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# HOW TO FIND REMOVABLE DISCONTINUITY AT THE POINT
## About "How to Find Removable Discontinuity At The Point"
How to Find Removable Discontinuity At The Point :
Here we are going to see how to test if the given function has removable discontinuity at the given point.
The function f(x) is defined at all points of the real line except x = 0. That is, f(0) is undefined, but lim x -> 0 sin x/x = 1. If we redefine the function f(x) as
h is defined at all points of the real line including x = 0. Moreover, h is continuous at x = 0 since
lim x -> 0 h(x) = lim x -> 0 (sin x / x) = 1 = h(0)
Note that h(x) = f(x) for all x ≠ 0. Even though the original function f(x) fails to be continuous at x = 0, the redefined function became continuous at 0.
That is, we could remove the discontinuity by redefining the function. Such discontinuous points are called removable discontinuities. This example leads us to have the following.
Definition of removable discontinuity :
A function f defined on an interval I ⊆ R is said to have removable discontinuity at x0 ∈ I if there is a function h :
I -> R such that h(x) =
## Finding Removable Discontinuity At the given point - Examples
Question 1 :
Which of the following functions f has a removable discontinuity at x = x0? If the discontinuity is removable, find a function g that agrees with f for x ≠ x0 and is continuous on R.
(i) f(x) = (x3 + 64)/(x + 4), x0 = -4
Solution :
In order to check if the given function is continuous at the point x0 = -4, let us apply -4
f(x) = ((-4)3 + 64)/(-4 + 4),
= (-64 + 64)/0
= 0/0
The given function is not continuous at x = -4. In order to redefine the function, we have to simplify f(x).
f(x) = (x + 4)(x2 - 4x + 16)/(x + 4)
f(x) = (x2 - 4x + 16)
f(-4) = ((-4)2 - 4(-4) + 16)
= 16 + 16 + 16
= 48
Hence it has removable discontinuity at x = -4. By redefining the function, we get
(iii) f(x) = (3 - √x)/(9 - x), x0 = 9
Solution :
In order to check if the given function is continuous at the point x0 = 9, let us apply 9
f(x) = (3 - √x)/(9 - x),
= (3 - √9)/(9 - 9)
= (3 – 3) / 0
= 0/0
The given function is not continuous at x = 9. In order to redefine the function, we have to simplify f(x).
f(x) = (3 - √x)/(32 – (√x)2)
f(x) = (3 - √x)/(3 + √x) (3 – √x)
= 1/(3 + √x)
f(9) = 1/(3 + √9)
= 1/(3 + 3)
= 1/6
Hence it has removable discontinuity at x = 9. By redefining the function, we get
After having gone through the stuff given above, we hope that the students would have understood, "How to Find Removable Discontinuity At The Point"
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# Equation of a Line Parallel to a Line
We will learn how to find the equation of a line parallel to a line.
Prove that the equation of a line parallel to a given line ax + by + λ = 0, where λ is a constant.
Let, ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.
Now, convert the equation ax + by + c = 0 to its slope-intercept form.
ax + by+ c = 0
⇒ by = - ax - c
Dividing both sides by b, [b ≠ 0] we get,
y = -$$\frac{a}{b}$$ x - $$\frac{c}{b}$$, which is the slope-intercept form.
Now comparing the above equation to slope-intercept form (y = mx + b) we get,
The slope of the line ax + by + c = 0 is (- $$\frac{a}{b}$$).
Since the required line is parallel to the given line, the slope of the required line is also (- $$\frac{a}{b}$$).
Let k (an arbitrary constant) be the intercept of the required straight line. Then the equation of the straight line is
y = - $$\frac{a}{b}$$ x + k
by = - ax + bk
ax + by = λ, Where λ = bk = another arbitrary constant.
Note: (i) Assigning different values to λ in ax + by = λ we shall get different straight lines each of which is parallel to the line ax + by + c = 0. Thus, we can have a family of straight lines parallel to a given line.
(ii) To write a line parallel to a given line we keep the expression containing x and y same and simply replace the given constant by a new constant λ. The value of λ can be determined by some given condition.
To get it more clear let us compare the equation ax + by = λ with equation ax + by + c = 0. It follows that to write the equation of a line parallel to a given straight line we simply need to replace the given constant by an arbitrary constant, the terms with x and y remain unaltered. For example, the equation of a straight line parallel to the straight line 7x - 5y + 9 = 0 is 7x - 5y + λ = 0 where λ is an arbitrary constant.
### Solved examples to find the equations of straight lines parallel to a given line:
1. Find the equation of the straight line which is parallel to 5x - 7y = 0 and passing through the point (2, - 3).
Solution:
The equation of any straight line parallel to the line 5x - 7y = 0 is 5x - 7y + λ = 0 …………… (i) [Where λ is an arbitrary constant].
If the line (i) passes through the point (2, - 3) then we shall have,
5 ∙ 2 - 7 ∙ (-3) + λ = 0
10 + 21 + λ = 0
31 + λ = 0
λ = -31
Therefore, the equation of the required straight line is 5x - 7y - 31 = 0.
2. Find the equation of the straight line passing through the point (5, - 6) and parallel to the straight line 3x - 2y + 10 = 0.
Solution:
The equation of any straight line parallel to the line 3x - 2y + 10 = 0 is 3x - 2y + k = 0 …………… (i) [Where k is an arbitrary constant].
According to the problem, the line (i) passes through the point (5, - 6) then we shall have,
3 ∙ 5 - 2 ∙ (-6) + k = 0
15 + 21 + k = 0
36 + k = 0
k = -36
Therefore, the equation of the required straight line is 3x - 2y - 36 = 0.
The Straight Line
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The relationship between function and curve and equation The curve must be an equation? The equation must be a curve? A function must be an equation? A function must be an equation?
The relationship between function and curve and equation The curve must be an equation? The equation must be a curve? A function must be an equation? A function must be an equation?
(1) Curve and equation are two different concepts. Curve is a geometric concept, equation is a mathematical concept, and curve can be expressed by equation. Therefore, it can not be said that curve must be equation or equation must be curve. (2) the image of function is not necessarily curve, or straight line, and curve does not necessarily correspond to function
Given that the abscissa of the intersection of line L and line y = 2x + 1 is 2, and the ordinate of the intersection of line y = x + 1 is 2, find the expression of line L. (note the format)
Let l be y = KX + B
The abscissa is 2, where y = 2x + 1
y=2*2+1=5
So l passed (2,5)
The ordinate is 2, where y = x + 1
2=x+1
x=1
So l passed (1,2)
So 5 = 2K + B
2=k+b
So k = 3, B = - 1
So l is y = 3x-1
Periodic function If the image of the function f (x) defined on R is symmetrical about the point (- 3 / 4,0), and f (x) = - f (x + 3 / 2), f (1) = 1, f (0) = 2, then the value of F (1) + f (2) + F (3) +... + F (2011) is __
F (x) = - f (x + 3 / 2), i.e. f (x + 3 / 2) = - f (x + 3 / 2) + 3 / 2] = - f (x + 3 / 2) = f (x), so the function period is 3. Because the image of function f (x) is symmetrical about point (- 3 / 4,0), then f (x) + F (- 3 / 2-x) = 0, and f (x + 3 / 2) = f (x), so f (- 3 / 2-x) = f (x
Given that the function f (x) satisfies f (x + 2) = f (x), f (2 + x) = f (2-x) and X ∈ [2,3], f (x) = (X-2) 2, find the analytical formula of F (x) on [4,6]
F (2 + x) = f (2-x), it is obtained that the f (x) image is symmetrical about x = 2;
F (x) = (X-2) 2 on X ∈ [2,3] is symmetric about x = 2, and the analytical formula of function f (x) on [1,2] is also f (x) = (X-2) 2;
That is, the function f (x) = (X-2) 2 on [1,3];
From F (x + 2) = f (x), we know that the period of F (x) is 2, and [1,3] is exactly a period of F (x). The image of translating one period to the right and then another period is as follows:
It can be seen from the figure that the analytical formula F (x) = (x-2-2) 2 = (x-4) 2 on [4,5], and the analytical formula on [5,6] is f (x) = (x-2-2) 2 = (X-6) 2;
That is, the analytical formula of F (x) on [4,6] is: F (x) =
(x−4)2 x∈[4，5]
(x−6)2 x∈(5，6] ．
Comprehensive application of function Let the function f (x) = (x-1) ^ 2 (x + b) e ^ x, if x = 1, be a maximum point of F (x), then the value range of real number B is______
Firstly, we can see that when x = 1, no matter how much B is taken, f (x) = 0, which means that near x = 1, f (x) is always negative, so that x = 1 is a maximum point
1： When B is not equal to - 1, (x-1) ^ 2 is always greater than or equal to 0, and e ^ x is always greater than 0, so in the neighborhood of x = 1 (x + b)
Mathematical problems of function equation Log3 (x ^ 2-10) = 1 + log3 (x) the 3 after solving the log is the subscript. I can't type it!
Move log3 (x) on the right to the left to get log3 (x ^ 2-10) - log3 (x) = 1 + log3 (x)
So log3 [(x ^ 2-10) / x] = 1, so [(x ^ 2-10) / x = 0, so x = √ 10 or - √ 10. Also, because X in log3 (x) cannot be negative, x = + √ 10
Note: √ is the root sign |
# Multiplication Table of 1
Repeated addition by 1’s means the multiplication table of 1.
(i) When 3 candle-stand having 1 candle each.
By repeated addition we can show 1 + 1 + 1 = 3
Then, one 3 times or 3 one’s
3 × 1 = 3
Therefore, there are 3 candles.
(ii) When 8 groups having 1 lollipop each.
By repeated addition we can show 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8
Then, one 8 times or 8 one’s
8 × 1 = 8
Therefore, there are 8 lollipops.
We will learn how to use the number line for counting the multiplication table of 1.
(i) Start at 0. Hop 1, five times. Stop at 5.
5 one’s are 5 5 × 1 = 5
(ii) Start at 0. Hop 1, nine times.
Stop at ____. Thus, it will be 9
9 one’s are 9 9 × 1 = 9
(iii) Start at 0. Hop 1, eleven times.
Stop at ____. Thus, it will be 11
11 one’s are 253 11 × 1 = 11
How to read and write the table of 1?
The above chart will help us to read and write the 1 times table.
Read 1 one is 1 2 one’s are 2 3 one’s are 3 4 one’s are 4 5 one’s are 5 6 one’s are 6 7 one’s are 7 8 one’s are 8 9 one’s are 9 10 one’s are 10 11 one’s are 11 12 one’s are 12 Write 1 × 1 = 1 2 × 1 = 2 3 × 1 = 3 4 × 1 = 4 5 × 1 = 5 6 × 1 = 6 7 × 1 = 7 8 × 1 = 8 9 × 1 = 9 10 × 1 = 10 11 × 1 = 11 12 × 1 = 12
Now we will learn how to do forward counting and backward counting by 1’s.
Forward counting by 1’s: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, ……
Backward counting by 1’s: ……, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0
Multiplication Table
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# Important Statistics Equations You Need To Be Aware Of
Karl Pearson has beautifully said – “Statistics Is The Grammar Of Science”
Although statistics is said to be a part of mathematics. There are various application of mathematics in statistics.
The mathematics of working with numeric data and analyzing it is called statistics. More formally, it is the study of data collection, interpretation, organization, and presentation. According to statistical theory, a statistic is a function of a sample that doesn’t depend on the sample’s distribution. Statistics equations enable you to measure the tendencies of a population in various ways and work out the deviation of values.
This branch of math enables you to draw insights from data and even predict patterns and possibilities by working with the facts of the numeric data. Here is a collection of important statistics math equations that students, statisticians, marketing analysts, financial analysts, and a wide array of professionals commonly use.
## Statistics Probability Equations
Here are seven common statistics equations:
See also How to become the experts in social science in class seventh?
### #1 Population Mean
This term refers to the average of a characteristic in a population. The formula is:
μ = ( Σ Xi ) / N
Here, “μ” is the population mean, “Σ Xi” refers to the sum of the characteristics in the population, and “N” is the total size of the population.
### #2 Population Standard Deviation
Standard deviation is a measure of the “spread” or the variability of scores in a data set, represented by:
σ = sqrt [ Σ ( Xi – μ )2 / N ]
As you may have figured out, the term is represented by “σ.” “Sqrt” denotes the square root, and Σ ( Xi – μ )2 is the sum of the squared deviations of the values minus the population mean.
### #3 Population Variance
The variance shows us the spread of the values in a population and is the square of the standard deviation. The formula is:
σ2 = Σ ( Xi – μ )2 / N
### #4 Sample Mean
The average score of a sample on a specific variable is its sample mean, and its formula is:
x_bar = ( Σ xi ) / n
It is the sum of the scores in the sample divided by the total number of data points in the sample.
### #5 Sample Standard Deviation
The term represents the spread of the scores in a sample for a specific variable. Here’s its formula:
s = sqrt [ Σ ( xi – x_bar )2 / ( n – 1 ) ]
In this formula, the “Σ ( xi – x_bar )2” represents the sum of squared deviations of scores from the sample mean.
### #6 Sample Variance
It is the square of the sample standard deviation and is denoted by s2, like so:
s2 = Σ ( xi – x_bar )2 / ( n – 1 )
## Probability Statistics Equations
The probability of something represents the possibility of an outcome of an event. It is a test of the likelihood of something happening. The most common way of explaining this is the coin toss – the possibility of the coin showing heads or tails is half and half.
The formula for the probability of an event E is:
P(E) = Number of positive outcomes/Number of total outcomes = n(E)/n(S)
There are many statistics equations that are used by students and professionals alike. If A and B are two events, then:
## Regression Equations Statistics
Regression equations allow us to determine whether there is any relationship between two data sets. The equations facilitate regression analysis, one of the most popular methods to determine the connection between dependent and independent variables.
The equations help assess the strength of the connection between two variables and depending on the outcome; the equations may be used to model the relationship between the variables later.
A simple regression analysis is measuring the height growth of a child. Let’s say you measure the child’s height annually and determine that they grow three inches every year. This trend of growth may be modeled using a regression equation.
Real-world regression analysis examples include calculating the odds of natural disasters and price changes in commodities of all kinds.
The formula for regression is:
Y = a + bX + ∈
The “Y” is the dependent variable, and as you can guess, “X” is the independent variable. The “a” represents the intercept, “b” is the slope, and “∈” is the error.
Working out the values of the intercept and slope is as simple as applying the following formula:
a = (Σy)(Σx2) – (Σx)(Σxy)/ n(Σx2) – (Σx)2
b = n (Σxy) – (Σx)(Σy) /n(Σx2) – (Σx)2
## Conclusion
Statistics is considered to be a game changer, and statistical techniques are created to examine the key factors of vast data. Various organisations and governments utilise statistical methods to compute a collective characteristic about workers or people. This property then influences the decisions that the organisations and governments make.
Today, we have discussed all the important statistics equations that are useful to improve your statistics knowledge. Try to remember these equations as, without these, you can not solve any of the statistics-related problems.
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# Heart of Algebra
Related Topics:
More Lessons for SAT Math
More Resources for SAT
We have lots of free resources and videos to help you prepare for the SAT. These materials are for the redesigned SAT which is for you if you are taking the SAT in March 2016 and beyond.
Heart of Algebra consists of questions that will test your ability to analyze and and solve linear equations and systems of linear equations. You need to be able to create linear equations and inequalities to represent relationships between quantities and to solve problems. You need to understand and use the relationship between linear equations and inequalities and their graphs to solve problems.
Many Heart of Algebra questions can be answered using the following steps:
1. Define the variables to represent the quantities in the question.
2. Write the equations, expressions, inequalities, or functions to represent the relationships described in the question.
3. Solve the equations or inequalities and interpret the solution in terms of what is required in the question.
4. You may need to expand, simplify or rearrange your solution to match an answer choice.
There are many ways that you can be tested and practicing different types of questions will help you to be prepared for the SAT.
The following video lessons will show you how to solve a variety of Heart of Algebra questions in different situations.
Linear Equations, Linear Inequalities and Linear Functions
Absolute Value
Heart of Algebra includes absolute value expressions, inequalities, and equations.
The absolute value of any real number is nonnegative. Therefore, |-x| = |x|, for any real number x.
For any real number a and b, |a - b| is the distance between a and b on the number line.
You will need to know how to solve absolute value inequalities and to find the absolute value inequality when given the interval or range of values.
Systems of Linear Equations and Inequalities
You can use either the elimination or substitution method when solving a system of linear equations. Look at the equations closely because sometimes one method may get you the answer quicker that the other.
You will need to know the conditions when a systems of linear equations has no solution, one solution, and infinite solutions.
Lines in the Coordinate Plane
A system of two linear equations in two variables can be solved by graphing the lines in the coordinate plane. The point of inter section gives the solution to the system.
There are three possibilities:
1. The lines intersect at one point. Therefore, the system has a unique solution.
2. The lines are parallel and do not meet. Therefore, the system has no solution.
3. The lines are identical. Therefore, the system has infinitely many solutions.
If the slopes of the line l and line k are defined (i.e. neither line is a vertical line) then
1. line l and line k are parallel if and only if they have the same slope.
2. line l and line k are perpendicular if and only if the product of their slopes is -1.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
McGraw Hill My Math Grade 1 Chapter 3 Lesson 5 Answer Key Use Near Doubles to Add
All the solutions provided in McGraw Hill My Math Grade 1 Answer Key PDF Chapter 3 Lesson 5 Use Near Doubles to Add will give you a clear idea of the concepts.
Explore and Explain
Explanation:
I used to model the problem
I shown the doubles fact 3 + 3
Then i took away 1 cube
3 + 2 = 5
So, the addition number sentence is 3 + 2 = 5.
Teacher Directions: Use to model. Show the doubles fact 3 + 3 on the billboard. Add or take away one cube from one of the groups of cubes. Trace the cubes. Write the addition number sentence.
See and Show
You can use near doubles facts to find a sum. If you know 5 + 5 = 10, you can find 5 + 6 and 5 + 4.
Use to model. Write the addition number sentence.
Question 1.
Explanation:
The doubles fact is 4 + 4 = 8
The doubles plus 1 is 4 + 5 = 9.
Question 2.
Explanation:
The doubles fact is 3 + 3 = 6
The doubles minus 1 is 3 + 2 = 5.
Talk Math How do doubles facts help you learn near doubles facts?
The doubles facts helps us to learn the near doubles either by adding or subtracting 1 from the doubles.
On My Own
Use to model. Find each sum.
Question 3.
2 + 2 = ____
2 + 3 = ____
Explanation:
The doubles fact is 2 + 2 = 4
The doubles plus 1 is 2 + 3 = 5.
Question 4.
3 + 3 = ____
3 + 2 = ____
Explanation:
The doubles fact is 3 + 3 = 6
The doubles minus 1 is 3 + 2 = 5.
Question 5.
8 + 7 = ____
Explanation:
The doubles fact is 7 + 7 = 14
I drew doubles plus 1
The addition number sentence is 8 + 7 = 15.
Question 6.
7 + 7 = ____
Explanation:
The doubles fact is 7 + 7 = 14.
Question 7.
6 + 6 = ____
Explanation:
The doubles fact is 6 + 6 = 12.
Question 8.
6 + 5 = ____
Explanation:
The doubles fact is 5 + 5 = 10
Doubles plus 1 is 6 + 5 = 11.
Question 9.
Explanation:
7 + 7 = 14
7 + 6 = 13
So, the sum of 7 and 6 is 13.
Question 10.
Explanation:
7 + 7 = 14
7 + 8 = 15
So, the sum of 7 and 8 is 15.
Question 11.
Explanation:
4 + 4 = 8
4 + 3 = 7
So, the sum of 4 and 3 is 7.
Question 12.
Explanation:
5 + 5 = 10
5 + 4 = 9
So, the sum of 5 and 4 is 9.
Question 13.
Explanation:
5 + 5 = 10
So, the sum of 5 and 5 is 10.
Question 14.
Explanation:
5 + 5 = 10
5 + 6 = 11
So, the sum of 5 and 6 is 11.
Question 15.
Explanation:
The sum of 7 and 3 is 10.
Question 16.
Explanation:
The sum of 7 and 5 is 12.
Question 17.
Explanation:
The sum of 7 and 4 is 11.
Problem Solving
Solve. Write the doubles foot that helped you solve the problem.
Question 18.
Tyra sees 5 beetles. Sara sees 6 beetles. How many beetles do they see in all?
____ beetles
____ + ___ = ____
Explanation:
Tyra sees 5 beetles
Sara sees 6 beetles
The doubles fact is 5 + 5 = 10
Doubles fact plus 1 is 5 + 6 = 11
So, they saw 11 beetles in all.
Question 19.
Adam has 9 red flowers and 8 purple flowers. How many flowers does he have in all?
_____ flowers
____ + ___ = ____
Explanation:
8 purple flowers
Doubles fact is 9 + 9 = 18
Doubles fact minus 1 is 9 + 8 = 17
So, Adam ha s17 flowers in all.
HOT Problem Devon has 6 pets. Maya has 7 pets. The answer is 13 pets. What is the question?
________________________
________________________
________________________
Devon has 6 pets. Maya has 7 pets. How many pets do they have in all?
McGraw Hill My Math Grade 1 Chapter 3 Lesson 5 My Homework Answer Key
Practice
Find each sum.
Question 1.
4 + 4 = _____
4 + 5 = ____
Explanation:
The doubles fact is 4 + 4 = 8
The doubles plus 1 is 4 + 5 = 9.
Question 2.
8 + 8 = ____
8 + 7 = ____
Explanation:
The doubles fact is 8 + 8 = 7
The doubles minus 1 is 8 + 7 = 15.
Question 3.
6 + 5 = ____
Explanation:
The doubles fact is 5 + 5 = 10
The doubles plus 1 is 6 + 5 = 11.
Question 4.
6 + 6 = ____
Explanation:
The doubles fact is 6 + 6 = 12.
Question 5.
Explanation:
7 + 7 = 14
7 + 6 = 15
So, the sum of 7 and 6 is 13.
Question 6.
Explanation:
The doubles fact is 7 + 7 = 14.
Question 7.
Explanation:
7 + 7 = 14
7 + 8 = 15
So, the sum of 7 and 8 is 15.
Solve. Write the doubles fact that helped you solve the problem.
Question 8.
Paul is a dog walker. He walks 7 small dogs. He walks 6 large dogs. How many dogs does he walk in all?
____ dogs
____ + ____ = ____
Explanation:
Paul is a dog walker
He walks 7 small dogs
He walks 6 large dogs
Doubles fact is 7 + 7 = 14
Doubles minus 1 is 7 + 6 = 13
So, Paul walks 13 dogs in all.
Question 9.
There are 9 girls and 8 boys in Maria’s class who walk to school. How many children in her class walk to school in all?
____ children
____ + ____ = ____
Explanation:
There are 9 girls and 8 boys in Maria’s class who walk to school
The doubles fact is 8 + 8= 16
The doubles plus 1 is 9 + 8 = 17
So, 17 children walk to school.
Vocabulary Check
Circle the missing word.
doubles plus 1 doubles minus 1
Question 10.
If you know that 8 + 8 = 16, then you can use ____ to find 8 + 7. |
# How the dot product measures similarity
The dot product is one of the most fundamental concepts in machine learning, making appearances almost everywhere. In introductory linear algebra classes, we learn that in the vector space $\mathbb{R}^n$, it is defined by the formula
$\langle x, y \rangle = \sum_{i=1}^{n} x_i y_i, \quad x = (x_1, \dots, x_n), \quad y = (y_1, \dots, y_n).$
One of its most important applications is to measure similarity between feature vectors.
But how are similarity and inner product related? The definition doesn't reveal much. In this post, our goal is to unravel the dot product and provide a simple geometric explanation!
## The fundamental properties of the dot product
To see what the dot product has to do with similarity, we have three key observations. First, we can see that it is linear in both variables. This property is called bilinearity:
\begin{aligned} \langle ax + by, z \rangle &= a \langle x, z \rangle + b \langle y, z \rangle \\ \langle x, ay + bz \rangle &= a \langle x, y \rangle + b \langle x, z \rangle \\ x, y, z \in &\mathbb{R}^n, \quad a, b \in \mathbb{R}. \end{aligned}
Second, the dot product of orthogonal vectors is zero.
Third, the dot product of a vector with itself equals the square of its magnitude:
$\langle x, x \rangle = \sum_{k=1}^{n} (x_i)^2 = | x |^2.$
Armed with these, we are ready to explore how similarity is measured!
## Dot product as similarity
Suppose that we have two vectors, $\textstyle x$ and $\textstyle y$. To see the geometric interpretation of their dot product, we first note that $\textstyle x$ can be decomposed into the sum of two components: one is parallel to $\textstyle y$, while the other is orthogonal.
So, the dot product $\langle x, y \rangle$ equals to $\langle x_y, y \rangle$. If we write $x_y$ as $x_y = ry$, a scalar multiple of $\textstyle y$, we can simplify the dot product:
\begin{align*} \langle x, y \rangle &= \langle x_y, y \rangle \\ &= \langle ry, y \rangle \\ &= r | y |^2. \end{align*}
We can go even one step further. If we assume that both $\textstyle x$ and $\textstyle y$ have a magnitude of one, the dot product equals the scaling factor!
\begin{align*} \langle x, y \rangle &= \langle x_y, y \rangle \\ &= \langle ry, y \rangle \\ &= r \end{align*} Note that this scaling factor is in the interval $[-1, 1]$. It can be negative if the directions of $x_y$ and $\textstyle y$ are the opposite.
Now comes the really interesting part! $\textstyle r$ has a simple geometric meaning. To see this, let's illustrate what is happening. (Recall we assumed that $\textstyle x$ and $\textstyle y$ both have a magnitude of one.)
Since cosine is defined by the ratio of the adjacent side and the hypotenuse, it turns out that the scaling factor $\textstyle r$ also equals the cosine of the angle between $\textstyle x$ and $\textstyle y$.
It is the reason why cosine similarity is defined this way:
$\cos \alpha = \bigg\langle \frac{x}{|x|}, \frac{y}{|y|} \bigg\rangle.$
I hope that this short post helps you make sense of this concept, and armed with this knowledge, you'll be more confident when dealing with it! |
# Induction and Recursion
Mathematical Induction
Overview
Mathematical induction can be used to prove statements that assert that P(n) is true for all positive integers n, where P(n) is a propositional function
PRINCIPLE OF MATHEMATICAL INDUCTION
To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, we complete two steps :
BASIC STEP:
We verify that P(1) is true.
INDUCTIVE STEP:
We show that the conditional statement P(k)P(k + 1) is true for all positive integers k.
Expressed as a rule of inference, this proof technique can be stated as
(P(1) ∧ ∀k(P(k)P(k + 1))) → ∀nP(n),
when the domain is the set of positive integers.
Why Mathematical Induction is Valid
The reason comes from the well-ordering property.
To show: Mathematical Induction is valid.
Proved Facts: P(1) is true and P(k) implies P(k + 1) is true for all positive integers.
Assumption: There is at least one positive integer for which P(n) is false.
Proof:
Then the set S of positive integers for which P(n) is false is nonempty. Thus, by the well-ordering property, S has a least element, which will be denoted by m.
m cannot be 1, because P(1) is true
Because m is positive and greater than 1, m − 1 is a positive integer.
Furthermore, because m − 1 is less than m, it is not in S, so P(m − 1) must be true.
Because the conditional statement P(m − 1) → P(m) is also true, it must be the case that P(m) is true. This contradicts the choice of m).
Hence, P(n) must be true for every positive integer n.
The Good and the Bad of Mathematical Induction
Good: it can be used to prove a conjecture once it is has been made (and is true).
Bad: It cannot be be used to find new theorems. Mathematicians sometimes find proofs by mathematical induction unsatisfying because they do not provide insights as to why theorems are true. Many theorems can be proved in many ways, including by mathematical induction. Proofs of these theorems by methods other than mathematical induction are often preferred because of the insights they bring.
Examples of Proofs by Mathematical Induction
Proving Summation Formulae
Show that if n is a positive integer, then 1 + 2 + . . . + n = n(n+1) 2
BASIS STEP:
P(1) is true, because 1 = 1(1+1) 2
INDUCTIVE STEP:
P(k) holds for an arbitrary positive integer k. we assume that 1 + 2 + . . . + k = k(k+1) 2
Under this assumption, it must be shown that P(k + 1) is true, namely, that
1 + 2 + . . . + k + (k + 1) = (k + 1)[(k + 1) + 1] 2 (k + 1)(k + 2) 2
is also true. When we add k + 1 to both sides of the equation in P(k), we obtain
1 + 2+. . .+ k + (k + 1) = k(k+1) 2 + (k + 1)
= k(k + 1) + 2(k + 1) 2
= (k + 1)(k + 2) 2
This last equation shows that P(k + 1) is true under the assumption that P(k) is true.
Proving Inequalities
Use mathematical induction to prove the inequality
n < 2n
for all positive integers n.
BASIS STEP:
P(1) is true, because 1 < 21 = 2. This completes the basis step.
INDUCTIVE STEP:
We first assume the inductive hypothesis that P(k) is true for an arbitrary positive integer k. That is, the inductive hypothesis P(k) is the statement that k < 2k.
We need to show that if P(k) is true, then P(k + 1), which is the statement that k + 1 < 2k+1, is true. That is, we need to show that if k < 2k, then k + 1 < 2k+1.
k + 1 < 2k + 1 ≤ 2k + 2k = 2 . 2k = 2k + 1.
This shows that P(k + 1) is true, namely, that k + 1 < 2k+1, based on the assumption that P(k) is true.
Proving Divisibility Results
Use mathematical induction to prove that n3n is divisible by 3 whenever n is a positive integer.
BASIS STEP:
The statement P(1) is true because 13 − 1 = 0 is divisible by 3.
INDUCTIVE STEP:
Assume that P(k) is true; that is, we assume that k3k is divisible by 3 for an arbitrary positive integer k
(k + 1)3 − (k + 1) = (k3 + 3k2 + 3k + 1) − (k + 1)
= (k3k) + 3(k2 + k). Using the inductive hypothesis, we conclude that the first term k3k is divisible by 3.
The second term is divisible by 3 because it is 3 times an integer.
So, (k + 1)3 − (k + 1) is also divisible by 3.
The Number of Subsets of a Finite Set Use mathematical induction to show that if S is a finite set with n elements, where n is a nonnegative integer, then S has 2n subsets.
Let P(n) be the proposition that a set with n elements has 2n subsets.
BASIS STEP:
P(0) is true, because a set with zero elements, the empty set, has exactly 20 = 1 subset, namely, itself.
INDUCTIVE STEP:
P(k) is true for an arbitrary nonnegative integer k, that is, we assume that every set with k elements has 2k subsets
Let T be a set with k + 1 elements.
It is possible to write T = S ∪ {a}, where a is one of the elements of T and S = T − {a} (and hence |S| = k).
The subsets of T can be obtained in the following way. For each subset X of S there are exactly two subsets of T , namely, X and X ∪ {a}.
These constitute all the subsets of T and are all distinct.
S has 2k subsets, because it has k elements. We also know that there are two subsets of T for each subset of S.
Therefore, there are 2 . 2k = 2k+1 subsets of T. This finishes the inductive argument.
P(n) is true for all nonnegative integers n. That is, we have proved that a set with n elements has 2n subsets whenever n is a nonnegative integer.
NOTE: This is Pascal's Triangle showing up again!
Here we will discuss the greedy scheduling algorithm.
Mistaken Proofs by Mathematical Induction
Not covered for Fall 2017
Guidelines for Proofs by Mathematical Induction
Template for Proofs by Mathematical Induction
1. Express the statement that is to be proved in the form "for all nb, P(n)" for a fixed integer b.
2. Write out the words "Basis Step." Then show that P(b) is true, taking care that the correct value of b is used. This completes the first part of the proof.
3. Write out the words "Inductive Step."
4. State, and clearly identify, the inductive hypothesis, in the form "assume that P(k) is true for an arbitrary fixed integer k ≥ b."
5. State what needs to be proved under the assumption that the inductive hypothesis is true. That is, write out what P(k + 1) says.
6. Prove the statement P(k + 1) making use the assumption P(k). Be sure that your proof is valid for all integers k with k ≥ b, taking care that the proof works for small values of k, including k = b.
7. Clearly identify the conclusion of the inductive step, such as by saying "this completes the inductive step."
8. After completing the basis step and the inductive step, state the conclusion, namely that by mathematical induction, P(n) is true for all integers n with n ≥ b.
Test Yourself!
1. Which of the following is not correct of mathematical induction?
1. They provide insights about why a theorem is true.
2. It can be used to prove a conjecture once it has been made.
3. It can not be used to find new theorems.
4. All the statements are correct.
2. Given f(0) = 0 and f(1) = 3 , Find the value of f(5) for f(n) = f(n - 2) + 3.
1. 6
2. 3
3. 12
4. 9
3. Given f(0) = 2 and f(1) = 2 , Find the recursive definition for f.
1. f(n) = f(n - 1) - 2 , n > 0
2. f(n) = 2f(n - 1) - 2, n > 0
3. f(n) = f(n - 1) + 2 , n > 0
4. f(n) = 3f(n - 1) - 3 , n > 0
1. a; 2. d; 3. b;
Strong Induction and Well-Ordering
#### Strong Induction
To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, we complete two steps:
BASIC STEP:
We verify that the proposition P(1) is true.
INDUCTIVE STEP:
We show that the conditional statement [P(1) ∨ P(2) ∨ . . . ∨ P(k)] → P(k + 1) is true for all positive integers k.
Example
Show that if n is an integer greater than 1, then n can be written as the product of primes.
BASIC STEP:
P(2) is true, because 2 can be written as the product of one prime, itself.
INDUCTIVE STEP:
The inductive hypothesis is the assumption that P(j) is true for all integers j with 2 ≤ jk, that is, the assumption that j can be written as the product of primes whenever j is a positive integer at least 2 and not exceeding k.
There are two cases to consider, namely, when k + 1 is prime and when k + 1 is composite. If k + 1 is prime, we immediately see that P(k + 1) is true. Otherwise, k + 1 is composite and can be written as the product of two positive integers a and b with 2 ≤ ab < k + 1.
Because both a and b are integers at least 2 and not exceeding k, we can use the inductive hypothesis to write both a and b as the product of primes. Thus, if k + 1 is composite, it can be written as the product of primes, namely, those primes in the factorization of a and those in the factorization of b.
Hence, every nonnegative integer can be written uniquely as the product of primes in nondecreasing order.
Using Strong Induction in Computational Geometry
Not Covered Spring 2018
Not covered for Spring 2018
Proofs Using the Well-Ordering Property
Not Covered Spring 2018
Not covered for Spring 2018
Recursive Definitions and Structural Induction
Recursively Defined Functions
We use two steps to define a function with the set of nonnegative integers as its domain.
BASIC STEP:
Specify the value of the function at zero.
RECURSIVE STEP:
Give a rule for finding its value at an integer from its values at smaller integers.
Example
Suppose that f is defined recursively by
f(0) = 3,
f(n + 1) = 2f(n) + 3.
Find f(1), f(2), f(3), and f(4).
Solution: From the recursive definition it follows that
f(1) = 2f(0) + 3 = 2 * 3 + 3 = 9,
f(2) = 2f(1) + 3 = 2 * 9 + 3 = 21,
f(3) = 2f(2) + 3 = 2 * 21 + 3 = 45,
f(4) = 2f(3) + 3 = 2 * 45 + 3 = 93.
Recursively Defined Sets and Structures
In the basis step, an initial collection of elements is specified.
In the recursive step, rules for forming new elements in the set from those already known to be in the set are provided.
Recursive definitions may also include an exclusion rule, which specifies that a recursively defined set contains nothing other than those elements specified in the basis step or generated by applications of the recursive step.
DEFINITION 1
The set ∑* of strings over the alphabet ∑ is defined recursively by
BASIS STEP:
λ ∈ ∑* (where λ is the empty string containing no symbols).
RECURSIVE STEP:
If w ∈ ∑* and x ∈ ∑, then wx ∈ ∑*.
DEFINITION 2
Two strings can be combined via the operation of concatenation. Let ∑ be a set of symbols and ∑* the set of strings formed from symbols in ∑. We can define the concatenation of two strings, denoted by ·, recursively as follows.
BASIS STEP:
If w ∈ ∑*, then w · λ = w, where λ is the empty string.
RECURSIVE STEP:
If w1 ∈ ∑* and w2 ∈ ∑* and x ∈ ∑, then w1 · (w2x) = (w1 · w2)x.
Example: Give a recursive definition of l(w), the length of the string w.
The length of a string can be recursively defined by
l(λ) = 0;
l(wx) = l(w) + 1 if w ∈ ∑* and x ∈ ∑.
DEFINITION 3
The set of rooted trees, where a rooted tree consists of a set of vertices containing a distinguished vertex called the root, and edges connecting these vertices, can be defined recursively by these steps:
BASIC STEP:
A single vertex r is a rooted tree.
RECURSIVE STEP:
Suppose that T1, T2, . . ., Tn are disjoint rooted trees with roots r1, r2, . . . , rn, respectively. Then the graph formed by starting with a root r, which is not in any of the rooted trees T1, T2, . . . , Tn, and adding an edge from r to each of the vertices r1, r2, . . . , rn, is also a rooted tree.
DEFINITION 4
The set of extended binary trees can be defined recursively by these steps:
BASIC STEP:
The empty set is an extended binary tree.
RECURSIVE STEP:
If T1 and T2 are disjoint extended binary trees, there is an extended binary tree, denoted by T1 · T2, consisting of a root r together with edges connecting the root to each of the roots of the left subtree T1 and the right subtree T2 when these trees are nonempty.
DEFINITION 5
The set of full binary trees can be defined recursively by these steps:
BASIC STEP:
There is a full binary tree consisting only of a single vertex r.
RECURSIVE STEP:
If T1 and T2 are disjoint full binary trees, there is a full binary tree, denoted by T1 * T2, consisting of a root r together with edges connecting the root to each of the roots of the left subtree T1 and the right subtree T2.
Structural Induction
DEFINITION 6
We define the height h(T) of a full binary tree T recursively.
BASIC STEP:
The height of the full binary tree T consisting of only a root r is h(T) = 0.
RECURSIVE STEP:
If T1 and T2 are full binary trees, then the full binary tree v = T1 * T2 has height h(T ) = 1 + max(h(T1), h(T2)).
Recursive Algorithms
DEFINITION 1
An algorithm is called recursive if it solves a problem by reducing it to an instance of the same problem with smaller input.
Example 1
Give a recursive algorithm for computing n!, where n is a nonnegative integer.
Example 2
Give a recursive algorithm for computing an, where a is a nonzero real number and n is a nonnegative integer.
Example 3
Give a recursive algorithm for computing the greatest common divisor of two nonnegative integers a and b with a < b.
Example 4
Devise a recursive algorithm for computing bn mod m, where b, n, and m are integers with m ≤ 2, n ≥ 0, and 1 ≤ b < m.
Example 5
Express the linear search algorithm as a recursive procedure.
Example 6
Construct a recursive version of a binary search algorithm.
Recursion and Iteration
Instead of successively reducing the computation to the evaluation of the function at smaller integers, we can start with the value of the function at one or more integers, the base cases, and successively apply the recursive definition to find the values of the function at successive larger integers. Such a procedure is called iterative.
A Recursive Algorithm for Fibonacci Numbers.
An Iterative Algorithm for Fibonacci Numbers.
Merge Sort
A Recursive Merge Sort.
Merging Two lists
Program Correctness
Not Covered Spring 2018
Program Verification
A program is said to be correct if it produces the correct output for every possible input. A proof that a program is correct consists of two parts.
The first part shows that the correct answer is obtained if the program terminates. This part of the proof establishes the partial correctness of the program.
The second part of the proof shows that the program always terminates.
To specify what it means for a program to produce the correct output, two propositions are used.
The first is the initial assertion, which gives the properties that the input values must have.
The second is the final assertion, which gives the properties that the output of the program should have, if the program did what was intended.
Definition 1
A program, or program segment, S is said to be partially correct with respect to the initial assertion p and the final assertion q if whenever p is true for the input values of S and S terminates, then q is true for the output values of S. The notation p{S}q indicates that the program, or program segment, S is partially correct with respect to the initial assertion p and the final assertion q.
#### Conditional Statements
Not covered for Spring 2018
#### Loop Invariants
Not covered for Spring 2018 |
Mathematics NCERT Grade 9, Chapter 10: Circles- We all see many objects in our daily life which are round in shape.
In this chapter, students will study circles, its related terms and some properties of a circle
• A circle is a collection of all points in a plane, which is equidistant from a fixed point in the plane
• In this chapter we will study some terms related to circles like centre of circle, radius, diameter, arcs and its types- major arc and minor arc, semicircle, segments and its two types major segment and minor segment.
• The next section is about Angle Subtended by a Chord at a point. 2 theorems are discussed in this section, equal chords of a circle subtend equal angles at the centre and its converse
This is followed by the theorems that perpendicular from the centre of a circle to a chord bisects the chord and its converse.
Next we learn about circumcircle. According to which through three given non-collinear points there is one circle that can be drawn.The centre and radius are respectively circumcentre and circumradius respectively.
The next section covers the topics equal chords of a circle are equidistant from the centre and its converse.
This topic is followed by Angle Subtended by an Arc of a Circle where three theorems are discussed.
• The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
• Angles in the same segment of a circle are equal.
• If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).
Cyclic Quadrilaterals is another important topic from this chapter.
• A quadrilateral is cyclic if all the vertices of the quadrilateral lie on a circle.
• The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.
• If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.
The chapter contains 5 unsolved exercises, and also an optional exercise is given.
The summary of the chapter-Circles is provided in the end.
#### Question 1:
Fill in the blanks
(i) The centre of a circle lies in __________ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a __________ of the circle.
(iv) An arc is a __________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and __________ of the circle.
(vi) A circle divides the plane, on which it lies, in __________ parts.
(i) The centre of a circle lies in interior of the circle.
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semi-circle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.
##### Video Solution for circles (Page: 171 , Q.No.: 1)
NCERT Solution for Class 9 math - circles 171 , Question 1
#### Question 2:
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.
(i) True. All the points on the circle are at equal distances from the centre of the circle, and this equal distance is called as radius of the circle.
(ii) False. There are infinite points on a circle. Therefore, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords.
(iii) False. Consider three arcs of same length as AB, BC, and CA. It can be observed that for minor arc BDC, CAB is a major arc. Therefore, AB, BC, and CA are minor arcs of the circle.
(iv) True. Let AB be a chord which is twice as long as its radius. It can be observed that in this situation, our chord will be passing through the centre of the circle. Therefore, it will be the diameter of the circle.
(v) False. Sector is the region between an arc and two radii joining the centre to the end points of the arc. For example, in the given figure, OAB is the sector of the circle.
(vi) True. A circle is a two-dimensional figure and it can also be referred to as a plane figure.
##### Video Solution for circles (Page: 171 , Q.No.: 2)
NCERT Solution for Class 9 math - circles 171 , Question 2
#### Question 1:
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle. And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose two circles of equal radius, then both circles will cover each other. Therefore, two circles are congruent if they have equal radius.
Consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths.
In ΔAOB and ΔCO'D,
AB = CD (Chords of same length)
OA = O'C (Radii of congruent circles)
OB = O'D (Radii of congruent circles)
∴ ΔAOB ≅ ΔCO'D (SSS congruence rule)
⇒ ∠AOB = ∠CO'D (By CPCT)
Hence, equal chords of congruent circles subtend equal angles at their centres.
##### Video Solution for circles (Page: 173 , Q.No.: 1)
NCERT Solution for Class 9 math - circles 173 , Question 1
#### Question 2:
Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Let us consider two congruent circles (circles of same radius) with centres as O and O'.
In ΔAOB and ΔCO'D,
∠AOB = ∠CO'D (Given)
OA = O'C (Radii of congruent circles)
OB = O'D (Radii of congruent circles)
∴ ΔAOB ≅ ΔCO'D (SAS congruence rule)
⇒ AB = CD (By CPCT)
Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
##### Video Solution for circles (Page: 173 , Q.No.: 2)
NCERT Solution for Class 9 math - circles 173 , Question 2
#### Question 1:
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Consider the following pair of circles.
The above circles do not intersect each other at any point. Therefore, they do not have any point in common.
The above circles touch each other only at one point Y. Therefore, there is 1 point in common.
The above circles touch each other at 1 point X only. Therefore, the circles have 1 point in common.
These circles intersect each other at two points G and H. Therefore, the circles have two points in common. It can be observed that there can be a maximum of 2 points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.
##### Video Solution for circles (Page: 176 , Q.No.: 1)
NCERT Solution for Class 9 math - circles 176 , Question 1
#### Question 2:
Suppose you are given a circle. Give a construction to find its centre.
The below given steps will be followed to find the centre of the given circle.
Step1. Take the given circle.
Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these chords.
Step3. Let these perpendicular bisectors meet at point O. Hence, O is the centre of the given circle.
##### Video Solution for circles (Page: 176 , Q.No.: 2)
NCERT Solution for Class 9 math - circles 176 , Question 2
#### Question 3:
If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.
Consider two circles centered at point O and O’, intersecting each other at point A and B respectively.
Join AB. AB is the chord of the circle centered at O. Therefore, perpendicular bisector of AB will pass through O.
Again, AB is also the chord of the circle centered at O’. Therefore, perpendicular bisector of AB will also pass through O’.
Clearly, the centres of these circles lie on the perpendicular bisector of the common chord.
##### Video Solution for circles (Page: 176 , Q.No.: 3)
NCERT Solution for Class 9 math - circles 176 , Question 3
#### Question 1:
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.
OA = OB = 5 cm
O'A = O'B = 3 cm
OO' will be the perpendicular bisector of chord AB.
∴ AC = CB
It is given that, OO' = 4 cm
Let OC be x. Therefore, O'C will be x − 4
In ΔOAC,
OA2 = AC2 + OC2
⇒ 52 = AC2 + x2
⇒ 25 − x2 = AC2 ... (1)
In ΔO'AC,
O'A2 = AC2 + O'C2
⇒ 32 = AC2 + (x − 4)2
⇒ 9 = AC2 + x2 + 16 − 8x
⇒ AC2 = − x2 − 7 + 8x ... (2)
From equations (1) and (2), we obtain
25 − x2 = − x2 − 7 + 8x
8x = 32
x = 4
Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.
Length of the common chord AB = 2 O'A = (2 × 3) cm = 6 cm
##### Video Solution for circles (Page: 179 , Q.No.: 1)
NCERT Solution for Class 9 math - circles 179 , Question 1
#### Question 2:
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In ΔOVT and ΔOUT,
OV = OU (Equal chords of a circle are equidistant from the centre)
∠OVT = ∠OUT (Each 90°)
OT = OT (Common)
∴ ΔOVT ≅ ΔOUT (RHS congruence rule)
∴ VT = UT (By CPCT) ... (1)
It is given that,
PQ = RS ... (2)
⇒ PV = RU ... (3)
On adding equations (1) and (3), we obtain
PV + VT = RU + UT
⇒ PT = RT ... (4)
On subtracting equation (4) from equation (2), we obtain
PQ − PT = RS − RT
⇒ QT = ST ... (5)
Equations (4) and (5) indicate that the corresponding segments of chords PQ and RS are congruent to each other.
##### Video Solution for circles (Page: 179 , Q.No.: 2)
NCERT Solution for Class 9 math - circles 179 , Question 2
#### Question 3:
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In ΔOVT and ΔOUT,
OV = OU (Equal chords of a circle are equidistant from the centre)
∠OVT = ∠OUT (Each 90°)
OT = OT (Common)
∴ ΔOVT ≅ ΔOUT (RHS congruence rule)
∴ ∠OTV = ∠OTU (By CPCT)
Therefore, it is proved that the line joining the point of intersection to the centre makes equal angles with the chords.
##### Video Solution for circles (Page: 179 , Q.No.: 3)
NCERT Solution for Class 9 math - circles 179 , Question 3
#### Question 4:
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see figure 10.25).
Let us draw a perpendicular OM on line AD.
It can be observed that BC is the chord of the smaller circle and AD is the chord of the bigger circle.
We know that perpendicular drawn from the centre of the circle bisects the chord.
∴ BM = MC ... (1)
And, AM = MD ... (2)
On subtracting equation (2) from (1), we obtain
AM − BM = MD − MC
⇒ AB = CD
##### Video Solution for circles (Page: 179 , Q.No.: 4)
NCERT Solution for Class 9 math - circles 179 , Question 4
#### Question 5:
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Draw perpendiculars OA and OB on RS and SM respectively.
OR = OS = OM = 5 m. (Radii of the circle)
In ΔOAR,
OA2 + AR2 = OR2
OA2 + (3 m)2 = (5 m)2
OA2 = (25 − 9) m2 = 16 m2
OA = 4 m
ORSM will be a kite (OR = OM and RS = SM). We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.
∴∠RCS will be of 90° and RC = CM
Area of ΔORS =
Therefore, the distance between Reshma and Mandip is 9.6 m.
##### Video Solution for circles (Page: 179 , Q.No.: 5)
NCERT Solution for Class 9 math - circles 179 , Question 5
#### Question 6:
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
It is given that AS = SD = DA
Therefore, ΔASD is an equilateral triangle.
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ASD. We also know that medians intersect each other in the ratio 2: 1. As AB is the median of equilateral triangle ASD, we can write
∴ AB = OA + OB = (20 + 10) m = 30 m
In ΔABD,
Therefore, the length of the string of each phone will be m.
##### Video Solution for circles (Page: 179 , Q.No.: 6)
NCERT Solution for Class 9 math - circles 179 , Question 6
#### Question 1:
In the given figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
It can be observed that
∠AOC = ∠AOB + ∠BOC
= 60° + 30°
= 90°
We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
##### Video Solution for circles (Page: 184 , Q.No.: 1)
NCERT Solution for Class 9 math - circles 184 , Question 1
#### Question 2:
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
In ΔOAB,
AB = OA = OB = radius
∴ ΔOAB is an equilateral triangle.
Therefore, each interior angle of this triangle will be of 60°.
∴ ∠AOB = 60°
⇒ ∠ADB = 180° − 30° = 150°
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.
##### Video Solution for circles (Page: 185 , Q.No.: 2)
NCERT Solution for Class 9 math - circles 185 , Question 2
#### Question 3:
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Consider PR as a chord of the circle.
Take any point S on the major arc of the circle.
∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠PSR = 180° − 100° = 80°
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠POR = 2∠PSR = 2 (80°) = 160°
In ΔPOR,
OP = OR (Radii of the same circle)
∴ ∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)
∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)
2 ∠OPR + 160° = 180°
2 ∠OPR = 180° − 160° = 20º
∠OPR = 10°
##### Video Solution for circles (Page: 185 , Q.No.: 3)
NCERT Solution for Class 9 math - circles 185 , Question 3
#### Question 4:
In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property of a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° − 100º
⇒ ∠BAC = 80°
∠BDC = ∠BAC = 80° (Angles in the same segment of a circle are equal)
##### Video Solution for circles (Page: 185 , Q.No.: 4)
NCERT Solution for Class 9 math - circles 185 , Question 4
#### Question 5:
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
In ΔCDE,
∠CDE + ∠DCE = ∠CEB (Exterior angle)
⇒ ∠CDE + 20° = 130°
⇒ ∠CDE = 110°
However, ∠BAC = ∠CDE (Angles in the same segment of a circle)
⇒ ∠BAC = 110°
##### Video Solution for circles (Page: 185 , Q.No.: 5)
NCERT Solution for Class 9 math - circles 185 , Question 5
#### Question 6:
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
For chord CD,
∠CBD = ∠CAD (Angles in the same segment)
∠BCD + 100° = 180°
∠BCD = 80°
In ΔABC,
AB = BC (Given)
∴ ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
⇒ ∠BCA = 30°
We have, ∠BCD = 80°
⇒ ∠BCA + ∠ACD = 80°
30° + ∠ACD = 80°
⇒ ∠ACD = 50°
⇒ ∠ECD = 50°
##### Video Solution for circles (Page: 185 , Q.No.: 6)
NCERT Solution for Class 9 math - circles 185 , Question 6
#### Question 7:
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.
(Consider BD as a chord)
∠BCD = 180° − 90° = 90°
(Considering AC as a chord)
90° + ∠ABC = 180°
∠ABC = 90°
Each interior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.
##### Video Solution for circles (Page: 185 , Q.No.: 7)
NCERT Solution for Class 9 math - circles 185 , Question 7
#### Question 8:
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Consider a trapezium ABCD with AB | |CD and BC = AD.
Draw AM ⊥ CD and BN ⊥ CD.
In ΔAMD and ΔBNC,
∠AMD = ∠BNC (By construction, each is 90°)
AM = BN (Perpendicular distance between two parallel lines is same)
∴ ΔAMD ≅ ΔBNC (RHS congruence rule)
∴ ∠ADC = ∠BCD (CPCT) ... (1)
∠BAD + ∠BCD = 180° [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.
##### Video Solution for circles (Page: 185 , Q.No.: 8)
NCERT Solution for Class 9 math - circles 185 , Question 8
#### Question 9:
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
Join chords AP and DQ.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) ... (1)
For chord DQ,
∠DBQ = ∠QCD (Angles in the same segment) ... (2)
ABD and PBQ are line segments intersecting at B.
∴ ∠PBA = ∠DBQ (Vertically opposite angles) ... (3)
From equations (1), (2), and (3), we obtain
∠ACP = ∠QCD
##### Video Solution for circles (Page: 186 , Q.No.: 9)
NCERT Solution for Class 9 math - circles 186 , Question 9
#### Question 10:
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Consider a ΔABC.
Two circles are drawn while taking AB and AC as the diameter.
Let they intersect each other at D and let D not lie on BC.
∠ADB = 90° (Angle subtended by semi-circle)
∠ADC = 90° (Angle subtended by semi-circle)
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, Point D lies on third side BC of ΔABC.
##### Video Solution for circles (Page: 186 , Q.No.: 10)
NCERT Solution for Class 9 math - circles 186 , Question 10
#### Question 11:
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
In ΔABC,
∠ABC + ∠BCA + ∠CAB = 180° (Angle sum property of a triangle)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90° ... (1)
∠CDA + ∠ACD + ∠DAC = 180° (Angle sum property of a triangle)
⇒ 90° + ∠ACD + ∠DAC = 180°
⇒ ∠ACD + ∠DAC = 90° ... (2)
Adding equations (1) and (2), we obtain
∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°
⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°
∠BCD + ∠DAB = 180° ... (3)
However, it is given that
∠B + ∠D = 90° + 90° = 180° ... (4)
From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.
Consider chord CD.
∠CAD = ∠CBD (Angles in the same segment)
##### Video Solution for circles (Page: 186 , Q.No.: 11)
NCERT Solution for Class 9 math - circles 186 , Question 11
#### Question 12:
Prove that a cyclic parallelogram is a rectangle.
Let ABCD be a cyclic parallelogram.
∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral) ... (1)
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C and ∠B = ∠D
From equation (1),
∠A + ∠C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
Parallelogram ABCD has one of its interior angles as 90°. Therefore, it is a rectangle.
##### Video Solution for circles (Page: 186 , Q.No.: 12)
NCERT Solution for Class 9 math - circles 186 , Question 12
#### Question 1:
Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Let two circles having their centres as O and intersect each other at point A and B respectively. Let us join O.
In ΔAO and BO,
OA = OB (Radius of circle 1)
A = B (Radius of circle 2)
O = O (Common)
ΔAO ≅ ΔBO (By SSS congruence rule)
∠OA = ∠OB (By CPCT)
Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.
##### Video Solution for circles (Page: 186 , Q.No.: 1)
NCERT Solution for Class 9 math - circles 186 , Question 1
#### Question 2:
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.
(Perpendicular from the centre bisects the chord)
Let ON be x. Therefore, OM will be 6− x.
In ΔMOB,
In ΔNOD,
We have OB = OD (Radii of the same circle)
Therefore, from equation (1) and (2),
From equation (2),
Therefore, the radius of the circle iscm.
##### Video Solution for circles (Page: 186 , Q.No.: 2)
NCERT Solution for Class 9 math - circles 186 , Question 2
#### Question 3:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.
Distance of smaller chord AB from the centre of the circle = 4 cm
OM = 4 cm
MB =
In ΔOMB,
In ΔOND,
Therefore, the distance of the bigger chord from the centre is 3 cm.
##### Video Solution for circles (Page: 186 , Q.No.: 3)
NCERT Solution for Class 9 math - circles 186 , Question 3
#### Question 4:
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
In ΔAOD and ΔCOE,
OA = OC (Radii of the same circle)
OD = OE (Radii of the same circle)
∴ ΔAOD ≅ ΔCOE (SSS congruence rule)
∠OAD = ∠OCE (By CPCT) ... (1)
∠ODA = ∠OEC (By CPCT) ... (2)
Also,
∠OAD = ∠ODA (As OA = OD) ... (3)
From equations (1), (2), and (3), we obtain
∠OAD = ∠OCE = ∠ODA = ∠OEC
Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x
In Δ OAC,
OA = OC
∴ ∠OCA = ∠OAC (Let a)
In Δ ODE,
OD = OE
∠OED = ∠ODE (Let y)
∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)
x + a + x + y = 180°
2x + a + y = 180°
y = 180º − 2xa ... (4)
However, ∠DOE = 180º − 2y
And, ∠AOC = 180º − 2a
∠DOE − ∠AOC = 2a − 2y = 2a − 2 (180º − 2xa)
= 4a + 4x − 360° ... (5)
∠BAC + ∠CAD = 180º (Linear pair)
⇒ ∠BAC = 180º − ∠CAD = 180º − (a + x)
Similarly, ∠ACB = 180º − (a + x)
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180º (Angle sum property of a triangle)
∠ABC = 180º − ∠BAC − ∠ACB
= 180º − (180º − ax) − (180º − ax)
= 2a + 2x − 180º
= [4a + 4x − 360°]
∠ABC = [∠DOE − ∠ AOC] [Using equation (5)]
##### Video Solution for circles (Page: 186 , Q.No.: 4)
NCERT Solution for Class 9 math - circles 186 , Question 4
#### Question 5:
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.
∴ ∠COD = 90°
Also, in rhombus, the diagonals intersect each other at 90°.
∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Clearly, point O has to lie on the circle.
##### Video Solution for circles (Page: 186 , Q.No.: 5)
NCERT Solution for Class 9 math - circles 186 , Question 5
#### Question 6:
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.
∠AEC + ∠CBA = 180°
∠AEC + ∠AED = 180° (Linear pair)
∠AED = ∠CBA ... (1)
For a parallelogram, opposite angles are equal.
From (1) and (2),
AD = AE (Angles opposite to equal sides of a triangle)
##### Video Solution for circles (Page: 186 , Q.No.: 6)
NCERT Solution for Class 9 math - circles 186 , Question 6
#### Question 7:
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
Let two chords AB and CD are intersecting each other at point O.
In ΔAOB and ΔCOD,
OA = OC (Given)
OB = OD (Given)
∠AOB = ∠COD (Vertically opposite angles)
ΔAOB ≅ ΔCOD (SAS congruence rule)
AB = CD (By CPCT)
Similarly, it can be proved that ΔAOD ≅ ΔCOB
∴ AD = CB (By CPCT)
Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C
However, ∠A + ∠C = 180° (ABCD is a cyclic quadrilateral)
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
As ACBD is a parallelogram and one of its interior angles is 90°, therefore, it is a rectangle.
∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.
##### Video Solution for circles (Page: 186 , Q.No.: 7)
NCERT Solution for Class 9 math - circles 186 , Question 7
#### Question 8:
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° .
It is given that BE is the bisector of ∠B.
∴ ∠ABE =
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
Similarly, ∠ACF = ∠ADF = (Angle in the same segment for chord AF)
Similarly, it can be proved that
##### Video Solution for circles (Page: 186 , Q.No.: 8)
NCERT Solution for Class 9 math - circles 186 , Question 8
#### Question 9:
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
AB is the common chord in both the congruent circles.
∴ ∠APB = ∠AQB
In ΔBPQ,
∠APB = ∠AQB
∴ BQ = BP (Angles opposite to equal sides of a triangle)
##### Video Solution for circles (Page: 187 , Q.No.: 9)
NCERT Solution for Class 9 math - circles 187 , Question 9
#### Question 10:
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.
Let perpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of side BC intersect it at E.
Perpendicular bisector of side BC will pass through circumcentre O of the circle. ∠BOC and ∠BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively. We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠BOC = 2 ∠BAC = 2 ∠A ... (1)
In ΔBOE and ΔCOE,
OE = OE (Common)
OB = OC (Radii of same circle)
∠OEB = ∠OEC (Each 90° as OD ⊥ BC)
∴ ΔBOE ≅ ∠COE (RHS congruence rule)
∠BOE = ∠COE (By CPCT) ... (2)
However, ∠BOE + ∠COE = ∠BOC
⇒ ∠BOE +∠BOE = 2 ∠A [Using equations (1) and (2)]
⇒ 2 ∠BOE = 2 ∠A
⇒ ∠BOE = ∠A
∴ ∠BOE = ∠COE = ∠A
The perpendicular bisector of side BC and angle bisector of ∠A meet at point D.
∴ ∠BOD = ∠BOE = ∠A ... (3)
Since AD is the bisector of angle ∠A,
⇒ 2 ∠BAD = ∠A ... (4)
From equations (3) and (4), we obtain |
## Section3.5Solving Combinatorial Problems Recursively
In this section, we present examples of combinatorial problems for which solutions can be computed recursively. In Chapter 9, we return to these problems and obtain even more compact solutions. Our first problem is one discussed in our introductory chapter.
###### Example3.3.
A family of $$n$$ lines is drawn in the plane with (1) each pair of lines crossing and (2) no three lines crossing in the same point. Let $$r(n)$$ denote the number of regions into which the plane is partitioned by these lines. Evidently, $$r(1)=2\text{,}$$ $$r(2)=4\text{,}$$ $$r(3)=7$$ and $$r(4)=11\text{.}$$ To determine $$r(n)$$ for all positive integers, it is enough to note that $$r(1)=2\text{,}$$ and when $$n>1\text{,}$$ $$r(n)=n+r(n-1)\text{.}$$ This formula follows from the observation that if we label the lines as $$L_1\text{,}$$ $$L_2, \dots, L_n\text{,}$$ then the $$n-1$$ points on line $$L_n$$ where it crosses the other lines in the family divide $$L_n$$ into $$n$$ segments, two of which are infinite. Each of these segments is associated with a region determined by the first $$n-1$$ lines that has now been subdivided into two, giving us $$n$$ more regions than were determined by $$n-1$$ lines. This situation is illustrated in Figure 3.4, where the line containing the three dots is $$L_4\text{.}$$ The other lines divide it into four segments, which then divide larger regions to create regions $$1$$ and $$5\text{,}$$ $$2$$ and $$6\text{,}$$ $$7$$ and $$8\text{,}$$ and $$4$$ and $$9\text{.}$$
With the recursive formula, we thus have $$r(5)=5+11=16\text{,}$$ $$r(6)=6+16=22$$ and $$r(7)=7+22=29\text{.}$$ Even by hand, it wouldn't be all that much trouble to calculate $$r(100)\text{.}$$ We could do it before lunch.
###### Example3.5.
A $$2\times n$$ checkerboard will be tiled with rectangles of size $$2\times1$$ and $$1\times2\text{.}$$ Find a recursive formula for the number $$t(n)$$ of tilings. Clearly, $$t(1)=1$$ and $$t(2)=2\text{.}$$ When $$n>2\text{,}$$ consider the rectangle that covers the square in the upper right corner. If it is vertical, then preceding it, we have a tiling of the first $$n-1$$ columns. If it is horizontal, then so is the rectangle immediately underneath it, and proceeding them is a tiling of the first $$n-2$$ columns. This shows that $$t(n)=t(n-1)+t(n-2)\text{.}$$ In particular, $$t(3)=1+2=3\text{,}$$ $$t(4)=2+3=5$$ and $$t(5)= 3+5=8\text{.}$$
Again, if compelled, we could get $$t(100)$$ by hand, and a computer algebra system could get $$t(1000)\text{.}$$
###### Example3.6.
Call a ternary string good if it never contains a $$2$$ followed immediately by a $$0\text{;}$$ otherwise, call it bad. Let $$g(n)$$ be the number of good strings of length $$n\text{.}$$ Obviously $$g(1)=3\text{,}$$ since all strings of length $$1$$ are good. Also, $$g(2)=8$$ since the only bad string of length $$2$$ is $$(2,0)\text{.}$$ Now consider a value of $$n$$ larger than $$2\text{.}$$
Partition the set of good strings of length $$n$$ into three parts, according to the last character. Good strings ending in $$1$$ can be preceded by any good string of length $$n-1\text{,}$$ so there are $$g(n-1)$$ such strings. The same applies for good strings ending in $$2\text{.}$$ For good strings ending in $$0\text{,}$$ however, we have to be more careful. We can precede the $$0$$ by a good string of length $$n-1$$ provided that the string does not end in $$2\text{.}$$ There are $$g(n-1)$$ good strings of length $$n-1$$ and of these, exactly $$g(n-2)$$ end in a $$2\text{.}$$ Therefore there are $$g(n-1)-g(n-2)$$ good strings of length $$n$$ that end in a $$0\text{.}$$ Hence the total number of good strings of length $$n$$ satisfies the recursive formula $$g(n) = 3g(n-1) - g(n-2)\text{.}$$ Thus $$g(3) = 3\cdot8 -3= 21$$ and $$g(4)= 3\cdot21-8= 55\text{.}$$
Once more, $$g(100)$$ is doable by hand, while even a modest computer can be coaxed into giving us $$g(5000)\text{.}$$
### Subsection3.5.1Finding Greatest Common Divisors
There is more meat than you might think to the following elementary theorem, which seems to simply state a fact that you've known since second grade.
We settle the claim for existence. The uniqueness part is just high-school algebra. If the theorem fails to hold, then let $$t$$ be the least positive integer for which there are integers $$m$$ and $$n$$ with $$m+n=t\text{,}$$ but there do not exist integers $$q$$ and $$r$$ with $$m=qn+r$$ and $$0\le r\lt n\text{.}$$
First, we note that $$n\neq 1\text{,}$$ for if $$n=1\text{,}$$ then we could take $$q=m$$ and $$r=0\text{.}$$ Also, we cannot have $$m=1\text{,}$$ for if $$m=1\text{,}$$ then we can take $$q=0$$ and $$r=1\text{.}$$ Now the statement holds for the pair $$m-1\text{,}$$ $$n$$ so there are integers $$q$$ and $$r$$ so that
\begin{equation*} m-1 = q\cdot n+r\quad\text{and} \quad 0 \le r \lt n. \end{equation*}
Since $$r\lt n\text{,}$$ we know that $$r+1\le n\text{.}$$ If $$r+1\lt n\text{,}$$ then
\begin{equation*} m = q\cdot n+(r+1)\quad\text{and} \quad 0 \le r+1 \lt n. \end{equation*}
On the other hand, if $$r+1=n\text{,}$$ then
\begin{equation*} m = q\cdot n+(r+1)=nq+n=(q+1)n=(q+1)n+0. \end{equation*}
Recall that an integer $$n$$ is a divisor of an integer $$m$$ if there is an integer $$q$$ such that $$m=qn\text{.}$$ (We write $$n\mid m$$ and read “$$n$$ divides $$m$$”.) An integer $$d$$ is a common divisor of integers $$m$$ and $$n$$ if $$d$$ is a divisor of both $$m$$ and $$n\text{.}$$ The greatest common divisor of $$m$$ and $$n\text{,}$$ written $$\gcd(m,n)\text{,}$$ is the largest of all the common divisors of $$m$$ and $$n\text{.}$$
Here's a particularly elegant application of the preceding basic theorem:
Consider the expression $$m=q\cdot n+r\text{,}$$ which is equivalent to $$m-q\cdot n = r\text{.}$$ If a number $$d$$ is a divisor of $$m$$ and $$n\text{,}$$ then $$d$$ must also divide $$r\text{.}$$ Similarly, if $$d$$ is a divisor of $$n$$ and $$r\text{,}$$ then $$d$$ must also divide $$m\text{.}$$
Here is a code snippet that computes the greatest common divisor of $$m$$ and $$n$$ when $$m$$ and $$n$$ are positive integers with $$m\ge n\text{.}$$ We use the familiar notation m%n to denote the remainder $$r$$ in the expression $$m=q\cdot n+r\text{,}$$ with $$0\le r \lt n\text{.}$$
Feel free to change the values 12 and 5 above in the SageMath cell in the HTML version of the text to calculate the greatest common divisor of some other integers. Just remember that the code assumes $$m\geq n$$ when you do so!
