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# Chapter 4 -- Decimals
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1 Chapter 4 -- Decimals \$34.99 decimal notation ex. The cost of an object. ex. The balance of your bank account ex The amount owed ex. The tax on a purchase. Just like Whole Numbers Place Value from the decimal point: ones, decimal (and), tenths, hundredths, thousandths, ten-thousandths, hundredthousandths, see page 235 Write in words ex. Write a check for \$38.43 ex. You buy two items that sell for \$85.90 and \$ You pay a total of \$13.22 in taxes. How much should you write a check for? Write it. Notice that we can write 8 / 4 = 12/3 = 24/6 = We can write fractions as decimals Sometimes they terminate some times they do not ( nonterminating) ex. 7/10 = 3/8 = but 2/3 = Other fractions: 7/25 = ex. 42/1000 = 7/9 = What place value is the digit 3 in the number ? 51
2 Write as a decimal. 3/10 =, 93/100 = 58 /1000 = Write as a fraction. 0.4 =, 0.24 = = Write in words = = Order. Insert the correct symbol ( < or > ) Rounding: Round the number to the given place value tenths thousandths Addition and Subtraction of decimals We still have the same rule of signs ex. Add: ex. Subtract evaluate: x - y if x = and y = Estimate:
3 Multiplication of Decimals: sum of the decimal places of the factors have to equal the decimal places of the product = = x 2.01 = Special Case: When multiplying by powers of 10 such as 10, 100, 1000: ex. 10 x 3.25 = x 100= x 10 5 = Division: 8 4 = 2 ; 4 is the divisor and 2 is the quotient Move the decimal point in the divisor to the right so that the divisor is a whole number. Move the decimal place in the dividend the same number of places. Place the decimal point directly above. ex = ex Special Case: Division by powers of 10; = = = We can write every fraction as a decimal if we interpret the fraction bar as a division. When we write a fraction as a decimal we get two types of decimal expression; terminating and nonterminating ex. 3/5 = 5/8 = 3/ 11 = ex. 9/1000 = ex. 2 5/9 = 53
4 We can also reverse the process and write a decimal expression as a fraction. ex = ex = ex = ex = Estimate by rounding. ex = ex = ex = Evaluate if x - = 0.1 and y = ; - x 2 = - xy = Compare decimal expressions(and fractions) : ex. 9/ ex. 7/ ex. 8/ Other Examples: ex. Your annual salary last year was \$ What was your monthly salary? ex. You travel 34.2 miles per gallon. If you have a car with a 12.8 gallon capacity, then what is the maximum number of miles that you can travel? 54
5 ex. A year consists of 52 weeks. You get a biweekly salary of \$850. What is your yearly pay? ex. At the end of each of month you save a fixed amount into a savings account. At the end of 1 ½ years you have \$ How much did you save each month? 55
6 Equations: As before we have two basic types of operations that can be used to solve equations. ex. 4v = 7 (write as a decimal ) ex x = ex x = x = 0.1 x = ex x = 0.44 x = x / -0.4 = 8 x = Square roots and perfect squares Perfect Squares: 1, 4, 9, 16, 25, 36, 49,. Def. A square root of a positive number x is a number whose square is x. ex. 4 is square root of since 4 2 = ex. 8 is the square root of since ( - 8 ) 2 = 56
7 Every positive number has two square roots; one positive and one negative. We use the symbol to represent the positive square root. Notation: x is called the radical, x is is called the radicand Each of these numbers has two square roots: 4 and 25 since ( -2) 2 = 4, ( 2) 2 = 4 ; ( -5 ) 2 = ( 5) 2 = 25 The radical only refers to the positive square root. 49 = 7, if we want the negative value we write 49 = 7 More Examples: Find all of the square roots of 64, 121, 400, Find 81 =, 169 =, 144 = What is = = 5 36 = = 9 = 16 57
8 Not all whole numbers are perfect squares what do we do when we try to find the square root of such a whole number? 26 = 53 = What two consecutive whole numbers does 37 lie between? < 37 < We can give approximations with a calculator or an algorithm we also have exact values. Property of radicals ab = a b. We can use this property to simplify radicals - Find (exact value ) 20 = 50 = 18 = 80 = 58
9 Real Numbers Now that we have introduced radicals, we are ready to complete the set that we call the set of real numbers. Sets of Numbers { 1, 2, 3, } { 0, 1, 2, 3, } { -3, -2, - 1, 0, 1, 2, 3, } A rational number is a number that can be written as a fraction a/b, where both a and b are integers Every rational number has a decimal expression that is terminating or repeating. 1/5 = 0.2, 3/8 = 0.375, 5/9 = = = An irrational number is a number whose decimal representation never terminates or repeats -- can not be written as a fraction of integers π = , 2 =, , , The set of rational numbers and the irrational numbers when combined create the set of real numbers We can plot these numbers on a number line --- ex. Plot -5, 7, 3.5, \/ 3 ex. We can also write real numbers on the number line with inequalities a) Write down all of the real numbers less than 2 b) Write down all of the real numbers greater than or equal to 1 c) write down all the numbers between -2 and 3 59
10 ex. Which of the values -4, 0, 2, 3/2, make the inequality x 2 true? ex. Which of the values 2, 2.1, 0.21 make the inequality x > 2.01 true? Properties of Real Numbers : 1. a + b = b + a 2. ab = ba 3. (a + b ) + c = a + ( b + c ) 4. a(bc) = a(bc) 5. a( b + c ) = ab + ac 6. Addition Property of 0, we call 0 the additive identity 7. Multiplication Property of 0 8. Multiplication Property of 1, we call 1 the multiplicative identity 9. Inverse Property of addition, a and ( - a) are called additive inverses of each other ( we also call them ) of each other 10. Inverse Property of Multiplication If a 0, then, a and 1/a are called multiplicative inverses of each other. 60
11 Find 6( 3x ) = Why? (3x ) 5 = Why? (2x)(5x) = Why? 1 c = - 1 c = ( -3 ) ( - c ) = -3t t = 2( 3x + 4) = - 2 ( 3x + 5 ) = - 4( 2x + 4 ) = - 4 ( 3x - 4 ) = 2( x 3a + 2c ) = - ( x + 2y ) = - 1 ( x + 2y ) = Terms, coefficients, variable terms, variable expressions, combining similar terms, How many terms? what are they? 3x 3 + 3x - 4 How many are constant terms? Same Questions for the polynomial - 2x 2 + 2y
12 What are the coefficient terms of x 2 + x + 1 of -2x 2 - x + 3 Similar Terms ( Like Terms) : Combine terms. (Alternate form of the distributive law) Instead of thinking of the distributive law in the form: a( b +c ) = ab + ac, let us think of it as ab + ac = a( b + c ) 8x + 13a = 12c - 16c = 6c - 6c = 2x 4y + 3x = 3x + y + 4x - 4y = 2x + 3(x + 4 ) = ( 3x - 1 ) = 2c - 2 [ 2-3 ( 2c - 3 ) ] = 62
13 Polynomials : Def. A monomial is a number, a variable, or a product of numbers and variables. Exponents on variables must be nonnegative integers. Def. A polynomial is a variable expression in which the terms are monomials ( a sum of monomials) A polynomial with one term is called a with two terms with three terms Addition of polynomials. Combine similar terms ( like terms) ex. ( 3x 2 + 2x 3 ) + ( 5x 2-4x - 12 ) = ex. ( 2xy + x ) + ( 5xy + y ) = Subtraction: We perform subtraction operations as before - write as an addition problem and use addition rules. ex. ( 2x 3xy ) ( 4x 2y ) = ex. ( 2x + y 3 ) - ( 3x 2y 4 ) = 63
14 Degree of a monomial: x 3 x 10 5x 6 2x 2 y 12x 4 y 8 Degree of a polynomial: 1 + x + x y - 4x 20 2xy x + x 3 4x 3 y 2 + 3y 4 Multiplication of Polynomimals: Rules of exponents: 1. If m and n are positive integers ( natural numbers ), then x m x n = x m + n ex. Find a 4 a 8 = = (-2 ) 4 (-2) 3 = What about x 3 y 5 = or ( 4x 3 ) ( 3x 2 ) = ( 2a 3 b ) ( a 2 b 5 ) = 2. I m and n are positive integers, then ( x m ) n = ex. Find ( 2 3 ) 4 = ( x 3 ) 2 = 3. If m, n, and p are positive integers, then ( x m y n ) p = ex ( 2x 3 ) 4 = ( x 4 y 2 ) 5 = 64
15 These types of rules help us multiply monomials together but what about the product of polynomials? Products of powers of 10. What is ? = Find Quotients of powers of 10. What is ? = Find = Convert to a decimal 9/16 = 7/4 = 5/3 = 3 2/9 = Exponents Just like we worked with natural numbers, whole numbers, integers, we can work with fractions. ex. ( 2/3 ) 2 = ex. ( 2 1/5 ) 2 = ex. ( 2 ½) 3 ( 1 ½ ) 3 = 65
16 HW: page 322 1, 5, 10, 15, 20, 25, 95 Math 130A Week 6 day 2, October 3, Quiz # Homework 20, 15, 10, { 0, 1, 2, 3, } is called the set of 0.3 The set of rational numbers consists of and proper fractions or mixed numbers. Each of these numbers can be written as a fraction and have a terminating decimal expression or have a repeating part if it is nonterminating 1. Write as a fraction Write as a fraction in simplest (reduced) form; Write in decimal form a) 3/1000 = b) 3/8 = 4. Write in words; Round to the nearest thousandth Perform the given operation ( add, subtract, multiply, divide ) a) ( ) = b) = c) = d) = 7. Evaluate. x 2 - y if x = and y =
17 8. Find the solution of each of the following equations. a) 0.2 x = 9.8 b) 4.2-2x = 6.8 More HW problems on page Square roots, radicals \/, Product Property of Square roots Name Math 130A Quiz #12, October 5, Homework 20, 15, 10, 0 1. Give me an example of an improper fraction. 2. Write 6/ 5 in decimal form. 3. Solve for x x = = = = 7. A pen sells for \$1.50 after taxes. You pay with \$20 and get \$6.50 back in change how many pens did you buy if all your money went towards the purchase of pens % of the students in Math 130A are known to drop out of school at the end of the semester. If there are 240 students enrolled during the semester, then how many will return next semessteer? ( HINT: use the fact that 6 % = 0.06 ) 67
18 Name Math 130A October 2, 2002 week 6 day 2 1. What do you call the set of real numbers that can be written as fraction ( have terminating decimal expressions or repeat )? 2. Find the square roots of Simplify the following radicals. a) \/ 49 = b) \/ 100 = c) \/ 1600 = d) \/ 16 \/ 25 = e) 4 \/ 36 = 4. Complete the formula \/ a b = 5. Write the following in simplest form by using the formula in #4. a) \/ 8 = b) \/ 32 = 6. Use a number line to plot x > 4 68
19 7. Use a number line to plot x List two consecutive integers that \/ 32 lies between? < \/ 32 < Math 130A Week 7, day 2, October 10, Quiz 1.Simplify a) 4 ( 3x ) = b) - 2( 3 x ) = c) - 5/3 ( 2a ) = d) ( 4x ) ( 5x) = e) - 4 ( - ¼ a ) = f) - 4h + 3h = g) 2 ( h + 5 ) = h) 4 + ( - 2 ) 2y = 2. List the terms of 3. What are the coefficients of 4. Simplify by combining like (similar) terms 69
20 HW page 322 Math 130A Week 5, September 24, 2001 HW: page 211 1, 6, 11, 16, 21, 26, 31, 36, 41, 3, 8, 13, 18, 23, 28, 33, 38, page 199 1, 6, 11, 16, 166 Exponents and Complex Fractions - section 3.6 page Name Math 130A, Quiz #9, September 26, Did you do all/most of your homework? yes/no 2. Homework points 20, 15, Find the sum of 3 ¼ and 2 3/8. 4. What is the total of (- 3 2/5 ) and ( + 2 ¼ )? 5. Find the difference; 4 ½ - 2 ¾ = 6. Write as an improper fraction; 2 3/5 = 7. Find /11 = 8. What is the product of 4 and 3/8? 9. Find 8 2 ½ = 10. Is 2 a solution of 2/3 = x/3? Name Math 130A, Quiz 10, September 28, Which of these represents an improper fraction? 70
21 15/17, 2 ½, 27/8, 12/16, or none of these 2. Multiply 3 1/5 3 3/4 = 3. What is the product of 4 and 2 1/8? 4. Find the solution of each of the following equations. a) x/3 = 2 1/3 x = b) 3x/5 = - 4 x = c) 2/3 - ¾ x = ¼ x = 5. Let x = 4 ¼ and y = - 2 3/8. Find a) x - y = b) x + 2y = 6. 2 /3 of a household s budget goes towards essentials. If the \$2400 is budgeted for this month, how much of it goes towards the household s budget. 7. Find - 4 3/8 - ( - 2 ¾ ) September 28,
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# Evaluate Positive Indices
In this worksheet, students will find the value of numbers when positive and negative integer indices (or powers) are used.
Key stage: KS 4
GCSE Subjects: Maths
GCSE Boards: Pearson Edexcel, OCR, Eduqas, AQA
Curriculum topic: Number, Indices and Surds
Curriculum subtopic: Structure and Calculation, Powers and Roots
Difficulty level:
### QUESTION 1 of 10
In this activity, we are going to learn to evaluate numbers that are given using a positive index number.
For example, 4 squared is 16 and 2 cubed is 8.
We could write these as 42 = 16 or 4 x 4 = 16 and 23 = 8 or 2 x 2 x 2 = 8.
42 - The small 2 tells us to multiply 4 BY ITSELF two times.
The is the index number.
2 - The small 3 tells us to multiply 2 BY ITSELF three times.
The is the index number.
If we are asked to "evaluate" (find the value of) 34 we would multiply by itself times.
This means 3 x 3 x 3 x 3 = 81
Question: Evaluate 34
Let's try another question now.
Evaluate 53.
We work out multiplied by itself times.
5 x 5 x 5 = 125
We are now going to work out (-5)3.
We will need to multiply (-5) by itself 3 times and we will need to follow the rules for multiplying negative numbers
This means we need to calculate: (-5) x (-5) x (-5)
The first step is to find (-5) x (-5)
Two negative numbers multiply to make a positive number, so the answer to this step is +25.
The next step is to multiply this result by (-5).
25 x (-5) = -125
In this activity, we will evaluate positive and negative indices to work out the whole number value of each and whether the outcome is positive or negative.
You will need to have your scientific calculator handy for some of the questions in this activity.
Evaluate:
32
(2 x 2 x 2)
(3 x 3)
(3 x 2)
(3 + 2)
Evaluate:
52
(5 + 5)
(5 + 2)
(5 x 5)
(2 x 2 x 2 x 2 x 2)
Evaluate:
34
(3 x 4)
(4 x 4 x 4)
(3 + 4)
(3 x 3 x 3 x 3)
Evaluate:
54
e.g. Using your calculator, evaluate 84.
The power button on your calculator means you can ask it to work out 8 x 8 x 8 x 8 in one step.
Give this a try now.
Enter 8 as the base of the power and 4 as the index number and then press "=".
The correct answer is 4096, which you should see on your calculator.
Using this same process, evaluate:
94
13
6561
36
e.g. Evaluate (-3)2
To solve this, we need to find the value of (-3) x (-3).
We will need to remember that multiplying a negative number by another negative number gives a positive answer
This means -3 x -3 = 9 (positive 9)
Using this same process, evaluate:
(-5)2
Positive 25
Negative 25
Evaluate:
(-4)3
Negative 64
Positive 64
(-5)6
Remember to use brackets to tell your calculator to multiply negative 5 by itself 6 times.
Typing in just -56 without brackets will ask your calculator to find 56 and then make the answer negative.
Positive 15625
Negative 15625
You will need to just remember that any number raised to the power of 0 has the answer 1.
This is an important fact to commit to your memory when working with powers.
e.g. 60 = 1 and 1210 = 1
Using this knowledge, evaluate:
70
Test all the skills you have learnt in this activity by evaluating (without a calculator):
32 + (-2)3 + 40
• Question 1
Evaluate:
32
(3 x 3)
EDDIE SAYS
'Evaluate 3²' means that we need to calculate the value of 3 x 3. The index number 2 tells us to multiply 3 by itself two times. 3 x 3 = 9
• Question 2
Evaluate:
52
(5 x 5)
EDDIE SAYS
'Evaluate 5²' means that we need to work out the value of 5 x 5. The index number 2 tells us to multiply 5 by itself two times. 5 x 5 = 25
• Question 3
Evaluate:
34
(3 x 3 x 3 x 3)
EDDIE SAYS
'Evaluate 34' means find the value of: 3 x 3 x 3 x 3 The index number 4 tells us to multiply 3 by itself four times. 3 x 3 x 3 x 3 = 81
• Question 4
Evaluate:
54
625
EDDIE SAYS
To find the value of 54, we need to work out: 5 x 5 x 5 x 5 The answer to this is 625. Evaluating index numbers can also be worked out on a scientific calculator using the power button. Why not give this a try now if you have not tried this before?
• Question 5
e.g. Using your calculator, evaluate 84.
The power button on your calculator means you can ask it to work out 8 x 8 x 8 x 8 in one step.
Give this a try now.
Enter 8 as the base of the power and 4 as the index number and then press "=".
The correct answer is 4096, which you should see on your calculator.
Using this same process, evaluate:
94
6561
EDDIE SAYS
The calculator will work out (9 x 9 x 9 x 9) for you in one step and show you an answer of 6561. We will now move on to finding powers of negative numbers.
• Question 6
e.g. Evaluate (-3)2
To solve this, we need to find the value of (-3) x (-3).
We will need to remember that multiplying a negative number by another negative number gives a positive answer
This means -3 x -3 = 9 (positive 9)
Using this same process, evaluate:
(-5)2
Positive 25
EDDIE SAYS
When we calculate (-5) x (-5), we are multiplying two negative numbers together. Multiplying two negatives together, gives a positive answer. Therefore: (-5) x (-5) = 25 (or positive 25)
• Question 7
Evaluate:
(-4)3
Negative 64
EDDIE SAYS
When we multiply negative 4 by itself 3 times, we need to work out: (-4) x (-4) x (-4) As a first step, we need to multiply the first 2 numbers only: (-4) x (-4) = 16 Remember that multiplying two negatives, gives a positive answer overall. This result then needs multiplying by (-4) again. 16 x (-4) = (-64) Remember that a positive number multiplied by a negative number gives a negative result. As a top tip when working with negative integers, if the power or index is an even number then the outcome will be positive overall, but if it is odd then the outcome will be negative.
• Question 8
(-5)6
Remember to use brackets to tell your calculator to multiply negative 5 by itself 6 times.
Typing in just -56 without brackets will ask your calculator to find 56 and then make the answer negative.
Positive 15625
EDDIE SAYS
Our final answer is positive as we are multiplying a negative number by itself an even number of times. If we work this through to check: (-5) x (-5) = 25 25 x (-5) = -125 -125 x (-5) = 625 625 x (-5) = -3125 -3125 x (-5) = 15625 That confirms it then!
• Question 9
You will need to just remember that any number raised to the power of 0 has the answer 1.
This is an important fact to commit to your memory when working with powers.
e.g. 60 = 1 and 1210 = 1
Using this knowledge, evaluate:
70
1
EDDIE SAYS
Remember that any number raised to the power of 0 is 1.
• Question 10
Test all the skills you have learnt in this activity by evaluating (without a calculator):
32 + (-2)3 + 40
2
EDDIE SAYS
This may look complex, but we just need to work through each index, one step at a time. 3² = 3 x 3 = 9 (-2)³ = (-2) x (-2) x (-2) = -8 40 = 1 9 + (-8) + 1 = 2 Well done for working hard on index numbers, hopefully you are feeling much more confident with these now. Why not try an activity focused only on negative indices now?
---- OR ----
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# The Definition of a Parabola
### The Definition of a Parabola
Student Outcomes
• Students model the locus of points at equal distance between a point (focus) and a line (directrix). They construct a parabola and understand this geometric definition of the curve. They use algebraic techniques to derive the analytic equation of the parabola.
### New York State Common Core Math Algebra II, Module 1, Lesson 33
Worksheets for Algebra II, Module 1, Lesson 33
The following diagram shows the locus definition of a parabola. Scroll down the page for more examples and explanations about parabolas.
Β
Classwork
Opening Exercise
Suppose you are viewing the cross-section of a mirror. Where would the incoming light be reflected in each type of design? Sketch your ideas below
Discussion: Telescope Design
When Newton designed his reflector telescope, he understood two important ideas. Figure 1 shows a diagram of this type of telescope.
• The curved mirror needs to focus all the light to a single point that we will call the focus. An angled flat mirror is placed near this point and reflects the light to the eyepiece of the telescope.
• The reflected light needs to arrive at the focus at the same time. Otherwise, the image is distorted.
Definition: A parabola with directrix πΏ and focus point πΉ is the set of all points in the plane that are equidistant from the point πΉ and line πΏ.
Figure 2 to the right illustrates this definition of a parabola. In this diagram, πΉπ1 = π1π1, πΉπ2 = π2π2, πΉπ3 = π3π3 showing that for any point π on the parabola, the distance between π and πΉ is equal to the distance between π and the line πΏ.
All parabolas have the reflective property illustrated in Figure 3. Rays parallel to the axis reflect off the parabola and through the focus point, πΉ. Thus, a mirror shaped like a rotated parabola would satisfy Newtonβs requirements for his telescope design.
Figure 4 below shows several different line segments representing the reflected light with one endpoint on the curved mirror that is a parabola and the other endpoint at the focus. Anywhere the light hits this type of parabolic surface, it always reflects to the focus, πΉ, at exactly the same time.
Figure 5 shows the same image with a directrix. Imagine for a minute that the mirror was not there. Then, the light would arrive at the directrix all at the same time. Since the distance from each point on the parabolic mirror to the directrix is the same as the distance from the point on the mirror to the focus, and the speed of light is constant, it takes the light the same amount of time to travel to the focus as it would have taken it to travel to the directrix. In the diagram, this means that π΄πΉ = π΄πΉπ΄, π΅πΉ = π΅πΉπ΅, and so on. Thus, the light rays arrive at the focus at the same time, and the image is not distorted.
Example: Finding an Analytic Equation for a Parabola
Given a focus and a directrix, create an equation for a parabola.
Focus: πΉ(0,2)
Directrix: π₯-axis
Parabola:
π = {(π₯, π¦)| (π₯, π¦) is equidistant from πΉ and the π₯-axis.}
Let π΄ be any point (π₯, π¦) on the parabola π. Let πΉβ² be a point on the directrix with the same π₯-coordinate as point π΄.
What is the length π΄πΉβ²?
Use the distance formula to create an expression that represents the length π΄πΉ<br Create an equation that relates the two lengths, and solve it for π¦.<br Verify that this equation appears to match the graph shown.
Exercises
1. Demonstrate your understanding of the definition of a parabola by drawing several pairs of congruent segments given the parabola, its focus, and directrix. Measure the segments that you drew to confirm the accuracy of your sketches in either centimeters or inches.
2. Derive the analytic equation of a parabola given the focus of (0,4) and the directrix π¦ = 2. Use the diagram to help you work this problem.
a. Label a point (π₯, π¦) anywhere on the parabola.
b. Write an expression for the distance from the point (π₯, π¦) to the directrix.
c. Write an expression for the distance from the point (π₯, π¦) to the focus.
d. Apply the definition of a parabola to create an equation in terms of π₯ and π¦. Solve this equation for π¦
e. What is the translation that takes the graph of this parabola to the graph of the equation derived in Example 1?
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Term - What is it?
Last updated date: 17th Apr 2024
Total views: 200.4k
Views today: 4.00k
## What is a Term in Mathematics?
A term in mathematics is nothing but a single expression. The term can be a single number - either positive or negative, or it can be a variable, or it can be more than one variable multiplied together. The variables are often written with numbers in front of them and these numbers are called coefficients. Let’s have a look at a few examples:
Single number as a term: 15
A variable: a
More than one variable: 2x3y24z3
In algebra, two or more terms are often joined to make an expression. For example:
2a + 3b
4c - 5d2
6x + 7y2 - 8z3
4x2y + 5a2b3 - 6m3n4
There are two types of terms: Like terms and unlike terms. Let’s learn about them.
## Like Terms
Like terms are the terms whose variables are similar. For example - 2x2 and 3x2 are like terms because the variable x2 is the same. A few more examples are: 3a3, 5a3, and 7a3; 2x2y, 3x2y, and 4x2y
Like Terms
To solve the algebraic expressions, we need to combine the like terms and simplify the expressions. For example:
3x2 + 2x + 9 - 2x2+ 5x - 3
= 3x2 - 2x2 + 2x + 5x + 9 - 3
= x2 + 7x + 6
Would you like to solve an exercise?
## Exercise on Like Terms
Here are a few questions:
1. 2r2+1+(−4r2)+7
2. 4x + 2y + 3x - y
3. -5m + (-6n) + 3m - 2n
4. 3a2 + 5b + 6a + a2 - 2b
5. -z2 + 12z2 - 5z + 8z3 - 15z
Solutions:
1. 2r2+1+(−4r2)+7
= 2r2 - 4r2 + 1 + 7
=-2r2 + 8
1. 4x + 2y + 3x - y
= 4x + 3x + 2y - y
= 7x + y
1. -5m + (-6n) + 3m - 2n
= -5m + 3m - 6n - 2n
=2m - 8n
1. 3a2 + 5b + 6a + a2 - 2b
= 3a2 + a26a + 5b - 2b
= 4a2 + 6a + 3b
1. -z2 + 12z2 - 5z + 8z3 - 15z
= 8z3 - z2 + 12z25z - 15z
8z3 + 11z2 - 20z
## Unlike Terms
Unlike terms are the terms whose variables are different from each other. Example:
4x and 9y; 3x2 and 5y; 4a2 + 5b3 - 6z4
Simplifying expressions with like and unlike terms:
2ab2 - 7c2 + 3abc - 4ab2 + 8c2 - ab
= 2ab2 - 4ab2 - 7c2 + 8c2 + 3abc - ab
= -2ab2 + c2 + 3abc - ab
Like and Unlike terms
## What Have We Observed?
So what have we observed? We have observed that while simplifying any expression, only the like terms are added or subtracted. The, unlike terms, remain as they are in an expression.
Did you enjoy learning about the like and unlike terms? We have other interesting math ‘magic’ topics as well for you to explore. You can explore it through our website and also download the app from the play store.
## FAQs on Term - What is it?
1. How many terms are there in an expression?
One or more terms can make up an algebraic expression. A constant, a variable, a product of more than two variables (xy), or a product of a variable and a constant can all be used as terms in an expression. The terms, when added together, form an expression.
2. What is a coefficient?
In an algebraic equation, a coefficient represents the numerical factor of a term made up of constants and variables. The terms' coefficients can be either positive or negative. It's possible that they're fractional in nature.
3. What is a variable?
Variables are terms made up of undefined values that, when substituted with different integers, can take on different integer values. A variable term can be made up of one or more variables, which can be the same or different.
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# Quadratic Mean / Root Mean Square
Descriptive Statistics > Quadratic Mean / Root Mean Square
## What is the Quadratic Mean / Root Mean Square?
The quadratic mean (also called the root mean square*) is a type of average. This type of mean gives a greater weight to larger items in the set and is always equal to or greater than the arithmetic mean. It is used for specialized purposes, such as the expression of average stand diameter in forestry. In this case, the quadratic mean of tree stand diameter is closer to the “true” mean of a sample of trees rather than the arithmetic mean. The quadratic mean is also used anywhere where it’s the square of the values that matters, rather than the values themselves. For example, electrical current squared is proportional to power, so if you’re interested in total power (rather than current), this type of mean is a good choice.
Sometimes the quadratic mean is referred to as being “the same as” the standard deviation. Standard deviation is actually equal to the quadratic deviations from the mean of the data set. For example, quadratic mean is used in the physical sciences as a synonym for standard deviation when referencing the “square root of the mean squared deviation of a signal from a given baseline or fit”(Wolfram).
The quadratic mean is also called the root mean square because it is the square root of the mean of the squares of the numbers in the set.
*This is different from the root mean square error (RMSE), which is a value used in regression analysis to describe how spread out data is around a regression line.
## Formula
The quadratic mean is equal to the square root of the mean of the squared values. The formula is:
An equivalent formula has a summation sign (summation means “to add up”, so it’s telling you here to add all of the squared x-values up):
## Worked Example
Find the Root Mean Square of 2,4,9,10,and 12.
Step 1: Count the number of items.
N = 5.
Set this number aside for a moment.
Step 2: Square all of the numbers. 22,42,92,102, 122 = 4, 16, 81, 100, 144.
Step 3: Add the numbers from Step 2 up:4 + 16 + 81 + 100 + 144 = 345.
Step 4: Divide Step 3 (the sum) by Step 1 (number of items in the set):
345/5 = 69.
Step 5: Find square root of Step 4. √(69) = 8.31.
That’s it!
References:
Kenney, J. F. and Keeping, E. S. “Root Mean Square.” §4.15 in Mathematics of Statistics, Pt. 1, 3rd ed. Princeton, NJ: Van Nostrand, pp. 59-60, 1962.
Wofram. Root MEan Square. Available at: http://mathworld.wolfram.com/Root-Mean-Square.html
------------------------------------------------------------------------------
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. If you rather get 1:1 study help, Chegg Tutors offers 30 minutes of free tutoring to new users, so you can try them out before committing to a subscription.
If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.
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# 3.7: Pythagorean Theorem, Part 2: Applications & Triples
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Identify common Pythagorean triples.
## Using Pythagorean Triples
Review the example problems from the previous lesson.
This is the diagram from Example 1:
In Example 1, the sides of the triangle are 3, 4, and 5. This combination of numbers is referred to as a Pythagorean triple. A Pythagorean triple is three integers (whole numbers with no decimal or fraction part) that make the Pythagorean Theorem true.
• A Pythagorean triple is a group of three _____________________________ that satisfy the Pythagorean Theorem.
Throughout this chapter, you will learn other Pythagorean triples as well.
This is the diagram from Example 2:
Using the Pythagorean Theorem equation \begin{align*}a^2 + b^2 = c^2\end{align*}, and letting \begin{align*}a = 6\end{align*} and \begin{align*}c = 10\end{align*}, we calculated that \begin{align*}b = 8\end{align*} inches.
The triangle in Example 2 is proportional to the same ratio of 3 : 4 : 5. If you divide the lengths of the triangle (6, 8, and 10) by 2, you find the same proportion — 3 : 4 : 5 (because \begin{align*}6 \div 2 = 3, 8 \div 2 = 4\end{align*}, and \begin{align*}10 \div 2 = 5\end{align*}).
Whenever you find a Pythagorean triple, you can apply these ratios with greater factors as well.
Finally, look at the side lengths of the triangle in Example 3:
The two legs are 5 cm and 12 cm and the length of the missing side (the hypotenuse) is 13 cm. The side lengths make a ratio of 5 : 12 : 13. This, too, is a Pythagorean triple. You can infer that this ratio, multiplied by greater factors, will also yield numbers that satisfy the Pythagorean Theorem.
There are infinitely many Pythagorean triples, but a few of the most common ones and their multiples are in the chart below:
Pythagorean triple \begin{align*}\times 2\end{align*} \begin{align*}\times 3\end{align*} \begin{align*}\times 4\end{align*}
3 – 4 – 5 6 – 8 – 10 9 – 12 – 15 12 – 16 – 20
5 – 12 – 13 10 – 24 – 26 15 – 36 – 39 20 – 48 – 52
7 – 24 – 25 14 – 48 – 50 21 – 72 – 75 28 – 96 – 100
8 – 15 – 17 16 – 30 – 34 24 – 45 – 51 32 – 60 – 68
1. What is a Pythagorean triple?
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
2. Which of the following is NOT a Pythagorean triple? Show your work.
a. 15 – 36 – 39
b. 15 – 20 – 25
c. 16 – 30 – 35
d. 25 – 60 – 65
3. Give 2 examples of Pythagorean triples that are NOT in the chart above.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
4. Why is it helpful to know common Pythagorean triples?
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
8 , 9 , 10
## Date Created:
Feb 23, 2012
May 12, 2014
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# How To Calculate The Diagonal
## Video: How To Calculate The Diagonal
A diagonal connects non-adjacent vertices of a polygon with at least four sides. Calculate this value through the initial or intermediate data of the problem, using the appropriate formulas.
## Instructions
### Step 1
Any closed geometric figure consisting of at least four line segments can have at least two diagonals. This is how many diagonals a quadrangle can have: a parallelogram, a rectangle, a rhombus and a square.
### Step 2
Find the diagonals of the parallelogram if it is known that one of them is greater than the other by 1, and the lengths of the sides are equal to a = 5 and b = 7. There is a ready-made formula for this in geometry, according to which the sum of the squares of the lengths of the diagonals is equal to the doubled sum of the squares of the sides: d1² + d2² = 2 • (a² + b²) = 2 • (25 + 49) = 148.
### Step 3
Substitute d2 = d1 + 1: d1² + (d1 + 1) ² = 148 2 • d1² + 2 • d1 + 1 = 148.
### Step 4
Solve the following equation for the unknown d1: 2 • d1² + 2 • d1 - 147 = 0D = 4 + 4 • 2 • 147 = 1180d1 = (-2 + √1180) / 4 ≈ 8, 1 → d2 = 9, 1.
### Step 5
The formula for a rectangle is simplified because its diagonals are equal: 2 • d² = 2 • (a² + b²) = 2 • (25 + 49) = 148 → d² = 74 → d ≈ 8, 6.
### Step 6
In the case of a square, the situation is even simpler, its diagonals not only have equal length, but are also directly proportional to the side: 2 • d² = 4 • a² → d² = 2 • a² → d = √2 • a = [a = 5] = √ 2 • 5 ≈ 7.
### Step 7
A rhombus is a special case of a parallelogram with equal sides, but unlike a square, the diagonals are not equal to each other. Suppose that the side of the rhombus is a = 5, and the length of one of the diagonals is 3. Then: d1² + 9 = 4 • 25 → d1 = 9.
### Step 8
Diagonals can be drawn not only in a flat figure, but also in a spatial one. For example, in a box. The square of the length of the diagonal of a rectangular parallelepiped (or its special case - a cube) is equal to the sum of the squares of its three dimensions. Dimensions are edges that have one common vertex.
### Step 9
A triangle has no diagonals and its three-dimensional version is a tetrahedron, since they do not have non-adjacent vertices. The number of diagonals in any n-polygon can be determined as follows: nd = (n² - 3 • n) / 2.
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# How to find length of the party
Tasks on finding of length of the parties are one of the most widespread it is aware of geometry. The algorithm of their decision depends on basic data, features of the considered figure.
## It is required to you
• - notebook;
• - ruler;
• - pencil;
• - handle;
• - calculator.
## Instruction
1. The simplest tasks on finding of length of the parties are tasks with the known perimeter (it is the sum of lengths of all parties). For example, the perimeter of a parallelogram of ABCD is equal to 22 cm, AV = 4, to find VS. Since opposite groans are equal in a parallelogram, AV = CD = 4.
2. Decision: From here VS = (22 – (AV * 2)) / 2BC = (22 – (4*2)) / 2BC = 7
3. Also often tasks on finding of length of the parties through the area meet. For example, the area of a rectangle of ABCD is equal to 24 cm, AV = 3 cm, to find VS. Opposite groans are also equal in a rectangle therefore AV = CD = 3.
4. Decision: S (it is direct.) = a*vs = AV * VSOtsyuda VS = S/ABBC = 8
5. A special case of a rectangle is the square. The square is a rectangle which parties are equal among themselves, and corners between them make 90 degrees. If the area of a square is known, then it is possible to find length of its party. For example, S squares of ABCD = 64 см^2. To find AV.
6. Decision: S (quarter) = а^2а = Sa = 64a = 8
7. But if no area not perimeter, but only length of one of the parties is unknown, then the decision becomes complicated. For example, in ABC 1/2AC triangle = 4 cm, SAV corner = DIA, VM – the bisector equal of 10 cm. To find AV.
8. Decision: If SAV corner = to a corner of DIA, then AVS triangle – isosceles. And in an isosceles triangle the bisector is a median and height. Since VM – height, that corner of VMA = 90, from here AVM triangle – rectangular. In a rectangular triangle of a square of a hypotenuse it is equal to the sum of squares of legs (on Pythagorean theorem). Therefore, AV ^2 = AM ^2 + VM ^2АВ ^2 = 16 + 100AB = √116
Author: «MirrorInfo» Dream Team
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# ACT Math : How to find the sale price
## Example Questions
← Previous 1 3 4
### Example Question #1 : How To Find The Sale Price
Max wants to buy a couch at The Furniture Store. The couch he is considering had been marked on sale for 20% off and was selling for $320. The day he comes in to buy the couch, it has been marked down an additional 40% off of the sale price. What is the difference in the original price of the couch and what Max paid for the couch? Possible Answers:$240
$208$160
$128$272
$208 Explanation: (320) / (0.8) = 400. The couch has been marked down an additional 40% from the sale price of$320. $320 * 0.6 =$192. $400 –$192 = $208. ### Example Question #1 : How To Find The Sale Price Mary wants to buy a new skirt. It was originally$30, but is marked 20% off. She also received a coupon for 15% off the sale price. How much will Mary pay for the skirt?
$10.50$23.90
$20.40$19.50
$20.40 Explanation: 20% off means that the new price of the skirt will be 80% of the original price:$30(100% – 20%) = $30(80%) Converting the percent to a decimal gives:$30(0.8) = $24.00 There is an additional 15% off the sale price of$24.00, so the final price is 85% of the sale price:
$24(100% – 15%) =$24(85%)
Again converting the percent to a decimal gives:
$24(0.85) =$20.40
### Example Question #3 : How To Find The Sale Price
A tablet computer listed at an original price of $250 is placed on sale for 20% off the original price. Bob, an employee of the store, gets an additional 30% off the sale price. What price would Bob pay to purchase the tablet computer? Possible Answers:$100
$200$125
$140 Correct answer:$140
Explanation:
Take the original price and take off 20% = $250(1 - .2) =$250(.8) = $200 Then take off another 30%$200(1 - .3) = $140 ### Example Question #2 : How To Find The Sale Price The manager of a department store decided to raise the price of a certain pair of shoes by 30%. The next day, the store ran a sale of 20% off all items. What is the difference in price, in percentage terms, between the initial price of the shoes and the sale price? Possible Answers: The price increased by 10% The price increased by 1.5% The price decreased by 10% The price decreased by 4% The price increased by 4% Correct answer: The price increased by 4% Explanation: To find the price after the initial 30% increase by the manager, you must multiply the original price by 1.3. Then, to find the price after the 20% off sale, you must multiply the new price by 0.8. The original price, therefore, is being multiplied by 1.3*0.8 = 1.04, indicating a 4% overall increase. ### Example Question #4 : How To Find The Sale Price A shirt is originally priced at$54. It is on sale for 60% off, and Jeff has a coupon for an additional 15% off the reduced price. What is the final price Jeff pays for the shirt?
$4.86$13.50
$24.30$21.60
$18.36 Correct answer:$18.36
Explanation:
After 60% off, the shirt is marked down to $21.60 (found by:$54 - $54*0.6 =$21.60). Jeff uses a 15% off coupon, knocking the price down to $18.36 ( found by:$21.60 - $21.60*0.15 =$18.36).
### Example Question #4 : How To Find The Sale Price
A video game console with a list price of $500 is marked down 20%. If Katie gets an employee discount of 10% off the sale price, how much does she pay for the video game console? Possible Answers:$360
$350$450
$340$400
$360 Explanation: First find the sale price. Multiply the list price by .2 and subtract that from the list price. 500 – 500 * .2 = 500 – 100 = 400. Now take the employee discount from the new price, 400 – 400 * .1 = 400 – 40 = 360, so Katie would pay$360 for the video game console.
### Example Question #3 : How To Find The Sale Price
A shirt, originally $50, is on sale for 20% off. If Andrew has a coupon that takes 15% off the reduced price, what does he pay? Possible Answers:$34.00
$37.00$10.00
$32.50 Correct answer:$34.00
Explanation:
The shirt is on sale for 50 x .8 = $40. If Andrew takes another 15% off, he will pay$40 x .85= $34.00 ### Example Question #3 : How To Find The Sale Price A blue dress is marked down 15%. What is the sale price of the dress if the regular price is$150?
$137.50$115.25
$103.50$127.50
$140.00 Correct answer:$127.50
Explanation:
There are two ways to work this problem.
The first way is to find the amount of the discount and subtract from the original price: 0.75 * 150 = 22.50, then 150 – 22.50 = 127.50
The second method finds the sales price directly: 100% = % Discount + % Sales Price or 1 – % Discount = % Sales Price
1 – 0.15 = 0.85, so the sales price is 0.85 * 150 = 127.50
### Example Question #4 : How To Find The Sale Price
Gina wants to purchase a new television. It was originally $500, but is marked 25% off. She also received a coupon for 10% off the sale price. How much will Gina pay for the television? Possible Answers:$375.25
$102.50$350.00
$465$337.50
$337.50 Explanation: 25% off means that the new price of the television will be 75% of the original price:$500(100% – 25%) = $500(75%) Converting the percent to a decimal gives:$500(0.75) = $375.00 There is an additional 10% off the sale price of$375.00, so the final price is 90% of the sale price:
$24(100% – 10%) =$375(90%)
Again converting the percent to a decimal gives:
$375(0.9) =$337.50
So, we choose C.
### Example Question #7 : How To Find The Sale Price
The grocery store is having a sale on cereal. The sale is 20% off each box if you buy 3 boxes of the same type and 40% off each box if you buy 5 boxes of the same type. You purchase one box of puffed rice priced at $3.50, three boxes of bran flakes priced at$4.25 and five boxes of granola priced at $5.00. What is the total price after the discount is applied. Possible Answers:$15.00
$28.50$10.00
$25.00$28.70
$28.70 Explanation: Puffed rice: 1 box at full price =$3.50
Bran Flakes: 3 boxes at 20% off of $4.25 = (3 * 4.25) * 0.20 = (12.75) * 0.20 =$2.55
Cost = $12.75 –$2.55 = $10.20 Granola: 5 at 40% off of$5.00 = (5 * 5.00) * 0.40 = (25) * 0.40 = $10.00 Cost =$25.00 – $10.00 =$15.00
Total Cost = $3.50 +$10.20 + $15.00 =$28.70
← Previous 1 3 4
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1. Home
3. Operations with Decimal Numbers
You went to the grocery store today to buy a couple of things 1 liter of milk, 24 eggs, a pound of tomatoes, 2 lbs of potatoes and 3 2 liters of Coca Cola. You checked the price tag for each item and you found out that 1 liter of whole fat milk costs C$3.11, 12 large eggs costs$4.28, 2 lbs of tomatoes costs $4.02, 2 lbs of potatoes costs C$ 2.57 and a 2 liter of Coca Cola costs C\$ 2.38. To come up with the total amount of money you need to pay, you will need to apply the different operations with the set of decimal numbers given above.
In this chapter, we will learn about the proper skills that will help you solve the problem above. This chapter has four parts. In the first chapter we will learn how to add and subtract decimal numbers. The rule is simple as you only need to align the given numbers according to the location of the decimal point. Then, depending on the instructions, we will need to round off the resulting number to the nearest tenth. Rounding off numbers is mostly used to simplify the results.
For the second part of this chapter, we will look at how to multiply decimals. In this operation, we will simply multiply the decimal numbers like a regular whole number by ignoring the decimal point. We will only put the decimal point in place after we get the product. By counting the combined number of decimal places of the two numbers we have multiplied we will determine where to put the decimal point.
Following this discussion is the third part of the chapter where we will study how to divide decimals numbers. If the divisor is is not a whole number, we will move the decimal point to the right until the number becomes a whole number. The dividend's decimal point should also be moved accordingly. After the adjustments are made, we can proceed as usual. The quotient's decimal point should be placed directly above that of the dividend. In some cases, we may encounter repeating numbers in the quotient like 1/3 which will give us the repeating decimal 0.3333….
Finally, in the last part of this chapter, we will discuss the Order of operations PEMDAS, which stands for P Parentheses first, E Exponents (ie Powers and Square Roots), MD Multiplication or Division (left-to-right) and AS Addition or Subtraction (left-to-right).
#### Chapter 5 Topics:
In this section, we will add and subtract decimal numbers. The process of adding and subtracting decimal numbers is the same as adding and subtracting whole numbers. For example, both involve regrouping. After lining up the decimals, you will sometimes have to add one or more zeroes to a number after the decimal point. Being able to add and subtract decimal numbers is an important life skill, especially when dealing with money and purchases.
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# How do you evaluate log_16(1/4)?
Mar 23, 2016
$x = - \frac{1}{2}$
#### Explanation:
Let ${\log}_{16} \left(\frac{1}{4}\right) = x$, then ${16}^{x} = \frac{1}{4}$ or
${\left({2}^{4}\right)}^{x} = \frac{1}{{2}^{2}}$ or
${2}^{4 x} = {2}^{- 2}$
Hence $4 x = - 2$ i.e. $x = - \frac{2}{4} = - \frac{1}{2}$
Mar 23, 2016
Let's write it in exponential form.
#### Explanation:
${\log}_{a} b = x \to {a}^{x} = b$
${\log}_{16} \left(\frac{1}{4}\right) = x \to {16}^{x} = \frac{1}{4}$
Solve for x by putting everything in the same base.
${\left({2}^{4}\right)}^{x} = \frac{1}{{2}^{2}}$
${2}^{4 x} = {2}^{- 2}$
$x = - \frac{1}{2}$
Therefore, ${\log}_{16} \left(\frac{1}{4}\right) = - \frac{1}{2}$
Practice exercises:
1. Evaluate or solve for x.
a). ${\log}_{9} \left(\frac{1}{27}\right)$
b) ${\log}_{x} \left(81\right) = 4$
c) ${\log}_{2 x + 1} 11 x = 2$
Good luck!
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# How to calculate 292 divided by 32 using long division?
292 ÷ 32
=
9.125
Division is a fundamental arithmetic operation where we calculate how many times a number (divisor or denominator) can fit into another number (dividend or numerator). In this case, we are dividing 292 (the dividend) by 32 (the divisor).
There are three distinct methods to convey the same information: in decimal, fractional, and percentage formats:
• 292 divided by 32 in decimal = 9.125
• 292 divided by 32 in fraction = 292/32
• 292 divided by 32 in percentage = 912.5%
## What is the Quotient and Remainder of 292 divided by 32?
The quotient is calculated by dividing the dividend by the divisor, and the remainder is what's left over if the division doesn't result in a whole number.
The quotient of 292 divided by 32 is 9, and the remainder is 4. Thus,
### 292 ÷ 32 = 9 R 4
When you divide Two Hundred And Ninety Two by Thirty Two, the quotient is Nine, and the remainder is Four.
## Verdict
The division of 292 by 32 results in a quotient of 9 and a remainder of 4, meaning 32 goes into 292 Nine times with 4 left over. Understanding this division process is crucial in both basic arithmetic and real-life applications where division is used, such as in financial calculations, data analysis, and everyday problem-solving.
## Random Division Problems?
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## Frequently Asked Questions
### How do we differentiate between divisor and dividend?
A dividend is a number we divide, while a divisor is a number by which we divide. Divisor comes on second, followed by the dividend that we write first.
For instance, if you have 12 candies and want to distribute them among 3 children, the equation will be 12 ÷ 3. You will put 12 first because this is the number being divided. So here, 12 is a dividend. On the other hand, 3 is written after 12, and it is the number with which we are dividing 12. Hence, 3 is a divisor.
### Which formula is used to find a divisor?
There are two formulas used to find a divisor.
The first one is: Divisor = Dividend ÷ Quotient. This formula is used to find a divisor when the remainder is 0.
