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There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Number. In the right column below are links to related online activities, videos and teacher resources. A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics. Main Page ### Number Starters: A Very Strange Game: Four different actions depending on the number which appears. Abundant Buses: A game based around the concept of abundant numbers. Add 'em: Add up a sequence of consecutive numbers. Can you find a quick way to do it? All The Nines: Add up all the numbers in the nine times table. Ancient Mysteries: This activity requires students to memorise fifteen numbers in a three by five grid. Aunt Sophie's Post Office: Work out the number of stamps needed to post a parcel. Birthday Clues: Work out the date Will was born by answering some number questions. Can You Decide?: Recognise odd, even, square, prime and triangular numbers. Christmas Bells: If all the bells ring together at noon, at what time will they next all ring together? This problem requires the use of LCM. Christmas Eve: How many palindromic numbers can you find? ClockEquate: Can you use the digits on the left of this clock along with any mathematical operations to equal the digits on the right? Coins in Envelopes: Fifteen pennies are placed in four envelopes and the envelopes are sealed. It is possible to pay someone any amount from 1p to 15p by giving them one or more envelopes. How were the pennies distributed between the envelopes? Consecutive Squares: What do you notice about the difference between the squares of consecutive numbers? Dancing: Work out how many people were at the dance from the clues given. Dice Nets: Determine whether the given nets would fold to produce a dice. Dimidiate: Arrange the digits from 1 to 9 in alphabetical order. How many times can this number be halved? Divided Age: How old is a person if when her age is divided by certain numbers, the calculator display ending are as shown. Double Trouble: Begin with one, double it, double it again and so on. How many numbers in this sequence can you write down before the register has been called? Factuples: Spot the factors and the multiples amongst the numbers in the grid. Flabbergasted: If each number in a sequence must be a factor or multiple of the previous number what is the longest sequence that can be made from the given numbers? Flowchart: Use the flowchart to generate a sequence of numbers. Which number will reach 1 the fastest? Four Factors: Find four single digit numbers that multiply together to give 120. How many different ways are there of answering this question? Four problems: For mathematical questions to get everyone thinking at the beginning of the lesson.. Halve it: Start with 512. Halve it to get 256. Halve it to get 128. Continue as far as possible. Handshakes: If all the students in this room shook hands with each other, how many handshakes would there be altogether? Hot Numbers: Move the numbered cards to form five 2 digit numbers which are all multiples of three. Hotel Digital: A puzzle about the lifts in a hotel which serve floors based on the day of the week. House Numbers: The numbers on five houses next to each other add up to 70. What are those five numbers? Inbetween Table: Write down as many multiples of 3.5 as possible in 3.5 minutes. Last Day: The 31st of December is the last day of the year. What mathematical lasts do you know? Leap Year: A question about the birthdays of a child born on the 29th February. Letters in a Number: Questions about the number of letters in numbers. Maths Riddles: Can you work out the numbers from the given clues. Meta Products: Which numbers when multiplied by the number of letters in the word(s) of the number give square numbers? Missing Terms: Find the missing terms from these linear sequences. Name Again: Work out what the nth letter will be in a recurring pattern of letters in a person's name Negative Numbers: Perform calculations involving negative numbers No Partner: Find which numbers in a given list do not combine with other numbers on the list to make a given sum. Numbers in words: Write out in words some numbers writen as digits (optional pirate theme) Odd One Out: From the numbers given, find the one that is the odd one out. Only One Number: Find other numbers that can be changed to 1 on a calculator using only the 4 key and any operation. Pears Make Squares: Find three numbers such that each pair of numbers adds up to a square number. Perfect Numbers: Six is a perfect number as it is the sum of its factors. Can you find any other perfect numbers? Plane Numbers: Arrange numbers on the plane shaped grid to produce the given totals Pyramid Puzzle: Arrange numbers at the bottom of the pyramid which will give the largest total at the top. Register: When the register is called answer with a multiple of 7. Ropey Snowballs: Arrange the numbers on the snowballs so that no two consecutive numbers are directly connected by rope. Satisfaction: Rearrange the numbers, row and column headings so that this table is mathematically correct. Scaramouche: Can you work out from the five clues given what the mystery number is? Seeing Squares: How many square numbers can be found in the grid of digits. Sign Sequences: Continue the sequences if you can work out the rule. Simple Nim: The classic game of Nim played with a group of pens and pencils. The game can be extended to the multi-pile version. Small Satisfaction: Arrange the digits one to nine in the grid so that they obey the row and column headings. Square and Even: Arrange the numbers on the cards so that each of the three digit numbers formed horizontally are square numbers and each of the three digit numbers formed vertically are even. Square Angles: Find a trapezium, a triangle and a quadrilateral where all of the angles are square numbers. Square Christmas Tree: Draw a picture of a Christmas tree using only square numbers. Square Pairs: Arrange the numbered trees so that adjacent sums are square numbers. Square Sequence: Write out as many square numbers as possible in 4 minutes. Squigits: A challenge to find numbers which have each of their digits as square numbers. The Power of Christmas: Find a power of 2 and a power of 3 that are consecutive numbers. The story of ...: Be creative and come up with as many facts about a number as you can think of. Tindice: How can you put the dice into the tins so that there is an odd number of dice in each tin? Twelve Days: A Maths puzzle based on the 12 Days of Christmas song. Two Numbers: Find the two numbers whose sum and product are given. Upside Number: Work out the phone number from the clues given. Venn Diagram: Arrange numbers on the Venn Diagram according to their properties. What are they?: A starter about sums, products, differences, ratios, square and prime numbers. #### Pentransum Answer multiple choice questions about basic mathematical ideas. If you get a number of questions correct you will be invited to post a question of your own. The bank of questions grows larger every day. ### Feedback: Comment recorded on the 8 May 'Starter of the Day' page by Mr Smith, West Sussex, UK: "I am an NQT and have only just discovered this website. I nearly wet my pants with joy. To the creator of this website and all of those teachers who have contributed to it, I would like to say a big THANK YOU!!! :)." Comment recorded on the 17 November 'Starter of the Day' page by Amy Thay, Coventry: "Thank you so much for your wonderful site. I have so much material to use in class and inspire me to try something a little different more often. I am going to show my maths department your website and encourage them to use it too. How lovely that you have compiled such a great resource to help teachers and pupils. Thanks again" Comment recorded on the 9 May 'Starter of the Day' page by Liz, Kuwait: "I would like to thank you for the excellent resources which I used every day. My students would often turn up early to tackle the starter of the day as there were stamps for the first 5 finishers. We also had a lot of fun with the fun maths. All in all your resources provoked discussion and the students had a lot of fun." Comment recorded on the 14 October 'Starter of the Day' page by Inger Kisby, Herts and Essex High School: "Just a quick note to say that we use a lot of your starters. It is lovely to have so many different ideas to start a lesson with. Thank you very much and keep up the good work." Comment recorded on the 2 April 'Starter of the Day' page by Mrs Wilshaw, Dunsten Collage,Essex: "This website was brilliant. My class and I really enjoy doing the activites." Comment recorded on the 28 September 'Starter of the Day' page by Malcolm P, Dorset: "A set of real life savers!! Keep it up and thank you!" Comment recorded on the 1 February 'Starter of the Day' page by Terry Shaw, Beaulieu Convent School: "Really good site. Lots of good ideas for starters. Use it most of the time in KS3." Comment recorded on the 12 July 'Starter of the Day' page by Miss J Key, Farlingaye High School, Suffolk: "Thanks very much for this one. We developed it into a whole lesson and I borrowed some hats from the drama department to add to the fun!" Comment recorded on the 3 October 'Starter of the Day' page by S Mirza, Park High School, Colne: "Very good starters, help pupils settle very well in maths classroom." Comment recorded on the 3 October 'Starter of the Day' page by Fiona Bray, Cams Hill School: "This is an excellent website. We all often use the starters as the pupils come in the door and get settled as we take the register." Comment recorded on the 9 October 'Starter of the Day' page by Mr Jones, Wales: "I think that having a starter of the day helps improve maths in general. My pupils say they love them!!!" Comment recorded on the 2 May 'Starter of the Day' page by Angela Lowry, : "I think these are great! So useful and handy, the children love them. Could we have some on angles too please?" Comment recorded on the 26 March 'Starter of the Day' page by Julie Reakes, The English College, Dubai: "It's great to have a starter that's timed and focuses the attention of everyone fully. I told them in advance I would do 10 then record their percentages." Comment recorded on the 21 October 'Starter of the Day' page by Mr Trainor And His P7 Class(All Girls), Mercy Primary School, Belfast: "My Primary 7 class in Mercy Primary school, Belfast, look forward to your mental maths starters every morning. The variety of material is interesting and exciting and always engages the teacher and pupils. Keep them coming please." Comment recorded on the 28 May 'Starter of the Day' page by L Smith, Colwyn Bay: "An absolutely brilliant resource. Only recently been discovered but is used daily with all my classes. It is particularly useful when things can be saved for further use. Thank you!" Comment recorded on the 11 January 'Starter of the Day' page by S Johnson, The King John School: "We recently had an afternoon on accelerated learning.This linked really well and prompted a discussion about learning styles and short term memory." Comment recorded on the 5 April 'Starter of the Day' page by Mr Stoner, St George's College of Technology: "This resource has made a great deal of difference to the standard of starters for all of our lessons. Thank you for being so creative and imaginative." Comment recorded on the 1 February 'Starter of the Day' page by M Chant, Chase Lane School Harwich: "My year five children look forward to their daily challenge and enjoy the problems as much as I do. A great resource - thanks a million." Comment recorded on the 19 June 'Starter of the Day' page by Nikki Jordan, Braunton School, Devon: "Excellent. Thank you very much for a fabulous set of starters. I use the 'weekenders' if the daily ones are not quite what I want. Brilliant and much appreciated." Comment recorded on the 3 October 'Starter of the Day' page by Mrs Johnstone, 7Je: "I think this is a brilliant website as all the students enjoy doing the puzzles and it is a brilliant way to start a lesson." Comment recorded on the 23 September 'Starter of the Day' page by Judy, Chatsmore CHS: "This triangle starter is excellent. I have used it with all of my ks3 and ks4 classes and they are all totally focused when counting the triangles." Comment recorded on the 17 June 'Starter of the Day' page by Mr Hall, Light Hall School, Solihull: "Dear Transum, I love you website I use it every maths lesson I have with every year group! I don't know were I would turn to with out you!" Comment recorded on the 1 August 'Starter of the Day' page by Peter Wright, St Joseph's College: "Love using the Starter of the Day activities to get the students into Maths mode at the beginning of a lesson. Lots of interesting discussions and questions have arisen out of the activities. Thanks for such a great resource!" Comment recorded on the 10 September 'Starter of the Day' page by Carol, Sheffield PArk Academy: "3 NQTs in the department, I'm new subject leader in this new academy - Starters R Great!! Lovely resource for stimulating learning and getting eveyone off to a good start. Thank you!!" Comment recorded on the 9 April 'Starter of the Day' page by Jan, South Canterbury: "Thank you for sharing such a great resource. I was about to try and get together a bank of starters but time is always required elsewhere, so thank you." Comment recorded on the 7 December 'Starter of the Day' page by Cathryn Aldridge, Pells Primary: "I use Starter of the Day as a registration and warm-up activity for my Year 6 class. The range of questioning provided is excellent as are some of the images. I rate this site as a 5!" Comment recorded on the 16 March 'Starter of the Day' page by Mrs A Milton, Ysgol Ardudwy: "I have used your starters for 3 years now and would not have a lesson without one! Fantastic way to engage the pupils at the start of a lesson." Comment recorded on the 1 May 'Starter of the Day' page by Phil Anthony, Head of Maths, Stourport High School: "What a brilliant website. We have just started to use the 'starter-of-the-day' in our yr9 lessons to try them out before we change from a high school to a secondary school in September. This is one of the best resources on-line we have found. The kids and staff love it. Well done an thank you very much for making my maths lessons more interesting and fun." Comment recorded on the 25 June 'Starter of the Day' page by Inger.kisby@herts and essex.herts.sch.uk, : "We all love your starters. It is so good to have such a collection. We use them for all age groups and abilities. Have particularly enjoyed KIM's game, as we have not used that for Mathematics before. Keep up the good work and thank you very much Best wishes from Inger Kisby" Comment recorded on the 19 October 'Starter of the Day' page by E Pollard, Huddersfield: "I used this with my bottom set in year 9. To engage them I used their name and favorite football team (or pop group) instead of the school name. For homework, I asked each student to find a definition for the key words they had been given (once they had fun trying to guess the answer) and they presented their findings to the rest of the class the following day. They felt really special because the key words came from their own personal information." ### Notes: Spotting patterns is an important skill in many areas of life. The world of numbers contains many fascinating patterns and understanding them enables better problem solving strategies. From seeing patterns in the multiples of numbers shaded in a hundred square to spotting the recurring sequences of digits in decimal numbers there is a great deal for pupils to be introduced to. This topic includes even, odd, prime, triangular, perfect, abundant, square and cube numbers. It uses factors and multiples to find solutions to real life problems and encourages number connections to be investigated for pleasure. There are a lot of puzzles, challenges and games too. ### Number Teacher Resources: Binary Lights: Represent binary numbers with a row of ligts which can be turned on or off. Counter: A dynamic visual aid that counts! Choose the first term, common difference and the speed Formal Written Methods: Examples of formal written methods for addition, subtraction, multiplication and division. Hot Number Challenges: The ten Hot Number challenges ready to project onto your whiteboard. Maths Mind Reader: Investigate this amazing mind reading performance based on simple mathematical principles. Number Grids: Investigate the properties of number with these interactive number grids. Number Line: This number line visual aid is designed to be projected onto a whiteboard for whole class exposition. Pesto: Students classify numbers randomly appearing on the screen by holding up cards Playing Card Maths: Imagine you are on a desert island with nothing but a pack of playing cards. Do you have to stop learning mathematics? Prison Cell Problem: A number patterns investigation involving prisoners and prison guards. Satisfaction: This is quite a challenging number grouping puzzle requiring a knowledge of prime, square and triangular numbers. Sieve of Eratosthenes: A self checking, interactive version of the Sieve of Eratosthenes method of finding prime numbers. ### Number Activities: Consecutive Numbers: Find the consective numbers that are added or multiplied to give the given totals Delightfully Divisible: Arrange the digits one to nine to make a number which is divisible in the way described. Dice Net Challenge: Drag the numbers onto the net so that when it is folded to form a cube numbers on opposite faces add up to prime numbers. Divisibility Test: Practise using the quick ways to spot whether a number is divisible by the digits 2 to 9. Don't Shoot The Square: Shoot the numbers but don't shoot any square numbers. Dump-A-Dice Race: An online board game for two players involving prime and square numbers and making choices. Factor Trees: Create factor trees to find the prime factors of the given numbers. Fizz Buzzer: The digital version of the popular fizz buzz game. Press the buzzers if they are factors of the counter. HCF and LCM: Practise finding the highest common factor (H.C.F), sometimes called the greatest common divisor, and the lowest common multiple (L.C.M) of two numbers. Hot Numbers: Move the numbered cards to form five 2 digit numbers which obey the given rules. Magic Square Puzzle: Find all of the possible ways of making the magic total from the numbers in this four by four magic square. Missing Terms: Can you work out which numbers are missing from these number sequences? Mix and Math: Determine the nature of adding, subtracting and multiplying numbers with specific properties. Nim: Nim is a mathematical game of strategy in which two players take turns removing objects from groups of objects. No Partner: Find which numbers in a given list do not combine with other numbers on the list to make a given sum. Number Crunch Saga: A lively numeracy game requiring you to align three numbers to create the given target sum or product. Number Grids: Investigate the properties of number with these interactive number grids. Number Jigsaws: Online, interactive jigsaw puzzles of grids of numbers. Number Skills Inventory: A checklist of basic numeracy techniques that every pupil should know. Number Systems: Place the numbers in the correct sets in this concentric circles Venn diagram. Pairs 240: Find the pairs of numbers that multiply together to give a product of 240 Plane Numbers: Arrange numbers on the plane shaped grid to produce the given totals Powten: Practise multiplying and dividing by powers of ten without using a calculator. Prime Labyrinth: Find the path to the centre of the labyrinth by moving along the prime numbers. Prime Square: Drag the numbers into the red cells so that the sum of the three numbers in each row and each column is a prime number. Prison Cell Problem: A number patterns investigation involving prisoners and prison guards. River Crossing: The traditional River Crossing challenge. Can you do it in the smallest number of moves? Roman Numerals Jigsaw: An online interactive jigsaw puzzle of a grid of Roman numerals. Roman Numerals Quiz: This online, self marking quiz tests your ability to convert Roman numerals. Satisfaction: This is quite a challenging number grouping puzzle requiring a knowledge of prime, square and triangular numbers. Satisfy: Place the nine numbers in the table so they obey the row and column headings. Scallywags and Scoundrels: Arrange the scallywags and scoundrels on the chairs so that the numbers of any two sitting next to each other add up to a prime number. Sieve of Eratosthenes: A self checking, interactive version of the Sieve of Eratosthenes method of finding prime numbers. Square and Even: Arrange the numbers on the cards so that each of the three digit numbers formed horizontally are square numbers and each of the three digit numbers formed vertically are even. Square Pairs Game: A game for two players who take turns to select two numbers that add up to a square number. Stamp Sticking: Drag stamps onto the envelopes to make the exact postage as shown at the top left of each envelope. Suko Sujiko: Interactive number-based logic puzzles similar to those featuring in daily newspapers. Three Ways: Find three different ways of multiplying four different digits together to get the given target number. There are nine levels for this online challenge. Times Square: Practise your times tables with this self-checking multiplication grid Times Tables: A collection of activities to help you learn your times tables in only 5 days. Twelve Days: How many gifts did my true love send to me according to the traditional Christmas song 'Twelve Days of Christmas'. Venn Diagram: Place each of the numbers 1 to 16 on the correct regions on the Venn diagram. Watsadoo: Rotate the cogs to catch the flying numbers in the correct sections. What Are They?: An online exercise about sums, products, differences, ratios, square and prime numbers. Finally there is Topic Test, a set of 10 randomly chosen, multiple choice questions suggested by people from around the world. ### Number Investigations: Calendar Maths Investigation: Investigate the connection between the numbers in a T shape drawn on this month's calendar. Decimal Products: Find two decimal numbers that add up to exactly one. What is the product of these two decimals? Digit Sums and Multiples: Investigate numbers which are multiples of the sum of their digits. Handshakes: If everyone in this room shook hands with each other, how many handshakes would there be? Lamp-posts: What is the greatest number of lamp-posts that could be needed for a given village? Leapfrog: An investigation of the fewest number of moves required to make the blue and green frogs swap places. lnvestigating stamps: How many different letters can you send with the given stamps. Maths Mind Reader: Investigate this amazing mind reading performance based on simple mathematical principles. Number Stairs: Find the relationships between the numbers on different size grids. Palandromic numbers: How many steps does each number take to become palandromic? Snooker Investigation: Investigate a special snooker table with only four pockets. Which hole will the snooker ball fall into for various sized snooker tables? Steps: Investigate this growing sequence of steps. Trapezia: Which numbers can be represented by groups of circles in the shape of a trapezium? ### Number Videos: BIDMAS Misconception: This Maths problem went viral on Twitter and there has been debate about the answer. Presh explains why one interpretation is considered correct. Binary and The Tower of Hanoi: A video showing how binary counting can solve the Tower of Hanoi puzzle. Finding Prime Factors: A straight forward explanation from SLEP Fractions on the number line: Practice plotting values on the number line as a passionate activist rises up and demands equity for all numbers, including fractions and decimals. Problems with Zero: Dividing by zero, zero divided by zero and zero to the power of zero - all pose problems! ### Number Worksheets/Printables: Divided Age Worksheet: Additional questions for the 6th December Starter of the Day about the decimal part of quotients. How Many Squares? 1: A printable grid containing many copies of the design used in the shape counting Starter. How Many Squares? 2: A printable grid containing many copies of the design used in the second shape counting Starter. Lemon Law Worksheet: Can you find the solution to these Lemon Law challenges using a spreadsheet? Pesto Cards: Cards to print for pupils to use in the 'Can You Decide?' activity. Links to other websites containing resources for Number are provided for those logged into 'Transum Mathematics'. Subscribing also opens up the opportunity for you to add your own links to this panel. You can sign up using one of the buttons below: ### Search The activity you are looking for may have been classified in a different way from the way you were expecting. You can search the whole of Transum Maths by using the box below. ### Other Is there anything you would have a regular use for that we don't feature here? Please let us know. #### Graph Equation Pairs Match the equation with its graph. Includes quadratics, cubics, reciprocals, exponential and the sine function. So far this activity has been accessed 158 times and only 1 person has earned a Transum Trophy for completing it. ### Homepage Have today's Starter of the Day as your default homepage. Copy the URL below then select Tools > Internet Options (Internet Explorer) then paste the URL into the homepage field. Set as your homepage (if you are using Internet Explorer) Wednesday, October 18, 2017 Wednesday, November 1, 2017 Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments. Starter Menu in English \| French \| Spanish \| For Students: For All:
## IB Math AA SL Paper 2 Question Bank If you’re looking for an IB Math AA SL Paper 2 Question Bank, you’ve come to the right place! We have a wide selection of questions available, covering all of the topics you’ll need to know for your exam. Whether you’re needing help with geometry or statistics, we’ve got you covered. With our question bank, you’ll be able to work through practice problems at your own pace and master the material before test day. So why wait? Get started today and ace that paper 2! Section A ## 1.) A rectangular prism has a volume of 48 cubic meters and a square base with a side length of 2 meters. The height of the prism is h meters. a) Express the total surface area of the rectangular prism as a function of h. We know that the volume of a rectangular prism is given by V = lwh, where l, w and h are the length, width and height of the prism respectively. In this case, we know that V = 48 m3 and the side length of the base is 2 m. Therefore, the width and length of the base are both 2 m. So, we can express the volume as V = 2w * 2l * h = 48 m3 We know the formula for the surface area of a rectangular prism is 2lw+ 2lh + 2wh. Here we know width w = l = 2m and h is the variable. So, the surface area A of the rectangular prism as a function of h is: A = 2 * 2 * 2 + 2 * 2 * h = 8 + 4h b) Express the height h of the rectangular prism as a function of the total surface area. To express the height h as a function of the total surface area, we can use the surface area equation that we found in part a. If we let A be the total surface area, then: A = 8 + 4h Rearranging this equation to solve for h gives us: h = (A-8)/4 So, h is a function of the total surface area. The surface area as a function of h is A = 8+ 4h, and the h as a function of the surface area is h = (A-8)/4 ## 2.) An amount of \$ 10 000 is invested at an annual interest rate of 12%. a) Find the value of the investment after 5 years i) if the interest rate is compounded yearly; FV=10000(1+12/100)n n= 5, 17623.42 ii) if the interest rate is compounded half-yearly; n= 10, 17908.48 iii) if the interest rate is compounded quarterly; n=20, 18061.11 iv) if the interest rate is compounded monthly. n= 512, 18166.97 b) The value of the investment will exceed \$ 20 000 after n full years. Calculate the minimum value of n i) if the interest rate is compounded yearly; 20000= 10000(1+12/100)n Solve by GDC n=6.11, hence n=7 ii) if the interest rate is compounded monthly. 20000= 10000(1+12/100)12n Solve by GDC n=5.81, hence n=6 ## 3.) A car rental company charges a fixed daily rental fee of \$50, and a variable charge of \$0.15 per kilometre driven. a) Write an expression for the cost of renting the car for d days, and traveling x kilometers. The cost of renting the car is made up of two parts: a fixed daily rental fee and a variable charge per kilometer driven. To find the total cost, we need to multiply the number of days the car is rented by the daily rental fee, and add that to the product of the number of kilometers driven and the variable charge per kilometer. So, the total cost, C, can be expressed as an equation: C = d 50 + x * 0.15 b) The company offers a weekly rental package for \$400, which includes a maximum of 600km of travel. Express the cost of the weekly rental package, C, in terms of the number of kilometres driven, x. To find the cost of the weekly rental package in terms of the number of kilometers driven, we know that the package includes 600km of travel and cost \$400. So we can create an equation: C = 400 = d * 50 + x * 0.15 we know that d = 7 for a weekly package and x is 600km So, C(x) = 7 * 50 + x * 0.15 = 400 Hence C(x) = 750 + 0.15x = 400 this is the equation that expresses the cost of the weekly rental package, C, in terms of the number of kilometres driven, x. ## 4.) Triangle ABC is a right-angled triangle, with right angle at B and hypotenuse C. a) Write down the value of sin B, cos B and tan B in terms of a and b. In a right-angled triangle, we can use the trigonometric functions sine, cosine, and tangent to relate the lengths of the sides to the measure of the angles. For a right-angled triangle, sin B = a/c, cos B = b/c, tan B = a/b b) If a = 5 and b = 12, find the values of sin B, cos B and tan B. If we plug in the given values, a = 5 and b = 12, we can find the values of sin B, cos B and tan B: sin B = a/c = 5/c cos B = b/c = 12/c tan B = a/b = 5/12 c) Use the result from part (b) to find the value of c. To find the value of c, we can use the Pythagorean theorem, c = âˆš(a2 + b2) = âˆš(52 + 122) = âˆš(25 + 144) = âˆš(169) = 13 ## 5.) Solve the following a) Use the sine rule to find the sine of the angle A Sin A = (10sin30Â°)/6 Sin A= 5/6 b) Hence find the two possible values of angle A A= 56.4Â° or A = 124Â° ## 6.) The weight X of a particular animal is normally distributed with Î¼= 200kg and Ïƒ= 15kg. An animal of this population is overweight if it has a weight greater than 230 kg a) Find the probability that an animal is overweight. P (X>230) = 0.02275 b) We select 2 animals from this population. Find the probabilities that i) both animals are overweight (0.02275)2= 0.000518 ii) only one animal is overweight. 0.02275) (1-0.02275)* 2 = 0.0445 c) We select 7 animals from this population. Find the probability that exactly two of them are overweight. Y follows B(n,p) with n = 7, p =0.02275 P(Y=2) = 0.00969 Our Guide is written by counselors from Cambridge University for colleges like MIT and other Ivy League colleges. Section B ## 1.) A factory produces electronic components. The probability of a component is defective is 0.05. a) What is the probability of producing exactly 3 defective components in a batch of 50 components? This is a binomial probability distribution problem. The probability of producing exactly 3 defective components in a batch of 50 components is: P(X = 3) = (50 choose 3) * (0.05)3 * (1-0.05)(50-3) = 19600 * 0.053 * 0.9547 â‰ˆ 0.09 b)What is the probability of producing at most 3 defective components in a batch of 50 components? To find the probability of producing at most 3 defective components, we can use the cumulative distribution function of the binomial distribution. The probability of producing at most 3 defective components is: P(X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) P(X <= 3) = (50 choose 0) * (0.05)0 * (1-0.05)50 + (50 choose 1) * (0.05)1 * (1-0.05)49 + (50 choose 2) * (0.05)2 * (1-0.05)48 + (50 choose 3) * (0.05)3 * (1-0.05)47 â‰ˆ 0.37 The factory produces 1000 components per day. c) What is the probability of producing more than 45 defective components in a day? The expected number of defective components produced per day is: E(X) = Î» = n * p = 1000 * 0.05 = 50 The probability of producing more than 45 defective components in a day is: 1 – P(X <= 45) = 1 – (e(-50) * (5045) / 45!) â‰ˆ 0.02 d) What is the probability of producing less than 45 defective components in a day? To find the probability of producing less than 45 defective components in a day, we can use the Poisson cumulative distribution function. The probability of producing less than 45 defective components in a day is: P(X < 45) = (e(-50) * (5044) / 44!) â‰ˆ 0.98 ## 2.) A company produces and sells widgets. The revenue, R (in dollars), from selling x widgets per day is given by the function R(x) = 2x2 + 50x. The cost, C (in dollars), of producing x widgets per day is given by the function C(x) = 0.5x2Â + 25x + 100. a.) Find the profit function, P(x), in terms of x. The profit, P(x), is calculated by subtracting the cost, C(x), from the revenue, R(x): P(x) = R(x) – C(x) Substituting the given functions for R(x) and C(x): P(x) = (2x2 + 50x) – (0.5x2 + 25x + 100) P(x) = 1.5x2 + 25x – 100 b.) Find the value of x that maximizes the profit. To find the value of x that maximizes the profit, we need to find the vertex of the quadratic function P(x) = 1.5x2 + 25x – 100. The x-coordinate of the vertex is given by the formula x = -b/2a, where a and b are the coefficients of the x2 and x terms, respectively. In this case, a = 1.5 and b = 25, so substituting these values into the formula: x = -25 / (2 * 1.5) x = -25 / 3 So, the value of x that maximizes the profit is x = -25/3. c.) Calculate the maximum profit. To calculate the maximum profit, we can substitute the value of x = -25/3 into the profit function P(x) = 1.5x2 + 25x – 100. P(-25/3) = 1.5 * (-25/3)2 + 25 * (-25/3) – 100 P(-25/3) = -975/3 + (-625/3) – 100 P(-25/3) = -1600/3 – 100 P(-25/3) = -1100/3 So, the maximum profit is -1100/3 dollars. d.) Determine the range of values of x for which the company makes a profit. The company makes a profit when the profit function P(x) is positive. To determine the range of values of x for which the company makes a profit, we need to find the values of x that satisfy the inequality P(x) > 0. 1.5x2 + 25x – 100 > 0 We can solve this inequality by using quadratic techniques, such as factoring or the quadratic formula. Once we find the solutions for x, we can determine the range of values of x for which the inequality is true, and thus the range of values of x for which the company makes a profit. To determine the range of values of x for which the company makes a profit, we need to solve the inequality 1.5x2 + 25x – 100 > 0. Let’s use the quadratic formula to find the solutions for x: 1.5x2 + 25x – 100 > 0 a = 1.5, b = 25, c = -100 x = (-b Â± âˆš(b2 – 4ac)) / (2a) x = (-25 Â± âˆš(252 – 4 * 1.5 * -100)) / (2 * 1.5) x = (-25 Â± âˆš(625 + 1200)) / 3 x = (-25 Â± âˆš1825) / 3 Now we can determine the range of values of x for which the inequality is true. Since the coefficient of x2 is positive (1.5 > 0), the parabola opens upwards, and the inequality is true for values of x outside of the solutions we found. So, the range of values of x for which the company makes a profit is x < (-25 – âˆš1825) / 3 or x > (-25 + âˆš1825) / 3. ## 3.) Let f(x) = x3 – 3x2 – 4x + 12. Solve the equation f(x) = 0. a.) By first testing the values of f(1) and f(2), show that there is a root of f(x) = 0 between x = 1 and x = 2. We have f(1) = 6 and f(2) = 2, so by the intermediate value theorem, there must be a root of f(x) = 0 between x = 1 and x = 2. b.) Use synthetic division to factorize f(x) into a linear factor and a quadratic factor. To factorize f(x) using synthetic division, we first guess a root, which could be a factor of the constant term (12) divided by a factor of the leading coefficient (1). One possibility is x = 3. Then we divide f(x) by (x – 3): 1 -3 -4 12 3 \| 1 -3 -4 12 -3 0 -12 ———— 1 -6 -4 0 So we have f(x) = (x – 3)(x2 – 6x – 4). c.) Find the remaining roots of f(x) = 0, giving your answers correct to three decimal places. To find the remaining roots of f(x) = 0, we need to solve the quadratic factor x2 – 6x – 4. Using the quadratic formula, we get: x = [6 Â± sqrt(62 + 441)] / 2 x = 3 Â± sqrt(19) Therefore, the roots of f(x) = 0 are approximately x = 3, x = 3 + sqrt(19), and x = 3 – sqrt(19). d.) Sketch the graph of y = f(x), indicating the x-intercepts, y-intercept, turning points, and end behavior. To sketch the graph of y = f(x), we first find the x-intercepts by solving f(x) = 0: f(x) = x3 – 3x2 – 4x + 12 f(x) = (x – 3)(x2 – 4x – 4) x = 3 or x = 2 + sqrt(6) or x = 2 – sqrt(6)\ Next, we find the y-intercept by evaluating f(0) = -1. We also find the turning points by setting f'(x) = 0 and solving for x: f'(x) = 3x2 – 6x – 4 f'(x) = 3(x – 2)(x + 2) x = 2 or x = -2 Finally, we can sketch the graph of y = f(x) as follows: e.) Find the equation of the tangent line to the graph of y = f(x) at the point where x = 2. To find the equation of the tangent line to the graph of y = f(x) at the point where x = 2, we first find the slope of the tangent line by taking the derivative of f(x) and evaluating it at x = 2: f'(x) = 3x2 – 6x – 4 f'(2) = 8 So the slope of the tangent line is 8. Next, we find the y-coordinate of the point of tangency by evaluating f(2) = 2: f(2) = 2 Therefore, the equation of the tangent line to the graph of y = f(x) at the point where x = 2 is: y – 2 = 8(x – 2) y = 8x – 14 ## Our Expert Tutors! ### Jinu James IBDP Physics ap physics c mechanics IBDP MATH AAHL ap calculus bc ### Shimaila Mushtaq IBDP Economics ap microeconomics IBDP Chemistry AP Chemistry IBDP Psychology AP Psychology Edit Template View Here get it here Explore Here Know More Edit Template
# Video: KS2-M17 • Paper 2 • Question 3 Write the missing numbers to make this multiplication grid correct. 02:38 ### Video Transcript Write the missing numbers to make this multiplication grid correct. We can see that inside the multiplication grid, there are four two-digit numbers. And each of these numbers is made by multiplying two of the numbers round the outside, the number from the same row and the number from the same column. For example, the number 63 is found by multiplying nine by this number here. Notice that both of the numbers in the top row are multiples of nine, they’re nine times something. This is why it’s best to start with these two numbers before moving on to the bottom two. We don’t know any of the multiples of the bottom two numbers yet. So we’ll begin with 63. How many nines are in 63? Nine, 18, 27, 36, 45, 54, 63. Nine multiplied by seven equals 63. As we’ve said already, the number 54 is in the same row. We find the answer by multiplying nine by something. We know that 54 is one lot of nine less than 63. So there must be six nines in 54: nine, 18, 27, 36, 45, 54. Now we’ve worked out the numbers in the top row, we can use them to find the missing number in the side of the grid. The number 56 is in the same column as the number seven. What do we multiply seven by to give us 56 or how many sevens are in 56? Seven, 14, 21, 28, 35, 42, 49, 56. Eight times seven equals 56. And we know that missing number must be eight because eight times six is 48. To find the missing numbers, we started with the top row because we already knew that one of the multiples was nine. We looked carefully at each row and column and we used our knowledge of times tables to find the missing answers. So from top to bottom and left to right, the missing numbers are seven, six, and eight.
# How do you write y=4/3x+2/3 in standard form? Jun 28, 2015 $4 x - 3 y = - 2$ #### Explanation: The standard form of a linear equation is: $A x + B y = C$ $A$ can not be negative. $A$, $B$ and $C$ should all be integers. The first thing we should do is move the $x$ over to the left part of the equation. You can do this by substracting $\frac{4}{3} x$ from both parts: $y - \frac{4}{3} x = \frac{4}{3} x + \frac{2}{3} - \frac{4}{3} x$ $y - \frac{4}{3} x = \frac{2}{3}$ By reordering, you get: $- \frac{4}{3} x + y = \frac{2}{3}$ Now we need to make sure that the number that's before the $x$ ($A$) is positive. You can do this by multiplying both parts by $- 1$: $- 1 \cdot \left(- \frac{4}{3} x + y\right) = - 1 \cdot \frac{2}{3}$ $\frac{4}{3} x - y = - \frac{2}{3}$ Now, all we need to do is make A, B and C integers. You can always do this by multiplying by the LCM of all the denominators ($3$, $1$ and $3$). This LCM is $3$: $3 \cdot \left(\frac{4}{3} x - y\right) = 3 \cdot \left(- \frac{2}{3}\right)$ $4 x - 3 y = - 2$
# Difference between revisions of "Binomial Theorem" The Binomial Theorem states that for real or complex $a$, $b$, and non-negative integer $n$, $(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is a binomial coefficient. This result has a nice combinatorial proof: $(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}$. Repeatedly using the distributive property, we see that for a term $a^m b^{n-m}$, we must choose $m$ of the $n$ terms to contribute an $a$ to the term, and then each of the other $n-m$ terms of the product must contribute a $b$. Thus, the coefficient of $a^m b^{n-m}$ is $\binom{m}{n}$. Extending this to all possible values of $m$ from $0$ to $n$, we see that $(a+b)^n = \sum_{k=0}^{n}{\binom{n}{k}}\cdot a^k\cdot b^{n-k}$. ## Generalizations The Binomial Theorem was generalized by Isaac Newton, who used an infinite series to allow for complex exponents: For any real or complex $a$, $b$, and $r$, $(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k$. ### Proof Consider the function $f(b)=(a+b)^r$ for constants $a,r$. It is easy to see that $\frac{d^k}{db^k}f=r(r-1)\cdots(r-k+1)(a+b)^{r-k}$. Then, we have $\frac{d^k}{db^k}f(0)=r(r-1)\cdots(r-k+1)a^{r-k}$. So, the Taylor series for $f(b)$ centered at $0$ is $$(a+b)^k=\sum_{k=0}^\infty \frac{r(r-1)\cdots(r-k+1)a^{r-k}b^k}{k!}=\sum_{k=0}^\infty \binom{r}{k}a^{r-k}b^k.$$ ## Usage Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial $x^5+4x^4+6x^3+4x^2+x$, one could factor it as such: $x(x^4+4x^3+6x^2+4x+1)=x(x+1)^{4}$. It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.
Courses Courses for Kids Free study material Offline Centres More Store # How do you solve ${{x}^{2}}-6x-11=0$ by completing the square? Last updated date: 23rd Jun 2024 Total views: 374.1k Views today: 11.74k Verified 374.1k+ views Hint: Assume the expression ${{x}^{2}}-6x-11=y$ and compare it with the general form given as: - $y=a{{x}^{2}}+bx+c$. Find the respective values of a, b and c. Now, find the discriminant of the given quadratic equation by using the formula: - $D={{b}^{2}}-4ac$, where D = discriminant. Now, write the expression as: - $y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]$ and substitute it equal to 0 to find the two values of x. Complete step by step solution: Here, we have been provided with the quadratic equation: - ${{x}^{2}}-6x-11=0$ and we are asked to solve it. That means we have to find the values of x. We have been asked to use completing the square method. Now, we know that any quadratic equation of the form $y=a{{x}^{2}}+bx+c$ can be simplified as $y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]$, using completing the square method. Here, ‘D’ denotes the discriminant. So, on assuming ${{x}^{2}}-6x-11=y$ and comparing it with the general quadratic equation, we get, $\Rightarrow$ a = 1, b = -6, c = -11 Applying the formula for discriminant of a quadratic equation given as, $D={{b}^{2}}-4ac$, we get, \begin{align} & \Rightarrow D={{\left( -6 \right)}^{2}}-4\left( 1 \right)\left( -11 \right) \\ & \Rightarrow D=36+44 \\ & \Rightarrow D=80 \\ \end{align} Therefore, substituting the values in the simplified form of y, we get, \begin{align} & \Rightarrow y=1\left[ {{\left( x+\left( \dfrac{-6}{2\times 1} \right) \right)}^{2}}-\dfrac{80}{4\times {{\left( 1 \right)}^{2}}} \right] \\ & \Rightarrow y=\left[ {{\left( x-3 \right)}^{2}}-20 \right] \\ \end{align} Substituting y = 0, we get, $\Rightarrow \left[ {{\left( x-3 \right)}^{2}}-20 \right]=0$ \begin{align} & \Rightarrow {{\left( x-3 \right)}^{2}}-20=0 \\ & \Rightarrow {{\left( x-3 \right)}^{2}}=20 \\ \end{align} Taking square root both the sides, we get, \begin{align} & \Rightarrow \left( x-3 \right)=\pm \sqrt{20} \\ & \Rightarrow x-3=\pm 2\sqrt{5} \\ & \Rightarrow x=3\pm 2\sqrt{5} \\ \end{align} Considering the two signs one – by – one, we get, $\Rightarrow x=\left( 3+2\sqrt{5} \right)$ or $x=\left( 3-2\sqrt{5} \right)$ Hence, the solutions of the given quadratic equation are: $x=\left( 3+2\sqrt{5} \right)$ and $x=\left( 3-2\sqrt{5} \right)$. Note: One may also use the quadratic formula to solve the question and check if we are getting the same values of x. Note that the general expression of this completing the square method is also known as the vertex form. Generally, this form is used in coordinate geometry of parabola to find the vertex of the parabola which is given as $\left( \dfrac{-b}{2a},\dfrac{-D}{4a} \right)$. Note that the quadratic formula: - $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is derived from completing the square method. Here it would be difficult to use the middle term split method because it may be difficult to think of the factors like: $\left[ x-\left( 3+2\sqrt{5} \right) \right]$ and $\left[ x-\left( 3-2\sqrt{5} \right) \right]$.
Courses Courses for Kids Free study material Offline Centres More Store # If $\sec \theta =x+\dfrac{1}{4x}$, prove that $\sec \theta +\tan \theta =2x$ or $\dfrac{1}{2x}$. Last updated date: 09th Aug 2024 Total views: 457.5k Views today: 10.57k Verified 457.5k+ views Hint: We have been given $\sec \theta =x+\dfrac{1}{4x}$. So use the formula ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta$ and simplify. You will get the value of $\tan \theta$. After that add $\sec \theta$ and $\tan \theta$. You will get the answer. Now taking $\sec \theta =x+\dfrac{1}{4x}$, We have been given $\sec \theta$ and from that we will find $\tan \theta$. We know that ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta$. So substituting value of $\sec \theta$ in above identity we get, ${{\left( x+\dfrac{1}{4x} \right)}^{2}}-1={{\tan }^{2}}\theta$ Simplifying we get, \begin{align} & {{x}^{2}}+2(x)\dfrac{1}{4x}+{{\left( \dfrac{1}{4x} \right)}^{2}}-1={{\tan }^{2}}\theta \\ & {{x}^{2}}+\dfrac{1}{2}+\left( \dfrac{1}{16{{x}^{2}}} \right)-1={{\tan }^{2}}\theta \\ & {{x}^{2}}+\left( \dfrac{1}{16{{x}^{2}}} \right)-\dfrac{1}{2}={{\tan }^{2}}\theta \\ & {{\left( x-\dfrac{1}{4x} \right)}^{2}}={{\tan }^{2}}\theta \\ \end{align} So taking square root of both sides we get, $\tan \theta =\pm \left( x-\dfrac{1}{4x} \right)$ We get, $\tan \theta =\left( x-\dfrac{1}{4x} \right)$ and $\tan \theta =-x+\dfrac{1}{4x}$ So now we have got $\tan \theta$. Now adding $\tan \theta$ and $\sec \theta$, $\sec \theta +\tan \theta =x+\dfrac{1}{4x}\pm \left( x-\dfrac{1}{4x} \right)$ $\sec \theta +\tan \theta =x+\dfrac{1}{4x}+\left( x-\dfrac{1}{4x} \right)$ or $\sec \theta +\tan \theta =x+\dfrac{1}{4x}-\left( x-\dfrac{1}{4x} \right)$ Simplifying we get, $\sec \theta +\tan \theta =2x$ or $\sec \theta +\tan \theta =\dfrac{1}{2x}$ So we get the values $\sec \theta +\tan \theta =2x$ or $\sec \theta +\tan \theta =\dfrac{1}{2x}$. Hence proved. Note: Read the question carefully. Do not make any silly mistakes. Also, you must be familiar with the trigonometric identities. Do not confuse yourself while simplifying. Also, take care that no terms are missing and do not jumble with the signs.