The disadvantage of this approach is the somewhat wasteful use of memory due to recursive function calls. It is not difficult to develop code for computing the greatest common divisor of $$m$$ and $$n$$ using only a loop, i.e., there are no recursive calls. With minimal extra work, such code can also be designed to solve the following diophantine equation problem:
Let's see how the Euclidean algorithm can be used to write $$\gcd(m,n)$$ in the form $$am+bn$$ with $$a,b\in\ints$$ with the following example.
###### Example3.10.
Find the greatest common divisor $$d$$ of $$3920$$ and $$252$$ and find integers $$a$$ and $$b$$ such that $$d=3920a+252b\text{.}$$
Solution.
In solving the problem, we demonstrate how to perform the Euclidean algorithm so that we can find $$a$$ and $$b$$ by working backward. First, we note that
\begin{equation*} 3920 = 15\cdot 252 + 140. \end{equation*}
Now the Euclidean algorithm tells us that $$\gcd(3920,252)=\gcd(252,140)\text{,}$$ so we write
\begin{equation*} 252 = 1\cdot 140 + 112. \end{equation*}
Continuing, we have $$140= 1\cdot 112 + 28$$ and $$112 = 4\cdot 28+0\text{,}$$ so $$d=28\text{.}$$
To find $$a$$ and $$b\text{,}$$ we now work backward through the equations we found earlier, “solving” them for the remainder term and then substituting. We begin with
\begin{equation*} 28 = 140-1\cdot 112. \end{equation*}
But we know that $$112=252-1\cdot 140\text{,}$$ so
\begin{equation*} 28=140-1(252-1\cdot 140) = 2\cdot 140 - 1\cdot 252. \end{equation*}
Finally, $$140 = 3920-15\cdot 252\text{,}$$ so now we have
\begin{equation*} 28= 2(3920-15\cdot 252) - 1\cdot 252 = 2\cdot 3920-31\cdot 252. \end{equation*}
Therefore $$a=2$$ and $$b=-31\text{.}$$
### Subsection3.5.2Sorting
One of the most common and most basic computing problems is sorting: Given a sequence $$a_1,a_2,\dots,a_n$$ of $$n$$ distinct integers, rearrange them so that they are in increasing order. We describe here an easy recursive strategy for accomplishing this task. This strategy is known as Merge Sort, and it is one of several optimal algorithms for sorting. Introductory computer science courses treat this topic in greater depth. In our course, we simply need some good strategy and merge sort works fine for our purposes.
To present merge sort, must first develop a strategy for solving a special case of the sorting problem. Suppose we have $$s+t$$ distinct integers
\begin{equation*} \{u_0,u_1,\dots,u_{s-1},v_0,v_1,\dots,v_{t-1}\} \end{equation*}
arranged as two lists with $$u_0\lt u_1\lt \dots\lt u_{s-1}$$ and $$v_0\lt v_1\lt \dots\lt v_{t-1}\text{.}$$ How do we merge these two sequences into a single increasing sequence of length $$s+t\text{.}$$ Imagine the two sequences placed on two horizontal lines, one immediately under the other. Then let $$u$$ be the least integer in the first sequence and $$v$$ the least integer in the second. At the moment, this implies that $$u=u_0$$ and $$v=v_0\text{,}$$ but integers will be deleted from the two sequences as the process is carried out. Regardless, the meaning of $$u$$ and $$v$$ will be preserved. Also, set $$i=0\text{.}$$ Then take $$a_i$$ as the minimum of $$u$$ and $$v$$ and delete $$a_i$$ from the sequence in which it occurs. Then increase $$i$$ by $$1$$ and repeat. Here is a code snippet for accomplishing a merge operation, with $$u_p$$ now written as u[p] and $$v_q$$ now written as v[q].
Now that we have a good strategy for merging, it is easy to develop a recursive strategy for sorting. Given a sequence $$a_1,a_2,\dots,a_n$$ of $$n$$ distinct integers, we set $$s=\lceil n/2\rceil$$ and $$t=\lfloor n/2\rfloor\text{.}$$ Then let $$u_i=a_i$$ for $$i=1,2,\dots,s$$ and $$v_j=a_{s+j}\text{,}$$ for $$j=1,2,\dots,t\text{.}$$ Sort the two subsequences and then merge them. For a concrete example, given the sequence $$(2,8,5,9,3,7,4,1,6)\text{,}$$ we split into $$(2,8,5,9,3)$$ and $$(7,4,1,6)\text{.}$$ These subsequences are sorted (by a recursive call) into $$(2,3,5,8,9)$$ and $$(1,4,6,7)\text{,}$$ and then these two sorted sequences are merged.
For running time, if $$S(n)$$ is the number of operations it takes to sort a sequence of $$n$$ distinct integers, then $$S(2n)\le2 S(n) + 2n\text{,}$$ since it clearly takes $$2n$$ steps to merge two sorted sequences of length $$n\text{.}$$ This leads to the bound $$S(n) \lt C n\log n$$ for some positive constant $$C\text{,}$$ and in computer science courses, you will learn (here it is an exercise) that this is optimal. |
Question
1. # Line Qr Contains (2, 8) And (3, 10) Line St Contains Points (0, 6) And (−2, 2). Lines Qr And St Are?
Do you remember your high school algebra lessons? If so, you’re in luck! This blog post focuses on the basics of linear equations and how to determine if two lines intersect. We’ll also discuss the concept of parallel and perpendicular lines, as well as how to use these concepts to solve for points of intersection. By the end of this article, you should have a better understanding of linear equations and be able to answer the question posed in the title: “Lines QR and ST are what?” Let’s get started!
## What is a line?
A line is a straight path between two points. In geometry, we use the term “line” to refer to a line segment, which is a part of a line that has two endpoints. A line segment is different from a ray, which is a part of a line that has one endpoint and goes on forever in one direction.
## What is a point?
A point is an exact location in space. It has no size, and it is not affected by translation or rotation. A point is represented by a set of coordinates, which are the numbers that describe its location. The most common coordinate system is the Cartesian coordinate system, which uses a set of perpendicular axes to define a location.
## What are the properties of a line?
A line is a one-dimensional object that is defined by two points. It has no width or thickness, and extends infinitely in both directions. A line can be straight or curved. The properties of a line include its length, direction, and location.
## What is the equation of a line?
Assuming you are referring to the equation of a line in slope-intercept form, the equation of a line is:
y = mx + b
Where “m” is the slope of the line and “b” is the y-intercept.
## How to find the equation of a line given two points?
There are several ways to find the equation of a line given two points. One way is to use the slope intercept form of the equation, which is y = mx + b. To use this method, you first need to calculate the slope of the line using the formula m = (y2-y1)/(x2-x1). Then, plug in the values for x and y for one of the points into the equation to solve for b. Once you have both m and b, you can plug them into the slope intercept form to get the equation of the line.
Another way to find the equation of a line given two points is to use the point-slope form of the equation, which is y – y1 = m(x – x1). To use this method, you again need to calculate the slope using the same formula as before. Once you have the slope, plug in the values for x and y for one of the points into the point-slope form to solve for y1. Then, plug in all of your values into point-slope form to get the equation of your line.
## What is the slope of a line?
The slope of a line is the measure of how steep the line is. It is the ratio of the rise to the run. The rise is the vertical distance between two points on the line, and the run is the horizontal distance between those same two points.
## How to find the slope of a line given two points?
Slope is defined as the change in y divided by the change in x. You can calculate the slope of a line using this formula:
Slope = (y2 – y1) / (x2 – x1)
where (x1, y1) and (x2, y2) are points on the line.
For example, to find the slope of the line that contains points (-2, 1) and (4, 3), you would use this formula:
Slope = (3 – 1) / (4 – (-2)) = 4/6 = 2/3
## What is the intercept of a line?
Assuming that you are referring to the equation of a line, the intercept is the point at which the line crosses the y-axis. In other words, it is the value of y when x = 0. For example, in the equation y = 2x + 3, the intercept is 3.
## How to find the intercept of a line given two
To find the intercept of a line, you need to know two things: the slope of the line and the y-intercept. The slope is the number that tells you how steep the line is, and the y-intercept is the point where the line crosses the y-axis. To find these values, you can use either a graph or some algebra.
If you’re using a graph, you can find the slope by looking at how many units the line rises for every unit it runs. For instance, if the line rises 3 units for every 2 units it runs, then its slope is 3/2. To find the y-intercept, simply look at where the line crosses the y-axis. In this case, it would be at (0,3).
If you’re using algebra, you can find the slope by solving for m in the equation y=mx+b. The y-intercept can be found by plugging in one of your points and solving for b. For instance, if your line was represented by the equation y=2x+5, you could plug in one of your points to solve for b. Let’s say your point was (4,10). You would plug this into your equation as 10=2(4)+5 and solve from there. This would give you a value of b=3, which means that your y-intercept is (0,3).
2. Are you trying to solve a geometry problem? If so, the lines Qr and St are both lines in the same plane.
The line Qr contains the points (2, 8) and (3, 10). This means that the line goes from point (2, 8) to point (3, 10). To find the equation for the line, we need to use the point-slope formula.
The slope of the line is (10 – 8)/(3 – 2) which is 2/1, or 2. The equation for this line is y = 2x + b. If we plug in the point (2, 8) for x and y, we get 8 = 2(2) + b. Solving for b, we get b = 4. So the equation of the line Qr is y = 2x + 4.
The line St contains the points (0, 6) and (−2, 2). This means that the line goes from point (0, 6) to point (−2, 2). To find the equation for the line, we need to use the point-slope formula.
The slope of the line is (2 – 6)/(-2 – 0) which is -4/2, or -2. The equation for this line is y = -2x + b. If we plug in the point (0, 6) for x and y, we get 6 = -2(0) + b. Solving for b, we get b = 6. So the equation of the line St is y = -2x + 6.
So, the lines Qr and St are the lines y = 2x + 4 and y = -2x + 6. |
# Order of operations
Order of operations
In mathematics and computer programming, the order of operations (sometimes called operator precedence) is a rule used to clarify unambiguously which procedures should be performed first in a given mathematical expression.
For example, in mathematics and most computer languages multiplication is done before addition; in the expression 2 + 3 × 4, the answer is 14. Brackets, which have their own rules, may be used to avoid confusion, thus the preceding expression may also be rendered 2 + (3 × 4), but the brackets are not required as multiplication still has precedence without them.
From the introduction of modern algebraic notation, where juxtaposition indicates multiplication of variables, multiplication took precedence over addition, whichever side of a number it appeared on.[1] Thus 3 + 4 × 5 = 4 × 5 + 3 = 23. When exponents were first introduced, in the 16th and 17th centuries, exponents took precedence over both addition and multiplication, and could be placed only as a superscript to the right of their base. Thus 3 + 52 = 28 and 3 × 52 = 75. To change the order of operations, originally a vinculum (an overline or underline) was used. Today we use brackets. Thus, to force addition to precede multiplication, we write (2 + 3) × 4 = 20, and to force addition to precede exponentiation, we write (3 + 5)2 = 64.
## The standard order of operations
The order of operations, or precedence, used in mathematics and many programming languages is expressed here:[citation needed]
terms inside parentheses or brackets
exponents and roots
multiplication and division As they appear left to right
addition and subtraction As they appear left to right
This means that if a mathematical expression is preceded by one operator and followed by another, the operator higher on the list should be applied first. The commutative and associative laws of addition and multiplication allow terms to be added in any order and factors to be multiplied in any order, but mixed operations must obey the standard order of operations.
It is helpful to treat division as multiplication by the reciprocal (multiplicative inverse) and subtraction as addition of the opposite (additive inverse). Thus 3/4 = 3 ÷ 4 = 3 • ¼; in other words the quotient of 3 and 4 equals the product of 3 and ¼. Also 3 − 4 = 3 + (−4); in other words the difference of 3 and 4 equals the sum of positive three and negative four. With this understanding, we can think of 1 - 2 + 3 as the sum of 1, negative 2, and 3, and add in any order: (1 - 2) + 3 = -1 + 3 = 2 and in reverse order (3 - 2) + 1 = 1 + 1 = 2. The important thing is to keep the negative sign with the 2.
The root symbol, √, requires a symbol of grouping around the radicand. The usual symbol of grouping is a bar (called vinculum) over the radicand. For the application of other functions, such as logarithm or cosine, the grouping is advised and denoted by parenthesis or brackets.
Stacked exponents are applied from the top down.
Symbols of grouping can be used to override the usual order of operations. Grouped symbols can be treated as a single expression. Symbols of grouping can be removed using the associative and distributive laws.
### Examples
$\sqrt{1+3}+5=\sqrt4+5=2+5=7.\,$
A horizontal fractional line also acts as a symbol of grouping:
$\frac{1+2}{3+4}+5=\frac37+5.$
For ease in reading, other grouping symbols such as braces, sometimes called curly braces { }, or brackets, sometimes called square brackets [ ], are often used along with parentheses ( ). For example,
$[(1+2)-3]-(4-5) = [3-3]-(-1) = 1. \,$
### Gaps in the standard
There exist differing conventions concerning the unary operator − (usually read "minus"). In written or printed mathematics, the expression −32 is interpreted to mean −(32) = −9,[citation needed] but in some applications and programming languages, notably the application Microsoft Office Excel and the programming language bc, unary operators have a higher priority than binary operators, that is, the unary minus (negation) has higher precedence than exponentiation, so in those languages −32 will be interpreted as (−3)2 = 9. [1]. In any case where there is a possibility that the notation might be misinterpreted, it is advisable to use brackets to clarify which interpretation is intended.
Similarly, care must be exercised when using the slash ('/') symbol. The string of characters "1/2x" is interpreted by the above conventions as (1/2)x. The contrary interpretation should be written explicitly as 1/(2x). Again, the use of brackets will clarify the meaning and should be used if there is any chance of misinterpretation.
## Mnemonics
Mnemonics are often used to help students remember the rules, but the rules taught by the use of acronyms can be misleading. In the United States the acronym PEMDAS is common. It stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. In other English speaking countries, Parentheses may be called Brackets, or symbols of inclusion and Exponentiation may be called either Indices, Powers or Orders, and since multiplication and division are of equal precedence, M and D are often interchanged, leading to such acronyms as BEDMAS, BIDMAS, BODMAS, BERDMAS, PERDMAS, and BPODMAS.
These mnemonics may be misleading when written this way, especially if the user is not aware that multiplication and division are of equal precedence, as are addition and subtraction. Using any of the above rules in the order "addition first, subtraction afterward" would also give the wrong answer.
$10 - 3 + 2 \,$
The correct answer is 9 (and not 5, which we get when we add 3 and 2 first to get 5,and then subtract it from 10 to get the final ans as 5), which is best understood by thinking of the problem as the sum of positive ten, negative three, and positive two.
$10 + (-3) + 2 \,$
An alternative way to write the mnemonic is:
P
E
MD
AS
Or, simply as BEMA, where it is taught that multiplication and divsion inherently share the same precedence; and that addition and subtraction inherently share the same precedence. BEMA is one of the mnemonics taught in New Zealand.
This makes the equivalence of multiplication and division, and addition and subtraction, clear.
Once logs are introduced the mnemonic can be written as:
P
EL
MD
AS
## Special cases
An exclamation mark indicates that one should compute the factorial of the term immediately to its left, before computing any of the lower-precedence operations, unless grouping symbols dictate otherwise. But 23! means (23)! = 8! = 40320 while 23! = 26 = 64; a factorial in an exponent applies to the exponent, while a factorial not in the exponent applies to the entire power.
If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$a^{b^c} = a^{(b^c)} \ne (a^b)^c. \,$
A function name usually applies to the monomial following the name, thus "sin xy" means sin (xy) but sin x + y means (sin x) + y. Calculators usually require brackets, and brackets should be used in complicated expressions to prevent misunderstanding.
Sometimes a dash or a heavy dot is used as a multiplication sign which has higher precedence than division.
## Calculators
Different calculators follow different orders of operations. Most non-scientific calculators without a stack work left to right without any priority given to different operators, for example giving
$1 + 2 \times 3 = 9, \;$
while more sophisticated calculators will use a more standard priority, for example giving
$1 + 2 \times 3 = 7. \;$
The Microsoft Calculator program uses the former in its standard view and the latter in its scientific view.
The non-scientific calculator expects two operands and an operator. When the next operator is pressed, the expression is immediately evaluated and the answer becomes the left hand of the next operator. Advanced calculators allow entry of the whole expression, grouped as necessary, and only evaluates when the user uses the equals sign.
Calculators may associate exponents to the left or to the right depending on the model. For example, the expression a ^ b ^ c on the TI-92 and TI-30XII (both Texas Instruments calculators) associates two different ways:
The TI-92 associates to the right, that is
a ^ b ^ c = a ^ (b ^ c) = $a^{(b^c)} = a^{b^c}$
whereas, the TI-30XII associates to the left, that is
a ^ b ^ c = (a ^ b) ^ c = (ab)c.
An expression like 1/2x is interpreted as 1/(2x) by TI-82, but as (1/2)x by TI-83.[2] While the first interpretation may be expected by some users, only the latter is in agreement with the standard rules stated above.
## Programming languages
Many programming languages use precedence levels that conform to the order commonly used in mathematics, though some, such as APL and Smalltalk, have no operator precedence rules (in APL evaluation is strictly right to left, in Smalltalk it's strictly left to right).
The logical bitwise operators in C (and all programming languages that borrowed precedence rules from C, for example, C++, Perl and PHP) have a precedence level that the creator of the C language considers to be unsatisfactory.[3] However, many programmers have become accustomed to this order. The relative precedence levels of operators found in many C-style languages are as follows:
1 () [] -> . :: Grouping, scope, array/member access 2 ! ~ - + * & sizeof type cast ++x --x (most) unary operations, sizeof and type casts 3 * / % Multiplication, division, modulo 4 + - Addition and subtraction 5 << >> Bitwise shift left and right 6 < <= > >= Comparisons: less-than, ... 7 == != Comparisons: equal and not equal 8 & Bitwise AND 9 ^ Bitwise exclusive OR 10 | Bitwise inclusive (normal) OR 11 && Logical AND 12 || Logical OR 13 ?: Conditional expression (ternary operator) 14 = += -= *= /= %= &= |= ^= <<= >>= Assignment operators 15 , Comma operator
Examples:
• !A + !B $\equiv$ (!A) + (!B)
• ++A + !B $\equiv$ (++A) + (!B)
• A * B + C $\equiv$ (A * B) + C
• A || B && C $\equiv$ A || (B && C)
• ( A && B == C ) $\equiv$ ( A && ( B == C ) )
The accuracy of software developer knowledge about binary operator precedence has been found to closely follow their frequency of occurrence in source code.[4]
## References
1. ^ http://mathforum.org/library/drmath/view/52582.html
2. ^ "Implied Multiplication Versus Explicit Multiplication on TI Graphing Calculators". Texas Instruments Incorporated. 16 January 2011. Retrieved 29 April 2011.
3. ^ Dennis M. Ritchie: The Development of the C Language. In History of Programming Languages, 2nd ed., ACM Press 1996.
4. ^ "Developer beliefs about binary operator precedence" Derek M. Jones, CVu 18(4):14--21
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# Internal angle
Internal angle
.
If every internal angle of a polygon is at most 180 degrees, the polygon is called convex.
In contrast, an exterior angle (or external angle) is an angle formed by one side of a simple polygon and a line extended from an adjacent side.
Interior angle measures of regular polygons
To find the total measure of degrees in a regular polygon, (regular meaning all sides and angles are equal) you must take the number of sides the polygon has, "n", subtract 2 from it, then multiply that number by 180°.
Example:
A decagon, a polygon with 10 sides, is a simple shape to figure the total measure of
: $\left(n-2\right) imes 180^circ !$
= measure in degrees, when "n" = number of sides
Solution to the decagon:
: $\left(10-2\right) imes 180^circ =1440^circ. !$
The total measure of the decagon is 1440°.
Divide that number by the number of sides, in this case, 10, to find the measure of each angle.
Each interior angle of a regular decagon is 144°.
It is easier to use measure of an exterior angle. Since every regular polygon can be built from "n" isosceles triangles,to get the measure of an internal angle simply subtract measure of exterior angle (see below) from 180°
For decagon this gives us:
: $180^circ - frac\left\{360^circ\right\}\left\{10\right\} = 180^circ - 36^circ = 144^circ$
For pentagon:: $180^circ - frac\left\{360^circ\right\}\left\{5\right\} = 180^circ - 72^circ = 108^circ$
Finding the exterior angles on a regular polygon
To find the measure of a regular decagon's exterior angles, divide 360° by the number of sides the polygon has, in this case, 10.
: $frac\left\{360^circ\right\}\left\{10\right\} = 36^circ.$
So all the exterior angles in a regular decagon are 36°. |
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Slope Intercept Form Answers – Among the many forms that are used to represent a linear equation, one of the most commonly seen is the slope intercept form. You can use the formula of the slope-intercept to identify a line equation when you have the straight line’s slope , and the y-intercept, which is the point’s y-coordinate where the y-axis intersects the line. Read more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three primary forms of linear equations: standard, slope-intercept, and point-slope. While they all provide similar results when used however, you can get the information line produced faster by using the slope-intercept form. It is a form that, as the name suggests, this form uses a sloped line in which it is the “steepness” of the line reflects its value.
This formula can be utilized to calculate the slope of a straight line, the y-intercept or x-intercept where you can apply different formulas that are available. The line equation of this specific formula is y = mx + b. The straight line’s slope is represented by “m”, while its y-intercept is indicated with “b”. Each point of the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” need to remain variables.
## An Example of Applied Slope Intercept Form in Problems
When it comes to the actual world in the real world, the slope-intercept form is commonly used to show how an item or issue evolves over its course. The value provided by the vertical axis represents how the equation deals with the extent of changes over what is represented via the horizontal axis (typically in the form of time).
A simple example of the use of this formula is to figure out how many people live in a particular area as the years go by. Based on the assumption that the area’s population increases yearly by a fixed amount, the point values of the horizontal axis will grow one point at a time each year and the value of the vertical axis will rise in proportion to the population growth by the fixed amount.
Also, you can note the starting point of a question. The beginning value is located at the y-value in the y-intercept. The Y-intercept marks the point where x is zero. If we take the example of the above problem the beginning point could be the time when the reading of population begins or when the time tracking begins , along with the associated changes.
So, the y-intercept is the location where the population starts to be monitored to the researchers. Let’s assume that the researcher is beginning to calculate or measure in the year 1995. In this case, 1995 will represent”the “base” year, and the x = 0 points will occur in 1995. Therefore, you can say that the population in 1995 will be the “y-intercept.
Linear equation problems that use straight-line formulas are almost always solved this way. The starting value is depicted by the y-intercept and the change rate is expressed through the slope. The main issue with the slope intercept form typically lies in the interpretation of horizontal variables particularly when the variable is accorded to the specific year (or any type of unit). The most important thing to do is to make sure you understand the variables’ definitions clearly. |
# Ken Ward's Mathematics Pages
## Arithmetic Algebra: Cube Roots
Main Arithmetic Algebra Page
## Rationale
Noting that:[2.1]
We try to understand how to extract the cube root:
We seek: [2.2]
We know the answers in the following and it is the method which is important. Also, this gives a vocabulary when considering applications in algebra and arithmetic.
a+b Answer Cube a3 We choose a as the cube root a 3a2b 3a2b+3ab2+b3 Remainder To find b, we can use (3a2+3ab+b2)b, to divide into the remainder to find b. Or assuming that 3a2b is the largest term and the rest is small in comparison, then we can divide by 3a2, to find b. a+b (3a2+3ab+b2)b Remainder after using (a+b) as the root so far: (3a2+3ab+b2)b 0
Observations:
1. First we seek a cube root, a.
2. Having chosen it, we subtract a3 from the cube
3. With the remainder, we need to find a number, b, such that 3a2b+3ab2+b3≤remainder
4. We can estimate b by dividing the remainder by 3a2.
5. This division gives us b+b/a+b3/3a2, if b is small in respect to a, then we get a fair estimate of b, otherwise, not.
6. In arithmetic, where a and b are numbers between 0 and 10, we actually have b+b/(10a)+b3/300a2, which makes it more likely that the b estimated by dividing by 3a2 is close to the real b.
## Application to algebra
Answer: Answer so far 1+x Find a a=1 1 1 Subtract a3 from the cube 3·(x/3) x Remainder after subtracting a3 Find b, so 3b≤-x b=x/3 1+x/3 x+x2/3+x3/27 3a2b+3ab2+b3 a=1, b=x/3 =x+x2/3+x3/27 (3+2x+x2/3)(-x2/9) -x2/3-x3/27 Remainder. New a=1+x/3 a2=1+2x/3+x2/9 3a2=3+2x+x2/3 b=-x2/27 1+x/3-x2/9 (3+2x+x2/3)·(-x2/9)+ (1+x/3)·(x4/81)+ x6/729 Subtract: 3a2b+3ab2+b3= (3+2x+x2/3)·(-x2/9)+ (1+x/3)·(x4/81)+ x6/729 3·5x3/81... (-x3/27)+2x3/9.... =5x3/27... Remainder. New a=(1+x/3-x2/9) ...
Below is the expansion of (1+x)1/3, to term 10:
[2.3]
As with the square root, the expansion of the cube root gives us a pre-Binomial way of expanding expressions.
## Application to Arithmetic
In applying the method to arithmetic, we note that instead of our remainder being 3a2b+3ab2+b3, it is:
300a2b+30ab2+b3
Where a and b are numbers between 0 and 10.
So to estimate b, we divide the remainder:
300a2
Using this approximation there are sometimes problems with the second figure of the cube, and fine tuning is required, see the second example. This occurs when the first figure is low, 1 for example, and the next one is high, 9, for example.
If this estimate ( 300a2) does not give a number between 0 and 10, we can fine tune it to:
300b+30b2+b3 ≤ remainder
Divide throughout by 300:
b+b2/10+b3/300 ≤ remainder/300
This is the exact expression, but the arithmetic is very much easier after the division.
Also, because a number between 0 and 10, when cubed, is less than 1000, we divide the cube into groups of 3.
## An Example Which Works As Expected.
We seek the cube root of 362467097
This example, has been carefully engineered to work as expected. That is, the second figure of the cube is small compared with the first.
Answer: 713 Answer so far 362 467 097 We divide the number in groups of three. 7 343 We find the nearest cube less than 362, which is 73=343. Subtract this from the cube. 14 700·b 19 467 Remainder. a=7, 300a2=14 700 Estimate b as 19467/14700, say 1 71 14 911 Calculate 300a2b+30ab2+b3 300a2b=14700 30ab2=210 b3=1 1512300·b 4 556 097 Subtract 300a2b+30ab2+b3, and bring down the rest of the number a=71, 300a2=300.712=1512300 Estimate b as: 4556097/1512300 Estimate is 3. 713 4556097 Compute 300a2b+30ab2+b3= 1512300+ 19170+ 27 The remainder is zero, so we have our cube root.
## An Example Which Needs Fine Tuning
Sometimes, it isn't easy to find the second figure. The problem and solution are explained below in the example.
We seek the cube root of 6859.
Answer 1 9 Answer so far 6 859 1 a=1, we subtract a3 1 5 859 b is estimated as: 5859/300 Giving b as approximately 19. (b must be between 0 and 10). We need to be more precise to find b. 300b+30b2+b3≤5859 Divide throughout by 300: b+b2/10+b3/300≤19.53 Try b=9, approximately: 9+81/10+729/300≏19 As this is very close, we will work this out precisely: 9+8.1+2.43=19.53 As this is exactly the figure we require, we have finished. 19
## A Longer Example Which Requires Fine Tuning
We seek the cube root of 5 079 577 959
Answer 1719 Answer so far 5 079 577 959 1 The nearest cube less than 5 is 1 1 4 079 300b≤4079, from this, b is approximately 13, but it must be between 0 and 10.So we have no idea what b might be!We need to fine tune:300b+30b2+b3≤4079Divide throughout by 300:b+b2/10+b3/300≤13.593..Try b=9: (working approximately)9+8+2=19 This is too big, so try:b=8:8+6+1=15Also too big.Try b=77+5+1=13Working precisely now:7+49/10+342/300=12.006...So b=7 is correct. 17 3913 300·7+30·72+73=3913We subtract this number: 166 577 a=17, 300a2=8670086700b≤166577b≅1, 171 87211 Take b=1, so300·172+30·17+1=86700+510+1=87211Subtract this from the remainder so far. 7 9366 959 Bring down the next group of 3.a=171300·1712=87723008772300b≅79366959 79366959 b≏9300a2b=7895070030ab2=415530b3=729Subtract the total 1719 0 As this leaves zero, we are done.
It seems that when there is a problem with this method, it is with the second figure, when it isn't small in comparison to the first one. That is, b isn't small compared with a.
By being more precise, and working approximately, it is much easier to find the value for b. After this, the normal rule of 300a2b as an approximation for b seems to work fine, because the new a's are always large compared with the new b's and the longer the computation, the more this is true, so b becomes smaller and smaller in relation to a.
Ken Ward's Mathematics Pages
# Faster Arithmetic - by Ken Ward
Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle:
Buy now at Amazon.com |
# Common Core: 4th Grade Math : Multiply or Divide to Solve Word Problems Involving Multiplicative Comparison: CCSS.Math.Content.4.OA.A.2
## Example Questions
← Previous 1
### Example Question #1 : Multiply Or Divide To Solve Word Problems Involving Multiplicative Comparison: Ccss.Math.Content.4.Oa.A.2
It takes Hannah minutes to get dressed in the morning. It takes her older sister, Sara, times as long. How long does it take Sara to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Sara to get dressed.
Think: minutes 4 times would be how many minutes?
### Example Question #2 : Multiply Or Divide To Solve Word Problems Involving Multiplicative Comparison: Ccss.Math.Content.4.Oa.A.2
It takes Charlie minutes to get dressed in the morning. It takes her older sister, Alison, times as long. How long does it take Alison to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Alison to get dressed.
Think: minutes times would be how many minutes?
### Example Question #3 : Multiply Or Divide To Solve Word Problems Involving Multiplicative Comparison: Ccss.Math.Content.4.Oa.A.2
It takes Mark minutes to get dressed in the morning. It takes her older sister, Shelly, times as long. How long does it take Shelly to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Shelly to get dressed.
Think: minutes times would be how many minutes?
### Example Question #4 : Multiply Or Divide To Solve Word Problems Involving Multiplicative Comparison: Ccss.Math.Content.4.Oa.A.2
It takes Cassie minutes to get dressed in the morning. It takes her older sister, Linden, times as long. How long does it take Linden to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Linden to get dressed.
Think: minutes times would be how many minutes?
### Example Question #5 : Multiply Or Divide To Solve Word Problems Involving Multiplicative Comparison: Ccss.Math.Content.4.Oa.A.2
It takes Sam minutes to get dressed in the morning. It takes her older sister, Kate, times as long. How long does it take Kate to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Kate to get dressed.
Think: minutes times would be how many minutes?
### Example Question #6 : Multiply Or Divide To Solve Word Problems Involving Multiplicative Comparison: Ccss.Math.Content.4.Oa.A.2
It takes Maggie minutes to get dressed in the morning. It takes her older sister, Alex, times as long. How long does it take Alex to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Alex to get dressed.
Think: minutes times would be how many minutes?
### Example Question #7 : Multiply Or Divide To Solve Word Problems Involving Multiplicative Comparison: Ccss.Math.Content.4.Oa.A.2
It takes Tim to get dressed in the morning. It takes his older sister, Lisa, times as long. How long does it take Lisa to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Lisa to get dressed.
Think: minutes times would be how many minutes?
### Example Question #8 : Multiply Or Divide To Solve Word Problems Involving Multiplicative Comparison: Ccss.Math.Content.4.Oa.A.2
It takes Molly to get dressed in the morning. It takes her older sister, Emily, times as long. How long does it take Emily to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Emily to get dressed.
Think: minutes times would be how many minutes?
### Example Question #572 : How To Multiply
It takes Nancy to get dressed in the morning. It takes her older sister, Jessica, times as long. How long does it take Jessica to get dressed?
Explanation:
The phrase " times as long" tells you that we are going to multiply by
Let's have represent how long it takes Jessica to get dressed.
Think: minutes times would be how many minutes?
### Example Question #573 : How To Multiply
It takes David to get dressed in the morning. It takes her older sister, Melissa, times as long. How long does it take Melissa to get dressed? |
# Three Meanings of the Minus Sign
We often see bright kids in 5th or 6th grade getting the deer-in-the-headlights look when it comes to dealing with equations with multiple negative or minus signs. Often it is good to stop and think about the different roles the humble minus sign take on in integer arithmetic. In fact, at this level, there are three meanings of the minus sign we should be aware of, and be able to tell which role is applicable for each part of the math we’re trying to solve.
## 1. As “Take-away” or Difference
This is the first use of the minus sign we all learn in kindergarten or first grade – i.e. to denote the difference or how far a value is from another.
As a difference operator, the minus sign takes in two arguments (10 and 7) and produces one output (3). In this sense, the use of a dash symbol is most obvious, as we would often define source-destination with a dash in between.
## 2. As a Label for Negative Numbers
The second time we see the minus sign is when we learn about negative numbers or numbers that are on the left side of the zero on the number line.
Here, we are not viewing the negative numbers as “reflection” of the positive numbers yet. They are just numbers that extends the positive number line we were familiar with and which are progressing leftwards. Here the minus signs are understood as labels to be used as part of the symbol, e.g. “-3”.
## 3. As a “Flip” Operation
Soon after we learn about negative numbers, we would learn about the third meaning of the minus sign – the “flip” over zero. For a positive quantity or expression, the minus sign flips it across zero on the number line.
This time, as a flip operator, the minus sign takes in only one argument (10-7) and produces one result -(10-7). We can concatenate or nest the operation multiple times, each time just flipping the quantity over zero.
In summary, the minus sign has multiple roles to play and often takes on different meanings in the same equation. It helps to keep these roles in our mind when mixing them in a mathematical expression.
• Minus sign as a difference operator
• Minus sign as part of the extended number line labeling
• Minus sign as a “flip” over zero operator
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Mass
# Convert Stone to Gram
Written by Rahul Lath
Updated on: 24 May 2023
### Convert Stone to Gram
Conversion stone to gram: Stones and grams are used to measure the mass. A mass of 6350.29 Grams makes a stone. This article will help you to convert stones to grams.
Stone
st
Grams
g
## What Is Stone?
A Stone is an English and Imperial unit of mass that is used to measure the weight of dry products and it is equal to 14 pounds. It’s denoted as ‘st.’
Common Usage:
• Used to measure the weight of people.
• Used to measure the weight of large animals.
• Used to measure the weight of a sack of wool.
## What Is Gram?
In the International System of Standards (SI), a gram is a unit used for measuring the mass which is equal to one thousand of a kilogram. It’s denoted by ‘g.’
1 g = 1/1000 kg
Common Usage:
• Used for measuring the mass of an apple.
• Used for measuring the mass of peas in a small bowl.
• Used for measuring the mass of a newspaper.
## How to Convert Stone to Gram?
To convert stones to grams, multiply the given value in stones by to get the value in grams. As
1 Stone = 6350.29 Grams
The formula to convert stones to grams is given below:
X st = X × 6350.29 g
where X is the value in stones.
## Examples of the Conversion of Stone to Kilogram.
Problem 1: Convert 10 stones to grams.
Solution 1:
Step 1: The given value is 10 stones.
Step 2: To convert stones to grams, substitute the given values at the required places in the conversion formula.
X st = X × 6350.29 g
Hence,
10 st = 10 × 6350.29 g
= 63502.9 g
Therefore, 10 stones are 63502.9 grams.
Problem 2: Convert 30 stones to grams.
Solution 2:
Step 1: The given value is 30 stones.
Step 2: To convert stones to grams, substitute the given values at the required places in the conversion formula.
X st = X × 6350.29 g
Hence,
30 st = 30 × 6350.29 g
= 190509 g
Therefore, 30 stones are 190509 grams.
Problem 3: Convert 50 stones to grams.
Solution 3:
Step 1: The given value is 50 stones.
Step 2: To convert stones to grams, substitute the given values at the required places in the conversion formula.
X st = X × 6350.29 g
Hence,
50 st = 50 × 6350.29 g
= 317515 g
Therefore, 50 stones are 317515 grams.
Problem 4: Convert 440 stones to grams.
Solution 4:
Step 1: The given value is 440 stones.
Step 2: To convert stones to grams, substitute the given values at the required places in the conversion formula.
X st = X × 6350.29 g
Hence,
440 st = 440 × 6350.29 g
= 2.794e+6 g
Therefore, 440 stones are 2.794e+6 grams.
Problem 5: Convert 32 stones to grams.
Solution 5:
Step 1: The given value is 32 stones.
Step 2: To convert stones to grams, substitute the given values at the required places in the conversion formula.
X st = X × 6350.29 g
Hence,
32 st = 32 × 6350.29 g
= 203209 g
Therefore, 32 stones are 203209 grams.
## FAQs on the Conversion of Stone to Gram
How many grams are in a stone?
1 Stone is 6350.29 Grams.
Convert 20 stones to grams.
Solution:
Step 1: The given value is 20 stones.
Step 2: To convert stones to grams, substitute the given values at the required places in the conversion formula.
X st = X × 6350.29 g
Hence,
20 st = 20 × 6350.29 g
= 127006 g
Therefore, 20 stones are 127006 grams.
Is 1 stone the same as 14 pounds?
Yes, 1 stone is 14 pounds.
Convert 5 stones to grams.
Solution:
Step 1: The given value is 20 stones.
Step 2: To convert stones to grams, substitute the given values at the required places in the conversion formula.
X st = X × 6350.29 g
Hence,
5 st = 5 × 6350.29 g
= 31751.5 g
Therefore, 5 stones are 31751.5 grams.
What is a stone?
A Stone is an English and Imperial unit of mass that is used to measure the weight of dry products and it is equal to 14 pounds. It’s denoted as ‘st.’
References
The SI Base Units
Written by by
Rahul Lath
Reviewed by by
Arpit Rankwar
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# A can do a work in 6 days. B takes 8 days to complete the work. C takes the time equal to the time taken by A and B to work together to complete that task. If B and C work together, how much time will they take to complete the work?
1. $$\frac{{13}}{5}$$ days
2. $$\frac{{14}}{5}$$ days
3. $$\frac{{12}}{5}$$ days
4. $$\frac{{11}}{5}$$days
Option 3 : $$\frac{{12}}{5}$$ days
Free
AFCAT 2/2021 All India Mock Test
14842
100 Questions 300 Marks 120 Mins
## Detailed Solution
Given:
Time taken by A to complete the work = 6 days
Time taken by B to complete the work = 8 days
Formula used:
Efficiency = Total work/Time taken
Calculation:
LCM of 6 and 8 = 24 = Total work
Efficiency of A = 24/6 = 4 units/day
Efficiency of B = 24/8 = 3 units/day
Total effiiciency of A and B = (4 + 3) units/day
⇒ 7 units/day
So, Efficiency of C will also be 7 units/day as it is taking same number of days as A and B together
So, Total efficiency of B and C = (3 + 7) units/day
⇒ 10 units/day
Time taken by B and C to complete the work = 24/10 days
⇒ 12/5 days
∴ They will take 12/5 days to complete the work |
Prime Factors of 1134
Prime Factors of 1134 are 2, 3, 3, 3, 3, and 7
How to find prime factors of a number
1. Prime Factorization of 1134 by Division Method 2. Prime Factorization of 1134 by Factor Tree Method 3. Definition of Prime Factors 4. Frequently Asked Questions
Steps to find Prime Factors of 1134 by Division Method
To find the primefactors of 1134 using the division method, follow these steps:
• Step 1. Start dividing 1134 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 1134, which is 2. Divide 1134 by 2 to obtain the quotient (567).
1134 ÷ 2 = 567
• Step 3. Repeat step 1 with the obtained quotient (567).
567 ÷ 3 = 189
189 ÷ 3 = 63
63 ÷ 3 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1
So, the prime factorization of 1134 is, 1134 = 2 x 3 x 3 x 3 x 3 x 7.
Steps to find Prime Factors of 1134 by Factor Tree Method
We can follow the same procedure using the factor tree of 1134 as shown below:
So, the prime factorization of 1134 is, 1134 = 2 x 3 x 3 x 3 x 3 x 7.
What does Prime factor of a number mean?
A prime number in mathematics is defined as any natural number greater than 1, that is not divisible by any number except 1 and the number itself. When we express any number as the product of these prime numbers than these prime numbers become prime factors of that number. Eg- Prime Factors of 1134 are 2 x 3 x 3 x 3 x 3 x 7.
Properties of Prime Factors
• Two prime factors of a given number are always coprime to each other.
• 2 is the only even prime factor any number can have.
• Two prime factors are always coprime to each other.
• 1 is neither a prime number nor a composite number and also 1 is the factor of every given number. So, 1 is the factor of 1134 but not a prime factor of 1134.
• Which is the smallest prime factor of 1134?
Smallest prime factor of 1134 is 2.
• Is 1134 a perfect square?
No 1134 is not a perfect square.
• What is the prime factorization of 1134?
Prime factorization of 1134 is 2 x 3 x 3 x 3 x 3 x 7.
• What are the factors of 1134?
Factors of 1134 are 1 , 2 , 3 , 6 , 7 , 9 , 14 , 18 , 21 , 27 , 42 , 54 , 63 , 81 , 126 , 162 , 189 , 378 , 567 , 1134.
• What is the product of all prime factors of 1134?
Prime factors of 1134 are 2 x 3 x 3 x 3 x 3 x 7. Therefore, their product is 1134.
• What numbers are the prime factors of 1134?
Prime factors of 1134 are 2 , 3 , 3 , 3 , 3 , 7.
• What is the sum of all odd prime factors of 1134?
Prime factors of 1134 are 2 , 3 , 3 , 3 , 3 , 7, out of which 3 , 3 , 3 , 3 , 7 are odd numbers. So, the sum of odd prime factors of 1134 is 3 + 3 + 3 + 3 + 7 = 19.
• Is there any even prime factor of 1134?
Yes there is a even prime factor of 1134.
• What is the product of all odd prime factors of 1134?
Prime factors of 1134 are 2 , 3 , 3 , 3 , 3 , 7, out of which 3 , 3 , 3 , 3 , 7 are odd numbers. So, the product of odd prime factors of 1134 is 3 x 3 x 3 x 3 x 7 = 567.
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# Problem Solving on Subtraction
Problem solving on subtraction will help us to get the idea on how to solve the basic subtraction statement problems.
1. Eight birds sat on a wire. Three birds flew away. How many were left?
Total number of birds sat on a wire = 8
Number of birds flew away = 3
Therefore, number of birds left = 8 - 3 = 5
2. Sam had 7 dollars. He spent 4 dollars. How many dollars is he left with?
Total amount of money Sam had = $7 He spent =$4
Therefore, amount of money left with him = $7 -$4 = \$3
3. Five boats were tied up. Four of the boats sailed away. How many were left?
Total number of boats tied up = 5
Number of boats sailed away = 4
Therefore, number of boats were left = 5 - 4 = 1
4. Ron had 10 stamps. His father took 2 stamps. How many stamps does Ron have now?
Total number of stamps Ron had = 10
Number of stamps his father took = 2
Therefore, number of stamps he have now = 10 - 2 = 8
5. Diana had 18 toffees. She gave 5 toffees to her friend. How many toffees left with her?
Total number of toffees Diana had = 18
Number of toffees she gave to her friend = 5
Therefore, number of toffees left = 18 - 5 = 13
More examples on statement problem solving on subtraction:
6. Mr. Daniel had 39 goats in a pasture. When he opened the pasture gate, 13 goats went out. How many goats remained in?
Total number of goats in a pasture Mr. Daniel had = 39
Number of goats went out = 13
Therefore, number of goats remained in = 39 - 13 = 26
7. Derek’s father is 47 years old. His mother is 35 years old. What is the difference of their ages?
Age of Derek’s father = 47 years
Age of his mother = 35 years
Therefore, difference of their ages = 47 - 35 = 12 years
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What Are The Factors Of 10000: Prime Factors Of 10000
Below I have discussed all the factors of 10000. This is the most important part of basic mathematics. This article will clear your all basic concepts regarding factors. In this article, you will be able to find information about the Prime factors for 10000, factors for 10000, pair factors for 10000, and the factors of other numbers as well. One by one I am going to explain in a very simple format. Now we are going to see the list of factors of 10000.
What Are The Factors Of 10000
The factors are the number that divides the number without leaving any remainder.
So the all factors of 10000 are 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 125, 200, 250, 400, 500, 225, 1000, 1250, 2000, 2500, 5000, 10000.
• 1(10000➗1=1000)
• 2(10000➗2=500)
• 4(10000➗4=250)
• 5(10000➗5=200)
• 8(10000➗8=125)
• 10(10000➗10=100)
• 16(10000➗20=50)
• 20(10000➗25=40)
• 25(10000➗40=25)
• 40(10000➗50=20)
Continue in the same manner…
1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 125, 200, 250, 400, 500, 625, 1000, 1250, 2000, 2500, 5000, 10000 are the factors for 10000.
Prime Factors Of 10000
For the calculation of Prime factors of 10000, you must know the prime numbers up to 10000. After that, we have to divide 10000 with all prime numbers. We are going to calculate the prime factors by factorization of the 10000 methods.
• 10000➗2 = 500
• 10000➗3 = 333.33
• 10000➗5 = 2000
• 10000➗7 = 1428.5
• 10000➗11 = 909.0
• 10000➗13 = 769.2
• 10000➗17 = 588.2
• 10000➗19 = 526.3
• 10000➗23 = 434.7
• 10000➗29 = 344.8
Continue with the same manner…
So the prime factors for the number 1000 are 2 and 5.
Factors Of 10000 In Pairs
Let’s see the pair factors of 10000.
• 1 x 10000 = 10000
• 2 x 5000 = 10000
• 4 x 2500 = 10000
• 5 x 2000 = 10000
• 8 x 1250 = 10000
• 10 x 1000 = 10000
• 16 x 625 = 10000
• 20 x 500 = 10000
• 25 x 400 = 10000
• 40 x 250 = 10000
• 50 x 200 = 10000
Continue with same manner…
So the pair factors for 10000 are 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 125 200, 250, 400, 500, 625, 1000, 1250, 2000, 2500, 5000, 10000.
Factors Of -10000
Here You can see the factors of -10000.
• -1(-10000➗-1=-1000)
• -2(-10000➗-2=-500)
• -4(-10000➗-4=-250)
• -5(-10000➗-5=-200)
• -8(-10000➗-8=-125)
• -10(-10000➗-10=-100)
• -16(-10000➗-20=-50)
• -20(-10000➗-25=-40)
• -25(-10000➗-40=-25)
• -40(-10000➗-50=-20)
Continue in the same manner…
-1, -2,- 4, -5, -8, -10, -16, -20, -25, -40, -50, -80, -100, -125, -200, -250, -400, -500, -625, -1000, -1250, -2000, -2500, -5000, -10000 factors for negative 10000.
Factors Of 10000 And Other Number
Below are the common factors of 10000 and other numbers.
The Factors Of 10000 And 64
The factors for 10000 and 64 are 1, 2, 4, 8, 16
• 10000 factors = 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 125, 200, 250, 400, 500, 225, 1000, 1250, 2000, 2500, 5000, 10000.
• 64 factors are = 1, 2, 4, 8, 16, 32, 64
HCF of 10000 and 64 = 1, 2, 4, 8, 16.
All factors for the numbers can be seen here on Factorsweb
Click here – What Are The Factors Of 17: Prime Factors Of 17
FAQ
What Are The Factors Of 10000?
Here is the list of all Postive Factors of 10000 in numerical order: 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 125, 200, 250, 400, 500, 625, 1000, 1250, 2000, 2500, 5000, and 10000. Factors of 10000 include negative numbers.
What Is The 10000 Th Prime Number?
104729
The ten-thousandth prime, prime(10000) , is 104729. We obtained these as shown below.
How Do I Get To The Root Of 10000?
The square root of 10000 is 100. This means when we multiply 100 twice with itself, the answer is 10000. Thus, The square of 100 is 10000, and the square root of 10000 is ±100.
How Many Primes Are There Between 1000 And 10000?
Well, there are 1061 Prime Numbers between 1,000 and 10,000.
How Many Factors Does 10000 Have?
The number 10000 has 23 factor not counting 1 or itself makeing it a square and maybe composite. Prime Factors for the number 10000 = (24 × 54) or(2 × 2 × 2 × 2 × 5 × 5 × 5 × 5).
Which Number Under 10000 Has The Most Factors?
2310 is the lowest number below 10000 with the most distinct prime factors, 9240 the highest, but there are many.
How Do You Find Prime Factors?
The simplest algorithm to find the prime factors of a number is to keep on dividing the original number by prime factors until we get the remainder equal to 1. For example, prime factorizing the number 30 we get, 30/2 = 15, 15/3 = 5, 5/5 = 1. Since we received the remainder, it cannot be further factorized.
Conclusion
From this article, you will be able to know about the factors of 10000. In addition to that, I have also explained the prime factors for 10000, factors for 10000 in pairs, factors for -10000, and factor numbers of 10000 with another number. This article will definitely help to clear the basic concepts regarding factors. Now you know all about the factors of 10000.
What are the prime factors of 10000 |
# How To Multiply Fractions By Whole Numbers
How To Multiply Fractions By Whole Numbers – Multiplying a fraction by a whole number is a simple operation. This is one of the basic concepts taught in elementary school. Students are taught to increase their arithmetic skills. Students often make mistakes when multiplying and dividing fractions. This article discusses methods of multiplying fractions by whole numbers with some examples.
Before learning how to multiply fractions by whole numbers, let’s review the basic terms used in multiplication. Do you know what a fraction is? Fractions are generally numbers expressed as p/q. For example, 2/3, 9/2, etc.
## How To Multiply Fractions By Whole Numbers
You need to understand that fractions have two parts. The part above the “dash” is called the numerator, and the number below the “dash” is called the divisor. All kinds of mathematical operations can be done with fractions.
#### Multiplying Fraction By Whole Number Crack The Code
If you divide the numerator by the denominator, you get a whole number or a decimal number. It no longer remains a faction. So a value can never be a fraction unless the form is p/q.
To multiply a fraction by a whole number, you must first learn how to multiply a fraction by another fraction. Let us know about it in the article.
Suppose we have two fractions, d/c and j/k. To multiply these fractions, you need to write them correctly on the opposite side. Then look at the numerator and denominator of the fraction. Multiply the numerators together (d x j) = x. Write the result of multiplying the numerator by another fraction and say “x”.
Similarly, multiply the denominators of fractions. Write the result of multiplying the divisor under the resulting numerator, (c x k) = y. Therefore, the product of two fractions can be expressed as:
#### Multiply Fractions By Whole Numbers Activity Pbl
After multiplying 4/5 by 3/7, the fraction is 12/35. If the number in the fraction is a multiple of some small number, you can easily reduce the fraction to simpler form. The only condition is that if you divide the numerator by any number ‘a’, you must divide the denominator by the same number ‘a’.
Solution: We see that numerator = 12 and denominator = 9 are both multiples of 3. So we divide the denominator by 3 and the denominator by 3.
Now we have learned how to multiply two fractions. In this section, you will learn how to multiply fractions by whole numbers. Suppose we have a d/c fraction and an integer ‘k’. The first step in multiplying fractions and whole numbers begins by writing them correctly opposite each other using the multiplication sign. For integers, we don’t have the form “p/q”. So first we convert the number to fractional form. To do this, put a line under the whole number whose scale is 1. Now we have an integer of the form “p/q”.
Now we follow similar steps mentioned in multiplying fractions. Now look at the denominator of the fraction and the whole number. Multiplying numerator by integer ‘k’ gives (d x k) = x. Now multiply the denominator of the fraction by 1 (since the denominator of the whole number is 1). This gives (c x 1 = c). Therefore, the product of two fractions can be expressed as:
## How To Teach Multiplying Fractions With Pattern Blocks — Mix And Math
When multiplying a fraction by a whole number, the denominator of the fraction is preserved. So whenever you find this type of multiplication, place the denominator as the divisor of the fraction and multiply the number.
Step 2: Knowing that the denominator remains the same in such cases, we need to multiply the denominator.
I hope the question of how to multiply a fraction by a whole number is now simple. Let’s look at a very interesting idea further.
But you can see that if we subtract 11 from the beginning, we still get the answer 3.
### Decomposing Fractions To Multiply
As shown in the example above. Suppose a fraction is multiplied by a product of the denominator or an integer equal to the denominator. In this case, you can subtract or reduce the number before the actual multiplication. Understand this from the examples below.
We see that the whole number and the denominator are the same. So we can eliminate them and get the number of fractions as the answer.
We observe that the denominator and the whole number are multiples of 7. Therefore, we reduced them to a minimum form. If you divide 7 by itself, you get 1, and if you divide 21 by 7, you get 3. Now there is 3×3 left in the numerator, so the answer to this question is 9.
As we know, integers are the set of real numbers starting from zero and extending to the positive limit. We have already seen all the cases of what happens when a fraction is multiplied by a whole number. If you multiply a fraction by 1, the number comes out by itself. But what happens when a fraction is multiplied by zero?
## How To Multiply Proper Fractions By A Whole Number
We know that everything multiplied by zero is zero. Therefore, when a fraction is multiplied by zero, the resulting fraction is zero or 0/1.
A number reversal is when the numerator and denominator switch positions. Since 0 = 0/1, its opposite is 1/0. Assuming we have the fraction “a/b”. Using the steps we’ve learned so far, a x 1 = a and ab x 0 = 0. So the resulting fraction is a/0. In mathematics, 1/0 or something divided by zero is not defined. Therefore, we cannot multiply a number by the opposite of zero.
Word Problem: Jill makes homemade hot chocolate. A quarter teaspoon of hot chocolate is used to make 1 cup of hot chocolate. Calculate how many spoons are needed to make 10 cups of hot chocolate.
Answer. Fractions can be multiplied by multiplying the numerator (the top number) and then the denominator (the bottom number).
### Divide Whole Numbers By Fractions
Answer. There are many ways to multiply fractions. The easiest way is to find the lowest common multiple of the denominator, multiply by the denominator of one fraction, and then multiply by the denominator of the other fraction. If you have a calculator, this can be done very quickly.
Answer. A general rule for multiplying fractions is to multiply the numerators, multiply the denominators, and then simplify.
Answer. To multiply fractions, you need to switch to fraction mode on your calculator. Then you enter the first fraction (at the top) and press the multiply button. The second fraction is automatically entered and multiplied by the first fraction. Then click right away to get the answer!
Composite figures consist of simple geometric figures. It is a 2D representation of basic 2D shapes like square, triangle, rectangle, circle etc. The field has different forms that differ from each other. From laptops to books, everything has its own section. To understand the dynamics of composite […]
## Free} Fraction Of A Whole Number Worksheets
Learn all about special right triangles – their types, formulas and examples are explained in detail for better understanding. What is the shorter ratio of the side lengths of the special right triangles 30 60 90 and 45 45 90? How do these ratios relate to the Pythagorean theorem? Right triangle Ninety degree triangle […]
Simplifying algebraic expressions in mathematics is a collection of various numerical expressions that have been refuted by several philosophers and historians. Speaking of algebra, this branch of mathematics deals with the oldest concepts of mathematics, geometry, and number theory. It is one of the oldest branches in the history of mathematics. Mathematical research […]
In this article, we will learn how to solve right triangles. But first, familiarize yourself with the triangle. A triangle consists of three line segments. These three segments join together to form three angles. Side length and angle measure are related. If you know the size (length) […]We use cookies to make them awesome. By using our site, you agree to our cookie policy. Cookie settings
This article was co-authored by Professor Mario Banuelos, Ph.D., and author Jessica Gibson. Mario Banuelos is an assistant professor of mathematics at California State University, Fresno. With more than eight years of teaching experience, Mario specializes in mathematical biology, optimization, statistical modeling of genome evolution, and data science. Mario received his BA and Ph.D. in mathematics from California State University, Fresno. in applied mathematics from the University of California, Merced. Mario has taught in both high school and college.
## Multiply Fractions By Whole Number Worksheet
Fractions are easy to multiply by mixed fractions or whole numbers. Begin by changing mixed fractions or whole numbers to improper fractions. Then multiply the number of two improper fractions. Multiply the times and simplify the result.
This article was co-authored by Professor Mario Banuelos, Ph.D., and author Jessica Gibson. Mario Banuelos is an assistant professor of mathematics at California State University, Fresno. With over eight years of teaching experience, Mario is qualified |
# 'Lesson 4 standard 1' presentation slideshows
## Splash Screen
Chapter 1 Place Value and Number Sense Click the mouse or press the space bar to continue. Splash Screen. Place Value and Number Sense. 1. Lesson 1-1 Number Patterns Lesson 1-2 Problem-Solving Strategy: Use the Four-Step Plan Lesson 1-3 Place Value Through 1,000
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## Lesson 1 MI/Vocab
Probability and Outcomes. 16-1. I will describe probability. outcome probability. Lesson 1 MI/Vocab. Probability and Outcomes. 16-1. Standard 4SDAP2.2 Express outcomes of experimental probability situations verbally and numerically (e.g., 3 out of 4; ). Lesson 1 Standard 1.
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## Lesson 4-1
Lesson 4-1. Maximum and Minimum Values. Quiz. Homework Problem: Related Rates 3-10 Two cars start moving from the same point. One travels south at 60 mph and the other travels west at 25 mph. At what rate is the distance between the cars increasing two hours later. Reading questions:
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## Lesson 4-1
Lesson 4-1. Parallel Lines and Planes. Ohio Content Standards:. Ohio Content Standards:. Formally define geometric figures. Ohio Content Standards:. Recognize and apply angle relationships in situations involving intersecting lines, perpendicular lines and parallel lines. Parallel Lines.
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## Lesson 1-4
Angles. Lesson 1-4. Angle and Points. An Angle is a figure formed by two rays with a common endpoint, called the vertex . ray. vertex. ray. Angles can have points in the interior, in the exterior or on the angle. A. E. D. B. C.
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## LESSON 4-1
LESSON 4-1. Responsibility Accounting for a Merchandising Business. TERMS REVIEW. page 91. fiscal period – length of time in which a business reports financial information
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## Lesson 1-4
Angles. Lesson 1-4. RA : RA and all points Y such that A is between R and Y. ( the symbol RA is read as “ray RA” ). Ray. Definition:. How to sketch:. How to name:. Opposite Rays. Definition :. If A is between X and Y, AX and AY are opposite rays.
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## LESSON 4-1
LESSON 4-1. Responsibility Accounting for a Merchandising Business. RECORDING A DIRECT EXPENSE. page 90. RECORDING AN INDIRECT EXPENSE. page 90. TERMS REVIEW. page 91. fiscal period responsibility accounting direct expense indirect expense departmental margin
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## LESSON 4-1
LESSON 4-1. DIVISIBILITY. 4-1. Divisibility. Course 1. Warm Up 1. 20 2. 48 3. 16. Write each number as a product of two whole numbers in as many ways as possible. 1 20, 2 10, 4 5. 1 48, 2 24, 3 16, 4 12, 6 8. 1 16, 2 8, 4 4. 4-1. Divisibility. Course 1. |
# How to Find the Mean: A Step-by-Step Guide
Baca Cepat
## Introduction
Welcome to our guide on how to find the mean! If you’ve ever wondered how to calculate the average value of a set of numbers, you’ve come to the right place. The mean is a fundamental statistical measure that is used in a wide variety of applications, from calculating grades to analyzing stock market trends. In this article, we’ll provide you with a detailed explanation of what the mean is, how to calculate it, and some practical examples to help you understand the concept better. Let’s get started!
### What is the Mean?
The mean, also known as the arithmetic mean or average, is a measure of central tendency that represents the sum of a set of values divided by the total number of values. It is a common way to describe the typical value of a set of numbers, and it is often used in statistics and other fields that involve quantitative analysis. The formula for calculating the mean is as follows:
Notation Formula Description
`x̄` `x̄ = (x1 + x2 + ... + xn) / n` Arithmetic Mean
where `x1, x2, ..., xn` are the individual values, and `n` is the total number of values in the set.
### How to Find the Mean
Now that you know what the mean is, let’s dive into how to calculate it. Here are the steps to follow:
#### Step 1: Add up all the values
The first step is to add up all the values in the set. For example, if you have a set of five numbers {3, 5, 7, 9, 11}, you would add them up as follows:
`3 + 5 + 7 + 9 + 11 = 35`
#### Step 2: Count the number of values
The next step is to count the number of values in the set. In our example, there are five numbers, so `n = 5`.
#### Step 3: Divide the sum by the number of values
The final step is to divide the sum of the values by the number of values in the set. In our example, we would divide 35 by 5:
`35 / 5 = 7`
Therefore, the mean of the set {3, 5, 7, 9, 11} is 7.
### Examples of Finding the Mean
Let’s look at some more examples to solidify your understanding of how to find the mean.
#### Example 1
Find the mean of the following set of numbers:
{2, 4, 6, 8, 10}
Solution:
`(2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6`
The mean of the set {2, 4, 6, 8, 10} is 6.
#### Example 2
Find the mean of the following set of numbers:
{1.5, 2.7, 3.9, 4.3, 5.1, 6.2, 7.4}
Solution:
`(1.5 + 2.7 + 3.9 + 4.3 + 5.1 + 6.2 + 7.4) / 7 ≈ 4.23`
The mean of the set {1.5, 2.7, 3.9, 4.3, 5.1, 6.2, 7.4} is approximately 4.23.
### FAQs
#### Q1: What is the difference between the mean and the median?
A: The mean is a measure of central tendency that represents the average value of a set of numbers, while the median is the middle value in a set of numbers.
#### Q2: What is the mode?
A: The mode is the most frequently occurring value in a set of numbers.
#### Q3: Can the mean be negative?
A: Yes, the mean can be negative if the set of numbers contains negative values.
#### Q4: What is an outlier?
A: An outlier is a data point that is significantly different from the others in a set of numbers, and it can affect the calculation of the mean.
#### Q5: Can the mean be used for qualitative data?
A: No, the mean is typically used for quantitative data, while other measures such as the mode and median are used for qualitative data.
#### Q6: What is the weighted mean?
A: The weighted mean is a type of mean that takes into account the relative importance or weight of each value in the set.
#### Q7: How is the mean used in finance?
A: The mean is used in finance to calculate average returns on investments and to analyze market trends over time.
### Conclusion
Congratulations! You’ve now learned how to find the mean and everything you need to know about this fundamental statistical measure. Remember to follow the steps we outlined to calculate the mean accurately and practice with different sets of numbers to solidify your understanding. Whether you’re a student, a researcher, or an analyst, the mean is an essential tool that you’ll use time and time again. Don’t forget to put this knowledge into practice and keep exploring the fascinating world of statistics!
#### Still have questions?
If you have any questions or comments on this guide, please don’t hesitate to reach out to us. We’re here to help you succeed!
## Closing Disclaimer
The information contained in this article is for educational and informational purposes only and should not be used as a substitute for professional advice. We make no representations or warranties of any kind, express or implied, about the completeness, accuracy, reliability, suitability, or availability with respect to the information contained in this article. Any reliance you place on such information is therefore strictly at your own risk.
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# Inverse for Complex Multiplication
## Theorem
Each element $z = x + i y$ of the set of non-zero complex numbers $\C_{\ne 0}$ has an inverse element $z^{-1}$ under the operation of complex multiplication:
$\forall z \in \C_{\ne 0}: \exists z^{-1} \in \C_{\ne 0}: z \times z^{-1} = 1 + 0 i = z^{-1} \times z$
This inverse can be expressed as:
$\dfrac 1 z := \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$
where $\overline z$ is the complex conjugate of $z$.
## Proof
$\ds \paren {x + i y} \frac {x - i y} {x^2 + y^2}$ $=$ $\ds \frac {\paren {x \cdot x - y \cdot \paren {-y} } + i \paren {x \cdot \paren {-y} + x \cdot y} } {x^2 + y^2}$ $\ds$ $=$ $\ds \frac {\paren {x^2 + y^2} + 0 i} {x^2 + y^2}$ $\ds$ $=$ $\ds 1 + 0 i$
Similarly for $\dfrac {x - i y} {x^2 + y^2} \paren {x + i y}$.
So the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is $\dfrac {x - i y} {x^2 + y^2}$.
As $x^2 + y^2 > 0 \iff x, y \ne 0$ the inverse is defined for all $z \in \C: z \ne 0 + 0 i$.
$\Box$
From the definition, the complex conjugate $\overline z$ of $z = x + i y$ is $x - i y$.
From the definition of the modulus of a complex number, we have:
$\cmod z = \sqrt {a^2 + b^2}$
From Modulus in Terms of Conjugate, we have that:
$\cmod z^2 = z \overline z$
Hence the result:
$\dfrac 1 z = \dfrac {\overline z} {z \overline z}$
$\blacksquare$
## Examples
### Example: $\dfrac 1 {1 + i}$
$\dfrac 1 {1 + i} = \dfrac {1 - i} 2$
### Example: $\dfrac 1 {3 + 2 i}$
$\dfrac 1 {3 + 2 i} = \dfrac 3 {13} + \dfrac {2 i} {13}$ |
# How To Figure Out Ratios With Fractions
How To Figure Out Ratios With Fractions – Here we learn about ratios to fractions, including using ratios to find fractions and using fractions to find ratios.
Worksheets based on Edexcel, AQA and OCR exam questions, plus further guidance on where to go next if you’re still stuck.
## How To Figure Out Ratios With Fractions
A ratio to a fraction is a way of writing a ratio differently. A ratio compares how much one thing is compared to another. It can be written using the word “:” or “to” or as a decimal.
#### Fractions, Ratios, And Percent
When two values are parts of a whole we have the ratio a:b, we can say the ratio frac and frac to convert the ratios into fractions.
The chart below has a bar pattern showing a blue:red ratio of 3:2 (3 to 2). There are 3 blue blocks and 2 red blocks, so there are 5 blocks in total.
Here, the total number of shares in the ratio is equal to a+b (the denominator of each fraction), and the numerator is the part of the ratio we are interested in.
#### Module 1: Ratios And Unit Rates
Note: If the fraction (or ratio) can be simplified, simplify it, but do not forget to use whole numbers (whole numbers).
Ratio to Fraction is part of our series of lessons to support ratio revision. It may be helpful to start with a basic ratio lesson to get a brief idea of what to expect, or use the step-by-step guides below for more detailed information on individual elements. Other lessons in this series:
Ann and Bob share a box of cookies in the ratio 3:4. What fraction of the cookies does Bob get?
## Vector Problems With Ratio
Since there are 5 yellow beads and 3 red beads in each repetition, the ratio of yellow to red beads is 5:3.
A group of friends want to watch a movie at a movie theater. Below is a bar model representing the choice of friends. Friends who want to watch a romantic comedy are in red, those who want to watch a sci-fi movie are in yellow, and the rest want to watch the latest action blockbuster.
What part of a group of friends doesn’t want to watch a sci-fi movie? Write your answer in simplest form.
### Ways To Tell If Two Ratios Are In Proportion
Since the question was asked for the number of people who would not want to see a science fiction movie, it is 3+5=8 people.
frac A male school of fish. The rest are women. Write the female to male ratio in the school.
A cat spends almost every day sleeping. State the ratio of waking hours to sleeping hours for Pepper.
### Everything You Need To Know About Ratio Tables
The dental practice sells 3 different packages of dental care plans: Basic, Premium and Family. frac have their members basic plan and frac premium plan. Basic : Premium : State the family ratio.
Before we can calculate this value, we need to know the sum of the two fractions we have. A common denominator also helps to tell the sum of the parts in a ratio
This ratio asks us Basic : Premium : Family. The three fractions we use are frac, frac and frac. Since fractions have the same denominator, we can use the numerators to write the ratio.
#### Question Video: Finding The Sum Of An Infinite Geometric Series
For example, the ratio 2:3 is expressed as a fraction frac rather than frac. This is a misunderstanding of the sum of the parts ratio. Be careful what you ask for as the denominator can be a part or a whole.