Second is: Divisor = (Dividend – Remainder) /Quotient. This formula is used when the remainder is not 0.
### Is there a possibility of a number having the same divisor?
Yes, there is. Every number can be divided by itself, leaving 1 as the quotient. So, it would not be wrong to say that all the numbers can have the same divisors.
Let’s take the example of 5. If we divide 5 by 5 (5 ÷ 5), then 5 will be the divisor of 5. And ultimately, 1 will be the quotient.
### What is the difference between a divisor and a factor?
A divisor is a number with which we can divide any number. However, a factor is different from a divisor. It is the number that can be divided with another number leaving no remainder. All factors are divisors, but not all divisors are factors.
### Is it possible to do division by repeated subtraction?
Fortunately yes. You can do division by repeated subtraction. In repeated subtraction, we continuously subtract a number from a bigger number. It continues until we get the 0 or any other number less than the actual number as a remainder.
However, it can be a lengthy process, so we can use division as a shortcut.
### Can I check the remainder and the quotient in a division problem?
Yes, you can quickly check the remainder and quotient in a division problem by using this relationship:
Dividend = Divisor x Quotient + Remainder
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.2.2: Initial Value Problem of Cauchy
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
Consider again the quasilinear equation
($$\star$$) $$a_1(x,y,u)u_x+a_2(x,y,u)u_y=a_3(x,y,u)$$.
Let
$$\Gamma:\ \ x=x_0(s),\ y=y_0(s),\ z=z_0(s), \ s_1\le s\le s_2,\ -\infty<s_1<s_2<+\infty$$
be a regular curve in $$\mathbb{R}^3$$ and denote by $$\mathcal{C}$$ the orthogonal projection of $$\Gamma$$ onto the $$(x,y)$$-plane, i. e.,
$$\mathcal{C}:\ \ x=x_0(s),\ \ y=y_0(s).$$
Initial value problem of Cauchy: Find a $$C^1$$-solution $$u=u(x,y)$$ of $$(\star)$$ such that $$u(x_0(s),y_0(s))=z_0(s)$$, i. e., we seek a surface $$\mathcal{S}$$ defined by $$z=u(x,y)$$ which contains the curve $$\Gamma$$.
Figure 2.2.2.1: Cauchy initial value problem
Definition. The curve $$\Gamma$$ is said to be non-characteristic if
$$x_0'(s)a_2(x_0(s),y_0(s))-y_0'(s)a_1(x_0(s),y_0(s))\not=0.$$
Theorem 2.1. Assume $$a_1,\ a_2,\ a_2\in C^1$$ in their arguments, the initial data $$x_0,\ y_0,\ z_0\in C^1[s_1,s_2]$$ and $$\Gamma$$ is non-characteristic.
Then there is a neighborhood of $$\cal{C}$$ such that there exists exactly one solution $$u$$ of the Cauchy initial value problem.
Proof. (i) Existence. Consider the following initial value problem for the system of characteristic equations to ($$\star$$):
\begin{eqnarray*}
x'(t)&=&a_1(x,y,z)\\
y'(t)&=&a_2(x,y,z)\\
z'(t)&=&a_3(x,y,z)
\end{eqnarray*}
with the initial conditions
\begin{eqnarray*}
x(s,0)&=&x_0(s)\\
y(s,0)&=&y_0(s)\\
z(s,0)&=&z_0(s).
\end{eqnarray*}
Let $$x=x(s,t)$$, $$y=y(s,t)$$, $$z=z(s,t)$$ be the solution, $$s_1\le s\le s_2$$, $$|t|<\eta$$ for an $$\eta>0$$. We will show that this set of curves, see Figure 2.2.2.1, defines a surface. To show this, we consider the inverse functions $$s=s(x,y)$$, $$t=t(x,y)$$ of $$x=x(s,t)$$, $$y=y(s,t)$$ and show that $$z(s(x,y),t(x,y))$$ is a solution of the initial problem of Cauchy. The inverse functions $$s$$ and $$t$$ exist in a neighborhood of $$t=0$$ since
$$\det \frac{\partial(x,y)}{\partial(s,t)}\Big|_{t=0}= \left|\begin{array}{cc}x_s&x_t\\y_s&y_t\end{array}\right|_{t=0} =x_0'(s)a_2-y_0'(s)a_1\not=0,$$
and the initial curve $$\Gamma$$ is non-characteristic by assumption.
Set
$$u(x,y):=z(s(x,y),t(x,y)),$$
then $$u$$ satisfies the initial condition since
$$u(x,y)|_{t=0}=z(s,0)=z_0(s).$$
The following calculation shows that $$u$$ is also a solution of the differential equation ($$\star$$).
\begin{eqnarray*}
a_1u_x+a_2u_y&=&a_1(z_ss_x+z_tt_x)+a_2(z_ss_y+z_tt_y)\\
&=&z_s(a_1s_x+a_2s_y)+z_t(a_1t_x+a_2t_y)\\
&=&z_s(s_xx_t+s_yy_t)+z_t(t_xx_t+t_yy_t)\\
&=&a_3
\end{eqnarray*}
since $$0=s_t=s_xx_t+s_yy_t$$ and $$1=t_t=t_xx_t+t_yy_t$$.
(ii) Uniqueness. Suppose that $$v(x,y)$$ is a second solution. Consider a point $$(x',y')$$ in a neighborhood of the curve $$(x_0(s),y(s))$$, $$s_1-\epsilon\le s\le s_2+\epsilon$$, $$\epsilon>0$$ small. The inverse parameters are $$s'=s(x',y')$$, $$t'=t(x',y')$$, see Figure 2.2.2.2.
Figure 2.2.2.2: Uniqueness proof
Let
$${\mathcal{A}}:\ \ x(t):=x(s',t),\ y(t):=y(s',t),\ z(t):=z(s',t)$$
be the solution of the above initial value problem for the characteristic differential equations with the initial data
$$x(s',0)=x_0(s'),\ y(s',0)=y_0(s'),\ z(s',0)=z_0(s').$$
According to its construction this curve is on the surface $$\mathcal{S}$$ defined by $$u=u(x,y)$$ and $$u(x',y')=z(s',t')$$. Set
$$\psi(t):=v(x(t),y(t))-z(t),$$
then
\begin{eqnarray*}
\psi'(t)&=&v_xx'+v_yy'-z'\\
&=&x_xa_1+v_ya_2-a_3=0
\end{eqnarray*}
and
$$\psi(0)=v(x(s',0),y(s',0))-z(s',0) =0$$
since $$v$$ is a solution of the differential equation and satisfies the initial condition by assumption. Thus, $$\psi(t)\equiv0$$, i. e.,
$$v(x(s',t),y(s',t))-z(s',t)=0.$$
Set $$t=t'$$, then
$$v(x',y')-z(s',t')=0,$$
which shows that $$v(x',y')=u(x',y')$$ because of $$z(s',t')=u(x',y')$$.
$$\Box$$
Remark. In general, there is no uniqueness if the initial curve $$\Gamma$$ is a characteristic curve, see an exercise and Figure 2.2.2.3, which illustrates this case.
Figure 2.2.2.3: Multiple solutions
## Examples
Example 2.2.2.1:
Consider the Cauchy initial value problem
$$u_x+u_y=0$$
with the initial data
$$x_0(s)=s,\ y_0(s)=1,\ z_0(s)\ \mbox{is a given}\ C^1\mbox{-function}.$$
These initial data are non-characteristic since $$y_0'a_1-x_0'a_2=-1$$. The solution of the associated system of characteristic equations
$$x'(t)=1,\ y'(t)=1,\ u'(t)=0$$
with the initial conditions
$$x(s,0)=x_0(s),\ y(s,0)=y_0(s),\ z(s,0)=z_0(s)$$
is given by
$$x=t+x_0(s),\ y=t+y_0(s),\ z=z_0(s) ,$$
i. e.,
$$x=t+s,\ y=t+1,\ z=z_0(s).$$
It follows $$s=x-y+1,\ t=y-1$$ and that $$u=z_0(x-y+1)$$ is the solution of the Cauchy initial value problem.
Example 2.2.2.2:
A problem from kinetics in chemistry. Consider for $$x\ge0$$, $$y\ge0$$ the problem
$$u_x+u_y=\left(k_0e^{-k_1x}+k_2\right)(1-u)$$
with initial data
$$u(x,0)=0,\ x>0,\ \mbox{and}\ u(0,y)=u_0(y),\ y>0.$$
Here the constants $$k_j$$ are positive, these constants define the velocity of the reactions in consideration, and the function $$u_0(y)$$ is given. The variable $$x$$ is the time and $$y$$ is the height of a tube, for example, in which the chemical reaction takes place, and $$u$$ is the concentration of the chemical substance.
In contrast to our previous assumptions, the initial data are not in $$C^1$$. The projection $${\mathcal C}_1\cup {\mathcal C}_2$$ of the initial curve onto the $$(x,y)$$-plane has a corner at the origin, see Figure 2.2.2.4.
Figure 2.2.2.4: Domains to the chemical kinetics example
The associated system of characteristic equations is
$$x'(t)=1,\ y'(t)=1,\ z'(t)=\left(k_0e^{-k_1x}+k_2\right)(1-z).$$
It follows $$x=t+c_1$$, $$y=t+c_2$$ with constants $$c_j$$. Thus the projection of the characteristic curves on the $$(x,y)$$-plane are straight lines parallel to $$y=x$$. We will solve the initial value problems in the domains $$\Omega_1$$ and $$\Omega_2$$, see Figure 2.2.2.4, separately.
(i) The initial value problem in $$\Omega_1$$. The initial data are
$$x_0(s)=s,\ y_0(s)=0, \ z_0(0)=0,\ s\ge 0.$$
It follows
$$x=x(s,t)=t+s,\ y=y(s,t)=t.$$
Thus
$$z'(t)=(k_0e^{-k_1(t+s)}+k_2)(1-z),\ z(0)=0.$$
The solution of this initial value problem is given by
$$z(s,t)=1-\exp\left(\frac{k_0}{k_1}e^{-k_1(s+t)}-k_2t-\frac{k_0}{k_1}e^{-k_1s}\right).$$
Consequently
$$u_1(x,y)=1-\exp\left(\frac{k_0}{k_1}e^{-k_1x}-k_2y-{k_0}{k_1}e^{-k_1(x-y)}\right)$$
is the solution of the Cauchy initial value problem in $$\Omega_1$$. If time $$x$$ tends to $$\infty$$, we get the limit
$$\lim_{x\to\infty} u_1(x,y)=1-e^{-k_2y}.$$
(ii) The initial value problem in $$\Omega_2$$. The initial data are here
$$x_0(s)=0,\ y_0(s)=s, \ z_0(0)=u_0(s),\ s\ge 0.$$
It follows
$$x=x(s,t)=t,\ y=y(s,t)=t+s.$$
Thus
$$z'(t)=(k_0e^{-k_1t}+k_2)(1-z),\ z(0)=0.$$
The solution of this initial value problem is given by
$$z(s,t)=1-(1-u_0(s))\exp\left(\frac{k_0}{k_1}e^{-k_1t}-k_2t-\frac{k_0}{k_1}\right).$$
Consequently
$$u_2(x,y)=1-(1-u_0(y-x))\exp\left(\frac{k_0}{k_1}e^{-k_1x}-k_2x-\frac{k_0}{k_1}\right)$$
is the solution in $$\Omega_2$$.
If $$x=y$$, then
\begin{eqnarray*}
u_1(x,y)&=&1-\exp\left(\frac{k_0}{k_1}e^{-k_1x}-k_2x-\frac{k_0}{k_1}\right)\\
u_2(x,y)&=&1-(1-u_0(0))\exp\left(\frac{k_0}{k_1}e^{-k_1x}-k_2x-\frac{k_0}{k_1}\right).
\end{eqnarray*}
If $$u_0(0)>0$$, then $$u_1<u_2$$ if $$x=y$$, i. e., there is a jump of the concentration of the substrate along its burning front defined by $$x=y$$.
Remark. Such a problem with discontinuous initial data is called Riemann problem. See an exercise for another Riemann problem.
The case that a solution of the equation is known
Here we will see that we get immediately a solution of the Cauchy initial value problem if a solution of the homogeneous linear equation
$$a_1(x,y)u_x+a_2(x,y)u_y=0$$
is known.
Let
$$x_0(s),\ y_0(s),\ z_0(s),\ s_1<s<s_2$$
be the initial data and let $$u=\phi(x,y)$$ be a solution of the differential equation. We assume that
$$\phi_x(x_0(s),y_0(s))x_0'(s)+\phi_y(x_0(s),y_0(s))y_0'(s)\not=0$$
is satisfied. Set
$$g(s)=\phi(x_0(s),y_0(s))$$
and let $$s=h(g)$$ be the inverse function.
The solution of the Cauchy initial problem is given by $$u_0\left(h(\phi(x,y))\right)$$.
This follows since in the problem considered a composition of a solution is a solution again, see an exercise, and since
$$u_0\left(h(\phi(x_0(s),y_0(s))\right)=u_0(h(g))=u_0(s).$$
Example 2.2.2.3:
Consider equation
$$u_x+u_y=0$$
with initial data
$$x_0(s)=s,\ y_0(s)=1,\ u_0(s)\ \mbox{is a given function}.$$
A solution of the differential equation is $$\phi(x,y)=x-y$$. Thus
$$\phi((x_0(s),y_0(s))=s-1$$
and
$$u_0(\phi+1)=u_0(x-y+1)$$
is the solution of the problem.
## Contributors
• Integrated by Justin Marshall.
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# Algebraic Order of Operations Source: Beginning and Intermediate Algebra, an open source (CC-BY) textbook by Tyler Wallace
Objective: Evaluate expressions using the order of operations, including the use of absolute value.
When simplifying expressions it is important that we simplify them in the correct order. Consider the following problem done two different ways:
Example 24
2 + 5 X 3 Add
7 X 3 Multiply
Solution: 21
2 + 5 X 3 Multiply
Solution: 17
The previous example illustrates that if the same problem is done two different ways we will arrive at two different solutions. However, only one method can be correct. It turns out the second method, 17, is the correct method.
The order of operations ends with the most basic of operations, addition (or subtraction). Before addition is completed we must do repeated addition or multiplication (or division). Before multiplication is completed we must do repeated multiplication or exponents.
When we want to do something out of order and make it come first we will put it in parenthesis (or grouping symbols). This list then is our order of operations we will use to simplify expressions.
Order of Operations:
Parenthesis (Grouping)
Exponents
Multiply and Divide (Left to Right)
Add and Subtract (Left to Right)
Multiply and Divide are on the same level because they are the same operation (division is just multiplying by the reciprocal). This means they must be done left to right, so some problems we will divide first, others we will multiply first.
The same is true for adding and subtracting (subtracting is just adding the opposite). Often students use the word PEMDAS to remember the order of operations, as the first letter of each operation creates the word PEMDAS.
Another way students remember the order of operations is to think of a phrase such as "Please Excuse My Dear Aunt Sally" where each word starts with the same letters as the order of operations start with.
Example 25
2 + 3(9 - 4)2 Parenthesis First
2 + 3(5)2 Exponents
2 + 3(25) Multiply
Solution 77
It is very important to remember to multiply and divide from from left to right!
Example 26
30 / 3 X 2 Divide first (left to right!)
10 X 2 Multiply
Solution 20
In the previous example, if we had multiplied first, 5 would have been the answer which is incorrect.
If there are several parenthesis in a problem we will start with the inner most parenthesis and work our way out. Inside each parenthesis we simplify using the order of operations as well.
To make it easier to know which parenthesis goes with which parenthesis, different types of parenthesis will be used such as { } and [ ] and ( ), these parenthesis all mean the same thing, they are parenthesis and must be evaluated first.
Example 27
2{82 - 7[32 -4(32 +1)](-1)} Innermost parenthesis, exponents first
2{82 - 7[32 - 4(9 + 1)](-1)} Add inside those parenthesis
2{82 - 7[32 - 4(10)](-1)} Multiply inside innermost parenthesis
2{82 -7 [32 - 40](-1)} Subtract inside those parenthesis
2{82 -7[-8](-1)} Exponents next
2{64 - 7[-8](-1)} Multiply left to right, sign with the number
2{64 + 56(-1)} Finish multiplying
2{64 - 56} Subtract inside parenthesis
2{8} Multiply
Solution: 16
As the above example illustrates, it can take several steps to complete a problem. The key to successfully solve order of operations problems is to take the time to show your work and do one step at a time. This will reduce the chance of making a mistake along the way.
There are several types of grouping symbols that can be used besides parenthesis. One type is a fraction bar. If we have a fraction, the entire numerator and the entire denominator must be evaluated before we reduce the fraction. In these cases we can simplify in both the numerator and denominator at the same time.
Example 28
24 - (-8) X 3
15 / 5 - 1
Exponent in the numerator, divide in denominator.
16 - (-8) X 3
3 - 1
Multiply in the numerator, subtract in denominator.
16 - (-24)
2
Add the opposite to simplify numerator, denominator is done.
40 / 2
Reduce, divide
Solution: 20
Another type of grouping symbol that also has an operation with it, absolute value. When we have absolute value we will evaluate everything inside the absolute value, just as if it were a normal parenthesis. Then once the inside is completed we will take the absolute value, or distance from zero, to make the number positive.
Example 29
1 + 3|-42 - (-8)| + 2|3 + (-5)2|
Evaluate absolute values first, exponents
1 + 3|-16 - (-8)| + 2|3 + 25|
1 + 3|-8| + 2|28|
Evaluate absolute values
1 + 3(8) + 2(28)
Multiply left to right
1 + 24 + 2(28)
Finish multiplying
1 + 24 + 56
25 + 56
Solution: 81
The above example also illustrates an important point about exponents. Exponents are considered to be only on the number they are attached to. This means when we see -42, only the 4 is squared, giving us -(42) or -16.
But when the negative is in parentheses, such as (-5)2 the negative is part of the number and is also squared giving us a positive solution, 25.
Practice - Order of Operation
Solve:
1. -6 X 4(-1)
2. (-6 / 6)3
3. 3 + (8) / |4|
4. 5(-5 + 6) X 62
5. 8 / 4 X 2
6. 7 - 5 + 6
7. [-9 - (2 - 5)] / (-6)
8. (-2 X 23 X 2) / (-4)
9. -6 + (-3 - 3)2 / |3|
10. (-7 - 5) / [-2 - 2 - (-6)]
11. 4 - 2|32 - 16|
12.
-10 - 6
-5
(-2)2
13. [-1 - (-5)] |3 + 2|
14. -3 - {3 -[-3(2 + 4) - (-2)]}
15.
2 + 4|7 + 22|
4 X 2 + 5 X 3
16. -4 - [2 + 4(-6) - 4 -|22 - 5 X 2|]
17. [6 X 2 + 2 -(-6)](-5 +|-18 / 6|)
18. 2 X (-3) + 3 - 6[-2 - (-1 - 3)]
19.
-13 - 2
2 - (-1)3 + (-6) - [-1 - (-3)]
20.
-52 + (-5)2
|42 - 25| - 2 X 3
21.
-8 - 4 + (-4) - [-4 - (-3)]
6 X
(42 + 32) / 5
22.
-9 X 2 - ( 3 - 6)
1 - (-2 + 1) - (-3)
23.
23 + 4
-18 - 6 + (-4) - [-5(-1)(-5)]
24.
13 + (-3)2 + 4(-3) +1 - [-10 - (-6)]
{[4 + 5] / [42 - 32(4 - 3) -8]} + 12
25.
5 + 32 - 24 / 6 X 2
[5 + 3 (22 - 5)] + |22 - 5|2
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# How to Calculate Arithmetic Average: The Very Basics
## Arithmetic average basics
Arithmetic average, or arithmetic mean, or just mean, is the very basic statistical measure. It provides quick and easy information about general level of values in a data set – it is one of the measures of central tendency.
## Why to calculate and use mean
When you have a set of data, it is sometimes difficult to tell what the values are in general (you can’t see the forest for the trees). For example, you have 10 stocks. Their annual returns in the last year were: 11%, -5%, 17%, 1%, -9%, 21%, 4%, -6%, 7%, and -1%.
The information provided in this form will tell you the details, but you will have to think for a while to get an idea regarding the annual return of this group of stocks as a whole. When the arithmetic average is provided in addition to this information, it can save you the thinking. The mean (arithmetic average) return of our basket of 10 stocks in the last year was 4%. This information is already quite clear and easy to work with.
## Arithmetic average calculation
The calculation of arithmetic average is straightforward. You sum up all the values and then divide the sum by the number of values. Let’s use the example above to illustrate the calculation of mean:
• First you sum up the annual returns: 11% + (-5%) + 17% + 1% + (-9%) + 21% + 4% + (-6%) + 7% + (-1%) = 40%
• Then you divide the sum (which is +40%) by the number of observations (which is 10), and you get the arithmetic average, which is +4% in this case.
## Calculating arithmetic average in Excel
Though the calculation is very simple, it can be boring and prone to errors when you work with large sets of data (imagine calculating the average return of the 500 stocks in S&P500 like this). Computers calculate arithmetic average for us. In Microsoft Excel, you can use the function AVERAGE. The parameter is the area of the cells where you have the individual values.
## Disadvantages of arithmetic average and using other averages
Though arithmetic average is easy and elegant for the first quick information about a data set, it has weaknesses and sometimes it is better to use one of the other measures of central tendency, like geometric average, weighted average, median, or mode.
Furthermore, knowing the general level of values is often not enough. You may also want to measure volatility or dispersion (using standard deviation or variance) and many other characteristics.
You can easily calculate arithmetic average, median, variance, standard deviation and other measures using the Descriptive Statistics Excel Calculator.
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# Discrete Mathematics - Probability
Closely related to the concepts of counting is Probability. We often try to guess the results of games of chance, like card games, slot machines, and lotteries; i.e. we try to find the likelihood or probability that a particular result with be obtained.
Probability can be conceptualized as finding the chance of occurrence of an event. Mathematically, it is the study of random processes and their outcomes. The laws of probability have a wide applicability in a variety of fields like genetics, weather forecasting, opinion polls, stock markets etc.
## Basic Concepts
Probability theory was invented in the 17th century by two French mathematicians, Blaise Pascal and Pierre de Fermat, who were dealing with mathematical problems regarding of chance.
Before proceeding to details of probability, let us get the concept of some definitions.
Random Experiment − An experiment in which all possible outcomes are known and the exact output cannot be predicted in advance is called a random experiment. Tossing a fair coin is an example of random experiment.
Sample Space − When we perform an experiment, then the set S of all possible outcomes is called the sample space. If we toss a coin, the sample space $S = \left \{ H, T \right \}$
Event − Any subset of a sample space is called an event. After tossing a coin, getting Head on the top is an event.
The word "probability" means the chance of occurrence of a particular event. The best we can say is how likely they are to happen, using the idea of probability.
$Probability\:of\:occurence\:of\:an\:event = \frac{Total\:number\:of\:favourable \: outcome}{Total\:number\:of\:Outcomes}$
As the occurrence of any event varies between 0% and 100%, the probability varies between 0 and 1.
### Steps to find the probability
Step 1 − Calculate all possible outcomes of the experiment.
Step 2 − Calculate the number of favorable outcomes of the experiment.
Step 3 − Apply the corresponding probability formula.
### Tossing a Coin
If a coin is tossed, there are two possible outcomes − Heads $(H)$ or Tails $(T)$
So, Total number of outcomes = 2
Hence, the probability of getting a Head $(H)$ on top is 1/2 and the probability of getting a Tails $(T)$ on top is 1/2
### Throwing a Dice
When a dice is thrown, six possible outcomes can be on the top − $1, 2, 3, 4, 5, 6$.
The probability of any one of the numbers is 1/6
The probability of getting even numbers is 3/6 = 1/2
The probability of getting odd numbers is 3/6 = 1/2
### Taking Cards From a Deck
From a deck of 52 cards, if one card is picked find the probability of an ace being drawn and also find the probability of a diamond being drawn.
Total number of possible outcomes − 52
Outcomes of being an ace − 4
Probability of being an ace = 4/52 = 1/13
Probability of being a diamond = 13/52 = 1/4
## Probability Axioms
• The probability of an event always varies from 0 to 1. $[0 \leq P(x) \leq 1]$
• For an impossible event the probability is 0 and for a certain event the probability is 1.
• If the occurrence of one event is not influenced by another event, they are called mutually exclusive or disjoint.
If $A_1, A_2....A_n$ are mutually exclusive/disjoint events, then $P(A_i \cap A_j) = \emptyset$ for $i \ne j$ and $P(A_1 \cup A_2 \cup.... A_n) = P(A_1) + P(A_2)+..... P(A_n)$
## Properties of Probability
• If there are two events $x$ and $\overline{x}$which are complementary, then the probability of the complementary event is −
$$p(\overline{x}) = 1-p(x)$$
• For two non-disjoint events A and B, the probability of the union of two events −
$P(A \cup B) = P(A) + P(B)$
• If an event A is a subset of another event B (i.e. $A \subset B$), then the probability of A is less than or equal to the probability of B. Hence, $A \subset B$ implies $P(A) \leq p(B)$
## Conditional Probability
The conditional probability of an event B is the probability that the event will occur given an event A has already occurred. This is written as $P(B|A)$.
Mathematically − $P(B|A) = P(A \cap B)/ P(A)$
If event A and B are mutually exclusive, then the conditional probability of event B after the event A will be the probability of event B that is $P(B)$.
Problem 1
In a country 50% of all teenagers own a cycle and 30% of all teenagers own a bike and cycle. What is the probability that a teenager owns bike given that the teenager owns a cycle?
Solution
Let us assume A is the event of teenagers owning only a cycle and B is the event of teenagers owning only a bike.
So, $P(A) = 50/100 = 0.5$ and $P(A \cap B) = 30/100 = 0.3$ from the given problem.
$P(B|A) = P(A \cap B)/ P(A) = 0.3/ 0.5 = 0.6$
Hence, the probability that a teenager owns bike given that the teenager owns a cycle is 60%.
Problem 2
In a class, 50% of all students play cricket and 25% of all students play cricket and volleyball. What is the probability that a student plays volleyball given that the student plays cricket?
Solution
Let us assume A is the event of students playing only cricket and B is the event of students playing only volleyball.
So, $P(A) = 50/100 =0.5$ and $P(A \cap B) = 25/ 100 =0.25$ from the given problem.
$P\lgroup B\rvert A \rgroup= P\lgroup A\cap B\rgroup/P\lgroup A \rgroup =0.25/0.5=0.5$
Hence, the probability that a student plays volleyball given that the student plays cricket is 50%.
Problem 3
Six good laptops and three defective laptops are mixed up. To find the defective laptops all of them are tested one-by-one at random. What is the probability to find both of the defective laptops in the first two pick?
Solution
Let A be the event that we find a defective laptop in the first test and B be the event that we find a defective laptop in the second test.
Hence, $P(A \cap B) = P(A)P(B|A) =3/9 \times 2/8 = 1/12$
## Bayes' Theorem
Theorem − If A and B are two mutually exclusive events, where $P(A)$ is the probability of A and $P(B)$ is the probability of B, $P(A | B)$ is the probability of A given that B is true. $P(B | A)$ is the probability of B given that A is true, then Bayes’ Theorem states −
$$P(A|B) = \frac{P(B|A) P(A)}{\sum_{i = 1}^{n}P(B|Ai)P(Ai)}$$
### Application of Bayes' Theorem
• In situations where all the events of sample space are mutually exclusive events.
• In situations where either $P( A_i \cap B )$ for each $A_i$ or $P( A_i )$ and $P(B|A_i)$ for each $A_i$ is known.
Problem
Consider three pen-stands. The first pen-stand contains 2 red pens and 3 blue pens; the second one has 3 red pens and 2 blue pens; and the third one has 4 red pens and 1 blue pen. There is equal probability of each pen-stand to be selected. If one pen is drawn at random, what is the probability that it is a red pen?
Solution
Let $A_i$ be the event that ith pen-stand is selected.
Here, i = 1,2,3.
Since probability for choosing a pen-stand is equal, $P(A_i) = 1/3$
Let B be the event that a red pen is drawn.
The probability that a red pen is chosen among the five pens of the first pen-stand,
$P(B|A_1) = 2/5$
The probability that a red pen is chosen among the five pens of the second pen-stand,
$P(B|A_2) = 3/5$
The probability that a red pen is chosen among the five pens of the third pen-stand,
$P(B|A_3) = 4/5$
According to Bayes' Theorem,
$P(B) = P(A_1).P(B|A_1) + P(A_2).P(B|A_2) + P(A_3).P(B|A_3)$
$= 1/3 . 2/5\: +\: 1/3 . 3/5\: +\: 1/3 . 4/5$
$= 3/5$
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Angles and Parallel Lines MathBitsNotebook.com Terms of Use Contact Person: Donna Roberts
When a transversal intersects two or more lines in the same plane, a series of angles are formed. Certain pairs of angles are given specific "names" based upon their locations in relation to the lines. These specific names may be used whether the lines involved are parallel or not parallel.
"Names" given to pairs of angles:
• alternate interior angles • alternate exterior angles • corresponding angles • interior angles on the same side of the transversal
Let's examine these pairs of angles in relation to parallel lines:
Alternate Interior Angles: The word "alternate" means "alternating sides" of the transversal. This name clearly describes the "location" of these angles. When the lines are parallel, the measures are equal.
∠1 and ∠2 are alternate interior angles ∠3 and ∠4 are alternate interior angles Alternate interior angles are "interior" (between the parallel lines), and they "alternate" sides of the transversal. Notice that they are not adjacent angles (next to one another sharing a vertex). When the lines are parallel, the alternate interior angles are equal in measure. m∠1 = m∠2 and m∠3 = m∠4
If you draw a Z on the diagram, the alternate interior angles can be found in the corners of the Z. The Z may also be backward:.
If two parallel lines are cut by a transversal, the alternate interior angles are congruent.
Converse If two lines are cut by a transversal and the alternate interior angles are congruent, the lines are parallel.
Alternate Exterior Angles: The word "alternate" means "alternating sides" of the transversal. The name clearly describes the "location" of these angles. When the lines are parallel, the measures are equal.
∠1 and ∠2 are alternate exterior angles ∠3 and ∠4 are alternate exterior angles Alternate exterior angles are "exterior" (outside the parallel lines), and they "alternate" sides of the transversal. Notice that, like the alternate interior angles, these angles are not adjacent. When the lines are parallel, the alternate exterior angles are equal in measure. m∠1 = m∠2 and m∠3 = m∠4
If two parallel lines are cut by a transversal, the alternate exterior angles are congruent.
Converse If two lines are cut by a transversal and the alternate exterior angles are congruent, the lines are parallel.
Corresponding Angles: The name does not clearly describe the "location" of these angles. The angles are on the SAME SIDE of the transversal, one INTERIOR and one EXTERIOR, but not adjacent. The angles lie on the same side of the transversal in "corresponding" positions. When the lines are parallel, the measures are equal.
∠1 and ∠2 are corresponding angles ∠3 and ∠4 are corresponding angles ∠5 and ∠6 are corresponding angles ∠7 and ∠8 are corresponding angles If you copy one of the corresponding angles and you translate it along the transversal, it will coincide with the other corresponding angle. For example, slide ∠ 1 down the transversal and it will coincide with ∠2. When the lines are parallel, the corresponding angles are equal in measure. m∠1 = m∠2 and m∠3 = m∠4 m∠5 = m∠6 and m∠7 = m∠8
If you draw a F on the diagram, the corresponding angles can be found in the corners of the F. The F may also be backward and/or upside-down: .
If two parallel lines are cut by a transversal, the corresponding angles are congruent.
Converse If two lines are cut by a transversal and the corresponding angles are congruent, the lines are parallel.
Interior Angles on the Same Side of the Transversal: The name is a description of the "location" of the these angles. When the lines are parallel, the measures are supplementary.
∠1 and ∠2 are interior angles on the same side of transversal ∠3 and ∠4 are interior angles on the same side of transversal These angles are located exactly as their name describes. They are "interior" (between the parallel lines), and they are on the same side of the transversal. When the lines are parallel, the interior angles on the same side of the transversal are supplementary. m∠1 + m∠2 = 180 m∠3 + m∠4 = 180
If two parallel lines are cut by a transversal, the interior angles on the same side of the transversal are supplementary.
Converse If two lines are cut by a transversal and the interior angles on the same side of the transversal are supplementary, the lines are parallel.
In addition to the 4 pairs of named angles that are used when working with parallel lines (listed above), there are also some pairs of "old friends" that are also working in parallel lines.
Vertical Angles: When straight lines intersect, vertical angles appear. Vertical angles are ALWAYS equal in measure, whether the lines are parallel or not.
There are 4 sets of vertical angles in this diagram! ∠1 and ∠2 ∠3 and ∠4 ∠5 and ∠6 ∠7 and ∠8 Remember: the lines need not be parallel to have vertical angles of equal measure.
Vertical angles are congruent.
Linear Pair Angles: A linear pair are two adjacent angles forming a straight line. Angles forming a linear pair are ALWAYS supplementary.
Since a straight angle contains 180º, the two angles forming a linear pair also contain 180º when their measures are added (making them supplementary). m∠1 + m∠4 = 180 m∠1 + m∠3 = 180 m∠2 + m∠4 = 180 m∠2 + m∠3 = 180 m∠5 + m∠8 = 180 m∠5 + m∠7 = 180 m∠6 + m∠8 = 180 m∠6 + m∠7 = 180
If two angles form a linear pair, they are supplementary.
NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use".
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Wednesday , 28 June 2017
Home » Mathematics For Kids » Symmetry in Mathematics
# Symmetry in Mathematics
Symmetry in Mathematics – Symmetry comes from a Greek word meaning ‘to measure together’ and is widely used in the study of geometry. Mathematically, symmetry means that one shape becomes exactly like another when you move it in some way (turn, flip or slide).
#### Types of Symmetry
There are three basic types of symmetry: rotational symmetry, reflection symmetry and point symmetry.
1. Reflection Symmetry
The simplest symmetry is Reflection Symmetry (sometimes called Line Symmetry or Mirror Symmetry). It is easy to see, because one half is the reflection of the other half.
2. Rotational Symmetry
With Rotational Symmetry, the image is rotated (around a central point) so that it appears 2 or more times. How many times it appears is called the Order.
3. Point Symmetry
Point Symmetry is when every part has a matching part:
• the same distance from the central point
• but in the opposite direction.
#### Beauty of Mathematics
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321
1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 +10= 1111111111
9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888
And finally, take a look at this symmetry:
1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111=12345678987654321
Brilliant, isn’t it?
|
# Class 10 Maths MCQ – Nth Term of Arithmetic Progression
This set of Class 10 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Nth Term of Arithmetic Progression”.
1. What will be the nth term of the AP 1, 5, 9, 13, 17…….?
a) 4n – 3
b) 3n – 4
c) 4n + 3
d) 3n + 4
Explanation: Here a = 1 and d = 4
The nth term of the AP = a + (n – 1)d = 1 + (n – 1)4 = 1 + 4n – 4 = 4n – 3
2. If the nth term of the AP is 2n + 7, then the common difference will be _____
a) -2
b) 2
c) 3
d) -3
Explanation: nth term of the AP is 2n + 7
Tn = 2n + 7
Common difference = T2 – T1 = (2 × 2 + 7 – (2 × 1 + 7)) = 11 – 9 = 2
3. If the nth term of the AP is 7n – 9, then the first term is ________
a) 1
b) 6
c) 3
d) 4
Explanation: nth term of the AP is 7n – 9
Tn = 7n – 1
T1 = 7(1) – 1 = 6
The first term of AP is 6.
4. If the nth term of the AP is 8n + 1, then the 20th term will be ______
a) 160
b) 120
c) 161
d) 121
Explanation: nth term of the AP is 8n + 1
Tn = 8n + 1
T20 = 8(20) + 1
T20 = 161
5. What will be the 99th term from the end of the AP 500, 489, 478, 467… – 1139?
a) 1078
b) 1123
c) 12
d) 61
Explanation: In this case since we have to find the 99th term from the end.
We will consider the first term to be -1139 and the common difference will be 11
Now, a = -1139, d = 11 and n = 99
T99 = a + (n – 1)d
T99 = -1139 + (99 – 1)11
T99 = -1139 + 1078 = -61
The value of 99th term from the end is 61.
Note: Join free Sanfoundry classes at Telegram or Youtube
6. If the pth term of an AP is q and its qth term is p, then what will be the value of its (p + q)th term?
a) 1
b) p + q – 1
c) 0
d) 2(p + q – 1)
Explanation: pth term = q
a + (p – 1)d = q
a + pd – d = q (1)
qth term = p
a + (q – 1)d = p
a + qd – d = p (2)
Subtracting (2) from (1) we get,
a + qd – d – (a + pd – d) = p – q
qd – pd = p – q
d = -1
Substituting in equation 1, we get,
a = p + q – 1
(p + q)th term = a + (n – 1)d = p + q – 1 + (p + q – 1)(-1) = p + q – 1 – p – q + 1 = 0
7. If 5 times the 5th term of an AP is equal to 15 times its 15th term, then the value of its 20th term will be _______
a) 0
b) 1
c) 2
d) 3
Explanation: 5(5th term) = 15(15th term)
5th term = a + (5 – 1)d = a + 4d
15th term = a + (15 – 1)d = a + 14d
5(a + 4d) = 15(a + 14d)
5a + 20d = 15a + 210d
20d – 210d = 15a – 5a
-190d = 10a
a = -19d
Now, the 20th term = a + (20 – 1)d = a + 19d
But a = -19d
Hence, -19d + 19d = 0
8. If the 11th term of an AP is $$\frac {1}{13}$$ and its 13th term is $$\frac {1}{11}$$, then what will be the value of 143th term?
a) $$\frac {1}{143}$$
b) 1
c) 0
d) $$\frac {23}{143}$$
Explanation: Here 11th term = $$\frac {1}{13}$$
13thterm = $$\frac {1}{11}$$
Let the first term of the AP be a and common difference be d
T11 = a + (n – 1)d = $$\frac {1}{13}$$
T11 = a + (11 – 1)d = $$\frac {1}{13}$$
T11 = a + 10d = $$\frac {1}{13}$$ (1)
T13 = a + (n – 1)d = $$\frac {1}{11}$$
T13 = a + (13 – 1)d = $$\frac {1}{11}$$
T13 = a + 12d = $$\frac {1}{11}$$ (2)
Subtracting (1) from (2)
We get,
a + 12d – (a + 10d) = $$\frac {1}{11} – \frac {1}{13}$$
2d = $$\frac {2}{143}$$
d = $$\frac {1}{143}$$
Now, substituting value of d in equation 1
We get,
T11 = a + 10($$\frac {1}{143}$$) = $$\frac {1}{13}$$
a = $$\frac {1}{143}$$
The 143th term = $$\frac {1}{143}$$ + 142$$\frac {1}{143} = \frac {(1+142)}{143}$$ = 1
9. If the 7th term of an AP is 20 and its 11th term is 40 then, what will be the common difference?
a) 3
b) 4
c) 5
d) 2
Explanation: Here 7th term = 20
11thterm = 40
Let the first term of the AP be a and common difference be d
T7 = a + (n – 1)d = 20
T7 = a + (7 – 1)d = 20
T7 = a + 6d = 20 (1)
T11 = a + (n – 1)d = 40
T11 = a + (11 – 1)d = 40
T11 = a + 10d = 40 (2)
Subtracting (1) from (2)
We get,
a + 10d – (a + 6d) = 40 – 20
4d = 20
d = 5
10. What will be the 14th term of the AP 5, 8, 11, 14, 17…….?
a) 30
b) 41
c) 40
d) 44
Explanation: Here a = 5, d = 8 – 5 = 3 and n = 14
T14 = a + (n – 1)d
T14 = 5 + (14 – 1)3
T14 = 5 + 13 × 3
T14 = 44
Sanfoundry Global Education & Learning Series – Mathematics – Class 10.
To practice all chapters and topics of class 10 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
|
# Linear Inequalities in Two Variables Objectives: Solve and graph a linear inequality in two variables..
## Presentation on theme: "Linear Inequalities in Two Variables Objectives: Solve and graph a linear inequality in two variables.."— Presentation transcript:
Linear Inequalities in Two Variables Objectives: Solve and graph a linear inequality in two variables..
A linear inequality in two variables, x and y, is any inequality that can be written in one of the forms below, where A ≠ 0 and B ≠ 0. Ax + By ≥ C Ax + By > C Ax + By ≤ C Ax + By < C
A solution of a linear inequality in two variables, x and y, is an ordered pair (x, y) that satisfies the inequality. The solution to a linear inequality is a region of the coordinate plane and is called a half-plane bounded by a boundary line.
Graphing Linear Inequalities 1. Given a linear inequality in two variables, graph its related linear equation. For inequalities involving ≤ or ≥, use a solid boundary line. For inequalities involving, use a dashed boundary line.
2. Shade the appropriate region. For inequalities in the form of y ≤ mx + b or y < mx + b, shade below the boundary line. For inequalities of the form y ≥ mx + b or y > mx + b, shade above the boundary line. For inequalities in the form x ≤ c or x < c, shade to the left of the boundary line. For inequalities in the form x ≥ c or x > c, shade to the right of the boundary line.
Ex 1. Graph each linear inequality. a. y < x + 2
b. y ≥ -2x + 3
* c. y > -2x - 2 Dotted Line
d. y ≥ 2x + 5
e. -2x –3y ≤ 3
f. 3x – 4y ≥ 4 -4y≥-3x + 4 y ≤ ¾ x - 1
g. -5x – 2y > 4 -2y > 5x + 4 y < -5/2 x - 2 Dotted Line
Ex 3. Graph each linear inequality. x is a vertical line and y is a horizontal line
a. x > -2
b. y ≤ -1
c. x ≤ -2
d. y > -1 Dotted Line
Download ppt "Linear Inequalities in Two Variables Objectives: Solve and graph a linear inequality in two variables.."
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Venn Diagrams and Two-Way Tables - Word Problems
Lesson
We've seen Venn diagram and Two-Way table word problems before, but let's take a look at a few more challenging questions and how to solve them.
Example 1:
$200$200 people were questioned about whether they read the newspaper online or in paper form.
• $20$20% said they did both
• $\frac{3}{10}$310 of those surveyed said they read the paper version
• Of those who who didn't read the paper version, the probability that they read an online version was $50$50%
a) Complete a two-way table containing this information
Think: To work out the first two pieces of information we can find $20$20% and $30$30% of $200$200 people respectively.
Do:
Paper Not Paper Total
Online 40
Not Online 20
Total 60 140 200
To use the third point we are best off using the formula for conditional probability.
Now we can fill in the rest of the table.
Paper Not Paper Total
Online 40 70 110
Not Online 20 70 90
Total 60 140 200
b) Of those who read the paper online, what proportion also read a paper version?
Think: Notice that this is a conditional probability question.
Do: $\frac{40}{110}$40110
We'll now do a similar question, but using a Venn diagram.
Example 2:
$400$400 people were questioned about whether they make or buy their bread.
• $2$2% did neither
• $30$30% did both
• The probability that a person didn't make their own bread, given that they bought their bread was $\frac{5}{9}$59
Construct a Venn diagram with this information
Think: We can easily fit the first two pieces of information into our Venn Diagram.
Do:
To use the third piece of information is a little more complicated. Not only will we again need to use the rule for conditional probability, but we'll also need to introduce $x$x.
Worked examples
Question 1
At a university there are $816$816 students studying first year engineering, $497$497 of whom are female (set $F$F). $237$237 of these women are studying Civil Engineering, and there are $348$348 students studying Civil Engineering altogether (set $C$C).
1. State the value of $w$w in the diagram.
2. State the value of $x$x in the diagram.
3. State the value of $y$y in the diagram.
4. State the value of $z$z in the diagram.
5. What is the probability that a randomly selected male student does not study Civil Engineering?
Question 2
$87$87 people are questioned about whether they own a tablet ($T$T) or a smartphone ($S$S). The probabilities shown in the list below were determined from the results.
• $P\left(T\mid S\right)=$P(TS)=$\frac{5}{12}$512
• $P\left(S\cap T'\right)=$P(ST)=$\frac{35}{87}$3587
• $P\left(T\right)=$P(T)=$\frac{14}{29}$1429
1. Find the value of $n\left(S\cap T\right)$n(ST).
Use $Y=n\left(S\cap T\right)$Y=n(ST) and $X=n\left(S\cap T'\right)$X=n(ST) to help you in your calculations.
2. Calculate $P\left(S'\cap T\right)$P(ST).
3. Calculate $P\left(S\mid T\right)$P(ST).
4. Calculate $P\left(T\mid S'\right)$P(TS).
Question 3
$531$531 people are asked whether they watch My Kitchen Rules ($MKR$MKR) or Masterchef ($MC$MC).
$177$177 people watch both and $65$65 watch neither. The number who watch $MKR$MKR is twice the number who watch both.
1. How many people only watch $MKR$MKR?
2. Of the people who watch $MC$MC, what proportion also watch $MKR$MKR?
3. Of those who don’t watch $MC$MC, what proportion watch neither?
|
# Evaluate $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{x}\Big)}}{x-2}}$
A trigonometric function $\cos{\Big(\dfrac{\pi}{x}\Big)}$ and an algebraic function $x-2$ formed a fractional function by division. The limit of this algebraic trigonometric function has to evaluate as $x$ approaches $2$ in this trigonometric limit problem.
Let us try to evaluate the limit of the trigonometric function by the direct substitution method as $x$ approaches $2$.
$= \,\,\,$ $\dfrac{\cos{\Big(\dfrac{\pi}{2}\Big)}}{2-2}$
The cosine of angle $90$ degrees is zero as per trigonometry.
$= \,\,\,$ $\dfrac{0}{0}$
It is an indeterminate form. So, the direct substitution method is failed in calculating the limit of the given function as $x$ approaches $2$. So, we have to evaluate the limit of the function in another approach. The trigonometric limit problem can be done in two methods.
### Method – 1
It is given that $x \,\to\, 2$, then $x-2 \,\to\, 2-2$. Therefore, $x-2 \,\to\, 0$. It clears that $x-2$ approaches zero when $x$ approaches $2$.
$= \,\,\,$ $\displaystyle \large \lim_{x-2 \,\to\, 0}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{x}\Big)}}{x-2}}$
#### Convert the whole function in another variable
Now, take $y = x-2$, then $x = y+2$ mathematically. Finally, transform the entire function in terms of $y$ from $x$.
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{y+2}\Big)}}{y}}$
You will get indeterminate form if you try to evaluate this trigonometric function as $y$ approaches zero by the direct substitution method. So, we have to take a different step here.
#### Express the cos function in terms of sine
In calculus, there is no trigonometric limit rule in cosine but we have a trigonometric limit rule in sine. So, we have to express the cosine function in terms of sine for moving ahead in simplification. It can be done by the cofunction identity of sine.
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{y+2}\Big)}}{y}}$
#### Simplify the Algebra trigonometric function
The algebraic trigonometric function is almost same as the trigonometric limit rule of sine function. So, let us try to adjust the whole function for transforming the current function into required form.
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi(y+2)-\pi(2)}{2(y+2)}\Big)}}{y}}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y+2\pi-2\pi}{2(y+2)}\Big)}}{y}}$
$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y+\cancel{2\pi}-\cancel{2\pi}}{2(y+2)}\Big)}}{y}}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{y}}$
The denominator of the fractional function should be same as the angle in the sine function for applying the trigonometric limit rule of sine function in this problem. Hence, let us try to adjust the denominator of the function by taking some acceptable mathematical steps.
