle \int \arcsin(x)\,\mathrm {d} x=x\arcsin x-\int {\frac {x}{\sqrt {1-x^{2}}}}\,\mathrm {d} x}$
${\displaystyle k=1-x^{2}.\,}$
${\displaystyle \mathrm {d} k=-2x\,\mathrm {d} x}$
${\displaystyle \int {\frac {x}{\sqrt {1-x^{2}}}}\,\mathrm {d} x=-{\frac {1}{2}}\int {\frac {\mathrm {d} k}{\sqrt {k}}}=-{\sqrt {k}}}$
${\displaystyle \int \arcsin(x)\,\mathrm {d} x=x\arcsin x+{\sqrt {1-x^{2}}}+C}$
加法公式和減法公式
${\displaystyle \arcsin x+\arcsin y}$
${\displaystyle \arcsin x+\arcsin y=\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),xy\leq 0\lor x^{2}+y^{2}\leq 1}$
${\displaystyle \arcsin x+\arcsin y=\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),x>0,y>0,x^{2}+y^{2}>1}$
${\displaystyle \arcsin x+\arcsin y=-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),x<0,y<0,x^{2}+y^{2}>1}$
${\displaystyle \arcsin x-\arcsin y}$
${\displaystyle \arcsin x-\arcsin y=\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right),xy\geq 0\lor x^{2}+y^{2}\leq 1}$
${\displaystyle \arcsin x-\arcsin y=\pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right),x>0,y<0,x^{2}+y^{2}>1}$
${\displaystyle \arcsin x-\arcsin y=-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),x<0,y>0,x^{2}+y^{2}>1}$
${\displaystyle \arccos x+\arccos y}$
${\displaystyle \arccos x+\arccos y=\arccos \left(xy-{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),x+y\geq 0}$
${\displaystyle \arccos x+\arccos y=2\pi -\arccos \left(xy-{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),x+y<0}$
${\displaystyle \arccos x-\arccos y}$
${\displaystyle \arccos x-\arccos y=-\arccos \left(xy+{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),x\geq y}$
${\displaystyle \arccos x-\arccos y=\arccos \left(xy+{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),x
${\displaystyle \arctan x+\arctan y}$
${\displaystyle \arctan \,x+\arctan \,y=\arctan \,{\frac {x+y}{1-xy}},xy<1}$
${\displaystyle \arctan \,x+\arctan \,y=\pi +\arctan \,{\frac {x+y}{1-xy}},x>0,xy>1}$
${\displaystyle \arctan \,x+\arctan \,y=-\pi +\arctan \,{\frac {x+y}{1-xy}},x<0,xy>1}$
${\displaystyle \arctan x-\arctan y}$
${\displaystyle \arctan x-\arctan y=\arctan {\frac {x-y}{1+xy}},xy>-1}$
${\displaystyle \arctan x-\arctan y=\pi +\arctan {\frac {x-y}{1+xy}},x>0,xy<-1}$
${\displaystyle \arctan x-\arctan y=-\pi +\arctan {\frac {x-y}{1+xy}},x<0,xy<-1}$
${\displaystyle \operatorname {arccot} x+\operatorname {arccot} y}$
${\displaystyle \operatorname {arccot} x+\operatorname {arccot} y=\operatorname {arccot} {\frac {xy-1}{x+y}},x>-y}$
${\displaystyle \operatorname {arccot} x+\operatorname {arccot} y=\operatorname {arccot} {\frac {xy-1}{x+y}}+\pi ,x<-y}$
${\displaystyle \arcsin x+\arccos x}$
${\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}},|x|\leq 1}$
${\displaystyle \arctan x+\operatorname {arccot} x}$
${\displaystyle \arctan x+\operatorname {arccot} x={\frac {\pi }{2}}}$
註釋與引用
1. ^ ${\displaystyle \theta =\arccos x}$，得到：
${\displaystyle {\frac {d\arccos x}{dx}}={\frac {d\theta }{d\cos \theta }}={\frac {-1}{\sin \theta }}={\frac {1}{\sqrt {1-\cos ^{2}\theta }}}={\frac {-1}{\sqrt {1-x^{2}}}}}$
因為要使根號內部恆為正，所以在條件加上${\displaystyle |x|<1}$
${\displaystyle \theta =\arctan x}$，得到：
${\displaystyle {\frac {d\arctan x}{dx}}={\frac {d\theta }{d\tan \theta }}={\frac {1}{\sec ^{2}\theta }}={\frac {1}{1+\tan ^{2}\theta }}={\frac {1}{1+x^{2}}}}$
${\displaystyle \theta =\operatorname {arccot} x}$，得到：
${\displaystyle {\frac {d\operatorname {arccot} x}{dx}}={\frac {d\theta }{d\cot \theta }}={\frac {-1}{\csc ^{2}\theta }}={\frac {1}{1+\cot ^{2}\theta }}={\frac {-1}{1+x^{2}}}}$
${\displaystyle \theta =\operatorname {arcsec} x}$，得到：
${\displaystyle {\frac {d\operatorname {arcsec} x}{dx}}={\frac {d\theta }{d\sec \theta }}={\frac {1}{\sec \theta \tan \theta }}={\frac {1}{\left|x\right|{\sqrt {x^{2}-1}}}}}$
因為要使根號內部恆為正，所以在條件加上${\displaystyle |x|>1}$，比較容易被忽略是${\displaystyle \sec \theta }$產生的絕對值${\displaystyle \sec ^{-1}\theta }$的定義域是${\displaystyle 0\leq \theta \leq \pi ,\theta \neq {\frac {\pi }{2}}}$，所以${\displaystyle \tan \theta =\pm {\sqrt {x^{2}-1}}}$，所以${\displaystyle x}$要加绝对值。
${\displaystyle \theta =\operatorname {arccsc} x}$，得到：
${\displaystyle {\frac {d\operatorname {arccsc} x}{dx}}={\frac {d\theta }{d\csc \theta }}={\frac {-1}{\csc \theta \cot \theta }}={\frac {-1}{\left|x\right|{\sqrt {x^{2}-1}}}}}$
因為要使根號內部恆為正，所以在條件加上${\displaystyle |x|>1}$，比較容易被忽略是${\displaystyle \csc \theta }$產生的絕對值${\displaystyle \csc ^{-1}\theta }$的定義域是${\displaystyle -{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}},\theta \neq 0}$ |
Mensuration
Mensuration You'll learn here the basic concepts of Mensuration. Space Figure A space figure, also called three-dimensional figure is a figure which has depth in addition to width and height. Everyday Objects in every day life like a tennis ball, a box, and a redwood tree are all examples of space figures. Some common simple space figures include cubes, spheres, cylinders, prisms, cones, and pyramids etc. A space figure having all flat faces is called a polyhedron. A cube and a pyramid are both polyhedrons; but a sphere, cylinder, and cone are not. Solids Solids are three-dimensional objects, bound by one or more surfaces. When plane surfaces bind a solid, they are called its faces. The lines of intersection of adjacent faces are called edges of the solid. For any regular solid with plane surfaces: Remember: Number of faces + Number of vertices = Number of edges + 2 This formula is called Euler's formula. Volume Volume of solid figure is the amount of space enclosed by its bounding surfaces. Volume is measured in cubic units. Weight of a solid = Volume x Density 1 cubic meter = 10 x 10 x 10 cubic cm = 1000 cubic cm.
People who saw this lesson also found the following lessons useful: Arithmetic Mean of Individual Observations Angle of Elevation and Depression Trigonometric Ratios Coordinate Geometry: Determination of Section Ratio
Cube
A cube is a special case of a parallelepiped in which length, breadth and height are equal i.e. it is bound by six square faces.
Thus it has 6 equal-area faces and 12 equal-length edges. If the length of one edge is l, the volume (V) of the cube is given by: V = l3 Its surface area (A) is calculated by finding the area of one square: l x l = l 2; and multiplying it by 6: 6 x l 2 A = 6l2
Example: Three cubes whose edges measure 4 cm, 5 cm and 6 cm respectively to form a single cube. Find its edge. Also, find the surface area of the new cube.
Solution Let x cm be the edge of the new cube. Then,
Volume of the new cube = Sum of the volumes of three cubes
x3 = 43 + 53 + 63 = 64 + 125 + 216
x3 = (405)3
x = 315
Edge of the new cube is 315 cm long.
Surface area of the new cube = 6 x2 = 6 * (315)2 cm2 = 810 cm2.
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# NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.4
## Chapter 1 Ex.1.4 Question 1
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) \begin{align}\frac{13}{3125} \end{align}
(ii) \begin{align}\frac{{17}}{8}\end{align}
(iii) \begin{align}\frac{64}{455}\end{align}
(iv) \begin{align}\frac{15}{1600} \end{align}
(v) \begin{align}\frac{29}{343}\end{align}
(vi) \begin{align}\frac{{23}}{{{2^3}{5^2}}}\end{align}
(vii) \begin{align}\frac{{129}}{{{2^2}{5^7}{7^5}}}\end{align}
(viii) \begin{align}\frac{6}{{15}}\end{align}
(ix) \begin{align}\frac{{35}}{{50}} \end{align}
(x)\begin{align}\frac{{77}}{{210}}\end{align}
### Solution
Reasoning:
Let $$x = \frac{p}{q}$$ be a rational number, such that the prime factorization of $$q$$ is of the form $${2^n} \times {5^m}$$, where $$n,\; m$$ are non-negative integers. Then $$x$$ has a decimal expansion which terminates.
Steps:
(i) \begin{align}\;\frac{{13}}{{3125}}\end{align}
The denominator is of the form $${5^5}.$$
Hence, the decimal expansion of \begin{align}\,\frac{{13}}{{3125}}\end{align}is terminating.
(ii) \begin{align}\;\frac{{17}}{8}\end{align}
The denominator is of the form $${2^3}.$$
Hence, the decimal expansion of \begin{align}\,\frac{{17}}{8}\end{align} is terminating.
(iii) \begin{align}\;\frac{{64}}{{455}}\end{align}
$455 = 5 \times 7 \times 13$
Since the denominator is not in the form $${2^m} \times {5^n},$$ and it also contains $$7$$ and $$13$$ as its factors, its decimal expansion will be non-terminating repeating.
(iv) \begin{align}\;\frac{{15}}{{1600}}\end{align}
$1600 = {2^6} \times {5^2}$
The denominator is of the form $${2^m} \times {5^n}.$$
Hence, the decimal expansion of \begin{align}\frac{{15}}{{1600}}\end{align}is terminating.
(v) \begin{align}\;\frac{{29}}{{343}}\end{align}
$343 = {7^3}$
Since the denominator is not in the form $$\,{2^m} \times {5^n},$$ and it has $$7$$ as its factor, the decimal expansion of \begin{align}\frac{{29}}{{343}}\end{align} is non-terminating repeating.
(vi) \begin{align}\;\frac{23}{{{2}^{3}}\times {{5}^{2}}}\end{align}
The denominator is of the form $${2^m} \times {5^n}.$$
Hence, the decimal expansion of \begin{align}\frac{{23}}{{{2^3} \times {5^2}}}\end{align}is terminating.
(vii) \begin{align}\;\frac{{129}}{{{2^2} \times {5^7} \times {7^5}}}\end{align}
Since the denominator is not of the form $$2^{m} \times 5^{n},$$ and it also has $$7$$ as its factor, the decimal expansion of \begin{align}\frac{{129}}{{{2^2} \times {5^7} \times {7^5}}}\end{align} is non-terminating repeating.
(viii) \begin{align}\frac{6}{{15}}\end{align}
\begin{align}\frac{6}{{15}} &= \frac{{2 \times 3}}{{3 \times 5}}\\&= \frac{2}{5}\end{align}
The denominator is of the form $${5^{n}}.$$
Hence, the decimal expansion of \begin{align}\frac{6}{{15}}\end{align} is terminating.
(ix) \begin{align}\,\frac{{35}}{{50}}\end{align}
\begin{align}\,\frac{{35}}{{50}} &= \frac{{7 \times 5}}{{10 \times 5}}\\&= \frac{7}{{10}}\end{align}
$10 = 2 \times 5$
The denominator is of the form $${2^m} \times {5^n}.$$
Hence, the decimal expansion of \begin{align}\frac{{35}}{{50}}\end{align} is terminating.
(x) \begin{align}\,\frac{{77}}{{210}}\end{align}
\begin{align}\,\,\,\,\frac{{77}}{{210}} &= \frac{{11 \times 7}}{{30 \times 7}}\\ &= \frac{{11}}{{30}}\end{align}
$30 = 2 \times 3 \times 5$
Since the denominator is not of the form $${2^m} \times {5^n},$$ and it also has $$3$$ as its factors, the decimal expansion of \begin{align}\frac{{77}}{{210}}\end{align} is non-terminating repeating.
## Chapter 1 Ex.1.4 Question 2
Write down the decimal expansions of those rational numbers in Question $$1$$ above which have terminating decimal expansions.
### Solution
Steps:
(i) \begin{align}\;\;\frac{13}{3125}=0.00416\end{align}
(ii) \begin{align}\,\,\frac{{17}}{8} = 2.125\end{align}
(iii) \begin{align}\,\frac{{64}}{{455}}\end{align} is non - terminating
(iv) \begin{align}\,\frac{{15}}{{1600}} = 0.009375\end{align}
(v) \begin{align}\frac{{29}}{{343}}\end{align} is non - terminating
(vi) \begin{align}\; \frac{23}{2^{3} \times 5^{2}} &=\frac{23}{200} \\ &=0.115 \end{align}
(vii) \begin{align}\,\frac{129}{{{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}}\end{align} it is non-terminating
(viii) \begin{align}\frac{6}{{15}} \end{align}
\begin{align}\frac{6}{{15}} &= \frac{{2 \times 3}}{{3 \times 5}}\\&= \frac{2}{5}\\ &= 0.4\end{align}
(ix) \begin{align}\,\frac{{35}}{{50}} = 0.7\end{align}
(x) \begin{align}\,\frac{77}{210}\end{align} it is non-terminating
## Chapter 1 Ex.1.4 Question 3
The following real numbers have decimal expansions as given below.
In each case, decide whether they are rational or not. If they are rational, and of the form \begin{align}\frac{p}{q},\end{align} what can you say about the prime factor of $$q$$ ?
(i) $$\,43.123456789$$
(ii) $$\,0.120120012000120000 \dots \dots$$
(iii) $$\,43.\mathop {123456789}\limits^{\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$$
### Solution
Reasoning:
Let $$x$$ be a rational number whose decimal expansion terminates.
Then $$x$$ can be expressed in the form \begin{align}\frac{p}{q},\end{align} where $$p$$ and $$q$$ are coprime, and the prime factorisation of $$q$$ is of the form $${2^n}\, \times \,{5^m}$$, where $$n, \;m$$ are non-negative integers.
Steps:
(i) $$\,43.123456789$$
Since this number has a terminating decimal expansion, it is a rational number of the form \begin{align} \frac{p}{q} \end{align} and $$q$$ is of the form $$2^{m} \times \,5^{n}$$
i.e., the prime factors of $$q$$ will be either $$2$$ or $$5$$ or both.
(ii) $$\,0.120120012000120000 \dots \dots$$
The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.
(iii) $$\,43.\mathop {123456789}\limits^{\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$$
Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form \begin{align} \frac{p}{q} \end{align} and $$q$$ is not of the form $$2^{m}\times 5^{n}$$
i.e., the prime factors of $$q$$ will also have a factor other than $$2$$ or $$5.$$
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# Sum to Product identity of Sine functions
## Formula
$\sin{\alpha}+\sin{\beta}$ $\,=\,$ $2\sin{\Big(\dfrac{\alpha+\beta}{2}\Big)}\cos{\Big(\dfrac{\alpha-\beta}{2}\Big)}$
An identity that expresses the transformation of sum of sine functions into product form is called the sum to product identity of sine functions.
### Introduction
Let $\alpha$ and $\beta$ be two angles of right triangles. The sine functions with the two angles are written as $\sin{\alpha}$ and $\sin{\beta}$ mathematically. The sum of the two sine functions is written mathematically in the following form.
$\sin{\alpha}+\sin{\beta}$
The sum of sine functions can be transformed into the product of the trigonometric functions as follows.
$\implies$ $\sin{\alpha}+\sin{\beta}$ $\,=\,$ $2\sin{\Big(\dfrac{\alpha+\beta}{2}\Big)}\cos{\Big(\dfrac{\alpha-\beta}{2}\Big)}$
#### Other forms
The sum to product transformation rule of sin functions is popular written in two forms.
$(1). \,\,\,$ $\sin{x}+\sin{y}$ $\,=\,$ $2\sin{\Big(\dfrac{x+y}{2}\Big)}\cos{\Big(\dfrac{x-y}{2}\Big)}$
$(2). \,\,\,$ $\sin{C}+\sin{D}$ $\,=\,$ $2\sin{\Big(\dfrac{C+D}{2}\Big)}\cos{\Big(\dfrac{C-D}{2}\Big)}$
In the same way, you can write the sum to product transformation formula of sine functions in terms of any two angles.
#### Proof
Learn how to prove the sum to product transformation identity of sine functions in trigonometry.
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# Quadratic Function (Vertex/ Graph/ Absolute value)
(CIE 0606/2021/m/12/Q4)
(a) Show that $2x^2+5x-3$ can be written in the form $a(x+b)^2+c$, where $a, b$ and $c$ are constants. [3]
(b) Hence write down the coordinates of the stationary point on the curve with equation $y=2x^2+5x-3.$ [2]
(c) On the axes below, sketch the graph of $y=|2x^2+5x-3|,$ stating the coordinates of the intercepts with the axes.[3]
(d) Write down the value of $k$ for which the equation $|2x^2+5x-3|=k$ has exactly 3 distinct solutions. [1]
*******\\
*********math solution*************
$\begin{array}{ll} \text { (a) } 2 x^{2}+5 x-3 &=2\left(x^{2}+\frac{5}{2} x\right)-3 \\&=2\left(x^{2}+\dfrac{5}{2} x+\left(\dfrac{5}{4}\right)^{2}-\left(\dfrac{5}{4}\right)^{2}\right)-3 \\&=2\left(\left(x+\dfrac{5}{4}\right)^{2}\right)-2 \times \dfrac{25}{16}-3 \\&=2\left(x+\dfrac{5}{4}\right)^{2}-\dfrac{49}{8} \\\text { (b) Stationary point}&=\left(-\dfrac{5}{4}\text{ , }-\dfrac{49}{8}\right) \end{array}$
(c)
(d) $k=\dfrac{49}{8}$
**********end math solution******************** |
# Fraction calculator
This calculator divides fractions. The first step makes the reciprocal value of the second fraction - exchange numerator and denominator of 2nd fraction. Then multiply both numerators and place the result over the product of both denominators. Then simplify the result to the lowest terms or a mixed number.
## The result:
### 3/4/5/6 = 1/40 = 0.025
The spelled result in words is one fortieth.
### How do we solve fractions step by step?
1. Divide: 3/4 : 5 = 3/4 · 1/5 = 3 · 1/4 · 5 = 3/20
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 5/1 is 1/5) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - three quarters divided by five is three twentieths.
2. Divide: the result of step No. 1 : 6 = 3/20 : 6 = 3/20 · 1/6 = 3 · 1/20 · 6 = 3/120 = 3 · 1 /3 · 40 = 1/40
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 6/1 is 1/6) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the following intermediate step, cancel by a common factor of 3 gives 1/40.
In other words - three twentieths divided by six is one fortieth.
### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
OpSymbol nameSymbol MeaningExample
+plus signaddition 1/2 + 1/3
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4) |
# GRE Math : How to find circumference
## Example Questions
### Example Question #1 : Circles
What is the circumference of a circle with an area of 36π?
12π
32
15π
12π
Explanation:
We know that the area of a circle can be expressed: a = πr2
If we know that the area is 36π, we can substitute this into said equation and get: 36π = πr2
Solving for r, we get: 36 = r2; (after taking the square root of both sides:) 6 = r
Now, we know that the circuference of a circle is expressed: c = πd. Since we know that d = 2r (two radii, placed one after the other, make a diameter), we can rewrite the circumference equation to be: c = 2πr
Since we have r, we can rewrite this to be: c = 2π*6 = 12π
### Example Question #1 : How To Find Circumference
Which is greater: the circumference of a circle with an area of , or the perimeter of a square with side length inches?
The relationship cannot be determined from the information given.
The circumference of the circle is greater.
The perimeter of the square is greater.
The two quantities are equal.
The circumference of the circle is greater.
Explanation:
Starting with the circle, we need to find the radius in order to get the circumference. Find by plugging our given area into the equation for the area of a circle:
Then calculate circumference:
(approximating as 3.14)
To find the perimeter of the square, we can use , where is the perimeter and is the side length:
, so the circle's circumference is greater.
### Example Question #2 : How To Find Circumference
Circle A has an area of . What is the perimeter of an enclosed semi-circle with half the radius of circle A?
Explanation:
Based on our information, we know that the 121π = πr2; 121 = r2; r = 11.
Our other circle with half the radius of A has a diameter equal to the radius of A. Therefore, the circumference of this circle is 11π. Half of this is 5.5π. However, since this is a semi circle, it is enclosed and looks like this:
Therefore, we have to include the diameter in the perimeter. Therefore, the total perimeter of the semi-circle is 5.5π + 11.
### Example Question #3 : How To Find Circumference
Quantity A: The circumference of a circle with radius
Quantity B: The area of a circle with a diameter one fourth the radius of the circle in Quantity A
Which of the following is true?
The two quantities are equal.
Quantity B is larger.
Quantity A is larger.
The relationship between the two values cannot be determined.
The relationship between the two values cannot be determined.
Explanation:
Let's compute each value separately. We know that the radii are positive numbers that are greater than or equal to . This means that we do not need to worry about the fact that the area could represent a square of a decimal value like .
Quantity A
Since , we know:
Quantity B
If the diameter is one-fourth the radius of A, we know:
Thus, the radius must be half of that, or .
Now, we need to compute the area of this circle. We know:
Therefore,
Now, notice that if , Quantity A is larger.
However, if we choose a value like , we have:
Quantity A:
Quantity B:
Therefore, the relation cannot be determined!
### Example Question #1 : Plane Geometry
Circle has a center in the center of Square .
The area of Square is .
What is the circumference of Circle ?
Explanation:
Since we know that the area of Square is , we know , where is the length of one of its sides. From this, we can solve for by taking the square root of both sides. You will have to do this by estimating upward. Therefore, you know that is . By careful guessing, you can quickly see that is . From this, you know that the diameter of your circle must be half of , or (because it is circumscribed). Therefore, you can draw:
The circumference of this circle is defined as: |
# Introduction to Basic Probability #1
In this post we will get an introduction to basic Probability and what it entails. Statistics and Probability are two topics that go hand in hand. We always need to study and understand probability before venturing into statistics.
## What is Probability?
Probability is defined as the likelihood of occurrence of a specified (desired or undesired) event. The most common usage of probability that can be told globally is the chance of rainfall on any given day. You will notice in your weather app that the chance of rainfall is given as a percentage value, anywhere between 0% to 100%. If you ask someone a question, such as ” are you attending the class tomorrow?” and they answer “probably”, then there is a chance that the person may not attend the class. If we try to calculate this mathematically, then we are solving his probability of attending the class.
Probability of any event will always have its value between 0(Zero) and 1, Or 0% and 100% if you convert the probability into percentages. 0 and 1 are the extreme cases of probability. If an event has 0 probability. then it will not occur. and if it has 1.0 as probability, it is certain to occur.
It becomes easy to explain probability with the use of working examples.
## Examples on basic probability
### The Classic Coin Toss
You will notice that most available books and online courses go with the coin toss, because it is the easiest of the probability examples. Where you can have only 2 outcomes. Heads and Tails. each outcome has an equal likelihood of occurring. So P(heads) = 0.50 and P(tails) = 0.50. This may also be expressed as a percentage. P(heads) = 50% and P(tails) = 50%
### Simple Dice.
A normal board game dice has 6 sides, showing the numbers 1 to 6 on the faces. A balanced die will show each number with equal likelihood. Since there are 6 outcomes of the dice throw, the probability of each event is
### School Students
Here is another classic problem given for probability basics. There is a school with a certain number of students and they have a certain characteristic of interest to the observer. lets look at the given data.
The data shows that the event of interest is Chicken Pox. If you were to pick a student at random, what is the probability that they had chicken pox? In this problem. the answer will be P(yes) = 155/210 = 0.738 (or 73.8%). We see that there is 73% chance that the random student we pick will have had chicken pox.
What is the probability that a randomly picked student is a boy? P(boy) = 100/210 = 0.476 (or 47.6%).
What is the probability that you pick a girl who did not have chicken pox? P (G no CP) = 30/210 = 0.143 (or 14.3%)
Notice that the number of girls that did not get chicken pox is very small as compared to the total number of students. So the likelihood of randomly picking out one that fits is also low.
This tells us that Probability values that are closer to 0.0 are less likely to occur. Probability values closer to 1.0 are more likely to occur.
## Categories of Basic Probability
We can divide basic probability in two categories. 1) unconditional probability and 2) conditional Probability.
### Unconditional Probability
When we compare our event of interest in the probability experiment, with the whole population, it is known as unconditional probability. We worked with unconditional probability in the above school example. The denominator in every equation is the total population (210).
### Conditional Probability
In many experiments we need specialized probability values based on a condition occurring first. It is know as Conditional Probability and is denoted as P(A|B). this means we are looking for probability of A given B has occurred.
What is the Probability of picking a boy given he did not have chickenpox. P(Boy|No CP) = 25/55 = 0.454 (or 45.4%). Note that the event that has occurred is now in the denominator in this case.
What is the probability that you picking someone who had chicken pox given its a girl?
P(CP|girl) = 80/110 = 0.727 (or 72.7%)
You can find more problems on basic Probability here. |
# Combinatorics (Part I)
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1 Combinatorics (Part I) BEGINNERS 02/01/2015 Combinatorics is the study of the methods of counting. Part I: Multiplication Principle 1. If you have two pairs of shorts, one blue and one black, and three T-shirts, red, white and yellow, how many outfits can you make? Let s see how many different combinations of T-shirts and shorts we can have. Fill in the blanks. Shorts T-shirts Red Blue Yellow Red White We see that we can have 1. Blue shorts and T-shirt 2. Blue shorts and T-shirt 3. Blue shorts and T-shirt 4. Black shorts and T-shirt 5. Black shorts and T-shirt 6. Black shorts and T-shirt This is 6 outfits, or 6 combinations, in all. But sometimes, we can t list out all the possible ways. Let us look at a similar problem. 1
2 2. If there are four chairs and four students Ashley, Benjamin, Caitlyn and Daniel in a classroom, in how many ways can the students be seated on the chairs? i. How many different students have the option of sitting on the first chair? ii. If one of them sits on the first chair, how many can sit on the second? iii. Now, how many can sit on the third chair? iv. How many can sit on the fourth chair? So, let s see how many total ways of arranging the students on the chairs there are. For each choice of the student who sits on the first chair (4 in total), we can then consider all the choices of who can sit on the second (3 in total). This gives us 4 3 = different arrangements for the first two chairs. Then, for all the possible choices of who can sit on the first and the second chairs, we can consider all the choices of who can sit on the third chair (2 in total). So, we can have = arrangements for the first three chairs. Then, for the different possible arrangements of students sitting on the first three chairs, we can consider how many students can sit on the last chair. This gives us total arrangements for the four chairs. So, there are ways in which the all students can be seated on the four chairs. It looks like we just multiplied the different number of ways to fill each individual other. Now go back to the first question and see what we did there. What can we multiply together to get the number of all possible outfits? We can generalize: If a process can be broken down into a number of steps, and all the steps are independent of each other like we saw above (choosing a particular pair of shorts does not affect your choice of T-shirt), then the total number of outcomes of the process is the number of possibilities in the first step times the number of possibilities in the second step times the number of possibilities in each step until the last one. This is called the Multiplication Principle of combinatorics. 2
3 Practice Problems 1. There are four fiction books and seven science books in a bookstore. In how many ways can you buy a fiction and a science book? 2. A sweater has to be knitted with three different colors of yarn: one for the left sleeve, one for the right sleeve and one for the body. If we have nine differently colored yarns available, how many different kinds of sweaters can be made? 3. A store carries 7 styles of pants. For each style, there are 10 different waist sizes, 8 different length options and 2 color choices. How many different pairs of pants does the store sell? 4. There are three ways of going from New York to Los Angeles: by train, by plane and by teleportation. And there are three ways of going from Los Angeles to Hawaii: by plane, by teleportation and by sea. How many ways are there to go from New York to Hawaii via Los Angeles? 3
4 Part II: Addition Principle Let us look at a new idea now. Max wants to go from New York to Hawaii. He has two options. One, he can go from New York to LA and then to Hawaii, like we did above. Or, he can go from New York to Chicago, where his grandmother lives, and then to Hawaii. There are four ways to get from New York to Chicago: by train, by car, by teleportation or by plane. And there are two ways to get from Chicago to Hawaii: by plane and by teleportation. In how many ways can Max get to Hawaii from New York? We can break this down as follows: How many ways are there to go from New York to Hawaii via Chicago? How many ways are there to go from New York to Hawaii via LA? How many ways are there to go from New York to Hawaii in all? It is easy to see that you add the number of ways in both travel plans to get the total number of ways. This is the Addition Principle of combinatorics: When you have a choice of methods for performing a procedure (like, via Chicago or via Los Angeles), then the number of ways of performing the procedure is found by adding the number of ways for all the methods individually. 4
5 Practice Problems 1. Angie has to draw one card from a deck with 52 playing cards. In how many ways can she choose a king or a queen? 2. If a bookstore sells five fiction books, four science books and three children's books, in how many ways can you buy two books of different kinds? 3. A college acting troupe has 6 junior women, 8 junior men, 5 senior women and 4 senior men. In how many ways can the teacher select a senior couple or a junior couple to play the roles of Romeo and Juliet? 4. In a movie rental store, there are five different English movies, three French movies, two Russian movies and four Hindi movies. How many ways are there to rent three movies in different languages? 5
6 Part III: Multiple Independent Events A natural number is odd-looking if all its digits are odd. How many odd-looking sixdigit numbers are there? i. How many different odd digits are there? ii. iii. iv. How many different digits can appear in the first box? How many different digits can appear in the second box? How many different digits can appear in the third box? v. How many different digits can appear in the fourth box? vi. vii. How many different digits can appear in the fifth box? How many different digits can appear in the sixth box? How many odd-looking six-digit numbers are possible? Write the answer as a product. Which principle of combinatorics did you use? Let s see if there is another way to write the answer can also be written as 5 6, which is read as 5 to the 6 th power. It means that we are multiplying 5 by itself six times. Therefore, there are 5 6 = 15,625 odd-looking six-digit numbers possible. That s a big number! 6
7 Practice Problems 1. If you toss a coin, how many different outcomes can you get? 2. If you now toss two coins a nickel and a dime how many different outcomes can you get? Remember that a heads on the nickel is different from a heads on the dime. 3. If you then toss three coins, a penny, a nickel and a dime, how many different outcomes can you get? 4. Each box in a 3 3 square can be colored using one out of three colors. How many different colorings of the grid are there? 5. The alphabet of a tribal language contains 4 letters. A word in this language can be formed by any combination of these letters and the words can have between 1 and 5 letters. How many words are there in this language? You may leave your answer in the form of powers. (Hint: Calculate separately the number of one-letter, two-letter, three- letter, four-letter and five-letter words in this language.) 7
8 Permutations are arrangements that can be made by placing different objects in a row. Like we did for the question of students on the chairs in the beginning, the order in which you place these objects is important. 1. How many three-digit numbers can be made using the digits 2, 4 and 6 only once each? In the case of numbers, the order of how the digits are arranged matters. Of course, the number 24 is different from 42; 246 is different from 264 and 426, and so on. i. In how many ways can you fill the hundred s place box? ii. In how many ways can you fill the ten s place box? iii. In how many ways can you fill the one s place box? How many permutations are possible in all? Which principle of combinatorics did you use? 2. In how many ways can we award gold, silver and bronze medals among eight contestants? The order matters. One of the permutations gold, bronze, silver is different from the other permutation of gold, silver, bronze. i. How many students can you give the gold medal to? ii. iii. How many students are left to receive the silver medal? How many students are left to receive the bronze medal? How many permutations are possible in all? Which principle of combinatorics did you use? 8
9 What is different between example 1 and example 2? How many four-digit numbers can you make using the digits 2, 4, 6 and 8? _ _ _ _ How many five-digit numbers can you make using the digits 0, 2, 4, 6 and 8? _ _ _ _ _ Both of these involve the product of decreasing whole numbers. These are known as factorials, written in math as n! It is generally helpful to write these out in descending order. 8! = ! = ! = 5! = Practice Problems 1. How many permutations are there of the set {red, blue, yellow, green, purple, orange}? 2. How many permutations are there of a set of n distinct elements? 9
10 Challenge Yourself! 1. There are five books on a shelf. In how many ways can you create unique piles of books using some (or all) books? Remember that you need at least two books to create a pile. 2. Cory has 24 students out of which he needs to call out 6, one after the other to solve 6 questions on the board. In how many ways can he do that? 3. License plates contain one digit, then three letters, then three more digits. What is the maximum number of cars that can be registered? 4. (Math Kangaroo) A pizza parlor sells small, medium, and large pizzas. Each pizza is made with cheese, tomatoes, and at least one of the following toppings: mushroom, onion, peppers and olives. How many different pizzas are possible? 10
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If you want to learn more, like how to simplify your fraction when you're finished, keep reading the article! In this case, to change the mixed number to an improper fraction, you need to divide 52 by 6. Find the product. Now try our lesson on Converting Fractions to Decimals where we learn how to write a fraction as a decimal. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, … Just follow the instructions. To turn a whole number into a fraction, make the numerator the whole number and make the denominator one. Ratios can be written and calculated in fractions and in percentages. Dividing fractions: 2/5 ÷ 7/3. It's already simplified. So for example, if we've got: 1 2 ÷ 2 \frac{1}{2} \div 2 2 1 ÷ 2, take the 2 from the bottom of 1 2 \frac{1}{2} 2 1 and multiply it with the 2 that's on the right. Check it out: How about another one? Dividing whole numbers by unit fractions. Flip the last number. Now, you have: 5/1× 4/3 Step 3:Multiply the two, and you will get your answer. Your support helps wikiHow to create more in-depth illustrated articles and videos and to share our trusted brand of instructional content with millions of people all over the world. change the division sign into a multiplication sign. Free help from wikiHow. We divide the numerator and leave the denominator the same. Your reciprocal should look like 1 over the whole number. Change to multiplication. We will multiply 7 by 2. Dividing fractions by a whole number isn't as hard as it looks. When teaching dividing fractions by whole numbers, it is important to remember that increasing the number on the top of the fraction makes the fraction bigger but increasing the number on the bottom of the fraction makes it smaller. So, let’s change that fraction into a division. One block was one fifth of each. Write down the fraction on a piece of paper. We now have 4 out of 15 parts which is a smaller amount that 4 out of 5 parts. When you come across dividing a fraction by a whole number, the steps to doing this is pretty simple. Dividing a whole number by a fraction. Thanks to all authors for creating a page that has been read 737,380 times. Next, reverse the numerator and denominator of the fraction you’re dividing the whole number with. Keep the first number. 2 × 2 = 4 and 4 ÷ 1 = 4 2 ÷ 1 / … We keep 2 as 2. Let’s take the same two numbers, but reverse the order. Then, switch to a multiplication problem by multiply by the reciprocal of the divisor. Before we get to the good stuff, let's look at the four step process that we will use every ti… 5 ÷ 3/4 Step 1:Convert the whole number into a fraction. Remember the following: "Dividing fractions is easy as pie, flip the second number and multiply!". Displaying top 8 worksheets found for - Divide Fractions With Whole Numbers. You want it? Turn the second fraction (the one you want to divide by) upside down (this is now a reciprocal). If not, multiply the bottom of the fraction by the whole number instead. 1 / 2 means that we have 1 out of 2 equal parts. We can see in the visual model that final shaded area is a smaller fraction than we started with but the number at the bottom of the fraction increased from 2 to 6. Dividing a whole number by a fraction. Next lesson. Video transcript - [Instructor] So let's see if we can figure out what 1/3 divided by five is. To turn a whole number into a fraction, make the numerator the whole number and make the denominator one. Head on over to the next page to try some fraction practice problems... 1; 2; Continue > Fractions Games. Exponents. Below are six versions of our grade 5 fractions worksheet on dividing proper fractions by whole numbers. Flip the fraction you are dividing by. Khan Academy is a 501(c)(3) nonprofit organization. Page 1 of 2. Step 2. To divide a fraction by a whole number, we can either multiply the denominator by the whole number or we can divide the numerator by the whole number. Doing so gives you 8 with 4 remaining. The resources below which includes Fractions Worksheets will help your KS2 maths lessons that involve dividing fractions become more engaging for the children you teach. Dividing Fractions with Whole and Mixed Numbers 2 - Cool Math has free online cool math lessons, cool math games and fun math activities. Problem 2 : Simplify : 7/5 ÷ 3. Divide the numerator by the denominator, multiply by 100, and add the % sign. Well, remember, a fraction is just another way of writing a “division” question. 1/2 ÷ 1/3 = 1/2 × 3/1 = (1×3) / (2×1) = 3 / 2 = 1 1/2 . To divide fractions by whole numbers use the following steps: If you used step 2, you may need to simplify your answer by dividing the top and bottom of the fraction by the same number. We do not multiply the 10 on the bottom by three because we have already done our division by dividing the 9 by 3. Khan Academy is a 501(c)(3) nonprofit organization. But, fractions are not integers and are represented in ratios, where both the numerator and denominator are integers and denominator is … That reciprocal Step 3: multiply the bottom number of the fraction by 2, change... Also means to reverse the numerator can be divided exactly by 3 you do when --... The special name given to an equivalent multiplication to do so, 15 is the same two numbers but! The division we have already done our division by dividing the whole number instead the. Simplifying at the end can Easily solve can multiply the bottom of the fraction by the whole.... 1 to get our answer of 3 / 7 ÷ 2 = shown with a contribution wikiHow!, flip the second number and make the numerator can be written and calculated in fractions and whole division! 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A regular polygon, remember, is a polygon whose sides and interior angles are all congruent. To understand the formula for the area of such a polygon, some new vocabulary is necessary.
The center of a regular polygon is the point from which all the vertices are equidistant. The radius of a regular polygon is a segment with one endpoint at the center and the other endpoint at one of the vertices. Thus, there are n radii in an n-sided regular polygon. The center and radius of a regular polygon are the same as the center and radius of a circle circumscribed about that regular polygon.
An apothem of a regular polygon is a segment with one endpoint at the center and the other endpoint at the midpoint of one of the sides. The apothem of a regular polygon is the perpendicular bisector of whichever side on which it has its endpoint. A central angle of a regular polygon is an angle whose vertex is the center and whose rays, or sides, contain the endpoints of a side of the regular polygon. Thus, an n-sided regular polygon has n apothems and n central angles, each of whose measure is 360/n degrees. Every apothem is the angle bisector of the central angle that contains the side to which the apothem extends. Below are pictured these characteristics of a regular polygon.
Once you have mastered these new definitions, the formula for the area of a regular polygon is an easy one. The area of a regular polygon is one-half the product of its apothem and its perimeter. Often the formula is written like this: Area=1/2(ap), where a denotes the length of an apothem, and p denotes the perimeter.
When an n-sided polygon is split up into n triangles, its area is equal to the sum of the areas of the triangles. Can you see how 1/2(ap) is equal to the sum of the areas of the triangles that make up a regular polygon? The apothem is equal to the altitude, and the perimeter is equal to the sum of the bases. So 1/2(ap) is only a slightly simpler way to express the sum of the areas of the n triangles that make up an n-sided regular polygon. |
# Geometric Distribution
In a series of trials, if you assume that the probability of either success or failure of a random variable in each trial is the same, geometric distribution gives the probability of achieving success after N number of failures. The distribution is essentially a set of probabilities that presents the chance of success after zero failures, one failure, two failures and so on.
## Geometric Distribution Definition
A geometric distribution is defined as a discrete probability distribution of a random variable “x” which satisfies some of the conditions. The geometric distribution conditions are
• A phenomenon that has a series of trials
• Each trial has only two possible outcomes – either success or failure
• The probability of success is the same for each trial
## Geometric Distribution Formula
In probability and statistics, geometric distribution defines the probability that first success occurs after k number of trials. If p is the probability of success or failure of each trial, then the probability that success occurs on the
$$\begin{array}{l}k^{th}\end{array}$$
trial is given by the formula
$$\begin{array}{l}Pr (X = k) = (1-p)^{k-1}p\end{array}$$
### Examples
• Consider a couple who are planning to have a child and they will continue to babies until it is a girl. What is the probability that they have zero boys, one boy and two boys and so on until a girl is born?
• A person is seeking new employment that is both challenging and fulfilling. What is the probability that he will quit zero times, one time, two times so on until he finds his ideal job?
• A pharmaceutical company is designing a new drug to treat a certain disease that will have minimal side effects. What is the probability that zero drugs fail the test, one drug fails the test, two drugs fail the test and so on until they have designed the ideal drug?
### Solved Problem
Let us consider one of the situations where geometric distribution can be applied. Let us say a person is throwing dice and will stop once he gets 5. Since there are 6 possible outcomes, the probability of success p =
$$\begin{array}{l}\frac{1}{6}\end{array}$$
= 0.17. Therefore the probability of failure
q = 1 – p = 1 – 0.17 = .83
Let us calculate the probability of the first three trials.
The person gets number 5 for the first time. The number of failures before the first success is zero. Therefore X = 0, k = 1
Substituting the values of X, k, p and q in distribution, we have
$$\begin{array}{l}Pr(X=0)= (0.83^{0})\times 0.17\end{array}$$
= 0.17
The person gets number 5 for the second time. The number of failures before the first success is 1. Therefore X = 1 and k = 2.
Substituting the values of X, k, p and q in distribution, we have
$$\begin{array}{l}Pr(X=1)= (0.83^{2-1})\times 0.17\end{array}$$
= 0.83 x 0.17 = 0.14
This way we can construct a series of geometric distribution for a series of trials.
### Related Topics
Learn more about geometric distribution and related topics in a simple and easy way. Register with BYJU’S – The Learning App today. |
# 6.1.1B Equivalence & Representations
6
Subject:
Math
Strand:
Number & Operation
Standard 6.1.1
Read, write, represent and compare positive rational numbers expressed as fractions, decimals, percents and ratios; write positive integers as products of factors; use these representations in real-world and mathematical situations.
Benchmark: 6.1.1.3 Percents
Understand that percent represents parts out of 100 and ratios to 100.
For example: 75% corresponds to the ratio 75 to 100, which is equivalent to the ratio 3 to 4.