Make sure all the units in the ratio are the same. For example, in Example 6, all units in the ratio are in milliliters. We have not mixed the ratio of ml and L.
The numerator is out of proportion. For example, the number of mirrors in the kitchen is written in the ratio of 4:3. Write the proportion of cups in the kitchen. The correct answer is frac. The value of “mugs” is the first number in the ratio, m:g = 4:3, because we use the same order as the written sentence.
### Ratio To Fraction Calculator
The parts of the ratio are written in the wrong order. For example, the ratio of dogs to cats is given as 12:13, but the solution is written as 13:12.
1. The ratio of Tyrannosaurus rex and Velociraptor fossils is 3:8. What part of the fossil record does Tyrannosaurus rex belong to? Give your answer in simpler form as a fraction.
2. A tailor sells silk and polyester ties in the ratio of 8:7. Calculate the percentage of bundles containing silk.
### Part:whole Ratios (video)
3. HEX color used for websites #428715 is based on 66:135:21 ratio of red and green to blue respectively. What part is green? Simplify your answer.
4. There are 52 cards in a deck. frac cards are picture cards. State the ratio of picture cards to non-picture cards in simplest form.
5. The football team won their league games. They did not draw any matches. Write the win:loss ratio.
#### Fractions Decimals Percents Worksheets
6. The number of beetles (B), slugs (S) and worms (W) eaten by a hedgehog was recorded in one night. frac Edible insects are beetles and frac slugs. Give the ratio of insects eaten by a hedgehog in simple form, B:S:W.
1. A market stall sells only apples, pears and bananas. The ratio of apples to pears is given by the composite numbers 5:2:1. Write the percentage of apples sold.
2. (a) The school orchestra is playing in a concert. 30% of ticket sales went to the public (P), \ frac sales went to musicians’ family and friends (F), and the rest to school staff (S).
### Chapter 6 Objectives Convert Among Fractions, Decimal Numbers, Ratios, And Percents Learn The Relationships Among Fractions, Decimal Numbers, Ratios, And.
B) If 440 tickets were sold to the public and employees, how many tickets were sold to members of the public only?
(b) Write the proportion of people wearing spectacles to the total number in simple form.
Prepare your KS4 students for Maths GCSE success with Third Space Learning. Review weekly online GCSE maths lessons from expert maths tutors.
## Average Ratios Of The Studied Metals In The Different Fractions Of The…
An essential guide for all SLTs and subject leaders
### Ratio And Proportion Applications
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This article was co-authored by staff. Our team of trained editors and researchers verify articles for accuracy and completeness. Our content management team closely monitors our editorial work to ensure that each article is supported by credible research and meets our high quality standards.
#### Fibonacci Numbers, Continued Fractions, And The Golden Ratio
You have already encountered fractions like 12}}. A ratio is a pair of fractions that are equal to each other, for example 12=24}=}}. There are many ways to solve proportion problems that ask you to find the missing number x, and you don’t need to learn them today. If you took pre-algebra and started using ratios, read on until you find a method that makes sense to you. If you’re doing algebra and working on more advanced ratio problems, you’ll need to move on to the next methods.
This article was co-authored by staff. Our team of trained editors and researchers verify articles for accuracy and completeness. Our content management team closely monitors our editorial work to ensure that each article is supported by credible research and meets our high quality standards. The article has been read 30,537 times.
To solve ratios, take the denominator or top number of the fraction you know and multiply it by the denominator or bottom number of the fraction you don’t know. |
# Points Defined by Adjacent Pairs of Digits of Reciprocal of 7 lie on Ellipse
## Theorem
Consider the digits that form the recurring part of the reciprocal of $7$:
$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$
Take the digits in ordered pairs, and treat them as coordinates of a Cartesian plane.
It will be found that they all lie on an ellipse:
## Proof
Let the points be labelled to simplify:
$A := \left({1, 4}\right)$
$B := \left({2, 8}\right)$
$C := \left({4, 2}\right)$
$D := \left({8, 5}\right)$
$E := \left({7, 1}\right)$
$F := \left({5, 7}\right)$
Let $ABCDEF$ be considered as a hexagon.
We join the opposite points of $ABCDEF$:
$AF: \left({1, 4}\right) \to \left({5, 7}\right)$
$BC: \left({2, 8}\right) \to \left({4, 2}\right)$
$BE: \left({2, 8}\right) \to \left({7, 1}\right)$
$AD: \left({1, 4}\right) \to \left({8, 5}\right)$
$CD: \left({4, 2}\right) \to \left({8, 5}\right)$
$EF: \left({7, 1}\right) \to \left({5, 7}\right)$
It is to be shown that the intersections of:
$AF$ and $BC$
$BE$ and $AD$
$CD$ and $EF$
all lie on the same straight line.
The result then follows from Pascal's Mystic Hexagram.
$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$
Thus:
$\text {(AF)}: \quad$ $\displaystyle \frac {y - 4} {x - 1}$ $=$ $\displaystyle \frac {7 - 4} {5 - 1}$ $\displaystyle$ $=$ $\displaystyle \frac 3 4$ $\displaystyle \leadsto \ \$ $\displaystyle 4 \left({y - 4}\right)$ $=$ $\displaystyle 3 \left({x - 1}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac 3 4 x + \dfrac {13} 4$
$\text {(BC)}: \quad$ $\displaystyle \frac {y - 8} {x - 2}$ $=$ $\displaystyle \frac {2 - 8} {4 - 2}$ $\displaystyle$ $=$ $\displaystyle -3$ $\displaystyle \leadsto \ \$ $\displaystyle y - 8$ $=$ $\displaystyle -3 \left({x - 2}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -3 x + 14$
$\text {(BE)}: \quad$ $\displaystyle \frac {y - 8} {x - 2}$ $=$ $\displaystyle \frac {1 - 8} {7 - 2}$ $\displaystyle$ $=$ $\displaystyle -\frac 7 5$ $\displaystyle \leadsto \ \$ $\displaystyle 5 \left({y - 8}\right)$ $=$ $\displaystyle -7 \left({x - 2}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle 5 y - 40$ $=$ $\displaystyle -7 x + 14$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -\frac 7 5 x + \frac {54} 5$
$\text {(AD)}: \quad$ $\displaystyle \frac {y - 4} {x - 1}$ $=$ $\displaystyle \frac {5 - 4} {8 - 1}$ $\displaystyle$ $=$ $\displaystyle \frac 1 7$ $\displaystyle \leadsto \ \$ $\displaystyle 7 \left({y - 4}\right)$ $=$ $\displaystyle x - 1$ $\displaystyle \leadsto \ \$ $\displaystyle 7 y - 28$ $=$ $\displaystyle x - 1$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \frac x 7 + \frac {27} 7$
$\text {(CD)}: \quad$ $\displaystyle \frac {y - 2} {x - 4}$ $=$ $\displaystyle \frac {5 - 2} {8 - 4}$ $\displaystyle$ $=$ $\displaystyle \frac 3 4$ $\displaystyle \leadsto \ \$ $\displaystyle 4 \left({y - 2}\right)$ $=$ $\displaystyle 3 \left({x - 4}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle 4 y - 8$ $=$ $\displaystyle 3 x - 12$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \frac 3 4 x - 1$
$\text {(EF)}: \quad$ $\displaystyle \frac {y - 1} {x - 7}$ $=$ $\displaystyle \frac {7 - 1} {5 - 7}$ $\displaystyle$ $=$ $\displaystyle -3$ $\displaystyle \leadsto \ \$ $\displaystyle y - 1$ $=$ $\displaystyle -3 x + 21$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -3 x + 22$
Evaluate the intersection of $AF$ and $BC$:
$\displaystyle y$ $=$ $\displaystyle \dfrac 3 4 x + \dfrac {13} 4$ $\displaystyle y$ $=$ $\displaystyle -3 x + 14$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac 3 4 x + \dfrac {13} 4$ $=$ $\displaystyle -3 x + 14$ $\displaystyle \leadsto \ \$ $\displaystyle 3 x + 13$ $=$ $\displaystyle -12 x + 56$ $\displaystyle \leadsto \ \$ $\displaystyle 15 x$ $=$ $\displaystyle 43$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \dfrac {43} {15}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -3 \left({\dfrac {43} {15} }\right) + 14$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {-129 + 210} {15}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {81} {15}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {27} 5$
So $AF$ and $BC$ intersect at $\left({\dfrac {43} {15}, \dfrac {27} 5}\right)$.
Evaluate the intersection of $BE$ and $AD$:
$\displaystyle y$ $=$ $\displaystyle -\frac 7 5 x + \frac {54} 5$ $\displaystyle y$ $=$ $\displaystyle \frac x 7 + \frac {27} 7$ $\displaystyle \leadsto \ \$ $\displaystyle -\frac 7 5 x + \frac {54} 5$ $=$ $\displaystyle \frac x 7 + \frac {27} 7$ $\displaystyle \leadsto \ \$ $\displaystyle -49 x + 378$ $=$ $\displaystyle 5 x + 135$ $\displaystyle \leadsto \ \$ $\displaystyle 54 x$ $=$ $\displaystyle 243$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \dfrac 9 2$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \frac 1 7 \left({\frac 9 2}\right) + \frac {27} 7$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {9 + 54} {14}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {63} {14}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac 9 2$
So $BE$ and $AD$ intersect at $\left({\dfrac 9 2, \dfrac 9 2}\right)$.
Evaluate the intersection of $CD$ and $EF$:
$\displaystyle y$ $=$ $\displaystyle \frac 3 4 x - 1$ $\displaystyle y$ $=$ $\displaystyle -3 x + 22$ $\displaystyle \leadsto \ \$ $\displaystyle \frac 3 4 x - 1$ $=$ $\displaystyle -3 x + 22$ $\displaystyle \leadsto \ \$ $\displaystyle \frac 3 x - 4$ $=$ $\displaystyle -12 x + 88$ $\displaystyle \leadsto \ \$ $\displaystyle 15 x$ $=$ $\displaystyle 92$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \dfrac {92} {15}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -3 \left({\dfrac {92} {15} }\right) + 22$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {-92 + 110} 5$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {18} 5$
So $CD$ and $EF$ intersect at $\left({\dfrac {92} {15}, \dfrac {18} 5}\right)$.
It remains to be shown that those points of intersection:
$\left({\dfrac {43} {15}, \dfrac {27} 5}\right)$, $\left({\dfrac 9 2, \dfrac 9 2}\right)$, $\left({\dfrac {92} {15}, \dfrac {18} 5}\right)$
all lie on the same straight line.
$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$
Thus:
$\displaystyle \frac {y - \frac {27} 5} {x - \frac {43} {15} }$ $=$ $\displaystyle \frac {\frac {18} 5 - \frac {27} 5} {\frac {92} {15} - \frac {43} {15} }$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {5 y - 27} {15 x - 43}$ $=$ $\displaystyle \frac {18 - 27} {92 - 43}$ simplifying $\displaystyle \leadsto \ \$ $\displaystyle \frac {5 y - 27} {15 x - 43}$ $=$ $\displaystyle -\frac 9 {49}$ $\displaystyle \leadsto \ \$ $\displaystyle 245 y - 1323$ $=$ $\displaystyle - 135 x + 387$ $\displaystyle \leadsto \ \$ $\displaystyle 245 y$ $=$ $\displaystyle - 135 x + 1710$ $\displaystyle \leadsto \ \$ $\displaystyle 49 y + 27 x$ $=$ $\displaystyle 342$
It remains to demonstrate that $\left({\dfrac 9 2, \dfrac 9 2}\right)$ lies on this line:
$\displaystyle 49 \dfrac 9 2 + 27 \dfrac 9 2$ $=$ $\displaystyle \dfrac {441 + 243} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {684} 2$ $\displaystyle$ $=$ $\displaystyle 342$
Bingo.
$\blacksquare$ |
# Lesson Video: Place Value of Three-Digit Numbers Mathematics • 2nd Grade
In this video, we will learn how to recognize numbers from a model of hundreds, tens, and ones and state the place value of digits.
13:57
### Video Transcript
Place Value of Three-Digit Numbers
In this video, we’re going to learn how to recognize numbers from a model of hundreds, tens, and ones. And we’re also going to learn how to state the place value of digits in a three-digit number. Let’s make a three-digit number out of place value blocks. To make our number, we’re going to need three different types of blocks: some hundreds, some tens, and some ones. Let’s take three hundreds, five tens, and we’ll have eight ones. What number have we made? We’ve got three hundreds blocks. So we know that these have a value of 300. Our five tens are worth 10, 20, 30, 40, 50. And of course, our eight ones are worth eight. So we’ve made the number 358.
Now, how can we write the number 358 using digits, like this? Can you see what we’ve done here? We’ve written the numbers 300, 50, and eight all in a long row. This isn’t correct though. It’s not how numbers work. A number that contains hundreds, tens, and ones only needs three digits. Remember the title of this video, Place Value of Three-Digit Numbers. We only need a digit to represent the number of hundreds, one to represent the number of tens, and one for the ones.
Now, we have three hundreds, five tens, and eight ones. So we can write the number 358 using the digits three, five, and eight. This is where these place value arrow cards come in really useful. When they’re split up like this, we can see the value of each digit. We have 300, 50, and eight. But if we push them together, we can see how to write the number using three digits. Three hundreds, five tens, and eight ones are 358.
Now, it’s really important that we understand the value of each digit when we’re working with three-digit numbers. Which of these digits is worth the most? Well, if we took these three cards and put them anywhere on the page, all we can see are the digits two, four, and nine separately. And of course, we’d say nine is the largest digit, isn’t it? But as soon as we put these digits together to make a three-digit number, the position or the place of each digit really matters.
Imagine for a moment that you’ve seen this lady. But the position or the place that you’ve seen her is sitting next to you on the bus with her shopping on her lap. You probably wouldn’t think she was a queen, would you? She looks like a queen, but she’s not in the right place or position. But if you saw her on a golden throne, there’s a different story. Now, her position tells you that she probably is the queen.
So to find out which of our three digits is worth the most, we don’t just look at the digits, but we look at what position they’re in. Out of the digits two, four, and nine, nine is the largest digit. But where is it in the number? It’s sitting in the ones place, isn’t it? Its value is nine ones. Now, on its own, the digit four is less than nine. But in this number, it’s sitting in the tens place. It has a value of four tens or 40.
So is the digit four worth the most? No, because although the digit two looks like the smallest digit if we looked at it on its own, it’s the king or queen of our number, isn’t it? We’ve put it in the most important position. It’s in the hundreds place. And so our tiny little digit two has a value of 200.
To find the digit with the largest value in this number, we need to look at the place that has the largest value. So we can say that in the number 249, the digit that’s worth the most is the two. The lowest, the largest digit on its own, the digit that’s worth the least in this number, is the nine, nine ones.
Let’s try answering some questions now where we have to think carefully about the place value of three-digit numbers. And remember, it’s all about the position we put our digits in.
Using a place value grid, we can figure out the value of each digit. To find the value of the three in 435, I can write the digits into the grid: four hundreds, three tens, five ones. The three is in the tens column. So I know the three has the value of three tens, which is the same as 30.
Use a grid to help you find the value of the six in 126.
We’re given a three-digit number in this problem, aren’t we? It’s the number 126. And of course, it’s made up out of the digits one, two, and six. Now, we’re being asked to find the value of the six in this number. What is this six digit worth? But we know that the value of a digit in a three-digit number depends on where it is in the number. Its position really matters.
Let’s draw a grid to help us, exactly the same as the one we’ve just looked at. As it’s a three-digit number, we’re going to need three columns for our hundreds, tens, and ones. And to help us remember what each one of these is worth, we could pop a hundreds block in the hundreds place, a tens block in the tens place, and a one in the ones place. Now, we can write our three-digit number in the grid, 126.
Rather than having just one block in each column, should we try to make the number? We already have one hundred, so that can stay the same. But we’ll put in one more tens so that we’ve got two tens. And five more ones will give us six ones. By using that place value grid to help us, we can see where the digit six is. It’s in the ones place, isn’t it? And because it’s in this position, we know what the six is worth. The value of the six in 126 is six ones.
Select the number that has a two in the hundreds place.
We’re given four numbers as possible answers to this question. Can you see? They’re all made up of three digits. And our job is to select or to choose the number that has the digit two in the hundreds place. Now, we know that the digits in three-digit numbers like this all have a different value. If we read each number from left to right, the digits are worth hundreds, tens, and ones.
Let’s look at the first number to begin with. It’s made up of a six followed by a two and then a three. This number has a two in it. But when we look at our grid, we can see that the two is in the tens place. This isn’t the number we’re looking for, is it? The digit two has a value of two tens or 20. And we read this number as 623.
Our second number starts off with a seven. So that’s a seven in the hundreds place. We can see straight away that this isn’t the number we’re looking for, is it? It has five tens. And this number also includes a digit two. But this time, it’s in the ones place. In this number, the digit two has a value of just two. And we’d read this number as 752.
Now, this next number is the number we’re looking for. But can you see why? Maybe if we write it on our grid, you’ll be able to see. We have two in the hundreds place. The rest of the number has a four in the tens place, which is the same as 40, and a five in the ones place, 245. This is the number that has a two in the hundreds place. And if we just check the last number, we can see this contains two twos. But neither of them are in the hundreds place, are they? We know that when we read a three-digit number from left to right, the first digit we see is the hundreds place. And the only one of these numbers that has a two in the hundreds place is 245.
Find the value of the nine in the number 192.
We’re given a three-digit number in this question. And it’s made up of the digits one, nine, and two. Now, we know that a digit’s position in a number gives it its value. And in this question, we’re thinking about the digit nine. It’s in the middle of our number, isn’t it? But what’s its value? Let’s use a place value grid to help us.
We know that three-digit numbers are made up of hundreds, tens, and ones. Now, our number, which is made up of a one and nine and a two, has a one in the hundreds place and nine in the tens place and the digit two in the ones place. We could also use place value equipment to model this number: one hundred, nine tens, and two ones.
Now, if we want to think about the digit nine in our number, we need to think about the middle column, don’t we? The nine is worth nine tens. Now, what’s the value of this? 10, 20, 30, 40, 50, 60, 70, 80, 90. We read our number as 192. And because the digit nine is in the tens place, we know that it has a value of 90.
In which of the following numbers does the digit nine have the smallest value?
What do you notice about the four numbers that we’re given in this problem? Well, you might notice that they’re all made up of three digits. They’re all three-digit numbers. But if we look more closely, what else can we see? They’re all made up of the same three digits, aren’t they? They all contain a one, a nine, and a two. And you know, we could make some of these numbers by taking digit cards that have a one, a nine, and a two on them and just swapping them around. Here’s our second number, our third number, and our last number.
So we can see that all four possible answers contain a one, a nine, and a two. But it’s the nine that we really want to be thinking about because our question asks us, in which of the numbers does the digit nine have the smallest value? And to be able to answer this question, we need to think about what the value of each place in a three-digit number is worth.
Now, we know that the digits in a three-digit number are worth hundreds, tens, and ones. Here’s a hundred. This is what a ten looks like. And here’s a one. Now, if we look at our four numbers, we can see that the digit nine sometimes appears in the hundreds place, sometimes in the middle of the number, which is the tens place, and sometimes in the ones place. Now, which is the smallest value, hundreds, tens, or ones?
Well, we know it’s the ones, don’t we? This is the digit on the right of the number. And can you see which of our numbers has a nine in the ones place? The number where the digit nine has a smallest value is this number here. It’s 129. We know that the smallest value place in a three-digit number is the ones place. And so we were looking for the number that had nine in the ones place. Instead of being worth 900 or 90, the nine in this number is just worth nine. The digit nine has the smallest value in the number 129.
What have we learned in this video? We’ve learned how to recognize numbers from a model of hundreds, tens, and ones. We’ve also learned how to recognize the place value of digits. |
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# Help with (b) I halfway got it correct but I think I am calculating it wrong. :/
0
193
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Help with (b) I halfway got it correct but I think I am calculating it wrong. :/
Sep 23, 2018
#1
+2340
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a)
Since these are composite figures, it is probably best to think about each composite figure as each individual shape that it is comprised of. Let's first discern the perimeter of the trapezoid.
We know three of the four sides, and the fourth side is of equal length to its opposite side. The fourth side is not exposed, so we can ignore it from the calculation. Just add these lengths together to obtain the perimeter. Multiply this perimeter by two because there are two trapezoids in this diagram.
$$P_{\text{trpzd}}=(13+9+7)\text{mm}\\ P_{\text{trpzd}}=29\text{mm}\\ 2P_{\text{trpzd}}=58\text{mm}$$
The only length left to determine is the arc length of the circular arc. The circumference of the circle is $$C=2\pi r$$, but we only want 44 of the 360 degrees of the circle. Add these perimeters together to find the final perimeter.
$$\text{Arclength}=2\pi r*\frac{44}{360}; r=9\text{mm}\\ \text{Arclength}=\frac{99\pi}{45}\text{mm}\approx6.9115\text{mm}$$
Add these perimeters together to find the final perimeter:
$$2P_{\text{trpzd}}+\text{Arclength}=58\text{mm}+6.9115\text{mm}\\ 2P_{\text{trpzd}}+\text{Arclength}=64.9115\text{mm}\approx64.91\text{mm}$$
And that's that!
b)
The base of the equilateral triangle is 7cm. The collinear radii of the circular arc are also 7cm. This means that
$$P_{\triangle\text{ & radii}}=(7+7+7)\text{cm}\\ P_{\triangle\text{ & radii}}=21\text{cm}$$
The only thing left to do is find the arc length, just like the previous problem. There are two arcs this time. Make sure to take that into account.
$$\text{Arclength}=2\pi r*\frac{120}{360}; r=7\text{cm}\\ \text{Arclength}=\frac{14\pi}{3}\text{cm}\\ 2\text{Arclength}=\frac{28\pi}{3}\text{cm}\approx29.3215\text{cm}$$
Find the sum of these perimeters again:
$$P_{\triangle\text{ & radii}}+\text{Arclength}=21\text{cm}+29.3215\text{cm}\\ P_{\triangle\text{ & radii}}+\text{Arclength}=50.3215\text{cm}\approx50.32\text{cm}$$
.
Sep 24, 2018 |
# How to Simplify a Square Root
Simplifying a square root isn't as hard as it looks. To simplify a square root, you just have to factor the number and pull the roots of any perfect squares you find out of the radical sign. Once you've memorized a few common perfect squares and know how to factor a number, you'll be well on your way to simplifying the square root.
Method 1
Method 1 of 3:
### Simplifying a Square Root by Factoring
1. 1
Understand factoring. The goal of simplifying a square root is to rewrite it in a form that is easy to understand and to use in math problems. Factoring breaks down a large number into two or more smaller factors, for instance turning 9 into 3 x 3. Once we find these factors, we can rewrite the square root in simpler form, sometimes even turning it into a normal integer. For example, √9 = √(3x3) = 3. Follow the steps below to learn this process for more complicated square roots.[1]
2. 2
Divide by the smallest prime number possible. If the number under the square root is even, divide it by 2. If your number is odd, try dividing it by 3 instead. If neither of these gives you a whole number, move down this list, testing the other primes until you get a whole number result. You only need to test the prime numbers, since all other numbers have prime numbers as their factors. For example, you don't need to test 4, because any number divisible by 4 is also divisible by 2, which you already tried.[2]
• 2
• 3
• 5
• 7
• 11
• 13
• 17
3. 3
Rewrite the square root as a multiplication problem. Keep everything underneath the square root sign, and don't forget to include both factors. For example, if you're trying to simplify √98, follow the step above to discover that 98 ÷ 2 = 49, so 98 = 2 x 49. Rewrite the "98" in the original square root using this information: √98 = √(2 x 49).[3]
4. 4
Repeat with one of the remaining numbers. Before we can simplify the square root, we keep factoring it until we've broken it down into two identical parts. This makes sense if you think about what a square root means: the term √(2 x 2) means "the number you can multiply with itself to equal 2 x 2." Obviously, this number is 2! With this goal in mind, let's repeat the steps above for our example problem, √(2 x 49):
• 2 is already factored as low as it will go. (In other words, it's one of those prime numbers on the list above.) We'll ignore this for now and try to divide 49 instead.
• 49 can't be evenly divided by 2, or by 3, or by 5. You can test this yourself using a calculator or long division. Because these don't give us nice, whole number results, we'll ignore them and keep trying.
• 49 can be evenly divided by seven. 49 ÷ 7 = 7, so 49 = 7 x 7.
• Rewrite the problem: √(2 x 49) = √(2 x 7 x 7).
5. 5
Finish simplifying by "pulling out" an integer. Once you've broken the problem down into two identical factors, you can turn that into a regular integer outside the square root. Leave all other factors inside the square root. For example, √(2 x 7 x 7) = √(2)√(7 x 7) = √(2) x 7 = 7√(2).[4]
• Even if it's possible to keep factoring, you don't need to once you've found two identical factors. For example, √(16) = √(4 x 4) = 4. If we kept on factoring, we'd end up with the same answer but have to do more work: √(16) = √(4 x 4) = √(2 x 2 x 2 x 2) = √(2 x 2)√(2 x 2) = 2 x 2 = 4.
6. 6
Multiply integers together if there are more than one. With some large square roots, you can simplify more than once. If this happens, multiply the integers together to get your final problem. Here's an example:
• √180 = √(2 x 90)
• √180 = √(2 x 2 x 45)
• √180 = 2√45, but this can still be simplified further.
• √180 = 2√(3 x 15)
• √180 = 2√(3 x 3 x 5)
• √180 = (2)(3√5)
• √180 = 6√5
7. 7
Write "cannot be simplified" if there are no two identical factors. Some square roots are already in simplest form. If you keep factoring until every term under the square root is a prime number (listed in one of the steps above), and no two are the same, then there's nothing you can do. You might have been given a trick question! For example, let's try to simplify √70:[5]
• 70 = 35 x 2, so √70 = √(35 x 2)
• 35 = 7 x 5, so √(35 x 2) = √(7 x 5 x 2)
• All three of these numbers are prime, so they cannot be factored further. They're all different, so there's no way to "pull out" an integer. √70 cannot be simplified.
Method 2
Method 2 of 3:
### Knowing the Perfect Squares
1. 1
Memorize a few perfect squares. Squaring a number, or multiplying it by itself, creates a perfect square. For example, 25 is a perfect square because 5 x 5, or 52, equals 25. Memorizing at least the first ten perfect squares can help you recognize and quickly simplify perfect square roots. Here are the first ten perfect squares:
• 12 = 1
• 22 = 4
• 32 = 9
• 42 = 16
• 52 = 25
• 62 = 36
• 72 = 49
• 82 = 64
• 92 = 81
• 102 = 100
2. 2
Find the square root of a perfect square. If you recognize a perfect square under a square root symbol, you can immediately turn it into its square root and get rid of the radical sign (√). For example, if you see the number 25 under the square root sign, you know that the answer is 5 because 25 is a perfect square. Here's the same list as above, going from the square root to the answer:[6]
• √1 = 1
• √4 = 2
• √9 = 3
• √16 = 4
• √25 = 5
• √36 = 6
• √49 = 7
• √64 = 8
• √81 = 9
• √100 = 10
3. 3
Factor numbers into perfect squares. Use the perfect squares to your advantage when following the factor method of simplifying square roots. If you notice a way to factor out a perfect square, it can save you time and effort. Here are some tips:[7]
• √50 = √(25 x 2) = 5√2. If the last two digits of a number end in 25, 50, or 75, you can always factor out 25.
• √1700 = √(100 x 17) = 10√17. If the last two digits end in 00, you can always factor out 100.
• √72 = √(9 x 8) = 3√8. Recognizing multiples of nine is often helpful. There's a trick to it: if all digits in a number add up to nine, then nine is always a factor.
• √12 = √(4 x 3) = 2√3. There's no special trick here, but it's usually easy to check whether a small number is divisible by 4. Keep this in mind when looking for factors.
4. 4
Factor a number with more than one perfect square. If the number's factors contain more than one perfect square, move them all outside the radical symbol. If you found multiple perfect squares during your simplification process, move all of their square roots to the outside of the √ symbol and multiply them together. For example, let's simplify √72:
• √72 = √(9 x 8)
• √72 = √(9 x 4 x 2)
• √72 = √(9) x √(4) x √(2)
• √72 = 3 x 2 x √2
• √72 = 6√2
Method 3
Method 3 of 3:
### Knowing the Terminology
1. 1
Know that the radical symbol (√) is the square root symbol. For example, in the problem, √25, "√" is the radical symbol.[8]
2. 2
Know that the radicand is the number inside the radical symbol. You will need to find the square root of this number. For example, in the problem √25, "25" is the radicand.[9]
3. 3
Know that the coefficient is the number outside the radical symbol. This is the number that the square root is being multiplied by; this sits to the left of the √ symbol. For example, in the problem, 7√2, "7" is the coefficient.
4. 4
Know that a factor is a number that can be evenly divided out of another number. For example, 2 is a factor of 8 because 8 ÷ 4 = 2, but 3 is not a factor of 8 because 8÷3 doesn’t result in a whole number. As another example, 5 is a factor of 25 because 5 x 5 = 25.
5. 5
Understand the meaning of simplifying a square root. Simplifying a square root just means factoring out any perfect squares from the radicand, moving them to the left of the radical symbol, and leaving the other factor inside the radical symbol. If the number is a perfect square, then the radical sign will disappear once you write down its root. For example, √98 can be simplified to 7√2.
## Community Q&A
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• Question
What is the simplest radical form?
wikiHow Staff Editor
This answer was written by one of our trained team of researchers who validated it for accuracy and comprehensiveness.
wikiHow Staff Editor
If something is written in its simplest radical form, that means that you have already found all possible roots and eliminated any radicals from the denominator of a fraction. A square root can’t be simplified any further if there are no 2 identical factors remaining and every term under the radical symbol is a prime number.
• Question
How do you simplify radical fractions?
wikiHow Staff Editor
This answer was written by one of our trained team of researchers who validated it for accuracy and comprehensiveness.
wikiHow Staff Editor
Start by multiplying the numerator and denominator by a radical that will eliminate the radical in the denominator. For example, if your fraction is 2/√7, multiply by √7/√7 to get 2√7/√49. This will simplify to 2√7/7. If possible, simplify the fraction by dividing out any common factors in the numerator and denominator.
• Question
How do you simplify square roots with variables?
wikiHow Staff Editor
This answer was written by one of our trained team of researchers who validated it for accuracy and comprehensiveness.
wikiHow Staff Editor
Use the same procedure you would for simplifying square roots with numerical radicands. Bring any identical pairs of factors out of the radical to become the coefficient. For example, if you need to simplify √25a^3, change it to √5×5×a×a×a. Factor out 5 and a^2 to get 5a^2√a.
200 characters left
## Tips
• One way to find perfect squares that factor into a number is to look through the list of perfect squares, beginning with the one that is the next smallest compared to your radicand, or the number under the square root sign. For example, when looking for the perfect square that goes into 27, you might start at 25 and go down the list to 16 and stop at 9, when you found the one that divides into 27.
⧼thumbs_response⧽
## Warnings
• Calculators can be useful for large numbers, but the more you practice working this out on your own, the easier this will get.
⧼thumbs_response⧽
• Simplifying is not the same as evaluating. At no point in this process should you get a number with a decimal point in it!
⧼thumbs_response⧽
Co-authored by:
Co-authors: 81
Updated: December 2, 2022
Views: 1,364,551
Article SummaryX
To simplify a square root, start by dividing the square root by the smallest prime number possible. For example, if you're trying to find the square root of 98, the smallest prime number possible is 2. If you divide 98 by 2, you get 49. Then, rewrite the square root as a multiplication problem under the square root sign. In this case, you'd rewrite the square root as 2 × 49 under the square root sign. From there, keep factoring the numbers until you have 2 identical factors. In this example, 49 divided by 7 is 7. Rewrite the square root as 2 × 7 × 7. Finally, once you have two identical numbers, move them outside of the square root to make them a regular integer. So the simplified square root of 98 is 7 × the square root of 2. However, if you factor the numbers inside the square root as much as you can without getting two identical numbers, then your square root can't be simplified! To learn other ways you can simplify a square root, keep reading!
Thanks to all authors for creating a page that has been read 1,364,551 times. |
Base Systems
Table of contents Introduction Conversions From One Base System to Another Arithmetic Operations While Operating in a Specific Base System Important Concepts of Base System Previous Year Questions based on Base System
## Introduction
• The numbers that we commonly use are the decimal number. The system is called the decimal system. Why is it called the decimal system? It is because it has 10 symbols : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Example: 85 means: 8 * 101 + 5 * 100
• Here, 10 is known as the base of the decimal system.
• Base: It is the number of distinct symbols used in a particular number system.
So, depending on the number of digits in the base system, there are many other systems possible. Have a look at the following table:
Classification of Number System
• In the questions involving base systems, we are expected to figure out the different numbers of digits in that system of counting.
• Example: If we had 7 digits instead of 10 i.e. 0, 1, 2, 3, 4, 5, 6, we would have a system of counting as follows: 0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 13, 14… .
In above system , 45 will be expanded as = 4 * 91 + 5 * 90
Questions in MBA entrance exams are based on two concepts:
1. Conversion of numbers from one base system to another.
2. Arithmetic Operations of numbers in any base system.
## Conversions From One Base System to Another
### 1. Decimal to Binary
(a) (10.25)10
Note: Keep multiplying the fractional part with 2 until decimal part 0.00 is obtained.
(0.25)10 = (0.01)2
(b) (28.3125)10
The given number has 2 parts:
(i) Conversion of an integral part:
(28)10 = (11100)2
(ii) Conversion of the fractional part:
• Multiply the decimal part with 2 successively and take the integral part of all the products starting from the first.
• 0.3125*2 = 0.6250 → (0)
• 0.6250*2 = 1.2500 → (1)
• 0.2500*2 = 0.500 → (0)
• 0.500*2 = 1 → (1)
• Therefore,
• (0.3125)10 = (0.0101)2
• (28.3125)10 = (11100.0101)2
Note: We should stop multiplying the factorial part by 2, once we get 0 as a fraction or the fractional part is non-terminating. It can be decided depending on the number of digits in the fractional part required.
### 2. Binary to Decimal
• (110101)2 = 1*25+ 1*24+ 0*23+ 1*22+ 0*21+ 1*20 = (53)10
• (0.11001)= 1*2-1+ 1*2-2+ 0*2-3+ 0*2-4+ 1*2-5 = (0.78125)10
Question for Base Systems
Try yourself:Convert the binary equivalent 10101 to its decimal equivalent.
### 3. Decimal to Octal
(a) (4324.235)10
(i) Integral Part
(4324)10 = (10344)8
(ii) Fractional Part
• 0.235*8 = 1.88 → (1)
• 0.88*8 = 7.04 → (7)
• 0.04*8 = 0.32 → (0)
• We can stop here as the fraction is non-terminating.
• (4324.235)10 = (10344.170)8
Question for Base Systems
Try yourself:What could be the maximum value of a single digit in an octal number system?
### 4. Octal to Decimal
• (677.47)8
= 6*82+ 7*81+ 7*80+ 4*8-1+ 7*8-2 = (447.609375)10
• (23524)10
(23524)10 = (5BE4)16
• (62A)16 = 6*162+ 2*161+ 10*160 = (1578)10
Now, we will look at conversions which do not include base 10. We can observe that the following cases include 2 bases (let them be a and b), where a=bn.
### 7. Binary to Octal
• (octal) is a cube of 2 (binary). So, to convert binary to octal, we just need to club three digits of binary number starting from unit digit and write the decimal equivalent of each group.
• (1101001011)2 = (001 101 001 011)2
(We can introduce two zeroes to form groups of 3 without changing the magnitude of the number)
(001 101 001 011)= (1513)8
### 8. Octal to Binary
Here, we need to express every digit of octal number into its binary form comprising of 3 digits.
• (645)8
• 6 = (110)2
• 4 = (100)2
• 5 = (101)2
• (645)8 = (110 100 101)2
### 9. Binary to Hexa-Decimal
• It is similar to the method discussed in the 7th point. Instead of clubbing 3, we club 4 digits.
• (10101110)2 = (1010 1110)2 = (AE)16
### 10. Hexa-Decimal to Binary
• We need to express every digit of Hexa-decimal into its binary form comprising of 4 digits.
• (3A91)16
• 3 = (0011)2
• A = (1010)2
• 9 = (1001)2
• 1 = (0001)2
• (3A91)16 = (11101010010001)2
Note: Using the logic discussed in points 7 to 10, we can do direct conversions in any two bases a and b such that a=bn where we will form blocks of n digits when the number is in base b and then write its decimal equivalent.
Example: For conversion from base 3 to base 9, we need to make blocks of 2 digits as 9 = 32, for instance- (22112)3 = (02 21 12)3 = (275)9
## Arithmetic Operations While Operating in a Specific Base System
• One method of performing arithmetic operations is to first convert the numbers to the decimal system, perform the required operations and then convert the numbers back to the required base.
• However, arithmetic operations can also be directly done given that numbers are expressed in the same base.
(a) (8358)10 + (5684)10
(8358)10 + (5684)10 = (14042)10
➢ Logic
• Start from a unit position; 8+4=12, and 12, when divided by 10 (base), gives us remainder as 2, which is the total for that column. 12, when divided by 10, gives us quotient as 1, which is carried over to the next column.
• In the second column, 1+5+8=14. When 14 is divided by 10 (base), we get 4 as remainder (total for that column) and 1 as quotient which is carried over to next column.
• In third column, 1 + 3 + 6 = 10. 10, when divided by 10, gives us 0 as remainder and 1 as quotient. We proceed in similar and get the required sum.
(b) (3542)6 + (4124)6
(3542)+ (4124)6 = (12110)6
### 2. Subtraction
(a) (237)10 – (199)10
(237)10 – (199)10 = (38)10
➢ Logic
• In units place 7 < 9, hence we borrow 10 (base) from tens place. 17-9=8
• Tens place reduces from 3 to 2. Still 2<9, hence we borrow 10 (base) from hundreds place. 12-9=3
• Hundreds place reduces to 1 (as we borrowed from it in an earlier step). 1-1=0
(b) (422)5 – (243)5
(422)5 – (243)5 = (124)5
### 3. Multiplication
For multiplying numbers in any base system, multiply them as we normally do for decimal numbers and while writing, write each number in the given base system
Example: Calculate (52)8 × (6)8
Solution:
Given, (52)8 × (6)8
We first multiply 6 with 2 i.e., 2 × 6 = 12 and write it in base 8 = (14)8.
Now, 4 will be the unit's digit of the final answer and 1 will be carried forward.
Now we multiply 6 with 5 and add any carry forward i.e., 5 × 6 + 1 = 31 and write it in base 8 = (37)6
Now, 37 will be the leftmost digit of the final answer.
∴ (52)8 × (6)8 = (374)8
### 4. Division
• Division of a number in base n by (n - 1)
(x)n is divisible by (n - 1) if the sum of all the digits of (x)n is divisible by (n - 1)
• Division of a number in base n by (n + 1)
(x)n is divisible by (n + 1) if the difference of the sums of alternate digits of (x)is either 0 or divisible by (n + 1)
• Number of zeroes at the end of a number in base n
For a number in base n, if there are k zeroes in the end then it is divisible by nk. Also, k is the highest power of n in the number.
(a) Is (7364)divisible by 8?
• The logic behind this question is same as checking the divisibility of any number (in decimal system) by 9. We add the digits and then check the divisibility.
• 7 + 3 + 6 + 4 = 20 which is not divisible by 8. Hence, the given number is not divisible by 8.
Rule: (x)b is divisible by (b-1) if all the digits of (x)b add up to be divisible by (b-1).
(b) Is (5236)9 divisible by 10?
• The logic behind this question is same as checking the divisibility of any number (in decimal system) by 11.
• We first find the sum of alternate digits and then find the difference of the sums obtained. This difference should either be divisible by 0 or divisible by 11(or 10 in the case of this question).
• 5 + 3 = 8; 2 + 6 = 8
• 8 - 8 = 0. Hence, the number is divisible by 10.
Rule: (x)b is divisible by (b+1) if the difference of the sums of alternate digits of (x)b is either 0 or divisible by (b+1).
(c) What is the IGP (Index of Greatest Power) of 9 in (780)9?
• (780)9 = 7*92+ 8*91+ 0*90 = 9(7*9+ 8)
• Thus IGP= 1. As the highest power of 9 with which the number is divisible is 1.
Rule: For a number in base b, if there are k zeroes in the end then it is divisible by bk. Also, k is the IGP of b in the number.
## Important Concepts of Base System
Example 1. Find the fifth root of (15AA51)19
• (15AA51)19 = 1*195+ 5*194+ 10*193+ 10*192+ 5*191+ 1*19= (19+1)5 = 20(Using binomial theorem)
• Therefore, the fifth root is 20.
Other examples of similar kind are:
• (121)n = n2+ 2n+ 1= (n+1)(n>2)
• (1331)= n3+ 3n2+ 3n+ 1= (n+1)3 (n>3)
• (14641)n, (15AA51)n and so on (the digits of the numbers used in the above examples form Pascal’s triangle)
Example 2. How many 4-digit numbers in base 9 are perfect squares?
• First, we need to know the range of 4-digit numbers in base 9
• Least 4 digit number possible= (1000)= 93 =729
• Observation: Lowest n digit number in base k = k(n-1)
• Highest 4 digit number possible= (8888)9 = 94-1= 6560
• Observation: Highest n digit number in base k = kn-1
• From 729 to 6560, the squares vary from 272 to 802.
• Number of perfect squares present = 80 - 26 = 54.
## Previous Year Questions based on Base System
Q.1. Let a, b, c be distinct digits. Consider a two-digit number ‘ab’ and a three-digit number ‘ccb’, both defined under the usual decimal number system, if (ab)2 = ccb > 300, then the value of b is? (CAT 1999)
(a) 1
(b) 0
(c) 5
(d) 6
• (ab)2 = ccb, the greatest possible value of ‘ab’ can be 31, since 312 = 961 (and since ccb > 300), 300 < ccb < 961, so 18 < ab < 31.
• So the possible value of ab which satisfies (ab)2 = ccb is 21.
• So 212 = 441, a = 2, b = 1, c = 4.
Q.2. Convert the number 1982 from base 10 to base 12. The result is? (CAT 2000)
(a) 1182
(b) 1912
(c) 1192
(d) 1292
Q.3. In a number system, the product of 44 and 11 is 1034. The number 3111 of this system, when converted to the decimal number system, becomes? (XAT 2001)
(a) 406
(b) 1086
(c) 213
(d) 691
(e) None of the above
• Let the base be n
• (4n+4)(n+1) = n3+3n+4
• n3-4n2-5n = 0
• n(n-5)(n+1) = 0
• n = 5
• (3111)= (406)10
Q.4.The product of two numbers 231 and ABA is BA4AA in a certain base system (where the base is less than 10), where A and B are distinct digits. What is the base of that system? (CAT 2010)
(a) 5
(b) 6
(c) 7
(d) 8
(e) 4
• Hence ,we can write (b+4+a)-2a = 0 or (base)+ 1 (let’s take (base+1))
i.e. b = base + a – 3
• 231 * aba = 2a(base)4+(2b+3a)(base)3+(3a+3b)(base)2+(3a+b)(base)+a
• Now put b = base + a – 3 , in above equation and Compare it with ba4aa, We get:
• 2a + 2 = b
• 4a – 3 = a
• Solving them gives a = 1 , b = 4
• Hence, base = b - a + 3 = 6
The document Base Systems | Quantitative Aptitude (Quant) - CAT is a part of the CAT Course Quantitative Aptitude (Quant).
All you need of CAT at this link: CAT
## Quantitative Aptitude (Quant)
185 videos|158 docs|113 tests
## FAQs on Base Systems - Quantitative Aptitude (Quant) - CAT
1. What is a base system?
Ans. A base system, also known as a number system, is a way of representing numbers using a specific set of digits. The most common base system is the decimal system, which uses 10 digits (0-9). Other popular base systems include binary (base 2), octal (base 8), and hexadecimal (base 16).
2. How do you convert a number from one base system to another?
Ans. To convert a number from one base system to another, you can use the concept of place value. First, express the number in the original base system as a sum of powers of the base. Then, rewrite each power of the base in the new base system. Finally, sum up the new representations to get the converted number.
3. How do you perform arithmetic operations in a specific base system?
Ans. To perform arithmetic operations in a specific base system, you need to follow the rules of that base system. For addition and subtraction, you can simply carry or borrow as needed. For multiplication and division, you can use the standard algorithms but make sure to perform the operations in the given base system.
4. What are some important concepts of the base system?
Ans. Some important concepts of the base system include the concept of place value, where the position of a digit determines its value, and the concept of carrying or borrowing when performing arithmetic operations. Understanding these concepts is crucial for working with different base systems.
5. Can you provide some previous year questions based on the base system?
Ans. Here are some previous year questions based on the base system: 1. Convert the binary number 1011 to its decimal equivalent. 2. Perform the multiplication 43 * 12 in base 5. 3. What is the hexadecimal representation of the decimal number 255? 4. Subtract the octal number 75 from the decimal number 100. 5. Divide the binary number 1101 by the decimal number 5.
## Quantitative Aptitude (Quant)
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# Angle Properties, Postulates, and Theorems
In order to study geometry
in a logical way, it will be important to understand key mathematical properties
and to know how to apply useful postulates and theorems. A postulate is a
proposition that has not been proven true, but is considered to be true on the basis
for mathematical reasoning. Theorems, on the other hand, are statements that
have been proven to be true with the use of other theorems or statements. While
some postulates and theorems have been introduced in the previous sections, others
are new to our study of geometry. We will apply these properties, postulates, and
theorems to help drive our mathematical proofs in a very logical, reason-based way.
Before we begin, we must introduce the concept of congruency. Angles are congruent
if their measures, in degrees, are equal. Note: “congruent” does not
mean “equal.” While they seem quite similar, congruent angles do not have to point
in the same direction. The only way to get equal angles is by piling two angles
of equal measure on top of each other.
## Properties
We will utilize the following properties to help us reason through several geometric
proofs.
### Reflexive Property
A quantity is equal to itself.
### Symmetric Property
If A = B, then B = A.
### Transitive Property
If A = B and B = C, then A = C.
If A = B, then A + C = B + C.
## Angle Postulates
If a point lies on the interior of an angle, that angle is the sum of two smaller
angles with legs that go through the given point.
Consider the figure below in which point T lies on the interior of
?QRS. By this postulate, we have that ?QRS = ?QRT + ?TRS.
We have actually applied this postulate when we practiced finding the complements
and supplements of angles in the previous section.
### Corresponding Angles Postulate
If a transversal intersects two parallel lines, the pairs of corresponding
angles are congruent.
Converse also true: If a transversal intersects two lines and the corresponding
angles are congruent, then the lines are parallel.
The figure above yields four pairs of corresponding angles.
### Parallel Postulate
Given a line and a point not on that line, there exists a unique line through the
point parallel to the given line.
The parallel postulate is what sets Euclidean geometry apart from non-Euclidean geometry.
There are an infinite number of lines that pass through point E, but only
the red line runs parallel to line CD. Any other line through E will
eventually intersect line CD.
## Angle Theorems
### Alternate Exterior Angles Theorem
If a transversal intersects two parallel lines, then the alternate exterior
angles are congruent.
Converse also true: If a transversal intersects two lines and the alternate
exterior angles are congruent, then the lines are parallel.
The alternate exterior angles have the same degree measures because the lines are
parallel to each other.
### Alternate Interior Angles Theorem
If a transversal intersects two parallel lines, then the alternate interior
angles are congruent.
Converse also true: If a transversal intersects two lines and the alternate
interior angles are congruent, then the lines are parallel.
The alternate interior angles have the same degree measures because the lines are
parallel to each other.
### Congruent Complements Theorem
If two angles are complements of the same angle (or of congruent angles), then the
two angles are congruent.
### Congruent Supplements Theorem
If two angles are supplements of the same angle (or of congruent angles), then the
two angles are congruent.
### Right Angles Theorem
All right angles are congruent.
### Same-Side Interior Angles Theorem
If a transversal intersects two parallel lines, then the interior angles
on the same side of the transversal are supplementary.
Converse also true: If a transversal intersects two lines and the interior
angles on the same side of the transversal are supplementary, then the lines are
parallel.
The sum of the degree measures of the same-side interior angles is 180°.
### Vertical Angles Theorem
If two angles are vertical angles, then they have equal measures.
The vertical angles have equal degree measures. There are two pairs of vertical angles.
## Exercises
(1) Given: m?DGH = 131
Find: m?GHK
First, we must rely on the information we are given to begin our proof. In this
exercise, we note that the measure of ?DGH is 131°.
From the illustration provided, we also see that lines DJ and EK
are parallel to each other. Therefore, we can utilize some of the angle theorems
above in order to find the measure of ?GHK.
We realize that there exists a relationship between ?DGH and ?EHI:
they are corresponding angles. Thus, we can utilize the Corresponding Angles Postulate
to determine that ?DGH??EHI.
Directly opposite from ?EHI is ?GHK. Since they are
vertical angles, we can use the Vertical Angles Theorem, to see that ?EHI??GHK.
Now, by transitivity, we have that ?DGH??GHK.
Congruent angles have equal degree measures, so the measure of ?DGH
is equal to the measure of ?GHK.
Finally, we use substitution to conclude that the measure of ?GHK
is 131°. This argument is organized in two-column proof form below.
(2) Given: m?1 = m?3
Prove: m?PTR = m?STQ
We begin our proof with the fact that the measures of ?1 and ?3
are equal.
In our second step, we use the Reflexive Property to show that ?2
is equal to itself.
Though trivial, the previous step was necessary because it set us up to use the
Addition Property of Equality by showing that adding the measure of ?2
to two equal angles preserves equality.
Then, by the Angle Addition Postulate we see that ?PTR is the
sum of ?1 and ?2, whereas ?STQ is the
sum of ?3 and ?2.
Ultimately, through substitution, it is clear that the measures of ?PTR
and ?STQ are equal. The two-column proof for this exercise is shown
below.
(3) Given: m?DCJ = 71, m?GFJ = 46
Prove: m?AJH = 117
We are given the measure of ?DCJ and ?GFJ to begin the
exercise. Also, notice that the three lines that run horizontally in the illustration
are parallel to each other. The diagram also shows us that the final steps of our
proof may require us to add up the two angles that compose ?AJH.
We find that there exists a relationship between ?DCJ and ?AJI:
they are alternate interior angles. Thus, we can use the Alternate Interior Angles
Theorem
to claim that they are congruent to each other.
By the definition of congruence, their angles have the same measures, so
they are equal.
Now, we substitute the measure of ?DCJ with 71
since we were given that quantity. This tells us that ?AJI is also
71°.
Since ?GFJ and ?HJI are also alternate interior angles,
we claim congruence between them by the Alternate Interior Angles Theorem.
The definition of congruent angles once again proves that the angles have equal
measures. Since we knew the measure of ?GFJ, we just substitute
to show that 46 is the degree measure of ?HJI.
As predicted above, we can use the Angle Addition Postulate to get the sum
of ?AJI and ?HJI since they compose ?AJH.
Ultimately, we see that the sum of these two angles gives us 117°.
The two-column proof for this exercise is shown below.
(4) Given: m?1 = 4x + 9, m?2 = 7(x + 4)
Find: m?3
In this exercise, we are not given specific degree measures for the angles shown.
Rather, we must use some algebra
to help us determine the measure of ?3. As always, we begin with the
information given in the problem. In this case, we are given equations for the measures
of ?1 and ?2. Also, we note that there exists two pairs
of parallel lines in the diagram.
By the Same-Side Interior Angles Theorem, we know that that sum of ?1
and ?2 is 180 because they are supplementary.
After substituting these angles by the measures given to us and simplifying,
we have 11x + 37 = 180. In order to solve for x, we
first subtract both sides of the equation by 37, and then divide both sides by 11.
Once we have determined that the value of x is 13, we plug it back in to the equation for the measure
of ?2 with the intention of eventually using the Corresponding Angles
Postulate
. Plugging 13 in for x gives us a measure of
119 for ?2.
Finally, we conclude that ?3
must have this degree measure as well since ?2 and ?3
are congruent. The two-column proof that shows this argument is shown below.
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# Find a point which is equidistant from
Question:
Find a point which is equidistant from the points A (- 5, 4) and B (- 1, 6). How many such points are there?
Solution:
Let P (h, k) be the point which is equidistant from the points A (- 5, 4) and B (-1, 6).
$\therefore \quad P A=P B \quad\left[\because\right.$ by distance formula, distance $=\sqrt{\left.\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}$
$\Rightarrow \quad(P A)^{2}=(P B)^{2}$
$\Rightarrow \quad(-5-h)^{2}+(4-k)^{2}=(-1-h)^{2}+(6-k)^{2}$
$\Rightarrow \quad 25+h^{2}+10 h+16+k^{2}-8 k=1+h^{2}+2 h+36+k^{2}-12 k$
$\Rightarrow \quad 25+10 h+16-8 k=1+2 h+36-12 k$
$\Rightarrow \quad 8 h+4 k+41-37=0$
$\Rightarrow \quad 8 h+4 k+4=0$
$\Rightarrow \quad 2 h+k+1=0$ $\ldots$ (i)
Mid-point of $A B=\left(\frac{-5-1}{2}, \frac{4+6}{2}\right)=(-3,5)$
$\left[\because\right.$ mid-point $\left.=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\right]$
At point $(-3,5)$, from Eq. (i),
$2 h+k=2(-3)+5$
$=-6+5=-1$
$\Rightarrow \quad 2 h+k+1=0$
So, the mid-point of AB satisfy the Eq. (i). Hence, infinite number of points, in fact all points which are solution of the equation 2h + k +1 = 0, are
equidistant from the points A and B.
Replacing h, k by x, y in above equation, we have 2 x +y+1= 0
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# Find the eigenvalues, eigenvectors and determinant of a matrix with unknowns along diagonal, first row and first column?
How do I find the eigenvalues, eigenvectors and determinant of the matrix
$$\begin{matrix} a & b & b & ... & b \\ c & a & 0 & ... & 0 \\ c & 0 & a & ... & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ c & 0 & 0 & ... & a \\ \end{matrix}$$
?
I think it has something to do with adding a matrix of form $cI$, which alters the eigenvalues and makes it easier to find the eigenvalues. But I'm not sure how to do this. Would really appreciate any help!
Let $A$ be your given matrix. Then $A=B+aI$ where $$B=\pmatrix{0 & b & b & \dots & b \\ c & 0 & 0 & \dots & 0 \\ c & 0 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots &\vdots \\ c & 0 & 0 & \dots & 0}$$ If you can find the eigenvalues and eigenvectors of this simpler matrix $B$, then you can easily find the eigenvalues and eigenvectors of $A$. Namely, the eigenvectors of $A$ will be identical to those of $B$, while the eigenvalues of $B$ will be $\lambda_1+a,\dots,\lambda_n+a$ where $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $B$.
To see why this works, suppose that $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, so $Av=\lambda v$. Then $$Bv = (A+aI)v = Av + aIv = \lambda v + av = (\lambda +a)v$$ so $v$ is an eigenvector of $B$ with eigenvalue $\lambda+a$.
From here, to get the eigenvalues and eigenvectors of $B$, you could probably just do it directly (i.e., compute the characteristic polynomial, etc.). Another approach though, which might be easier, is to first apply a similarity transformation: Define $C=SBS^{-1}$, where $$S=\pmatrix{1 & 0 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & 0& \dots & 0 \\ 0 & -1 & 1 & 0 & \dots & 0 \\0 & -1 & 0 & 1&\dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots&\vdots \\ 0 & -1 & 0 & 0 & \dots & 1}$$ The idea is that $S$ performs the necessary row operations to $B$ in order to cancel out all but one of instances of $c$. You should then be able to find the characteristic polynomial of $C$ directly and find its eigenvalues and eigenvectors of $C$. The eigenvalues of $B$ will be the same, and the eigenvectors of $B$ will have the form $S^{-1}v$, as $v$ ranges over the eigenvectors of $C$.
• yes I got this, but I was wondering how do you find the eigenvalues and eigenvectors of matrix B? – thbcm Feb 27 '15 at 21:08
• I've added a bit more to my answer now to explain how this can be done. – Brent Kerby Feb 27 '15 at 21:45
I can offer some help for the determinant (sorry not high enough to just comment). Use Block matrices!
Let: $A = a$ (a $1x1$ matrix)
$B = [b \ b \ b \ ...]$ (your top row less the first $a$)
$C = \begin{bmatrix} c\\ c\\ c\\ .\\ .\\ \end{bmatrix}$
and $D = aI$
Then from a formula in the above link:
$Det\begin{pmatrix} A & B\\ C & D \end{pmatrix} = Det(A)Det(D-CA^{-1}B)$
Hope this helps (with the determinant at least). |
## How do you find the point slope equation?
What does it stand for?(x1, y1) is a known point.m is the slope of the line.(x, y) is any other point on the line.Slope m = change in y change in x = y − y1 x − x1So point (x1, y1) is actually at (0, b)
## What is point Slope example?
Examples of Applying the Concept of Point-Slope Form of a Line. Example 1: Write the point-slope form of the line with a slope of 3 which passes through the point (2,5). The slope is given as m = 3 m = 3 m=3, and the point (2,5) has coordinates of x 1 = 2 {x_1} = 2 x1=2 and y 1 = 5 {y_1} = 5 y1=5.
## What is standard form slope?
Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3. A, B, C are integers (positive or negative whole numbers) No fractions nor decimals in standard form.
## What is the Y intercept formula?
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.
## How do I find the slope of a line?
The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .
## What are the 3 slope formulas?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form.
## How do you do slope intercept form?
1) start with the point slope form y-y1=m(x-x1) and solve for y. y-8=2(x-6) so y-8=2x-12, add 8 to end up with y=2x-4. 2) start with slope intercept form and solve for b, then substitute that back in: y=mx + b, so 8=2(6)+b, 8=12+b, subtract 12 to get b=-4, therefore equation is y=2x-4.
## How do you write an equation given two points?
Find the Equation of a Line Given That You Know Two Points it Passes Through. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
## Is C the slope in standard form?
Lesson Summary When we have a linear equation in slope-intercept form, it’s easy to identify the slope of the line as the number in front of x. The standard form of a linear equation is Ax + By = C.
### Releated
#### Depreciation equation
What are the 3 depreciation methods? There are three methods for depreciation: straight line, declining balance, sum-of-the-years’ digits, and units of production. What do you mean by depreciation? Definition: The monetary value of an asset decreases over time due to use, wear and tear or obsolescence. This decrease is measured as depreciation. How do you […]
#### Polar to cartesian equation calculator wolfram
How do you convert polar to Cartesian? Summary: to convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) 😡 = r × cos( θ )y = r × sin( θ ) How do you find the polar Cartesian equation? Convert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding […] |
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# Module 1 lesson 4
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### Module 1 lesson 4
1. 1. Module 1 Lesson 4.notebook 1 September 16, 2013 Homework: Lesson 4 Problem Set and CRS + J #1 Sept. 16th Aim: Identifying Proportional and Non-proportional relationships in Tables Module 1, Lesson 4 DO NOW: Check HW Answers 1.) Yes, Proportional 2.) Not Proportional 3.) Yes Proportional 4.) Yes Proportional 5.) Yes Proportional 6.) Yes Proportional 7.) Not Proportional 8.) Not Proportional 9.) Answers will vary QUIZ 1 FRIDAY
2. 2. Module 1 Lesson 4.notebook 2 September 16, 2013 Problem Set 3 Solutions
3. 3. Module 1 Lesson 4.notebook 3 September 16, 2013 Problem Set 3 Solutions
4. 4. Module 1 Lesson 4.notebook 4 September 16, 2013 Problem Set 3 Solutions
5. 5. Module 1 Lesson 4.notebook 5 September 16, 2013
6. 6. Module 1 Lesson 4.notebook 6 September 16, 2013 Complete the following class work independently. Which Team Will Win The Race?? • You have decided to run in a long distance race. There are two teams that you can join. • Team A runs at a constant rate of 2.5 miles per hour. • Team B runs 4 miles the first hour and then 2 miles per hour after that. Task: Create a table for each team showing the distances that would be run for times of 1, 2, 3, 4, ,5 and 6 hours.
7. 7. Module 1 Lesson 4.notebook 7 September 16, 2013
8. 8. Module 1 Lesson 4.notebook 8 September 16, 2013 a. For which team is distance proportional to time? Explain your reasoning. b. Explain how you know distance for the other team is not proportional to time.
9. 9. Module 1 Lesson 4.notebook 9 September 16, 2013 c. If the race were 2.5 miles long, which team would win? Explain. If the race were 3.5 miles long, which team would win? Explain. If the race were 4.5 miles long, which team would win? Explain.
10. 10. Module 1 Lesson 4.notebook 10 September 16, 2013 d. For what length race would it be better to be on Team B than Team A? Explain e. Using this relationship, if the members on the team ran for 10 hours, how far would each member run on each team?
11. 11. Module 1 Lesson 4.notebook 11 September 16, 2013 f. Will there always be a winning team, no matter what the length of the course? Why or why not? g. If the race is 12 miles long, which team should you choose to be on if you wish to win? Why would you choose this team? h. How much sooner would you finish on that team compared to the other team?
12. 12. Module 1 Lesson 4.notebook 12 September 16, 2013 Closing How does knowing two quantities are proportional help answer questions about the quantities? For example, if we know 1 cup = 8 oz, what does that allow us to do?
13. 13. Module 1 Lesson 4.notebook 13 September 16, 2013 EXIT TICKET Complete the table. If Gabby wants to make an octagon with a side length of 20 inches using wire, how much wire does she need? Justify your reasoning with an explanation of whether perimeter is proportional to the side length. The table below shows the relationship between the side lengths of a regular octagon and its perimeter.
14. 14. Module 1 Lesson 4.notebook 14 September 16, 2013 |
# Regrouping in Subtraction
This is a complete lesson about how to teach regrouping in subtraction (borrowing) step-by-step with 2-digit numbers, meant for 2nd grade. The lesson contains a teaching video, instruction with visual models, and many exercises.
In the video below, I explain an idea of breaking down the concept of regrouping (borrowing) so that students can see what actually happens in it. For example, to subtract 52 − 38, we write 52 as 50 + 2 (breaking it down into its tens and ones). Then, regrouping means that 50 + 2 becomes 40 + 12. This makes the process totally transparent.
We will now study regrouping
(also called "borrowing") in subtraction.
As a first step, we study breaking
a ten-pillar into ten little cubes.
This is called regrouping,
because one ten "changes
groups" from the tens group
into the ones.
Breaka ten. 4 tens 5 ones 3 tens 15 ones First we have 45. We "break" one ten-pillar into little cubes. Now we have 3 tens and 15 ones. It is still 45, but written in a different way.
Here is another example. First
we have 5 tens 3 ones. We
"break" one ten-pillar into
10 little cubes. We end up with
4 tens 13 ones.
Breaka ten. 5 tens 3 ones 4 tens 13 ones
1. Break a ten into 10 ones. What do you get? Draw or use manipulatives to help.
a. 3 tens 0 ones ___tens ____ones
b. ___ tens ____ones ___ tens ____ones
c. ___ tens ____ones ___tens ____ones
d. ___ tens ____ones ___ tens ____ones
e. ___ tens ____ones ___tens ____ones
f. ___ tens ____ones ___ tens ____ones
Let's study subtraction. The pictures
on the right illustrate 45 − 17.
First, a ten is broken into 10 ones.
So, 4 tens 5 ones becomes
3 tens 15 ones.
After that, cross out (subtract)
1 ten 7 ones.
Breaka ten. 4 tens 5 ones 3 tens 15 ones
Cross out 1 ten 7 ones (from the second picture).
What is left? ____ tens ____ ones
The pictures on the right
illustrate 52 − 39.
First, a ten is broken into 10 ones.
So, 5 tens 2 ones becomes
4 tens 12 ones.
After that, cross out (subtract)
3 tens 9 ones.
Breaka ten. 5 tens 2 ones 4 tens 12 ones
Cross out 3 tens 9 ones (from the second picture).
What is left? ____ tens ____ ones
2. Fill in. Always subtract (cross out some) from the second picture.
Breaka ten. 3 tens 6 ones 2 tens 16 ones
a. Subtract 8 ones (from the second picture).
What is left? ____ tens ____ ones
Break a ten. ___ tens ___ ones ___ tens ___ ones
b. Subtract 2 tens 7 ones.
What is left? ____ tens ____ ones
Break a ten. ___ tens ___ ones ___ tens ___ ones
c. Cross out 2 tens 5 ones.
What is left? ____ tens ____ ones
Break a ten. ___ tens ___ ones ___ tens ___ ones
d. Cross out 4 tens 4 ones.
What is left? ____ tens ____ ones
3. First, break a ten. Then subtract ones and tens separately. Look at the example.
a.
5 tens 5 ones
4 tens 15 ones − 3 tens 7 ones 1 ten 8 ones
b.
7 tens 2 ones
___ tens ___ ones − 3 tens 5 ones ___ tens ___ ones
c.
6 tens 0 ones
___ tens ___ ones − 2 tens 7 ones ___ tens ___ ones
d.
6 tens 4 ones
___ tens ___ ones − 3 tens 8 ones ___ tens ___ ones
e.
7 tens 6 ones
___ tens ___ ones − 4 tens 7 ones ___ tens ___ ones
f.
5 tens 0 ones
___ tens ___ ones − 2 tens 2 ones ___ tens ___ ones
g.
8 tens 1 one
___ tens ___ ones − 6 tens 5 ones ___ tens ___ ones
h.
6 tens 3 ones
___ tens ___ ones − 2 tens 8 ones ___ tens ___ ones
4. Jessica had 27 colored pencils and her brother and sister had none. Then Jessica gave
10 of them to her brother, and four to her sister.
a.How many pencils does Jessica have now?
b. How many more pencils does Jessica have than her brother?
c. How many more pencils does Jessica have than her sister?
This lesson is taken from my book Math Mammoth Add & Subtract 2B.
#### Math Mammoth Add & Subtract 2-B
A self-teaching worktext for 2nd grade that covers mental addition and subtraction with two-digit numbers, and regrouping in addition and subtraction (carrying & borrowing). |
# As the Wheel Turns
Alignments to Content Standards: F-TF.B.5 F-IF.B.4
A wheel of radius 0.2 meters begins to move along a flat surface so that the center of the wheel moves forward at a constant speed of 2.4 meters per second. At the moment the wheel begins to turn, a marked point $P$ on the wheel is touching the flat surface.
1. Write an algebraic expression for the function $y$ that gives the height (in meters) of the point $P$, measured from the flat surface, as a function of $t$, the number of seconds after the wheel begins moving.
2. Sketch a graph of the function $y$ for $t>0$. What do you notice about the graph? Explain your observations in terms of the real-world context given in this problem.
3. We define the horizontal position of the point $P$ to be the number of meters the point has traveled forward from its starting position, disregarding any vertical movement the point has made. Write an algebraic expression for the function $x$ that gives the horizontal position (in meters) of the point $P$ as a function of $t$, the number of seconds after the wheel begins moving.
4. Sketch a graph of the function $x$ for $t>0.$ Is there a time when the point $P$ is moving backwards? Use your graph to justify your answer.
## IM Commentary
In this task, students use trigonometric functions to model the movement of a point around a wheel and, in the case of part (c), through space (F-TF.5). Students also interpret features of graphs in terms of the given real-world context (F-IF.4).
In order to complete part (a), students must use the linear speed of the wheel to determine its angular speed, and use the unit circle definitions of trigonometric functions (or judicious guessing) to construct a function that gives the vertical position of the point P. In part (c), students must use similar ideas to write a function for the horizontal position of the point P with respect to the center of the wheel, and then combine this with the horizontal position of the center of the wheel to obtain the point's horizontal position with respect to the starting point. The difficulty of the task likely makes it more appropriate for groupwork than as an individual exercise.