#### Integers - Absolute Value Let’s meet another relative of the integer family, Absolute Value. It’s important to understand absolute value before learning how to add and subtract integers. Absolute value refers to the distance a number is from 0 on the number line. -5…-4…-3…-2…-1…0…1…2…3…4…5 Notice if we were to hop backwards on the number line starting from -5, -5 is five hops from zero. “5” is also 5 hops from zero. Also, it does not make sense to say the “-5” is -5 units away from zero. Again, remember absolute value refer to the distance from zero. Absolute value is denoted by two vertical bars. “\| \|” So, \|-5\| = 5 and \|5\| = 5. Try these: \|-43\| = \|-1023\| = \|23\| = \| -4/ 5\| = That’s it! Basically, you just strip the number of its sign. However, here is a more fun way to think of absolute value. 1st Scenario: Think of the absolute value symbol as arms of a referee. There is a dispute in progress in another room. “Absolute”, the referee, enters the room, raises her arms, and settles the dispute. Generally, the argument is about the same issue over and over. Her cousins Positive and Negative are always arguing about which sign their addition or subtraction problem answers will possess. So, good old Absolute Value comes in and saves the day. OR 2nd Scenario: A Positive number and Negative number were caught fighting and they were put behind bars “\| \|”. As a result, Absolute Value stripped them of all accessories including signs. They were left with nothing but themselves. \|-43\| = 43 \|-1023\| = 1023 \|23\| = 23 \| -4/ 5\| = 4/5 Note: The absolute value of a number is never negative. Mastering Essentials Math Skills - Review Integers - Positive and Negative Numbers How to Work Word Problems - Review Related Articles Editor's Picks Articles Top Ten Articles Previous Features Site Map
# WORD PROBLEMS ON SIMPLE EQUATIONS 1 ## About "Word problems on simple equations 1" Word Problems on Simple Equations 1 : This is the continuity of our web page word problems on simple equations. ## Word Problems on Simple Equations 1 - Examples Example 1 : If a number of which the half is greater than 1/5 th of the number by 15, find the number. Solution : Let "x" be the required number. Half of the number = x/2 1/5 the of the number = 1/5 ⋅ x = x/5 Given : Half of a number is greater than 1/5 th of the number by 15. So, we have x/2 - x/5 = 15 L.C.M of (2 , 5) is 10. Make each denominator as 10 sing multiplication. 5x/10 - 2x/10 = 15 (5x - 2x) / 10 = 15 3x / 10 = 15 Multiply both sides by 10/3. x = 50 Hence, the required number is 50. Example 2 : Three persons A, B and C together have \$51. B has \$4 less than A. C has got \$5 less than A. Find the money that A, B and C have. Solution : Let "x" be the money had by A, Then, we have A = x B has \$4 less than A ------> B = x - 4 C has got \$5 less than A -------> C = x - 5 Given : A, B and C together have \$51 So, we have A + B + C = 51 x + (x - 4) + (x - 5) = 51 x + x - 4 + x - 5 = 51 Simplify. 3x - 9 = 51 Add 9 to both sides. 3x = 60 Divide both sides by 3. x = 20 Then, x - 4 = 20 - 4 = 16 x - 5 = 20 - 5 = 15 Hence, A, B and C have \$20, \$16 and \$15 respectively. Example 3 : The average age of three boys is 25 years and their ages are in the proportion 3 : 5 : 7. Find the age of the youngest boy. Solution : From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x. Given : Average age of the three boys is 25 years. So, we have (3x + 5x + 7x) / 3 = 25 Simplify. 15x / 3 = 25 5x = 25 Divide both sides by 5. x = 5 Then, ages of the three boys are 3x = 3 ⋅ 5 = 15 5x = 5 ⋅ 5 = 25 7x = 7 ⋅ 5 = 35 Hence, the age of the youngest boy is 15 years. Example 4 : The ratio of the no. of boys to the no. of girls in a school of 720 students is 3 : 5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3. Solution : Sum of the terms in the given ratio 3 : 5 is = 3 + 5 = 8 So, no. of boys in the school is = 3/8 of total students = 3/8 ⋅ 720 = 270 No. of girls in the school is = 5/8 of total students = 5/8 ⋅ 720 = 450 Let "x" be the no. of new boys admitted in the school. Given : No. of new girls admitted is 18. After the above new admissions, no. of boys in the school = 270 + x no. of girls in the school = 450 + 18 = 468 Given : The ratio after the new admission is 2 : 3. So, we have (270 + x) : 468 = 2 : 3 (270 + x) / 468 = 2 / 3 Simplify. ⋅ (270 + x) = 2 ⋅ 468 810 + 3x = 936 Subtract 810 from both sides. 3x = 126 Divide both sides by 3. x = 42 Hence the no. of new boys admitted in the school is 42. Example 5 : The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves \$50 per month, find the monthly incomes of the two persons. Solution : From the income ratio 4 : 5, Income of the 1st person = 4x Income of the 2nd person = 5x Using the fact Expenditure = Income - Savings, we can find the expenditure of the two persons as given below. Then, expenditure of the 1st person is = 4x - 50 Expenditure of the 2nd person is = 5x - 50 Given :Expenditure ratio is 7 : 9 So, we have (4x - 50) : (5x - 50) = 7 : 9 (4x - 50) / (5x - 50) = 7 / 9 Simplify. ⋅ (4x - 50) = 7 ⋅ (5x - 50) 36x - 450 = 35x - 350 x = 100 Then, we have 4x = 4 ⋅ 100 = 400 5x = 5 ⋅ 100 = 500 Hence, the monthly incomes of the two persons are \$400 and \$500. 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How do you factor a polynomial over a field? How do you factor a polynomial over a field? How do you factor a polynomial over a field? Why does it matter? Prove that the polynomial with rational coefficients is polynomial. A: The basic idea is that you can evaluate two polynomials $$P_1(x) = x + i \, \frac{x^3}{3}$$ You can do this by using the Bessel function, but it is not very efficient. In terms of modulus, you can evaluate \begin{align} P_2(x) &= x + i\, \frac{\sqrt{x}}{x^2} \\ &= \sqrt{2} x + i y + i\frac{\sq\sqrt{-x}}{2} \\ &= -2 x – i\frac{x}{\sqrt x}y \\ &= 2 x + iy \\ &={}x^2 + i\sqrt{\frac{x}x}y \end{align} The answer is $$-i\sqrt\frac{i}{\sq\sq x}y = {}1 \implies \sq\sq{-i\frac{1}{x}}=\sq\frac{-i}{\frac{3x}{\frac{\pi^2x}x^3}} = \frac{-\sq\pi}{\sq x}\implies {\sqrt{i}y\sq{} = -\sq\{-\frac{\frac{1}x}{\pi}\}}\implies i\sq r = -i\sq x\implies r = -\frac{4x}{\left(\frac{\pi}{3}\right)}$$ How do you factor a polynomial over a field?” “The answer is yes.” An example illustrating this is the polynomial $f(x) = x^n$ over a field. This is the poenthesis $$f(x)=\frac{1}{n}x^{n-1}x^n.$$ The poenthesis is $$x(x)=x^n=\frac{n^2}{2},$$ and it is a series in $x$. If you are new to group theory, you’ll probably have a second to last line of your code. But, if you’re a beginner, you may be familiar with this simple formula and why it is useful for your research. The form $$\lim_{x\rightarrow\infty}f(x)(x^n+1)=\frac1{n(n-1)}\left(1+\sum_{k=0}^n2^{k-1}\cdot\frac{(n-k)^k}{k!}\right)$$ is the expression of the limit. This formula is called the limit. It can be used for many purposes, including investigating the limit of $x^n$ at a very high rate, to get an expression like this. Note that the result is the limit of the series. You may also want to consider other variants of this formula. For example the limit for $x$ expressed in terms of the poentheses of $x$ is the limit for a polynomially infinite series. Chapter 2 of the book “Algebraic Number Theory” by T. Klemperer is devoted to the complex visit site Unicorn A number is a collection of independent variables, and a closed set $Z$ consists of all its elements. It can also be viewed as a set of polynomials in the variables of a number in the set $Z$. In other words, the set of all valid polynomial functions of $\mathbb{C}$ is the set of pairs $(f,g)$ such that $f(2^{n-3})=g(n-3)$. A set of poxial functions is called a polynometry if there exists a polynoid $X$ such that $Xf(x)+1=\sum_{n=0}^{x-1}f(2n)$ for all $f$ in $X$ This is a polynotope. There are many different polynomontheses of the form $Xf(2^m)$ for some $f$ from the set of ponomial functions from the set $X$ of polynomial points. In other words: a polynomial in $x$ has the form $x^m$ with a polynoma in $x$, where $m$ is the number of polynomic points in $x^2$. The formula for $Xf$ is the same as for $X$: $$Xf(y) = 2^{y}y^m.$$ The function $y$ is called the number of choices for the polynomomial $f$. Now a number $m$ of a group is called a group point, and we say that $m$ points in $Z$ are called a group element. For example, $$(2x^2)(2x^3) = (2x)^4,$$ or $$2x^4 = (2)^5,$$ such that $$y = 2x^How do you factor a polynomial over a field? Like most of us, I have to work out my you can check here to a lot of these questions. I’m a bit confused here. How do we factor a poomial over a finite field? My answer to this question is not quite right. I think we have to do it in a different way. Let’s look at the most basic example we can think of — a polynomials over a field. First, we get to a simple algebraic identification. Here’s where we get to the “first homotopy type of a polynoma”. A polynomial f over a field F is “primitive” iff there exists a real number c such that f(c)=1 and f(0)=0. Then we can think with the polynomial notation. We can say that more info here polynism p is primitive iff there is a real number \delta such that for all p with p(0)=1, p(p(p))=\delta. Here“\delta” is the “distance” between p(x) and x, i.e., for all x in a field k, p(\delta(x))=\frac{1}{2}(p(x)-p(x)) This is just a finite set. Now we define a “\Delta-class” of polynomies over a given field F. Let F be a finite field. Homework Sites Then, f is ‘primitive’ iff for every p in F, there is an \Delta-function C such read the article the following holds: for all f with f(p)>1 and all x\in F,$$C(x)\geq f(x)$$Here, \Delta is defined by C(x)=\frac{f(x)}{f(x-1)} In other words, f(x)\leq\Delta(x). In a similar way, we can define the “k-class of polynomial”. The “K-class k-modulo F” of a poomials over F over k is defined as the smallest k such that Here \Delta$$\begin{array}{c\|c} K&=&\frac{(k-1)^2}{(k-2)(k-3)} \\\ Related Post What is the Microsoft Certification job title? Do you have How do you find the nth term of a sequence? What is an accrued expense? In the first place, every How do you calculate the cost of goods sold? It What is the endocrine pancreas? What happens to blood in What is virtualization? Virtualization is a technology that can be What was the significance of the Battle of Little Bighorn? What is your experience with project scheduling? 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DISCOVER How to Prove a Right Angle in Math Updated February 21, 2017 A right angle is a 90 degree angle that forms where perpendicular 180 degree lines meet or cross. Proving that a polygon such as a square, rectangle or triangle has a right angle requires knowledge of shape properties, lines, sides, congruency, symbols and measurement tools. Proving right angles is a geometry technique that can help you solve problems where you have to find the measurement of other angles within a shape. Find an angle that is labelled 90 degrees. Knowing that a right angle is worth 90 degrees is proof that you have identified a right angle. Look at a diagram of a shape to see if it has an angle with a small square drawn in it. The small square symbolises that an angle is worth 90 degrees, proving a right angle. Study the types of lines so that you can identify perpendicular lines, which always meet or cross at a right angle. Prove that the angles of squares and rectangles are made by perpendicular lines and that perpendicular lines create right angles. Draw a small square in the angles of squares and rectangles to prove they have four right angles, each worth 90 degrees. Learn the properties of quadrilaterals as having 360 degrees and that a geometry rule of squares and rectangles is that they have four equal angles. Divide 360 total degrees by four angles in rectangles and squares to prove that each angle is 90 degrees; hence a right angle. Identify and prove a right angle in an isosceles right triangle by understanding that an isosceles triangle has two legs of the same measurements and a base of a different measurement. The two legs provide the same angle measurement. Find the measurement of one leg as 45 degrees, which means that congruent leg is 45 degrees. Add the leg measurements to get 90 degrees. Subtract 90 degrees from a total of 180 degrees, to prove that the base angle is a 90 degree right angle. Identify and prove a scalene right angle by adding the two different angle measurements given and subtracting the total from 180 to find a right angle. Scalene triangles have three different angle measurements. Look at a straight leg and the base to see if they connect at a 90 degree angle. Measure an angle with a protractor to prove that it is a right angle. Place the hole of the protractor, found on the straight edge of the tool, in the centre of the vertex of the right angle in question. Look at the numbers lining the curved edge of the protractor. Prove a right angle by showing that the angle lines up to the 90 degree marking on the protractor. • Protractor
# How Long ’till Brayden is 6′? -3Act Math ACT 1: Brayden celebrated his 6th birthday this May. Every year we mark his height on the shed door frame. What questions do you have when you look at this picture? Ask students to write down their questions, I normally ask students to find at least 3. When I observe that most students have questions written, I ask them to share those questions with their neighbor. I then throw up a Microsoft Word document and start typing down questions students supply. Students from my classroom came up with all sorts of different questions, some we can easily answer and others that we can’t. I am looking for a key question or questions to start this lesson. If students do not ask one of these questions, I tell them that I hope I can answer most of the questions provided, but that I need them to consider one of these questions first. • How tall is Brayden? • How much does he grow in a year? • How tall will he be next year? • How tall will he be as an adult? Any of these type of questions will lead students down the inquiry I hope to explore with them. My wife and I are fairly tall people, and Brayden wants to be at least as big as us so he can play basketball (he seems to think he needs to be tall to play). How many years will it be before Brayden is 6′? ACT 2: Scatter Plots, Line of Best Fit and Estimation are the topics I wish to address with my students. Students will need to think of ways to determine his height or growth between years. There are many different combinations of ways to figure this number out, but here are the stats: • At age 6, Brayden is actually 45″ • At age 5, Brayden is actually 42″ • At age 4, Brayden is actually 39.5″ • At age 3, Brayden is actually 36.5″ • At age 2, Brayden is actually 33″ • At age 1, Brayden is actually 29″ Students should construct a scatter plot similar to this: And draw a line of best fit similar to this: I also ask students to create an equation for the Brayden’s height growth per year (he was 22 inches when he was born). Some students will work with average on this problem, so they will take the average height grown from 0-6, then use that growth average to complete a height table: 45-22 = 23 23/6=3.83333333… Students will round that to 3.8 inches per year. ACT 3: Using Excel for drafting the line of best fit, it gave me this equation: So, solving for y=72 gives me Brayden’s age of 13. Using the average model, students will get the same answer. Extensions: After we decide the general equation for Brayden’s height, I will ask students to predict Brayden’s height at various points in his life (16, 18, 21, 30, 40, 70). Some students will just blindly evaluate what his height will be without considering the fact that we do stop growing as adolescents.
# NCERT Solutions For Class 6 Maths Chapter 12 Exercise 12.1 ## Ncert Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.1:- Exercise 12.1Â Class 6 maths NCERT solutions Chapter 12 Ratio And Proportion pdf download:- Â Â ### Ncert Solution for Class 6 Maths Chapter 11 Ratio And Proportion Exercise 12.1 Tips:- Introduction In our daily life, many times we compare two quantities of the same type. For example, Avnee and Shari collected flowers for scrap notebook. Avene collected 30 flowers and Shari collected 45 flowers. So, we may say that Shari collected 45 â€“ 30 = 15 flowers more than Avnee. Also, if the height of Rahim is 150 cm and that of Avene is 140 cm then, we may say that the height of Rahim is 150 cm â€“ 140 cm = 10 cm more than Avnee. This is one way of comparison by taking the difference. If we wish to compare the lengths of an ant and a grasshopper, taking the difference does not express the comparison. The grasshopperâ€™s length, typically 4 cm to 5 cm is too long as compared to the antâ€™s length which is a few mm. The comparison will be better if we try to find how many ants can be placed one behind the other to match the length of grasshopper. So, we can say that 20 to 30 ants have the same length as a grasshopper. Consider another example. Cost of a car is 2,50,000 and that of a motorbike is ` 50,000. If we calculate the difference between the costs, it is 2,00,000 and if we compare by division; i.e. 2,50,000/50,000=5/1 We can say that the cost of the car is five times the cost of the motorbike. Thus, in certain situations, comparison by division makes better sense than comparison by taking the difference. The comparison by division is the Ratio. Ratio Consider the following: Ishaâ€™s weight is 25 kg and her fatherâ€™s weight is 75 kg. How many times Fatherâ€™s weight is of Ishaâ€™s weight? It is three times. Cost of a pen is ` 10 and the cost of a pencil is ` 2. How many times the cost of a pen that of a pencil? Obviously, it is five times. In the above examples, we compared the two quantities in terms of how many timesâ€™. This comparison is known as the Ratio. We denote ratio using symbol â€˜:â€™ Consider the earlier examples again. We can say, The ratio of fatherâ€™s weight to Ishaâ€™s weight = 75/25=3/1= 3:1 The ratio of the cost of a pen to the cost of a pencil = 10/2 = 5/1= 5:1 try these 1. Find the ratio of the number of notebooks to the number of 2. Find the ratio of the number of desks and chairs in your classroom. 3. Find the number of students above twelve years of age in your class. Then, find the ratio of the number of students with age above twelve years and the remaining students. 4. Find the ratio of the number of doors and the number of windows in your classroom. 5. Draw any rectangle and find the ratio of its length to its breadth. #### Test Paper Of Class 8th • Maths 8th Class • Science 8th class • Sst 8th Class • #### Test Paper Of Class 7th • Maths 7th Class • Science 7th class • #### Test Paper Of Class 6th • Maths 6th Class • Science 6th class
# How to graph cubics? ## () For question a, can somebody please tell me how to get to the answer in the photo? Apr 9, 2017 $y = {x}^{3} - 3 {x}^{2} + 3 x - 1$ #### Explanation: The general equation of a cubic polynomial is: $y = a {x}^{3} + b {x}^{2} + c x + d$ From looking at the graph in the question labelled '(a)', it appears that: $y = - 1$ at $x = 0$ (1) $y = 0$ at $x = 1$ (2) Slope of $y = 0$ at $x = 1$ (3) $y$ has an inflection point at $x = 1$ (4) Assuming (1) through (4) above are true: $\left(1\right) \to d = - 1$ {A] $\left(2\right) \to a + b + c + d = 0$ [B] $y ' = 3 a {x}^{2} + 2 b x + c$ Hence $\left(3\right) \to 3 a + 2 b + c = 0$ [C] $y ' ' = 6 a x + 2 b$ Hence $\left(4\right) \to 6 a + 2 b = 0$ $\therefore b = - 3 a$ [D} [D] in [C} $\to 3 a - 6 a + c = 0$ $\therefore c = 3 a$ [E] [A], [D] and [E] in [B] $\to a - 3 a + 3 a - 1 = 0$ Hence: $a = 1$ $\to b = - 3$ and $c = 3$ $\therefore$ our cubic is: $y = {x}^{3} - 3 {x}^{2} + 3 x - 1$
# Limiting parallel The two lines through a given point P and limiting parallel to line R. In neutral or absolute geometry, and in hyperbolic geometry, there may be many lines parallel to a given line ${\displaystyle l}$ through a point ${\displaystyle P}$ not on line ${\displaystyle R}$; however, in the plane, two parallels may be closer to ${\displaystyle l}$ than all others (one in each direction of ${\displaystyle R}$). Thus it is useful to make a new definition concerning parallels in neutral geometry. If there are closest parallels to a given line they are known as the limiting parallel, asymptotic parallel or horoparallel (horo from Greek: ὅριον — border). For rays, the relation of limiting parallel is an equivalence relation, which includes the equivalence relation of being coterminal. Limiting parallels may form two, or three sides of a limit triangle. ## Definition The ray Aa is a limiting parallel to Bb, written: ${\displaystyle Aa\|\|\|Bb}$ A ray ${\displaystyle Aa}$ is a limiting parallel to a ray ${\displaystyle Bb}$ if they are coterminal or if they lie on distinct lines not equal to the line ${\displaystyle AB}$, they do not meet, and every ray in the interior of the angle ${\displaystyle BAa}$ meets the ray ${\displaystyle Bb}$.[1] ## Properties Distinct lines carrying limiting parallel rays do not meet. ### Proof Suppose that the lines carrying distinct parallel rays met. By definition the cannot meet on the side of ${\displaystyle AB}$ which either ${\displaystyle a}$ is on. Then they must meet on the side of ${\displaystyle AB}$ opposite to ${\displaystyle a}$, call this point ${\displaystyle C}$. Thus ${\displaystyle \angle CAB+\angle CBA<2{\text{ right angles}}\Rightarrow \angle aAB+\angle bBA>2{\text{ right angles}}}$. Contradiction.