Benchmark: 6.1.1.4 Fractions, Decimals, Percents: Equivalence
Determine equivalences among fractions, decimals and percents; select among these representations to solve problems.
For example: If a woman making $25 an hour gets a 10% raise, she will make an additional$2.50 an hour, because $2.50 is $\frac{1}{10}$ or 10% of$25.
Benchmark: 6.1.1.7 Positive Rational Numbers: Equivalent Representations
Convert between equivalent representations of positive rational numbers.
For example: Express $\frac{10}{7}$ as $\frac{7+3}{7}=\frac{7}{7}+\frac{3}{7}=1\frac{3}{7}$.
## Overview
Big Ideas and Essential Understandings
Relationships of equivalence with different forms of rational numbers can be illustrated in a variety of representations. Students use fractions, decimals, and percents to describe equivalent positive rational numbers. Clement (2004) suggests five different kinds of representations for teaching students the concepts of fractions, decimals and percents. These five representations are pictures, manipulatives, spoken language, written symbols, and relevant situations. Conceptual understanding of these equivalencies is developed as students describe these rational numbers in concrete representational forms (fraction strips, Cuisenaire rods, pattern blocks, etc.); visual representational forms (grids, diagrams, pictures, etc.); and abstract symbolic form. Using these same structures, students expand their understanding of equivalence with rational numbers to making comparisons between them. Students' learning experiences with these different forms of representation guide them in identifying and selecting appropriate forms for making comparisons and conversions in a particular situation.
The skills of prime factorization, least common multiple, and greatest common factor become tools for students in their formation of equivalent fractional numbers. Student understanding of representing whole numbers as a product of factors with exponents is aided by their previous work with whole numbers, multiples, factors and exponents.
All Standard Benchmarks
• 6.1.1.1 Locate positive rational numbers on a number line and plot pairs of positive rational numbers on a coordinate grid.
• 6.1.1.2 Compare positive rational numbers represented in various forms. Use the symbols <, =, and >.
• 6.1.1.3 Understand that percent represents parts out of 100 and ratios to 100.
• 6.1.1.4 Determine equivalences among fractions, decimals, and percents: select among these representations to solve problems.
• 6.1.1.5 Factor whole numbers; express a whole number as a product of prime factors with exponents.
• 6.1.1.6 Determine greatest common factors and least common multiples. Use common factors and common multiples to calculate with fractions and find equivalent fractions.
• 6.1.1.7 Convert between equivalent representations of positive rational numbers.
Benchmark Cluster
Benchmark Group B
• 6.1.1.3 Understand that percent represents parts out of 100 and ratios to 100.
• 6.1.1.4 Determine equivalences among fractions, decimals, and percents; select among these representations to solve problems.
• 6.1.1.7 Convert between equivalent representations of positive rational numbers.
What students should know and be able to do [at a mastery level] related to these benchmarks.
• Demonstrate knowledge of percent as representing part per 100 or ratios to 100;
• Understand that 100% is equivalent to 1;
• Determine equivalencies between fractions and decimals, decimals and percents, and fractions and percents;
• Recognize common benchmarks; e.g., $\frac{1}{4}$ = 0.25 = 25%
• Understand that percents must be converted to fractions or decimals in order to multiply or divide;
• Choose among representations of rational numbers (fractions, decimals, and percents) to solve problems.
• Work from previous grades that supports this new learning includes:
Recognize and generate equivalent decimals, fractions, mixed numbers and improper fractions in various contexts.
• Divide multi-digit numbers using efficient and generalizable procedures, based on knowledge of place value, including standard algorithms. Recognize that quotients can be represented in a variety of ways, including a whole number with a remainder, a fraction or mixed number, or a decimal. Divide numerator by denominator to produce a decimal.
• Estimate solutions to arithmetic problems in order to access the reasonableness of results.
• Visualize or use area models to determine equivalency
Correlations
NCTM Standards
Understand numbers, ways of representing numbers, relationships among numbers, and number systems:
• work flexibly with fractions, decimals, and percents to solve problems;
• compare and order fractions, decimals, and percents efficiently and find their approximate locations on a number line;
• develop meaning for percents greater than 100 and less than 1;
• understand and use ratios and proportions to represent quantitative relationships;
• use factors, multiples, prime factorization, and relatively prime numbers to solve problems.
Common Core State Standards
4NF (NUMBER AND OPERATIONS - FRACTIONS) Extend understanding of fraction equivalence and ordering.
• 4NF.7 Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual model.
4OA. (OPERATIONS AND ALGEBRAIC THINKING) Gain familiarity with factors and multiples.
• 4OA.4 Find all factor pairs for a whole number in the range 1-100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1-100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1-100 is prime or composite.
5NBT (NUMBER AND OPERATIONS IN BASE TEN) Understand the place value system.
• 5NBT.3b Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.
6NS (NUMBER SYSTEM) Apply and extend previous understandings of multiplication and division to divide fractions by fractions.
• 6NS.4 Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor.
• 6NS.6 Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates.
• Recognize opposite signs of numbers as indicating locations on opposite sides of 0 on the number line; recognize that the opposite of the opposite of a number is the number itself, e.g., -(-3) = 3, and that 0 is its own opposite.
• Understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane; recognize that when two ordered pairs differ only by signs, the locations of the points are related by reflections across one or both axes.
• Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane.
• 6NS.7 Understand ordering and absolute value of rational numbers.
• 6NS.7.b Write, interpret, and explain statements of order for rational numbers in real-world contexts. For example, write -30C > -70C to express the fact that -30C is warmer than -70C.
6RP (RATIOS AND PROPORTIONAL RELATIONSHIPS) Understand ratio concepts and use ratio reasoning to solve problems.
• 6RP.3Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
• 6RP3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
## Misconceptions
Student Misconceptions
• Students may have difficulty dividing the numerator by the denominator to identify a decimal equivalency to be translated to a percent. Creating a percent from a decimal number requires students to understand the meaning of the percent symbol and the procedure of moving the decimal point two place values to the left of the current decimal point. Some students leave the decimal in its exact place while creating an equivalent percent from a decimal.
Example: 0.12 = 12% not .12%
• When looking at a 100-grid containing $28\frac{1}{2}$ colored squares, students may not understand how to write $\frac{28.5}{100}$ as 0.285 since $\frac{1}{2}$ of $\frac{1}{100}$ is 0.005.
• Students may not recognize that, in a problem solving situation, they must change a percent into a fraction or decimal in order to work with it.
• Students may have difficulty understanding the value of fractions when comparing fractions with the same numerators and different denominators.
Example: $\frac{3}{5}$ is less than $\frac{3}{8}$ because 5 is less than 8.
• Students may think that a number containing a "longer" portion (more digits) to the right of the decimal point is actually "greater" than a "shorter" number.
Example, 3.12345 is NOT greater than 3.2.
• Students may assume that $4\frac{7}{4}$ is smaller than $5\frac{1}{4}$ because the whole unit, 4, is smaller than the 5. They often do not consider the improper portion of a mixed number.
• Students might not use benchmark numbers like 0, $\frac{1}{2}$, and 1 to compare fractions because their understanding reflects only part-whole situations and they do not think of the fractions as numbers.
• Students may think that multiplying the numerator and the denominator by the same number increases the value of the fraction.
• Students may write the percent as a fraction using the wrong denominator, or try to remove the decimal point when writing it over 100.
Example: a student may write $\frac{775}{100}$ instead of$\frac{77.5}{100}$.
• When changing a decimal to a percent, students may only add the percent symbol without considering parts per hundred.
Example: 0.12 may be written as .12%.
## Vignette
In the Classroom
In this vignette, Mr. Olson takes advantage of his students' interest watching TV to collect and represent data as fractions, decimals, and percents.
Teacher: "How many of you enjoy homework?"
(Audible groans are heard from all parts of the room.)
Teacher: Today we're going to try a different kind of homework assignment that you may find more exciting. Instead of giving you a worksheet to practice translating between fractions, decimals, and percents, I'm going to 'require' you to watch 30 minutes of television.
Teacher: Recently I've heard some of you complaining about how often commercials interrupt your favorite TV shows. Your homework is to collect some data to investigate whether your complaints are justified. Here's how it works. Tonight, you must watch 30 minutes of a program your family allows you to see. For 30 minutes, you need to keep track of whether you're watching a commercial or viewing an actual show. I want you to record data every minute, using the words either "Show" or "Commercial" to indicate what you're watching.
Student: Is it all right if I watch more than 30 minutes?
Teacher: Class, what do you think?
Student: You mean some people want to do more homework?
Student: I don't want to do more homework. I just want to watch a show that lasts an hour.
Teacher: I'm asking you to collect data for 30 minutes. You'll have to discuss how much TV you watch tonight with your family.
Student: What if you don't have a TV?
Teacher: Then I'll arrange for you to have some class time tomorrow for data collection. Let's talk about how we might organize our data collection. Any ideas?
Student: It would be easy to record the data in our notebook. We could number the lines 1-30 to show the minutes, and then write either "Show" or "Commercial" on each line
Teacher: Is this what you mean?
1.
2.
3.
4.
Student: Yes, only I'd number all the way to 30.
Teacher: Yes, that's one way to organize your data collection. Any other ideas?
Student: I was thinking about making two columns - one for "Show" and one for "Commercial." Then I would use tally marks to show what was happening each minute.
Teacher: Show me what that would look like.
Student: Maybe like this.
Teacher: That's another strategy that could be used. Just to be sure that you understand what the assignment is, we're going to watch 3 minutes of a TV and practice recording the data as either "Show" or "Commercial." Take a moment now to create a table or chart that you can use to collect 3 minutes of data, and then we'll practice.
(Class watches 3 minutes TV show; teacher models both strategies offered by students for data collection.)
Teacher: Are there any questions about tonight's homework assignment? If not, I look forward to seeing your data tomorrow.
The next day...
Mr. Olson has used the whiteboard at the front of the room to create a table for each student to record a summary of the data.
Teacher: Who got their homework done last night?
(Every hand is raised. Students are smiling and sharing their "homework" experiences.
Teacher: That's great! I'm curious to see your data. I've created a table on the white board where I'd like each of you to write the name of the show you watched, as well as the fraction, decimal, and percent that represent what part of the total time was spent on the actual show and what part was spent on commercials.
Note: This chart can provide an opportunity for students to see that the sum of the parts equals the whole; i.e., actual show time + commercials = total time. Students can also represent this data in a circle graph.
## Resources
Instructional Notes
Teacher Notes
• Give students enough time and multiple opportunities to construct and process the concepts of representing and relating equivalents in their minds. Students learn "in a halting, back-and-forth sort of way." The process of learning is not just cumulative, but students are "disconnecting, connecting, and reorganizing" the knowledge they have acquired.
• Students have typically been taught about fractions, decimals, and percents in isolation. A conscious effort needs to be spent identifying numbers in a multitude of ways simultaneously to encourage flexible thinking. Students will benefit from using 100-grids as visual representations to connect fractions, decimals, and percent equivalencies.
• Students may have difficulty understanding decimal movement and placement with representing equivalent values from decimal to percent.
• Students may have difficulty understanding that the percent symbol implies a ratio of parts per hundred.
• By repeated exposure in multiple situations and representations, students should be developing a quick recall of common benchmark fractions and percents: halves, fourths, fifths, and tenths.
• Students have little difficulty writing a percent as a fraction. Writing fractions as percents is more challenging. Ask students how many percents are in one whole (100). Remind students that the denominator will always be 100. Also tell students that the number to the left of the percent symbol will always be the numerator.
• Students may not know when it is appropriate to use an equivalent fraction to write a fraction as a percent. While there is no error in writing the fraction as a decimal first, sometimes it is easier to just writing an equivalent fraction. Brainstorm some denominators that are easily rewritten as denominators of 100, such as 4, 5, 10, 20, 25, etc.
• Students have difficulty with percents greater than 100% or less than 1% (i.e., 150%, 0.015%). They want to quickly move the decimal point without thinking first. Reinforce the meaning behind moving the decimal point. Practice converting with greater and lesser percents when students are comfortable with percents between 1 and 100.
• 15% is 15 per one hundred or 0.15;
• 150% is 150 per one hundred or 1.5;
• 0.15% is 15 hundredths per one hundred or 0.0015;
• 0.015% is 15 thousandths per one hundred or 0.00015.
• Many struggle with the meaning of the percent symbol. Some may believe that 2.4% is the same as 2.4, while others think that 0.4 is the same as 4% Remind students that when the percent is given, for instance 40%, the percent symbol (%) means writing the number as a fraction with a denominator of 100. Therefore, 40% is $\frac{40}{100}$ (which can be simplified to $\frac{2}{5}$) = 0.40. Using the same idea, 2.4% is the same as $\frac{2.4}{100}= \frac{2.4}{100} \times \frac{10}{10}= 2.4 \times \frac{10}{100} \times 10 = \frac{24}{1000}$ (which can be simplified to $\frac{3}{125}$) = 0.024.
• Students may move the decimal point the wrong way, forget to place zeroes as placeholders, or move the decimal point too many places (especially when the percent is greater than 100). As a class, gather some helpful information for remembering how to convert decimals and percents.
• Students may struggle with the idea that three different forms of a number are useful. Have a classroom discussion where students brainstorm when to use fractions (construction, recipes), when to use decimals (money, science), and when to use percents (shopping, chance of rain).
Instructional Resources
Asks students to fill in missing portion of the following three; part, whole or percent.
Explore different representations for fractions including improper fractions, mixed numbers, decimals, and percentages. Additionally, there are length, area, region, and set models. Adjust numerators and denominators to see how they alter the representations and models. Use the table to keep track of interesting fractions.
In this lesson, students use real-world models to develop an understanding of fractions, decimals, unit rates, proportions, and problem solving.
This article uses current research to assist with relating decimals, fractions and percents.
Allows practice with percentage, fraction to decimal, fraction to percentage, algebra, comparing fractions, reducing fractions, adding & subtracting fraction, multiplying & dividing fractions, rounding decimals, and improper fractions to mixed numbers.
New Vocabulary
per: for each, or for everyExample: If apples cost $1.99 per pound, then each pound of apples costs$1.99.
percent: per hundred, or out of 100Example: "12% of sixth grade students are left-handed" means that out every 100 sixth grade students, 12 are left-handed.
rational number: any number that can be expressed in the form $\frac{a}{b}$ where a and b are integers and b≠0. A rational number can always be represented by either a terminating or a repeating decimal. Examples: $\frac{2}{3}$, 4 (which can be expressed as $\frac{4}{1}$); 2.25 (which can be expressed as $\frac{225}{100}$).
simplify (fraction): to express in simplest form, or lowest terms. The numerator and denominator of proper fractions in simplest form have no common factor other than 1. Improper fractions and mixed numbers are in simplest form when the fraction part is proper and in simplest form. Examples: The numerator and denominator of $\frac{4}{8}$ share the common factor 4, so must be rewritten as $\frac{1}{2}$ to be in simplest form; $\frac{10}{4}$ written in simplest form is $\frac{5}{2}$ or $2\frac{1}{2}$.
translate: change from one form of a number to another. Example: 50% can be translated to a decimal (0.5) or fraction ($\frac{1}{2}$).
Professional Learning Communities
Professional Learning Communities
Reflection - Critical Questions regarding the teaching and learning of these benchmarks:
How has the students' ability to conceptualize fractions, decimals, and percents been strengthened by today's lesson?
• What evidence exists to show that students understand percent as a ratio to 100?
• What evidence exists to show that students recognize fraction, decimal, and percent equivalencies?
• What everyday student experiences can be used as problem-solving opportunities?
• How was prior knowledge used to connect different representations of equivalencies?
Materials
Authors: Jenny K. Tsankova and Karmen Pjanic
The algorithm for multiplying proper fractions is often taught by gently leading students to notice patterns when finding part of a fractional part. An area-model approach will extend students' understanding.
References
Clement, L. L. (2004). A model for understanding, using, and connecting representations. Teaching Children Mathematics, 11 (2), 97 - 102.
Keeley, P., & Rose, C. (2006). Mathematics Curriculum Topic Study. Thousand Oaks, CA: Corwin Press.
Kilpatrick, J., Martin, W., & Schifter, D. (Eds.). (2003). A Research Companion to Principles and Standards for School Mathematics. Reston, VA: National Council of Teachers of Mathematics, Inc.
Mathematics Curriculum Framework. (2000). Malden, MA: Massachusetts Department of Education.
Minnesota's K-12 Mathematics Frameworks. (1998). St. Paul, MN: SciMathMN.
Mathematics Framework for the 2009 National Assessment of Educational Progress. (2009). Washington, D.C.: National Assessment Governing Board U.S. Department of Education.
Mathematics 6-8. GaDOE:Georgia Department of Education, n.d. Web. 29 Mar. 2011.
The Rational Number Project (choose RNP: Initial Fraction Ideas)Cramer, K., Behr, M., Post T., Lesh, R., (2009). The Rational Number Project (RNP) advocates teaching fractions using a model that emphasizes multiple representations and connections among different representations.
Behr, M. & Post, T. (1992). Teaching rational number and decimal concepts. In T. Post (Ed.), Teaching mathematics in grades K-8: Research-based methods (2nd ed.) (pp. 201-248). Boston: Allyn and Bacon.
## Assessment
(DOK: Level 1)
1. What percent of the figure is shaded red?
(DOK Level 1)
2. Write $8\frac{11}{20}$ as a decimal.
(DOK Level 2)
3. A baseball player's batting average is 0.625. Write this batting average as a fraction.
Answer: $\frac{5}{8}$
(DOK Level 2)
4. A 200-pound man lost 10 pounds. Which of the following represents the portion of body weight lost?
a. 0.10 b. $\frac{1}{200}$ c. 5% d. $\frac{1}{10}$
(DOK Level 2)
5. Ernest got a box of chocolates for his birthday. He has eaten 6 of the 20 pieces. What percent has he eaten?
(DOK: Level 3)
6. A species of cicada has a life cycle of 17 years. A parasite that harms the cicada has a life cycle of 4 years. If the last known encounter between these two was 1980. When can the cicada expect the next encounter with this parasite?
(DOK: Level 3)
30% were black
25% were red
15% were white
The rest of the beads were blue. How many beads were blue?
A. 25
B. 65
C. 90
D. 130
(DOK Level 4)
8. A sixth grade class was surveyed to find out whether like ice cream or cake better. The survey results are shown in the table below. Use the table to answer the following questions.
a) What fraction of the boys like cake better?
b) What percent of total students like cake better?
c) How does the fraction of boys that like cake better compare to the percent of total students that like cake better? Explain how you know.
a) $\frac{15}{27}=\frac{5}{9}$
b) $\frac{45}{75}=0.6=60%$
c) The fraction of boys that like cake better is less than the percent of total students. $\frac{5}{9}<60%$; I know because $\frac{5}{9}$ written as a decimal is 0.555...., which is the same as 55.5%. 55.5% is less than 60%.
## Differentiation
Struggling Learners
Struggling Students
● Encourage student use of concrete and pictorial models to increase the understanding of abstract ideas. Fraction bars, fraction circles, area models, 10 by 10 grids, graph paper, number lines, coordinate grids, and Cuisenaire rods can be used to develop order relations among fractions, decimals, and percents.
Fraction Circles are commonly used for fraction instruction; however, these can also be used for decimal and percent concepts.
● Provide a fraction-stick chart that shows common equivalent fractions.
● Use concrete tools, such as percent circles, 100 grids, and meter sticks to aid in solving percent problems;
● Suggest websites containing interactive opportunities.
Example: Interactive Tutorial on Percentage. This applet contains a 100-grid, percent, and fraction interactive components.Given a partially completed chart of fractions, decimals, and percents, after students have completed the chart it could be used as a tool to assist them in daily work;
• Have students provide fraction, decimal and percent forms of the same number whether a specific problem requires it or not.
• Provide simpler problems when working with a more difficulty concept.
English Language Learners
English Language Learners
• Connect the word "percent" to "per cent," as in cents or pennies;
• Post charts that shows how to convert among fractions, decimals, and percents;
• Pair students with others that have strong English language skills;
• Model alternate ways of solving problems; and, model more than once;
• Encourage parents to teach the same math skills in their native tongue to build a more cohesive understanding of concepts.
• Model the strategies needed to convert equivalencies for the student while coaching the student through the process. Repeat and reverse the roles.
• Use graphic organizers such as the Frayer model shown below, for vocabulary development.
Extending the Learning
Extending the Learning
● Ask students to design their own problem using a circle graph; e.g., construct a circle graph that includes 4 or more fractions that are not equivalent with different denominators. Students can then share their problems and possible solutions.
● Pose problems involving percents greater than 100% or less than 1%.
● Ask students to predict whether increasing a number by 10%, then decreasing by 10% results in the original number and justify their answer.
Classroom Observation
Students are: (descriptive list) Teachers are: (descriptive list) using concrete models and pictures to develop an understanding of equivalencies among fractions, decimals, and percents. providing sufficient practice with the concrete before moving toward the abstract. recognizing equivalencies among fractions, decimals and percents; and then choosing the best representation of the solution from among the representations. allowing students time to recognize equivalencies among different representations and choose among them to solve problems. converting among fractions, decimals, and percents. posing questions that require students to convert among different representations. discussing and writing about the processes used to determine equivalencies and justifying their results. actively listening, questioning, and providing encouragement for student talk regarding percents and equivalency relationships. becoming fluent at converting among the fractions, decimals, and percents of common benchmarks such as $\frac{1}{4}=0.25=25%$. asking students to estimate solutions to problems by using common benchmarks. solving real-world problems that require conversion among fractions, decimals, and percents. using students' everyday experiences as a source of rich problems.
Parents
● Interactive Tutorial on Percentage This applet contains a 100-grid, percent, and fraction interactive components.
● Gifford, S., & Thaler S. (2003). Piece=Part=Portion. Berkley, CA: Tricycle Press. Print. Explains how in the language of mathematics, fractions, decimals and percents are three different ways of describing the same parts of things
● Percent Uses grids, fractions and decimals to identify percents. It also shares how to identify percent when 100 items are not available. |
# Slope review
CCSS Math: HSF.LE.A.2
The slope of a line is a measure of its steepness. Mathematically, slope is calculated as "rise over run" (change in y divided by change in x).
## What is slope?
Slope is a measure of the steepness of a line.
$\text{Slope} = \dfrac{\text{rise}}{\text{run}}=\dfrac{\Delta y}{\Delta x}$
Want an in-depth introduction to slope? Check out this video.
### Example: Slope from graph
We're given the graph of a line and asked to find its slope.
The line appears to go through the points $(0,5)$ and $(4,2)$.
$\text{Slope}=\dfrac{\Delta y}{\Delta x}=\dfrac{2-5}{4-0}=\dfrac{-3}{4}$
In other words, for every three units we move vertically down the line, we move four units horizontally to the right.
### Example: Slope from two points
We're told that a certain linear equation has the following two solutions:
Solution: $x=11.4 ~~~ y=11.5$
Solution: $x=12.7 ~~~ y=15.4$
And we're asked to find the slope of the graph of that equation.
The first thing to realize is that each solution is a point on the line. So, all we need to do is find the slope of the line through the points $(11.4,11.5)$ and $(12.7,15.4)$.
\begin{aligned} \text{Slope}=\dfrac{\Delta y}{\Delta x}&=\dfrac{15.4-11.5}{12.7-11.4}\\\\ &=\dfrac{3.9}{1.3}\\\\ &=\dfrac{39}{13}\\\\ &=3\end{aligned}
The slope of the line is $3$.
## Practice
Problem 1
What is the slope of the line below?
Give an exact number.
Want more practice? Check out this Slope from graphs exercise and this Slope from points exercise. |
# Discrete Mathematics: An Active Approach to Mathematical Reasoning
## Section7.2One-to-One, Onto, Inverse Functions
In this section we will look at specific properties of functions. We will learn how to prove a function is one-to-one and/or onto its codomain. These properies are important as they are the exact properties we need in order for a function to have an inverse function.
### Definition7.2.1.
A function, $$f:X\rightarrow Y\text{,}$$ is one-to-one or injective if for all $$x_1, x_2\in X\text{,}$$ if $$f(x_1)=f(x_2)$$ then $$x_1=x_2\text{.}$$
Equivalently, $$f$$ is one-to-one if $$x_1\neq x_2$$ implies $$f(x_1)\neq f(x_2)\text{.}$$ We note, this is just the contrapositive of the definition.
Although it is easier to prove a function is one-to-one using the definition, the contrapositive can be helpful for deciding if a function is one-to-one.
### Proving a Function is One-to-One.
To prove $$f:X\rightarrow Y$$ is one-to-one:
• Assume $$f(x_1)=f(x_2).$$
• Show $$x_1=x_2.$$
To prove $$f:X\rightarrow Y$$ is not one-to-one:
• Find a counterexample.
• In particular, find $$x_1, x_2\in X$$ with $$x_1\neq x_2$$ and $$f(x_1)=f(x_2)\text{.}$$
The function given in Figure 7.2.3 in not one-to-one since $$c$$ and $$d$$ both map to the same value in $$Y\text{.}$$
Let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be given by $$f(x)=3x+2\text{.}$$ Prove $$f$$ is one-to-one.
Assume $$f(x_1)=f(x_2)\text{.}$$ Then
\begin{align*} 3x_1+2&=3x_2+2\\ 3x_1&=3x_2\\ x_1&=x_2 \end{align*}
which is what we wanted to show.
Let $$f:\mathbb{Z}\rightarrow \mathbb{Z}$$ be given by $$f(x)=x^2-1\text{.}$$ Disprove $$f$$ is one-to-one.
We need a counterexample with $$x_1\neq x_2$$ and $$f(x_1)=f(x_2)\text{.}$$ Let $$x_1=2, x_2=-2\text{.}$$
Then $$f(2)=4-1=3$$ and $$f(-2)=4-1=3\text{.}$$ So $$f(x_1)=f(x_2)\text{,}$$ but $$2\neq-2\text{.}$$
### Definition7.2.6.
A function, $$f:X\rightarrow Y\text{,}$$ is onto $$Y$$ or surjective if for all $$y\in Y$$ there exists $$x\in X$$ such that $$f(x)=y\text{.}$$
Although we need the definition for onto to be able to write a proof, the concept of onto is easier to understand without the definition. Basically, we need every $$y\in Y$$ to get mapped to by some $$x\in X\text{.}$$ We can also think about onto in terms of sets. A function is onto $$Y$$ if $$Y$$ is the range of $$f\text{.}$$
### Proving a Function is Onto.
To prove $$f:X\rightarrow Y$$ is onto $$Y\text{:}$$
• Let $$y$$ be a general element of $$Y\text{.}$$ You should not be using any information about the function at this point.
• Find $$x\in X$$ such that $$f(x)=y\text{.}$$ Finding $$x$$ may involve scratchwork.
• In your proof, state $$x\text{,}$$ show $$x\in X\text{,}$$ and show $$f(x)=y\text{.}$$
To prove $$f:X\rightarrow Y$$ is not onto $$Y\text{:}$$
• Find a counterexample.
• In particular, find $$y\in Y$$ such that no $$x\in X$$ will map to $$y\text{.}$$
The function given in Figure 7.2.8 in not onto $$Y$$ since 2 is not mapped to by any value in $$X\text{.}$$
Let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be given by $$f(x)=3x+2\text{.}$$ Prove $$f$$ is onto $$\mathbb{R}\text{.}$$
Let $$y\in\mathbb{R}\text{.}$$
[Scratchwork: we want to find $$x$$ so that $$f(x)=y\text{.}$$ So we want $$3x+2=y\text{,}$$ or $$x=\frac{y-2}{3}\text{.}$$]
Let $$x=\frac{y-2}{3}\text{.}$$ Then since $$y\in \mathbb{R}, x\in\mathbb{R}\text{.}$$ Furthermore,
\begin{equation*} f(x)=f(\frac{y-2}{3})=3(\frac{y-2}{3})+2=y-2+2=y, \end{equation*}
which is what we wanted to show.
Let $$f:\mathbb{Z}\rightarrow \mathbb{Z}$$ be given by $$f(x)=3x+2\text{.}$$ Prove $$f$$ is not onto $$\mathbb{Z}\text{.}$$
Let $$y\in\mathbb{Z}\text{.}$$
We saw in the previous example $$x=\frac{y-2}{3}\text{.}$$ But $$x$$ is not necessarily in $$\mathbb{Z}\text{.}$$ So for our counterexample, let $$y=1\text{.}$$ Then we would need $$x=\frac{-1}{3}\notin \mathbb{Z}\text{.}$$
Hence no element in $$\mathbb{Z}$$ will map to $$y=1\text{.}$$ Therefore, $$f$$ is not onto $$\mathbb{Z}\text{.}$$
Let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be given by $$f(x)=x^2-1\text{.}$$ Prove or disprove $$f$$ is onto $$\mathbb{R}\text{.}$$
Let $$y=-2\text{.}$$ Then if $$f$$ is onto $$\mathbb{R}\text{,}$$ we could find $$x$$ with $$f(x)=-2\text{.}$$
But if $$f(x)=-2\text{,}$$ then $$x^2-1=-2\text{,}$$ or $$x^2=-1\text{.}$$ We know there are no real solutions to this equation. Hence no element in $$\mathbb{R}$$ will map to $$y=-2\text{.}$$ Therefore, $$f$$ is not onto $$\mathbb{R}\text{.}$$
### Activity7.2.1.
Define $$f:\mathbb{R}\rightarrow\mathbb{R}$$ by $$f(x)=5x\text{.}$$
#### (a)
Prove or disprove $$f$$ is one-to-one.
#### (b)
Prove or disprove $$f$$ is onto $$\mathbb{R}\text{.}$$
### Activity7.2.2.
Define $$f:\mathbb{Z}\rightarrow\mathbb{Z}$$ by $$f(n)=5n\text{.}$$
#### (a)
Prove or disprove $$f$$ is one-to-one.
#### (b)
Prove or disprove $$f$$ is onto $$\mathbb{Z}\text{.}$$
### Activity7.2.3.
Define $$g:\mathbb{Z}\rightarrow\{0, 1, 2\}$$ by $$g(n)=r$$ where $$r$$ is the remainder when $$n$$ is divided by 3.
#### (a)
Prove or disprove $$g$$ is one-to-one.
#### (b)
Prove or disprove $$g$$ is onto $$\{0, 1, 2\}\text{.}$$
### Activity7.2.4.
Define $$h:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}$$ by $$h(a, b)=a-b\text{.}$$
#### (a)
Prove or disprove $$h$$ is one-to-one.
#### (b)
Prove or disprove $$h$$ is onto $$\mathbb{Z}\text{.}$$
### Activity7.2.5.
Define $$d:\mathbb{Z}\rightarrow\mathbb{Z}\times\mathbb{Z}$$ by $$d(a)=(a, a)\text{.}$$
#### (a)
Prove or disprove $$d$$ is one-to-one.
#### (b)
Prove or disprove $$d$$ is onto $$\mathbb{Z}\times\mathbb{Z}\text{.}$$
### Definition7.2.12.
A function $$f:X\rightarrow Y$$ is a one-to-one correspondence or bijection if $$f$$ is one-to-one and onto $$Y\text{.}$$
We showed in the above examples that $$f:\mathbb{R}\rightarrow\mathbb{R}$$ given by $$f(x)=3x+2$$ is one-to-one and onto $$\mathbb{R}\text{.}$$ Thus, it is an example of a one-to-one correspondence.
If it exists, we say $$f^{-1}$$ is the inverse of $$f\text{.}$$
Since $$f:\mathbb{R}\rightarrow\mathbb{R}$$ given by $$f(x)=3x+2$$ is one-to-one and onto, it has an inverse. We can find the inverse as we did in calculus.
Let $$y=3x+2\text{,}$$ solve for $$x\text{.}$$
We get $$x=\frac{y-2}{3}\text{.}$$ Thus $$f^{-1}(x)=\frac{x-2}{3}\text{.}$$
Show $$f^{-1}$$ is one-to-one: assume $$f^{-1}(y_1)=f^{-1}(y_2)\text{.}$$ Then $$f^{-1}(y_1)=f^{-1}(y_2)=x$$ for some $$x\in X\text{.}$$ Thus, $$f(x)=y_1$$ and $$f(x)=y_2\text{.}$$ Since $$f$$ is a function, $$y_1=y_2\text{.}$$
Show $$f^{-1}$$ is onto $$X\text{.}$$ Let $$x\in X\text{.}$$ Then there exists $$y\in Y$$ such that $$f(x)=y$$ since $$f$$ is a function from $$X\text{.}$$ Now, $$f^{-1}(y)=x\text{.}$$ Therefore, there exists $$y\in Y$$ such that $$f^{-1}(y)=x\text{.}$$
#### 1.
True or false: $$f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=5x-2$$ is one-to-one.
• True.
• False.
#### 2.
True or false: $$f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=5x-2$$ is onto $$\mathbb{R}\text{.}$$
• True.
• False.
#### 3.
True or false: $$f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=5x-2$$ is one-to-one.
• True.
• False.
#### 4.
True or false: $$f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=5x-2$$ is onto $$\mathbb{Z}\text{.}$$
• True.
• False.
#### 5.
True or false: $$f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=x^3-3$$ is one-to-one.
• True.
• False.
#### 6.
True or false: $$f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=x^3-3$$ is onto $$\mathbb{R}\text{.}$$
• True.
• False.
#### 7.
True or false: $$f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=x^3-3$$ is one-to-one.
• True.
• False.
#### 8.
True or false: $$f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=x^3-3$$ is onto $$\mathbb{Z}\text{.}$$
• True.
• False.
#### 9.
True or false: $$f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}, f(x, y)=x$$ is one-to-one.
• True.
• False.
#### 10.
True or false: $$f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}, f(x, y)=x$$ is onto $$\mathbb{Z}\text{.}$$
• True.
• False.
#### 11.
True or false: $$f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}, f(x, y)=(y, x)$$ is one-to-one.
• True.
• False.
#### 12.
True or false: $$f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}, f(x, y)=(y, x)$$ is onto $$\mathbb{Z}\times\mathbb{Z}\text{.}$$
• True.
• False.
### ExercisesExercises
#### 1.
All but two of the following are correct ways to express the fact that a function $$f$$ is onto. Find the two that are incorrect.
1. $$f$$ is onto if and only if every element in its codomain is the image of some element in its domain.
2. $$f$$ is onto if and only if every element in its domain has a corresponding image in its codomain.
3. $$f$$ is onto if and only if $$\forall y\in Y, \exists x\in X$$ such that $$f(x)=y\text{.}$$
4. $$f$$ is onto if and only if $$\forall x\in X, \exists y\in Y$$ such that $$f(x)=y\text{.}$$
5. $$f$$ is onto if and only if the range of $$f$$ is the same as the codomain of $$f\text{.}$$
#### 2.
Let $$X=\{1, 5, 9\}$$ and $$Y=\{3, 4, 7\}\text{.}$$
1. Define $$f:X\rightarrow Y$$ by specifying that $$f(1)=4, f(5)=7, f(9)=4\text{.}$$
Is $$f$$ one-to-one? Is $$f$$ onto? Explain your answers.
2. Define $$g:X\rightarrow Y$$ by specifying that $$g(1)=7, g(5)=3, g(9)=4\text{.}$$
Is $$g$$ one-to-one? Is $$g$$ onto? Explain your answers.
#### 3.
Let $$X=\{1, 2, 3\}\text{,}$$ $$Y=\{1, 2, 3, 4\}$$ and $$Z=\{1, 2\}\text{.}$$
1. Define a function $$f:X\rightarrow Y$$ that is one-to-one but not onto.
2. Define a function $$g:X\rightarrow Z$$ that is onto but not one-to-one.
3. Define a function $$h:X\rightarrow X$$ that is neither one-to-one nor onto.
4. Define a function $$k:X\rightarrow X$$ that is one-to-one and onto, but is not the identity function.
#### 4.
Define $$f:\mathbb{Z}\rightarrow\mathbb{Z}$$ by $$f(n)=4n-5\text{.}$$
1. Prove or disprove $$f$$ is one-to-one.
2. Prove or disprove $$f$$ is onto $$\mathbb{Z}\text{.}$$
#### 5.
Define $$g:\mathbb{R}\rightarrow\mathbb{R}$$ by $$g(x)=4x-5\text{.}$$
1. Prove or disprove $$g$$ is one-to-one.
2. Prove or disprove $$g$$ is onto $$\mathbb{R}\text{.}$$
#### 6.
Define $$F:{\cal P}(\{a, b, c\})\rightarrow\mathbb{Z}$$ by
$$F(A)=$$ the number of elements in $$A\text{.}$$
1. Prove or disprove $$F$$ is one-to-one.
2. Prove or disprove $$F$$ is onto $$\mathbb{Z}\text{.}$$
#### 7.
Define $$G:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\times\mathbb{R}$$ by
$$G(x, y)=(2y, -x)\text{.}$$
1. Prove or disprove $$G$$ is one-to-one.
2. Prove or disprove $$G$$ is onto $$\mathbb{R}\times\mathbb{R}\text{.}$$
#### 8.
Define $$H:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\times\mathbb{R}$$ by
$$H(x, y)=(x+1, 2-y)\text{.}$$
1. Prove or disprove $$H$$ is one-to-one.
2. Prove or disprove $$H$$ is onto $$\mathbb{R}\times\mathbb{R}\text{.}$$ |
# NCERT Solutions for Class 10 Maths Chapter 10 Circles
Written by Team Trustudies
Updated at 2021-05-07
## NCERT solutions for class 10 Maths Chapter 10 Circles Exercise - 10.1
Q1 ) How many tangents can a circle have ?
NCERT Solutions for Class 10 Maths Chapter 10 Circles
A circle can have an infinite number of tangents.
Q2 ) Fill in the blanks :
(i) A tangent to a circle intersects it in _________ point(s).
(ii) A line intersecting a circle in two points is called a _______.
(iii) A circle can have _______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ______.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
(i) exactly one
(ii) secant
(iii) two
(iv) point of contact
Q3 ) A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length of PQ is :
(A) $12$ cm
(B) $13$ cm
(C) $8.5$ cm
(D) $\sqrt{119}$ cm
NCERT Solutions for Class 10 Maths Chapter 10 Circles
cm
So, (D) option is correct
Q4 ) Draw a circle and two lines parallel to a given line such that one is tangent and the other, a secant to the circle.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
## NCERT solutions for class 10 Maths Chapter 10 Circles Exercise - 10.2
Q1 ) From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) $7$ cm
(B) $12$ cm
(C) $15$ cm
(D) $24.5$ cm
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Since QT is a tangent to the circle at T and OT is radius,
It is given that OQ = 25 cm and QT = 24 cm,
$?$ By Pythagoras theorem, we have
$?$ radius of the circle is 7 cm
$?$ Option (A) is correct.
Q2 ) In figure, if TP and TQ are the two tangents to a circle with centre O so that , then is equal to
(A) $60°$
(B) $70°$
(C) $80°$
(D) $90°$
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Since TP and TQ are tangents to a circle with centre O so that ,
$?$ and
and
In the quadrilateral TPOQ, we have
[ \therefore Sum of all angles in a quadrilateral = 360° ]
$?$ Option (B) is correct.
Q3 ) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then is equal to
(A) $50°$
(B) $60°$
(C) $70°$
(D) $80°$
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Since PA and PB are tangents to a circle with centre O,
, and
and
In the quadrilateral PAOB, we have
In and $OBP$, we have
[Common]
[Each = 90°]
$?$ ( By SAS Criterion)
[C.P.C.T.]
$?$
? The (A) is the correct option.
Q4 ) Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Let PQ be a diameter of the given circle with centre O.
Let AB and CD be the tangents drawn to the circle at the end points of the diameter PQ respectively.
$?$ Tangent at a point to a circle is perpendicular to the radius through the point,
$?$ and
[$?$ and are alternate angles]
Hence, the tangents drawn at the ends of a diameter of a circle are parallel.
Q5 ) Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Let AB be the tangent drawn at the point P on the circle with O.
If possible, let PQ be perpendicular to AB, not passing through O. Join OP.
$?$ Tangent at a point to a circle is perpendicular to the radius through the point,
$?$
(Construction)
, which is not possible.
$?$ It contradicts our supposition.
Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Q6 ) The Length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Since tangent to a circle is perpendicular to the radius through the point of contact,
$?$
In , we have
$?$ Radius of the circle is 3 cm.
Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Let O be the common centre of two concentric circles, and let AB be a chord of the larger circle touching the smaller circle at P. Join OP.
? OP is the radius of the smaller circle and AB is tangent to this circle at P,
? .
We know that, the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.
So, and .
In , we have
Now, ???? [? ]
? The length of the chord of the larger circle which touches the smaller circle is 8 cm.
Q8 ) A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
? Lengths of two tangents drawn from an external point of circle are equal,
, , , and
Hence proved
Q9 ) In figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that .
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Since tangents drawn from an external point to a circle are equal,
$?$ .
Thus in and , we have
[Common]
[ By SSS-criterion of congruence]
Similarly, we can prove that
Now,
[Sum of the interior angles on the same side of transversal is 180°]
[$?$ ]
Q10 ) Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Let PA and PB be two tangents drawn from an external point P to a circle with centre O. We have to prove that-
In right and $OBP$, we have
[Tangents drawn from an external point are equal]
[Common]
[by SSS - criterion of congruence]
and,
and,
But,
[ is right triangle]
Hence proved that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Q11 ) Prove that the parallelogram circumscribing a circle is a rhombus.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Let ABCD be a parallelogram such that its sides touch a circle with centre O.
We know that the tangents to a circle from an exterior point are equal in length.
, , , and
[$?$ ABCD is a parallelogram ]
and ]
$?$
$?$ ABCD is a rhombus.
Q12 ) A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Let us join AO, OC, and OB.
It is given that BD = 8cm, CD = 6 cm.
$?$ Lengths of two tangents drawn from an external point of circle are equal.
$?$
cm.
Then, the sides of the triangle are 14 cm, (x+6) cm and (x+8) cm.
,
and,
$?$ Area of
Also, Area of
Squaring, we get
But x cannot be negative,
Thus, AB = x + 8 = 7 + 8 = 15 cm and AC = x + 6 = 6 + 7 = 13 cm.
Hence, the sides AB and AC are 15 cm and 13 cm respectively.
Q13 ) Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Let a circle with centre O touch the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.
We have to prove that-
and,
Join OP, OQ, OR and OS.
$?$ The two tangents drawn from an external point to a circle subtend equal angles at the centre.
$?$
Now, [$?$ Sum of all the ? s around a point is 360°]
and,
and
Now since $?2+?3=?AOB$ ,
and |
This is an interesting problem I bumped into the other day. What’s so interesting about it is that, while it requires only basic knowledge of probability to solve, it is far from trivial and the answer may be quite surprising! Here’s the problem:
A line of 100 passengers is waiting to board a 100-seat airplane. However, the first passenger in line forgot his designated seat and will sit in a random seat. The following passengers will sit in their assigned seats unless their seat is already taken, in which case they will sit in a random available seat.
What is the probability the 100th passenger will manage to sit in her designated seat?