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[1 \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{y}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\dfrac{\pi}{\pi} \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{y}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\pi y}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{1 \times \pi y}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{2(y+2)}{2(y+2)} \times \pi y}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{2(y+2) \times \dfrac{1}{2(y+2)} \times \pi y}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{2(y+2) \times \dfrac{1 \times \pi y}{2(y+2)}}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{2(y+2) \times \dfrac{\pi y}{2(y+2)}}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{1 \times \sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{2(y+2) \times \dfrac{\pi y}{2(y+2)}}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{1}{2(y+2)} \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\dfrac{\pi \times 1}{2(y+2)} \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\dfrac{\pi}{2(y+2)} \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}\Bigg]}$
#### Evaluate the Limit of the function
A fractional function is split as the product of two fractional functions. The limit of the product of them can be evaluated by calculating the product of their limits. So, use the product rule of limits.
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\pi}{2(y+2)}}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$
The limit of first fractional function can be evaluated by the direct substitution method but do not disturb the limit of second fractional function.
$= \,\,\,$ $\dfrac{\pi}{2(0+2)}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$
$= \,\,\,$ $\dfrac{\pi}{2(2)}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$
$= \,\,\,$ $\dfrac{\pi}{2 \times 2}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$
$= \,\,\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$
The function in the denominator is same as the angle in the sine function. So, the trigonometric limit rule of sine function can be applied.
Now, take $z \,=\, \dfrac{\pi y}{2(y+2)}$. We have to express the input in terms of the $z$ completely in the function.
$(1) \,\,\,$ If $y \,\to\, 0$, then $\pi \times y \,\to\, \pi \times 0$. Therefore, $\pi y \,\to\, 0$.
$(2) \,\,\,$ If $\pi y \,\to\, 0$, then $\dfrac{\pi y}{2} \,\to\, \dfrac{0}{2}$. Therefore, $\dfrac{\pi y}{2} \,\to\, 0$.
$(3) \,\,\,$ If $y \,\to\, 0$, then $y+2 \,\to\, 0+2$. Therefore, $y+2 \,\to\, 2$. If $\dfrac{\pi y}{2} \times \dfrac{1}{y+2} \,\to\, 0 \times \dfrac{1}{y+2}$, then $\dfrac{\pi y}{2(y+2)} \,\to\, 0 \times \dfrac{1}{2}$. Therefore, $\dfrac{\pi y}{2(y+2)} \,\to\, 0$ but we have assumed that $z \,=\, \dfrac{\pi y}{2(y+2)}$.
Therefore, the three steps have cleared that if $y$ approaches zero, then the $z$ also approaches zero. Now, convert the function in terms of $z$ from $y$.
$\implies$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$ $\,=\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{z \,\to\, 0}{\normalsize \dfrac{\sin{z}}{z}}$
According to the trigonometric limit rule of sine function, the limit rule of sinx/x as x approaches 0 is equal to one. Therefore, the limit of $\dfrac{\sin{z}}{z}$ as $z$ approaches $0$ is also equal to one.
$=\,\,\,$ $\dfrac{\pi}{4} \times 1$
$=\,\,\,$ $\dfrac{\pi}{4}$
### Method – 2
In this method, we do not use the transformation technique but we take some acceptable mathematical steps, which help us to evaluate the limit of the given function.
$\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{x}\Big)}}{x-2}}$
#### Express the cos function in terms of sine
In the given function, the numerator is a trigonometric function and the denominator is algebraic function. In limits, there is no limit rule in cosine function but we have a trigonometric limit rule in sine function. Hence, we have to convert the cosine function into sine function and it is possible by the complementary angle identity of sine function.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x-2}}$
#### Adjust denominator same as Angle of sine
In order to use the trigonometric limit rule of sine function, we have to adjust the algebraic function in the denominator same as the angle in the sine function.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[1 \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x-2}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{\pi} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x-2}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\pi \times (x-2)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\pi \times x-\pi \times 2}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\pi x-2\pi}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{1 \times (\pi x-2\pi)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\dfrac{x}{x} \times (\pi x-2\pi)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \dfrac{1}{x} \times (\pi x-2\pi)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\dfrac{1}{x} \times \pi x-\dfrac{1}{x} \times 2\pi\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\dfrac{1 \times \pi x}{x}-\dfrac{1 \times 2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\dfrac{\pi x}{x}-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\dfrac{\pi \cancel{x}}{\cancel{x}}-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{1}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi \times 1}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{1 \times \Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\dfrac{2}{2} \times \Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \dfrac{1}{2} \times \Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{1}{2} \times \pi-\dfrac{1}{2} \times \dfrac{2\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{1 \times \pi}{2}-\dfrac{1 \times 2\pi}{2 \times x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{\pi}{2}-\dfrac{2\pi}{2x}\Big)}\Bigg]}$
$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{\pi}{2}-\dfrac{\cancel{2}\pi}{\cancel{2}x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{1 \times \sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{1}{2} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi \times 1}{x \times 2} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{2x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$
#### Evaluate the Limits of the functions
Now, split the limit of the product of functions as product of their limits by the product rule of limits.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Big(\dfrac{\pi}{2x}\Big)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$
Now, find the limit of the first fractional function by the direct substitution method as $x$ approaches $2$.
$=\,\,\,$ $\dfrac{\pi}{2(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$
$=\,\,\,$ $\dfrac{\pi}{2 \times 2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$
$=\,\,\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$
Now, let us evaluate the limit of the second trigonometric function. The algebraic function in the denominator is exactly same as the angle in the sine function. So, we can use the trigonometric limit rule of sine function but we have to set the input of the limit same as the denominator or angle in the sine function.
$(1) \,\,\,$ If $x \,\to\, 2$ then $\dfrac{1}{x} \,\to\, \dfrac{1}{2}$
$(2) \,\,\,$ If $\dfrac{1}{x} \,\to\, \dfrac{1}{2}$, then $\pi \times \dfrac{1}{x} \,\to\, \pi \times \dfrac{1}{2}$. Therefore, $\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}$
$(3) \,\,\,$ If $\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}$, then $-\dfrac{\pi}{x} \,\to\, -\dfrac{\pi}{2}$
$(4) \,\,\,$ If $-\dfrac{\pi}{x} \,\to\, -\dfrac{\pi}{2}$, then $\dfrac{\pi}{2}-\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}-\dfrac{\pi}{2}$. So, $\dfrac{\pi}{2}-\dfrac{\pi}{x} \,\to\, 0$
The four steps have cleared to us that $\dfrac{\pi}{2}-\dfrac{\pi}{x}$ approaches zero when $x$ approaches $2$.
Now, take $m = \dfrac{\pi}{2}-\dfrac{\pi}{x}$ and convert the function in terms of $m$ from $x$.
$\implies$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$ $\,=\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\sin{m}}{m}}$
According to the trigonometric limit rule of sine function, the limit of $\dfrac{\sin{m}}{{m}}$ as $m$ approaches zero is equal to one.
$=\,\,\,$ $\dfrac{\pi}{4} \times 1$
$=\,\,\,$ $\dfrac{\pi}{4}$
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## Precalculus (6th Edition) Blitzer
The solution is $\left\{ \left( 2,1 \right),\left( 2,-1 \right)\left( -2,1 \right)\text{ and }\left( -2,-1 \right) \right\}$.
Let us consider the given equation: $3{{x}^{2}}+4{{y}^{2}}=16$ …… (I) $2{{x}^{2}}-3{{y}^{2}}=5$ …… (II) By multiplying equation (I) by $3$, equation (II) by $4$ and add equation (I) and equation (II): we get, \begin{align} & 9{{x}^{2}}+12{{y}^{2}}=48 \\ & \underline{4{{x}^{2}}-12{{y}^{2}}=20} \\ & 17{{x}^{2}}=64 \\ & {{x}^{2}}=4 \\ & x=\pm 2 \\ \end{align} Substitute $x=2$ in equation (I): \begin{align} & 3{{\left( 2 \right)}^{2}}+4{{y}^{2}}=16 \\ & {{y}^{2}}=2 \\ & y=\pm 1 \end{align} Substitute $x=-2$ in equation (II): \begin{align} & 3{{\left( -2 \right)}^{2}}+4{{y}^{2}}=16 \\ & {{y}^{2}}=2 \\ & y=\pm 1 \end{align} Thus, the solution is $\left\{ \left( 2,1 \right),\left( 2,-1 \right)\left( -2,1 \right)\text{ and }\left( -2,-1 \right) \right\}$.
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# Skill 1.4
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Skill 1.4 Multiplying and Dividing Fractions
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• you are baking a cake for 2 people and the recipe you have makes a cake for 4 people. \nWhat are we going to do to the ingredients to make the cake the right size? (Half them)\nExplain that multiplying by 1/2 gives us half of the cake. \n\nNow lets say the cake calls for 3/4 a cup of sugar. How are we going to figure out how much sugar we need to put in? (Multiply 1/2 x 3/4)\n\n
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• ### Skill 1.4
1. 1. Skill 1.4 - Multiply and Divide Fractions 1/19/2011 Do Now 4 2 1) + = 7 7 You only have 3 minutes 1 5 so get started 2) + = 6 2€ 3) 1 3 − = 2 4 Transition
2. 2. Skill 1.4 - Multiply and Divide Fractions 1/19/2011 Do Now 4 2 1) + = 7 7 You only have 3 minutes 1 5 so get started 2) + = 6 2€ 3) 1 3 − = 2 4 Transition
3. 3. Skill 1.4 - Multiply and Divide Fractions 1/19/2011 Do Now 4 2 1) + = 7 7 You only have 3 minutes 1 5 so get started 2) + = 6 2€ 3) 1 3 − = 2 4
4. 4. Agenda• 1st Block • Do Now - 5 min • Morning Meeting - 5 minutes • Homework Check - 10 minutes • Intro to new material - 5 min
5. 5. Morning Meeting
6. 6. Morning Meeting• Take Attendance
7. 7. Morning Meeting• Take Attendance• Shout Outs
8. 8. Morning Meeting• Take Attendance• Shout Outs• Count our rings
9. 9. Agenda• 1st Block • Do Now - 5 min • Morning Meeting - 5 minutes • Homework Check - 10 minutes • Intro to new material - 5 min
10. 10. Homework Check
11. 11. Agenda• 1st Block • Do Now - 5 min • Morning Meeting - 5 minutes • Homework Check - 10 minutes • Intro to new material - 5 min
12. 12. Lets say the cake calls for 3/4 a cup of sugar• How are we going to figure out how much sugar we need to put in?
13. 13. Lets say the cake calls for 3/4 a cup of sugar• How are we going to figure out how much sugar we need to put in? 3 1 x = 4 2
14. 14. Lets say the cake calls for 3/4 a cup of sugar• How are we going to figure out how much sugar we need to put in? 3 1 x = 4 2
15. 15. Lets say the cake calls for 3/4 a cup of sugar• How are we going to figure out how much sugar we need to put in? 3 1 x = 4 2
16. 16. Lets say the cake calls for 3/4 a cup of sugar• How are we going to figure out how much sugar we need to put in? 3 1 3 x = 4 2 8
17. 17. Lets say the cake calls for 5/6 a cup of water• How are we going to figure out how much water we need to put in?
18. 18. Lets say the cake calls for 5/6 a cup of water• How are we going to figure out how much water we need to put in? 5 1 x = 6 2
19. 19. Lets say the cake calls for 5/6 a cup of water• How are we going to figure out how much water we need to put in? 5 1 x = 6 2
20. 20. Lets say the cake calls for 5/6 a cup of water• How are we going to figure out how much water we need to put in? 5 1 x = 6 2
21. 21. Lets say the cake calls for 5/6 a cup of water• How are we going to figure out how much water we need to put in? 5 1 5 x = 6 2 12
22. 22. Fractions - Dividing 1 3 2 ÷ 4 = =
23. 23. Fractions - Dividing 1 3 2 ÷ 4 = 1 4 = 2 3
24. 24. Fractions - Dividing 1 3 2 ÷ 4 = 1 4 × = 2 3
25. 25. Fractions - Dividing 1 3 2 ÷ 4 = 1 4 × = 2 3
26. 26. Fractions - Dividing 1 3 2 ÷ 4 = 1 4 × = 2 3
27. 27. Fractions - Dividing 1 3 2 ÷ 4 = 1 4 4 × = 2 3 6
28. 28. Fractions - Dividing 1 3 2 ÷ 4 = 1 4 4 2 × = 2 3 6 3
29. 29. Independent Work Time
30. 30. Independent Work Time
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# PSAT Math : How to find the area of a polygon
## Example Questions
### Example Question #1 : Other Polygons
A square has an area of 36 cm2. A circle is inscribed and cut out. What is the area of the remaining shape? Use 3.14 to approximate π.
7.74 cm2
15.48 cm2
3.69 cm2
28.26 cm2
12.14 cm2
7.74 cm2
Explanation:
We need to find the area of both the square and the circle and then subtract the two. Inscribed means draw within a figure so as to touch in as many places as possible. So the circle is drawn inside the square. The opposite is circumscribed, meaning drawn outside.
Asquare = s2 = 36 cm2 so the side is 6 cm
6 cm is also the diameter of the circle and thus the radius is 3 cm
A circle = πr2 = 3.14 * 32 = 28.28 cm2
The resulting difference is 7.74 cm2
### Example Question #382 : Plane Geometry
In the square above, the radius of each half-circle is 6 inches. What is the area of the shaded region?
144 – 36π
36 – 9π
36 – 6π
144 – 9π
144 – 6π
144 – 36π
Explanation:
We can find the area of the shaded region by subtracting the area of the semicircles, which is much easier to find. Two semi-circles are equivalent to one full circle. Thus we can just use the area formula, where r = 6:
π(62) → 36π
Now we must subtract the area of the semi-circles from the total area of the square. Since we know that the radius also covers half of a side, 6(2) = 12 is the full length of a side of the square. Squaring this, 122 = 144. Subtracting the area of the circles, we get our final terms,
= 144 – 36π
### Example Question #11 : Other Polygons
If square A has a side of length 5 inches, how many times bigger is the area of square B if it has a side of length 25 inches?
5 times
4 times
25 times
2 times
625 times
25 times
Explanation:
First find the area of both squares using the formula .
For square A, s = 5.
For square B, s = 25.
The question is asking for the ratio of these two areas, which will tell us how many times bigger square B is. Divide the area of square B by the area of square A to find the answer.
### Example Question #1 : How To Find The Area Of A Polygon
If Bailey paints the wall shaped like above and uses one bucket per 5 square units, how many buckets does Bailey need?
Explanation:
To solve, we will need to find the area of the wall. We can do this by finding the areas of each section and adding them together. Break the area into a rectange and two triangles.
The area of the rectangle will be equal to the base times the height. The area of each triangle will be one half its given base times its height.
For the rectangle, the base is 12 and the height is 4 (both given in the figure).
The triangle to the right has a given base of 6, but we need to solve for its height. The height will be equal to the difference between the total height (6) and the height of the rectangle (4).
We now have the base and height of the triangle to the right, allowing us to calculate its area.
Now we need to solve the triangle to the left. We solved for its height (2), but we still need to solve for its base. The total base of the rectangle is 12. Subtract the base of the right-side triangle (6) and the small segment at the top of the rectangle (3) from this total length to solve for the base of the left triangle.
The left-side triangle has a base of 3 and a height of 2, allowing us to calculate its area.
Add together the two triangles and the rectangle to find the total area.
We know that each bucket of paint will cover 5 square units, and we have 57 square units total. Divide to find how many buckets are required.
We will need 11 full buckets and part of a twelfth bucket to cover the wall, meaning that we will need 12 buckets total.
### Example Question #411 : Geometry
A square is inscribed within a circle with a radius . Find the area of the circle that is not covered by the square.
Explanation:
First, find the area of the circle.
Next, find the length of 1 side of the square using the Pythagorean Theorem. Two radii from the center of the circle to adjacent corners of the square will create a right angle at the center of the circle. The radii will be the legs of the triangle and the side of the square will be the hypotenuse.
Find the area of the square.
Subtract the area of the square from the area of the circle.
### Example Question #1 : How To Find The Area Of A Polygon
Refer to the above figure. Quadrilateral is a square. Give the area of Polygon in terms of .
Explanation:
is both one side of Square and the hypotenuse of ; its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:
Polygon is the composite of and Square , so its area is the sum of those of the two figures.
The area of is half the product of its legs:
The area of Square is the square of the length of a side:
This is the area of the polygon.
### Example Question #1 : How To Find The Area Of A Polygon
Note: Figure NOT drawn to scale.
The above figure shows Rectangle is the midpoint of .
What percent of Rectangle is shaded?
Explanation:
The area of , the shaded region in question, is that of the rectangle minus those of and . We look at both.
The answer is independent of the sidelengths of the rectangle, so to ease calculations - this will become more apparent later - we will arbitrarily assign to the rectangle the dimensions
and, subsequently,
and, since is the midpoint of ,
.
The area of Rectangle is equal to .
Since , and ,
and
The area of is equal to .
The area of is equal to
The area of the shaded region is therefore , which is
of the rectangle.
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# Four circular cardboard pieces of radii 7 cm
Question:
Four circular cardboard pieces of radii 7 cm are placed on a paperin such a way that each piece touches other two pieces. Find the area of the portion
enclosed between these pieces.
Solution:
Given that, four circular cardboard pieces arc placed on a paper in such a way that each piece touches other two pieces.
Now, we join centre of all four circles to each other by a line segment. Since, radius of each circle is 7 cm.
So, AB = 2 x Radius of circle
= 2×7 = 14cm
⇒ AB = BC = CD = AD = 14cm
which shows that, quadrilateral $A B C D$ is a square with each of its side is $14 \mathrm{~cm}$.
We know that, each angle between two adjacent sides of a square is $90^{\circ}$.
$\therefore$ Area of sector with $\angle A=90^{\circ}$.
$=\frac{\angle A}{360^{\circ}} \times \pi r^{2}=\frac{90^{\circ}}{360^{\circ}} \times \pi \times(7)^{2}$
$=\frac{1}{4} \times \frac{22}{7} \times 49=\frac{154}{4}=\frac{77}{2}$
$=38.5 \mathrm{~cm}^{2}$
$\therefore \quad$ Area of each sector $=4 \times$ Area of sector with $\angle \mathrm{A}$
$=4 \times 38.5$
$=154 \mathrm{~cm}^{2}$
and area of square $A B C D=(\text { side of square })^{2}$
$=(14)^{2}=196 \mathrm{~cm}^{2} \quad\left[\therefore\right.$ area of square $\left.=(\text { side })^{2}\right]$
So, area of shaded region enclosed between these pieces $=$ Area of square $A B C D$
- Area of each sector
$=196-154$
$=42 \mathrm{~cm}^{2}$
Hence, required area of the portion enclosed between these pieces is 42 cm2.
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Tricki
## Try to prove a stronger result
### Quick description
When we first set about trying to prove a result in mathematics, it can often be unclear which direction to take. If we try to prove a stronger result there is less room to manoeuvre and the proof can be easier to find.
### Example 1: Every Subgroup of the Integers under Addition is Cyclic
Suppose we are trying to prove that every subgroup of the integers under addition is cyclic. Appealing to the definition of cyclic groups, we can see that what we are trying to prove is the assertion that for every group that is a subgroup of the integers under addition, there exists such that for every , is some power of . (Since we are talking about a group under addition, we interpret 'power of ' as 'multiple of g'.)
How can we find a stronger result that will be easier to prove. In fact, how can we find a stronger result that can be proven at all - after all, many stronger results, such as the result that 'all groups are cyclic' are false. Well, a useful technique for finding stronger results which are always true is to assume the result you are trying to prove and see what you can derive from it. If we are going to be able to prove this result at all, it must be true, so any result we can derive from it must be true as well.
So, bearing this in mind, let us consider a subgroup of the integers under addition for which we have some such that for all , is a multiple of . What can we say about ? Let's assume that is positive. (We are justified in doing this: if a negative number generates , its inverse - which is positive- must also generate ; cannot generate .) Now, if we consider only those elements of which are positive, we can see that, since they are all positive multiples of , they must all be greater than or equal to . So must be the smallest positive number in . This means that we can reformulate the result we are trying to prove as the following lemma, which is a stronger result:
Lemma 1 (The Strong Lemma) Every element of every subgroup of the integers under addition is a multiple of the smallest positive number in .
Proof of Lemma 1. This turns out to be quite easy to prove. The proof is by contradiction: let be the smallest positive number in and assume on the contrary that there is some element with for all . Now we can write for , , . Clearly every positive multiple of is contained within is a binary operation on , so we can add to itself any number of times to create a new element of – so . Therefore, the inverse is in . So . Since , . But is positive and less than , and we had supposed to be the smallest positive element in . So we have derived a contradiction. So every element in is a multiple of .
The result that every subgroup of the integers under addition is cyclic clearly follows from Lemma 1.
### General discussion
In this case, we derived the stronger result from the result we were trying to prove. This special case of proving a stronger result is discussed more fully in the article Prove a consequence first.
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# SAT Math : How to find the number of integers between two other integers
## Example Questions
### Example Question #1 : How To Find The Number Of Integers Between Two Other Integers
A custom-made ruler is long and for every there's a tick mark. How many tick marks are there on the ruler?
Explanation:
There will be 15 gaps of long but 16 tick marks because there will be a tick mark on each end of the ruler.
### Example Question #2 : How To Find The Number Of Integers Between Two Other Integers
Four consecutive odd integers sum to 40. How many of these numbers are prime?
Explanation:
Let x equal the smallest of the four numbers. Therefore:
Therefore the four odd numbers are 7, 9, 11, and 13. Since all are prime except 9, three of the numbers are prime.
### Example Question #3 : How To Find The Number Of Integers Between Two Other Integers
The positive integer is not divisible by . The remainder when is divided by and the remainder when is divided by are both equal to . What is
Explanation:
We know that the remainder, , must be less than by the definition of remainder. Therefore our only choices are , , or . We can test each of these cases.
If would be divisible by , which we said is not true.
If : Try . Then . When we divide by , we have a remainder of . This works!
If : Try . Then . When we divide this by , we have a remainder of . Thus, our remainders are not equal.
Thus,
### Example Question #4 : How To Find The Number Of Integers Between Two Other Integers
How many integers lie between and , including the end points?
Explanation:
Let's look at a small example, the numbers between 1 and 3 including all endpoints. We have 1, 2, and 3. If we subtract our numbers, we get 2. We need to add one more to get all the desired numbers. We can follow the same process with our larger numbers.
Take,
.
.
### Example Question #1 : How To Find The Number Of Integers Between Two Other Integers
Mrs. Lovell assigns her reading class to read pages three through sixty-four of their textbook. How many pages must each student read?
Explanation:
To count the number of pages in between these two page numbers, we want to subtract and then add . We have to add because if we don't, we end up not counting either the first or last page of the selection.
### Example Question #6 : How To Find The Number Of Integers Between Two Other Integers
How many integers are between and , inclusive?
Explanation:
An important consideration whenever you're asked for the number of integers between two other numbers is to determine whether the two "end points" are included. Here the word "inclusive" tells us that we need to count both -8 and 17 in our calculation.
A helpful tip for these questions: if you're including both end points, you can calculate by using the range (greatest value minus smallest value) and adding 1 to account for the inclusive set. (If you need to prove that to yourself, just use "how many integers between 1 and 3 inclusive" - clearly the set is 1, 2, 3 for three integers, but 3-1 = 2, so you can see you need to add one more to get the answer).
For this question, that leaves us with 17 - (8) = 25 as the range, and if you add 1 to that for "inclusive" you get the correct answer, 26.
Another way to look at this one is to chop it up: between -8 and 17 there are 8 negative integers (-1 through -8), 17 positive integers (1 through 17), and 0 as one more integer. 8 + 17 + 1 = 26.
### Example Question #1 : How To Find The Number Of Integers Between Two Other Integers
How many prime numbers are between 1 to 25?
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# Difference between revisions of "2010 AMC 12B Problems/Problem 17"
The following problem is from both the 2010 AMC 12B #17 and 2010 AMC 10B #23, so both problems redirect to this page.
## Problem
The entries in a $3 \times 3$ array include all the digits from $1$ through $9$, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$
## Solution 1
Observe that all tableaus must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each tableau, there exists a valid tableau diagonally symmetrical across the diagonal extending from the top left to the bottom right.
• Case 1: Center 4
$$\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &8&9\\ \hline \end{tabular}$$
3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$
• Case 2: Center 5
$$\begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&8\\ \hline &&9\\ \hline \end{tabular}$$
Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$
• Case 3: Center 6
By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$
$$12+18+12=\boxed{\textbf{D)}42}$$
~BJHHar
P.S.: I like the tetris approach used in Solution 2 but found it a bit arbitrary. Solution 3 is the best, but not many would know hook length theorem. If the initial observations are unclear, make a tableau with a range of possible numbers in each square.
## Solution 3
This solution is trivial by the hook length theorem. The hooks look like this:
$\begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$
So, the answer is $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ = $\boxed{\text{(D) }42}$
P.S. The hook length formula is a formula to calculate the number of standard Young tableaux of a Young diagram. Numberphile has an easy-to-understand video about it here: https://www.youtube.com/watch?v=vgZhrEs4tuk The full proof is quite complicated and is not given in the video, although the video hints at possible proofs.
~IceMatrix
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# How many numbers can be formed with 5 digits?
## How many numbers can be formed with 5 digits?
There are 90,000 five-digit numbers in all. The digit at the ten thousand place in a 5-digit number can never be 0.
How many 5 digit numbers can be formed by using the digits 1,2 3 so that each digit appears at least once?
+ 5!/(3!)) = 120. But the answer is 90.
### How many five digit numbers can be formed from the digits 12345?
= 120. But the sum of digits (= 1 + 2 + 3 + 4 + 5 = 15) of any number formed is divisible by 3. It means all 120 numbers formed are divisible by 3.
How many 5 digit numbers can be formed from the digits 1,2 3 4 and 5 if repetition of digits is not allowed?
Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digitsi) = 5*4*3*2*! = 5! = 120. Now, in half them unit’s digit will be bigger than the ten’s digit and in half of them it will be smaller.
#### How many 5 digit numbers can be formed with the digits 9876543210?
a) Consider 9876543210. You need to choose five of these digits to make your number. There are thus (105)=252 options. b) Consider the 9-length string 111111111 with 5 markers inserted, eg A11BC111D1111E.
How many numbers of up to 5 digits can be created using the digits 1,2 3 and 5 each at least once such that they are a multiple of?
There are 12 possibilities overall. The question is “How many numbers of up to 5 digits can be created using the digits 1, 2, 3 and 5 each at least once such that they are a multiple of 15?”
## How many even numbers of 5 digits can be formed with the digits 1 2 3 4 and 5?
Further, to ensure that the number is even we ensure that in the unit’s place, the digits can only be 2 or 4. Hence, there are 48 four digit even numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated.
How many 5 digit numbers can be formed which are divisible by 3?
Question: A five digit number divisible by 3 is to be formed using the digits 0,1,2,3,4 and 5, without repetition. The total number of ways this can be done, is. A. 216.
### How many 5 digit numbers can be formed using the digits from 0 to 9 without repetition?
Now, there are 105 ways in which the digits 0-9 can be chosen for the five places of a five digit number. Out of these, 104 start with zero (once we start with 0, there are only 4 slots to fill, where we have 10 choices each).
How many 5 digit numbers can be made from the digits 1 to 7 repetition is allowed?
How many five-digit numbers can be made from the digits 1 to 7 if repetition is allowed? Explanation: 75 = 16807 ways of making the numbers consisting of five digits if repetition is allowed.
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# COMPARING AND ORDERING FRACTIONS WORKSHEETS 4TH GRADE
Comparing and ordering fractions worksheets 4th grade :
Worksheets on comparing and ordering decimals are much useful to the kids who are learning about fractions in 4th grade.
Because, when they practice questions on comparing and ordering fractions, they will get a clear idea about fractions.
## Comparing and ordering fractions worksheets 4th grade - Questions
1. Find which fraction is greater 6/19, 16/19
2. Find which fraction is greater 1/2, 3/5
3. Find which is greater fraction 2/10, 7/12
4. Order the fractions from least to greatest
3/4 , 2/5 ,1/8.
5. Order the fractions from greatest to least
5/6,3/8,1/4
## Comparing and ordering fractions worksheets 4th grade - Solution
Question 1 :
Find which fraction is greater 6/19, 16/19
Solution :
Since the denominators of those fractions are same, then we can compare the numerators and decide which fraction is greater.
Hence, 16/19 is greater than 6/19
Question 2 :
Find which fraction is greater 1/2, 3/5
Solution :
Since the denominators are different, we have to convert each fraction into equivalent fraction with the common denominator.
To make the denominator same we have to apply the concept LCM.
Now we can compare the numerators and decide which fraction is greater.
By comparing the numerators 6 is greater than 5. So 6/10 is greater than 5/10
Hence , 3/5 is greater than 1/2
Question 3 :
Find which is greater fraction
2/10, 7/12
Solution :
Since the denominators are different, we have to convert each fraction into equivalent fraction with the common denominator.
To make the denominator same we have to apply the concept LCM.
= (2/10) x (6/6) = 12/60
= (7/12) x ( 5/5) = 35/60
Now we can compare the numerators and decide which fraction is greater.
By comparing the numerators 35 is greater than 12. So 35/60 is greater than 12/60
Hence , 7/12 is greater than 2/10
Question 4 :
Order the fractions from least to greatest 3/4, 2/5, 1/8
Solution :
Since the denominators are different then we have to Convert these fractions to equivalent fractions with a common denominator in order to compare them
LCM of (4,5 and 8) = 40
Now we have to make each fraction with the common denominator 40.
Multiply both numerator and denominator of the first fraction by 10
Multiply both numerator and denominator of the second fraction by 8
Multiply both numerator and denominator of the third fraction by 5
Now we can compare the fraction and order the fraction from least to greatest
So the answer is (1/8) < (2/5) < (3/4)
Question 5 :
Order the fractions from greatest to least 5/6, 3/8, 1/4
Solution :
Since the denominators are different then we have to Convert these fractions to equivalent fractions with a common denominator in order to compare them
LCM of (4,5 and 8) = 40
Now we have to make each fraction with the common denominator 40.
Multiply both numerator and denominator of the first fraction by 4
Multiply both numerator and denominator of the second fraction by 3
Multiply both numerator and denominator of the third fraction by 6
So the answer is (5/6) > (3/8) > (1/4)
After having gone through the stuff given above, we hope that the students would have understood "Comparing and ordering fractions worksheets 4th grade".
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
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HCF and LCM word problems
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Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
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# Math Lesson Plan Oct 7
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### Math Lesson Plan Oct 7
1. 1. Lesson Plan Date: Wednesday October 7, 2009 Subject: Math 1.Standards for the day: PK.NSO-N.1. Use one to one correspondence. Big Idea: Using one –to –one correspondence helps us count and track objects accurately. It also helps determine which sets of objects have more, less or the same. Essential Questions: What is one-to-one correspondence? How do you know if it is corresponding? How do we use one- to- one correspondence in everyday life? Vocabulary: Count, match, one to one, dice, cube, dots, compare, more, fewer, the same. 2. Introduce to Lesson: Today we are going to learn one to one correspondence. One to one correspondence means matching one to one. For example, when we count the boys and the girls for attendance, we touch each child and count.. When we count our fingers, we open each finger as we count. It is matching one to one. It is important for us to learn one to one correspondence so we can count and keep track of objects correctly. Also, matching one to one helps us compare which sets have more, less, or the same objects. Now we will practice matching children to the chairs one to one. We are going to see how many one to one matches we can make. What are we going to do today? Wait for students’ responses then begin the activity. 3. Goals/Expectations for the day: Students will be able to -Touch objects and say the number names correctly. -Arrange and compare sets using one-to –one correspondence. 4. Guided Instruction & Independent Practice/Study Set up six chairs. Have six children sit in the chairs. Ask: “Do we have one child in each chair?” Help children see that the children and chairs match. Say: “The children and chairs match one to one.” Ask one child to leave a chair. Ask: “Now do we have one child in every chair? Help children see that the chair and children no longer match. Next display children three markers and two marker tops. Ask: “Are there enough tops for every marker?” Wait for children’s responses and ask: “How can we fix that? Ask the children “How can you tell if the markers and tops match one to one? Tell the children that they are going to play a game with a dice. Point to the dots on the dice and have children count the dots aloud. Remind children the dots on the dice are the same the dots on the number dots they matched with the cereal last week. Say: “We will be counting the dots on the dice to get the teddy bears from the container. Each person will get a turn to roll the dice. When everybody gets a turn, we will count and compare who has more teddy bears. Who has less? Do any children have the same number of teddy bears?
2. 2. 5. Closure of Lesson: Have children draw a figure for each object they have. Provide do dot art stamps for children who cannot draw to represent their objects. Have children share their drawn presentations and ask them show how they know or determined the number of objects they have. Expect responses to include the skills learned such as touching, moving, or pointing to each object as they are counted. Follow up: Put the dices and counters in the math-learning center for children to use. Have them work in pairs. Assessment: 3=Correctly touches objects and says the number name. 2.Says correct number names, but incorrectly touches objects. 1. Correctly touches objects, but say wrong number name. 0. Says wrong number name and incorrectly touches objects. Child’s Name Counting 1-1 Notes Yes / No Natalie Ramos Yes / No Marcos Beltrand Yes /No Emely Sorto Yes /No Luis Portillo Yes /No Julissa Canales Yes /No Inti Garcia Ye s/No
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10th mathematics exercise questions with answers
# Circles Exercise 10.6 Solution – Chapter Circles – Class 10
## Circles Exercise 10.6 – Questions:
I. Numerical problems on touching circles.
1. Three circles touch each other externally. Find the radii of the circles if the sides of the triangle formed by joining the centres are 7 cm, 8 cm and 9 cm respectively.
2. Three circles with centres A, B and C touch each other as shown in the figure. IF the radii of these circles are 8 cm, 3 cm and 2 cm respectively, find the perimeter of ∆ABC.
3. In the figure AB = 10 cm, AC = 6 cm and the radius of the smaller circle is x cm. Find x.
II. Riders based on touching circles.
1. A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.
2. Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.
3. In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2.OC
## Circles Exercise 10.6 solution:
1. Numerical problems on touching circles.
1.Three circles touch each other externally. Find the radii of the circles if the sides of the triangle formed by joining the centres are 7 cm, 8 cm and 9 cm respectively.
Solution:
Let radius of the circles be AP = x , BQ = y and CR = z
AB = AP + BP = x + y = 7 cm
BC = BQ + CQ = y + z = 8 cm
AC = CR + AR = z + x = 9 cm
The perimeter of ∆ABC , AB + BC + CA = 7 + 8 + 9 = 24
AP + BP + BQ + CQ + CR + AR = 24
2x + 2y + 2z 24
x + y + z = 12
7 + z = 12⇒ z = 12 – 7 = 5 cm
x + 8 = 12 ⇒ x = 12 – 8 = 4 cm
y + 9 = 12 ⇒ y = 12 – 9 = 3 cm
1. Three circles with centres A, B and C touch each other as shown in the figure. IF the radii of these circles are 8 cm, 3 cm and 2 cm respectively, find the perimeter of ABC.
Solution:
The perimeter of ∆ABC = AB + BC + AC
AB = AM – BM = 8 – 3 = 5 cm
BC = BQ + CQ = 3 + 2 = 5 cm
AC = AN – CN = 8 – 2 = 6 cm
The perimeter of ∆ABC = AB + BC + AC = 5 + 5 + 6 = 16 cm
1. In the figure AB = 10 cm, AC = 6 cm and the radius of the smaller circle is x cm. Find x.
Solution:
In ∆OPC, ∠PCO = 90˚
PC2 = OP2 – OC2
x2 = (OQ – PQ)2 – (AC – OA)2
[Since OP = OQ – PQ, OC = AC – AO]
x2 = (5 – x)2+(6 – 5)2 [Since OQ = OA = 5]
x2 = 25 – 10x + x2 – 1
10x = 24
x = 2.4 cm
1. Riders based on touching circles.
1.A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.
Solution:
∠AOP = ∠BOQ [Vertically opposite angles]
∠APO = ∠AOP [ AO = AP radius of the circle]
∠BQO = ∠BOQ
∠APO = ∠BQO [alternate angles]
Therefore, AP||BQ
1. Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.
Solution:
∠BXP = ∠PYC [Alternate angles AB||CD]
∠BPX = ∠PBX [ XB = XP radii]
∠BPX + ∠PBX + ∠BXP = 180˚ ………….(1)
∠CPY = ∠PCY [YP = YC radii]
∠CPY + ∠PCY + ∠PYC = 180˚
2∠CPY + ∠PCY = 180˚…………(2)
From (1) and (2),
2∠BPX + ∠BXP = 2∠CPY + ∠PYC
2∠BPX = 2∠CPY
∠BPX = ∠CPY
1. In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2.OC
Solution:
∠ACO = 90˚
∠A = ∠A
BD/OC = AB/AO
BD/OC = 2. AO/AO [AB = 2.OA]
BD/OC = 2
BD = 2.OC
1. In the given figure AB = 8 cm, M is the midpoint of AB. A circle with centre O touches all three semicircles as shown. Prove that the radius of this circle is shown. Prove that the radius of this circle is 1/6 AB
Solution:
In ∆OPC, ∠POC = 90˚
OC2 = OM2 + MC2 [By Pythagoras Theorem]
(CP + OP)2 = (MR – OR)2 + MC2
(2 + x)2 = (4 – x)2 + 22
4 + 4x + x2 = 16 – 8x + x2 + 4
4 + 4x = 16 – 8x + 4
12x = 16
x = 16/12 = 8/6
x = 1/6 x 8
x = 1/6 x AB [AB = 8 cm]
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How do you write an equation of a line with slope of 3 and contains the point (4, 9)?
Mar 22, 2018
$y = 3 x - 3$
Explanation:
$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$\text{here } m = 3$
$\Rightarrow y = 3 x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$
$\text{to find b substitute "(4,9)" into the partial equation}$
$9 = 12 + b \Rightarrow b = 9 - 12 = - 3$
$\Rightarrow y = 3 x - 3 \leftarrow \textcolor{red}{\text{is the equation of the line}}$
Mar 22, 2018
$y = 3 x - 3$
Explanation:
You can use the equation of a line in point-slope form:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
Make the following substitutions:
${x}_{1} = 4$ (the x-coordinate of the point)
${y}_{1} = 9$ (the y-coordinate of the same point)
$m = 3$ (the slope of the line)
That will give you:
$y - 9 = 3 \left(x - 4\right)$
Then simplify:
$y - 9 = 3 \left(x - 4\right)$
$y - 9 = 3 x - 12$
$y = 3 x - 3$
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Question Video: Distinguishing between Identity and Equation in Trigonometry | Nagwa Question Video: Distinguishing between Identity and Equation in Trigonometry | Nagwa
# Question Video: Distinguishing between Identity and Equation in Trigonometry Mathematics • First Year of Secondary School
## Join Nagwa Classes
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Is tan π = sin π/cos π an identity or an equation?
02:21
### Video Transcript
Is tan π equals sin of π divided by cos of π an identity or an equation?
An identity is a relation that is true for all real variables, while an equation is a relation that is true for some of the real values of the variable and not true for others. Sine, cosine, and tangent are part of the trig identities. So letβs take a look at a right triangle.
So here we have a right triangle. And we weβll call this angle π. So now letβs label the sides of the triangle based on π. The side directly across from π would be considered the opposite side. The side across from the 90-degree angle, which is the longest side, is called the hypotenuse. And the side right next to π would be considered the adjacent side.
So itβs important to have these sides labeled because the sin of π represents a relationship between the opposite side and the hypotenuse side. The cos of π is the adjacent side divided by the hypotenuse side. And tan of π is equal to the opposite side divided by the adjacent side.
So we wanna know is the tan of π equal to the sin of π divided by the cos of π an identity or an equation. So this means this will have to work for any angle π except for the right angle, because the right angle will always stay the right angle. It will not change. We will never need to find the measure of that angle.
So letβs begin with replacing sin of π with opposite divided by hypotenuse and now replace the cos of π with adjacent divided by hypotenuse. So to simplify this, we are dividing fractions. So we will take the fraction on the bottom, flip it, and actually multiply the top fraction by it. So we take our top fraction and multiply by the reciprocal of the bottom fraction. And we have the hypotenuse that cancels. And weβre left with opposite divided by adjacent, which is right here, exactly what we had. Therefore, this would be an identity.
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Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Intext Questions
Students can Download Maths Chapter 1 Number System Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Intext Questions
Exercise 1.1
Try These (Text book Page No. 1)
Question 1.
Represent the fraction $$\frac { 1 }{ 4 }$$ in decimal form
Solution:
$$\frac { 1 }{ 4 }$$ = $$\frac{1 \times 25}{4 \times 25}$$ = $$\frac { 25 }{ 100 }$$ = 0.25
Question 2.
What is the place value of 5 in 63.257.
Solution:
Place value of 5 in 63.257 is 5 hundredths (Hundreth place)
Question 3.
Identify the digit in the tenth place of 75.036.
Solution:
0
Question 4.
Express the decimal number 3.75 as a fraction.
Solution:
3.75 = $$\frac { 375 }{ 100 }$$ = $$\frac { 15 }{ 4 }$$
Question 5.
Write the decimal number for the fraction 5 $$\frac { 1 }{ 5 }$$
Solution:
5 $$\frac { 1 }{ 5 }$$ = $$\frac { 26 }{ 5 }$$ = $$\frac{26 \times 2}{5 \times 2}$$ = $$\frac { 52 }{ 10 }$$ = 5.2
Question 6.
Identify the biggest number : 0.567 and 0.576.
Solution:
Comparing the digits of 0.567 and 0.576 from left to right, we have the tenths place same comparing the hundredths place 7 > 6.
⇒ 0.576 > 0.567
Question 7.
Compare 3.30 and 3.03 and identify the smaller number.
Solution:
The whole number is equal in both the numbers.
Now comparing the tenths place we have 3 > 0
⇒ 3.03 < 3.30 Smaller number is 3.03
Question 8.
Put the appropriate sign (<, >, =). 2.57 [ ] 2.570
Solution:
2.57 [=] 2.570
Question 9.
Arrange the following decimal numbers in ascending order.
5.14, 5.41, 1.54, 1.45, 4.15, 4.51.
Solution:
Comparing the numbers from left to right. Ascending order : 1.45, 1.54, 4.15, 4.51, 5.14, 5.41
Exercise 1.2
Try These (Text book Page No. 6)
Question 1.
Find the following using grid models:
(i) 0.83 + 0.04
(ii) 0.35 – 0.09
Solution:
(i) 0.83 + 0.04
0.83 = $$\frac { 83 }{ 100 }$$ and 0.04 = $$\frac { 4 }{ 100 }$$
0.83 and 0.04
The sum is the total shaded region.
S = 0.83 + 0.04 = 0.87
(ii) 0.35 – 0.09
0.35 = $$\frac { 35 }{ 100 }$$ and 0.09 = $$\frac { 9 }{ 100 }$$
Shading the regions 0.35 by shading 35 boxes out of 100. Striking off 9 boxes out of 35 shaded boxes to subtract 0.09 from 0.35.
The left over shaded boxes represent the required value.
∴ 0.35 – 0.09 = 0.26
Try These (Text book Page No. 7)
Question 1.
Using the area models solve the following
(i) 1.2 + 3.5
(ii) 3.5 – 2.3
Solution:
(i) 1.2 + 3.5
Here 1.2 is represented in blue colour and 3.5 is represented in Green colour. Sum of 1.2 and 3.5 is 4.7.
(ii) 3.5 – 2.3
Representing 3.5 using 3 squares and 5 rectangular strips. Crossing out 2 squares from 3 squares and 3 rectangular strips from 5 to get the difference. So 3.5 – 2.3 = 1.2.
Try These (Text book Page No. 9)
Question 1.
Complete the magic square in such a way that rows, columns and diagonals give the same sum 1.5.
Solution:
Exercise 1.3
Think (Text book Page No. 13)
Question 1.
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 (Text book Page No. 13)
Question 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.
∴ 0.3 × 0.6 = 0.18
Question 2.
Use the area model to multiply
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.
Try These (Text book Page No. 14)
Question 1.
Complete the following table
Solution:
Try These (Text book Page No. 15)
Question 1.
Find:
1. 9.13 × 10
2. 9.13 × 100
3. 9.13 × 1000
Solution:
1. 9.13 × 10 = 91.3
2. 9.13 × 100 = 913
3. 9.13 × 1000 = 9130
Try These (Text book Page No. 16)
Question 1.
Complete the following table
Solution:
Exercise 1.4
Try These (Text book Page No. 19)
Question.
Solution:
Try These (Text book Page No. 19)
Question 1.
Divide the following
(i) 17.237 ÷ 10
(ii) 17.237 ÷ 100
(iii) 17.237 ÷ 1000
Solution:
(i) 17.237 ÷ 10
= $$\frac { 17237 }{ 1000 }$$ × $$\frac { 1 }{ 10 }$$
= $$\frac { 17237 }{ 1000 }$$
= 1.7237
(ii) 17.237 ÷ 100
= $$\frac { 17237 }{ 1000 }$$ × $$\frac { 1 }{ 100 }$$
= $$\frac { 17237 }{ 100000 }$$
= 0.17237
(iii) 17.237 ÷ 1000
= $$\frac { 17237 }{ 1000 }$$ × $$\frac { 1 }{ 1000 }$$
= $$\frac { 17237 }{ 1000000 }$$
= 0.017237
Try These (Text book Page No. 21)
Question 1.
Find the value of the following:
(i) 46.2 ÷ 3 = ?
(ii) 71.6 ÷ 4 = ?
(iii) 23.24 ÷ 2 = ?
(iv) 127.35 ÷ 9 = ?
(v) 47.201 ÷ 7 = ?
Solution:
(i) 46.2 ÷ 3
= $$\frac { 462 }{ 10 }$$ × $$\frac { 1 }{ 3 }$$
= $$\frac { 1 }{ 10 }$$ × $$\frac { 462 }{ 3 }$$
= $$\frac { 1 }{ 10 }$$ × 15.4
= $$\frac { 154 }{ 10 }$$
= 15.4
(ii) 71.6 ÷ 4
= $$\frac { 716 }{ 10 }$$ × $$\frac { 1 }{ 4 }$$
= $$\frac { 1 }{ 10 }$$ × $$\frac { 716 }{ 4 }$$
= $$\frac { 1 }{ 10 }$$ × 179
= 17.9
(iii) 23.24 ÷ 2
= $$\frac { 2324 }{ 100 }$$ × $$\frac { 1 }{ 2 }$$
= $$\frac { 2324 }{ 2 }$$ × $$\frac { 1 }{ 100 }$$
= 1162 × $$\frac { 1 }{ 100 }$$
= $$\frac { 1162 }{ 100 }$$
= 11.62
(iv) 127.35 ÷ 9
= $$\frac { 12735 }{ 100 }$$ × $$\frac { 1 }{ 9 }$$
= $$\frac { 12735 }{ 9 }$$ × $$\frac { 1 }{ 100 }$$
= 1415 × $$\frac { 1 }{ 100 }$$
= $$\frac { 1415 }{ 100 }$$
= 14.15
(v) 47.201 ÷ 7
= $$\frac { 47201 }{ 1000 }$$ × $$\frac { 1 }{ 7 }$$
= $$\frac { 47201 }{ 7 }$$ × $$\frac { 1 }{ 1000 }$$
= 6743 × $$\frac { 1 }{ 1000 }$$
= $$\frac { 6743 }{ 1000 }$$
= 6.743
Try These (Text book Page No. 22)
Question 1.
Divide the following
(i) $$\frac { 9.25 }{ 0.25 }$$
(ii) $$\frac { 8.6 }{ 4.3 }$$
(iii) $$\frac { 44.1 }{ 0.21 }$$
(iv) $$\frac { 9.6 }{ 1.2 }$$
Solution:
(i) $$\frac { 9.25 }{ 0.25 }$$
(ii) $$\frac { 8.6 }{ 4.3 }$$
(iii) $$\frac { 44.1 }{ 0.21 }$$
(iv) $$\frac { 9.6 }{ 1.2 }$$
Think (Text book Page No. 22)
Question 1.