## Solution
1. Since we are currently interested only in the vertical position of the point $P$, we can ignore the fact that the wheel is moving horizontally and pretend that the center of the wheel is stationary. To find an expression for $y(t)$, we start by defining $\theta$ to be the angle shown below:
Since the wheel moves at a speed of 2.4 m/s, and the circumference of the wheel is $2\pi r = 0.4 \pi$ meters, the wheel completes $\frac{2.4}{0.4 \pi} = \frac{6}{\pi}$ rotations per second. This means that $\theta$ is increasing at a rate of $\frac{6}{\pi} \cdot 2\pi = 12$ radians per second. So $\theta = 12t$, where $t$ is the time in seconds after the wheel begins to move.
We observe that the height of the point P is equal to 0.2 meters, the height of the center of the wheel, plus or minus the vertical part of the radius from the center of the wheel to P. The vertical part of the radius is $0.2 \cos \theta = 0.2 \cos (12t)$, so the height of the point P is given by $$y(t) = 0.2 - 0.2 \cos(12t).$$
2. Graphing the function $y(t)$, we notice that the graph of the function touches the line $y = 0$ but does not go below this line. This makes sense in terms of the real-world context, because the height of the point $P$ reaches zero but does not go below zero, since the point $P$ never goes beneath the surface.
3. If we pretend temporarily that the center of the wheel is stationary, we can use our finding from part (a) that $\theta = 12t$ to show that the horizontal position of the point $P$, with respect to the center of the circle, is $- 0.2\sin(12 t)$. (This is because the length of the horizontal part of the radius from the center to point $P$ is $0.2 \sin(12 t)$, and this part initially points to the left.)
Now we take into account the horizontal motion of the center of the wheel. Since the center is advancing at a rate of 2.4 meters per second, the horizontal position of the point $P$ is equal to $$x(t) = 2.4 t - 0.2 \sin(12 t).$$
4. Note that the scale for this graph is not the same as the scale for the graph given in part (b).
To determine whether the point $P$ ever moves backwards, we look at the graph of $x(t)$ to see whether $x$ ever decreases. The graph suggests that $x$ does not decrease, though there are points at which the graph is momentarily horizontal. At the times corresponding to these points, the horizontal movement of $P$ has momentarily slowed to a halt. If we put the graphs of $x(t)$ and $y(t)$ on the same set of axes, we can also observe that the times when $P$ stops advancing horizontally are the same times when $P\,$ touches the surface. |
# How do you factor 6x^6-3x^4-9x^2?
Jul 26, 2016
$3 {x}^{2} \left(2 {x}^{2} - 3\right) \left({x}^{2} + 1\right)$.
#### Explanation:
The Expression$= 6 {x}^{6} - 3 {x}^{4} - 9 {x}^{2}$
$= 3 {x}^{2} \left(2 {x}^{4} - {x}^{2} - 3\right)$
Now, to factorise $= 2 {x}^{4} - {x}^{2} - 3$, let us put ${x}^{2} = t$ so that the
poly. will become,
$2 {t}^{2} - t - 3$
$= 2 {t}^{2} - 3 t + 2 t - 3$
$= t \left(2 t - 3\right) + 1 \left(2 t - 3\right)$
$= \left(2 t - 3\right) \left(t + 1\right)$
$= \left(2 {x}^{2} - 3\right) \left({x}^{2} + 1\right)$..................[ as, $t = {x}^{2}$]
Therefore, the Exp.$= 3 {x}^{2} \left(2 {x}^{2} - 3\right) \left({x}^{2} + 1\right)$. |
# Statistical Methods 1 - PowerPoint PPT Presentation
Statistical Methods 1
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Statistical Methods 1
## Statistical Methods 1
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##### Presentation Transcript
1. Statistical Methods 1 Lecture Notes Chapter 7: Inferences Based on a Single Sample: Estimation with Confidence Intervals
2. Introductory Example • A survey by the Roper Organization found that 45% of the people who were offended by a television program would change the channel, while 15% would turn off their television sets. The survey further stated that the margin of error is 3 percentage points and 4000 adults were introduced. • How do these estimates compare with the true population percentages? • Is the sample of 4000 large enough to represent the population of all adults who watch television in the United States.
3. objectives • Find the confidence interval for the mean when σ is known or n 30. • Determine the minimum sample size for finding a confidence interval of the mean • Find the confidence interval for the mean when σ is unknown and n< 30. • Find the confidence interval for a proportion • Determine the minimum sample size for finding a confidence interval of the proportion
4. 7.1 Identifying and Estimating the Target Parameter • Target Parameter Key words associated with parameters:
5. Confidence Interval • Recall that a statistic such as the sample mean is a point estimator of the population mean .
6. 7.2 Confidence Interval for a Population Mean: Normal (z) Statistic • The goal is to determine how to estimate the population mean and assess the estimate’s reliability. • Example: Find such that . Solution:
7. Example (cont.) • According to the Central Limit Theorem, the distribution of will be approximately a normal and 95% of all from a sample of size n lie within of the mean. • There is a probability of .95 that will lie in the interval . • The interval is called a large-sample 95% confidence interval for the population mean . • Large sample means the sample must be large enough so that the Central Limit Theorem can be applied. (Rule of thumb, n ≥ 30)
8. Example (cont.) • Calculating knowing , . We usually don’t know , but with a large sample s is a good estimator of . • The interval being called the 95% confidence interval for the population mean means that if a large number of samples were taken and this interval calculated each time, 95% would contain . • The probability, .95, that measures the confidence we can place in the interval estimate is called a confidence coefficient. The percentage, 95%, is called the confidence level for the interval estimate. • The Margin of error= (This is the maximum error of estimate for a 95% confidence interval)
9. Definition • Confidence coefficient – probability that a randomly selected confidence interval encloses the population parameter • Confidence level – Confidence coefficient expressed as a percentage
10. The confidence coefficient • The confidence coefficient is equal to 1- , where is called the significance level and it is the amount of area assigned to the tails of the sampling distribution, and is split between the two tails of the distribution.
11. Example: 90% confidence interval • John says he is 90% confident that the population mean is contained within the interval I when the values of the population are normally distributed.
12. Example: A publishing company has just published has just a new textbook. Before the company decides the price at which to sell this textbook, it wants to know the average price of all such textbooks in the market. The research department at the company took a sample of 36 comparable textbooks and collected information on their prices. This information produces a mean of \$70.50 for this sample. It is known that the standard deviation of the prices of all such textbooks is \$4.50. A) What is the point estimate of the mean price of all such college textbooks? B) What is the margin error of this estimate? C) Construct the 90% confidence interval for the mean price of all such college textbooks.
13. Large-Sample 100(1 - )% Confidence Interval for Conditions required for a Valid Large-Sample Confidence Interval for 1. A random sample is selected from the target population. 2. The sample size n is large, n 30. p. 306 # 11
14. Example: p. 306 #11. A random sample of 100 observations from a normally distributed population possesses a mean equal to 83.2 and standard deviation equal to 6.4. a. Find a 95% confidence interval for μ. b. What do you mean when you say that a confidence coefficient is .95? c. Find a 99% confidence interval for μ. d. What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held fixed? e. Would your confidence intervals of parts a and c be valid if the distribution of the original population were not normal? Explain.
15. Interpretation
16. 7.3 Confidence Interval for a Population Mean: Student's t-Statistic • Many times, inferences must be made from small samples. But The Central Limit Theorem does not guarantee that sampling distribution of will be normal for small sample sizes. The sampling distribution of will be normal if the population is normal. • Also, the population standard deviation may not be known and the sample standard deviation s may not provide a good approximation for . Instead of using the statistic , which requires a good approximation of , the statistic is used.
17. t-statistic vs z-statistic • The t-statistic is very much like the z-statistic. It is mound shaped, symmetric, and has mean 0. The t-statistic is different in that it has two random quantities ( and s) while the z-statistic only has one ( ). • The variability of t depends on the sample size n. Variability is expressed as (n-1) degrees of freedom (df). As df gets smaller, variability increases.
18. t-statistic vs z-statistic
19. Table for t-distribution: Table IV • Table for t-distribution contains t-value for various combinations of degrees of freedom and tα. tα is the point where the upper tail of the t-distribution contains an area of . • The last row, where df = , contains the standard normal z-values.
20. Small-Sample 100(1 - )% Confidence Interval for Conditions required for a Valid Large-Sample Confidence Interval for 1. A random sample is selected from the target population. • 2. The population has a relative frequency distribution that • is approximately normal.
21. Example: • The data below represent a sample of the number of homes fires started by candles for the past several years. Find the 99% confidence interval for the mean of homes started by candle each year. p. 317 # 33ab,39
22. Example: p. 317 #33 The following random sample was selected from a normal distribution: 4, 6, 3, 5, 9, 3. a. Construct a 90% confidence interval for the population mean μ. b. Construct a 95% confidence interval for the population mean μ. c. Construct a 99% confidence interval for the population mean μ. d. Assume that the sample mean and sample standard deviation s remain exactly the same as those you just calculated, but that they are based on a sample of n = 25 observations rather than n = 6 observations. Repeat parts a-c . What is the effect of increasing the sample size on the width of the confidence intervals?
23. Example: p. 318 #39 • Radioactive lichen. Refer to the Lichen Radionuclide Baseline Research project at the University of Alaska, presented in Exercise 2.36 (p. 47). Recall that the researchers collected 9 lichen specimens and measured the amount (in microcuries per milliliter) of the radioactive element cesium-137 for each. (The natural logarithms of the data values are saved in the LICHEN file.) A MINITAB printout with summary statistics for the actual data is shown below.
24. a. Give a point estimate for the mean amount of cesium in lichen specimens collected in Alaska. • b. Give the t-value used in a small-sample 95% confidence interval for the true mean amount of cesium in Alaskan lichen specimens. • c. Use the result you obtained in part b and the values of and s shown on the MINITAB printout to form a 95% confidence interval for the true mean amount of cesium in Alaskan lichen specimens. • d. Check the interval you found in part c with the 95% confidence interval shown on the MINITAB printout. • e. Give a practical interpretation for the interval you obtained in part c .
25. 7.4 Large-Sample Confidence Interval for a Population Proportion • Confidence intervals around a proportion are confidence intervals around the probability of success in a binomial experiment. • Sample statistic of interest is , where .
26. Sampling Distribution of
27. Large Sample Confidence Interval of p. Conditions required for a Valid Large-Sample Confidence Interval of p • A random sample is selected from the target population. • The sample size n is large. (A sample size is considered large if both and . and are the number of success and the number of failures in the sample.)
28. Example • A sample of 500 nursing applications included 60 from men. Find the 90% confidence interval of the true proportion of men who applied to the nursing program. p. 325 # 51,55
29. Example: p. 325#51 • A random sample of size n = 196 yielded = .64. • a. Is the sample size large enough to use the methods of this section to construct a confidence interval for p? Explain. • b. Construct a 95% confidence interval for p. • c. Interpret the 95% confidence interval. • d. Explain what is meant by the phrase “95% confidence interval.”
30. Example: p. 325#55 • Is Starbucks coffee overpriced? The Minneapolis Star Tribune (August 12, 2008) reported that 73% of Americans say that Starbucks coffee is overpriced. The source of this information was a national telephone survey of 1,000 American adults conducted by Rasmussen Reports. • Identify the population of interest in this study. • Identify the sample for the study. • Identify the parameter of interest in the study. • Find and interpret a 95% confidence interval for the parameter of interest.
31. Summary of Confidence Intervals A. Large-Sample Confidence Interval for a Population Mean For random samples of size 30, the confidence interval is expressed as B. Small-Sample Confidence Interval for a Population Mean The small sample confidence interval will be Where is based on (n – 1) degrees of freedom. C. Large-Sample Confidence Interval for a Population Proportion Large-Sample Confidence Interval for p Where and . • A sample size is considered large if both and .
32. 7.5 Determining the Sample Size A. Estimating a Population Mean • The width of a confidence interval depends on the sample size: • As the sample size increases, the width of the interval decreases for any given confidence coefficient. • When we want to estimate to within a given number of units with a (1- ) level of confidence, we can calculate the sample size needed by solving the equation = interval width for n.
33. Sampling Error The reliability associated with a confidence interval for the population mean is expressed using the sampling error within which we want to estimate with 100(1- )% confidence.
34. Example: p. 332#71 • If you wish to estimate a population mean to within .2 with a 95% confidence interval and you know from previous sampling that σ2 is approximately equal to 5.4, how many observations would you have to include in your sample?
35. sampling error SE of a confidence interval for a population proportion p • The Sampling Error (SE) is half the width of the confidence interval. #71 p.
36. B. Estimating a Population Proportion #78, 81 p.
37. Example: p. 332#81 • Scanning errors at Wal-Mart. Refer to the National Institute for Standards and Technology (NIST) study of the accuracy of checkout scanners at Wal-Mart stores in California, presented in Exercise 3.52 (p. 132). NIST sets standards so that no more than 2 of every 100 items scanned through an electronic checkout scanner can have an inaccurate price. Recall that in a sample of 60 Wal-Mart stores, 52 violated the NIST scanner accuracy standard (Tampa Tribune, Nov. 22, 2005). Suppose you want to estimate the true proportion of Wal-Mart stores in California that violate the NIST standard.
38. a. Explain why the large-sample methodology of Section 7.4 is inappropriate for this study. • b. Determine the number of Wal-Mart stores that must be sampled in order to estimate the true proportion to within .05 with 90% confidence, using the large-sample method. |
# AP Board 6th Class Maths Notes Chapter 2 Whole Numbers
Students can go through AP Board 6th Class Maths Notes Chapter 2 Whole Numbers to understand and remember the concepts easily.
## AP State Board Syllabus 6th Class Maths Notes Chapter 2 Whole Numbers
→ Natural numbers: The numbers which we use for counting are called Natural numbers N= {1, 2, 3, 4, 5, 6,…} .
→ Successor: Every natural number has a successor, which is one more than it. Example : Successor of 15 is 15 + 1 = 16
→ Predecessor: Every natural number has a predecessor except 0. Predecessor of a number is one less than it.
Example: Predecessor of 56 is 56 – 1 = 55
→ Whole numbers: The natural numbers along with zero forms the set of Whole numbers.
W = {0, 1, 2, 3, 4, 5, 6,…}
• Every whole number has a successor and every whole number has a predecessor except zero.
• Every natural number is a whole number and every whole number is a natural number except zero.
• The smallest natural number is 1.
• The smallest whole number is 0.
• Addition, subtraction and multiplication can be represented on a number line.
→ Closure property: Sum of any two whole numbers is always a whole number.
Example : 5 + 3 = 8, a whole number.
This is called closure property of whole numbers w.r.t. addition.
→ Closure property: Product of any two whole numbers is always a whole number. Example : 5 × 3 = 15, a whole number.
This is called closure property of whole numbers w.r.t. multiplication.
In other words whole numbers are closed under addition and multiplication.
But whole numbers are not closed under subtraction and division.
Example: 7 – 12, is not a whole number.
9 ÷ 14, is not a whole number.
Division by zero is not defined.
Example: 5 ÷ 0, is not defined
→ Commutative property: Sum of any two whole numbers taken in any order is always same.
Example: 5 + 3 = 3 + 5 = 8
This is called commutative property of whole numbers w.r.t. addition.
→ Commutative property: Product of any two whole numbers taken in any order is always same.
Example: 5 × 3 = 3 × 5 = 15
This is called commutative property of whole numbers w.r.t. addition.
In other words whole numbers are commutative w.r.t. addition and multiplication. But whole numbers are not commutative w.r.t. subtraction and division.
Example: 4 – 9 ≠ 9 – 4 and 8 ÷ 11 ≠ 11 ÷ 8
→ Associative property: The sum of any three whole numbers taken in any or der is always same.
Example : 5 + (8 + 3) = (5 + 8) + 3 =16
This is called associative property of whole numbers w.r.t. addition.
→ Associative property: The product of any three whole numbers taken in any order is always same.
Example : 5 × (8 × 3) = (5 × 8) × 3 = 120
This is called associative property of whole numbers w.r.t. multiplication.
In other words whole numbers are associative w.r.t. addition and multiplication.
But whole numbers are not associative w.r.t. subtraction and division.
Example : 5 – (8 – 3) ≠ (5 – 8) – 3
: 5 ÷ (8 ÷ 3) ≠ (5 ÷ 8) ÷ 3
→ Distributive property of multiplication over addition:
(Example: 5 × (8 + 3) = (5 × 8) + (5 × 3) = 55
→ Additive identity: If zero is added to any whole number, then the result is the number itself.- Here zero is called the additive identity.
Example: 5 + 0 = 5, 7 + 0 = 7, 0 + 9 = 9 and so on.
→ Multiplicative identity: If any whole number is multiplied by 1, then the result is the number itself. Here one is called the multiplicative identity.
Example: 5 × 1 = 5, 7 × 1 = 7, 1 × 9 = 9 and so on.
If we represent the number 1 as a (.) dot, then a whole number can be represented either as an array of a triangle or a square.
The triangular numbers are 3, 6, 10, 15, 21, 28, ……. etc.
The square numbers are 4, 9, 16, 25, 36, …… etc.
→ Multiplication by 9/99/999/9999….etc.
74 × 99 = (74 – 1)/(9 – 7)(10 – 4) = 73/26
256 × 999 = (256 – 1)/(9 – 2)(9 – 5)(10 – 6) = 255/744
4267 × 9999 = (4267 – 1)/(9 – 4)(9 – 2)(9 – 6)(10 – 7) = 4266/5733
Here the number of 9’s in the multiplier is equal to number of digits in the multiplicand. |
# Equation of a Circle
Equation Of a Circle. There are three different forms of the equation of a circle that can be used to define its radius and centre or point of origin. Depending on the form you need, you can use either the distance formula or the midpoint formula to find the coordinates of the centre of your circle. The radius, r, can then be found with either the distance formula or by taking half of the length between (x1, y1) and (x2, y2).
## EQUATION OF A CIRCLE
Every circle has a radius, so r is the length from the centre to any point on the edge. The x-axis is drawn perpendicular to the radius and intersects with it at two points. One issue is where y=0 (the centre), and one end is where y=r (the edge).
The equation for a circle centred at (h,k) with radius r can be written as: (x-h)2+(y-k)2 = r2. This implies that all points in a circle are equidistant from the centre. The distance between the point’s location and the centre can be found by subtracting h or k from r.
When h=k, then x2+y2 = r2 becomes (x-h)2+(y-k)2 = 0 or (-h+k) = 0 or h=k. So all points on a circle centred at (h, k) are either on its edge or centre.
In general, given any point (x, y) on a circle of radius r and centre (h, k), it’s equation is x2+y2 = r2 - h2 + k2. The distance from that point to (h,k) can be found by subtracting h or k from r.
If we let C be any point on a circle with radius r, (h1, k1) and (h2, k2) are two different points on that circle. If h1 = h2 and k1 = k2, then C is on that circle. For example, let’s use (9,-12),(-3,-6),(-7,-8) to represent three distinct points which satisfy x2+y2=9.
## EQUATION OF A CIRCLE WITH CENTER AND RADIUS
A circle is defined by its centre point and radius. The equation for a process is r = √(x2 + y2), where x and y are the coordinates on the plane. For example, if the coordinate (x,y) lies in quadrant I, then
on the unit circle with a radius one centred at the origin, x=1, y=-1/√3=-√3, and we have r=√(-√3-1)=1
The equation of a circle is x2+ y2 = r2, and The general equation for a loop with a centre point (x0, y0) and radius r is (x- x0)2 + (y- y0)2 = r2. Here, we want to find out how to determine if an arbitrary point(x,y) on a plane lies inside or outside a circle.
We define R_min and R_max as follows : Then, if x- x0 2 + y- y0 2 > R_max, then (x,y) lies outside our circle. Similarly, if x- x0 2 + y- y0 2 < R_min, then (x,y) lies inside our circle.
Now, we can re-write our earlier equation as x - x0 2 + y - y0 2 = (R_max - r2)/(2R_min) and compare it with the cosecant function to find out if (x,y) lies inside or outside the circle.
If x0 is not equal to 0, then the cosecant function will have values less than 0 for (x - x0)/(x0), meaning our point lies outside the circle.
Similarly, if y0 is not equal to 0, then the cosecant function will have values greater than 0 for (y - y0)/(y0), meaning our point lies inside the circle.
## EQUATION OF A CIRCLE GIVEN TWO POINTS
The equation of the circle with centre at (h,k) and radius r is given by the following formula:
x^2 + y^2 = r^2 . Let’s explore some examples. Knowing the coordinates for two points on the plane can determine what kind of curve they form.
In this example, let’s say that one point has coordinates (-5,-5), and another has coordinates (4,-4). We can use these points to create a circle with a centre at (-5,-5) and a radius of 2.
Linear equations connect the coordinates for points on a circle. Let’s say that we have three points that make up our curve, each with its own distinct set of x and y coordinates. We know that these are points on a circle when they fall within an x-y coordinate plane.
The equation to describe such a circle is given by x^2 + y^2 - 4x - 4y + 16 = 0. If you simplify the equation using the quadratic formula, it becomes 2(x+1)(x-1) - 4(x+1)(y+1)(x+1) -4(x+1)(y-1)(x-1)=0.
We can solve for (x-1) and (y+1), so that we have: x^2 + y^2 - 4x - 4y + 16 = 0. Now, let’s put our values in (-1)^2 + y^2 - 4(-1) - 4(4) + 16 = 0. We now know that our circle has the equation: (-1)-4•4+(16)=0.
Let’s start by using points (1,1) and (-2,-2). We know that: 1^2 + (-2)^2 - 4(1)(-2) - 4(-2)(1) + 16 = 0. With these numbers plugged in, we learn that our equation is: 1-4•-2+16=0. Next, if we take (3,7) and (-8,-8), then the result will be 3^2 + (-8)^2 - 4(3)(-8) - 4(-8)(3) + 16 = 0.
## EQUATION OF A CIRCLE DESMOS
When discussing the equation for a circle, we assume that the centre is the point (h, k) with coordinates (h, 0) and that the radius is r.
We also assume that our circle has no vertical or horizontal tangents. Any line drawn through the centre will intersect with only one point on each side.
For an ellipse, we do not assume that there are no tangents, which means that any line drawn through either centre will intersect with at least one point on each side. This means that our formula for r is a function of both t and φ
You can see from the figure that formula_5 is parallel to formula_6 and perpendicular to formula_7.
This means that because they are perpendicular to each other, we can use properties of right triangles to solve for our unknowns.
Because both sides of our triangle are radii, we know they are congruent by SSS (Side-Side-Side) Postulate. We see that side adjacent = adjacent over hypotenuse and opposite = opposite over hypotenuse.
We know that because both sides are radii, we can use Pythagorean Theorem to solve for formula_10 and formula_11, which means that we now have enough information to write our final equation. From here, we can find r, t, and φ.
## EQUATION OF A CIRCLE EXPLAINED
The equation for a circle is (x-h)2+(y-k)2=r2. It has a centre point (h,k) and a radius, r. This formula can be used to find the equation for any point on the circle by plugging in the coordinates for that point into the recipe.
In contrast, (x-h)2+(y-k)2 is just equal to (x-h+r)2+(y-k+r)2; r is cancelled out because each side has it and thus simplifies to x2 + y 2 = r 2. Now we can put in any values for x and y that we please and get our circle.
We take square roots on both sides to get our centre point, h and k. This will give us (h+r/2) and (k+r/2). We then add r / 2 to each side to ensure that our circle is centred at our coordinate.
We can use our equation to create any point on our circle by plugging in coordinates. A good example is (3,5).
Plugging those into our equation gives us: 32 + 52 = 25 + 25 = 50. Thus, (3,5) is an equation for our circle. To put that another way, 3 and 5 are X and Y coordinates for a point in our process, and 50 is its radius.
## HOW TO FIND THE EQUATION OF A CIRCLE
To find the equation of a circle, use the radius and any point on the process as your coordinates. Here’s how it works for a loop with centre (h, k) and radius r: x^2 + y^2 = h^2 + k^2 - 2rhk + r^2 FINDING THE EQUATION OF A CIRCLE WITHOUT CENTER POINT:
If you don’t have an intersection point or if the intersection is outside of the circle, we can use (0, 0). In this case, you would still need to know the radius to calculate other information, such as circumference or area. Here’s what that looks like x^2 + y^2 = 0
If you already have an intersection point, that’s awesome! Here’s what it looks like: (x1, y1) ^2 + (x2, y2) ^2 = r^2. So using our same example where the equation of a circle with centre (h, k) and radius r is x^2 + y^2 = h^2 + k^2 - 2rhk + r^2, our values are replaced by x1= 4, y1= 8, h= 4, k=-6.
And our equation is 4^2 + 8^2 = 16 - 2(4)(-6) + 9^2 = 0. So the equation of our circle would be (x, y) = (-5, 5). That’s it! Now you know how to find an equation for any circle!
So how do you use that equation to find things like area or circumference? Let’s walk through it. First, we must replace x and y with our centre point (h, k). Here’s what that looks like: h^2 + k^2 - 2rhk + r^2 = (4)^2 + (-6)^2 - 2(4)(-6) + 9^2 = 36.
## EQUATION OF A CIRCLE EXAMPLES WITH ANSWERS
The equation for the circle with centre at (h, k) and radius r is given by the following equation: x^2 + y^2 = r^2. For example, if we want to find the equation for a circle with a centre at (-1,-1) and radius 3, we would plug in h=-1 and k=-1.
Another way to view circle equations is as circles pass through (h, k). In other words, if h and k are real numbers, we can write an equation for every point on that circle.
By plugging in (-1,-1) into our original equation, we get x^2 + y^2 = r^2. This shows that all points along our circle satisfy the equation x^2 + y^2 = r^2.
Using x = (h - k)^2 and y = 2k, we get a circle that passes through (-1,-1). In other words, all points on our circle are given by x^2 + y^2 = (-1-0)^2+ 2*(-1)*0 = 1. So plugging in (0,-3), for example, we see that its coordinates satisfy our equation.
With any value we plug in for h and k, both x and y satisfy our equation, as do all coordinates. For example, if we plug (1,-2) into our original equation x^2 + y^2 = r^2 we find that (1,-2) satisfies both x^2+y^2=r^2 and 2(1)(-3) + (-12) = -13.
In contrast, (x-h)2+(y-k)2 is just equal to (x-h+r)2+(y-k+r)2; We take square roots on both sides to get our centre point, h and k.
Cancelled out because each side has it and thus simplifies to x2 + y 2 = r 2. This will give us (h+r/2) and (k+r/2). We then add r / 2 to each side.
Now we can put in any values for x and y that we please and get our circle. To ensure that our circle is centred at our coordinate.
## GENERAL EQUATION OF A CIRCLE CALCULATOR
We must first find the radius to find the equation for a circle. To do this, use the following formula:
r = √(x^2 + y^2) Once you’ve found the radius, plug it into one of these two equations: (x- h)^2+(y-k)^2=r^2 or (x-h)^2+ (y+k) ^ 2=r ^ 2.
(x-h)^2+(y-k)^2=r^2 or (x-h)^2+ (y+k) ^ 2=r ^ 2 If you aren’t sure which equation to use, plug in each set of variables and compare the results. If one of them works, you’ve got your equation for a circle!
The circle equation (x-h)^2+(y-k)^2=r^2 is often referred to as quadrantal, and (x-h)^2+ (y+k) ^ 2=r ^ 2 is referred to as polar.
Which one you use doesn’t really matter. Some equations are more accessible to memorize, but either way will get you a circle!
If you prefer using degrees, we have an alternative equation for you! You may be more familiar with (x-h)^2+(y-k)^2=(r * 2 * pi)^2 or (x-h)^2+(y-k)^2=(r * 2pi) ^ 2, and that’s totally fine. It’ll get you a circle no matter what! Just make sure to plug in radians for r and your variables.
## EQUATION OF A CIRCLE GIVEN CENTER AND POINT
The equation for a circle with center (h, k) and point (x, y) on the circle is (x^2 + y^2 = h^2) The equation can also be written as:
(x^2 + y^2 = r^2) Where r is the radius. You can also find this relationship by looking at the distance formula to solve for x. First, we substitute (x, y) for each variable. (d = \sqrt{(x - h)^2 + (y - k)^2}) Now we square both sides.
(d^2 = \sqrt{(x - h)^2 + (y - k)^2}) Then take square root on both sides. (d = \sqrt{r^2}) Multiply both sides by r.
(r^2 = d) Solve for r. (r = \sqrt{d}) And that’s how to find x when given center and point. To use these methods, you must already know how to find the centre and radius in a circle.
The video tutorial below covers finding these values in detail. The methods are discussed in detail below as well.
Finding Center and Radius using Pythagorean Theorem The video below is a detailed step-by-step tutorial on how to find the centre and radius for any circle, given its centre and point.
If you don’t know how to do that, check out that video before attempting these methods.
Get to our equation for finding x when given the centre and point. As we saw in that video, there are two ways you can use the distance formula to solve for x.
## Summary
The equation for a circle is (x-h)2+(y-k)2=r2 where x, y, and r are the coordinates. The centre (h,k) is found by solving for either x or y first. A circle can also represent an equation with the form (x+w)2+(y+h)2=r2.
Some Important Questions For Your Knowledge:
### 1) How do we find the equation of a circle?
A circle is the set of points at a given distance from one point. The equation for the circle with centre (0,0) at the origin and radius r is x2 + y2 = r2. This can be rewritten as x2 + y2 - 2rx = 0. We can also write this as x2 + y2 = ±r2. These two equations represent the total possible solutions for this problem!
### 2) What is the equation of a circle GCSE?
The equation for a circle with center ((h, k)) and radius (r) is:$$x^2 + y^2 = r^2$$ \begin{aligned}&x = \sqrt{h^2+k^2-r^2} \&y = \sqrt{h^2+k^2-r^2}\end{aligned} The two points on the circumference are given by (\left(\frac{\pi}{2}, 0\right)) and (\left(\frac{3\pi}{2}, 0\right)).
### 3) What are the three circle formulas?
The formula for the circumference is 2πr. The formula for the area is πr2. And the formula for the radius is r. A circle with an equation x2+y2=1 has both of these circles formulas because it can be graphed in two different ways: from x=0 to x=-1 and y=-1 to y=0 or from x=-1 to x=0 and y=-1 to y=0, depending on which one you want to use as the outer circle or inner circle.
### 4) What is the standard form of the equation of a circle?
The equation for the equation of a circle is: (x-h)2+(y-k)2=r^2. The graph shows the x, y, and r coordinates, and h and k are the x and y intercepts. It is important to note that this expression holds for any point on the chart with an x value greater than or equal to h and less than or equal to k.
### 5) How do we find the radius of a circle?
To see the radius of a process, we use the equation for the circumference of the process and divide it by 2π. The radius is halfway between the centre and any point on the edge. To find the centre, we draw two perpendicular lines from any point on the border to either side of our circle.
### 6) How do you find the equation of a circle given two points?
The equation for a circle is x2 + y2 = r2. To find the equation of a circle given two points, you need to solve for r in the following equation: x - (y/r) = 0 and y - (x/r) = 0. These equations can be solved using substitution and linear algebra. For example, if we have x = 3 and y = 4, then r would equal 9.
### 7) What is the general form of an equation?
The equation for the equation of a circle is as follows: (x-h)2+(y-k)2=r2. It is called the general form because it can be used to find any point on the circle, not just the centre. To find the distance between two points, use the following formula: d=sqrt((x1-x2)^2+(y1-y2)^2).
### 8) How do you form equations?
The equation for a circle with a centre is the following: X^2 + Y^2 = R^2 where R = distance from the centre to any point on the circle’s circumference. The radius (R) can be either positive or negative. WOW, that was easy! To get an equation in standard form, you only need to solve it and put it into parentheses! For example, let’s say we have 4x - 3y = 2, so what we do is multiplie diply by y, then add 3 to both sides, so we get 4x - 3y + 3y = 6y, which then equals y(4x-3).
### 9) What are the two equations of a line?
The equation for the circle’s centre is ‘x^2+y^2=r^2’ where ‘r’ is the radius, and the equation for a point on the circumference of the circle is (x-h)^2+(y-k)^2=(r+h)^2. If you know any two points on the process, you can find its radius theiry taking their difference squared.
### 10) What is an equation for a line?
An equation for a line is y = MX + b, where x and y are the coordinates of the line. The slope, or steepness, of the line is represented by m, and b is the y-intercept.
## Conclusion
The equation for a circle is found by finding the centre, then plugging it into the radius and drawing a line from the centre to any point on the circumference. The radius of the circle is equal to its distance from the centre. A circle has an infinite number of points on its circumference, but only one point t called this e centre. |
# The Free Throws Problem Part 2
Here’s an extension to the problem in my previous post. Time has run out, and a player is at the free throw line. If he makes the first shot, he gets a second try. If he makes both shots, his team wins; if he misses the first, his team loses. Otherwise, it’s a tie game, and it goes into overtime.
1. Dave tracks his averages for his first and second free throws separately. He says that, using those numbers, the probability of a tie game in the scenario is greater than either of the other two possibilities. Does he do better or worse on his second free throw?
If we used the visual area model, we could see that this scenario is now possible because the green area can be any rectangle, so the congruence discussed in the previous post no longer necessarily holds. But it doesn’t tell us anything immediate about the relationship between the first and second free throw percentages.
Let’s use a completely algebraic method. Let p and q be the probability of hitting the first and second free throws, respectively, and w, t, and l against be the respective probabilities of winning, tying, and losing. Then $w = pq \\ t = p(1-q) = p – pq \\ l = 1 – p$
We want t to be greater than both l and w. Let’s solve those individually.
In the first case, $$t > l$$ so $$p – pq > 1 – p \Rightarrow -pq > 1 – 2p \Rightarrow q < (2p – 1)/p$$.
In the second case, $$t > w$$ so $$p – pq > pq \Rightarrow p > 2pq \Rightarrow 1/2 > q \Rightarrow q < 1/2$$.
Since $$q > 0$$, $$(2p – 1)/p > 0 \Rightarrow 2p – 1 > 0 \Rightarrow p > 1/2$$.
So t is only greater than both l and w when $$p > 1/2$$ and $$q < 1/2$$, meaning that $$q < p$$. That is, Dave has a worse average on his second free throws than on his first.
We can see the exact domain for p and q by graphing the two inequalities for q in Desmos. The overlapped area are the cases where p (x-axis) and q (y-axis) make the probability of tying the highest. It’s not sufficient that q merely be below 0.5 and p be above 0.5. We can calculate the intersection point for the red and blue dotted lines: $$(2p – 1)/p = 1/2 \Rightarrow 2(2p – 1) = p \Rightarrow 4p-2 = p \Rightarrow 3p = 2$$. So when $$p = 2/3$$ and $$q = 1/2$$, the three values (w, l, and t) ought to be the same.
Let’s check that that’s true. $$w = pq = (2/3)(1/2) = 1/3$$; $$t = p – pq = 2/3 – 1/3 = 1/3$$; and $$l = 1 – p = 1 – 2/3 = 1/3$$. This is the only case where all three probabilities are the same.
One more question: What is the minimum difference between p and q? Clearly, for values $$p > 2/3$$, $$q < 1/2$$, so any value of p above 2/3 has a greater minimal difference with q than $$p = 2/3$$. At $$p = 2/3$$, the difference is $$2/3 – 1/2 = 1/6$$. At $$p = 1/2$$, the difference is $$1/2 – 0 = 1/2$$.
Adding the line $$q = p – 1/6$$, we can easily see that this intersection point represents the minimal difference.
So if Dave’s free throw averages are such that the chance of a tie in the scenario given is greater than the other two possibilities, his second throw average has to be at least 1/6 less than his first throw average. |
# Place Value and Face Value – Definition, Examples | Difference Between Place Value and Face Value
Place Value of a Number deals majorly with the position of the digit i.e. where does it live in the number. Whereas face value deals with the actual digit and how much is the digit worth. Face Value remains the same irrespective of the place it is present in the number. Check out the Definition of Place Value and Face Value, Key Differences that Separate Place Value and Face Value, Writing the Expanded Form of a Given Number, etc.
## Definition of Place Value and Face Value
Place Value: Place Value defines the value of a digit depending on the position in a number. It changes according to the digit’s place. For Example: In 3456 place value of 5 is 5*10 i.e. 50
Face Value: Face Value of a Number is the Actual Value of a Digit in a Number. It is definite and cannot be changed. For Example: In 456 the face value of 6 is 6 the digit itself.
### How to find the Place Value?
Follow the simple steps listed below to determine the Place Value of a Number. They are along the lines
• Note down the digit for which you want to determine the place value.
• Count the number of digits next to the digit you want to find the place value.
• Place as many zeros as the number of the digits after the number you want to calculate the place value and that is the actual place value of the number.
You can better understand this by considering an example.
Find the place value of 8 in 5536891?
Solution:
• Firstly find the number you want to determine the place value. i.e. 8
• Count the number of digits after 8.
• As there are 2 digits place 2 zeros after 8 to obtain the place value.
• By doing so, we will get 800.
Therefore, the place value of 8 in the 5536891 is 800.
### Difference Between Place Value and Face Value
One can find the key differences that separate the Place Value and Face Value in the below sections. In the case of place value, the number system starts from 0 to tens, hundreds, thousands, and so on. Refer to the following table to know about the major differences between place value and face value.
Place value
Face value
Place Value denotes the position of a digit in the given number. Face Value represents the value of the digit itself.
The Value of Each digit in the number depends on its position. The value of the digit is independent of the position or place of the digit in a number.
Place Value of digit can be obtained using the formula Place Value = (Face value) × (Numerical Value of Place) Face Value of the digit = Numerical Value of the Digit Itself.
Place Value of 0 is always 0 no matter it is present in any position of a number. Face Value of 0 is also 0.
Example: The place value of digit 3 in 7,831 = 3 × 10 = 30. Example: The face value of digit 3 in 7831 is 3
### Place Value and Face Value of Digits Examples
Go through the below provided examples on determining the face and place value to understand the concept better. We have provided the expanded form of a given number, how to express them in words, as well as place value, face value for the respective number.
Number Expanded Form in Words Place Value Face Value
12 10+2 Twelve 10 1
35 30+5 Thirty Five 30 3
74 70+4 Seventy Four 70 7
46 40+6 Forty Six 40 4
68 60+8 Sixty Eight 60 6
83 80+3 Eighty Three 80 8
29 20+9 Twenty Nine 20 2
58 50+8 Fifty Eight 50 5
123 100+20+3 One Hundred Twenty Three 20 2
134 100+30+4 One Hundred Thirty Four 30 3
175 100+70+5 One Hundred Seventy Five 70 7
164 100+60+4 One Hundred Sixty Four 60 6
193 100+90+3 One Hundred Ninty Three 90 9
278 200+70+8 Two Hundred Seventy Eight 200 2
281 200+80+1 Two Hundred Eighty One 80 8
387 300+80+7 Three Hundred Eighty Seven 300 3
394 300+90+4 Three Hundred Ninty Four 90 9
423 400+20+3 Four Hundred Twenty Three 400 4
569 500+6+9 Five Hundred Sixty Nine 60 6
673 600+70+3 Six Hundred Seventy Three 600 6
749 700+40+9 Seven Hundred Forty Nine 40 4
836 800+30+6 Eight Hundred Thirty Six 800 8
932 900+30+2 Nine Hundred Thirty Two 30 3
1234 1000+200+30+4 One Thousand Two Hundred Thirty Four 200 2
1576 1000+500+70+6 One Thousand Five Hundred Seventy Six 1000 1
2345 2000+300+40+5 Two Thousand Three Hundred Forty Five 40 4
4567 4000+500+60+7 Four Thousand Five Hundred Sixty Seven 500 5
5678 5000+600+70+8 Five Thousand Six Hundred Seventy Eight 5000 5
6789 6000+700+80+9 Six Thousand Seven Hundred Eighty Nine 9 9
87654 80000+7000+600+50+4 Eighty Seven Thousand Six Hundred Fifty Four 7000 7
65434 60000+5000+400+30+4 Sixty Five Thousand Four Hundred Thirty Four 60000 6
76543 70000+6000+500+40+3 Seventy Six Thousand Five Hundred Forty Three 40 4
65432 60000+5000+400+30+2 Sixty Five Thousand Four Hundred Thirty Two 5000 5
543256 500000+40000+3000+200+50+6 Five Lakh Forty Thousand Three Thousand Two Hundred Fifty-Six 3000 3
765434 700000+60000+5000+400+30+4 Seven Lakh Sixty Thousand Five Thousand Four Hundred Thirty-Four 400 4
765813 700000+60000+5000+800+10+3 Seven Lakh Sixty Thousand Five Thousand Eight Hundred Thirteen 60000 6
345656 300000+40000+5000+600+50+6 Three Lakh Forty-Five Thousand Six Hundred Fifty Six 5000 5
776543 700000+70000+6000+500+40+3 Seven Lakh Seventy Thousand Six Thousand Five Hundred Forty Three 70000 7
345643 300000+40000+5000+600+40+3 Three Lakh Forty Five Thousand Six Hundred Forty Three 40 4
5678987 5000000+600000+70000+8000+900+80+7 Fifty Six Lakh Seventy Eighty Thousand Nine Hundred Eighty Seven 8000 8
6543566 6000000+500000+30000+2000+500+60+6 Sixty Five Lakh Forty Three Thousand Five Hundred Sixty-Six 3000 3
1356 1000+300+50+6 One Thousand Three Hundred Fifty Six 50 5
12567 10000+2000+500+60+7 Twelve Thousand Five Hundred Sixty Seven 2000 2
45362 40000+5000+300+60+2 Forty Five Thousand Three Hundred Sixty Two 60 6
92725 90000+2000+700+20+5 Ninty Two Thousand Seven Hundred Twenty Five 700 7
658493 600000+50000+8000+400+90+3 Six Lakh Fifty Eight Thousand Four Hundred Ninty Three 8000 8
884635 800000+80000+4000+600+30+5 Eight Lakh Eighty Four Thousand Six Hundred Thirty Five 80000 8
749352 700000+40000+9000+300+50+2 Seven Lakh Forty Nine Thousand Three Hundred Fifty Two 9000 9
926455 900000+20000+6000+400+50+5 Nine Lakh Twenty Six Thousand Four Hundred Fifty Five 50 5
### FAQs on Place Value and Face Value
1. What is Place Value?
Place Value is defined as the value represented by the digit in a number depending on the position in the number.
2. What is Face Value?
Face Value is defined as the digit itself within a number.
3. What is the face value of 4 in 94206?
Face Value of 4 in 94206 is 4.
4. When will Place Value and Face Value be the same?
Place Value and Face Value will be the same if both are at one’s digit place.
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# Mathematical Induction
Mathematical Induction is the process by which a certain formula or expression is proved to be true for an infinite set of integers. An example of such a formula would be
$Latex formula$
which may be proven true using Mathematical Induction. The process of Mathematical Induction simply involves assuming the formula true for some integer $Latex formula$ and then proving that if the formula is true for $Latex formula$ then the formula is true for $Latex formula$. From this we may show that the formula is true, if and only if there is a base case. That is, if there exists some integer for which the formula is true. After this, the formula is true for all integers larger than that integral value, since if it is true for some integer, then it is true for the integer plus one and then that integer plus one and so on. The idea is set out below,
STEP 1 (Base Case)Show that the statement/formula is true for some integral value, $Latex formula$ say.STEP 2 (Assumption)Assume that the statement/formula is true for some integral value, $Latex formula$ say. STEP 3 (Inductive Step) Show that if the assumption is true then the statement is true for $Latex formula$. STEP 4 (Conclusion) End the proof by writing “Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer”, and $Latex formula$ was the integer the base case is true for.
Before we move on, we shall review some basic notation and terminology which becomes useful in this topic.
• An integer is a whole number.
• A number greater than zero which is an integer is called a positive integer.
• A number greater than or equal to zero which is an integer is called a non-negative integer.
• Be familiar with the summation and product notation
$Latex formula$
and
$Latex formula$
We shall now illustrate the method of mathematical induction by proving the formula for the sum of the first positive integers.
### Example 1
Prove that for all positive integers $Latex formula$,
$Latex formula$
### Solution 1
STEP 1:
We prove that the statement is true for $Latex formula$. (Since we are required to prove the statement for all positive integers).
For $Latex formula$
LHS $Latex formula$
RHS $Latex formula$$Latex formula$$Latex formula$LHS
Hence as RHS $Latex formula$ LHS then the statement is true for $Latex formula$
STEP 2:
We assume that the statement is true for some positive integer $Latex formula$. That is,
$Latex formula$
STEP 3:
We now must prove that the statement is true for $Latex formula$. That is, we must prove,
$Latex formula$
LHS $Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$ RHS
Hence as the statement is true for $Latex formula$, it follows that the statement is true for $Latex formula$.
STEP 4:
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
This illustrates the technique of mathematical induction. We shall now move on to an important category of questions requiring mathematical induction.
## Proving Summation Statements using Mathematical Induction
We will study some further examples of summation problems in mathematical induction. In general, the three main types of mathematical induction problems are classified into summation, division or inequality problems. Some problems fall outside these categories, and we shall study them to encourage a more holistic view of Mathematical Induction.
### Example 2
Prove by mathematical induction that for all positive integral values of $Latex formula$,
$Latex formula$
### Solution 2
This is exactly the same as proving
$Latex formula$
In this example we shall not write down the steps being taken as they should be obvious by the reasoning provided.
Firstly, for $Latex formula$
LHS $Latex formula$
RHS $Latex formula$$Latex formula$$Latex formula$ LHS
Hence as RHS $Latex formula$ LHS, the statement is true for $Latex formula$.
Assume the statement is true for $Latex formula$, where $Latex formula$ is a positive integer. i.e.
$Latex formula$
We are now, R.T.P. (required to prove) that the statement is true for $Latex formula$. i.e.
$Latex formula$
$Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$Hence as the statement is true for $Latex formula$, it follows that the statement is true for $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
Note: A very good technique with these types of questions is to remove as a factor anything that occurs in the final statement. For e.g. in the simplification of
$Latex formula$
Every student will remove a factor of $Latex formula$. It is also a good idea to remove the $Latex formula$ out as a factor and this greatly simplifies ones working out. Be careful to account for the factor of $Latex formula$ though in every term. For e.g. we have that,
$Latex formula$
Many students however incorrectly factor out the $Latex formula$ and have the incorrect result,
$Latex formula$
### Example 3
Prove by mathematical induction that for all positive integers $Latex formula$,
$Latex formula$
### Solution 3
This is exactly the same as proving,
$Latex formula$
So, for $Latex formula$,
$Latex formula$
$Latex formula$
Hence as $Latex formula$ it thus follows that the statement is true for $Latex formula$.
Assume that the statement is true for $Latex formula$ where $Latex formula$ is a positive integer. i.e.
$Latex formula$
(Note here that we have used $Latex formula$ instead of $Latex formula$ to avoid confusion, since the dummy variable within the summation is $Latex formula$)
We are now R.T.P. that the statement is true for $Latex formula$. That is,
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
Hence as the statement is true for $Latex formula$, it follows that the statement is true for $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
## Proving divisibility statements using mathematical induction
Mathematical Induction is also very useful in proving that a certain expression is always divisible by another, given that the expressions have integers as there input. An example question would be,
“Prove that $Latex formula$ is divisible by 4 for all integers, $Latex formula$
Such a question is proved using the exact same steps as those used previously, except that at the assumption step, we assume that the quotient is equal to some integer if numbers are being divided, or a polynomial if polynomials are being divided. This is due to divisibility being defined as thus. Also, at the inductive step, we must rearrange the expression so as to be able to substitute the assumed expression in. We shall now illustrate this by completing the above question as an example.
### Example 4
Prove by mathematical induction that for all positive integers $Latex formula$, $Latex formula$ is divisible by $Latex formula$.
### Solution 4
We shall firstly consider the base case.For $Latex formula$, $Latex formula$, which is clearly divisible by $Latex formula$. Hence the statement is true for $Latex formula$.
We now assume that the statement is true for $Latex formula$, where $Latex formula$ is a positive integer. i.e.
$Latex formula$
where $Latex formula$ is some integer.
We may write this in a more useful form, namely that
$Latex formula$
(Note this step!)
We now are R.T.P. that the statement is true for $Latex formula$. That is, $Latex formula$ is divisible by $Latex formula$.
We have that
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
where $Latex formula$ is an integer. Since $Latex formula$ is divisible by $Latex formula$ it thus follows that $Latex formula$ is divisible by 4.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
Note: Notice that as in all induction proofs, we must use the assumed statement to prove the case for $Latex formula$. Now, the step used, (that is the process of multiplying through by the divisor) is the key step which allows one to prove the result using induction. Consider another example to clarify the process.
### Example 5
Prove that $Latex formula$ is an even number for all positive integral $Latex formula$.
### Solution 5
An even number is any number that is divisible by $Latex formula$. Hence in this question we are proving that the required expression is divisible by $Latex formula$.So, for $Latex formula$ (the first positive integer),
$Latex formula$
Which is divisible by $Latex formula$. Hence the statement is true for $Latex formula$.
Assume that the statement is true for $Latex formula$ where $Latex formula$ is a positive integer. i.e.
$Latex formula$
where $Latex formula$ is an integer. i.e.
$Latex formula$
We now are R.T.P. that the statement is true for $Latex formula$. i.e. we are required to prove that $Latex formula$ is divisible by $Latex formula$.
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
Where $Latex formula$ is an integer. Hence as $Latex formula$ is divisible by $Latex formula$ then it follows that $Latex formula$ is divisible by $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
Note: At this point it should be mentioned that some teachers insist that there students write out a conclusion to their mathematical induction proofs on the lines of “Since the statement is true for $Latex formula$ it thus follows that the statement is true for $Latex formula$, and hence for $Latex formula$ and so on. Hence it follows that the statement is true for all integers $Latex formula$.” Not only is this excessive and unnecessary, it also is incorrect because it has not mentioned once that the proof was by the method of mathematical induction. It is said in the Mathematics Extension 1 examiners comments that students should only write down a statement such as “Hence the statement is true for integers $Latex formula$, by mathematical induction.” This is because Mathematical Induction is an axiom upon which Mathematics is built, not a theory that has a reasoning or proof behind it. Hence any type of explanation of Mathematical Induction from a heuristic approach is deemed to be incorrect, and students should keep to a simple conclusion as given in these notes. Students should note though that the values for which the statement is true is important and should be stated clearly in the conclusion.
We shall now consider another example;
### Example 6
Prove by mathematical induction that $Latex formula$ is divisible by $Latex formula$ for all integers $Latex formula$.
### Solution 6
We now prove the base case, where $Latex formula$. (Note the difference in the values for which we are required to prove the statement is true).For $Latex formula$,
$Latex formula$
which is clearly divisible by $Latex formula$. Hence the statement is true for $Latex formula$.
Assume that the statement is true for $Latex formula$, where $Latex formula$ is an integer and $Latex formula$. i.e.
$Latex formula$
where $Latex formula$ is an integer. i.e.
$Latex formula$
i.e.
$Latex formula$
(Note this technique!)
We are R.T.P. that the statement is true for $Latex formula$. i.e. that $Latex formula$ is divisible by $Latex formula$.
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
where $Latex formula$ is an integer. Since $Latex formula$ is divisible by $Latex formula$, it thus follows that $Latex formula$ is divisible by $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
Note:The technique used in the assumption in the above example is useful in simplifying ones working out.
We shall now consider an example of polynomial division.
### Example 7
Prove by mathematical induction that $Latex formula$ Is divisible by $Latex formula$ for all positive integral $Latex formula$. (You may suppose that $Latex formula$)
### Solution 7
For $Latex formula$, we have that
$Latex formula$
which is clearly divisible by $Latex formula$. Hence the statement is true for $Latex formula$.
Assume that the statement holds for $Latex formula$. That is,
$Latex formula$
Where $Latex formula$ is some polynomial in $Latex formula$. (This is the definition of polynomial division)
That is,
$Latex formula$
We are now required to prove the statement true for $Latex formula$. i.e. we are required to prove that $Latex formula$ is divisible by $Latex formula$.
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
where [latext]p(x)[/latext] is some polynomial in $Latex formula$. Hence as $Latex formula$ is clearly divisible by $Latex formula$, it thus follows that $Latex formula$ is also divisible by $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
We shall now move on to induction problems involving inequality statements.
## Proving Inequality Statements using Mathematical Induction
Inequality statements are not at all times straightforward to answer, and it is ones experience in answering such questions that proves to be the largest factor in solving these questions. Questions involving inequalities are typically questions such as,
“Prove that $Latex formula$ for all integral $Latex formula$ where $Latex formula$
These types of questions are solved using the exact same technique as in the summation and divisibility questions. In these questions however, one proves the inductive step by showing that RHS and LHS satisfy the inequality required, after using the assumed statement. The below example will clarify this point.
### Example 8
Prove by mathematical induction that $Latex formula$ for $Latex formula$ where $Latex formula$ is an integer.
### Solution 8
For $Latex formula$,LHS $Latex formula$
RHS $Latex formula$
Obviously we have that LHS $Latex formula$ RHS and thus the statement is true for $Latex formula$.
Assume that the statement is true for $Latex formula$, where $Latex formula$ is a positive integer. i.e.
$Latex formula$
R.T.P. true that the statement is true for $Latex formula$. i.e.
$Latex formula$
LHS $Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$RHS
Hence we have that LHS $Latex formula$ RHS and that the statement is true for $Latex formula$, if the statement is true for $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
We shall now investigate one more example to illustrate some of the techniques involved in solving these problems.
### Example 9
Prove that $Latex formula$ for $Latex formula$ where $Latex formula$ is an integer.
### Solution 9
For $Latex formula$,LHS $Latex formula$
RHS $Latex formula$
Obviously LHS $Latex formula$ RHS and thus we have that the statement is true for $Latex formula$.
Assume that the statement is true for $Latex formula$ where $Latex formula$ is a positive integer. i.e.
$Latex formula$
R.T.P. that the statement is true for $Latex formula$. i.e.
$Latex formula$
LHS $Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$ RHS
Hence as we have LHS $Latex formula$ RHS, then it follows that the statement is true for $Latex formula$, if the statement is true for $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
Note: With inequality problems, a standard technique to use is to bring all terms onto one side to obtain a zero on another side. After this the problem boils down to proving that one side is always positive, negative or non-negative. Solving the problem using this method is rarely the best way to do so, but it is included so that the student may add this into his or her arsenal.
## Proving miscellaneous problems using Mathematical Induction
We shall now investigate problems that can only come under the appropriately named category “miscellaneous”. These problems require deeper understanding and a more perseverance then most induction questions. We shall investigate some of these to show the method of solving these questions.
### Example 10
Assuming that the product rule for differentiation holds, show that
$Latex formula$
For all positive integers .
### Solution 10
For $Latex formula$,
LHS $Latex formula$ $Latex formula$(Since it equals the gradient of $Latex formula$ at the point $Latex formula$ which is $Latex formula$)
RHS $Latex formula$ LHS
Hence the statement is true for $Latex formula$.
Assume that the statement is true for $Latex formula$ where $Latex formula$ is a positive integer. i.e.
$Latex formula$
R.T.P. true that the statement is true for $Latex formula$. i.e.
$Latex formula$
LHS $Latex formula$ $Latex formula$ $Latex formula$
$Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$$Latex formula$ RHS
Hence as the statement is true for $Latex formula$ it thus is true for $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer.
The above problem was in fact the induction question from the 2009 Mathematics Extension 1 HSC. The next example shows the method of “Strong Induction” where you assume the statement to be true for more than one value. This is not included in the Mathematics Extension 1 syllabus, but is the focus of Mathematics Extension 2 induction. Also, it is another example of polynomial division.
### Example 11
Prove by induction that $Latex formula$ is divisible by $Latex formula$ for all $Latex formula$ where $Latex formula$ is integral and $Latex formula$.
### Solution 11
Since we are using strong induction in this case, we then must prove that the statement is true for two base cases, namely $Latex formula$ and $Latex formula$. (Since we assume the statement to be true for two values that are consecutive)For $Latex formula$;
$Latex formula$
which is clearly divisible by $Latex formula$. Hence the statement is true for $Latex formula$.
For $Latex formula$;
$Latex formula$
Which is clearly divisible by $Latex formula$. Hence the statement is true for $Latex formula$ and $Latex formula$.
Assume the statement is true for $Latex formula$ and $Latex formula$ where $Latex formula$ is a positive integer greater than $Latex formula$. That is,
$Latex formula$ (*)
$Latex formula$ (#)
Note here that $Latex formula$ represents a polynomial in two variables $Latex formula$ and $Latex formula$. An example of such a polynomial would be $Latex formula$.
R.T.P. that the statement is true for $Latex formula$. i.e. that $Latex formula$ is divisible by $Latex formula$.
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$ (Using (*) and (#))
$Latex formula$
$Latex formula$
where $Latex formula$ is a polynomial in two variables $Latex formula$ and $Latex formula$. Obviously, as $Latex formula$ is divisible by $Latex formula$ it thus follows that $Latex formula$ is divisible by $Latex formula$.
Hence by Mathematical Induction the statement is true for $Latex formula$ where $Latex formula$ is an integer. |
# Photography Tutorial:
## Where do those "F/stop" numbers come from? (detailed explanation)
The camera lens transmits a circular image that completely covers the camera sensor. The area of a circle is calculated by taking the pure number Pi (approximately 3.14 and pronounced like cherry 'pie') times the radius squared. When we square a number, that just means to take the number and multiply it by itself. Four squared means 4 * 4, which equals 16. 10 squared means 10 * 10, which equals 100. The number Pi is a pure number (without any units) and never changes.
The radius of a circle is the length from the center of a circle to the circle itself. The diameter of a circle is the length from one side of the circle to the other side, passing through the center. Obviously, the diameter of a circle equals exactly twice the radius.
So, let's calculate the area of a circle with a radius of 1 unit (millimeters, feet, miles, doesn't matter):
Putting in our numbers, this gives us:
Area = (3.14) * (1 unit) * (1 unit) giving us 3.14 square units. This will be our starting point.
Now, if we want to double the area of a circle, what can we change? The number Pi never changes so the only thing we can change is the radius of the circle. Our first thought might be to simply double the length of the radius to double the area of the circle. Let's see how those numbers come out:
Area = (3.14) * (2 units) * (2 units) - giving us 12.56 square units. Wow! That's 4 times the starting area, not twice. Now, we could try guessing a few times using a calculator to get close to a final area of 6.28 (twice our starting point area) or we can figure it out mathematically. Or I can simply tell you that the answer is the square root of 2. What's a square root? Earlier, we said that a number squared means to take a number and multiply it by itself. A square root is going the other direction. It means you start with a number and figure out what other number, when multiplied by itself gives you the starting number. So the square root of 100 would be 10 (because 10 * 10 = 100) and the square root of 25 would be 5 (because 5 * 5 = 25).
So what is the square root of 2? It is approximately 1.414. The number actually continues on forever past that (1.414213562...) but let's just use the 1.414 part.
Go ahead and get out your calculator. I know you're thinking you've got to check it yourself. That's ok, I'll wait.
Check out ok? Thought so.
Area = (3.14) * (1.414 units) * (1.414 units) gives us 6.28 square units. Now THIS is twice the area of our first circle. So what happened to the diameter in this exercise? In the starting circle, the radius was 1 unit which makes the diameter 2 units. In our second circle, the radius was 1.414 units so the diameter would be 2.828 units. The diameter increased by the same factor as the radius, ie, radius 1 to 1.414 and diameter 2 to 2.828. Increasing the diameter by the same square root of 2 value doubles the area of the resulting circle. The diameter of the opening through which the light is transmitted is the actual Aperture measurement.
So, what can we conclude? We can conclude that each time we want to double the area of a circle, we need to increase the radius (or diameter) of the circle by a factor of the square root of two (multiply by 1.414...).
What happens if the focal length of the lens and the aperture values are the same? Let's start with a 100mm focal length lens and give it a 100 mm aperture.
As we know, F/stop = (focal length) / (aperture) so the F/stop in this example would be:
F/stop = (100 mm) / (100 mm) - which gives us that magic value, F/stop = '1'.
The area of this aperture opening equals (Pi) * (radius) * (radius) or:
(3.14) * (50 mm) * (50 mm) = 7853.98 sq mm
Now let's start decreasing the diameter (aperture) by that square root of two value (to cut the area of the circle in half) with this lens and see what happens.
Remember, each step will divide the diameter by 1.414 to get the next diameter or Aperture value:
Aperture Area Radius (focal length) (mm) (sq mm) (mm) (aperture) 100.00 7853.98 50.00 1 70.71 3926.99 35.35 1.41 50.00 1963.50 25.00 2.00 35.35 981.75 17.68 2.83 25.00 490.87 12.50 4.00 17.68 245.44 8.84 5.66 12.50 122.72 6.25 8.00 8.84 61.36 4.42 11.31 6.25 30.68 3.13 16.00 4.42 15.34 2.21 22.63 3.13 7.67 1.56 32.00
So, do the numbers in that last column look familiar yet? Yup, those are the F/stop values on your lens. Each change above represents cutting the area of the resulting circle in half, thereby cutting the amount of light allowed through the lens in half for the same shutter speed.
As you change the F/stop from a setting of 1 to 1.4 you are cutting the amount of light reaching the sensor in half. As you change the F/stop from a setting of 1.4 to 2 you cut the amount of light in half again. This means that the amount of light reaching the sensor at an F/stop setting of 1 is four times the amount of light reaching the sensor at an F/stop setting of 2.0. Remember, we cut the amount of light in half twice (1 to 1.4 and again from 1.4 to 2.0) and 2 * 2 equals 4.
### Ramifications:
The kit lens I bought with my Nikon D80 has a variable zoom range from 18mm to 135mm. This is a wonderful working range for focal lengths, covering the moderate wide angle to medium zoom range. It's also a variable aperture lens. This means that as I change the focal length, the maximum aperture value I can use (largest opening) also changes. Let's look at how the maximum aperture changes as I vary the focal length for this lens:
FocalLength MaximumAperture 18mm F/3.5 19mm F/3.8 24mm F/4.0 31mm F/4.2 35mm F/4.5 44mm F/4.8 50mm F/5.0 58mm F/5.3 70 - 135mm F/5.6
Once I get past 70mm focal length, the amount of light reaching my camera's sensor is half the amount it received when the focal length was between 24 and 30mm (F/5.6 vs F/4.0). This is an excellent "walkaround" lens. Yes, there are distortion and vignetting issues at specific settings, but once you understand how to work around these points the lens performs very well. It is very sharp throughout the F/stop range. Unfortunately, the variable F/stop range available with this lens means it cannot be used in poor light without using a flash or very slow shutter speeds.
I also have a Tamron 17-50mm constant F/2.8 and a Nikon 50mm F/1.8 lens. At 50mm, the Tamron can let in more than 3 times as much light as the variable aperture kit lens and the 50mm Nikon can let in more than 7 times as much light! This has several effects. First, I can shoot the same shot at much higher shutter speeds and have the same exposure. Second, I have a greater range of apertures to choose from so I can control the depth of field to a tighter range. Third, the viewfinder is much brighter, making it easier to see what you're shooting and fourth, the camera is able to focus better in low light.
How can the camera focus better in low light with a larger F/stop? When using your camera to set up a shot, the aperture remains wide open (lowest F/stop number, largest opening). It is only when you actually take the shot that the aperture closes down to the selected position. [Note: If your camera has a "Depth of Field Preview" button, you can push this to manually close down the aperture to the selected position and see how much of your picture will be in focus.] With your camera using the largest opening possible for the given lens, the autofocus circuitry has the maximum amount of light (and thereby contrast) available to control the focus. When lens A lets in 2 or 4 or 7 times as much light as lens B, the camera can control focus MUCH easier with lens A.
Now you see why that Nikon 70-200mm, constant F/2.8 VR lens costs four times as much as my 70-300mm, F/4.5-5.6 VR lens.
Happy shooting,
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# The following equation has: ${{x}^{\left[ {{\left( {{\log }_{5}}x \right)}^{2}}-\left( 9/2 \right){{\log }_{5}}x+5 \right]}}=5\sqrt{5}$A. exactly three real solutionsB. at least one real solutionsC. exactly one irrational solution D. complex roots
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Hint: Here we have to substitute ${{\log }_{5}}x=t$ and $x={{5}^{t}}$ and apply rule that when bases are same then exponents will be equal and then factorize the equation to get values of t and put it back to get values of x.
Complete step-by-step solution:
In the question we are given the function,
${{x}^{\left[ {{\left( {{\log }_{5}}x \right)}^{2}}-\left( 9/2 \right){{\log }_{5}}x+5 \right]}}=5\sqrt{5}\ldots \ldots (1)$
Now for solving the equation (1) we will consider ${{\log }_{5}}x=t$ then the equation (1) can be represented as
${{x}^{\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}=5\sqrt{5}\ldots \ldots (2)$
Here in the equation (2) we can replace x by ${{5}^{t}}$ as ${{\log }_{5}}x=t$ so it can be written as,
${{5}^{t\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}=5\sqrt{5}\ldots \ldots (3)$
In the right hand side of equation (3) $5\sqrt{5}$ can be represented as ${{5}^{\dfrac{3}{2}}}$ to make the bases same so that we can apply the rule of indices which is when bases are same then exponents will be equal and vice-versa hence equation (3) can be represented as,
${{5}^{t\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}={{5}^{\dfrac{3}{2}}}\ldots \ldots (4)$
Now by applying the rule of indices which we get,
${{t}^{3}}-\dfrac{9}{2}{{t}^{2}}+5t=\dfrac{3}{2}\ldots \ldots (5)$
Now we are moving $\dfrac{3}{2}$ from right hand side to left hand side of equation (5) so we get,
${{t}^{3}}-\dfrac{9}{2}{{t}^{2}}+5t-\dfrac{3}{2}=0\ldots \ldots (6)$
Now we will factorize the left hand side of equation (6) to find the value of t so it goes like,
${{t}^{3}}-{{t}^{2}}-\dfrac{7}{2}{{t}^{2}}+\dfrac{7}{2}t+\dfrac{3}{2}t-\dfrac{3}{2}=0$
which can be further written as,
${{t}^{2}}\left( t-1 \right)-\dfrac{7}{2}t\left( t-1 \right)+\dfrac{3}{2}\left( t-1 \right)=0$
Taking $\left( t-1 \right)$ common we will get,
$\left( t-1 \right)\left\{ {{t}^{2}}-\dfrac{7}{2}t+\dfrac{3}{2} \right\}=0$
Further factoring $\left( {{t}^{2}}-\dfrac{7}{2}t+\dfrac{3}{2} \right)$ we get,
$\left( t-1 \right)\left\{ {{t}^{2}}-\dfrac{1}{2}t-3t+\dfrac{3}{2} \right\}=0$
which can be further written as,
$\left( t-1 \right)\left\{ t\left( t-\dfrac{1}{2} \right)-3\left( t-\dfrac{1}{2} \right) \right\}=0$
which can finally written as,
$\left( t-1 \right)\left( t-3 \right)\left( t-\dfrac{1}{2} \right)=0$
We can say that for values t=1, 3, $\dfrac{1}{2}$ the equation (2) satisfies.
As we know the value of t we can find the value of x by using $x={{5}^{t}}$ which we took earlier for solving.
So, the value of x is ${{5}^{1}},{{5}^{3}},{{5}^{\dfrac{1}{2}}}$ which can be written as 5, 125, $\sqrt{5}$ respectively.
Now analysing the option we can say that equation A. holds as there are exactly three real solutions. For option B. it also holds as it has minimum one real solution to satisfy. Now for option C. it holds as only one irrational solution $\sqrt{5}$ satisfy the equation (1). But it does not hold for option D. as there are no complex solutions.
Hence the correct options are ‘A’, ‘B’, ‘C’.
Note: Student while taking ${{\log }_{5}}x=t$ they confuse how to replace with x so they can use ${{\log }_{a}}b=c$ can also be represented as ${{a}^{c}}=b$ hence it can be used here by ${{\log }_{5}}x=t$ as $x={{5}^{t}}$. And after finding values of t, the students forget to put it back to x as hence get mistakes while choosing options. |
# Numbers Types of Numbers | Number names 1 to 100
Here we will discuss Numbers and its type also see Number names 1 to 10.
## What are Numbers?
Within our everyday routines, we use numerals. They are frequently referred to as numbers.
We can’t count objects, calendars, age, cash, or anything else without numerals. These digits are often used for measuring and several other situations for labeling.
Numbers have features that allow them to conduct numerical calculations on them.
These figures are given both numerically and in utterances. For instance, 4 is written as four, 98 is written as ninety-eight, and so on.