Tricky, huh? Try to work out the answer to that in your head and then come back here later to find out if you got it right!
Okay, let’s try and solve this now. For starters, let’s investigate what happens in a much smaller plane, of only 2 seats, and with only 2 passengers. We’ll use the notation $n = 2$ for this case, meaning the number of both seats and passengers is 2. Now we’ll introduce the notation $P_n$ which is the probability the $n$-th passenger will find her seat available. So right now let’s work on finding out $P_2$, the solution to the 2-seat problem.
Turns out, $P_2$ is really easy: either the first passenger sat in his assigned seat, in which case the second passenger will find her seat available; or he didn’t, in which case the second passenger’s seat will be occupied. We know from basic prob that this is basically a coin toss, with 50% chance for each. So our answer is that passenger 2 has 50% chance to find her seat available. Using our notation:
Now let’s work with a 3-seat, 3-passenger plane. The first passenger can sit in any of three seats, at random: his own, passenger 2’s, and passenger 3’s. The chance for each is exactly one third.
In the first case, we know that passenger 3 will find her seat unoccupied, and in the third case she won’t. Let’s introduce some notation to represent this: $P_{n|x}$, meaning the probability that the $n$-th passenger will find her seat available given that the first passenger sat in passenger $x$’s seat. So, for the 3-seat case we found out that:
What about $P_{3|2}$? Let’s think about it. If the first passenger sits in passenger 2’s seat, now passenger 2 will have to sit in a random seat: either passenger 1’s or passenger 3’s. Now, haven’t we seen that one before? It looks exactly like the 2-seat problem, in which passenger 3 will have a 50% chance to find her seat available.
This is not a coincidence: when passenger 1 sat in passenger 2’s seat, we can say that passenger 2’s assigned seat is now passenger 1’s, and solve this exactly like the 2-seat problem. In notation:
Now, remembering that the chance of passenger 1 to sit in any given seat was exactly one third, so it follows that:
Substituting the probabilities we already calculated:
Hmmmmmmm… so again the probability is 50%. Interesting. Let’s investigate the 4-seat problem to see if a pattern emerges, then. Starting with the basics:
Just like with 3 seats, we know the trivial cases. If passenger 1 sits in his assigned seat, passenger 4 will manage to do so; and if passenger 1 sits in passenger 4’s seat right away, then she’ll never manage to do it:
For the other two cases, we can do exactly like we did before, and “downgrade” our 4-seat problem to a smaller problem. If passenger 1 sits in passenger 2’s seat, we can say that passenger 2’s assigned seat is now passenger 1’s, and treat it like a 3-seat problem!
And in case he sits in passenger 3’s seat, then we can reassign passenger 3 to passenger 1’s seat and treat it like a 2-seat problem:
Putting it all together:
Huh! 50% again! And, if we think back to how we solved $P_{3|2}$, $P_{4|2}$ and $P_{4|3}$, a pattern did emerge: if the first passenger takes a seat that is neither his nor the last passenger’s, the passenger who lost his seat can swap assignment with passenger 1 and we can treat this like a smaller problem. Let’s try to put that into notation.
Considering we’re dealing with a $k$-seat problem, and passenger 1 just sat in passenger $x$’s seat, then we can say that they swap seat assignments and we’ll treat this as a problem of size $(k - x + 1)$:
Putting it all together to solve $P_k$:
Now if we consider that all problems smaller than $k$ had probability of one half:
Then we can solve $P_k$ by substituting all the smaller problem probabilities by one half, as follows:
Considering we already know that $P_2$ is one half, we can solve $P_k$ for any $k > 2$, and the result will always be one half. Thus, we have proven by induction that:
That is the surprising (at least for me) part of the problem: no matter how many seats (either 2, 200 or 2 million), the answer always will be 50%.
Evidently, this means that $P_{100}$ of our original problem will be 50%, the same as with any other $n$, so we can state the following answer:
The probability that the 100th passenger will find her seat free is 50%. |
## Lesson 11: Volume of Solids (I)
May 31, 2012
Vocabulary on Volume of Solids
Volume Worksheet
Volume Word Problems
I Have to Know by the End of this Lesson
### 1.1. Volume units
Volume is length by length by length, so the basic unit of volume is a cube with edge length one metre. Its volume is 1 metre × 1 metre × 1 metre, which is written m3 (cubic metre).
The basic unit of volume is the cubic metre, which is written m3.
Multiples and submultiples of cubic metre are the following:
unit it is the volume of a it equals symbol multiples cubic kilometre cube whose edges measure one kilometre 1000000000 m3 km3 cubic hectometre cube whose edges measure one hectometre 1000000 m3 hm3 cubic decametre cube whose edges measure one decametre 1000 m3 dam3 base unit cubic metre cube whose edges measure one metre m3 submultiples cubic decimetre cube whose edges measure one decimetre 0.001 m3 dm3 cubic centimetre cube whose edges measure one centimetre 0.000001 m3 cm3 cubic millimetre cube whose edges measure one millimetre 0.000000001 m3 mm3
You have to bear in mind that volume units are cubes and three-dimensional. The conversion of a unit into another one is done by dividing or multiplying both the length, the height and the width, that is dividing and multiplying by ten three times; in other words, dividing and multiplying by 1000.
The volume units go up by a factor of 1000 at the time.
Each jump to a smaller unit is equivalent to multiply by 1000, you have to move the decimal point three places to the right. Each jump to a larger unit is equivalent to divide by 1000, you have to move the decimal point three places to the left. Here you are a sketch:
Now you can practice on the following links:
1 – Volume unit conversion. (with hints)
### 1.2 Volume of an object
Volume of an object is the amount of space it occupies.
We will begin by a basic solid, the cube.
### 2.1 Volume and capacity
The capacity of a container is known as the volume of the liquid or gas that it can hold.
Capacity and volume have equivalent meanings. Establishing a special unit to measure volume isn’t necessary, so using the cubic metre would be enough, but for practical use the litre was established as a unit. If you pour one litre in a cube with an edge of 1 dm it will fit in the cube exactly.
The litre is identical to the cubic decimetre (dm³), although it wasn’t always so . Recall
The litre is the capacity of a cubic decimetre.
We know 1 dm is 0·1 m or 10 cm, so 1 dm³ is 10×10×10 cm:
This, of course, means that there are 1,000 cm³ in a litre, or that 1 cm³ is equal to 1 mL.
The following table shows the equivalence between volume units and capacity units.
Volume units m3 dm3 cm3 Capacity units kl hl dal l dl cl ml
2.2. Volume, mass and capacity
A recipient contains a litre of pure water, which occupies 1 dm3. We weight it and it weights 1 kilogram exactly.
One kilogram is the weight of 1 dm3 of pure water.
If we weigh a container with 1 ml of pure water with, which occupies 1 cm3, it weights 1 gram.
One gram is the weight of 1 cm3 of water
The following table shows the equivalence among volume units, capacity units and mass units for pure water.
Volume units m3 dm3 cm3 Capacity units kl hl dal l dl cl ml Mass units t q mag kg dag hg g
1 l =1 dm3 = 1 kg of pure water
NOTE: If we have other substance different from pure water one litre doesn’t weight one kilo.
In the following video you can find a long explanations about these topics:
## Lesson 10: Solids. Surface Areas. An Extra Question with a Prize (IV)
May 23, 2012
### Are you able to prove it?
Think of the net of a cone.
If the base of the cone has radius r and the cone has generatrix g , I told you the lateral area (the area of the circle sector) was πrg
Could you prove it?
There is an extra 0.5 in this lesson if you give the right proof. Anyway your interest and effort will be prized in the global mark of this term.
## Lesson 10: Solids. Surface Areas. Revolution solids (III)
May 22, 2012
### 5. Solids of revolution
A solid of revolution is a solid figure obtained by rotating a flat figure around a line that lies on the same plane. The axis of revolution is the line about which the revolution takes place.
### 5.1. Cylinder
A cylinder is a solid of revolution generated by a rectangle that rotates around one of its sides.
Elements of a cylinder
Surface area of a cylinder
The net of a cylinder is formed by:
• A rectangle whose length is the length of the circumference of the bases and its width is the height of the cylinder.
• Two equal circles that are its bases.
The surface area of a cylinder can be calculated from its net. So it will be the sum of the lateral area and the areas of the two equal bases:
• Lateral area AL: It is the area of the rectangle whose length is the length of the circumference of the bases with radius r, 2πr and its width is the height of the cylinder,h.
AL = 2πr · h
• Area of the bases AB: This is the sum of the areas of the bases with radius .
AB = πr2
The surface area of a cylinder is:
AT = AL + 2·AB = 2πr · h + 2· πr2 = 2πr (h + r)
### 5.2. Cone
The cone is a solid of revolution generated by a right triangle that rotates around one of its legs.
Elements of a cone
Calculation of the generatrix of a cone
By the Pythagorean Theorem, the slant height or generatrix (or slant height, s) of the cone is equal to:
Surface area of a cone
The net of a cone is formed by:
• Circular sector of length2πr · h, being r the radius of the base, and radius g the generatrix of the cone.
• A circle that is the base
As in other cases we work out the surface area from its net. So it will be the sum of the lateral area and the area of the base:
• Lateral area AL: It is the area of the circular sector whose length is 2πr and its radius is,g.
AL = πr · g
• Area of the base AB: This is the area of the base with radius .
AB = πr2
The area of a cone is:
AT = AL + AB = πr · g + · πr2 = πr (g + r)
You can get the net of a cone and a cylinder: http://www.senteacher.org/wk/3dshape.php
### 5.3. Sphere
A spherical surface is the surface generated by rotating a circle about its diameter.
A sphere is the region of the space that is inside a spherical surface.
A sphere can also be seen as the region of the space obtained by rotating a semicircle around its diameter.
Elements of a sphere
The centre is the interior point that is equidistant to all points on the surface of the sphere.
Radius is the distance from the centre to any point on the surface of the sphere, .
The diameter is the distance from one point through the centre to another point . A diameter is twice the radius.
Surface area of a sphere
The surface area of sphere of radius is :
A = 4 πr2
In the following links you will find exercises about all the 3-D shapes we studied:
Finally, the following video will show you how to find the areas of a cylinder, a cone and a sphere. Improve your listening!
## Lesson 10: Solids. Surface Areas. Prisms and Pyramids (II)
May 22, 2012
### 3. Prisms
Prisms are polyhedrons that have two parallel, equally sized faces called bases and their lateral faces are parallelograms.
prism is a 3D shape which has a constant cross-section – both ends of the solid are the same shape and anywhere you cut parallel to these ends gives you the same shape too.
### 3.1.Types of prisms
To name a prism we use to refer to the base polygon. This way we can name them by their bases:
Right prisms are those that have lateral faces rectangles or squares, i.e. basic edges are perpendicular to the lateral edges
Oblique prisms are those whose lateral faces are rhomboids.
Oblique prism
Regular prisms are right prism whose bases are regular polygons.
Irregular prisms are prisms whose bases are irregular polygons.
Parallelepipeds are prisms whose bases are parallelograms.
Cuboids are right parallelepipeds, i.e. their faces are rectangular. Cuboids are very common in daily life. They are rectangular prisms.
3.2. Elements of a prism
3.3. Surface area of a prism
The net of a right prism is formed by:
• A rectangle of length all the sides of the base, this is the perimeter of the base, and width the height of the prism.
• Two equal polygons, the ones that are the bases.
It is easy to deduce the surface area of a prism from its net. So it will be the sum of the lateral area and the areas of the two equal basis:
• Lateral area AL: It is the sum of the areas of the lateral faces but, if we look at the net, the lateral surface is a rectangle of length the perimeter of the base PB, and width the height of the prism h.
• Area of the bases AB: This is the sum of the areas of the bases.
The surface area of a right prism is:
AT = AL + 2·AB = PB · h + 2·AB
Now, you can see a video about how to find the surface area of a prism.
### 4. Pyramids
A pyramid is a polyhedron whose base can be any polygon and whose lateral faces are triangles with a common vertex (apex of the pyramid).
pyramid has sloping sides that meet at a point.
4.1. Elements of a pyramid
4.2. Types of pyramids
To name a pyramid we use to refer to its base polygon. This way we can name them by their bases:
A right pyramid has isosceles triangles as its lateral faces and its apex lies directly above the midpoint of the base.
An oblique pyramid does not have all isosceles triangles as its sides.
Oblique pyramid
A regular pyramid is right pyramid whose base is a regular polygon and its lateral faces are equally sized. In other case is named irregular pyramid.
4.3 Surface area of a regular pyramid
The net of a regular pyramid is formed by:
• As many isosceles triangles as sides the base has.
• The base polygon
Let the number of sides of the base of the regular polygon that is the base of the pyramid.
We deduce the surface area of a prism from its net. So it will be the sum of the lateral area and the area of the base:
• Lateral area AL: It is the sum of the areas of the lateral faces but, if we look at the net, the lateral face is n isosceles triangles whose base is the side of the polygon, b, of the base and height the apothem of the pyramid. If we add them up, we have:
AL = n · (b · a) = (PB · a)/2
where PB the perimeter of the base, and a the apothem of the pyramid.
• Area of the bases AB: As the base is a regular polygon:
AB = (PB · a’)/2
where a’ the apothem of the base.
The surface area of a regular pyramid is:
AT = AL + AB = (PB · a)/2 + PB · a’)/2
Now, you can see a video about how to find the surface area of a pyramid, in this case cuadrangular).
The following links will provide you enough exercises
## Lesson 10: Solids.Surface Areas (I)
May 16, 2012
Vocabulary on Solids
Lesson 10: Solids. Surface Areas (Notes)
Practice Problems on Classifying Solids
Practice problems on Surface Area of Solids
I Have to Know by the End of this Lesson
Three-D shapes have 3-dimensions- length, width and depth. We are going to study some of them: polyhedrons and revolution solids.
### 1. Polyhedrons
A polyhedron is a three-dimensional region of the space bounded by polygons.
Some solids have curved surfaces or a mix of curved and flat surfaces (so they aren’t polyhedrons).
If you click on the links you will learn more about these 3-D shapes in the web where it is taken from the table below, www.mathisfun.com. You can also find questions with answers.
Polyhedrons :
(they must have flat faces)
Platonic Solids Prisms Pyramids
Non-Polyhedra:
(if any surface is not flat)
Sphere Torus Cylinder Cone
Elements of a polyhedron
Faces: These polygons that limit the polyhedrons.
Edges: Line segments where two faces of a polyhedron meet. They are sides of the faces.
Vertices of the polyhedron: Points at which three or more polyhedron edges of a polyhedron meet.
Diagonals of a polyhedron: Segments, joining two vertices, which are not placed on the same face, are called diagonals of polyhedron. The tetrahedron has no diagonals.
Dihedral angle (also called the face angle) is the internal angle at which two adjacent faces meet. All dihedral angles between the edges are ≤ 180º.
Polyhedral angle: is the portion of space limited by tree or more faces which meet at a vertex. All polyhedral angles between the edges are ≤ 360º .
Net: It is an arrangement of edge-joined polygons in the plane which can be folded (along edges) to become the faces of the polyhedron. There are several possibilities for a net of a polyhedron.
Tetrahedrom Net for a tetrahedron
You can find the net of the main polyhedron and their different possibilities on
http://gwydir.demon.co.uk/jo/solid/index.htm
### 2.1. Types of polyhedrons
A convex polyhedron is defined as follows: no line segment joining two of its points contains a point belonging to its exterior. There are many examples known by you: the cube, prisms, pyramids….
A concave polyhedron, on the other hand, will have line segments that join two of its points with all but the two points lying in its exterior.
Below is an example of a concave polyhedron.
The study of polyhedrons was a popular study item in Greek geometry even before the time of Plato (427 – 347 B.C.E.) In 1640, Rene Descartes, a French philosopher, mathematician, and scientist, observed the following formula. In 1752, Leonhard Euler, a Swiss mathematician, rediscovered and used it.
In a convex polyhedron:
If F = number of faces , V = number of vertices and E = number of edges, then
F + V = E +2
This formula is named Euler’s Formula.
All the convex polyhedrons verify this formula. There are some concave polyhedrons that verify it. However, there are concave polyhedrons that don’t verify this formula.
### 2.2. Regular polyhedrons
A regular polyhedron is a polyhedron where:
• each face is the same regular polygon
• the same number of faces (polygons) meet at each vertex (corner)
They are also called Platonic solids.
There are five Platonic solids:
1. Tetrahedron, which has three equilateral triangles at each corner. 2. Cube, which has three squares at each corner. 3. Octahedron, which has four equilateral triangles at each corner. 4. dodecahedron, which has three regular pentagons at each corner. 5. Icosahedron, which has five equilateral triangles at each corner.
Exercise: Fill in this table and check they satisfy Euler’s formula. Why?
``` faces edges vertices
tetrahedron ___ ___ ___
cube ___ ___ ___
octahedron ___ ___ ___
dodecahedron ___ ___ ___
icosahedron ___ ___ ___``` |
Find the solutions in the app
##### Sections
Exercise name Free?
###### Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Monitoring Progress 6
Monitoring Progress 7
Monitoring Progress 8
Monitoring Progress 9
Monitoring Progress 10
Monitoring Progress 11
Monitoring Progress 12
Monitoring Progress 13
Monitoring Progress 14
Monitoring Progress 15
Monitoring Progress 16
Monitoring Progress 17
Monitoring Progress 18
Monitoring Progress 19
Monitoring Progress 20
Monitoring Progress 21
Monitoring Progress 22
Monitoring Progress 23
Monitoring Progress 24
Monitoring Progress 25
Monitoring Progress 26
Monitoring Progress 27
Monitoring Progress 28
###### Exercises
Exercise name Free?
Exercises 1 When we have a radical expression in the denominator of a fraction whose radicand is not a perfect square — for example, 21 — we can multiply by the appropriate form of 1; in this case, 22=1. This way, we won't change the value of the fraction and we will get rid of the radical term in the denominator. 21ba=b⋅2a⋅22⋅21⋅2a⋅a=a2⋅21⋅2Multiply121a=a2 Notice that the denominator changed from 2 to 1. That is, it changed from an irrational number to a rational number. That is why we call this process rationalizing the denominator. With this in mind, we can complete the exercise sentence.The process of eliminating a radical from the denominator of a radical expression is called rationalizing the denominator.
Exercises 2 Let's start by reviewing what the conjugate of a binomial is. The binomials ab+cd and ab−cd where a, b, c, and d are rational numbers are called conjugates. Let's take a look at the expression given in the exercise. 6+4⇔16+41 As we can see, this expression is in the format mentioned above, ab+cd, so we can write its conjugate by changing the positive sign to a negative one. Original binomialConjugate binomial6+46−4
Exercises 3 The exercise asks us to tell if the expressions given are equivalent. 92x=?312x For this it will be useful to remember the property for dividing square roots.The square root of a quotient equals the quotient of the square roots of the numerator and denominator. ba=ba Now, let's see if we can use this to start from one of the given expressions and rewrite it to obtain the other one. 92xba=ba92xCalculate root32xba=b1⋅a312x We have found that 92x=312x. Therefore, the expressions given are equivalent.
Exercises 4 Let's start by reviewing what a like radical is. Like radicals are those which have the same index and radicals. Let's now take a look at the radicals given in the exercise. -316 6361 613 -3361 Notice that all the radical expressions involve square roots having 3 as radicand except for 316. This means that all the expressions are like radicals, except for this one. Therefore, this is the odd one out.
Exercises 5 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 19 to ensure it's not evenly divisible by any perfect squares by noticing that it is prime. 19=1⋅19 Since it contains no repeating roots, we cannot form any perfect squares. Therefore, the first rule is satisfied. The second and third rules are about fractions, but our expression does not contain any fractions. Therefore, all three rules are satisfied which proves that 19 is in simplest form because it cannot be simplified further.
Exercises 6 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 7 to ensure it's not evenly divisible by any perfect squares by noticing that it is prime. 7=1⋅7 Since it contains no repeating roots, we cannot form any perfect squares. Therefore, the first rule is satisfied. The second and third rules are about fractions, and our expression has a radical in the denominator of a fraction. Therefore, all three rules are not satisfied which proves that 71 is not in simplest form because it could be simplified further.
Exercises 7 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 48 to ensure it's not evenly divisible by any perfect squares by showing its prime factorization. 48=2⋅2⋅2⋅2⋅3=22⋅22⋅3 Do you see how there are repeating primes allowing us to form perfect squares? This means that 48 is divisible by some perfect square which tells us that this expression could be simplified further and that 48 is not in simplest form.
Exercises 8 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 34 to ensure it's not evenly divisible by any perfect squares by showing its prime factorization. 34=2⋅17 Since there are no repeating primes, the expression satisfies the first rule. The second and third rules are about fractions, but our expression does not contain any fractions. Therefore, all three rules are satisfied which proves that 34 is in simplest form because it cannot be simplified further.
Exercises 9 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 2 to ensure it's not evenly divisible by any perfect squares by showing its prime factorization. 2=2⋅1 Since there are no repeating primes, the expression satisfies the first rule. The second rule is about roots being in the denominator. Since our denominator is 2, the rule is not met. Therefore, 25 is not in simplest form because it can be simplified further.
Exercises 10 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 2 to ensure it's not evenly divisible by any perfect squares by showing its prime factorization. 10=2⋅5 Since there are no repeating primes, the expression satisfies the first rule. The second and third rules are about fractions. Our expression does does not contain any fractions within the radicand or any radicals within the denominator. Therefore, all rules are satisfied which proves that 4310 is in simplest form because it cannot be simplified further.
Exercises 11 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a cube root, for the first rule, n=3. We can check 2 to ensure it's not evenly divisible by any perfect cubes by showing its prime factorization. 2=2⋅1 Since there are no repeating primes, the expression satisfies the first rule. The second and third rules are about fractions, our expression does does not contain any fractions within the radicand. However, it does contain a radical in the denominator. Therefore, all rules are not satisfied which proves that 2+321 is not in simplest form because it can be simplified further.
Exercises 12 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a cube root, for the first rule, n=3. We can check 54 to ensure it's not evenly divisible by any perfect cubes by showing its prime factorization. 54=2⋅3⋅3⋅3=2⋅33 Do you see how there are repeating primes allowing us to form perfect cubes? This means the expression is divisible by some perfect cube which tells us the expression does not satisfy the first rule. Therefore, all rules are not satisfied which proves that 6−354 is not in simplest form because it can be simplified further.
Exercises 13 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 20Split into factors4⋅5a⋅b=a⋅b4⋅5Calculate root25
Exercises 14 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 32Split into factors16⋅2a⋅b=a⋅b16⋅2Calculate root42
Exercises 15 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 128Split into factors64⋅2a⋅b=a⋅b64⋅2Calculate root82
Exercises 16 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. -72Split into factors-36⋅2a⋅b=a⋅b-36⋅2Calculate root-62
Exercises 17 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 125bSplit into factors25⋅5ba⋅b=a⋅b25⋅5bCalculate root55b
Exercises 18 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 4x2Split into factors4⋅x2a⋅b=a⋅b4⋅x2Calculate root2x
Exercises 19 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. -81m3Split into factors-81⋅m2⋅ma⋅b=a⋅b-81⋅m2⋅mCalculate root-9mm
Exercises 20 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 48n5Split into factors16⋅3⋅n4⋅nCommutative Property of Multiplication16⋅n4⋅3⋅na⋅b=a⋅b16⋅n4⋅3nWrite as a power16⋅(n2)2⋅3nCalculate root4n23n
Exercises 21 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 494ba=ba494Calculate root72
Exercises 22 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. -817ba=ba-817Calculate root-97
Exercises 23 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. -6423ba=ba-6423Calculate root-823
Exercises 24 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 12165ba=ba12165Calculate root1165
Exercises 25 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 49a3ba=ba49a3Split into factors49a2⋅aa⋅b=a⋅b49a2⋅aCalculate root7aa
Exercises 26 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. k2144ba=bak2144Calculate rootk12
Exercises 27 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 4x2100ba=ba4x2100Calculate root2x10ba=b/2a/2x5
Exercises 28 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 3625v2ba=ba3625v2a⋅b=a⋅b3625⋅v2Calculate root65v
Exercises 29 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 316Split into factors38⋅23a⋅b=3a⋅3b38⋅32Calculate root232
Exercises 30 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3-108Split into factors3-27⋅43a⋅b=3a⋅3b3-27⋅34Calculate root-334
Exercises 31 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3-64x5Split into factors3-64⋅x3⋅x23a⋅b=3a⋅3b3-64⋅3x3⋅3x2Calculate root-4x3x2
Exercises 32 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. -3343n2Split into factors-3343⋅n23a⋅b=3a⋅3b-3343⋅3n2Calculate root-73n2
Exercises 33 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3-1256c3ba=3b3a3-12536cCalculate root-536cPut minus sign in front of fraction-536c
Exercises 34 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3278h43ba=3b3a32738h4Split into factors32738⋅h3⋅h3a⋅b=3a⋅3b32738⋅3h3⋅3hCalculate root32h3h
Exercises 35 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. -31000x381y23ba=3b3a-31000x3381y2Split into factors-31000⋅x3327⋅3⋅y23a⋅b=3a⋅3b-31000⋅3x3327⋅33⋅y2Calculate root-10x333y2
Exercises 36 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3-64a3b6213ba=3b3a3-64a3b6321Split into factors3-64⋅a3⋅b63213a⋅b=3a⋅3b3-64⋅3a3⋅3b6321Calculate root-4ab2321Put minus sign in front of fraction-4ab2321
Exercises 37 Let's simplify the expression and compare our answer to given work. 72Split into factors4⋅9⋅2a⋅b=a⋅b4⋅9⋅2Calculate root2⋅3⋅2Multiply62 Comparing our work, we can see that they didn't factor completely to find all the perfect squares.
Exercises 38 Let's simplify the expression and compare our answer to given work. 3125128y3ba=ba31253128y3Split into factors3125364⋅2⋅y3a⋅b=a⋅b3125364⋅32⋅3y3Calculate root54⋅32⋅3y3nan=a{n}54y32 Comparing our work, we can see that the quotient property of radicals wasn't used correctly.
Exercises 39 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 64ba=b⋅6a⋅66⋅646a⋅b=a⋅b6⋅646a⋅a=a26246a2=a646 To rationalize the denominator, we have expanded the expression by a factor of 66.
Exercises 40 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 13z1ba=b⋅13za⋅13z13z⋅13z13za⋅b=a⋅b13z⋅13z13za⋅a=a2(13z)213za2=a13z13z To rationalize the denominator, we have expanded the expression by a factor of 13z13z.
Exercises 41 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a cube root, we need to multiply it by a cube root that will give us a perfect cube under the radical. 3x22ba=b⋅3xa⋅3x3x2⋅3x23x3a⋅3b=3a⋅b3x2⋅x23xa⋅am=a1+m3x323xnan=a{n}x23x To rationalize the denominator, we have expanded the expression by a factor of 3x3x.
Exercises 42 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a cube root, we need to multiply it by a cube root that will give us a perfect cube under the radical. 343mba=b⋅316a⋅31634⋅3163m3163a⋅3b=3a⋅b34⋅163m316Multiply3643m316Calculate root43m316 To rationalize the denominator, we have expanded the expression by a factor of 316316.
Exercises 43 To rationalize a fraction with a binomial denominator, we multiply the numerator and denominator of the fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the second term of the expression.BinomialConjugate a + ba − b a − ba + b In this case, because the expression is 5−82 the conjugate of the denominator is 5+8. Therefore, we have to multiply by 5+85+8.
Exercises 44 To rationalize a fraction with a binomial denominator, we multiply the numerator and denominator of the fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the second term of the expression.BinomialConjugate a + ba − b a − ba + b In this case, because the expression is 3+75 the conjugate of the denominator is 3−7. Therefore, we have to multiply by 3−73−7.
Exercises 45 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 22ba=b⋅2a⋅22⋅222a⋅b=a⋅b2⋅222a⋅a=a22222a2=a222ba=b/2a/22
Exercises 46 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 34ba=b⋅3a⋅33⋅343a⋅b=a⋅b3⋅343a⋅a=a23243a2=a343
Exercises 47 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 485ba=b⋅48a⋅4848⋅485⋅48a⋅b=a⋅b48⋅485⋅48a⋅a=a24825⋅48Multiply482240a2=a48240 We know that we have successfully rationalized the denominator because the radical has been eliminated. However, our fraction can still be simplified a bit further. 48240Split into factors4816⋅15a⋅b=a⋅b4816⋅15Calculate root48415ba=b/4a/41215
Exercises 48 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 524ba=b/4a/4131ba=ba131ba=b⋅13a⋅1313⋅1313a⋅b=a⋅b13⋅1313a⋅a=a213213a2=a1313
Exercises 49 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. a3ba=b⋅aa⋅aa⋅a3aa⋅b=a⋅ba⋅a3aa⋅a=a2a23aa2=aa3a
Exercises 50 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 2x1ba=b⋅2xa⋅2x2x⋅2x2xa⋅b=a⋅b2x⋅2x2xa⋅a=a2(2x)22xa2=a2x2x
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First half of Exercise 3.1 …
Question 12: Find the rationalising factors of the following:
(i) $\sqrt{108}$
$\sqrt{108} = \sqrt{2^2 \times 3^3} = 6 \sqrt{3}$.
We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt{3} =$ $3^{\frac{1}{2}}$ the rationalizing factor should be $3^{(1-\frac{1}{2})} = 3^{\frac{1}{2}} = \sqrt{3}$
(ii) $\sqrt[5]{486}$
$\sqrt[5]{486} = \sqrt[5]{2 \times 3^5} = 3 \sqrt[5]{2}$.
We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt[5]{2} =$ $2^{\frac{1}{5}}$ the rationalizing factor should be $2^{(1-\frac{1}{5})} = 2^{\frac{4}{5}} = \sqrt[5]{16}$
(iii) $\sqrt{2}+\sqrt{3}+\sqrt{5}$
We find that
$( \sqrt{2}+\sqrt{3}+\sqrt{5}) ( \sqrt{2}+\sqrt{3}-\sqrt{5}) = ( \sqrt{2}+\sqrt{3})^2 - (\sqrt{5})^2 = 2 + 3 + 2 \sqrt{6} - 5 = 2 \sqrt{6}$
Rationalizing factor of $\sqrt{6}$ is $\sqrt{6}$.
Hence $\sqrt{6}( \sqrt{2}+\sqrt{3}-\sqrt{5})$ is the rationalizing factor of $\sqrt{2}+\sqrt{3}+\sqrt{5}$
(iv) $\sqrt{3}+ \sqrt{8+2\sqrt{15}}$
We have $\sqrt{3}+ \sqrt{8+2\sqrt{15}} = \sqrt{3}+ \sqrt{5 + 3+2\sqrt{5 \times 3}} = \sqrt{3}+ \sqrt{(\sqrt{5} + \sqrt{3})^2}$
$= \sqrt{3} +\sqrt{5}+\sqrt{3} = 2 \sqrt{3}+ \sqrt{5}$
The conjugate of $(2 \sqrt{3}+ \sqrt{5})$ is $(2 \sqrt{3}- \sqrt{5})$
Therefore the rationalizing factor of $\sqrt{3}+ \sqrt{8+2\sqrt{15}}$ is $(2\sqrt{3}-\sqrt{5})$.
(v) $\sqrt[3]{5}$
We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt[3]{5} =$ $5^{\frac{1}{3}}$ the rationalizing factor should be $5^{(1-\frac{1}{3})} = 5^{\frac{2}{3}} = \sqrt[3]{25}$
Question 13: Rationalize the denominator and simplify
(i) $\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$
$=$ $\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$ $\times$ $\frac{6-4\sqrt{2}}{6-4\sqrt{2}}$
$=$ $\frac{36 + 32 -48\sqrt{2}}{36-32}$
$=$ $\frac{78-48\sqrt{2}}{4}$
$= 17 - 12\sqrt{2}$
(ii) $\frac{b^2}{\sqrt{a^2+b^2} +a}$
$=$ $\frac{b^2}{\sqrt{a^2+b^2} +a}$ $\times$ $\frac{\sqrt{a^2+b^2} -a}{\sqrt{a^2+b^2} -a}$
$=$ $\frac{b^2(\sqrt{a^2+b^2)}}{a^2+b^2-a^2}$
$=$ $\sqrt{a^2+b^2} - a$
(iii) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+ \sqrt{2}}$
$=$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+ \sqrt{2}}$ $\times$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}- \sqrt{2}}$
$=$ $\frac{3+2-2\sqrt{6}}{3-2}$
$=$ $5 - 2\sqrt{6}$
(iv) $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$
$=$ $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$ $\times$ $\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}$
$=$ $\frac{6\sqrt{30} - 15 + 24 - 2 \sqrt{30}}{45-24}$
$=$ $\frac{4\sqrt{3}+9}{21}$
(v) $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$
$=$ $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$ $\times$ $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}}$
$=$ $\frac{a+b+\sqrt{a^2-b^2} + \sqrt{a^2-b^2}+a-b}{a+b-a+b}$
$=$ $\frac{2(a+\sqrt{a^2-b^2})}{2b}$
$=$ $\frac{a+\sqrt{a^2-b^2}}{b}$
Question 14: Simplify
(i) $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$ $+$ $\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$
$=$ $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$ $\times$ $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$ $+$ $\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$ $\times$ $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=$ $\frac{18-6\sqrt{6} - 6\sqrt{6}+12}{18-12}$ $+$ $\frac{\sqrt{12}(\sqrt{3}+\sqrt{2})}{1}$
$=$ $\frac{30-12\sqrt{6}}{6}$ $+$ $\sqrt{12} (\sqrt{3}+\sqrt{2})$
$=$ $5 - 2\sqrt{6} + 6 + 2\sqrt{6}$
$=$ $11$
(ii) $\frac{1}{2+\sqrt{3}}$ $+$ $\frac{2}{\sqrt{5} - \sqrt{3}}$ $+$ $\frac{1}{2-\sqrt{5}}$
$=$ $\frac{1}{2+\sqrt{3}}$ $\times$ $\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $+$ $\frac{2}{\sqrt{5} - \sqrt{3}}$ $\times$ $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ $+$ $\frac{1}{2-\sqrt{5}}$ $\times$ $\frac{2+\sqrt{5}}{2+\sqrt{5}}$
$=$ $\frac{2-\sqrt{3}}{4-3}$ $+$ $\frac{2(\sqrt{5}+\sqrt{3})}{5-3}$ $+$ $\frac{2+\sqrt{5}}{-1}$
$= 2 -\sqrt{3}+\sqrt{5}+\sqrt{3}-2-\sqrt{5}$
$= 0$
(iii) $\frac{2}{\sqrt{5}+\sqrt{3}}$ $+$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $-$ $\frac{3} {\sqrt{5}+\sqrt{2}}$
$=$ $\frac{2}{\sqrt{5}+\sqrt{3}}$ $\times$ $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ $+$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $\times$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $+$ $\frac{3} {\sqrt{5}+\sqrt{2}}$ $\times$ $\frac{\sqrt{5}-\sqrt{2}} {\sqrt{5}-\sqrt{2}}$
$=$ $\frac{2(\sqrt{5}-\sqrt{3})}{2}$ $+$ $\frac{\sqrt{3}-\sqrt{2}}{1}$ $-$ $\frac{3(\sqrt{5}-\sqrt{2})}{3}$
$=$ $\sqrt{5} - \sqrt{3} + \sqrt{3} - \sqrt{2} - \sqrt{5} + \sqrt{2}$
$= 0$
(iv) $\frac{1}{2-\sqrt{3}}$ $-$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $+$ $\frac{5}{3-\sqrt{2}}$
$=$ $\frac{1}{2-\sqrt{3}}$ $\times$ $\frac{2+\sqrt{3}}{2+\sqrt{3}}$ $-$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $\times$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $+$ $\frac{5}{3-\sqrt{2}}$ $\times$ $\frac{3+\sqrt{2}}{3+\sqrt{2}}$
$=$ $\frac{2+\sqrt{3}}{4-3}$ $-$ $\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{5(\sqrt{3}+\sqrt{2})}{9-2}$
$=$ $2 + \sqrt{3}-\sqrt{3}+\sqrt{2}$ $+$ $\frac{15+5\sqrt{2}}{7}$
$=$ $\frac{14+7\sqrt{2}+15+5\sqrt{2}}{7}$
$=$ $\frac{29+12\sqrt{2}}{7}$
(v) $\frac{4\sqrt{3}}{2+\sqrt{3}}$ $-$ $\frac{30}{4\sqrt{3}-3\sqrt{2}}$ $-$ $\frac{3\sqrt{2}}{3+2\sqrt{2}}$
$=$ $\frac{4\sqrt{3}}{2+\sqrt{3}}$ $\times$ $\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $-$ $\frac{30}{4\sqrt{3}-3\sqrt{2}}$ $\times$ $\frac{4\sqrt{3}+3\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$ $-$ $\frac{3\sqrt{2}}{3+2\sqrt{2}}$ $\times$ $\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$
$=8\sqrt{3} - 12 - 4\sqrt{3}-3\sqrt{2}-9\sqrt{2}+12$
$= 4\sqrt{3}-12\sqrt{2}$
$= 4(\sqrt{3}-3\sqrt{2})$
Question 15: Determine rational numbers $a$ and $b$
(i) $\frac{3+\sqrt{2}}{3-\sqrt{2}}$ $=$ $a + b\sqrt{2}$
$\Rightarrow$ $\frac{3+\sqrt{2}}{3-\sqrt{2}}$ $\times$ $\frac{3+\sqrt{2}}{3+\sqrt{2}}$ $=$ $a + b\sqrt{2}$
$\Rightarrow$ $\frac{9+2+6\sqrt{2}}{9-2}$ $=$ $a + b\sqrt{2}$
$\Rightarrow$ $\frac{11}{7} + \frac{6}{11}$ $\sqrt{2}$ $=$ $a + b\sqrt{2}$
$\Rightarrow$ $a =$ $\frac{11}{7}$ and $b =$ $\frac{6}{11}$
(ii) $\frac{5+3\sqrt{3}}{7+4\sqrt{3}}$ $=$ $a + b\sqrt{3}$
$\Rightarrow$ $\frac{5+3\sqrt{2}}{7+3\sqrt{3}}$ $\times$ $\frac{7-4\sqrt{3}}{7-4\sqrt{3}}$ $=$ $a + b\sqrt{3}$
$\Rightarrow$ $\frac{35+21\sqrt{3}-20\sqrt{3}-36}{49-48}$ $=$ $a + b\sqrt{3}$
$\Rightarrow$ $-1+\sqrt{3}$ $=$ $a + b\sqrt{3}$
$\Rightarrow$ $a = -1$ and $b = 1$
(iii) $\frac{1+\sqrt{48}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{72} - \sqrt{108}+\sqrt{8}+2}$ $=$ $a + b\sqrt{3}$
Let’s first simplify
$\frac{1+\sqrt{48}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{72} - \sqrt{108}+\sqrt{8}+2}$
$=$ $\frac{1+\sqrt{2^4 \times 3}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{2^3 \times 3^2} - \sqrt{2^2 \times 3^3}+\sqrt{2^3}+2}$
$=$ $\frac{1+3\sqrt{3}}{5\sqrt{3} + 4\sqrt{2} - 6\sqrt{2} - 6\sqrt{3}+2\sqrt{2}+2}$
$=$ $\frac{1+4\sqrt{3}}{2-\sqrt{3}}$
$=$ $\frac{1+4\sqrt{3}}{2-\sqrt{3}}$ $\times$ $\frac{2+\sqrt{3}}{2+\sqrt{3}}$
$=$ $\frac{2+8\sqrt{3}+\sqrt{3}+12}{4-3}$
$= 14+9\sqrt{3}$
Now comparing, $14+9\sqrt{3} = a + b\sqrt{3}$
$\Rightarrow a = 14 \ and \ b = 9$
Question 16: If $x$ $=$ $\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $y$ $=$ $\frac{\sqrt{2}-1}{\sqrt{2}+1}$, find $x^2+xy+y^2$
$x^2+xy+y^2$
$=$ $(\frac{\sqrt{2}+1}{\sqrt{2}-1})^2$ $+$ $(\frac{\sqrt{2}+1}{\sqrt{2}-1})(\frac{\sqrt{2}-1}{\sqrt{2}+1})$ $+$ $(\frac{\sqrt{2}-1}{\sqrt{2}+1})^2$
$=$ $\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$ $+ 1 +$ $\frac{2+1-2\sqrt{2}}{2+1+2\sqrt{2}}$
$=$ $\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$ $+ 1 +$ $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$
$=$ $\frac{9+9+12\sqrt{2} + 9 + 8 - 12\sqrt{2}}{9-8}$ $+1$
$= 34+1 = 35$
Question 17: If $x$ $=$ $\frac{1}{3-2\sqrt{2}}$ and $y$ $=$ $\frac{1}{3+2\sqrt{2}}$, find $xy^2+x^2y$
$xy^2+x^2y$
$= ($ $\frac{1}{3-2\sqrt{2}})(\frac{1}{3+2\sqrt{2}})^2$ $+ ($ $\frac{1}{3-2\sqrt{2}})^2(\frac{1}{3+2\sqrt{2}})$
$=$ $\frac{1}{9-8}(\frac{1}{3+2\sqrt{2}})$ $+$ $\frac{1}{9-8}(\frac{1}{3-2\sqrt{2}})$
$=$ $\frac{1}{3+2\sqrt{2}}$ $+$ $\frac{1}{3-2\sqrt{2}}$
$=$ $\frac{3-2\sqrt{2}+3+2\sqrt{2}}{9-8}$
$= 6$
Question 18: If $x = 2 + \sqrt{3}$, find the value of $x^3 +$ $\frac{1}{x^3}$
$x = 2 + \sqrt{3}$
Therefore $\frac{1}{x}$ $=$ $\frac{1}{2 + \sqrt{3}}$ $\times$ $\frac{2 - \sqrt{3}}{2 - \sqrt{3}}$ $=$ $2-\sqrt{3}$
Hence $x +$ $\frac{1}{x}$ $= 2 + \sqrt{3} + 2-\sqrt{3} = 4$
$x^3 +$ $\frac{1}{x^3}$ $=$ $($ $x+$ $\frac{1}{x})^3$ $- 3(x+$ $\frac{1}{x}$ $) = 4^3 - 3 \times 4 = 52$
Question 19: If $x = 3 + \sqrt{8}$, find the value of $x^2 +$ $\frac{1}{x^2}$
$x = 3 + \sqrt{8}$
Therefore $\frac{1}{x}$ $=$ $\frac{1}{3 + \sqrt{8}}$ $\times$ $\frac{3 - \sqrt{8}}{3 - \sqrt{8}}$ $=$ $3-\sqrt{8}$
Hence $x +$ $\frac{1}{x}$ $= 3 + \sqrt{8} + 3-\sqrt{8} = 6$
$x^2 +$ $\frac{1}{x^2}$ $=$ $($ $x+$ $\frac{1}{x})^2$ $- 2$ $= 6^2 - 2 = 34$
Question 20: If $x =$ $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $y =$ $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ , find the value of $3x^2+4xy-3y^2$
$3x^2+4xy-3y^2$
$= 3($ $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}}$ $)^2 + 4 ($ $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}}$ $)($ $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ $) - 3 ($ $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ $)^2$
$= 3 ($ $\frac{7+\sqrt{2}}{7-\sqrt{2}}$ $) + 4 - 3 ($ $\frac{7-\sqrt{2}}{7+\sqrt{2}}$ $)$
$= 3 ($ $\frac{49+40+28\sqrt{10}-49-40+28\sqrt{10}}{49-40}$ $)$
$=$ $\frac{56\sqrt{10}+12}{3}$
Question 21: If $x =$ $\frac{\sqrt{3}+1}{2}$, find the value of $4x^3+2x^2-8x+7$
$4x^3+2x^2-8x+7$
$= 4 \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^3 + 2 \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^2 - 8\Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big] + 7$
$= \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^2 \Big[ ($ $\frac{4\sqrt{3}+4}{2}$ $) + 2 \Big] - 8\Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big] + 7$
$=$ $(\frac{3+1+2\sqrt{3}}{2})(\frac{4\sqrt{3}+4+4}{2})$ $- 8$ $(\frac{\sqrt{3}+1}{2})$ $+ 7$
$= (4+2\sqrt{3})$ $(\frac{\sqrt{3}+2}{2})$ $- 8$ $(\frac{\sqrt{3}+1}{2})$ $+ 7$
$=$ $\frac{4\sqrt{3}+6+8+4\sqrt{3}-8\sqrt{3}-8+14}{2}$
$= 10$
Question 22: Given $\sqrt{3}= 1.732, \sqrt{5}= 2.236, \sqrt{2}=1.4142, \sqrt{6}=2.4495$ and $\sqrt{10}=3.162$, find
(i) $\frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}$ $+$ $\frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}$
$=$ $\frac{\sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} + \sqrt{5} + \sqrt{3} -\sqrt{10} -\sqrt{6}}{5-3}$
$=$ $\frac{2\sqrt{5}-2\sqrt{6}}{2}$
$= \sqrt{5}-\sqrt{6}$
$= 2.236 - 2.4495 = -0.213$
(ii) $\frac{6}{\sqrt{5}-\sqrt{3}}$
$=$ $\frac{6}{\sqrt{5}-\sqrt{3}}$ $\times$ $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
$=$ $\frac{6(\sqrt{5}+\sqrt{3})}{5-3}$
$= 3 (\sqrt{5}+\sqrt{3})$
$= 3(2.236+1.732) = 11.904$
(iii) $\frac{2+\sqrt{3}}{2-\sqrt{3}}$ $+$ $\frac{2-\sqrt{3}}{2+\sqrt{3}}$ $+$ $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=$ $\frac{2+\sqrt{3}}{2-\sqrt{3}}$ $\times$ $\frac{2+\sqrt{3}}{2+\sqrt{3}}$ $+$ $\frac{2-\sqrt{3}}{2+\sqrt{3}}$ $\times$ $\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $\times$ $\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $\times$ $\frac{\sqrt{3}-1}{\sqrt{3}-1}$
$= 4+3+4\sqrt{3}+4+3-4\sqrt{3} +$ $\frac{3+1-2\sqrt{3}}{2}$
$= 14 + (2-\sqrt{3})$
$= 14 + (2-1.732) = 14.268$
Question 23: Rationalize and simplify:
(i) $\frac{1}{1+\sqrt{2}-\sqrt{3}}$
$=$ $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ $\times$ $\frac{1+\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}$
$=$ $\frac{1+\sqrt{2}+\sqrt{3}}{1+2+2\sqrt{2}-3}$
$=$ $\frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$ $\times$ $\frac{\sqrt{2}}{\sqrt{2}}$
$=$ $\frac{\sqrt{2}+2+\sqrt{6}}{4}$
(ii) $\frac{1}{1+\sqrt{5}+\sqrt{3}}$
$=$ $\frac{1}{1+\sqrt{5}+\sqrt{3}}$ $\times$ $\frac{1+\sqrt{5} - \sqrt{3}}{1+\sqrt{5} -\sqrt{3}}$
$=$ $\frac{1+\sqrt{5}-\sqrt{3}}{1+5+2\sqrt{5}-3}$
$=$ $\frac{1+\sqrt{5}-\sqrt{3}}{3+2\sqrt{5}}$ $\times$ $\frac{3-2\sqrt{5}}{3-2\sqrt{5}}$
$=$ $\frac{3+3\sqrt{5}-3\sqrt{3}-2\sqrt{5}-10+2\sqrt{15} }{9-20}$
$=$ $\frac{7-\sqrt{5}+3\sqrt{3}-2\sqrt{15}}{11}$
(iii) $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$
$=$ $\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$ $\times$ $\frac{{\sqrt{2}+\sqrt{3}+\sqrt{5}}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$
$=$ $\frac{2+\sqrt{6}+\sqrt{10}}{2+3+2\sqrt{6}-5}$
$=$ $\frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}}$
$=$ $\frac{2\sqrt{6}+6+\sqrt{60}}{12}$
$=$ $\frac{\sqrt{6}+3+\sqrt{15}}{6}$ |
# S Chand ICSE Mathematics Class 9 Solution Sixteenth Chapter Area of Plane Figures Exercise 16C (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)
### S Chand ICSE Mathematics Class 9 Solution Sixteenth Chapter Area of Plane Figures Exercise 16C (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)
Parallelogram
(1) Find the area of a parallelogram whose base and height are as given below:
(i) (ii) (iii) (iv) Base 8 cm 2.8 cm 12 mm 6.5 m Height 3 cm 5 cm 8.7 mm 4.8 m
Solution: (1) (i) Area = Base x height
= 8 x 3
= 24 cm2
(ii) Area = (2.8 x 5) cm2
= 14 cm2
(iii) Area = 12 x 8.7 mm2
= 104.4 mm2
(iv) Area = (6.5 x 4.8) m2
= 31.20 m2
(2) The area of a parallelogram is 1 ½ ares. Its base is 20 m. Find its height. (1 arc = 100 m2)
Solution: Area = (3/2 x 100) m
= ( 3 x 50) m2
= 150 m2
∴ 150 = 20 x h
h = 7.5 m
(3) In a parallelogram ABCD, AB = 8 cm, BC = 5 cm, perp. From A to DC = 3 cm. Find the length of the perp. drawn from B to AD.