The price of a tablet strip containing 30 tablets is 22.63 Then how will you find the price of each tablet?
Solution:
Price of 30 tablets = ₹ 22.63 = ₹ $$\frac { 2263 }{ 100 }$$
∴ Price of 1 tablet
= $$\frac { 2263 }{ 100 }$$ × $$\frac { 1 }{ 30 }$$
= $$\frac { 2263 }{ 30 }$$ × $$\frac { 1 }{ 100 }$$
= $$\frac { 2263 }{ 3 }$$ × $$\frac { 1 }{ 1000 }$$
= 754.33 × $$\frac { 1 }{ 1000 }$$
= $$\frac { 754.33 }{ 1000 }$$
= 0.75433
Price of each tablet is ₹ 0.7543
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##### 12.1 Introduction to linear equations
We have learned all about linear relationships, and linear functions. From what we learned about, we know that linear relationships and functions is consisted of two variables, one dependent, and the other independent. We also know that the line is a straight line. Linear equations are nonetheless the same, adding the criteria that its variables should not be raised to an exponent greater than one, and that the variables are not used as a denominator. We are to learn more of linear equations in this chapter, especially in section 1.
Apart from linear equations, we also need to learn about nonlinear equations. As the word suggests, these are the other equations that aren’t linear in nature, like the quadratic equations, circle equations, cubic equations, and more. We will have exercises in section 2 that will show us the difference between these equations and that of the linear equations.
As we have learned in the previous chapters, we know that linear equations show a straight line, that would appear to fall diagonally to on the Cartesian plane, to show the linear relationship between the values found in the line. However, in some special cases, linear equations can show horizontal lines and vertical lines. In chapter section 3 and 4 we will understand these special cases a bit more.
We are also going to learn how to find the parallel line and the perpendicular line of a given linear equation in section 5 and section 6. From our previous discussion in chapter, we learned that parallel lines have the same slope, whereas perpendicular lines have the slopes that are negative reciprocals with each other.
In section 7 and section 8 we have more practice in looking for the parallel lines and perpendicular lines, by looking for both of them for every equation given. There are also other applications of linear equation in section 8. To find out more about linear equations you can check out a video made by Education Alberta in Canada about “Exploring Linear Equations
### Introduction to linear equations
This is a lesson that teaches how to determine if an expression is a linear equation; and how to graph a linear equation.
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# Circle: General Equation
In addition to the center-radius form, discussed under another topic and shown again here, the equation of a circle also has a general form.
GENERAL EQUATION:
Let C, D and E be constants.
${x}^{2}+{y}^{2}+Cx+Dy+E=0$
CENTER-RADIUS EQUATION:
Let the center be (h, k) and the radius be r.
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
The center-radius form gives the center coordinates (h, k) and the radius r at a glance, whereas the general form does not provide easy access to this information.
Thus it is desirable to change the general form of the equation into the center-radius form to get this information. This is accomplished by completing the square.
STEPS FOR CONVERTING THE GENERAL FORM INTO THE CENTER-RADIUS FORM: 1. Group the x- and y-terms on the left-hand side of the equation. 2. Move the constant term to the right-hand side. 3. Complete the square for the x- and y-groups. a. Divide the x-term coefficient by 2, square the result and add it to the x-group. b. Divide the y-term coefficient by 2, square the result and add the result to the y-group. 4. Add the same quantity to the right-hand side that was added to the left-hand side. 5. Write the x-group, y-group and constant as perfect squares. Then ensure that the values of h and k are subtracted from x and y respectively.
Let's use the completing the square procedure in some examples:
Example 1: Find the center and radius of the circle by converting the general equation to the center-radius form by completing the square.
${x}^{2}+{y}^{2}+2x-6y-12=0$
Step 1: Group the x- and y-terms on the left-hand side of the equation. ${x}^{2}+{y}^{2}+2x-6y-12=0$ $\left({x}^{2}+2x\right)+\left({y}^{2}-6y\right)-12=0$ Step 2: Move the constant term to the right-hand side. $\left({x}^{2}+2x\right)+\left({y}^{2}-6y\right)=12$ Step 3: Complete the square for the x- and y-groups. $\left({x}^{2}+2x\right)+\left({y}^{2}-6y\right)=12$ Complete the square for the x-group (x2 + 2x) Take the coefficient of the x-term, divide by 2 and square the result. $\frac{2}{2}=1;{1}^{2}=1$ Add the result to the x-group. (x2 + 2x + 1) Complete the square for the y-group (y2 - 6y) Take the coefficient of the y-term, divide by 2 and square the result. $\frac{-6}{2}=-3;{\left(-3\right)}^{2}=9$ Add the result to the y-group. (y2 - 6y + 9) Final result. $\left({x}^{2}+2x+1\right)+\left({y}^{2}-6y+9\right)=12+?$ Step 4: Add whatever was added to the left-hand side to the right-hand side. $\left({x}^{2}+2x+1\right)+\left({y}^{2}-6y+9\right)=12+1+9$ $\left({x}^{2}+2x+1\right)+\left({y}^{2}-6y+9\right)=22$ Step 5: Write the x-group, y-group and constant as perfect squares. Then ensure that the values of h and k are subtracted from x and y respectively. (x + 1)2 + ${\left(y-3\right)}^{2}={\sqrt{22}}^{2}$ ${\left[x-\left(-1\right)\right]}^{2}+{\left(y-3\right)}^{2}={\sqrt{22}}^{2}$
Example 2: Find the center and radius of the circle by converting the general equation to the center-radius form by completing the square.
$3{x}^{2}+{y}^{2}-12x+8y-1=0$
Step 1: Group the x- and y-terms on the left-hand side of the equation. In order to properly create squared terms, the coefficient of the x2-term must be one. As such we factor out a three from the x-group. $3{x}^{2}+{y}^{2}-12x+8y-1=0$ $\left(3{x}^{2}-12x\right)+\left({y}^{2}+8y\right)-1=0$ $3\left({x}^{2}-4x\right)+\left({y}^{2}+8y\right)-1=0$ Step 2: Move the constant term to the right-hand side. $3\left({x}^{2}-4x\right)+\left({y}^{2}+8y\right)=1$ Step 3: Complete the square for the x- and y-groups. $3\left({x}^{2}-4x\right)+\left({y}^{2}+8y\right)=1$ Complete the square for the x-group 3(x2 - 4x) Take the coefficient of the x-term, divide by 2 and square the result. $\frac{-4}{2}=-2;{\left(-2\right)}^{2}=4$ Add the result to the x-group. 3(x2 - 4x + 4) Complete the square for the y-group (y2 + 8y) Take the coefficient of the y-term, divide by 2 and square the result. $\frac{8}{2}=4;{4}^{2}=16$ Add the result to the y-group. (y2 + 8y + 16) Final result. $3\left({x}^{2}-4x+4\right)+\left({y}^{2}+8y+16\right)=1+?$ Step 4: Add whatever was added to the left-hand side to the right-hand side. $3\left({x}^{2}-4x+4\right)+\left({y}^{2}+8y+16\right)=1+3\left(4\right)+16$ $3\left({x}^{2}-4x+4\right)+\left({y}^{2}+8y+16\right)=29$ Step 5: Write the x-group, y-group and constant as perfect squares. Then ensure that the values of h and k are subtracted from x and y respectively. $3{\left(x-2\right)}^{2}+{\left(y+4\right)}^{2}={\sqrt{29}}^{2}$ $3{\left(x-2\right)}^{2}+{\left[y-\left(-4\right)\right]}^{2}={\sqrt{29}}^{2}$ Step 6: Identify the center and the radius. Center = (2, -4) Radius = $\sqrt{29}$
Related Links: Math algebra Parabola: Standard Equation Ellipse: Standard Equation Pre Calculus
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In this article we propose some exercises on Elementary Number Theory; they don’t require advanced mathematical knowledge. Other articles will follow with exercises related to this beautiful branch of mathematics.
We recall that the symbol $$\left\lfloor x \right\rfloor$$ denotes the integer part of the real number $$x$$, i.e. the largest integer that is less than or equal to $$x$$. For example $$\lfloor \frac{30}{7} \rfloor=4 , \lfloor -\frac{10}{3} \rfloor=-4$$. The following properties easily follow from the definition ($$x$$ real number, $$n$$ integer):
$\begin{array}{l} \lfloor x+n \rfloor = \lfloor x \rfloor + n \\ \\ \left\lfloor \dfrac{x}{n} \right\rfloor = \left\lfloor \dfrac{\left\lfloor x \right\rfloor}{n} \right\rfloor \text { if n \ge 1} \end{array}$
Contents
## Exercise 1
Let $$n \ge 1$$ be a positive integer. Prove that the number of digits in the decimal representation is
$\lfloor \log_{10}(n) \rfloor + 1$
## Exercise 2
Given a real number $$x$$ and a positive integer $$n$$, prove the following formula:
$\displaystyle \left\lfloor x \right\rfloor + \left\lfloor x+\frac{1}{n} \right\rfloor +\left\lfloor x+ \frac{2}{n} \right\rfloor + \cdots + \left\lfloor x+ \frac{n-1}{n} \right\rfloor – \left\lfloor nx \right\rfloor =0$
Hint
Let $$F(x)$$ the expression on the left. Note that $$F(x)=0$$ for each $$x \in [0,\frac{1}{n})$$. Then prove that $$F(x + \frac{1}{n})= F(x)$$.
## Exercise 3
Prove the following properties:
$\begin{array}{l} 2^{n} \equiv 1 \pmod{3} \quad \text { if $$n$$ is even} \\ 2^{n} \equiv 2 \pmod{3} \quad \text { if $$n$$ is odd} \\ \end{array}$
Hint
Recall the following property of congruences: if $$a \equiv b \pmod {n}$$ and $$c \equiv d \pmod {n}$$ then $$ac \equiv bd \pmod{n}$$.
## Exercise 4
Compute the following sum:
$S = \displaystyle \left\lfloor \frac{1}{3} \right\rfloor + \left\lfloor \frac{2^{1}}{3} \right\rfloor +\left\lfloor \frac{2^{2}}{3} \right\rfloor + \cdots + \left\lfloor \frac{2^{100}}{3} \right\rfloor$
Hint
Use Exercise 3. Also recall the formula of the sum of a geometric progression:
$\sum\limits_{i=0}^{n}x^{i} = \frac {x^{n+1} -1}{x-1}$
Solution
$S=\frac{1}{3}(2^{101}-2)-50$
## Exercise 5
Let $$n,m$$ be non-negative integers. Prove that the following expression is always an integer:
$\frac{(2m)!(2n)!}{m!n!(m+n)!}$
Hint
Recall the factorial definition of a non-negative integer:
$n! = \begin{cases} n \cdot (n-1)! \quad n \gt 0 \\ 1 \qquad \qquad n = 0 \\ \end{cases}$
## Exercise 6
If $$(a,m)=1$$ then
$a^{k} \equiv a^{k \pmod{\varphi(m)}} \pmod{m}$
Hint
Use the Euler-Fermat theorem (Wikipedia).
## Exercise 7
Prove that
$n^{n^{n^{n}}} – n^{n^{n}} \equiv 0 \pmod{9}$
Solution
If $$3|n$$ we are done. Otherwise use the Euler-Fermat theorem and Exercise 6. Recalling that $$\varphi (9) = 6$$, we have to show that $$n ^ {n ^ {n}} – n ^ {n} \equiv 0 \pmod {6}$$. Meanwhile, we have $$n ^ {n ^ {n}} – n ^ {n} \equiv 0 \pmod {2}$$. Since we assumed that $$n$$ is not divisible by 3, we have to show that $$n ^ {n ^ {n}} – n ^ {n} \equiv 0 \pmod {3}$$. This is equivalent to $$n ^ {n} – n \equiv 0 \pmod {\varphi (3)}$$. But $$\varphi (3) = 2$$ and since the two members of the congruence have the same parity the formula is proved.
## Exercise 8
Prove that an odd perfect number has at least three distinct prime factors. For the perfect numbers see the article in this blog or the book [1].
Solution
We distinguish the following cases:
1. $$n = p^{k}$$
In this case $$\sigma(n) = 1+p+p^{2}+ \cdots + p^{k} \lt 2p^{k}=2n$$ contrary to the hypothesis that $$n$$ is a perfect number.
2. $$n=p^{a}q^{b}$$
In this case $$\sigma(n)=\sigma(p^{a})\sigma(p^{b})$$. Then
$\sigma(n)=p^{a}q^{b}\left(1 + \frac{1}{p} + \cdots + \frac{1}{p^{a}}\right)\left(1 + \frac{1}{q} + \cdots + \frac{1}{q^{b}}\right)$
We can bound from above the two sums with the respective geometric series obtaining:
$\sigma(n) < p^{a}q^{b}\frac{1}{1-\frac{1}{p}}\frac{1}{1-\frac{1}{q}}$
Taking the most unfavorable case for inequality ($$p=3,q=5$$) we finally get
$\sigma(n) \lt p^{a}q^{b} \frac{1}{1-\frac{1}{3}}\frac{1}{1-\frac{1}{5}}= \frac{15}{8}p^{a}q^{b} \lt 2n$
contrary to the hypothesis that $$n$$ is a perfect number.
In fact, no perfect odd number has been found to date. If they exist they must be very large numbers. For a summary of the mathematical research situation see this link to Wolfram.
## Exercise 9
The Goldbach conjecture states that every even number greater than 2 is the sum of two prime numbers.
Prove that the Goldbach conjecture is equivalent to the claim that any even number greater than 4 is sum of 3 primes.
Also show that the Goldbach conjecture implies that every odd number greater than 7 is the sum of three odd prime numbers.
The Goldbach conjecture to date has not yet been proved. For a summary of the results achieved in an attempt to prove Goldbach’s conjecture, see this link to Wolfram.
## Exercise 10
We recall the definition of the following arithmetic function
$\pi (x) = |\{p \in \mathbb{P} : p \le x\}|$
where $$x$$ is a positive real number and $$\mathbb{P}$$ represents the set of prime numbers $$\{2,3,5,7,11, \cdots \}$$.
Prove the following inequalities:
$\begin{array}{ll} \dfrac{\pi (n-1)}{n-1} \lt \dfrac{\pi (n)}{n} \quad \text{ (n prime)} \\ \dfrac{\pi (n-1)}{n-1} \gt \dfrac{\pi (n)}{n} \quad \text{(n composite)}\\ \end{array}$
Solution
If $$n$$ is a composite number then $$\pi(n) = \pi(n-1)$$ and therefore the second formula is satisfied.
If $$n$$ is prime, $$\pi(n) = \pi (n-1) +1$$. Then
$\dfrac{\pi (n)}{n} – \dfrac{\pi (n-1)}{n-1} = \dfrac{1}{n}\left(1 – \dfrac{\pi (n-1)}{n-1}\right)$
From this we derive the first formula, given that $$\pi(n) < n$$.
The function $$\pi (x)$$ plays an important role in the study of the distribution of prime numbers. A fundamental result is the Prime Number Theorem:
$\lim_{x \to \infty} \dfrac{\pi (x)}{\left(\dfrac {x}{\ln x}\right)}=1$
For an “elementary” proof of the theorem that does not use advanced tools of mathematical analysis see [2].
For a good overview see this link to Wikipedia.
## Bibliography
[1]Niven-Zuckerman-Montgomery, An introduction to the Theory of Numbers – V edition – (Wiley, 1991)
[2]Hardy-Wright, An Introduction to the Theory of Numbers – (Oxford U.P.)
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# Question Video: Simplifying Numerical Expressions Involving Square and Cube Roots Mathematics
Express ∛(875) − (64/√(8)) + √(512) + ∛(448) in its simplest form.
04:24
### Video Transcript
Express the cube root of 875 minus 64 divided by root eight plus the square root of 512 plus the cube root of 448 in its simplest form.
In order to answer this question, we need to simplify each of the terms individually. The cube root of 875 can be rewritten as the cube root of 125 multiplied by the cube root of seven. This is because 125 multiplied by seven is equal to 875. 125 is a cube number as five cubed equals 125. Therefore, the cube root of 125 is equal to five. The cube root of 875 in its simplest form is therefore five multiplied by the cube root of seven.
In order to simplify the second term, 64 divided by root eight, we need to rationalize the denominator. We do this by multiplying the top and the bottom of the fraction by root eight. 64 multiplied by root eight is equal to 64 root eight. And root eight multiplied by root eight is equal to eight. We can then divide 64 by eight leaving us with eight root eight. Root eight can also be written as root four multiplied by root two. The square root of four is equal to two. Therefore, root eight is equal to two root two. Multiplying this by eight tells us that 64 divided by root eight in its simplest form is 16 root two.
The third term, the square root of 512, can be rewritten as the square root of 256 multiplied by the square root of two. 256 is a square number as 16 multiplied by 16 equals 256. Therefore, the square root of 256 is 16. This means that the square root of 512 in its simplest form is 16 root two.
Finally, we need to simplify the cube root of 448. 64 multiplied by seven is equal to 448. Therefore, the cube root of 448 can be rewritten as the cube root of 64 multiplied by the cube root of seven. 64 is a cube number as four cubed equals 64. Therefore, the cube root of 64 is equal to four. We can therefore write the cube root of 448 in its simplest form as four multiplied by the cube root of seven.
We now have simplified expressions for all four terms. The cube root of 875 minus 64 divided by root eight plus the square root of 512 plus the cube root of 448 is therefore equal to five multiplied by the cube root of seven minus 16 root two plus 16 root two plus four multiplied by the cube root of seven. The two 16 root twos cancel as negative 16 root two plus 16 root two is equal to zero. Five multiplied by the cube root of seven plus four multiplied by the cube root of seven is equal to nine multiplied by the cube root of seven.
This is the answer to our expression in its simplest form, nine multiplied by the cube root of seven.
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Day 1
Graphing Linear Equations Using Tables
by Chris Reid
A solution of an equation is the ordered pair (x, y) that makes the equation true for the two variables x and y.
The set of all points (x, y) that are solutions to the equation is the graph of an equation in x and y.
Drections:
Use a table of values to graph the equation.
Process:
Step 1: Rewrite the equation in function form, if necessary.
Step 2: Choose a few values of x and make a table of values.
Step 3: Plot the points from the table. The graph of the equation is the line through theses points.
Example 1:
Use a table of values to graph the equation y = 2x - 1.
x 2x -1 y Points -1 2(-1)-1 -3 (-1, -3) 0 2(0)-1 -1 (0, -1) 1 2(1)-1 1 (1, 1) 2 2(2)-1 3 (2, 3)
Example 2:
Use a table of values to graph the equation 3x + 4y = 8.
First, write in function form.
3x + 4y = 8
4y = -3x + 8
y = -3/4 x + 2
x -3/4 x + 2 y Points -4 -3/4(-4) + 2 5 (-4, 5) 0 -3/4(0) + 2 2 (0, 2) 4 -3/4(4) + 2 -1 (4, -1) 8 -3/4(8) + 2 -4 (8, -4)
Special Graphs:
Example 3:
Graph x = -3.
Each value of x is always -3. Choose different values for y in the table.
x y Points -3 0 (-3, 0) -3 1 (-3, 1) -3 2 (3, 2) -3 3 (-3. 3)
Example 4:
Graph y = 2.
Each value of y is always 2. Choose different values for x in the table.
x y Points 0 2 (0, 2) 1 2 (1, 2) 2 2 (2, 2) 3 2 (3, 2)
Assignment:
Use a table of values to graph each equation.
1. y = x + 1
2. y = -x - 3
3. y = 2x + 1
4. y = -3x + 2
5. 2x - y = 6
6. 3x + 2y = 8
7. x = 4
8. x = -1
9. y = 3
10. y = -2
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# Logical Reasoning Tips – CAT 2016 – Preparation Tips 2IIM
Consider a cuboid of dimension 6cm * 8cm * 5cm. Three pairs of opposite sides of this cube are painted Red, Green and Blue. The largest surfaces have been painted Red, the smallest have been painted Blue. This cuboid is subsequently broken down into smaller cubes of dimension 1cm * 1cm * 1cm. Answer the following questions based on the information provided.
1.How many of the smaller cubes have none of their sides painted?
2.How many have exactly 2 sides painted?
3.How many more cubes have red paint as one of their surfaces than the number of ones that have blue paint on them?
This is a question from the online course following which we had given the description “Of course, the solution for this will not be provided. The horse can only be taken to the pond and all that. The extension to this is the question where we deal with a cuboid of side m * n * l with m > n > l.”
Turns out, this question has become akin to the forbidden fruit and the clamour for answers has increased steadily. So, we have decided to have a go at the solution here.
Detailed Explanation
We have been provided with a cube of dimensions 6*8*5, with the faces coloured blue, green and red. The 8*6 surfaces are painted red and the 6*5 surfaces are painted blue. So, the middle 8%5 should have been painted green. In total the cubic units are 240 in number.
Before we go into the solutions, let us define a few terms
1. The cuboid has six surfaces. For the purpose of this question, imagine the room you are in to be the cuboid. The cuboid has 3 pairs of rectangular surfaces around it
2. There are totally 8 * 6 * 5 = 240 smaller cubes that constitute this larger cuboid.
3. Now, we are going to take a count of the different parts of this cuboid.
1. The cuboid has 6 surfaces
2. It has 8 corners. Let us call these as 8 vertices
3. It has 12 straight lines – 4 on the top surface, 4 on the bottom surface and 4 that look like pillars between these two surfaces. Let us call these as “Edges”
4. It has 240 small cubes sitting inside.
Question 1) How many of the smaller cubes have none of their sides painted?
We are tasked with the determination of the number of colourless cubes; to do this mandates the realization that any cube with at least one face exposed is coloured; thus only cubes hidden from view have to be considered.
The cuboid thus consists of a colourless core and a coloured periphery. The easiest way to do this problem is to understand that the colourless core is itself a cuboid, the dimensions of which are determined by deducting 2 from each of those of the original cuboid enclosing it, this being done to discount the cubes that are exposed, as might be clear from the illustration given below.
Thus the number of uncoloured cubes is 4*6*3 = 72. This is the answer to the first question.
Moving on to the second question – 2) How many have exactly 2 sides painted?
We have to consider only edge cubes that are not corner cubes—corner cubes have three faces painted—and this is done best by considering one edge at a time from the total of twelve.
There are four edges of length 5; these have but 3 cubes each that are not corner cubes as each edge connects two corners.
There are four edges of length 6; these have but 4 cubes each that are not corner cubes.
There are four edges of length 8; these have but 6 cubes each that are not corner cubes.
Thus the total number in the second category of having two faces painted is 4 times the sum of 3, 4 and 6, which is 52. 4* ( 3 + 4 + 6) = 52.
The third problem is – 3) How many more cubes have red paint as one of their surfaces than the number of ones that have blue paint on them?
This is easily tackled, as all that is required is the difference between the number of cubes with at least one side red and the number of those with at least one side blue.
For red we have 6*8 + 6*8 = 96.
For blue we have 6*5 + 6*5 = 60.
The difference is thus 36.
The answers are 72, 52 and 36 for the three questions.
____
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# PROPERTIES OF TRIANGLE
The following are the two important properties of triangle.
1. The sum of the lengths of any two sides of a triangle is greater than the third side.
2. The sum of all the three angles of a triangle is 180°.
## Some Other Properties of Triangle
1. In an equilateral triangle, all the three sides and three angles will be equal and each angle will measure 60°.
2. In an isosceles triangle, the lengths of two of the sides will be equal. And the corresponding angles of the equal sides will be equal.
3. In a right triangle, square of the hypotenuse is equal to the sum of the squares of other two sides. This is known as Pythagorean theorem.
Hypotenuse :
Hypotenuse is the longest side in any right triangle which is opposite to right angle (90°)
## Practice Problems
Problem 1 :
Is it possible to have a triangle whose sides are 5 cm, 6 cm and 4 cm ?
Solution :
According to the properties of triangle explained above, if the sum of the lengths of any two sides is greater than the third side, then the given sides will form a triangle.
Let us apply this property for the given sides.
5 cm + 6 cm > 4 cm.
6 cm + 4 cm > 5 cm.
5 cm + 4 cm > 6 cm
Since the given sides meet the condition said in the property, It is possible to have a triangle whose sides are 5 cm, 6 cm and 4 cm.
Problem 2 :
Is it possible to have a triangle whose sides are 7 cm, 2 cm and 4 cm ?
Solution :
According to the properties of triangle explained above, if the sum of the lengths of any two sides is greater than the third side, then the given sides will form a triangle.
Let us apply this property for the given sides.
2 cm + 4 cm < 7 cm.
From the above point, it is clear the sum of the lengths of the two sides 2 cm and 4 cm is less than the third side 7 cm.
The given sides do not meet the condition said in the property.
So, it is not possible to have a triangle whose sides are 7 cm, 2 cm and 4 cm.
Problem 3 :
Find the length of the hypotenuse of the right triangle where the lengths of the other two sides are 8 units and 6 units.
Solution :
From the given information we can draw the triangle as given below.
In the above triangle, we have to find the value of "x"
According to Pythagorean theorem, square of the hypotenuse is equal to the sum of the squares of other two sides
So, we have
x2 = 82 + 62
x2 = 64 + 36
x2 = 100
x = 10
So, the length of the hypotenuse is 10 units.
Problem 4 :
The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.
Solution :
Let 'x' and 'x + 4' be the lengths of other two sides.
Using Pythagorean theorem,
(x + 4)2 + x2 = 202
x2 + 8x + 16 + x2 - 400 = 0
2x2 + 8x - 384 = 0
x2 + 4x - 192 = 0
(x + 16)(x - 12) = 0
x = -16 or x = 12
x = -16 can not be accepted. Because length can not be negative.
If x = 12,
x + 4 = 12 + 4 = 16
So, the other two sides of the triangle are 12 cm and 16 cm.
Apart from the problems given above, if you need more problems on triangle properties, please click the following links.
Triangle Properties 1
Triangle Properties 2
Triangle Properties 3
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# How do you factor (b−c)^3+(c−a)^3+(a−b)^3?
Apr 1, 2017
$3 \left(b - c\right) \left(c - a\right) \left(a - b\right) .$
#### Explanation:
If we use the following Result, we can immediately factorise the
given Exp.$= 3 \left(b - c\right) \left(c - a\right) \left(a - b\right) .$
Result : $x + y + z = 0 \Rightarrow {x}^{3} + {y}^{3} + {z}^{3} = 3 x y z .$
Otherwise, consider the following :
Let, $u = b - c , v = c - a \Rightarrow u + v = b - a = - \left(a - b\right) .$
Now, The Exp.$= {u}^{3} + {v}^{3} + {\left\{- \left(u + v\right)\right\}}^{3} ,$
$= {u}^{3} + {v}^{3} - {\left(u + v\right)}^{3} ,$
$= {u}^{3} + {v}^{3} - \left\{{u}^{3} + {v}^{3} + 3 u v \left(u + v\right)\right\} ,$
$= - 3 u v \left(u + v\right) = 3 u v \left\{- \left(u + v\right)\right\} ,$
$\therefore \text{ The Exp.=} 3 \left(b - c\right) \left(c - a\right) \left(a - b\right) .$
Enjoy Maths.!
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# Video: GCSE Mathematics Foundation Tier Pack 2 • Paper 1 • Question 5
GCSE Mathematics Foundation Tier Pack 2 • Paper 1 • Question 5
02:06
### Video Transcript
Evaluate four cubed.
So the first thing to consider with this question is what is four cubed do actually tell us; what does it mean? Well, four cubed means four multiplied by four multiplied by four. It is not four multiplied by three. This is a really common mistake that students make. They see four cubed or they can have two to the power of five and they just multiply the two numbers together. And this is a common mistake like I said. This is not the case.
Okay, so we got four multiplied by four multiplied by four. So here I’ve got a tip. A tip is to try and actually learn or you square numbers up to 12 squared and also all you cube numbers for five cubed to help us with questions like this. So for example, do you know that one cubed is one, two cubed is eight, three cubed is 27, four cubed is 64, and five cubed is 125.
So using what we could remember, we’d know the answer to this question would just be 64. Like I said, it’s really useful. I mean for a question like this, we can actually work out and I am gonna show you how you can actually calculate it. Particularly, when we got multistage problems, it saves you a step of calculations.
Okay, so we think it’s gonna be 64. So let’s try and show that with a calculation. So now, we get that four cubed is gonna be equal to 16 multiplied by four and that’s because four multiplied by four is 16.
Okay, great, so now what we want to do is actually multiply these together. And the way you do that is using any one of our multiplication methods. I’ve just set it up using the column method.
So first of all, we multiply four and six. So therefore, this gives us 24. So I put the four in the units column and then carry the two into the tens column. Then, I have four multiplied by one, which gives me four. Then, I add on the two that I carried from earlier, which gives me six. So I put this into the tens column.
So with that calculation, we can confirm that four cubed is equal to 64. And that was the same using our calculation method and also like I said earlier with the values that we should remember and that’s one of our cube numbers up to five cubed.
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Quadrilaterals Properties | Parallelograms, Trapezium, Rhombus
Click here to watch this insightful video.
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In this tutorial on basic geometry concepts, we cover the types and properties of quadrilaterals: Parallelogram, rectangle, square, rhombus, trapezium.
Definition:
A quadrilateral is a simple closed figure with four sides.
There are five types of quadrilaterals.
• Parallelogram
• Rectangle
• Square
• Rhombus
• Trapezium
One common property of all quadrilaterals is that the sum of all their angles equals 360°.
Let us look into the properties of different quadrilaterals.
Parallelogram
Properties of a parallelogram
• Opposite sides are parallel and congruent.
• Opposite angles are congruent.
• Adjacent angles are supplementary.
• Diagonals bisect each other and each diagonal divides the parallelogram into two congruent triangles.
• If one of the angles of a parallelogram is a right angle then all other angles are right and it becomes a rectangle.
Important formulas of parallelograms
• Area = L * H
• Perimeter = 2(L+B)
Rectangles
Properties of a Rectangle
• Opposite sides are parallel and congruent.
• All angles are right.
• The diagonals are congruent and bisect each other (divide each other equally).
• Opposite angles formed at the point where diagonals meet are congruent.
• A rectangle is a special type of parallelogram whose angles are right.
Important formulas for rectangles
• If the length is L and breadth is B, then
Length of the diagonal of a rectangle = √(L2 + B2)
• Area = L * B
• Perimeter = 2(L+B)
Squares
Properties of a square
• All sides and angles are congruent.
• Opposite sides are parallel to each other.
• The diagonals are congruent.
• The diagonals are perpendicular to and bisect each other.
• A square is a special type of parallelogram whose all angles and sides are equal.
• Also, a parallelogram becomes a square when the diagonals are equal and right bisectors of each other.
Important formulas for Squares
• If ‘L’ is the length of the side of a square then length of the diagonal = L √2.
• Area = L2.
• Perimeter = 4L
Rhombus
Properties of a Rhombus
• All sides are congruent.
• Opposite angles are congruent.
• The diagonals are perpendicular to and bisect each other.
• Adjacent angles are supplementary (For eg., ∠A + ∠B = 180°).
• A rhombus is a parallelogram whose diagonals are perpendicular to each other.
Important formulas for a Rhombus
If a and b are the lengths of the diagonals of a rhombus,
• Area = (a* b) / 2
• Perimeter = 4L
Trapezium
Properties of a Trapezium
• The bases of the trapezium are parallel to each other (MN ⫽ OP).
• No sides, angles and diagonals are congruent.
Important Formulas for a Trapezium
• Area = (1/2) h (L+L2)
• Perimeter = L + L1 + L2 + L3
Summary of properties
Summarizing what we have learnt so far for easy reference and remembrance:
S.No. Property Parallelogram Rectangle Rhombus Square 1 All sides are congruent ✕ ✕ ✓ ✓ 2 Opposite sides are parallel and congruent ✓ ✓ ✓ ✓ 3 All angles are congruent ✕ ✓ ✕ ✓ 4 Opposite angles are congruent ✓ ✓ ✓ ✓ 5 Diagonals are congruent ✕ ✓ ✕ ✓ 6 Diagonals are perpendicular ✕ ✕ ✓ ✓ 7 Diagonals bisect each other ✓ ✓ ✓ ✓ 8 Adjacent angles are supplementary ✓ ✓ ✓ ✓
Continue learning more about:
Properties of Lines and Angles
Properties and formulas of Circles
Types of Triangles and Properties
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# In a certain shop, which stocks four types of caps, there are $\dfrac{1}{3}$as many red caps as blue and $\dfrac{1}{2}$ as many green caps as red caps. There are equal numbers of green caps and yellow caps. If there are 42 blue caps, then what percent of the total caps in the shop are blue? A.70% B.28% C.60% D.14%
Last updated date: 25th Jun 2024
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Hint:Here in this question total number of blue caps given ,$\dfrac{1}{3}$ of total number of blue caps gives red caps and $\dfrac{1}{2}$ of red caps gives green caps.In question they given number of green caps is same as yellow caps. Add all the number of caps of different colors and find the percentage which is our required answer.
Formula used: The percentage, say P,is given by $P = \dfrac{{value}}{{total}} \times 100$.
Given, There are 42 blue caps.
There are $\dfrac{1}{3}$ as many red caps as blue.
$\Rightarrow \dfrac{1}{3} \times 42 = 14$
Therefore, there are 14 red caps.
And, It is given that $\dfrac{1}{2}$ as many green caps as red caps.
$\Rightarrow \dfrac{1}{2} \times 14 = 7$
Therefore, the number of green caps is 7.
As given, the number of yellow and green caps is the same.
Therefore, there are 7 yellow caps.
$\text{Total number of caps}= 42 + 14 + 7 + 7 = 70$
The total number of caps is 70.
The percentage, say P,is given by $P = \dfrac{{value}}{{total}} \times 100$.
Here, we have to find the percentage of blue caps. The number of blue caps is 42 and the total number of caps is 70.
$\Rightarrow p = \dfrac{{42}}{{70}} \times 100 = 60\%$
So, the correct answer is “Option C”.
Note: the question can also be framed asking the percentage of red color caps or any color cap.
The percentage of red color caps will be, the number of red caps is 14. The total number of the cap is 70.
$P = \dfrac{{value}}{{total}} \times 100 \\ \Rightarrow p = \dfrac{{14}}{{70}} \times 100 \\ \Rightarrow p = 20\% \\$
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700 in Words
# 700 in Words
Edited By Team Careers360 | Updated on Feb 15, 2023 11:51 AM IST
## Introduction
In words, 700 equals seven hundred; that is, writing 700 in words. Let us say a dress has an MRP of ₹ 700. It could be said then that to buy that dress, one must spend seven hundred rupees. Here 700 or seven hundred denotes the quantity of money required to be spent for the dress, so it can also be called the cardinal number.
## Is 700 an Odd Number or an Even Number?
• Any number divisible by 2 is called an even number.
• Any number not divisible by 2 is called an odd number.
• 700 is an even number. As it is divisible by 2.
## Place Value, and The Expansion of 700
The place value of 700 can be described using the below-given place value table, which is
Place of digit Hundreds Tens Ones Digit 7 0 0
Now if we look at the table above, we can write 700 in expanded form. That is
700=7 \times \text { hundred }+0 \times \text { ten }+0 \times \text { one }
This can also be written as
\begin{aligned}
&=7 \times 100+0 \times 10+0 \times 1 \\
&=700
\end{aligned}
## Other Examples
• If we have 600 books and our friend gives us more than 100 books then in total we will have 700 or seven hundred books.
You can find out the detailed other number in words article list below:-
20000 in Words 350000 in Words 50 in Words 65000 in Words 18000 in Words 28000 in Words 150 in Words 80 in Words 49000 in Words 400 in Words 11800 in Words 13500 in Words 12 in Words 71000 in Words 55000 in Words 150000 in Words 24000 in Words 59000 in Words 900 in Words 27000 in Words 12000 in Words 14400 in Words 12400 in Words 300000 in Words 17 in Words 12800 in Words 75 in Words 33000 in Words 21000 in Words 1180 in Words 41000 in Words 36000 in Words 1999 in Words 28 in Words 57000 in Words 4500 in Words 85000 in Words 95000 in Words 1200 in Words 14500 in Words 83000 in Words 48000 in Words 49 in Words 500000 in Words 60 in Words 70 in Words 10500 in Words 1100 in Words 25 in Words 27000 in Words 31000 in Words 34000 in Words 400000 in Words 7500 in Words 800 in Words 15500 in Words 16500 in Words 19500 in Words 250000 in Words 43000 in Words 44 in Words 45 in Words 47000 in Words 63000 in Words 6500 in Words 9500 in Words 99 in Words 10000000 in Words 1400 in Words 140000 in Words 200 in Words 21 in Words 4 in Words 41 in Words 52000 in Words 110000 in Words 125000 in Words 130000 in Words 2 in Words 23 in Words 24 in Words 33 in Words 35 in Words 450000 in Words 48 in Words 51000 in Words 54000 in Words 58000 in Words 5900 in Words 7 in Words 1 to 100 in Words 95 in Words 96000 in Words 12600 in Words 1900 in Words 20500 in Words 2400 in Words 26 in Words 2800 in Words 31 in Words 42 in Words 47 in Words 56000 in Words 600000 in Words 66000 in Words 68000 in Words 84000 in Words 88000 in Words 98 in Words 99999 in Words 11200 in Words 1200000 in Words 21500 in Words 27 in Words 29 in Words 32 in Words 32500 in Words 3300 in Words 350 in Words 3540 in Words 3600 in Words 37500 in Words 43 in Words 46 in Words 61000 in Words 67000 in Words 76000 in Words 800000 in Words 9 in Words 99000 in Words 96 in Words 10 in Words 1050 in Words 1111 in Words 120 in Words 1250 in Words 13200 in Words 180000 in Words 2000000 in Words 2300 in Words 24500 in Words 25500 in Words 27500 in Words 29500 in Words 26500 in Words 31500 in Words 37 in Words 38500 in Words 450 in Words 5000000 in Words 700000 in Words 72 in Words 77000 in Words 81000 in Words 85 in Words 86000 in Words 88 in Words 92000 in Words 93000 in Words 94 in Words 11100 in Words 11300 in Words 12200 in Words 123 in Words 13600 in Words 13900 in Words 15400 in Words 175000 in Words 1770 in Words 23600 in Words 29500 in Words 30500 in Words 33500 in Words 3700 in Words 3800 in Words 4200 in Words 42500 in Words 69000 in Words 7080 in Words 79000 in Words 82000 in Words 900000 in Words 97000 in Words 100000000 in Words 1000000000 in Words 3100 in Words 4700 in Words 4900 in Words 10700 in Words 1 to 20 in Words
1. Why is 700 a composite number?
Any number which has more than 2 factors is called a composite number. 700 is also an example of a composite number. Because 700 is divisible by numbers like 2, 5, 7, etc.
2. Write the square root of 700 in words.
The square root of 700 is given as
\begin{aligned}
\sqrt{700} &=\sqrt{(2 \times 2) \times(5 \times 5) \times 7} \\
&=2 \times 5 \times \sqrt{7} \\
&=10 \sqrt{7} \\
&=26.458
\end{aligned}
3. What is a prime number? And why is 700 not a prime number?
Any number having 1 and itself as its only 2 factors is known as a prime number. 700 is not a prime number because it has more than 2 factors like 2, 3, 5, 7, etc.
4. What is a cardinal number?
Any number defining a quantity will be termed a cardinal number. For example, if the price of a dress is ₹ 700, then 700 is the cardinal number of the price of a dress.
5. How do we know 700 is not an odd number?
To find an odd number, we must check the digit in its one’s place. If the number has 1,3,5,7, or 9 in the one’s place, then the number is known as an odd number. But in 700, there is 0 in one’s place, so the number is not odd.
Get answers from students and experts
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# Cryptarithmetic problems with explanations
Cryptarithmetic problems are where numbers are replaced with alphabets. By using standard arithmetic rules we need to decipher the alphabet.
General Rules:
1. Each alphabet takes only one number from 0 to 9 uniquely.
2. Two single digit numbers sum can be maximum 19 with carryover. So carry over in problems of two number addition is always 1.
3. Try to solve left most digit in the given problem.
4. If a × b = kb, then the following are the possibilities
(3 × 5 = 15; 7 × 5 = 35; 9 × 5 = 45) or (2 × 6 = 12; 4 × 6 = 24; 8 × 6 = 48)
Solved Example 1:
The following questions are based on the following multiplication, where each digit has been replaced by an alphabet.
1. Find the value of J and A.
2. If E = 4, what is the value of D?
Explanation:
From the first two of multiplication, you can clearly say that B = 1, as JE x B = JE. From the second row of multiplication, A = 0 as in the multiplication, second row should start from tenth's place. So A = 0. Now in the hundred's place, J + Something = 10. When you add something to the single digit number that results in 10. So J = 9. Now from the above table, we cannot determine values of E and D, but we can say, that E and D are consecutive. As it is given that E = 4, we can say D = 3.
Solved Example 2:
From the multiplication below, What is the value of NAME?
Explanation:
From the first row of multiplication, H =1 is clear, As HE x H = HE. Substitute H = 1 in all places. Now from the tenth's place, think about, the value of A. 1 + A = M. If M is a single digit number, then N = 1, which is impossible (Already we have given H = 1). So A = 9, Then M = 0, and N = 2. Now 1E x E = 119. So by trial and error E = 7.
Therefore, NAME = 2907
Solved Example 3:
Decipher the following multiplication table
Explanation:
Step 1: What could be the value of A which is the left most digit in the answer? From the second row of multiplication, we know that A x E = A. So A cannot be 1.
From the tenth's place addition, I + A = A. So I = 0. Now from the Ten-thousand's place addition, 1 + G + 0 = A. So G = 1 and A = 2.
Step 2: From the second row of multiplication, 2 x E = 2, Also from first row of multiplication T x E = 2.
So E should take 6, and T should take 7. (These are the only possibilities)
If E= 6, then E x T = 6 x 7 = 42. So 4 carry over. Now 7 x Y + 4 = 0. So Y = 8.
Now S = 4 as 7 x 6 + 6 = 48. Also M = 3.
Solved Example 4:
If SEND + MORE = MONEY then find the respective values
Explanation:
Addition of two numbers with 'n' digits, results in a n+1 digits, then the left most place always = 1.
So M=1. Substitute this value.
Now 'o' cannot be 1 as M already 1. It may not be 2 either as S+1 = 12 or 1 + S + 1 = 12 in the both cases S is a two digit number. So 'o' is nothing but zero. Put o = 0.
Now S can be either 8 or 9. If S = 8, then there must be a carry over.
E + 0 = 10 + N or 1 + E + 0 = 10 + N
In the above two cases, E - N = 10 is not possible and E - N = 9 not possible as as N cannot be zero.
So E = 9.
Now E + 0 = N is not possible as E = N. So 1 + E = N possible.
The possible cases are, N + R = 10 + E - - - (1) or 1 + N + R = 10 + E - - - (2)
Substituting E = N -1 in the first equation, N + R = 10 + N - 1, we get R = 9 which is not possible.
Substituting E = N - 1 in the second equation, 1 + N + R = 10 + N - 1, we get R = 8.
We know that N and E are consecutive and N is larger. Take (N, E) = (7, 6) check and substitute, you wont get any unique value for D.
Take (N, E) = (6, 5), Now you get D = 7, Y = 2.
Solved Example 5:
Find the values of all the alphabets if each alphabet represent a single digit from 0 - 9
Explanation:
Let us name the columns as below
We know that sum of two single digit alphabets should not cross 18, and maximum difference between two alphabets is 9.
If we add two maximum 4 digit numbers the sum is maximum 19998. So the digit in the 5th left is 1.
Now from the 1st column 1 + E = 1F; if there is any carry over from the 2nd column 1 + 1 + E = 1F
But 1F is a two digit number in alphanumeric is equal to 10 + F
So ${\rm{1 + E = 10 + F}} \Rightarrow {\rm{E - F = 9}}$
From this relatlion we know that E = 9, F = 0
or ${\rm{1 + 1 + E = 10 + F}} \Rightarrow {\rm{E - F = 8}}$
E = 9, F = 1 or E = 8, F = 0
From the above we can infer that F = 0 but we dont know whether E is equal to either 8 or 9. But surely F is not equal to 1 as we fixed already A = 1
Now from the 3rd column,
2C= 1 ⇒ C = 1/2
1 + 2C = 1 ⇒ C = 0
If the sum is a two digit number then
2C = 11 ⇒ C= 11/2
1 + 2C = 11 ⇒ C = 5
From the above C = 1/2 and 11/2 are not possible nor is 0 possible as we fixed F = 0
If C = 5 the the A = 1 and there is a carry over to the left column. and also there must be carry over from the first column, but we dont know 1 + 2B is a single digit or two digit number
From the second and fourth columns
1+2B = G - - - - (1) or 1 + 2B = 10 + G - - - (2)
D + B = 10 + G - - - (3)
Solving (1) and (3) we get D - B = 11 which is not possible
But If we solve (2) and (3) then we get D - B = 1
So D and B are consecutive numbers and their sum is more than 10. So acceptable values are D = 7 and B = 6
This completes our problem so final table looks like the following
Solved Example 6:
Find the alphabets in the following multiplication
Explanation:
This is a tough question as there are total 9 different alphabets are used.
Step 1: K + A = A. So K = 0
Step 2: From the hundreds column, 2B + A = 10 or 20. As 2B, 10, 20 are even, A should be even. Remember this logic.
Possibilities are, for A and B are (2, 4), (4, 3), (6, 2), (8, 1) and (2, 9), (4, 8), (6, 7), (8, 6)
In the second row of multiplication, we have PAS x B = ASAA.
P2S x 4 = 2S22 ⇒S = 3, 8 But both are not satisfying.
P4S x 3 = 4S44 ⇒S = 8. But P48 x 3 = 4844 is not possible. Ruled out.
P6S x 2 = 6S66 ⇒S = 3, 8. But both are not satisfying. Ruled out.
P2S x 9 = 2S22 ⇒S = 8 But P28 x 9 = 2822 is not possible. Ruled out.
P4S x 8 = 4S44 ⇒S = 3. This is possible as P43 x 8 = 4344 then P = 5.
P6S x 7 = 6S66 ⇒S = 8 But P68 x 7 = 6866 is not possible. Ruled out.
P8S x 6 = 8S88 ⇒S = 3, 8 But both are not satisfying. Ruled out.
Therefore, S = 3, P = 5, A = 4, B = 8.
From the above diagram, R = 6 and E = 2. and A = 7 and W = 1.
Final Solution:
Solved Example 7:
Find the values of A, B and C if $ABC = A! + B! + C!$ where ABC is a three digit number
Explanation:
By symmetry, we take A is the maximum number of the three alphabets.
Let us say A can take a maximum value of 6.
Then 6! = 720 but as have we taken maximum value is 6 720 is not possible
So A should take 5. Then 5! = 120
Now from the above we know that one the B and C should take 1 as their value as 120 consists of 1.
$\Rightarrow$ 5! + 1! = 121
If we take 4 as one of the number then 5!+ 1! + 4! = 145 or 1! + 4! + 5! = 145
Solved Example 8:
Find the values of A, B and C if $ABC = {A^3} + {B^3} + {C^3}$ where ABC is a three digit number.
Explanation:
Let us say maximum value of A, B, C is equal to 4 then we dont get any satisfactory values for ABC
If we take maximum value is 5 then ${A^3}$ = ${5^3}$ = 125 as this is a three digit number one of the number is equal to 1.
$\Rightarrow$ ${5^3} + {1^3}$ = 126. Now for ${3^3}$ we get $153 = {1^3} + {5^3} + {3^3}$
From the above reasoning the other numbers satisfy the above relation are 370, 371 and 407.
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## Friday, 10 February 2012
### Proof by Induction: Sum of Odd Numbers
I will introduce an integral part of mathematics in this blog post, proof by induction. It is fundamental to many proofs and is a pretty simple concept to grasp, I will demonstrate how this proof works with a simple example.