To understand further, students must learn how to write the numerals from 1 to 100 in syllables and by spelling.
There are several sorts of numbers that we study in Mathematics. Natural and whole numerals, odd and even numerals, rational and irrational numerals, etc., are all examples.
In this post, we’ll go through all of the various sorts. Aside from that, numbers are utilized in a variety of industries, including sequence, arithmetic formulas, and so on.
## Types of Numbers
The number system is a way of categorizing numerals into groups. In math, there are several distinct sorts of numbers:
Natural Numerals: Natural numbers are accounting numbers made up of positive numerals ranging from 1 until ∞. The group containing natural numbers is symbolized by the letter “N,” and it consists of N = 6,7,8,9…….
Whole Numerals: Whole numbers are non-negative numbers that have no fractional or decimal components. It is represented by the letter “W,” and the group of whole numbers contains W = 0, 1, 2, 3, 4, 5,………..
Integers: The collection of all whole numerals, as well as a negative group of natural numerals, is known as integers. Integers are represented by the letter “Z,” and the group of integers can be represented as Z = -2,-1,0,1,2.
Real Numbers: Real numbers are all positive and negative numerals, fractions, and decimal values that do not contain imaginary values. The letter “R” is used to signify it.
Rational Numbers: Rational numbers are any numerals that may be represented as a relation of one value to another value. Any integer that may be expressed in terms of p/q complies with the requirement to be a rational number. The rational number is represented by the letter “Q.”
Irrational Numbers: Irrational numbers are those which can be stated as a proportion of one to another and are denoted by the letter “P.”
Complex Numbers: Complex numbers “C” are values that may be represented as a+bi, in which “both a,b” are real values and “i” is an arbitrary value.
Imaginary Numbers: Imaginary numbers are complex numbers that may be expressed as a combination of a real value and the imaginary component “I.”
Whole number All natural numbers including zero are called whole numbers.
Properties of the whole number:-
Addition– closure property- the sum of whole numbers is always a whole number.
Commutative property- a+b = b+a
Associative property- (a+b)+c = a+(b+c)
Identity element- if zero is added to any whole number the sum is the number itself.
0+a = a+0 = a
subtraction – closure property- the difference of two whole numbers is not necessary a whole number a-b ≠ b-a not necessary a whole number.
Commutative property a-b ≠ b-a is not defined.
12-4 ≠ 4-12 is not defined.
Property of zero- 0-a ≠ 0-b is not defined.
Associative property- (a-b)-c ≠ a-(b-c)
Division– Dividend- the number which is to be divided is called a dividend.
Divisor– the number by which dividend is divided is called the divisor.
Quotient– the number of times the divisor is contained in the dividend is called the quotient.
Reminder– the leftover number after division is called the reminder.
Thus, the relationship between these terms.
Fractions The numbers in the form of a/b where a and b are whole numbers and
Rational numbers The numbers in the form of a/b a where a and b are integers.
Fractions are also rational numbers a is called the numerator and b is called denominator
equivalent rational number .
Relational numbers between two relational numbers
If given two relational numbers are a and b then is relational number between a and b.
Example
and
More that one relational numbers between two rational numbers
Write and rational numbers between and
Convert two given rational number into equivalent form with common denominator
To obtain 4 Relational numbers between them multiply numerator and denominators with (4+1=5) to make equivalent relational numbers(Multiply n+1 times where n is required number of relational number between two relational number)
and
then
and
where the denominator of Relational numbers is greater.
Relational numbers multiplication
Relational numbers division
Factor– A factor of a number is an exact divisor of the number. Itself in other word the number that are multiplied to get a product are called the factor . for e.g.- 1,2,3,4,6&12 are the factors of 12.
Multiple- a multiple of a number is a number obtained by multiplying it by a natural number. If we multiply 4 by 1,2,3 we get 4,8, 12 which we all multiple of 4.
Even number– A natural number which is exactly divided by 2 is called an even number.
Odd number– A natural number which is not exactly divisible by 2 is called an odd number.
Prime number– A natural number which is greater than 1 & whose only factors are 1 and the number itself is a prime number.
Composite number– The number having more than two factors are called composite numbers.
Co-prime numbers– The two numbers which has no common factor other than one are called co-prime number (2,3) (3,4)
Further research into numbers shows that some more numerals exist in addition to the ones listed previously, such as even and odd digits, prime numerals, and composite numerals. They are discussed below:
• Even Numbers: Even numbers are numerical values that are perfectly divisible by two. Positive or negative numbers, like -92,-38,6,18,24, 28, etc., are some examples of even numbers.
• Odd Numbers: Odd numbers are numerical values that are not perfectly divisible by two. Positive and negative numbers, like -1,-17,13,15,21,23, are some of the examples of odd numbers.
• Prime Numbers: The numerals with only two factors are known as prime numerals. 1 and the digit itself, to be precise. In other terms, prime numbers are the result of dividing a value by one and the value itself. 13,19,53,61,89 are all prime numbers within 100 number.
• Composite Numerals: A composite value consists of 3 or more factors. For instance, the value 15 is a composite number since it is divisible by 1,3,5 and 15. 44,66,91,82,36 are some of the composite numbers under 100 digit value.
There are some special numbers in mathematics as well,
Cardinal Numbers: A cardinal value means the number of units in a sequence, like fifty, thirteen,sixty, seventy, etc., are some of the examples of cardinal values under 100 digit value.
Ordinal Numbers: To describe the place of anything in a list, we use ordinal numbers. For example, 20th will be written as twentieth.
Nominal Numbers: The term “nominal number” refers to a number that is solely referred to as a name. It has no bearing on a thing’s real value or placement.
Pi: Pi is a unique number that is roughly equivalent to 3.14. The proportion of the circumference of a circle reduced by the diameter of a circle is known as Pi ().
Circumference/Diameter = 3.14.
Euler’s Number (e): Euler’s number is roughly equivalent to 2.71, and it is among the most important metrics in mathematics. It is the root of the natural log and is an irrational numeral.
Golden Ratio: The golden ratio is really a unique number that is roughly equivalent to 1.618. It’s an irrational numerical with no discernible pattern among the numbers.
Various properties of numbers
For real values, the characteristics of numbers are mentioned clearly. The following are some of the shared properties:
Commutative Property: If a and b represent two actual values, then the commutative principle states that their sum or multiplication is equal.
a + b = b+a
b.a = a.b
4+5 = 5+4 and 4.5 = 5.4 are two examples.
Associative Property: If a, b, and c represent three real values, then the associative principle states that
(a+b)+c = a+(b+c)
(a.b).c is the same as a.(b.c)
(4+5)+6 = 4+(5+6), (4.5).6=4.(5.6) are two examples
Distributive Property:If a, b, and c represent 3 real values, then the distributive principle states that a
a*(b+c)=a*c+a*c
We can verify the above principle as,
4*(5+6)=4*5+4*6
4*11=20+24
44=44
L.H.S=R.H.S
Closure Property: If one value is added to some other, the output is a single number, as in a+b = c, where a, b, and c represent 3 real values.
5+6=11 is an example.
Identity Property:The identity property states that if we multiply a number by one or we add zero with it, the result will be the same.
x+0= x
x.1=x
7+0 Equals 7 and 7 x 1 = 7 are two examples.
When a positive number is added by the negative value, the outcome is zero.
0 = x+(-x)
5+(-5) = 5-5 = 0 as an example
Inverse Multiplication: When a number other than 0 is multiplied by its own reciprocal, the outcome is 1.
1 = x * (1/x)
50 x (1/50) = 1 as an example
Property of Zero Product: When x.y = 0, either x or b must be 0.
For instance, 9 x 0 = 0 as well as 0 x 10 = 0
Reflexive Property: The value mirrors the digit itself.
x=1
12=12 is an example.
Indian Numeral System
Consider integer 111 as an example. It’s worth noting that the digit 1 appears thrice in this integer. They each have a distinct worth. We distinguish them by mentioning their positioned value, which can be specified as a digit’s numerical value based on its location in a number. As a result, the leftmost one has a place value of hundreds, whereas the one in the middle has a place value of tens, and the one on the right side has a place value of ones.
Returning to the Indian numeral concept, digits have place values of Ones, Tens, Hundreds, Thousands, Ten Thousand, Lakhs, Ten Lakhs, Crores, etc.
The positional meanings of each digit in the integer 12,48,94,231 are:
• 1 – Ones
• 3 – Tens
• 2 – Hundreds
• 4 – Thousands
• 9 – Ten Thousand
• 8 – Lakhs
• 4 – Ten Lakhs
• 2 – Crores
• 1 – Ten Crores
International Numeral System
As in international numeral standard, digit place values are assigned to Ones, Tens, Hundreds, Thousands, Ten Thousand, Hundred Thousands, Millions, Ten Million, and so on. The position values of every digit in the number 30,264,019 are:
• 9– Ones
• 1– Tens
• 0 – Hundreds
• 4 – Thousands
• 6 – Ten Thousand
• 2 – Hundred Thousands
• 0 – Millions
• 3 – Ten Million
How was Zero Number discovered?
0 (zero) is an integer as well as a numeric value used to depict it in numbers. Zero is a number that denotes the lack of all other numbers. It is the identity component of fractions, real values, and several other arithmetic frameworks and hence plays a crucial role in maths. In position value systems, zero is often used as a temporary replacement, just like a digit.
Aryabhata was a famous Mathematician-Astronomer during the ancient period. He is recognized as one of the finest mathematicians of all periods, having been born in Pataliputra, Magadha. Aryabhata became eternal after giving the public the numeral “0” (zero). The Aryabhatiya, his treatise, contained astronomical and mathematical ideas in which the Globe was assumed to be rotating on its axis, and the planets’ cycles were calculated in relation to the rotation of the sun.
The guidelines for working with integer zero are as follows. Unless otherwise indicated, these laws apply to any complex number x.
• Multiplication: a + 0 = 0 + a = a (In other words, when it comes to addition, 0 is the identity component.)
• Subtraction rules: a-0 = a and 0-a = -a.
• Multiplication: a.0=0.a=0
• Division: For nonzero a, divide by 0 / a = 0. However, a/ 0 is undefined since, as a result of the preceding criterion, 0 doesn’t have a multiplicative inverse. For positive a the quotient rises toward positive infinity as b in a / b tends to zero from positive numbers, while when b tends to zero with negative numbers, the quotient rises toward -∞. Division with zero is indeterminate, as evidenced by the various quotients. Exponentiation: a.0 = 1, with the exception that in rare cases, the condition a = 0 may indeed be left undetermined. 0.a Equals 0 for any positive number a.
• Note:The multiplication of 0 integers equals 1, while the total of 0 numbers equals 0.
## Number names 1 to 100
Here are the Names spellings of numbers from 1 to 100
## Why is it important to name the numbers?
Each student must be familiar with the titles of the numerals. These are essential math skills that will assist children inappropriately spelling numbers. They can also readily write such values when they participate in lessons, and their professors spell them out.
In math’s, integers are extremely important. These numbers provide the foundation for all algebraic and numeric processes, not just in basic school but also in secondary ed.
## How can I learn Number Names 1 to 100 Easily?
All you have to do is assist your children in memorizing the digits up to 20, and then a sequence is established from Thirty to 100, as shown below:
Properties of Whole Numbers:
•The digit 0 is the lowest, and initializing whole numbers, as well as all-natural numbers, including zero, are referred to as whole numbers.
• There is no such thing as a culminating or biggest whole number.
• Because whole numerals are indefinite, there is no greatest whole numeral.
• There is an unlimited amount or an unimaginable range of whole numerals.
• Every Natural Number is a whole numeral.
• Each digit is one higher than the one before it.
• Each Whole number is not a Natural number.
## Example 1: Spell 44 digit.
Solution:
In digit 44, one’s place is taken by 4, and tens place is also taken by 4. So 4 tens and 4 ones are equal to 44, and we can spell 44 as ‘forty-four.’
## Example 2: Write, 18, 20, 14, 200, 5000, 18000, and 60000 in words.
Solution:
• 18 in words – Eighteen
• 20 in words – Twenty
• 14 in words – Fourteen
• 200 in words – Two Hundred
• 5000 in words – Five thousand
• 18000 in words – Eighteen thousand
• 60000 in words – Sixty thousand
## Example 3:By using Number system represent 8,299,213,811,552
Solution:
In the starting, the first digit is 8. It is present in the place of a trillion – eight trillion
The next comma to the right side is billion – two hundred ninety-nine billion
The next comma to the right side is million – two hundred thirteen million
The next comma to the right side is thousands –eight hundred eleven thousand
The comma present on the rightmost side represents – five hundred and fifty-two
8,299,213,811,552 in number system format is eight trillion, two hundred ninety-nine billion, two hundred thirteen million, eight hundred eleven thousand, five hundred and fifty-two.
Example 4:A student did a study and concluded that the number of users of laptops in Russia in a week in the year 2015 was 456643218. Write it in the number system.
Solution:
After applying the commas, the numeral is represented by 456,643,218.
Millions– four hundred and fifty-six million
Thousands– six hundred and forty-three thousand
One’s Place – two hundred and eighteen
The number of users of laptops in Russia in a week in the year 2015 was four hundred and fifty-six million, six hundred and forty-three thousand, two hundred and eighteen(456,643,218).
Example 5: Following statements of numbers are given write their numeric value
a] Ninety-four million, five hundred seven thousand, nine hundred thirty-one
b] Eight billion, one hundred thirty-five million, eighty-two thousand, three hundred forty-five
Solution:
a] Ninety-four million, five hundred seven thousand, nine hundred thirty-one
The Numeric Value is 94,507,931.
b] Eight billion, one hundred thirty-five million, eighty-two thousand, three hundred forty-five
The Numeric Value is 8,135,082,945.
Example 6:
Choose the correct option:
Ninty billion two hundred thirty-five million four hundred thirteen thousand four hundred five
a] 90,235,413,405
b] 9,023,514,301 ,405
c]90,235,403,405
Solution:
The correct option is 90,235,413,405
Example 7: Give the numerical value of the following statements:
a] The distance between Earth and Sun is around 58 billion km.
b] The net weight of a navy ship is 204 km pounds.
Solution:
The situations can be represented according to the number system as:
a] The distance between Earth and Sun is approximately 58 billion km.
According to the number system, it is 58,000,000,000 km
b] The net weight of a navy ship is 305 km pounds.
According to the number system, it is 305,000,000 km
Example 8: New York’s reserve fund is around 555 trillion dollars. Write the reserve fund according to the number system?
Solution:
New York’s reserve fund is around 555 trillion dollars.
Be sure to take care of commas or periods; three digits are given in billions.
To express the value in the number system, the rest values of millions and others are made zero.
555 billion
Millions – 000
Thousands – 000
Ones – 000
The budget announced was about 555,000,000,000 dollars.
Check these most asked questions on the webpage:
Q1) How do you write 450 in words?
A1) 450 is written as Four Hundred Fifty.
Q2) How to write 9 in English?
A2) Nine
Q3) How can I spell 15?
A3) 15 is spelled as Fif+teen= Fifteen
Q4) Is there any use of learning number system?
A4) Digits can be used to quantify, weigh, classify, and organize items, as well as for barcodes and license plates. Even primitive civilizations placed graphs or charts on items to indicate that they were counting things, if it was weeks, recording transactions, or anything else. There is also indications that the digits were utilized by early Bablyonians and Romans for different uses.
We can use the numbers practically in following forms-
• A laptop translates any characters, phrases, or passwords into a defined numerical notation, which is incredibly useful.
• Digits assist us in expressing or displaying the accurate estimation of any thing.
• A quantitative data collection method is used to arrange or display distinct elements in a specific sequence.
## Q: The sum of a number and twenty-one is sixty-four. what is the value of the number? the number equals.
Let us the number is x
the sum of a number and twenty-one is sixty-four
x+21=64
so x= 64-21
x=43
so the number is 43 |
Home | | Maths 7th Std | Multiplication of Decimal Numbers
# Multiplication of Decimal Numbers
(i) Decimal multiplication through models (ii) Multiplication of Decimal Numbers by 10, 100 and 1000
Operations on Decimal Numbers
Already we are familiar with decimal numbers. We know how to represent a decimal number as a fraction and the place values of digits. Now, let us learn the operations on decimal numbers.
Multiplication of Decimal Numbers
Mathan wants to buy a shirt material which costs ₹ 75.50 per metre. He needs 1.5 metre to stitch a shirt. How much does he have to pay? Here we need to multiply 75.50 and 1.5. In real life, there are many situations where we need to multiply decimal numbers.
(i) Decimal multiplication through models
Let us try to understand decimal multiplication using grid model.
Let us find 0.1 × 0.1.
0.1= 1/10 . Therefore, 0.1× 0.1= 1/10× 1/10
That is, 1/10 th of 1/10 .
Shade horizontally 1/10 by blue colour (Fig. 1.7). Shade vertically 1/10 by green colour (Fig. 1.8).
Then 1/10 th of 1/10 is the common portion, that is 1/100 th .
Therefore, 1/10 × 1/10 = 1/100 = 0.01
Hence, 0.1 × 0.1 = 0.01.
Example 1.16
Find 0.3 × 0.4
Solution
First shade 4 rows of the grid in blue colour to represent 0.4. Shade 3 columns of the grid in green colour to represent 0.3 of 0.4. Now 12 squares represents the common portion. This represents 12 hundredth or 0.12. Hence 0.3 × 0.4 = 0.12.
Note
The number of decimal digits in 0.12 is two. So, we can conclude that the number of decimal digits in the product of two decimal numbers is equal to the sum of decimal digits that are multiplied.
Area model
We have already learnt about the area model in the addition and subtraction of decimal numbers. In the same way we are going to multiply the decimal numbers. Now we shall see an example.
Example 1.17
Multiply 3.2 × 2.1
Solution
Let us try to represent the product of decimal numbers (3.2 × 2.1) as the area of a rectangle. Let us consider a rectangular portion as shown in Fig. 1.9.
The rectangular portion is split into 3 wholes and 2 tenth along it’s length (Fig. 1.10).
Since, 3 wholes and 2 tenth is multiplied with 2.1, we split the same area into 2 wholes and 1 tenth along its breadth (Fig. 1.11).
Here, each row contains 3 wholes and 2 tenth. Each column contains 2 wholes and 1 tenth. The entire area model represents 6 wholes, 7 tenth and 2 hundredth.
Therfore, 3.2 × 2.1 = 6.72
Think
How are the products 2.1 × 3.2 and 21 × 32 alike? How are they different.
Solution:
2.1 × 3.2 = 6.72 and 21 × 32 = 672.
In both the cases the digits ambers are the same. But the place value differs.
Try these
(1) Shade the grid to multiply 0.3 × 0.6.
Solution:
3 rows of Yellow represent 0.3, 6 columns of Red colour represent 0.6 Double shaded 18 squares of orange colour represent.
0.30 × 0.6 = 0.18
(2) Use the area model to multiply 1.2 × 2.5
Solution:
Here each row contains 1 whole and 2 tenths. Each column contains 2 wholes and 5 tenths. The entire area model represents 2 wholes 9 tenths and 10 hundredths ( = 1 tenths). So 1.2 × 2.5 = 3.
Example 1.18
Multiply the following
(i) 2.3 × 1.4
(ii) 5.6 × 3.2
Solution
(i) 2.3 × 1.4
First, let us multiply 23 × 14
23×14=322
Now 2.3 × 1.4 = 3.22
(ii) 5.6 × 3.2
First, let us multiply 56 × 32
56 × 32 =1792
Now 5.6 × 3.2 = 17.92
Example 1.19
Latha purchased a churidhar material of 3.75 m at the rate of ₹ 62.50 per metre. Find the amount to be paid.
Solution
Cost of churidhar material = ₹ 62.50 per metre
Length of churidhar material = 3.75 m
Amount to be paid = 3.75 × 62.50
= ₹ 234.3750
= ₹ 234.38 (rounded to two decimals).
Example 1.20
The length and breadth of a rectangle is 23.5 cm and 1.5 cm respectively. Find the area of the rectangle.
Solution
Area of a rectangle = l × b sq. units
Here, l =23.5 cm, b=1.5 cm
Area of the rectangle = 23.5 × 1.5
= 35.25 sq.cm.
(ii) Multiplication of Decimal Numbers by 10, 100 and 1000
We have studied about conversion of decimals into fractions in the first term. Consider 45.6 and 4.56.
Expressing these decimal numbers into fractions, we get
45.6 = 40 + 5 + 6/10 = 45 + 6/10 = 456/10
Now, 4.56 = 4 + 5/10 + 6/100= 456/100
Comparing the two fractions, we see that if there is one digit after the decimal point, then the denominator is 10 and if there are two digits after the decimal point, then the denominator is 100.
Let us see what happens if a decimal number is multiplied by 10, 100 and 1000.
Observe the following table and complete it.
Try these
Can you observe any pattern in the above table? There is a pattern in the shift of decimal points of the products in the table. In 2.35 × 10 = 23.5 the digits are the same, that is 2, 3, 5. Observe 2.35 and 23.5. To which side has the decimal point been shifted, right or left? The decimal point is shifted to the right by one place. Also note that 10 is one zero followed by 1.
In 2.35 × 100 = 235.0, observe 2.35 and 235. To which side and by how many digits has the decimal point been shifted? The decimal point has been shifted to the right by two places. Note that 100 is two zeros followed by 1.
Similarly in 2.35 × 1000 = 2350.0, we see that the decimal point has been shifted to the right by three places by adding one more digit 0 to the number 235. Note that 100 is three zeros followed by 1.
So we conclude that when a decimal number is multiplied by 10, 100 or 1000, the digits in the product are same as in the decimal number but the decimal point in the product is shifted to the right by as many places as there are zeros followed by 1.
Based on these observations we can now say,
0.02 × 10 = 0.2; 0.02 × 100 = 2 and 0.02 × 1000 = 20.
Can you find the following?
2.76×10=?
2.76 × 100 = ?
2.76 × 1000 = ?
Try these
Find
(i) 9.13×10 = 91.3
(ii) 9.13×100 = 913
(iii) 9.13×1000 = 9130
Example 1.21
Find the value of the following
(i) 3.26×10
(ii) 3.26×100
(iii) 3.26×1000
(iv) 7.01×10
(v) 7.01×100
(vi) 7.01×1000
Solution
(i) 3.26 × 10 = 32.6
(ii) 3.26 × 100 = 326.0
(iii) 3.26 × 1000 = 3260.0
(iv) 7.01 × 10 = 70.1
(v) 7.01 × 100 = 701.0
(vi) 7.01 × 1000 = 7010.0
Example 1.22
A concessional entrance ticket for students to visit a zoo is ₹ 12.50. How much has to be paid for 20 tickets?
Solution
Cost of one ticket = ₹ 12.50
Amount to be paid for 20 students = 12.50 × 20 = ₹ 250.00
We have already discussed about the multiplication of decimal numbers by 10, 100 and 1000. In the same way we can find patterns for multiplying decimal numbers by 0.1, 0.01 and 0.001. Observe the following.
From the above multiplication we can conclude that, when multiplying by
• 0.1, the decimal point moves one place left.
• 0.01, the decimal point moves two places left.
• 0.001, the decimal point moves three places left.
Zeros may be needed after the decimal point as they are needed before the decimal point of numbers when multiplied 0.1, 0.01 and 0.0001.
Try this
Complete the following table :
Tags : Number System | Term 3 Chapter 1 | 7th Maths , 7th Maths : Term 3 Unit 1 : Number System
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 3 Unit 1 : Number System : Multiplication of Decimal Numbers | Number System | Term 3 Chapter 1 | 7th Maths |
# 7.05 Powers of 10
Lesson
In previous years we have learned how to multiply whole numbers by 10 and can relate the multiplication to place value. Let's review.
### Examples
#### Example 1
Complete these number sentences.
a
9\times10=⬚
Worked Solution
Create a strategy
Put the number in a place value table and move the digit one column to the left.
Apply the idea
Here is 9 in a place value table.
To multiply by 10, move each digit one column to the left. Use 0 as a placeholder.
9\times 10 = 90
b
90\times10=⬚
Worked Solution
Create a strategy
Put the number in a place value table and move each digit one column to the left.
Apply the idea
Here is 90 in a place value table.
To multiply by 10, move each digit one column to the left. Use 0 as a placeholder.
90\times 10 = 900
Idea summary
When we multiply a number by 10, each digit increases in place value by 1 place value.
We can use a place value table to multiply numbers by powers of 10.
## Multiply decimals by 10 or 100
This video looks at what happens to numbers with decimals when we multiply by 10 or 100.
### Examples
#### Example 2
Find the value of 29.34\times100.
Worked Solution
Create a strategy
Put the decimal number in a place value table and move each digit two columns to the left.
Apply the idea
Here is 29.34 in a place value table.
To multiply by 100, move each digit two columns to the left.
29.34\times100= 2934
Idea summary
We can use a place value table to multiply decimals by powers of 10.
When multiplying by 10 each digit moves one place value to the left.
When multiplying by 100 each digit moves two place values to the left.
## Divide decimals by 10 or 100
This video looks at what happens to numbers with decimals when we divide by 10 or 100.
### Examples
#### Example 3
Find the value of 15.8 \div 10.
Worked Solution
Create a strategy
Put the decimal number in a place value table and move each digit one column to the right.
Apply the idea
Here is 15.8 in a place value table.
To divide by 10, move each digit one column to the right.
15.8 \div 10 =1.58
Idea summary
We can use a place value table to multiply decimals by powers of 10.
When multiplying by 10 each digit moves one place value to the left.
When multiplying by 100 each digit moves two place values to the left.
## Multiply and divide decimals by larger powers of 10
This video looks at what we can do when multiplying and dividing by 10\,000.
### Examples
#### Example 4
Find the value of 63.05\times10\,000.
Worked Solution
Create a strategy
Put the decimal number in a place value table and move each digit four columns to the right.
Apply the idea
Here is 63.05 in a place value table.
To multiply by 10\,000, move each digit four columns to the left. Use 0s as placeholders.
63.05\times10\,000=630\,500
Idea summary
For each power of 10: \, 10,\,100,\,1000,\,10\,000, when you multiply or divide by these values, each digit of your number is increased or decreased by that value, and moves up or down place value positions.
The number of place value positions each digit moves is equal to the number of zeros in the power of 10.
### Outcomes
#### MA3-7NA
compares, orders and calculates with fractions, decimals and percentages |
## Math-Linux.com
Knowledge base dedicated to Linux and applied mathematics.
Home > Mathematics > Trigonometry > Demonstration / proof of cos²x + sin²x=1
# Demonstration / proof of cos²x + sin²x=1
All the versions of this article: <English> <français>
We will demonstrate the following equality cos²x + sin²x=1 in several ways. By using the notion of derivative, addition formula and then geometrically by using the trigonometric circle.
Let’s prove the following equality:
$$\forall x\in \mathbb{R}, \quad\cos^2 x+ \sin^2 x =1$$
## Proof/Demonstration using the derivative
Let $f$ be the function defined as follows:
$$\forall x\in \mathbb{R}, \quad f(x)=\cos^2 x+ \sin^2 x$$
\begin{aligned} f’(x)&=(\cos^2 x+ \sin^2 x)’ \\ &= 2 \cos x (-\sin x) + 2 \sin x \cos x\\ &= - 2 \cos x \sin x + 2 \sin x \cos x\\ & =0 \end{aligned}
This means that $f$ is constant on $\mathbb{R}$:
$$\exists C \in \mathbb{R},\forall x\in \mathbb{R}, \quad f(x)=C$$
Let’s take $x=0$:
$$f(x=0)=\cos^2 0+ \sin^2 0=1$$
We conclude:
$$\forall x\in \mathbb{R}, \quad f(x)=\cos^2 x+ \sin^2 x =1$$
## Proof/Demonstration using addition formulas
We had previously demonstrated the addition formula
$$\cos(a+b)=\cos a \cos b - \sin a \sin b$$
Let $a=x\in \mathbb{R}$. Let $b=-a=-x$, we have, since cosine is an even function and sine an odd function:
\begin{aligned} \cos(x-x)&=\cos x \cos (-x) - \sin x \sin (-x)\\ &= \cos x \cos x - \sin x (-\sin x)\\ &= \cos^2 x + \sin^2 x\\ &=\cos 0\\ &=1 \end{aligned}
We conclude then:
$$\forall x\in \mathbb{R}, \quad f(x)=\cos^2 x+ \sin^2 x=1$$
## Proof/Demonstration using the trigonometric circle
Consider the trigonometric circle of radius $r=1$. In the following triangle, we can apply the Pythagorean theorem:
$x= \cos \theta$, $y= \sin \theta$. The hypotenuse $r= 1$, we then have:
\begin{aligned} x^2+y^2=r^2&=1 \\ \cos^2 \theta+ \sin^2 \theta&=1 \end{aligned}
Then we have:
$$\forall \theta\in [0, 2\pi ], \quad \cos^2 \theta+ \sin^2 \theta =1$$
We conclude that:
$$\forall x\in \mathbb{R}, \quad\cos^2 x+ \sin^2 x =1$$
The conversion from radians $\theta$ to degrees $x$ is done as follows:
$$x=\theta \times \frac{180}{\pi}$$
Example: $\theta=\pi/4$
$$x= \frac{\pi}{4} \times \frac{180 }{\pi}= \frac{180 }{4}=45^\circ$$ |
### Let $a$ and $b$ be positive real numbers such that$\dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0$ Find the value of $\left(\dfrac{b}{a} + \dfrac{a}{b} \right)^2$.
$5$
Step by Step Explanation:
1. We are given,
$\dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0$
Let $x = \dfrac{b}{a}$
$\implies b = ax$
2. Now,
\begin{align} & \implies \dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{ a + b } = 0 \\ & \implies \dfrac{1}{a} - \dfrac{1}{ ax } - \dfrac{1}{ a + ax } = 0 \\ & \implies \dfrac{1}{a}\left(1 - \dfrac{1}{ x } - \dfrac{1}{ 1 + x } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x(1 + x) - (1 + x) - x }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x + x^2 - 1 - x - x }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{1}{a}\left( \dfrac{ x^2 - x - 1 }{ x(1 + x) } \right) = 0 \\ & \implies \dfrac{ x^2 - x - 1 }{ x(1 + x) } = 0 && (\because a > 0) \\ & \implies x^2 - x - 1 = 0 \\ & \implies x = \dfrac{ 1 \pm \sqrt{ 1 + 4 } }{ 2(1) } \\ & \implies x = \dfrac{ 1 \pm \sqrt{5} }{ 2 } \\ & \text{ Since a and b are positive, x > 0 } \\ & \implies x = \dfrac{ 1 + \sqrt{ 5 } }{ 2 } \end{align}
3. Now,
\begin{align} \left( \dfrac{b}{a} + \dfrac{a}{b} \right)& = \left(x + \dfrac{1}{x} \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 }{ 1 + \sqrt{ 5 } } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 }{ 1 + \sqrt{ 5 } } \times \dfrac{ 1 - \sqrt{ 5 } }{ 1 - \sqrt{ 5 } } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } + \dfrac{ 2 - 2\sqrt{5} }{ -4 } \right)^2 \\ & = \left(\dfrac{ 1 + \sqrt{ 5 } }{ 2 } - \dfrac{1}{2} + \dfrac{ \sqrt{5} }{ 2 } \right)^2 \\ & = \left( \sqrt{5} \right)^2 \\ & = 5 \end{align}
4. Hence, the value of $\left(\dfrac{b}{a} + \dfrac{a}{b} \right)^2$ is $5$. |
Update all PDFs
# Log Ride
Alignments to Content Standards: 6.EE.B.5
A theme park has a log ride that can hold 12 people. They also have a weight limit of 1500 lbs per log for safety reasons. If the average adult weighs 150 lbs, the average child weighs 100 lbs and the log itself weighs 200, the ride can operate safely if the inequality
$$150A + 100C + 200 \leq 1500$$
is satisfied ($A$ is the number of adults and $C$ is the number of children in the log ride together). There are several groups of children of differing numbers waiting to ride. Group one has 4 children, group two has 3 children, group three has 9 children, group four 6 children while group five has 5 children.
If 4 adults are already seated in the log, which groups of children can safely ride with them?
## IM Commentary
In this instructional task students are given two inequalities, one as a formula and one in words, and a set of possible solutions. They have to decide which of the given numbers actually solve the inequalities. There are two solutions presented, the first is the more natural one, where we simply substitute the given values and check if the inequality is satisfied. In the second solution, we actually solve the inequality first and then see if the given values are in the solution set. While this seems unnecessary, it is the approach many students will use because they have often been "conditioned" to solve any inequality or equality they encounter. A class discussion could clear up when it is necessary to actually solve an inequality and when it is easier to use substitution. If we were given a large number of possible solutions to check, solving would be the preferred solution method, while for a small number, substitution is faster.
This task provides a way to see if students understand the meaning of a solution to an inequality (or equation), namely values of the variables that make the inequality (equation) true. Often students see solutions only as the result of solving an inequality or equation. This idea is an important one, since for some equations it is possible to check if a given number is a solution, while it would be hard or impossible to solve, e.g. higher degree polynomials, equations involving trigonometric or exponential functions.
A natural extensions for this task would be to ask students to find other groups of adults and children that can or cannot ride together, either because of the number of people allowed or because of the weight limit. Some combinations are not possible. For example, it is not possible to find a group of children that satisfies the number requirement but does not satisfy the weight requirement.
## Solutions
Solution: By Substitution
As there are five different groups waiting to ride, we can substitute each group’s number of children into the weight equation for $C$, using $A = 4$ as there are already 4 adults seated in the log. For group one, we get
$$150(4) + 100(4) + 200 = 1200 \leq 1500$$
So, group one can safely ride the ride. Using this technique with the other four groups, we find that group two, group four, and group five can all safely ride the ride both in weight and in number-of-riders. We find that group three exceeds the number-of-riders limit, as there will then be 13 people on the log, as well as the weight limit.
$$150(4) + 100(9) + 200 = 1700 \not \leq 1500$$
We can also observe that group one together with group two would be able to ride the log ride.
$$150(4) + 100(4 + 3) + 200 = 1500$$ and $$7 + 4 \lt 12$$
Solution: Solving the Inequality
While this approach is not expected of 6th graders, students who are ready for this approach may try it in 6th grade. This solution approach is expected in 7th grade.
First, we have that $A = 4$, as there are 4 adults already seated. By substitution, we can solve for $C$ and see how many children can safely ride based solely on the weight limit.
\begin{align} 150(4) + 100C + 200 &\leq 1500 \\ 600 + 100C + 200 &\leq 1500 \\ 100C &\leq 700 \\ C&\leq7 \end{align}
So, up to 7 children will satisfy the weight limit inequality. Since $7 + 4 \lt 12$, the number-of-riders limit is also satisfied, and so we find that 7 is the maximum number of children that can safely ride the ride with the 4 adults. Thus, group three is the only group that cannot safely ride in the log. Note: We can observe that group one and group two would be able to ride the log ride together with the 4 adults.
#### kassymov says:
over 3 years
The cluster is "Reason about and solve one-variable equations and inequalities." Will it be better to combine the adults and the log into one number (for example, having one or two park people guiding the log and providing their weight + the weight of the log)?
If the tasks stays the way it is, in the part "Solution: Solving the Inequality" it's hard to make sense of the phrase "using the given equation".
#### Kristin says:
over 3 years
If students think to do that, then it certainly makes sense; as long as the reasoning is correct, I think it fits this cluster.
I agreed about the awkwardness of the phrase "using the given equation" and removed it. Thanks!
#### NEB says:
almost 5 years
I understand the Substitution Method for solving at Grade 6. I interpreted the standards as reserving solving inequalities for 7th grade. In sixth grade I only see '1-step' inequalities written and the recognition that they have multiple solutions. Actually solving is in later grades, as is having multi-step inequalities. Please correct me if I have misinterpreted this progression.
#### Kristin says:
almost 5 years
I think you are right. I've added a caveat to the third solution--do you think this is sufficient or would it be better to move this to the commentary of the task so that there is no question about the difference in expectation from 6th to 7th grade? |
# Constructing a 60°, 30° or 15° Angle
Rule
### InstructionsforConstructinga$\text{}60\text{}\text{°}$Angle
1.
Draw a straight line $l$.
2.
Mark a point called $P$ on the line $l$.
3.
Put the point of your draft compass on $P$ and make an arc that intersects $l$ and passes above $P$. Mark a point $A$ where the arc intersects with $l$.
4.
Without changing the distance between the legs of your draft compass, put the point on $A$ and make a small arc that intersects with the larger arc you made previously.
5.
Draw a line from the point $P$ through the intersection of the two arcs. Ta-da, you’ve got a $60$° angle.
## Constructing a $30\text{°}$ Angle
If you need to find an angle that is the half the size of another angle, you use the bisection technique. So, if you want to construct a $30$° angle, you just construct a $60$° angle, and then bisect it. You just make a cross midway between the two sides of the angle and draw a line from the vertex through this cross.
To do this, set the point of your draft compass at $A$ and draw an arc to the right of your existing $60$° arc. Keep the distance of the legs of your draft compass, and set the point of the compass at the intersection of the two crossing arcs and the $60$° line. From this point, draw a new arc intersecting the arc you marked when the point was set at $A$. These intersecting arcs should make a cross. Finally, to bisect the $60$° angle, draw a straight line from $P$ until this line intersects the newly created cross.
The two new angles are both $30$°, because $30+30=60$.
## Constructing a $15\text{°}$ Angle
To make a $15$° angle, you’ll once again need to construct an angle that is half the angle you have just constructed, so you only need to use the bisection technique again. In other words, all you need to do to construct a $15$° angle is bisect a $30$° angle.
That means you just make a cross midway between the two sides of the angle using a drafting compass and draw a line from the vertex through this cross, using the same instructions as before. The two new angles are both $15$°, because $15+15=30$. |
# Simple Ratios problem from old SAT book
• Ognerok
In summary, the conversation discusses the problem of finding the cent value of a set of 20 coins with a ratio of 2:3 between nickels and dimes. Through solving the equations, it is determined that there are 12 nickels and 8 dimes in the set. The final step is to calculate the total value of the coins.
Ognerok
## Homework Statement
Aggregate # of nickels and dimes = 20 coins. If the ratio between nickels and dimes is 2:3, what is cent value of 20-coin set?
2x = 3y
## The Attempt at a Solution
20 = nickels(x)+dimes(y); ratio = 2x:3y; 2x = 3y; y = 2x/3
20 = x + 2x/3; x = 12 nickels; 6 dimes? ...confused from there.
Ognerok said:
## Homework Statement
Aggregate # of nickels and dimes = 20 coins. If the ratio between nickels and dimes is 2:3, what is cent value of 20-coin set?
2x = 3y
## The Attempt at a Solution
20 = nickels(x)+dimes(y); ratio = 2x:3y; 2x = 3y; y = 2x/3
20 = x + 2x/3; x = 12 nickels; 6 dimes? ...confused from there.
2x=3y => 24=3y => y=8, not 6!
N:D = 2:3
total ratio =5
thus, number of N = 2/5 * 20. I think you can do the rest.
## What is a simple ratio problem?
A simple ratio problem is a type of mathematical problem that involves finding the relationship between two quantities by expressing them in the form of a ratio.
## How is a simple ratio problem typically presented?
A simple ratio problem is usually presented as a word problem, where two quantities are compared using words such as "per", "to", or "for every".
## What are the steps for solving a simple ratio problem?
The steps for solving a simple ratio problem include: 1) Identifying the two quantities being compared, 2) Expressing the two quantities in the form of a ratio, 3) Simplifying the ratio if possible, and 4) Solving for the unknown quantity using cross-multiplication.
## What are some common mistakes when solving simple ratio problems?
Some common mistakes when solving simple ratio problems include: 1) Not correctly identifying the two quantities being compared, 2) Using the wrong numbers or units in the ratio, 3) Forgetting to simplify the ratio before solving, and 4) Making errors in cross-multiplication.
## How can simple ratio problems be applied in real-life situations?
Simple ratio problems can be used to solve everyday problems, such as finding the cost per unit of a product, determining the ratio of ingredients in a recipe, or calculating the speed of an object based on distance and time. They are also commonly used in business and finance to analyze financial statements and make investment decisions.
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CBSE NOTES CLASS 9 MATHEMATICS CHATER 2
POLYNOMIALS
Basic Concepts and Important Formulae
• A polynomial p(x) in one variable x is an algebraic expression in x of the form
p(x) = anxn + an–1xn–1 + . . . + a2x2 + a1x + a0,
where a0, a1, a2, . . ., an are real constants and an ≠ 0.
• The powers of x in a polynomial can only be positive integers.
• The constants a0, a1, a2, . . ., an are respectively the coefficients of x0, x, x2, . . ., xn. A coefficient can be any real number.
• Each of anxn, an–1xn–1, …, a2x2, a1x, a0, with an ≠ 0, is called a term of the polynomial p(x).
• Polynomials are denoted by p(x), q(x), g(x), r(x) etc.
For example in p(x) = 2x2 + 3x - 1
Here the terms are 2x2, 3x and - 1
The coefficient of x2 is 2, of x if 3 and constant term is-1
• Variable names can be x, y, t u, v etc.
• A polynomial of one term is called a monomial.
A polynomial of two terms is called a binomial.
A polynomial of three terms is called a trinomial.
• The highest power of the variable x (that is n) is called degree of the polynomial.
A polynomial with degree 0 has only a non-zero constant term and it is called a constant polynomial, i.e. p(x) = 2.
Only a constant term with value 0 is called zero polynomial. The degree of zero polynomial is not defined.
A polynomial of degree one is called a linear polynomial, i.e.,
p(x) = 3x + 1
A polynomial of degree two is called a quadratic polynomial, i.e.,
p(x) = 2x2 + 3x - 1.
A polynomial of degree three is called a cubic polynomial, i.e.,
p(x) = 4x3+2x2+3x-1
• In standard form of a polynomial, the terms are arranged from highest power of variable to lowest power.
For a quadratic polynomial the standard form is ax2+bx+c, a ≠ 0
Value of a polynomial
By putting x = a, we get the value of polynomial at x = a.
For example, if p(x) = 5x3 - 2x2 + 3x – 4, then,
p(0) = value of polynomial at (x=0)
= 5×03 - 2×02 + 3×0 - 4 = -4
p(1) = value of polynomial at (x=1)
= 5×13 - 2×12 + 3×1 - 4
= 5 – 2 + 3 - 4 = 2
And so on.
Zeroes of a polynomial
• A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. ‘a’ is also called a root of the equation p(x) = 0.
• Every linear polynomial in one variable has a unique zero
• A non-zero constant polynomial has no zero.
• Every real number is a zero of the zero polynomial.
• For any polynomial, number of zeroes ≤ degree of polynomial
• For finding zeroes of a polynomial, put p(x) = 0 and solve for x.
For example, to find zero of 3x – 2, we put,
3x - 2 = 0 x = $\frac{2}{3}$.
Hence $\frac{2}{3}$ is zero of 3x -2.
Dividing one polynomial by another
Example
Divide p(x) = x + 3x21 by g(x) =1 + x
Step 1: Write the dividend x + 3x2 – 1 and the divisor 1 + x in the standard form, i.e., after arranging the terms in the descending order of their degrees. So, the dividend is 3x2 + x –1 and divisor is x + 1.
Step 2: Divide the first term of the dividend by the first term of the divisor, i.e., we divide
3x2 by x, and get 3x. This gives the first term of the quotient.
Step 3: Mmultiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply x + 1 by 3x and subtract the product 3x2 + 3x from the dividend 3x2 + x – 1. This gives us the remainder as –2x – 1.
Step 4: Treat the remainder –2x – 1 as the new dividend. The divisor remains the same. Repeat Step 2 to get the next term of the quotient, i.e., we divide the first term – 2x of the (new) dividend by the first term x of the divisor and obtain – 2. Thus, – 2 is the second term in the quotient.
Step 5: Multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is, we multiply x + 1 by – 2 and subtract the product – 2x – 2 from the dividend – 2x – 1. This gives us 1 as the remainder.
• This process continues till the remainder is 0 or the degree of the new dividend is less than the degree of the divisor. At this stage, this new dividend becomes the remainder and the sum of the quotients gives us the whole quotient.
Step 6: Thus, the quotient in full is 3x – 2 and the remainder is 1.
The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomials q(x) and r(x) such that
dividend = divisor × quotient + remainder
p(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree r(x) < degree g(x).
Factor
If r(x) = 0, then p(x) = g(x) × q(x) ⇒q(x) and g(x) are called factors of p(x).
If g(x) is a factor of p(x) then, we put g(x) = 0 and solve for zeroes of g(x). If we put the value of this zero in p(x) we should get r(x) = 0
Remainder Theorem
If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial x a, then the remainder is p(a).
Proof: Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x a, the quotient is q(x) and the remainder is r(x), i.e.,
p(x) = (x a) q(x) + r(x)
Since the degree of x a is 1 and the degree of r(x) is less than the degree of x a, the degree of r(x) = 0. This means that r(x) is a constant, say r.
So, for every value of x, r(x) = r.
Therefore, p(x) = (x a) q(x) + r
In particular, if x = a, this equation gives us
p(a) = (a a) q(a) + r = r
• We can verify the result by dividing p(x) by x-a by long division method. We see that remainder = p(a).
Factor Theorem: x a is a factor of the polynomial p(x), if p(a) = 0. Also, if x a is a factor of p(x), then p(a) = 0.
Proof: By the Remainder Theorem,
p(x) = (xa) q(x) + p(a).
• If p(a) = 0, then p(x) = (xa) q(x), which shows that xa is a factor of p(x).
• Since xa is a factor of p(x),
p(x) = (xa) g(x) for same polynomial g(x).
In this case, p(a) = (aa) g(a) = 0.
• Factorization by mid-term splitting
Find a and b such that
(x + a) (x + b) = x2 + (a + b)x + ab
is satisfied.
• Factorization by factor theorem
• Factorization by mixed combination
Algebraic identities for factorization
(x + y)2 = x2 + 2xy + y2 ⇒ x2 + y2 = (x + y)2 - 2xy (x – y)2 = x2 – 2xy + y2 ⇒ x2 + y2 = (x - y)2 + 2xy (x + y)2 = (x - y)2 + 4xy ⇒ (x - y)2 = (x + y)2 - 2xy x2 – y2 = (x + y) (x – y) (x + a) (x + b) = x2 + (a + b)x + ab (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx (x + y)3 = x3 + y3 + 3xy(x + y) = x3 + 3x2y + 3xy2 + y3 (x – y)3 = x3 – y3 – 3xy(x – y) = x3 – 3x2y + 3xy2 – y3 x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) x3 + y3 = (x + y) (x2 + y2 - xy) x3 - y3 = (x - y) (x2 + y2 + xy) If x + y + z = 0 then x3 + y3 + z3 = 3xyz |
# Math:Resource Preservation
Let's say you have 120 bars, 80% resource preservation, and 0% doubling chance. If a dagger takes one bar to create, how many daggers would you create on average out of 120 bars?
The answer is $\displaystyle{ \frac{120}{1 - 80/100} = \frac{120}{0.2} = 600. }$
The main idea behind resource preservation is that you can reuse the preserved resources to make more items, but those resources when used will themselves be preserved also.
Using the same example with 120 bars, you take 120 actions producing 120 daggers, but with 80% preservation you are left with $\displaystyle{ 120 \times 0.8 = 96 }$ bars.
Now you can use those 96 bars to do the same thing, and you will be left with $\displaystyle{ 96 \times 0.8 = 76.8 }$ bars on average.
Repeat that to infinity, and you will get the average number of daggers that you will produce from 120 bars.
But how do we calculate this number?
So the number of daggers produced is $\displaystyle{ 120 + (120 \times 0.8) + (120 \times 0.8 \times 0.8) +\cdots + (120 \times 0.8^n) }$ where $\displaystyle{ n }$ goes to infinity.
If we generalize and represent the number of starting resources ($\displaystyle{ 120 }$) with $\displaystyle{ a }$ and the probability ($\displaystyle{ 0.8 }$) with $\displaystyle{ r }$, the number of actions performed with $\displaystyle{ a }$ action's worth of starting resources and a resource preservation of $\displaystyle{ r }$ (where $\displaystyle{ 0 \le r \lt 1 }$) can be expressed as follows:
$\displaystyle{ a + (a \times r) + (a \times r \times r) + \cdots + (a \times r^n) = \sum\limits^{\infty}_{n=0} ar^n }$
This expression is an infinite geometric series, which can simply be expressed as $\displaystyle{ \displaystyle \lim_{n \to \infty}\tfrac{a(1 - r^n)}{1 - r} = \frac{a}{1 - r} }$.
So, coming back to our example where $\displaystyle{ a = 120 }$ and $\displaystyle{ r = 0.8 }$, we get $\displaystyle{ \frac{120}{1 - 0.8} = 600 }$. |
# Math explainer
Looking for Math explainer? Look no further! We can solve math word problems.
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Best of all, Math explainer is free to use, so there's no reason not to give it a try! In addition, the built-in practice exercises further reinforce the lesson material. As a result, Think Through Math is an extremely effective way for students to learn and improve their math skills. Best of all, the app is available for free, so there's no excuse not to give it a try!
Solving for a side in a right triangle can be done using the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This theorem can be represented by the equation: a^2+b^2=c^2. In this equation, c is the hypotenuse and a and b are the other two sides. To solve for a side, one would rearrange this equation to isolate the desired variable. For example, to solve for c, one would rearrange the equation to get c^2=a^2+b^2. To solve for a, one would rearrange the equation to get a^2=c^2-b^2. Once the equation is rearranged, one can then use basic algebraic techniques to solve for the desired variable. In this way, the Pythagorean theorem can be used to solve for any side in a right triangle.
To find the domain and range of a given function, we can use a graph. For example, consider the function f(x) = 2x + 1. We can plot this function on a coordinate plane: As we can see, the function produces valid y-values for all real numbers x. Therefore, the domain of this function is all real numbers. The range of this function is also all real numbers, since the function produces valid y-values for all real numbers x. To find the domain and range of a given function, we simply need to examine its graph and look for any restrictions on the input (domain) or output (range).
Then, take the square root of this number to find the length of the hypotenuse. For example, if you know that one side is 3 feet long and another side is 4 feet long, you would first square these numbers to get 9 and 16. Then, you would add these numbers together to get 25. Taking the square root of 25 gives you 5, so you know that the hypotenuse is 5 feet long. Solving for x in a right triangle is a simple matter of using the Pythagorean theorem. With a little practice, you'll be able to do it in your sleep! |
# Table of 126
Created by: Team Maths - Examples.com, Last Updated: August 12, 2024
## Table of 126
The multiplication table of 126 is a helpful reference for understanding the results of multiplying 126 by whole numbers. Starting with 126 × 1 = 126, 126 × 2 = 252, and continuing sequentially, this table is ideal for students, educators, and professionals needing quick access to multiplication results for various applications.
## What is the Multiplication Table of 126?
The multiplication table of 126 is a sequential representation showing the results of multiplying 126 by whole numbers, starting from 1 and increasing incrementally. For example, 126 multiplied by 1 equals 126, 126 multiplied by 2 equals 252, 126 multiplied by 3 equals 378, and so on. This table continues with 126 multiplied by each subsequent whole number, providing a quick reference for calculations. It’s a valuable tool for students, educators, and professionals who need to perform or verify multiplication operations efficiently in various academic and practical contexts.
## 126ᵗʰ Table
The 126ᵗʰ table lists products of multiplying 126 by whole numbers: 126, 252, 378, 504, 630, 756, 882, 1008, 1134, and 1260, aiding in quick and easy calculations.
## Tips for 126ᵗʰ Table
### Memorize Key Multiples:
Focus on memorizing key multiples such as 126×1=126, 126×2=252, 126×5=630, and 126×10=1260 for quick reference.
### Use Patterns:
Notice patterns in the results, such as the increment of 126 in each step, to help predict the next value.
### Practice Regularly:
Regularly practice multiplying 126 by different numbers to reinforce memory and improve speed.
### Break Down Multiplications:
For larger numbers, break down the multiplication into smaller, more manageable parts. For example, 126×4 can be calculated as (126×2)×2.
### Apply in Real-Life Scenarios:
Use the 126ᵗʰ table in practical situations, like budgeting or measuring, to become more familiar with it.
## How to Read 126ᵗʰ Tables?
• One time 126 is 126
• Two times 126 is 252
• Three times 126 is 378
• Four times 126 is 504
• Five times 126 is 630
• Six times 126 is 756
• Seven times 126 is 882
• Eight times 126 is 1008
• Nine times 126 is 1134
• Ten times 126 is 1260
To read the 126th multiplication table, start at 126 and sequentially multiply it by numbers (1, 2, 3, etc.) to find the product for each respective multiple.
## Example 1: Simple Multiplication
Question: What is 126 times 2?
Solution: To find 126 times 2, you multiply 126 by 2.
Calculation: 126 x 2 = 252
Answer: 126 times 2 equals 252.
Question: How much is 126 added to itself 3 times?
Solution: Adding 126 to itself 3 times is the same as multiplying 126 by 3.
Calculation: 126 x 3 = 378
## Example 3: Real-world Application
Question: If a shirt costs \$126, how much will 4 shirts cost?
Solution: Multiply the cost of one shirt by the number of shirts.
Calculation: 126 x 4 = 504
## Example 4: Doubling
Question: What is the double of 126?
Solution: Doubling a number is the same as multiplying it by 2.
Calculation: 126 x 2 = 252
Answer: The double of 126 is 252.
## Example 5: Multiplication by Ten
Question: What is 126 times 10?
Solution: To find 126 times 10, you multiply 126 by 10.
Calculation: 126 x 10 = 1260
Answer: 126 times 10 equals 1260.
## Example 6: Cost Analysis
Question: If one lamp costs \$126, how much do 7 lamps cost?
Solution: Multiply the cost of one lamp by the total number of lamps.
Calculation: 126 x 7 = 882
## Example 7: Division into Groups
Question: How many groups of 126 can you form with 1008 items?
Solution: Divide the total number of items by 126.
Calculation: 1008 ÷ 126 = 8
Answer: You can form 8 groups of 126 with 1008 items.
## Example 8: Multiplying by a Hundred
Question: What is 126 times 100?
Solution: To find 126 times 100, you multiply 126 by 100.
Calculation: 126 x 100 = 12600
Answer: 126 times 100 equals 12600.
## Example 9: Increment by Quarters
Question: What is 126 increased by a quarter of its value?
Solution: To increase 126 by a quarter, multiply 126 by 0.25 and add it to 126.
Calculation: 126 + (126 x 0.25) = 126 + 31.5 = 157.5
Answer: 126 increased by a quarter equals 157.5.
## Example 10: Triple Amount
Question: What is three times 126?
Solution: Tripling a number is the same as multiplying it by 3.
Calculation: 126 x 3 = 378
Answer: Three times 126 is 378.
Text prompt |
You are on page 1of 7
# NAME:___________________________
Mr. Rogove
Date:__________
## Learning Objective: We will solve linear equations using the standard
form of a linear equation. (G8M4L11)
Exploration:
Brandon tells you he scored 32 points in a basketball game with ONLY two- and
three- point shots (no free throws). How many of each type of basket did he score?
Use the table to organize your work.
Number of two-pointers Number of three-pointers
Let x be the number of two-pointers he scored, and y be the number of threepointers he scored. Write an equation to represent the situation.
Concept Development:
Standard Form of a Linear Equation
!" + !" = !
!, !, and ! are constants and at least one of ! or ! does not equal zero
Example:
50! + ! = 15
A solution to a linear equation is an ordered pair of numbers (!, !) so that x and y
make the equation a true statement.
We can find solutions by fixing a number for x or y and solving for the other
variable.
## G8M4L11: Standard Form of a Linear Equation
NAME:___________________________
## Math _______ , Period ____________
Mr. Rogove
Date:__________
Guided Practice:
Steps for Finding Solutions to Linear Equations Written in Standard Form
1. Create a table.
2. Fix a number for x and solve for y (or vice versa).
3. Plot each point on a graph.
4. List the solutions as ordered pairs
Find five solutions for the linear equation ! + ! = 3 and plot the solutions as points
on a coordinate plane.
x
Linear equation
y
Solution
! + ! = !
(!, !)
## G8M4L11: Standard Form of a Linear Equation
NAME:___________________________
## Math _______ , Period ____________
Mr. Rogove
Date:__________
Find five solutions to the linear equation 2! ! = 10 and then plot the solutions as
points on a coordinate plane.
x
Linear equation
y
Solution
!" ! = !"
(!, !)
## G8M4L11: Standard Form of a Linear Equation
NAME:___________________________
## Math _______ , Period ____________
Mr. Rogove
Date:__________
At the store, you can buy a bag of candy for \$2 and a drink for \$1. Assume you have a
total of \$35 to spend and you are feeling generous and want to buy some snacks for
your friends. Write an equation in standard form to represent the number of bags of
candy, x, and number of drinks, y, you can buy with your \$35.
Find five solutions to the linear equation and plot the solutions as points on the
coordinate plane.
x
Linear
y
Solution
equation
(candy)
(drinks)
(!, !)
## G8M4L11: Standard Form of a Linear Equation
NAME:___________________________
## Math _______ , Period ____________
Mr. Rogove
Date:__________
!
Find five solutions to the linear equation ! ! + ! = 11 and then plot the solutions as
points on a coordinate plane.
x
Linear equation
y
Solution
!
! + ! = !!
!
(!, !)
## G8M4L11: Standard Form of a Linear Equation
NAME:___________________________
## Math _______ , Period ____________
Mr. Rogove
Date:__________
!
Find five solutions to the linear equation ! ! ! = 2 and then plot the solutions as
points on a coordinate plane.
x
Linear equation
y
Solution (!, !)
!
! ! = !
!
## G8M4L11: Standard Form of a Linear Equation
NAME:___________________________
## Math _______ , Period ____________
Mr. Rogove
Date:__________
Independent Practice:
Hand out Problem Setwhatever doesnt get done is HW.
Closure:
## Hand out exit ticket for lesson 12
Teacher Notes:
Matches lesson 12 from mod 4, grade 8 |
### You are Visitor Number:
2 2 6 1 0 1
Q.1 A gardener increased the area of his rectangular garden by increasing its length by 40% & decreasing its width by 20%.The area of the new garden
A) has increased by 20%
B) has increased by 12%
C) has increased by 8%
D) Same as the old area
Ans. B
Explanation:
The area of the new garden=
=(40-20- 40*20/100)%
=(20-80)%
=12%
Q.2 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?
A) 274 & 100 m
B) 275 & 200 m
C) 275 & 100 m
D) None of the above
Ans. C
Let the common ratio between the length and breadth of the rectangular plot be x. Hence, the length and breadth of the rectangular plot will be 11x m and 4x m respectively.
Perimeter of the plot = 2(Length + Breadth)
It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75, 000.
? 100 × Perimeter = 75000
100 × 30x = 75000
3000x = 75000
Dividing both sides by 3000, we obtain
x = 25
Length = 11x m = (11 × 25) m = 275 m
Breadth = 4x m = (4 × 25) m = 100 m
Hence, the dimensions of the plot are 275 m and 100 m respectively.
Q.3 A Village having a population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20m x 15m x 6m. The water of this tank will last for
A) 2 days
B) 3 days
C) 4 days
D) 5 days
Ans. B
Total amount of water for 4000 people=
4000*150=6,00,000 ltrs.
=600 m3
Volume of tank= 20m * 15m * 6m
=1800 m3
Therefore,required no. of days=
1800/600 =3
Q.4 Cost of carpeting a room 8m long & 4m wide with a carpet 2m wide is Rs.50 per metre. Find the cost of carpeting the floor?
A) Rs. 650
B) Rs. 700
C) Rs. 750
D) Rs. 800
Ans. D
Length = L * B/ Width of carpet
Or 8 * 4 / 2
=16m
Now, Cost = 16*50 = Rs.800
Q.5 A garden is 112m long & 78 m wide. It has walking path 25m wide all around it on inside. Find the area of path ?
A) 925 m2
B) 927 m2
C) 929 m2
D) 934 m2
Ans. A
If path is within the garden then use this formula:
2 * B * (L + B – 2 * B)
2 * 25 (112 + 78 – 2 * 25)
= 5 * 185 = 925 m2 |
# Tangents and Normals
Contents
## Summary
• A tangent is a line that just touches the curve but doesn’t go through it.
The gradient of a tangent = Gradient of a curve at that point
• A normal is a straight line perpendicular (at right angle 90°) to a curve.
$Gradient\quad of\quad a\quad normal\quad =\quad \frac { -1 }{ (Gradient\quad of\quad a\quad curve\quad or\quad tangent\quad at\quad that\quad point) }$
We know that differentiation is the process that we use to find the gradient of a point on the curve. However, we can also find the gradient of a curve at a given point by drawing a tangent at that point (which is a straight line) and finding the gradient of that line which is equal to the gradient/derivative of the curve as well.
#### What is a Tangent?
As shown in Fig 1, a tangent just touches the curve but doesn’t go through it. The line intersects the curve at a point and the gradient of the line is equal to the derivative of the curve at the point.
#### Example #1
Q. Find the equation of tangent to the curve y = (6 – x)(x + 3) at the point (3, 5)
Solution:
Write the equation in the form that we can differentiate:
$y\quad =\quad (6\quad -\quad x)(x\quad +\quad 3)$
$=\quad { -x }^{ 2 }\quad +\quad 3x\quad +\quad 18$
Now differentiate it:
$\frac { dy }{ dx } \quad =\quad -2x\quad +\quad 3$
The gradient of the tangent is equal to gradient of the curve at point (3, 5):
At x=3
$\frac { dy }{ dx } \quad =\quad -3$
Hence m(gradient) = -3 as well.
Now to find the equation of the tangent:
$y\quad -\quad { y }_{ 1 }\quad =\quad m(x\quad -\quad { x }_{ 1 })$
Point $({ x }_{ 1 },\quad { y }_{ 1 })$ is (3, 5) and m = -3
$y\quad -\quad 5\quad =\quad -3(x\quad -\quad 3)$
$y\quad =\quad -3x\quad +\quad 14\quad \quad \quad \longrightarrow$ equation of tangent at point (3, 5)
#### What is a Normal?
As shown in Fig 2, A normal is a straight line perpendicular (at right angle 90°) to a curve or to the tangent of the curve at that point.
If two lines, with gradients m1 and m2 are at right angles then $m1\quad \times \quad m2\quad =\quad -1$.
Therefore if the gradient of the tangent is m1 than the gradient of the normal is $m2\quad =\quad \frac { -1 }{ m1 }$
#### Example #2
Q. Find the equation of the normal to the curve $y\quad =\quad { x }^{ 3 }\quad -\quad 4x\quad +\quad 2$ at the point where x = 2.
Solution:
Firstly find the y coordinate at x=2:
$y\quad =\quad { (2) }^{ 3 }\quad -\quad 4(2)\quad +\quad 2$
$y\quad =\quad 2$
Now differentiate the equation of the curve to find the gradient of the tangent at:
$\frac { dy }{ dx } \quad =\quad { 3x }^{ 2 }\quad -\quad 4$
At x=2
$\frac { dy }{ dx } \quad =\quad 8$
Hence m1 = 8
Find m2:
$m2\quad =\quad \frac { -1 }{ 8 }$
Equation of normal is $y\quad -\quad { y }_{ 1 }\quad =\quad m(x\quad -\quad { x }_{ 1 })$ where Point $({ x }_{ 1 },\quad { y }_{ 1 })$ is (2, 2) and $m\quad =\quad \frac { -1 }{ 8 }$.
$y\quad -\quad 2\quad =\quad \frac { -1 }{ 8 } (x\quad -\quad 2)$
$y\quad =\quad \frac { -1 }{ 8 } x\quad +\quad \frac { 9 }{ 4 } \quad \quad \quad \longrightarrow$ equation of normal at point (2, 2)
##### Reference
1. CGP AS-Level Edexcel Mathematics complete revision and practice book |
calculation of probability
Este contenido está disponible en los siguientes idiomas: Español | Inglés
Refers to measure the degree of likelihood of a certain outcome when performing a random experiment or one that we cannot predict the outcome.
The likelihood values between 0 and 1 (or expressed as a percentage, between 0% and 100%):
- The zero value corresponds to the impossible event: throw a dice and the probability for number 7 is zero.
- The value corresponds to one certain event: throw a dice and the probability of getting any number from 1 to 6 is equal to one (100%).
All other events will be likely between zero and one: it will be greater the more likely it is that the event occurs. How to measure the probability? Favorable cases are divided among the possible cases. P (A) = favorable cases / possible cases. Sticking with the dice case, we have the following examples:
1. Probability of rolling a dice to get number 2: the favorable case is just one (which comes up the number 2) while there are six possible cases (can come up any number of 1 to 6). Therefore: Probability = 1/6 = 0.166 (or what is the same, 16.6%).
2. Probability of rolling a dice to get an even number: here are three favorable cases (comes up 2, 4 or 6), while six cases are still possible. Therefore: Probability = 3/6 = 0.50 (or what is the same, 50%).
3. Probability of rolling a dice to get fewer than 5: in this case we have four cases favorable (to get 1, 2, 3 or 4), compared with six cases. Therefore: Probability = 4/6 = 0.666 (or what is the same, 66.6%).
Sometimes, the probability that a given event occurs depends on whether another event has occurred previously or not. This is sometimes the fact that there is a certain phenomenon that can do more or less likely another fact to occur next. Such probabilities are called conditional probabilities, and are denoted by P (A / B) to the conditional probability of event A assuming that the event B has already occurred. The multiplicative law of probability indicates that the probability that two events occur simultaneously A and B equals: Probability (A y B) = P (A / B). P (B). The above multiplication law is also used to determine conditional probability P (A / B) from the values P (A and B) and P (B): Probability (A / B) = P (A and B) / P (B) where: - P (A / B) is the probability of that event A has been conditioned to the event given B. - P (B and A) is the probability of simultaneous occurrence of A and B - P (B) is the prior probability of the event B
Examples:
1. Someone rolls a dice and we know that the probability of rolling a 2 is 1/6 (prior probability). If we incorporate new information (for example, someone tells us that the result has been an even number) then the probability changes: P (A and B) = 1/6 P (B) = 1/2 P (A / B) = (1 / 6) / (1/2) = 1/3 then the probability that the number 2 if we know that has come up an even number is 1/3 (greater than the prior probability of 1/6).
2. A medical study has concluded that the probability that a person suffers coronary problems (event B) is 0.10 (a priori probability). Furthermore, the probability that a person suffering from obesity problems (event A) is 0.25 and the probability that a person suffers both obesity and coronary problems (event intersection of A and B) is 0.05. Calculate the probability that a person suffers heart problems if that person is obese (conditional probability P (B / A)). P (BYA) = 0.05 P (A) = 0.25 P (B / A) = 0.05 / 0.25 = 0.20 |
This is a free lesson from our course in Geometry
In this lesson youll learn the basics and concepts of Area of Polygons and Circles, an important part of the geometric applications. First the related and earlier learnt basics will be reviewed, which will follow on the important relationship of parts of the polygons/circle and angular measurements. and that help to apply them to solve real world application problems. The presentation covering such content will be done by the instructor in own handwriting, using video and with the help of several examples and solution. This will help you understand important trigonometric relationships to solve home work problems and also use them for real life applications. (More text below video...)
Other useful lessons:
Area of a Rectangle Area of a Triangle - Areas of Polygons and Circles Area of a Square Area of a Parallelogram Area of a Trapezoid Area of a Circle Effect of dimension changes on Area Real World Applications - Area of Polygons and Circles Midpoint of a Line Segment
(Continued from above) Area is a measure of the amount of space contained inside a closed figure i. e. polygon and circle. The area of a regular polygon may involve relatively simple figures such as an equilateral triangle, square or complex figures combining different shapes. To find the area of say an equilateral triangle, you can use the formula:1/2 * (bh), where b is the base and h, altitude or height of the triangle. Similarly area of a square with side 's', shall be s2 sq units.