Solution:
(4) A parallelogram has side 34 cm and 20 cm. One of its diagonals is 42 cm. Calculate its area.
Solution:
(5) ABCD is a parallelogram with side AB = 12 cm. Its diagonals AC and BD are the lengths 20 cm and 16 cm respectively. Find the area of || gm ABCD
Solution:
Rhombus
(6) What is the area of a rhombus which has diagonals of 8 cm and 10 cm.
Solution: Area of rhombus = ½ x 8 x 10 cm2
= 40 cm2
(7) The area of a rhombus is 98 cm2. If one of its diagonals is 14 cm, what is the length of the other diagonal?
Solution: Area of rhombus = ½ x Product of diagonals
98 = ½ x 14 x a
Or, a = 14 cm
(8) PQRS is a rhombus.
(i) If it is given that PQ = 3 cm, calculate the perimeter of PQRS
(ii) If the height of the rhombus is 2.5 cm, calculate its area,
(iii) The diagonals of a rhombus are 8 cm and 6 cm respectively. Find its perimeter.
Solution:
(i) PQ = 3 cm
∴ Perimeter = 4 x 3 cm
= 12 cm
(9) The sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate,
(i) The length of the other diagonal and
(ii) The area of the rhombus.
Solution:
(i) Let, length of other diagonal is l cm
(10) In the given figure, ABCX is a rhombus of side 5 cm. angles BAD and ADC are right angles. If DC = 8 cm, calculate the area of ABCX.
Solution:
Updated: October 21, 2019 — 1:00 pm |
Class 10 Maths
Theorems on Tangent to Circle
Tangent to a circle: A line which intersects a circle at any one point is called the tangent.
1. There is only one tangent at a point of the circle.
2. The tangent to a circle is perpendicular to the radius through the point of contact.
3. The lengths of the two tangents from an external point to a circle are equal.
Theorem 1
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Construction: Draw a circle with centre O. Draw a tangent XY which touches point P at the circle.
To Prove: OP is perpendicular to XY.
Draw a point Q on XY; other than O and join OQ. Here OQ is longer than the radius OP.
OQ > OP
For every point on the line XY other than O, like Q1, Q2, Q3, ……….Qn;
OQ_1 > OP
OQ_2> OP
OQ_3 > OP
OQ_4 > OP
Since OP is the shortest line
Hence, OP ⊥ XY proved
Theorem 2
The lengths of tangents drawn from an external point to a circle are equal.
Construction: Draw a circle with centre O. From a point P outside the circle, draw two tangents P and R.
To Prove: PQ = PR
Proof: In Δ POQ and Δ POR
OQ = OR (radii)
PO = PO (common side)
∠PQO = ∠PRO (Right angle)
Hence; Δ POQ ≅ Δ POR proved |
PDF chapter test
Here are the degrees of prime numbers that are often used:
$$n$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ ${2}^{n}$ $$2$$ $$4$$ $$8$$ $$16$$ $$32$$ $$64$$ $$128$$ $$256$$ $$512$$ $$1024$$ ${3}^{n}$ $$3$$ $$9$$ $$27$$ $$81$$ $$243$$ $$729$$ - - - - ${5}^{n}$ $$5$$ $$25$$ $$125$$ $$625$$ - - - - - - ${7}^{n}$ $$7$$ $$49$$ $$343$$ - - - - - - -
Example:
Calculate ${7}^{2}-{2}^{5}$.
The first step is always exponentiation.
$\begin{array}{l}{7}^{2}=49;\\ {2}^{5}=32.\end{array}$
Substituting the found values, we obtain:
${7}^{2}-{2}^{5}=49-32=17$.
Consider examples of degrees with negative bases:
$\begin{array}{l}{\left(-3\right)}^{2}=\left(-3\right)\cdot \left(-3\right)=9;\\ {\left(-3\right)}^{3}=\left(-3\right)\cdot \left(-3\right)\cdot \left(-3\right)=-27;\\ {\left(-3\right)}^{4}=\left(-3\right)\cdot \left(-3\right)\cdot \left(-3\right)\cdot \left(-3\right)=81.\end{array}$
Important!
An even degree of a negative number is positive; an odd one is negative. |
# How to Solve Algebra Equations: Step by Step Instructions
Page content
Need a refresher on algebra equations to complete your homework? An algebraic equation is a statement of mathematical equality between two algebraic expressions. The goal is to find the unknown in the equation, and the core to solving algebra equations is preserving the equality.
Let’s take a sample problem and break it down step by step.
## Sample Problem
Suppose you have a word problem and that you have finally succeeded in turning it into an equation (a fairly tricky step by itself; you can read more about constructing word problems here). Your equation is 5t + 7 = -23 + 2t.
Obviously, the variable here is “t” and we should try to determine its value. As we said earlier, when one wants to know how to solve algebra equations, he or she should keep in mind that it is possible to apply almost ANY operation, provided it is applied to BOTH sides of the equation – so the equilibrium is preserved.
## Gather Variables
As the first step, let’s try to gather all the variable-related members in one side of the equation. As we have “2t” member at the right, we can apply the operation of “-2t” to both sides:
5t + 7 – 2t = -23 + 2t – 2t. The equilibrium is preserved.
Let’s simplify by subtracting 2t from both 5t on the left and from 2t on the right. You’ll have:
3t + 7 = -23
Ok, first stage completed, all the variables are on the same side (we could alternatively, of course, apply “-5t” to both sides, getting, after simplification, 7 = -23-3t. You are invited to check this one yourself.)
## Group Free Members
Now let’s gather all the “free members” (those without the variable) into the other side of the equation. To eliminate “7” from the left side, we apply “-7” operation to both sides:
3t + 7 – 7 = -23 – 7
Simplifying to 3t = -30. Mission accomplished. (If you were going the second way around, working on the 7 = -23-3t example, you should apply +23 to both sides, getting, after simplification, 30 = -3t).
## The Final Step
Ok, we now know that 3t equals -30. To determine the value of a single t, we should divide both parts by 3:
3t / 3 = -30 /3; which is, of course, t = -10. Problem solved! (Same would, of course, happen with the 30 = -3t example: we would divide both parts by -3 (important not to forget the “-” sign), getting -10 = t, same answer, of course).
There is only one thing to memorize – preserve the equilibrium while applying similar operations to both parts of the equation. Gather variables to one side, free members to another side and get the solution. Enjoy!
## References and Image Credits
https://encyclopedia2.thefreedictionary.com/algebraic+equation
https://www.algebrahelp.com/lessons/equationbasics/
Image Credits:
https://www.algebra-class.com/solving-algebra-equations.html
https://www.algebrahelp.com/lessons/equationbasics/ |
Chapter 1 Class 12 Relation and Functions
Class 12
Important Questions for exams Class 12
Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
### Transcript
Example 23 (Method 1) Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse. Checking inverse Step 1 f(x) = 4x + 3 Let f(x) = y y = 4x + 3 y – 3 = 4x 4x = y – 3 x = (𝑦 − 3)/4 Rough Checking inverse of f:X → Y Step 1: Calculate g: Y → X Step 2: Prove gof = IX Step 3: Prove fog = IY g is the inverse of f Let g(y) = (𝑦 − 3)/4 where g: Y → N Step 2: gof = g(f(x)) = g(4x + 3) = ((4𝑥 + 3) − 3)/4 = (4𝑥 + 3 − 3)/4 = 4𝑥/4 = x = IN Rough Checking inverse of f:X → Y Step 1: Calculate g: Y → X Step 2: Prove gof = IX Step 3: Prove fog = IY g is the inverse of f Step 3: fog = f(g(y)) = f((𝑦 − 3)/4) = 4 ((𝑦 − 3)/4) + 3 = y – 3 + 3 = y + 0 = y = IY Since gof = IN and fog = IY, f is invertible & Inverse of f = g(y) = (𝒚 − 𝟑)/𝟒 Rough Checking inverse of f:X → Y Step 1: Calculate g: Y → X Step 2: Prove gof = IX Step 3: Prove fog = IY g is the inverse of f Example 23 (Method 2) Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse. f is invertible if f is one-one and onto Checking one-one f(x1) = 4x1 + 3 f(x2) = 4x2 + 3 Putting f(x1) = f(x2) 4x1 + 3 = 4x2 + 3 4x1 = 4x2 x1 = x2 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 If f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Checking onto f(x) = 4x + 3 Let f(x) = y, where y ∈ Y y = 4x + 3 y – 3 = 4x 4x = y – 3 x = (𝑦 − 3)/4 Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((𝑦 − 3)/4) = 4((𝑦 − 3)/4) + 3 = y − 3 + 3 = y For every y in Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. There is a value of x which is a natural number such that f(x) = y Thus, f is onto Since f is one-one and onto f is invertible Finding inverse Inverse of x = 𝑓^(−1) (𝑦) = (𝑦 − 3)/4 ∴ Inverse of f = g(y) = (𝒚 − 𝟑)/𝟒 where g: Y → N |
# How do you find the limit of (sqrt(x+6)-x)/(x^3-3x^2) as x->-oo?
Feb 6, 2018
$0$
#### Explanation:
$\frac{\sqrt{x + 6} - x}{{x}^{3} - 3 {x}^{2}}$
Divide by ${x}^{3}$:
$\frac{\frac{\sqrt{x + 6}}{x} ^ 3 - \frac{x}{x} ^ 3}{{x}^{3} / {x}^{3} - 3 {x}^{2} / {x}^{3}} = \frac{\frac{\sqrt{x + 6}}{x} ^ 3 - \frac{1}{x} ^ 2}{1 - \frac{3}{x}}$
lim_(x->-oo)((sqrt(x+6))/x^3-1/x^2)/(1-3/x)=(lim_(x->-oo)((sqrt(x+6))/x^3-1/x^2))/(lim_(x->-oo)(1-3/x)
${\lim}_{x \to - \infty} \left(\frac{\sqrt{x + 6}}{x} ^ 3 - \frac{1}{x} ^ 2\right) = {\lim}_{x \to - \infty} \frac{\sqrt{x + 6}}{x} ^ 3 - {\lim}_{x \to - \infty} \left(\frac{1}{x} ^ 2\right)$
${\lim}_{x \to - \infty} \frac{\sqrt{x + 6}}{x} ^ 3 = 0$
${\lim}_{x \to - \infty} \left(\frac{1}{x} ^ 2\right) = 0$
$0 - 0 = 0$
${\lim}_{x \to - \infty} \left(1 - \frac{3}{x}\right) = {\lim}_{x \to - \infty} \left(1\right) - {\lim}_{x \to - \infty} \left(\frac{3}{x}\right)$
${\lim}_{x \to - \infty} \left(1\right) = 1$
${\lim}_{x \to - \infty} \left(\frac{3}{x}\right) = 0$
$\therefore$
$1 - 0 = 1$
$\frac{0}{1} = 0$
Hence:
${\lim}_{x \to - \infty} \frac{\sqrt{x + 6} - x}{{x}^{3} - 3 {x}^{2}} = 0$ |
# If cos α and cos β (0 < α < β < π) are the roots of the quadratic 4x2 – 3 = 0, then what is the value of sec α × sec β?
This question was previously asked in
NDA (Held On: 9 Sept 2018) Maths Previous Year paper
View all NDA Papers >
1. $$- \frac{4}{3}$$
2. $$\frac{4}{3}$$
3. $$\frac{3}{4}$$
4. $$- \frac{3}{4}$$
Option 1 : $$- \frac{4}{3}$$
## Detailed Solution
Concept:
Let us consider the standard form of a quadratic equation, ax2 + bx + c =0
Where a, b and c are constants with a ≠ 0
Let α and β be the two roots of the above quadratic equation.
• The sum of the roots of a quadratic equation are:
$${\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}$$
• The product of the roots is given by:
$${\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}$$
Calculation:
Given: cos α and cos β (0 < α < β < π) are the roots of the quadratic equation.
4x2 – 3 = 0
Product of the roots = cos α × cos β = - 3/4
$$\Rightarrow {\rm{\;}}\frac{1}{{\sec {\rm{\alpha \;}}\sec {\rm{\beta }}}} = {\rm{\;}}\frac{{ - 3}}{4}$$
∴ sec α sec β = -4/3
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There are four suits in a deck of cards.
# Verbal Reasoning - Complete the Sum 1
This Math quiz is called 'Verbal Reasoning - Complete the Sum 1' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is an enjoyable way to learn if you are in the 3rd, 4th or 5th grade - aged 8 to 11.
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In this quiz you have to find the number that will complete the sum correctly. The only operations that will be used are multiply, divide, add and subtract. Choose the correct answer from the four choices available.
Sometimes you have to work out the sum to find the last number, and sometimes you have to find a number in the middle of the sum. An example has been done for you (see below and read it carefully so you understand before playing the quiz).
Example:
26 + 7 = 18 + ?
13
14
15
16
This is because 26 + 7 = 33. 18 + ? = 33. This makes 15 the only possible answer.
1.
Find the number that will complete the sum below correctly.
17 + ? = 49 - 13.
18
19
17
21
This is because 49 - 13 is 36. The answer must be 19 as 17 + 19 = 36. 19 is a prime number. A prime number is a number which is only divisible by itself and 1
2.
Find the number that will complete the sum below correctly.
? + 26 = 30 + 4.
8
9
10
11
This is because 30 + 4 = 34. To turn ? + 26 into 34 we need to add 8. Here's an interesting fact for you: a number is always divisible by 8 if its last three digits are also divisible by 8, for example 6,240
3.
Find the number that will complete the sum below correctly.
? ÷ 9 = 15 - 7
76
64
58
72
This is because 15 - 7 = 8 and 72 ÷ 9 = 8
4.
Find the number that will complete the sum below correctly.
35 - ? = 4 x 4
16
17
18
19
This is because 4 x 4 = 16. This is equal to 35 - 19. When we multiply a number by itself, the product is known as the square of that number. So 16 is the square of 4. Another example is 25 which is the square of 5
5.
Find the number that will complete the sum below correctly.
19 x 4 = ? + 52
25
24
23
26
This is because 24 + 52 = 76. 19 x 4 also gives the answer of 76. Many things are arranged in fours. A deck of cards has four suits, there are four points of a compass and, if you're lucky, there are four leaves on a clover!
6.
Find the number that will complete the sum below correctly.
18 - 5 = 78 ÷ ?
6
5
7
4
This is because 18 - 5 = 13. To get from 78 to 13 we need to divide it by 6
7.
Find the number that will complete the sum below correctly.
56 x 2 = 127 - ?
12
13
14
15
This is because 127 - 15 = 112 which is equal to 56 x 2
8.
Find the number that will complete the sum below correctly.
12 x 4 = 52 - ?
6
3
4
8
This is because 12 x 4 = 48 and 52 - 4 is also 48
9.
Find the number that will complete the sum below correctly.
14 ÷ 2 = 63 ÷ ?
7
5
9
6
This is because 14 ÷ 2 = 7 and 63 ÷ 9 is also equal to 7. Can you remember what a prime number is? Good. Well, 2 is the only prime number which is also an even number
10.
Find the number that will complete the sum below correctly.
45 - ? = 13 x 3
7
6
5
4
This is because 45 - 6 = 39, which is the same as = 13 x 3. Some people consider 13 to be an unlucky number and Friday 13th is thought to be an unlucky date. Any month's 13th day will fall on a Friday if the month begins with a Sunday
Author: Stephen O'Hara |
# How do you solve | 2x - 5 | + 3 = 12?
Mar 3, 2018
$x = 7 , - 2$
#### Explanation:
$| 2 x - 5 | + 3 = 12$ Given Equation
$| 2 x - 5 | = 9$ Subtracted $3$ from both sides
$| 2 x | = 14$ Added $5$ to each side.
$x = 7$ Divided 2 by both sides
Since this is absolute value, you are also required to also solve the equation where in this case $x$ could also be negative aswell as positive.
$- | 2 x - 5 | + 3 = 12$
$- 2 x + 5 + 3 = 12$ Distributed the negative sign.
$- 2 x + 8 = 12$ Added 5 and 3.
$- 2 x = 4$ Subtracted 8 from 12.
$x = - 2$ Divided -2 to 4.
Mar 3, 2018
$x = - 2 , 7$
#### Explanation:
Solve:
$\left\mid 2 x - 5 \right\mid + 3 = 12$
Since $\left\mid a \right\mid = a \mathmr{and} - a$, the equation can be broken into two equations:
$2 x - 5 + 3 = 12$ and $- \left(2 x - 5\right) + 3 = 12$
Solve the first equation:
$2 x - 5 + 3 = 12$
Simplify $\left(- 5 + 3\right)$ to $- 2$.
$2 x - 2 = 12$
Add $2$ to both sides of the equation.
$2 x = 12 + 2$
Simplify.
$2 x = 14$
Divide both sides by $2$.
$x = \frac{14}{2}$
$x = 7$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solve the second equation:
$- \left(2 x - 5\right) + 3 = 12$
Expand.
$- 2 x + 5 + 3 = 12$
Simplify $\left(5 + 3\right)$ to $8$.
$- 2 x + 8 = 12$
Subtract $8$ from $12$
$- 2 x = 12 - 8$
Simplify.
$- 2 x = 4$
Divide both sides by $- 2$.
$x = \frac{4}{- 2}$
$x = - 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$x = - 2 , 7$ |
# 1 - 1.Introduction Cyclic codes form an important subclass...
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Unformatted text preview: 1.Introduction: Cyclic codes form an important subclass of linear codes. These codes are attractive for two reasons: first, encoding and decoding can be implemented easily by employing shift-registers with feedback connections (as will be seen later); and second, because they have considerable inherent algebric structure, it is possible to find various practical methods for decoding them. In this section the reader can find a general overview of binary cyclic codes followed by their mathematical representation and their main algebric properties (section 2); The third section deals with the encoding procedure and finally, the structure of the generator and parity-check matrices will be shown. In order to fully understand the cyclic codes material, it is advised to comprehend thoroughly the linear block codes issue first. 2.Description of cyclic codes: 2.1 Cyclic code definition: Lets start with the definition of cyclic shift : If the components of an n-tuple are cyclically shifted one place to the right, we obtain another n-tuple: which is called a cyclic shift of . If the components of are cyclically shifted i places to the right, the resultant n-tuple would be : . Clearly, cyclically shifting i places to the right is equivalent to cyclically shifting n-i places to the left. Cyclic code definition: An (n,k) linear code C is called a cyclic code if every cyclic shift of a code vector in C is also a code vector in C. 2.2 Polynomial representation: In order to develop the algebric properties of a cyclic code, we treat the components of a code vector as the coefficients of a polynomial as follows: . Thus, each code vector corresponds to a polynomial of degree n-1 or less. We shall call the code polynomial of . Before looking further into cyclic codes, we'll go over some polynomial basics first:...
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# PSEB 12th Class Physics Solutions Chapter 9 Ray Optics and Optical Instruments
Punjab State Board PSEB 12th Class Physics Book Solutions Chapter 9 Ray Optics and Optical Instruments Textbook Exercise Questions and Answers.
## PSEB Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments
PSEB 12th Class Physics Guide Ray Optics and Optical Instruments Textbook Questions and Answers
Question 1.
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = -27 cm
Radius of curvature of the concave mirror, R = -36 cm
Focal length of the concave mirror, f = $$\frac{R}{2}=\frac{-36}{2}$$ = -18 cm
Image distance = v
The image distance can be obtained using the mirror formula
The negative sign shows that the image is formed in front of the mirror i.e., on the side of the object itself. Thus the screen must be placed at a distance of 54 cm in front of the mirror.
The magnification of the image is given as
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.
Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Given u = -12 cm, f = +15 cm. (convex mirror)
That is image is formed at a distance of 6.67 cm behind the mirror.
Magnification m = $$-\frac{v}{u}=-\frac{\frac{20}{3}}{-12}$$ = $$\frac{5}{9}$$
Size of image I = mO = $$\frac{5}{9}$$ x 4.5 = 2.5 cm
The image is erect, virtual and has a size 2.5 cm.
Its position is 6.67 cm behind the mirror when needle is moved farther, the image moves towards the focus and its size goes on decreasing.
Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Case I:
When tank is filled with water Actual depth of the needle in water, h1 = 12.5cm
Apparent depth of the needle in water, h2 =9.4cm
Refractive index of water = μ
The value μ can be obtained as follows
μ = $$\frac{\text { Actual depth }}{\text { Apparent depth }}$$
= $$\frac{h_{1}}{h_{2}}=\frac{12.5}{9.4}$$ ≈ 1.33
Hence, the refractive index of water is about 1.33
Case II: When tank is filled with liquid
Water is replaced by a liquid of refractive index, μ’ = 1.63
The actual depth of the needle remains the same, but its apparent depth changes.
Let y be the new apparent depth of the needle. Hence, we can write the relation
μ’ = $$\frac{h_{1}}{y}$$
y = $$\frac{h_{1}}{\mu^{\prime}}=\frac{12.5}{1.63}$$ = 7.67 cm
Hence, the new apparent depth of the needle is 7.67cm. It is less than h2 Therefore, to focus the needle again, the microscope should be moved up. Distance by which the microscope should be moved up =9.4-7.67 = 1.73 cm.
Question 4.
Figures 9.34 (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.34 (c)]
As per the given figure, for the glass-air interface
Angle of incidence, i = 60°
Angle of refraction, r = 35°
The relative refractive index of glass with respect to air is given by Snell’s law as
aμg = $$\frac{\sin i}{\sin r}$$
= $$\frac{\sin 60^{\circ}}{\sin 35^{\circ}}=\frac{0.8660}{0.5736}$$ = 1.51 …………………….. (1)
As per the given figure, for the air-water interface
Angle of incidence, j = 600
Angle of refraction, r = 470
The relative refractive index of water with respect to air is given by Snell’s law as
wμw = $$\frac{\sin i}{\sin r}$$
= $$\frac{\sin 60^{\circ}}{\sin 47^{\circ}}=\frac{0.8660}{0.7314}$$ = 1.184 …………………………… (2)
Using equations (1) and (2), the relative refractive index of glass with respect to water can be obtained as
wμg = $$\frac{a_{g}}{a_{w_{w}}}$$
= $$\frac{1.51}{1.184}$$ = 1.275
The following figure shows the situation involving the glass-water interface
Angle of incidence, i = 45
Angle of reflection = r
From Snell’s law, r can be calculated as, $$\frac{\sin i}{\sin r}$$ = wμg
$$\frac{\sin 45^{\circ}}{\sin r}$$ = 1.275
sin r = $$\frac{\frac{1}{\sqrt{2}}}{1.275}=\frac{0.707}{1.275}$$ = 0.5546
r = sin-1(0.5546) = 38.68°
Hence, the angle of refraction at the water-glass interface is 38.68°
Question 5.
A small bulb is placed at the bottom of a tank containirg water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33 (Consider the bulb to be a point source.)
Actual depth of the bulb in water, d1 = 80 cm = 0.8 m
Refractive index of water, μ = 1.33
The given situation is shown in the following figure
where,
i = Angle of Incidence
r = Angle of Refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle of radius,
R = $$\frac{A C}{2}$$ = AO = OC
Using Snell’s law, we can write the relation for the refractive index of water as
μ = $$\frac{\sin r}{\sin i}$$
1.33 = $$\frac{\sin 90^{\circ}}{\sin i}$$
i = sin-1$$\left(\frac{1}{1.33}\right)$$ = 48.75°
Using the given figure, we have the relation
tan i = $$\frac{O C}{O B}=\frac{R}{d_{1}}$$
∴R = tan 48.75° x 0.8 = 0.91 m
∴ Area of the surface of water = πR2
= π(0.91)2
= 2.61 m2
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2.
Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Angle of minimum deviation, δm = 40 °
Refracting angle of the prism, A = 60°
Refractive index of water, μ = 1.33
Let μ’ be the refractive index of the material of the prism.
The angle of deviation and refracting angle of the prism are related to refractive index (μ’) as
μ’ = $$\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$
= $$\frac{\sin \left(\frac{60^{\circ}+40^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)}=\frac{\sin 50^{\circ}}{\sin 30^{\circ}}=\frac{0.766}{0.5}$$
= 1.532
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let 8 ^ be the new angle of minimum deviation for the same prism.
The refractive index of glass with respect to water is given by the relation
Hence, the new minimum angle of deviation is 10.32°.
Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Lens maker formula is
$$\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$ …………………………………… (1)
If R is radius of curvature of double convex lens, then,
∴ R = 2(n-1)f
Here, n =1.55, f = +20 cm
∴ R = 2 (1.55 -1) x 20 = 22 cm
Question 8.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
In the given situation, the object is virtual and the image formed is real.
Object distance, u = +12cm
(a) Focal length of the convex lens, f = 20 cm
Image distance = v
According to the lens formula, we have the relation
∴ v = $$\frac{60}{8}$$ = 7.5cm
Hence, the image is formed 7.5cm away from the lens, toward its right.
(b) Focal length of the concave lens, f = -16 cm
Image distance = v
According to the lens formula, we have the relation
∴ v = 48 cm
Hence, the image is formed 48 cm away from the lens, toward its right.
Question 9.
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Size of object O = 3.0 cm
u = -14 cm, f = -21 cm (concave lens)
∴ Formula $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$
⇒ $$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$$
or $$\frac{1}{v}=\frac{1}{-21}+\frac{1}{-14}=-\frac{2+3}{42}$$
or v = $$-\frac{42}{5}$$ = -8.4 cm
Size of image I = $$\frac{v}{u}$$ O
= $$\frac{-8.4}{-14}$$ x 3.0 cm = 1.8 cm
That is, image is formed at a distance of 8.4 cm in front of lens. The image is virtual, erect and of size 1.8 cm. As the object is moved farther from the lens, the image goes on shifting towards focus and its size goes on decreasing. The image is never formed beyond the focus of the concave lens.
Question 10.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system
a converging or a diverging lens? Ignore thickness of the lenses.
Given f1 = +30 cm, f2 = -20 cm
The focal length (F) of combination is given by
$$\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$$
⇒ F = $$\frac{f_{1} f_{2}}{f_{1}+f_{2}}$$
= $$\frac{30 \times(-20)}{30-20}$$ = -60 cm
That is, the focal length of combination is 60 cm and it acts like a diverging lens.
Question 11.
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Focal length of the objective lens, f0 = 2.0 cm
Focal length of the eyepiece, fe = 6.25cm
Distance between the objective lens and the eyepiece, d = 15cm
(a) Least distance of distinct vision, d’ = 25cm
∴ Image distance for the eyepiece, ve = -25cm
Object distance for the eyepiece = ue
According to the lens formula, we have the relation
$$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$$
or $$\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}$$
= $$\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$$
∴ ue = -5cm
Image distance for the objective lens, v0 = d + ue =15-5 = 10 cm
Object distance for the objective lens = u0
According to the lens formula, we have the relation
$$\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}$$
$$\frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$$
∴ u0=-2.5cm
Magnitude of the object distance, |u0| = 2.5 cm
The magnifying power of a compound microscope is given by the relation
m = $$\frac{v_{o}}{\left|u_{o}\right|}\left(1+\frac{d^{\prime}}{f_{e}}\right)$$
= $$\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)$$ = 4(1+4) = 20
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
∴ Image distance for the eyepiece, ve = ∞
Object distance for the eyepiece = ue
According to the lens formula, we have the relation
$$\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}$$
$$\frac{1}{u_{o}}=\frac{1}{v_{o}}-\frac{1}{f_{o}}=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$$
∴ u0 = $$\frac{17.5}{6.75}$$ = -2.59 cm
Magnitude of the object distance, |u0| = 2.59 cm
The magnifying power of a compound microscope is given by the relation
m = $$\frac{v_{o}}{\left|u_{o}\right|}\left(1+\frac{d^{\prime}}{f_{e}}\right)$$
= $$\frac{8.75}{2.59} \times\left(1+\frac{25}{6.25}\right)$$ = 13.51
Hence, the magnifying power of the microscope is 13.51.
Question 12.
A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Focal length of the objective lens, f0= 8 mm = 0.8cm
Focal length of the eyepiece, fe = 2.5 cm
Object distance for the objective lens, u0 = -9.0 mm = -0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, ve = -d = -25 cm
Object distance for the eyepiece = ue
Using the lens formula, we can obtain the value of ue as
∴ ue = $$-\frac{25}{11}$$ = -2.27 cm
We can also obtain the value of the image distance for the objective lens (v0) using the lens formula.
∴ v0 = 7.2 cm
The distance between the objective lens and the eyepiece = |ue|+v0
= 2.27+ 7.2 = 9.47cm
The magnifying power of the microscope is calculated as $$\frac{v_{o}}{\left|u_{o}\right|}\left(1+\frac{d}{f_{e}}\right)$$
= $$\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)$$
= 8(1 +10) = 88
Hence, the magnifying power of the microscope is 88.
Question 13.
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Focal length of the objective lens, f0 = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power of the telescope is given as, m = $$\frac{f_{o}}{f_{e}}=\frac{144}{6}$$ = 24
The separation between the objective lens and the eyepiece is calculated as
= fo + fe
= 144 + 6 = 150 cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.
Question 14.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 x 106 m, and the radius of lunar orbit is 3.8 x 108 m.
(a) Given f0 = 15 m,
fe = 1.0 cm = 1.0 x 10-2 m
Angular magnification of telescope,
m = $$-\frac{f_{o}}{f_{e}}=-\frac{15}{1.0 \times 10^{-2}}$$ = -1500
Negative sign shows that the final image is inverted.
(b) Let D be diameter of moon, d diameter of image of moon formed by objective and r be the distance of moon from objective lens, then
Question 15.
Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished In size and is located between the focus and the
pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
(a) For a concave mirror, the focal length (f) is negative
∴ f<o
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u<O
For image distance v, we can write the mirror formula
$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$$ …………………………………… (1)
The object lies between f and 2f.
∴ 2f < u < f (∵ u and f are negative) ∴ $$\frac{1}{2 f}>\frac{1}{u}>\frac{1}{f}$$
$$-\frac{1}{2 f}<-\frac{1}{u}<-\frac{1}{f}$$
$$\frac{1}{f}-\frac{1}{2 f}<\frac{1}{f}-\frac{1}{u}<0$$ ………………………………… (2)
Using equation (1), we get
$$\frac{1}{2 f}<\frac{1}{v}<0$$
∴ $$\frac{1}{v}$$ is negative, i.e., v is negative.
$$\frac{1}{2 f}<\frac{1}{v}$$ 2f > v
-v > -2 f
Therefore, the image lies beyond 2f.
(b) For a convex mirror, the focal length (f) is positive.
∴ f>o
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u<O
For image distance y, we have the mirror formula
$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$$
Using equation (2), we can conclude that
$$\frac{1}{\nu}$$ < 0 v v> 0
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
(c) For a convex mirror, the focal length (f) is positive.
∴ f> 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u< 0
For image distance v, we have the mirror formula
$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$$
But we have u < 0 ∴ $$\frac{1}{v}>\frac{1}{f}$$
v < f
Hence, the image formed is diminished and is located between the focus (f) and the pole.
(d) For a concave mirror, the focal length (f) is negative.
∴ f< 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u< 0 It is placed between the focus (f) and the pole. ∴f > u > 0
$$\frac{1}{f}<\frac{1}{u}$$ < 0 $$\frac{1}{f}-\frac{1}{u}$$ > 0
For image distance v, we have the mirror formula
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write
$$\frac{1}{u}>\frac{1}{v}$$
v > u
Magnification, m = $$\frac{v}{u}$$ > 1 u
Hence, the formed image is enlarged.
Question 16.
A small pin fixed on a tabletop is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Actual depth of the pin, d = 15cm
Apparent depth of the pin = d’
Refractive index of glass, µ = 1.5
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
µ = $$\frac{d}{d^{\prime}}$$
∴ d’ = $$\frac{d}{\mu}$$
= $$\frac{15}{1.5}$$ = 10 cm
The distance at which the pin appears to be raised = d-d’=15-10 = 5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.
Question 17.
(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?
(a) Refractive index of the glass fibre, µ2 = 1.68
Refractive index of the outer covering of the pipe, µ1 =1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index (µ) of the inner core-outer core interface is given as
µ = $$\frac{\mu_{2}}{\mu_{1}}=\frac{1}{\sin i^{\prime}}$$
sin i’ = $$\frac{\mu_{1}}{\mu_{2}}=\frac{1.44}{1.68}$$ = 0.8571
∴ i’ = 59°
For the critical angle, total internal reflection (TIR) takes place only when i > i’. i.e., i > 59°
Maximum angle of reflection, rmax = 90°-i’ = 90°-59°= 31°
Let, imax be the maximum angle of incidence.
The refractive index at the air – glass interface, µ2 =1.68
µ2 = $$\frac{\sin i_{\max }}{\sin r_{\max }}$$
sin imax = µ2 sin rmax = 1.68 sin31°
= 1.68 x 0.5150
= 0.8652
∴imax = sin-1 (0.8652) ≈ 60°
Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.
(b) If the outer covering of the pipe is not present, then
Refractive index of the outer pipe, µ1 = Refractive index of air = 1
For the angle of incidence i =90°, we can write Snell’s law at the air-pipe interface as
$$\frac{\sin i}{\sin r}$$ = µ2 = 1.68
sin r = $$\frac{\sin 90^{\circ}}{1.68}=\frac{1}{1.68}$$
r = sin-1(0.5952)
∴ i’ = 90°-36.5°= 53.5°
Since i’ > r, all incident rays will suffer total internal reflection.
Question 18.
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it onto the ‘screen’(i.e., the retina) of our eye. Is there a contradiction?
(c) A diver underwater, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that – of ordinary glass. Is this fact of some use to a diamond cutter?
(a) Yes, they produce real images under some circumstances. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.
(b) No, there is no contradiction. A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.
(c) The diver is in the water and the fisherman is on land (i.e., in the air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are traveling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.
(d) Yes, the apparent depth of a tank of water changes when viewed obliquely. This is because light bends on traveling from one medium to another. The apparent depth of the tank, when viewed obliquely, is less than the near-normal viewing.
(e) Yes, the refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.
Question 19.
The image of a small electric bulb on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose? ’’
Here, u + v = 3 m, :.v = 3 -u
From lens formula,
or u = $$\frac{3 \pm \sqrt{3^{2}-4.3 f}}{2}$$
For real solution, 9 -12, f should be positive.
It., 9 -12f > 0
or 9 >12f.
or f < $$\frac{9}{12}$$ < $$\frac{3}{4}$$ m
∴ The maximum focal length of the lens required for the purpose is $$\frac{3}{4}$$ m
i.e, fmax = 0.7 m
Question 20.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Here, O is a position of object and I is position of image (screen).
Distance OI = 90 cm
L1 and L2 are the two positions of the lens.
∴ Distance between L1 and L2 = O1 O2 = 20 cm
For Position L1 of the Lens: Let x be the distance of the object from the lens.
∴ u1 = -x
∴ Distance of the image from the lens, v1 = +(90 – x)
If f be the focal length of the lens, then using lens formula,
$$-\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$$ we get
$$-\frac{1}{-x}+\frac{1}{90-x}=\frac{1}{f}$$
or $$\frac{1}{f}=\frac{1}{x}+\frac{1}{90-x}$$ ……………………………….. (1)
For Position L2 of the Lens : Let u2 and v2 be the distances of the object and image from the lens in this position.
∴ u2=-(X + 20),
v2 = +[90-(x+20)] = +(70-x)
∴ Using lens formula,
Question 21.
(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system and the size of the image.
Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens,f2 = -20 cm
Distance between the two lenses, d = 8.0 cm
(a)
(i) When the parallel beam of light is incident on the convex lens first.
According to the lens formula, we have
$$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$$
where, μ1 = Object distance = ∞, v1 = Image distance = ?
$$\frac{1}{v_{1}}=\frac{1}{30}-\frac{1}{\infty}=\frac{1}{30}$$
∴ v1 = 30 cm
The image will act as a virtual object for the concave lens. Applying lens formula to the concave lens, we have
$$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$$
where, u2 = Object distance = (30 – d) = 30 – 8 = 22 cm,
v2 = Image distance=?
$$\frac{1}{v_{2}}=\frac{1}{22}-\frac{1}{20}=\frac{10-11}{220}=\frac{-1}{220}$$
∴ v2 = -220 cm
The parallel incident beam appears to diverge from a point that is $$\left(220-\frac{d}{2}=220-\frac{8}{2}=220-4=216 \mathrm{~cm}\right)$$ from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, on the concave lens first. According to the lens formula, we have
$$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$$
$$\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}$$
where, u2 = Object distance = -∞, v2 = Image distance = ?
$$\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{-\infty}=-\frac{1}{20}$$
∴ v2 = -20 cm
The image will act as a real object for the .convex lens.