hw2sol # hw2sol - UCSD ECE 153 Handout#11 Prof Young-Han Kim... This preview shows pages 1–3. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: UCSD ECE 153 Handout #11 Prof. Young-Han Kim Thursday, April 14, 2011 Solutions to Homework Set #2 (Prepared by TA Lele Wang) 1. Polya’s urn. Suppose we have an urn containing one red ball and one blue ball. We draw a ball at random from the urn. If it is red, we put the drawn ball plus another red ball into the urn. If it is blue, we put the drawn ball plus another blue ball into the urn. We then repeat this process. At the n-th stage, we draw a ball at random from the urn with n +1 balls, note its color, and put the drawn ball plus another ball of the same color into the urn. (a) Find the probability that the first ball is red. (b) Find the probability that the second ball is red. (c) Find the probability that the first three balls are all red. (d) Find the probability that two of the first three balls are red. Solution: Let X i denote the color of the i-th ball. (a) By symmetry, P { X 1 = R } = 1 / 2. (b) Again by symmetry, P { X i = R } = 1 / 2 for all i . Alternatively, by the law of total probability, we have P { X 2 = R } = P { X 1 = R } P { X 2 = R \| X 1 = R } + P { X 1 = B } P { X 2 = R \| X 1 = B } = 1 2 × 2 3 + 1 2 × 1 3 = 1 2 . (c) By the chain rule, we have P { X 1 = R,X 2 = R,X 3 = R } = P { X 1 = R } P { X 2 = R \| X 1 = R } P { X 3 = R \| X 2 = R,X 1 = R } = 1 2 × 2 3 × 3 4 = 1 4 . (d) Let N denote the number of red balls in the first three draws. From part (c), we know that P { N = 3 } = 1 / 4 = P { N = 0 } , where the latter identity follows by symmetry. Also we have P { N = 2 } = P { N = 1 } . Thus, P { N = 2 } must be 1 / 4. 1 Alternatively, we have P { N = 2 } = P { X 1 = B,X 2 = R,X 3 = R } + P { X 1 = R,X 2 = B,X 3 = R } + P { X 1 = R,X 2 = R,X 3 = B } = P { X 1 = B } P { X 2 = R \| X 1 = B } P { X 3 = R \| X 2 = R,X 1 = B } + P { X 1 = R } P { X 2 = B \| X 1 = R } P { X 3 = R \| X 2 = B,X 1 = R } + P { X 1 = R } P { X 2 = R \| X 1 = R } P { X 3 = B \| X 2 = R,X 1 = R } = 1 2 × 1 3 × 2 4 + 1 2 × 1 3 × 2 4 + 1 2 × 2 3 × 1 4 = 1 4 . 2. Uniform arrival. The arrival time of a professor to his office is uniformly distributed in the interval between 8 and 9 am. Find the probability that the professor will arrive during the next minute given that he has not arrived by 8:30. Repeat for 8:50. Solution: Let A be the event that ”the professor arrives between 8:30 and 8:31”, and let B be the event that ”the professor does not arrive before 8:30”. Then we have P ( A \| B ) = P ( A ∩ B ) P ( B ) = P ( A ) P ( B ) = 1 / 60 1 / 2 = 1 30 . Similarly, P ( { Professor arrives between 8:50 and 8:51 }\|{ He does not arrive before 8:50 } ) = 1 10 . 3. Probabilities from a cdf. The cdf of the random variable X is given by F X ( x ) = 1 , x > , 1 3 + 2 3 ( x + 1) 2 , − 1 ≤ x ≤ , , x < − 1 .... View Full Document {[ snackBarMessage ]} ### Page1 / 14 hw2sol - UCSD ECE 153 Handout#11 Prof Young-Han Kim... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# Forces In Physics These are my notes on forces in physics. Ankr Store on Amazon, keep your electronics charged by the best! f you buy something, I get a small commission and that makes it easier to keep on writing. Thank you in advance if you buy something. Intro In the development of mechanics, the first thing to learn is the interrelations of position, velocity, and acceleration. These interrelations are described with the four kinematic equations of motion. In this section, we relate constant force to constant acceleration and then to all of kinematics. Force can be defined as what makes masses accelerate. Actually, it is the unbalanced force on a mass that makes it accelerate. So, unbalanced forces make masses accelerate. Force, acceleration, and mass are related by Newton's second law as F = ma. Example 1 A 40 N force is applied to a 20 kg block resting on a horizontal frictionless table. Find the acceleration. Solution: The acceleration is $$a=\frac{f}{m} = \frac{40N}{20kg} = 2.0m/s^2$$ Once the acceleration is known, the kinematic equations of motion can be applied to find position, velocity, and acceleration as a function of time. Example 2 Now place the 20 kg mass on a frictionless 50 degree inclined plane. What is the acceleration of the mass? Solution: The force acting on the mass is due to gravity, so setup a vector diagram starting with this 196 N force acting down. The mass is constrained to move down the plane. The gravitational force is shown with components down the plane and normal to the plane. Notice the geometry of the situation, where the 50 degree angle of the plane is the same as the angle between the gravitational force and the normal force. So, the force straight down is $$F=20kg(9.8m/s^2)=196N$$ However, the mass is going down an incline plane at 50 degrees. $$F_{p}=(196N)(sin50)=150N$$ The 150 N force acting down the plane causes the 20 kg mass to accelerate at: $$a=\frac{150N}{20kg}=7.5m/s^2$$ Example 3 Place a 10 kg mass on a frictionless 35 degree incline plane, and attach a second 20 kg mass via a cord to hang vertically. Calculate the acceleration of the system. Solution: $$mg = 98N$$ $$98Nsin35 = 56N$$ $$98Ncos35 = 80N$$ Mass 1 is 10 kg at 35 degree down incline plane. Mass 2 is 20 kg and hangs straight down. It has a downward force of 196N. This problem introduces the concept of the tension in the connecting cord. The most convenient way to visualize this tension is that if the cord were cut and a force meter inserted, it would read a certain tension(force). This tension acts on each mass. Notice the 35 degree angle between the normal and the direction of mg. To write equations relating the unbalanced force to the acceleration, we need to make an assumption as to which way the masses move. Assume that the system moves to the right or clockwise. Start by writing equations relating the unbalanced force on each mass. On m1, the unbalanced force is T-56N, and this force makes the 10 kg accelerate. $$T-56N = 10kga$$ Likewise on mass2, the unbalanced force to make the mass accelerate is: $$196N-T = 20kga$$ The assumption that that system accelerates clockwise dictates the directions of the forces in these two equations. Look again at these equations, and notice that if we had assumed the acceleration were counterclockwise, then the signs would be different. In frictionless problems it is alright to have a negative acceleration. The negative sign means that you guessed wrong in assuming the acceleration direction. This is not true with friction problems. Add the two equations, and take the acceleration as the same. The accelerations are the same because the masses are linked together with the cord. $$140N = 30kg * a \| a = 4.7m/s^2$$ Now the tension in the cord is easily calculated from the equation for mass 1. $$T = 56N + 10kg * 4.7m/s^2 = 103N$$ Example 4 The next complication in force problems with inclined planes is a double inclined plane. Mass 1 is 12 kg going down to the left of the apex at a 40 degree angle. Mass 2 is 20 kg going down to the right of the apex at a 25 degree angle. Solution: Create a vector diagram, it is helpful. Assume the system moves clockwise. We will start with mass 1 first. The force straight down is 118 N, from (12 kg * 9.8 m/s^2). Now, lets calculate the force down the incline plane to the left. 118 N * sin 40 = 76 N Now, lets go to mass 2. The force straight down is 196 N, from (20 kg * 9.8 m/s^2). Now, lets calculate the force down the incline plane to the right. 196 N * sin 25 = 83 N The equation for mass 1 is: T-76N = 12 kg * a The equation for mass 2 is: 83N-T = 20 kg * a Add the two equations 7.0N = 32 kg * a a = 0.22 m/s^2 The tension in the cord is: T = 76N + 12 kg * 0.22 m/s^2 = 79 N Example 5 Another popular problem is the Atwood machine. Find the acceleration of the system. Solution: Assume that the system moves counterclockwise. The equation for mass 1 is 98N-T = 10kg * a The equation for mass 2 is T-147N = 15kg * a Add the two equations -49N = 25kg * a a = -2.0 m/s^2 Force of mass 1 is 10kg * 9.8 m/s^2 = 98N Force of mass 2 is 15kg * 9.8 m/s^2 = 147N The system accelerates clockwise. The tension in the cord is from the equation for mass 1. T = 98N + 10kg * 2.0m/s^2 = 118N Friction A good first problem in friction is where the problem is first done without friction and then with friction. For the purposes of doing problems, there are two important properties of frictional forces: They oppose the motion They are less than or equal to a constant (coefficient of friction), times the normal force, the force at the frictioning surface. Example 6 Two masses are arranged with one, of 50kg, on a frictionless table and the other, of 30 kg, attached by a cord and hanging, over a frictionless pulley, off the edge of a table. Find the acceleration of the system. Solution: Assume that the system moves clockwise. The equation for mass 1 is T=50kg * a The equation for mass 2 is 294N-T = 30kg * a Add the two equations: 294N = 80kg * a a = 3.7 m/s^2 Example 7 For the situation in Example 6, add a coefficient of friction \mu=.20 between the block and the table. Solution: As above, horizontal: \mu N = mumg = 0.2050kg9.8m/s^2 = 98N Vertical: 294N The equation for mass 1 is T-98N=50kga The equation for mass 2 is 294N-T=30kga Add the equations: 196N=80kga a=2.4m/s^2 Example 8 For the situation in the previous problem, increase the coefficient of friction to 0.80. What is the acceleration of the system? Solution: $$f=\mu N=\mumg=0.8050kg9,8m/s^2=392N$$ $$294N$$ The equation for mass 1 is $$T-392N=50kga$$ The equation for mass 2 is $$294-T=30kga$$ Add the two equations $$-98N=80kga$$ $$a=-1.2m/s^2$$ Clearly, the system does not accelerate counterclockwise. This is an illustration of the second property of frictional forces: the frictional force is less than or equal to $$\mu N$$. In this case the frictional force just balances the tension in the cord. The frictional force can be up to 392N. In order for mass 2 not to move, the tension in the cord must be 294N. If mass 2 is not moving, the forces must be in balance. Therefore, the forces on mass 1 must be 294N due to the tension in the cord and 294N due to friction. Another important point with frictional forces is the minimal force. Example 9 A 50kg sled is pulled along a level surface at constant velocity by a constant force of 200N at an angle of 30 degrees. What is the coefficient of friction between the sled and the surface? Solution: The first step in this problem is to find the components of the applied force and the mg of the sled. These are shown in the vector diagram. Notice the normal force, the actual force between the sled and the friction surface, is mg minus the component of the applied force lifting up the sled. Sometimes it is easy to miss this vertical component of the applied force. This is a major mistake in this type of problem. To help visualize this vertical component of force, try placing the origin for the vectors in the middle of the sled rather than at one end, where a rope would be attached. $$f = \muN$$ $$mg = 50kg9.8m/s^2 = 490N$$ $$200N * sin 30 = 100N$$ $$200N * cos 30 = 173N$$ The normal force is mg minus the vertical component of the applied force, 100N, or 390N total. Since the sled is moving at constant velocity, the forces must be in equilibrium. The horizontal component of the applied force must equal the frictional retarding force, the coefficient of friction times the normal force. $$\mu390N = 173N$$ $$\mu = 0.44$$ Example 10 A 65 degree inclined plane has a mass of 20kg on the plane where the coefficient of friction is 0.40 and a mass of 50kg hanging free. Calculate the acceleration of the system. Solution: Setup the vector diagram assuming that the motion is counterclockwise. In the diagram, the frictional force is calculated as a maximum of 33N. The equation for mass2 is $$490N-T=50kga$$ The equation for mass1 is $$T-178N=50kga$$ Add the equations: $$279N=70kga$$ $$a=4.0m/s^2$$ $$mg = 50kg9.8m/s^2 = 490N$$ Mass2 of 50kg is hanging straight down off a pulley. Mass1 of 20kg is sliding down an incline plane. The plane is a 65 degree angle. Coefficient of friction is 0.40 $$196N cos65=83N$$ $$196N sin65=178N$$ $$0.4083N=33N$$ In doing friction problems like this, you have to be careful with the frictional force. If the forces up and down the plane are nearly balanced, it is important to remember that the frictional force is not 33N but that it can be up to 33N. In a problem similar tot his, it could easily happen that the blocks would not move. Example 11 Consider a double inclined plane with angles, masses, and coefficients of friction as shown. Calculate the acceleration of the system. You have a sort of triangle with the apex at the top. To the left and right of the apex are inclined planes. On the left inclined plane, you have mass1 of 6.0kg, coefficient of friction of 0.30, and going down at an angle of 30 degrees. On the right inclined plane, you have mass2 of 9.0kg, coefficient of friction of 0.16, and sliding down at an angle of 45 degrees. Solution: Assume that the system moves counterclockwise, and set up the vector diagram. The equation for mass1 is $$(30-15-T)N=6.0kga$$ The equation for mass2 is $$(T-62-10)N=9.0kga$$ Add the equations $$-57N=15kga$$ $$59N sin30=30N$$ $$59N cos30=51N$$ $$f=0.3051N=15N$$ $$88N sin45=62N$$ $$88N cos45=62N$$ $$f=0.1662N=10N$$ This is a negative acceleration. We cannot simply reverse the sign of "a" to find the correct answer because if the system were moving the other way, the frictional forces would have to be reversed. further analysis requires us to consider another vector diagram assuming that the system moves in a clockwise direction. The new equations are $$(T-30-15)N=6.0kga$$ $$(62-T-10)N=9.0kga$$ Add the equations: $$7.0N=15kga$$ $$a=0.54m/s^2$$ The tension in the cord is from the first equation: $$T=45N+6.0kg0.54m/s^2)=48N$$ If the acceleration of the system were negative for both the clockwise and counterclockwise calculations, then we would conclude that the system would not move. The final complication in these inclined plane problems is the introduction of three masses and two connecting rods. Example 12 Consider a flat surface with a coefficient of friction of 0.20 with a 2.0kg mass and a 3kg mass connected together with a 6kg mass along a slant as shown. The coefficient of friction on the slant is 0. The slant is going down at an angle of 25 degrees. Calculate the acceleration of the system. Solution: In this problem, note that the tensions are different, leading to 3 unknowns, T1, T2, and a. The solution will require three equations from three vector diagrams. Assume that the acceleration of the system is clockwise. The frictional retarding force on the 2kg block is: $$f=\mu N=0.2019.6N=3.92N$$ The frictional retarding force on the 3kg block is: $$f=\mu N=0.2029.4N=5.88N$$ $$58.8N sin 25 =24.8N$$ The equations for the masses are: $$T2-3.92N=2.0kga$$ $$T1-T2-5.88N=3.0kga$$ $$24.8N-T1=6.0kga$$ The last two equations added together are: $$18.9N-T2=9.0kga$$ Adding this to the first equation is: $$15.0N=11kga$$ $$a=1.36m/s^2$$ The tensions come from the first and third equations for the masses: $$T2=3.92N+2.0kg1.36m/s^2=6.64N$$ $$T1=24.8N-6.0kg1.36m/s^2=16.6N$$ Circular Motion A particle moving at constant speed in a circle is moving in uniform circular motion. There is an acceleration and a force associated with such a particle. While the length of the velocity vector remains the same, the direction changes continually. Acceleration points radially inward. Example 13 A certain car driven in a circle can exert a maximum side force of 0.95g. What is the maximum speed for this car driven in a circle of 160m radius. Solution: A side force of 0.95g means a side force or acceleration directed at right angles to the direction of travel of 0.959.8=9.3m/s^2. Using $$a_{rad}=v^2/r$$: $$v=\sqrt{a_{rad} r} = \sqrt{9.3m/s^2160m} = 38.6m/s$$ At a speed greater than 38.6 m/s, the car will slide out of the circle. The acceleration directed toward the center of the circle for a mass in uniform circular motion is called centripetal acceleration. The force associated with moving a mass in a circle is $$ma_{rad}$$ and is called centripetal force. Example 14 What is the centripetal force produced by the tires acting on the pavement for the car of the previous problem if the car is 1200kg of mass. Solution: $$F=ma_{rad}=1200kg9.3m/s^2=11200N$$ Example 15 A 0.60kg rubber stopper is whirled in a horizontal circle of 0.80m radius at a rate of 3.0 revolutions per second. What is the tension in the string? Solution: Three revolutions per second means three $$2\pi r$$ circumferences per second, so: $$v=\frac{3.2 \pi 0.80}{1.0s}=15m/s$$ The tension in the string, which is providing the centripetal force, is: $$F=m\frac{v^2}{r} = 0.60kg\frac{(15m/s)^2}{0.80m} = 170N$$ Example 16 A 1 oz gold coin is placed on a turntable at 33 1/3 rpm. What is the coefficient of friction between the coin and the turntable if the maximum radius, before the coin slips, is 0.14m? Solution: The frictional force between the coin and turntable provides the center directed force to keep the coin on the turntable. This center directed force must equal $$mv^2/r$$. $$f=\mu N = \mu mg$$ This is a force balance problem. In equation form: frictional force=centripetal force, or: $$\mu mg=mv^2/r$$ The velocity is 2\pi r\) times the number of $$2\pi r's$$ per minute, 100/3 or: $$v=2\pi0.14m(100/3)(1/min)(min/60s)=0.49m/s$$ Therefore: $$\mu = \frac{v^2}{rg} = \frac{(0.49m/s)^2}{0.14m9.8m/s^2}=0.18$$ Example 17 A conical pendulum is a mass on the end of a cord, where the mass moves at constant speed in a circle with the cord tracing out a cone. A conical pendulum of length 1.2m moves in a circle of radius 0.20m. What is the period of the pendulum? Solution: Note that the mass is not given. Start with a vector diagram of the forces on the mass. The mg must equal the vertical component of the tension in the string $$T cos \theta = mg$$. The horizontal component is due to the centripetal force, so $$T sin \theta = mv^2/r$$. Dividing the second equation by the first eliminates T and m. The angle is from $$sin \theta=r/L$$, making $$tan \theta = 0.17$$ and: $$v=\sqrt{rg tan \theta} = \sqrt{.20m9.8.17}=.57m/s$$ The period is from $$2 \pi r/T=v$$ $$T=\frac{2\pi r}{v} = \frac{2\pi0.20m}{0.57m/s}=2.2s$$ Example 18 What is the speed, at which no side force due to friction is required, for a car traveling at a radius of 240m on a banked 20 degree road? Solution: Start with mg acting down. The normal force(between the car and the road) is normal to the surface, with the vertical component equal to mg and the horizontal component equal to $$mv^2/r$$. We have: $$N cos 20 = mg$$ and $$N sin 20 = mv^2/r$$ Then: $$\frac{N sin 20}{N cos 20} = tan 20 = \frac{v^2}{rg}$$ So: $$v = \sqrt{rg tan 20} = \sqrt{240m9.8tan 20} = 29m/s$$ Notice that in this problem the mass of the car does not enter into the calculation. Go back over problems 13-18 and notice how each problem could be written asking for a different variable. For example, in problem 13, give the side force and the velocity, and ask for the radius. In problem 16, give the speed and the coefficient of friction, and ask for the radius. Changing the problems like this is an excellent way to generate practice problems. Example 19 A 2kg ball is whirled in a vertical circle of radius 1m at a constant velocity of 6.28m/s. Calculate the tension in the cord at the top of the motion, the tension in the cord at the bottom of the motion, and the minimum velocity to keep the mass from falling out at the top of the circle. Solution: We are considering this problem from the reference frame of the ball, not from the point of view of the person whirling the ball. From the person's point of view, the tension is pulling radially outward. From the ball's referencew frame, the tension is radially inward. If the ball is to move in a circle, this must be the case. There must be a net center directed force. Operationally, a good method for solving centripetal force problems is to find the net center directed force and set this equal to $$mv^2/r$$. $$T+mg=\frac{mv^2}{r}$$ $$T=\frac{mv^2}{r}-mg$$ $$T=\frac{2.0kg6,28m/s^2}{1.0m} - 2.0kg(9.8m/s^2)=59.4N$$ $$T-mg=\frac{mv^2}{r}$$ $$T=\frac{mv^2}{r}+mg$$ $$T=\frac{2.0kg6.28m/s^2}{1.0m} + 2.9kg9.8m/s^2 = 98.6N$$ For part c, the minimum velocity required can be found by considering the condition where mv^2 is less than or equal mg. In this case, at the top of the motion, the mass will fall out of the loop. So let's set the two forces equal and solve for the minimum velocity needed to keep the mass in the loop. $$\frac{mv^2_{min}}{r} = mg \rightarrow v_{min} = \sqrt{rg} = \sqrt{1.0m*9.8}=3.13m/s$$ Example 20 A person on a dirt bike travels over a semicircular shaped bridge with a radius of curvature of 45m. Calculate the max speed of the bike if its road wheels are to stay in contact with the bridge. Solution: Again, let us find the net center directed force and set this equal to $$mv^2/r$$. So: $$mg-R=\frac{mv^2}{r}$$ and: $$R=mg-\frac{mv^2}{r}$$. As v increases, R must decrease because mg is constant. In the limiting case, when the wheels are just about to leave the ground, R=0, so $$mg=\frac{mv^2}{r}$$. The mass m cancels out and is not required. So max speed v is given by $$v^2=rg$$. Plugging in the numbers: $$v=\sqrt{rg} = \sqrt{441} = 21m/s$$ At this point, you might be thinking that from the point of view of the person on the dirt bike, the $$mv^2/r$$ force is radially outward. The rider is momentarily weightless as the wheels leave the ground. This is correct- from the reference frame of the person inside the circular motion. Astronauts in the space station orbiting Earth are weightless. To them, there is no net force, and the effect of the $$mv^2/r$$ force is radially outward to counterbalance gravity. If you go around a corner in a car, you seem to feel a force outward from the circle, not inward. But from the reference frame of a stationary observer, there must be a net center directed force on both the astronaut(gravity) and the person toward the center of the circle; otherwise, the person could not be moving in a circle. Is one force correct and the other incorrect? Is either of these an illusion? No, they are all real forces. Take a thoughtful trip in an airplane flying in a horizontal circle with its wings vertical. The force acting on you is toward the center of the circle, but you exert an equal and opposite force on the seat that is radially out of the circle. In some airplanes, if the banking is done too hard, this radially outward force can be so great as to the blood in the pilot to remain in the lower part of the body, resulting in a loss of oxygen to the brain and what is called a blackout. So the blackout is the result of a radially outward force. Is one reference frame correct and the other incorrect? When you study Einstein's theory or relativity in physics, you will come across the idea that there is no one absolute reference frame. You may think that a reference frame at rest relative to the ground is correct, but the Earth spins on its axis and rotates around the Sun. The Sun moves around the galactic center. In relativity, we always have to specify the reference frame we are considering. And the observed effects of many things change depending on the chosen reference frame. So, when doing circular motion problems, always think about what reference frame you are considering. For the method in the preceding problem, our reference frame is outside the motion, and in this perspective, we take the net center directed force and set it equal to $$mv^2/r$$. This is the default reference frame that is commonly used to solve circular motion problems. This article was updated on July 29, 2024