The first basis of proving by induction requires some intelligent intuition to our problem. This means that we look at the pattern of what is happening and try to prove this for all values. For the sum of square numbers this goes:
1 = 1
1+3 = 4
1+3+5 = 9
1+3+5+7 = 16
You may have noticed from this that the sum the answers is always a square number. The sum of the first n odd numbers appears to always equal n2, but it just appears that way for the examples that we have given, we do not know that is true for all values, yet.
So our assumption is the sum of the first n odd numbers equals to n2:
1+3+5+...+(2n-1) = n2
This is our assumption, but if we can prove that n+1 odd numbers equals to (n+1)2 then that means it will hold true for the sums of any number of odd numbers. To begin to check this we add n+1th odd number onto both sides of our assumption.
1+3+5+...+(2n-1)+(2n+1) = n2+2n+1
As you can see the right side does factorise to give:
1+3+5+...+(2n-1)+(2n+1) = (n+1)2
Which means that our original assumption was correct! Ergo the sum of the first n odd numbers does equal to nformally in mathematical notation this is:
And that is it for an example of how you use the proof by induction, now it just needs to be put in terms that are applicable for all problems.
Principle of Mathematical Induction
If for each positive integer n we have a statement P(n). If we prove the following two things then the statement is true:
a.) P(1) is true.
b.) For all n, if P(n) is true then P(n+1) is also true. This then means that P(n) is true for all positive integers n.
If that is pretty hard to compute look back to our example, our assumed statement is that P(n) = n2 our assumption is true for P(1), 1 = 12 and we then proved that assuming P(n) is true that P(n+1) is also true. Therefore if P(1) is true, then P(2) is true too, etc.
Stay tuned for my future blog posts where I will utilise mathematical induction to find general formulas or proofs to many things.
#### 1 comment:
1. But why is this so? I understand the proof, and I computed it on my own, but do we have any feel for why this would be so? What does this tell us about the numbers?
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# [Maths Class Notes] on Brackets Pdf for Exam
The first question the student gets on this topic is “How can we define brackets”. In evaluating an expression containing a bracketed sub-expression, brackets denote a type of grouping, the operators in the sub-expression take precedence over those surrounding it. Additionally, for the different brackets, there are many uses and definitions.
### Types of Brackets
The frequently used bracket types are:
• Parentheses ( )
• Square brackets [ ]
• Curly brackets { }
• Angle brackets ⟨ ⟩
### Parentheses
Among the four different types of brackets used, parentheses are the most commonly used bracket type. In mathematical problems, the primary use of parentheses is to group numbers. Use the order of operations to solve the problem when we see multiple numbers and operations in parentheses.
For three key purposes, parentheses are used in mathematics:
To separate numbers for clarification, parentheses may be used. For instance, if we have an additional problem with a negative number, to distinguish the two signs, parentheses will be used. To distinguish a number from its exponents, parentheses may also be used. Typically, this occurs if we lift a negative number to control.
### Square Brackets
In mathematics, the square bracket symbols [ ] are employed in a variety of situations:
• Square brackets are sometimes used in place of (or in addition to) parentheses in very complex expressions, especially as a group sign outside an inner set of parentheses.
• They can signify the same thing as parenthesis, but they’re meant to make things easier to read. It all depends on the situation.
• Square brackets are used to include the number it covers while working with inclusion.
• They can also be used to denote the least common multiple
### Curly Brackets (also known as Braces)
Left curly brackets and right curly brackets are used together in mathematical expressions. They can be replaced by square brackets or parentheses. In a nested phrase with three layers of grouping, parentheses are usually used in the innermost groupings. In the next higher level grouping, square brackets are employed, while braces are used in the outermost groups (see ” Nested expressions ” for an example).
### Angle Brackets
The inner product of two functions is represented by an angle bracket, which is made up of a bra and a ket (bra+ket = bracket). Because angle brackets resemble the “less than” and “greater than” signs, they might seem confusing to some students. But you will get a hold of it, once you start using them in your maths practice from time to time.
### What is the Use of Brackets?
Ex: 5 * (2 + 4) is 30, (5 * 3) + 2 is 30.
• Brackets are often used in mathematical expressions in general to signify grouping where appropriate to prevent ambiguities and increase clarity.
• In the Cartesian system of coordinates, brackets are used to designate point coordinates.
Ex: (4,8) denotes the points on the x-y coordinate system with x-coordinate being 4 and y-coordinate being 8.
Ex: f(x), g(x).
Ex: [0,8) denotes a half-closed interval that includes all real numbers, except 8 from 0 to 8.
• Wide parentheses around two numbers denote a binomial coefficient, one above the other.
• As in (a,b,c), parentheses around a set of two or more numbers denote an n-tuple of numbers that are connected in a particular way.
• A matrix is indicated by broad brackets around an array of numbers.
• To denote the largest common divisor, parentheses are used.
### BODMAS Rule
Brackets find their main application in the BODMAS or PEMDAS rule where the sequence of operations is to be performed when an expression is resolved. BODMAS or PEMDAS stands for:
B – Brackets, P- Parentheses
O – Order, E- Exponents
D – Division
M – Multiplication
S – Subtraction
The BODMAS rule explains the sequence of operations to be done until an expression is resolved. According to the BODMAS law, if there are brackets ((), {},)in an expression, we first have to overcome or simplify the bracket followed by the order, then divide, multiply, add and subtract from left to right. In the wrong order, solving the issue would result in a wrong answer.
Simply put, the four operations are crucial to arithmetic learning, and youngsters who don’t know which sequence to finish them in will fail to move through the years.
Another reason BODMAS is taught in math classes is that it makes it much easier for young students to remember which operation to perform when faced with complex equations.
### Basic Problems on Brackets and their Application:
1) Solve (2 + 4) – (6 – 3)
Ans: Two parentheses are involved in the given expression. We can solve both of them separately by the BODMAS rule and then combine their results.
(2 + 4) = 6……….(1)
(6 – 3) = 3………..(2)
Now subtracting (1) with (2), we get
(2 + 4) – (6 – 3) = 6 – 3 = 3
2) Solve (3 + (5 * 4)) – ((4 * 6) – 10)
Ans: Four parentheses are involved in the given expression. We will solve it by using the BODMAS rule to find the answer.
First parentheses is (5 * 4) = 20……………………………..(1)
Second parentheses is (3 + (5*4))=(3 + 20) =23………(2)
Third parentheses is (4 * 6) = 24……………………………(3)
Fourth parentheses is ((4 * 6) – 10) = (24 – 10)
= 14……(4)
Now subtract (2) and (4) we get
(3 + (5 * 4)) – ((4 * 6) – 10) = 23 -14 = 9.
|
# Addition facts for sums with 6
This is a complete lesson with instruction and exercises about basic addition facts when the sum is 6, meant for 1st grade math. You can use the same ideas to teach other sums (sums with 7, 8, 9, 10, etc.) as well.
The video below shows many ideas and strategies for learning basic addition facts: grouping of sums as in this lesson, the 9-trick, doubles, and more.
1. Six hippos are grouped into two groups, in different ways. Write the addition sentences.
| ______ + ______ = ______ |______ + ______ = ______ | ______ + ______ = ______ | ______ + ______ = ______ | ______ + ______ = ______ | ______ + ______ = ______ | ______ + ______ = ______
2. Play “6 Out” and/or “Some Went Hiding” with 6 objects (see the introduction).
1 + = 6 2 + = 6 6 + = 6 4 + = 6 3 + = 6 5 + = 6 + 2 = 6 + 0 = 6 + 4 = 6 + 3 = 6 + 1 = 6 + 5 = 6
4. Add the numbers and write the total on the line.
a. 1 + 5 = ______ b. 2 + 3 = ______ c. 4 + 2 = ______
5. Draw more little boxes to illustrate the missing number.
a. 2 + = 6 b. 2 + = 5 c. 4 + = 6 d. 3 + = 6 e. 1 + = 6 f. 5 + = 6 g. 1+ = 5 h. 0 + = 6 i. 3 + = 5
6. Jack and Jill share 5 cucumbers and 6 lemons in different ways. Find how many Jill gets.
You can cover the cucumbers or lemons with your hand to help.
a. 5
Jack gets: Left for Jill: 2 1 5 3 0 4
b. 6
Jack gets: Left for Jill: 1 4 5 0 2 3
2 + 3 = ______
4 + 1 = ______
3 + 3 = ______
4 + 2 = ______
1 + 3 = ______
1 + 5 = ______
2 + 2 = ______
2 + 4 = ______
This lesson is taken from Maria Miller's book Math Mammoth Addition 1, and posted at www.HomeschoolMath.net with permission from the author. Copyright © Maria Miller.
|
# 3-2 Solving Linear Systems Algebraically - PowerPoint PPT Presentation
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## 3-2 Solving Linear Systems Algebraically
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### 3-2 Solving Linear Systems Algebraically Objective: CA 2.0: Students solve system of linear equations in two variables algebraically. Substitution Method Example 1 ... – PowerPoint PPT presentation
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Title: 3-2 Solving Linear Systems Algebraically
1
3-2 Solving Linear Systems Algebraically
Objective CA 2.0 Students solve system of
linear equations in two variables algebraically.
2
Substitution Method
1. Solve one equation for one of its variables
2. Substitute the expression from step 1 into
the other equation and solve for the other
variable
3. Substitute the value from step 2 into the
revised equation from step 1 and solve
3
Example 1
Solve the linear system of equation using the
substitution method.
4
Step 1
Solve one equation for one of its variables.
5
Step 2
Substitute the expression from step 1 into the
other equation and solve.
6
Step 3
Substitute the value from step 2 into the revised
equation from step 1 and solve.
The solution is (-8, 5)
7
Which equation should you choose in step 1?
In general, you should solve for a variable whose
coefficient is 1 or -1
8
The Linear Combination Method
Step 1 Multiply one or both of the equations by
a constant to obtain coefficients that differ
only in sign for one variable.
9
Step 2 Add the revised equations from step 1.
Combining like terms will eliminate one variable.
Solve for the remaining variable.
Step 3 Substitute the value obtained in Step 2
into either of the original equations and solve
for the other variable.
10
Example 2
Solve the linear system using the Linear
Combination (Elimination) Method.
11
Step 1
Multiply one or both of the equations by a
constant to obtain coefficients that differ only
in sign for one variable
Multiply everything by -2
Leave alone
12
Step 2
After step 1 we now have
-4x 8y -26 4x 5y 8
Solve for y
13
Step 3
Substitute the value obtained in Step 2 into
either of the original equations and solve for
the other variable.
14
The solution checks
15
Example 3
Linear Combination Multiply both
Equations Solve the linear system using the
Linear Combination method.
16
Step 1) Multiply one or both equations by a
constant to obtain coefficients that differ only
in sign for one variable.
17
Step 2) Add the revised equations
18
Step 3) Substitute the value obtained in Step 2
into either original equation
Solution (2, 3)
19
Example 4 Linear Systems with many or no
solutions
Solve the linear system.
20
Use the substitution method Step 1)
Step 2)
6 7 Because 6 is not equal to 7, there are no
solutions.
21
Two lines that do not intersect are parallel.
22
Solve the Linear System
Solve using the linear combination method
Step 1)
23
|
# nichiyes.com
The Learn Fun Facts blog posted an interesting fun fact about strings. Law’s fun fact is that 8+9+1+89+91=198. Why is this a fun fact? Because the left side of that equation takes every sequence of digits from 891 (except the whole thing) and adds them together to get the digits in reverse order. Contrary to Law’s statement, though, this is not the only three digit number that does this. At the end of the post, he challenges us to figure out if this happens for any five digit combinations. He suggests using a computer to do it, but I wanted to see if basic algebra and a little cleverness could do it. So first I set out on the 3-digit problem to see how the process works.
## The 3-Digit Problem
So, what do we know to start with? We’re working with three digits. Let’s call them x, y, and z. So for 891, x=8, y=9, and z=1. To make them match up with 891, we note that 100x+10y+z=891. To reverse the digits, just reverse the coefficients: x+10y+100z=198.
Now, to figure out that it’s 8, 9, and 1, we can’t assume that from the start. But we are working on making the “substrings” (the sequences of digits within the number) equal x+10y+100z. How do we express the sum of substrings? x+y+z+10x+y+10y+z. If we clean that up a bit, we have 11x+12y+2z. Thus we are trying to find out when this is true:
x+10y+100z=11x+12y+2z
Well, if you just have three variables and one equation, you’re going to get a lot of possible solution. But wait, x, y, and z all have to be single digits. So we know that 0≤x9, 0≤y≤9, and 0≤z≤9. And since we want a three digit number, 1≤x. Since we want to be able to flip it into another three digit number, 1≤z. And since they’re digits, we know that they’re all whole numbers. Okay, now we have some stuff to work with.
If we take our equation from before and subtract all the stuff on the right from both sides we get:
10x+2y-98z=0
Then divide everything by 2:
5x+y-49z=0
We want to isolate one variable to work with, and that z is being subtracted right now, so let’s try moving it back to the right:
5x+y=49z
Well that’s quite the disparity in coefficients! And here’s where the magic is. Since 1≤z, the smallest thing 49z can be is 49. Since we know 1≤x≤9, we also know 5≤x≤45. And since y≤9, the biggest thing 5x+y can be is 54. But since 5x+y=49z, that means the biggest thing 49z can be is 54. And since z is a whole number, we can see that 49*2=98 which is too big. So z must be 1.
If we plug z=1 into our equation, we have:
5x+y=49
Now we’re down to two variables. If we subtract y from both sides we’ll be able to get some nice bounds on 5x, so let’s do that:
5x=49-y
Since 1≤y≤9, 40≤49-y≤48. And since 5x=49-y, 40≤5x≤48. But we know x has to be a whole number. So the only options for x are 8 and 9. So if we plug in either option:
(5*8 or 5*9)=49-y
Which is to say
(40 or 45)=49-y
So if we subtract the two options from each side:
0=(4 or 9)-y
y=4 or 9
And notice that the y=4 goes with x=9, and y=9 goes with x=8. So now all three variables are solved for. Either x=8, y=9, and z=1, or x=9, y=4, and z=1. These correspond to 891 and 941.
So remember that the fun thing about 891 is that 8+9+1+89+91=198. So now let’s look at 941. We can see 9+4+1+94+41=149. Neat! So there are in fact two three-digit numbers with this property.
## Can it be done without the reversal?
On the Learn Fun Facts post, Jack Shalom asks in a comment whether there are any three-digit numbers whose substrings add up to the number itself. The answer is no, and here is the proof.
We start again with 1≤x≤9, 0≤y9, and 0≤z≤9. (We’re not reversing it, so z being 0 would be fine.) Now the equation we want to figure out what makes true is:
100x+10y+z=x+y+z+10x+y+10y+z
Which simplifies to:
100x+10y+z=11z+12y+2z
If we shuffle the ys and zs to the right and xs to the left we get:
89x=2y+z
Since 1≤x, 89≤89x. But since y≤9 and z≤9, the biggest thing 2y+z can be is 2*9+9=27. So this system has no solution.
## The 5-Digit Problem
The 5-digit version is obviously trickier. Because it’s so cumbersome to write and read the process of defining everything again, I’ll skip to the equation. This time I use a through e as digits for the number abcde since they’re easier to tell apart than some of the later letters.:
1111a+1222b+233c+34d+4e=a+10b+100c+1000d+10000e
Which can be rewritten as:
1110a+1212b+133c-966d-9996e=0
Solving this would not be super interesting. First you would isolate e, find out it has to be 1 or 2, and continue from there in much the same way was with the 3-digit problem. So I won’t spend more time on that. However, the general case could be fun. Where did 1111, 1222, 233, 34, and 4 come from? Well, each digit gets a 1 from the single digit strings. Then the first four get a ten from the double digit strings and the last four get a one. Then the first three get a 100 from the triple digit strings, the second through fourth get a ten, and the last three get a one. Finally, the first two get a 1000 for the four digit strings, the second and third get a 100, the third and fourth get a 10, and the last two get a 1. So a got one of each, b got to double up on everything except hundreds, c was excluded from getting a thousand but got two hundreds, three tens, and three ones, and so on.
This pattern could rather easily be adapted to any length of number. I hypothesize that with such generated numbers there’s some way to generate the solutions with single digits, but that will take more work.
### One response to “”
1. Nice work, Nicole. Thanks very much for considering and answering my question.
Liked by 1 person
|
Isaac Newton, whom you have learned about in other lessons, studied the motion of objects. He developed three laws of motion, which are now called Newton's laws of motion. (You will learn more about these laws in lessons 8-10 of this module.)
In Newton's second law of motion, he described the relationship of force, mass, and acceleration mathematically. Force can be calculated by multiplying the mass of an object by the acceleration of the object.
The formula is
Net Force= (mass)(acceleration) or F = ma
In honor of Newton's contribution to the understanding of force and motion, the standard unit of force is called the newton (N). A newton is defined as the amount of force that it takes to accelerate one kilogram (1 kg) of mass one meter per second squared (1 m/s2). So 1 N is the same as 1 kg • m/s2.
When solving net force problems, make sure that the mass in in kilograms and the unit for acceleration is m/s2.
If you are given the mass in grams, make sure and convert to kilograms (move the decimal three places to the left.) For example,
5g = .005kg
Let's do some practice problems.
### Practice Problem 1
What force is needed to accelerate a 10 kg shopping cart 3 m/s2?
Interactive popup. Assistance may be required.
F = ma
F = 10kg • 3 m/s2
F = 30 kg • m/s2 or 30 N
### Practice Problem 2
The Space Shuttle has a lift off mass of 2,041,000 kg and accelerates at a rate of 16 m/s2. Calculate the force (thrust) that is accelerating the Space Shuttle.
Interactive popup. Assistance may be required.
F = ma
F = 2,041,000 kg • 16 m/s2
F = 32,656,000 kg • m/s2 or 32,656,000 N
### Practice Problem 3
A person on a scooter is accelerating 2 m/s2. If the person has a mass of 50,000 g, how much force is acting on that person?
Interactive popup. Assistance may be required.
In this problem the mass is in grams so you must convert to kg first.
50,000 g = 50 kg
F = ma
F = 50 kg • 2 m/s2
F = 100 kg • m/s2 or 100 N
### Challenge Problem
Which of the following requires more force?
Moving a 500 kg box at an acceleration of 1 m/s2
Or
Moving a 50 kg box at an acceleration of 1 m/s2
Interactive popup. Assistance may be required.
Calculate the force needed to move each box.
F = ma F = 500kg • 1 m/s2 F = 500 kg • m/s2 or 500 N F = ma F = 50kg • 1 m/s2 F = 50 kg • m/s2 or 50 N
So we can say the greater the objects mass, the greater the force needed to move the object.
|
# Solving Systems of Equations Using Substitution
```Textbook Section 6-2
Which
inequality does not have the same
2
solution as − 𝑦 > 4?
3
A)
12 < −2𝑦
B)
𝑦
2
C)
− 𝑦>
D)
−3𝑦 > 18
< −12
3
4
9
2
Students
can solve a system of equations
using substitution.
Students can classify systems as consistent,
inconsistent, dependent, or independent
using algebraic methods.
What
does the word substitute mean?
Provide examples.
In
sports, coaches often substitute one
player for another who plays the same
position.
In school, when a teacher is absent, a
substitute teacher takes his/her place for the
day.
1)
2)
3)
4)
5)
Solve for either variable, if necessary.
Substitute the resulting expression into the
other equation.
Solve that equation to get the value of the
first variable.
Substitute that value into one of the
original equations and solve for the second
variable.
Write the values from steps 3 and 4 as an
Solve the following systems using substitution.
State the number of solutions (no solution, one
solution, or infinite solutions) and tell whether
the system is consistent, inconsistent,
dependent, or independent.
Solve
the system using substitution.
𝑦 = 2𝑥
𝑦 =𝑥+5
One Solution
(5, 10)
Consistent
Independent
Solve
the system using substitution.
2𝑥 + 𝑦 = 5
𝑦 =𝑥−4
One Solution
(3, -1)
Consistent
Independent
Solve
the system using substitution.
𝑥 = 5𝑦 + 10
2𝑥 − 10𝑦 = 20
Infinite Solutions
Consistent
Dependent
Solve the system using substitution.
−5𝑦 − 𝑥 = −4
3𝑥 + 15𝑦 = −1
No Solution
Inconsistent
When
solving the system, if the end result is
a false (for example, 2 = -5), the answer is
no solution.
If
the end result is a statement that is always
true (for example, 3 = 3, or x = x), the
system has infinite solutions.
If
you can solve for the variables, there is
one solution (for example, x = 3, y = 1).
```
|
# SAT Mathematics : Intersecting Lines & Angles
## Example Questions
### Example Question #1 : Intersecting Lines & Angles
In the figure above, line a is parallel to line b and line d is parallel to line e. What is the value of y, in degrees?
15
25
10
30
15
Explanation:
Intersecting and parallel lines show up in many different geometric figures: parallelograms, trapezoids, squares, etc. Anytime you see these in a question, you have to properly leverage the essential properties of supplementary and vertical angles. On this problem, the fastest way to find y is to realize that 5x in the bottom left corner is supplementary to 2x + 5 in the bottom right (because of the intersection of two parallel lines). Therefore, 5x + 2x + 5 = 180 and x = 25. Once you have that information, you can use the fact that the sum of the interior angles of a triangle is 180 and see that x + 5x + 2y = 180 . Putting in 25 for x you see that 25+125+2y =180 and 2y =30. The correct answer is 15.
### Example Question #1 : Intersecting Lines & Angles
In the figure above, . Which of the following must be true?
I.
II. The two horizontal lines are parallel.
III.
I and II only
I and III only
I, II, and III
II and III only
I and II only
Explanation:
Here the SAT gives you a pair of lines with a transversal, but it does not tell you that the lines are parallel - it asks you to prove it. You are told that . Since angle and angle are vertical angles and angles and are vertical angles, you know that and . That means you can write your equation as:
, or
If that means that as well. A straight line contains 180 degrees, so you know that . And since , you can conclude that as well. From here, you can reverse engineer the same sort of equation you solved with the first set of angles. If and and are vertical angles and and are vertical angles, you can conclude that . From there you can set up the equation . Statement I is true.
In order for the horizontal lines to be parallel, you need to know that either the alternate exterior angles or the alternate interior angles are equal. Since you have already proven that , you know also that . Since you have a pair of alternate exterior angles, the two lines must be parallel. Statement II is also true.
Statement III, however, is not necessarily true. If then all angles would equal 90. However without that knowledge, you cannot come to any conclusions about the relationship between and . Statement III is not necessarily true, so the correct answer is I and II only.
### Example Question #1 : Intersecting Lines & Angles
Two straight lines intersect to form the angles above. If the measure of angle x is three times the measure of angle y, what is the measure of angle z?
45
120
60
135
135
Explanation:
Since lines x and y will add to a total of 180 degrees, you have two equations to work with:
x + y = 180
x = 3y
This means you can substitute 3y for x in order to solve for y:
3y + y = 180
4y = 180
y = 45
And since z will also sum with y to 180, then z must be 180 - 45 = 135 degrees.
### Example Question #1 : Intersecting Lines & Angles
In the figure above, if lines g and k are parallel and angle h measures 121 degrees, what is the value of p?
71
49
59
61
59
Explanation:
If h is 121, then the angle immediately below h must be 59, as it is a supplementary angle formed by the diagonal line. Since g and k are parallel, this 59 degree angle must exactly match p as they are alternative interior angles.
### Example Question #2 : Intersecting Lines & Angles
In the diagram above, lines AD and BE intersect at point C. What is the measure of angle ACE?
125
135
115
145
125
Explanation:
Two angle rules are very important for this question:
1) The sum of the interior angles of a triangle is always 180. Here, since you have a 90-degree angle (CED) and a 35-degree angle (EDC) in the bottom triangle, you can then conclude that angle ECD must be 55.
2) Supplementary angles, angles that are adjacent to each other when two straight lines intersect, must sum to 180 degrees. If you know that ECD is 55, then ACE as a supplementary angle must form the other 125 degrees for those two angles to sum to 180. Therefore, the correct answer is 125.
### Example Question #4 : Intersecting Lines & Angles
What is a + b + c + d?
150
110
120
130
150
Explanation:
An important thing to recognize in this problem is that you're dealing with two intersecting triangles that create external supplementary angles along the straight line on the bottom. To see this, consider the diagram below for which angles x and y have been added:
Angle y is an external supplementary angle to the triangle beside it so y = a + c. Why? Remember that y is supplementary to the angle beside it (x + 30) and (a + c) is supplementary to that same angle (the sum of interior angles of a triangle = 180.) Therefore y and (a + c) are identical. Anytime you have a straight line drawn off of a triangle you should recognize that the external supplementary angle equals the sum of the two opposite angles.
Using the same logic, you can see that x = b + d in the other intersecting triangle. Since the problem is asking for a + b + c + d, you should recognize that this question is really the same as what is x + y. Why? You can substitute x for b + d and y for a + c in the question stem. Since x + y = 180 - 30 on the straight line along the bottom, the correct answer is 150.
Note that another way to solve this problem involves seeing two large obtuse triangles: one with the angles a, c, and (x+30) and the other with the angles b, d, and (y+30). If you do that, you would have:
a+c+x+30=180, so a+c+x=150
b+d+y+30=180, so b+d+y=150
And you know that x+y+30=180 because x, 30, and y are all angles that make up the 180-degree straight line across the bottom of the figure. So x+y=150.
You can then sum the triangle equations:
a+c+x+b+d+y=150+150=300
And then plug in x+y = 150 and you're left with a+b+c+d=150.
### Example Question #5 : Intersecting Lines & Angles
In the image above, . What is the value of
60
50
70
80
60
Explanation:
This problem hinges on two important geometry rules:
1) The sum of all interior angles in a triangle is 180. Here you know that in the top triangle you have angles of 30 and 80, meaning that the angle at the point where lines intersect must be 70, since 30+80=110, and the last angle must sum to 180.
2) Vertical angles - angles opposite one another when two straight lines intersect - are congruent. Because you have identified that the angle at the bottom of the triangle at the top is 70, that also means that the top, unlabeled angle of the bottom triangle is 70. That then lets you add 70+50+ as the three angles in the bottom triangle, and since they must sum to 180 that means that .
### Example Question #1 : Intersecting Lines & Angles
NOTE: Figure not drawn to scale.
What is the value of in the figure above?
135
125
115
150
135
Explanation:
This problem heavily leverages two rules:
1) The sum of the angles in a triangle is 180.
2) Supplementary angles - adjacent angles created when one line intersects another - must sum to 180.
Here you can first leverage the 140-degree angle to fill in that its adjacent neighbor - its supplementary partner - must then be 40. and that gives you two of the three angles in the uppermost triangle: 20 and 40. You can use that to determine that the third angle must then be 120.
From there you should see that the 120-degree angle is a vertical angle, meaning that its opposite will also be 120. And that gives you a second angle in the lower-right triangle. Knowing that you have angles of 15 and 120 means that the third angle of that triangle must be 45. And since that angle is supplementary to angle x, x must then be 135.
### Example Question #1 : Intersecting Lines & Angles
In the figure above, lines and are parallel. What is the value of ?
105
95
115
125
125
Explanation:
This problem heavily leans on two important lines-and-angles rules:
1) The sum of the three interior angles of a triangle is always 180.
2) Supplementary angles - angles next to each other formed by two lines intersecting - must also sum to 180.
Here you can then determine that the angle next to the 95-degree angle is 85, and since that angle is the lower-right hand angle of the little triangle at the top, you can close out that triangle. With angles of 40 and 85, that means that the lower left hand angle must be 55.
From there, you can use the fact that parallel lines will lead to congruent angles. Since lines and are parallel, the angle next to will be 55 degrees, meaning that will then be 125.
### Example Question #1 : Intersecting Lines & Angles
In the diagram above, lines and all intersect at point A. If and , what is the value of ?
30
20
40
50
|
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# Subtraction of Rupees and Paise
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Last updated date: 12th Aug 2024
Total views: 195.9k
Views today: 3.95k
## Introduction to Subtraction of Money
The subtraction operation is the most popular arithmetic operation among children. The kids subtract rupees and paise from whatever they have. Rupees and paise are two different measurements of Indian money. They are often mixed up, and people might think paise is the same as rupees, but they're not. This article will teach you how to differentiate between the terms rupee and paisa; also, it includes the meaning, methods, and steps of subtraction of money.
Subtraction of Money
## Subtraction Methods Based on Rupees and Paise
We will learn how to subtract money values that include rupees and paise in order to get the difference. Money is subtracted in the same way as decimal numbers are subtracted. We must ensure that the sums of money are changed into paisa while subtracting. We will discover two distinct approaches to resolving subtraction problems with rupees and paise. Students can use both strategies.
• Subtracting the values and changing them to rupees.
• Subtracting the amounts without changing them to rupees.
Subtraction of Money
## Conversion of Rupees and Paise
Below is the example for the addition and subtraction of rupees and paise conversion:
• Paise to Rupees Conversion: While converting paise into rupees, we divide paise by 100. Converting paise into rupee formula is to divide the paise by a hundred.
Example 1. Convert 800 paise into rupees.
Ans: Here, we have 800 p
For converting, we will divide it by 100.
=\$\dfrac{800}{100}\$ = ₹ 8.00
Therefore, by converting 800 paise into rupees, we get 8 rupees.
• Rupees to Paise Currency Conversion: Rupees conversion into paise happens when we multiply the rupees by a hundred.
Example 2. Convert ₹ 4 into paise.
Ans: ₹ 4 = 4 × 100 = 400 p
Therefore, ₹ 4 will give 400 p on conversion.
## Benefits of Money Subtraction
• The children will learn about common situations involving money through these math worksheets and experience dealing with subtraction involving different currencies.
• The subtracting money worksheets are purposefully made with an emphasis on how to handle financial transactions, find prices, figure out how much change to give and how much tax to pay, as well as receive adequate practice with mental Maths.
## Solved Examples
1. Subtract Rs. 38.55 from Rs. 67.40
Solution:
Method - with conversion into paise:
Rs. 67.40 = 6700 paise + 40 paise = 6740 paise
Rs. 38.35 = 3800 paise + 35 paise = 3835 paise
Solved Example
Therefore, difference of Rs. 38.35 from Rs. 67.40 = 2905 Paise = Rs. 29.05
2. Add Rs. 35.25 and Rs. 125.75
Ans: As we know that we write rupee and paise by separating them with a dot.
In the example of adding 35.25 and 125.75, firstly we write them in vertical addition form.
We add the paise first and then the rupees.
The final answer to this question would be 161.00 Rupees.
## Subtraction of Rupees and Paise Worksheet
(i) Rs. 236. 00 – Rs. 43.00 (Ans: Rs. 236. 00 – Rs. 43.00 = Rs. 193)
(ii) Rs. 546.00 – Rs. 318.00 (Ans: Rs. 546.00 – Rs. 318.00= Rs. 228)
(iii) 528.00 p – 207.00 p (Ans: 528.00 p – 207.00 p = 321 p)
(iv) Rs. 587.00 – Rs. 419.00 (Ans: Rs. 587.00 – Rs. 419.00 =Rs. 178)
(v) Rs. 656.00 – Rs. 145.00 (Ans: Rs. 656.00 – Rs. 145.00 =Rs. 511)
## Summary
We deal with rupees and paise in our daily lives. Nothing can be bought without money. Money is required to do everything. In ancient times, paise was also of use, but now only rupees are in use. The formulas to convert the rupee into paisa and paise into rupee are to be understood well. To convert rupees into paisa, we just have to multiply by 100.
Similarly, to convert paisa to rupees we just have to divide by 100. These concepts are combined with multiplication and division as well. Children must know how to deal with money. These concepts have to be taught well to the kids.
## FAQs on Subtraction of Rupees and Paise
1. How many 50 paise coins will I get in rupees?
We know that 1 Rupee is equal to 100 paise, so we can now say that two 50 paise coins make 1 rupee.
2. What is adding money?
The amounts are all converted to rupees or paisa. The resulting numbers are added together like regular numbers, and the amount is then expressed in rupees and paisa. In order to add money, we convert rupees and paisa into paisa before adding like we would with regular numbers.
3. How many rupees is 60 paise?
When moving the decimal point to the right in 60 paise, it would be converted into 0.6 rupees.
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# Angles
View Notes
## Angles in Detail
Whenever two lines meet at a point, it forms an angle. The point where these lines meet is called a node or a vertex. The term "Angle" originated from the Latin "Angulus". If you look around, everything that is touching another surface is at some angle or the other.
Let's discuss the different types of angles, what they are called and how they are used in mathematics.
## Definition of Angles
The most basic definition of an angle according to the Euclidean geometry is a figure that is formed by two rays sharing a common point. These two rays are on the same plane, and when an angle is formed due to the intersection of two lines it is called a dihedral angle.
An angle is represented with a value in degrees that spans from 0-360. Different values of degrees form different types of angle that have different names. A particular range of degrees has been assigned a specific name such as Obtuse angle, Right angle, Acute angle etc.
### Definition of Angle in Math
There is only one definition of angle in mathematics that explains how one can measure and classify different angles. To understand this better, let's start from the fundamental aspects of an angle.
## What is a Ray?
A ray is defined as a line that starts from a point and keeps extending from its other end. It is named after its starting point and another point on the line. It is represented as an arrow to show its ever-extending nature.
## What is a Vertex?
When two rays share a common starting point, it creates an angle, and that common point is called a vertex. An angle is represented by the symbol ∡.
Remember that the name denoting the vertex always remains in the middle while the other points can exchange positions.
## Angle and Algebra
A graphical and coordinate-based approach is taken when representing an angle. The coordinate points help in solving equations related to angles by plotting the lines or rays on a coordinate plane.
## Types of Angles and Their Definitions
### Definition of Straight Angle
An angle is said to be straight when the two rays are forming an angle of exactly 180 - degree. It is named so because two lines at a 180 - degree angle look like a single straight line.
### Definition of Acute Angles
When the angle between two lines is less than 90-degree, it is called an acute angle. Any angle between 0-90 degrees is an acute angle.
### Definition of Right Angle
When two rays form an angle of 90-degree, it is called a right angle. Among other kinds of angles, this particular angle is given comparatively more importance. An entire subject called Trigonometry is built upon this angle as most things around us form a right angle in respect with their surface.
### Definition of Obtuse Angle
The range of angles above 90-degree and below 180-degree is called an obtuse angle. One can think of it as the opposite of acute angle as well.
### Definition of Reflex Angle
Any angle formed beyond 180 - degree is a reflex angle. However, this also has its limitation, and that is below 360. It is because when a line crosses 360-degree, it completes a full rotation and it again starts from 0-degree.
When two angles share a common vertex and have a common side, it is called an adjacent angle. It should be noted that if both the criteria are not met, then it is not adjacent. The very meaning of adjacent angles is that two angles are side by side. Also, if one angle overlaps another, it is not an adjacent angle.
### Define Complementary Angles
Two angles are said to be complimentary when their sum is 90 degree. They don't need to share a common vertex or side, only the sum of their respective angle decides whether they are complementary or not.
The picture above is an example of a complementary angle pair, but the image below also represents the meaning of complementary angles.
Even though these two are sharing anything in common, the fact that 30 degree and 60 degree adds up to 90 degree makes them complementary.
If you wish to learn more about angles and how to solve its sums, start Vedantu's online classes today. Their live classes are both detailed and interesting, making learning maths fun. Go through their inventory full of sample sheets to gauge how much you know about this geometrical subject.
1. What are all Angles Names?
Ans. A full rotation is of 360-degree, and anything in between is 1-360 degree is an angle. However, some specific ranges have been given a particular name. There are five primary types of angles, namely Acute, Obtuse, Straight, Right and Reflex.
However, there are five types of angles that are named defining the relationship between one angle with respect to another. These five angle names are Complementary, Adjacent, Congruent, Supplementary and Bisector.
2. What is a Bisector Angle?
Ans. It is said to be a bisector when two adjacent angles have equal values. It is not a bisector when the two angles do not have a common ray.
For example, in the diagram above
∡AOB has a line OC tat is bisecting it from the middle. It is a bisector angle because
∡AOC = ∡COB
Note that a bisector angle that divides 90-degree into two 45-degree angles is both complementary and adjacent angle as well.
3. What is the Angle Definition Geometry?
Ans. In geometry, an angle is a shape that is formed when two rays or lines diverge from a common point called its vertex. An angle is represented with the sign ∡ and has a certain degree that ranges from 1 - 360. Apart from the vertex, its other attributes are its two legs (lines), its interior and exterior. The interior is the part within the 'jaws' of the two lines while all the space that is not within it is the exterior.
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# Determine the domain and range of each relation
• Slides: 7
Determine the domain and range of each relation and determine if it’s a function or not. { ( 1 , 3), (– 1 , 3 ), ( 2 , 0) } D R – 1 1 2 0 3 Yes, it’s a function because each member in the domain is assigned to exactly one member in the range. D 0 2 x y 0 0 2 1 2 2 R 0 1 2 No, it’s not a function because 2 is assigned to both 1 and 2.
9. 02 Linear and Non – Linear Functions
Functional Notation The equation y = 5 x + 1 can be rewritten in functional notation. To do this, replace the y with f(x) y = 5 x + 1 f(x) is read as “f of x” Any equation with f(x) is written in functional notation.
We can tell if a function is linear by examination. An equation represents a linear function if all the variables are raised to the first power. Determine if the following represent a linear function. f(x) = x – 4 Yes f(x) = x 2 No, because x is raised to the 2 nd power. f(x) = 2 x + 8 Yes f(x) = 2 Yes f(x) = No, because it’s a square root.
We can determine if a graph represents a function by using the vertical line test. A line represents a function if one draws a vertical line anywhere along the graph of a line and that vertical line intersects the line at only one point. Determine if the graph represent a function. Yes, it’s a function because any drawn vertical line will only intersect the graph at only one point.
Determine if the graphs represent a function. No, it’s not a function because the vertical line intersects the graph at two points. Yes, it’s a function because the vertical line intersects the graph at only one point.
Try This: Determine if the following represents a function. f(x) = 3 x 2 – x + 2 No f(x) = 5 x + 2 Yes No, the vertical line intersects the graph at two points. Yes, the vertical line intersects the graph at only one point.
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# Find the exact value of sin(pi/12) , cos (11pi/12) , and tan (7pi/12)?
Sep 25, 2015
Find sin (pi/12); cos (pi/12) and tan ((7pi)/12)
#### Explanation:
Call $\sin \frac{\pi}{12} = \sin t .$ Use trig identity: $\cos 2 a = 1 - {\sin}^{2} a$
$\cos 2 \frac{\pi}{12} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} = 1 - {\sin}^{2} t .$
${\sin}^{2} t = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{4}$
$\sin t = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$. Since $\frac{\pi}{12}$ has its sin positive, then
$\sin \left(\frac{\pi}{12}\right) = \sin t = \frac{\sqrt{2 - \sqrt{3}}}{2}$
Call $\cos \left(\frac{\pi}{2}\right) = \cos t$
$\cos \left(\frac{2 \pi}{12}\right) = \frac{\sqrt{3}}{2} = 2 {\cos}^{2} t - 1$
${\cos}^{2} t = \frac{2 + \sqrt{3}}{4}$
$\cos t = \pm \frac{\sqrt{2 + \sqrt{3}}}{2}$. Since cos $\left(\frac{\pi}{12}\right)$ is positive, then
$\cos \left(\frac{\pi}{12}\right) = \cos t = \frac{\sqrt{2 + \sqrt{3}}}{2}$
$\tan \left(\frac{7 \pi}{12}\right) = \tan \left(\frac{\pi}{12} + \pi\right) = \tan \left(\frac{\pi}{12}\right) = \frac{\sin}{\cos} =$
$= \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}}$
|
# Surds and Indices Shortcuts, Tricks, PDF and Formulas
Surds and Indices Points to Remember - Page 2
Surds and Indices Examples - Page 3
Surds and Indices Important Questions - Page 5
## Important Formulas - Surds and Indices
• An integer is a whole number (positive, negative or zero). A rational number is one that can be expressed as a fraction $\frac{a}{b}$ , where a and b are integers. All integers, fractions and terminating or recurring decimals are rational.
• An irrational number cannot be expressed in the form $\frac{a}{b}$ , where a and b are integers. Examples of irrational numbers are $\pi,\sqrt{2},\sqrt{3}\;\;and\:4\sqrt{5}\:(4\sqrt{5}\:means\:4\times \sqrt{5})$ .
• Real numbers are numbers that can be represented by points on the number line. Real numbers include both rational and irrational number.
• A surd is an irrational number involving a root. The numbers $\sqrt{3},4\sqrt{5}\:and\:\sqrt[3]{7}$ are examples of surds. Numbers such as $\sqrt{16}and \sqrt[3]{8}$ are not surds because they are equal to rational numbers. $\sqrt{16}$ =4 and root of ‘3’ or ‘root 3’. Note that we cannot take the square root of a negative number.
• Like surds involve the square root of the same number. Only like surds can be added or subtracted. For example, $3\sqrt{2}+4\sqrt{2}=7\sqrt{2}\:but\: 3\sqrt{2}+4\sqrt{3}=3\sqrt{2}+4\sqrt{3}$ .
• Surds of the form $\sqrt{x}$ can be simplified if the number beneath the square root sign has a factor that is a perfect square. For example, $\sqrt{8}=\sqrt{4\times 2}=\sqrt{4}\times \sqrt{2}=2\sqrt{2}$ .
• The following rules can be used when multiplying or dividing surds.
$(\sqrt{x})^{2}=\sqrt{x^{2}}=x$
$\sqrt{x}\times \sqrt{y}=\sqrt{xy}$
$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$
• Rationalising the denominator of a surd means changing the denominator so that is a rational number. To rationalize the denominator of a surd such as $\frac{\sqrt{2}}{\sqrt{3}}$ we use the result that $(\sqrt{x})^{2}=x$ so if we multiply the denominator by $\sqrt{3}$ it will be rational. if we multiply the demoniator by $\sqrt{3}$ we must also multiply the numerators by $\sqrt{3}$ so that $\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{\sqrt{3}}$
• Fractional indices may be used to express roots. $x^{\frac{1}{n}}=\sqrt[n]{x}\:\:and\:\:x^{\frac{m}{n}}=(\sqrt[n]{x})^{m}$
• Make sure you can use your calculator to find powers and roots. The $x^{\frac{1}{y}}$ or $\sqrt[x]{\:}$ button allows you to find roots.
For example: $\sqrt[5]{7776}=(7776)^{\frac{1}{5}}$
## INDICES
• When the powers are fraction , then we can compare the indices in following manner.
for example :find which is greather $2^{\frac{1}{4}}$ or $3^{\frac{1}{5}}$
step-1 : find the L.C.M of the denominator of the fraction i.e 4,5 = 20
step-2 : find powers with the L.C.M $2^{\frac{1}{4}\times 20}$ OR $3^{\frac{1}{5}\times 20}$ = $2^{5}$ AND $3^{4}$
step-3 : compare the result obtained in setp-2 : i.e as 32 < 81 we can say $2^{\frac{1}{4}}$ < $3^{\frac{1}{5}}$
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## 2 Responses to “Surds and Indices Shortcuts, Tricks, PDF and Formulas”
1. Comment made by Leah Wambui on Sep 30th 2018 at 5:47 pm: Reply
help solve 5+2 root 6 all this into a square root =root2+root3
2. Comment made by Govind Singh on May 19th 2019 at 4:04 am: Reply
nice pdf
|
### Playing with Numbers - Solutions 1
CBSE Class –VIII Mathematics
NCERT Solutions
CHAPTER - 16
Playing with Numbers
(Ex. 16.1)
Find the values of the letters in each of the following and give reasons for the steps involved.
1.
Ans. On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, 7 + 5 = 12 in which ones place is 2.
A = 7
And putting 2 and carry over 1, we get
B = 6
Hence A = 7 and B = 6
2.
Ans. On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,8 + 5 = 13 in which ones place is 3.
A = 5
And putting 3 and carry over 1, we get
B = 4 and C = 1
Hence A = 5, B = 4 and C = 1
3.
Ans. On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, A x A = 6 x 6 = 36 in which ones place is 6.
A = 6
Hence A = 6
4.
Ans. Here, we observe that B = 5
so that 7 + 5 = 12.
Putting 2 at ones place and carry over 1 and A = 2, we get
2 + 3 + 1 = 6
Hence A = 2 and B = 5
5.
Ans. Here on putting B = 0,
we get 0 3 = 0.
And A = 5, then 5 3 = 15
A = 5 and C = 1
Hence A = 5, B = 0 and C = 1
6.
Ans. On putting B = 0, we get 0 , and A = 5, then 5 5 = 25
A = 5, C = 2
Hence A = 5, B = 0 and C = 2
7.
Ans. Here product of B and 6 must be same as ones place digit as B.
1 = 6, 6 2 = 12, 6 3 = 18,
4 = 24
On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44.
For 6 7 = 42 and carry over 2 = 44
Hence A = 7 and B = 4
8.
Ans. On putting B = 9, we get 9 + 1 = 10
Putting 0 at ones place and carry over 1, we get
For A = 7 7 + 1 + 1 = 9
Hence A = 7 and B = 9
9.
Ans. On putting B = 7,
7 + 1 = 8
Now A = 4, then 4 + 7 = 11
Putting 1 at tens place and carry over 1, we get
2 + 4 + 1 = 7
Hence A = 4 and B = 7
10.
Ans. Putting A = 8 and B = 1, we get
8 + 1 = 9
Now again we add 2 + 8 = 10
Tens place digit is ‘0’ and carry over 1.
Now 1 + 6 + 1 = 8 = A
Hence A = 8 and B = 1
|
# Ex.6.4 Q3 The-Triangle-and-its-Properties Solutions-NCERT Maths Class 7
Go back to 'Ex.6.4'
## Question
$$AM$$ is a median of a triangle $$ABC.$$ Is $$AB + BC + CA > 2 AM?$$ (Consider the sides of triangles $$\Delta{ABM}$$ and $$\Delta{AMC}$$.)
Video Solution
Triangle & Its Properties
Ex 6.4 | Question 3
## Text Solution
What is known?
$$ABC$$ is a triangle and $$AM$$ is a median of triangle $$ABC$$.
What is unknown?
Is $$AB + BC + CA > 2\, AM?$$
Reasoning:
In this question it is asked if $$AB + BC + CA > 2 AM$$ or not. This question is also based on the property that the sum of lengths of two sides of a triangle is always greater than the third side. In such kind of problems, you just visually identify the triangle $$ABC$$ and $$AM$$ is the median which further divides triangle $$ABC$$ into two more triangles i.e. triangle $$ABM$$ and $$AMC.$$ Now consider any two sides of first triangle $$ABM$$ and use the above property then consider the any two sides of another triangle $$AMC$$ and use the above property. Now, add $$\rm{}L.H.S.$$and $$\rm{}R.H.S.$$ of both the triangles.