Applying lens formula to the convex lens, we have
$$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$$
where, u1 = Object distance = -(20 + d) = -(20 + 8) = -28 cm v1 = Image distance = ?
$$\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-28}=\frac{14-15}{420}=\frac{-1}{420}$$
∴ v1 = -420 cm
Hence, the parallel incident beam appear to diverge from a point that is (420 – 4 = 416 cm) from the left of the centre of the combination of the two lenses. The answer depends on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.
(b) Height of the object, h1 =1.5 cm
Object distance from the side of the convex lens, u1 = -40 cm
|ui| = 40 cm
According to the lens formula
$$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$$
where, v1 = Image distance =?
$$\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-40}=\frac{4-3}{120}=\frac{1}{120}$$
∴ v1 = 120 cm
Magnification, m= $$\frac{v_{1}}{\left|u_{1}\right|}=\frac{120}{40}$$ = 3
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula
$$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$$
where,
u2 = Object distance = +(120 —8)=112 cm
v2= Image distance =?
Magnification, m’ = $$\left|\frac{v_{2}}{u_{2}}\right|=\frac{2240}{92} \times \frac{1}{112}=\frac{20}{92}$$
Hence, the magnification due to the concave lens is $$\frac{20}{92}$$
The magnification produced by the combination of the two lenses is calculated as m x m’ = $$3 \times \frac{20}{92}=\frac{60}{92}$$ = 0.652
The magnification of the combination is given as
$$\frac{h_{2}}{h_{1}}$$ = 0.652
h2 = 0.652 x h1
where, h1 = Object size = 1.5 cm,
h2 = Size of the image
∴ h2 = 0.652 x 1.5 = 0.98 cm
Hence, the height of the image is 0.98 cm.
Question 22.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure
Angle of prism, A = 60°
Refractive index of the prism, μ = 1.524
i1 = Incident angle
r2 = Refracted angle
r2 = Angle of incidence at the face
AC = e = Emergent angle = 90°
According to Snell’s law, for face AC, we can have sine
$$\frac{\sin e}{\sin r_{2}}$$ = μ
It is clear from the figure that angle A = r1 + r2
According to Snell’s law, we have the relation
μ = $$\frac{\sin i_{1}}{\sin r_{1}}$$
sin i1 = μ sin r1
= 1.524 x sin19°= 0.496
∴ i1= 29.75°
Hence, the angle of incidence is 29.75°.
Question 23.
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
(a) deviate a pencil of white light without much dispersion,
(b) disperse (and displace) a pencil of white light without much deviation.
(a) Place the two prisms beside each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism, it will recombine and white light will emerge from the combination of the two prisms.
(b) Take the system of the two prisms as suggested in answer (a). Adjust (increase) the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will disperse the pencil of white light without much deviation.
Question 24.
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of . the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Least distance of distinct vision, d = 25 cm
Far point of a normal eye, d’ = ∞
Converging power of the cornea, Pc = 40 D
Least converging power of the eye- lens, Pe = 20 D
To see the objects at infinity, the eye uses its least converging power.
Power of the eye-lens, P = Pc + Pe =40+20 = 60 D
Power of the eye-lens is given as
P = $$\frac{1}{\text { Focal length of the eye lens }(f)}$$
f = $$=\frac{1}{P}=\frac{1}{60 D}=\frac{100}{60}=\frac{5}{3}$$ cm
To focus an object at the near point, object distance (u) = -d = -25 cm
Focal length of the eye-lens = Distance between the cornea and the retina = Image distance
Hence, image distance, v = $$\frac{5}{3}$$ cm
According to the lens formula, we can write
$$\frac{1}{f^{\prime}}=\frac{1}{v}-\frac{1}{u}$$
Where f’ = Focal length
Power of the eye-lens = 64-40 = 24 D
Hence, the range of accommodation of the eye-lens is from 20 D to24D.
Question 25.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?
A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eyeballs get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened.
When the eye-lens loses its ability of accommodation, the defect is called presbyopia.
Question 26.
A myopic person has been using spectacles of power -1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power +2.0 dioptres. Explain what may have happened.
The power of the spectacles used by the myopic person, P = -1.0 D
Focal length of the spectacles, f = $$\frac{1}{P}=\frac{1}{-1 \times 10^{-2}}$$ = -100 cm
Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm.
He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.
During old age, the person uses reading glasses of power, P’ = +2D The ability of accommodation is lost in old age.
This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.
Question 27.
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is tailed astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.
Question 28.
A man with normal near point (25cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
(a) Focal length of the magnifying glass, f = 5 cm
Least distance of distinct vision, d = 25 cm
Closest object distance = u
Image distance, v = -d = -25 cm
According to the lens formula, we have
Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distance (u’), the image distance (v’) = ∞
According to the lens formula, we have
∴ u’ = -5 cm
Hence, the farthest distance at which the person can read the book is 5 cm.
(b) Maximum angular magnification is given by the relation
αmax= $$\frac{d}{|u|}=\frac{25}{\frac{25}{6}}$$ = 6
Minimum angular magnification is given by the relation
αmin = $$\frac{d}{\left|u^{\prime}\right|}=\frac{25}{5}$$ = 5.
Question 29.
A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)?
Explain.
(a) Area of each square, A = 1 mm2
Object distance, u = -9 cm
Focal length of the converging lens, f = 10 cm
For image distance v, the lens formula can be written as
∴ v = -90 cm
Magnification, m = $$=\frac{v}{u}=\frac{-90}{-9}$$ =10
∴ Area of each square in the virtual image = (10)2A
= 102 x 1 =100 mm2 = 1 cm2
(b) Magnifying power of the lens = $$\frac{d}{|u|}=\frac{25}{9}$$ = 2.8
(c) The magnification in (a) is not the same as the magnifying power in(b).
The magnification magnitude is $$\left(\left|\frac{v}{u}\right|\right)$$ and the magnifying power is $$\left(\frac{d}{|u|}\right)$$ .
The two quantities will be equal when the image is formed at the near point (25 cm).
Question 30.
(a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
(a) The maximum possible magnification is obtained when the image is formed at the near point (d = 25cm).
Image distance, v = -d = -25 cm
Focal length, f = 10 cm
Object distance = u
According to the lens formula, we have
∴ u = $$-\frac{50}{7}$$ = -7.14 cm
Hence, to view the squares distinctly, the, lens should be kept 7.14 cm away from them. .
(b) Magnifying = $$\left|\frac{v}{u}\right|=\frac{25}{50}$$ =3.5
(c) Magnifying power = $$\frac{d}{u}=\frac{25}{\frac{50}{7}}$$ = 3.5
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.
Question 31.
What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Area of the virtual image of each square, A = 6.25 mm
Area of each square, A0 = 1 mm2
Hence, the linear magnification of the object can be calculated as
m = $$\sqrt{\frac{A}{A_{0}}}=\sqrt{\frac{6.25}{1}}$$ = 2.5
But m = $$\frac{\text { Image distance }(v)}{\text { Object distance }(u)}$$
∴ v = mu = 2.5 u
Focal length of the magnifying glass, f = 10 cm
According to the lens formula, we have the relation
∴ u = $$-\frac{1.5 \times 10}{2.5}$$ = -6 cm
and v = 2.5 u = 2.5 x 6 = -15 cm
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.
Question 32.
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
(a) Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
(b) Yes, the angular magnification changes when the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.
(c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.
(d) The ang lar magificarin produced by’the eyepiece of a compound microscope is $$\left[\left(\frac{25}{f_{e}}\right)+1\right]$$
Where fe = Focal length of the eyepiece
It can be inferred that fe is small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as
$$\frac{1}{\left(\left|u_{o}\right| f_{o}\right)}$$
Where, u0 = Object distance for the objective lens, f0 = Focal length of the objective
The magnification is large when u0> f0 . In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little.
Since u0 is small, f0 will be even smaller. Therefore, fe and f0 are both small in the given condition.
(e) When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much-refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.
Question 33.
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Focal length of the objective lens, f0 = 1.25 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation
me = $$\left(1+\frac{d}{f_{e}}\right)=\left(1+\frac{25}{5}\right)$$ = 1+5 = 6
The angular magnification of the objective lens (m0) is related to me as
mome=m
or m0 = $$\frac{m}{m_{e}}=\frac{30}{6}$$ = 5
We also have the relation
m = $$\frac{\text { Image distance for the objective lens }\left(v_{o}\right)}{\text { Object distance for the objective lens }\left(-u_{0}\right)}$$
5 = $$\frac{v_{o}}{-u_{o}}$$
∴ v0 = -5u0 …………………………….. (1)
Applying the lens formula for the objective lens
and v0 = -5u0
= -5 x (-1.5) = 7.5 cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece
$$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$$
where,
ve = Image distance for the eyepiece = -d = -25 cm
ue = Object distance for the eyepiece
$$\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}=\frac{-1}{25}-\frac{1}{5}=-\frac{6}{25}$$
ue =-4.17 cm
Separation between the objective lens and the eyepiece = $$\left|u_{e}\right|+\left|v_{o}\right|$$
= 4.17 + 7.5 = 11.67 cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
Question 34.
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?
Focal length of the objective lens, f0 =140 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
(a) When the telescope is in normal adjustment, its magnifying power is given as
m = $$\frac{f_{o}}{f_{e}}=\frac{140}{5}$$ = 28
(b) When the final image is formed at d, the magnifying power of the telescope is given as
$$\frac{f_{o}}{f_{e}}\left[1+\frac{f_{e}}{d}\right]=\frac{140}{5}\left[1+\frac{5}{25}\right]$$
= 28[1 +0.2] = 28×1.2 = 33.6
Question 35.
(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?
Focal length of the objective lens, f0 =140 cm
Focal length of the eyepiece, fe= 5 cm
(a) In normal adjustment, the separation between the objective lens and the eyepiece = f0 + fe = 140 + 5 = 145 cm
(b) Height of the tower, h1 = 100 m
Distance of the tower (object) from the telescope, u = 3 km = 3000 m
The angle subtended by the tower at the telescope is given as
θ’ = $$\frac{h_{2}}{f_{o}}=\frac{h_{2}}{140}$$ rad
where,
h2 = Height of the image of the tower formed by the objective lens
$$\frac{1}{30}=\frac{h_{2}}{140}$$ (∵θ=θ’)
∴ h2 = $$\frac{140}{30}$$ = 4.7 cm
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
(c) Image is formed at a distance, d = 25 cm
The magnification of the eyepiece is given by the relation
m = 1 + $$\frac{d}{f_{e}}$$
= 1+ $$\frac{25}{5}$$ =1 + 5 = 6
Height of the final image = mh2 = 6 x 4.7 = 28.2 cm
Hence, the height of the final image of the tower is 28.2 cm.
Question 36.
A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart.
If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
Given, r1 = 220 mm, f1 = $$\frac{r_{1}}{2}$$ = 110 mm = 11 cm
r2 = 140 mm, f2 = $$\frac{r_{2}}{2}$$ = 70 mm = 7.0 cm
Distance between mirrors, d = 20 mm = 2.0 cm
The parallel incident rays coming from distant objects fall on the concave mirror and try to be focused at the principal focus of concave lens, i. e., v1 = -f1 = -11 cm
But in the path of rays reflected from concave mirror, a convex mirror is placed. Therefore the image formed by the concave mirror acts as a virtual object for convex mirror.
For convex mirror f2 = -7.0 cm, u2 = -(11 -2) = -9 cm
v2 = $$-\frac{63}{2}$$ cm = -31.5 cm
This is the distance of the final image formed by the convex mirror. Thus, the final image is formed at a distance of 31.5 cm from the smaller (convex) mirror behind the bigger mirror.
Question 37.
Light incident normally on a plane mirror attached to a galvanometer coil retraces backward as shown in Fig. 9.36. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
Angle of deflection, θ = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ = 2 x 3.5 = 7.0°
The displacement (d) of the reflected spot of light on the screen is given as
tan 2θ = $$\frac{d}{1.5}$$ d =1.5 x tan7°= 0.184 m = 18.4 cm
Hence, the displacement of the reflected spot of light is 18.4 cm.
Question 38.
Figure 9.37 shows an biconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror.
A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to he 30.0 cm. What is the refractive index of the liquid?
Focal length of the convex lens, f1 = 30 cm
The liquid acts as a mirror. Focal length of the liquid = f2
Focal length of the system (convex lens + liquid), f = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as
$$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$$
$$\frac{1}{f_{2}}=\frac{1}{f}-\frac{1}{f_{1}}$$
= $$\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}$$
∴ f2 = -90 cm
Let the refractive index of the lens be μ1 and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is R.
R can be obtained using the relation $$\frac{1}{f_{1}}=\left(\mu_{1}-1\right)\left(\frac{1}{R}+\frac{1}{-R}\right)$$
$$\frac{1}{30}=(1.5-1)\left(\frac{2}{R}\right)$$
∴ R = $$\frac{30}{0.5 \times 2}$$ = 30 cm
Let μ2 be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane minor = ∞
Radius of curvature of the liquid on the side of the lens, R = -30 cm
The value of μ2, can be calculated using the relation
$$\frac{1}{f_{2}}=\left(\mu_{2}-1\right)\left[\frac{1}{-R}-\frac{1}{\infty}\right]$$
$$\frac{-1}{90}=\left(\mu_{2}-1\right)\left[\frac{1}{+30}-0\right]$$
μ2 – 1 = $$\frac{1}{3}$$
∴ μ2 = $$\frac{4}{3}$$ = 133
Hence, the refractive index of the liquid is 1.33. |
# How do you solve 2x^2+3x+5=0 using the quadratic formula?
##### 1 Answer
Nov 13, 2016
Please see the explanation for steps leading to:
$x = - \frac{3}{4} + \frac{\sqrt{31} i}{4}$ and $x = - \frac{3}{4} - \frac{\sqrt{31} i}{4}$
#### Explanation:
Comparing with the standard form, $a {x}^{2} + b x + c$, we see that:
$a = 2$
$b = 3$
$c = 5$
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$
Substitute in the values of a, b, and c:
$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(2\right) \left(5\right)}}{2 \left(2\right)}$
Simplify:
$x = \frac{- 3 \pm \sqrt{- 31}}{4}$
We must factor out $\sqrt{- 1}$ and write it as $i$
$x = - \frac{3}{4} + \frac{\sqrt{31} i}{4}$ and $x = - \frac{3}{4} - \frac{\sqrt{31} i}{4}$ |
# How do you simplify x^2/(x^2-4) = x/(x+2)-2/(2-x) ?
Apr 14, 2017
The given equation represents an impossible relation,
...unless (see below)
#### Explanation:
If we attempt to simplify the given equation (by multiplying both sides by $\left({x}^{2} - 4\right)$ with the assumption $x \notin \left\{- 2 , + 2\right\}$
x^2=x(x-2)+(2(x+2)
$\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 2 x}}} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{+ 2 x}}} + 4$
$0 = 4 \textcolor{w h i t e}{\text{XXXX}}$Impossible!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As an alternative answer to the one given above, it is possible to claim a pair of solutions: $\textcolor{m a \ge n t a}{x = - 2}$ or $\textcolor{m a \ge n t a}{x = + 2}$
and if we look at the graphs for the left and right sides of the given equation this (sort of) makes sense:
The limits at $x = \pm 2$ when approached from the same sides are equal. |
Browse Questions
# Solve the system of inequalities : $x+y \leq 9; y > x;x \geq 0$
Toolbox:
• To represent the solution of linear inequality of one or two variable in a plane if the inequality involves $'\geq'$ or $' \leq$ we draw the graph of the line as a thick line to indicate the line is included in the solution set.
• If the inequality involves $'>'$ as $'<'$ we draw the graph of the line using is not included in the solution set.
• To solve an inequality $ax+by > c \qquad a \neq 0, b \neq 0 ( or \;> )$
• We consider the corresponding equation $ax+by =c$ which represents a straight line This line divides the plane into two half planes I and II
• We take any point in I half plane and check if it satisfies the given inequality will be one half plane (called solution region ) Containing the point satisfying the inequality
Step 1:
The first inequality is $x+y \leq 9$----(1)
Consider the equation $x +y =9$
The points (9,0) and (0,9) satisfy the equation.
The graph of the line is drawn as shown.
The line divides the xy -plane into two half planes .
Consider the point (0,0) We see that
$0+0 \leq 9$
$0 \leq 9$ is true.
Thus the inequality $x+y \leq 9$ represents the region below the line $x+y =9$ containing the point (0,0) including the line .
Step 2:
The second inequality $y > x$ ----(2)
consider the equation $y=x$
The points $(1,1) ,(0,0)$ satisfy the equation.
The graph of the line $y=x$ is drawn using dotted line as shown .
The graph divides the xy plane into two half planes.
consider the point $(1,0)$ we see that
$0 > 1$ is false.
Thus the region represented by the inequality $y >x$ is above the line not containing point (1,0) (excluding the line y=x)
Step 3:
The third inequality $x \geq 0$
The inequality represents the region right of the y-axis (including y axis )
Step 4:
The solution of the system of inequality is represented by the common shaded region including the points on the line $x+y =9$ and $x=0$ and excluding the points on the line $y=x$. |
Why negative and fractional powers work the way they do
Most of the students I help have a pretty good grasp of the three straightforward power laws:
$(x^a)^b = x^{ab}$
$x^a \times x^b = x^{a+b}$
$x^a \div x^b = x^{a-b}$
So far, so dandy - and usually good enough if you're hoping for a B at GCSE. The trouble comes when they start throwing strange things in: what's $3^{-2}$? Or $81^{\frac14}$? Or, for the love of all that's holy, $16^{-\frac32}$? How on earth do you multiply something by itself negative two times? Or a quarter of a time?
Non-positive powers
Non-positive numbers are probably the easier of the two to get to grips with, and I have two ways to explain them. The first involves making a list:
$10^3 = 1,000$
$10^2 = 100$
$10^1 = 10$ ... you see how it's dividing by 10 each time? That pattern continues:
$10^0 = 1$
$10^{-1} = \frac{1}{10}$
$10^{-2} = \frac{1}{100}$
... and so on. In general, $x^{-k} = \frac{1}{x^k}$ - the negative power just 'flips' whatever you're working with and turns it into a fraction.
That means $3^{-2} = \frac{1}{3^2} = \frac19$; similarly, $2^{-6} = \frac{1}{2^6} = \frac{1}{64}$.
The second argument is that $3^{-2}$ must be the same as $3^{0-2} = 3^0 \div 3^2 = \frac{1}{9}$. Easy!
Non-integer powers
Fractional powers are a bit harder to get your head around, but they do make sense - fractions, remember are really division sums. Division sums are the opposite of multiplications.
Remember that $x^{ab} = (x^{a})^b$? Well, it stands to reason - since roots are the opposites of powers - that $x^{\frac ab}$ is the same as $\sqrt[b]{x^a}$.
So, to work out $81^\frac14$, you need to work out the fourth root of 81. 81 is $9^2$, or $3^4$, so $81^\frac14 = 3$.
In the same vein, $8^\frac23 = \sqrt[3]{8}^2 = 2^2 = 4$.
Combining the two
And how about when they're combined? Well, you break it down into small steps. If you've got $16^{-\frac32}$, you deal with the ugliest thing first: the bottom of the fraction. That means 'square root', so you're left with $4^{-3}$. Already looking better! $4^3 = 64$, so you've got $64^{-1}$; the power of negative one is just the reciprocal - so your answer is $\frac{1}{64}$.
Colin
Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.
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# Exponents and powers class 8 worksheets with answers
In this page we have exponents and powers class 8 worksheets with answers .This worksheet has all format of questions covering the whole chapter. Questions format is Multiple choice questions, Match the column, Fill in the blank, short answer type. Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
Evaluate
1. 2-2
2. (-2)-2
3. (3/2)-5
As we know that
b-n= 1/bn
(i) 2-2 = 1/ 22 =1/4
(ii) (-2)-2 = 1/(-2)2= 1/4
(iii) (3/2)-5 =3-5/ 2-5 =25/35 = 32/243
Question 2
Simplify and express the result in power notation with positive exponent.
(i) (-2)5 ÷ (-2)4
(ii) (1/2)2 × (2/5)2
(iii) (-5)2 × (3/5)
(i) (-2)5 ÷ (-2)4
= (-2)5 / (-2)4
= (-2)5-4
=-2
(ii) (1/2)2 × (2/5)2
= (1/4) X (4/25)
= 1/25
(iii) (-5)2 × (3/5)
=25 × (3/5) =15
Question 3
Find the value of.
(i)(40 + 4-1) × 22
(ii)(3-1 × 9-1) ÷ 3-2
(iii)(11-1 + 12-1 + 13-1)0
(i)(40 + 4-1) × 22 = (1+1/4) × 4
= 4+1=5
(ii)(3-1 × 9-1) ÷ 3-2 = [(1/3) × (1/9)] ÷ (1/9)
=1/3
(iii)(11-1 + 12-1 + 13-1)0
=1 as a0=1
Question 4
Find the value of x here
$( \frac {11}{9})^3 \times (\frac {9}{11})^6 = (\frac {11}{9})^{2x-1}$
$( \frac {11}{9})^3 \times (\frac {9}{11})^6 = (\frac {11}{9})^{2x-1}$
$(\frac {11}{9})^{3-6} = (\frac {11}{9})^{2x-1}$
-3=2x-1
Or
x=-1
Question 5
Find the value of m for which 2m ÷ 2-4 = 45
2m ÷ 2-4 = 45
2m × (1/2-4) = 210
2m+4 =210
So m+4=10
m=6
Question 6
Express the following numbers in standard form.
(i) 0.0000000015
(ii) 0.00000001425
(iii) 102000000000000000
(i)0.0000000015
=1.5 ×10-9
(ii)0.00000001425
=1.425×10-8
(iii)102000000000000000
=1.02×1017
Question 7
Express the following numbers in usual form.
(i) 34.02 x 10-5
(ii) 9.5 x 105
(iii) 9 x 10-4
(iv) 2.0001 x 108
(i)34.02 x 10-5 = .0003402
(ii)9.5 x 105
=950000
(iii) 9 x 10-4
=.0009
(iv)2.0001 x 108
=200010000
Question 8
Solve for the variables
(i) $\frac {1}{y^{-5}} = 32$
(ii) $(\frac {1}{7})^3 \div (\frac {1}{7})^8 = 7^n$
(iii) $\frac {3^x .3}{3^{2x+1}} = 27$
(i) $\frac {1}{y^{-5}} = 32$
$y^5 = 32$
$y^5 = 2^5$
Therefore y=2
(ii) $(\frac {1}{7})^3 \div (\frac {1}{7})^8 = 7^n$
$(\frac {1}{7})^3 \times 7^8 = 7^n$
$7^5 = 7^n$
Hence n=5
(iii) $\frac {3^x .3}{3^{2x+1}} = 27$
$\frac {3^{x+1}}{3^{2x+1}} = 27$
$3^{x + 1 - 2x -1} =27$
$3^{-x}= 27$
$3^{-x}=3^3$
Hence x =-3
Question 9
Find the Multiplicative inverse of
(i) $3^{25}$
(ii) $4^{-3}$
(iii) $(\frac {2}{3})^{-2}$
(i) $3^{-25}$
(ii) $4^{3}$
(iii) $(\frac {2}{3})^{2}$
Question 10
Simplify the following
(i) $1 + \left [ \left ( \frac{1}{3} \right )^{-3} - \left ( \frac{1}{2} \right )^{-3} \right ] \div 38$
(ii) $\frac {1}{1 + p^{a-b}} + \frac {1}{1 + p^{b-a}}$
(i) $1 + \left [ \left ( \frac{1}{3} \right )^{-3} - \left ( \frac{1}{2} \right )^{-3} \right ] \div 38$
=$1 + (3^3 - 2^3) \div 38$
=$1 + 19 \div 38$
=$\frac {3}{2}$
(ii) $\frac {1}{1 + p^{a-b}} + \frac {1}{1 + p^{b-a}}$
=$\frac {p^b}{p^b + p^a} + \frac {p^a}{p^a + p^b}$
=$\frac {p^b + p^a}{p^b + p^a}$
=1
## True and false
Question 11
True and False
(i) Very small numbers can be expressed in standard form using negative exponents.
(ii) $a^p \times b^q= (ab)^{pq}$
(iii) $.00567 = 5.67 \times 10^{-3}$
(iv) $\frac {1}{(8)^{-3}}= 2^9$
(v) $4^0 =4$
(vi)$(-1)^3=1$
(vii) The multiplicative inverse of $(–2)^{–2}$ is $(2)^2$.
(i)True
(ii) False
(iii) True
(iv) True as $\frac {1}{(8)^{-3}}=8^3=(2^3)^3= 2^9$
(v) False
(vi) False
(vii) True as $(–2)^{–2}= \frac {1}{4}$
## Fill in the blanks
Question 12
Fill in the blank
(i) The multiplicative inverse of $(3)^{–2}$ is _____
(ii) $9^7 \times 3^{-2}$=_____
(iii) .0000067 in exponent form _____
(iv) Very large numbers can be expressed in standard form by using_________ exponents
(v) $5^0$ =____
(vi)$(-1)^100$=____
(vii) The value of $[4^{–1} +3^{–1} + 6^{–2}]^{–1}$ is ____.
(i)$(3)^{2}$
(ii) $3^{11}$
(iii) $6.7 \times 10^6$
(iv) positive
(v) 1
(vi) 1
(vii) 18/11
Question 13
(p) -> (ii)
(q) -> (iii)
(r) -> (iv)
(s) -> (i)
## Multiple choice questions
Question 14
If a be any non-zero integer, then $a^{–1}$ is equal to
(a) a
(b) -a
(c) 1/a
(d) -1/a
Question 15
The multiplicative inverse of $(–7)^{–2} \div (90)^{–1}$ is
(a) $-(7)^2 \times (90)^{–1}$
(b) $-(7)^{-}2 \times (90)^{–1}$
(c) $-(7)^2 \div (90)^{–1}$
(d) $-(7)^2 \times (90)^{1}$
Question 16
If $5^{3x–1} \div 25 = 125$, Then the value x is
(a) 1
(b) 2
(c) 3
(d) 4
(13) (c)
(14) $(–7)^{–2} \div (90)^{–1}= - (1/7)^2 \times 90^1$. So Multiplication inverse is $-(7)^2 \times (90)^{–1}$. Hence Option (a)
(15) $5^{3x–1} \div 25 = 125$
$5^{3x–1} \times 5^2=5^3$
$5^{3x-3}=5^3$
Hence x=2
Option(b) is correct
## Summary
This exponents and powers class 8 worksheets with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail. |
Understanding the Concept: Combining Like Terms in Mathematics
In the vast and intricate world of algebra, mastering the concept of combining like terms stands as a crucial foundational skill. This technique, simple in its essence, dramatically simplifies algebraic expressions and equations, making complex mathematical problem-solving more accessible. By identifying and merging terms with common variables and powers, mathematicians can condense equations to their most succinct form.
This article will provide a comprehensive understanding of what it means to combine like terms, how to do it effectively, and the importance of this process in mathematics, all illustrated with relevant examples.
What are Like Terms?
Before we dive into combining like terms, let’s start with combining like terms definition:
In mathematics, like terms, refer to algebraic expressions that share the same variable to the same power. This means that both the variable and its exponent must be identical. These shared components can be multiplied by any constant, which could be different for different terms.
Example: Consider the following terms: 5x, 3x, and 7x2.
• The terms 5x and 3x are like terms because they share the same variable (x) raised to the same power (1, in this case).
• However, 7x2 isn’t a like term to 5x and 3x because, though it has the same variable (x), the power (2) is different.
Process of Combining Like Terms
Combining like terms involves adding or subtracting the numerical coefficients of like terms to simplify the expression or equation. When the coefficients are added together or subtracted, the resulting term remains a ‘like term’ because it has the same variable and the same power.
Steps to Combine Like Terms
1. Identify like terms: Look for terms with the same variables raised to the same power.
2. Addition or subtraction: Combine the coefficients of the like terms by addition or subtraction, depending on the operation sign in front of the terms.
3. Write the resulting term: The combined term will have the same variable and exponent.
Example of Combining Like Terms
Consider the expression: 5x + 3x – 7x2 + 9x2 – 2x + 8 – 4.
First, identify the like terms:
• Like terms with x: 5x, 3x, and -2x
• Like terms with x2: -7x2 and 9x2
• Constants: 8 and -4
Next, add or subtract the coefficients of the like terms:
• Combine coefficients of x: 5x + 3x – 2x = 6x
• Combine coefficients of x2: -7x2 + 9x2 = 2x2
• Combine constants: 8 – 4 = 4
The simplified form of the original expression is: 2x2 + 6x + 4.
The Importance of Combining Like Terms
Combining like terms plays a crucial role in many areas of mathematics. It is employed in solving linear and quadratic equations, simplifying polynomial expressions, and in various multiplier situations. It is a basic tactic in algebra that allows for brevity and clarity, generating mathematically crisp and less complex solutions.
Quick Tips for Combining Like Terms
• Consider Sign Changes: Always pay attention to the operation signs before each term. The sign before a term belongs to that term. Therefore, a change in sign means a change in operation.
• Practice Regularly: By combining like terms and other algebraic simplification techniques, proficiency comes with practice. Regularly working with various expressions will make identifying and combining like terms almost instinctive.
Conclusion
In essence, combining like terms is an indispensable tool in the world of algebra. It condenses our expressions and equations into their simplest forms, allowing for enhanced understanding and easier problem-solving. By practicing the identification and merging of like terms, one can significantly improve their mathematical dexterity, thus paving the way for more complex and fascinating algebraic explorations. |
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# Mixed Numbers: Subtraction
When a number is expressed as a combination of a whole number and a fraction, such as $4\frac{2}{3}$ , we call that a mixed number. It would be difficult to add and subtract mixed numbers as they are written.
Imagine the problem $5\frac{2}{3}+3\frac{5}{6}$ . You probably know how to subtract whole numbers and how to subtract fractions, but how does it work when the fraction you are subtracting is larger than the fraction you are subtracting it from? There is a more straightforward way to subtract mixed numbers, and that is to write them as improper fractions.
## Subtracting mixed numbers with like denominators
As with subtracting fractions, subtracting mixed numbers with like denominators is easier than subtracting those with unlike denominators. Let's try a problem.
Subtract the following:
$5\frac{3}{4}-3\frac{1}{4}$
The first step is to write the mixed numbers as improper fractions.
$5$ and $\frac{3}{4}$ is equivalent to $\frac{23}{4}$
$3$ and $\frac{1}{4}$ is equivalent to $\frac{13}{4}$
Because the denominators are already the same, all you need to do is subtract the numerators.
$\frac{23}{4}-\frac{13}{4}=\frac{10}{4}$
Then turn the answer back into a mixed number and simplify it.
$\frac{10}{4}$ simplifies to $2\frac{1}{2}$ .
## Subtracting mixed numbers with unlike denominators
Subtracting mixed numbers with unlike denominators includes an extra step, the same as subtracting fractions with unlike denominators does. Let's try a problem to see how it works.
Subtract the following:
$9\frac{1}{2}-5\frac{3}{4}$
For the first step, we write the mixed numbers as improper fractions.
$9\frac{1}{2}$ is equivalent to $\frac{19}{2}$
$5\frac{3}{4}$ is equivalent to $\frac{23}{4}$
Now the problem is:
$\frac{19}{2}-\frac{23}{4}$
To find the difference, the next step is to find the least common denominator (LCD) of the fractions.
The least common multiple (LCM) of the two denominators 2 and 4 is 4. Therefore, the least common denominator (LCD) of the fractions is 4.
First, we rewrite $\frac{19}{2}$ using the LCD of 4.
$\frac{19}{2}$ is equivalent to $\frac{19}{2}$ times $\frac{2}{2}$ which is $\frac{38}{4}$
So the subtraction now is:
$\frac{19}{2}-\frac{23}{4}$
$\frac{38}{4}-\frac{23}{4}$
Since the denominators are now the same, we can subtract the numerators to find the answer.
$\frac{38}{4}-\frac{23}{4}=\frac{15}{4}$
Write the improper fraction as a mixed number and you have the solution.
$\frac{15}{4}$ simplifies to $3\frac{15}{4}$ .
## Get help learning about subtracting mixed numbers
If your student is struggling with subtracting mixed numbers, it might be a good idea to have them get together with an expert math tutor. A tutor can figure out your student's learning style and use tutoring methods that are most effective for them. They can work at your student's pace so your student can clearly understand each concept before moving on to the next. To find out more about how tutoring can help your student with subtracting mixed numbers, contact the Educational Directors at Varsity Tutors today.
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# How do you find cos(x/2) given sin(x)=1/4?
Oct 17, 2017
$\cos \left(\frac{x}{2}\right) = 0.99$
$\cos \left(\frac{x}{2}\right) = 0.127$
#### Explanation:
$\sin x = \frac{1}{4}$ . Find cos x
${\cos}^{2} x = 1 - {\sin}^{2} x = 1 - \frac{1}{16} = \frac{15}{16}$
$\cos x = \pm \frac{\sqrt{15}}{4}$
There are 2 values of cos x because if $\sin x = \frac{1}{4}$, x could either be in Quadrant 1 or in Quadrant 2
Use trig identity:
$2 {\cos}^{2} \left(\frac{x}{2}\right) = 1 - \cos 2 a$
In this case:
${\cos}^{2} \left(\frac{x}{2}\right) = \frac{1}{2} \pm \frac{\sqrt{15}}{8} = \frac{1}{2} \pm 0.484$
a. ${\cos}^{2} \left(\frac{x}{2}\right) = 0.984$
b. ${\cos}^{2} \left(\frac{x}{2}\right) = 0.016$
a. $\cos \left(\frac{x}{2}\right) = \sqrt{0.984} = 0.99$
b. $\cos \left(\frac{x}{2}\right) = \sqrt{0.016} = 0.127$
Check with calculator.
a. $\cos \left(\frac{x}{2}\right) = 0.99$ --> $\frac{x}{2} = {7}^{\circ} 27$ --> $x = {14}^{\circ} 53$
$\sin \left({14}^{\circ} 52\right) = 0.25$. Proved
b. $\cos \left(\frac{x}{2}\right) = 0.127$ --> $\frac{x}{2} = {82}^{\circ} 70$ --> $x = {165}^{\circ} 41$
$\sin \left({165}^{\circ} 41\right) = 0.25$. Proved |
# Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.
Question:
Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.
Solution:
Here, we are given an A.P., whose $n^{\text {th }}$ term is given by the following expression, $a_{n}=5-6 n$
So, here we can find the sum of the $n$ terms of the given A.P., using the formula, $S_{n}=\left(\frac{n}{2}\right)(a+l)$
Where, a = the first term
l = the last term
So, for the given A.P,
The first term (a) will be calculated using in the given equation for nth term of A.P.
$a=5-6(1)$
$=5-6$
$=-1$
Now, the last term (l) or the nth term is given
$a_{e}=5-6 n$
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
$S_{n}=\left(\frac{n}{2}\right)[(-1)+5-6 n]$
$=\left(\frac{n}{2}\right)[4-6 n]$
$=\left(\frac{n}{2}\right)(2)[2-3 n]$
$=(n)(2-3 n)$
Therefore, the sum of the $n$ terms of the given A.P. is $(n)(2-3 n)$. |
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## Lesson Tutor : Algebra Lesson : Balancing Equations to Solve Variables (Finding ‘x’)
/ Lesson Tutor : Algebra Lesson : Balancing Equations to Solve Variables (Finding ‘x’)
Basic Algebra – Lesson 6
Balancing Equations to Solve Variables
by Elaine Ernst Schneider
Objective(s): By the end of this lesson the student will be able to:
balance equations to solve for variables (solve for ‘x’)
Pre Class Assignment: Completion or review of Basic Algebra – Lesson 5
Resources/Equipment/Time Required:
Outline:
In the last lesson, you learned to write expressions and to find the values of expressions.
When two expressions can be written to balance or equal one another, it is called an
equation. In other words, whatever is expressed on one side of the equal sign is calculated to be the exact same value as whatever is on the other side of the equal sign. IN other words, the left member and the right member of the equation “balance.”
Examples of equations:
5 + 7 – 2 = 2(5) (the answer on the left side is 10 and the answer on the right side is 10)
100/25 = 347 – 343 (the answer is 4 on the left of the equal sign and 4 on the right also)
Equations involving unknown variables are solved by balancing the left member and the right member. For example, in the equation y + 5 = 12, I know that the left member must equal 12 for it to balance with the right member. This means y would be 7 because 7 + 5 = 12. That one is easy enough to do the math without a formal procedure. But because algebraic equations can become much longer and more calculated, a system is necessary to solve for variable and balance equations.
Let’s take the above equation:
y + 5 = 12
The proper procedure in solving for y is to “isolate” y. This means we want y to stand by itself on one side of the equal sign.
Now, keep in mind that we want every thing to “balance” on either side of the equal sign. This means whatever I do to the left member, I must do to the right member. So, to isolate y and solve the equation, I must “move” the 5 to the other side of the equal sign. To do this, I must make it zero on the left side of the equation by subtracting 5:
y + 5 = 12
y + 5 – 5 = 12 – 5
y + 0 = 12 – 5
y = 7
Try solving for the variable by keeping each side equal. Remember, the idea is to “balance” the equation, so what you do to one side, you must do to the other.
100/5 = y + 2
100/5 -2 = y +2 – 2
20 – 2 = y
18 = y **
**The answer displayed prior to June 10/01 was incorrect and unreported. Thanks to the class of L. Crawford for the notice. JM
Now it’s time for you to try a few…
EXERCISE:
1. y – 10 = 17
2. c + 4 = 29
3. 5y = 90
4. 1/3 t = 29
5. 6y = 72
6. 1/9 g = 58
7. 4y = 100
8. 12a = 132
9. r + 9 = 48
10. x + 79 = 422
Cut Here —————————————————–
1. 27
2. 25
3. 18
4. 87
5. 12
6. 522
7. 25
8. 11
9. 39
10. 343
Pre-Requisite To: Basic Algebra Lesson 7 |
# By using properties of determinants, show that:
Question:
By using properties of determinants, show that:
(i) $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$
(ii) $\left|\begin{array}{llr}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$
Solution:
(i) $\Delta=\left|\begin{array}{ccr}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we have:
$\Delta=\left|\begin{array}{lll}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{lll}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$, we have:
$\Delta=(a+b+c)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c)\end{array}\right|$
$=(a+b+c)^{3}\left|\begin{array}{lll}1 & 0 & 0 \\ 2 b & -1 & 0 \\ 2 c & 0 & -1\end{array}\right|$
Expanding along C3, we have:
$\Delta=(a+b+c)^{3}(-1)(-1)=(a+b+c)^{3}$
Hence, the given result is proved.
(ii) $\Delta=\left|\begin{array}{clr}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|$
Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$, we have:
\begin{aligned} \Delta &=\left|\begin{array}{lll}2(x+y+z) & x & y \\ 2(x+y+z) & \quad y+z+2 x & y \\ 2(x+y+z) & x & z+x+2 y\end{array}\right| \\ &=2(x+y+z)\left|\begin{array}{lll}1 & x & y \\ 1 & y+z+2 x & y \\ 1 & x & z+x+2 y\end{array}\right| \end{aligned}
Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we have:
\begin{aligned} \Delta &=2(x+y+z)\left|\begin{array}{llll}1 & & x & y \\ 0 & & x+y+z & 0 \\ 0 & & 0 & x+y+z\end{array}\right| \\ &=2(x+y+z)\left|\begin{array}{lll}1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right| \end{aligned}
Expanding along R3, we have:
$\Delta=2(x+y+z)^{3}(1)(1-0)=2(x+y+z)^{3}$
Hence, the given result is proved. |
# Find the equation of the line passing through
Question:
Find the equation of the line passing through ( - 3, 5) and perpendicular to the line through the points (2, 5) and ( - 3, 6).
Solution:
Given: The line perpendicular to the line passing through $(2,5)$ and $(-3,6)$ passes through $(-3,5)$.
Formula to be used: If $(a, b)$ and $(c, d)$ are two points then the equation of the line passing through them is
$\frac{y-b}{x-a}=\frac{b-d}{a-c}$
Product of slopes of two perpendicular lines $=-1$
The equation of the line joining points $(2,5)$ and $(-3,6)$ is
$\frac{y-5}{x-2}=\frac{5-6}{2-(-3)}$
or, $\frac{y-5}{x-2}=\frac{-1}{5}$
Or, $5 y-25=-x+2$
i.e. the given line is x + 5y = 27.
The slope of this line is $-1 / 5$.
$\therefore$ the slope of the perpendicular line =
$\frac{-1}{-1 / 5}=5$
The equation of the line can be written in the form y = 5x + c
(c is the y - intercept)
This line passes through ( - 3,5).
Hence, 5 = 5x( - 3) + c or, c = 20
The required equation of the line will be y = 5x + 20
i.e. 5x - y + 20 = 0 |
## Probability and Statistics Background
Statistics – A subject which most statisticians find difficult, but in which nearly all physicians are expert. – Stephen S. Senn
## Introduction
For us, we will regard probability theory as a way of logically reasoning about uncertainty. I realize that this is not a precise mathematical definition, but neither is ‘probability theory is the mathematics arising from studying non-negative numbers which add up to 1’, which is at least partially accurate.
Some additional material is covered elsewhere:
* Statistical inference.
To get well-grounded let’s begin with a sequence of definitions.
### First definitions
#### Definition
A probability space is a measure space $D$ with measure $P$ such that $P(D)=1$. The space $D$ is also sometimes called the sample space and the measurable subsets of $D$ are called events1.
#### Remark
The definition of probability space is sufficiently general to include lots of degenerate examples. For example, we can take any set $S$ and make it into a probability space by decreeing that the only measurable subsets are $S$ and $\emptyset$ with $P(S)=1$. Although we will try to make this explicit, we will almost always want that singleton sets, i.e., sets with just a single element, are measurable. When a probability space has this property and every measurable subset is a countable union of singleton sets we will call the probability space discrete.
#### Exercise
Make the positive integers into a discrete probability space where every point has non-zero probability.
#### Definition
The probability of an event $E$ is $P(E)=:\int_E dP$. For discrete probability spaces we can also write $\int_E dP=\sum_{x\in E} P(x)$2.
#### Construction
Given two probability spaces $D_1$ and $D_2$ with respective probability measures $P_1$ and $P_2$. We can define a probability space $D_1\times D_2$ by:
1. The underlying set is the cartesian product $D_1\times D_2$.
2. The measurable subsets are generated under countable unions and complements by the products sets $I_1\times I_2$, where $I_1\subseteq D_1$ and $I_2\subseteq D_2$ are measurable subsets.
3. The probability measure is determined by $P(I_1\times I_2)=P_1(I_1)\cdot P(I_2)$, where $I_1$ and $I_2$ are as in the previous statement.