Notice that in cases where the number of sides of the regular polygon is 3 or 4, it may be easy to calculate the area. Similarly area of a regular hexagon also is simple to calculate, as it can be broken into 6 equilateral triangles. Then sum up for all the equilateral triangles. Now look into a case of a regular polygon with n number of sides. You may recall the required trigonometric relationships, while working out the formula for determining the area:
In the figure on right, look at the part of the regular polygon with side s and n number of sides. Point O is the center of the polygon and r, the distance from the center to a vertex i.e. radius of the polygon. With this information, formula for the area can be derived: POQ has a measure of (360/n), and MOQ has a measure of (180/n). POQ is a central triangle, and there shall be n such triangles in this polygon. Since PQ is side with measure s, MQ = s/2. (basic right triangle trig)
Now that,
h = r cos(180/ n) and b = r sin(180/ n)
and MQ = s/2
Thus area of POQ = 1/2 (b* h)
= s/2 * h = sin(180/ n) * cos(180/ n) r2
This is the area of POQ.
Notice that the polygon has n number of such triangles, area of the polygon shall be: The area of a regular polygon of n sides and radius r = n * sin(180/ n) * cos(180/ n) r2
Common Formulas for Area
Wikipedia
Shape Formula Variables Square s is the length of one side of the square. Rectangle l and w are the length and width of the rectangle. Rhombus a and b are the lengths of the diagonals of the rhombus. Parallelogram b is the length of the base and h is the perpendicular height. Equilateral triangle s is the length of one side of the triangle. Triangle a, b and c are the length of each side, and s is half the perimeter. Triangle a and b are any two sides, and is the angle between them. Triangle b and h are the base and altitude (measured perpendicular to the base), respectively. Trapezoid a and b are the parallel sides and h the distance (height) between the parallel sides. Regular hexagon s is the length of one side of the hexagon. Regular Polygon a is the apothem, or the radius of an inscribed circle in the polygon, and p is the perimeter of the polygon. Circle r is the radius and d the diameter. Circular sector r and are the radius and angle (in radians), respectively. Ellipse a and b are the semi-major and semi-minor axes, respectively.
Further on, youll explore more details about the area of polygons and circles looking at some of the illustrative examples to have better understanding of the concepts and relationships:
Polygon
Regular A polygon with all sides and interior angles the same. Regular polygons are always convex i.e. all interior angles less than 180°, and all vertices 'point outwards' .
Properties Summary:
Sum of Interior Angles [180 (n-2)], where n is the number of sides
For a regular polygon, all the interior angles have the same values. E.g. the interior angles of a hexagon always add up to 720°, so in a regular hexagon, each one is one sixth of that, or 120°. It can be thus said: each interior angle of a regular polygon is given by: [180(n-2)/n], where n is the number of sides.
The apothem of a polygon is a line from the center to the midpoint of a side. This is also the, radius of the incircle.
The radius of a regular polygon is a line from the center to any vertex. It is also the radius of the circumcircle of the polygon.
It is a polygon with four 'sides' or edges and four vertices or corners.
Parallelogram- opposite sides parallel.
Rectangle- opposite sides equal, all angles 90°.
Square- all sides equal, all angles 90°.
Trapezoid- two sides parallel.
Rhombus- opposite sides parallel and equal.
Parallelogram
It is a quadrilateral with two pairs of parallel sides.
Properties and important points to remember-
Opposite sides of a parallelogram are equal in length.
Opposite angles of a parallelogram are equal in measure.
Opposite sides of a parallelogram can never intersect.
The diagonals of a parallelogram bisect each other.
Consecutive angles are supplementary, add to 180°.
Area (A), of a parallelogram is A = bh, where is the base and is height of parallelogram.
The area of a parallelogram is twice the area of a triangle created by one of its diagonals.
A parallelogram is a quadrilateral with opposite sides parallel and congruent.
Notice that:
A rectangle is a parallelogram, but with each angle as 90°.
A rhombus is a parallelogram, but with all sides equal in length.
A square is a parallelogram but with all sides equal in length, and each angle at 90°
Example 1: Given- let ABCDEF be a regular hexagon, as shown in the figure below (NTS). What is the ratio of the area of the triangle ACE to that of the hexagon ABCDEF?
Step 1: Let the side of a regular hexagon be 'a'
Step 2: The area of hexagon ABCDEF = ((33)/2)a2
Notice and consider the rhombus COED,
Step 3: Area of Rhombus COED = (1/2) times product of the length of its diagonal = (1/2) * CE * OD
= (1/2) * CE * a .................(1)
Also, Area of Rhombus COED = Area of COD + Area of DOE
= (3/4)a2 + (3/4)a2
= (3/2)a2.................(2)
Step 4: Comparing (1) and (2), gives, (1/2) * CE * a = (3/2)a2 i.e. CE = (3)a
Step 5: Area of ACE = (3/4)CE2 = (3/4)((3)a)2 = ((33)/4)a2
The ratio of the area of the triangle ACE to that of the hexagon ABCDEF is =((33)/4)a2 / ((33)/2)a2 = 1/2, as the final answer.
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# Percentage of a quantity
Lesson
Percentages are used for a variety of things, usually when we want to describe how much of something there is. For example, perhaps you only want $50%$50% of the juice in your cup or when the car dashboard says that the fuel tank is only $20%$20% full. However, $50%$50% of the water in a $100$100 L swimming pool is obviously very different to $50%$50% of the $2$2 L milk in your fridge. Let's take a look at how we can figure out how much there ACTUALLY is when we hear about percentages.
## Finding Percentages of a Quantity
We already know how to find a fraction of a quantity through multiplication. For example, we know to find $\frac{2}{3}$23 of $60$60 all we do is multiply the two numbers together, so $\frac{2}{3}\times60=40$23×60=40 is our answer. We can do the same with percentages as we know how to turn them into fractions with $100$100 as the denominator.
For example, we want to find what $71%$71% of $526$526 is, so let's multiply them together.
$71%$71% of $526$526 $=$= $71%\times526$71%×526 can be rewritten as $=$= $\frac{71}{100}\times526$71100×526 $=$= $\frac{71\times526}{100}$71×526100 Get out the calculator! $=$= $\frac{37346}{100}$37346100 simplify $=$= $\frac{18673}{50}$1867350
Sometimes such large messy improper fractions are easier to understand as mixed number, so in this case we can evaluate $\frac{18673}{50}$1867350 as $373\frac{23}{50}$3732350
Can you see we can easily estimate this to $373\frac{1}{2}$37312?
So much simpler!
#### Worked Examples
##### QUESTION 1
By converting the percentage to a decimal, find $74%$74% of $4600$4600 kilometres.
##### QUESTION 2
Evaluate $24%$24% of $272$272. Leave your answer as a fraction.
##### QUESTION 3
When tickets to a football match went on sale, $29%$29% of the tickets were purchased in the first hour. If the stadium seats $58000$58000 people, what was the number of seats still available after the first hour?
### Outcomes
#### NA4-3
Find fractions, decimals, and percentages of amounts expressed as whole numbers, simple fractions, and decimals |
### Expression and Equation
This is a continuation about algebra. We have covered definitions of the various terms used in algebra such as variables, like terms etc. We discussed about fundamental concepts and differences between expressions and equations.
Go through the site Study Guides and Strategies and attempt the following questions.
1. What is the difference between an equation and an expression?
2. What are the characteristics of a linear expression?
3. What does it mean to solve an equation?
4. Can an equation have multiple solutions?
5. Can an equation has no solution?
Solve the following equations for the value/s of x.
1. 8x - 46 = 2
2. 8x + 46 = 2
3. -8x +46 = 2
4. 8x + 46 = 46
5. 8x + 46 = 8x
1. What is the significance of the solutions of the above equations?
2. Can an equation have multiple solutions or no solution? If so, what is the implication.
1. a)8x - 46 = 2
x=6
b)8x + 46 = 2
x=-6
c)-8x +46 = 2
x=-6
d)8x + 46 = 46
x=0
e)8x + 46 = 8x
x= not possible to deduce.
a) If 8x - 46 = 2,
2 + 46 = 48
8x = 48
x = 48 / 8 = 6
b) If 8x + 46 = 2
46 - 2 = 44
-44 + 46 = 2
8x = -44
x = -44 / 8 = -5.5
c) If -8x + 46 = 2
46 - 2 = 44
-44 + 46 = 2
-8x = (-44)
x = (-44) / (-8) = 5.5
d) If 8x + 46 = 46
46 - 46 = 0
8x = 0
x = 0 / 8 = 0
e) If 8x + 46 = 8x
The following question cannot be define
3. This comment has been removed by the author.
1. 8x - 46 = 2
2 + 46 = 48
8x = 48
x = 6
2. 8x + 46 = 2
2 - 46 = (-44)
8x = (-44)
x = (-5.5)
3. -8x +46 = 2
2 - 46 = (-44)
-8x = (-44)
x = 5.5
4. 8x + 46 = 46
46 - 46 = 0
8x = 0
x = 0
5. 8x + 46 = 8x
The question is undefined as 8x + 46 could never be equal to 8x as 8x must be added with a 0 to give the possible outcome thus this question cannot be solved.
5. 1)8x - 46 = 2
8x=48
x=6
2)8x + 46 = 2
8x=-44
x=-5.5
3)-8x +46 = 2
-8x=-44
x=5.5
4)8x + 46 = 46
x=0
8x + 46 = 8x
0x=-46
This Q cannot be defined.
6. 1) 8x - 46 = 2
8x = 48
x = 6
2) 8x + 46 = 2
46-2 = 8x
44 = 8x
5.5 = x
3) -8x +46 = 2
46-2 = 8x
44 = 8x
5.5 = x
4) 8x + 46 = 46
8x = 46 - 46
8x = 0
x = 0/8 which is undefined.
5) 8x + 46 = 8x
This question is not definable as 8x will always be = to 8x. So having 46 added to one side will cause the whole statement to be unbalanced thus making it a mathematical error.
7. Ques 1. 8x - 46 = 2
8x = 2 + 46
8x = 48
x = 6
Ques 2, 8x + 46 = 2
8x = 2 - 46
8x = -44
x = -5.5
Ques 3. -8x +46 = 2
-8x = 2 - 46
-8x = -44
-x = -5.5
x = 5.5
Ques 4. 8x + 46 = 46
8x = 46 - 46
8x = 0
x = 0
Ques 5. 8x + 46 = 8x
8x = 8x - 46
(Which is impossible)
What is the significance of the solutions of the above equations?
- It shows that the value of the variable is dependent on the answer of the equation.
Can an equation have multiple solutions or no solution? If so, what is the implication.
- A equation can only have one solution, eg. 8x - 46 = 2. There is no other value of x except 6. But, an equation with a variable within can have various type of answers as the final product.
8. 1) 8x - 46 = 2
8x=2+46
8x=48
x=6
2)8x + 46 = 2
8x=-46+2
8x=-44
x=-5.5
3)-8x +46 = 2
-8x = 2 - 46
-8x = -44
-x = -5.5
x = 5.5
4)8x + 46 = 46
8x = 46 - 46
8x = 0
x = 0
5)8x+46=8x
8x-8x=-46????
Frankly speaking, I really think this is the way to do it...but it's different from other people's.
9. a)8x - 46 = 2(ans.)
x Will equal to 6
b)8x + 46 = 2 (ans.)
x Will equal to
-6
c)-8x +46 = 2 (ans.)
x Will equal to
-6
d)8x + 46 = 46 (ans.)
x Will equal to 0
e)8x + 46 = 8x (ans.)
X is Impossible to deduce as the before and after have the same values with an added 46.
1. Equation can be answered but expression can only be simplified.
e.g (Equation - 10=x+15) (Expression - 10x-5y)
2. a mathematical statement that performs functions of addition, subtraction, multiplication, and division
These are examples of linear expressions:
x + 4
2x + 4
2x + 4y
These are not linear expressions:
x2 (no exponents on variables)
2xy + 4 (can't multiply two variables)
2x / 4y (can't divide two variables)
Ö x (no square root sign on variables)
Source:http://www.studygs.net/equations.htm
3. No. It will be a definite answer.
e.g 10=x-5 (x=15)
4. Yes, if only the expression is false.
e.g 3>x = 5<y
11. 1. 8x - 46 = 2
46＋2＝48
48／8＝6
x=6
2. 8x + 46 = 2
2-46=-44
-44/8=-11/2=-5.5
x=5.5
3. -8x +46 = 2
2-46=-44
-44/-8=11/2=5.5
x=5.5
4. 8x + 46 = 46
46-46=0
0/8=0
x=0
5. 8x + 46 = 8x
cannot be done because 46 is a constant so it cannot be ignored
12. 1) 8x - 46 = 2
x = 6
2) 8x + 46 = 2
x = -6
3) 8x + 46 = 2
x = -6
4) 8x + 46 = 46
x = 0
5) 8x + 46 = 8x
x = Undefined.
13. This comment has been removed by the author.
14. What is the difference between an equation and an expression?
Equations has an answer and no variable. An Expression is solved by simplifying and may have a variable.
What are the characteristics of a linear expression?
The linear expression must have a variable.
What does it mean to solve an equation?
To simplify the variable.
Can an equation have multiple solutions?
No. It cannot have multiple solutions.
Can an equation has no solution?
No. It must have one solution.
-Jasper
15. Difference between Equation and Expression
- Equation has an equal sign while expression doesn’t
- Equation can be solved but and expression can only be simplified
Characteristics of Linear Equation
- Has no operations other than addition, subtraction and multiplication of a variable by a constant
- Variables may not be multiplied together or appear in a denominator
- A linear equation does not contain variable with exponents other than 1
- The graph of a linear equation is ALWAYS a line
Can an equation have multiple solution
- An equation cannot have multiple solution
Can an equation has no solution
- An equation must have one solution if not it would not be an equation
16. An equation has a equal sign but the expression does not have an equal sign.
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.
The left hand and the right hand of the equation must be simplified and after simplification, the both sides of the equation must be having the same value
No because there is only one solution and answer to the equation in order to meet the exact value.
No. An equation has one solution.
17. 1) 8x - 46 = 2
x = 6
2) 8x + 46 = 2
x = -6
3) 8x + 46 = 2
x = -6
4) 8x + 46 = 46
x = 0
5) 8x + 46 = 8x
x = Cannot be defined
18. 1)
8x - 46 = 2
8x = 48
x = 6
2)
8x + 46 = 2
8x = (-44)
x = -5.5
3)
-8x +46 = 2
-8x =(-44)
x = 5.5
4)
8x + 46 = 46
8x = 0
x = 0
5)
8x + 46 = 8x
x = not possible
19. This comment has been removed by a blog administrator.
20. What is the difference between an equation and an expression?
An equation must be solved, while an expression can only be simplified
What are the characteristics of a linear expression?
It must have a variable.
What does it mean to solve an equation?
To find the value of a variable
Can an equation have multiple solutions?
No, an equation cannot have more than one solution
Can an equation has no solution?
No, an equation must have only one solution
21. This comment has been removed by the author.
1. Equation can be answered but expression can only be simplified.
2. A mathematical statement that performs functions of addition, subtraction, multiplication, and division.
3. Simplifying the terms.
4. no an equation cannot have multiple solutions.
5. No, equations must have a solution.
a)8x - 46 = 2
8x=2+46
8x=48
x=6
b)8x + 46 = 2
8x=2-46
8x=44
x=5.5
c)-8x +46 = 2
-8x=2-46
-8x=-44
-x=-5.5
x=5.5
d)8x + 46 = 46
8x=0
x=0
e)8x + 46 = 8x
8x-8x=-46
x=0/undefined
23. 1. What is the difference between an equation and an expression?
~An expression can only be simplified while an equation can be solved.
2. What are the characteristics of a linear expression?
~ Is an equation between 2 variables that gives a straight line.
3. What does it mean to solve an equation?
~ To find the value of something. e.g. x
4. Can an equation have multiple solutions?
~ No. An equation cannot have multiple solutions.
5. Can an equation has no solution?
~No. An equation will have a solution no matter what.
1. What is the difference between an equation and an expression?
An equation has an answer which can be solved while an expression cannot be solved .
2. What are the characteristics of a linear expression?
In a linear expression , there should be a variable .
3. What does it mean to solve an equation?
To solve an equation means simplifying the statement and the variable .
4. Can an equation have multiple solutions?
No . It cannot have multipe solutions .
5. Can an equation has no solution?
An equation should atleast have a single solution in order to be an equation .
1) 8x - 46 = 2
8x = 2 + 46
x = 6
2) 8x + 46 = 2
8x = 2 - 46
= -44
x = -5.5
3) -8x + 46 = 2
-8x = 2 - 46
= -44
x = 5.5
4) 8x + 46 = 46
8x = 46 - 46
= 0
x = 0
5) 8x + 46 = 8x
Ans = Undefined as x does not have any vlue .
25. 1) 8x - 46 = 2
8x = 46 + 2
8x = 48
x = 48 / 8
x = 6
2) 8x + 46 = 2
8x = 2 - 46
8x = -44
x = -44 / 8
x = -5.5
3) -8x + 46 = 2
-8x = 2 - 46
-8x = -44
x = -44 / -8
x = 5.5
4) 8x + 46 = 46
8x = 46 + 46
8x = 92
x = 92 / 8
x = 11.5
5) 8x + 46 = 8x
This solution cannot be defined
What is the difference between an equation and an expression?
An equation will have a solution but an expression will not.
What are the characteristics of a linear expression?
It has a variable.
What does it mean to solve an equation?
It is to find out the value of the equation.
Can an equation have multiple solutions?
No. An equation only has 1 solution.
Can an equation has no solution?
Yes. The answer can be undefined.
8x - 46 = 2
8x=48
x=6
8x + 46 = 2
8x=-44
x=-5.5
-8x +46 = 2
-8x=-44
x=5.5
8x + 46 = 46
8x=0
x=0
8x + 46 = 8x
8x=8x-46
27. 1. 8x - 46 = 2
x = 6
2. 8x + 46 = 2
x = -6
3. -8x +46 = 2
x = 5.5
4. 8x + 46 = 46
x = 0
5. 8x + 46 = 8x
cannot be solved
What is the difference between an equation and an expression?
An equation is a combination of two expressions with an equal sign. Eg. 2x-3=5. However, an expression does not have an equal sign. Eg. 2x
What are the characteristics of a linear expression?
Linear expressions can have more than two expressions and must have at least one variable. Eg, y=3x+4y
What does it mean to solve an equation?
To solve an equation means to simplify it into one or more expressions.
Can an equation have multiple solutions?
No it cannot have multiple solutions.
Can an equation has no solution?
No, it must have a solution, unless the equation is somehow wrong.
1) 8x - 46 = 2
8x=2 +46
x=6
2)8x + 46 = 2
8x=2-46
x=-5.5
3)-8x +46 = 2
-8x=2-46
-x=-44÷8
x=-5.5
4)8x + 46 = 46
8x=46-46
x=0
5)8x + 46 = 8x
8x=8x-46
That is impossible to find as we do not know the value of x. as the two 8x are the same.
1) An equation has a defined answer that can be solved without a variable. An expression can be solved by simplifying, expansion or factorisation that the answer comes out in variables.
2) A linear expression has a term with a variable in it.
3) To solve an equation, we need to find the variable(e.g x) into a number.
4) No. It only can have one answer.
5) Yes. When nothing else can be done if the solution cannot be solved.
1)
8x-46=2
8x =48
48÷8=6
x=6
2)
8x+46=2
8x =-44
x=-5.5
3)
-8x+46 = 2
-8x =-44
x=5.5
4)
8x+46=46
46-46=0
x=0
5)
8x+46=8x
x=undefined variable
Go through the site Study Guides and Strategies and attempt the following questions.
What is the difference between an equation and an expression?
An expression requires us to simplify or factories but not solve while an equation requires us to find a value of a variable.
What are the characteristics of a linear expression?
- Has no operations other than addition, subtraction and multiplication of a variable by a constant
- Variables may not be multiplied together or appear in a denominator
- A linear equation does not contain variable with exponents other than 1
- The graph of a linear equation is always a line
What does it mean to solve an equation?
To find out the value of the variable.
Can an equation have multiple solutions?
No.
Can an equation has no solution?
Yes.
Solve the following equations for the value/s of x.
8x - 46 = 2
8x - 48 = 0
8x = 48
1x= 6
8x + 46 = 2
8x + 44 = 0
8x = -44
1x = -5.5
-8x + 46 = 2
-8x + 44 = 0
-8x = -44
8x = 44
1x = 5.5
8x + 46 = 46
8x + 0 = 0
8x = 0
1x = undefined
8x + 46 = 8x
1x = undefined or infinity
What is the significance of the solutions of the above equations?
They all only have one solution and has no operations other than addition and subtraction.
Can an equation have multiple solutions or no solution? If so, what is the implication.
No.
If the question have an undefined answer, there is no solution to that question.
-Christopher Nah |
# Modeling, Functions, and Graphs
## Section9.4Infinite Geometric Series
### SubsectionSummation Notation
There is a convenient notation for representing a series. For example, consider the sum of the first fifteen terms of the sequence
\begin{equation*} 4, 7, 10, \cdots, 3n+1, \cdots \end{equation*}
To find the terms of the series we replace $$n$$ in the general term $$3n+1$$ by the numbers 1 through 15. So, instead of writing out all the terms, we might express the sum as
\begin{equation*} \text{The sum, as}~ n~ \text{runs from 1 to 15, of}~ 3n+1 \end{equation*}
We use the Greek letter $$\Sigma$$ (called "sigma") to stand for "the sum," and we show the first and last values of $$n$$ below and above the summation symbol $$\Sigma\text{,}$$ like this:
\begin{equation*} \sum_{n=1}^{15} 3n+1 \end{equation*}
Writing such an expression is sometimes called "using sigma notation." The letter $$n$$ is called the index of summation; it is like a variable because it represents numerical values.
#### Note9.53.
Any letter can be used for the index of summation; $$i,~j,$$ and $$k$$ are other common choices. Of course, the letter used for the index of summation does not affect the sum.
#### Example9.54.
Use sigma notation to represent the sum of the first 20 terms of the sequence
\begin{equation*} -1, 2, 7, \cdots, k^2-2, \cdots \end{equation*}
Solution.
The general term of the sequence is $$k^2-2\text{,}$$ and the first term is $$-1\text{,}$$ which we find by letting $$k=1$$ in the formula for the general term. Thus,
\begin{equation*} \sum_{k=1}^{20} (k^2-2) \end{equation*}
#### Checkpoint9.55.Practice 1.
Use sigma notation to represent the sum of the first 20 terms of the sequence
\begin{equation*} \begin{gathered} 5, 8, 11, \ldots, 3k+2, \ldots \end{gathered} \end{equation*}
\begin{equation*} \sum_{k=1}^{b} a_k =5+ 8+11+\cdots+ (3k+2)+\cdots \end{equation*}
where $$b=$$ and $$a_k=$$
$$20$$
$$3k+2$$
Solution.
$$\displaystyle\sum_{k=1}^{20} (3k+2)$$
#### Checkpoint9.56.QuickCheck 1.
What does the notation $$\displaystyle{\sum_{j=1}^{8} \frac{1}{j}}$$ mean?
• List 8 terms of the sequence $$a_j=\displaystyle\frac{1}{j}\text{.}$$
• Add the numbers from 1 to 8 and take the reciprocal of the sum.
• Add the reciprocals of the numbers from 1 to 8.
• Add the first and eighth term of the sequence described.
$$\text{Choice 3}$$
Solution.
Add the reciprocals of the numbers from 1 to 8.
The expanded form of a series written in sigma notation is obtained by writing out all the terms of the series. For example,
\begin{equation*} \sum_{m=4}^{8} \dfrac{3}{m} = \dfrac{3}{4} + \dfrac{3}{5} + \dfrac{3}{6} + \dfrac{3}{7} + \dfrac{3}{8} \end{equation*}
#### Note9.57.
Notice that the series above has five terms, which is one more than the difference of the upper and lower limits of summation, $$8 - 4\text{.}$$
Recall that the general term of an arithmetic series is a linear function of the index, and the general term of a geometric series is an exponential function. If we recognize a given series as one of these two types, we can use the formulas developed in the last section to evaluate the sum.
#### Example9.58.
Compute the value of each series.
1. $$\displaystyle \displaystyle{\sum_{i=1}^{13} (90-5i)}$$
2. $$\displaystyle \displaystyle{\sum_{k=0}^{9} 2^k}$$
Solution.
1. Because the general term $$90-5i$$ is linear, this is an arithmetic series. By writing out the first few terms of the series,
\begin{equation*} 85+80+75+70+\cdots \end{equation*}
we can verify that the first term of the series is $$a_1=85$$ and the common difference is $$d=-5\text{.}$$ We also need to know the last term of the series, so we substitute $$n=13$$ in the general term to find $$a_{13} = 90-5(13) = 25$$ Thus,
\begin{equation*} \sum_{i=1}^{13} (90-5i) = \dfrac{13}{2}(85+25) = 715 \end{equation*}
2. The general term of this series is exponential, so the series is geometric. By writing out a few terms of the series,
\begin{equation*} 1+2+4+8+\cdots \end{equation*}
we confirm that the first term of the series is $$a_1=1$$ and the common ratio is $$r=2\text{.}$$ The series has 10 terms, from $$k=0$$ to $$k=9\text{,}$$ so $$n=10\text{.}$$ Finally, we substitute these values into the formula for geometric series to obtain
\begin{equation*} \sum_{k=0}^{9} 2^k = \dfrac{1(1-2^{10})}{1-2} = 1023 \end{equation*}
#### Checkpoint9.59.Practice 2.
Compute the value of each series.
1. $$\displaystyle{\sum_{k=1}^{50} (3k+2)}=$$
2. $$\displaystyle{\sum_{n=1}^{8} 10^n}=$$
$$3925$$
$$1.11111\times 10^{8}$$
Solution.
1. 3925
2. 111,111,110
If the series is not arithmetic or geometric, we don’t have a formula for the sum, so we must compute the sum directly.
#### Example9.60.
Compute the value of each series.
1. $$\displaystyle \displaystyle{\sum_{m=1}^{5} m^2}$$
2. $$\displaystyle \displaystyle{\sum_{p=1}^{800} 5}$$
Solution.
1. Because the general term, $$m^2\text{,}$$ is neither linear nor exponential, we know that the series is not arithmetic or geometric. However, we can expand the series and evaluate it directly.
\begin{equation*} \begin{aligned}[t] \displaystyle{\sum_{m=1}^{5} m^2}\amp = 1^2+2^2+3^2+4^2+5^2\\ \amp = 1+4+9+16+25=55\\ \end{aligned} \end{equation*}
2. The general term is the number 5. Because the index $$p$$ runs from 1 to 800, we are adding 800 terms, each of which is 5. Thus
\begin{equation*} \begin{aligned}[t] \displaystyle{\sum_{p=1}^{800} 5}\amp = 5+5+5+\cdots+5+5\\ \amp = 800(5) = 4000\\ \end{aligned} \end{equation*}
#### Checkpoint9.61.Practice 3.
Compute the value of each series.
1. $$\displaystyle{\sum_{k=0}^{20} \dfrac{1}{3}}=$$
2. $$\displaystyle{\sum_{k=0}^{3} \dfrac{k}{k+1}}=$$
$$7$$
$$\frac{23}{12}$$
Solution.
1. 7
2. $$\displaystyle \dfrac{23}{12}$$
#### Checkpoint9.62.QuickCheck 2.
Which of these series is neither arithmetic nor geometric?
• $$\displaystyle \displaystyle\sum_{k=5}^{15} \left(\frac{1}{2}k-4\right)$$
• $$\displaystyle \displaystyle\sum_{m=1}^{50} 0.8^m$$
• $$\displaystyle \displaystyle\sum_{n=2}^{12} (n^3+2)$$
• $$\displaystyle \displaystyle\sum_{p=1}^{5} 5(1.5)^{2p}$$
$$\text{Choice 3}$$
Solution.
$$\displaystyle\sum_{n=2}^{12} (n^3+2)$$ is neither arithmetic nor geometric. (The first series is arithmetic; the second and fourth series are both geometric.
### SubsectionInfinite Series
A series with infinitely many terms is called an infinite series. Is it possible to add infinitely many terms and arrive at a finite sum? In some cases, if the terms added are small enough, the answer is yes.
Consider the infinite geometric series
\begin{equation*} \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\cdots \end{equation*}
The $$n^{\text{th}}$$ partial sum of the series is the sum of its first $$n$$ terms, and is denoted by $$S_n\text{.}$$ Thus,
\begin{align*} S_1\amp = \dfrac{1}{2}\\ S_2 \amp = \dfrac{1}{2}+\dfrac{1}{4} = \dfrac{3}{4}\\ S_3 \amp = \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} = \dfrac{7}{8}\\ S_4 \amp = \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16} = \dfrac{15}{16} \end{align*}
Notice that for each partial sum, the new term added gets smaller and smaller: $$\dfrac{1}{2}\text{,}$$ $$\dfrac{1}{4}\text{,}$$ $$\dfrac{1}{8}\text{,}$$ $$\dfrac{1}{16},$$ and so on. You can also see that as $$n$$ increases and we add more and more terms of the series, the partial sums are getting closer to 1. In fact, as $$n$$ becomes very large, $$S_n$$ gets very close to 1. It seems reasonable that the sum of all the terms of the series is 1.
#### Note9.63.
Will the partial sums ever be greater than 1? No, because each new term $$a_n$$ added is half the difference between $$S_n$$ and 1.
We can make this conjecture more plausible by examining the formula for the $$n^{\text{th}}$$ partial sum of a geometric series,
\begin{align*} S_n\amp = \dfrac{a_{n+1} - a_1}{r-1}\\ \amp = \dfrac{ar^n - a}{r-1} = \dfrac{a - ar^n}{1-r} \amp\amp (r \not= 1) \end{align*}
Look at the second term of the numerator, $$ar^n\text{.}$$ This term is the only part of the formula that depends on $$n\text{.}$$ What happens to $$ar^n$$ as $$n$$ increases? Consider two examples:
• If $$r = \dfrac{1}{2}\text{,}$$ then
\begin{equation*} r^2 = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4},\qquad r^3 = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8},\qquad r^4 = \left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}, \end{equation*}
and so on, with $$\left(\dfrac{1}{2}\right)^n$$ becoming smaller and smaller for larger values of $$n\text{.}$$ Each time we multiply by another factor of $$r=\dfrac{1}{2}\text{,}$$ the product gets smaller, because $$\dfrac{1}{2} \lt 1\text{.}$$
• On the other hand, if $$r \gt 1\text{,}$$ multiplying by $$r$$ makes the product larger. If $$r = \dfrac{3}{2}$$ for example, then
\begin{equation*} r^2 = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4},\qquad r^3 = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8},\qquad r^4 = \left(\dfrac{3}{2}\right)^4 = \dfrac{81}{16}, \end{equation*}
and so on. In this case the powers of $$r$$ are increasing.
In general, we have the following result.
1. If $$0 \lt \abs{r} \lt 1\text{,}$$ then $$r^n$$ gets closer to zero as $$n$$ increases.
2. If $$\abs{r} \gt 1\text{,}$$ then $$r^n$$ does not approach a finite number as $$n$$ increases.
Now let’s return to the formula for $$S_n$$ and write it in the form
\begin{equation*} S_n = \dfrac{a}{1-r} (1-r^n) \end{equation*}
where we have factored $$a$$ from the numerator. There are two cases to consider. If $$\abs{r} \lt 1\text{,}$$ then $$r^n$$ approaches 0, and the factor $$1-r^n$$ gets closer and closer to 1 as $$n$$ grows larger. Consequently, if we compute $$S_n$$ for larger and larger values of $$n\text{,}$$ the sum approaches the value
\begin{equation*} \dfrac{a}{1-r} \end{equation*}
This analysis motivates us to define the sum of an infinite geometric series as follows.
#### Sum of an Infinite Geometric Series.
The sum of an infinite geometric series $$~~\displaystyle{\sum_{k=0}^{\infty} ar^{k-1}}~~$$ is
\begin{equation*} S_{\infty} = \dfrac{a}{1-r}~~~~~~\text{if}~~~~-1 \lt r \lt 1 \end{equation*}
In the second case, if $$\abs{r} \gt 1\text{,}$$ as in the infinite series
\begin{equation*} 3 + 6 + 12 + \cdots \end{equation*}
where $$r = 2\text{,}$$ the terms become larger as $$n$$ increases, and the sum of the series is not a finite number. In this case the series does not have a sum.
#### Checkpoint9.64.QuickCheck 3.
Under what circumstances does the series $$\displaystyle\sum_{k=0}^{\infty} ar^k$$ have a finite sum?
• If $$a \lt 1$$
• If $$a \lt 1$$
• If $$| r | \lt 1$$
• If $$ar \lt r$$
$$\text{Choice 3}$$
Solution.
If $$\abs{r} \lt 1$$
#### Example9.65.
Make a table showing the first five partial sums of each series. Then use the formula to find the sum, if it exists.
1. $$\displaystyle ~~\displaystyle{\sum_{j=0}^{\infty} 30(0.8)^j}~~$$
2. $$\displaystyle ~~\displaystyle{\sum_{m=0}^{\infty} 3(\dfrac{4}{3})^m}~~$$
Solution.
1. First, we evaluate the general term $$30(0.8)^j$$ for $$j=1, 2, \cdots, 5$$ and compute the partial sums. For example,
\begin{equation*} \begin{aligned}[t] S_1\amp = 30(0.8)^1 = 24\\ S_2\amp = 30(0.8)^1+30(0.8)^2 = 43.2\\ \end{aligned} \end{equation*}
and so on. The results, rounded to hundredths, are shown in the table.
$$n$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$S_n$$ $$24$$ $$43.2$$ $$58.56$$ $$70.85$$ $$80.68$$
This is an infinite geometric series with $$a=30$$ and $$r=0.8\text{.}$$ The series has a sum because $$\abs{r} \lt 1\text{.}$$ Thus,
\begin{equation*} S_{\infty}=\dfrac{a}{1-r} = \dfrac{30}{1-0.8} = 150 \end{equation*}
2. We evaluate the general term $$3(\dfrac{4}{3})^m$$ for $$m=1, 2, \cdots, 5$$ and compute the partial sums. For example,
\begin{equation*} \begin{aligned}[t] S_1\amp = 3(\dfrac{4}{3})^1 = 4\\ S_2\amp = 3(\dfrac{4}{3})^1 + 3(\dfrac{4}{3})^2 = \dfrac{28}{3}\\ \end{aligned} \end{equation*}
and so on. The results, rounded to hundredths, are shown in the table.
$$n$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$S_n$$ $$4$$ $$9.33$$ $$16.44$$ $$25.93$$ $$38.57$$
This is a geometric series with $$a=3$$ and $$r = \dfrac{4}{3}\text{.}$$ The infinite series does not have a sum because $$\abs{r} \gt 1\text{.}$$
#### Checkpoint9.66.Practice 4.
Find the sum, if it exists. If the series does not have a sum, enter “no sum”.
1. $$\displaystyle{\sum_{j=0}^{\infty} 13\left(\dfrac{7}{6}\right)^j}=$$
2. $$\displaystyle{\sum_{m=0}^{\infty} 5.9(0.9)^m}=$$
$$\text{no sum}\hbox{, }\text{dne}\hbox{, or }\text{diverges}$$
$$\text{59}$$
Solution.
1. No sum
2. 59
### SubsectionRepeating Decimals
An interesting application of geometric series involves repeating decimals. Recall that the decimal representation of a rational number either terminates, as does $$0.75,$$ or repeats a pattern of digits. For example, you probably recognize $$0.333\overline{3}$$ as the decimal representation of $$\dfrac{1}{3}\text{.}$$ It is easy to find the decimal form of a fraction: we just divide the denominator into the numerator. Is there a way to find the fractional form of a repeating decimal?
Consider the repeating decimal
\begin{equation*} 0.2121\overline{21} \end{equation*}
We can write this number as an infinite geometric series:
\begin{equation*} 0.21 + 0.0021 + 0.000021 + \cdots \end{equation*}
The first term of this series is $$0.21\text{,}$$ or $$\dfrac{21}{100}\text{,}$$ and its common ratio is $$r=0.01\text{,}$$ or $$\dfrac{1}{100}\text{.}$$ Because $$\abs{r} \lt 1\text{,}$$ the series has a sum given by
\begin{equation*} \begin{aligned}[t] S_{\infty}\amp = \dfrac{a}{1-r} = \dfrac{\dfrac{21}{100}}{1-\dfrac{1}{100}}\\ \amp = \dfrac{\dfrac{21}{100}}{\dfrac{99}{100}} = \dfrac{21}{99} = \dfrac{7}{33}\\ \end{aligned} \end{equation*}
Thus, the decimal number $$0.2121\overline{21}$$ is equal to the fraction $$\dfrac{7}{33}\text{.}$$
#### Example9.67.
Find a common fraction equivalent to $$0.3\overline{7}\text{.}$$
Solution.
The decimal can be written as $$0.3 + 0.0\overline{7}\text{.}$$ We will find a fraction equivalent to the repeating decimal $$0.0\overline{7}$$ and add that to $$0.3\text{,}$$ or $$\dfrac{3}{10}\text{.}$$ We write $$0.0\overline{7}$$ as a series:
\begin{equation*} 0.0\overline{7} = \dfrac{7}{100} + \dfrac{7}{1000} + \dfrac{7}{10,000} + \cdots \end{equation*}
This is an infinite geometric series with first term $$\dfrac{7}{100}$$ and common ratio $$\dfrac{1}{10}\text{.}$$ The sum of the series is given by
\begin{equation*} \begin{aligned}[t] S_{\infty}\amp = \dfrac{a}{1-r} = \dfrac{\dfrac{7}{100}}{1-\dfrac{1}{10}}\\ \amp = \dfrac{\dfrac{7}{100}}{\dfrac{9}{10}} = \dfrac{7}{100} \cdot \dfrac{9}{10} = \dfrac{7}{90}\\ \end{aligned} \end{equation*}
Finally, we add $$\dfrac{7}{90}$$ and $$\dfrac{3}{10}$$ to find
\begin{equation*} 0.3\overline{7} = \dfrac{3}{10}+\dfrac{7}{90} = \dfrac{34}{90} \end{equation*}
#### Checkpoint9.68.Practice 5.
Find a common fraction equivalent to $$0.\overline{8}\text{:}$$
$$\frac{8}{9}$$
Solution.
$$\dfrac{8}{9}$$
### SubsectionSection Summary
#### SubsubsectionVocabulary
• Sigma notation
• Index of summation
• Infinite series
• Partial sum
#### SubsubsectionCONCEPTS
1. We can use sigma notation to denote a series.
2. It is possible to add infinitely many terms and arrive at a finite sum if the terms are small enough.
3. The sum of an infinite geometric series $$~~\displaystyle{\sum_{k=0}^{\infty} ar^{k-1}}~~$$ is
\begin{equation*} S_{\infty} = \dfrac{a}{1-r}~~~~~~\text{if}~~~~-1 \lt r \lt 1 \end{equation*}
4. If $$\abs{r} \gt 1$$ in an infinite geometric series, the series does not have a sum.
#### SubsubsectionSTUDY QUESTIONS
1. Explain how to use sigma notation to define a series.
2. What is an infinite series?
3. What is a partial sum of an infinite series?
4. State a formula for evaluating an infinite geometric series. Under what conditions is the formula valid?
5. A repeating decimal can be rewritten as what kind of series?
#### SubsubsectionSKILLS
Practice each skill in the Homework Problems listed.
1. Express sigma notation in expanded form: #1-8
2. Write a series in sigma notation: #12-20
3. Identify a series as arithmetic, geometric, or neither and evaluate: #21-40
4. Evaluate an infinite geometric series: #41-48, #57-60
5. Write a repeating decimal as a common fraction: #49-56
### ExercisesHomework 9.4
#### Exercise Group.
For Problems 1–8, write the sum in expanded form.
##### 1.
$$\displaystyle{\sum_{i=1}^{4} i^2}$$
##### 2.
$$\displaystyle{\sum_{i=1}^{3} (3i-2)}$$
##### 3.
$$\displaystyle{\sum_{j=5}^{7} (j-2)}$$
##### 4.
$$\displaystyle{\sum_{j=2}^{6} (j^2+1)}$$
##### 5.
$$\displaystyle{\sum_{k=1}^{4} k(k+1)}$$
##### 6.
$$\displaystyle{\sum_{k=2}^{6} \dfrac{k}{2}(k+1)}$$
##### 7.
$$\displaystyle{\sum_{m=1}^{4} \dfrac{(-1)^m}{2^m}}$$
##### 8.
$$\displaystyle{\sum_{m=3}^{5} \dfrac{(-1)^{m+1}}{m-2}}$$
#### Exercise Group.
For Problems 9–20, write the series using sigma notation.
##### 9.
$$1+3+5+7$$
##### 10.
$$2+4+6+8$$
##### 11.
$$5+5^3+5^5+5^7$$
##### 12.
$$4^3+4^5+4^7+4^9+4^{11}$$
##### 13.
$$1+4+9+16+25$$
##### 14.
$$1+8+27+64+125$$
##### 15.
$$\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+\dfrac{5}{6}$$
##### 16.
$$\dfrac{2}{1}+\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}+\dfrac{6}{5}$$
##### 17.
$$\dfrac{1}{1}+\dfrac{2}{3}+\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{5}{9}+\dfrac{6}{11}$$
##### 18.
$$\dfrac{3}{1}+\dfrac{5}{3}+\dfrac{7}{5}+\dfrac{9}{7}+\dfrac{11}{9}$$
##### 19.
$$\dfrac{1}{1}+\dfrac{2}{2}+\dfrac{4}{3}+\dfrac{8}{4}+\cdots$$
##### 20.
$$\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{9}{6}+\dfrac{27}{8}+\cdots$$
#### Exercise Group.
For Problems 21–40, identify the series as arithmetic, geometric or neither, then evaluate it.
##### 21.
$$\displaystyle{\sum_{i=1}^{6} (i^2+1)}$$
##### 22.
$$\displaystyle{\sum_{i=1}^{5} 3i^2}$$
##### 23.
$$\displaystyle{\sum_{j=1}^{4} \dfrac{1}{j}}$$
##### 24.
$$\displaystyle{\sum_{j=0}^{4} \dfrac{2}{j+1}}$$
##### 25.
$$\displaystyle{\sum_{k=1}^{100} 1}$$
##### 26.
$$\displaystyle{\sum_{k=1}^{300} 3}$$
##### 27.
$$\displaystyle{\sum_{q=1}^{20} 3^q}$$
##### 28.
$$\displaystyle{\sum_{p=1}^{30} 2^p}$$
##### 29.
$$\displaystyle{\sum_{k=1}^{200} k}$$
##### 30.
$$\displaystyle{\sum_{k=1}^{150} k}$$
##### 31.
$$\displaystyle{\sum_{n=1}^{6} n^3}$$
##### 32.
$$\displaystyle{\sum_{n=1}^{7} n^2}$$
##### 33.
$$\displaystyle{\sum_{n=0}^{30} (3n-1)}$$
##### 34.
$$\displaystyle{\sum_{k=0}^{20} (5k+2)}$$
##### 35.
$$\displaystyle{\sum_{k=0}^{25} (5-2k)}$$
##### 36.
$$\displaystyle{\sum_{p=0}^{15} (2-3p)}$$
##### 37.
$$\displaystyle{\sum_{j=0}^{10} 5\cdot 2^j}$$
##### 38.
$$\displaystyle{\sum_{j=0}^{10} 3\cdot 2^j}$$
##### 39.
$$\displaystyle{\sum_{m=0}^{12} 50(1.08)^m}$$
##### 40.
$$\displaystyle{\sum_{m=0}^{18} 300(1.12)^m}$$
#### Exercise Group.
For Problems 41–48,
1. Make a table of values showing the first five partial sums of the series.
2. Evaluate the series algebraically.
##### 41.
$$\displaystyle{\sum_{n=1}^{\infty} \left(\dfrac{1}{2}\right)^n}$$
##### 42.
$$\displaystyle{\sum_{n=1}^{\infty} \left(\dfrac{1}{3}\right)^n}$$
##### 43.
$$\displaystyle{\sum_{k=0}^{\infty} 12(0.15)^{k-1}}$$
##### 44.
$$\displaystyle{\sum_{k=0}^{\infty} 25(0.08)^{k-1}}$$
##### 45.
$$\displaystyle{\sum_{j=0}^{\infty} 4\left(\dfrac{-3}{5}\right)^{j}}$$
##### 46.
$$\displaystyle{\sum_{j=0}^{\infty} 6\left(\dfrac{-2}{5}\right)^{j}}$$
##### 47.
$$\displaystyle{\sum_{n=4}^{\infty} 3\left(\dfrac{1}{2}\right)^{n}}$$
##### 48.
$$\displaystyle{\sum_{n=3}^{\infty} 2\left(\dfrac{1}{3}\right)^{n}}$$
#### Exercise Group.
For Problems 49–56, find a fraction equivalent to the repeating decimal.
##### 49.
$$0.\overline{4}$$
##### 50.
$$0.\overline{6}$$
##### 51.
$$0.\overline{31}$$
##### 52.
$$0.\overline{45}$$
##### 53.
$$0.\overline{410}$$
##### 54.
$$0.\overline{027}$$
##### 55.
$$0.12\overline{8}$$
##### 56.
$$0.8\overline{3}$$
#### 57.
The arc length through which the bob of a pendulum moves is nine-tenths of its preceding arc length. Approximately how far will the bob move before coming to rest if the first arc length is $$12$$ inches?
#### 58.
A force is applied to a particle moving in a straight line in such a fashion that each second it moves only one-half of the distance it moved the preceding second. If the particle moves $$10$$ centimeters the first second, approximately how far will it move before coming to rest?
#### 59.
A ball returns two-thirds of its preceding height on each bounce. If the ball is dropped from a height of $$6$$ feet, approximately what is the total distance the ball travels before coming to rest? (Hint: Compute separately the total distance the ball falls from the total distance it moves upwards.)
#### 60.
If a ball is dropped from a height of $$10$$ feet and returns three-fifths of its preceding height on each bounce, approximately what is the total distance the ball travels before coming to rest? (See the Hint for Problem 59.) |
# Table of 47
0
70
The “47 times table” or the multiplication table for 47 shows the results of multiplying 47 by whole numbers. By using this table, you can see a pattern of adding 47 repeatedly when multiplying by different whole numbers. For example, when you multiply 47 by 3, you add 47 to itself three times, resulting in 141 (47 x 3 = 47+47+47 = 141). As a result, the products obtained from multiplying 47 by different whole numbers are diverse. To create the table of 47, you can simply multiply 47 by successive positive integers.
Here is the multiplication table for 47, which can assist students in making rapid calculations, particularly in competitive exams where efficient time management is vital. The table has been presented in various layouts up to 20 times to facilitate the memorization of the 47 times table.
## Multiplication Table of 47
Visual aids such as tables and charts are often utilized to improve children’s learning since they can be more effective. Therefore, a visual representation of the 47 times table has been provided below to assist children in understanding and memorizing the information more quickly.
47 Table
47 X 1 = 47
47 X 2 = 94
47 X 3 = 141
47 X 4 = 188
47 X 5 = 235
47 X 6 = 282
47 X 7 = 329
47 X 8 = 376
47 X 9 = 423
47 X 10 = 470
47 X 11 = 517
47 X 12 = 564
47 X 13 = 611
47 X 14 = 658
47 X 15 = 705
47 X 16 = 752
47 X 17 = 799
47 X 18 = 846
47 X 19 = 893
47 X 20 = 940
## Table of 47 PDF
The PDF version of the Table of 47 is available for download below, which enables students to learn offline at their convenience.
## How to Read 47 Times Table?
The 47 times table is simple to learn, but it can take some practice to memorize it. Students can quickly memorize the table, thereby enabling them to solve problems that depend on it. Now, let us explore how to read the table of 47.
One time 47 is 47
Two times 47 is 94
Three times 47 is 141
Four times 47 is 188
Five times 47 is 235
Six times 47 is 282
Seven times 47 is 329
Eight times 47 is 376
Nine times 47 is 423
Ten times 47 is 470
## Tips to Memorize Table of 47
Memorizing the 47 times table is a straightforward process. By following the tips below, you can quickly and easily memorize the table.
To get the next multiple of 47, simply add 47 to the current multiple consecutively.
The table of 47 can be generated by performing skip counting by 47.
Practice the table of 47 by writing it down multiple times and reciting it aloud.
Related Articles:-
## FAQs on 47 Tables
### What is the result of multiplying 47 by 6?
Referring to the Table of 47, we can find that 47 times 6 equals 47 multiplied by 6, which is equal to 282.
47 x 6 = 282
### What is the sum of 47 times 5 and 47 times 8?
To find the sum of 47 times 5 and 47 times 8, we first need to find the products of each multiplication. Using the table of 47:
47 times 5 is equal to 235 (47 x 5 = 235)
47 times 8 is equal to 376 (47 x 8 = 376)
To get the sum, we need to add these two products:
235 + 376 = 611
Therefore, the sum of 47 times 5 and 47 times 8 is 611. |
# 2.7 Linear inequalities and absolute value inequalities (Page 4/11)
Page 4 / 11
$\begin{array}{c}-200\le x-600\le 200\\ -200+600\le x-600+600\le 200+600\\ 400\le x\le 800\end{array}$
This means our returns would be between $400 and$800.
To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.
## Absolute value inequalities
For an algebraic expression X, and $\text{\hspace{0.17em}}k>0,$ an absolute value inequality is an inequality of the form
These statements also apply to $\text{\hspace{0.17em}}|X|\le k\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}|X|\ge k.$
## Determining a number within a prescribed distance
Describe all values $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ within a distance of 4 from the number 5.
We want the distance between $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and 5 to be less than or equal to 4. We can draw a number line, such as in [link] , to represent the condition to be satisfied.
The distance from $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to 5 can be represented using an absolute value symbol, $\text{\hspace{0.17em}}|x-5|.\text{\hspace{0.17em}}$ Write the values of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ that satisfy the condition as an absolute value inequality.
$|x-5|\le 4$
We need to write two inequalities as there are always two solutions to an absolute value equation.
$\begin{array}{lll}x-5\le 4\hfill & \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\hfill & x-5\ge -4\hfill \\ \phantom{\rule{1.8em}{0ex}}x\le 9\hfill & \hfill & \phantom{\rule{1.8em}{0ex}}x\ge 1\hfill \end{array}$
If the solution set is $\text{\hspace{0.17em}}x\le 9\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x\ge 1,$ then the solution set is an interval including all real numbers between and including 1 and 9.
So $\text{\hspace{0.17em}}|x-5|\le 4\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}\left[1,9\right]\text{\hspace{0.17em}}$ in interval notation.
Describe all x- values within a distance of 3 from the number 2.
$|x-2|\le 3$
## Solving an absolute value inequality
Solve $|x-1|\le 3$ .
$\begin{array}{l}|x-1|\le 3\hfill \\ \hfill \\ -3\le x-1\le 3\hfill \\ \hfill \\ -2\le x\le 4\hfill \\ \hfill \\ \left[-2,4\right]\hfill \end{array}$
## Using a graphical approach to solve absolute value inequalities
Given the equation $y=-\frac{1}{2}|4x-5|+3,$ determine the x -values for which the y -values are negative.
We are trying to determine where $\text{\hspace{0.17em}}y<0,$ which is when $\text{\hspace{0.17em}}-\frac{1}{2}|4x-5|+3<0.\text{\hspace{0.17em}}$ We begin by isolating the absolute value.
Next, we solve for the equality $|4x-5|=6.$
$\begin{array}{lll}4x-5=6\hfill & \hfill & 4x-5=-6\hfill \\ \phantom{\rule{1.9em}{0ex}}4x=11\hfill & \phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}\hfill & \phantom{\rule{1.9em}{0ex}}4x=-1\hfill \\ \phantom{\rule{2em}{0ex}}x=\frac{11}{4}\hfill & \hfill & \phantom{\rule{2em}{0ex}}x=-\frac{1}{4}\hfill \end{array}$
Now, we can examine the graph to observe where the y- values are negative. We observe where the branches are below the x- axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at $\text{\hspace{0.17em}}x=-\frac{1}{4}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=\frac{11}{4},$ and that the graph opens downward. See [link] .
Solve $\text{\hspace{0.17em}}-2|k-4|\le -6.$
$k\le 1\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}k\ge 7;$ in interval notation, this would be $\text{\hspace{0.17em}}\left(-\infty ,1\right]\cup \left[7,\infty \right).$
Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities.
## Key concepts
• Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See [link] and [link] .
• Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See [link] , [link] , [link] , and [link] .
• Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See [link] and [link] .
• Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See [link] and [link] .
• Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See [link] .
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when can I use sin, cos tan in a giving question
depending on the question
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maybe provide us videos
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a
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nothing. I accidentally press it
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b=P-4a-2c
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like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
yah
immy
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
how would this look as an equation?
Hayden
5x+x=45
Khay |
# Trigonometry/Law of Sines
For any triangle with vertices A, B, and C, corresponding angles A, B, and C, and corresponding opposite side lengths a, b, and c, the Law of Sines states that
${a \over \sin A} = {b \over \sin B} = {c \over \sin C}.$
Each of these expressions is also equal to the diameter of the triangle's circumcircle (the circle that passes through the points A, B, and C). The law can also be written in terms of the reciprocals:
${\sin A \over a} = {\sin B \over b} = {\sin C \over c}.$
## Proof
Dropping a perpendicular OC from vertex C to intersect AB (or AB extended) at O splits this triangle into two right-angled triangles AOC and BOC. We can calculate the length h of the altitude OC in two different ways:
• Using the triangle AOC gives
$\displaystyle h=b \sin A ;$
• and using the triangle BOC gives
$\displaystyle h=a \sin B .$
• Eliminate h from these two equations:
$\displaystyle a \sin B =b \sin A .$
• Rearrange to obtain
$\displaystyle {a \over \sin A} = {b \over \sin B}.$
By using the other two perpendiculars the full law of sines can be proved. QED.
## Application
This formula can be used to find the other two sides of a triangle when one side and the three angles are known. (If two angles are known, the third is easily found since the sum of the angles is 180º.) See Solving Triangles Given ASA. It can also be used to find an angle when two sides and the angle opposite one side are known.
## Area of a triangle
The area of a triangle may be found in various ways. If all three sides are known, use Heron's theorem.
If two sides and the included angle are known, consider the second diagram above. Let the sides b and c, and the angle between them α be known. From triangle ACO, the altitude h = CO is bsin(α) so the area is 12cbsin(α).
If two angles and the included side are known, again consider the second diagram above. Let the side c and the angles α and γ be known. Let AO =x. Then
$\displaystyle \frac{x}{h} = \cot(\alpha) \text {; } \frac{c-x}{h} = \cot(\gamma) \text {; adding these, } \frac{c}{h} = \cot(\alpha) + \cot(\gamma)$
Thus
$\displaystyle h = \frac{c}{\cot(\alpha) + \cot(\gamma)} \text { so area } = \frac{c^2}{2(\cot(\alpha) + \cot(\gamma))}$. |
# Linear Problem - Slope of Curve Through Point
• Sep 13th 2011, 09:40 AM
celtic1234
Linear Problem - Slope of Curve Through Point
Hi all,
I have started on differential equations today and need some help (as usual).
From what i have read so far.
The solution to a differential equation will be another function or differential equation.
For example the slope of a curve is a differential equation. The solution of this differential equation will actually give you the curve of the function or the equation of the function itself.
Now i have this problem to solve.
"find the curve whose slope at any point (x,y) is equal to 5y and which passes through the point (1,-2)".
Now the slope is given as 5y;
$slope=\frac{dy}{dx}=5y$
Now this is a differential equation in itself and the solution to the differential equation will give me the curve of the function itself.
rearranging the slope term
multiply across by dx:
$\frac{dy}{dx}.dx=5y.dx$
$dy=5y.dx$
To solve the differential equation i need to integrate both sides;
$\int dy=\int 5y.dx$
$x+C1= \frac{5y^2}{2} +C2$
rearranging similar terms:
$x- \frac{5y^2}{2} =C2-C1$
$x- \frac{5y^2}{2} =C.........equation (1)$
Now we are given the point (1,-2) so substitute into equation (1) for x and y
$(1)- \frac{5(-2)^2}{2} =C.........equation (1)$
$C=-9$
Now the solution to the problem is C inserted into equation (1)
$x- \frac{5y^2}{2} =-9$
or
$x- \frac{5y^2}{2} +9=0$
Now to check this if i substitute the solution back into the original equation for y and if i differentiate y with respect to x i should get the slope of 5y?
I think i understand it-some of the text overcomplicates this if you ask me(or more probably i have not got the brains to understand it),
can anybody give me a more basic explanation if i have not got it correct,
Is my method/solution correct?
Thanks
John
• Sep 13th 2011, 09:51 AM
HallsofIvy
re: Linear Problem - Slope of Curve Through Point
Quote:
Originally Posted by celtic1234
Hi all,
I have started on differential equations today and need some help (as usual).
From what i have read so far.
The solution to a differential equation will be another function or differential equation.
For example the slope of a curve is a differential equation. The solution of this differential equation will actually give you the curve of the function or the equation of the function itself.
Now i have this problem to solve.
"find the curve whose slope at any point (x,y) is equal to 5y and which passes through the point (1,-2)".
Now the slope is given as 5y;
$slope=\frac{dy}{dx}=5y$
Now this is a differential equation in itself and the solution to the differential equation will give me the curve of the function itself.
rearranging the slope term
multiply across by dx:
$\frac{dy}{dx}.dx=5y.dx$
$dy=5y.dx$
To solve the differential equation i need to integrate both sides;
$\int dy=\int 5y.dx$
$x+C1= \frac{5y^2}{2} +C2$
NO!! $\int 5y dy= 5y^2/2+ C$ but you cannot integrate y with respect to x! y is an unknown function of x.
Instead, you need to separate the x and y terms:
$\frac{dy}{y}= 5 dx$
Now integrate
$\int \frac{dy}{y}= \int 5 dx$
Quote:
rearranging similar terms:
$x- \frac{5y^2}{2} =C2-C1$
$x- \frac{5y^2}{2} =C.........equation (1)$
Now we are given the point (1,-2) so substitute into equation (1) for x and y
$(1)- \frac{5(-2)^2}{2} =C.........equation (1)$
$C=-9$
Now the solution to the problem is C inserted into equation (1)
$x- \frac{5y^2}{2} =-9$
or
$x- \frac{5y^2}{2} +9=0$
Now to check this if i substitute the solution back into the original equation for y and if i differentiate y with respect to x i should get the slope of 5y?
I think i understand it-some of the text overcomplicates this if you ask me(or more probably i have not got the brains to understand it),
can anybody give me a more basic explanation if i have not got it correct,
Is my method/solution correct?
Thanks
John
No, your method is not correct. In particular, you cannot integrate y, an unknown function of x, with respect to x. $\int y dx$ is NOT the same as $\int y dy$.
• Sep 13th 2011, 10:56 AM
celtic1234
Re: Linear Problem - Slope of Curve Through Point
ok,
so if i am getting what you re saying;
$\frac{1}{y}.dy=5.dx$
integrate both sides:
$\int \frac{1}{dy}.dy=\int 5.dx$
$ln y + C1= 5x + C2$
$ln y-5x=C$
At point (1,-2)
$-ln (-2)-5(1)=C$
$C =-5.69315$
Substituting into the solution for C
$ln y-5x=-5.69315$
$ln y-5x+5.69315 =0$
Is this it?
John |
# Introduction to Hypothesis Testing
A statistical hypothesis is an assumption about a population parameter. For example, we may assume that the mean height of a male in the U.S. is 70 inches. The assumption about the height is the statistical hypothesis and the true mean height of a male in the U.S. is the population parameter.
A hypothesis test is a formal statistical test we use to reject or fail to reject a statistical hypothesis.
## The Two Types of Statistical Hypotheses
To test whether a statistical hypothesis about a population parameter is true, we obtain a random sample from the population and perform a hypothesis test on the sample data.
There are two types of statistical hypotheses:
The null hypothesis, denoted as H0, is the hypothesis that the sample data occurs purely from chance.
The alternative hypothesis, denoted as H1 or Ha, is the hypothesis that the sample data is influenced by some non-random cause.
## Hypothesis Tests
A hypothesis test consists of five steps:
1. State the hypotheses.
State the null and alternative hypotheses. These two hypotheses need to be mutually exclusive, so if one is true then the other must be false.
2. Determine a significance level to use for the hypothesis.
Decide on a significance level. Common choices are .01, .05, and .1.
3. Find the test statistic.
Find the test statistic and the corresponding p-value. Often we are analyzing a population mean or proportion and the general formula to find the test statistic is: (sample statistic – population parameter) / (standard deviation of statistic)
4. Reject or fail to reject the null hypothesis.
Using the test statistic or the p-value, determine if you can reject or fail to reject the null hypothesis based on the significance level.
The p-value tells us the strength of evidence in support of a null hypothesis. If the p-value is less than the significance level, we reject the null hypothesis.
5. Interpret the results.
Interpret the results of the hypothesis test in the context of the question being asked.
## The Two Types of Decision Errors
There are two types of decision errors that one can make when doing a hypothesis test:
Type I error: You reject the null hypothesis when it is actually true. The probability of committing a Type I error is equal to the significance level, often called alpha, and denoted as α.
Type II error: You fail to reject the null hypothesis when it is actually false. The probability of committing a Type II error is called the Power of the test or Beta, denoted as β.
## One-Tailed and Two-Tailed Tests
A statistical hypothesis can be one-tailed or two-tailed.
A one-tailed hypothesis involves making a “greater than” or “less than ” statement. For example, suppose we assume the mean height of a male in the U.S. is greater than or equal to 70 inches. The null hypothesis would be H0: µ ≥ 70 inches and the alternative hypothesis would be Ha: µ < 70 inches.
A two-tailed hypothesis involves making an “equal to” or “not equal to” statement. For example, suppose we assume the mean height of a male in the U.S. is equal to 70 inches. The null hypothesis would be H0: µ = 70 inches and the alternative hypothesis would be Ha: µ ≠ 70 inches.
Note: The “equal” sign is always included in the null hypothesis, whether it is =, ≥, or ≤. |
# Chapter 2 – Properties of Real Numbers
## Presentation on theme: "Chapter 2 – Properties of Real Numbers"— Presentation transcript:
Chapter 2 – Properties of Real Numbers
2.3 – Subtraction of Real Numbers
2.3 – Subtraction of Real Numbers
Today we will be learning how to: Subtract real numbers using the subtraction rule Using subtraction of real numbers to solve real-life problems
2.3 – Subtraction of Real Numbers
Some addition expressions can be evaluated using subtraction. ADDITION PROBLEM EQUIVALENT SUBTRACTION PROBLEM 5 + (-3) = 2 5 – 3 = 2 9 + (-6) = 3 9 – 6 = 3
2.3 – Subtraction of Real Numbers
Adding the opposite of a number is equivalent to subtracting the number. SUBTRACT RULE To subtract b from a, add the opposite of b to a a – b = a + (-b) The result is the difference of a and b Example: 6 – 4 = 6 + (-4)
2.3 – Subtraction of Real Numbers
Example 1 Find the difference -6 – 3 1/2 – 2/3 -15 – (-16) -3 – 6
2.3 – Subtraction of Real Numbers
Look at the first and fourth example. In addition, we have the commutative property. In subtraction, the order in which we subtract matters.
2.3 – Subtraction of Real Numbers
Example 2 Evaluate the expression 10 – – (-18)
2.3 – Subtraction of Real Numbers
When an expression is written as a sum, the parts that are added are the TERMS of the expression. For example, you can write the expression 8 – y as 8 + (-y). The terms are 8 and –y.
2.3 – Subtraction of Real Numbers
Example 3 Find the terms of 3x – 5.
2.3 – Subtraction of Real Numbers
Example 4 Evaluate the function y = -x – 7 for these values of x: -2, -1, 0, 1. Organize your results in a table & describe the pattern. Input Function Output x = -2 y = -(-2) - 7 -5 x = -1 x = 0 x = 1
2.3 – Subtraction of Real Numbers
HOMEWORK Page 82 #16 – 50 even |
# Lowest Common Denominator (LCD) of 1/1 and 1/87
So you wanna find the lowest common denominator of 1/1 and 1/87? Lucky for you that's exactly what this page is here to help you with! In this quick guide, we'll walk you through how to calculate the lowest common denominator for any fractions you need to check. Here we go!
In a rush and just want the answer? No worries! The LCD of 1/1 and 1/87 is 87. In fraction form it's 1/87.
LCD(1/1, 1/87) = 87
In fraction form:
1 / 87
Read on to find out how we worked that out!
As we always do in these articles, it's worth a very quick recap on fraction terminology. The number above the line is called the numerator and the number about the line is called the denominator. So, in this example, our numerators are 1 and 1, while our denominators are 1 and 87.
## What is the Lowest Common Denominator?
To calculate the lowest common denominator, the easiest way is to look at the factors of those numbers and find the lowest common multiple. Here's how that looks for 1 and 87:
• Factors for 1: 1
• Factors for 87: 1, 3, 29, and 87
The next step is to calculate the numerator and complete our fraction. To do this, we need to find the greatest common factor of the numerators, which are 1 and 1.
• Factors for 1: 1
In this case there is only one factor, 1, so this lucky little number becomes the numerator in our fraction by default:
1 / 87
Hopefully this article has helped you to understand how to calculate the lowest common denominator of two fractions. Isn't math fun? If you want to challenge yourself, try to work out the least common denominator of some fractions by yourself and use our LCD calculator to check your answers! |
The simple binomial theorem of degree 2 can be written as:
${(x+a)}^2=x^2+2xa+a^2 \ \ldots (1)$
Replacing $a$ by $(n+a)$ where $x, n, a \in \mathbb{R}$ , we can have
${(x+(n+a))}^2= x^2+2x(n+a)+{(n+a)}^2$
or, ${(x+n+a)}^2 =x^2+2xn+2ax+{(n+a)}^2$
Arranging terms in a way that
$${(x+n+a)}^2 =ax+{(n+a)}^2+x^2+2xn+ax$$
$$=ax+{(n+a)}^2+x(x+2n+a)$$
Taking Square-root of both sides
$$x+n+a=\sqrt{ax+{(n+a)}^2+x(x+2n+a)} \ \ldots (2)$$
Take a break. And now think about $(x+2n+a)$ in the same way, as:
$x+2n+a =(x+n)+n+a$ .
Therefore, in equation (2), if we replace $x$ by $x+n$ , we get
$$x+2n+a=(x+n)+n+a$$
$$=\sqrt{a(x+n)+{(n+a)}^2+(x+n)((x+n)+2n+a)}$$
or, $x+2n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)} \ \ldots (3)$
Similarly, $x+3n+a=\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)} \ \ldots (4)$
and also, $x+4n+a=\sqrt{a(x+3n)+{(n+a)}^2+(x+3n)(x+5n+a)} \ \ldots (5)$
Similarly,
$x+kn+a=\sqrt{a(x+(k-1)n)+{(n+a)}^2+(x+(k-1)n)(x+(k+1)n+a)} \ \ldots (6)$
where, $k \in \mathbb{N}$
Putting the value of $x+2n+a$ from equation (3) in equation (2), we get:
$x+n+a=\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)}} \ \ldots (7)$
Again, putting the value of $x+3n+a$ from equation (4) in equation (7), we get
$$x+n+a =\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n) \sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)}}} \ \ldots (8)$$
Generalizing the result for $k$ -nested radicals:
$$x+n+a$$
$$=\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+ \\ (x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots +(x+(k-2)n)\sqrt{a(x+(k-1)n)+ \\ {(n+a)}^2+x(x+(k+1)n+a)}}}}} \ \ldots (9)$$
This is the general formula of Ramanujan Nested Radicals up-to $k$ roots.
### Some interesting points
As $x,n$ and $a$ all are real numbers, thus they can be interchanged with each other.
i.e.,
$$x+n+a$$
$$=\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots+(x+(k-2)n)\\ \sqrt{a(x+(k-1)n)+{(n+a)}^2+x(x+(k+1)n+a)}}}}})$$
$$=\sqrt{an+{(x+a)}^2+n\sqrt{a(n+x)+{(x+a)}^2+(n+x)\sqrt{a(n+2x)+{(x+a)}^2+(n+2x)\sqrt{\ldots+(n+(k-2)x) \\ \sqrt{a(n+(k-1)x)+{(x+a)}^2+n(n+(k+1)x+a)}}}}})$$
$$=\sqrt{xa+{(n+x)}^2+a\sqrt{x(a+n)+{(n+x)}^2+(a+n)\sqrt{x(a+2n)+{(n+x)}^2+(a+2n)\sqrt{\ldots+(a+(k-2)n) \\ \sqrt{ x(a+(k-1)n)+{(n+x)}^2+a(a+(k+1)n+x)}}}}} \ \ldots (10)$$
etc.