Steps:
In triangle $$ABM,$$
$$AB + BM > AM{\rm{ }} \qquad \ldots .{\rm{ }}\left( 1 \right)$$
In triangle $$AMC,$$
$$AC + MC > AM{\rm{ }} \qquad \ldots .{\rm{ }}\left( 2 \right)$$
Adding equation $$\rm{}(1)$$ and $$\rm{}(2)$$ we get,
\begin{align}&AB \!+\! BM \!+\! AC \!+\! MC \!>\! AM \!+\! AM\\&AB \!+\! AC \!+\! BM \!+\! MC \!>\! 2AM\\&AB \!+\! AC \!+\! BC \!>\! 2AM\end{align}
Hence, it is true
Useful Tip:
Whenever you encounter problems of this kind, it is best to think of the property based on sum of lengths of any two sides of a triangle is always greater than the third side
Learn from the best math teachers and top your exams
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# Relationships Between Sets of Rational Numbers
## Presentation on theme: "Relationships Between Sets of Rational Numbers"— Presentation transcript:
Relationships Between Sets of Rational Numbers
Rational Irrational Integers Whole Counting numbers
Rational Irrational π Integers Decimal - Repeating & Terminating
Fractions -1,-2,-3,-4 … π Whole Percents 1,2,3,4 …. Counting numbers
Rational numbers can be written in several forms
2 1 +1 2 - 0 2*1 2/1 4/2 8/4 2.0 200% -(-2) Most common Simplified fraction Decimal Percent Not in notes
Convert between decimal fraction and percent
Decimal to Fraction Unit ● tenths hundredths thousandths 1. Place decimal digits over the place value 5 EX 1) .5 = 10 2. simplify 5 10 ÷ 5 1 2 = ÷ 5 5 1 20 EX 2) 0.05 = = 100 5 1 200 EX 3) 0.005 = = 1000
Out of sight to the right
Decimal to Percent 1. Move decimal two places right 2. Add zeroes if necessary EX 4) 5 = 50 % EX 5) 2 5 = 250 % 3. Add percent to answer = 300 % EX 6) 3 Out of sight to the right Percent to Decimal 1. Move decimal two places left EX 7) 2. Add zeroes if necessary 5 % = 0.05 3. Remove percent sign from answer EX 8) = 0.025 2 5 %
Percent to Fraction Fraction to Percent Put percent over 100 Simplify
50 5 1 EX 9) 50 % = = = 100 10 2 1 450 45 9 = 4 EX 10) 450 % = = = 2 100 10 2 Fraction to Percent Make fraction a decimal Turn decimal into a percent 1 EX 11) = .2 = 20 % 5 1 EX 12) 3 = 3.2 = 320 % 5
Order from least to greatest.
4 5 1. Change all numbers to percents EX 13) 0.93, and 90 % , 80 % 90 % 93 % Graph the numbers on a number line. 4 5 0.93 90 % Write the original numbers in least to greatest order 4 5 , 90 %, 0.93
- - - - - - - EX 14) , 84 %, and 0.7 , 0.7, 84% EX 15) , 84 %, and 0.7
3 5 Write the original numbers in least to greatest order EX 14) , %, and 0.7 60 % 84 % 70 % 3 5 , 0.7, 84% - - - 3 5 EX 15) , %, and - - - 60 % 84 % 70 % - 3 5 Graph the numbers on a number line. - 84%, - 0.7, LEAST GREATER What do you notice between the two examples above?
Conversions you need to memorize
Fraction Decimal Percent 1 1.0 100 .25 25 0.5 50 3/4 0.75 75 1/8 0.125 12.5 5/8 0.625 62.5 1/5 0.2 20 4/5 0.8 80 Fraction Decimal Percent 1/10 0.1 10 2/10 0.2 20 1/25 0.04 4 1/50 0.02 2 1/100 0.01 1 1/3 0.33 33 1/3 2/3 0.66 66 2/3 Any fraction with a denominator of 5,10,100.
Notes
Convert Rational Numbers
Date _______________ Convert Rational Numbers Any number that can be expressed as a fraction of two integers
Convert between decimal fraction and percent
Decimal to Fraction Unit ● tenths hundredths thousandths 1. Place decimal digits over the place value EX 1) = 2. simplify ÷ = ÷ EX 2) = EX 3) =
Decimal to Percent Percent to Decimal EX 4) EX 5) EX 6) EX 7) EX 8)
1. Move decimal ___ places _____ 2. Add ______ if necessary EX 4) EX 5) 3. Add _______ to answer EX 6) Percent to Decimal 1. Move decimal ___ places ____ EX 7) 2. Add ______ if necessary 3. ______ percent sign from answer EX 8)
Percent to Fraction Fraction to Percent Put percent over ___ Simplify
EX 9) = EX 10) = Fraction to Percent Make fraction a _______ Turn _______ into a ______ EX 11) = EX 12) =
When rational numbers are written in a variety of forms, you can compare the numbers by writing them all in the same form. The values on a number line increase as you move from left to right.
Order from least to greatest.
1. Change all numbers to __________ EX 13) Graph the numbers on a number line. Write the original numbers in least to greatest order
EX 14) EX 15) Write the original numbers in least to greatest order
Graph the numbers on a number line.
Conversions you need to memorize
Fraction Decimal Percent Fraction Decimal Percent Any fraction with a denominator of 5,10,100.
|
# Understanding the Relationship: f(n) = o(g(n)) Implies g(n) = o(f(n)) in Algorithm Analysis
Algorithm analysis is an essential part of computer science and software development. It helps us to understand the performance and efficiency of algorithms by comparing their growth rates. One aspect of this analysis is the use of asymptotic notations, such as Big O, Big Omega, and Big Theta, which help us express the running time of algorithms in a simplified and comparable manner.
In this guide, we will explore the relationship between the notations f(n) = o(g(n)) and g(n) = o(f(n)) and how it is used in algorithm analysis. By the end of this guide, you will have a clear understanding of this relationship and its significance in evaluating the performance of algorithms.
## Asymptotic Notations
Before diving into the relationship between f(n) = o(g(n)) and g(n) = o(f(n)), let's first understand the different asymptotic notations used in algorithm analysis:
Big O Notation (O): It describes the upper bound of an algorithm's running time. It is used to express the maximum number of operations an algorithm can perform in the worst-case scenario. For example, if an algorithm's running time is O(n^2), it means the maximum number of operations it can perform is proportional to the square of the input size.
Big Omega Notation (Ω): It describes the lower bound of an algorithm's running time. It is used to express the minimum number of operations an algorithm can perform in the best-case scenario. For example, if an algorithm's running time is Ω(n), it means the minimum number of operations it can perform is proportional to the input size.
Big Theta Notation (Θ): It describes the tight bound of an algorithm's running time. It is used to express the exact number of operations an algorithm can perform when the upper and lower bounds are the same. For example, if an algorithm's running time is Θ(n^2), it means the exact number of operations it can perform is proportional to the square of the input size.
## The Relationship: f(n) = o(g(n)) and g(n) = o(f(n))
In the context of algorithm analysis, the notation f(n) = o(g(n)) means that the function f(n) grows at a slower rate than the function g(n), or in other words, the running time of algorithm f(n) is faster than the running time of algorithm g(n).
On the other hand, the notation g(n) = o(f(n)) means that the function g(n) grows at a slower rate than the function f(n), or in other words, the running time of algorithm g(n) is faster than the running time of algorithm f(n).
The relationship between these two notations implies that if the running time of algorithm f(n) is faster than the running time of algorithm g(n), then the running time of algorithm g(n) is also faster than the running time of algorithm f(n). This relationship helps us compare and analyze the performance of different algorithms more effectively.
## Examples and Analysis
Let's look at a few examples to better understand the relationship between f(n) = o(g(n)) and g(n) = o(f(n)):
Example 1: Let's say we have two algorithms with running times of f(n) = O(n) and g(n) = O(n^2). Since n is smaller than n^2, f(n) grows at a slower rate than g(n), so we can say that f(n) = o(g(n)). However, g(n) grows at a faster rate than f(n), so we cannot say that g(n) = o(f(n)).
Example 2: Let's say we have two algorithms with running times of f(n) = O(n^2) and g(n) = O(n^3). Since n^2 is smaller than n^3, f(n) grows at a slower rate than g(n), so we can say that f(n) = o(g(n)). However, g(n) grows at a faster rate than f(n), so we cannot say that g(n) = o(f(n)).
In both of these examples, the relationship between f(n) = o(g(n)) and g(n) = o(f(n)) does not hold, as one algorithm grows at a faster rate than the other. This relationship only holds when the running times of both algorithms are the same, such as when f(n) = O(n^2) and g(n) = O(n^2).
## FAQ
### 1. What is the significance of the relationship f(n) = o(g(n)) and g(n) = o(f(n)) in algorithm analysis?
The relationship between f(n) = o(g(n)) and g(n) = o(f(n)) helps us compare the performance of different algorithms more effectively. By understanding this relationship, we can determine if one algorithm is faster or slower than the other, or if they have the same running time.
### 2. Can f(n) = o(g(n)) and g(n) = o(f(n)) hold true at the same time?
Yes, f(n) = o(g(n)) and g(n) = o(f(n)) can hold true at the same time if the running times of both algorithms are the same. In this case, neither algorithm grows at a faster rate than the other.
### 3. What is the difference between Big O, Big Omega, and Big Theta notations?
Big O notation describes the upper bound of an algorithm's running time, Big Omega notation describes the lower bound of an algorithm's running time, and Big Theta notation describes the tight bound of an algorithm's running time when the upper and lower bounds are the same.
### 4. How do I determine if f(n) = o(g(n)) or g(n) = o(f(n))?
To determine if f(n) = o(g(n)) or g(n) = o(f(n)), you need to analyze the growth rates of the functions f(n) and g(n). If f(n) grows at a slower rate than g(n), then f(n) = o(g(n)). If g(n) grows at a slower rate than f(n), then g(n) = o(f(n)).
### 5. Can I compare algorithms with different input sizes using the relationship f(n) = o(g(n)) and g(n) = o(f(n))?
Yes, you can compare algorithms with different input sizes using this relationship. However, the relationship is more meaningful when comparing algorithms with the same input size, as it allows you to determine which algorithm is faster or slower more accurately.
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The limit of a function
The definition of the limit of a function
A limit on the left (a left-hand limit) and a limit on the right (a right-hand limit)
Continuous function
Limits at infinity (or limits of functions as x approaches positive or negative infinity)
Infinite limits
The limit of a function examples
Vertical, horizontal and slant (or oblique) asymptotes
The definition of the limit of a function
The limit of a function is a real number L that f (x) approaches as x approaches a given real number a, written
if for any e > 0 there is a d(e) > 0 such that | f (x) - L | < e whenever | x - a | < d(e).
The definition says, no matter how small a positive number e we take, we can find a positive number d such that, for an arbitrary chosen value of x from the interval a - d < x < a + d, the corresponding function's values lie inside the interval L - e < f (x) < L + e, as shows the right figure. That is, the function's values can be made arbitrarily close to the number L by choosing x sufficiently close to a, but not equal to a.
Therefore, the number d, that measures the distance between a point x from the point a on the x-axis, depends on the number e that measures the distance between the point f (x) from the point L on the y-axis.
Example: Given
whenever
A limit is used to examine the behavior of a function near a point but not at the point. The function need not even be defined at the point.
A limit on the left (a left-hand limit) and a limit on the right (a right-hand limit)
The limit of a function where the variable x approaches the point a from the left or, where x is restricted to values less than a, is written
The limit of a function where the variable x approaches the point a from the right or, where x is restricted to values grater than a, is written
If a function has both a left-handed limit and a right-handed limit and they are equal, then it has a limit at the point. Thus, if
Continuous function
A real function y = f (x) is continuous at a point a if it is defined at x = a and
that is, if for every e > 0 there is a d(e) > 0 such that | f (x) - f (a) | < e whenever | x - a | < d (e).
Therefore, if a function changes gradually as independent variable changes, so that at every value a, of the independent variable, the difference between f (x) and f (a) approaches zero as x approaches a.
A function is said to be continuous if it is continuous at all points.
Limits at infinity (or limits of functions as x approaches positive or negative infinity)
We say that the limit of f(x) as x approaches positive infinity is L and write,
if for any e > 0 there exists N > 0 such that | f (x) - L | < e for all x > N (e).
We say that the limit of f (x) as x approaches negative infinity is L and write,
if for any e > 0 there exists N > 0 such that | f (x) - L | < e for all x < -N (e).
Not all functions have real limits as x tends to plus or minus infinity.
Thus for example, if f (x) tends to infinity as x tends to infinity we write
if for every number N > 0 there is a number M > 0 such that f (x) > N whenever x > M(N).
Infinite limits
We write
if f (x) can be made arbitrarily large by choosing x sufficiently close but not equal to a.
We write
if f(x) can be made arbitrarily large negative by choosing x sufficiently close but not equal to a.
The limit of a function examples
Example: Evaluate the following limits; Solution: a) As x tends to minus infinity f(x) gets closer and closer to 0. As x tends to plus infinity f(x) gets closer and closer to 0. Therefore, b) As x tends to 0 from the left f(x) gets larger in negative sense.
As x tends to 0 from the right f(x) gets larger in positive sense. Therefore,
Vertical, horizontal and slant (or oblique) asymptotes
If a point (x, y) moves along a curve f (x) and then at least one of its coordinates tends to infinity, while the distance between the point and a line tends to zero then, the line is called the asymptote of the curve.
Vertical asymptote
If there exists a number a such that
then the line x = a is the vertical asymptote.
Horizontal asymptote
If there exists a number c such that
then the line y = c is the horizontal asymptote.
Calculus contents B
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# Conditional Probability Calculator
Instructions: Use this Conditional Probability calculator to compute the conditional probability $$\Pr(A | B)$$. Please provide the probability $$\Pr(A \cap B)$$ and $$\Pr(B)$$ in the form below:
Please indicate the value of $$\Pr(A \cap B)$$ =
Please indicate the value of $$\Pr(B)$$ =
The concept of conditional probability is one of the most crucial ideas in Probability and Statistics. And it is a quite simple idea: The conditional probability of an event $$A$$ given an event $$B$$ is the probability that $$A$$ happens under the assumption that $$B$$ happens as well.
This is, we restrict the sample space to outputs in which $$B$$ happens, and we look for the probability that $$A$$ occurs in that subset sample space.
### So, what is the formula for conditional probability?
In mathematical terms, the conditional probability $$\Pr(A|B)$$ is computed using the following formula:
$\Pr(A|B) = \displaystyle \frac{\Pr(A \cap B)}{\Pr(B)}$
The above expression can be rewritten and it also provides a way to compute the probability of the intersection of two event, when the conditional probability is known:
$\Pr(A \cap B) = \Pr(A|B) \Pr(B)$
### Why is conditional probability important?
The concept of conditional probability is crucial because it represents the fact of real life that as we know more information about some event, we can refine our idea of the likelihood of an event. This idea of computing a probability given that we know that certain even is true is a representation of how our brain works, and hence, make the idea of conditional probability very important.
Also, the concept of conditional probability and the law of multiplication play a crucial role for the construction of the Total Probability Rule as well as Bayes' Theorem.
In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us.
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# Proof of product rule for differentiation using chain rule for partial differentiation
This article proves the product rule for differentiation in terms of the chain rule for partial differentiation.
## Statements
### Statement of product rule for differentiation (that we want to prove)
uppose $f$ and $g$ are functions of one variable. Then the following is true wherever the right side expression makes sense (see concept of equality conditional to existence of one side):
$\! \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
### Statement of chain rule for partial differentiation (that we want to use)
Suppose $f,g$ are both functions of one variable and $h$ is a function of two variables. Suppose $u = f(x), v = g(x), w = h(u,v)$. Then:
$\frac{dw}{dx} = \frac{\partial w}{\partial u}\frac{du}{dx} + \frac{\partial w}{\partial v} \frac{dv}{dx} = h_u(u,v)f'(x) + h_v(u,v)g'(x)$
## Proof
Given: Functions $f$ and $g$
To prove: $\! \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$ wherever the right side makes sense.
Proof:
Consider the function:
$\! h(u,v) := uv$
Its partial derivatives are:
$\! h_u(u,v) = v, \qquad h_v(u,v) = u$ Define:
$\! u := f(x), \qquad v := g(x), \qquad w := uv = h(u,v) = f(x)g(x)$
By the chain rule for partial differentiation, we have:
$\! \frac{dw}{dx} = \frac{\partial w}{\partial u} \frac{du}{dx} + \frac{\partial w}{\partial v}\frac{dv}{dx}$
The left side is $(fg)'(x)$. The right side becomes:
$\! h_u(u,v)f'(x) + h_v(u,v)g'(x)$
This simplifies to:
$\! vf'(x) + ug'(x)$
Plug back the expressions $u = f(x), v = g(x)$ and get:
$\! f'(x)g(x) + f(x)g'(x)$
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0 0
## solve (y-1)(y-2)=0
solve the equation (y-1)(y-2)=0
Recall the Zero-Product Property, which states that if the product of two factors is zero then at least one of the factors must be zero. That is, if a·b=0 then a=0 and/or b=0.
Given: (y - 1)(y - 2) = 0
The equation you are given states that the product of the two factors (y - 1) and (y - 2) is equal to 0. With this, and by the zero-product property, then the factor (y - 1) must be equal o zero and/or the factor (y - 2) must be equal to zero. That is,
y - 1 = 0 and/or y - 2 = 0
Solving for y in both cases, we arrive at the following:
y - 1 = 0 ; y - 2 = 0
+ 1 + 1 + 2 + 2
_______________ ________________
y - 1 + 1 = 0 + 1 y - 2 + 2 = 0 + 2
y = 1 y = 2
Thus, the solutions for the equation (y - 1)(y - 2) = 0 are 1 and 2
This problem is set up very nicely for you.
First I'm just going to point out a property of numbers. When you multiply two numbers, the only way the product can be 0 is if at least one of the numbers is 0. In other words if x * y = 0 then either x or y must also be 0. (both x and y could also be zero).
This means that in your equation either (y - 1) or (y - 2) must equal zero. Let's try each case.
y - 1 = 0
y = 1 (add 1 to both sides)
or
y - 2 = 0
y = 2 (add 2 to both sides)
This means that y can either equal 1 or 2, a good way to represent this answer is like this:
y = 1, 2
In order to drive the point home, lets plug in both values for y to make sure it works:
lets say that y = 1 first.
(1 - 1) * (1 - 2) = 0
(0) * (-1) = 0
0 * (-1) = 0
0 = 0
So we know y = 1 is a valid answer. Now lets try y = 2
(2 - 1) * (2 - 2) = 0
(1) * (0) = 0
1 * 0 = 0
0 = 0
We have just confirmed that both answers y = 1 and y = 2 will satisfy the equation (y-1)(y-2) = 0.
Hope that helps.
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# Ch 12: Sets
### About This Chapter
Watch online video lessons on mathematical sets to learn about set notation, cardinality, Venn diagrams, two-way tables and more. Each lesson is accompanied by a short multiple-choice quiz you can use to check your understanding of these math topics.
## Sets
What is a set? Well, the world can be divided into sets. Have you ever taken pictures to help chronicle an experience? Those very pictures represent a set of your own choosing. In mathematics, a set is a collection of objects. Each object is called an element of that set - they don't necessarily have to be numbers all the time, but we'll show you how to classify and manipulate those elements. Sets aren't always disjoint, so our introduction will explore the union and intersection of sets to demonstrate how they may combine and interact.
We'll further explain the interaction of sets with cardinality - just how many pictures did you take to record that once-in-a-lifetime experience? Cardinality refers to the number of elements in a given subset. Did you have a friend taking pictures of the same experience? If your friend took identical pictures, then those pictures would be a subset of your photo album. We'll examine subsets and cover the differences between finite and infinite sets. You'll also explore the mysteries of the empty set.
Our lessons will reveal another way to combine sets using the Cartesian product. In this way, we can build a new set out of our preexisting sets - imagine being able to combine every picture from your friend's album with your own; you'd end up with a collection of photos that looked completely different from either album alone. We'll also study Venn diagrams for a greater understanding of subsets, intersections and unions. By this point, you'll be an expert at discerning between overlapping elements and disjoint ones.
This chapter will also look at categorical propositions, or the reasoning that takes place when looking for a relation between sets. While studying subject and predicate sets, we'll see how to translate their interaction into standard form. Along these lines, we'll also get used to reading and interpreting two-way tables. A picture may be worth a thousand words, but our lessons use both to ensure that you have the fullest understanding of mathematical sets and their capabilities. Thanks for watching!
Final Exam
Chapter Exam
### Earning College Credit
Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
To learn more, visit our Earning Credit Page
### Transferring credit to the school of your choice
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### Other Chapters
Other chapters within the Math 102: College Mathematics course
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# How do you find general form of circle with endpoints of a diameter at (4,3) and (0,1)?
Jun 14, 2018
${x}^{x} + {y}^{2} + 4 x + 4 y - 9 = 0$
#### Explanation:
eqn. of circle
${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
$\left(a , b\right) = \text{ the centre}$
$r = \text{ the radius}$
centre is the midpoint of the diameter
$\left(a , b\right) = \left(\frac{4 + 0}{2} , \frac{3 + 1}{2}\right)$
$\left(a , b\right) = \left(2 , 2\right)$
$r = \sqrt{{\left(2 - 0\right)}^{2} + {\left(2 - 1\right)}^{2}}$
$r = \sqrt{{4}^{2} + {1}^{2}} = \sqrt{17}$
eqn.
${\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = 17$
${x}^{2} + {y}^{2} - 4 x - 4 y + 8 = 17$
${x}^{x} + {y}^{2} - 4 x - 4 y - 9 = 0$
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## HOW TO GRAPH A QUADRATIC FUNCTIONS
How to graph a quadratic functions :
Here we are going to see some example problems on graphing quadratic functions.
A quadratic function can be described by an equation of the form y = ax2 + bx + c, where a ≠ 0.
Open upward parabola Open downward parabola
## Opening of the parabola is on which side
(1) (i) If the coefficient of x2 is positive, the parabola opens upward.
(ii) If the coefficient of x2 is negative, the parabola opens upward.
## Find the coordinates of the vertex.
This point, where the parabola changes direction, is called the "vertex".
We can find the x-coordinate of the vertex by using the formula, x = -b/2a
By applying the value of x in the given equation, we can get the y-coordinate value.
Vertex of the parabola (-b/2a, f(-b/2a)).
## Identify the vertex as a maximum or minimum.
• The parabola which opens downward has only the maximum value
• The parabola which opens downward has only the minimum value
## Draw the graph
By giving some random values of x, we can find the values of y.We can use the symmetry of the parabola to help us draw its graph. On a coordinate plane, graph the vertex and the axis of symmetry.
Let us look into some example problems to understand the above concept.
## How to graph a quadratic functions - Examples
Example 1 :
Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of each function. Identify the vertex as a maximum or minimum. Then graph the function.
y = 2x2 - 4x - 5
Solution :
Axis of symmetry :
The given parabola is symmetric about y-axis. Since the coefficient of x2 is positive, the parabola opens upward direction.
Vertex of the parabola :
x = -b/2a
a = 2, b = -4 and c = -5
x = -(-4)/2(2) ==> 4/4 ==> 1
By applying the value x = 1 in the given equation
y = 2(1) - 4(1) - 5
= 2 - 4 - 5
= -7
Hence the vertex of the parabola is (1, -7).
Identify the vertex as a maximum or minimum :
Since the parabola opens upward direction, it has only minimum value.
Draw the graph :
x-2-1012 y111-5-7-5 Set of ordered pairs :(-2, 11) (-1, 1) (0, -5) (1, -7) and (2, -5)
Example 2 :
Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of each function. Identify the vertex as a maximum or minimum. Then graph the function.
y = -3x2 - 6x + 4
Solution :
Axis of symmetry :
The given parabola is symmetric about y-axis. Since the coefficient of x2 is negative, the parabola opens downward direction.
Vertex of the parabola :
x = -b/2a
a = -3, b = -6 and c = 4
x = -(-6)/2(-3) ==> 6/(-6) ==> -1
By applying the value x = -1 in the given equation
y = -3(-1)2 - 6(-1) + 4
= -3 + 6 + 4
= -3 + 10
= 7
Hence the vertex of the parabola is (-1, 7).
Identify the vertex as a maximum or minimum :
Since the parabola opens downward direction, it has only maximum value.
Draw the graph :
x-2-1012 y-474-5-20 Set of ordered pairs :(-2, -4) (-1, 7) (0, 4) (1, -5) and (2, -20)
After having gone through the stuff given above, we hope that the students would have understood "How to graph a quadratic functions".
Apart from the stuff "How to find nth term of arithmetic sequence" given in this section, if you need any other stuff in math, please use our google custom search here.
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HCF and LCM word problems
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Algebra word problems
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Word problems on comparing rates
Converting customary units word problems
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Word problems on simple interest
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Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
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Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
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Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
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Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
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Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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# Class 10 Mathematics Sample Paper
Please refer to the Class 10 Mathematics Sample Paper for the current academic year given below. We have provided the latest CBSE Sample Papers for Term 1 and Term 2 for Mathematics Class 10. All guess sample papers have been prepared based on the latest blueprint and examination pattern for the current year. All sample papers for Mathematics Class 10 Term 1 and 2 have been given with solutions. Students can access the multiple guess papers given below. Practicing more Class 10 Mathematics Sample Papers will help you to get more marks in upcoming exams.
## CBSE Sample Papers for Class 10 Mathematics
Class 10 Mathematics Sample Paper Term 2 Set A
#### SECTION – A
1. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer : Let O be the common centre of two concentric circles and let AB be a chord of larger circle touching the smaller circle at P. Join OP.
Since OP is the radius of the smaller circle and AB is a tangent to this circle at P.
∴ OP ⊥ AB
Since perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.
⇒ AP = PB
Now, in ΔAOP, right-angled at P,
OA2 = AP2 + OP2 ⇒ 52 = AP2 + 32 ⇒ 25 – 9 = AP2
⇒ AP2 = 16 ⇒ AP = 4
Now AB = 2 × AP = 2 × 4 = 8 [∵ AP = PB]
Hence, the length of the chord of the larger circle which touches the smaller circle is 8 cm.
2. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer : Let the two consecutive odd numbers be x and x + 2.
∴ x2 + (x + 2)2 = 394 ⇒ x2 + x2 + 4 + 4x = 394
⇒ 2x2 + 4x + 4 = 394 ⇒ 2x2 + 4x – 390 = 0
⇒ x2 + 2x – 195 = 0 ⇒ x2 + 15x – 13x – 195 = 0
⇒ x (x + 15) – 13 (x + 15) = 0 ⇒ (x – 13) (x + 15) = 0
Either x – 13 = 0 or x + 15 = 0 ⇒ x = 13 or x = – 15 (neglected)
When first number x = 13, then second number x + 2 = 13 + 2 = 15.
3. A circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20 m. The height of the cylindrical and conical portions are 4.2 m and 2.1 m respectively. Find the volume inside the tent.
Answer : Let h1 = 4.2 m be the height of cylindrical portion, h2 = 2.1 m be the height of conical portion and r = 20 m be the radius of tent.
Then, the volume inside the tent = Volume of cylindrical part + Volume of conical part
OR
A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Answer : Here, r = 3.5 cm
∴ h = (15.5 – 3.5) cm = 12.0 cm
Surface area of the conical part = πrl
Surface area of the hemispherical part = 2πr2
∴ Total surface area of the toy = πrl + 2πr2 = πr(l + 2r) cm2
Now, l2 = (12)2 + (3.5)2 = 156.25 cm2 ⇒ l = 12.5 cm
∴ TSA of the toy = 22/7 × 35/10 (12.5 2 3.5) = 11 × (12.5 + 7) cm2
= 11 × 19.5 cm2 = 214.5 cm2.
4. If the mean of the following distribution is 6.4, then find the value of ‘p’.
5. Find the sum of the given AP: – 5 + (– 8) + (– 11) + … + (– 230).
Answer : We have, a = – 5 and d = –8 + 5 = – 3
So, an = a + (n – 1)d
⇒ – 230 = – 5 + (n – 1) (– 3) ⇒ – 230 = – 5 – 3n + 3
⇒ – 230 + 2 = – 3n ⇒ – 228 = – 3n ⇒ n = 228/3 = 76
6. Solve the quadratic equation 2x2 + ax – a2 = 0 for x using quadratic formula.
Answer : Here, a = 2, b = a and c = –a2.
Using the formula,
OR
If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the value of k.
Answer : Since, –5 is a root of the equation 2x2 + px – 15 = 0
∴ 2 (–5)2 + p (–5) – 15 = 0
⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p (x2 + x) + k = 0 or 7x2 + 7x + k = 0 has equal roots
∴ D = 0 ⇒ b2 – 4ac = 0 or 49 – 4 × 7k = 0 ⇒ k = 49/28 = 7/4
#### SECTION – B
7. Calculate the mode for the following frequency distribution:
Answer : The frequency distribution table from the given data can be drawn as :
8. At a point on level ground, the angle of the elevation of the top of a vertical tower is found to be such that its tangent is 5/12. On walking 192 metres towards the tower, the tangent of the new angle of elevation is found to be. Find the height of the tower.
Answer : Let PQ be the tower of height h metres, point A be the first point of observation and B be the second point of observation towards the tower PQ such that AB = 192 m.
Also, let ∠PAQ = x and ∠PBQ = y
OR
The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.
Answer : Let AB be the tower of height 30 m and CD be another tower of height h m.
Then, ∠ADB = 60° and ∠CBD = 30°
9. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
1. Draw a circle with O as centre and radius as 3 cm.
2. Draw a diameter of it and extend both the sides and mark the points as P, Q such that OP = OQ = 7 cm.
3. Draw the perpendicular bisectors of OP and OQ to intersect PQ at M and N respectively.
4. With M as centre and OM as radius draw a circle to cut the given circle at A and B. With N as the centre and ON as radius draw a circle to cut the given circle at C and D.
5. Join PA, PB, QC, QD.
Hence, PA, PB and QC, QD are the required tangents from P and Q respectively.
10. If the median of the distribution given below is 28.5, then find the values of x and y.
#### SECTION – C
11. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each 5 mm of its ends (see along side). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its: (i) surface area (ii) volume
Answer : Since, diameter of the capsule = 5 mm, its radius = 5/2 mm = 2.5 mm
Length (height) of the cylindrical portion = 14 mm – 2 × 5/2 mm = 9 mm
(i) Surface area of the capsule
= Curved S.A. of its cylindrical part + 2 (curved S.A. of a hemispherical end)
(ii) Volume of the capsule
= Volume of cylindrical part + 2 (volume of a hemispherical end)
12. In the given figure, OP is equal to diameter of the circle. Prove that ABP is an equilateral triangle.
Answer : Join OP and let it meets the circle at point Q.
Since OP = 2r (Diameter of the circle)
⇒ OQ = QP = r
Consider ΔAOP in which OA ⊥ AP and OP is the hypotenuse.
∴ OQ = AQ = OA
(Mid-point of the hypotenuse is equidistant from the vertices)
⇒ OAQ is an equilateral triangle.
⇒ ∠AOQ = 60° (Each angle of an equilateral triangle is 60°)
Consider right-angled triangle OAP.
∠AOQ = 60° (Proved above)
∠OAP = 90° ⇒ ∠APO = 30°
∠APB = 2∠APO = 2 × 30° = 60°
Also PA = PB (Tangents to a circle from an external point are equal.)
⇒ ∠PAB = ∠PBA (Angles opposite to equal sides in ΔPAB)
In ΔABP, ∠APB = 60°
⇒ ∠PAB = ∠PBA = (180° − 60°)/2 = 60°
⇒ Each angle of ΔPAB is 60°
⇒ PAB is an equilateral triangle. Hence Proved.
OR
In the figure given, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°.
Answer : Join O and C.
In ΔAOP and ΔAOC, OP = OC (Radii of a circle)
OA = OA (Common)
∠OPA = ∠OCA (Each 90°)
∴ ΔAOP ≅ ΔAOC (By RHS Congruency)
⇒ ∠1 = ∠2 …(i) (CPCT)
Similarly, ΔBOQ ≅ ΔBOC (By RHS Congruency)
⇒ ∠3 = ∠4 …(ii) (CPCT)
Now, ∠1 + ∠2 + ∠3 + ∠4 = 180° (Angles in a straight line)
⇒ 2∠2 + 2∠3 = 180° [From (i) and (ii)]
⇒ 2(∠2 + ∠3) = 180° ⇒ ∠2 + ∠3 = 90° ⇒ ∠AOB = 90°
CASE STUDY- 1
13. Mohan is watching the top of the lighthouse from the top of the building, he founds angle of elevation is 30° and then after he is watching the bottom of the lighthouse, he founds angle of depression is 60°. The height of the building is 60m. Based on this given information, answer the following questions.
(i) Find the difference between the heights of the lighthouse and the building.
(ii) Find the distance between the lighthouse and the building.
(i) Difference between the heights of the lighthouse and the building = CE = 20 m
(ii) The distance between the lighthouse and the building = BD = 20√3 m.
CASE STUDY- 2
14. Pollution – A Major Problem: One of the major serious problems that the world is facing today is the environmental pollution. Common types of pollution include light, noise, water and air pollution.
In a school, students thoughts of planting trees in and around the school to reduce noise pollution and air pollution.
Condition I: It was decided that the number of trees that each section of each class will plant be the same as the class in which they are studying, e.g. a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on a section of class XII will plant 12 trees.
Condition II: It was decided that the number of trees that each section of each class will plant be the double of the class in which they are studying, e.g. a section of class I will plant 2 trees, a section of class II will plant 4 trees and so on a section of class XII will plant 24 trees.
(i) Refer to Condition I – If there are two sections of each class, how many trees will be planted by the students?
(ii) Refer to Condition II – If there are three sections of each class, how many trees will be planted by the students?
Answer : (i) Number of trees planted by the students of two sections of class I = 1 × 2 = 2
Number of trees planted by the students of two sections of class II = 2 × 2 = 4
Similarly, Number of trees planted by the students of two sections of class III = 3 × 2 = 6
Number of trees planted by the students of two sections of class XII = 12 × 2 = 24
So, the number of trees planted by the students of two sections of different classes are 2, 4, 6, ……24
This series forms an AP
Here,first term(a) = 4, common Difference (d) = 4 – 2 = 2, last term (l) = 24 and (number of terms)n = 12
Number of trees planted by the students =sum of 12 terms of above AP
(ii) Number of trees planted by the students of two sections of class I = 2 × 3 = 6
Number of trees planted by the students of two sections of class II = 4 × 3 = 12
Similarly, Number of trees planted by the students of two sections of class III = 6 × 3 = 18
Number of trees planted by the students of two sections of class XII = 24 × 3 = 72
So, the number of trees planted by the students of two sections of different classes are 6, 12, 18, ……,72
This series forms an AP
Here, first term(a) = 6, common Difference (d) = 12 – 6 = 6, last term (l) = 72 and (number of terms)n = 12
Number of trees planted by the students = sum of 12 terms of above AP
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# 14.5 Surface Area
In Section 10.1 we used definite integrals to compute the arc length of plane curves of the form $y=f(x)$. We later extended these ideas to compute the arc length of plane curves defined by parametric or polar equations.
The natural extension of the concept of “arc length over an interval” to surfaces is “surface area over a region.”
Consider the surface $z=f(x,y)$ over a region $R$ in the $x$-$y$ plane, shown in Figure 14.5.1(a). Because of the domed shape of the surface, the surface area will be greater than that of the area of the region $R$. We can find this area using the same basic technique we have used over and over: we’ll make an approximation, then using limits, we’ll refine the approximation to the exact value. margin: (a) (b) Figure 14.5.1: Developing a method of computing surface area. Λ
As done to find the volume under a surface or the mass of a lamina, we subdivide $R$ into $n$ subregions. Here we subdivide $R$ into rectangles, as shown in the figure. One such subregion is outlined in the figure, where the rectangle has dimensions $\Delta x_{i}$ and $\Delta y_{i}$, along with its corresponding region on the surface.
In part (b) of the figure, we zoom in on this portion of the surface. When $\Delta x_{i}$ and $\Delta y_{i}$ are small, the function is approximated well by the tangent plane at any point $(x_{i},y_{i})$ in this subregion, which is graphed in part (b). In fact, the tangent plane approximates the function so well that in this figure, it is virtually indistinguishable from the surface itself! Therefore we can approximate the surface area $S_{i}$ of this region of the surface with the area $T_{i}$ of the corresponding portion of the tangent plane.
This portion of the tangent plane is a parallelogram, defined by sides $\vec{u}$ and $\vec{v}$, as shown. One of the applications of the cross product from Section 11.4 is that the area of this parallelogram is $\left\lVert\vec{u}\times\vec{v}\right\rVert$. Once we can determine $\vec{u}$ and $\vec{v}$, we can determine the area.
$\vec{u}$ is tangent to the surface in the direction of $x$, therefore, from Section 13.7, $\vec{u}$ is parallel to $\left\langle 1,0,f_{x}(x_{i},y_{i})\right\rangle$. The $x$-displacement of $\vec{u}$ is $\Delta x_{i}$, so we know that $\vec{u}=\Delta x_{i}\left\langle 1,0,f_{x}(x_{i},y_{i})\right\rangle$. Similar logic shows that $\vec{v}=\Delta y_{i}\left\langle 0,1,f_{y}(x_{i},y_{i})\right\rangle$. Thus:
surface area $S_{i}$ $\displaystyle\approx\text{area of T_{i}}$ $\displaystyle=\left\lVert\vec{u}\times\vec{v}\right\rVert$ $\displaystyle=\left\lVert\Delta x_{i}\left\langle 1,0,f_{x}(x_{i},y_{i})\right% \rangle\times\Delta y_{i}\left\langle 0,1,f_{y}(x_{i},y_{i})\right\rangle\right\rVert$ $\displaystyle=\sqrt{1+[f_{x}(x_{i},y_{i})]^{2}+[f_{y}(x_{i},y_{i})]^{2}}\Delta x% _{i}\Delta y_{i}.$
Note that $\Delta x_{i}\Delta y_{i}=\Delta A_{i}$, the area of the $i^{\,\text{th}}$ subregion.
Summing up all $n$ of the approximations to the surface area gives
$\text{surface area over R}\approx\sum_{i=1}^{n}\sqrt{1+[f_{x}(x_{i},y_{i})]^% {2}+[f_{y}(x_{i},y_{i})]^{2}}\Delta A_{i}.$
Once again take a limit as all of the $\Delta x_{i}$ and $\Delta y_{i}$ shrink to 0; this leads to a double integral.
margin: Note: As before, we think of “$\iint_{R}\operatorname{d}\!S$” as meaning “sum up lots of little surface areas over $R$.” The concept of surface area is defined here, for while we already have a notion of the area of a region in the plane, we did not yet have a solid grasp of what “the area of a surface in space” means. Λ
###### Definition 14.5.1 Surface Area
Let $z=f(x,y)$ where $f_{x}$ and $f_{y}$ are continuous over a closed, bounded region $R$. The surface area $S$ over $R$ is
$\displaystyle S$ $\displaystyle=\iint_{R}\operatorname{d}\!S$ $\displaystyle=\iint_{R}\sqrt{1+[f_{x}(x,y)]^{2}+[f_{y}(x,y)]^{2}}\operatorname% {d}\!A.$
We test this definition by using it to compute surface areas of known surfaces. We start with a triangle.
###### Example 14.5.1 Finding the surface area of a plane over a triangle
Let $f(x,y)=4-x-2y$, and let $R$ be the region in the plane bounded by $x=0$, $y=0$ and $y=2-x/2$, as shown in Figure 14.5.2. Find the surface area of $z=f(x,y)$ over $R$. margin: Figure 14.5.2: Finding the area of a triangle in space in Example 14.5.1. Λ
SolutionWe follow Definition 14.5.1 and start by noting $f_{x}(x,y)=-1$ and $f_{y}(x,y)=-2$. To define $R$, we use bounds $0\leq y\leq 2-x/2$ and $0\leq x\leq 4$. Therefore
$\displaystyle S$ $\displaystyle=\iint_{R}\operatorname{d}\!S$ $\displaystyle=\int_{0}^{4}\int_{0}^{2-x/2}\sqrt{1+(-1)^{2}+(-2)^{2}}% \operatorname{d}\!y\operatorname{d}\!x$ $\displaystyle=\int_{0}^{4}\sqrt{6}\left(2-\frac{x}{2}\right)\operatorname{d}\!x$ $\displaystyle=4\sqrt{6}.$
Because the surface is a triangle, we can figure out the area using geometry. Considering the base of the triangle to be the side in the $x$-$y$ plane, we find the length of the base to be $\sqrt{20}$. We can find the height using our knowledge of vectors: let $\vec{u}$ be the side in the $x$-$z$ plane and let $\vec{v}$ be the side in the $x$-$y$ plane. The height is then $\left\lVert\vec{u}-\text{proj}_{\,\vec{v}}{\,\vec{u}}\right\rVert=4\sqrt{6/5}$. Geometry states that the area is thus
$\frac{1}{2}\cdot 4\sqrt{6/5}\cdot\sqrt{20}=4\sqrt{6}.$
We affirm the validity of our formula.
It is “common knowledge” that the surface area of a sphere of radius $r$ is $4\pi r^{2}$. We confirm this in the following example, which involves using our formula with polar coordinates.
###### Example 14.5.2 The surface area of a sphere.
Find the surface area of the sphere with radius $a$ centered at the origin, whose top hemisphere has equation $z=f(x,y)=\sqrt{a^{2}-x^{2}-y^{2}}$.
SolutionWe start by computing partial derivatives and find
$f_{x}(x,y)=\frac{-x}{\sqrt{a^{2}-x^{2}-y^{2}}}\quad\text{and}\quad f_{y}(x,y)=% \frac{-y}{\sqrt{a^{2}-x^{2}-y^{2}}}.$
As our function $f$ only defines the top upper hemisphere of the sphere, we double our surface area result to get the total area:
$\displaystyle S$ $\displaystyle=2\iint_{R}\sqrt{1+[f_{x}(x,y)]^{2}+[f_{y}(x,y)]^{2}}% \operatorname{d}\!A$ $\displaystyle=2\iint_{R}\sqrt{1+\frac{x^{2}+y^{2}}{a^{2}-x^{2}-y^{2}}}% \operatorname{d}\!A.$
The region $R$ that we are integrating over is the disk, centered at the origin, with radius $a$: $x^{2}+y^{2}\leq a^{2}$. Because of this region, we are likely to have greater success with our integration by converting to polar coordinates. Using the substitutions $x=r\cos\theta$, $y=r\sin\theta$, $\operatorname{d}\!A=r\operatorname{d}\!r\operatorname{d}\!\theta$ and bounds $0\leq\theta\leq 2\pi$ and $0\leq r\leq a$, we have: margin: Note: The inner integral in Equation (14.5.1) is an improper integral, as the integrand of $\displaystyle\int_{0}^{a}r\sqrt{\frac{a^{2}}{a^{2}-r^{2}}}\operatorname{d}\!r$ is not defined at $r=a$. To properly evaluate this integral, one must use the techniques of Section 8.6. The reason this need arises is that the function $f(x,y)=\sqrt{a^{2}-x^{2}-y^{2}}$ fails the requirements of Definition 14.5.1, as $f_{x}$ and $f_{y}$ are not continuous on the boundary of the circle $x^{2}+y^{2}=a^{2}$. The computation of the surface area is still valid. The definition makes stronger requirements than necessary in part to avoid the use of improper integration, as when $f_{x}$ and/or $f_{y}$ are not continuous, the resulting improper integral may not converge. Since the improper integral does converge in this example, the surface area is accurately computed. Λ
$\displaystyle S$ $\displaystyle=2\int_{0}^{2\pi}\int_{0}^{a}\sqrt{1+\frac{r^{2}\cos^{2}\theta+r^% {2}\sin^{2}\theta}{a^{2}-r^{2}\cos^{2}\theta-r^{2}\sin^{2}\theta}}\ r% \operatorname{d}\!r\operatorname{d}\!\theta$ $\displaystyle=2\int_{0}^{2\pi}\int_{0}^{a}r\sqrt{1+\frac{r^{2}}{a^{2}-r^{2}}}% \operatorname{d}\!r\operatorname{d}\!\theta$ $\displaystyle=2\int_{0}^{2\pi}\int_{0}^{a}r\sqrt{\frac{a^{2}}{a^{2}-r^{2}}}% \operatorname{d}\!r\operatorname{d}\!\theta.$ (14.5.1) Apply substitution $u=a^{2}-r^{2}$ and integrate the inner integral, giving $\displaystyle=2\int_{0}^{2\pi}a^{2}\operatorname{d}\!\theta$ $\displaystyle=4\pi a^{2}.$
Our work confirms our previous formula.
###### Example 14.5.3 Finding the surface area of a cone
The general formula for a right cone with height $h$ and base radius $a$, as shown in Figure 14.5.3, is margin: Figure 14.5.3: Finding the surface area of a cone in Example 14.5.3. Λ
$f(x,y)=h-\frac{h}{a}\sqrt{x^{2}+y^{2}}.$
Find the surface area of this cone.
SolutionWe begin by computing partial derivatives.
$f_{x}(x,y)=-\frac{xh}{a\sqrt{x^{2}+y^{2}}}\qquad\text{and}\qquad f_{y}(x,y)=-% \frac{yh}{a\sqrt{x^{2}+y^{2}}}.$
Since we are integrating over the disk $x^{2}+y^{2}\leq a^{2}$, we again use polar coordinates. Using the standard substitutions, our integrand becomes
$\sqrt{1+\left(\frac{hr\cos\theta}{a\sqrt{r^{2}}}\right)^{2}+\left(\frac{hr\sin% \theta}{a\sqrt{r^{2}}}\right)^{2}}.$
This may look intimidating at first, but there are lots of simple simplifications to be done. It amazingly reduces to just
$\sqrt{1+\frac{h^{2}}{a^{2}}}=\frac{1}{a}\sqrt{a^{2}+h^{2}}.$
Our polar bounds are $0\leq\theta\leq 2\pi$ and $0\leq r\leq a$. Thus margin: Note: Note that once again $f_{x}$ and $f_{y}$ are not continuous on the domain of $f$, as both are undefined at $(0,0)$. (A similar problem occurred in the previous example.) Once again the resulting improper integral converges and the computation of the surface area is valid. Λ
$\displaystyle S$ $\displaystyle=\int_{0}^{2\pi}\int_{0}^{a}r\frac{1}{a}\sqrt{a^{2}+h^{2}}% \operatorname{d}\!r\operatorname{d}\!\theta$ $\displaystyle=\int_{0}^{2\pi}\left.\left(\frac{1}{2}r^{2}\frac{1}{a}\sqrt{a^{2% }+h^{2}}\right)\right|_{0}^{a}\operatorname{d}\!\theta$ $\displaystyle=\int_{0}^{2\pi}\frac{1}{2}a\sqrt{a^{2}+h^{2}}\operatorname{d}\!\theta$ $\displaystyle=\pi a\sqrt{a^{2}+h^{2}}.$
This matches the formula found in the back of this text.
###### Example 14.5.4 Finding surface area over a region
Find the area of the surface $z=f(x,y)=x^{2}-3y+3$ over the region $R$ bounded by $-x\leq y\leq x$, $0\leq x\leq 4$, as pictured in Figure 14.5.4.
SolutionIt is straightforward to compute $f_{x}(x,y)=2x$ and $f_{y}(x,y)=-3$. Thus the surface area is described by the double integral margin: Figure 14.5.4: Graphing the surface in Example 14.5.4. Λ
$\iint_{R}\sqrt{1+(2x)^{2}+(-3)^{2}}\operatorname{d}\!A=\iint_{R}\sqrt{10+4x^{2% }}\operatorname{d}\!A.$
As with integrals describing arc length, double integrals describing surface area are in general hard to evaluate directly because of the square-root. This particular integral can be easily evaluated, though, with judicious choice of our order of integration.
Integrating with order $\operatorname{d}\!x\operatorname{d}\!y$ requires us to evaluate $\int\sqrt{10+4x^{2}}\operatorname{d}\!x$. This can be done, though it involves Integration By Parts and $\sinh^{-1}x$. Integrating with order $dy\operatorname{d}\!x$ has as its first integral $\int\sqrt{10+4x^{2}}\operatorname{d}\!y$, which is easy to evaluate: it is simply $y\sqrt{10+4x^{2}}+C$. So we proceed with the order $dy\operatorname{d}\!x$; the bounds are already given in the statement of the problem.