#### Example
Suppose we have a fair coin that we flip twice. Each of the four possible outcomes $D=\{HH,HT,TH,TT\}$ are equally likely and form a discrete probability space such that $P(x)=1/4$ for all $x\in D$. The probability of the event $E$, where we get precisely one head, is $P(E)=P(HT)+P(TH)=1/2$.
#### Definition
A random variable $X\colon D\to T$ is a measurable function from a probability space $D$ to a measure space $T$.
We can associate to each such $X$ a probability measure $P_X$ on $T$ by assigning to each measurable subset $U\subset T$, $P_X(U)=P(X^{-1}(U))$. Indeed it is clear that $P_X(T)=1$ and that for the measure of a countable disjoint union is $$P_X(\coprod U_i)=P(X^{-1}(\coprod U_i))=P(\coprod(X^{-1}U_i))=\sum P(U_i).$$
#### Remark
There is an unfortunate clash in the language of probability theory and standard english usage. For example, imagine that we have a box with a single button on it and a numerical display. Every time we push the button the screen displays a number between 1 and 10. In common usage we say that these values are random if there is no way to know which number will appear on the screen every time we push the button.
It is important to know that mathematics/probability theory/statistics do not provide any such mechanism. There is no function whose values are “randomly” chosen given a particular input. In particular, mathematics does not provide a method of randomly choosing objects.
One should keep this in mind when talking about random variables. Random variables are not objects with random values; they are functions. The additional data that a random variable $X$ does define are numbers associated to the preimages $X^{-1}(I)$ (for measurable subsets $I$), which we can use to weight the values of $X$.
This can also be used to shed light on statistical mechanics, which uses probability theory to model situations arising in physics. The fact that such models have been extremely successful in the field of quantum mechanics does not necessarily mean there is something random, in the common usage sense, about the universe; we are not claiming that “God plays dice with the universe”. It is just that our best mathematical models for these phenomena are constructed using the language of probability theory.
Finally, we should remark that the closest mathematical object to a random number generator in the sense of english is a pseudorandom number generator. These are deterministic functions which output sequences of numbers which attempt to model our intuition of what a random number generator should produce. Although not truly random, these are heavily used in simulations and Monte Carlo methods.
#### Conventions
If we are regarding $\Bbb R$ as a measure space and do not specify an alternative measure, we will mean that it is equipped with its standard Borel measurable subsets and the Borel measure3 $E\mapsto \int_E dx$. If we regard a discrete finite set $S$ or any interval $[a,b]$ (with $a<b$) as a probability space and do not specify the measure then we will mean that it is equipped with a uniform measure. In other words, $P(s)=1/|S|$ for all $s\in S$ and for all measurable $E\subset I$ we have $P(E)=P_{\Bbb R}(E)/(b-a)$.
#### Remarks4:
If the measure $P_X$ from the previous definition is absolutely continuous with respect to the standard Borel measure (i.e., the preimage of every measure 0 set with respect to the standard Borel measure is of measure 0), then there is a measurable function $dP_X/dx \colon T\to \Bbb R$ such that for all measurable $E\subset T$, $$P_X(E) := \int_{X^{-1} E} dP := \int_E dP_X = \int_{E} \frac{dP_X}{dx} dx.$$ All of these integrals are Lebesgue integrals.
The measurable function $dP_X/dx$ is called a Radon-Nikodym derivative and any two such derivatives disagree on a set of measure 0, i.e., they agree almost everywhere. Without the absolute continuity hypothesis there is only a distribution satisfying this property. Having a measure defined in such a way obvious implies absolute continuity, so the first sentence can be formulated as an if and only if statement. This is the Radon-Nikodym theorem.
#### Definition
For a discrete probability space $D$ the function $p\colon D\to [0,1]$, defined by $d\mapsto p(d):=P(d)$ is called the probability mass function (PMF) of $D$.
Note that the measure $P$ on $D$ is uniquely determined by the associated probability mass function.
#### Definition
Suppose that $\Bbb R$ is equipped with a probability measure $P$ and the cumulative distribution function (cdf) $F(a)=P(x\leq a)$ is a continuously differentiable function of $a$, then $F(x)=\int_{-\infty}^x F'(x) dx$ and $F'(x)$ is called the probability density function (pdf) of $F$ (or $P$).
Note that the probability measure $P$ is determined by $F$ and hence the probability density function $F’$. This can lead to some confusing abuses of language.
#### Example
Let $D$ be the probability space from the first example. Let $X\colon D\to \mathbb{R}$ be the random variable which counts the number of occurrences of heads in a given event. Then the cumulative density function of $P_X$ is $F_X(x)=0$ if $x<0$, $F_X(x)=1/4$ if $0\leq x < 1$, $F_X(x)=3/4$ if $1\leq x < 2$ and $F_X(x)=1$ if $x\geq 2$. This function is discontinuous and hence the probability density function is not defined5.
### Moments of distributions
Typically when we are handed a probability space $D$, we analyze it by constructing a random variable $X\colon D\to T$ where $T$ is either a countable subset of $\Bbb R$ or to $\Bbb R$. Using the procedure of the previous section we obtain a probability measure $P_X$ on $T$ and we now study this probability space. Usually a great deal of information is lost about $D$ during this process, but it allows us to focus our energies and work in the more tractable and explicit space $T\subset\Bbb R$.
So, we know focus on such probability spaces. This is usually decomposed into two cases, when $T$ is discrete (e.g., a subset of $\Bbb N$) and when $T$ is $\Bbb R$ (or some interval in $\Bbb R$). We could study the first case as a special case of the latter and just studying probability measures on $\Bbb R$, but that would require throwing in a lot of Dirac delta distributions at some point and I sense that you may not like that. We will seek a compromise and still use the integral notation to cover both cases although integrals in the discrete case can be expressed as sums.
There are two special properties of this situation that we will end up using:
1. It makes sense to multiply elements of $T$ with real valued functions.
2. There is a natural ordering on $T$ (so we can define a cdf).
3. We can now meaningfully compare random variables with values in $\Bbb R$ which are defined on different probability spaces, by comparing their associated probability measures on $\Bbb R$ (or their cdfs/pdfs when these exist).
For example the first property allows us to make sense of:
#### Definition
1. The expected value or mean of a random variable $X\colon D\to T\subset \Bbb R$, is $$\mu_X:=E(X)= \int_{x\in T} x\cdot dP_X = \int_{d\in D} X(d) dP.$$
2. Let $F_X$ denote the cdf of $X$. The median of $X$ is those $t\in \Bbb R$ such that $F_X(t)=0.5$.
3. Suppose that $X$ admits a pdf $f_X$. The modes of $X$ are those $t\in \Bbb R$ such that $f_X(t)$ is maximal.
#### Example
In our coin flipping example, the expected value of the random variable $X$ which counts the heads is $$\int_D X dP = \sum_{d\in D} X(d)p(d) = 2/4+1/4+1/4+0/4=1,$$
as expected.
The third property lets us make sense of:
#### Definition
Two random variables $X\colon D_1\to T$ and $Y\colon D_2\to T$ are identically distributed if they define the same probability measure on $T$, i.e., $P_X(I)=P_Y(I)$ for all measurable subsets $I\subseteq T$. In this case, we write $X\sim Y$.
#### Definition
We associate to two random variables $X,Y\colon D\to T$ a random variable $X\times Y\colon D \to T^2$ by $X\times Y(x)=(X(x),Y(x))$. This induces a probability measure $P_{X,Y}$ on $T^2.$ When $T=\Bbb R$ we can then define an associated joint cdf, $F_{X,Y}\colon \Bbb R^2\to [0,1]$ defined by $F_{X,Y}(a,b)=P_{X,Y}(x\leq a, y\leq b)$, which when $X\times Y$ is absolutely continuous with respect to the Lebesgue measure admits a joint pdf. Similarly, we can extend this to joint probability distributions of any number of random variables.
#### Definition
Two random variables $X,Y\colon D\to T$ with corresponding probability measures $P_X$ and $P_Y$ on $T$ are independent if the associated joint probability measure $P_{X,Y}$ on $T^2$ satisfies $P_{X,Y}(I_1\times I_2)=P_X(I_1)P_Y(I_2)$ for all measurable subsets $I_1, I_2\subseteq T$. When two variables are both independent and identically distributed then we abbreviate this to iid.
#### Definition
Suppose that $T$ is a probability space whose singleton sets are measurable. A random sample of size $n$ from $T$ will be any point of the associated product probability space $T^n$.
#### Exercise
1. Show that if $X$ and $Y$ are two $\Bbb R$ random variables then they are independent if and only if their joint cdf is the product of their individual cdfs.
2. Suppose that moreover $X$ and $Y$ and the joint distribution admit pdfs $p_X, p_Y,$ and $p_{X,Y}$ respectively, then show that the $f_{X,Y}=f_X f_Y$ if and only if the distributions are independent.
#### Definition
1. The $k$th moment of a random variable $X\colon D \to \Bbb R$ is $E(X^k)$.
2. The variance of a random variable $X\colon D\to \Bbb R$ is
$$\sigma_X^2=E((X-\mu_X)^2)=\int_D (X-\mu_X)^2 dP$$.
3. The standard deviation of $X$ is $\sigma_X=\sqrt{\sigma_X^2}$.
4. The covariance of a pair of random variables $X,Y\colon D\to \Bbb R$ is
$$Cov(X,Y) = E((X-\mu_X)(Y-\mu_Y)).$$
5. The correlation coefficient of a pair of random variables $X,Y\colon D\to \Bbb R$ is $$\frac{Cov(X,Y)}{\sigma_X \sigma_Y}.$$
#### Exercise
Suppose that $X$ and $Y$ are two independent $\Bbb R$-valued random variables with finite means $\mu_X$ and $\mu_Y$ and finite variances $\sigma^2_X$ and $\sigma^2_Y$ respectively.
1. Show that, for $a,b\in \Bbb R$ the mean of $aX+bY$ is $a\mu_X + b\mu_Y$.
1. Show that the variance of $aX+bY$ is $a^2 \sigma^2_X + b^2 \sigma^2_Y$.
1. Show that $E(XY)$ is $\mu_X\cdot \mu_Y$.
1. Show that $E(X^2)=\sigma_X^2+\mu_X^2$.
#### Definition
The characteristic function of a random variable $X\colon D\to \Bbb R$ is the complex function $$\varphi_X(t)=E(e^{itX})=\int_{x\in \Bbb R} e^{itx} dP_X = \int_{d\in D} e^{itX(d)} dP.$$
#### Remarks
1. The characteristic function is always defined (because we are integrating an absolutely bounded function over a finite measure space).
2. When $X$ admits a pdf $p_X$, then up to a reparametrization the characteristic function is the Fourier transform of $p_X$: $F(p_X)(X)=\varphi_X(-2\pi t)$.
3. Two random variables have the same characteristic functions if and only if they are identically distributed6.
## Some Important Results
1. The Law of Large Numbers, which essentially says that the average $S_n$ of a sum of $n$ iif random variables with finite mean $\mu$ “converges” to the common mean.
2. The Central Limit Theorem, which says that under the above hypotheses plus the assumption that the random variables have a finite variance $\sigma^2$, the random variable $\sqrt{n}(S_n-\mu)$ converges in distribution to the normal distribution with mean $0$ and variance $\sigma^2$. This result is the basis behind many normality assumptions and is critical to hypothesis testing which is used throughout the sciences.
## Conditional probability
Suppose we have a probability space $P\colon D\to [0,1]$ and two events $A,B\in D$. Then we write $P(A,B)=P(A\cap B)$. Suppose that $P(B)>0$, then define the conditional probability of $A$ given $B$ as $$P(A|B)=P(A,B)/P(B).$$ A similar definition is also given for the conditional pdf of two random variables $X$ and $Y$: $f_{X,Y}(x|y)=f_{X,Y}(x,y)/f_Y(y)$ where $f_Y(y)=\int_{x \in \Bbb R} f(x,y) dx$ is the marginal density.
#### Bayes Rule
Let $A$ be an event in a probability space $P\colon D\to [0,1]$ and that $\{B_i\}_{i=1}^n$ is a disjoint union of events which cover $D$ and all of these events have non-zero probability. Then
There is also the pdf form:
$$f_{X,Y}(x|y)=\frac{f_{X,Y}(y|x)f_X(x)}{\int f_{X,Y}(y|x) f_X(x) dx}.$$
The usefulness of Bayes rule is that it allows us to write a conditional probability that we do not understand (the dependence of $X$ on $Y$) in terms that we might understand (the dependence of $Y$ on $X$).
1. If you don’t know what a measurable subset is then you don’t know what a measure space is and you should consult the references. If you don’t consult the references and you just believe that all subsets of $D$ are measurable, it will take a long time to find out that you are wrong.
2. Here and elsewhere we will abuse notation and denote measure of a singleton set ${{x}}$ by $P(x)$.
3. For $\Bbb R^n$ and $n>1$ we should probably use the Lebesgue measure, which is a completion of the product Borel measure.
4. This material is quite advanced, so don’t worry if it goes over your head. The notation here is chosen so that in special cases where the [fundamental theorem of calculus] (https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus) applies the Radon-Nikodym derivative can be chosen to be an actual derivative.
5. Although we could define the “pdf” as a linear combination of Dirac delta distributions, but then it wouldn’t be a function (no matter what a physicist tells you).
6. For a proof see these notes
## Supervised Learning
A big computer, a complex algorithm, and a long time does not equal science. – Robert Gentleman
## Examples
Before getting into what supervised learning precisely is, let’s look at some examples of supervised learning tasks:
#### Definition
Supervised learning is concerned with the construction of machine learning algorithms $f\colon 2^{D\times T}\times D\to T$. The subsets of $D\times T$ are referred to as subsets of labeled examples.
We often assume that the labeled examples arise from restricting some presumed function $F\colon D\to T$ whose values we know on $E\subset D$. We can then train $f$ on pairs $\{s, F(s) | s\in E\}$. We can then devise a cost function which measures the distance from the learned $f$ and the presumed $F$ (e.g., $L^2$-distance).
Supervised learning has been the most successful of the three branches to real world problems. The existence of labeled examples usually leads to well-defined performance metrics that transform supervised learning tasks into two other tasks finding an appropriate parametrized class of functions to choose $f$ from and an optimization problem of finding the best function in that class.
Typically $D$ will be some subset of $\Bbb R^n$ and we will refer to the components of $\Bbb R^n$ as features, dependent variables, or attributes (many concepts in machine learning have many names). Sometimes $D$ will be discrete, in which case we refer to these as categorical variables. There are a few simple tricks to map categorical variables into $\Bbb R^n$ (such as one-hot encoding), so it usually does not hurt to think of $D$ as a subset of $\Bbb R^n$.
When $T$ is discrete (typically finite), then the supervised learning problem is called a classification problem. When it is continuous (e.g, $\Bbb R$) then it is called a regression problem.
### Examples
Let us consider the following sequence of supervised learning methods in turn.
1. Linear Regression (Regression problems).
2. $k$-nearest Neighbors (Classification or Regression problems).
3. Logistic Regression (Classification problems1).
4. Naive Bayes Classifier (Classification problems).
5. Linear Discriminant Analysis (Classification problems).
6. Support Vector Machines (Classification problems).
7. Decision trees (Primarily Classification problems).
8. Neural networks (Classification or Regression problems).
1. I know the name is confusing.
## Machine Learning Overview
Science is knowledge which we understand so well that we can teach it to a computer; and if we don’t fully understand something, it is an art to deal with it. Donald Knuth
## Introduction
#### First Attempt at a Definition
One says that an algorithm learns if its performance improves with additional experience.
This is not a precise definition, so let’s try to make it so.
#### Definition
An algorithm will be a function1 $f\colon D\to T$ and whose individual values can be calculated by a computer (i.e., a Turing machine) in a finite amount of time using a finite amount of memory. We can refer to $D$ as the space of inputs or samples or examples and $T$ as the outputs or labels.
For the following definition we will want to be able to form a sum indexed by the elements of $D$. For this purpose we could suppose that our input is finite or that the domain is a measure space and that the algorithm will be a measurable function. For simplicity and convenience, we will pretend that $D$ is finite. That is if $D$ is infinite, then we have actually implicitly restricted to to a finite subset (e.g., $D=\Bbb R$ should really be interpreted as the finite subset of floating point numbers expressed in some fixed number of bits). This poses no real restriction in applications (all real world problem domains are effectively finite).
#### Definition
A performance measure of an algorithm will be a function $P\colon D\times T \to \Bbb R$. For two algorithms $f_1$ and $f_2$, we will say that $f_1$ performs better than $f_2$ (relative to $P$) if $$\sum_{d\in D} P(d,f_1(d)) > \sum_{d\in D} P(d, f_2(d)).$$ We similarly define ‘performs worse than’ and ‘performs at least as well as’, etc.
Alternatively, we may be presented a cost function $C\colon D\times T\to \Bbb R$ which we want to minimize. One can postcompose any performance measure (resp. cost function) with an order reversing map to obtain a cost function (resp. performance measure). This allows us to pass between the two notions.
#### Definition
A machine learning algorithm is an algorithm $f_-\colon 2^E \times D\to T$, that learns with respect to a specified performance measure $P$ (defined below). Here $2^E$ is the set of all subsets of $E$. We call an element $S\in 2^E$ (i.e., a subset $S\subseteq E$) a set of training data. For such an $S$ and $f_-$, let $f_S=f_-(S,-) \colon D\to T$ denote the resulting algorithm (which we will say is trained on $S$).
#### Example
Consider the problem of finding a linear function $f\colon D\subset \Bbb R\to \Bbb R$ which closely approximates a fixed function $F\colon D\subset \Bbb R\to \Bbb R$. We can regard the function $F$ as a subset $S$ of $E=D\times \Bbb R$ by $F\mapsto S= \{(s,F(s)) | s\in D\}$2. We can define a cost function $C\colon D\times \Bbb R\to \Bbb R$ by $$C(d,t)=(F(d)-t)^2.$$ We can reformulate problem of minimizing the cost of $C(d,f(d))$ as maximizing the performance function $P(d,t)=\frac{1}{1+C(d,t)}$. The method of Linear Regression will define a machine learning algorithm $f_-\colon 2^E\times D\to \Bbb R$.
One can play with this problem using this demo.
#### Second Attempt at a Definition
A machine learning algorithm $f\colon 2^E\times D\to T$ learns (relative to a performance measure $P$) if for all proper subset inclusions $S_1\subset S_2\subseteq E$, $f_{S_2}$ performs better than $f_{S_1}$.
This is at least clearly defined, but it does not cover many examples of learning algorithms. For example, the method of linear regression will perform poorly when given a small training sample containing outliers (those $d\in D$ such that $F(d)$ takes on an unlikely value under some assumed probability distribution associated to $F$). So, the act of adjoining outliers to our training data will typically decrease the performance of this machine learning algorithm.
#### Final Definition
A machine learning algorithm $f\colon 2^E\times D\to T$ learns (relative to a performance measure $P$) if for each pair of natural numbers $m<n$, the average of the performances of $f_{S}$ over the subsets $S\subseteq E$ of cardinality $n$ is better than the average of the performances over the subsets of cardinality $m$.
#### Crucial remark
Without a specified performance measure, we have no obvious goal in machine learning. We can only evaluate machine learning algorithms relative to a performance measure3 and an algorithm that performs well with respect to one measure may perform poorly with respect to another. So the field of machine learning depends on first finding a good performance measure for the task at hand and then finding a machine learning algorithm that performs well with respect to this measure.
#### Remarks
Now that we have come to a definition of a learning algorithm that learns that seems to cover at least one standard example, I will unveil some unfortunate truths:
• In practice, we often do not have a performance measure that is defined over all of $D\times T$, instead we only have it over a subset.
• Machine learning algorithms are often presented without proofs that they learn in the sense above.
## Taxonomy of machine learning
We can roughly divide up the field of machine learning into 3 branches:
1. Supervised Learning
2. Unsupervised Learning
3. Reinforcement learning
## Supervised learning
Before getting into what supervised learning precisely is, let’s look at some examples of supervised learning tasks:
#### Definition
Supervised learning is concerned with the construction of machine learning algorithms $f\colon 2^{D\times T}\times D\to T$. The subsets of $D\times T$ are referred to as subsets of labeled examples.
We often assume that the labeled examples arise from restricting some presumed function $F\colon D\to T$ whose values we know on $E\subset D$. We can then train $f$ on pairs $\{s, F(s) | s\in E\}$. We can then devise a cost function which measures the distance from the learned $f$ and the presumed $F$ (e.g., $L^2$-distance).
Supervised learning has been the most successful of the three branches to real world problems. The existence of labeled examples usually leads to well-defined performance metrics that transform supervised learning tasks into two other tasks:
1. Find an appropriate parametrized class of functions to choose $f$ from.
2. Solve the optimization problem of finding the best function in that class.
#### Example
The first example of a supervised learning task is the linear regression problem mentioned above.
## Unsupervised learning
First some examples:
1. Some sample techniques in unsupervised learning can be seen at this beautiful blog post.
2. Word vector embeddings.
3. Independent components analysis
#### Definition
Unsupervised learning is concerned with the construction of machine learning algorithms of the form $f\colon 2^D\times D\to T$.
In this case, there is no obvious measure of performance for such an algorithm. The choice of performance measure essentially defines the unsupervised learning task.
Often, one can vaguely state that the goal of an unsupervised learning task is to summarize the data or determine its essential features. In other words, suppose that we have a function $i\colon T\to D$, then we can state the goal is to find an $f\colon D\to T$ such that $i\circ f$ approximates the identity function. That is our unsupervised learning task is to find a compression algorithm! In this case, we have transformed the unsupervised learning task into a supervised learning task (because we have as many labels as we like of the identity function).
The only task remaining (and it is the hard task) is to make sense of the word `approximates’ above by choosing an appropriate measure. For example, how do you quantitatively measure the quality of a video compression algorithm? Somehow we can usually see the difference between a terrible algorithm and a good one, but it can be difficult to turn this intuition into a well-defined measure.
#### Example
One of the typical tasks in unsupervised learning is to identify ‘clusters’ of data points in a high dimensional space. We may be looking for difficult to identify relationships between data points. Once we have identified a classification that we deem to be ‘meaningful’, then our algorithm should be able to sort new data points into the correct classes. For example, can try to look at demographic data to identify clusters of voters with similar voting patterns and then try to identify the correct approach to appeal to this group.
## Reinforcement learning
Reinforcement learning is applied to tasks where one must choose actions whose outcomes eventually lead to various rewards/punishments which are used to encourage or discourage certain choices:
1. Breakout
2. One of the more difficult challenges is Montezuma’s revenge where there is very little feedback and one of the implicit motivations is curiousity.
3. Another truly amazing achievement is AlphaGo, or its successor AlphaGo Zero.
4. Self-driving cars
To construct a reasonably precise definition, let’s describe a typical reinforcement learning task4. In this case the algorithm is to construct an agent (the name comes from the field of artificial intelligence). The agent learns a policy which is a mapping from perceived states to actions. Paired with the agent is an environment. The environment maintains an environment state and when given an action returns an observation and a reward to the agent, which it then uses to update its policy and perceived state and choose an action. This creates a feedback loop:
We can break up the agent’s task into two parts:
1. It must take in an observation and reward and update its perceived state. To update its perceived state it might maintain a model of its environment and use the observation and reward to update this model.
2. It must take in a sequence of observations and rewards and update its policy. The policy is supposed to choose an allowable action and this decision could be made by defining a value function, which assigns to each possible action given our state a value, and choosing the action that maximizes the value. Since the agent is operating with imperfect information, the value function will only be an estimate, and it may choose to alternatively explore and choose a lower valued action in the hopes that such a state is being undervalued. We can model this exploration possibility, by saying that the policy assigns a probability to each possible action from a fixed perceived state.
Writing the general task out as a learning algorithm will get quite cumbersome. So let’s just hit the main points.
• The environment will be encoded as a probability space $S_e \times A\times S_e$, which encodes the probability of transitioning from one environment state to another given a fixed action.
• The policy will be encoded as a probability space $S_p \times A$, which encodes the probability of choosing an action from a given probability.
• We will also require some type of model, which will be a probability space $S_p\times A\times O\times R\times S_p$ which will encode the probability of moving from one perceived state to another given an action, an observation, and a reward.
• We will be a given a environmental feedback, which will be a random variable $F\colon S_e\times A\times S_e \to O\times R$.
All of these components can then be combined and we can form the expected reward of all sequences of length $n$ of tuples of the above variables5. The reinforcement learning task is to learn the policy function and the interpreter function so that we can maximize the expected reward over sequences of length $n$ as $n$ goes to $\infty$.
In this generality, the task is extremely difficult. We could simplify things by assuming the environment provides complete feedback about the perceived state, so that the observation is the perceived state (for example, a game with perfect information). But even then this task can be extremely difficult if there are many potential perceived states and many potential actions. For example, in the game Go there are two players (black and white) which take alternating turns placing pieces on a 19×19 board, occasionally removing captured pieces. As an approximate upper bound on the perceived state space we can see that there are no more than $3^{19\cdot 19}\approx 10^{172}$ possible board configurations (each square is either black, white, or blank), from each of these positions we have no more than $19^2=361$ possible actions choose from. Searching through all possible chains of configurations and actions is completely intractable. Even encoding an optimal policy in tabular form is impossible. This is part of the reason that performing well on this game was such a high-hurdle for the AI community.
1. We could also allow partially defined functions.
2. Indeed if you look under the hood, this is how functions are actually defined.
3. Okay, runtime and memory usage are also important too.
4. I must admit this first attempt at finding a definition of reinforcement learning that fits into the above framework is a bit clumsy and accurately reflects my limited knowledge in this area.
5. As a non-trivial exercise write this down for sequences of length 1 and 2. |
# The Geometry Rule That's More Important Than You Think
This tip for improving your GMAT score was provided by David Newland at Veritas Prep.
There is one very simple rule that students seem to have trouble remembering and even more difficulty applying: the Third-Side Rule.
The essence of the rule is that to have an actual triangle, no one side of the triangle can be longer than the other two sides put together. Imagine a triangle with one side length 10 and the other two sides have lengths of 3 and 5. There is simply no way that the two shorter sides could connect. You would not have a triangle.
The actual rule is stated in terms of the third side (which is why it is called the third-side rule): “The third side of a triangle must be smaller than the sum of the other two sides and must be greater than the difference between the other two sides.”
Use the third side rule to solve the following question quickly and efficiently from the GMATPrep software: “If two sides of a triangle have lengths of 2 and 5, which of the following could be the perimeter of the triangle?
I. 9
II. 15
III. 19
A) None
B) I only
C) II only
D) II and III
E) I, II, and III”
Apply the first part of the rule to determine the maximum value that the perimeter could equal. Given that the third side—in this case, the unknown side—must be smaller than the sum of the two known sides, and since that sum is 7 (2 + 5), the third side of the triangle must be smaller than 7. Therefore the perimeter must be less than 14.
The second part of the rule establishes the minimum for the unknown third side. Since the third side must be greater than the difference between the two known sides, and since that difference is 3 (5 – 2), the third side must be bigger than 3. Therefore the perimeter must be more than 10.
None of the available options is between 10 and 14, so none could be the third side of this triangle. Therefore the correct answer is A: “none.”
A surprising number of GMAT questions rely on the third-side rule. Some questions, such as the one we just solved, clearly ask for the third side, while others are cleverly disguised. An excellent example of a question that is made so much easier by the third-side rule is this question, which also appears courtesy the GMATPrep software: “The perimeter of a certain right isosceles triangle is 16 + 16 √2. What is the length of the hypotenuse?
A) 8
B) 16
C) 4 √2
D) 8 √2
E) 16 √2”
Most people can set up the algebra on this problem but often do not know where to go from there. If you look at this one logically from the start, you can see where the third-side rule applies.
An isosceles right triangle has two equal sides and a hypotenuse that is √2 larger. Use x to represent each of the equal sides, and you get the equation, “2x + x √2 = 16 + 16 √2.” Many test takers look at this equation and think that the hypotenuse must have the square root; most people choose 16 for the sides and 16 √2 for the hypotenuse. Unfortunately, choice E is the trap answer on this question.
Using the third-side rule could help you quickly find the correct answer. Begin by recognizing that the √2 can be attached to either the sides or the hypotenuse but not both. Either the sides are an integer, such as 10, and the hypotenuse is 10 √2, or the sides have the square root, such as 2 √2, and the hypotenuse is the integer 4.
Now apply the rule: The sum of the two sides must be greater than the length of the third side – the hypotenuse. So the 16 √2 (which you can estimate to be approximately 23) must be the sum of the two sides and 16 must be the length of the hypotenuse. Otherwise, that ~23 would be too large for the two smaller sides, since they’d add up to 16. Therefore, the correct answer is B.
The third-side rule is one of the simplest rules to memorize, but it can also be one of the most important.
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Stem and Leaf Plots, Range of a Data Set
Range, median and distribution from the plot.
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Stem-and-Leaf Plots, Range of a Data Set
Remember Julie from the Stem-and-Leaf Plots Concept? Have you ever wondered how to calculate range?
Well, here is the stem-and-leaf plot that Julie created in the previous Concept.
This Concept is all about how you can use a stem-and-leaf plot to determine the range of a set of data. Pay attention and by the end of the Concept you will know how Julie can figure out the range of the data.
Guidance
Previously we worked on how to create a stem-and-leaf plot for a set of data. Once you have the stem-and-leaf plot made, you can use it to figure out the range of a set of data.
What is the range?
The range is the difference between the maximum score and the minimum score.
The smallest number in the stem-and-leaf plot is 22. You can see that by looking at the first stem and the first leaf. The greatest number is the last stem and the last leaf on the chart. In this case, the largest number is 55. To find the range, we subtract the smallest number from the largest number. This difference will give us the range.
55 - 22 = 33
The range is 33 for this set of data.
Look at the following stem-and-leaf plot and answer these questions.
Example A
What is the range for this data set?
Solution: 35
Example B
What is the smallest interval?
Solution: 12 - 14
Example C
What is the greatest interval?
Solution: 42 - 47
Now back to Julie and the stem-and-leaf plot. Here it is once again.
What is the range of Julie's data?
To figure this out, we can find the difference between the greatest value in the data set and the smallest value in the data set.
\begin{align*}86 - 52 = 34\end{align*}
The range of the data is 34.
Vocabulary
Stem-and-leaf plot
a way of organizing numbers in a data set from least to greatest using place value to organize.
Data
information that has been collected to represent real life information
Ascending
from smallest to largest
Descending
from largest to smallest
Interval
a specific period or arrangement of data
Range
the difference from the largest value to the smallest value
Guided Practice
Here is one for you to try on your own.
What is the range of this data set?
To figure this out, we find the difference between the largest value in the data set and the smallest value in the data set.
The largest value is 68.
The smallest value is 33.
\begin{align*}68 - 33 = 35\end{align*}
The range of the data set is 35.
Video Review
Great video on organizing, building and interpreting a stem and leaf plot.
Practice
Directions: Use each stem - and - leaf plot to answer the following questions.
Stem Leaf
6 8
7 5 7 9
8 0 2
9 2 6 6 7
1. What is the smallest value in the plot?
2. What is the greatest value in the plot?
3. What is the range of the data?
Stem Leaf
0 8
1 2 7 8 9
2 2 3
3 1 5
4 0
4. What is the smallest value in the data set?
5. What is the greatest value in the data set?
6. What is the range of the data?
7. Name the first interval.
8. Name the second interval.
9. Name the third interval.
10. Name the fourth interval.
11. Which interval has the greatest number of values in it?
12. Which interval has the smallest number of values in it?
13. Because the 4 stem has a zero in it, does that mean that there aren't any values in it?
14. True or false. You can create a stem - and - leaf plot with a range in the hundreds.
15. True or false. You can create a stem - and - leaf plot without using intervals.
Vocabulary Language: English
Ascending
Ascending
Ascending order indicates that values are arranged from smallest to largest (to ascend means to move upward).
Data
Data
Data is information that has been collected to represent real life situations, usually in number form.
Descending
Descending
A descending pattern indicates that values in the pattern are arranged from greatest to least (to descend means to move downward).
Interval
Interval
An interval is a range of data in a data set.
Range
Range
The range of a data set is the difference between the smallest value and the greatest value in the data set.
Stem-and-leaf plot
Stem-and-leaf plot
A stem-and-leaf plot is a way of organizing data values from least to greatest using place value. Usually, the last digit of each data value becomes the "leaf" and the other digits become the "stem". |
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# Pythagorean triples
A Pythagorean triple is a triple of positive integers ${\displaystyle \scriptstyle (a,\,b,\,c)\,}$ which represent the side lengths of a Pythagorean triangle, i.e.
${\displaystyle a^{2}+b^{2}=c^{2},\quad a
where ${\displaystyle a}$ and ${\displaystyle b}$ are the leg (short and long, respectively) lengths of the right triangle and ${\displaystyle c}$ is the hypotenuse length.
Pythagorean triples ${\displaystyle \scriptstyle (a,\,b,\,c)\,}$ such that GCD${\displaystyle \scriptstyle (a,\,b,\,c)\,}$ = 1 are called primitive Pythagorean triples. If a Pythagorean triple ${\displaystyle \scriptstyle (a,\,b,\,c)\,}$ is not primitive, it is possible to use it to find a primitive triple ${\displaystyle \scriptstyle (a',\,b',\,c')\,}$ through division of ${\displaystyle \scriptstyle (a,\,b,\,c)\,}$ by GCD${\displaystyle \scriptstyle (a,\,b,\,c)\,}$. For example, 24, 32, 40 is not a primitive triple, but dividing each number by 8 it leads to the primitive triple 3, 4, 5.
## Hypotenuse numbers
Hypotenuse numbers are positive integers such their square is the sum of 2 distinct nonzero squares, hence the hypotenuse of a Pythagorean triangle.
## Formulae
${\displaystyle a^{2}+b^{2}=c^{2},\,a
if and only if
${\displaystyle c^{2}={\frac {m^{2}+n^{2}}{2}},\,m=b-a,\,n=b+a.\,}$
This provides a way to obtain all Pythagorean triples, primitive and otherwise, by iterating through pairs of integers. To obtain just the primitive Pythagorean triples requires just a few restrictions on the pairs of integers.
Theorem PYT.
In order for positive integers ${\displaystyle r}$ and ${\displaystyle s}$ to give ${\displaystyle x=r^{2}-s^{2}}$, ${\displaystyle y=2rs}$, ${\displaystyle z=r^{2}+s^{2}}$ that form a primitive solution to ${\displaystyle x^{2}+y^{2}=z^{2}}$, it is necessary that ${\displaystyle \gcd(r,s)=1}$ and that one of ${\displaystyle r}$ and ${\displaystyle s}$ be even.
Proof. First we verify that ${\displaystyle r}$ and ${\displaystyle s}$ give a solution as prescribed by expanding ${\displaystyle x^{2}+y^{2}=z^{2}}$ thus: ${\displaystyle (r^{2}-s^{2})^{2}+(2rs)^{2}=(r^{2}+s^{2})^{2}}$ and then ${\displaystyle (r^{4}+s^{4}-2r^{2}s^{2})+4r^{2}s^{2}=r^{4}+s^{4}+2r^{2}s^{2}}$. If ${\displaystyle \gcd(r,s)>1}$, that means there is a prime ${\displaystyle p}$ such that ${\displaystyle p|r}$ and ${\displaystyle p|s}$. Then ${\displaystyle x=(pa)^{2}-(pb)^{2}}$, ${\displaystyle y=2p^{2}ab}$, ${\displaystyle z=(pa)^{2}+(pb)^{2}}$. Dividing out ${\displaystyle p^{2}}$, we obtain ${\displaystyle \scriptstyle u\,=\,{\frac {x}{p^{2}}}\,=\,a^{2}-b^{2}}$, ${\displaystyle \scriptstyle v\,=\,{\frac {y}{p^{2}}}\,=\,2ab}$ and ${\displaystyle \scriptstyle w\,=\,{\frac {z}{p^{2}}}\,=\,a^{2}+b^{2}}$, and therefore ${\displaystyle u^{2}+v^{2}=w^{2}=a^{4}+b^{4}+2a^{2}b^{2}}$, which means ${\displaystyle x,y,z}$ is not a primitive solution.
If ${\displaystyle \gcd(r,s)=1}$ and both ${\displaystyle r}$ and ${\displaystyle s}$ are odd, then, since the difference of two odd numbers is even, ${\displaystyle \gcd(x,y)=\gcd(r^{2}-s^{2},2rs)=2}$, and also ${\displaystyle \gcd(x,z)=\gcd(r^{2}-s^{2},r^{2}+s^{2})=2}$, which means that ${\displaystyle x,y,z}$ are all even and we can divide out ${\displaystyle p=2}$. That leaves us just the case ${\displaystyle \gcd(r,s)=1}$ with either ${\displaystyle r}$ or ${\displaystyle s}$ even and the other odd. Now we can be certain that ${\displaystyle x=r^{2}-s^{2}}$ is odd while ${\displaystyle y=2rs}$ is at least doubly even, regardless of which of ${\displaystyle r}$ or ${\displaystyle s}$ is even. Furthermore, ${\displaystyle \gcd(x,y)=1}$ because ${\displaystyle x}$ is divisible by neither ${\displaystyle r}$ nor ${\displaystyle s}$, while ${\displaystyle y}$ is divisible by both. Likewise with ${\displaystyle z=r^{2}+s^{2}}$, we see that it is coprime to ${\displaystyle x}$ since ${\displaystyle x=z-2s^{2}}$ or ${\displaystyle z=x+2s^{2}}$, and ${\displaystyle z}$ is also coprime to ${\displaystyle y}$, which is even and divisible by both ${\displaystyle r}$ and ${\displaystyle s}$, confirming that ${\displaystyle x,y,z}$ is indeed a primitive Pythagorean triple, and that it could only be obtained with coprime ${\displaystyle r}$ and ${\displaystyle s}$, one of which is even, as specified by the theorem.
So, for example, the pair 5, 2 will give the primitive triple 21, 20, 29, while 5, 3 gives the triple 16, 30, 34, which can be 'reduced' to the primitive triple 8, 15, 17.
## Sequences
### Sequences (legs)
A118905 Sum of legs of Pythagorean triangles (without multiple entries).
{7, 14, 17, 21, 23, 28, 31, 34, 35, 41, 42, 46, 47, 49, 51, 56, 62, 63, 68, 69, 70, 71, 73, 77, 79, 82, 84, 85, 89, 91, 92, 93, 94, 97, 98, 102, 103, 105, 112, 113, 115, 119, 123, 124, 126, 127, 133, 136, ...}
#### Sequences (short legs)
A020884 Ordered short legs of primitive Pythagorean triangles.
{3, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 20, 21, 23, 24, 25, 27, 28, 28, 29, 31, 32, 33, 33, 35, 36, 36, 37, 39, 39, 40, 41, 43, 44, 44, 45, 47, 48, 48, 49, 51, 51, 52, 52, 53, 55, 56, 57, 57, 59, 60, ...}
A009004 Ordered short legs of Pythagorean triangles.
{3, 5, 6, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 15, 16, 16, 17, 18, 18, 19, 20, 20, 20, 21, 21, 21, 22, 23, 24, 24, 24, 24, 25, 25, 26, 27, 27, 27, 28, 28, 28, 29, 30, 30, 30, 31, 32, 32, 32, 33, 33, 33, ...}
A046083 The smallest member ${\displaystyle \scriptstyle a\,}$ of the Pythagorean triples ${\displaystyle \scriptstyle (a,\,b,\,c)\,}$ ordered by increasing ${\displaystyle \scriptstyle c\,}$.
{3, 6, 5, 9, 8, 12, 15, 7, 10, 20, 18, 16, 21, 12, 15, 24, 9, 27, 30, 14, 24, 20, 28, 33, 40, 36, 11, 39, 33, 25, 16, 32, 42, 48, 24, 45, 21, 30, 48, 18, 51, 40, 36, 13, 60, 39, 54, 35, 57, 65, 60, 28, 20, 48, ...}
#### Sequences (long legs)
A020883 Ordered long legs of primitive Pythagorean triangles.
{4, 12, 15, 21, 24, 35, 40, 45, 55, 56, 60, 63, 72, 77, 80, 84, 91, 99, 105, 112, 117, 120, 132, 140, 143, 144, 153, 156, 165, 168, 171, 176, 180, 187, 195, 208, 209, 220, 221, 224, 231, 240, 247, 252, 253, ...}
A009012 Ordered long legs of Pythagorean triangles.
{4, 8, 12, 12, 15, 16, 20, 21, 24, 24, 24, 28, 30, 32, 35, 36, 36, 40, 40, 42, 44, 45, 45, 48, 48, 48, 52, 55, 56, 56, 60, 60, 60, 60, 63, 63, 64, 68, 70, 72, 72, 72, 72, 75, 76, 77, 80, 80, 80, 84, 84, 84, 84, ...}
A046084 The middle member ${\displaystyle \scriptstyle b\,}$ of the Pythagorean triples ${\displaystyle \scriptstyle (a,\,b,\,c)\,}$ ordered by increasing ${\displaystyle \scriptstyle c\,}$.
{4, 8, 12, 12, 15, 16, 20, 24, 24, 21, 24, 30, 28, 35, 36, 32, 40, 36, 40, 48, 45, 48, 45, 44, 42, 48, 60, 52, 56, 60, 63, 60, 56, 55, 70, 60, 72, 72, 64, 80, 68, 75, 77, 84, 63, 80, 72, 84, 76, 72, 80, 96, 99, ...}
### Sequences (hypotenuse)
A020882 Ordered hypotenuse numbers of primitive Pythagorean triangles (squares are sums of 2 distinct nonzero squares and GCD[a,b,c] = 1).
{5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 65, 73, 85, 85, 89, 97, 101, 109, 113, 125, 137, 145, 145, 149, 157, 169, 173, 181, 185, 185, 193, 197, 205, 205, 221, 221, 229, 233, 241, 257, 265, 265, 269, 277, ...}
A009003 Hypotenuse numbers (squares are sums of 2 distinct nonzero squares).
{5, 10, 13, 15, 17, 20, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45, 50, 51, 52, 53, 55, 58, 60, 61, 65, 68, 70, 73, 74, 75, 78, 80, 82, 85, 87, 89, 90, 91, 95, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, ...}
A009000 Ordered hypotenuse numbers (squares are sums of 2 distinct nonzero squares). The largest member ${\displaystyle \scriptstyle c\,}$ of the Pythagorean triples ${\displaystyle \scriptstyle (a,\,b,\,c)\,}$ ordered by increasing ${\displaystyle \scriptstyle c\,}$.
{5, 10, 13, 15, 17, 20, 25, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45, 50, 50, 51, 52, 53, 55, 58, 60, 61, 65, 65, 65, 65, 68, 70, 73, 74, 75, 75, 78, 80, 82, 85, 85, 85, 85, 87, 89, 90, 91, 95, 97, 100, 100, ...} |
# If sum of the 3rd and the 8th terms of an AP is 7 and the sum of the 7th and 14th terms is -3, then find the 10th term.