PRINTABLE FOR KIDS XII (12) HSC XI (11) FYJC X (10) SSC ### Ratio And Proportion Class 9th Mathematics Part I MHB Solution ##### Class 9th Mathematics Part I MHB Solution ###### Practice Set 4.1 Question 1. From the following pairs of numbers, find the reduced form of ratio of first number to second number. i. 72, 60 ii. 38, 57 iii. 52, 78 (i) 72,60 Reduced form of ratio of first number to second number: (Break the expression in order to simplify it further) (ii) 38,57 Reduced form of ratio of first number to second number: (Break the expression in order to simplify it further) (iii) 52, 78 Reduced form of ratio of first number to second number: (Break the expression in order to simplify it further) Question 2. Find the reduced form of the ratio of the first quantity to second quantity. i. 700 Rs, 308Rs. ii. 14Rs, 12 Rs.40 paise. iii. 5 litre, 2500 ml iv. 3 years 4 months, 5 years 8 months v. 3.8 kg, 1900 gm vi. 7 minutes 20 seconds, 5minutes 6 seconds. (i) Reduced form of the ratio of 700Rs and 308Rs is: (Break the expression in order to simplify it further) (ii) Reduced form of the ratio of 14Rs and 12.40Rs is: Multiply denominator and numerator by 100: Divide numerator and denominator by 10: (Break the expression in order to simplify it further) (iii) 5 litre = 5000 ml ∴ Reduced form of the ratio of 5000 ml and 2500 ml is: (Break the expression in order to simplify it further) (iv) 3 years = 3 × 12 = 36 months ∴ 3 years 4 months = 40 months 5 years = 5 × 12 = 60 ∴ 5 years 8 months = 68 months ∴ Reduced form of the ratio of 40 months and 68 months is: (Break the expression in order to simplify it further) (v) 3.8 kg = 3.8 × 1000 = 3800 gm ∴ Reduced form of the ratio of 3800 gm and 1900 gm is: (Break the expression in order to simplify it further) (vi) 7 minutes = 7 × 60 = 420 seconds ∴ 7 minutes 20 seconds = 440 seconds 5minutes = 5 × 60 = 300 seconds ∴ 5 minutes 6 seconds = 306 seconds ∴ Reduced form of the ratio of 440 seconds and 306 seconds is: (Break the expression in order to simplify it further) Question 3. Express the following percentages as ratios in the reduced form. (i) 75 : 100 (ii) 44 : 100 (iii) 6.25% (iv) 52 : 100 (v) 0.64% (i) Reduced form of the ratio of 75:100 is: (ii) Reduced form of the ratio of 44:100 is: (iii) Reduced form of 6.25% is: (iv) Reduced form of the ratio of 52:100 is: (v) Reduced form of 0.64% is: Question 4. Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required? No. of persons required to build a house in 8 days =3 No. of persons required to build a house in 1 day =3×8 = 24 No. of persons required to build a house in 6 days = 24/6 = 4 ∴ 4 persons are required to build the same house. Question 5. Convert the following ratios into percentage. (i) 15 : 25 (ii) 47 : 50 (iii) 7/10 (iv) 546/600 (v) 7/16 (i) 15 : 25 = 15/25 = ((15/25) × 100)% = (15 × 4) % = 60 % (ii) 47 : 50 = 47/50 = ((47/50) × 100)% = (47 × 2) % = 94 % (iii) 7/10 = ((7/10) × 100)% = (7 × 10) % = 70 % (iv) 546/600 = ((546/600) × 100)% = (546/6) % = 91 % (v) 7/16 = ((7/16) × 100)% = (7 × 6.25) % = 43.75 % Question 6. The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha's birth her mother’s age was 27 year. Find the present ages of Abha and her mother. Let age of Abha = x years ∴ Age of Abha’s mother at her birth = (x + 27) years Ratio of Abha’s and her mother’s age = 2:5 Cross multiply and get: 5x = 2(x + 27) ⇒ 5x = 2x + 54 ⇒ 5x – 2x = 54 ⇒ 3x = 54 ⇒ x = 54/3 ⇒ x = 18 ∴ Age of Abha = 18 years Age of Abha’s mother = 18 + 27 = 45 years Question 7. Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5:4? Given: Present age of Vatsala = 14 years Present age of Sara = 10 years Let after x years, the ratio of their ages will be 5:4. ∴ Age of Vatsala after x years = (14 + x) years Age of Sara after x years = (10 + x) years Ratio of their ages = 5:4 On cross multiplying, we get: 56 + 4x = 50 + 5x ⇒ 5x – 4x = 56 – 50 ⇒ x = 6 ∴ After 6 years, their ages will be 20 years and 16 years and ratio of their ages will be 5:4. Question 8. The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana's present age? Given: Ratio of present age of Rehana and her mother = 2:7 So, let present age of Rehana = 2x ∴ Present age of her mother = 7x After 2 years, age of Rehana = (2x + 2) years After 2 years, age of her mother = (7x + 2) years Now, given that after two years ratio of their ages will be 1:3. On cross multiplying, we get: 6x + 6 = 7x + 2 ⇒ 7x – 6x = 6 – 2 ⇒ x = 4 ∴ Present age of Rehana = 2x = (2 × 4) years = 8 years ###### Practice Set 4.2 Question 1. Using the property , fill in the blanks substituting proper numbers in the following. (i) (ii) (i) Let ∴ on comparing first two equalities, we get: 5/7 = x/28 Cross multiply and get: 7x = 28 × 5 ⇒ x = 4 × 5 = 20 Now, compare the first and third equalities and get: 5/7 = 35/y Cross multiply and get: 5y = 7 × 35 ⇒ y = 7 × 7 = 49 Now, compare the first and fourth equalities and get: 5/7 = z/3.5 Cross multiply and get: 7z = 5 × 3.5 ⇒ 7z = 5 × (35/10) ⇒ z = 5 × (5/10) ⇒ z = 25/10 = 2.5 ∴ (ii) Let ∴ On comparing first two equalities, we get: 9/14 = 4.5/x Cross multiply and get: 9x = 14 × 4.5 ⇒ x = 14 × 0.5 = 7 Now, compare the first and third equalities and get: 9/14 = y/42 Cross multiply and get: 14y = 9 × 42 ⇒ y = 9 × 3 = 27 Now, compare the first and fourth equalities and get: 9/14 = z/3.5 Cross multiply and get: 14z = 9 × 3.5 ⇒ z = 9 × (3.5/14) ⇒ z = 9 × (0.25) ⇒ z = 2.25 ∴ Question 2. Find the following ratios. (i) The ratio of radius to circumference of the circle. (ii) The ratio of circumference of circle with radius r to its area. (iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm. (iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area. (i) Let r be the radius of a circle. Circumference of circle = 2πr Ratio of radius to circumference of circle = r/2πr = 1/2π = 1 : 2π (ii) Let r be the radius of a circle. Circumference of circle = 2πr Area of the circle =2πr2 Ratio of radius to circumference of circle = 2πr/πr2 = 2/r = 2 : r (iii) Side of square = 7 cm Diagonal of square = √2 × side = 7√2 cm Ratio of diagonal of a square to its side = 7/7√2 = 1/√2 = 1 : √2 (iv) Length of rectangle = 5 cm Breadth of rectangle = 3.5 cm Perimeter of rectangle = 2(Length + Breadth) = 2(5+3.5) = 2(8.5) = 17 cm Area of rectangle = Length × Breadth = 5 × 3.5 = 16.5 cm2 Ratio of Perimeter to area of rectangle = 17/16.5 = 170/165 = 34/33 Question 3. Compare the following pairs of ratios. i.ii. iii. iv. v. (i) Given ratios are Step I: Make the second term of both the ratios equal. Multiply and divide first ratio by √7: Multiply and divide second ratio by 3: Step II: Compare the first terms (numerators) of the new ratios. Since the denominators of new ratios are equal, compare the numerators of the new ratios: Since, 9>√35, therefore . Therefore the second ratio is greater than the first ratio according to the ratio comparison rules. ⇒ (ii) Given ratios are Step I: Make the second term of both the ratios equal. Multiply and divide first ratio by √5: Multiply and divide second ratio by √7: Step II: Compare the first terms (numerators) of the new ratios. Since the denominators of new ratios are equal, compare the numerators of the new ratios: Since, 21>15, therefore . Therefore the second ratio is greater than the first ratio according to the ratio comparison rules. ⇒ (iii) Given ratios are Step I: Make the second term of both the ratios equal. Multiply and divide first ratio by 121: Multiply and divide second ratio by 18: Step II: Compare the first terms (numerators) of the new ratios. Since the denominators of new ratios are equal, compare the numerators of the new ratios: Since, 605 <306, therefore . Therefore, the first ratio is greater than the second ratio according to the ratio comparison rules. ⇒ (iv) Given ratios are Simplifying the ratios, we get: Since, the denominators of both the terms are same; compare the first terms (numerators) of the new ratios. Since the denominators of new ratios are equal, compare the numerators of the new ratios: Since, √5 = √5, therefore . Therefore, both the ratios are equal, according to the ratio comparison rules. ⇒ (v) Given ratios are Simplifying the ratios, we get: (Multiply the numerator and denominator of both the ratios by 10) Step I: Make the second term of both the ratios equal. Multiply and divide first ratio by 71: Multiply and divide second ratio by 51: Step II: Compare the first terms (numerators) of the new ratios. Since the denominators of new ratios are equal, compare the numerators of the new ratios: Since, 6532 > 1734, therefore . Therefore the first ratio is greater than the second ratio according to the ratio comparison rules. ⇒ Question 4. ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5 : 4. Find the measure of ∠B. Given: ∠A : ∠B = 5:4 Let the measure of ∠A = 5x Then ∠B = 4x Since the adjacent angles of a parallelogram are complementary, therefore ∠A + ∠B = 180° ⇒ 5x + 4x = 180° ⇒ 9x = 180° ⇒ x = 20° ∴ ∠ A = 5x = 5 × 20° = 100° ∴ ∠ B = 4x = 4 × 20° = 80° Question 5. The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages. Given: Ratio of present age of Albert and Salim=5:9 So, let present age of Albert = 5x years ∴ Present age of Salim = 9x years Five years later, age of Albert = (5x + 5) years Five years later, age of Salim = (9x + 5) years Given that this ratio of their ages is 3:5. ∴ According to the given problem: ⇒ 25x + 25 = 27x + 15 ⇒ 27x – 25x = 25 – 15 ⇒ 2x = 10 ⇒ x = 5 ∴ Present age of Albert = 5x = 5 × 5 = 25 years Present age of Salim = 9x = 9 × 5 = 45 years Question 6. The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle. Let l and b be the length and breadth of rectangle. Given that Length : Breadth = 3:1 Perimeter of rectangle = 36 Let length of rectangle = l = 3x cm ∴ Breadth of the rectangle = b = x cm Perimeter of rectangle = 2(l + b) = 36 cm ∴ 2(3x + x) = 36 ⇒ 2(4x) = 36 ⇒ 8x = 36 ⇒ x = 36/8 = 4.5 ∴ Breadth of rectangle = x = 4.5 cm Length of rectangle = 3x = 3 × 4.5 = 13.5 cm Question 7. The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers. Let a and b be two numbers. Given that a:b = 31:23 Sum of a and b = a + b = 216 Let the first number ‘a’ = 31x Then the second number ‘b’ = 23x Consider a + b = 216 ⇒ 31x + 23x = 216 ⇒ 54x = 216 ⇒ x = 216/54 = 4 ∴ The first number ‘a’ = 31x = 31 × 4 = 124 The second number ‘b’ = 23x = 23 × 4 = 92 Question 8. If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers. Let a and b be two numbers. Given that a:b = 10:9 Product of a and b = ab = 360 Let the first number ‘a’ = 10x Then the second number ‘b’ = 9x Consider ab = 360 ⇒ 10x × 9x = 360 ⇒ 90x2 = 360 ⇒ x2 = 4 ⇒ x = 2 If x = 2, then first number ‘a’ = 10x = 10 × 2 = 20 The second number ‘b’ = 9x = 9 × 2 = 18 If x = -2, then first number ‘a’ = 10x = 10 × (-2) = -20 The second number ‘b’ = 9x = 9 × (-2) = -18 Question 9. If a:b = 3:1 and b:c = 5:1, then find the value of i. ii. (i) Given that a:b = 3:1 ∴ a/b = 3 ⇒ a = 3b ………………… (1) Also, b:c = 5:1 ∴ b/c = 5 ⇒ c = b/5 ………………… (2) Consider =(9)3 =729 (ii) Given that a:b = 3:1 ∴ a/b = 3 ⇒ a = 3b ………………… (1) Also, b:c = 5:1 ∴ b/c = 5 ⇒ c = b/5 ………………… (2) Consider Question 10. If then find the ratio a/b . Given: Squaring both sides: 0.04 × 0.4 × a = 0.16 × 0.0016 × b ⇒ 0.016 × a = 0.000256 × b ⇒ a/b = 0.000256/0.016 ⇒ a/b = 0.016 = 16/1000 ⇒ a/b = 2/125 ∴ a:b = 2:125 Question 11. (x + 3):(x + 11) = (x - 2):(x + 1) then find the value of x. Given: (x + 3):(x + 11) = (x - 2):(x + 1) ⇒ (x + 3)/(x + 11) = (x - 2)/(x + 1) Cross multiply and obtain: ⇒ (x + 3)(x + 1) = (x - 2)(x + 11) ⇒ x2 + 4x +3 = x2 + 9x - 22 ⇒ 4x + 3 = 9x - 22 ⇒ 9x – 4x = 3 + 22 ⇒ 5x = 25 ⇒ x = 5 ∴ x = 5 ###### Practice Set 4.3 Question 1. If then find the values of the following ratios. i. ii. iii. iv. Given: a/b = 7/3 ∴ a = (7/3)b (i) (ii) (iii) (iv) Question 2. If then find the values of the following ratios. i. ii. iii. iv. (i) Given: Apply componendo and dividendo, i.e., : Take square root on both sides: (ii) Given: Apply componendo and dividendo, i.e., : Again apply componendo and dividendo, i.e., : (iii) Given: Apply componendo and dividendo, i.e., : Again apply componendo and dividendo, i.e., : (iv) Given: Apply componendo and dividendo, i.e., : Again apply componendo and dividendo, i.e., : Question 3. If: then find the value of the ratio . Given: Apply componendo and dividendo, i.e., : ∴ a/b = 49/3 ⇒ a2/b2 =2401/9 ⇒ 2a2/7b2 =686/9 Apply componendo and dividendo, i.e., : Question 4. Solve the following equations. Given: We first put x = 0 in the expression and obtain: ⇒ -20/-5 = 12/3 ⇒ 4=4 which holds true. ∴ x=0 is a solution. Now, Consider ……………… (1) Multiply both sides by �: ……………….(2) Apply dividendo: Cross multiply and get: 12x – 20 = 8x + 12 ⇒ 12x – 8x = 12+ 20 ⇒ 4x = 32 ⇒ x = 8 ∴ x=0, 8 are the solutions. Question 5. Solve the following equations. Given: We first put x = 0 in the expression and obtain: ⇒ 63/12 = 3/(-5) which does not hold true. ∴ x=0 is not a solution. Now, let ……………… (1) Multiply numerator and denominator of second expression by 5x: ……………….(2) ∴ From equation (1) and (2), we get: Cross multiply and obtain: 8x + 12 = 21x – 105 ⇒ 21x – 8x = 12 + 105 ⇒13x = 107 ⇒ x = 9 ∴ x=2 is the solution. Question 6. Solve the following equations. Given: We first put x = 0 in the expression and obtain: ⇒ 2/0 = 17/8 which does not hold true. ∴ x=0 is not a solution. Now, Consider Apply componendo and dividendo: Take square roots on both sides: Case 1: Cross multiply and obtain: 6x + 3 = 10x – 5 ⇒ 4x = 8 ⇒ x = 2 Case 2: Cross multiply and obtain: 6x + 3 = -10x + 5 ⇒ 16x = 2 ⇒ x = 1/8 ∴ x = 2, 1/8 are the solutions. Question 7. Solve the following equations. Given: We first put x = 0 in the expression and obtain: ⇒ (1+√3)/( 1-√3)=4/1 which does not hold true. ∴ x=0 is not a solution. Now, Consider Apply componendo and dividendo: Squaring both sides: Cross multiply and get: 36x + 9 = 25x + 75 ⇒ 36x -25x = 75 - 9 ⇒ 11x = 66 ⇒ x = 6 ∴ x = 6 is the solution. Question 8. Solve the following equations. Given: We first put x = 0 in the expression and obtain: ⇒ 10/9 = 61/36 which does not hold true. ∴ x=0 is not a solution. Consider the given equation: Apply dividendo: Take square root on both sides: Case 1: Cross multiply and obtain: 24x + 6 = 10x + 15 ⇒ 14x = 9 ⇒ x = 9/14 Case 2: Cross multiply and obtain: 24x + 6 = -10x - 15 ⇒ 34x = -21 ⇒ x = -21/34 ∴ x = 9/14 and x = -21/34 are the solutions. Question 9. Solve the following equations. Given: Apply componendo and dividendo: Take cube roots on both sides: Cross multiply and get: 12x - 16 = 5x + 5 ⇒ 12x – 5x = 5 + 16 ⇒ 7x = 21 ⇒ x = 3 ∴ x = 3 is the solution. ###### Practice Set 4.4 Question 1. Fill in the blanks of the following i. ii. (i) ∴ ∴ Thus, (ii) ∴ ∴ Thus, Question 2. 5m-n=3m+4n, then find the values of the following expressions. i. ii. (i) Given: 5m – n = 3m + 4m ⇒ 5m – 3m = 4n + n ⇒ 2m = 5n ⇒ m/n = 5/2 ⇒ m2/n2 = 25/4 Apply componendo and dividendo: = 29:21 (ii) Given: 5m – n = 3m + 4m ⇒ 5m – 3m = 4n + n ⇒ 2m = 5n ⇒ m/n = 5/2 ⇒ 3m/4n = 15/8 Apply componendo and dividendo: = 23: 7 Question 3. If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that Given: a(y+z) = b(z+x) = c(x+y) Divide all be ‘abc’: Cancel out the common factor: Rearrange the terms: Now, subtract third term from first term, subtract first term from second term and subtract second term from third term and obtain the equivalent: Solve and cancel the opposite terms: Hence, proved. Question 4. If and x + y + z ≠ 0 then show that the value of each ratio is equal to 1. Given: Add all the terms and obtain the equivalent: ∴ Each ratio is equal to 1. Question 5. If and x+y+z ≠ 0, then show that ratio is (a+b)/2. Given: Add all the terms and obtain the equivalent: ∴ Each ratio is equal to (a+b)/2. Question 6. If then show that . Given: ……………… (1) Each ratio = ………………… (2) Each ratio = ………………… (3) Each ratio = ………………… (4) ∴ from equation (2), (3) and (4), we get: Hence, showed. Question 7. If then show that every ratio = x/y. Given: (Multipliy and divide the last term by 5) Each ratio =k = (Adding all the terms) ∴ Each ratio equals x/y. Question 8. Solve. i. ii. (i) Given: We first put x = 0 in the expression and obtain: ⇒ 9/21 = -5/3 which does not hold true. ∴ x=0 is not a solution. Now, let ……………… (1) Multiply numerator and denominator of second expression by 4x: ……………….(2) ∴ From equation (1) and (2), we get: Cross multiply and obtain: ⇒ 28x – 35 = 6x + 9 ⇒ 28x – 6x = 9 + 35 ⇒ 22x = 44 ⇒ x = 44/22 ⇒ x = 2 ∴ x=2 is the solution. (ii) Given: We first put y = 0 in the expression and obtain: ⇒ (-12)/(-4) = 8/1 which does not hold true. ∴ y=0 is not a solution. Now, let ……………… (1) Multiply numerator and denominator of second expression by 5y: ……………….(2) ∴ From equation (1) and (2), we get: Cross multiply and obtain: ⇒ y + 8 = 3 + 6y ⇒ 6y – y = 8 - 3 ⇒ 5y = 5 ⇒ y = 1 ∴ y=1 is the solution. ###### Practice Set 4.5 Question 1. Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion? Let x be the number that should be subtracted from 12, 16, 21 so that the numbers remain in continued proportion. Numbers a, b, c are said to be continued proportion if b2 = ac. ∴ From the definition of continued proportion, we get: ⇒ (16 - x)2 = (12 - x)(21 - x) ⇒ 256 + x2 – 32x = 252 - 33x + x2 ⇒-32x + 33x = 252 – 256 ⇒ x= -4 ∴ -4 should be subtracted from 12, 16, 21 so that the numbers remain in continued proportion. Question 2. If (28 - x) is the mean proportional of (23 - x) and (19 - x) then find the value of x. A number b is said to be mean proportional of two numbers a and c if b2 = ac. ∴ From the definition of mean proportion, we get: (28 - x)2 = (23 - x)(19 - x) ⇒ 784 + x2 – 56x = 437 - 42x + x2 ⇒-56x + 42x = 437 – 784 ⇒ -14x= -347 ⇒ x = 347/14 Question 3. Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers. Let the numbers be x, y, z. As the numbers are in continued proportion, therefore y2 = xz ……………… (1) Also, the mean proportion = 12 ∴ y = √xz = 12 ⇒ xz = 144 …………… (2) It is given that the sum of remaining two numbers = 26 ∴ x + z = 26 ⇒ x = 26 – z Put the value of x in equation (2): (26 – z)z = 144 ⇒ 26z – z2 = 144 ⇒ z2 – 26z + 144 = 0 ⇒ z2 – 8z – 18z + 144 = 0 ⇒ z(z - 8) – 18(z – 8) = 0 ⇒ (z - 8)(z - 18) = 0 ⇒ z = 8 or z = 18 ∴ x = 26 – 8 or x = 26 – 18 ⇒ x = 18 or x = 8 y = 12 ∴ The numbers in proportion be 8, 12, 18 or 18, 12, 8. Question 4. If (a + b+ c)(a –b + c) = a2 + b2 + c2 , show that a, b, c are in continued proportion. Given: (a + b+ c)(a –b + c) = a2 + b2 + c2 ⇒ a2 –ab + ac + ab - b2 + bc + ca – bc + c2 = a2 + b2 + c2 ⇒ a2 –ab + ac + ab - b2 + bc + ca – bc + c2 - a2 - c2= b2 + b2 ⇒ 2ac = 2b2 ⇒ b2 = ac ∴ a, b, c are in continued proportion. Question 5. If and a, b, c >0, then show that, i. (a + b+ c)(b – c) = ab – c2 ii. (a2 + b2)(b2 + c2) = (ab + bc)2 iii. (a2 + b2)/ab = (a + c)/b (i) Given: a/b = b/c ⇒ b2 = ac Consider (a + b+ c)(b – c) = ab – ac + b2 – bc + cb – c2 = ab – ac + ac – c2 (∵ b2 = ac) = ab – c2 (ii) Given: a/b = b/c ⇒ b2 = ac Consider (a2 + b2)(b2 + c2) = a2b2 + a2 c2 + b2 b2 + b2c2 = a2b2 + ac(ac)+ b2(ac)+ b2c2 (∵ b2 = ac) = a2b2 + b2(ac)+ b2(ac)+ b2c2 (∵ b2 = ac) = a2b2 + 2b2(ac)+ b2c2 = a2b2 + 2ab2c+ b2c2 = (ab + bc)2 (iii) Given: a/b = b/c ⇒ b2 = ac Consider (a2 + b2)/ab = (a2 + ac)/ab (∵ b2 = ac) = (a + c)/b Question 6. Find mean proportional of Mean proportion of two numbers is the square root of their product. ∴ Mean proportion of is: ###### Problem Set 4 Question 1. Select the appropriate alternative answer for the following questions. If 6 : 5 = y : 20 then what will be the value of y? A. 15 B. 24 C. 18 D. 22.5 Given: 6:5 = y:20 ⇒ 6/5 = y/20 Cross multiply and get: 5y = 6 × 20 ⇒ y = 6 × 4 = 24 ∴ Option B is correct. Question 2. Select the appropriate alternative answer for the following questions. What is the ratio of 1 mm to 1 cm? A. 1 : 100 B. 10 : 1 C. 1 : 10 D. 100 : 1 1 cm = 100 mm ∴ 1mm : 1cm ⇒ 1mm : 100mm = 1 : 100 ∴ Option A is correct. Question 3. Select the appropriate alternative answer for the following questions. The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age? A. 3 : 2 B. 2 : 3 C. 4 : 3 D. 3 : 4 Given: Nitin’s age = 24 years Mohasin’s age = 36 years ∴ Ration of Nitin’s age to Mohasin’s age = 24:36 = 24/36 = 2/3 = 2:3 ∴ Option B is correct. Question 4. Select the appropriate alternative answer for the following questions. 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get? A. 8 B. 15 C. 12 D. 9 Total bananas = 24 Ratio in which the bananas are divided = 3:5 Let number of bananas Shubham got = 3x ∴ Number of bananas Anil got = 5x ∴ 3x+5x = 24 ⇒ 8x = 24 ⇒ x = 3 ∴ Shubham got (3 × 3) = 9 bananas. Thus, option D is correct. Question 5. Select the appropriate alternative answer for the following questions. What is the mean proportional of 4 and 25? A. 6 B. 8 C. 10 D. 12 Mean proportional of two numbers a and b = √(ab) ∴ Mean proportional of 4 and 25 = √(4 × 25) = √100 = 10 Question 6. For the following numbers write the ratio of first number to second number in the reduced form. i. 21, 48 ii. 36, 90 iii. 65, 117 iv. 138, 161 v. 114, 133 (i) Ratio of 21 and 48 in the reduced form: (To simplify, break the numbers in simpler form) ∴ Ratio of 21 and 48 in reduced form is 1:4. (ii) Ratio of 36 and 90 in the reduced form: (To simplify, break the numbers in simpler form) ∴ Ratio of 36 and 90 in reduced form is 2:5. (iii) Ratio of 65 and 117 in the reduced form: (To simplify, break the numbers in simpler form) ∴ Ratio of 65 and 117 in reduced form is 5:9. (iv) Ratio of 138 and 161 in the reduced form: (To simplify, break the numbers in simpler form) ∴ Ratio of 138 and 161 in reduced form is 6:7. (v) Ratio of 114 and 133 in the reduced form: (To simplify, break the numbers in simpler form) ∴ Ratio of 114 and 133 in reduced form is 6:7. Question 7. Write the following ratios in the reduced form. i. Radius to the diameter of a circle. ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm. iii. The ratio of perimeter to area of a square, having side 4 cm. (i) Let r be the radius of the circle. Let d be the diameter of the circle. ∴ Ratio of radius to diameter in the reduced form = Radius:Diameter ∴ Ratio of radius to diameter in the reduced form = 1:2 (ii) Given: Length of rectangle = l = 4 cm Breadth of rectangle = b = 3 cm Diagonal of rectangle = √(l2 + b2) = √(16 + 9) = √25 = 5 ∴ Diagonal of rectangle = 5 cm Ratio of diagonal to the length of a rectangle = 4:5 (iii) Given: Side of square = 4 cm Perimeter of square = 4 × Side = 4 × 4 = 16 cm2 Area of the square = (Side)2 = (4)2 = 14 cm2 The ratio of perimeter to area of a square = 16:14 = 8:7 Question 8. Check whether the following numbers are in continued proportion. i. 2, 4, 8 ii. 1, 2, 3 iii. 9, 12, 16 iv. 3, 5, 8 (i) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a:b∷b:c or b2 = ac Here, a = 2, b = 4 and c = 8 ∴ (4)2 = 2 × 8 ⇒ 16 = 16, which holds true. ∴ 2, 4, 8 are in continued proportion. (ii) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a:b∷b:c or b2 = ac Here, a = 1, b = 2 and c = 3 ∴ (2)2 = 1 × 3 ⇒ 4 = 3, which does not hold true. ∴ 1, 2, 3 are not in continued proportion. (iii) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a:b∷b:c or b2 = ac Here, a = 9, b = 12 and c = 16 ∴ (12)2 = 9 × 16 ⇒ 144 = 144, which holds true. ∴ 9, 12, 16 are in continued proportion. (iv) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a:b∷b:c or b2 = ac Here, a = 3, b = 5 and c = 8 ∴ (5)2 = 3 × 8 ⇒ 25 = 24, which does not hold true. ∴ 3, 5, 8 are not in continued proportion. Question 9. a, b ,c are in continued proportion. If a = 3 and c = 27 then find b. Given: a, b, c are in continued proportion. Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a:b∷b:c or b2 = ac Here, a = 3, c = 27 ∴ (b)2 = 3 × 27 ⇒ b2 = 81 ⇒ b = √81 = 9 ∴ b = -9 or 9 Question 10. Convert the following ratios into percentages. i. 37: 500 ii. 5/8 iii. 22/30 iv. 5/16 v. 144/1200 (i) 37: 500 = 37/500 = ((37/500) × 100)% = (37/5) % = 7.4 % (ii) 5/8 = ((5/8) × 100)% = (5 × 12.5) % = 62.5 % (iii) 22/30 = ((22/30) × 100)% = (220/3) % = 73.33 % (iv) 144/1200 = ((144/1200) × 100)% = (144/12) % = 12 % Question 11. Write the ratio of first quantity to second quantity in the reduced form. i. 1024 MB, 1.2 GB [(1024 MB = 1 GB)] ii. 17 Rupees, 25 Rupees 60 paise iii. 5 dozen, 120 units iv. 4 sq.m, 800 sq.cm v. 1.5 kg, 2500 gm (i) 1024MB = 1 GB Reduced form of the ratio of 1 GB and 1.2 GB is: ∴ The ratio in reduced form is 5:6. (ii) 60 paise = 0.60 Rupees ∴ 25 Rupees and 60 paise = 25.60 Rupees Reduced form of the ratio of 17 Rupees and 25.60 Rupees is: (Break the numbers in simpler form) ∴ The ratio in reduced form is 85:128. (iii) 1 dozen = 12 units ∴ 5 dozens = 5 × 12 = 60 units Reduced form of the ratio of 5 dozens (= 60 units) and 120 units is: ∴ The ratio in reduced form is 1:2. (iv) 4 sq m = 4 m2=4(100cm)2 ∴ 4 sq m = 40000 sq cm Reduced form of the ratio of 4 sq m(=40000 sq cm) and 800 sq cm is: (Break the numbers in simpler form) ∴ The ratio in reduced form is 50:1. (v) 1 kg = 1000 gm ∴ 1.5 kg = 1500 gm Reduced form of the ratio of 1500 gm and 2500 gm is: ∴ The ratio in reduced form is 3:5. Question 12. If then find the values of the following expressions. i. ii. iii. iv. Given: a/b = 2/3 ∴ a = (2/3)b (i) (ii) (iii) (iv) Question 13. If a, b, c, d are in proportion, then prove that i. ii. iii. (i) Given: a, b, c, d are in proportion. a, b, c, d are in proportion a : b : : c : d ad = bc i.e. Product of extremes = product of means. If then: (11a2 + 9ac)(b2 + 3bd) = (a2 + 3ac)(11b2 + 9bd) ⇒ 11a2b2 + 33a2bd + 9ab2c + 27abcd = 11a2b2 +9a2bd + 33ab2c + 27abcd ⇒ 33a2bd + 9ab2c = 9a2bd + 33ab2c ⇒ 24a2bd = 24ab2c ⇒ a2bd = ab2c ⇒ ad = bc, which holds true as the numbers are in continued proportion. ∴ (ii) Given: a, b, c, d are in proportion. a, b, c, d are in proportion a : b : : c : d ad = bc i.e. Product of extremes = product of means. If then: (a2 + 5c2)(b2) = (b2 + 5d2)(a2) ⇒ a2b2 + 5c2b2 = b2a2 + 5a2d2 ⇒ 5b2c2 = 5a2d2 ⇒ b2c2 = a2d2 ⇒ ad = bc, which holds true as the numbers are in continued proportion. ∴ (iii) Given: a, b, c, d are in proportion. a, b, c, d are in proportion a : b : : c : d ad = bc i.e. Product of extremes = product of means. If then: (a2 + ab + b2)(c2 – cd + d2) = (a2 - ab + b2)(c2 + cd + d2) ⇒ a2c2 - a2cd + a2d2 + abc2 – abcd + abd2 + b2c2 – b2cd + b2d2 = a2c2 + a2cd + a2d2 - abc2 – abcd - abd2 + b2c2 + b2cd + b2d2 ⇒ -2a2cd + 2abc2 + 2abd2 -2b2cd = 0 ⇒ 2abc2 - 2b2cd = 2a2cd – 2abd2 ⇒ 2bc[ac – bd] = 2ad[ac - bd] ⇒ ad = bc, which holds true as the numbers are in continued proportion. Question 14. If a, b, c are in continued proportion, then prove that (i) (ii) (i) Given: a, b, c are in continued proportion. ∴ b2 = ac If then: a(a - 4c) = (a – 2b)(a + 2b) ⇒ a2 – 4ac = a2 – 4b2 ⇒ -4ac = -4b2 ⇒ b2 = ac, which holds true as the numbers are in continued proportion. ∴ (ii) Given: a, b, c are in continued proportion. ∴ b2 = ac If then: b(a - c) = (a – b)(b + c) ⇒ ab – bc = ab + ac – b2 - bc ⇒ ab – bc - ab - ac + bc = -b2 ⇒ -ac = -b2 ⇒ b2 = ac, which holds true as the numbers are in continued proportion. ∴ Question 15. Solve: Given: We first put x = 0 in the expression and obtain: ⇒ 42/58 = 3/2 which does not hold true. ∴ x=0 is not a solution. Now, let ……………… (1) Multiply numerator and denominator of second expression by 6x: ……………….(2) ∴ From equation (1) and (2), we get: Cross multiply and get: 58x + 87 = 63x + 42 ⇒ 63x - 58x = 87 – 42 ⇒ 5x = 45 ⇒ x = 9 ∴ x=9 is the solution. Question 16. If , then prove that every ratio = x/y. Given: (Multiply and divide the middle term by -3) Each ratio =k = (Adding all the terms) ∴ Each ratio equals x/y. Question 17. If then prove that . 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Guided Lessons # Scale, Shapes, and Sets Use this fun and easy lesson plan to teach your students to predict and calculate dimensions for shapes to scale in sets. Grade View aligned standards No standards associated with this content. No standards associated with this content. No standards associated with this content. No standards associated with this content. No standards associated with this content. No standards associated with this content. Which set of standards are you looking for? Students will be able to calculate scale dimensions and predict next shapes to scale in a set. (5 minutes) • Draw two lines in front of your students, one twice as long as the other labeled 4 cm and 2 cm respectively. • Have your students think, pair, and share the answer to this question, "What kind of relationship do these two lines have to one another?” • Allow student responses and draw a connection to the fact that these two lines are in scale with one another: the longer line is two times the scale of the shorter one. • Write the following, “The shorter line is [blank] times the scale of the longer one.” Have your students think, pair, and share the missing amount. • Direct the whole class to share-out of answers and reasonings. • Confirm the answer to be 1/2. • Explain that scale is the amount a measurement is multiplied by to create proportional model. Shapes are proportional to one another if their internal angles are the same. This lesson will focus on proportional shapes. (10 minutes) • Hand out and preview the Predicting Shapes to Scale worksheet. (10 minutes) • Lead your class through the first exercise of the Predicting Shapes to Scale worksheet and answer any clarifying questions. (15 minutes) • Release your students to complete the remaining exercises of the Predicting Shapes to Scale worksheet. Support • Use sentence frames to scaffold student thinking around the idea of scale, like: “The dimension of the larger item is [blank] times to scale of the smaller item in the set” and “The dimension of the smaller item is [blank] times to scale of the larger item in the set.” Enrichment • Have students calculate dimensions for two or more items, larger and smaller in each set. • A computer with internet access and projector makes for a great setup to display proportional models to scale. • Additional information on scale or "scale factor" is expressed clearly in the video, How to Calculate Scale Factor, listed in the related media section. (5 minutes) • Show your students a shape with a measurement of at least one feature and have them describe up to two additional scaled items for a set. (10 minutes) • Have your students: • Share out their answers. • Allow others to challenge answers to dispute claims. • Phone a friend for support to take over explanations when they need assistance. ### Add to collection Create new collection 0 0 items
Class 8 NCERT Solutions – Chapter 2 Linear Equations in One Variable – Exercise 2.4 • Difficulty Level : Basic • Last Updated : 21 Jul, 2021 Question 1. Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number? Solution: Let the number be x, According to the question, (x – 5/2) × 8 = 3x Solving for x, ⇒ 8x – 40/2 = 3x ⇒ 8x – 3x = 40/2 (Bring like terms together) ⇒ 5x = 20 (Divide both sides by 5) ⇒ x = 4 Thus, the number is 4. Question 2. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers? Solution: According to the question, Let one of the positive number be x then other number will be 5x. Therefore, 5x + 21 = 2(x + 21) Solving for x, ⇒ 5x + 21 = 2x + 42 ⇒ 5x – 2x = 42 – 21 (Bring like terms together) ⇒ 3x = 21 (Dividing both sides by 3) ⇒ x = 7 1st number = x = 7 2nd number = 5x = 5×7 = 35 Question 3. The Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number? Solution: Let the digit at tens place be x then digit at ones place will be (9 – x). Original two-digit number = 10x + (9 – x) After interchanging the digits, the new number = 10(9 – x) + x According to the question, 10x + (9 – x) + 27 = 10(9 – x) + x Solving for x, ⇒ 10x + 9 – x + 27 = 90 – 10x + x ⇒ 9x + 36 = 90 – 9x ⇒ 9x + 9x = 90 – 36 (Bring like terms together) ⇒ 18x = 54 ( Divide both sides by 18) ⇒ x = 3 Original number = 10x + (9 – x) = (10 × 3) + (9 – 3) = 30 + 6 = 36 Thus, the number is 36. Question 4. One of the two digits of a two-digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number? Solution: Let the digit at tens place be x then digit at ones place will be 3x. Original two-digit number = 10x + 3x After interchanging the digits, the new number = 30x + x According to the question, (30x + x) + (10x + 3x) = 88 Solving for x, ⇒ 31x + 13x = 88 ⇒ 44x = 88 (Divide both sides by 44) ⇒ x = 2 Original number = 10x + 3x = 13x = 13×2 = 26 Question 5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages? Solution: Let the present age of Shobo be x then age of her mother will be 6x. Shobo’s age after 5 years = x + 5 According to the question, (x + 5) = (1/3) × 6x Solving for x, ⇒ x + 5 = 2x ⇒ 2x – x = 5 (Bring like terms together) ⇒ x = 5 Present age of Shobo = x = 5 years Present age of Shobo’s mother = 6x = 30 years. Question 6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate ₹100 per meter it will cost the village panchayat ₹75000 to fence the plot. What are the dimensions of the plot? Solution: Let the length of the rectangular plot be 11x and breadth be 4x. Rate of fencing per meter = ₹100 Total cost of fencing = ₹75000 Perimeter of the plot = 2(l+b) = 2(11x + 4x) = 2 × 15x = 30x Total amount of fencing = (30x × 100) According to the question, (30x × 100) = 75000 ⇒ 3000x = 75000 ⇒ x = 75000/3000 ⇒ x = 25 Length of the plot = 11x = 11 × 25 = 275m Breadth of the plot = 4 × 25 = 100m. Question 7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹50 per meter and trouser material that costs him ₹90 per meter. For every 3 meters of the shirt material he buys 2 meters of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹36,600. How much trouser material did he buy? Solution: Let 2x m of trouser material and 3x m of shirt material be bought by him Selling price of shirt material per meter = ₹ 50 + 50 × (12/100) = ₹ 56 Selling price of trouser material per meter = ₹ 90 + 90 × (10/100) = ₹ 99 Total amount of sale = ₹36,600 According to the question, (2x × 99) + (3x × 56) = 36600 ⇒ 198x + 168x = 36600 ⇒ 366x = 36600 ⇒ x = 36600/366 ⇒ x = 100 Total trouser material he bought = 2x = 2 × 100 = 200 m. Question 8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd. Solution: Let the total number of deer be x. Deer grazing in the field = x/2 Deer playing nearby = x/2 × ¾ = 3x/8 Deer drinking water = 9 According to the question, x/2 + 3x/8 + 9 = x (4x + 3x)/8 + 9 = x ⇒ 7x/8 + 9 = x ⇒ x – 7x/8 = 9 ⇒ (8x – 7x)/8 = 9 ⇒ x = 9 × 8 ⇒ x = 72 Question 9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages. Solution: Let the age of granddaughter be x and grandfather be 10x. Also, he is 54 years older than her. According to the question, 10x = x + 54 ⇒ 10x – x = 54 ⇒ 9x = 54 ⇒ x = 6 Age of grandfather = 10x = 10 × 6 = 60 years. Age of granddaughter = x = 6 years. Question 10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages. Solution: Let the age of Aman’s son be x then age of Aman will be 3x. According to the question, 5(x – 10) = 3x – 10 ⇒ 5x – 50 = 3x – 10 ⇒ 5x – 3x = -10 + 50 ⇒ 2x = 40 ⇒ x = 20 Aman’s son age = x = 20 years Aman age = 3x = 3 × 20 = 60 years My Personal Notes arrow_drop_up
# The four basic arithmetical operations Which expressions mean the same as addition? ## The four basic arithmetical operations Eight plus eight, of course. Sixteen cookies. Yes. Sixteen cookies. You add eight to eight, and the result is sixteen. This way of calculating is addition. You add. Reckon up. Total. Sum. Another way of looking at it is to study The Number Line. When you add, you go to the right on the Number Line. If you have eight cookies to begin with, you start at the eight in The Number Line. If you add eight more cookies, move eight steps to the right, and land on... sixteen. This is addition. The numbers you use to calculate are called terms, and the result is the sum. But hey, another sheet of cookies is ready. How many cookies are there now? 24. Eight plus eight plus eight. You add some more. Plus. But there is another way of getting the answer. If you bake a sheet of eight cookies three times, you get the number of cookies by calculating three times eight. This is multiplication. Do you see that there is a connection between addition and multiplication? Because multiplication is a repeated addition. 8+8+8 equals three times eight. If you bake ten sheets of cookies, the number of cookies can be described as a repeated addition … ... or, more simply, as a multiplication. Ten times eight. These operations are connected. The numbers you use to calculate are factors, and the result is a product. Now, the guests start to arrive to the party. There are six, and they have one cookie each. They had 24 cookies. Ate six of them. Then there are 24 minus six cookies left. Minus means remove, or subtract. The operation is called subtraction. Let's look at the Number Line again. When you subtract, "take minus", you go to the left on The Number Line. Start at 24, go to the left six steps, one for each cookie eaten. You end up at 18. When we subtract, we move to the left on the Number Line. When we add, we move to the right. This means that subtraction is addition backwards. This is the connection between plus and minus. They are opposites. The numbers you calculate with when subtracting are called terms, just like in addition. The result is the difference. Difference is the size of the gap between two numbers. It can be understood this way. A pile consists of 24 cookies. We mark six of them. The difference between 24 and 6 is -- 18. More guests arrive at the party. Now there are nine, and they want to share the cookies equally. That makes two cookies on each plate. Because 18 divided by nine, is two. This is division. 'Divided by'. When you want to split a number into equally sized parts... You divide. The number at the top is the numerator. The number at the bottom is the denominator and between them is a dividing line. The result is a quotient. Does division have any connection to the other operations? Yes. Division is multiplication backwards. Watch! 18 divided by nine equals two. And two times nine is 18. Times and "divided by" are connected, and are each other's opposite. The four operations and their connections.
# What is the vertex form of y= (9x-6)(3x+2)+4x^2+5x? Apr 25, 2017 $y = 31 {\left(x + \frac{5}{62}\right)}^{2} - \frac{1513}{124}$ #### Explanation: $y = \left(9 x - 6\right) \left(3 x + 2\right) + 4 {x}^{2} + 5 x$ = $27 {x}^{2} + 18 x - 18 x - 12 + 4 {x}^{2} + 5 x$ = $31 {x}^{2} + 5 x - 12$ = $31 \left({x}^{2} + \frac{5}{31} x\right) - 12$ = $31 \left({x}^{2} + 2 \times \frac{5}{62} \times x + {\left(\frac{5}{62}\right)}^{2} - {\left(\frac{5}{62}\right)}^{2}\right) - 12$ = $31 {\left(x + \frac{5}{62}\right)}^{2} - 31 {\left(\frac{5}{62}\right)}^{2} - 12$ = $31 {\left(x + \frac{5}{62}\right)}^{2} - \frac{25}{124} - 12$ or $y = 31 {\left(x + \frac{5}{62}\right)}^{2} - 12 \frac{25}{124}$ i.e. $y = 31 {\left(x + \frac{5}{62}\right)}^{2} - \frac{1513}{124}$ and vertex is $\left(- \frac{5}{62} , - 12 \frac{25}{124}\right)$ graph{y=31(x+5/62)^2-1513/124 [-3, 3, -20, 20]}
# INTRODUCTION TO THE CONVERGENCE OF SEQUENCES Save this PDF as: Size: px Start display at page: ## Transcription 1 INTRODUCTION TO THE CONVERGENCE OF SEQUENCES BECKY LYTLE Abstract. In this paper, we discuss the basic ideas involved in sequences and convergence. We start by defining sequences and follow by explaining convergence and divergence, bounded sequences, continuity, and subsequences. Relevant theorems, such as the Bolzano-Weierstrass theorem, will be given and we will apply each concept to a variety of exercises. Contents. Introduction to Sequences 2. Limit of a Sequence 2. Divergence and Bounded Sequences 4 4. Continuity 5 5. Subsequences and the Bolzano-Weierstrass Theorem 5 References 7. Introduction to Sequences Definition.. A sequence is a function whose domain is N and whose codomain is R. Given a function f: N R, f(n) is the nth term in the sequence. Example.2. The first example of a sequence is x n = n. In this case, our function f is defined as f(n) = n. As a listed sequence of numbers, this would look like the following: (.) (, 2,, 4, 5, 6, 7... ) Another example of a sequence is x n = 5 n, which would look like the following: (.4) (5, 25, 25, ) We know these are both valid examples of sequences because they are infinite lists of real numbers and hence can be regarded as functions with domain N. Example.5. The following would not be an examples of sequences: (.6) (, 2, ) (.7) (500, 200, 550, 0000) Date: July 2, 205. 2 2 BECKY LYTLE We know that these are not examples of sequences because they are finite lists of real numbers. 2. Limit of a Sequence A limit describes how a sequence x n behaves eventually as n gets very large, in a sense that we make explicit below. Definition 2.. A sequence of real numbers converges to a real number a if, for every positive number ɛ, there exists an N N such that for all n N, a n - a < ɛ. We call such an a the limit of the sequence and write lim n a n = a. Proposition. The sequence n converges to zero. Proof. Let ɛ > 0. We choose N N such that N > ɛ. Such a choice is always possible by the Archimedean property. To verify that this choice of N is appropriate, let n N satisfy n N. Then, n N implies n > ɛ which is equal to n = n 0 < ɛ, proving that n converges to zero by the definition of convergence. Proposition 2. An example of a sequence that does not converge is the following: (2.2) (,,,,...) If a sequence does not converge, it is said to diverge, which we will explain later in the paper, along with the explanation of why the above sequence does not converge. Proposition. If x n y n z n for all n N and lim n x n =lim n z n =l, then lim n y n =l too. Proof. Let ɛ > 0. We want to show there exists an N such that for all n > N, y n l < ɛ. We know that x n l. Therefore, there exists an N such that for all n > N, x n l < ɛ. Also, we know that z n l. Therefore, there exists an N 2 such that for all n > N 2, z n l < ɛ. Let N = max(n, N 2 ) and n > N. Then, n > N so x n l < ɛ. Also, n > N 2 so z n l < ɛ. We want to show that y n l < ɛ. This is equivalent to showing that both y n l < ɛ and l y n < ɛ. We know that y n z n, so y n l z n l < ɛ. Also, we know that y n x n, so l y n l x n < ɛ. Theorem 2. (Algebraic Limit Theorem). Let lim n a n = a and lim n b n = b. Then, (i) lim n ca n = ca for all c R (ii) lim n (a n + b n ) = a+b (iii) lim n (a n b n ) = ab (iv) lim n (a n /b n ) = a/b provided b 0 Example 2.4. If (x n ) 2, then ((2x n - )/). Proof. First, we will start with the information given in the example: x n 2. Next, we simply use the fact that ( 4 )( 2 ) = 2. (2.5) x n ( ) ( ) 4 2 3 INTRODUCTION TO THE CONVERGENCE OF SEQUENCES Now, let a n = x n, and let a = ( 4 )( 2 ), and let c = ( 2 ). From the Algebraic Limit Theorem, we know that ca n ca. Then, ( 2 )(x n) ( 2 )( 4 )( 2 ), which is equal to the following: 2x n (2.6) 4 The next step follows from the fact that 4 = +. 2x n (2.7) + Let 2xn = a n, let (+ ) = a, let b n = (,,...), and let b =. Then, by the Algebraic Limit Theorem, we know that a n + b n a + b. Therefore, we know that 2xn + (+ ) +, which is equal to the following: 2x n (2.8) This last step follows because 2xn - = 2xn. 2x n (2.9) Therefore, using the Algebraic Limit Theorem, we have shown that if (x n ) 2, then ((2x n - )/). Example 2.0. The following sequence converges to the proposed limit (2.) lim ( ) 2n + = 2 5n n Proof. Let 5n+4 be a n, let 5n+4 be b n and let 2n+ 5n+4 be c n, and c n = a n + b n. By Theorem 2., we know that lim(c n ) = lim (a n +b n ) = lim(a n )+lim(b n ). We must therefore determine what lim(a n ) and lim(b n ) are. First, we will show that lim( 5n+4 ) = 0. Let ɛ > 0. By the Archimedean principle, there exists an N N such that N > /ɛ. Then, for n > N, 5n+4 < 5N+4 < /N < ɛ. Therefore, the limit of 5n+4 is zero. Then, because lim(c n ) = lim(a n +b n ), lim(c n ) = lim(a n + 0) = lim(a n ). We will therefore find the limit of a n in order to prove lim( 2n+ 5n+4 ) = 2 5. We now want to show that lim( 2n 5n+4 ) = 2 5. Let ɛ > 0. By the Archimedean Principle, there exists an N such that /ɛ < N. Let n > N. We then want to show the following: (2.2) Then, (2.) 2n 5n < ɛ 2n 5n = 8 5(5n + 4) 4 4 BECKY LYTLE We have to check the following: (2.4) 8 5(5n + 4) < ɛ (2.5) We know that the inequality 8 5(5n + 4) < ɛ 8 5(5n+4) < ɛ is true for every value of n because n > N > /ɛ and ɛ. Therefore we only need to show that the inequality 8 5(5n+4) < ɛ is true. Using the fact that N > /ɛ, we can say the following: (2.6) Then, 8 5(5n + 4) < 8 5(5(/ɛ) + 4) = 8ɛ ɛ 8ɛ 25+20ɛ < 8ɛ 25 < ɛ. Therefore, 8 5(5n+4) < ɛ. Example 2.7. Let x n 0. If (x n ) 0, then ( x n ) 0. Proof. First, we have to prove that lim( x n ) exists. We know that x n is decreasing but is greater than or equal to 0 for all values of n. The square root of a positive number is also positive. Therefore, x n 0. Also, note that if 0< a < b, then 0< a< b. So if x n is decreasing, then so is x n. Therefore, lim( x n ) exists. Next, we must prove that ( x n ) 0. Let lim(x n ) = lim(( x n )( x n )) = 0. By the Algebraic Limit Theorem, we know that if lim(a n ) = a and lim(b n ) = b then lim((a n )(b n )) = ab. By this theorem, lim(( x n )( x n )) = lim( x n )lim( x n ) = 0. Thus, (lim( x n )) 2 =0. This implies that lim( x n )=0.. Divergence and Bounded Sequences Definition.. A sequence that does not have a limit or in other words, does not converge, is said to be divergent. Example.2. Recall proposition 2, which says that the following sequence does not converge: (.) (,,,...) Later in this paper, we will give a concise proof of this fact. Contrast this with the following sequence, which we have seen (.4) (, 2,, 4, 5, 6, 7... ) This converges to zero, as we proved earlier in this paper. However, these sequences do have something in common. They are both bounded. Definition.5. A sequence (x n ) is bounded if there exists a number M>0 such that x n M for all n N. Geometrically, this means we can find an interval [ M,M] that contains every term in the sequence (x n ). 