Putting $n=0$ in equation (9) we have
$x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots+x\sqrt{ax+{a}^2+x(x+a)}}}}} \ \ldots (11)$
or just, $x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots}}}} \ \ldots (12)$
Again putting $x=1 \ a=0$ in (9)
$1+n =\sqrt{{n}^2+\sqrt{n^2+(1+n)\sqrt{{n}^2+(1+2n)\sqrt{\ldots+(1+(k-2)n)\sqrt{{n}^2+1+(k+1)n}}}}} \ldots (13)$
Putting $x=1 \ a=0$ in equation (8)
$1+n =\sqrt{{n}^2+\sqrt{{n}^2+(1+n)\sqrt{{n}^2+(1+2n)(1+4n)}}} \ \ldots (14)$
Again putting $x=a=n$ =n(say) then
$3n=\sqrt{n^2+4{n}^2+n\sqrt{2n^2+4{n}^2+2n\sqrt{3n^2+4{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn^2+4{n}^2+(k+3)n^2}}}}}$
or, $3n=\sqrt{5{n}^2+n\sqrt{6{n}^2+2n\sqrt{7{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+4)n^2+(k+3)n^2}}}}} \ \ldots (15)$
Putting $n=1$ in (15)
$3=\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{\ldots+(k-1)\sqrt{(2k+7)}}}}} \ \ldots (16)$
Putting $x=n \in \mathbb{N}$ and $a=0$ in (9) we get even numbers
$2n =\sqrt{{n}^2+n\sqrt{{n}^2+2n\sqrt{{n}^2+3n)\sqrt{\ldots+(k-1)n\sqrt{(k-1)n)+{n}^2+(k+2)n^2}}}}} \ \ldots (17)$
Similarly putting $x=n \in \mathbb{N}$ and $a=1$ in (9) we get a formula for odd numbers:
$$2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn+{(n+1)}^2+(k+2)n^2+n}}}}} \ \ldots (18)$$
or,
$$2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$$ |
Videos, examples, and solutions to aid Kindergarten and also Grade 1 kids learn to include and subtract within 20, demonstrating fluency for addition and subtraction within 10 through using techniques such as making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14).
You are watching: Use make a 10 to add
Related PagesMake Ten SubtractionLesson Plans and also Worksheets because that Grade 1More Lessons because that Grade 1Common core For grade 1
Common Core: 1.OA.6
### Learning Targets:
I can add and subtract within 10 easily.I can add and subtract by do 10. (e.g. 7 + 3 = 10, 10 - 6 = 4)
The adhering to diagrams show how to add using the Make-Ten strategy.
Step 1: The first addend and what make ten?Step 2: create the number listed below the second addendStep 3: The number below the 2nd addend and also what do the 2nd addend?Step 4: add the rest to 10
Example:6 + 7 = 6 + 4 + 3 = 10 + 3 = 135 + 8 = 5 + 5 + 3 = 10 + 3 = 13
Mental math Strategies: make 10
This video clip explains how to use mental strategies to include within 20 utilizing the “make 10” strategy. This is among the strategies said in the usual Core State criter for Mathematics, 1.OA.6 and also 2.OA.2.
With practice, this strategy can assist students obtain fluency when including within 20. Ultimately students will memorize this facts, i beg your pardon is meant by the finish of grade 2.7 + 5 = 7 + 3 + 2 = 10 + 2 = 124 + 9 = 4 + 6 + 3 = 10 + 3 = 138 + 7 = 8 + 2 + 5 = 10 + 5 = 15
Make 10 Strategy because that Addition7 + 9 = 7 + 3 + 6 = 10 + 6 = 168 + 6 = 8 + 2 + 4 = 10 + 4 = 1475 + 7 = 75 + 5 + 2 = 80 + 2 = 8254 + 8 = 54 + 6 + 2 = 60 + 2 = 62
Grade 1 Make-a-Ten Strategy because that Unknown Total
Example:9 + 5 = 9 + 1 + 4 = 10 + 4 = 14
Grade 1 Make-a-Ten Strategy because that Unknown Partner
Example:9 + _ = 149 + 1 + 4 = 14
Make 10 enhancement Strategy
Example:9 + 4 = 9 + 1 + 3 = 10 + 3 = 139 + 6 = 9 + 1 + 5 = 10 + 5 = 159 + 5 = 9 + 1 + 4 = 10 + 4 = 149 + 3 = 9 + 1 + 2 = 10 + 2 = 12 9 + 8 = 9 + 1 + 7 = 10 + 7 = 178 + 5 = 8 + 2 + 3 = 10 + 3 = 138 + 4 = 8 + 2 + 2 = 10 + 2 = 12
Try the free Mathway calculator and also problem solver listed below to practice miscellaneous math topics.
See more: A Biologist Took A Count Of The Number Of Fish In A Particular Lake And Recounted The Lakes
Shot the offered examples, or kind in your very own problem and also check your answer v the step-by-step explanations. |
30/08/2022
## What does 20 simplify to?
The square root of 20 is represented as √20, where √ is the radical sign and 20 is said to be radicand. In the simplest form, the √20 can be written as 2√5.
### What is the simplified fraction of 2 8?
It is 14 because you find a common number that both the numerator and the denominator can be divided by. In this case, it is 2 . You can’t divide 14 anymore so that is the final answer.
#### What is the decimal for 2 20?
As you can see, in one quick calculation, we’ve converted the fraction 220 into it’s decimal expression, 0.1.
What is 2 20 as a percent?
We can see that this gives us the exact same answer as the first method: 2/20 as a percentage is 10%.
What is the reduced fraction 4 10?
1 Answer. 410 in lowest terms is 25 .
## What is the fraction in simplest form 9 12?
The factors of 12 are 1, 2, 3, 4, 6, and 12. So the greatest common factor of 9 and 12 is 3. So you divide both sides by 3. So the simplest form of 912 is 34 .
### What is the simplified form of 2 4?
2 and 4 share a common factor: 2. We can reduce this fraction by dividing both the numerator and denominator by their common factor, 2. 1 and 2 have no common factor other than 1, so the fraction is in lowest terms.
#### What is a 2 out of 20?
10.00%
The percentage score for 2 out of 20 is 10.00%.
What is .60 as a fraction?
In the decimal form, the fraction can be written as 0.6. The fraction can be written as 35.
What is the fraction of 2 out of 20?
10/100
We can either find the decimal for 2/20, or make 2/20 a fraction with 100 in the denominator. In this case, it is simpler to do the latter. We need to multiply 20 by 5 to get 100, thus we multiply 2 by 5 to get 10. Then, we have that 2/20 is equivalent to 10/100.
## What is the fraction of 3 divided by 12?
The number 3 is called the numerator or dividend, and the number 12 is called the denominator or divisor. The quotient of 3 and 12, the ratio of 3 and 12, as well as the fraction of 3 and 12 all mean (almost) the same: 3 divided by 12, often written as 3/12.
### What is 4 10as a decimal?
Answer: The fraction 4/10 written as a decimal number is equal to 0.4.
#### What is the fraction of 5 4?
5/4 = 54 = 1 14 = 1.25 Spelled result in words is five quarters (or one and one quarter). |
# Difference between revisions of "2010 AMC 10A Problems/Problem 25"
## Problem
Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$, then his sequence contains $5$ numbers:
$$\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}$$
Let $N$ be the smallest number for which Jim’s sequence has $8$ numbers. What is the units digit of $N$?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 3 \qquad \mathrm{(C)}\ 5 \qquad \mathrm{(D)}\ 7 \qquad \mathrm{(E)}\ 9$
## Solution
We can find the answer by working backwards. We begin with $1-1^2=0$ on the bottom row, then the $1$ goes to the right of the equal's sign in the row above. We find the smallest value $x$ for which $x-1^2=1$ and $x>1^2$, which is $x=2$.
We repeat the same procedure except with $x-1^2=1$ for the next row and $x-1^2=2$ for the row after that. However, at the fourth row, we see that solving $x-1^2=3$ yields $x=4$, in which case it would be incorrect since $1^2=1$ is not the greatest perfect square less than or equal to $x$ . So we make it a $2^2$ and solve $x-2^2=3$. We continue on using this same method where we increase the perfect square until $x$ can be made bigger than it. When we repeat this until we have $8$ rows, we get:
$$\begin{array}{ccccc}{}&{}&{}&{}&7223\\ 7223&-&84^{2}&=&167\\ 167&-&12^{2}&=&23\\ 23&-&4^{2}&=&7\\ 7&-&2^{2}&=&3\\ 3&-&1^{2}&=&2\\ 2&-&1^{2}&=&1\\ 1&-&1^{2}&=&0\\ \end{array}$$
Hence the solution is the last digit of $7223$, which is $\boxed{\textbf{(B)}\ 3}$.
Note: We can go up to $167$, and then notice the pattern of units digits alternating between $3$ and $7$, so we do not need to calculate $7223$.
## Solution 2
Notice that to get the previous term, we must add the smallest square number, (let's call it $n^2$) such that the sum is less than $(n+1)^2$. Otherwise, instead of subtracting $n^2$ from the previous term, we're subtracting a greater square number.
Remember that $(x+1)^2 = x^2 + x + (x+1)$. Recall that to find the previous term, we must add a square number such that it is less than the next square number. $a + n^2 < (n+1)^2$. For this to be true, $a < n + (n+1)$. What that means is that given a term $a$, we can find the previous term by adding $n^2$ where $n > \frac {a-1}{2}$.
For example, to find the term that precedes $167$, we know that $n>166/2 = 83$. Therefore, $n=84$ and the previous term is $167 + 84^2 = 7223$. The last digit of $7223$ is $3 \Rightarrow \boxed{\textbf{(B)}\ 3}$ |
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Arc (geometry) - Wikipedia, the free encyclopedia
Page 1 of 3
Arc (geometry)
In geometry, an arc is a closed segment of a differentiable curve in the two-dimensional plane; for example, a circular arc is a segment of the circumference of a circle. If the arc occupies a great circle (or great ellipse), it is called a great-arc.
Contents
A circular sector is shaded in green. Its curved boundary of length L is a circular arc.
Arc Length
The length of an arc of a circle with radius r and subtending an angle circle center i.e., the central angle equals . This is because
yields
and so the arc length equals
http://en.wikipedia.org/wiki/Arc_(geometry)
12/13/2010
Arc (geometry) - Wikipedia, the free encyclopedia
Page 2 of 3
A practical way to determine the length of an arc in a circle is to plot two lines from the arc's endpoints to the center of the circle, measure the angle where the two lines meet the center, then solve for L by cross-multiplying the statement: measure of angle/360 = L/Circumference. For example, if the measure of the angle is 60 degrees and the Circumference is 24", then 60/360 = L/24 360L=1440 L = 4". This is so due to the fact that the circumference of a circle and the degrees of a circle, of which there are always 360, are directly proportionate.
Arc Area
The area between an arc and the center of a circle is:
The area A has the same proportion to the circle area as the angle to a full circle:
By multiplying both sides by r , we get the final result:
Using the conversion described above, we find that the area of the sector for a central angle measured in degrees is:
Arc Segment Area
The area of the shape limited by the arc and a straight line between the two end points is:
http://en.wikipedia.org/wiki/Arc_(geometry)
12/13/2010
Arc (geometry) - Wikipedia, the free encyclopedia
Page 3 of 3
To get the area of the arc segment, we need to subtract the area of the triangle made up by the circle's center and the two end points of the arc from the area A. See Circular segment for details. |
# The sum of the first and third terms of a geometric sequence is 40 while the sum of its second and fourth terms is 96. How do you find the sixth term of the sequence?
Jan 15, 2016
Solve to find the initial term $a$, common ratio $r$ and hence:
${a}_{6} = a {r}^{5} = \frac{1990656}{4225}$
#### Explanation:
The general term of a geometric sequence is:
${a}_{n} = a {r}^{n - 1}$
where $a$ is the initial term and $r$ is the common ratio.
We are given:
$40 = {a}_{1} + {a}_{3} = a + a {r}^{2} = a \left(1 + {r}^{2}\right)$
$96 = {a}_{2} + {a}_{4} = a r + a {r}^{3} = a r \left(1 + {r}^{2}\right)$
So:
$r = \frac{a r \left(1 + {r}^{2}\right)}{a \left(1 + {r}^{2}\right)} = \frac{96}{40} = \frac{12}{5}$
$a = \frac{40}{1 + {r}^{2}}$
$= \frac{40}{1 + {\left(\frac{12}{5}\right)}^{2}}$
$= \frac{40}{1 + \frac{144}{25}}$
$= \frac{40}{\frac{169}{25}}$
$= \frac{40 \cdot 25}{169}$
$= \frac{1000}{169}$
Then:
${a}_{6} = a {r}^{5}$
$= \frac{1000}{169} \cdot {\left(\frac{12}{5}\right)}^{5}$
$= \frac{{2}^{3} \cdot {5}^{3} \cdot {2}^{10} \cdot {3}^{5}}{{13}^{2} \cdot {5}^{5}}$
$= \frac{{2}^{13} \cdot {3}^{5}}{{5}^{2} \cdot {13}^{2}}$
$= \frac{1990656}{4225}$ |
# How do you differentiate g(x) = (5x-2)(x^2+1) using the product rule?
Feb 6, 2016
$g ' \left(x\right) = 2 x \left(5 x - 2\right) + 5 \left({x}^{2} + 1\right)$
#### Explanation:
differentiation using the product rule
If g(x) = f(x).h(x)
then g'(x) = f(x).h'(x) + h(x).f'(x)................(A)
here f(x) =(5x - 2 ) and f'(x) = 5
$\textcolor{b l a c k}{\text{-------------------------------------}}$
h(x) $= \left({x}^{2} + 1\right)$ hence h'(x) = 2x
$\textcolor{b l a c k}{\text{--------------------------------}}$
substituting these values into (A)
g'(x) = (5x - 2).2x + $\left({x}^{2} + 1\right) .5$
$\Rightarrow g ' \left(x\right) = 2 x \left(5 x - 2\right) + 5 \left({x}^{2} + 1\right)$ |
Short Notes: Playing with Numbers
# Playing with Numbers Class 6 Notes Maths
Table of contents Factors and Multiples Prime and Composite Numbers Tests for Divisibility of Numbers Common Factors and Common Multiples Prime Factorisation Highest Common Factor (HCF) Lowest Common Multiple (LCM) Solved Examples
## Factors and Multiples
Factors
The numbers which exactly divides the given number are called the Factors of that number.
As we can see that we get the number 12 by
1 × 12, 2 × 6, 3 × 4, 4 × 3, 6 × 2 and 12 ×1
Hence,
1, 2, 3, 4, 6 and 12 are the factors of 12.
The factors are always less than or equal to the given number.
### Multiples
If we say that 4 and 5 are the factors of 20 then 20 is the multiple of 4 and 5 both.
For Example:
List the multiples of 3
Multiples are always more than or equal to the given number.
Some facts about Factors and Multiples
• 1 is the only number which is the factor of every number.
• Every number is the factor of itself.
• All the factors of any number are the exact divisor of that number.
• All the factors are less than or equal to the given number.
• There are limited numbers of factors of any given number.
• All the multiples of any number are greater than or equal to the given number.
• There are unlimited multiples of any given numbers.
• Every number is a multiple of itself.
Question for Short Notes: Playing with Numbers
Try yourself:What is a factor of a number?
### Perfect Number
If the sum of all the factors of any number is equal to the double of that number then that number is called a Perfect Number.
Perfect Number Factors Sum of all the factors 6 1, 2, 3, 6 12 28 1, 2, 4, 7, 14, 28 56 496 1, 2, 4, 8, 16, 31, 62, 124, 248, 496 992
## Prime and Composite Numbers
Prime Numbers
The numbers whose only factors are 1 and the number itself are called the Prime Numbers.
Like 2, 3, 5, 7, 11 etc.
### Composite Numbers
All the numbers with more than 2 factors are called composite numbers or you can say that the numbers which are not prime numbers are called Composite Numbers.
Like 4, 6, 8, 10, 12 etc.
Remark: 1 is neither a prime nor a composite number.
### Even and Odd Numbers
All the multiples of 2 are even numbers. To check whether the number is even or not, we can check the number at one's place. If the number at ones place is 0,2,4,6 and 8 then the number is even number.
The numbers which are not even are called Odd Numbers.
Remark: 2 is the smallest even prime number. All the prime numbers except 2 are odd numbers.
## Tests for Divisibility of Numbers
1. Divisibility by 2: If there are any of the even numbers i.e. 0, 2, 4, 6 and 8 at the end of the digit then it is divisible by 2.
Example: Check whether 63 is divisible by 2 or not.
Sol:
The last digit of 63 is 3 i.e. odd number so 63 is not divisible by 2.
2. Divisibility by 3: A given number will only be divisible by 3 if the total of all the digits of that number is multiple of 3.
Example: Check whether 2400 is divisible by 3 or not.
Sol: The sum of the digits of 2400 i.e. 2 + 4 + 0 + 0 = 6, which is the multiple of 3 so 2400 is divisible by 3.
3. Divisibility by 4: We have to check whether the last two digits of the given number are divisible by 4 or not. If it is divisible by 4 then the whole number will be divisible by 4.
Example Check the number 23436 is divisible by 4 or not.
Sol:
The last two digits of 23436 are 36 which are divisible by 4, so 23436 are divisible by 4.
4. Divisibility by 5: Any given number will be divisible by 5 if the last digit of that number is ‘0' or ‘5'.
Example: Check whether 6300 is divisible by 5 or not.
Sol:
The last digit of 6300 is 0 so it is divisible by 5.
5. Divisibility by 6: Any given number will be divisible by 6 if it is divisible by 2 and 3 both. So we should do the divisibility test of 2 and 3 with the number and if it is divisible by both then it is divisible by 6 also.
Example: Check the number 342341 is divisible by 6 or not.
Sol:
342341 is not divisible by 2 as the digit at ones place is odd and is also not divisible by 3 as the sum of its digits i.e. 3 + 4 + 2 + 3 + 4 + 1 = 17 is also not divisible by 3.Hence 342341 is not divisible by 6.
6. Divisibility by 7: Any given number will be divisible by 7 if we double the last digit of the number and then subtract the result from the rest of the digits and check whether the remainder is divisible by 7 or not. If there is a large number of digits then we have to repeat the process until we get the number which could be checked for the divisibility of 7.
Example: Check the number 2030 is divisible by 7 or not.
Sol: Given number is 2030
• Double the last digit, 0 × 2 = 0
• Subtract 0 from the remaining number 203 i.e. 203 – 0 = 203
• Double the last digit, 3 × 2 = 6
• Subtract 6 from the remaining number 20 i.e. 20 - 6 = 14
• The remainder 14 is divisible by 7 hence the number 203 is divisible by 7.
7. Divisibility by 8: We have to check whether the last three digits of the given number are divisible by 8 or not. If it is divisible by 8 then the whole number will be divisible by 8.
Example: Check whether the number 74640 is divisible by 8 or not.
Sol: The last three digit of the number 74640 is 640. As the number 640 is divisible by 8 hence the number 74640 is also divisible by 8.
8. Divisibility by 9: Any given number will be divisible by 9 if the total of all the digits of that number is divisible by 9.
Example: Check whether 6390 is divisible by 9 or not.
Sol: The sum of the digits of 6390 is 6 + 3 + 9 + 0 = 18 which is divisible by 9 so 6390 is divisible by 9.
9. Divisibility by 10: Any given number will be divisible by 10 if the last digit of that number is zero.
Example: Check the number 123 and 2630 are divisible by 10 or not.
Sol:
• The ones place digit is 3 in 123 so it is not divisible by 10.
• The ones place digit is 0 in 2630 so it is divisible by 10.
Question for Short Notes: Playing with Numbers
Try yourself:Which condition determines divisibility by 5?
## Common Factors and Common Multiples
Example 1: What are the common factors of 25 and 55?
Sol:
Factors of 25 are 1, 5.
Factors of 55 are 1, 5, 11.
Common factors of 25 and 55 are 1 and 5.
Example 2: Find the common multiples of 3 and 4.
Sol:
Common multiples of 3 and 4 are 0, 12, 24 and so on.
### Co-prime Numbers
If 1 is the only common factor between two numbers then they are said to be Co-prime Numbers.
Example: Check whether 7 and 15 are co-prime numbers or not.
Sol: Factors of 7 are 1 and 7.
Factors of 15 are 1, 3, 5 and 15.
The common factor of 7 and 15 is 1 only. Hence they are the co-prime numbers.
## Prime Factorisation
Prime Factorisation is the process of finding all the prime factors of a number.
There are two methods to find the prime factors of a number-
1. Prime factorisation using a factor tree
We can find the prime factors of 70 in two ways.
The prime factors of 70 are 2, 5 and 7 in both the cases.
2. Repeated Division Method
Find the prime factorisation of 64 and 80.
The prime factorisation of 64 is 2 × 2 × 2 × 2 × 2 × 2.
The prime factorisation of 80 is 2 × 2 × 2 × 2 × 5.
## Highest Common Factor (HCF)
The highest common factor (HCF) of two or more given numbers is the greatest of their common factors. Its other name is (GCD) Greatest Common Divisor.
### Method to find HCF
To find the HCF of given numbers, we have to find the prime factorisation of each number and then find the HCF.
Example: Find the HCF of 60 and 72.
Sol: First, we have to find the prime factorisation of 60 and 72.
Then encircle the common factors.
HCF of 60 and 72 is 2 × 2 × 3 = 12.
Question for Short Notes: Playing with Numbers
Try yourself:What is the highest common factor (HCF) of two or more given numbers?
## Lowest Common Multiple (LCM)
The lowest common multiple of two or more given number is the smallest of their common multiples.
### Methods to find LCM
1. Prime Factorisation Method: To find the LCM we have to find the prime factorisation of all the given numbers and then multiply all the prime factors which have occurred a maximum number of times.
Example: Find the LCM of 60 and 72.
Sol: First, we have to find the prime factorisation of 60 and 72.
Then encircle the common factors.
To find the LCM, we will count the common factors one time and multiply them with the other remaining factors.
LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 = 360
2. Repeated Division Method: If we have to find the LCM of so many numbers then we use this method.
Example: Find the LCM of 105, 216 and 314.
Sol:
Use the repeated division method on all the numbers together and divide until we get 1 in the last row.
LCM of 105,216 and 314 is 2 × 2 × 2 × 3 × 3 × 3 × 5 × 7 × 157 = 1186920
## Solved Examples
Example 1: There are two containers having 240 litres and 1024 litres of petrol respectively. Calculate the maximum capacity of a container which can measure the petrol of both the containers when used an exact number of times.
Sol:
As we have to find the capacity of the container which is the exact divisor of the capacities of both the containers, i. e. maximum capacity, so we need to calculate the HCF.
The common factors of 240 and 1024 are 2 × 2 × 2 × 2.
Thus, the HCF of 240 and 1024 is 16.
Therefore, the maximum capacity of the required container is 16 litres.
Example 2: What could be the least number which when we divide by 20, 25 and 30 leaves a remainder of 6 in every case?
Sol: As we have to find the least number so we will calculate the LCM first.
LCM of 20, 25 and 30 is 2 × 2 × 3 × 5 × 5 = 300.
Here 300 is the least number which when divided by 20, 25 and 30 then they will leave remainder 0 in each case. But we have to find the least number which leaves remainder 6 in all cases.
Hence, the required number is 6 more than 300.
The required least number = 300 + 6 = 306.
The document Playing with Numbers Class 6 Notes Maths is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6
## Mathematics (Maths) Class 6
94 videos|347 docs|54 tests
## FAQs on Playing with Numbers Class 6 Notes Maths
1. What are factors and multiples?
Ans. Factors are the numbers that divide a given number without leaving a remainder, while multiples are the numbers obtained by multiplying a given number by any whole number.
2. What is a perfect number?
Ans. A perfect number is a positive integer that is equal to the sum of its proper divisors, excluding the number itself. For example, 6 is a perfect number because its proper divisors (1, 2, and 3) add up to 6.
3. What are prime numbers and composite numbers?
Ans. Prime numbers are positive integers greater than 1 that have no other divisors except 1 and themselves. Composite numbers, on the other hand, have more than two divisors. For example, 5 is a prime number, while 8 is a composite number.
4. What are even and odd numbers?
Ans. Even numbers are integers that are divisible by 2 without leaving a remainder, while odd numbers are integers that are not divisible by 2. For example, 4 is an even number, while 7 is an odd number.
5. How can we test the divisibility of numbers?
Ans. There are several tests for divisibility, such as the divisibility rule for 2 (if the last digit of a number is even), the divisibility rule for 3 (if the sum of the digits is divisible by 3), and the divisibility rule for 5 (if the last digit is either 0 or 5). These rules help determine if a number is divisible by another number without performing the actual division.
## Mathematics (Maths) Class 6
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# Integrals
## Class 12 NCERT
### NCERT
1 Find an anti-derivative (or integral) of the function $\sin 2x$ by the method of inspection.
The anti-derivative of $\sin 2x$ is a function of $x$ whose derivative is $\sin 2x$. It is known that, $\\$ $\dfrac{d}{dx}(\cos2x) =-2\sin2x\\ \implies \sin 2x=-\dfrac{1}{2}\dfrac{d}{dx}(cos 2x)\\ \therefore \sin2x =\dfrac{d}{dx}(-\dfrac{1}{2}\cos 2x)$$\\ Therefore, the anti-derivative of \sin 2x is -\dfrac{1}{2}\cos2x. 2 Find an anti-derivative (or integral) of the function \cos 3x by the method of inspection. ##### Solution : The anti-derivative of \cos 3x is a function of x whose derivative is \cos 3x.$$\\$ It is known that,$\\$ $\dfrac{d}{dx}(\sin3x)=3\cos3x\\ \implies \cos3x =\dfrac{1}{3}\dfrac{d}{dx}(\sin3x)\\ \therefore \cos3x=\dfrac{d}{dx}(\dfrac{1}{3}\sin3x)$$\\ Therefore, the anti-derivative of \cos3x is \dfrac{1}{3}\sin3x 3 Find an anti-derivative (or integral) of the function e^{ 2x} by the method of inspection. ##### Solution : The anti–derivative of e^{ 2x} is the function of x whose derivative is e^{ 2x} It is known that,\\ \dfrac{d}{dx}(e^{2x}) =2e^{2x}\\ \implies e^{2x} = \dfrac{1}{2}\dfrac{d}{dx}(e^{2x})\\ \therefore e^{2x} =\dfrac{d}{dx}(\dfrac{1}{2}e^{2x}).$$\\$ Therefore, the anti-derivative of $e^{2x}$ is $\dfrac{1}{2}e^{2x}$
4 Find an anti-derivative (or integral) of the function $(ax + b)^ 2$ by the method of inspection.
The anti-derivative of $(ax + b)^ 2$ is the function of $x$ whose derivative is $(ax + b)^ 2$.$\\$ It is known that,$\\$ $\dfrac{d}{dx}(ax+b)^3=3a(ax+b)^2\\ \implies (ax+b)^2=\dfrac{1}{3a}\dfrac{d}{dx}(ax+b)^3\\ \therefore (ax+b)^2=\dfrac{d}{dx}(\dfrac{1}{3a}(ax+b)^3)$$\\ Therefore, the anti-derivative of (ax+b)^2 is \dfrac{1}{3a}(ax+b)^3 5 Find an anti-derivative (or integral) of the function \sin 2x - 4e^{ 3x} by the method of inspection. ##### Solution : The anti-derivative of \sin 2x - 4e^{ 3x} is the function of x whose derivative is \sin2x - 4e^{ 3x} It is known that,\\ \dfrac{d}{dx}(-\dfrac{1}{2}\cos2x-\dfrac{4}{3}e^{3x}) =\\ \sin2x-4e^{3x}$$\\$ Therefore, the anti derivative of $(\sin2x -4e^{3x})$ is $(-\dfrac{1}{2}\cos2x-\dfrac{4}{3}e^{3x})$
6 $\int (4e^{3x}+1)\mathrm dx$
##### Solution :
$\int (4e^{3x}+1)\mathrm dx $$\\ =4 \int e^{3x} \mathrm dx+\int 1 \mathrm dx\\ 4(\dfrac{e^{3x}}{3})+x+C\\ \dfrac{4}{3}e^{3x}+x+C$$\\$ where $C$ is an arbitrary constant.
7 $\int x^2(1-\dfrac{1}{x^2}) \mathrm dx$
##### Solution :
$\int x^2(1-\dfrac{1}{x^2}) \mathrm dx$$\\ =\int(x^2-1)\mathrm dx\\ \int x^2 \mathrm dx -\int 1 \mathrm dx \\ \dfrac{x^3}{3}-x+C$$\\$ where $C$ is an arbitrary constant.
8 $\int (ax^2+ bx+c)\mathrm dx$
##### Solution :
$\int (ax^2+ bx+c)\mathrm dx$$\\ =a \int x^2 \mathrm dx + b\int x \mathrm dx +c \int 1. \mathrm dx\\ a(\dfrac{x^3}{3})+b(\dfrac{x^2}{2})+cx + C\\ =\dfrac{ax^3}{3}+\dfrac{bx^2}{2} + cx + C$$\\$ where $C$ is an arbitrary constant.
9 $\int (2x^2+e^x) \mathrm dx$
##### Solution :
$\int (2x^2+e^x) \mathrm dx$$\\ =2 \int x^2 \mathrm dx+ \int e^x \mathrm dx\\ =2(\dfrac{x^3}{3})+e^x+C\\ =\dfrac{2}{3}x^3+ e^x+C$$\\$ where $C$ is an arbitrary constant.
10 $\int (\sqrt{x}-\dfrac{1}{\sqrt{x}})^2 \mathrm dx$
##### Solution :
$\int (\sqrt{x}-\dfrac{1}{\sqrt{x}})^2 \mathrm dx$$\\ =\int (x+\dfrac{1}{x}-2)\mathrm dx\\ =\int x \mathrm dx+\int \dfrac{1}{x} \mathrm dx -2\int 1.\mathrm dx \\ =\dfrac{x^2}{2}+\log|x|-2x+C$$\\$ where $C$ is an arbitrary constant.
11 $\int \dfrac{x^3+ 5x^2-4}{x^2} \mathrm dx$
##### Solution :
$\int \dfrac{x^3+ 5x^2-4}{x^2} \mathrm dx$$\\ =\int (x+5-4x^{-2})\mathrm dx\\ =\int x \mathrm dx + 5\int 1.\mathrm dx -4\int x^{-2}\mathrm dx\\ \dfrac{x^2}{2}+5x-4(\dfrac{x^{-1}}{-1})+C\\ =\dfrac{x^2}{2}+5x+\dfrac{4}{x}+C$$\\$ where $C$ is an arbitrary constant.
12 $\int \dfrac{x^3+3x+4}{\sqrt{x}}\mathrm dx$
##### Solution :
$\int \dfrac{x^3+3x+4}{\sqrt{x}}\mathrm dx$$\\ =\int(x^{\frac{5}{2}}+3x^{\frac{1}{2}}+4x^{-\frac{1}{2}})\mathrm dx\\ \dfrac{x^{\frac{7}{2}}}{\frac{7}{2}}\\ +\dfrac{3(x^{\frac{3}{2}})}{\frac{3}{2}}\\ +\dfrac{4(x^{\frac{1}{2}})}{\dfrac{1}{2}}+C\\ =\dfrac{2}{7}x^{\frac{7}{2}}+2x^{3}{2}+8x^{\frac{1}{2}}+C\\ =\dfrac{2}{7}x^(\frac{7}{2})+2x^{\frac{3}{2}}+8\sqrt{x}+C$$\\$ where $C$ is an arbitrary constant.
13 $\int \dfrac{x^3-x^2+x-1}{x-1} \mathrm dx$
##### Solution :
$\int \dfrac{x^3-x^2+x-1}{x-1} \mathrm dx $$\\ On factorising, we obtain\\ \int \dfrac{(x^2+1)(x-1)}{x-1} \mathrm dx\\ =\int (x^2+1) \mathrm dx\\ =\int x^2 \mathrm dx + \int 1 \mathrm dx \\ \dfrac{x^3}{3}+x+C$$\\$ where $C$ is an arbitrary constant.
14 $\int (1-x)\sqrt{x} \mathrm dx$
##### Solution :
$\int (1-x)\sqrt{x} \mathrm dx $$\\ =\int(\sqrt{x}-x^{\frac{3}{2}}) \mathrm dx\\ =\int x^{\frac{1}{2}}\mathrm dx -\int x^{\frac{3}{2}} \mathrm dx\\ =\dfrac{x^{\frac{3}{2}}}{\dfrac{3}{2}}-\dfrac{x^{\frac{5}{2}}}{\dfrac{5}{2}}+C\\ =\dfrac{2}{3}x^{\frac{3}{2}}-\dfrac{2}{5}x^{5}{2}+C$$\\$ where $C$ is an arbitrary constant.
15 $\int \sqrt{x}(3x^2+2x+3)\mathrm dx$
##### Solution :
$\int \sqrt{x}(3x^2+2x+3)\mathrm dx $$\\ =3\int x^{\frac{5}{2}}\mathrm dx+2\int x^{\frac{3}{2}}\mathrm dx+3\int x ^{\frac{1}{2}} \mathrm dx\\ =3(\dfrac{x^{\frac{7}{2}}}{\dfrac{7}{2}})+2(\dfrac{x^{\frac{5}{2}}}{\dfrac{5}{2}})\\ +3 \dfrac{(x^{\frac{3}{2}})}{\dfrac{3}{2}}+C\\ =\dfrac{6}{7}x^{\frac{7}{2}}+\dfrac{4}{5}x^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$$\\$ where $C$ is an arbitrary constant.
16 $\int (2x-3 \cos x + e^x) \mathrm dx$
##### Solution :
$\int (2x-3 \cos x + e^x) \mathrm dx$$\\ =2\int x \mathrm dx -3 \int \cos x \mathrm dx +\int e^x \mathrm dx\\ =\dfrac{2 x^2}{2}-3(\sin x)+e^x+C\\ =x^2-3 \sin x +e^x+C$$\\$ where $C$is an arbitrary constant.
17 $\int (2x^2-3 \sin x +5\sqrt{x})\mathrm dx$
##### Solution :
$\int (2x^2-3 \sin x +5\sqrt{x})\mathrm dx$$\\ =2\int x^2 \mathrm dx -3 \int \sin x \mathrm dx\\ +5\int x^{\frac{1}{2}}\mathrm dx \\ =\dfrac{2 x^3}{3}-3(-\cos x)+5(\dfrac{x^{\frac{3}{2}}}{\dfrac{3}{2}})+C\\ =\dfrac{2}{3}x^3+3 \cos x+\dfrac{10}{3}x^{\frac{3}{2}}+C$$\\$ where $C$ is an arbitrary constant.
18 $\int \sec x(\sec x+ \tan x)\mathrm dx$
##### Solution :
$\int \sec x(\sec x+ \tan x)\mathrm dx$$\\ =\int (\sec^2 x+\sec x \tan x)\mathrm dx\\ =\int \sec^2 x \mathrm dx +\int \sec x \tan x \mathrm dx\\ =\tan x + \sec x + C$$\\$ where $C$ is an arbitrary constant.
19 $\int \dfrac{\sec^2 x}{\csc^2 x}\mathrm dx$
##### Solution :
$\int \dfrac{\sec^2 x}{\csc^2 x}\mathrm dx$$\\ =\int \dfrac{\dfrac{1}{\cos ^2 x}}{\dfrac{1}{\sin^2 x}}\mathrm dx\\ =\int \dfrac{\sin^2 x}{\cos ^2 x}\mathrm dx\\ =\int \tan^2 x \mathrm dx\\ =\int (\sec^2 x-1) \mathrm dx\\ =\int \sec^2 x \mathrm dx -\int 1 \mathrm dx \\ =\tan x- x+C$$\\$ where $C$ is an arbitrary constant.
20 $\int \dfrac{2-3 \sin x}{\cos^2 x}\mathrm dx$
$\int \dfrac{2-3 \sin x}{\cos^2 x}\mathrm dx\\ =\int (\dfrac{2}{\cos^2 x}-\dfrac{3 \sin x}{\cos^2 x})\mathrm dx\\ =\int 2 \sec^2 x \mathrm dx -3 \int \tan x \sec x \mathrm dx\\ =2 \tan x -3\sec x +C$$\\ where C is an arbitrary constant. 21 The anti-derivative of (\sqrt{x}+\dfrac{1}{\sqrt{x}}) equals ##### Solution : \int \sqrt{x}+\dfrac{1}{\sqrt{x}}\mathrm dx \\ =\int x^{\frac{1}{2}} \mathrm dx + \int x^{\frac{1}{2}} \mathrm dx\\ =\dfrac{x^{\frac{3}{2}}}{\dfrac{3}{2}}\\ +\dfrac{x^{\frac{1}{2}}}{\dfrac{1}{2}}+C\\ =\dfrac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C,$$\\$ where $C$ is an arbitrary constant.$\\$ Hence, the correct Answer is C.
22 If $\dfrac{\mathrm d}{\mathrm dx} f(x)=4x^3-\dfrac{3}{x^4}$such that $f(2)=0,$ then $f(x)$ is
It is given that,$\dfrac{\mathrm d}{\mathrm d} f(x)=4x^3-\dfrac{3}{x^4}$$\\ Anti-derivative of 4 x^3-\dfrac{3}{x^4}=f(x)\\ \therefore f(x)=\int 4 x^3-\dfrac{3}{x^4}=f(x)\\ f(x)=4\int x^3 \mathrm dx -3 \int (x^{-4}) \mathrm dx\\ f(x)=4(\dfrac{x^4}{4})-3 (\dfrac{x^{-3}}{-3})+C\\ f(x) x^4 + \dfrac{1}{x^3}+C$$\\$ Also,$\\$ $f(2)=0\\ \therefore f(2) =(2)^4+\dfrac{1}{(2)^3}+C=0\\ \implies 16+\dfrac{1}{8}+C=0\\ \implies C=-(16+\dfrac{1}{8})\\ \implies C=\dfrac{-129}{8}\\ \therefore f(x)=x^4+ \dfrac{1}{x^3}-\dfrac{129}{8}$$\\ Hence, the correct Answer is A. 23 Integrate \dfrac{2x}{1+x^2} ##### Solution : Let 1+x^2 =t\\ \therefore 2x \mathrm dx =\mathrm dt\\ \implies \int \dfrac{2x}{1+x^2}\mathrm dx\\ =\int \dfrac{1}{t} \mathrm dt\\ =\log|t|+C\\ =\log|1+x^2|+C\\ \log(1+x^2)+C$$\\$ where C is an arbitrary constant.
24 Integrate$\dfrac{(\log x)^2}{x}$
Let $\log x=t\\ \therefore \dfrac{1}{x} \mathrm dx =\mathrm dt\\ \implies \int \dfrac{(\log|x|)^2}{x}\mathrm dx \\ =\int t^2 \mathrm dt \\ =\dfrac{t^3}{3} + C\\ =\dfrac{(\log|x|)^3}{3}+C$$\\where C is an arbitrary constant. 25 Integrate \dfrac{1}{x+ x \log x} ##### Solution : The given function can be rewritten as\\ \dfrac{1}{x+x \log x}=\dfrac{1}{x(1+ \log x)}$$\\$ Let $1+\log x =t \\ \therefore \dfrac{1}{x}\mathrm dx=\mathrm dt\\ \implies \int \dfrac{1}{x(1+\log x)}\mathrm dx\\ =\int \dfrac{1}{t}\mathrm dt\\ =\log|t|+C\\ =\log |1+\log x|+C$$\\where C is an arbitrary constant. 26 Integrate \sin x . \sin (\cos x) ##### Solution : Let \cos x = t\\ \therefore -\sin x \mathrm dx = dt\\ \implies \int \sin x .\sin ( \cos x ) \mathrm dx =-\int \sin t \mathrm dt\\ =-[- \cos t ]+ C\\ =\cos t + C\\ = \cos ( \cos x )+ C$$\\$ where C is an arbitrary constant.
27 Integrate $\sin (ax + b) \cos (ax + b)$
##### Solution :
The given function can be rewritten as$\\$ $\sin (ax + b) \cos (ax + b) =\\ \dfrac{2 \sin(ax + b)\cos (ax +b)}{2}\\ =\dfrac{\sin 2(ax+b)}{2}$$\\ Let 2(ax+b)=t\\ \therefore 2 a \mathrm dx = \mathrm dt\\ \implies \int \dfrac{\sin 2 (ax+b)}{2}\mathrm dx\\ =\dfrac{1}{2}\int \dfrac{\sin t \mathrm dt}{2a}\\ =\dfrac{1}{4a}[-\cos t]+C\\ =\dfrac{-1}{4a}\cos 2(ax+b)+C$$\\$ where C is an arbitrary constant.
28 Integrate $\sqrt{ax+b}$
Let $ax + b = t\\ \implies a \mathrm dx=\mathrm dt\\ \therefore \mathrm dx=\dfrac{1}{a}\mathrm dt\\ \implies \int(ax+b)^{\frac{1}{2}} \mathrm dx\\ =\dfrac{1}{a}\int t^{\frac{1}{2}} \mathrm dt\\ =\dfrac{1}{a}(\dfrac{t^{\frac{1}{2}}}{\dfrac{3}{2}})+C\\ =\dfrac{2}{3a}(ax+b^{\frac{3}{2}})+C$$\\ where C is an arbitrary constant. 29 Integrate x\sqrt{x+2} ##### Solution : Let x + 2 = t\\ \therefore \mathrm dx = \mathrm dt\\ \implies \int x \sqrt{x+2}\mathrm dx\\ =\int (t-2)\sqrt{t}\mathrm dt\\ =\int (t^{\frac{3}{2}}-2t^{\frac{1}{2}}) \mathrm dt\\ =\int t^{\frac{3}{2}} \mathrm dt - 2 \int t^{\frac{1}{2}} \mathrm dt\\ =\dfrac{t^{\frac{5}{2}}}{\dfrac{5}{2}}\\ -2 (\dfrac{t^{\frac{3}{2}}}{\dfrac{3}{2}})+C\\ =\dfrac{2}{5}t^{\frac{5}{2}}-\dfrac{4}{3}t^{\frac{3}{2}}+C\\ =\dfrac{2}{5}(x+2)^{\frac{5}{2}}\\ -\dfrac{4}{3}(x+2)^{\frac{3}{2}}+C$$\\$where C is an arbitrary constant.
30 $x\sqrt{1+2x^2}$
Let $1+2x^2=t\\ \therefore 4x \mathrm dx =\mathrm dt\\ \implies \int x\sqrt{1+2x^2} \mathrm dx\\ =\int \dfrac{\sqrt{t}}{4}\mathrm dt\\ =\dfrac{1}{4}\int t^{\frac{1}{2}} \mathrm dt\\ =\dfrac{1}{4}(\dfrac{t^{\frac{3}{2}}}{\dfrac{3}{4}})+C\\ =\dfrac{1}{6}(1+2x^2)^{\frac{3}{2}}+C$$\\ where C is an arbitrary constant. 31 Integrate(4x+2)\sqrt{x^2+x+1} ##### Solution : Let x^2+x+1=t\\ \therefore (2x+1)\mathrm dx=\mathrm dt\\ \int (4x+2)\sqrt{x^2+x+1 }\mathrm dx\\ =\int 2 \sqrt{t}\mathrm dt\\ =2\int \sqrt{t}\mathrm dt\\ =2(\dfrac{t^{\frac{3}{2}}}{\dfrac{3}{2}})+C\\ =\dfrac{4}{3}(x^2+x+1)^{\frac{3}{2}}+C$$\\$ where C is an arbitrary constant.
32 Integrate $\dfrac{1}{x-\sqrt{x}}$
##### Solution :
The given function can be rewritten as$\\$ $\dfrac{1}{x-\sqrt{x}}=\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}$$\\ Let (\sqrt{x}-1)=t\\ \therefore \dfrac{1}{2\sqrt{x}}\mathrm dx =\mathrm dt\\ \implies \int\dfrac{1}{\sqrt{x}(\sqrt{x-1})}\mathrm dx\\ =\int \dfrac{2}{t}\mathrm dt\\ =2 \log|t|+C\\ =2 \log|\sqrt{x}-1|+C$$\\$where C is an arbitrary constant.
33 Integrate $\dfrac{x}{\sqrt{x+4}},x> 0$
##### Solution :
Let $x+4=t$$\\ \therefore \mathrm dx=\mathrm dt\\ \int \dfrac{x}{\sqrt{x+4} } \mathrm dx\\ =\int \dfrac{(t-4)}{\sqrt{t}} \mathrm dt\\ =\int(\sqrt{t}-\dfrac{4}{\sqrt{t}})\mathrm dt$$\\$ $=\dfrac{t^{\frac{3}{2}}}{\dfrac{3}{2}}-4(\dfrac{t^{\frac{1}{2}}}{\dfrac{1}{2}})+C\\ =\dfrac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C\\ =\dfrac{2}{3}t.(t)^{\frac{1}{2}}-8(t)^{\frac{1}{2}}+C\\ =\dfrac{2}{3}t(t)^{\frac{1}{2}}(t-12)+C\\ =\dfrac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C\\ =\dfrac{2}{3}\sqrt{x+4}(x-8)+C$ $\\$ where C is an arbitrary constant.
34 Integrate $(x^3-1)^{\frac{1}{3}}x^5$
##### Solution :
Let $x^3 -1=t\\ \therefore 3 x^2 \mathrm dx =\mathrm dt\\ \implies \int(x^3-1)^{\frac{1}{3}} x^5 \mathrm dx\\ =\int(x^3-1)^{\frac{1}{3}} x^3.x^2 \mathrm dx\\ =\int t^{\frac{1}{3}}(t+1) \dfrac{\mathrm dt}{3}\\ =\dfrac{1}{3}\int(t^{\frac{4}{3}}+t^{\frac{1}{3}})\mathrm dt\\ =\dfrac{1}{3}[\dfrac{t^{\frac{7}{3}}}{\dfrac{7}{3}}+\dfrac{t^{\frac{4}{3}}}{\dfrac{4}{3}}] +C\\ =\dfrac{1}{3}[\dfrac{3}{7}t^{\frac{7}{3}}+\dfrac{3}{4}t^{\frac{4}{3}}]+C\\ =\dfrac{1}{7}(x^3-1)^{\frac{7}{3}}+\dfrac{1}{4}(x^3-1)^{\frac{4}{3}}+C$$\\$ where C is an arbitrary constant. |
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Difficulty Level: At Grade Created by: CK-12
This chapter has presented three methods to solve a quadratic equation:
• By graphing to find the zeros;
• By solving using square roots; and
• By using completing the square to find the solutions
This lesson will present a fourth way to solve a quadratic equation: using the Quadratic Formula.
## History of the Quadratic Formula
As early as 1200 BC, people were interested in solving quadratic equations. The Babylonians solved simultaneous equations involving quadratics. In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit formula to solve a quadratic equation. The Quadratic Formula was written as it is today by the Arabic mathematician Al-Khwarizmi. It is his name upon which the word “Algebra” is based.
The solution to any quadratic equation in standard form 0=ax2+bx+c\begin{align*}0=ax^2+bx+c\end{align*} is
x=b±b24ac2a\begin{align*}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{align*}
Example: Solve x2+10x+9=0\begin{align*}x^2+10x+9=0\end{align*} using the Quadratic Formula.
Solution: We know from the last lesson the answers are x=1\begin{align*}x=-1\end{align*} or x=9\begin{align*}x=-9\end{align*}.
By applying the Quadratic Formula and a=1,b=10\begin{align*}a=1, b=10\end{align*}, and c=9\begin{align*}c=9\end{align*}, we get:
xxxxxx=10±(10)24(1)(9)2(1)=10±100362=10±642=10±82=10+82 or x=1082=1 or x=9\begin{align*}x &= \frac{-10 \pm \sqrt{(10)^2-4(1)(9)}}{2(1)}\\ x &= \frac{-10 \pm \sqrt{100-36}}{2}\\ x &= \frac{-10 \pm \sqrt{64}}{2}\\ x &= \frac{-10 \pm 8}{2}\\ x &= \frac{-10 + 8}{2} \ or \ x=\frac{-10-8}{2}\\ x &= -1 \ or \ x=-9\end{align*}
Example 1: Solve 4x2+x+1=0\begin{align*}-4x^2+x+1=0\end{align*} using the Quadratic Formula.
Solution:
Quadratic formula:Plug in the values a=4,b=1,c=1.Simplify.Separate the two options.Solve.xxxxx=b±b24ac2a=1±(1)24(4)(1)2(4)=1±1+168=1±178=1+178 and x=1178.39 and x.64\begin{align*}\text{Quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \text{Plug in the values} \ a=-4, b=1, c=1. && x& =\frac{-1 \pm \sqrt{(1)^2-4(-4)(1)}}{2(-4)}\\ \text{Simplify.} && x & =\frac{-1 \pm \sqrt{1+16}}{-8}=\frac{-1 \pm \sqrt{17}}{-8}\\ \text{Separate the two options.} && x&=\frac{-1+\sqrt{17}}{-8} \ \text{and} \ x=\frac{-1-\sqrt{17}}{-8}\\ \text{Solve.} && x & \approx -.39 \ \text{and} \ x \approx .64\end{align*}
Figure 2 provides more examples of solving equations using the quadratic equation. This video is not necessarily different from the examples above, but it does help reinforce the procedure of using the Quadratic Formula to solve equations.
## Finding the Vertex of a Quadratic Equation in Standard Form
The x\begin{align*}x-\end{align*}coordinate of the vertex of 0=ax2+bx+c\begin{align*}0=ax^2+bx+c\end{align*} is x=ba\begin{align*}x=-\frac{b}{a}\end{align*}
## Which Method to Use?
Usually you will not be told which method to use. You will have to make that decision yourself. However, here are some guidelines to which methods are better in different situations.
• Graphing – a good method to visualize the parabola and easily see the intersections. Not always precise.
• Factoring – best if the quadratic expression is easily factorable
• Taking the square root – is best used of the form 0=ax2c\begin{align*}0=ax^2-c\end{align*}
• Completing the square – can be used to solve any quadratic equation. It is a very important method for rewriting a quadratic function in vertex form.
• Quadratic Formula – is the method that is used most often for solving a quadratic equation. If you are using factoring or the Quadratic Formula, make sure that the equation is in standard form.
Example: The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square meters. Find the dimensions of the pool.
Solution: Begin by drawing a sketch. The formula for the area of a rectangle is A=l(w)\begin{align*}A=l(w)\end{align*}.
A875=(x+10)(x)=x2+10x\begin{align*}A &= (x+10)(x)\\ 875 &= x^2+10x\end{align*}
Now solve for x\begin{align*}x\end{align*} using any method you prefer.
The result is x=25\begin{align*}x=25\end{align*}. So, the length of the pool is 35 meters and the width is 25 meters.
## Practice Set
The following video will guide you through a proof of the Quadratic Formula. CK-12 Basic Algebra: Proof of Quadratic Formula (7:44)
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Using the Quadratic Formula (16:32)
1. What is the Quadratic Formula? When is the most appropriate situation to use this formula?
2. When was the first known solution of a quadratic equation recorded?
Find the x\begin{align*}x-\end{align*}coordinate of the vertex of the following equations.
1. x214x+45=0\begin{align*}x^2-14x+45=0\end{align*}
2. 8x216x42=0\begin{align*}8x^2-16x-42=0\end{align*}
3. 4x2+16x+12=0\begin{align*}4x^2+16x+12=0\end{align*}
4. x2+2x15=0\begin{align*}x^2+2x-15=0\end{align*}
1. x2+4x21=0\begin{align*}x^2+4x-21=0\end{align*}
2. x26x=12\begin{align*}x^2-6x=12\end{align*}
3. 3x212x=38\begin{align*}3x^2-\frac{1}{2}x=\frac{3}{8}\end{align*}
4. 2x2+x3=0\begin{align*}2x^2+x-3=0\end{align*}
5. x27x+12=0\begin{align*}-x^2-7x+12=0\end{align*}
6. 3x2+5x=0\begin{align*}-3x^2+5x=0\end{align*}
7. 4x2=0\begin{align*}4x^2=0\end{align*}
8. x2+2x+6=0\begin{align*}x^2+2x+6=0\end{align*}
1. x2x=6\begin{align*}x^2-x=6\end{align*}
2. x212=0\begin{align*}x^2-12=0\end{align*}
3. 2x2+5x3=0\begin{align*}-2x^2+5x-3=0\end{align*}
4. x2+7x18=0\begin{align*}x^2+7x-18=0\end{align*}
5. 3x2+6x=10\begin{align*}3x^2+6x=-10\end{align*}
6. 4x2+4000x=0\begin{align*}-4x^2+4000x=0\end{align*}
7. 3x2+12x+1=0\begin{align*}-3x^2+12x+1=0\end{align*}
8. x2+6x+9=0\begin{align*}x^2+6x+9=0\end{align*}
9. 81x2+1=0\begin{align*}81x^2+1=0\end{align*}
10. 4x2+4x=9\begin{align*}-4x^2+4x=9\end{align*}
11. 36x221=0\begin{align*}36x^2-21=0\end{align*}
12. x2+2x3=0\begin{align*}x^2+2x-3=0\end{align*}
13. The product of two consecutive integers is 72. Find the two numbers.
14. The product of two consecutive odd integers is 11 less than 3 times their sum. Find the integers.
15. The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches. Find its dimensions.
16. Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is 200 square feet. How much fencing does Suzie need?
17. Angel wants to cut off a square piece from the corner of a rectangular piece of plywood. The larger piece of wood is 4 feet×8 feet\begin{align*}4 \ \text{feet} \times 8 \ \text{feet}\end{align*} and the cut off part is 13\begin{align*}\frac{1}{3}\end{align*} of the total area of the plywood sheet. What is the length of the side of the square?
18. Mike wants to fence three sides of a rectangular patio that is adjacent the back of his house. The area of the patio is 192 ft2\begin{align*}192 \ ft^2\end{align*} and the length is 4 feet longer than the width. Find how much fencing Mike will need.
Mixed Review
1. The theatre has three types of seating: balcony, box, and floor. There are four times as many floor seats as balcony. There are 200 more box seats than balcony seats. The theatre has a total of 1,100 seats. Determine the number of balcony, box, and floor seats in the theatre.
2. Write an equation in slope-intercept form containing (10, 65) and (5, 30).
3. 120% of what number is 60?
4. Name the set() of numbers to which 16\begin{align*}\sqrt{16}\end{align*} belongs.
5. Divide 617÷234\begin{align*}6 \frac{1}{7} \div - 2 \frac{3}{4}\end{align*}.
6. The set is the number of books in a library. Which of the following is the most appropriate domain for this set: all real numbers; positive real numbers; integers; or whole numbers? Explain your reasoning.
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# Ch 2.5: The Fundamental Theorem of Algebra
## Presentation on theme: "Ch 2.5: The Fundamental Theorem of Algebra"— Presentation transcript:
Ch 2.5: The Fundamental Theorem of Algebra
Theorem: In the complex number system, every nth degree polynomial has n zeros Complex system: Real AND Imaginary numbers Finding all zeros 1. Use Descartes to determine possible solutions 2. Use the calculator to find all rational zeros 3. Use synthetic division to get the problem to a quadratic equation 4. Use the quadratic formula to find the remaining zeros ***If a solution is imaginary, then its conjugate is a solution as well***
Ex: Total Solutions: 5 Positive: 2 or 0 Negative: 1 Imaginary: 4 or 2
1. Find possible number of positive and negative zeros 2. Graph 3. Find the zeros **x=1 has multiplicity of 2 to give 2 positive zeros** Continued…. Positive: 2 or 0 Negative: 1 Imaginary: 4 or 2 X = -2 X = 1 X = 1
ALL SOLUTIONS 5. Synthetic divide one zero at a time
6. Use the quadratic formula to find the final zeros ALL SOLUTIONS
Sometimes you must factor…
1. Graph give 2 irrational zeros, so try factoring! 2. Set each factor equal to zero ALL SOLUTIONS
Find all roots when given one complex root
1. Change it to a factor 2. Multiply it by its conjugate (because it is a root as well) 3. Use Long division to get the remaining quadratic equation 4. Factor the equation to find the zeros or use the quadratic formula
Ex: Find all the roots of if 1 + 3i is a zero
1. Turn it into a factor 2. Multiply by its conjugate 3. Divide by the new factor Continued…
(remember, imaginary and its conjugate)
4. Factor the quotient ALL SOLUTIONS (remember, imaginary and its conjugate)
Ex: Find the 4th degree polynomial with zeros of 1, 1, and 3i
Turn into factors Remember conjugate is a factor as well!! Multiply and Simplify!
Ex: Find the cubic with zeros, 2 and 1-i where f(1)=3
1. Write the factors, including the conjugate 2. To find f(1) = 3, take the function, put a(function)=3 and solve 3. Plug in 1 for x and find a 4. Distribute a |
# Video: US-SAT04S4-Q24-715101364742v2
In the figure shown, π΄π΅ = 12 and π΅πΆ = 5. If the area of the circle is π₯π, find the value of π₯.
03:58
### Video Transcript
In the figure shown, π΄π΅ equals 12 and π΅πΆ equals five. If the area of the circle is π₯π, find the value of π₯.
Weβre told in the question that the length of π΄π΅ is equal to 12 and the length of π΅πΆ is equal to five. Joining the points π΄πΆ, as shown, creates a right-angled triangle as the angle π΄π΅πΆ is equal to 90 degrees. One of our circle theorems states that the angle in a semicircle is equal to 90 degrees. This means that the line π΄πΆ is the diameter of the circle. If we let π be the center of the circle, then π is the midpoint of π΄πΆ. Therefore, π΄π is the radius of the circle.
We might remember at this point that there are some special right triangles that we need to know. For example, a three, four, five and five, 12, 13 right triangle. In our case, as the two shorter sides are lengths five and 12, then the longest side or hypotenuse will have length 13. The diameter of the circle π΄πΆ is length 13. As the radius is half the length of the diameter, this will have length 13 over two or 13 divided by two. We could write this as the decimal 6.5 as a half of 13 is 6.5.
If we hadnβt recognized that we have a special triangle, we could use Pythagorasβ theorem to calculate the length of the diameter. This states that π squared plus π squared is equal to π squared, where π is the longest side of the right triangle known as the hypotenuse. As the length π΄πΆ, the diameter of the circle, is the hypotenuse, we have π΄πΆ squared is equal to 12 squared plus five squared. 12 squared is equal to 144 as 12 multiplied by 12 is 144. Five squared is equal to 25. This means that π΄πΆ squared is equal to 144 plus 25. These two numbers sum to 169. The opposite or inverse of squaring is square rooting. Therefore, we need to square root both sides of this equation. The square root of 169 is equal to 13 as 13 multiplied by 13 is 169. We have once again proved that the length of the diameter π΄πΆ is equal to 13.
We were told that the area of our circle is π₯π. And the formula to calculate the area of a circle is ππ squared. As the radius is equal to 13 over two, the area will be equal to π multiplied by 13 over two squared. If we want to square any fraction, we can square the numerator and denominator separately. 13 squared is equal to 169, as we have already seen. And two squared is equal to four. Therefore, the area is equal to π multiplied by 169 over four. This can be rewritten as 169 over four π.
Our answer is now written in the same form as the question, where π₯ is equal to 169 over four.
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# Lattice multiplication
Lattice multiplication, also known as gelosia multiplication, sieve multiplication, shabakh, Venetian squares, or the Hindu lattice, is a method of multiplication that uses a lattice to multiply two multi-digit numbers. It is mathematically identical to the more commonly used long multiplication algorithm, but it breaks the process into smaller steps, which some practitioners find easier to use.[1]
The method had already arisen by medieval times, and has been used for centuries in many different cultures. It is still being taught in certain curricula today.[2][3]
## Description
A grid is drawn up, and each cell is split diagonally. The two multiplicands of the product to be calculated are written along the top and right side of the lattice, respectively, with one digit per column across the top for the first multiplicand, and one digit per row down the right side for the second multiplicand. Then each cell of the lattice is filled in with product of its column and row digit. For example, if the column digit is 5 and the row digit is 2, then 10 will be written in the cell, with the digit 1 above the diagonal and the digit 0 below the diagonal (see picture for Step 1).
If the simple product lacks a digit in the tens place, simply fill in the tens place with a 0.[1]
Step 1
After all the cells are filled in this manner, the digits in each diagonal are summed, working from the bottom right diagonal to the top left. Each diagonal sum is written where the diagonal ends. If the sum contains more than one digit, the value of the tens place is carried into the next diagonal (see Step 2).
Step 2
Numbers are filled to the left and to the bottom of the grid, and the answer is the numbers read off down (on the left) and across (on the bottom).
Step 3
## Multiplication of Decimals
The lattice technique can also be used to multiply decimal fractions. For instance, to multiply 5.8 by 2.13, a line could be drawn straight down from the decimal in 5.8, and a line straight out from the decimal in 2.13. The lines are extended until they reach each other, at which point they merge and follow the diagonal. The positioning of this diagonal line in the final result is the location of the decimal point.[1]
## History
Lattice multiplication has been used historically in many different cultures. It is not known where it arose first, nor whether it developed independently within more than one region of the world.[4] The earliest recorded use of lattice multiplication:[5]
- in Arab mathematics was by Ibn al-Banna' al-Marrakushi in his Talkhīṣ a‘māl al-ḥisāb, in the Maghreb in the late 13th century
- in European mathematics was by the unknown author of a Latin treatise in England, Tractatus de minutis philosophicis et vulgaribus, c. 1300
- in Chinese mathematics was by Wu Jing in his Jiuzhang suanfa bilei daquan, completed in 1450.
The mathematician and educator David Eugene Smith asserted that lattice multiplication was brought to Italy from the Middle East.[6] This is reinforced by noting that the Arabic term for the method, shabakh, has the same meaning as the Italian term for the method, gelosia, namely, the metal grille or grating (lattice) for a window.
It is sometimes erroneously stated that lattice multiplication was described by Muḥammad ibn Mūsā al-Khwārizmī (Baghdad, c. 825) or by Fibonacci in his Liber Abaci (Italy, 1202, 1228).[7] In fact, however, no use of lattice multiplication by either of these two authors has been found. In Chapter 3 of his Liber Abaci, Fibonacci does describe a related technique of multiplication by what he termed quadrilatero in forma scacherii (“rectangle in the form of a chessboard”). In this technique, the square cells are not subdivided diagonally; only the lowest-order digit is written in each cell, while any higher-order digit must be remembered or recorded elsewhere and then "carried” to be added to the next cell. This is in contrast to lattice multiplication, a distinctive feature of which is that the each cell of the rectangle has its own correct place for the carry digit; this also implies that the cells can be filled in any order desired. Swetz [8] compares and contrasts multiplication by gelosia (lattice), by scacherii (chessboard), and other tableau methods.
Other notable historical uses of lattice multiplication include:[5]
- Jamshīd al-Kāshī’s Miftāḥ al-ḥisāb (Samarqand, 1427), in which the numerals used are sexagesimal (base 60), and the grid is turned 45 degrees to a “diamond” orientation
- the Arte dell’Abbaco, an anonymous text published in the Venetian dialect in 1478, often called the Treviso Arithmetic because it was printed in Treviso, just inland from Venice, Italy
- Luca Pacioli’s Summa de arithmetica (Venice, 1494)
- the Indian astronomer Gaṇeśa’s commentary on Bhāskara II’s Lilāvati (16th century).
## Derivations
Derivations of this method also appeared in the 16th century in Matrakci Nasuh's Umdet-ul Hisab.[9] Matrakçı Nasuh's triangular version of the multiplication technique is seen in the example showing 155 x 525 on the right, and explained in the example showing 236 x 175 on the left figure.[10]
The same principle described by Matrakci Nasuh underlay the later development of the calculating rods known as Napier's bones (Scotland, 1617) and Genaille–Lucas rulers (France, late 1800s).
## References
1. ^ a b c Thomas, Vicki (2005). "Lattice Multiplication". Learn NC. UNC School of Education. Retrieved 4 July 2014.
2. ^ Boag, Elizabeth, “Lattice Multiplication,” BSHM Bulletin: Journal of the British Society for the History of Mathematics 22:3 (Nov. 2007), p. 182.
3. ^ Nugent, Patricia M., “Lattice Multiplication in a Preservice Classroom”, Mathematics Teaching in the Middle School 13:2 (Sept. 2007), pp. 110-113.
4. ^ Jean-Luc Chabert, ed., A History of Algorithms: From the Pebble to the Microchip (Berlin: Springer, 1999), p. 21.
5. ^ a b Jean-Luc Chabert, ed., A History of Algorithms: From the Pebble to the Microchip (Berlin: Springer, 1999), pp. 21-26.
6. ^ Smith, David Eugene, History of Mathematics, Vol. 2, “Special Topics of Elementary Mathematics” (New York: Dover, 1968).
7. ^ The original 1202 version of Liber Abaci is lost. The 1228 version was later published in its original Latin in Boncompagni, Baldassarre, Scritti di Leonardo Pisano, vol. 1 (Rome: Tipografia delle Scienze Matematiche e Fisiche, 1857); an English translation of the same was published by Sigler, Laurence E., Fibonacci’s Liber Abaci: A Translation into Modern English of Leonardo Pisano’s Book of Calculation (New York: Springer Verlag, 2002).
8. ^ Swetz, Frank J., Capitalism and Arithmetic: The New Math of the 15th Century, Including the Full Text of the Treviso Arithmetic of 1478, Translated by David Eugene Smith (La Salle, IL: Open Court, 1987), pp. 205-209.
9. ^ Corlu, M.S., Burlbaw, L.M., Capraro, R. M., Corlu, M.A.,& Han, S. (2010). "The Ottoman Palace School Enderun and The Man with Multiple Talents, Matrakçı Nasuh." Journal of the Korea Society of Mathematical Education, Series D: Research in Mathematical Education. 14(1), p 19-31. |
# Explain how to factor a general expression x^2 + bx + c.
rakesh05 | Certified Educator
To factor a general quadratic expression `x^2+bx+c` we will proceed as under -
Let `(x-alpha)` and `(x-beta)` are the factors of the above expression.
Then `x^2+bx+c=(x-alpha)(x-beta)`
or, `x^2+bx+c=x^2-(alpha+beta)x+alphabeta`
Comparing coefficients of powers of `x` from both sides we get
`alpha+beta=b` (1) and `alphabeta=c` (2).
Now `(alpha-beta)^2=(alpha+beta)^2-4alphabeta`
or, `(alpha-beta)^2=b^2-4c` (Using equations (1) and (2))
or, `(alpha-beta)=+-sqrt(b^2-4c)` (3)
Solving (1) and (3) we get (by taking the positive sign in (3))
`alpha=(-b+sqrt(b^2-4c))/2` ,
`beta=(-b-sqrt(b^2-4c))/2` .
If we take negative sin we get the same values provided the role of `alpha ` and `beta ` are changed.
So the factors are
`x-{(-b+sqrt(b^2-4c))/2}` and `x-{(-b-sqrt(b^2-4c))/2}` .
justaguide | Certified Educator
For a general expression x^2 + bx + c, it is not certain that it can be factored as that depends on what the values of b and c are.
One way of determining the factors is using the formula for the roots of a quadratic equation. The roots of x^2 + bx + c = 0 are given by the formula `(-b+-sqrt(b^2 - 4c))/2`
The factored form of x^2 + bx + c is:
`(x - (-b+sqrt(b^2 - 4c))/2)(x - (-b-sqrt(b^2 - 4c))/2)`
If the term `sqrt(b^2 - 4c)` yields an integer, the expression given was one that could be factored.
Any expression x^2 + bx + c can be factored as `(x - (-b+sqrt(b^2 - 4c))/2)(x - (-b-sqrt(b^2 - 4c))/2)` |
# 26.4 - Student's t Distribution
26.4 - Student's t Distribution
We have just one more topic to tackle in this lesson, namely, Student's t distribution. Let's just jump right in and define it!