Video Solution
Text Solution
Verified by Experts
## Let the first term and common difference of an AP are a and d, respectively. According to the question, ${a}_{3}+{a}_{8}=7$ and ${a}_{7}+{a}_{14}=-3$ $⇒a+\left(3-1\right)d+a+\left(8-1\right)d=7\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\left[\therefore {a}_{n}=a+\left(n-1\right)d\right]$ and $a+\left(7-1\right)d+a+\left(14-1\right)d=-3$ $a+2d+a+7d=7$ and $a+6d+a+13d=-3$ $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}2a+9d=7\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ ...(i) and $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}2a+19d=-3\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ ...(ii) On subtracting Eq. (i) from Eq. (ii), we get $10d=-10⇒d=-1\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ [ from Eq. (i)] $2a+9\left(-1\right)=7$ $⇒2a-9=7$ $2a=16⇒a=8$ $\therefore {a}_{10}=a+\left(10-1\right)d$ $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=8+9\left(-1\right)$ $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=8-9=-1$
|
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# Introduction
Math > Math Concepts > Algebra > Conic Sections > Introduction
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## Introduction to Conic Sections
Conic Sections
A conic section or more simply, conic, is the intersection of a plane and a right circular conical surface. This intersection is a curve (or curves) that can be specified as a quadratic equation in terms of x and y.
This surface is generated by rotating a line in the Y-Z plane (see next figure) about the Z axis with the vertex at the origin. The Z axis the axis of rotation and is the axis of the cones. Each cone is referred to as a nappe (pronounced: năp).
There are 7 possible ways to pass a plane through this conical surface:
The plane passes through the shared vertex of each nappe, that is the vertex of the cone. The plane lies along the sides of the nappes and passes through the vertex of the cone. The plane contains the axis of the conical surface. The plane passes through the conical surface normal to it axis. Normal means that the axis is at right angles no matter how the angle from the axis to the plane is measured. The green and blue angle measurements illustrate this requirement. (These three lines represent 3D axes for the 3 dimensional coordinate system we use where all three lines must be at right angles with each other.) Imagine tilting the plane a bit. When we do this we stretch the circle a bit from one end to the next, that is we elongate it. The resulting intersection is an ellipse. Let's tilt the plane further until it becomes parallel to the edge of a nappe. The intersection created is the parabola. Let's tilt the plane further. When we do this the plane will pass through both nappes. We get two intersections here, one in each nappe, that defines the hyperbola.
Before proceeding to studying each conic, we should appreciate the history behind them.
Euclid cataloged the conics, his work became lost over time. Apollonius of Perga, Greece, 3rd century wrote eight books on the conics. Not having algebra he had to argue his points using figures, i.e., diagrams.
Kepler used Pollonius' ideas to model planetary motion with the ellipse, and Newton took this model further with his theory of gravitation.
Now, let's talk about the names.
circle: the plane is perpendicular to the cone's axis.
ellipse (ĭ - lĭps '): (Greek: to leave out) the plane tilts further, leaving out one nappe.
parabola (pə - răb' - ə - lə): (Greek: beside, compare) the plane is parallel to an edge of the cone. Only one nappe can be cut in this case.
hyperbola (hī - pur ' bə - lə): (Greek: hyperbole: exceeds ) the plane exceeds parallel. Both nappes must be cut.
The right circular cone for review:
This is another FREE ALGEBRA PRINTABLE presented to you from the Algebra section of K12math.com |
# Worksheet on Dividend, Divisor, Quotient and Remainder
Practice the questions given in the worksheet on dividend, divisor, quotient and remainder. We know the number which we divide is called the dividend and the number by which we divide is called the divisor and their result obtained is called the quotient and the number left over is called the remainder.
I. Find the quotient: (i) 137 ÷ 1 (ii) 0 ÷ 23 (iii) 55 ÷ 0 (iv) 512 ÷ 512 (v) 444 ÷ 1 (vi) 348 ÷ 348 (vii) 0 ÷ 540 (viii) 175 ÷ 175
II. Find the quotient and remainder (if any):
(i) 245 ÷ 5 (ii) 335 ÷ 4 (iii) 243 ÷ 8 (iv) 495 ÷ 6 (v) 2744 ÷ 7 (vi) 1947 ÷ 3 (vii) 2870 ÷ 4 (viii) 6436 ÷ 7
III. Find the quotient and the remainder:
(i) 6781 ÷ 8 (ii) 2908 ÷ 6 (iii) 1125 ÷ 3 (iv) 2455 ÷ 6 (v) 2572 ÷ 5 (vi) 4945 ÷ 7 (vii) 2870 ÷ 4 (viii) 1638 ÷ 5
IV. Find the quotient and remainder (if any):
(i) 996 ÷ 12 (ii) 788 ÷ 11 (iii) 990 ÷ 15 (iv) 860 ÷ 15 (v) 1570 ÷ 14 (vi) 2880 ÷ 16 (vii) 5780 ÷ 26 (viii) 2135 ÷ 22 (ix) 2260 ÷ 19 (x) 1900 ÷ 23 (xi) 2700 ÷ 14 (xii) 5434 ÷ 24 (xiii) 29481 ÷ 12 (xiv) 54528 ÷ 16 (xv) 16544 ÷ 32 (xvi) 57298 ÷ 66
Answers for the worksheet on dividend, divisor, quotient and remainder are given below to check the exact answers of the above questions on division.
I. (i) 137 (ii) 0 (iii) Not possible (iv) 1 (v) 444 (vi) 1 (vii) 0 (viii) 1
II. (i) Q = 49, R = 0 (ii) Q = 83, R = 3 (iii) Q = 30, R = 3 (iv) Q = 82, R = 3 (v) Q = 392, R = 0 (vi) Q = 649, R = 0 (vii) Q = 717, R = 2 (viii) Q = 919, R = 3
III. (i) Q = 847, R = 5 (ii) Q = 484, R = 4 (iii) Q = 375, R = 0 (iv) Q = 409, R = 1 (v) Q = 514, R = 2 (vi) Q = 706, R = 3 (vii) Q = 717, R = 2 (viii) Q = 327, R = 3
IV. (i) Q = 83, R = 0 (ii) Q = 71, R = 7 (iii) Q = 66, R = 0 (iv) Q = 57, R = 5 (v) Q = 112, R = 2 (vi) Q = 180, R = 0 (vii) Q = 222, R = 8 (viii) Q = 97, R = 1 (ix) Q = 118, R = 18 (x) Q = 82, R = 4 (xi) Q = 192, R = 12 (xii) Q = 226, R = 10 (xiii) Q = 2456, R = 9 (xiv) Q = 3408, R = 0 (xv) Q = 517, R = 0 (xvi) Q = 868, R = 10
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# Problem Study: The Remainder Theorem and Division Algorithm
Given that x5 + ax3 + bx2 − 3 = (x2 − 1) Q(x) x − 2 where Q(x) is a polynomial. State the degree of Q(x) and find the value of a and b. Find also the remainder when Q(x) is divided by x + 2.
Solution
Since $\displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3$ is divided by (x2 − 1), Q(x) is a polynomial of degree 3.
$\displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2$
$\displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=(x-1)(x+1)Q(x)-x-2$
When x = − 1, $\displaystyle -1-a+b-3=(-1-1)(-1+1)Q(x)-(-1)-2$
$\displaystyle \therefore -a+b=3$ --------------(1)
When x = 1, $\displaystyle 1+a+b-3=(1-1)(1+1)Q(x)-1-2$
$\displaystyle \therefore a+b=-1$ --------------(2)
$\displaystyle (2)+(1)\Rightarrow 2b=2\Rightarrow b=1$
$\displaystyle (2)-(1)\Rightarrow 2a=-4\Rightarrow a=-2$
$\displaystyle \therefore {{x}^{5}}-2{{x}^{3}}+{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2$
$\displaystyle \therefore {{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1=({{x}^{2}}-1)Q(x)$
$\displaystyle \therefore Q(x)=\frac{{{{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1}}{{{{x}^{2}}-1}}$
When Q(x) is divided by x + 2,
the remainder $\displaystyle =Q(-2)$
$\displaystyle =\frac{{{{{(-2)}}^{5}}-2{{{(-2)}}^{3}}+{{{(-2)}}^{2}}+(-2)-1}}{{{{{(-2)}}^{2}}-1}}$
$\displaystyle =\frac{{-32+16+4-2-1}}{3}$
$\displaystyle =-\frac{{15}}{3}$
$\displaystyle =-5$ |
U.S. Customary System Measurement Task
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Objective
The students will be able to convert measurements of length, weight, and capacity within the U.S. Customary System to solve real world problems.
Big Idea
I guess this is practical!
Opener
15 minutes
In today’s lesson students review the U.S. Customary system. First, they perform a classroom sort of vocabulary associated with the units of measurement. Then, they work in groups to solve real world story problems that have multiple steps. The students close this lesson by presenting their solution to the story problems.
To prepare for this lesson I make copies of the classroom sort words and cut them out. I ensure I have enough words cut so that each student gets a word. I pass out the words as a folded piece of paper and instruct the students not to open it until told to do so. I divide the room into four sections and label each one length, weight, capacity, and temperature.
Alright, on the count of three you have to open your paper and then put yourself into the appropriate group. Once you have everyone in your group you need to present what you have and ask a question to your group. For example if you have feet, you could say 1 foot equals how many inches? Or you could ask, what is a unit bigger than the foot? Or, what is bigger, one foot or one mile?
Once students play one round of the sort I collect the papers and distribute them again. This time the students have to create a statement that helps group members determine which unit they have. For example if a student has an ounce, they could say there are 16 of me in one pound. Or, I am the smallest unit of weight in the U.S. Customary system.
Practice
30 minutes
After the second round of the sort I bring students back to the whole group and explain the next portion of today’s lesson.
Today we are going to be working with our groups to solve some real world problems just like you have been in your exit slips from the last few days. But, these story problems are a bit more complicated. You may need to preform several steps and operations in order to complete the problem. You need to show your work on a separate sheet of paper and then present your findings at the end.
There are five word problems so I place my students in five groups and pass out the papers. I allow students about 15 - 20 minutes to solve the problem and prepare their presentation. I circulate the room and listen to conversation without aiding students in solving the problem. I want them to struggle a bit because these problems do require some deep thought. My role as a facilitator in this activity is to make sure all members of the group are participating in conversation.
The word problems used in today’s activity are derived from Kahn Academy. There is a video of solutions for each one of the problems.
Closer
15 minutes
Once I see the majority of groups preparing their visual for the presentation I give them a five minute warning of time until remaining. I have students place their work on the document camera and stand at the board to present their thinking. At the end of each presentation I allow the other students to ask questions about the solution or work showed. I also ask students presenting some further questions to check for understanding. |
The Navigation Computer: Rates and Ratios
Let us reinforce the division process by doing another example. Placing it into an aviation related situation:-.
The average rate of climb of a particular aircraft is given as 700 feet per minute. How long will it take to climb from 3000 to 20000 feet?
The aircraft is to climb through 17000 feet at 700 feet per minute. Straightforward division of 17000 by 700 will determine the number of minutes for the climb.
What is the rough order of magnitude?
700 feet in one minute so ...
7000 feet in ten minutes, and ...
14000 feet in twenty minutes; therefore ...
17000 feet in a few more minutes than twenty ...
Alternatively, 700 fpm is not too far from 1000 fpm. So the sum is approximately 17000 / 1000, which will be a bit more than 17, i.e. somewhere in the low twenties.
Hold the computer with the 17000 (shown as 17) of the outer scale at the top.
Rotate the disc to bring the 700 (shown as 7) of the inner scale immediately opposite the outer scale 17.
Turn the instrument the shortest way (clockwise in this case) to bring 10 of the inner scale to the top.
Opposite this ‘10’ read off from the outer scale the answer digits 243.
Which leads us to a final answer of 24.3 minutes.
You may prefer a slightly different approach to this type of problem, solving in terms of proportion rather than division. Restating the problem we have:-
Aircraft climbs 700 feet in one minute.
It will climb 17000 feet in ‘t’ minutes (where ‘t’ is the answer we wish to discover).
This can be written as:-
700 / 1 = 17000 / t
Now if you arrange this equation on your circular slide rule with the numerators 700 and 17000 on the outer scale set respectively above the denominators 1 and ‘t’ on the inner scale, the problem is solved.
As shown in Figure 6, align the 700 (shown as 70) of the outer scale exactly over the 1 (shown as 10) on the inner scale. This sets up the left hand side of the equation.
Turn the instrument clockwise and look for the right hand side of the equation. Look for 17000 (shown as 17) on the outer scale.
Immediately under it, read off 24.3 from the inner scale.
This is the same answer that we had before. If you solve the problem in this manner, you still have to perform a rough mental calculation, as before, to position the decimal point correctly.
This proportioning, or ratio, technique is important. It is the basis of all distance, speed and time calculations and of fuel consumption calculations. |
# How do you describe percentages?
## How do you describe percentages?
In mathematics, a percentage (from Latin per centum “by a hundred”) is a number or ratio expressed as a fraction of 100. It is often denoted using the percent sign, “%”, although the abbreviations “pct.”, “pct” and sometimes “pc” are also used.
### How do you express percentages in words?
Words and Numbers
1. Numbers up to ten are usually written as words, while larger numbers are written as numerals.
2. When a percentage is written as a word, it should be followed by “percent”
3. When a percentage is written as a numeral, it should be followed by the “%” sign.
How do you explain differences in percentages?
First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100. % increase = Increase ÷ Original Number × 100. If your answer is a negative number, then this is a percentage decrease.
How do you use percent in a sentence?
Examples of percentage in a Sentence If a goalie saves 96 out of 100 shots, his save percentage is 96 percent. The percentages of women completing high school and college were 95 percent and 52 percent, respectively. What percentage of your income do you spend on rent?
The APA rule for numbers is that you should begin a sentence with a word even if the number is greater than nine, and the word “percent” should also be used. For example: Forty-eight percent of the sample showed an increase.
### What are the rules for percentages?
To write a percentage as a decimal, simply divide it by 100. For example, 50% becomes 0.5, 20% becomes 0.2, 1% becomes 0.01 and so on. We can calculate percentages using this knowledge. 50% is the same as a half, so 50% of 10 is 5, because five is half of 10 (10 ÷ 2).
What is the easiest way to calculate percentages?
To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1. Then, to calculate what 10 percent of is, say, 250 students, simply multiply the number of students by 0.1.
How do you calculate percentages easily?
A percentage is a way to express a number as a part of a whole. To calculate a percentage, we look at the whole as equal to 100%. For example, say you have 10 apples (=100%). If you eat 2 apples, then you have eaten 2/10 × 100% = 20% of your apples and you are left with 80% of your original apples.
## How do you solve percentages without a calculator?
If you need to find a percentage of a number, here’s what you do – for example, to find 35% of 240: Divide the number by 10 to find 10%. In this case, 10% is 24. Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72.
### How do I convert a number into a percentage?
Multiply by 100 to convert a number from decimal to percent then add a percent sign %.
1. Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
2. Example: 0.10 becomes 0.10 x 100 = 10%
3. Example: 0.675 becomes 0.675 x 100 = 67.5%
How do you find 20% of a number?
If you know what the whole number is and you know what percent of that number you are looking for, you multiply. For example, if you are looking for 20% of 100, you multiply 100 by 0.2. If you want to find what percent of 100 is equal to 20, you would divide 100 by 20.
How do you find 30% of a number?
Once you have the decimal figure, multiply it by the number for which you seek to calculate the percentage; i.e., if you need to know 30 percent of 100, you convert 30 percent to a decimal (0.30) and multiply it by 100 (0.30 x 100, which equals 30).
## How do I use a calculator to find a percentage?
convert any number from its percentage form to its decimal form. Just enter the number and press the % button, and the calculator will show the decimal equivalent. To represent the number 4% on your calculator, just press the button 4 and then %.
### How do I figure out a percentage of two numbers?
If you want to know what percent A is of B, you simple divide A by B, then take that number and move the decimal place two spaces to the right. That’s your percentage! To use the calculator, enter two numbers to calculate the percentage the first is of the second by clicking Calculate Percentage.
How do I take a percentage off a number?
How to calculate percent off?
1. Divide the number by 100 (move the decimal place two places to the left).
2. Multiply this new number by the percentage you want to take off.
3. Subtract the number from step 2 from the original number. This is your percent off number.
What is \$20 with 10% off?
Percent Off Table For 20.00
1 percent off 20.00 is 19.80 The difference is 0.20
10 percent off 20.00 is 18.00 The difference is 2.00
11 percent off 20.00 is 17.80 The difference is 2.20
12 percent off 20.00 is 17.60 The difference is 2.40
13 percent off 20.00 is 17.40 The difference is 2.60
## What is 1/3 as a percent?
Example Values
Percent Decimal Fraction
331/3% 0.333… 1/3
50% 0.5 1/2
75% 0.75 3/4
80% 0.8 4/5
### What is 15% off?
Percent Off Table For 15.00
1 percent off 15.00 is 14.85 The difference is 0.15
14 percent off 15.00 is 12.90 The difference is 2.10
15 percent off 15.00 is 12.75 The difference is 2.25
16 percent off 15.00 is 12.60 The difference is 2.40
17 percent off 15.00 is 12.45 The difference is 2.55
What is 20% off?
First, convert the percentage discount to a decimal. A 20 percent discount is 0.20 in decimal format. Secondly, multiply the decimal discount by the price of the item to determine the savings in dollars. For example, if the original price of the item equals \$24, you would multiply 0.2 by \$24 to get \$4.80.
What is the percentage of 15%?
15% is 10% + 5% (or 0.15 = 0.1 + 0.05, dividing each percent by 100). Thinking about it this way is useful for two reasons. First, it’s easy to multiply any number by 0.1; just move the decimal point left one digit. For example, 75.00 x 0.1 = 7.50, or 346.43 x 0.1 = 34.64 (close enough).
## What number is 15% of 20?
75
### What number is 15% of 60?
9
What number is 15% of 50?
7.5
How do you find 40% 60?
Percentage Calculator: What is 40 percent of 60? = 24.
Begin typing your search term above and press enter to search. Press ESC to cancel. |
# Algebra/Arithmetic Progression (AP)
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
Example 1:
A frog jumps every 2 seconds. If we start from the very first second the frog jump we take it as $t = 0$. The next jump will be on $t = 2$. As the frog jumps every 2 seconds so; $t = 0 + 2 \rightarrow t = 2$
So can you tell when the next jump will be? If you said $t = 4$ then yes, you are correct. Do you remember how we got $t = 2$? Now how did you get $t = 4$? The same way right? Well what you did was you added the difference $d = 2$ to previous term (which was 2). So to get next term of AP you always add the common difference to current term, and to get previous term you always subtract the common difference from current term. A more general formula to get the ‘n’th term of an AP is:
$a_n = a_1 + (n - 1)d$
Where $a_1$ is the first term and d is the common difference of the AP. So the general terms of AP are:
$a_1,\; a_1 + d,\; a_1 + 2d,\; a_1 + 3d,\; \dots \; a_1 + (n - 1)d$
### §Finding the sum of ‘n’ terms of an arithmetic sequence:
When needed to find the sum of ‘n’ terms of an AP, you use:
$s_n=\frac{n}{2}(2 a_1 + (n - 1)d)$
!If you have the first and last term you can use a short-cut formula to obtain the sum.
$s_n=\frac{n(a_1 + a_n)}{2}$
Where $a_n$ is the last term.
## §Notes
1. The sum of infinite arithmetic sequence is essentially infinite, but if $a = b = 0$, then the sum is 0. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Graphs of Linear Systems
## Graph lines to identify intersection points
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Progress
Practice Graphs of Linear Systems
Progress
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Solving Systems with One Solution Using Graphing
Two of coin one plus four of coin two equals a total of 70 cents. One of coin one plus five of coin two equals a total of 50 cents. What is the value of each coin?
### Guidance
In this lesson we will be using various techniques to graph the pairs of lines in systems of linear equations to identify the point of intersection or the solution to the system. It is important to use graph paper and a straightedge to graph the lines accurately. Also, you are encouraged to check your answer algebraically as described in the previous lesson.
#### Example A
Graph and solve the system:
yy=x+1=12x2
Solution:
Since both of these equations are written in slope intercept form, we can graph them easily by plotting the y\begin{align*}y-\end{align*}intercept point and using the slope to locate additional points on each line.
The equation, y=x+1\begin{align*}y=-x+1\end{align*}, graphed in \begin{align*}{\color{blue}\mathbf{blue}}\end{align*}, has \begin{align*}y-\end{align*}intercept 1 and slope \begin{align*}- \frac{1}{1}\end{align*}.
The equation, \begin{align*}y=\frac{1}{2}x-2\end{align*}, graphed in \begin{align*}{\color{red}\mathbf{red}}\end{align*}, has \begin{align*}y-\end{align*}intercept -2 and slope \begin{align*}\frac{1}{2}\end{align*}.
Now that both lines have been graphed, the intersection is observed to be the point (2, -1).
Check this solution algebraically by substituting the point into both equations.
Equation 1: \begin{align*}y=-x+1\end{align*}, making the substitution gives: \begin{align*}(-1)=(-2)+1. \end{align*}
Equation 2: \begin{align*}y=\frac{1}{2}x-2\end{align*}, making the substitution gives: \begin{align*}-1=\frac{1}{2}(2)-2. \end{align*}
(2, -1) is the solution to the system.
#### Example B
Graph and solve the system:
Solution: This example is very similar to the first example. The only difference is that equation 1 is not in slope intercept form. We can either solve for \begin{align*}y\end{align*} to put it in slope intercept form or we can use the intercepts to graph the equation. To review using intercepts to graph lines, we will use the latter method.
Recall that the \begin{align*}x-\end{align*}intercept can be found by replacing \begin{align*}y\end{align*} with zero and solving for \begin{align*}x\end{align*}:
Similarly, the \begin{align*}y-\end{align*}intercept is found by replacing \begin{align*}x\end{align*} with zero and solving for \begin{align*}y\end{align*}:
We have two points, (2, 0) and (0, 3) to plot and graph this line. Equation 2 can be graphed using the \begin{align*}y-\end{align*}intercept and slope as shown in Example A.
Now that both lines are graphed we observe that their intersection is the point (4, -3).
Finally, check this solution by substituting it into each of the two equations.
Equation 1: \begin{align*}3x+2y =6; 3(4)+2(-3)=12-6=6 \end{align*}
Equation 2: \begin{align*}y = - \frac{1}{2}x-1; -3= - \frac{1}{2}(4)-1 \end{align*}
#### Example C
In this example we will use technology to solve the system:
This process may vary somewhat based on the technology you use. All directions here can be applied to the TI-83 or 84 (plus, silver, etc) calculators.
Solution: The first step is to graph these equations on the calculator. The first equation must be rearranged into slope intercept form to put in the calculator.
The graph obtained using the calculator should look like this:
The first equation, \begin{align*}y=\frac{2}{3}x-\frac{10}{3}\end{align*}, is graphed in \begin{align*}{\color{blue}\mathbf{blue}}\end{align*}. The second equation, \begin{align*}y=-\frac{2}{3}x+4\end{align*}, is graphed in \begin{align*}{\color{red}\mathbf{red}}\end{align*}.
The solution does not lie on the “grid” and is therefore difficult to observe visually. With technology we can calculate the intersection. If you have a TI-83 or 84, use the CALC menu, select INTERSECT. Then select each line by pressing ENTER on each one. The calculator will give you a “guess.” Press ENTER one more time and the calculator will then calculate the intersection of (5.5, .333...). We can also write this point as \begin{align*}\left(\frac{11}{2},\frac{1}{3}\right)\end{align*}. Check the solution algebraically.
Equation 1: \begin{align*}2x-3y=10; 2\left(\frac{11}{2}\right)-3\left(\frac{1}{3}\right)=11-1=10 \end{align*}
Equation 2: \begin{align*}y=- \frac{2}{3}x+4; - \frac{2}{3}\left(\frac{11}{2}\right)+4=-\frac{11}{3}+\frac{12}{3}=\frac{1}{3} \end{align*}
If you do not have a TI-83 or 84, the commands might be different. Check with your teacher.
Intro Problem Revisit The system of linear equations represented by this situation is:
If you plot both of these linear equations on the same graph, you find that the point of intersection is (25, 5). Therefore coin one has a value of 25 cents and coin two has a value of 5 cents.
### Guided Practice
Solve the following systems by graphing. Use technology for problem 3.
1.
2.
3.
1.
The first line is in slope intercept form and can be graphed accordingly.
The second line is a horizontal line through (0, 2).
The graph of the two equations is shown below. From this graph the solution appears to be (2, 2).
Checking this solution in each equation verifies that it is indeed correct.
Equation 1: \begin{align*}2=3(2)-4 \end{align*}
Equation 2: \begin{align*}2=2 \end{align*}
2.
Neither of these equations is in slope intercept form. The easiest way to graph them is to find their intercepts as shown in Example B.
Equation 1: Let \begin{align*}y=0\end{align*} to find the \begin{align*}x-\end{align*}intercept.
Now let \begin{align*}x=0\end{align*}, to find the \begin{align*}y-\end{align*}intercept.
Now we can use (-2, 0) and (0, 4) to graph the line as shown in the diagram. Using the same process, the intercepts for the second line can be found to be (-6, 0) and (0, -4).
Now the solution to the system can be observed to be (-3, -2). This solution can be verified algebraically as shown in the first problem.
3.
The first equation here must be rearranged to be \begin{align*}y=-5x+10\end{align*} before it can be entered into the calculator. The second equation can be entered as is.
Using the calculate menu on the calculator the solution is (3, -5).
Remember to verify this solution algebraically as a way to check your work.
### Explore More
Match the system of linear equations to its graph and state the solution.
Solve the following linear systems by graphing. Use graph paper and a straightedge to insure accuracy. You are encouraged to verify your answer algebraically.
1. .
1. .
1. .
1. .
1. .
1. .
Solve the following linear systems by graphing using technology. Solutions should be rounded to the nearest hundredth as necessary.
1. .
1. .
1. .
Use the following information to complete exercises 14-17.
Clara and her brother, Carl, are at the beach for vacation. They want to rent bikes to ride up and down the boardwalk. One rental shop, Bargain Bikes, advertises rates of $5 plus$1.50 per hour. A second shop, Frugal Wheels, advertises a rate of $6 plus$1.25 an hour.
1. How much does it cost to rent a bike for one hour from each shop? How about 10 hours?
2. Write equations to represent the cost of renting a bike from each shop. Let \begin{align*}x\end{align*} represent the number of hours and \begin{align*}y\end{align*} represent the total cost.
3. Solve your system to figure out when the cost is the same.
4. Clara and Carl want to rent the bikes for about 3 hours. Which shop should they use?
### Vocabulary Language: English
Consistent
Consistent
A system of equations is consistent if it has at least one solution.
Dependent
Dependent
A system of equations is dependent if every solution for one equation is a solution for the other(s).
Independent
Independent
A system of equations is independent if it has exactly one solution.
linear equation
linear equation
A linear equation is an equation between two variables that produces a straight line when graphed.
Linear Function
Linear Function
A linear function is a relation between two variables that produces a straight line when graphed. |
# Leave and Let Dice
Imagine a game with six players, numbered #1 to #6, and one six-sided die. Someone rolls the die and the player who matches the number wins the game. That is, if the die rolls 1, player #1 wins; if the die rolls 2, player #2 wins; and so on. With a fair die, this is a fair game, because each player has exactly a 1/6 chance of winning. You could call it a simultaneous game, because all players are playing at once. It has one rule:
• If the die rolls n, then player #n wins.
Now try a different game with six players and one die. Player #1 rolls the die. If he gets 1, he wins the game. If not, then he leaves the game and player #2 rolls the die. If he gets 2, he wins the game. If not, then he leaves the game and player #3 rolls the die. And so on. You could call this a sequential game, because the players are playing in sequence. It has two rules:
• If player #n rolls n on the die, then he wins.
• If player #n doesn’t roll n, then player n+1 rolls the die.
Is it a fair game? No, definitely not. Player #1 has the best chance of winning. 1/6 or 16.6% of the time he rolls 1 and wins the game. 5/6 of the time, he rolls 2, 3, 4, 5 or 6 and passes the die to player #2. Now player #2 has a 1/6 chance of rolling a 2 and winning. But he has the opportunity to roll the die only 5/6 of the time, so his chance of winning the game is 1/6 * 5/6 = 5/36 = 13.8%. However, if player #2 rolls a 1, 3, 4, 5 or 6, then he loses and player #3 rolls the die. But player #3 has that opportunity only 5/6 * 5/6 = 25/36 of the time. So his chance of winning is 1/6 * 25/36 = 11.57%. And so on.
To put it another way, if the six players play 46656 = 6^6 games under the sequential rules, then on average:
• Player #1 wins 7776 games
• Player #2 wins 6480 games
• Player #3 wins 5400 games
• Player #4 wins 4500 games
• Player #5 wins 3750 games
• Player #6 wins 3125 games
• 15625 games end without a winner.
In other words, player #1 is 20% more likely to win than player #2, 44% more likely than player #3, 72.8% more likely than player #4, 107% more likely than player #5, and 148.8% more likely than player #6. Furthermore, player #2 is 20% more likely to win than player #3, 44% more likely than player #4, 72.8% more likely than player #5, and so on.
But there is a simple way to make the sequential game perfectly fair, so long as it’s played with a fair die. At least, I’ve thought of a simple way, but there might be more than one.
To make the sequential game fair, you add an extra rule:
1. If player #n rolls n on the die, he wins the game.
2. If player #n rolls a number greater than n, he loses and the die passes to player n+1.
3. If player #n rolls a number less than n, then he rolls again.
Let’s run through a possible game to see that it’s fair. Player #1 rolls first. He has a 1/6 chance of rolling a 1 and winning the game. However, 5/6 of the time he loses and passes the die to player #2. If player #2 rolls a 1, he rolls again. In other words, player #2 is effectively playing with a five-sided die, because all rolls of 1 are ignored. Therefore, he has a 1/5 chance of winning the game at that stage.
But hold on: a 1/5 chance of winning is better than a 1/6 chance, which is what player #1 had. So how is the game fair? Well, note the qualifying phrase at the end of the previous paragraph: at that stage. The game doesn’t always reach that stage, because if player #1 rolls a 1, the game is over. Player #2 rolls only if player doesn’t roll 1, which is 5/6 of the time. Therefore player #2’s chance of winning is really 1/5 * 5/6 = 5/30 = 1/6.
However, 4/5 of the time player #2 rolls a 3, 4, 5 or 6 and the die passes to player #3. If player #3 rolls a 1 or 2, he rolls again. In other words, player #3 is effectively playing with a four-sided die, because all rolls of 1 and 2 are ignored. Therefore, he has a 1/4 chance of winning the game at that stage.
A 1/4 chance of winning is better than a 1/5 chance and a 1/6 chance, but the same reasoning applies as before. Player #3 rolls the die only 5/6 * 4/5 = 20/30 = 2/3 of the time, so his chance of winning is really 1/4 * 2/3 = 2/12 = 1/6.
However, 3/4 of the time player #2 rolls a 4, 5 or 6 and the die passes to player #4. If player #4 rolls a 1, 2 or 3, he rolls again. In other words, player #4 is effectively playing with a three-sided die, because all rolls of 1, 2 and 3 are ignored. Therefore, he has a 1/3 chance of winning the game at that stage. 1/3 > 1/4 > 1/5 > 1/6, but the same reasoning applies as before. Player #4 rolls the die only 5/6 * 4/5 * 3/4 = 60/120 = 1/2 of the time, so his chance of winning is really 1/3 * 1/2 = 1/6.
And so on. If the die reaches player #5 and he gets a 1, 2, 3 or 4, then he rolls again. He is effectively rolling with a two-sided die, so his chance of winning is 1/2 * 5/6 * 4/5 * 3/4 * 2/3 = 120/720 = 1/6. If player #5 rolls a 6, he loses and the die passes to player #6. But there’s no need for player #6 to roll the die, because he’s bound to win. He rolls again if he gets a 1, 2, 3, 4 or 5, so eventually he must get a 6 and win the game. If player #5 loses, then player #6 automatically wins.
It’s obvious that this form of the game will get slower as more players drop out, because later players will be rolling again more often. To speed the game up, you can refine the rules like this:
1. If Player #1 rolls a 1, he wins the game. Otherwise…
2. If player #2 rolls a 2, he wins the game. If he rolls a 1, he rolls again. Otherwise…
3. Player #3 rolls twice and adds his scores. If the total is 3, 4 or 5, he wins the game. Otherwise…
4. Player #4 rolls once. If he gets 1 or 2, he wins the game. Otherwise…
5. Player #5 rolls once. If he gets 1, 2 or 3, he wins the game. Otherwise…
6. Player #6 wins the game.
Only player #2 might have to roll more than twice. Player #3 has to roll twice because he needs a way to get a 1/4 chance of winning. If you roll two dice, there are:
• Two ways of getting a total of 3: roll #1 is 1 and roll #2 is 2, or vice versa.
• Three ways of getting a total of 4 = 1+3, 3+1, 2+2.
• Four ways of getting 5 = 1+4, 4+1, 2+3, 3+2.
This means player #3 has 2 + 3 + 4 = 9 ways of winning. But there are thirty-six ways of rolling one die twice. Therefore player #3 has a 9/36 = 1/4 chance of winning. Here are the thirty-six ways of rolling one die twice, with asterisks marking the winning totals for player #3:
01. (1,1)
02. (1,2)*
03. (2,1)*
04. (1,3)*
05. (3,1)*
06. (1,4)*
07. (4,1)*
08. (1,5)
09. (5,1)
10. (1,6)
11. (6,1)
12. (2,2)*
13. (2,3)*
14. (3,2)*
15. (2,4)
16. (4,2)
17. (2,5)
18. (5,2)
19. (2,6)
20. (6,2)
21. (3,3)
22. (3,4)
23. (4,3)
24. (3,5)
25. (5,3)
26. (3,6)
27. (6,3)
28. (4,4)
29. (4,5)
30. (5,4)
31. (4,6)
32. (6,4)
33. (5,5)
34. (5,6)
35. (6,5)
36. (6,6)
# Dice in the Witch House
“Who could associate mathematics with horror?”
John Buchan answered that question in “Space” (1911), long before H.P. Lovecraft wrote masterpieces like “The Call of Cthulhu” (1926) and “Dreams in the Witchhouse” (1933). But Lovecraft’s use of mathematics is central to his genius. So is his recognition of both the importance and the strangeness of mathematics. Weird fiction and maths go together very well.
But weird fiction is about the intrusion or eruption of the Other into the everyday. Maths can teach you that the everyday is already Other. In short, reality is weird — the World is a Witch House. Let’s start with a situation that isn’t obviously weird. Suppose you had three six-sided dice, A, B and C, each with different set of numbers, like this:
Die A = (1, 2, 3, 6, 6, 6)
Die B = (1, 2, 3, 4, 6, 6)
Die C = (1, 2, 3, 4, 5, 6)
If the dice are fair, i.e. each face has an equal chance of appearing, then it’s clear that, on average, die A will beat both die B and die C, while die B will beat die C. The reasoning is simple: if die A beats die B and die B beats die C, then surely die A will beat die C. It’s a transitive relationship: If Jack is taller than Jim and Jim is taller than John, then Jack is taller than John.
Now try another set of dice with different arrangements of digits:
Die A = (1, 2, 2, 5, 6, 6)
Die B = (1, 1, 4, 5, 5, 5)
Die C = (3, 3, 3, 3, 4, 6)
If you roll the dice, on average die A beats die B and die B beats die C. Clearly, then, die A will also beat die C. Or will it? In fact, it doesn’t: the dice are what is called non-transitive. Die A beats die B and die B beats die C, but die C beats die A.
But how does that work? To see a simpler example of non-transitivity, try a simpler set of random-number generators. Suppose you have a triangle with a short rod passing through its centre at right angles to the plane of the triangle. Now imagine numbering the edges of the triangles (1, 2, 3) and throwing it repeatedly so that it spins in the air before landing on a flat surface. It should be obvious that it will come to rest with one edge facing downward and that each edge has a 1/3 chance of landing like that.
In other words, you could use such a spiked triangle as a random-number generator — you could call it a “trie”, plural “trice”. Examine the set of three trice below. You’ll find that they have the same paradoxical property as the second set of six-sided dice above. Trie A beats trie B, trie B beats trie C, but trie C beats trie A:
Trie A = (1, 5, 8)
Trie B = (3, 4, 7)
Trie C = (2, 3, 9)
When you throw two of the trice, there are nine possible outcomes, because each of three edges on one trie can be matched with three possible edges on the other. The results look like this:
Trie A beats Trie B 5/9ths of the time.
Trie B beats Trie C 5/9ths of the time.
Trie C beats Trie A 5/9ths of the time.
To see how this works, here are the results throw-by-throw:
Trie A = (1, 5, 8)
Trie B = (3, 4, 7)
When Trie A rolls 1…
…and Trie B rolls 3, Trie B wins (Trie A has won 0 out of 1)
…and Trie B rolls 4, Trie B wins (0 out of 2)
…and Trie B rolls 7, Trie B wins (0 out of 3)
When Trie A rolls 5…
…and Trie B rolls 3, Trie A wins (1/4)
…and Trie B rolls 4, Trie A wins (2/5)
…and Trie B rolls 7, Trie B wins (2/6)
When Trie A rolls 8…
…and Trie B rolls 3, Trie A wins (3/7)
…and Trie B rolls 4, Trie A wins (4/8)
…and Trie B rolls 7, Trie A wins (5/9)
Trie B = (3, 4, 7)
Trie C = (2, 3, 9)
When Trie B rolls 3…
…and Trie C rolls 2, Trie B wins (Trie B has won 1 out of 1)
…and Trie C rolls 3, it’s a draw (1 out of 2)
…and Trie C rolls 9, Trie C wins (1 out of 3)
When Trie B rolls 4…
…and Trie C rolls 2, Trie B wins (2/4)
…and Trie C rolls 3, Trie B wins (3/5)
…and Trie C rolls 9, Trie C wins (3/6)
When Trie B rolls 7…
…and Trie C rolls 2, Trie B wins (4/7)
…and Trie C rolls 3, Trie B wins (5/8)
…and Trie C rolls 9, Trie C wins (5/9)
Trie C = (2, 3, 9)
Trie A = (1, 5, 8)
When Trie C rolls 2…
…and Trie A rolls 1, Trie C wins (Trie C has won 1 out of 1)
…and Trie A rolls 5, Trie A wins (1 out of 2)
…and Trie A rolls 8, Trie A wins (1 out of 3)
When Trie C rolls 3…
…and Trie A rolls 1, Trie C wins (2/4)
…and Trie A rolls 5, Trie A wins (2/5)
…and Trie A rolls 8, Trie A wins (2/6)
When Trie C rolls 9…
…and Trie A rolls 1, Trie C wins (3/7)
…and Trie A rolls 5, Trie C wins (4/8)
…and Trie A rolls 8, Trie C wins (5/9)
The same reasoning can be applied to the six-sided non-transitive dice, but there are 36 possible outcomes when two of the dice are thrown against each other, so I won’t list them.
Die A = (1, 2, 2, 5, 6, 6)
Die B = (1, 1, 4, 5, 5, 5)
Die C = (3, 3, 3, 3, 4, 6)
Elsewhere other-posted:
At the Mountains of Mathness
Simpson’s Paradox — a simple situation with a very weird outcome
# The Brain in Pain
You can stop reading now, if you want. Or can you? Are your decisions really your own, or are you and all other human beings merely spectators in the mind-arena, observing but neither influencing nor initiating what goes on there? Are all your apparent choices in your brain, but out of your hands, made by mechanisms beyond, or below, your conscious control?
In short, do you have free will? This is a big topic – one of the biggest. For me, the three most interesting things in the world are the Problem of Consciousness, the Problem of Existence and the Question of Free Will. I call consciousness and existence problems because I think they’re real. They’re actually there to be investigated and explained. I call free will a question because I don’t think it’s real. I don’t believe that human beings can choose freely or that any possible being, natural or supernatural, can do so. And I don’t believe we truly want free will: it’s an excuse for other things and something we gladly reject in certain circumstances.
Continue reading The Brain in Pain
# Flesh and Binary
It’s odd that probability theory is so counter-intuitive to human beings and so late-flowering in mathematics. Men have been gambling for thousands of years, but didn’t develop a good understanding of what happens when dice are rolled or coins are tossed until a few centuries ago. And an intuitive grasp of probability would have been useful long before gambling was invented. Our genes automatically equip us to speak, to walk and to throw, but they don’t equip us to understand by instinct why five-tails-in-a-row makes heads no more likely on the sixth coin-toss than it was on the first.
Dice and gambling tokens from ancient Rome
Or to understand why five-boys-in-a-row makes the birth of a girl next time no more likely than it was during the first pregnancy (at least in theory). Boy/girl, like heads/tails, is a binary choice, so binary numbers are useful for understanding the probabilities of birth or coin-tossing. Questions like these are often asked to test knowledge of elementary probability:
1. Suppose a family have two children and the elder is a boy. What is the probability that both are boys?
2. Suppose a family have two children and at least one is a boy. What is the probability that both are boys?
People sometimes assume that the two questions are equivalent, but binary makes it clear that they’re not. If 1 represents a boy, 0 represents a girl and digit-order represents birth-order, the first question covers these possibilities: 10, 11. So the chance of both children being boys is 1/2 or 50%. The second question covers these possibilities: 10, 01, 11. So the chance of both children being boys is 1/3 = 33·3%. But now examine this question:
3. Suppose a family have two children and only one is called John. What is the probability that both children are boys?
That might seem the equivalent of question 2, but it isn’t. The name “John” doesn’t just identify the child as a boy, it identifies him as a unique boy, distinct from any brother he happens to have. Binary isn’t sufficient any more. So, while boy = 1, John = 2. The possibilities are: 20, 21, 02, 12. The chance of both children being boys is then 1/2 = 50%.
The three questions above are very simple, but I don’t think Archimedes or Euclid ever addressed the mathematics behind them. Perhaps they would have made mistakes if they had. I hope I haven’t, more than two millennia later. Perhaps the difficulty of understanding probability relates to the fact that it involves movement and change. The Greeks developed a highly sophisticated mathematics of static geometry, but did not understand projectiles or falling objects. When mathematicians began understood those in Renaissance Italy, they also began to understand the behaviour of dice, coins and cards. Ideas were on the move then and this new mathematics was obviously related to the rise of science: Galileo (1564-1642) is an important figure in both fields. But the maths and science can be linked with apparently distinct phenomena like Protestantism and classical music. All of these things began to develop in a “band of genius” identified by the American researcher Charles Murray. It runs roughly from Italy through France and Germany to Scotland: from Galileo through Beethoven and Descartes to David Hume.