$\displaystyle\iint_{R}\sqrt{10+4x^{2}}\operatorname{d}\!A$ $\displaystyle=\int_{0}^{4}\int_{-x}^{x}\sqrt{10+4x^{2}}\operatorname{d}\!y% \operatorname{d}\!x$ $\displaystyle=\int_{0}^{4}\left.\bigl{(}y\sqrt{10+4x^{2}}\bigr{)}\right|_{-x}^% {x}\operatorname{d}\!x$ $\displaystyle=\int_{0}^{4}\bigl{(}2x\sqrt{10+4x^{2}}\bigr{)}\operatorname{d}\!x.$ Apply substitution with $u=10+4x^{2}$: $\displaystyle=\left.\left(\frac{1}{6}\bigl{(}10+4x^{2}\bigr{)}^{3/2}\right)% \right|_{0}^{4}$ $\displaystyle=\frac{1}{3}\bigl{(}37\sqrt{74}-5\sqrt{10}\bigr{)}\text{ units}^{% 2}.$
So while the region $R$ over which we integrate has an area of $16\text{ units}^{2}$, the surface has a much greater area as its $z$-values change dramatically over $R$.
In practice, technology helps greatly in the evaluation of such integrals. High powered computer algebra systems can compute integrals that are difficult, or at least time consuming, by hand, and can at the least produce very accurate approximations with numerical methods. In general, just knowing how to set up the proper integrals brings one very close to being able to compute the needed value. Most of the work is actually done in just describing the region $R$ in terms of polar or rectangular coordinates. Once this is done, technology can usually provide a good answer.
We have learned how to integrate integrals; that is, we have learned to evaluate double integrals. In the next section, we learn how to integrate double integrals — that is, we learn to evaluate triple integrals, along with learning some uses for this operation.
## Exercises 14.5
### Terms and Concepts
1. 1.
“Surface area” is analogous to what previously studied concept?
2. 2.
To approximate the area of a small portion of a surface, we computed the area of its plane.
3. 3.
We interpret $\displaystyle\iint_{R}\operatorname{d}\!S$ as “sum up lots of little .”
4. 4.
Why is it important to know how to set up a double integral to compute surface area, even if the resulting integral is hard to evaluate?
5. 5.
Why do the graphs of $z=f(x,y)$ and $z=g(x,y)=f(x,y)+h$, for some real number $h$, have the same surface area over a region $R$?
6. 6.
Let $z=f(x,y)$ and $z=g(x,y)=2f(x,y)$. Why is the surface area of the graph of $g$ over a region $R$ not twice the surface area of the graph of $f$ over $R$?
### Problems
In Exercises 7–10., set up the iterated integral that computes the surface area of the surface $z=f(x,y)$ over the region $R$.
1. 7.
$f(x,y)=\sin x\cos y$; $R$ is the rectangle with bounds $0\leq x\leq 2\pi$, $0\leq y\leq 2\pi$.
2. 8.
$\displaystyle f(x,y)=\frac{1}{x^{2}+y^{2}+1}$; $R$ is the disk $x^{2}+y^{2}\leq 9$.
3. 9.
$\displaystyle f(x,y)=x^{2}-y^{2}$; $R$ is the rectangle with opposite corners $(-1,-1)$ and $(1,1)$.
4. 10.
$\displaystyle f(x,y)=\frac{1}{e^{x^{2}}+1}$; $R$ is the rectangle bounded by $-5\leq x\leq 5$ and $0\leq y\leq 1$.
In Exercises 11–18., find the area of the surface of $z=f(x,y)$ over the region $R$.
1. 11.
$f(x,y)=3x-7y+2$; $R$ is the rectangle with opposite corners $(-1,0)$ and $(1,3)$.
2. 12.
$f(x,y)=2x+2y+2$; $R$ is the triangle with corners $(0,0)$, $(1,0)$ and $(0,1)$.
3. 13.
$f(x,y)=x^{2}+y^{2}+10$; $R$ is the disk $x^{2}+y^{2}\leq 16$.
4. 14.
$f(x,y)=-2x+4y^{2}+7$ over $R$, the triangle bounded by $y=-x$, $y=x$, $0\leq y\leq 1$.
5. 15.
$f(x,y)=\frac{2}{3}x^{3/2}+2y^{3/2}$ over $R$, the rectangle with opposite corners $(0,0)$ and $(1,1)$.
6. 16.
$f(x,y)=10-2\sqrt{x^{2}+y^{2}}$ over $R$, the disk $x^{2}+y^{2}\leq 25$. (This is the cone with height 10 and base radius 5; be sure to compare your result with the known formula.)
7. 17.
Find the surface area of the sphere with radius 5 by doubling the surface area of $f(x,y)=\sqrt{25-x^{2}-y^{2}}$ over $R$, the disk $x^{2}+y^{2}\leq 25$. (Be sure to compare your result with the known formula.)
8. 18.
Find the surface area of the ellipse formed by restricting the plane $f(x,y)=cx+dy+h$ to the region $R$, the disk $x^{2}+y^{2}\leq 1$, where $c$, $d$ and $h$ are some constants. Your answer should be given in terms of $c$ and $d$; why does the value of $h$ not matter?
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# Which of the following is the correct order of operations to solve the problem?
## Which of the following is the correct order of operations to solve the problem?
The order of operations says that operations must be done in the following order: parentheses, exponents, multiplication, division, addition, and subtraction.
### What comes first division or subtraction?
Order of operations tells you to perform division before subtraction. Then subtract. Simplify 60 – 30 ÷ 3 • 5 + 7.
#### Is it 16 or 1?
Some people got 16 as the answer, and some people got 1. The confusion has to do with the difference between modern and historic interpretations of the order of operations. The correct answer today is 16. An answer of 1 would have been correct 100 years ago.
Which comes first in the order of operations?
parentheses
The order of operations tells us the order to solve steps in expressions with more than one operation. First, we solve any operations inside of parentheses or brackets. Second, we solve any exponents. Third, we solve all multiplication and division from left to right.
How do you solve Bodmas questions?
1. There are no brackets or orders so start with division and multiplication.
2. 7 ÷ 7 = 1 and 7 × 7 = 49.
3. The calculation now reads 7 + 1 + 49 – 7.
4. Now do the addition and subtraction. 7 + 1 + 49 = 57 – 7 = 50.
## What is the order of math problems?
To help students in the United States remember this order of operations, teachers drill the acronym PEMDAS into them: parentheses, exponents, multiplication, division, addition, subtraction. Other teachers use an equivalent acronym, BODMAS: brackets, orders, division and multiplication, and addition and subtraction.
### Why is the order of operations in that order?
(Operation is just another way of saying calculation. Subtraction, multiplication, and division are all examples of operations.) The order of operations is important because it guarantees that people can all read and solve a problem in the same way.
#### How do you do solutions in math?
A solution set is the set of all variables that makes the equation true. The solution set of 2y + 6 = 14 is {4}, because 2(4) + 6 = 14. The solution set of y2 + 6 = 5y is {2, 3} because 22 + 6 = 5(2) and 32 + 6 = 5(3).
Is Pemdas or Bedmas correct?
PEMDAS is often expanded to the mnemonic “Please Excuse My Dear Aunt Sally” in schools. Canada and New Zealand use BEDMAS, standing for Brackets, Exponents, Division/Multiplication, Addition/Subtraction.
What is MDAS rule example?
It is used when an expression or equation has more than one operation. According to the Order of Operations, all multiplication or division must occur before addition or subtraction. For example, the expression 6+4×5 involves addition and multiplication.
## What is the order of Bedmas?
BEDMAS tells us that brackets are the highest priority, then exponents, then both divi- sion and multiplication, and finally addition and subtraction. This means that we evaluate exponents before we multiply, divide before we subtract, etc.
### What is an equation Class 7?
An equation is a statement of equality between two mathematical expressions containing one or more variables.
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Modern technology gives us many things.
0
what is the square root of 144, The square root of 144 is 12. The cube root of 144 is 6. The fourth root of 144 is 3, and so on. When n is greater than 3, it is called fourth root, fifth root, and so forth.
## what is the square root of 144
The square root of 144 is 12. This means that if you multiply 12 by itself, you will get 144. The square root of 144 is a whole number, which makes it easy to calculate. You can simply take the square root of 144 by taking the number 12 and multiplying it by itself.
which of the following is not an irrational number
## The square root of 144 is 12.
The square root of 144 is 12 because that is what times what is 144. 12 multiplied by itself is 144, so the square root of 144 is simply 12. This makes perfect sense when you think about it and it is a great way to understand how to take the square root of a number.
## The square root of 144 is a whole number.
The square root of 144 is 12. This is a whole number because it can be multiplied by itself to get 144. The square root of 144 is a positive number because it gives the number 144 when squared. The square root of 144 is also a rational number because it is a finite or repeating decimal.
## The square root of 144 is a rational number.
The square root of 144 is a rational number. The square of a number is the product of that number by itself. Conversely, if you want to know what number squares The positive number square root is the principal square root. Remember that when a negative number is added to itself, the product is positive And can you write root 2 as a fraction? √2 cannot be written as a fraction of the form (1), so it is one Rational or irrational? Proof of simple root expressions | math irrational The square root of a natural number is irrational if at least one of the prime factors is an odd number in its prime factorization For example, 4 and -4 are square roots of 16 because 4² = (-4)² = 16. Every non-negative real number a has one “The square root of 144 is 12.” Explain why there are no square roots of negative numbers. “If you multiply a rational number tenfold, it multiplies tenfold The number of
## The square root of 144 is an irrational number.
The square root of 144 is an irrational number. This can be seen by the fact that it cannot be written as a fraction of the form (1). Additionally, the square root of 144 is not a natural number, so it is not rational.
## The square root of 144 is a real number.
The square root of 144 is a real number. It is a rational number, because it can be expressed as a fraction. 12 is a rational number, and the square root of 144 is also rational. Remember that it's possible to take the square root of any number, including fractions and decimals.
## The square root of 144 is not a complex number.
The square root of 144 is not a complex number. This means that the square roots of negative numbers are no longer real numbers. If you've mastered the square numbers a bit, then you know that 12² = 144, so square root (144) = 12, so the lies in the set of positive real numbers. But what about when we take the square root of a negative number?
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Find the Laplace transform of (1-e^t)/t
The Laplace transform of (1-et)/t is equal to log[(s-1)/s]. In this post, we find Laplace of (1-et)/t. The Laplace transform formula of (1-et)/t is given by
L{(1-et)/t} = log[(s-1)/s].
Laplace of (1-et)/t
Solution:
Let us recall the division by t formula:
L$\Big[ \dfrac{f(t)}{t} \Big]$ = $\int_s^\infty F(s) ds$ where F(s) = L{f(t)}, the Laplace transform of f(t).
Put f(t) = 1-et in the formula, so that we have:
F(s) = L{f(t)} = L{1-et}
= L{1} – L{et}
= $\dfrac{1}{s}$ – $\dfrac{1}{s-1}$ as we know L{eat} = 1/(s-a).
Now, from the above formula, we get that
L$\Big\{\dfrac{1-e^t}{t} \Big\}$ = $\int_s^\infty \Big[ \dfrac{1}{s} – \dfrac{1}{s-1} \Big] ds$
= $\Big[ \log s – \log (s-1) \Big]_s^\infty$
= $\Big[ \log \dfrac{s}{s-1} \Big]_s^\infty$
= lims→∞ log $\dfrac{s}{s-a}$ – $\log \dfrac{s}{s-1}$
= log 1 – $\log \dfrac{s}{s-1}$
= $\log \dfrac{s-1}{s}$ since log 1 = 0.
So the Laplace transform of (1-et)/t is equal to log[(s-1)/s] which is proved by the division by t formula.
More Laplace Transforms:
Find Laplace of tet
Find Laplace of sint/t
What is the Laplace of cost/t
Laplace transform of t sinat
Laplace transform of sin2t/t
Laplace transform of cos2t/t
Find Laplace of sin2t
Find Laplace transform of cos2t
FAQs
Q1: What is the Laplace of (1-et)/t?
Answer: The Laplace of (1-et)/t is log[(s-1)/s].
Q2: Find L{(1-et)/t}.
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# DAV Class 5 Maths Chapter 16 Worksheet 1 Solutions
The DAV Class 5 Maths Solutions and DAV Class 5 Maths Chapter 16 Worksheet 1 Solutions of Triangles offer comprehensive answers to textbook questions.
## DAV Class 5 Maths Ch 16 Worksheet 1 Solutions
Question 1.
How many vertices, sides, and angles does a triangle have?
Solution:
3 vertices, 3 sides, and 3 angles.
Question 2.
Draw a triangle. Name its vertices, sides, and angles.
Solution:
A triangle has 3 vertices = A, B and C
A triangle has 3 sides = AB, BC, CA
A triangle has 3 angles = ∠BAC or ∠A, ∠CBA or ∠B, ∠ACB or ∠C
Question 3.
You are given some triangles. Name its vertices, sides, and angles.
Solution:
Question 4.
Count and write how many triangles are there in each figure.
Solution:
(a) 3
(b) 5
(c) 5
DAV Class 5 Maths Chapter 16 Worksheet 1 Notes
Triangle:
A triangle has three sides, three vertices, and three angles.
Line:
A line is an endless straight path that extends in both directions.
An infinite number of lines can be drawn to pass through a given point.
Acute Angle:
An angle that is less than 90° is called acute angle.
Obtuse Angle:
An angle more than 90° but less than 180° is called obtuse angle.
90° or Right Angle:
An angle that measures 90° is called a right angle.
Reflex Angle:
An angle that measures more than 180° is called a reflex angle.
A triangle has three angles ∠BAC or ∠A, ∠CBA or ∠B, ∠ACB or ∠C
A triangle has 3 comers or 3 vertices – A, B and C.
A triangle has 3 sides – AB, BC, CA.
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# Exercise Sheet 6
```CA659 Mathematical Models/Computational Science
In-Class Exercise 6
Problems On Differential Equations
1. Write down the equation for unconstrained (Malthusian) growth. Show that the
doubling time (i.e. the time for the population to double in size) is given by
π‘2 = log π 2/π, where π is the constant growth rate.
2. The spread of a rumour in a population can be modelled according to the equation:
ππ
= ππ [π β π ]
ππ‘
(1)
Where:
π(π‘), denotes those that have heard the rumour after time t;
π΄, denotes the number of individuals in the population;
π, is a constant dependent on how exciting/unexpected the rumour is.
If π = 1000 people, show that:
1
1
1
=
+
π (1000 β π ) 1000π 1000(1000 β π )
(2)
Let
1
π΄
π (1000βπ )
π΅
= π + (1000βπ )
Then
1
π΄
π (1000βπ )
π΅
= π + (1000βπ ) for constant A, B
So multiplying both sides by the denominator on the left hand side gives:
1 = π΄(1000 β π ) + π΅π
1
So, equating coefficients of π 0 gives π΄ = 1000
1
And equating coefficients of π 1 gives: 0 = βπ΄ + π΅ and hence π΅ = 1000
Therefore
1
s(1000βs)
1
1
= 1000s + 1000(1000βs) QED
Using the expression you have derived in Equation (2) above, and the initial
condition:
π (0) = 500 people
find the general solution to Equation (1) above, the number who have heard the
rumour after time t. If k=1 How long will it take for three quarters of the population
to have heard the rumour? (You may leave your answer in terms of logarithms).
π (π‘)
β«
500
π‘
1
ππ = β« πππ‘
π (1000 β π )
0
Therefore, from Equation (4.2)
π (π‘)
π (π‘)
π‘
1
ππ
ππ
[β«
+ β«
] = β« πππ‘
1000 500 π
500 (1000 β π )
0
And hence:
π (π‘)
1
π
= π[π‘]π‘0
[log π
]
1000
(1000 β π ) 500
Simplifying gives:
1
π
log π 1
= ππ‘
[log π
]β
1000
(1000 β π )
1000
With log_e (1)=0 this becomes, in terms of t:
1
π
π₯π¨π π
= ππ‘
1000
(ππππ β π)
Therefore, for π = 1 and π (π‘) = 750 people,
π‘=
1
750 log π 3
π₯π¨π π
=
1000
250 1000
3. Five mice in a stable population of 500 are intentionally infected with a contagious
disease to test a theory of epidemic spread that postulates that the rate of change in
the infected population is proportional to product of the number of mice who have
the disease with the number that are disease free. Assuming that the theory is
10β3
correct, and that π = hour find how long it will take half the population to contract
the disease.
Let N(t) denote the number of mice with the disease at time t. We are given that
N(0)=5 mice and it follows that 500 β π is the number of mice without the disease at
time t. The theory predicts that:
ππ
= ππ[500 β π]
ππ‘
Show that the general solution general solution for the number of mice infected with
the disease after time t is given by:
π
1 500ππ‘
=
π
500 β π 99
and hence or otherwise, determine π‘250 .
```
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# How do you solve (2a+4)/3=6/9?
Jul 14, 2016
$a = - 1$
#### Explanation:
Solve by cross multiplication ($\frac{a}{b} = \frac{c}{d}$ = $a d = b c$).
$\frac{2 a + 4}{3} = \frac{6}{9}$
$\left(2 a + 4\right) \left(9\right) = \left(3\right) \left(6\right)$
$18 a + 36 = 18$
$18 a = - 18$
$a = - 1$
Plug in $- 1$ for $a$ to check.
$\frac{2 \left(- 1\right) + 4}{3} = \frac{6}{9}$
$\frac{2}{3} = \frac{6}{9}$
$\frac{2}{3} = \frac{6 \div i \mathrm{de} 3}{9 \div i \mathrm{de} 3}$
$\frac{2}{3} = \frac{2}{3}$
|
14
Dec 21
## Sum of random variables and convolution
### Link between double and iterated integrals
Why do we need this link? For simplicity consider the rectangle $A=\left\{ a\leq x\leq b,c\leq y\leq d\right\} .$ The integrals
$I_{1}=\underset{A}{\int \int }f(x,y)dydx$
and
$I_{2}=\int_{a}^{b}\left( \int_{c}^{d}f(x,y)dy\right) dx$
both are taken over the rectangle $A$ but they are not the same. $I_{1}$ is a double (two-dimensional) integral, meaning that its definition uses elementary areas, while $I_{2}$ is an iterated integral, where each of the one-dimensional integrals uses elementary segments. To make sense of this, you need to consult an advanced text in calculus. The difference notwithstanding, in good cases their values are the same. Putting aside the question of what is a "good case", we concentrate on geometry: how a double integral can be expressed as an iterated integral.
It is enough to understand the idea in case of an oval $A$ on the plane. Let $y=l\left( x\right)$ be the function that describes the lower boundary of the oval and let $y=u\left( x\right)$ be the function that describes the upper part. Further, let the vertical lines $x=m$ and $x=M$ be the minimum and maximum values of $x$ in the oval (see Chart 1).
Chart 1. The boundary of the oval above the green line is described by u(x) and below - by l(x)
We can paint the oval with strokes along red lines from $y=l\left( x\right)$ to $y=u\left(x\right) .$ If we do this for all $x\in \left[ m,M\right] ,$ we'll have painted the whole oval. This corresponds to the representation of $A$ as the union of segments $\left\{ y:l\left( x\right) \leq y\leq u\left( x\right) \right\}$ with $x\in \left[ m,M\right] :$
$A=\bigcup\limits_{m\leq x\leq M}\left\{ y:l\left( x\right) \leq y\leq u\left( x\right) \right\}$
and to the equality of integrals
(double integral)$\underset{A}{\int \int }f(s,t)dsdt=\int_{m}^{M}\left( \int_{l\left( x\right) }^{u(x)}f(x,y)dy\right) dx$ (iterated integral)
### Density of a sum of two variables
Assumption 1 Suppose the random vector $\left( X,Y\right)$ has a density $f_{X,Y}$ and define $Z=X+Y$ (unlike the convolution theorem below, here $X,Y$ don't have to be independent).
From the definitions of the distribution function $F_{Z}\left( z\right)=P\left( Z\leq z\right)$ and probability
$P\left( A\right) =\underset{A}{\int \int }f_{X,Y}(x,y)dxdy$
we have
$F_{Z}\left( z\right) =P\left( Z\leq z\right) =P\left( X+Y\leq z\right) = \underset{x+y\leq z}{\int \int }f_{X,Y}(x,y)dxdy.$
The integral on the right is a double integral. The painting analogy (see Chart 2)
Chart 2. Integration for sum of two variables
suggests that
$\left\{ (x,y)\in R^{2}:x+y\leq z\right\} =\bigcup\limits_{-\infty
Hence,
$\int_{-\infty }^{z}f_{Z}\left( z\right) dz=F_{Z}\left( z\right) =\int_{R}\left( \int_{-\infty }^{z-x}f_{X,Y}(x,y)dy\right) dx.$
Differentiating both sides with respect to $z$ we get
$f_{Z}\left( z\right) =\int_{R}f_{X,Y}(x,z-x)dx.$
If we start with the inner integral that is with respect to $x$ and the outer integral $-$ with respect to $y,$ then similarly
$f_{Z}\left( z\right) =\int_{R}f_{X,Y}(z-y,y)dy.$
Exercise. Suppose the random vector $\left( X,Y\right)$ has a density $f_{X,Y}$ and define $Z=X-Y.$ Find $f_{Z}.$ Hint: review my post on Leibniz integral rule.
### Convolution theorem
In addition to Assumption 1, let $X,Y$ be independent. Then $f_{X,Y}(x,y)=f_{X}(x)f_{Y}\left( y\right)$ and the above formula gives
$f_{Z}\left( z\right) =\int_{R}f_{X}(x)f_{Y}\left( z-x\right) dx.$
This is denoted as $\left( f_{X}\ast f_{Y}\right) \left( z\right)$ and called a convolution.
The following may help to understand this formula. The function $g(x)=f_{Y}\left( -x\right)$ is a density (it is non-negative and integrates to 1). Its graph is a mirror image of that of $f_{Y}$ with respect to the vertical axis. The function $h_{z}(x)=f_{Y}\left( z-x\right)$ is a shift of $g$ by $z$ along the horizontal axis. For fixed $z,$ it is also a density. Thus in the definition of convolution we integrate the product of two densities $f_{X}(x)f_{Y}\left( z-x\right) .$ Further, to understand the asymptotic behavior of $\left( f_{X}\ast f_{Y}\right) \left( z\right)$ when $\left\vert z\right\vert \rightarrow \infty$ imagine two bell-shaped densities $f_{X}(x)$ and $f_{Y}\left( z-x\right) .$ When $z$ goes to, say, infinity, the humps of those densities are spread apart more and more. The hump of one of them gets multiplied by small values of the other. That's why $\left(f_{X}\ast f_{Y}\right) \left( z\right)$ goes to zero, in a certain sense.
The convolution of two densities is always a density because it is non-negative and integrates to one:
$\int_{R}f_{Z}\left( z\right) dz=\int_{R}\left( \int_{R}f_{X}(x)f_{Y}\left( z-x\right) dx\right) dz=\int_{R}f_{X}(x)\left( \int_{R}f_{Y}\left( z-x\right) dz\right) dx$
Replacing $z-x=y$ in the inner integral we see that this is
$\int_{R}f_{X}(x)dx\int_{R}f_{Y}\left( y\right) dy=1.$
|
Remember to find the conjugate all you have to do is change the sign between the two terms. Situation 3 – Working with a Reciprocal. Solved exercises of Rationalisation. The solution is similar to the above solution. Give the slope-intercept form of the equation of the line perpendicular to 5x-3y-8 and passing through (4,7) 6. Fixing it (by making the denominator rational) is called "Rationalizing the Denominator"Note: there is nothing wrong with an irrational denominator, it still works. An icon used to represent a menu that can be toggled by interacting with this icon. So, in order to rationalize the denominator, we have to get rid of all radicals that are in denominator. Rationalizing the denominator is the process of moving any root or irrational number (cube roots or square roots) out of the bottom of the fraction (denominator) and to top of the fraction (numerator).. To rationalize a denominator start by multiplying the numerator and denominator by the radical in the denominator. When we multiply the with √2, we get (1x√2), which further equals √2. yields the rational number with smallest denominator that lies within dx of x. arrow_forward. I'm not sure but is the primary reason we change $$\displaystyle \sqrt80$$ into $$\displaystyle 4\sqrt5$$ is to make it easier to calculate if you didn't have a calculator, so a relic of the days before calculators were commonplace? How to Rationalize a Radical Out of a Denominator - dummies Hot www.dummies.com. And since we had historically rationalized the denominators due to a lack of calculators this form became the standard one. The middle two terms always cancel each other, and the radicals disappear. The reciprocal is created by inverting the numerator and denominator of the starting expression. Rationalize the Denominator Calculator: Master your skills of solving any expression by rationalizing its denominator by using the handy calculator tool presented here. Step 2: Distribute (or FOIL) both the numerator and the denominator. BYJU’S online rationalize the denominator calculator tool makes the calculations faster and easier where it displays the result in a fraction of seconds. How to solve: Rationalize the numerator and simplify. This is OK. Rationalize the numerator of expression x + y x − y. x + y x − y = x + y x − y × x − y x − y = (x) 2 + (y) 2 (x − y) 2 Want to see the full answer? 105-181 19179 Blanco Rd #181 San Antonio, TX 78258 USA Calculation. 4. Rationalize the numerator and simplify: 2. Rationalize Denominator Widget. Step 1 : Multiply both numerator and denominator by a radical that will get rid of the radical in the denominator. Is there a general formula for rationalizing multiple terms with rational exponents? Knowledge-based, broadly deployed natural language. 349x7 16 O 7x2 31 2 V2x2 o 7x2 O 7x3 3 2-V14x o 7x3 3 2V14x2 . But if your question is how rationalize the numerator of $\frac{\sqrt{x+h}-\sqrt{x}}{h}$ then your answer $\frac{1}{\sqrt{x+h}+\sqrt{x}}$ is correct. This website uses cookies to ensure you get the best experience. The following steps are involved in rationalizing the denominator of rational expression. Thanks for the feedback. A historical reason: Before we had calculators that could compute radicals , we had to to calculate the value of radicals by hand, and it's much easier to do that when the radical is in the numerator. By using this website, you agree to our Cookie Policy. The bottom does NOT have a radical, which was our goal. But it is not "simplest form" and so can cost you marks.. And removing them may help you solve an equation, so you should learn how. In square root fractions, it is okay for the numerator to contain the square root, but it is subtle when the denominator contains the radical sign. To read our review of the Math way--which is what fuels this page's calculator, please go here. Therefore to rationalize this numerical expression we need to follow these steps:-Multiply both the numerator and denominator with √2. How to Use Rationalize the Denominator Calculator? Learn about how to rationalize the denominator of fraction manually. To rationalize frac sqrt 2 sqrt 3 1 sqrt 2 3 type r2 r3 for numerator and 1 r 2 3 for denominator. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. What we have here is a square root of an entire fraction. arrow_back. This part of the fraction can not have any irrational numbers. If so, would the same reason apply for why we rationalize the denominator? Preface. \sqrt {x^2 + 4} - x. Chapter 1, Problem 4BM. To add the first set of fractions together all we need to do is make a common denominator of 21 and then add like terms from the numerators. Rationalize the numerator. Combine terms. To rationalize a denominator, start by multiplying the numerator and denominator by the radical in the denominator. Multiply Both Top and Bottom by the Conjugate. The denominator is the bottom part of a fraction. Simply type into the app below and edit the expression. This process is called rationalizing the denominator. Math notebooks have been around for hundreds of years. This calculator eliminates radicals from a denominator. Central infrastructure for Wolfram's cloud products & services. When working with the reciprocal of an expression containing a radical, it may be necessary to rationalize the denominator. For better understanding, we have also given an example question. The process of directly calculating this solution is initiated by rearranging the resistance (R) and load (S) terms from Eq. The answer has a radical in the numerator. To rationalize the denominator, you must multiply both the numerator and the denominator by the conjugate of the denominator. This calculator eliminates radicals from a denominator. Assume all variables represent positive numbers. Then, simplify the fraction if necessary. Simplify and give your answer in simplified radical form: r3-8 3x +2x -8 3.Simplity. Lets take the example of a number such as 1/√2, which is an example of an irrational number. Rationalize the numerator: If a fraction has a numerator of the form A ± B C, rationalize numerator by multiplying numerator and denominator by the conjugate radical A ∓ B C. Calculation: The given expression is x 2 + 1 − x. Simplify: x2-34-x2-8x+16 X3-4--X4 +4 5. Get more help from Chegg. Example: to rationalize $\frac{\sqrt{2}-\sqrt{3}}{1-\sqrt{2/3}}$ type r2-r3 for numerator and 1-r(2/3) for denominator. Check out a sample textbook solution. share | cite | improve this answer | follow | edited Feb 5 '17 at 4:55. answered Feb 5 '17 at 4:38. Example 6: Write the reciprocal of 4 – √3 How do I rationalize the numerator? Rationalize the numerator. The first step is to apply the Quotient Rule of Square Roots. Juniven Juniven. Example 3: Rationalize \sqrt {{{27} \over {12}}}. Ste. If you're working with a fraction that has a binomial denominator, or two terms in the denominator, multiply the numerator and denominator by the conjugate of the denominator. You can visit this calculator on its own page here. This allows us to generate a fraction with a distinct numerator and denominator with radical symbols. To be in "simplest form" the denominator should not be irrational!. Rationalisation Calculator online with solution and steps. When calculating the square root, if the denominator contains the radical symbol, it should be removed. Is there something I can read or study to learn more about this? How To Rationalize Radical Numerator › how to move radicals from denominator › how to rationalize a radical › how to rationalize radical denominator › how to rationalize the numerator. By signing up, you'll get thousands of step-by-step solutions to your homework questions. To: mathgroup at smc.vnet.net; Subject: [mg18697] Re: [mg18633] Rationalizing the denominator (better solution!) Free rationalize numerator calculator - rationalize numerator of radical and complex fractions step-by-step This website uses cookies to ensure you get the best experience. See solution. The Math Way app will solve it form there. Free rationalize numerator calculator - rationalize numerator of radical and complex fractions step-by-step This website uses cookies to ensure you get the best experience. To rationalize a denominator, you need to find a quantity that, when multiplied by the denominator, will create a rational number (no radical terms) in the denominator. Make your calculations easy as the tool provides you the step by step explanation. Free rationalize numerator calculator - rationalize numerator of radical and complex fractions step-by-step This website uses cookies to ensure you get the best experience. Learn about rationalization with examples, rationalize calculator, and how to rationalize numerator in the concept of rationalization. Nowadays: There are two reasons why we still rationalize the denominator. = \frac{ 6 -3\sqrt{5} } {4 \red{+} 2\sqrt{5} \red{-} 2\sqrt{5}+ 5} = Simplifying rational expressions This calculator factor both the numerator and denominator completely then reduce the expression by canceling common factors. Detailed step by step solutions to your Rationalisation problems online with our math solver and calculator. Free rationalize numerator calculator - rationalize numerator of radical and complex fractions step-by-step This website uses cookies to ensure you get the best experience. The procedure to rationalize the denominator calculator is as follows: Step 1: Enter the numerator and the denominator value in the input field Want to see this answer and more? Check out the interactive simulations and rationalize using a calculator to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page. When the denominator contains a single term, as in $\frac{1}{\sqrt{5}}$, multiplying the fraction by $\frac{\sqrt{5}}{\sqrt{5}}$ will remove the radical from the denominator. Chapter 1, Problem 6BM. For this problem, you get x 2 – 2. (a) Rationalize the numerator Rationalize the numerator of Squareroot 2 - 6/5 + Squareroot 2. Around for hundreds of years ; Subject: [ mg18697 ] Re: [ mg18633 ] rationalizing denominator. By inverting the numerator and the denominator should not be irrational! rationalize frac sqrt 2 sqrt 3 sqrt! The radical symbol, it should be removed can read or study learn! We get ( 1x√2 ), which further equals √2 of an expression containing radical... Slope-Intercept form of the denominator calculator: Master your skills of solving any expression by rationalizing denominator! Represent a menu that can be toggled by interacting with this icon, please go here form!, and the denominator math solver and calculator 1 sqrt 2 sqrt 3 1 sqrt 3... To a lack of calculators this form became the standard one have a radical, which further √2! Better solution! mathgroup at smc.vnet.net ; Subject: [ mg18633 ] rationalizing the denominator all radicals that are denominator. Have here is a square root, if the denominator around for hundreds of.. Have also given an example of an entire fraction when calculating the root. Homework questions of rationalization 5 '17 at 4:38 better solution! on its own page here uses cookies to you. Of fraction manually been around for hundreds of years Rd # 181 San,. Page 's calculator, please go here icon used to represent a that... Change the sign between the two terms always cancel each other, and how solve... Complex fractions step-by-step this website uses cookies to ensure you get the best experience Wolfram 's cloud &! Terms from Eq an icon used to represent a menu that can toggled... Solving any expression by rationalizing its denominator by using the handy calculator tool presented here is example! Or study to learn more about this cloud products & services for denominator } } } had. This page 's calculator, please go here is change the sign between the terms! The standard one the example of an expression containing a radical, was. Wolfram 's cloud products & services this problem, you agree to our Policy. By signing up, you must multiply both the numerator of radical complex! Multiply the with √2, we have here is a square root of an number. 1/√2, which was our goal for denominator | cite | improve this answer | follow edited... Historically rationalized the denominators due to a lack of calculators this form the! From Eq 7x2 31 2 V2x2 O 7x2 31 2 V2x2 O 7x2 2... Solutions to your Rationalisation problems online with our math solver and calculator more about this calculator: Master skills... Line perpendicular to 5x-3y-8 and passing through ( 4,7 ) 6 of rationalization make your easy! Will solve it form rationalize the numerator calculator Rule of square Roots of directly calculating this solution is initiated by rearranging resistance! Math solver and calculator all radicals that are in denominator expression containing a radical that will rid! All you have to do is change the sign between the two terms 2! The with √2, we have also given an example question the rational with. To generate a fraction multiply the with √2 the reciprocal of 4 √3! By rearranging the resistance ( r ) and load ( S ) terms from Eq what have. By inverting the numerator and denominator with radical symbols give the slope-intercept form of the line to. Denominator, we have to do is change the sign between the two terms cancel... 3 2-V14x O 7x3 3 2V14x2 rationalize a denominator, start by multiplying the numerator and denominator a! To be in simplest form '' the denominator is the bottom not! Conjugate all you have to get rid of the equation of the math Way -- which is fuels... About rationalization with examples, rationalize calculator, please go here is there general... Take the example of a fraction with a distinct numerator and denominator by using handy! – √3 how to rationalize the numerator and simplify may be necessary to rationalize numerator! 'Ll get thousands of step-by-step solutions to your homework questions into the app below and the... Take the example of a number such as 1/√2, which is fuels..., it should be removed cancel each other, and how to rationalize the denominator by the symbol. Share | cite | improve this answer | follow | edited Feb '17! 7X2 O 7x3 3 2V14x2 improve this answer | follow | edited Feb 5 '17 at 4:38 any expression rationalizing... Irrational! problem, you 'll get thousands of step-by-step solutions to your Rationalisation problems online with our solver! To: mathgroup at smc.vnet.net ; Subject: [ mg18697 ] Re: [ ]... 3 2V14x2 tool presented here the same reason apply for why we still rationalize denominator... Get x 2 – 2 reasons why we rationalize the numerator of radical and complex fractions this! To 5x-3y-8 and passing through ( 4,7 ) 6 radical in the denominator edited Feb 5 '17 4:38... Quotient Rule of square Roots simplest form '' the denominator in denominator a denominator start by the! 'S cloud products & services, which is what fuels this page 's calculator, please go.! Learn about rationalization with examples, rationalize calculator, rationalize the numerator calculator the denominator of manually... Your answer in simplified radical form: r3-8 3x +2x -8 3.Simplity 2 - 6/5 + Squareroot 2 - +! Which further equals √2 sign between the two terms always cancel each other, and how rationalize. With examples, rationalize calculator, please go here rationalize the numerator calculator is initiated by rearranging resistance. To a lack of calculators this form became the standard one to frac... Blanco Rd # 181 San Antonio, TX 78258 USA Preface radical will! Complex fractions step-by-step this website uses cookies to ensure you get x 2 2... ; Subject: [ mg18697 ] Re: [ mg18633 ] rationalizing the denominator, you get. We multiply the with √2, we get ( 1x√2 ), further. Solution is initiated by rearranging the resistance ( r ) and load ( S ) terms from.! Solver and calculator # 181 San Antonio, TX 78258 USA Preface you can visit this on. Give the slope-intercept form of the fraction can not have any irrational numbers (! \Sqrt { { 27 } \over { 12 } } } } toggled by with... Step-By-Step solutions to your Rationalisation problems online with our math solver and calculator rationalize \sqrt { {! Denominator - dummies Hot www.dummies.com the resistance ( r ) and load ( S ) terms from Eq 5x-3y-8 passing. Historically rationalized the denominators due to a lack of calculators this form became the standard.... Step solutions to your Rationalisation problems online with our math solver and calculator with exponents. Complex fractions step-by-step this website, you get the best experience generate a fraction ( 1x√2 ) which... Tool provides you the step by step explanation using the handy calculator tool presented here calculator, please go.. Step-By-Step this website uses cookies to ensure you get the best experience for Wolfram 's cloud products &.... Rule of square Roots are involved in rationalizing the denominator radical Out of denominator... At 4:55. answered Feb 5 '17 at 4:38 and denominator with √2 we. Due to a lack of calculators this form became the standard one between the two terms cancel. Rational expression 7x2 O 7x3 3 2-V14x O 7x3 3 2V14x2 the Rule... 2 V2x2 O 7x2 O 7x3 3 2V14x2 Write the reciprocal of an expression containing a Out... Example 6: Write the reciprocal of 4 – √3 how to rationalize this numerical expression we to... Entire fraction math notebooks have been around for hundreds of years # 181 San Antonio, 78258! Steps are involved in rationalizing the denominator by a radical Out of a number such as 1/√2, which our... Can not have any irrational numbers a radical, which is an of... Handy calculator tool presented here to read our review of the fraction can not have a radical, which what... Denominator calculator: Master your skills of solving any expression by rationalizing its denominator the! Have a radical, which was our goal ) rationalize the denominator denominators. Answered Feb 5 '17 at 4:55. answered Feb 5 '17 at 4:38 the concept of rationalization take the example a. The radical in the concept of rationalization have to do is change the sign between two! -8 3.Simplity denominator - dummies Hot www.dummies.com this website uses cookies to ensure you get x 2 – 2 be! Step-By-Step solutions to your Rationalisation problems online with our math solver and calculator by rationalizing denominator. By rationalizing its denominator by a radical, it should be removed you must multiply numerator!: multiply both numerator and the radicals disappear rationalize \sqrt { { 27 } \over 12... Denominator by the radical in the denominator math Way app will solve it form there,! With this icon the same reason apply for why we rationalize the denominator the. A menu that can be toggled by interacting with this icon by conjugate. Make your calculations easy as the tool provides you the step by step explanation 5x-3y-8 and passing through 4,7! Rid of the radical in the denominator of fraction manually smc.vnet.net ;:. The following steps are involved in rationalizing the denominator rationalize numerator of radical and complex fractions step-by-step website! 1 sqrt 2 3 type r2 r3 for numerator and the denominator, start by multiplying numerator.
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# Measuring Area of Irregular Shapes
This instructable combines a few mathematical tricks to enable you to calculate the area of irregular shapes.
I come from a farming background. One of the things we often had to do, was measure the area of a section of land in order to calculate how much fertiliser was required.
This can be done from maps, (E.g. using Google Maps measure tools) but this doesn't take into account variations in height. Our farm was in a moderately hilly area, and often the real area of a paddock was 20% more than what the maps said.
This instructable could also be used to measure area of irregular plane shapes, as well as surface area of 3D shapes.
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## Step 1: Heron's Formula - Calculating the Area of a Single Triangle.
Heron's formula is a way to calculate the area of ANY triangle, knowing the lengths of it's three sides, a, b and c. More information can be found at http://en.wikipedia.org/wiki/Heron%27s_formula
Firstly, calculate the semiperimeter. This is just half of the triangle's perimeter.
s = (a + b + c) / 2
In our example,
s = (6.56 + 6.01 + 5.76)/2
s = 9.165
-----------------------------------------------------------------
Then, calculate the following differences:
s - a
s - b
s - c
E.g.
s - a = 9.165 - 6.56 = 2.605
s - b = 9.165 - 6.01 = 3.155
s - c = 9.165 - 5.76 = 3.405
------------------------------------------------------------------
A = sqrt( s * (s-a) * (s-b) * (s-c) )
(Multiply the three differences together, along with the semiperimeter, and then take the square root.)
A = sqrt( 9.165 * 2.605 * 3.155 * 3.405)
A = sqrt( 256.482)
A = 16.015
This gives the area of a single triangle.
For a point-and-click solution, Meander ( http://peacockmedia.co.uk/meander ) allows you to trace the perimeter of any shape you can see on your screen and read off the perimeter and area
Hi shieladixon. Does this measure projected area (I.e. treat the area as a flat plane), or does it take into account changes in elevation?
I come from a farming background in a hilly area. Topo maps and surveyed areas give you the first, but on the ground, a bit of undulation can take a 2ha paddock up to a 2.5ha paddock. Important if you're calculating seed/fertilizer quantities, etc.
2 months ago
Hi, thanks for your interest. At present it allows you to add in a hilliness factor for the line (eg walking route) but doesn't take this into account when calculating the area. Now that you've given me a practical application I'll update the app to do this.
Xellers4 years ago
This is OK for very simple approximations, but Green's Theorem and Surface Internals would be far more effective!
ancienthart (author) Xellers4 years ago
Yeah yeah. :D
The thought of keeping track of the line integral x*dy - y*dx as you walk around a paddock sounds reeeeellllllll fun! :P
mikchil4 years ago
'nother way is to divide the area into horizontal strips of equal width & measure the lengths of the strip centerlines from edge to edge of shape in question. Add the lengths & multiply by the strip width and you have a pretty good approximation of the area. The narrower the strips, the greater tha accuracy. It's sorta like integration iin calculus if memory serves. This works well for any kind of shape including those with curved edges. If there are holes in the middle, leave them out of the measurements.
ancienthart (author) mikchil4 years ago
This would definitely be a method you'd keep for moderate size areas. :D
I'd imagine this would take 4 hours on a large paddock, which tend to be straight edged any case. ;)
In the situations where this is necessary (E.g. measuring a curved lawn), the various forms of Simpson's Method may be a more accurate form of this approach. It gives approximately twice as much accuracy, allowing you to make the strips twice as big.
For gardens/areas that are almost circular, using Simpson's Method from the centre to measure out arcs (Keep the width of the ends constant) and divide the total by two is also a good approach.
4 years ago
I'm afraid my math education is far less comprehensive (and perhaps more ancient) than yours.
I looked up Simpson's rule and it made my head hurt. Didn't do very well in the calculus classes I attempted because of those lousey headaches.
My thoughts run more along the lines of the rectangular method using midpoint approximation. I often have to estimate irregular shapes on plans measuring from a few hundred square feet to hundreds of acres. After using this method for a number of years, I find that if I choose the width of the rectangles correctly, my results are close enough the output of high end digitizers or autocad and don't take much time to do using a simple measuring wheel and a straight edge. If the margins are made up of straight lines, subdividing into simple straight sided shapes can be a quicker approach if the original shape is regular enough but that's a judgement call.
I'm sure that the math approach will provide better accuracy if you have the right information but I don't find the error in the approach that I use to be significant in my world.
ancienthart (author) mikchil4 years ago
The simplest version is the 2/4 rule. Split your area up into an even number of sectors of equal widths. This will give you an odd number of measurements to make.
E.g. assume we split the area into six sectors, and seven measurements, which are [1,2,1,3,1,2,2].
Take the first and last value and add them together.
1 + 2 = 3
Now take each even value, add them together and multiply by four.
2 + 3 + 2 = 7
7 * 4 = 28
Take the remaining odd values, add them together and multiply by two.
1 + 1 = 2
2 * 2 = 4
Add all these results together and divide by 3.
3 + 28 + 4 = 35
35 / 3 = 11 2/3
Simpson's rule is based on taking polynomial approximations of each section.
Se1f_Destruct4 years ago
Perhaps someone could explain?
I thought that you find the area through:
area = (width * height) / 2
Is that any different?
Great job, by the way!
ancienthart (author) Se1f_Destruct4 years ago
This saves you ensuring that the height is at a perfect right angle. All you need to do is go for a walk around the paddock with a trundle wheel.
sonogo4 years ago
Thank you for the information and this method.
I would like to comment on the xls file you made. it is okay but I think you should have put some protection on the cells other than the value of a,b,and c
if some one write on the cell of S or a-S for example it would damage the file. some people doesn't know how to drag the formula to the other cells.
otherwise it is a great job you've done in there..
thanks.
ancienthart (author) sonogo4 years ago
Hi sonogo. I've uploaded some password-less, protected versions of the spreadsheet. The issue being that the version I've created has formulas that only go up to 30 triangles. I've left the unprotected versions in as well, if people want to adapt them.
4 years ago
Thank You very much for the very handy and useful method and file you gave in this instructable.
keep the good job.
rimar20004 years ago
Thank you for sharing this. Just I know this so useful formula.
gserrano7014 years ago
I enjoyed reading it. I'm sure I'll be using it some day. Thanks, well done.
kelseymh4 years ago
This is great! Well written, citations where appropriate, and clear diagrams. And it's got math :-) Too bad you didn't go in for non-Euclidean corrections, but I guess you're not trying to measure the area of a hyperboloid of revolution ;-> Featured and rated!
ancienthart (author) kelseymh4 years ago
I used geogebra ( http://www.geogebra.org/cms/ ) for the diagrams, and I'm not entirely certain how to do non-Euclidean corrections. Thank you for the featuring.
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# Square Root Of 3
The root of three is expressed as √3 within the radical type and as (3)½ or (3)0.5 within the exponent type. If the square root of 3 rounded up to 7 decimal places is 1.7320508.
• Square Root of 3: 1.7320508075688772
• Square Root of 3 in exponential form: (3)½ or (3)0.5
• Square Root of 3 in radical form: √3
## Square Root Of 3
If a number that when multiplied by itself, results in the original number then it’s said to be the square root of the given number.
For instance, the square root of 16 is 4, as 4 times 4 results in16.
There are square roots of some numbers that do not result in whole numbers, such as 3.
We can categorise the root of three in numerous ways that
• Decimal form= 1.732.
• Exponent form= 31/2
Can the square root of 3 be considered a rational or an irrational number?
• The decimal part of square root 3 is non-terminating.
• √3 = 1.732050807……. Is a never-ending decimal.
• Hence, the square root of 3 is an irrational number.
Finding the Square Root of 3?
The square root of 3 is never-ending and in such case, the long-division method is used to calculate its value.
• Step1: To begin, we write 3 as 3.000000, then from left to right, we group the 0s after the decimal point in pairs of two.
• Step 2: Conjure up a number that is less than or equal to 3 when multiplied by itself. That number would be 1 in this scenario.
• Step 3: We get a remainder of 2 by dividing 3 by 1 with a quotient of 1.
• Step 4: To make the dividend 200, drag a pair of 0s down and place them next to 2.
• Step 5: Divisor 1 is multiplied by itself, and the result is written below. We now have 2X as the new divisor, and we must find a number for X such that the product of 2X X is less than or equal to 200. The needed value is 27 in this case.
• Step 6: After a decimal place, the number 7 is added to the quotient. The subsequent division’s new divisor will be 2X + X, which in this case is 34.
We may calculate the rest of the decimals by repeating step 4 in the same manner.
NOTE:
• The real roots of √3 are ± 1.732.
• The Square root of other numbers is always irrational and a perfect square will always be rational. For example, √25 = 5, whereas √17 = 4.1231…
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# OPERATIONS WITH DECIMALS WORKSHEET
## About "Operations with decimals worksheet"
Operations with decimals worksheet :
Worksheet on operations with decimals is much useful to the students who would like to practice problems on decimals.
## Operations with decimals worksheet
2. Subtract 0.14 from 0.25
3. Multiply 1.5 and 4.3
4. Divide 4.2 by 3
5. Find 17.6 ÷ 0.4
6. Each part of a multipart question on a test is worth the same number of points. The whole question is worth 37.5 points. Daniel got 1/2 of the parts of a question correct. How many points did Daniel receive ?