Definition. If $$Z\sim N(0,1)$$ and $$U\sim \chi^2(r)$$ are independent, then the random variable:
$$T=\dfrac{Z}{\sqrt{U/r}}$$
follows a $$t$$-distribution with $$r$$ degrees of freedom. We write $$T\sim t(r)$$. The p.d.f. of T is:
$$f(t)=\dfrac{\Gamma((r+1)/2)}{\sqrt{\pi r} \Gamma(r/2)} \cdot \dfrac{1}{(1+t^2/r)^{(r+1)/2}}$$
for $$-\infty<t<\infty$$.
By the way, the $$t$$ distribution was first discovered by a man named W.S. Gosset. He discovered the distribution when working for an Irish brewery. Because he published under the pseudonym Student, the $$t$$ distribution is often called Student's $$t$$ distribution.
History aside, the above definition is probably not particularly enlightening. Let's try to get a feel for the $$t$$ distribution by way of simulation. Let's randomly generate 1000 standard normal values ($$Z$$) and 1000 chi-square(3) values ($$U$$). Then, the above definition tells us that, if we take those randomly generated values, calculate:
$$T=\dfrac{Z}{\sqrt{U/3}}$$
and create a histogram of the 1000 resulting $$T$$ values, we should get a histogram that looks like a $$t$$ distribution with 3 degrees of freedom. Well, here's a subset of the resulting values from one such simulation:
ROW Z CHISQ (3) T(3)
1 -2.60481 10.2497 -1.4092
2 2.92321 1.6517 3.9396
3 -0.48633 0.1757 -2.0099
4 -0.48212 3.8283 -0.4268
5 -0.04150 0.2422 -0.1461
6 -0.84225 -0.0903 -4.8544
7 -0.31205 1.6326 -0.4230
8 1.33068 5.2224 1.0086
9 -0.64104 0.9401 -1.1451
10 -0.05110 2.2632 -0.0588
11 1.61601 4.6566 1.2971
12 0.81522 2.1738 0.9577
13 0.38501 1.8404 0.4916
14 -1.63426 1.1265 -2.6669
...and so on...
994 -0.18942 3.5202 -0.1749
995 0.43078 3.3585 0.4071
996 -0.14068 0.6236 -0.3085
997 -1.76357 2.6188 -1.8876
998 -1.02310 3.2470 -0.9843
999 -0.93777 1.4991 -1.3266
1000 -0.37665 2.1231 -0.4477
Note, for example, in the first row:
$$T(3)=\dfrac{-2.60481}{\sqrt{10.2497/3}}=-1.4092$$
Here's what the resulting histogram of the 1000 randomly generated $$T(3)$$ values looks like, with a standard $$N(0,1)$$ curve superimposed:
Hmmm. The $$t$$-distribution seems to be quite similar to the standard normal distribution. Using the formula given above for the p.d.f. of $$T$$, we can plot the density curve of various $$t$$ random variables, say when $$r=1, r=4$$, and $$r=7$$, to see that that is indeed the case:
In fact, it looks as if, as the degrees of freedom $$r$$ increases, the $$t$$ density curve gets closer and closer to the standard normal curve. Let's summarize what we've learned in our little investigation about the characteristics of the t distribution:
1. The support appears to be $$-\infty<t<\infty$$. (It is!)
2. The probability distribution appears to be symmetric about $$t=0$$. (It is!)
3. The probability distribution appears to be bell-shaped. (It is!)
4. The density curve looks like a standard normal curve, but the tails of the $$t$$-distribution are "heavier" than the tails of the normal distribution. That is, we are more likely to get extreme $$t$$-values than extreme $$z$$-values.
5. As the degrees of freedom $$r$$ increases, the $$t$$-distribution appears to approach the standard normal $$z$$-distribution. (It does!)
As you'll soon see, we'll need to look up $$t$$-values, as well as probabilities concerning $$T$$ random variables, quite often in Stat 415. Therefore, we better make sure we know how to read a $$t$$ table.
## The $$t$$ Table
If you take a look at Table VI in the back of your textbook, you'll find what looks like a typical $$t$$ table. Here's what the top of Table VI looks like (well, minus the shading that I've added):
$P(T \leq t)=\int_{-\infty}^{t} \frac{\Gamma[(r+1) / 2]}{\sqrt{\pi r} \Gamma(r / 2)\left(1+w^{2} / r\right)^{(r+1) / 2}} d w$
$P(T \leq-t)=1-P(T \leq t)$
P(T≤ t) 0.60 0.75 0.90 0.95 0.975 0.99 0.995 r t0.40(r) t0.25(r) t0.10(r) t0.05(r) t0.025(r) t0.01(r) t0.005(r) 1 0.325 1.000 3.078 6.314 12.706 31.821 63.657 2 0.289 0.816 1.886 2.920 4.303 6.965 9.925 3 0.277 0.765 1.638 2.353 3.182 4.541 5.841 4 0.271 0.741 1.533 2.132 2.776 3.747 4.604 5 0.267 0.727 1.476 2.015 2.571 3.365 4.032 6 0.265 0.718 1.440 1.943 2.447 3.143 3.707 7 0.263 0.711 1.415 1.895 2.365 2.998 3.499 8 0.262 0.706 1.397 1.860 2.306 2.896 3.355 9 0.261 0.703 1.383 1.833 2.262 2.821 3.250 10 0.260 0.700 1.372 1.812 2.228 2.764 3.169
The $$t$$-table is similar to the chi-square table in that the inside of the $$t$$-table (shaded in purple) contains the $$t$$-values for various cumulative probabilities (shaded in red), such as 0.60, 0.75, 0.90, 0.95, 0.975, 0.99, and 0.995, and for various $$t$$ distributions with $$r$$ degrees of freedom (shaded in blue). The row shaded in green indicates the upper $$\alpha$$ probability that corresponds to the $$1-\alpha$$ cumulative probability. For example, if you're interested in either a cumulative probability of 0.60, or an upper probability of 0.40, you'll want to look for the $$t$$-value in the first column.
Let's use the $$t$$-table to read a few probabilities and $$t$$-values off of the table:
Let's take a look at a few more examples.
## Example 26-6
Let $$T$$ follow a $$t$$-distribution with $$r=8$$ df. What is the probability that the absolute value of $$T$$ is less than 2.306?
#### Solution
The probability calculation is quite similar to a calculation we'd have to make for a normal random variable. First, rewriting the probability in terms of $$T$$ instead of the absolute value of $$T$$, we get:
$$P(|T|<2.306)=P(-2.306<T<2.306)$$
Then, we have to rewrite the probability in terms of cumulative probabilities that we can actually find, that is:
$$P(|T|<2.306)=P(T<2.306)-P(T<-2.306)$$
Pictorially, the probability we are looking for looks something like this:
But the $$t$$-table doesn't contain negative $$t$$-values, so we'll have to take advantage of the symmetry of the $$T$$ distribution. That is:
>$$P(|T|<2.306)=P(T<2.306)-P(T>2.306)$$
Can you find the necessary $$t$$-values on the $$t$$-table?
P(T≤ t) 0.60 0.75 0.90 0.95 0.975 0.99 0.995 r t0.40(r) t0.25(r) t0.10(r) t0.05(r) t0.025(r) t0.01(r) t0.005(r) 1 0.325 1.000 3.078 6.314 12.706 31.821 63.657 2 0.289 0.816 1.886 2.920 4.303 6.965 9.925 3 0.277 0.765 1.638 2.353 3.182 4.541 5.841 4 0.271 0.741 1.533 2.132 2.776 3.747 4.604 5 0.267 0.727 1.476 2.015 2.571 3.365 4.032 6 0.265 0.718 1.440 1.943 2.447 3.143 3.707 7 0.263 0.711 1.415 1.895 2.365 2.998 3.499 8 0.262 0.706 1.397 1.860 2.306 2.896 3.355 9 0.261 0.703 1.383 1.833 2.262 2.821 3.250 10 0.260 0.700 1.372 1.812 2.228 2.764 3.169
P(T≤ t) 0.60 0.75 0.90 0.95 0.975 0.99 0.995 r t0.40(r) t0.25(r) t0.10(r) t0.05(r) t0.025(r) t0.01(r) t0.005(r) 1 0.325 1.000 3.078 6.314 12.706 31.821 63.657 2 0.289 0.816 1.886 2.920 4.303 6.965 9.925 3 0.277 0.765 1.638 2.353 3.182 4.541 5.841 4 0.271 0.741 1.533 2.132 2.776 3.747 4.604 5 0.267 0.727 1.476 2.015 2.571 3.365 4.032 6 0.265 0.718 1.440 1.943 2.447 3.143 3.707 7 0.263 0.711 1.415 1.895 2.365 2.998 3.499 8 0.262 0.706 1.397 1.860 2.306 2.896 3.355 9 0.261 0.703 1.383 1.833 2.262 2.821 3.250 10 0.260 0.700 1.372 1.812 2.228 2.764 3.169
The $$t$$-table tells us that $$P(T<2.306)=0.975$$ and $$P(T>2.306)=0.025$$. Therefore:
$$P(|T|>2.306)=0.975-0.025=0.95$$
What is $$t_{0.05}(8)$$?
#### Solution
The value $$t_{0.05}(8)$$ is the value $$t_{0.05}$$ such that the probability that a $$T$$ random variable with 8 degrees of freedom is greater than the value $$t_{0.05}$$ is 0.05. That is:
Can you find the value $$t_{0.05}$$ on the $$t$$-table?
P(T≤ t) 0.60 0.75 0.90 0.95 0.975 0.99 0.995 r t0.40(r) t0.25(r) t0.10(r) t0.05(r) t0.025(r) t0.01(r) t0.005(r) 1 0.325 1.000 3.078 6.314 12.706 31.821 63.657 2 0.289 0.816 1.886 2.920 4.303 6.965 9.925 3 0.277 0.765 1.638 2.353 3.182 4.541 5.841 4 0.271 0.741 1.533 2.132 2.776 3.747 4.604 5 0.267 0.727 1.476 2.015 2.571 3.365 4.032 6 0.265 0.718 1.440 1.943 2.447 3.143 3.707 7 0.263 0.711 1.415 1.895 2.365 2.998 3.499 8 0.262 0.706 1.397 1.860 2.306 2.896 3.355 9 0.261 0.703 1.383 1.833 2.262 2.821 3.250 10 0.260 0.700 1.372 1.812 2.228 2.764 3.169
P(T≤ t) 0.60 0.75 0.90 0.95 0.975 0.99 0.995 r t0.40(r) t0.25(r) t0.10(r) t0.05(r) t0.025(r) t0.01(r) t0.005(r) 1 0.325 1.000 3.078 6.314 12.706 31.821 63.657 2 0.289 0.816 1.886 2.920 4.303 6.965 9.925 3 0.277 0.765 1.638 2.353 3.182 4.541 5.841 4 0.271 0.741 1.533 2.132 2.776 3.747 4.604 5 0.267 0.727 1.476 2.015 2.571 3.365 4.032 6 0.265 0.718 1.440 1.943 2.447 3.143 3.707 7 0.263 0.711 1.415 1.895 2.365 2.998 3.499 8 0.262 0.706 1.397 1.860 2.306 2.896 3.355 9 0.261 0.703 1.383 1.833 2.262 2.821 3.250 10 0.260 0.700 1.372 1.812 2.228 2.764 3.169
We have determined that the probability that a $$T$$ random variable with 8 degrees of freedom is greater than the value 1.860 is 0.05.
## Why will we encounter a $$T$$ random variable?
Given a random sample $$X_1, X_2, \ldots, X_n$$ from a normal distribution, we know that:
$$Z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1)$$
Earlier in this lesson, we learned that:
$$U=\dfrac{(n-1)S^2}{\sigma^2}$$
follows a chi-square distribution with $$n-1$$ degrees of freedom. We also learned that $$Z$$ and $$U$$ are independent. Therefore, using the definition of a $$T$$ random variable, we get:
It is the resulting quantity, that is:
$$T=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}$$
that will help us, in Stat 415, to use a mean from a random sample, that is $$\bar{X}$$, to learn, with confidence, something about the population mean $$\mu$$.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility |
# Are all fractions are rational numbers?
## Are all fractions are rational numbers?
Each numerator and each denominator is an integer. We need to look at all the numbers we have used so far and verify that they are rational. The definition of rational numbers tells us that all fractions are rational.
## Is a fraction a rational or irrational?
Fractions are rational numbers so long as their bottom number (the denominator) is not zero, because dividing anything by zero is impossible.
Can a fraction be an irrational number?
An irrational number cannot be expressed as a ratio between two numbers and it cannot be written as a simple fraction because there is not a finite number of numbers when written as a decimal.
Can be written as a fraction?
An integer can be written as a fraction by giving it a denominator of one, so any integer is a rational number. A terminating decimal can be written as a fraction by using properties of place value. For example, 3.75 = three and seventy-five hundredths or 3 75 100 , which is equal to the improper fraction .
### Are all numbers fractions?
Every fraction is a rational number but a rational number need not be a fraction. Since every natural number is an integer. Therefore, a and b are integers. Thus, the fraction a/b is the quotient of two integers such that b ≠ 0.
### What makes a fraction rational?
A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. The number 8 is a rational number because it can be written as the fraction 8/1.
Are numbers rational?
Rational numbers are NOT only fractions but any number that can be expressed as fractions. Natural numbers, whole numbers, integers, fractions of integers, and terminating decimals are rational numbers. Non-terminating decimals with repeating patterns of decimals are also rational numbers.
How can I determine if a number is a rational number?
A rational number is one that can be expressed as the ratio (hence the name) of two integers. So if you can find any two integers p and q such that , then x is rational. Otherwise, x is irrational. A couple of rules: Any integer is rational. That’s because you can express any integer as that same integer divided by 1.
## Is 6.7 a rational number?
Answer and Explanation: Yes, 6/7 is a rational number. A rational number is any number that can be stated as p/q where p is an integer and q is a non-zero integer.
## Are fraction decimal and integers are rational numbers?
Rational Numbers . Rational numbers have integers AND fractions AND decimals. Now you can see that numbers can belong to more than one classification group.
Is a fraction considered a number?
Fraction, In arithmetic, a number expressed as a quotient, in which a numerator is divided by a denominator. In a simple fraction, both are integers. A complex fraction has a fraction in the numerator or denominator. |
# Order of Operations
While
solving the questions on order of operations we follow certain rules that
indicate the sequence for simplifying expressions that contain more than one fundamental
operation.
Steps to solve order of operations:
Step I: Simplify the operations inside grouping symbols.
Step II: Simplify the powers.
Step III: Solve multiplication and division from left to right.
Step IV: Solve addition and subtraction from left to right.
• In simplifying an expression, all the brackets must be removed first in the order and the grouping symbols are parentheses ( ), brackets [ ], braces or curly brackets { }.
• The other grouping symbols include fraction
bars, radical symbols, and absolute-value symbols.
• When we simplify expressions involving
more than one grouping symbol, first we need to simplify the innermost set.
Within each set, then follow the fundamental order of operations.
• Symbols of grouping can be used when
translating the expressions from words to math.
The product of 11 and the sum of 5, 8 and
13 is written as 11(5 + 8 + 13).
A. Worked-out problems on simplifying numerical expressions:
1. Evaluate
the expression:
(i) 27 ÷ 32 + 4 · 2 – 1
Solution:
27 ÷ 32 + 4 · 2 – 1
= 27 ÷ 9 + 4 · 2 – 1 = 3 + 4 · 2 – 1= 3 + 8 – 1 = 11 – 1 = 10 Evaluate powers.Divide 27 by 9.Multiply 4 by 2.Add 3 and 8.Subtract 1 from 11.
(ii) 27 – [5 + {28 – (29 – 7}]
Solution:
27 – [5 + {28 – (29 – 7}]= 27 – [5 + {28 – 22}]= 27 – [5 + 6]= 27 – 11= 16 Removing the parenthesis. Subtract 7 from 29.Removing the curly brackets. Subtract 22 from 28.Removing the brackets. Add 5 and 6.Subtract 11 from 27.
B. Worked-out problems on grouping symbols:
Evaluate
each expression:
(i) 6(12 – 8) – 3(3 + 1)
Solution:
6(12 – 8) – 3(3 + 1) = 6(4) – 3(4)= 24 – 12= 12 Evaluate inside grouping symbols.Multiply expressions left to right.Subtract 12 from 24.
(ii) 4[(24 ÷ 3) – (3 + 2)]
Solution:
4[(24 ÷ 3) – (3 + 2)]= 4[(8) – (5)]= 4[3]= 12 Evaluate innermost expression first.Evaluate expression in grouping symbol.Multiply.
(iii) (25 ÷ 2)/(15 – 23)
Solution:
(25 ÷ 2)/(15 – 23) means (25 ÷ 2) ÷ (15 – 23).
(25 ÷ 2)/(15 – 23)
= (32 ÷ 2)/(15 – 23)= 16/(15 – 23)= 16/(15 – 8)= 16/7 Evaluate the power in the numerator.Divide 32 by 2 in the numerator.Evaluate the power in the denominator.Subtract 8 from 15 in the denominator.
C. Worked-out problems on evaluating an algebraic expression:
Evaluate: (a2 – 2cb) + b3 if a = 8, b = 3 and c = 5.
Solution:
(a2 – 2cb) + b3
= (82 – 2 · 5 · 3) + 33= (64 – 2 · 5 · 3) + 33= (64 – 30) + 33= (34) + 33= 34 + 27= 61 Replace a with 8, b with 3 & c with 5.Evaluate 82.Multiply 2, 5, and 3.Subtract 30 from 64.Evaluate 33.Add 34 and 27.
D. Real-life word problem using algebraic expressions:
Ron has \$600 to invest for 5 years. She finds a bank that will invest her money at a simple interest rate of 5%. Interest I is equal to the principle P (amount invested) times the product of the rate r as a decimal and the time t in years.
a. Write an expression that represents simple interest.
Words VariablesExpression principle P = Principle,P
times t = time,× the product of rate and time r = rate(r × t)
b. Find the amount of interest earned after 5 years.
Evaluate Prt for P = 600, r = 0.05, and t = 5.
Prt = 600(0.05)(5) = 30(5) = 150 Replace P with 600, r with 0.05, and t with 5.Multiply 600 by 0.05.Multiply 30 by 5.
The amount of
interest Jamie will earn in 5 years will be \$150.
+ |
# How do you the equation of the line that is perpendicular to the line 2x - 3y = 3 and passes through the point (-8, 2)?
May 26, 2017
When given a line in form $a x - b y = c$, the equation of all lines that are perpendicular are of the form:
$b x + a y = k$
Use the given point to find the value of $k$
#### Explanation:
The equation of all lines that are perpendicular are:
$3 x + 2 y = k$
To find the value of k, substitute the point $\left(- 8 , 2\right)$
$3 \left(- 8\right) + 2 \left(2\right) = k$
$- 24 + 4 = k$
$k = - 20$
The equation of the line perpendicular to $2 x - 3 y = 3$ that passes through the point $\left(- 8 , 2\right)$ is:
$3 x + 2 y = - 20$ |
Improper Fractions to mixed numbers
The following instructions are the steps to change improper fractions to mixed numbers!! An improper fraction is a fraction that has a numerator that is LARGER than the denominator. 3/3 (three halves) is improper because 3 is bigger than 2.
One way you can change an improper fraction to a mixed number is by making a picture. For example, 3/2 (three halves) can change into the mixed number 1 1/2 (one and a half). They both mean the exact same amount! Here's a picture to help you see why.
= Three halves is the same as 1 whole and another half 3/2 = 1 1/2
Another way to change an improper fraction to a mixed number is to just use division. The first thing you need to do is divide the numerator by the denominator. Such as 3 divided by 2 = 1 remainder 1. So you take the quotient/answer (1) and use that as the whole number. Then, take the remainder (1) and use that as the numerator. Finally, take the number you divided by (2) and use that as the denominator.
Example:
3
2
3÷2= 1 remainder 1 , so the mixed number would be
1 1 2
Mixed Numbers to Improper fractions
The following directions are the steps to be able to change mixed numbers to improper fractions. First, you take the mixed number and multiply the denominator by the whole number. Then add the numerator to that answer. The denominator just stays the same.
For example:
1 3 4
4x1=4+3= 7
4
You can also make a picture to help you solve this type of problem. Just draw a picture of the mixed number and then count the number of individual pieces that make up the entire picture. Here is an example:
1 3 4
Equals one whole and three fourths is the same as seven fourths 1 3/4 = 7/4
There are really 7 fourths in one and three fourths.
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# State and prove the impulse-momentum theorem.
Last updated date: 02nd Aug 2024
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Hint: Impulse can be defined mathematically as the product of force and time. The Impulse momentum theorem can be gotten from Newton’s second law.
Formula used: In this solution we will be using the following formulae;
$F = \dfrac{{dp}}{{dt}}$ where $F$ is the force acting on a body, $p$ is the momentum of a body, and $t$ is time, and $\dfrac{{dp}}{{dt}}$ signifies instantaneous rate of change of momentum.
Complete Step-by-Step solution:
Generally, impulse is defined as the product of and time. It is generally used to quantify how long a force acts on a particular body. Its unit in Ns. 1 Ns is defined as the amount of impulse when 1 N of force acts on a body for one second. However, by relation, it is equal to the change in momentum of the body
The impulse – momentum theorem generally states that the impulse applied to a body is equal to the change in momentum of that body. This theorem can be proven from Newton’s law.
According to Newton’s second law, we have that
$F = \dfrac{{dp}}{{dt}}$ where $F$ is the force acting on a body, $p$ is the momentum of a body, and $t$ is time, and $\dfrac{{dp}}{{dt}}$ signifies instantaneous rate of change of momentum.
Hence, by cross multiplying, we have
$Fdt = dp$
Then, integrating both sides from initial point to final point for both momentum and time, we have
$\int_0^1 {Fdt} = \int_{{p_0}}^{{P_f}} {dp}$
Hence, by integrating, we have that
$Ft = {p_f} - {p_o}$
$\Rightarrow Ft = \Delta p$
Hence, the impulse is equal to change in momentum.
Note: Alternately, we can use the constant form of Newton's second law equation. Which can be given as,
$F = \dfrac{{mv - mu}}{t}$
Hence, simply by cross multiplying we have
$Ft = mv - mu$
Hence, we have that
$I = mv - mu$ which is the impulse-momentum theorem. |
##### Child pages
• Small Group Work
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# Small Group Work
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## Small Group Work
1. All Things Being Equal II - The equals sign signifies that amounts on each side are the same. The students will use Unifix blocks and the corresponding equations to represent equalities between additive amounts.
2. All Things Being Equal III - The students will write equations to represent verbal statements and successive transformations that maintain or do not maintain the equality.
3. Candy Boxes - This class centers on the possible amounts of candies two children, John and Maria, have. They each have the same, unspecified number of candies inside their own candy box. John has, in addition, one extra candy and Maria has three extra candies. What are the possible total candies they might have?
4. Comparing Different Functions - The students will discuss, represent, and solve a verbal problem involving the choice between two functions.
5. Comparing Discrete Quantities - Students compare amounts of tokens and unknown amounts of discrete quantities. In both cases they are guided to adopt line segments to represent discrete amounts and the differences between them. They are also asked to discuss composition of measures: "the difference plus the smaller amount is equal to the larger amount" and, "the larger amount minus the difference is equal to the smaller amount".
6. Comparing Heights I - Students compare the heights of two children, measure, compare, and represent one's own height in relation to a peer's height, and focus on the differences between heights.
7. Comparison Problems & Tables - This class will be used to review concepts and representations as applied to the solution of verbal comparison problems and to work on function tables.
8. Comparisons - Comparisons and comparison operators: =, ≠, <, >.
9. Comparisons and Attributes - Work with comparisons and comparison operators (=, ≠, <, >).
10. Dots Problem - We present to the students a problem dealing with a growing pattern over time. To begin, there is one dot. With each passing minute four more dots are drawn around the previous dot(s).
11. Formulas and Stories - The students will be required to work with the relation between different mathematical expressions (formulas) and stories.
12. Functioning Together - Students work together to develop multiple representations of a function. The students split up into groups of three with each student having a separate responsibility. When all the input values have been used up, the students are asked to, together, make up a story that describes their function.
13. Functions - Earning Money - The students will create tables and equations from given stories. The functions are additive and multiplicative.
14. Functions from Tables - Students work with a function, beginning with a table and then a formula, to generate ordered pairs that follow the rule of the function.
15. Heights as Functions - In this class children will work on the functional representation of two unknown heights and on the composition of the shorter height plus the difference between the heights as equal to the second height.
16. How Many Points? - Students work with: (a) a context — distance as a function of time; (b) generating coordinates.
17. Interpreting Graphs - Students will be given two linear distance-time graphs and asked to tell a story about each graph and to compare them. They will later explore comparisons between points in each line.
18. Linear vs Quadratic Functions - The students will use two functions (a linear and a quadratic) that are represented as a sequence of patterns and create a sequence of hops on the number line and an algebraic expression to express the functions.
19. Maps to Graphs - Students will be given two linear distance-time graphs and asked to tell a story about each graph and to compare them. They will later explore comparisons between points in each line.
20. Multiple Number Lines - Students continue to learn that two partial changes that start at different points on the number line are equivalent. At the end, they will work with notation for variables (N + 5 - 3 or N + 2).
21. N-Number Line I - Students work with the table they built in the previous class for multiple number lines, focusing on the notation for variables (N + 5 - 3 or N + 2).
22. N-Number Line II - Students use the N-Number line to make generalizations about an unknown amount of money in a piggy bank.
23. Number Line Shortcuts - The students will use a number line to see how two addends or subtrahends are equivalent to one single change once combined.
24. Part-Whole Relations - This class follows the discussion from the Candy Boxes I class. The challenge is to work with a visual representation of the relationships among the various quantities in the candy box problem and to relate the visual and numerical information contained in visual diagram(s) to verbal descriptions and to algorithms for finding unknown values.
25. Piggy Banks - The whole lesson revolves around a multipart story problem involving changes in two quantities over several days of a week. The initial quantities are equal yet unknown. Then transformations are applied to the quantities. Students are asked to compare the quantities throughout the week even though only their relative relationship can be determined.
26. Rates vs Totals - Students compare points on an hours/pay Cartesian space. The main challenge lies in recognizing that, although one student earned more, the other student was paid better, that is, at a higher rate of pay. They must indicate the difference in pay and the differences in amount worked.
27. Starting With A Rule - Students focus on whether given outputs are consistent with a given rule.
28. Symbols - Discussion about what symbols are; writing messages or "stories" with symbols; interpreting symbols.
29. Three Heights - In this class we will explore: (a) How the children deal with comparisons, (b) How they draw inferences from comparisons, and (c) How they represent comparisons between three unknown amounts.
30. Time and Time Lines - Students will discuss and learn about points and intervals on time lines of various sorts.
1. Cartesian Candy Bars II - Children work on sharing different amounts of candy bars among different numbers of people. They compare ratios (candy bars per person) and plot points in a Cartesian grid.
2. Comparing Functions - This lesson is split into two days. In the first class, the students will analyze eight basic graph shapes and will represent and solve a verbal problem involving the choice between two functions. In the second one they will be asked to choose, among the eight basic graph shapes, the ones that matches specific situations.
3. Equations and Inequalities - Students will work with equations and inequalities, first with simple ones and later with comparisons of two functions. The Wallet Problem, introduced in a previous lesson, will provide the background context.
4. Evaluation Problem - Students will be given a problem that asks about the amount of money each person has, based on the amount in a piggy bank. They will be given one graph and asked to draw the second graph.
5. Graphing A Story - A trip is described in miles, hours, and miles/hr. Students produce a graph from the description. They then produce a table from the graph and answer questions about the trip.
6. Graphing Halves and Doubles - Children work on a problem about distance and time and compare two rates: half a meter per second and two meters per second.
7. Graphing Thirds and Triples - Children work on a problem about distance and time and compare two rates: one third of a meter per second and three meters per second.
8. Intervals - Students reason about graphs showing growth over time. They compare heights of children and heights of two animals at different time intervals.
9. Multiplicative Candy Boxes II - This class is a continuation of the Multiplicative Candy Boxes I lesson. It centers on the possible amounts of candies two children, Juan and Marcia, have. Juan has a box of candy and Marcia has twice as much candy. What are the possible amounts of candies they might have?
10. The Better Paying Job I - Children work on a problem about rate of pay per hour of work. They compare ratios (dollars earned per hour of work) and discuss and plot points in a Cartesian plane.
11. The Better Paying Job II - Children work on a problem about rate of pay per hour of work. They compare ratios (dollars earned per hour of work) and discuss and plot points in a Cartesian plane.
12. Three Heights Review - In this class we will explore: (a) How children deal with comparisons, (b) How they draw inferences from comparisons, and (c) How they represent comparisons between three unknown amounts.
13. Three to One - Children discuss and produce verbal and mathematical statements on the proportion, S:L :: 1:3, that is, on the function f( x ) = 3x and on its inverse f -1( x ) = 1/3 x
14. Two Phone Plans I - Students compare two phone plans, one of which has a lower rate, but a monthly basic charge; the other has a higher rate but no basic charge.
15. Two Phone Plans II - Students will work on the comparison between two phone plans (also used in the lesson "Two Phone Plans I"), one of which has a lower rate, but a monthly basic charge, the other has a higher rate but no basic charge.
16. Varying Speed - Children are asked to tell a story about a trip depicted through a graph that has varying slopes/speeds.
17. Varying Velocity - Children are asked to tell a story about a trip depicted through a graph that has varying slopes/velocities.
18. Wallet Problem I - Students compare the amounts of money two students have. The amounts are described relationally but not through precise dollar amounts.
19. Wallet Problem II - Students will be given a wallet problem. They will be asked to compare the amounts of money two students have. The amounts are described relationally but not through precise dollar amounts.
20. Wallet Problem III - Students will continue working with the wallet problem. They will be shown a graph for Mike's amounts and asked to (a) determine whether it represents Robin's or Mike's money and (b) to predict where the line for Mike would fall. Later they will plot Mike's amounts and will discuss why the lines cross.
1. Basic Function Shapes - In this lesson, the students will (a) discuss, represent, and solve a verbal problem involving the choice between two functions; (b) choose, among 8 basic graphs (7 distinct shapes), the one that matches specific situations; and (c) write stories to match a specific graph shape.
2. Equations and Graphs - Students will further compare two linear functions in the context of evaluating two plans for shoveling snow. One plan has two parts: a basic charge plus a charge based on the number of square meters cleared. The other plan has no basic charge; it only charges according to the number of square meters cleared. However the per-meter charge is higher than in the other plan. Students are asked to determine the circumstances in which the bill from each plan would be the same. They then examine the graph of the two functions and discuss how equations and inequalities relate to the graph.
3. Equations in Groups - Students first discuss equality situations and how equal changes on both sides of the equality do not change the equality or the solution to the equation. In a second activity, A pair of students begins with a solved equation (e.g. N = 4) and passes the equation to their neighbor; the neighbor operates equally on each side of the equation and passes the equations to the following neighbor. They continue this process until the series of equations return to the first two students who, then, check whether the solution still holds. They also check the logic and correctness of their colleagues operations on the initial equation.
4. Equations in Groups II - A student (or a pair of students) begins with a solved equation (e.g. N = 4) and pass(es) the equation to neighbor (or pair of neighbors); the neighbor(s) operate(s) equally on each side of the equation. And so on, around the table. There should be at least three students or pair of students at each table. When the series of equations returns to the first students, each student (or pair of students) check whether the solution still holds for the solution they had proposed at the beginning. They also check the logic and correctness of the changes implemented by their classmates.
5. Phone Plans - Students will compare two linear functions in the context of evaluating phone plans. One plan has two parts: a basic charge plus a charge based upon the number of minutes used. The other plan has no basic charge; it only charges according to the minutes used. However the per-minute charge is higher than in the other plan. Students are asked to determine the circumstances in which the monthly bill from each plan would be the same. They then examine the graph of the two functions and discuss how equations and inequalities relate to the graph.
6. Solving Equations I - Students will be asked to use the syntactic rules of algebra to solve equations with variables on both sides of the equals sign.
7. Solving Equations II - Students will be asked to represent and solve verbal problems requiring algebra and to use the syntactic rules of algebra to solve equations with variables on both sides of the equals sign.
8. Wallet Review Problem - This activity is a review of the Wallet Problem done in fourth grade. It is intended to introduce new students to some of the concepts we have covered and to refresh the memories of our old students. Students compare the amounts of money two students have. The amounts are described relationally but not through specific dollar amounts.
###### Middle School Lessons
1. Biggest Output - Students will decide on what linear and quadratic functions will result in the greatest output, starting from an algebraic expression, and using tables and graphs to help them make these decisions.
2. Box of Clay Activity - Students will compare two cubic functions based on the context of the volumes of a box of clay.
3. Can We Predict Differences? - Students will predict, produce, and compare linear and non-linear function graphs used to represent the number of punches on a balloon.
4. Candy Experiment - Students will create their own data to construct a graph and equation of negative and fractional slope.
5. Compare and Contrast - Students will identify the y-intercept and slope using equations and then use that data to create corresponding tables and graphs.
6. Contrasting Equations - Students write equations for three graphs and examine their slopes by comparing and contrasting the graphs. Students also look at the same functions graphed on differently scaled coordinate planes.
7. Function Challenges - 20 Questions - Students will compete in a game to generate equations for functions that meet certain criteria, as given by the instructor.
8. Guess My Rule - Linear - Students will try to determine the equation to match their partner's created graph and work together to correct their own mistakes.
9. Guess My Rule - Non-Linear - Students will produce algebraic expressions starting from non-linear graphs produced by other students in the class.
10. Jason's Tree House - Students will extract data from a story and use tables and graphs to answers questions about proposed scenarios.
11. Same and Different - Students will compare graphs of linear functions, looking for similarities and differences, and will produce algebraic expressions, again identifying what is the same and what is different about each one.
12. Sound Loudness - Students will examine a non-linear function depicted in a graph and generate the corresponding function table and equation.
13. Who Shares My Function? - Linear with All Representations - Students will work in groups after finding other students who have the same linear function represented by a story, a table, a graph, or an equation. They will attempt to explain and discuss why the different representations refer to the same function.
14. Who Shares My Function? - Linear with Graphs and Stories - Students will make groups by finding other students who have the same quadratic or linear function in different representations.
15. Who Shares My Function? - Linear with Graphs, Tables, and Equations - Students will make groups by finding other students who have the same linear function, as shown in representations of graphs, tables, or equations. They will then generate a story to go with the function.
16. Who Shares My Function? - Linear with Negative and Fractional Slope - Students will find other functions that are the same as theirs, starting from a table, a graph, or an equation. Once they have identified the same function represented in a different way, they will create a story that describes all of the different representations of the same function.
17. Who Shares My Function? - Quadratics - Students will make groups by finding other students who have the same quadratic or linear function in different representations.
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# Solving an Equation
Solving an Equation is a way of finding the value of an unknown variable that the equation contains. An equation is said to be balanced if it has an ‘equal to’ sign. Thus, this equation represents that it has two quantities equal on both sides. The two sides of the equation are LHS (Left hand side) and RHS (Right hand side). Class 7 students can easily solve simple linear equations by reading this article here.
For example, x – 4 = 5 is an equation. It represents that x – 4 (LHS) is equal to 5 (RHS). x is an unknown quantity or variable here. Hence, we need to solve this equation here to get the value of x.
Fact: If we interchange LHS and RHS, then the equation remains the same.
Solving an equation means finding the solution of the equation. A solution is the value of an unknown variable that can be used to verify the given equation. Let us learn in this article how to solve an equation with examples.
## How to Solve an Equation?
What is an equation? An equation is a condition on a variable such that two expressions in the variable should have equal value. The value of the variable for which the equation is satisfied is called the solution of the equation.
To solve any equation we need to perform arithmetic operations, to separate the variable, such as:
• Adding the same number on both sides
• Subtracting same number on both sides
• Multiplying with the same number on both sides
• Dividing by same number on both the sides
Based on these operations, we can isolate the given variable in the equation and solve any equations. The solution so obtained will justify the equation. Let us solve some examples to understand better.
## Solved Examples
Example 1: Solve 3 + x = 4
Solution: Given, the equation is;
3 + x = 4
We can see, on the Left hand side, the variable x is present. Thus, we need to make the variable ‘x’ alone on LHS. Thus, by subtracting the 3 from LHS and RHS we get;
3 + x – 3 = 4 – 3
x = 1
Hence, the solution is x = 1.
Verification:
For x = 1
Taking LHS and proving it equal to RHS.
LHS = 3 + x = 3 + 1 = 4 = RHS
Hence, verified.
Example: 2: Solve x – 9 = 0.
Solution: Given, the equation is;
x – 9 = 0
On the left hand side of the equation, the variable x is present, for which we need to find the solution. Thus, adding 9 on both sides of the equation, we get;
x – 9 + 9 = 0 + 9
x = 9
Hence, the solution is x = 9.
Verification: To verify our solution, we have to put the value obtained for x into the given equation.
Taking LHS and proving it equal to RHS.
LHS = x – 9 = 9 – 9 = 0 = RHS
Hence, verified.
Note: Adding or subtracting a value on both the sides means the changing side of the value. Changing sides is called transposing. While transposing a number, we change its sign
Example 3: Solve 3x = 21.
Solution: Given, the equation is:
3x = 21
Since, x is multiplied by 3 in the left hand side of the equation, thus, we have to perform a division here to remove 3. Thus, dividing both sides by 3, we get;
(3x)/3 = 21/3
x = 7 (Since, 3 × 7 = 21)
Hence, the required solution is x = 7.
Verification: Put x = 7 on the LHS of the given equation.
LHS = 3x = 3 × 7 = 21 = RHS
Thus, verified.
Example 4: Solve the following equation: x/5 = 2.
Solution: Given, the equation is:
x/5 = 2
In this equation, the LHS side has the variable x which is divided by 5. Thus, we need to make the variable alone on the left hand side.
So, multiplying the equation by 5 on both sides, we get;
(x/5) × 5 = 2 × 5
x = 10
Hence, the required solution is x = 10.
Verification:
Put x = 10 in the LHS of the equation x/5 = 2, to get the value equal to RHS.
LHS = x/5 = 10/5 = 2 = RHS
Hence, verified.
Example 5: Solve 2x + 6 = 12 and verify your answer.
Solution: Given ,the equation is:
2x + 6 = 12
Subtract 6 on both sides of the equation.
2x + 6 – 6 = 12 – 6
2x = 6
Now, divide both sides by 2.
(2x/2) = 6/2
x = 3
Hence, the required solution is 6.
Verification:
Taking the LHS, we have
LHS = 2x + 6
Now putting the value of x = 6, we get;
LHS = 2 (3) + 6
= 6 + 6
= 12
= RHS
Hence, verified.
## Solving an Equation with More Than One Solution
In your higher classes, you will learn to solve equations with two or more solutions. Let us take an example of a simple equation.
Example: (x – 2) (x – 5) = 0
In this case, the variable x will have two solutions.
X – 2 = 0
X = 2
And
X – 5 = 0
X = 5
Thus, the two solutions are x = 2 and x = 5.
An equation having a degree of variable equal to 2 (variable raised to the power such as x²) is called a quadratic equation. We generally use the quadratic formula to solve such equations. Also, sometimes we can use the factorisation method to solve the equation with two solutions.
## Solving Equations With Fractions
If an equation has fractions, then we have to make the fractions as like fractions first and then solve it for variables. To convert fractions into like fractions means we need to make the denominators of the fractions the same, by using the LCM method. Let us solve an example to understand it.
Example: Solve ⅛ x +½ = ¼.
Solution: Given, the equation is:
⅛ x +½ = ¼
As we can see, in the given equation, both left and right hand sides have fractions present. We need to solve the equation for x.
Since, all the three fractions i.e., ⅛, ½ and ¼ are unlike (denominators are not same)
Thus, finding the LCM (Least common multiple) of denominators we have;
LCM (2, 4, 8) = 8
Now multiply both sides of the equation by 8, to get;
8 (⅛ x +½) = 8 (¼)
8.(⅛ x) + 8.(½) = 8.¼
X + 4 = 2
Now, we subtract 4 from both sides, to get;
X + 4 – 4 = 2 – 4
X = -2
Hence, the required solution is x = -2.
Verification:Let us check if the solution obtained is correct or not.
The equation is ⅛ x +½ = ¼.
Put x = -2, on the LHS.
LHS = ⅛ x +½
= ⅛ (-2) + ½
= -¼ +½
=-¼ + 2/4
= (-1+2)/4
= ¼
= RHS
## Word Problem On Solving an Equation
Problem: The sum of three times a number and 11 is 32. Find the number.
Solution: Let the number be x.
As per the given equation,
Sum of three times a number and 11 = 32
3x + 11 = 32
Now we need to solve the above equation for x.
Subtract the equation from 11 on both sides, to get;
3x + 11 – 11 = 32 – 11
3x = 21
Now divide the equation by 3 on both sides.
3x/3 = 21/3
x = 7
Hence, the required number is 7.
### Practice Questions
1. Solve the following equations:
• y – 4 = 4
• x/7 = 4
• 5b + 7 = 17
• 3s + 12 = 0
• -x/3 = 5
• -2(x+3) = 8
• 4(2 – x) = 8
• 7x + ½ = 12
• 2z/3 – 4 = 2
2. If Neha adds 19 to a number and divides the sum by 5, she will get 8. What is the number?
## Frequently Asked Questions on Solving an Equation
### What do you mean by solving an equation?
Solving an equation means finding the solution for the unknown variable present in the equation.
### What are the four steps involved in solving an equation?
To solve an equation, we can perform the four arithmetic operations, such as addition, subtraction, multiplication and division.
### What is the transposition of a number in an equation?
Transposing means moving and changing a sign from one side to the other (from LHS to RHS or vice versa). For example, addition on LHS transposes to subtraction on RHS. Similarly, multiplication on one side changes to division on the other. This process is called transposition.
### How to solve equations in a simple way?
While solving the equation we need to simply check if the equation is balanced or not. Equation is like a balanced scale that has both its sides equal. Thus, perform the operations in such a way that the balance of the equation is not disturbed.
### What is the solution of x + 5 = 2x?
If both the sides have a variable, then we need to bring the variable on one side of the equation. Thus, subtracting x from both sides, we get;
X + 5 – x = 2x – x
5 = x or x = 5
Thus, the required solution is x = 5. |
Verifying Trigonometric Identities
# Verifying Trigonometric Identities
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
• PRACTICE (online exercises and printable worksheets)
An identity is a mathematical sentence that is always true.
(A bit more precisely—everywhere it is defined, it is true.)
To verify (or prove) an identity means to prove that the sentence is always true.
This section discusses two common approaches for verifying (proving) identities.
Fundamental Trigonometric Identities talks about identities in general, and their usefulness.
It also introduces three important trigonometric identities (listed below). Review this earlier section as needed.
• the Pythagorean Identity: $\,\sin^2 t + \cos^2 t = 1\,$
• cosine is an even function: $\,\cos (-t) = \cos t$
• sine is an odd function: $\,\sin(-t) = -\sin t\,$
## Tools for Verifying Trigonometric Identities
Verifying identities is all about renaming—to show that two (initially different-looking) expressions are actually the same.
You'll use:
• familiar algebraic tools: e.g., factoring, distributive laws, getting common denominators
• definitions of the trigonometric functions $$\cssId{s17}{\text{For simplicity, inputs are not shown:}}\qquad \cssId{s18}{\tan = \frac{\sin}{\cos}}\qquad \qquad \cssId{s19}{\cot = \frac{\cos}{\sin} = \frac{1}{\tan}}\qquad \qquad \cssId{s20}{\sec = \frac{1}{\cos}}\qquad \qquad \cssId{s21}{\csc = \frac{1}{\sin}}$$
• trigonometric identities that have already been proved
This varies from teacher-to-teacher and course-to-course.
Always ask about which identities you're allowed to use when verifying new ones.
For the exercises in this section, you're only ‘allowed’ the three fundamental trigonometric identities cited above,
together with two that follow immediately from the Pythagorean Identity:
Start with: $\,\sin^2 + \cos^2 = 1\,$ (For simplicity, inputs are not shown.) divide both sides by $\,\cos^2\,$: $$\begin{gather} \cssId{s30}{\frac{\sin^2 + \cos^2}{\cos^2} = \frac{1}{\cos^2}}\cr\cr \cssId{s31}{\frac{\sin^2}{\cos^2} + \frac{\cos^2}{\cos^2} = \frac{1}{\cos^2}}\cr\cr \cssId{s32}{\left(\frac{\sin}{\cos}\right)^2 + 1 = \left(\frac{1}{\cos}\right)^2}\cr\cr\cr \cssId{s33}{\color{red}{\tan^2 + 1 = \sec^2}} \end{gather}$$ divide both sides by $\,\sin^2\,$: $$\begin{gather} \cssId{s35}{\frac{\sin^2 + \cos^2}{\sin^2} = \frac{1}{\sin^2}}\cr\cr \cssId{s36}{\frac{\sin^2}{\sin^2} + \frac{\cos^2}{\sin^2} = \frac{1}{\sin^2}}\cr\cr \cssId{s37}{1 + \left(\frac{\cos}{\sin}\right)^2 = \left(\frac{1}{\sin}\right)^2}\cr\cr\cr \cssId{s38}{\color{red}{1 + \cot^2 = \csc^2}} \end{gather}$$
In this lesson, the identities ‘$\,\color{red}{\tan^2 x + 1 = \sec^2 x}\,$’ and ‘$\,\color{red}{1 + \cot^2 x = \csc^2 x} \,$’
are referred to as alternate Pythagorean Identities.
You should be able to easily recognize equivalent versions of the five already-proved identities you're allowed to use in this section:
the Pythagorean Identity and its two alternate versions, cosine is even, and sine is odd.
For example, ‘$\,\sin^2 x = 1 - \cos^2 x\,$’ should be easily identified as an equivalent version of the Pythagorean Identity.
• Write Everything in Terms of Sines and Cosines:
If an expression involves a ‘mix’ of trigonometric functions,
it sometimes helps to write everything in terms of just sines and cosines.
• the Conjugate Technique:
By definition, $\,x + y\,$ and $\,x - y\,$ are called conjugates.
Thus, the conjugate of $\,x + y\,$ is $\,x - y\,$; the conjugate of $\,x - y\,$ is $\,x + y\,$.
When a numerator or denominator in a (suspected) trigonometric identity is of the form $\,1 \pm \sin x\,$ or $\,1\pm \cos x\,$,
then it often helps to multiply both numerator and denominator by the conjugate.
(Thus, you're multiplying the original expression by $\,1\,$ in the form $\,\frac{\text{conjugate}}{\text{conjugate}}\,$.)
Why? This reduces two terms to a single term, which is often easier to work with.
Below are a couple examples. (Review FOIL as needed.) $$\cssId{s57}{\overbrace{(1 - \sin x)}^{\text{original expression}}} \cssId{s58}{\cdot}\ \ \cssId{s59}{\overbrace{(1 + \sin x)}^{\text{conjugate}}} \qquad \cssId{s60}{\overbrace{=\strut}^{\text{FOIL}}}\qquad \cssId{s61}{1 - \sin^2 x} \cssId{s62}{\overbrace{=\strut}^{\text{Pythagorean Identity}}} \cssId{s63}{\cos^2 x}$$ $$\cssId{s64}{\overbrace{(1 + \cos x)}^{\text{original expression}}} \cssId{s65}{\cdot}\ \ \cssId{s66}{\overbrace{(1 - \cos x)}^{\text{conjugate}}} \qquad \cssId{s67}{\overbrace{=\strut}^{\text{FOIL}}}\qquad \cssId{s68}{1 - \cos^2 x} \cssId{s69}{\overbrace{=\strut}^{\text{Pythagorean Identity}}} \cssId{s70}{\sin^2 x}$$ As illustrated in an example below, the conjugate technique often helps out for other binomial (two-term) expressions, too!
## Two Basic Approaches for Verifying Identities
Equations of the form: COMPLICATED = SIMPLER COMPL1 = COMPL2 Often, one side of a (suspected) identity is more complicated than the other. In such cases, it is usually easiest to: start with the complicated side apply tools (one at a time) to rename it as the simpler side \cssId{s79}{ \begin{align} \text{C O M}&\text{P L I C A T E D}\cr &= \ \text{}\cr &= \ \text{}\cr &= \ \ldots\cr &= \ \text{SIMPLER} \end{align}} Often, both sides of a (suspected) identity are similarly complicated. In such cases, you can often work with each side separately, renaming each side as the same simpler expression: \cssId{s83}{ \begin{align} \text{COMPL1} &= \ \text{}\cr &= \ \ldots\cr &= \ \color{red}{\text{SIMPLER}}\cr\cr \text{COMPL2} &= \ \text{}\cr &= \ \ldots\cr &= \ \color{red}{\text{SIMPLER}} \end{align}} The two $\,\color{red}{\text{SIMPLER}}\,$ expressions are the same! (Sometimes, when renaming one side,you inadvertently end up with the other side. If so—you're done!) Notes: if a complicated side is long, put the first simplication on the next line, indented a bit for readability, line up all equal signs you may want to provide a reason for each step MANY IDENTITIES CAN BE VERIFIED IN MANY DIFFERENT WAYS! (In the examples and exercises, only one correct approach is shown.)
## EXAMPLE of ‘COMPLICATED = SIMPLER’:Turning a Complicated Side into the Simpler Side
Verify the identity: $\,\displaystyle \tan x + \frac{\cos x}{1 + \sin x} = \sec x\,$
SOLUTION:
The left side ($\,\tan x + \frac{\cos x}{1 + \sin x}\,$) is more complicated than the right side ($\,\sec x\,$).
rename it to match the expression on the right.
The solution shown here first rewrites everything in terms of sines and cosines, and then uses the conjugate technique: \begin{alignat}{2} &\cssId{s103}{\tan x + \frac{\cos x}{1 + \sin x}} &\qquad\qquad& \cssId{s104}{\text{(start with left side)}} \cr\cr &\qquad\cssId{s105}{= \ \frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}} &&\cssId{s106}{\text{(definition of tangent)}}\cr\cr &\qquad\cssId{s107}{= \ \frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}\cdot\frac{1 - \sin x}{1 - \sin x}} &&\cssId{s108}{\text{(conjugate technique)}}\cr\cr &\qquad\cssId{s109}{= \ \frac{\sin x}{\cos x} + \frac{(\cos x)(1 - \sin x)}{\cos^2 x}} &&\cssId{s110}{\text{(FOIL, Pythagorean Identity)}}\cr\cr &\qquad\cssId{s111}{= \ \frac{\sin x}{\cos x} + \frac{1 - \sin x}{\cos x}} &&\cssId{s112}{\text{(cancel)}}\cr\cr &\qquad\cssId{s113}{= \ \frac{1}{\cos x}} &&\cssId{s114}{\text{(add fractions)}}\cr\cr &\qquad\cssId{s115}{= \ \sec x} &&\cssId{s116}{\text{(definition of secant)}}\cr \end{alignat}
## EXAMPLE of ‘COMPL1 = COMPL2’:Working With Both Sides Separately
Verify the identity: $\,\displaystyle \frac{1 + \cos t}{\cos t} = \frac{\tan^2 t}{\sec t - 1}\,$
SOLUTION:
The left side ($\,\frac{1 + \cos t}{\cos t}\,$) and the right side ($\,\frac{\tan^2 t}{\sec t - 1}\,$) are similarly complicated.
Therefore, work with both sides separately, trying to reduce each to the same simpler expression.
Here, ‘LHS’ and ‘RHS’ stand for ‘left-hand side’ and ‘right-hand side’.
Note that the conjugate technique is successfully applied to an expression of the form $\,\sec t - 1\,$.
\begin{alignat}{2} &\cssId{s126}{\text{LHS}} \cssId{s127}{= \frac{1 + \cos t}{\cos t}} &\qquad\qquad& \cssId{s128}{\text{(start with LHS)}} \cr\cr &\qquad\cssId{s129}{= \ \frac{1}{\cos t} + \frac{\cos t}{\cos t}} &&\cssId{s130}{\text{(algebra)}}\cr\cr &\qquad\cssId{s131}{= \ \color{red}{\sec t + 1}} &&\cssId{s132}{\text{(definition of secant)}}\cr \end{alignat} \begin{alignat}{2} &\cssId{s133}{\text{RHS}} \cssId{s134}{= \frac{\tan^2 t}{\sec t - 1}} &\qquad\qquad& \cssId{s135}{\text{(start with RHS)}} \cr\cr &\qquad\cssId{s136}{= \ \frac{\tan^2 t}{\sec t - 1}\cdot\frac{\sec t + 1}{\sec t + 1}} &&\cssId{s137}{\text{(conjugate technique)}}\cr\cr &\qquad\cssId{s138}{= \ \frac{\tan^2 t(\sec t + 1)}{\sec^2 t - 1}} &&\cssId{s139}{\text{(FOIL)}}\cr\cr &\qquad\cssId{s140}{= \ \frac{\tan^2 t(\sec t + 1)}{\tan^2 t}} &&\cssId{s141}{\text{(alternate Pythagorean Identity)}}\cr\cr &\qquad\cssId{s142}{= \ \color{red}{\sec t + 1}} &&\cssId{s143}{\text{(cancel)}}\cr \end{alignat} Notice that both the LHS and RHS have been renamed as the same expression, $\,\sec t + 1\,$.
## Graphically Checking that an Equation is an Identity
For many years, I worked in a math lab. Any student could come in to get help.
I have a recollection of a time when I learned an important lesson.
I was helping someone verify an identity—but nothing was working! We tried oodles of things, for a very long time.
Finally, the thought occurred to me that perhaps it wasn't an identity after all—maybe it was just a mistake.
Indeed, it was a typo in the book.
If you ever find yourself in the same situation, check that you're really working with an identity.
If the LHS and RHS are always equal then they will have precisely the same graphs.
At WolframAlpha, if you type in two comma-separated expressions, then both will be graphed in the same coordinate system.
If you graph both sides of an identity, then it will look like a single graph, because the graphs will overlap perfectly!
For example, cut-and-paste the following into WolframAlpha:
tan(x) + cos(x)/(1 + sin(x)) , sec x
Here's what you'll see:
The blue and red graphs overlap perfectly, giving the appearance of a single graph. This graphically shows that ‘$\,\displaystyle \tan x + \frac{\cos x}{1 + \sin x} = \sec x\,$’ is an identity. If you prefer, you can get an equivalent equation with zero on one side: $$\cssId{s163}{\tan x + \frac{\cos x}{1 + \sin x} - \sec x = 0}$$ Since this is an identity, the left side graphs as the zero function. Not much to see!
## How to Prove that an Equation is NOT an Identity
If the graphs don't match up (and you've input both sides correctly), then you're not working with an identity.
To prove that a sentence is not an identity, you need only provide a single number for which the sentence is false.
For example, ‘$\,\displaystyle \tan x + \frac{\cos x}{1 + \sin x} = 1 + \sec x\,$’ is not an identity:
• when $\,x = 0\,$, the LHS is: $\displaystyle \,\tan 0 + \frac{\cos 0}{1 + \sin 0} = 0 + \frac{1}{1 + 0} = 1\,$
• when $\,x = 0\,$, the RHS is: $\displaystyle \,1 + \sec 0 = 1 + \frac{1}{\cos 0} = 1 + \frac{1}{1} = 2\,$
• $\,1\ne 2\,$
Master the ideas from this section |
# 6.3: Using the Normal Distribution
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The shaded area in the following graph indicates the area to the left of $$x$$. This area is represented by the probability $$P(X < x)$$. Normal tables, computers, and calculators provide or calculate the probability $$P(X < x)$$.
The area to the right is then $$P(X > x) = 1 – P(X < x)$$. Remember, $$P(X < x) =$$ Area to the left of the vertical line through $$x$$. $$P(X > x) = 1 – P(X < x) =$$ Area to the right of the vertical line through $$x$$. $$P(X < x)$$ is the same as $$P(X \leq x)$$ and $$P(X > x)$$ is the same as $$P(X \geq x)$$ for continuous distributions.
## Calculations of Probabilities
Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.To calculate the probability, use the probability tables provided in [link] without the use of technology. The tables include instructions for how to use them.
##### Example $$\PageIndex{1}$$
If the area to the left is 0.0228, then the area to the right is $$1 - 0.0228 = 0.9772$$.
##### Exercise $$\PageIndex{1}$$
If the area to the left of $$x$$ is $$0.012$$, then what is the area to the right?
$$1 - 0.012 = 0.988$$
##### Example $$\PageIndex{2}$$
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.
1. Find the probability that a randomly selected student scored more than 65 on the exam.
2. Find the probability that a randomly selected student scored less than 85.
3. Find the 90th percentile (that is, find the score $$k$$ that has 90% of the scores below k and 10% of the scores above $$k$$).
4. Find the 70th percentile (that is, find the score $$k$$ such that 70% of scores are below $$k$$ and 30% of the scores are above $$k$$).
a. Let $$X$$ = a score on the final exam. $$X \sim N(63, 5)$$, where $$\mu = 63$$ and $$\sigma = 5$$
Draw a graph.
Then, find $$P(x > 65)$$.
$P(x > 65) = 0.3446\nonumber$
The probability that any student selected at random scores more than 65 is 0.3446.
##### USING THE TI-83, 83+, 84, 84+ CALCULATOR
Go into 2nd DISTR.
After pressing 2nd DISTR, press 2:normalcdf.
The syntax for the instructions are as follows:
normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 1099) by pressing 1, the EE key (a 2nd key) and then 99. Or, you can enter 10^99instead. The number 1099 is way out in the right tail of the normal curve. We are calculating the area between 65 and 1099. In some instances, the lower number of the area might be –1E99 (= –1099). The number –1099 is way out in the left tail of the normal curve.
##### Historical Note
The TI probability program calculates a $$z$$-score and then the probability from the $$z$$-score. Before technology, the $$z$$-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the $$z$$-score was used. You calculate the $$z$$-score and look up the area to the left. The probability is the area to the right.
$z = 65 – 63565 – 635 = 0.4\nonumber$
Area to the left is 0.6554.
$P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446\nonumber$
##### USING THE TI-83, 83+, 84, 84+ CALCULATOR
Find the percentile for a student scoring 65:
*Press 2nd Distr
*Press 2:normalcdf(
*Enter lower bound, upper bound, mean, standard deviation followed by )
*Press ENTER.
For this Example, the steps are
2nd Distr
2:normalcdf(65,1,2nd EE,99,63,5) ENTER
The probability that a selected student scored more than 65 is 0.3446.
To find the probability that a selected student scored more than 65, subtract the percentile from 1.
b. Draw a graph.
Then find $$P(x < 85)$$, and shade the graph.
Using a computer or calculator, find $$P(x < 85) = 1$$.
$$\text{normalcdf}(0,85,63,5) = 1$$ (rounds to one)
The probability that one student scores less than 85 is approximately one (or 100%).
c. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the $$x$$-axis. Shade the area that corresponds to the 90th percentile.
Let $$k =$$ the 90th percentile. The variable $$k$$ is located on the $$x$$-axis. $$P(x < k)$$ is the area to the left of $$k$$. The 90th percentile $$k$$ separates the exam scores into those that are the same or lower than $$k$$ and those that are the same or higher. Ninety percent of the test scores are the same or lower than $$k$$, and ten percent are the same or higher. The variable $$k$$ is often called a critical value.
$$k = 69.4$$
The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:
invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation)
For this problem, $$\text{invNorm}(0.90,63,5) = 69.4$$
d. Find the 70th percentile.
Draw a new graph and label it appropriately. $$k = 65.6$$
The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.6 and 30% fall at or above.
$$\text{invNorm}(0.70,63,5) = 65.6$$
##### Exercise $$\PageIndex{2}$$
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65.
$$\text{normalcdf}(10^{99},65,68,3) = 0.1587$$
##### Example $$\PageIndex{3}$$
A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.
1. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.
2. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.
a. Let $$X =$$ the amount of time (in hours) a household personal computer is used for entertainment. $$X \sim N(2, 0.5)$$ where $$\mu = 2$$ and $$\sigma = 0.5$$.
Find $$P(1.8 < x < 2.75)$$.
The probability for which you are looking is the area between $$x = 1.8$$ and $$x = 2.75$$. $$P(1.8 < x < 2.75) = 0.5886$$
$\text{normalcdf}(1.8,2.75,2,0.5) = 0.5886\nonumber$
The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.
b.
To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, $$k$$, where $$P(x < k) = 0.25$$.
$\text{invNorm}(0.25,2,0.5) = 1.66\nonumber$
The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.
##### Exercise $$\PageIndex{3}$$
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.
$$\text{normalcdf}(66,70,68,3) = 0.4950$$
##### Example $$\PageIndex{4}$$
There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
1. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
2. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
3. Find the 80th percentile of this distribution, and interpret it in a complete sentence.
1. $$\text{normalcdf}(23,64.7,36.9,13.9) = 0.8186$$
2. $$\text{normalcdf}(-10^{99},50.8,36.9,13.9) = 0.8413$$
3. $$\text{invNorm}(0.80,36.9,13.9) = 48.6$$
The 80th percentile is 48.6 years.
80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.
Use the information in Example to answer the following questions.
##### Exercise $$\PageIndex{4}$$
1. Find the 30th percentile, and interpret it in a complete sentence.
2. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old and at least 0 years old?
70.
Let $$X =$$ a smart phone user whose age is 13 to 55+. $$X \sim N(36.9, 13.9)$$
To find the 30th percentile, find $$k$$ such that $$P(x < k) = 0.30$$.
$$\text{invNorm}(0.30, 36.9, 13.9) = 29.6$$ years
Thirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years. Find $$P(x < 27)$$
(Note that $$\text{normalcdf}(-10^{99},27,36.9,13.9) = 0.2382$$. The two answers differ only by 0.0040.)
$\text{normalcdf}(0,27,36.9,13.9) = 0.2342\nonumber$
##### Example $$\PageIndex{5}$$
In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).
1. Calculate the interquartile range ($$IQR$$).
2. Forty percent of the ages that range from 13 to 55+ are at least what age?
a.
$IQR = Q_{3} – Q_{1}\nonumber$
Calculate $$Q_{3} =$$ 75th percentile and $$Q_{1} =$$ 25th percentile.
\begin{align*} \text{invNorm}(0.75,36.9,13.9) &= Q_{3} = 46.2754 \\[4pt] \text{invNorm}(0.25,36.9,13.9) &= Q_{1} = 27.5246 \\[4pt] IQR &= Q_{3} - Q_{1} = 18.7508 \end{align*}
b.
Find $$k$$ where $$P(x > k) = 0.40$$ ("At least" translates to "greater than or equal to.")
$$0.40 =$$ the area to the right.
Area to the left $$= 1 – 0.40 = 0.60$$.
The area to the left of $$k = 0.60$$.
$$\text{invNorm}(0.60,36.9,13.9) = 40.4215$$.
$$k = 40.42$$.
Forty percent of the smartphone users from 13 to 55+ are at least 40.4 years.
##### Exercise $$\PageIndex{5}$$
Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean $$\mu = 81$$ points and standard deviation $$\sigma = 15$$ points.
1. Calculate the first- and third-quartile scores for this exam.
2. The middle 50% of the exam scores are between what two values?
1. $$Q_{1} =$$ 25th percentile $$= \text{invNorm}(0.25,81,15) = 70.9$$
$$Q_{3} =$$ 75th percentile $$= \text{invNorm}(0.75,81,15) = 91.1$$
2. The middle 50% of the scores are between 70.9 and 91.1.
##### Example $$\PageIndex{6}$$
A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.
1. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.
2. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.
3. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.
a. $$\text{normalcdf}(6,10^{99},5.85,0.24) = 0.2660$$
b.
$$1 – 0.20 = 0.80$$
The tails of the graph of the normal distribution each have an area of 0.40.
Find $$k1$$, the 40th percentile, and $$k2$$, the 60th percentile ($$0.40 + 0.20 = 0.60$$).
$$k1 = \text{invNorm}(0.40,5.85,0.24) = 5.79$$ cm
$$k2 = \text{invNorm}(0.60,5.85,0.24) = 5.91$$ cm
c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.
##### Exercise $$\PageIndex{6}$$
Using the information from Example, answer the following:
1. The middle 45% of mandarin oranges from this farm are between ______ and ______.
2. Find the 16th percentile and interpret it in a complete sentence.
The middle area $$= 0.40$$, so each tail has an area of 0.30.
$$– 0.40 = 0.60$$
The tails of the graph of the normal distribution each have an area of 0.30.
Find $$k1$$, the 30th percentile and $$k2$$, the 70th percentile ($$0.40 + 0.30 = 0.70$$).
$$k1 = \text{invNorm}(0.30,5.85,0.24) = 5.72$$ cm
$$k2 = \text{invNorm}(0.70,5.85,0.24) = 5.98$$ cm
$$\text{normalcdf}(5,10^{99},5.85,0.24) = 0.9998$$
## References
1. “Naegele’s rule.” Wikipedia. Available online at http://en.Wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013).
2. “403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at www.thisamericanlife.org/radi...sode/403/nummi (accessed May 14, 2013).
3. “Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at www.winatthelottery.com/publi...partment40.cfm (accessed May 14, 2013).
4. “Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013).
5. “Facebook Statistics.” Statistics Brain. Available online at http://www.statisticbrain.com/facebo...tics/(accessed May 14, 2013).
## Review
The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean $$\mu$$ and the standard deviation σ. A special normal distribution, called the standard normal distribution is the distribution of z-scores. Its mean is zero, and its standard deviation is one.
## Formula Review
• Normal Distribution: $$X \sim N(\mu, \sigma)$$ where $$\mu$$ is the mean and σ is the standard deviation.
• Standard Normal Distribution: $$Z \sim N(0, 1)$$.
• Calculator function for probability: normalcdf (lower $$x$$ value of the area, upper $$x$$ value of the area, mean, standard deviation)
• Calculator function for the $$k$$th percentile: $$k = \text{invNorm}$$ (area to the left of $$k$$, mean, standard deviation)
##### Exercise $$\PageIndex{7}$$
How would you represent the area to the left of one in a probability statement?
$$P(x < 1)$$
##### Exercise $$\PageIndex{8}$$
Is $$P(x < 1)$$ equal to $$P(x \leq 1)$$? Why?
Yes, because they are the same in a continuous distribution: $$P(x = 1) = 0$$
##### Exercise $$\PageIndex{9}$$
How would you represent the area to the left of three in a probability statement?
##### Exercise $$\PageIndex{10}$$
What is the area to the right of three?
$$1 – P(x < 3)$$ or $$P(x > 3)$$
##### Exercise $$\PageIndex{11}$$
If the area to the left of $$x$$ in a normal distribution is 0.123, what is the area to the right of $$x$$?
##### Exercise $$\PageIndex{12}$$
If the area to the right of $$x$$ in a normal distribution is 0.543, what is the area to the left of $$x$$?
$$1 - 0.543 = 0.457$$
Use the following information to answer the next four exercises:
$$X \sim N(54, 8)$$
##### Exercise $$\PageIndex{13}$$
Find the probability that $$x > 56$$.
##### Exercise $$\PageIndex{14}$$
Find the probability that $$x < 30$$.
0.0013
##### Exercise $$\PageIndex{15}$$
Find the 80th percentile.
##### Exercise $$\PageIndex{16}$$
Find the 60th percentile.
56.03
##### Exercise $$\PageIndex{17}$$
$$X \sim N(6, 2)$$
Find the probability that $$x$$ is between three and nine.
##### Exercise $$\PageIndex{18}$$
$$X \sim N(–3, 4)$$
Find the probability that $$x$$ is between one and four.
0.1186
##### Exercise $$\PageIndex{19}$$
$$X \sim N(4, 5)$$
Find the maximum of $$x$$ in the bottom quartile.
##### Exercise $$\PageIndex{20}$$
Use the following information to answer the next three exercise: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
Figure $$\PageIndex{12}$$.
$$P(0 < x <$$ ____________$$) =$$ ___________ (Use zero for the minimum value of $$x$$.)
1. Check student’s solution.
2. 3, 0.1979
##### Exercise $$\PageIndex{21}$$
Find the probability that a CD player will last between 2.8 and six years.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
Figure $$\PageIndex{13}$$.
$$P($$__________ $$< x <$$ __________$$)$$ = __________
##### Exercise $$\PageIndex{22}$$
Find the 70th percentile of the distribution for the time a CD player lasts.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70%.
Figure $$\PageIndex{14}$$.
$$P(x < k) =$$ __________ Therefore, $$k =$$ _________ |
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