Map of Europe from Mercator’s Atlas Cosmographicae (1596)
But how far is geography also biology? Having children is a form of gambling: the dice of DNA, shaken in testicle- and ovary-cups, are rolled in a casino run by Mother Nature. Or rather, in a series of casinos where different rules apply: the genetic bets placed in Africa or Europe or Asia haven’t paid off in the same way. In other words, what wins in one place may lose in another. Different environments have favoured different sets of genes with different effects on both bodies and brains. All human beings have many things in common, but saying that we all belong to the same race, the human race, is like saying that we all speak the same language, the human language. It’s a ludicrous and anti-scientific idea, however widely it may be accepted (and enforced) in the modern West.
Languages have fuzzy boundaries. So do races. Languages have dialects and accents, and so, in a sense, do races. The genius that unites Galileo, Beethoven and Hume may have been a particular genetic dialect spoken, as it were, in a particular area of Europe. Or perhaps it’s better to see European genius as a series of overlapping dialects. Testing that idea will involve mathematics and probability theory, and the computers that crunch the data about flesh will run on binary. Apparently disparate things are united by mathematics, but maths unites everything partly because it is everything. Understanding the behaviour of dice in the sixteenth century leads to understanding the behaviour of DNA in the twenty-first.
The next step will be to control the DNA-dice as they roll. China has already begun trying to do that using science first developed in the West. But the West itself is still in the thrall of crypto-religious ideas about equality and environment. These differences have biological causes: the way different races think about genetics, or persuade other races to think about genetics, is related to their genetics. You can’t escape genes any more than you can escape maths. But the latter is a ladder that allows us to see over the old genetic wall and glimpse the possibilities beyond it. The Chinese are trying to climb over the wall using super-computers; the West is still insisting that there’s nothing on the other side. Interesting times are ahead for both flesh and binary.
Appendix
1. Suppose a family have three children and the eldest is a girl. What is the probability that all three are girls?
2. Suppose a family have three children and at least one is a girl. What is the probability that all three are girls?
3. Suppose a family have three children and only one is called Joan. What is the probability that all three are girls?
The possibilities in the first case are: 000, 001, 010, 011. So the chance of three girls is 1/4 = 25%.
The possibilities in the second case are: 000, 001, 010, 011, 100, 101, 110. So the chance of three girls is 1/7 = 14·28%.
The possibilities in the third case are: 200, 201, 210, 211, 020, 021, 120, 121, 002, 012, 102, 112. So the chance of three girls is 3/12 = 1/4 = 25%.
# Live and Let Dice
How many ways are there to die? The answer is actually five, if by “die” you mean “roll a die” and by “rolled die” you mean “Platonic polyhedron”. The Platonic polyhedra are the solid shapes in which each polygonal face and each vertex (meeting-point of the edges) are the same. There are surprisingly few. Search as long and as far as you like: you’ll find only five of them in this or any other universe. The standard cubic die is the most familiar: each of its six faces is square and each of its eight vertices is the meeting-point of three edges. The other four Platonic polyhedra are the tetrahedron, with four triangular faces and four vertices; the octahedron, with eight triangular faces and six vertices; the dodecahedron, with twelve pentagonal faces and twenty vertices; and the icosahedron, with twenty triangular faces and twelve vertices. Note the symmetries of face- and vertex-number: the dodecahedron can be created inside the icosahedron, and vice versa. Similarly, the cube, or hexahedron, can be created inside the octahedron, and vice versa. The tetrahedron is self-spawning and pairs itself. Plato wrote about these shapes in his Timaeus (c. 360 B.C.) and based a mathemystical cosmology on them, which is why they are called the Platonic polyhedra.
Tetrahedron
Hexahedron
Octahedron
Dodecahedron
Icosahedron
They make good dice because they have no preferred way to fall: each face has the same relationship with the other faces and the centre of gravity, so no face is likelier to land uppermost. Or downmost, in the case of the tetrahedron, which is why it is the basis of the caltrop. This is a spiked weapon, used for many centuries, that always lands with a sharp point pointing upwards, ready to wound the feet of men and horses or damage tyres and tracks. The other four Platonic polyhedra don’t have a particular role in warfare, as far as I know, but all five might have a role in jurisprudence and might raise an interesting question about probability. Suppose, in some strange Tycholatric, or fortune-worshipping, nation, that one face of each Platonic die represents death. A criminal convicted of a serious offence has to choose one of the five dice. The die is then rolled f times, or as many times as it has faces. If the death-face is rolled, the criminal is executed; if not, he is imprisoned for life.
The question is: Which die should he choose to minimize, or maximize, his chance of getting the death-face? Or doesn’t it matter? After all, for each die, the odds of rolling the death-face are 1/f and the die is rolled f times. Each face of the tetrahedron has a 1/4 chance of being chosen, but the tetrahedron is rolled only four times. For the icosahedron, it’s a much smaller 1/20 chance, but the die is rolled twenty times. Well, it does matter which die is chosen. To see which offers the best odds, you have to raise the odds of not getting the death-face to the power of f, like this:
3/4 x 3/4 x 3/4 x 3/4 = 3/4 ^4 = 27/256 = 0·316…
5/6 ^6 = 15,625 / 46,656 = 0·335…
7/8 ^8 = 5,764,801 / 16,777,216 = 0·344…
11/12 ^12 = 3,138,428,376,721 / 8,916,100,448,256 = 0·352…
19/20 ^20 = 37,589,973,457,545,958,193,355,601 / 104,857,600,000,000,000,000,000,000 = 0·358…
Those represent the odds of avoiding the death-face. Criminals who want to avoid execution should choose the icosahedron. For the odds of rolling the death-face, simply subtract the avoidance-odds from 1, like this:
1 – 3/4 ^4 = 0·684…
1 – 5/6 ^6 = 0·665…
1 – 7/8 ^8 = 0·656…
1 – 11/12 ^12 = 0·648…
1 – 19/20 ^20 = 0·642…
So criminals who prefer execution to life-imprisonment should choose the tetrahedron. If the Tycholatric nation offers freedom to every criminal who rolls the same face of the die f times, then the tetrahedron is also clearly best. The odds of rolling a single specified face f times are 1/f ^f:
1/4 x 1/4 x 1/4 x 1/4 = 1/4^4 = 1 / 256
1/6^6 = 1 / 46,656
1/8^8 = 1 / 16,777,216
1/12^12 = 1 / 8,916,100,448,256
1/20^20 = 1 / 104,857,600,000,000,000,000,000,000
But there are f faces on each polyhedron, so the odds of rolling any face f times are 1/f ^(f-1). On average, of every sixty-four (256/4) criminals who choose to roll the tetrahedron, one will roll the same face four times and be reprieved. If a hundred criminals face the death-penalty each year and all choose to roll the tetrahedron, one criminal will be reprieved roughly every eight months. But if all criminals choose to roll the icosahedron and they have been rolling since the Big Bang, just under fourteen billion years ago, it is very, very, very unlikely that any have yet been reprieved. |
Polar Functions
collaborated by:
Laura Trkovsky
In this investigation we are going to look at polar functions and how these graphs change as we change different variables in the function, so let's get started.
The first function we are looking at is:
As we change different values for a, b, and k lets examine how the graph changes. When a=b and k is an integer we get the "n leaf rose" similar to this:
Here a=b=5 and k=8, we get a flower with eight petals. When a=b=k=1. we this pretty graph:
Now try playing with different values for a, b, and k. Graphing Calculator.
As the value for k increases so does the number of leaves. If it is not an integer then the leaves are not fully formed. As a and b are different the leaves do not touch. The closer they are to each other the closer they get to touching, but they also determine the length of the leaves. As a and b increase so does the length of the leaves. Also when b is larger than a then a second row of leaves are formed inside the larger ones.
Now we are going to look at if we change cosine to sine:
What happens to this as we change the a, b, and k values?
When k=1 this graph is only changed by the center looping get bigger as a and b get bigger. When k gets larger and a and b change the graph looks similar to the cosine graph lets check this out.
In the first one: a=6, b=8, k=3 And in the second: a=10 b=8, k=7.
We see that k still tells how many leaves, and when b is larger than a then the leaves repeat inside each other.
Second let's look at this:
We're gonna split these up and look at the first two first.
This is when a=1 and k=1 it looks like a normal circle.
But when we change the a and k values a little we might get something more like this.
when a=8 and k=6
The same trend is seen except the number of leaves is 2k instead of k.
This is just using cosine instead of sine. The circle has shifted, instead of the diameted being the y-axis it is now the x-axis. So what happens when we change a and k?
We see here that this looks similar to the sine graph except it is also rotated so that the x and y axes are bisecting the leaves, but there are still 2k leaves.
Now what about when we take the same equations and add b.
This is the same cosine equation but with b added. All a, b, and k equal one here. How do you think it will change as we change a, b, and k?
Here a=8, b=8 and k=6. We notice a difference this time. There are still 2k leaves but we have large and small leaves.
Here is a comparison of all the equations when a=b=k=1:
Right now they are all different but when we change the values for a, b, and k they look more alike.
If you look closely you can see how the red and purple are the same except for a rotation, the same with the blue and green. All four of these have four leaves, but the red and purple are all the same size and the blue and green have leaves of two different sizes. The black basically forms a line with point where it is undefined.
So we can see how we can change functions by just changing a few things like whether it is a sine or cosine function. The values being added, multiplied to the trig. function or being multiplied times the angle change each graph uniquely. |
# Deductive Reasoning, 2-3 9/1/02 …\sec2_3.ppt.
## Presentation on theme: "Deductive Reasoning, 2-3 9/1/02 …\sec2_3.ppt."— Presentation transcript:
Deductive Reasoning, 2-3 9/1/02 …\sec2_3.ppt
In this section we formalize their definition and use.
This section we will review the properties of equality that are useful for algebra in general and for geometric proofs specifically. We will specifically talk about the reflexive, symmetric, and transitive property of equality. In this section we formalize their definition and use. 4/15/2017 …\GeoSec02_04.ppt
Thus, the set of ALL cats does not have any dogs in that set.
Set theory deals with things (called elements or members) that we put in the same container. Thus, the set of ALL cats does not have any dogs in that set. The set of ALL house pets would include both cats and dogs, but not ALL cats and dogs are house pets. It is important to understand what is and is not in a set. 4/15/2017 …\GeoSec02_04.ppt
Deductive Reasoning, 2-3 9/1/02 Visualize the different sets as being separated, and that the relation between the sets (the operation that ties one element in A to an element in B) is represented by an arrow from set A to set B. The arrow is directional, so just because you can go from set A to set B, does not mean you can go from set B to set A. Once you have established sets, the next step is to build relations between any given two sets. These relations are called binary relations (meaning two) since we are dealing with two sets at any given time. Relation of A to B a2 a1 a3 b2 b1 b3 Once you have established sets, the next step is to build relations between any given two sets. These relations are called binary relations (meaning two) since we are dealing with two sets are any given time. Visualize the different sets as being separated, and that the relation between the sets (the operation that ties one element in A to an element in B) is represented by an arrow from set A to set B. The arrow is directional, so just because you can go from set A to set B, does not mean you can go from set B to set A. 4/15/2017 …\GeoSec02_04.ppt …\sec2_3.ppt
One of the most desirable relation is called an equivalence relation.
Mathematicians are always looking for new number systems that have specific relation properties. One of the most desirable relation is called an equivalence relation. An equivalence relation is defined are a relation that is reflexive, symmetric, and transitive. The concept of the equality relation (i.e., = symbol) is an equivalence relation and as we continue in this course we will find that congruence, , is an equivalence relation. 4/15/2017 …\GeoSec02_04.ppt
This property is called reflexive since it is a reflection of itself.
An element in set A has an equality relation with itself. Specifically, it is equal to itself. This property is called reflexive since it is a reflection of itself. This represents a relation. a2 a1 a3 A In the case to the left, a1 = a1. For example; 5 = 5 or x = x 4/15/2017 …\GeoSec02_04.ppt
If the order of the relation (in this case equality) does not matter, then the relation is called symmetric. You can think of this as looking the same on both sides of the equality symbol. In the case to the left, if a = b, then b = a For example; if 5 = 2 + 3, then = 5 b a c A 4/15/2017 …\GeoSec02_04.ppt
The third property requires three sets A, B, and C, and the relations A to B, and B to C. With these connections you can establish a new relation of A to C. This is the transitive property. Think of transitivity as moving through sets. Relation of A to C c2 c1 c3 a2 a1 a3 b2 b1 b3 Relation of A to B Relation of B to C The relation from A to B is different than the relation from B to C as well as the relation A to C. 4/15/2017 …\GeoSec02_04.ppt
Properties of Equality for Real Numbers
Reflexive Property For every number a, a = a. Symmetric Property For all numbers a and b, if a = b, then b = a. Transitive Property For all numbers a, b, and c, if a = b and b = c, then a = c Addition and Subtraction Properties For all numbers a, b, and c, if a = b, then a + c = b + c and a - c = b - c . Multiplication and Division Properties For all numbers a, b, and c, if a = b, then a • c = b • c and ,c 0. a/c = b/c Substitution Property For all numbers a and b, if a = b, then a may be replaced by b in any equation or expression. Distributive Property For all numbers a, b, and c, a(b + c) = ab + ac 4/15/2017 …\GeoSec02_04.ppt
Property Segments Angles
Reflexive PQ = PQ m 1 = m 1 Symmetric If AB = CD, then CD = AB. if m A = m B, then m B = m A Transitive If GH = JK, JK = LM, then if m 1 = m 2 and m 2 = m 3, GH = LM. then m 1 = m 3. 4/15/2017 …\GeoSec02_04.ppt
Summary In this section we established the reflexive, symmetric, and transitive properties of equality that you will use throughout the remainder of this course. We briefly established that segments and angles, from an equality perspective, are also reflexive, symmetric, and transitive. We will use these properties in the next section to show that congruence is reflexive, symmetric, and transitive. 4/15/2017 …\GeoSec02_04.ppt
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# Difference between revisions of "Newton's Sums"
Newton sums give us a clever and efficient way of finding the sums of roots of a polynomial raised to a power. They can also be used to derive several factoring identities.
## Basic Usage
Consider a polynomial:
$\displaystyle P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$
Let $P(x)=0$ have roots $x_1,x_2,\ldots,x_n$. Define the following sums:
$\displaystyle S_1 = x_1 + x_2 + \cdots + x_n$
$\displaystyle S_2 = x_1^2 + x_2^2 + \cdots + x_n^2$
$\vdots$
$\displaystyle S_k = x_1^k + x_2^k + \cdots + x_n^k$
$\vdots$
Newton sums tell us that,
$\displaystyle a_nS_1 + a_{n-1} = 0$
$\displaystyle a_nS_2 + a_{n-1}S_1 + 2a_{n-2}=0$
$\displaystyle a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3}=0$
$\vdots$
For a more concrete example, consider the polynomial $P(x) = x^3 + 3x^2 + 4x - 8$. Let the roots of $P(x)$ be $r, s$ and $t$. Find $r^2 + s^2 + t^2$ and $r^4 + s^4 + t^4$
Newton Sums tell us that:
$S_1 + 3 = 0$
$S_2 + 3S_1 + 8 = 0$
$S_3 + 3S_2 + 4S_1 - 24 = 0$
$S_4 + 3S_3 + 4S_2 - 8S_1 = 0$
Solving, first for $S_1$, and then for the other variables, yields,
$S_1 = r + s + t = -3$
$S_2 = r^2 + s^2 + t^2 = 1$
$S_3 = r^3 + s^3 + t^3 = 33$
$S_4 = r^4 + s^4 + t^4 = -127$
Which gives us our desired solutions, -127 and 1. |
# How do you find the determinant of |(4,-1,-2), (0, 2, 1), (2,1,3)|?
Jan 12, 2017
#### Answer:
The answer is $= 26$
#### Explanation:
One method is as follows :
$| \left(a , b , c\right) , \left(d , e , f\right) , \left(g , h , i\right) |$
$= a | \left(e , f\right) , \left(h , i\right) | - b | \left(d , f\right) , \left(g , i\right) | + c | \left(d , e\right) , \left(g , h\right) |$
$= a \left(e i - f h\right) - b \left(\mathrm{di} - g f\right) + c \left(\mathrm{dh} - e g\right)$
Therefore,
$| \left(4 , - 1 , - 2\right) , \left(0 , 2 , 1\right) , \left(2 , 1 , 3\right) |$
$= 4 \cdot | \left(2 , 1\right) , \left(1 , 3\right) | + 1 \cdot | \left(0 , 1\right) , \left(2 , 3\right) | - 2 \cdot | \left(0 , 2\right) , \left(2 , 1\right) |$
$= 4 \cdot \left(6 - 1\right) + 1 \cdot \left(0 - 2\right) - 2 \cdot \left(0 - 4\right)$
$= 4 \cdot 5 - 1 \cdot 2 - 2 \cdot - 4$
$= 20 - 2 + 8$
$= 26$ |
How do you solve the quadratic equation by completing the square: 2x^2 - 7x = 2?
Jul 14, 2015
The solution is $x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$.
Explanation:
$2 \left({x}^{2} - \frac{7}{2} x\right) = 2$
${x}^{2} - \frac{7}{2} x + \frac{49}{16} = 1 + \frac{49}{16}$
${\left(x - \frac{7}{4}\right)}^{2} = \frac{65}{16}$
$\left(x - \frac{7}{4}\right) = \pm \frac{\sqrt{65}}{4}$
$x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$
Jul 15, 2015
$x = \frac{7 + \sqrt{65}}{4} ,$ $\frac{7 - \sqrt{65}}{4}$
Explanation:
$2 {x}^{2} - 7 x = 2$
Divide both sides by $2$.
${x}^{2} - \frac{7}{2} x = 1$
To complete the square means to force a perfect square trinomial on the left side of the equation in the form ${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$.
Divide the coefficient of the $x$ term by $2$, square the result, and add to both sides of the equation.
$\frac{- 7}{2} \div 2 = \left(- \frac{7}{2}\right) \cdot \frac{1}{2} = - \frac{7}{4}$
${\left(- \frac{7}{4}\right)}^{2} = \frac{49}{16}$
${x}^{2} - \frac{7}{2} x + \frac{49}{16} = 1 + \frac{49}{16}$
The common denominator for $1$ and $\frac{49}{16}$ is $16$. Multiply $1$ times $\frac{16}{16}$, then add the two fractions.
${x}^{2} - \frac{7}{2} x + \frac{49}{16} = \frac{16}{16} + \frac{49}{16}$ =
${x}^{2} - \frac{7}{2} x + \frac{49}{16} = \frac{65}{16}$
We now have a perfect square trinomial on the left side, where $a = x$ and $b = \frac{7}{4}$.
${\left(x - \frac{7}{4}\right)}^{2} = \frac{65}{15}$
Take the square root of both sides.
$x - \frac{7}{4} = \pm \sqrt{\frac{65}{16}}$ =
$x - \frac{7}{4} = \pm \frac{\sqrt{65}}{4}$
Solve for $x$.
$x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$ =
$x = \frac{7 \pm \sqrt{65}}{4}$
$x = \frac{7 + \sqrt{65}}{4}$ =
$x = \frac{7 - \sqrt{65}}{4}$ |
## Presentation on theme: ""— Presentation transcript:
Finally a Formula for Derivative and Its Application
Can you see the pattern? And indeed, (Bring the exponent to the front as a coefficient, and reduce the exponent by 1, and use that as the new exponent.) This is our very first formula in derivatives without using the limit definition. It works for any base (as long as the base is the variable and nothing else) and any exponent (as long as the exponent is a real number). Technically, in mathematics, we use the letter n to denote a positive integer, but since this formula applies for any real number exponent, so it’s better to rewrite the formula as: Use the above formula to find the derivative of the following functions: For negative exponents: For rational exponents: For irrational exponents: However, the formula is NOT applicable for the following examples:
Derivative of a Constant Function and Some Derivative Formulas
Why the formula is NOT applicable for the following examples? Derivative of a Constant Function If f(x) = 7, what is f (x)? And in general, If f(x) = k where k is a constant, what is f (x)? Keep this in mind too: Some Fundamental Derivative Formulas In limits, we have laws. In derivatives we have formulas too. Limits Laws Derivatives Formulas 1. 2. 3. Applications: If f(x) = 2x3 – 4x2 + 5x – 6, what is f (x)? 2. Given g(t) = t6 + t–19 + t–2007, find g (t). 3.
Derivative of a Product
In the previous page, we showed you three formulas in derivatives, which really are complementary to three laws in limits. Limits Laws Derivatives Formulas In Words… 1. The limit/derivative of a constant times a function is the constant times the limit/derivative of the function. 2. The limit/derivative of a sum of two functions is the sum of the limit/ derivative of the two functions. 3. The limit/derivative of a difference of two functions is the difference of the limit/derivative of the two functions. In limits, we have the limit of a product of two functions is the product of the limit of the two functions: You might wonder: In derivatives, is the derivative of a product of two functions the product of the derivative of the two functions: f(x) g(x) If it is, then d/dx[x2x3] = But d/dx[(x2)(x3)] = If it is, then d/dx[2x3] = But d/dx[2x3] = f(x) g(x) The two examples above show that the derivative of a product ______ the product of the derivatives. That is,
The Product Rule If the derivative of a product is not the product of the derivatives, then what is the derivative of a product of two functions: It turns out the correct formula should be: and this is called the Product Rule. What did we just do: Differentiate the first factor function times Keep the second factor function plus Keep the first factor function times Differentiate the second factor function Why it works? Let’s see, with the product rule, we have: Check! You might say: Why do we need the product rule if it’s easier without it? Answer: Because we do need it sometimes, maybe a lot of the times, if not always. If you manage to get the derivative of the one above without using the product rule, let’s see how you handle this one:
If There Is a Product Rule Then There Is a …
If , then what is ? Ans: Definitely NOT . The correct formula is: and this is called the __________________. Let’s see how it works on some examples and compare it with the WRONG quotient rule too. The RIGHT Quotient Rule vs. The WRONG Quotient Rule How Do We Verify Which One Is Right?
Product Rule and Quotient Rule—In Depth
Sometimes the function might not be in terms of x, the variable we usually use and differentiate with respect to. For example, the function could be in terms of t, in that case, we would have written the product rule as: Regardless we are using d/dx when we differentiate with respect to x or d/dt when we differentiate with respect to t, we can always use the prime () notation for derivative. Hence, a shorthand for the product rule (by eliminating the x’s and t’s) is: (fg) = f g + fg and a shorthand for the quotient rule is: Déjà Vu? Example 1: Given f(x) = (2x + 3)(x2 – 4), find f (x). Example 2: Given f(x) = , find f (x). Different Versions of Product Rule and Quotient Rule: In some textbooks (including ours), the product rule can be written differently as: (fg) = f g + gf (fg) = gf + gf (fg) = gf + f g (fg) = f g + gf (fg) = gf + gf (fg) = gf + f g It’s because addition and multiplication are both commutative, i.e., a + b = b + a and ab = ba, hence we have these different versions. For quotient rule, you might see these versions in different textbooks: Our textbook Our textbook
Product Rule and Quotient Rule—To Use or Not To Use
My version of the product rule and quotient rule vs. The textbook’s version of the product rule and quotient rule (fg) = f g + fg (fg) = gf + gf Change “+” to “–” What is the relationship? Don’t see any. If you memorize my version of the product rule, for the quotient rule, all you need to do is to change the “+” sign the “–” sign, and put the whole thing over g2. If you memorize the textbook version of the product rule, it doesn’t help you to memorize its version of the quotient rule. That means you need to memorize the quotient rule too, which is obviously more complicated. When to use the product rule and quotient rule and when not to use them? Product Rule—Examples Use it / Don’t Use it / Doesn’t Matter Quotient Rule—Examples U / DU / DNM 1. f(x) = 2x3 2. f(x) = ¾(x2 + 5x – 6) 3. f(x) = (3x – 1)(2x + 4) 4. f(x) = (3x2 – 2x + 1)(4x2 + 5x – 4) |
# Math Snap
## The solution of the equation $\ln (2 z)=2-3 i$ is a. $\frac{1}{2} e^{2}(\cos 3+i \sin 3)$ b. $\frac{1}{2} e^{2}(\cos 3-i \sin 3)$ c. $\frac{1}{2} e^{2}(\cos 3-i \sin 3)$ d. $\frac{1}{2} e^{2}(\cos 3+i \sin 3)$
#### STEP 1
Assumptions 1. The given equation is $\ln(2z) = 2 - 3i$. 2. We need to solve for $z$. 3. The natural logarithm function $\ln$ has a complex argument. 4. The exponential form of a complex number $z$ is $z = re^{i\theta}$, where $r$ is the magnitude and $\theta$ is the argument. 5. The logarithm of a complex number $w$ can be expressed as $\ln(w) = \ln|w| + i\arg(w)$.
#### STEP 2
Rewrite the given equation $\ln(2z) = 2 - 3i$ in exponential form. $2z = e^{2 - 3i}$
#### STEP 3
Separate the real and imaginary parts in the exponent. $e^{2 - 3i} = e^2 \cdot e^{-3i}$
#### STEP 4
Recall Euler's formula $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. $e^{-3i} = \cos(-3) + i\sin(-3)$
#### STEP 5
Use the properties of trigonometric functions, noting that $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$. $e^{-3i} = \cos(3) - i\sin(3)$
#### STEP 6
Combine the results from STEP_3 and STEP_5. $e^{2 - 3i} = e^2 (\cos(3) - i\sin(3))$
#### STEP 7
Express $2z$ in terms of the exponential form. $2z = e^2 (\cos(3) - i\sin(3))$
#### STEP 8
Solve for $z$ by dividing both sides by 2. $z = \frac{1}{2} e^2 (\cos(3) - i\sin(3))$
##### SOLUTION
Compare the solution with the given options. The correct option is: $\boxed{\frac{1}{2} e^2 (\cos(3) - i\sin(3))}$ This matches option b. |
• Written By Anu_V
# Measures of Central Tendency: Mean, Median, Mode
A single number representing a set of data by identifying the central position within that set is a measure of central tendency. As a result, central tendency measures are also known as central location measures.
Central tendency focuses on the central distribution of data through a single value. Types of central tendency in real-life are average marks, rainfall, income, etc. Three commonly used measures of central tendency such as arithmetic mean, median, and mode.
This is the most important statistical method to represent the value of the dataset. We can choose the measures of central tendency based on the kind of data we have. Students learn this concept in order to solve mathematical problems in a fraction of seconds. You can check NCERT Solution for Class 11 Maths Chapter 15 for better understanding. We have provided detailed information on measure of central tendency in this article. Read on to find out about its definition, types, and examples.
## Measures of Central Tendency: Definition
The measure of central tendency is the summary of the data set in the form of a typical value. It focuses on providing an accurate description of the data. This is one of the important methods to solve statistical problems. The measures of central tendency provide a summary of the data set rather than the individual data.
### Understanding Mean, Median And Mode
There are several measures of central tendency, out of which the most commonly used are Mean, Median and Mode. Let us now focus on the most commonly used measures of central tendency as mentioned below:
#### Mean
Mean is defined as the sum of values in the dataset divided by the number of observations or values. This is the most commonly used measure of central tendency and is also called the arithmetic mean. It is calculated by using the formula:
X1 + X2+ X3 +……………….+ XN/ N
There are two types of arithmetic mean as the arithmetic means of ungrouped data and arithmetic mean for grouped data.
For example, A monthly income of 4 families is 1600, 1400, 1300, 1200. Find the average income of the family.
Solution: Given, 1600 + 1400 + 1300 +1200
= 1600 + 1400 + 1300 +1200 /4
= 1375
Therefore, the average monthly income of the family is Rs. 1375.
#### Median
It is defined as the value that divides the distribution into two halves. One part of the value is greater or equal to the median value and the other is less than or equal to it. You can calculate the median by arranging the data from the smallest value to the largest one.
For example, Arrange the data 5, 7, 6, 1, 8, 10, 12, 4, and 3 in ascending order and find out its median.
1, 3, 4, 5, 6, 7, 8, 10, 12
The middle value of the data is 6. Here, half of the numbers are greater than 6 and the other halves are smaller.
In case, the middle value has two numbers in the data, you calculate by using the formula mentioned below:
1, 3, 4, 5, 6, 7, 8, 10, 12 , 13
This can be calculated by adding two median values divided by the number of observations. For example, 6+7/ 2= 9.5
Therefore, the median value is 9.5.
#### Mode
Mode is the frequently observed data value. There are chances that we get repeated numbers in the data set.
For example, 1, 2, 3, 4, 4, 5, Here 4, 4 is the frequently observed value.
### Examples of Central Tendency
Some of the examples of measures of central tendency are given below:
Example 1: Find the mean deviation about the mean for the data 4,7,8,9,10,12,13,17
Solution: Step 1:The given data is 4,7,8,9,10,12,13,17
Mean of the data, ?̅= 4+7+8+9+10+12+13+17/ 8 = 80/ 8 = 10
The deviations of the respective observations from the mean ?̅, i.e. ?? − ?̅ are – 6, – 3, – 2, – 1,0,2,3,7
The absolute values of the deviations, i.e. |?? − ?̅|, are 6,3,2,1,0,2,3,7
The required mean deviation about the mean is
M.D. (?̅) = ∑ |?? − ?̅| / 10
12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 / 10 = 84 /10 = 8.4
Example 2: Find the mean deviation about the mean for the data 4,7,8,9,10,12,13,17
Solution: Step 1:The given data is 4,7,8,9,10,12,13,17
Mean of the data, ?̅= 4+7+8+9+10+12+13+17/ 8 = 80 /8 = 10
The deviations of the respective observations from the mean ?̅, i.e. ?? − ?̅ are – 6, – 3, – 2, – 1,0,2,3,7
The absolute values of the deviations, i.e. |?? − ?̅|, are 6,3,2,1,0,2,3,7
The required mean deviation about the mean is
M.D. (?̅) = ∑ |?? − ?̅|/ 8
= 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7/ 8 = 24 /8 = 3
### FAQs
The frequently asked questions on measures of central tendency are given below:
Now we have provided information on measures of central tendency in this article. So, study seriously and master every concept in the CBSE Class 11 syllabus. Make use of these NCERT solutions for Class 11. You can also solve CBSE Class 11 PCM questions on Embibe. These will help you in your Class 11 preparation as well as other competitive exams.
We hope this detailed article on measures of central tendency helps you. If you have any queries, feel free to ask in the comment section below. We will get back to you at the earliest.
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# NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
Written by Team Trustudies
Updated at 2021-05-07
## NCERT solutions for class 8 Maths Chapter 5 Data Handling Exercise 5.1
1.For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman’s bag.
(b) The height of competitors in an athletics meet.
(c) The number of cassettes produced by 5 companies.
(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m.. at a station.
Give reasons for each.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
In (b) and (d),histograms can be drawn because data can be divided into class intervals.
2.The shoppers who come to a departmental store are marked as: man (M), woman (W), boy (B) or Girl (G). The following list gives the shoppers who came during the first hour in the morning. W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
Arrangement of data in a table using tally marks is done as :
3.The weekly wages (in ?) of 30 workers in a factory are : 830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806,840.
Using tally marks make a frequency table with intervals as 800-810, 810-820 and so on.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
Grouped frequency table is made as follows:
4.Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions:
(i) Which group has the maximum number of workers?
(ii) How many workers earn ₹ 850 and more?
(iii) How many workers earn less than ₹ 850?
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
Histogram for the given data in question 3 is shown as:
From the above histogram, clearly we see that,
(i) The group 830-840 has the maximum number of workers i.e, 9.
(ii) The number of workers who earn equal and more than ₹ 850 are 1 + 3 + 1 + 1 + 4, i.e., 10.
(iii) The number of workers who earn less than ? 850 are 3 + 2 + 1 + 9 + 5, i.e., 20.
5.The number of hours for which students of a particular class watched television during holidays is shown through the given graph.
(i) For how many hours did the maximum number of students watch TV?
(ii) How many students watched TV for less than 4 hours?
(iii) How many students spent more than 5 hours watching TV?
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
We can observe the following from the given figure:
(i) The maximum number of students who watched TV for 4 to 5 hours is 32.
(ii) 4 + 8 + 22 = 34 students watched TV for less than 4 hours.
(iii)The number of students who watched TV for more than 5 hours is 8 + 6 = 14.
## NCERT solutions for class 8 Maths Chapter 5 Data Handling Exercise 5.2
1.A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey.
From this pie chart answer the following:
(i) If 20 people liked classical music, how many young people were surveyed?
(ii) Which type of music is liked by the maximum number of people?
(iii) If a cassette company were to make 1000 CDs. How many of each type would they make?
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
(i) let, the number of young people surveyed =x
∴10% of x=20
$$\Rightarrow \frac { 100\times x}{ 10 } = 20$$
$$\Rightarrow x=200$$
Thus, number of young people surveyed = 200
(ii) Light music is liked by the maximum people, i.e., 40%
(iii) Total number of CD = 1000
Number of CD’s in respect of:
classical music =$$\frac { 10\times 1000 }{ 100 } = 100$$
semi-classical music = $$\frac { 20\times 1000 }{ 100 } = 200$$
light music = $$\frac { 40\times 1000 }{ 100 } = 400$$
folk music = $$\frac { 30\times 1000 }{ 100 } = 300$$
2.A group of 360 people were asked to vote for their favourite season from the three seasons rainy, winter and summer.
(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii) Draw a pie chart to show this information.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
(i) Winter season got the maximum votes, i.e. 150.
(iii) Pie chart for the given data:
3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.
Find the proportion of each sector. For example, Blue is $$\frac { 18 }{ 36 } =\frac { 1 }{ 2 }$$ ; Green is $$\frac { 9 }{ 36 } =\frac { 1 }{ 4 }$$ and so on.
Use this to find the corresponding angles.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
Central angles with their color are:
Blue colour: $$=\left( \frac { 18 }{ 36 } \times 360 \right) ^{ \circ }=180^{ \circ }$$
Green colour: $$=\left( \frac { 9 }{ 36 } \times 360 \right) ^{ \circ }=90^{ \circ }$$
Red colour: $$=\left( \frac { 6 }{ 36 } \times 360 \right) ^{ \circ }=60^{ \circ }$$
Yellow colour: $$=\left( \frac { 3 }{ 36 } \times 360 \right) ^{ \circ }=30^{ \circ }$$
Now, various components is shown by the adjoining pie chart.
4. The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.
(i) In which subject did the student score 105 marks?
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
(i) For 540 marks, the central angle = $$360^{ \circ}$$
∴ the central angle for 1 mark = $$\left( \frac { 360 }{ 540 } \right) ^{ \circ }=\left( \frac { 2 }{ 3 } \right) ^{ \circ }$$
∴ the central angle for 105 marks =$$\left( \frac { 2 }{ 3 } \times 105 \right) ^{ \circ }=70^{ \circ }$$
Hence, from the given pie chart, the subject is Hindi.
(ii) Difference between the central angles made by the subject of Mathematics and Hindi= $$90^{\circ} – 70^{\circ} = 20^{\circ}$$.
Marks obtained for the central angle of $$360^{\circ}$$, = 540 Marks obtained for the central angle of $$1^{\circ}$$=$$\frac { 540 }{ 360 } =\frac { 3 }{ 2 }$$
Marks obtained for the central angle of $$20^{\circ}=\frac { 3 }{ 2 } \times 20=30$$
Hence, the student obtained 30 more marks in Mathematics than Hindi.
(iii) Sum of the central angles made by Social Science and Mathematics =$$65^{\circ}+ 90^{\circ} = 155^{\circ}$$.
And, the sum of the central angles made by Science and Hindi = $$80^{\circ}+ 70^{\circ} = 150^{\circ}$$.
Since $$155^{\circ} >150^{\circ}$$, therefore, the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
5.The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
Let us find the central angle of each sector with the total scale = 72. We thus have the following table :
And the pie chart for the given data is:
## NCERT solutions for class 8 Maths Chapter 5 Data Handling Exercise 5.3
1.List the outcomes you can see in these experiments.
(i) Spinning a wheel
(ii) Tossing two coins together
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
(i) We can get the possible outcomes on spinning the given wheel as B, C, D, E and A.
(ii) When two coins are tossed together, we get the following possible outcomes
HH, HT, TH, TT
.
2.When a die is thrown, list the outcomes of an event of getting
(i) (a) a prime number
(b) not a prime number
(ii) (a) a number greater than 5
(b) a number not greater than 5
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
In a throw of die, list of the outcomes of an event of getting
(i) (a) the prime number are 2, 3 and 5
(b) not a prime number are 1, 4 and 6
(ii) (a) a number greater than 5 = 6
(b) a number not greater than 5 are 1, 2, 3, 4, 5
3. Find the
(i) Probability of the pointer stopping on D in (Question 1-(a))?
(ii) Probability of getting an ace from a well-shuffled deck of 52 playing cards?
(iii) Probability of getting a red apple, (see figure below)
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
(i) Refer to fig. Question 1-(a)
Total number of sectors = 5
Number of sectors where the pointer stops at D = 1, because there is only one ‘D’ on the spinning wheel.
Probability of pointer stopping at D = $$\frac{1}5$$
(ii) Number of aces = 4 (one from each suit i.e. heart, diamond, club and spade)
Total number of playing cards = 52
Probability of getting an ace=$$\frac{Number\; of\; aces}{Total\; number\; of \;playing \;cards}$$
$$=\frac{4}{52}=\frac{1}{13}$$
(iii) Total number of apples = 7
Number of red apples = 4
Probability of getting red apples=$$\frac{Number \;of\; red \;apples}{Total \;number\; of \;apples}=\frac{4}7$$
4.Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of:
(i) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv) getting a 1-digit number?
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
(i)An event of getting a number 6, i. e., if we obtain a slip having number 6 as an outcome out of total 10 slips
So favourable number of outcomes = 1
∴ required probability of getting a number 6 = $$\frac{1}{10}$$
(ii) Probability of getting a number less than 6 = $$\frac{5}{10}= \frac{1}2$$ [∵ Numbers less than 6 are 1, 2, 3, 4, 5]
(iii) Probability of getting a number greater than 6 = $$\frac{4}{10}= \frac{2}5$$ [∵ Number greater than 6 are 7, 8, 9, 10]
(iv) Probability of getting a 1-digit number = $$\frac{9}{10}$$ [∵ 1-digit numbers are 9, i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9]
5.If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector?
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
Out of the total 5 sectors, the pointer can stop at any of the sectors in 5 ways.
∴ Total number of outcomes = 5
There are 3 green sectors in the spinning wheel, out of which one can be obtained in 3 ways. ∴ Favourable number of outcomes = 3
So, the required probability for green sector= $$\frac { 3 }{ 5 }$$
Further, there are 4 ways to obtain a non-blue sector out of 4 non-blue present in the spinning wheel.
So, the required probability for non- blue sector = $$\frac { 4 }{ 5 }$$
6.Find the probabilities of the events given in Question 2.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
In a single throw of a die, we can get any one of the six numbers marked on its six faces. Therefore, the total number of outcomes = 6
(i) Let A denote the event “getting a prime number”. So,event A occurs if we obtain 2, 3, 5 as an outcome.
∴ Favourable number of outcomes = 3
Hence, P$$\left( A \right) =\frac { 3 }{ 6 } =\frac { 1 }{ 2 }$$
(ii) Let A denote the event “not getting a prime number”. So, event A occurs if we obtain 1, 4, 6 as an outcome.
∴ Favourable number of outcomes = 3
Hence, P$$\left( A \right) =\frac { 3 }{ 6 } =\frac { 1 }{ 2 }$$
(iii) The event “getting a number greater than 5” will occur if we obtain the number 6.
∴ Favourable number of outcomes = 1
Hence, the required probability = $$\frac { 1}{ 6 }$$
(iv) The event “getting a number not greater than 5” will occur if we obtain one of the numbers 1, 2, 3, 4, 5.
∴ Favourable number of outcomes = 5
Hence, required probability = $$\frac { 5 }{ 6 }$$
##### FAQs Related to NCERT Solutions for Class 8 Maths Chapter 5 Data Handling
There are total 16 questions present in ncert solutions for class 8 maths chapter 5 data handling
There are total 2 long question/answers in ncert solutions for class 8 maths chapter 5 data handling
There are total 3 exercise present in ncert solutions for class 8 maths chapter 5 data handling |
# The population of Greenville is approximately 75 times the population of Fairview. There are 2.50 times 10^4 people living in Greenville. Approximately how many people are living in Fairview?
Nov 14, 2017
See a solution process below:
#### Explanation:
Let's call the population of Fairview: $f$
And, let's call the population of Greenville: $g$
Then we can write an equation:
$g = 75 f$ or $\frac{g}{75} = f$
Substituting $2.50 \times {10}^{4}$ for $g$ gives:
$\frac{2.50 \times {10}^{4}}{75} = f$
$\frac{2.50}{75} \times {10}^{4} = f$
$0.0 \overline{3} \times {10}^{4} = f$
We can move the decimal point two places to the left and subtract from the exponent for the 10s term giving:
$3. \overline{3} \times {10}^{2} = f$
The population of Fairview is approximately $3.33 \times {10}^{2}$ |
# 2020 AIME I Problems/Problem 12
## Problem
Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$
## Solution 1
Lifting the Exponent shows that $$v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1$$ so thus, $3^2$ divides $n$. It also shows that $$v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2$$ so thus, $7^5$ divides $n$.
Now, multiplying $n$ by $4$, we see $$v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})$$ and since $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4n}-2^{4n})=1+v_5(n)$ meaning that we have that by LTE, $4 \cdot 5^4$ divides $n$.
Since $3^2$, $7^5$ and $4\cdot 5^4$ all divide $n$, the smallest value of $n$ working is their LCM, also $3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.
~kevinmathz
## Solution 2 (Simpler, just basic mods and Fermat's theorem)
BY THE WAY, please feel free to correct my formatting. I don't know latex.
Note that for all $n$, $149^n - 2^n$ is divisible by $149-2 = 147$ because that is a factor. That is $3\cdot7^2$, so now we can clearly see that the smallest $n$ to make the expression divisible by $3^3$ is just $3^2$. Similarly, we can reason that the smallest $n$ to make the expression divisible by $7^7$ is just $7^5$.
Finally, for $5^5$, take $(\text{mod} 5)$ and $(\text{mod} 25)$ of each quantity (They happen to both be $-1$ and $2$ respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum $n$ for divisibility by $5$ is $4$, and other values are factors of $4$. Testing all of them(just $1$,$2$,$4$ using mods-not too bad), $4$ is indeed the smallest value to make the expression divisible by $5$, and this clearly is NOT divisible by $25$. Therefore, the smallest $n$ to make this expression divisible by $5^5$ is $2^2 \cdot 5^4$.
Calculating the LCM of all these, one gets $2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. Using the factor counting formula, the answer is $3\cdot3\cdot5\cdot6$ = $\boxed{270}$. Can someone please format better for me?
-Solution by thanosaops -formatted by MY-2 |
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