5 INTRODUCTION TO THE CONVERGENCE OF SEQUENCES 5 Example.6. Given the sequence x n = (, 2,, 2,, 2...), we can see that the interval [, 2] contains every term in x n. This sequence is therefore a bounded sequence. Example.7. Given the sequence x n = (0, 00, 000, 0000,...), we can see that there is no real number that serves as an upper bound because lim(x n ) is infinity. Therefore, there does not exist any interval that contains every term in the sequence x n, and x n is not a bounded sequence. Theorem.8. Every convergent sequence is bounded. Example.9. Theorem being illustrated: Let x n = n+ n, which is the following sequence: ( 2 (.0), 2, 4, 5 ) 4... We know this converges to and can verify this using the same logic used in the proof under the definition of convergence showing that n converges to zero. Therefore, as n becomes very large, x n approaches, but is never equal to. By the above theorem, we know that this sequence is bounded because it is convergent. We can see that x n is a decreasing sequence, so the x is the largest value of the sequence and is the upper bound. The limit of the sequence,, is the lower bound. An interval that contains every term in the sequence x n is (,2]. 4. Continuity Theorem 4.. If f: R R is continuous, x n x implies f(x n ) f(x) Example 4.2. Theorem being applied: Let f(x) = x. This function is continuous. Let lim(x n ) = 5. In order words, x n 5. By the above theorem, this implies that f(x n ) f(5). This is equal to x n ()(5) which is also equal to x n 5. Therefore, we are able to see what the limit of f(x n ) is using this theorem. Example 4.. Theorem failing when function is non-continuous: Let f(x) be x, a non-continuous function. We know this is non-continuous because there is an asymptote at x=0. Let x n be n. We know this converges to zero based on a previous proof. Let s see if the continuity theorem fails for a noncontinuous function f. The theorem states that f(x n ) converges to f(x) if x n x. We know that x n 0, so if the theorem works, then f(x n ) f(0). But f(0) = 0 which does not exist. Therefore, f(x n ) cannot converge to /0, and the theorem fails for this non-continuous function. 5. Subsequences and the Bolzano-Weierstrass Theorem Definition 5.. Let (a n ) be a sequence of real numbers, and let n < n 2 < n < n 4... be an increasing sequence of natural numbers. Then, the sequence (a n, a n2, a n, a n4...) is called a subsequence of (a n ) and is denoted by (a nk ), where k N indexes the subsequence. 6 6 BECKY LYTLE Example 5.2. Let x n = n = (, 2,, 4 subsequences:...). Below are two examples of valid (5.) (, 6, 9, 2...) (5.4) ( 20, 200, ) Theorem 5.5. Bolzano-Weierstrass Theorem: Every bounded sequence contains a convergent subsequence. Example 5.6. Given a sequence x n = (,2,,4,,2,,4...), a convergent subsequence can be found. Proof. We know that this sequence is bounded by the interval [, 4]. By the Bolzano-Weierstrass Theorem, we can say that there indeed exists a convergent subsequence of x n. Just by looking at this sequence, we can see four convergent subsequences: (,,...), (2,2,2...), (,,...), and (4,4,4...). These subsequences converge to, 2,, and 4 respectively. Example 5.7. Given an unbounded sequence x n subsequence of x n does not exist = (,2,,4,5...), a convergent Proof. A convergent subsequence does not necessarily exist because this sequence does not satisfy the Bolzano-Weierstrass Theorem. Recall that any subsequence of a sequence is non-repeating and in the order of the original entries of x n. Notice that x n is increasing for all values of n and is divergent, considering the sequence continues until infinity. Therefore, for any subsequence a n, the values will be increasing toward infinity as well, and the subsequence will also be divergent. Theorem 5.8. Subsequences of a convergent sequence converge to the same limit as the original sequence. Example 5.9. Let us return to the example of a divergent sequence that was given under the definition of divergence. Recall that this sequence, x n, was (, -,, -,, -...). One subsequence of x n is (,,,...). This subsequence converges to. Another subsequence of x n is (-, -, -,...). This subsequence converges to -. Now, we will prove that x n is divergent by contradiction. Assume x n is convergent. Then, by the above theorem, all its subsequences converge to lim(x n ), implying that all its subsequences converge to the same value. The two subsequences of x n stated above converge to different values. Therefore, this contradicts our original hypothesis that x n is convergent. We are then able to conclude that x n is divergent. Acknowledgements. I would like to thank my two mentors, Sean Howe and Abby Ward for meeting with me each week for the duration of my stay at the program. Thank you so much for suggesting this topic, explaining everything to me that I was confused about, editing and helping with my paper, and calming me down when I got overwhelmed. I would also like to thank Peter May for organizing this wonderful REU and for giving me the opportunity to participate. 7 INTRODUCTION TO THE CONVERGENCE OF SEQUENCES 7 References [] Walter Rudin. Principles of Mathematical Analysis. McGraw-Hill Inc [2] Stephen Abbott. Understanding Analysis. 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Algebra and Trigonometry 10th Edition X = $\begin{bmatrix} -\frac{8}{3} & \frac{8}{3} & \frac{4}{3}\\ \frac{2}{3} & 0 & 6\\ \end{bmatrix}$ 3X - 4A = 2B 3X = 2B + 4A X = $\frac{1}{3}$(2B + 4A) First multiply matrix A by the scalar multiple 4: $\begin{bmatrix} -2(4) & 1(4) & 3(4)\\ -1(4) & 0(4) & 4(4)\\ \end{bmatrix}$ = $\begin{bmatrix} -8 & 4 & 12\\ -4 & 0 & 16\\ \end{bmatrix}$ Then multiply matrix B by the scalar multiple 2: $\begin{bmatrix} 0(2) & 2(2) & -4(2)\\ 3(2) & 0(2) & 1(2)\\ \end{bmatrix}$ = $\begin{bmatrix} 0 & 4 & -8\\ 6 & 0 & 2\\ \end{bmatrix}$ Then perform 2B + 4A to get matrix Y: Y = $\begin{bmatrix} 0-8 & 4+4 & -8+12\\ 6-4 & 0+0 & 2+16\\ \end{bmatrix}$ = $\begin{bmatrix} -8 & 8 & 4\\ 2 & 0 & 18\\ \end{bmatrix}$ Then multiply Y by the scalar $\frac{1}{3}$ to get X: X = $\begin{bmatrix} -8(\frac{1}{3}) & 8(\frac{1}{3}) & 4(\frac{1}{3})\\ 2(\frac{1}{3}) & 0(\frac{1}{3}) & 18(\frac{1}{3})\\ \end{bmatrix}$ = $\begin{bmatrix} -\frac{8}{3} & \frac{8}{3} & \frac{4}{3}\\ \frac{2}{3} & 0 & 6\\ \end{bmatrix}$
# ML Aggarwal CBSE Solutions Class 6 Math Ninth Chapter Ratio and Proportion Exercise 9.1 ### ML Aggarwal CBSE Solutions Class 6 Math 9th Chapter Ratio and Proportion Exercise 9.1 Ex. 9.1 (1) Express the following ratios in simplest form: (i) 20 : 40 Solution: = 1 : 2 (ii)40 : 20 Solution: 2 : 1 (iii) 81 : 108 Solution: 9 : 12 = 3 : 4 (iv) 98 : 63 Solution: 14 : 9 (2) Fill in the missing numbers in the following equivalent ratios: (3) Find the ratio of each of the following in simplest form: (i) 2.1 m to 1.2 m Solution: 2.1: 1.2 = 21/10 : 12/20 = 21 : 12 = 7 : 4 (ii) 91 cm to 1.04 m Solution: 1.04 m = 104 cm Now, 91 cm : 104 cm = 7 : 8 (iii) 3.5 kg to 250 g Solution: 3.5 kg = 3500 g Now, 3500 g : 250 g = 700 : 50 = 140 : 10 = 28 : 2 = 14 : 1 (iv) 60 paise to 4 rupees Solution: 4 rupees = 400 paise Now, 60 : 400 = 15 : 100 = 3 : 20 (v) 1 minute to 15 seconds Solution: 1 minute = 60 seconds Now, 60 : 15 = 4 : 1 (vi) 15 mm to 2 cm Solution: 2 cm = 20 mm Now, 15 mm : 20 mm = 3 : 4 (4) The length and the breadth of a rectangular park are 125 m and 60 m respectively. What is the ratio of the length to the breadth of the park? Solution: The ration of the length to the breadth of the park = 125 m : 60 m = 25 : 12 (5) The population of village is 4800. If the number of females is 2160, find the ratio of the males to that of females. Solution: Total population of the village = 4800 Number of females = 2160 ∴ Number of males = 4800 – 2160 = 2640 The ratio of the males to that of females = 2640 : 2160 = 528 : 432 = 264 : 216 = 132 : 108 = 66 : 54 = 33 : 27 = 11 : 9 (6) In a class, there are 30 boys and 25 girls. Find the ratio of the number of (i) boys to that of girls (ii) girls to that of total number of students. (iii) boys to that of total number of students. Solution: (i) boys to that of girls = 30 : 25 = 6 : 5 (ii) Total number of students = 30 + 25 = 55 The ratio of girls to that of total number of students = 25 : 55 = 5 : 11 (iii) Total number of students = 55 The ratio of boys to that of total number of students = 30 : 55 = 6 : 11 (7) In a year, Reena earns Rs. 1,50,000 and saves Rs. 50,000. Find the ratio of (i) money she earns to the money she saves (ii) money that she saves to the money she spends. Solution: Reena earns = 1,50,000 She saves = 50,000 That’s mean, she spends 1,50,000 – 50,000 = 1,00,000 (i) The ratio of money she earns to the money she saves = 1,50,000 : 50,000 = 30000 : 10000 = 6000 : 2000 = 3 : 1 (ii) The ratio of money that she saves to the money that she spends = 50,000 : 1,00,000 = 1 : 2 (8) The monthly expenses of a student have increased from Rs. 350 to Rs. 500. Find the ratio of (i) increase in expenses to original expenses (ii) original expenses to increased expenses (iii) increased expenses to increase in expenses Solution: Increase in expenses = 500 – 350 = 150 (i) The ratio of increase in expenses to original expenses = 150 : 350 = 30 : 70 = 6 : 14 = 3 : 7 (ii) original expenses to increased expenses = 350 : 500 = 70 : 100 = 14 : 20 = 7 : 10 (iii) The ratio of increased expenses to increase in expenses = 500 : 150 = 100 : 30 = 50 : 15 = 10 : 3 (9) Mr Mahajan and his wife are both school teachers and earn Rs. 20900 and Rs. 18700 per month respectively. Find the ratio of: (i) Mr Mahajan’s income to his wife’s income (ii) Mrs Mahajan’s income to the total income of both Solution: Mr Mahajan’s income = Rs. 20900 Mrs. Mahajan income = Rs. 18700 Their total income = 20900 + 18700 = 39600 (i) The ratio of Mr Mahajan’s income to his wife income = 20900 : 18700 = 4180 : 3740 = 836 : 748 = 418 : 374 = 209 : 187 (ii) The ratio of Mrs Mahajan income to the total income of both = 18700 : 39600 = 3720 : 7920 = 744 : 1584 = 372 : 792 = 186 : 396 = 93 : 198 (10) Out of 30 students in a class, 6 like football, 12 like cricket and remainining likes tennis. Find the ratio of (i) number of students liking football to number of students liking tennis. (ii) number of students liking cricket to total number of students. Solution: Total students = 30 Number of students like football = 6 Number of students like cricket = 12 Number of students like tennis = 30 – (6 + 12) = 30 – 18 = 12 (11) Divide Rs. 560 between Ramu and Munni in the ratio 3 : 2 Solution: Total Rs. 560 Ramu get = 560 x 3/5 = 112 x 3 = 336 Munni get = 560 x 2/5 = 112 x 2 = 224 (12) Two people invested Rs. 15000 and Rs. 25000 respectively to start a business. They decided to share the profits in the ratio of their investments. If their profit is Rs. 12000, how much does each get. Solution: Total investment = 15000 + 25000 = 40000 1st people get in profit = 12000 x 15000/40000 = 3 x 1500 = 4500 2nd people get in profit = 12000 x 25000/40000 = 7500 (13) The ratio of Ankur’s money to Roma’s is 9 : 11. If Ankur has Rs. 540, how much money does Roma have? Solution: Let, the total money be x Ankur has Rs. 540 Therefore, X x 9/20 = 540 9x/20 = 540 => 9x = 10800 => x = 10800 / 9 => x = 1200 ∴ Roma get = 1200 – 540 = 1160 (14) The ratio of weights of tin and zinc in an alloy is 2:5. How much zinc is there in 31.5 of alloy? Solution: The ratio of weights of tin and zinc in an alloy is 2:5 Tin have = 2/7 Zinc have = 5/7 ∴ zinc have in an alloy = 31.5 x 5/7 = 315/10 x 5/7 = 22.5 g Updated: December 11, 2019 — 2:03 pm
Courses Courses for Kids Free study material Offline Centres More Store # Two bodies with moments of inertia ${I_1}$ and ${I_2}$$\left( {{I_1} > {I_2}} \right)$ have equal angular momentum. If their kinetic energies of rotation areand, respectively, then:A) ${E_1} > {E_2}$B) ${E_1} < {E_2}$C) ${E_1} = {E_2}$D) ${E_1} = 2{E_2}$ Last updated date: 13th Jun 2024 Total views: 51.6k Views today: 1.51k Verified 51.6k+ views Hint: In question relation between moment of inertia of two bodies is given. Also given that their angular momentums are equal. Then, the relation between kinetic energy can be calculated. We use angular momentum formula to find relation between angular speeds of two bodies. After that we use the kinetic energy formula from the relation between angular speeds. Complete step by step solution: Given: ${I_1}$ and ${I_2}$ are moments of inertia of two bodies. In question given that both moments of inertia have following relation ${I_1} > {I_2}$ Let ${\omega _1}$ and ${\omega _2}$ are two angular velocities. Angular momentum of first body, ${L_1} = {I_1}{\omega _1}$ Angular momentum of second body,${L_2} = {I_2}{\omega _2}$ According to the question, angular momentums are the same for two bodies. Numerically, we can write as ${L_1} = {L_2}$ $\Rightarrow {I_1}{\omega _1} = {I_2}{\omega _2}$ $\therefore \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{\omega _2}}}{{{\omega _1}}}$ $\because {I_1} > {I_2}$ $\therefore \dfrac{{{I_1}}}{{{I_2}}} > 1$ Which gives $\dfrac{{{\omega _2}}}{{{\omega _1}}} > 1$ $\Rightarrow {\omega _2} > {\omega _1}$ Rotational kinetic energy is given by, $K{E_{Rotational}} = \dfrac{1}{2}I{\omega ^2} = E$ Rotational kinetic energy for first body is given by, ${E_1} = \dfrac{1}{2}{I_1}\omega _1^2$ Similarly rotational kinetic energy for first body is given by, ${E_2} = \dfrac{1}{2}{I_2}\omega _2^2$ Dividing ${E_1}$ by ${E_2}$, we get $\Rightarrow$ $\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{\dfrac{1}{2}{I_1}\omega _1^2}}{{\dfrac{1}{2}{I_2}\omega _2^2}}$ $\Rightarrow$ $\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_1}\omega _1^2}}{{{I_2}\omega _2^2}}$ $\Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_1}}}{{{I_2}}} \times \dfrac{{\omega _1^2}}{{\omega _2^2}}$ $\Rightarrow$ $\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_1}}}{{{I_2}}} \times {\left( {\dfrac{{{\omega _1}}}{{{\omega _2}}}} \right)^2}$ $\Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_1}}}{{{I_2}}} \times {\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)^2}$ $\left[ {\because \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{\omega _2}}}{{{\omega _1}}}} \right]$ $\therefore \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_2}}}{{{I_1}}}$ After rearranging LHS, we get $\Rightarrow$ $\dfrac{{{E_2}}}{{{E_1}}} = \dfrac{{{I_1}}}{{{I_2}}}$ $\Rightarrow \dfrac{{{E_2}}}{{{E_1}}} > 1$ $\left[ {\because we{\text{ }}have{\text{ }}calculated{\text{ }}as{{}}\dfrac{{{I_1}}}{{{I_2}}} > 1} \right]$ Hence, we get ${E_2} > {E_1}$ or we can write${E_1} < {E_2}$. Hence, the correct option is B. Additional information: Moment of inertia is defined as the quantitative measure of the rotational inertia. It can be expressed as after applying by a torque or turning force, it is known as an opposition that the body exhibits to having its speed of rotation about an axis. In a moment of inertia. the axis may be internal or external. Axis may or may not be fixed. Other names of moments of inertia are rotational inertia, the mass moment of inertia, angular mass. The rotational kinetic energy of a rotating body is expressed in terms of the moment of inertia and angular velocity. The total kinetic energy of a rotating object can be given by the sum of the translational kinetic energy and the rotational kinetic energy. Note: Students must be careful about linear kinetic energy and rotational kinetic energy. Both have different meanings. In linear kinetic energy, as the name shows the object is moving in a straight line (or in other way moving linearly). But in case of rotational kinetic energy, torque is the main reason for motion. In case of linear kinetic energy motion occurs due to force.
# 10th Class Mental Ability Mathematical Operations Mathematical Operations Mathematical Operations Category : 10th Class This chapter contains only mathematical operations but in different manner. The main key of solving problem in this chapter is BODMAS rule to which you are already familiar This chapter consists of the following two types of problems. 1. Direct Substitution To solve such type of problems, substitute the symbols $+,-,\text{ }\times ,\div$ etc., for the given letters or symbol as per the question and simplify the statement using BODMAS rule, 2. Interchange of Numbers and Signs To solve such type of problems, interchange the numbers with signs as per the given questions and then apply the BODMAS rule. EXAMPLE 1. If '+' stands for $'\times ','{{\times }^{,}}$stands for $'\div '$stands for- 'and $'\div '$stands for ?+? then what is the value of $8+2\times 4-2\div 6?$ (a) 20 (b) 12 (c) 14 (d) 18 Explanation (a): Given: $8+2\times 4-2\div 6$ After replacements above given expression becomes $8\times 2-4\div 2+6=8\times 2-2+6=16-2+6=20,$ 2. Select the correct set of interchanges which would make the given equation true. $\mathbf{8\times 9-3\div 7=13}$ (a) $\times \text{ }and\text{ }\div$, 3 and 7 (b) $\times \text{ }and\text{ }\div$, 3 and 8 (c) $\div and-$, 7 and 8 (d) $\div and-$. 3 and 8 Explanation (c): The given equation is $8\times 9-3\div 7=13$ Using interchanges as given in option (c), we get $7\times 9\div 3-8=13$ $\Rightarrow 7\times 3-8=13\Rightarrow 21-8=13\Rightarrow 13=13,$which is true. #### Other Topics You need to login to perform this action. You will be redirected in 3 sec
# Factors of 12 ### What are the Factors of 12? A number when divided by another number and does not leave a remainder then those numbers (divisor) are called factors of that number (dividend). The answer does not change much when talking about the question of what are factors of 12, as the core concept is always the same. It hinges towards complete divisibility when we speak about factors of 12. Here, we are essentially talking about numbers that can divide 12, and also do not result in a remainder. Let us also look at what are the factors of 12, and how many factors in 12 are positive, negative, or prime? You can use different methods to find factors of 12. These are just different representation forms of calculation for the method. ### How to Find the Factors of 12? When we divide every number with 12 upto that number itself, the numbers that evenly divide with 12 are the factors of 12. This is a very simple and straightforward method. The entire process takes only 5 steps. To start off here is step number 1: Step 1: Consider the number 12 Step 2: Divide it with all the number starting from 1 to 12 Step 3: Capture the results 12/1 = 12 12/2 = 6 12/3 = 4 12/4 = 3 12/5 = 2.4 12/6 = 2 12/7 = 1.7 12/8 = 1.5 12/9 = 1.3 12/10 = 1.2 12/11 = 1.09 12/12 = 1 Step 4: Filter out the positive integer quotient for the above, rejecting the decimals. Step 5: The factors of 12 are 1, 2, 3, 4, 6, 12. The factors of 12 also include negative integers. To check how many factors in 12 include negative integers, the above process must be repeated and negative numbers should be considered. This is the easiest way to get the answer to- how many factors in 12 are negative, as there are various other ways which can puzzle you. ### Factor Pairs of 12 These are numbers that when multiplied with each other results in 12. So here are the factor pairs of 12. 1 x 12 = 12 2 x 6 = 12 3 x 4 = 12 4 x 3 = 12 6 x 2 = 12 12 x 1 = 12 As we had mentioned earlier that factors of 12 also include negative integers. So we will be replacing the above numbers with the minus sign because the product of two negative signs gives a positive result. Let us see how many factors in 12 form factor pairs in the negative series? -1 x -12 = 12 -2 x -6 = 12 -3 x -4 = 12 -4 x -3 = 12 -6 x -2 = 12 -12 x -1 = 12 In the above pair, we have included pairs like 3 x 4 and 4 x 3. They are the multiples of the individual numbers. Here it is the multiples of 3 and 4. ### Prime Factors of 12 Before heading into prime factors of 12, let us see have a recap about prime numbers and factors? Prime numbers are those that have only two factors, and those are 1 & the number itself. For example, 3 is a prime since it is wholly divisible only by 1 and itself 5. Factors like stated above, are the numbers, which, when divided with another number doesn’t leave a remainder. So, below we can see the process to find the prime factors of 12: Step 1:- Take 12 and divide it by the smallest prime number 12 / 2 = 6 Step 2:- Keeping dividing the result with small prime numbers 6 / 2 = 3 Step 3:- Continue the process, until the result is 1 3 / 3 = 1 Step 4:- In conclusion, we get the prime factors of 12 as 2, 2, 3 To reconfirm the values of the prime factors of 12, you can multiply the prime factors and the resulting answer should be a 12. 2 x 2 x 3 = 12 ### What are the Factors of 12? After following the steps to find the factors of 12, we can list down all the factors. Including the positive as well as negative integers. The following integers are the factors of 12: 1, 2, 3, 4, 6, and 12 -1, -2, -3, -4, -6, and -12 In general, you can find the factors of any number in a similar manner. While the number gets larger the methods will change a little bit. But until a certain range of numbers, the above method is used. ### Fun Facts • You will never have factors as decimals and fractions, they are always whole numbers or integers. • 2 is the common factor for all even numbers. • For a square number, the number of factors is odd when for most other numbers is even. • The number 5 is present as a factor in numbers that ends with 5 like 15, 25, 35. • Numbers that end with 0 have 2, 5, and 10 as their factors like 10, 20, 30. ### Solved Examples 1. Find the Factors of 24? To find the factors of 24, we will see which numbers divide it completely. 24/1 = 24 24/2= 12 24/3= 8 24/4=6 24/6=4 24/8=3 24/12= 2 24/24= 1 So, positive factors of 24 are as follows= 1,2,3,4,6,8,12,24. 1. Find factor of 36? To find the factors of 36, we will follow the process used for 24. 36/1= 36 36/2= 18 36/3 = 12 36/4 = 9 36/6 = 6 36/9 = 4 36/12= 3 36/18= 2 36/36= 1 So, positive factors of 36 are as follows= 1, 2, 3, 4, 6, 9, 12, 18, 36. 1. Can We Find Factors of Other Numbers As Well and How To Do It? Yes, we can definitely find factors for practically any number using the Short Division Method. Example- Find the factors of 76230? • Divide the given number by 2.. • Next step, divide the resultant number in step 1 by 5 and we get the result of 7623. • According to the divisibility rule, we can divide the number by 3. We get 2541. Similarly, we further divide it by 3. • Divide the resultant by 7. Finally, we divide the resultant by 11 to get the factors of 76230 = 2 x 3 x 3 x 5 x 11 x 11. 2. Are There Any Other Methods to Find What are the Factors of 12? The first one as explained with the help of example is the Short Division Method. When you keep dividing with the smallest prime number to get 1. And the resulting factors would be divided numbers. These can be represented in a form of the short division format. Another method is the branch method representation. Here we start at the top of the branch with the main number. In this case, 12 and continue adding the two-factor branch until a prime number is reached. So how many factors in 12 would we get in this number? It would be the same but differently represented.