7. The bill for a pizza was \$14.50. Charles paid for 3/5 of the bill. How much did he pay ?
## Operations with decimals worksheet - Solution
Problem 1 :
Solution :
Observe the figure given below.
0.24 and 0.15 are shaded in two different colors. Now the sum of 0.24 and 0.15 is 0.39
That is,
3 tenths and 9 hundredths
Problem 2 :
Subtract 0.14 from 0.25
Solution :
Method 1 :
Therefore,
0.25 - 0.14 = 0.11
Method 2 :
Not asaaaaaaaaaaaaaaaaa0 . 2 5 aaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaa- 0 . 1 4 aaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaa-----------aaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaa0 . 1 1 aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaa-----------
Therefore,
0.25 - 0.14 = 0.11
Note :
In both the methods, arrange the decimal numbers according to the place values and subtract them.
Problem 3 :
Multiply 1.5 and 4.3
Solution :
To multiply the decimals, first convert both of them into whole numbers.
Then, we have 15 and 43
Multiplying 15 and 43, we get 645.
Both, in 1.5 and 4.3, there is 1 digit to the right of the decimal point. So, count 2 digits from the right and put a decimal point.
So, we get 1.5 x 4.3 = 6.45
Problem 4 :
Divide 4.2 by 3
Solution :
Write the decimal number 4.2 as a fraction with multiples of 10.
Since there is only one digit after the decimal in 4.2, we can write 4.2 as 42/10.
(If there are two digits after the decimal, we have to take 100 as denominator)
Now, we have to divide 42/10 by 3 as given below.
4.2 ÷ 3 = 42/10 ÷ 3
4.2 ÷ 3 = 42/10 x 1/3
4.2 ÷ 3 = 14/10
4.2 ÷ 3 = 1.4
Problem 5 :
Find 17.6 ÷ 0.4
Solution :
First, let us write both the decimals 17.6 and 0.4 as 176/10 and 4/10.
Now, we can divide 176/10 by 4/10 as given below.
17.6 ÷ 0.4 = 176/10 ÷ 4/10
17.6 ÷ 0.4 = 176/10 x 10/4
17.6 ÷ 0.4 = 44
Problem 6 :
Each part of a multipart question on a test is worth the same number of points. The whole question is worth 37.5 points. Daniel got 1/2 of the parts of a question correct. How many points did Daniel receive ?
Solution :
To find the total points received by Daniel, we have to multiply 1/2 and 37.5
Step 1 :
Convert the fraction 1/2 as the decimal 0.5
(1/2) x 37.5 = 0.5 x 37.5
Step 2 :
Multiply.
0.5 x 37.5 = 18.75
Problem 7 :
The bill for a pizza was \$14.50. Charles paid for 3/5 of the bill. How much did he pay ?
Solution :
To find the amount paid by Charles, we have to multiply 3/5 and 14.50
Step 1 :
Convert the fraction 3/5 as the decimal 0.6
(3/5) x 14.50 = 0.6 x 14.50
Step 2 :
Multiply.
0.6 x 14.50 = 8.7
Hence, Charles paid \$8.7
After having gone through the stuff given above, we hope that the students would have understood "Operations with decimals worksheet".
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# Partial Products Calculator
Created by Anna Szczepanek, PhD
Reviewed by Wojciech Sas, PhD and Jack Bowater
Last updated: Jan 18, 2024
With the help of our partial products calculator, you can master the partial products algorithm of multiplication. If you're not yet familiar with the partial products method or need a refresher, scroll down to find a short article explaining what it means to multiply using partial products and see step-by-step how to do partial products multiplication using a table or column approach. We also include examples of partial products so that you can see this algorithm in action.
## What is the method of partial products in math?
The partial products algorithm allows pupils to quickly & easily multiply two long numbers by turning a more difficult product into a bunch of smaller problems. This strategy is based on the expanded form of a number and on the distributive property. For instance, we can transform 26 × 43 as follows:
26 × 43 = (20 + 6) × (40 + 3) = (20 × 40) + (20 × 3) + (6 × 40) + (6 × 3)
As you can see, first, we use the expanded form and then apply the distributive property.
Done with multiplication? Discover more about the other arithmetical operations with Omni's calculators:
## How do I do partial products by hand?
The partial products algorithm consists of two main steps:
1. Create partial products by multiplying ones times ones, tens times the ones, ones times tens, tens times tens, etc. For instance, 26 × 43 would give you the following partial products:
• 20 × 40 = 800;
• 20 × 3 = 60;
• 6 × 40 = 240; and
• 6 × 3 = 18.
2. Add partial products together to arrive at the final answer via the partial products method of multiplication. In our example of 26 × 43, we have
26 × 43 = 800 + 60 + 240 + 18 = 1118.
## How to write down the partial products algorithm?
There're two slightly different methods of writing down the partial products method: one that uses a box (table) and another that uses columns. We'll discuss them right now. Note, that our partial products calculator can use both methods - choose the one you prefer!
#### Box approach
When we use the box approach, we prepare a table with the number of rows and columns corresponding to the number of digits in the numbers we want to multiply. That is if we want to multiply:
• A 2-digit number by a 2-digit number, we need a 2 × 2 table;
• A 3-digit number by a 2-digit number, we need a 3 × 2 table;
• A 5-digit number by a 3-digit number, we need a 5 × 3 table; etc.
In the columns of our table, we put the ones, tens, hundreds, etc., of one number, and in the rows, we put the ones, tens, hundreds, etc., of the other number we want to multiply using partial products. For instance, to multiply 43 × 26, we prepare the following table:
$\begin{matrix} \times & 20 & 6 \\ 40 \\ 3 \end{matrix}$
Then we fill the table with the successive partial products:
$\begin{matrix} \times & 20 & 6 \\ 40 & 800 & 240 \\ 3 & 60 & 18 \end{matrix}$
The result of 43 × 26 is the sum of the four fields inside the table:
43 × 26 = 800 + 240 + 60 + 18 = 1118
#### Columns approach
When we use columns to calculate the product using partial products, we start in exactly the same way as in the long multiplication algorithm:
$\begin{array}{rr} & 43 \\ \times & 26 \\ \hline = & \end{array}$
Then we write down under the vertical bar the results of consecutive partial multiplications:
💡 It's crucial you remember to put a sufficient number of trailing zeroes, depending on whether you've performed the multiplication of ones or tens or hundreds, etc.
$\begin{array}{rr} & 43 \\ \times & 26 \\ \hline = & 18 \\ & 240 \\ & 60 \\ & 800 \\ \end{array}$
If you're already familiar with the standard multiplication algorithm, you can easily spot the difference: here, we write down the result (partial product) as it is and do not carry over anything to the next column.
Once we've got all the partial products, it remains to add them all together to get the final result:
$\begin{array}{rr} & 43 \\ \times & 26 \\ \hline = & 18 \\ & 240 \\ & 60 \\ & 800 \\ \hline = & 1118 \end{array}$
As you can see, there's nothing difficult in the partial products multiplication. However, it requires a bunch of calculations. Thankfully, Omni's partial products multiplication calculator can perform them for you!
## How to use this partial products calculator?
It's high time we discussed what is the most effective way of working with Omni's partial products calculator. To master it, follow these steps:
1. Input the numbers you want to multiply with the partial products method.
2. Choose the output method. As we've explained above, there are two popular approaches:
• Table (box); and
• Columns.
Try experimenting with both to choose the one that suits you best!
3. The result of the partial products multiplication will appear at the bottom of the partial products calculator.
## Partial products example
The best way to understand the definition of partial products is to experiment with bigger and bigger numbers. Here, as an example of the partial products algorithm, we'll discuss the partial products multiplication of 3-digit by 3-digit numbers. To see examples involving numbers with more digits, use our partial products multiplication calculator shamelessly!
Example. Let's multiply 432 × 118 using the table approach.
1. Prepare the table:
$\begin{array}{rrr} \times & 100 & 10 & 8 \\ 400 &&& \\ 30 &&& \\ 2 &&& \end{array}$
1. Next, we fill in the table with partial products:
$\begin{array}{rrrr} \times & 100 & 10 & 8 \\ 400 & 40000 & 4000 & 3200\\ 30 &3000 & 300 & 240\\ 2 &200&20&16 \end{array}$
1. Now, we add all the partial products together. There are nine of them, so we have to be careful not to forget anything. Let's go column by column:
(40000 + 3000 + 200) + (4000 + 300 + 20) + (3200 + 240 + 16) = 43200 + 4320 + 3456 = 50976
That's it! We've got the final result:
432 × 118 = 50976.
## FAQ
### What is a partial product in math?
We talk about the partial product when we multiply two numbers bit by bit. That is, instead of performing the whole multiplication all at once, we use the expanded form of the numbers and separately multiply tens by tens, tens by ones, ones by ones, etc. Each of these chunks is called a partial product. We then add them all together to get the final answer.
### How do I use an array to find partial products?
When you use the array (table, box) approach to the partial products multiplication, you prepare a table with rows corresponding to the hundreds, tens, ones of one number and columns corresponding to the hundreds, tens, ones of the other numbers. Then you fill in the table with the results of respective multiplications. These are the partial products. Then you sum these partial products to arrive at the final answer to your problem.
### What are the partial products of 123 × 45?
There are 2 × 3 = 6 partial products in 123 × 45, because you multiply a 3-digit number by a 2-digit number. The partial products are the following:
• 100 × 40 = 4000;
• 100 × 5 = 500;
• 20 × 40 = 800;
• 20 × 5 = 100;
• 3 × 40 = 120; and
• 3 × 5 = 15.
### How do I calculate 28 × 12 by partial products?
To multiply two numbers using partial products, you have to:
1. Compute one-by-one the partial products. In our case, these are:
• 20 × 10 = 200;
• 20 × 2 = 40;
• 8 × 10 = 80; and
• 8 × 2 = 16.
2. Add all the partial products together:
200 + 40 + 80 + 16 = 336,
which is the final answer to 28 × 12.
Anna Szczepanek, PhD
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Input two numbers
1st number
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# STRAIGHT LINE PROBLEMS WITH SOLUTIONS
Problem 1 :
A photocopy store charges \$1.50 per copy for the first 10 copies and \$1.00 per copy after the 10th copy. Let x be the number of copies, and let y be the total cost of photocopying (i) Draw graph of the cost as x goes from 0 to 50 copies (ii) Find the cost of making 40 copies.
Solution :
(i) :
(ii) :
x stands for number of copies and y stands for total cost.
y = 1.5x for 0 ≤ x ≤ 10
After 10th copy, cost per copy is \$1.00.
Let us construct a function for the number of copies which is greater than 10.
y = 1.5(10) + 1(x - 10)
y = 15 + x - 10
y = 5 + x for x > 10
Substitute 40 for x.
y = x + 5
= 40 + 5
= 45
The cost of making 40 copies is \$45.
Problem 2 :
Find at least two equations of the straight lines in the family of the lines y = 5x + b, for which b and the x-coordinate of the point of intersection of the lines with 3x - 4y = 6 are integers.
Solution :
Find the point of intersection of the above lines using substitution.
y = 5x + b
3x - 4y = 6
Substitute (5x + b) for y in 3x - 4y = 6.
3x - 4(5x + b) = 6
3x - 20x - 4b = 6
-17x - 4b = 6
-17x = 6 + 4b
x = -(6 + 4b)/17
Since x is an integer, (6 + 4b) has to be a multiple of 17.
6 + 4b = ±17, ±34,...............
6 + 4b = 17 ----> b = 11/4 (not an integer)
6 + 4b = -17 ----> b = -23/4 (not an integer)
6 + 4b = 34 ----> b = 7 (integer)
6 + 4b = -34 ----> b = -10 (integer)
The required equations are
y = 5x + 7
y = 5x - 10
Problem 3 :
Find all the equations of the straight lines in the family of the lines y = mx - 3, for which m and the x-coordinate of the point of intersection of the lines with x - y = 6 are integers.
Solution :
Find the point of intersection of the above lines using substitution.
y = mx − 3
x - y = 6
Substitute (mx - 3) for y in x - y = 6.
x - (mx - 3) = 6
x - mx + 3 = 6
x - mx = 3
x(1 - m) = 3
x = 3/(1 - m)
Find the possible integer values of m such that the value of x is also an integer.
m = -2 ----> x = 3/3 = 1
m = 0 ----> x = 3/1 = 3
m = 2 ----> x = 3/(-1) = -3
m = 4 ----> x = 3/(-3) = -1
The possible integer values of m are -2, 0, 2 and 4.
The required equations are
y = -2x - 3
y = -3
y = 2x - 3
y = 4x - 3
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1. ### SAT Math Videos
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2. ### Simplifying Algebraic Expressions with Fractional Coefficients
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Simplifying Algebraic Expressions with Fractional Coefficients
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# Percents
Math
Summary
## Computing Percents
Summary Computing Percents
### Finding a Percent of a Number
People deal with percents all the time in everyday life. When sales clerks compute sales tax, they compute "7% of your total purchase." Similarly, when diners are deciding what size tip to leave, they compute "15% of the cost of our food." How do they perform these calculations?
Just as with fractions, the word "of" in a percent problem indicates multiplication. The sales clerk multiplies 7% by the total purchase, and the diner multiplies 15% by the cost of her food.
To find a percent of a number, first change the percent into a decimal or a fraction. Next, multiply that decimal or fraction by the number.
Example 1: What is 15% of 220?
I. Using decimals.
15% = 0.15 and 0.15×220 = 33
II. Using fractions.
15% = 15/100 = 3/20 and 3/20×220/1 = 3/1×11/1 = 33
Example 2: What is 12.5% of 52?
I. Using decimals.
12.5% = 0.125 and 0.125×52 = 6.5
II. Using fractions.
12.5% = 0.125 = 125/1000 = 1/8 and 1/8×52/1 = 52/8 = 13/2 = 6 1/2
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# Question Video: Multiplying 2-Digit Numbers by 1-Digit Numbers with Base 10 Blocks Mathematics • 4th Grade
Find the product 43 × 2 by multiplying the tens followed by the ones and then adding.
02:20
### Video Transcript
Find the product 43 multiplied by two by multiplying the tens followed by the ones and then adding. 43 multiplied by two equals what?
In this question, we need to find the product, which is what we get when we multiply two numbers together, of 43 and two. And perhaps you know your two times table up to 10 times two, maybe even 12 times two. But what about 43 times two? This is quite a large two-digit number that we’re dealing with here. How can we multiply this two-digit number by two? Well, the question tells us. First, we’re to multiply the tens, then by the ones, and then add. What does this mean? Well, underneath the calculation, we can see a picture that can help us here. It’s the number 43 modeled out of place-value blocks. And place-value blocks are really useful because they help us see numbers in their different parts. 43 can be partitioned or split up into four tens or 40 and three ones, which of course have a value of three.
Now what this question is telling us to do is not to look at 43 all in one go, but to first think about the tens and multiply 40 by two, then think about the ones and multiply three by two. Now, we know that four times two is eight, and so four tens multiplied by two is worth eight tens or 80. So if we multiply the tens part of our number by two, we get the answer 80. Now let’s multiply the ones; three times two equals six. So we’ve got the answers 80 and six. And to find the overall answer, we just need to combine these two parts back together again. 80 plus six equals 86. We found the product of 43 and two by multiplying the tens, then the ones, and then adding the two answers together. 43 multiplied by two equals 86.
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# ISEE Lower Level Math : How to find the perimeter of a triangle
## Example Questions
← Previous 1 3 4 5 6
### Example Question #1 : How To Find The Perimeter Of A Triangle
The triangle below has a perimeter of . The lengths of two sides are shown below.
What is the length of the third side?
Explanation:
Remember, the perimeter of a triangle is the sum of its three sides. First, sum up the two side lengths that we are given in the question:
To find the third side, subtract the sides we know from the total perimeter:
### Example Question #2 : How To Find The Perimeter Of A Triangle
An equilateral triangle has a perimeter of . How long is each side?
Explanation:
An equilateral triangle has three congruent sides, i.e. the sides are all the same length.
The perimeter of an equilateral triangle is given by .
Thus, ; then divide each side by three to get .
### Example Question #1 : How To Find The Perimeter Of A Triangle
Use the triangle to answer the question.
What is the perimeter of the triangle?
Explanation:
To find the perimeter, add all the sides together.
### Example Question #4 : How To Find The Perimeter Of A Triangle
An equilateral triangle has a side length equal to .
What is the perimeter of the triangle?
Explanation:
Because it is an equilateral triangle, all sides have the same length .
The perimeter is the sum of the length of the sides.
### Example Question #5 : How To Find The Perimeter Of A Triangle
Figure not drawn to scale
The perimeter of isosceles triangle is .
If and are equal to each other and is , what is the length of ?
Explanation:
The perimeter of the triangle is equal to the sum of the lengths of all the sides.
Because the triangle is an isosceles triangle, the triangle has two equal sides, which we know are and
Since is the same as , we can say: .
We now solve for :
Subtract from both sides
Divide both sides by
### Example Question #1 : How To Find The Perimeter Of A Triangle
What is the name of the longest side of a right triangle?
Length
Base
Equilateral
Hypotenuse
Obtuse
Hypotenuse
Explanation:
The hypotenuse is the longest side of a triangle with a right angle.
All the other terms refer to something else. Obtuse refers to an angle over 90 degrees. Length is too general of a term, and refers to the measurement of any given side. Equilateral refers to a triangle in which all angles and side are of equal measurement. Base refers to the side that is perpendicular to the measure of the height, meaning that it will be one of the shorter sides.
### Example Question #1 : How To Find The Perimeter Of A Triangle
In an isosceles triangle, two side are 4 feet long. Which of the following is NOT a possible length of the third side?
8 feet
1 foot
7 feet
2.5 feet
5 feet
8 feet
Explanation:
For triangles, it is impossible for a third side to be equal to or greater than the sum of the two other sides.
Given that the sum of the two sides of this triangle is 8, it is impossible for the third side to be equal to 8 feet.
### Example Question #1271 : Isee Lower Level (Grades 5 6) Mathematics Achievement
In an equilateral triangle, the perimeter is 42. What is the length of 1 side?
Explanation:
In an equilateral triangle, all 3 sides are equal to one another. Therefore, if the perimeter is 42, the length of 1 side will be equal to 42 divided by 3. This results in 14, which is therefore the correct answer.
### Example Question #1 : How To Find The Perimeter Of A Triangle
If the side of an equilateral triangle is equal to and the perimeter is equal to 18, what is the value of ?
Explanation:
The three sides must sum to 18:
Combine like terms on the left:
Divide both sides by 9:
### Example Question #341 : Geometry
If an equilateral triangle has sidelengths of 3 inches, what is the perimeter?
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# How many terms are there in an algebraic expression?
An algebraic expression is an expression composed of various components, such as variables, constants, coefficients, and arithmetic operations. These components form various parts of the algebraic expressions. An algebraic expression is a linear equation composed of any number of variables. The highest power of the variable is known as the degree. An algebraic expression containing one variable is monomial, two variables is binomial, and so on. For instance, if we assume an expression to be,
2x+5
Here, the parts of the expression are as follows:
Coefficient of the expression is 2
Constant is 5
Variable is x
Terms are 2x and 5
Mathematical operators used are plus (+) and minus (-).
### What are Terms?
The terms of an algebraic expression are known as the components of the expression. An algebraic expression may be composed of one or more terms. A term of an expression may be a constant, a variable, a product of more than two variables (xy), or a product of a variable and a constant. Terms add up together to form an algebraic expression. For instance, in the algebraic expression 3x + y, the two terms are 3x and y. Terms can be further classified depending on the variables and the corresponding powers defining them. They may be divided into like and unlike terms:
• Like terms: The terms which are constituted by the same literal along with the same exponents. For example: 12x and -3x are like terms since both of them have the same power of the same variable.
• Unlike terms: The terms which are constituted by the same variable with different exponents or different variables with the same exponents. For example: x and x2 are unlike terms.
An algebraic expression can be composed of the following terms:
Coefficient
A coefficient is an integer that is the constant which accompanies the variable. To summarise, a coefficient in an algebraic expression is considered as the numerical factor of a term that is composed of constants and variables. Coefficients of the terms may be positive or negative in nature. They may be fractional in nature. For instance, in the term 2x, 2 is the coefficient for the variable x. The terms with no constant, that is with no numerical factor along with them have a unit coefficient. For instance, in the term z, +1 is the coefficient for the variable z. Similarly, we have, -5/2 as the coefficient of the term –5/2xy2
Constant
Constant is the term in the algebraic expression which is constituted of only integers. These terms contain variable counterparts. These values are fixed in nature since there is no variable accompanying them. Therefore, these terms have a fixed value throughout, since no change can occur in these. Terms may only be defined by constants. For instance, in the expression 7x2 + 3xy + 8, the constant term in this expression is 8.
Variable
Variables are terms composed of undefined values, which may assume different integer values on substituting them with different integers. A variable term can be composed of one or more variables, where the variables may or may not be the same. For example, we have, x3 which is a term composed of x raised to the power of 3, and xyz is composed of three different variables.
For instance, x3 can be 8 where the value of x = 2.
Some of the examples of terms are:
• 12x: Constant term = 12 ; Variable term = x
• 42a: Constant = 42 ; Variable = a
• xy: Variables = x and y
• 89: Constant = 89
• mn: Constant = 1 ; Variables = m and n
### Sample Questions
Question 1. Identify the terms, like terms, coefficients, and constants in the expression.
12m − 24n + 10 + m − 17
Solution:
Here, we have,
12m − 24n + 10 + m − 17
First, rewrite the subtractions as additions.
12m − 24n + 10 + m − 17 = 12m + (-24n) + 10 + m + (-17)
Therefore,
The terms: 12m, (−24n), m, 10, and (−17).
The like terms are the ones that contain the same variable
12m and m are a pair of like terms.
The constant terms 10 and −17 are like terms.
Coefficients: 12 is coefficient of m,
-24 is the coefficient of n
1 is the coefficient of m.
Therefore, the coefficients are 12, (−24), and 1.
Constants: 10, -17
Question 2. Differentiate between constants and variables.
Solution:
Question 3. Compute the value of x in the equation 2x + 20 = 40
Solution:
Given,
2x + 20 = 40
⇒ 2x + 20 = 40
⇒ 2x = 40 – 20
⇒ 2x = 20
⇒ x =
⇒ x = 10
Value of x in the equation 2x + 20 = 40 is 10.
Question 4. Find the Variable, coefficient, constant, and terms of the algebraic expression
90x + 22y – 31
Solution:
Here given algebraic expression
90x + 22y – 31
We have to find Variable, coefficient, constant, and terms
Variables: x and y
Terms: 90x, 22y and 31
Constant: 31
Coefficient: 90 of x and 22 of y.
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# Two Digit-Three Digit Numbers Product
5 Steps - 3 Clicks
# Two Digit-Three Digit Numbers Product
### Introduction
How to Quickly Calculate the Product of a 2-digit and 3-digit Number
In this article, we will be studying various tricks to calculate the Two Digit-Three Digit Numbers Product. We will understand this method using examples. We choose the base to be 100.
### Short Cuts
Example 1:
Find the product of 97 × 107 = ?
Step 1: Find the difference of the numbers with respect to 100.
Step 2: Add first number with the difference of the other. We get 97 + (+7) = 104.
Step 3: Multiply the result of Step 2 with 100. We get 104 × 100 = 10400.
Step 4: Multiply the differences obtained in Step 1.
Step 5: Add the results obtained in Step 3 and Step 4 to get the answer.
∴ The answer is 10400 + (-21) = 10379.
Example 2:
Find the product of 88 × 111 =?
Step 1: Find the difference of the numbers with respect to 100.
Step 2: Add first number with the difference of the other. We get 111 + (-12) = 99.
Step 3: Multiply the result of Step 2 with 100. We get 99 × 100 = 9900.
Step 4: Multiply the differences obtained in Step 1.
We get (+11) × (-12) = -132.
Step 5: Add the results obtained in Step 3 and Step 4 to get the answer.
∴ The answer is 9900 + (-132) = 9768.
### Exercises
1. Find the product of 88 × 106.
2. Find the product of 99 × 111.
3. Find the product of 109 × 91.
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# Fraction simplifier is a simplified way to unite numerator and denominator with a common factor
Fraction simplifier simplifies both proper and improper fractions into a smaller fraction or into a mixed number. When this simplifier gives an answer in both reduced fraction and a mixed number, some people call this a round down to lowest denominator
Fraction simplifier is a way of trying to put together some fractions into a whole number. Three (3) types of fraction are identified, they are:
• A proper fraction
• An improper fraction
• A mixed fraction
Proper fractions – this kind of fraction indicates a fraction whereby the numerator i.e. the number on top of the division line is less than the denominator i.e. the number below the division line. To further explain proper fractions, some fraction simplifier examples in form of images are given below:
• >5 8/ 5 6 = 324(5/8) ⋰ 424(5/6) =15/20 = 3/4
• ÷ 4
That is as far as we can go. The fraction simplifies to 23
It is also possible to convert either from an improper fraction to a mixed fraction or from a mixed fraction to an improper fraction. In any case, there certain steps or procedures we must follow
converting a mixed fraction into improper fraction – to convert a mixed fraction to an improper fraction, these steps must be followed:
• Multiply the whole number by the fraction’s bottom number, i.e. the denominator
• Add that the answer you get from step one above to the top number, i.e. the numerator
• Finally, write down the result on top of the bottom number (denominator)
Example: change 3(2/5) to an improper fraction using the steps described above:
• Multiply the whole number, i.e. 3 by the fraction’s denominator, i.e. 5:
3 × 5 = 15
• Add the fraction’s numerator, i.e. 2 to the answer in step one: 15 + 2 = 17
• Last step, put the answer in step over the denominator i.e.: 17/5
converting improper fractions to a mixed fraction – in order to convert an improper fraction into a mixed fraction, follow the steps below:
• Divide the numerator by the denominator
• Write down the whole number answer
• Write down any remainder above the denominator
Example: convert 11/4 to a mixed fraction
• Divide the top number by the bottom number: 11 ÷ 4 = 2 but with a remainder of 3
• Write down the 2 and then write down the remainder 3 over the denominator 4 i.e.: 2(4/3)
Fractions has for a very long time being part of everyday usage, it’s just that people do not really pay attention to it. A very good example of an everyday usage of fractions are given below:
• It is a lot easier to say I ate 2(1/2) loafs of bread, instead of saying I ate 5/2 loafs of bread
• It is also easier to say that John drank 3(1/2) of wine bottles, rather than saying He drank 7/2 of wine bottles
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# Five More Euler Problems
In this post I will be discussing five more Project Euler problems, with their solutions in Python, an explanation of the solutions and the background mathematics behind them. The first five I discussed here.
In this post, I’ve implemented spoiler tags that look like this: This is a spoiler!. Just mouse over or touch to reveal.
## Problem #6: Sum Square Difference
The sum of the squares of the first ten natural numbers is,
12 + 22 + … + 102 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + … + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural >numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
This is a fairly trivial problem. We just need to find the square of the sum of the numbers 1 – 100, the sum of the squares of the numbers 1 – 100, then find the difference between the two.
#### Solution
This gives the result 25164150.
#### Explanation
On line 3 we find the square of the sum of the natural numbers between 1 and 100. On line 4 we find the sum of the squares of the natural nubmers between 1 and 100. On line 5 we find the difference between the former and latter, then print the result.
This could be done as a one-liner such as follows:
## Problem #7: 10001st Prime
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
Now this is a little more like it. What we need is a way to generate the sequence of prime numbers up to the nth term, and a way to return the nth term in the sequence.
#### Solution
This gives us the result 104743.
### Explanation
This does require some explanation. In order to generate the sequence of primes, we use an implementation of an ancient algorithm known as the Sieve of Eratosthenes. This particular implementation is odds-only (which speeds up the process by about 25%1), along with with postponing the inclusion of the prime’s information in the sieve dictionary until that prime’s square is seen amongst the candidate numbers.
In terms of mathematics and complexity theory, this implementation is interesting, but all we care about is that it is a) it works and b) it is more efficient than a brute force approach. In a game like prime-number generation and indexing, optimisation is everything.
## Problem #8: Largest Product In A Series
The four adjacent digits in the 1000-digit number that have the greatest product are $9 × 9 × 8 × 9 = 5832$.
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
That’s a pretty huge number. The old wording of this problem was misleading, as it used consecutive instead of adjacent. The re-write also increased the number of adjacent digits from five to thirteen.
Essentailly, we need to find a way to iterate over the intger in chunks of 13, finding the products of said chunks and then finding the maximum.
#### Solution
This gives the result 23514624000.
#### Explanation
So what’s going on here? First of all we do some list trickery in the interests of formatting over performance (as horizontal scrolling is the devil).
Then, we have out integer ready to iterate over. We create an xrange object between 0 and the length of the large integer s, minus 13. This is because the length of the sequences we are finding the product of is 13. Otherwise, we would encounter a ListError as the list index would be out of range (looking for another 13-digit sequence that doesn’t exist).
It is then just a matter of finding the product of the 13-digit sequences p and if the product is greater than 0 (in the first loop) or greater than the previous product (in all other loops), set n as that product. Once we’ve iterated over the whole integer, we print n.
## Problem #9: Special Pythagorean Triplet
A Pythagorean triplet is a set of three natural numbers, $(a < b < c)$, for which, $$a^{2} + b^{2} = c^{2}$$ For example, $3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}$.
There exists exactly one Pythagorean triplet for which $a + b + c = 1000$.
Find the product $abc$.
A little bit of a breather after #8, as all we have to do now is some basic algebra.
This gives us the result: 31875000. The three values are: a: 200 b: 375 c: 425.
#### Explanation
We start out by defining our function, then we iterate over two ranges to get values for a and b. Why is the upper bound of the range 500? We know from the problem statement that a and b must be less than c, and if either one is 500, c must be 501 at a minimum, pushing us beyond the 1000 limit. If we wanted to really squeeze all of the performance out of this program, we’d set the upper bound of the ranges to 498 (again, as $a < b < c$).
Line 6 is our check to keep us within the 1000 limit. Line 7 just checks that the values for a, b and c we have generated satisfy Pythagoras’ theorem, then we just print the product of the three values.
## Problem #10: Summation of Primes
The sum of the primes below 10 is $2 + 3 + 5 + 7 = 17$.
Find the sum of all the primes below two million.
Fortunately, in the solution to #7, we have a very efficient way to generate prime numbers. All we need to do is use the Sieve of Eratosthenes to generate the primes up to $2\times10^{6}$. But how do we implement the bounding?
#### Solution
This gives us the result 142913828922
#### Explanation
It’s our old friend takewhile! We have the Sieve of Eratosthenes as our iterable, and lambda x: x < 2e6 as our predicate, so the sieve will generate prime numbers up until $2\times10^{6} – 1$, and we just sum the total of the primes. The sieve is an extremely useful tool for any problem that requires generating prime numbers, and it is remarkably efficient (this program only took ~0.5 seconds to run).
#### Conclusion
In this post I have discussed the Project Euler problems 6 – 10, along with their solutions written in Python. These are not definitive, nor elegant solutions. All solutions can be found as seperate .py files in the project_euler repository on my Github, and will be updatØed if/when the problems change parameters. I’m also including some timing code from the Python module time, to see how fast the solutions run in the interpreter.
#### What’s next?
An indepth discussion of a more complex Euler problem, more website tweaks and checks (implementing mathjax fully, along with some CSS tweaks for my purposes).
1. From this Python Cookbook excerpt.
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NEB Class 12 Questions: NEB 12 Model Questions 2080/2024
NEB Class 12 Routine 2080-2081: Class 12 Routine
# Elementary Group Theory - Exercise 3.3 : Class 12 Math
Complete Exercise 3.3 Elementary Group Theory. Exercise 3.2: Groups Complete Que Ans note, Class 12 Mathematics.
## Chapter 1: Elementary Group Theory.
### Exercise: 3.3
Complete Exercise of Elementary Group Theory - Exercise 3.3 : Class 12 Mathematics 2080 NEB.
Exercise 3.2 is about: Group and its types: Finite and infinite group, Trivial group, Abelian group Groups with elements other than numbers, Matrix Group are also include in this chapter.
Elementary Group Theory Exercise 3.3 Questions
#### Question no. 1
a.
Soln:
It is a false statement because the order of group G = Number elements in G = 4, which is finite.
b.
It is a true statement because the group contains two elements, so the order of given group is 2.
c.
Soln:
If is a false statement because it is not closed, for e.g.
(- 2).2 = - 4 ∉ {-2,-1,0,1,2}
d.
Soln:
It is false statement because it is not closed,
For e.g. 4 + 8 = 12 ∉ {2,4,6,8,10}
#### Question no. 2
Soln:
Here, 1 + 1 = 2 ∉ S where, 1 , 1 ԑ S.
So, S is not closed under addition.
Thus, S is not a group under addition.
#### Question no. 3
Soln:
Existence of Identity: Let a be any natural number and if it exists, let e be the identity element of a, then a + e = a. à e = a – a = 0 which is not a natural number,
So, identity element under addition doesn’t exist in the set of natural numbers,
Hence, the set of natural numbers does not form a group under the addition operation,
#### Question no. 4
Soln:
Closure property: From the multiplication table, we see that T is closed under multiplication, i.e.
Multiplication Table.
X -1 1 -1 1 -1 1 -1 1
(-1) * (-1) = 1, (-1) * 1 = -1
1 * (-1) = - 1, 1 * 1 = 1
So, a * b ԑ T for all a,b ԑ T.
Associative property: Here, the elements of T are – 1 and 1. We know that all integers under multiplication obey associative law. So, the element of T being integers satisfy associative law under multiplication.
i.e. a * b(b * c) = (a * b) * c for all a,b,c, ԑ T.
For e.g. (-1) * {1 * (-1)} = (-1) * (-1) = 1.
And {(-1) * 1} * (-1) = (-1) * (-1) = 1
So, (-1) * {1 * (-1} = {(-1) * 1} * (-1) and so on.
Existence of Identity element,
Since (-1) * 1 = (-1) = 1 * (-1)
And 1 * 1 = 1
So, 1 is the multiplicative identity in T.
Existence of inverse
Each element of T is an inverse of itself since,
1 * 1 = 1 and (-1) * (-1) = 1.
Hence, T = {-1,1} forms a group under multiplication.
#### Question no. 5
a.
Soln:
Multiplication Table.
X 1 -1 i -i 1 1 -1 i -i -1 -1 1 -i i i i -i -1 1 -i -i i 1 -1
From above table, we see that for al a,b ԑ G, a * b ԑ G. So G is closed under multiplication.
b.
Soln:
Since all the complex numbers satisfy associative law under multiplication. So all the elements of G being complex numbers also satisfy associative law.
Ie. a * (b * c) = (a * b) * c, for all a,b,c ԑ G.
c.
Soln:
For all a ԑ G, a * 1 = 1 * a = a.
So, 1 is the multiplication identity in G.
Since, 1 * 1 = 1
(-1) * (-1) = 1
i * (-i) = -i2 = 1
and (-i) * I = -i2 = 1
So, the inverse of 1, -1, i and – I and 1, - 1, - i and i , respectively. So ,every element of G possesses an inverse element in G. Hence, identity element and inverse exists.
d.
Soln:
Yes, G forms a group under multiplication as (G,x) is closed, associative, and the identity and inverse exist in G.
#### Question no. 6
Soln:
Let a,b ԑ Z.
Since, a,b ԑ Z à a + b ԑ Z.
So, closure property is satisfied.
If a,b,c ԑ Z, then
a + (b + c) = a + b + c ԑ Z.
Also, (a + b) + c = a + b + c ԑ Z.
So, a + (b + c) = (a + b) + c.
Hence, associative property is satisfied. 0 is an integer and
0 + a = a + 0 = a ԑ Z.
0 is an identity element.
Also, if a ԑ Z then – a ԑ Z.
Or, a + (- a) = (-a) + a = 0.
So, - a is the inverse element of a. Above relations are true for all elements of Z.
Hence, the set of integers Z forms a group under addition.
#### Question no. 7
Soln:
* a b c a a b c b b c a c c a b
From the table, we see that the operation defined on any two elements of G gives and element of G itself.
So, G is closed under the operation *.
a * (b * c) = a * a = a
(a * b) * c = b * c = a
So, a * (b * c) = (a *b) * c
So, * satisfies associative property.
Since, a * a = a, a * b = b * a= b
And a * c = c * a = c, so a is an identity element.
a * a = a, so a is the inverse of a,b * c = c * b = a, so b and c are the inverse elements of c and respectively,
So, (S,*) forms a group.
#### Question no. 8
Soln:
Composition table for G under the addition modulo r (+4) is presented below.
+4 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2
From the table, we see that sum of any two elements of G modulo 4 is an element of G. So, +4 satisfies closure property.
Again, 1 +4 (2 +4 3) = 1 +4 1 = 2.
And (1 +4 2) +4 3 = 3 +4 3 = 2.
This result is true for all elements of G. Hence, +4 satisfies associative property.
From the second row and second column of above table, 0 is the identity element.
Form the second row and second column of above table, 0 is the identity elemnt.
Since, 0 +4 0 = 0, 1 +4 3 = 3 +4 1 = 0
And 2 +4 2 = 0
So, the inverse elements of 0,1,2 and 3 are 0,3,2 and 1 respectively.
So, G forms a group under addition modulo 4.
#### Question no. 9
Soln:
Since a is an identity element, so,
a * a = a, a * b = b * a = a.
So, also a * a = a, a is the inverse of a.
And b * b = a so that b is the inverse of b.
Now the required composite table is given below.
* a b a a b b b a
### Elementary Group Theory - Exercise 3.3 : Class 12 Math PDF
Elementary Group Theory - Exercise 3.1 : Class 12 Math
Elementary Group Theory - Exercise 3.2 : Class 12 Math
Elementary Group Theory - Exercise 3.4 : Class 12 Math
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# Matrices4.5
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### Matrices4.5
1. 1. 4.5 Multiplication of two matrices<br />Name: Lee ShingKuan<br />MuhamadZaid<br />Chien Shin Yee<br />Lim Si Qi<br />
2. 2. Introduction<br />We can multiply two matrices if the number of columns in the first matric equal the number of rows in the second matric.<br />The product matric ’s demension are: row of first matric x columns of the second matric<br />Two matric can’t be multiply together because the number of column does not equal to the rows.In this case the multiplication of these two matric is not defined.<br />If we multiply a 2x2 matric with a 2x1 matric ,the product matric is 2x1 : (2x3) (3x1)=(2x1)<br />If we multiply a 2x2 matric with a 1x2 matric ,the matric can’t be multiply:(2x2) (1x2)=(2x1)<br />
3. 3. How to multiply two matrices?<br />·Matrixmultiplication falls into two general categories: Scalar in which a single number is multiplied with every entry of a matrix. <br />·Multiplication of an entire matrix by another entire matrix For the rest of the page, matrix multiplication will refer to this second category. <br />
4. 4. Example:<br />In the picture on the left, the matrices can be multiplied since the number of columns in the 1st one, matrix A, equals the number of rows in the 2nd, matrix B. <br />The dimensions of the product matrix <br /> - rows of 1st matrix ×columns of 2nd<br /> - 4 × 3<br />
5. 5. Matrix C and D below cannot be multiplied together because the number of columns in C does not equal the number of rows in D. In this case, the multiplication of these two matrices is not defined.<br />
6. 6. Matrix multiplication <br /><ul><li>Matrix multiplication is a binary operation that takes a pair of matrices, and produces another matrix.
7. 7. The process is the same for any size matrix. We multiply across rows of the first matrix and down columns of the second matrix, element by element. We then add the products:
8. 8. In this case, we multiply a 2 × 2 matrix by a 2 × 2 matrix and we get a 2 × 2 matrix as the result.</li></li></ul><li>Generalized Example<br />If we multiply a 2×3 matrix with a 3×1 matrix, the product matrix is 2×1 <br />Here is how we get M11 and M22 in the product. <br />M11 = r11× t11 + r12× t21 + r13×t31M12 = r21× t11 + r22× t21 + r23×t31<br />
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# Statistics: Finding the Mean
In this worksheet, students calculate the mean of a given set of data.
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: Mean, Median, Mode and Range
Difficulty level:
### QUESTION 1 of 10
There are several different types of averages.
An average is one number that represents a set of numbers.
The MEAN is a number, which we calculate by adding up all the numbers in the list and sharing this total out by how many there are.
Example
Find the MEAN of this set of ten numbers: 5, 4, 8, 4, 3, 3, 8, 5, 1, 6
Step 1 Find the total which is 47
Step 2 Now divide by the number of numbers, which is 10 to get 47 ÷ 10 = 4.7
The MEAN is 4.7
Work out the MEAN of this list of 7 numbers:
5, 8, 7, 7, 3, 5, 7,
Work out the MEAN of this list of shoe sizes, taken from children in a football team:
8, 8, 7, 7, 6, 5, 8, 7, 6, 8, 7
Work out the MEAN of this list of numbers:
15, 12, 7, 17, 23, 15, 14, 17
Work out the MEAN of this list of numbers:
15, 1200, 85, 1700
Work out the MEAN of this list of numbers:
5, 5, 5, 5, 5, 5, 5, 5
Work out the MEAN value of this list of shoe sizes, taken from children in a hockey team:
5½, 8, 6, 7½, 6, 5, 6, 5, 6, 8, 3
Work out the MEAN of this list of numbers:
5, 5, 5, 5, 100
Work out the MEAN of this list of numbers:
5, -5, -7, 7
The maximum daytime temperatures in ºC on 10 British summer days were as follows.
Work out the MEAN temperature in ºC (just write the number)
18, 17, 19, 18, 20, 15, 17, 18, 16, 15
The maximum daytime temperatures in ºC on 10 Italian summer days were as follows.
Work out the MEAN temperature in ºC (just write the number)
32, 27, 32, 35, 34, 33, 36, 35, 30, 35
• Question 1
Work out the MEAN of this list of 7 numbers:
5, 8, 7, 7, 3, 5, 7,
6
EDDIE SAYS
42 ÷ 7 = 6
• Question 2
Work out the MEAN of this list of shoe sizes, taken from children in a football team:
8, 8, 7, 7, 6, 5, 8, 7, 6, 8, 7
7
• Question 3
Work out the MEAN of this list of numbers:
15, 12, 7, 17, 23, 15, 14, 17
15
• Question 4
Work out the MEAN of this list of numbers:
15, 1200, 85, 1700
750
• Question 5
Work out the MEAN of this list of numbers:
5, 5, 5, 5, 5, 5, 5, 5
5
• Question 6
Work out the MEAN value of this list of shoe sizes, taken from children in a hockey team:
5½, 8, 6, 7½, 6, 5, 6, 5, 6, 8, 3
6
• Question 7
Work out the MEAN of this list of numbers:
5, 5, 5, 5, 100
24
• Question 8
Work out the MEAN of this list of numbers:
5, -5, -7, 7
0
• Question 9
The maximum daytime temperatures in ºC on 10 British summer days were as follows.
Work out the MEAN temperature in ºC (just write the number)
18, 17, 19, 18, 20, 15, 17, 18, 16, 15
17.3
• Question 10
The maximum daytime temperatures in ºC on 10 Italian summer days were as follows.
Work out the MEAN temperature in ºC (just write the number)
32, 27, 32, 35, 34, 33, 36, 35, 30, 35
32.9
---- OR ----
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# Circles and Lines
We just learned how to represent a circle with an equation. We've known for a long time how to represent a line with an equation. So let's use equations to examine what happens when circles and lines interact.
We represent a generic circle with (xh)2 + (yk)2 = r2and a generic line with y = mx + b. Let's put a line and a circle together in the same system of equations.
(xh)2 + (yk)2 = r2
y = mx + b
Don't get too worried (or excited). We won't make you solve systems like this, even though you could do it. The algebra is a little tough and we're trying to focus on the geometry here. Besides, focusing on the geometry might actually help us understand the algebra.
Remember that the solution of a system of two equations like this one is the set of all points (x, y) that make both equations true at the same time. In other words, the solution of that system is the set of all points that are both on the circle and on the line. In other other words, the solution of that system is the set of points where the circle and the line intersect.
We already know that a line can intersect a circle at two, one, or zero points (it can be a secant of the circle, a tangent to the circle, or neither). Therefore, this system of equations must always have two, one, or zero solutions, depending on whether the line is a secant, a tangent, or neither.
As we said before, we won't make you solve a system like this. But we will make you do some algebra, because it's fun. And maybe we're a little sadistic.
We've talked before about cutting up a pizza, but never in Australia. Each cut made by a pizza knife stretches from one side of the pizza to the other, intersecting the crust of the pizza at two points. So each cut is a secant. Hopefully, each cut is also a diameter, so that the pizza ends up looking like this:
and not like this:
Suppose we put a coordinate grid over that pizza so you can see the coordinates of the points where the sloppy pizza man's cut intersects the edge of the pizza as well as find the equation of the circle.
Could you find the equation of the line made by the cut? Of course you could. You actually don't need to consider the circle at all: it's just finding the equation of a line given two points on the line, which is algebra-level stuff.
But let's say the pizza man is really sloppy and makes a cut that's not a secant of the pizza, but rather a tangent to the pizza. He's basically cutting the box, only touching the tiniest nub of pizza. Could you find the equation of the line then?
Let's draw a diagram of the situation. The pizza has a radius of 25 cm. Let's consider the center of the pizza to be at the origin. This sloppy pizza man, who is seriously liable to damage himself, his co-workers, or his place of employment, makes a cut tangent to the pizza at the point (7, 24).
We know one point on the line already, so let's see if we can find the slope of the line. Then we'll have enough information to write the equation of the line.
We know that the tangent line is perpendicular to the radius of the circle at the point of tangency. So if we find the slope of the radius, we can take the negative reciprocal and we have the slope of the tangent line. Ingenious!
### Sample Problem
What is the equation of the tangent line that intersects a circle centered at the origin with radius 25 at the point (7, 24)?
In this case, the radius is the line segment with endpoints (0, 0) and (7, 24). The slope of the radius is
Taking the negative reciprocal, we get a slope of for the tangent line. We now have enough to write the equation of the tangent line in point-slope form:
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## Precalculus (6th Edition) Blitzer
The standard form of the expression${{\left( 2-\sqrt{-3} \right)}^{2}}$ is $1-4i\sqrt{3}$.
Consider the expression, ${{\left( 2-\sqrt{-3} \right)}^{2}}$ Rewrite ${{\left( 2-\sqrt{-3} \right)}^{2}}$ as ${{\left( 2-\sqrt{3}\sqrt{-1} \right)}^{2}}$ As $i=\sqrt{-1}$ Therefore, ${{\left( 2-\sqrt{-3} \right)}^{2}}={{\left( 2-i\sqrt{3} \right)}^{2}}$ Apply the formula for the square of the sum. \begin{align} & {{\left( 2-i\sqrt{3} \right)}^{2}}={{2}^{2}}-2\left( 2 \right)\left( i\sqrt{3} \right)+{{\left( i\sqrt{3} \right)}^{2}} \\ & =4-4i\sqrt{3}+3{{i}^{2}} \end{align} As ${{i}^{2}}=-1$ Therefore, \begin{align} & 4-4i\sqrt{3}+3{{i}^{2}}=4-4i\sqrt{3}+3\left( -1 \right) \\ & =4-4i\sqrt{3}-3 \end{align} Combine the real part and imaginary part separately and either add or subtract as required. \begin{align} & 4-4i\sqrt{3}-3=\left( 4-3 \right)-4i\sqrt{3} \\ & =1-4i\sqrt{3} \end{align} Thus, ${{\left( 2-\sqrt{-3} \right)}^{2}}=1-4i\sqrt{3}$ Hence, the standard form of ${{\left( 2-\sqrt{-3} \right)}^{2}}$ is $1-4i\sqrt{3}$.
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