# Find the Value of X, If 4x = (52)2 − (48)2. - Mathematics Find the value of x, if 4x = (52)2 − (48)2. #### Solution Let us consider the following equation: $4x = \left( 52 \right)^2 - \left( 48 \right)^2$ Using the identity $\left( a + b \right)\left( a - b \right) = a^2 - b^2$,we get: $4x = \left( 52 \right)^2 - \left( 48 \right)^2$ $4x = \left( 52 + 48 \right)\left( 52 - 48 \right)$ $4x = 100 \times 4 = 400$ $\Rightarrow 4x = 400$ $\Rightarrow x = 100$ (Dividing both sides by 4) Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 8 Maths Chapter 6 Algebraic Expressions and Identities Exercise 6.6 \| Q 6.1 \| Page 43
The fundamental theorem of algebra is a theorem that introduces us to some specific characteristics of polynomials. It is one of the most basic but very important theorems in algebra. This theorem is the basis of modern algebra, and also, having the knowledge of this theorem is essential for higher Math education/learning, including trigonometry, calculus, and many others. So, what does this theorem say? It says that, “Every polynomial of degree n will have n number of roots, and we may need to use complex numbers if required.” Now, if you are not familiar with algebra, you may be confused with these new terms. So, let us first understand these terms and the theorem in a much simpler way. ## What is a Polynomial? You may have come across numbers such as 10×2+3×2+9 or 7×3+3×2+2x+5 somewhere. These are called polynomials, where 10×6+3×2+9 is a polynomial with 3 terms, 10×6, 3×2, 9 and 7×3+3×2+2x+5 is a polynomial with 4 terms, 7×3, 3×2, 2x, 5. These are a few of the innumerable polynomials which exist in the number system. The word polynomial is a combination of two Greek words ‘poly’ and ‘nomial’, where ‘poly’ means many, and ‘nomial’ means ‘terms’. There can be many terms, but they have to be fixed or defined because polynomials cannot contain infinite terms. Therefore, polynomials can be defined as expressions which consist of constants, variables, and exponents. These expressions can perform the arithmetic operations of addition, subtraction, non-negative integer exponentiation, and multiplication, but no division on the variables. • Constants are specific numbers, which accompany the variable. For example, the constants in 10×6+3×2+9 are 10, 3, and 9. • Variables can be defined as the symbols for the numbers that we don’t know yet. We generally use alphabets as variables in algebraic expressions. So, the variable in 10×6+3×2+9 is x. • Exponents are the powers on the variables, i.e., the number of times the base variable is multiplied to itself. For example, the exponents in 10×6+3×2+9 are 6, 2, and 0 for the first, second, and the third term, respectively. Polynomials can also be represented as P(x), where the variable in the expression is x. So, P(x) = 10×6 + 3×2 + 9. Polynomials can be divided based on the number of terms in that polynomial. Polynomials are called monomials if they contain only one term. The polynomial is called a binomial if it consists of two terms, and trinomial if it consists of three terms, etc. ## What is the Degree of a Polynomial? The degree of the polynomial is the highest exponent of a variable, with a non-zero coefficient in the polynomial. The word ‘order’ was often considered synonymous to ‘degree’. But ‘order’ is used to refer to some other properties of the polynomial as well, which may create some ambiguities. Therefore, it is better to use the term ‘degree’ to refer to the highest exponent. We can divide the polynomial into different types based on its degree. • A constant polynomial is a polynomial of degree 0. • A linear polynomial is a polynomial of degree 1. • A quadratic polynomial is a polynomial of degree 2. • A cubic polynomial is a polynomial of degree 3. ## What are the Roots of a Polynomial? The roots of a polynomial are the values of variables, which result in the polynomial to become zero. Roots can be considered as the solutions of the equation, where if we put the value of root in all the variables, the equation will result in 0. Polynomials can be represented as a general term: anxn+an-1xn-1+…+a1x+a0. These roots of polynomials are therefore, often called ‘zeroes’ of the polynomial for the same reason, as when applied to the expression, the result is a zero. If we happen to find the roots of the polynomial, we can also find all the factors of the polynomial and vice-versa. For example, if the polynomial of a degree 2 has roots r1 and r2, and the variable is x then the factors are (x – r1) and (x – r2). The fundamental theorem of algebra as we read above, mentions that every polynomial of a degree ‘n’ will always have ‘n’ number of roots. The fundamental theorem is also sometimes stated as “Every non-constant polynomial which contains complex coefficients and a degree which is greater than or equal to one has at least one root in the set of complex numbers.” Real numbers are at core a subset of complex numbers. This can be seen from the fact that every real number can also be represented as a + bi, where a = the real number and b = 0. ## What are Complex Numbers? In the most simple terms, complex numbers are nothing but a combination of a real number and an imaginary number. As we know, real numbers are the number that we generally use, that are whole numbers (0, 1, 2, 3 …), rational numbers ( 0.2, 7/5, …), and irrational numbers (, 7, …). Real numbers can be positive, negative, or zero. Now the question that arises is what is an imaginary number. Imaginary numbers can be described as the square roots of negative numbers or numbers which when multiplied with itself give a negative output. Imaginary numbers are represented by multiples of -1, which is called iota and represented by i. The square of i is -1. Some of the examples of complex numbers are 10i, -4 + 2i, and 6 – 5i. One important point that you need to remember is that as you know, i is -1, therefore, i2 is -1. Similarly, i3 is i2i, which is -i and i4 is i2i2 which happens to be 1. As you can notice in the above examples, we combine natural numbers with imaginary numbers to form complex numbers. In 6 – 5i, the real number component is 6, and the imaginary component is -5. If the real part of the complex number is 0, the number is called purely imaginary. For example, 10i, where the real number component is 0, and therefore, the complex number is called purely imaginary. Whereas, if the imaginary part of the complex number is 0, we consider the complex number as purely real. For example, 12, where the imaginary component is 0, and therefore, the complex number is called purely real. As we read that roots of the polynomial can be real or complex, one fact that we should know here is that complex roots always occur in pairs. This means that if we have a complex number, it will be a conjugate pair. Conjugate pairs are the pair of complex numbers where we can change the sign in the middle to get the other one. For example, if one part of the conjugate pair happens to be 5 + 6i, then the other part of the conjugate pair is 5 – 6i. Here we just changed the sign in the middle to obtain the other part. The other part of the conjugate pair is called the conjugate inverse. Therefore, if one of the roots of a polynomial is a complex number, the other root will definitely be the conjugate inverse of that complex number. So, if you happen to find one of the roots to be a + bi, then the other root will be a – bi. ## Bottom line Now that you have a fair idea of what a polynomial is, the degree of a polynomial, the roots of a polynomial, and a broad idea of what complex numbers are, it will be easier for you to understand the fundamental theorem of Algebra and all other concepts related to it. The concepts you learned here will be the basis of your learning of algebra, and these concepts are essential for solving questions. You can try to solve questions based on the concepts you learn at Cuemath and know more about algebra and all other important topics under algebra. If you wish to visit Cuemath and try a free trial class to learn algebra and much more from all the other topics of Math you can visit their website. If you have any more suggestions or want to share your experience of using the fundamental theorem of algebra in your work or studies, please comment below and let us know your experience! Previous article5 Best Gaming Monitor under 150 In 2021
## Introductory Algebra for College Students (7th Edition) $9\sqrt{2}+8\sqrt{3}$ Factor the radicands so that one of the factors is a perfect square: $=4\sqrt{4(2)}-\sqrt{64(2)}+2\sqrt{16(3)}+3\sqrt{9(2)} \\=4\sqrt{2^2(2)}-\sqrt{8^2(2)}+2\sqrt{4^2(3)}+3\sqrt{3^2(2)}$ Simplify to obtain: $\\=4\cdot 2\sqrt{2}-8\sqrt{2}+2\cdot 2\sqrt{3}+3\cdot 3\sqrt{2} \\=8\sqrt{2}-8\sqrt{2}+8\sqrt{3}+9\sqrt{2}$ Combine like terms: $=(8\sqrt{2} - 8\sqrt{2}+9\sqrt{2}) + 8\sqrt{3} \\=(8-8+9)\sqrt{2} + 8\sqrt{3} \\=9\sqrt{2}+8\sqrt{3}$
## Tuesday, September 16, 2008 ### Determining the CG One of the easiest ways to understand weight and balance is to consider a board with weights placed at various locations. We can determine the CG of the board and observe the way the CG changes as the weights are moved. The CG of a board like the one in Figure 2-4 may be determined by using these four steps: 1. Measure the arm of each weight in inches from the datum. 2. Multiply each arm by its weight in pounds to determine the moment in pound-inches of each weight. 3. Determine the total of all weights and of all the moments. Disregard the weight of the board. 4. Divide the total moment by the total weight to determine the CG in inches from the datum. In Figure 2-4, the board has three weights, and the datum is located 50 inches to the left of the CG of weight A. Determine the CG by making a chart like the one in Figure 2-5. As noted in Figure 2-5, A weighs 100 pounds and is 50 inches from the datum: B weighs 100 pounds and is 90 inches from the datum; C weighs 200 pounds and is 150 inches from the datum. Thus the total of the three weights is 400 pounds, and the total moment is 44,000 lb-in. Determine the CG by dividing the total moment by the total weight. To prove this is the correct CG, move the datum to a location 110 to the right of the original datum and determine the arm of each weight from this new datum, as in Figure 2-6. Then make a new chart similar to the one in Figure 2-7. If the CG is correct, the sum of the moments will be zero. The new arm of weight A is 110 - 50 = 60 inches, and since this weight is to the left of the datum, its arm is negative, or -60 inches. The new arm of weight B is 110 - 90 = 20 inches, and it is also to the left of the datum, so it is - 20; the new arm of weight C is 150 - 110 = 40 inches. It is to the right of the datum and is therefore positive. The board is balanced when the sum of the moments is zero. The location of the datum used for determining the arms of the weights is not important; it can be anywhere. But all of the measurements must be made from the same datum location. Determining the CG of an airplane is done in the same way as determining the CG of the board in the previous example. [Figure 2-8] Prepare the airplane for weighing and place it on three scales. All tare weight, that is, the weight of any chocks or devices used to hold the aircraft on the scales, is subtracted from the scale reading, and the net weight from each wheel weigh point is entered on the chart like the one in Figure 2-9. The arms of the weighing points are specified in the Type Certificate Data Sheet (TCDS) for the airplane in terms of stations, which are distances in inches from the datum. Tare weight also includes items used to level the aircraft. The empty weight of this aircraft is 5,862 pounds. Its EWCG, determined by dividing the total moment by the total weight, is located at fuselage station 201.1. This is 201.1 inches behind the datum.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: Algebra basics>Unit 7 Factoring quadratics is very similar to multiplying binomials, just going the other way. For example, x^2+3x+2 factors to (x+1)(x+2) because (x+1)(x+2) multiplies to x^2+3x+2. This article reviews the basics of how to factor quadratics into the product of two binomials. ### Example Factor as the product of two binomials. ${x}^{2}+3x+2$ Our goal is to rewrite the expression in the form: $\left(x+a\right)\left(x+b\right)$ Expanding $\left(x+a\right)\left(x+b\right)$ gives us a clue. $\begin{array}{rl}{x}^{2}+3x+2& =\left(x+a\right)\left(x+b\right)\\ \\ & ={x}^{2}+ax+bx+ab\\ \\ & ={x}^{2}+\left(a+b\right)x+ab\end{array}$ So $\left(a+b\right)=3$ and $ab=2$. After playing around with different possibilities for $a$ and $b$, we discover that $a=1$, $b=2$ satisfies both conditions. Plugging these in, we get: $\left(x+1\right)\left(x+2\right)$ And we can multiply the binomials to check our solution if we'd like: Yep, we get our original expression back, so we know we factored correctly to get our answer: $\left(x+1\right)\left(x+2\right)$ Want to see another example? Check out this video. ## Practice Factor the quadratic expression as the product of two binomials. ${x}^{2}-x-42=$ Want more practice? Check out this exercise. ## Want to join the conversation? • What are you supposed to do when the exponent that is squared has an integer before it? • Here is an example: 3x^2 + 31x + 10 The way you solve this is pretty simple. Since the 3x is squared, you multiply 3x by x and you get 3x^2. The important thing to know about this is that 3x^2 does NOT equal 3x * 3x, because that would be 9x^2. So when you factor this quadratic, you need to find a way to factor the 3x^2 into 3x * x. The answer for the quadratic is (3x + 1)(x + 10) • I'm so confused... How do we find a and b? Do we just guess or something, because that's what it looks like to me.... I feel stupid because I don't understand this... • It takes a while to get used to factoring polynomials. You need to be familiar with multiplying polynomials before you attempt factoring. Factoring reverses the processes used to multiply polynomials. You also need to be good at find all factor pairs for a number. Consider what happens in this example... Multiply (x + 3)(x + 5). You would do this using FOIL or expanded distribution. As you multiply, here's what happens. x(x) + 5x + 3x + 3(5) x(x) + (5+3)x + 3(5) x^2 + 8x + 15 I wrote the above in great detail because you need to see what the X's, the 3 and the 5 create (where they go in the final result). Factoring reverses this process. Look at the 2nd row above: x(x) + (5+3)x + 3(5) The 1st term: x(x) comes for the 1st term in both binomials. So, you know your binomials will look like: "(x + _ ) (x + _ )". The 2nd and 3rd term are created using the 3 and the 5. In the 3rd term they are multiplied. In the 2nd term, they are added. So, we learn to factor polynomials by looking at the 3rd term (the 15 in the final result) and finding 2 factors of 15 that also add to the 2nd term (the 8) Factors of 15 are: 1, 15 3, 5 Only 3 and 5 add to 8, thus we know the binomials need to contain the 3 and 5. Thus, our factors become "(x+3)(x+5)" 2nd example: let's factor x^2 + 11x + 24 We know the lead terms in our binomials have to be "x" and "x". So, we have "(x + _) (x + _)" We start with the 24. We need to find 2 factors of 24 that add to 11 (the middle term) This is where you need your factoring skills. Factors of 24 are: 1, 24 2, 12 3, 8 4, 6 Which pair adds to 11? The 3 and 8 add to 11, so that is the pair we want. Thus, our factors become "(x + 3)(x + 8)" Hope this helps. • How would you do this if the leading coefficient isn't one? I can't seem to find an article for that on Khan. - Thanks • My question was to factor -3x^2+6x+9 completely, and after I did all my work I ended up with the answer "-3(x+1)(x-3)" which I thought was correct, but then it told me I was wrong, so I looked at how they did the problem, and I saw we did the exact same steps, and we got the exact same answer. I don't know if it was just a glitch or something, but id love to know if I got it incorrect, or if it was Khan Academy that made a mistake. • You can always check your factors - Do the multiplication and see if you get back to the original polynomial. -3(x+1)(x-3) = -3(x^2-3x+x-3) = -3x^2+9x-3x+9 = -3x^2+6x+9 Looks like your factors are correct. Note: If you find something like this when you are doing a KA exercise, use the "report a problem" link within the exercise to report it to KA. • how do i do this if an expression only has two terms • If you have 2 terms, then you have a binomial. Assuming you have a quadratic (highest exponent is 2 on the variable), then your options to factor are: 1) A greatest common factor that includes the variable. For example: 3x^2-15x factors into 3x(x-5) 2) The binomial is a difference of 2 squares like: x^2-9 which factors into (x-3)(x+3) Hope this helps. • how do i factorise the expression 18pq-7+4pq-4 • First, you should simplify the polynomial because in its current form, it is not factorable. Combine like terms. 22pq-11 Now, you have a binomial with a common factor. Factor out the common factor. Can you identify the factor and do the factoring from here? Comment back if you have questions. • is there a way to make negatives easier • If all of the terms are negative, you could factor a negative. Ex: -x^2-4x-3 = -1(x^2+4x+3). If they aren't all negative, you just have to do it normally.
# What Is The Least Common Multiple Of 12 and 8: What Is The Least Common Multiple Of 12 and 8: The least common multiple of 12 and 8 is 24. This is the smallest number that both 12 and 8 can be divided by without leaving a remainder. The other multiples of 12 and 8 are 36, 48, 60, 72, and 84. While there is no specific formula for finding the least common multiple, there are several methods that can be used to help find it. One such method is to list all of the multiples of each number and then find the smallest number that is listed in both lists. Another method is to use a calculator or a division algorithm to find the smallest number that both numbers can be divided by evenly. Whichever method is used, the final result should always be 24. The lowest common multiple (LCM) is the smallest number that is divisible by all of the given numbers and is often used to find multiples or factors. The highest common factor (HCF), also known as the greatest common divisor, is the largest number that can divide two or more numbers with no remainder. Another term for HCF is gcd, short for “greatest common divisor”. LCM and HCF can be different if one of the original numbers equals zero. For example, the LCM of 4 and 10 is 20, but their HCF equals 2 since they are both even integers. ### What is the least common multiple of 8 and 12: the least common multiple of 8 and 12 is the lowest number that can be divided into both 8 and 12 without leaving a remainder. `” least common multiple “=” LCM “=” lcm = lowest factor for two integers ` (i) 2 times 6 is less than or equal to 12 :- `12\div2=6` –> No good (ii) 2 times 7 is greater than or equal to 8 :- `8\div2=4` –> No good ### What is the least common multiple of 8 and 10: `” least common multiple “=” LCM “=” lcm = lowest factor for two integers ` (i) 2 times 5 is less than or equal to 10 :- `10\div2=5` –> No good (ii) 2 times 6 is greater than or equal to 8 :- `8\div2=4` –> No good ### What is the least common multiple of 12 and 14: `” least common multiple “=” LCM “=” lcm = lowest factor for two integers ` (i) 2 times 7 is less than or equal to 14 :- `14\div2=7` –> No good (ii) 2 times 8 is greater than or equal to 12 :- `12\div2=6` –> No good ### What is the least common multiple of 18 and 24: `” least common multiple “=” LCM “=” lcm = lowest factor for two integers ` (i) 3 times 6 is less than or equal to 24 : `24\div3=8` –> Yes! Good answer. (e) 3 is a good answer for the least common multiple of 18 and 24. what is the least common multiple of 30 and 36:- `” least common multiple “=” LCM “=” lcm = lowest factor for two integers ` (i) 2 times 15 is less than or equal to 36 :- `36\div2=18` –> Yes! Good answer. (e) 2 is a good answer for the least common multiple of 30 and 36. So, the least common multiple of 8 and 12 is 6, the least common multiple of 10 and 12 is 5, the least common multiple of 12 and 14 is 7, and the least common multiple of 18 and 24 is Lisa has been a freelance journalist who has worked for various print magazine online. After years of spent working in the field of journalism, she took a plunge and founded Asap Land sharing the latest news bulletins from the field of Business and Technology as well as general headlines. She writes mostly the General US Headlines and Business News.
# Strategies Using Doubles ## Presentation on theme: "Strategies Using Doubles"— Presentation transcript: Strategies Using Doubles Addition and subtraction with the same partners is called doubles. Odd and Even Doubles What do you know about even numbers? Odd and Even Doubles How can we decide if 6 is an even or odd number? Odd and Even Doubles How can we decide if 7 is an even or odd number? The Doubles Plus 1 and Doubles Minus One Strategy! How might we use doubles to solve this problem? = ___ 7 + 6 = ___ Using a Double Plus 1 or Using a Double Minus 1 7 + 8 = ___ Using a Double Plus 1 or Using a Double Minus 1 6 + 5 = ___ Using a Double Plus 1 or Using a Double Minus 1 8 + 9 = ___ Using a Double Plus 1 or Using a Double Minus 1 Let’s Practice Two-Step Story Problems Work in pairs. Retell the problem. What do you do first? What do you do second? Show your work. Use a variety of strategies. Aretha has 7 whistles. She has 4 more whistles than Jeremy Aretha has 7 whistles. She has 4 more whistles than Jeremy. How many whistles do Aretha and Jeremy have altogether? Mina buys 4 shirts and 5 skirts. She returns 2 skirts to the store Mina buys 4 shirts and 5 skirts. She returns 2 skirts to the store. How many new pieces of clothing does Mina keep? Ed bought 9 bananas at one fruit stand and 3 at another Ed bought 9 bananas at one fruit stand and 3 at another. He bought 8 more bananas than Gail. How many did Gail buy? Felix had 7 pounds of white potatoes and 8 pounds of red potatoes to cook for the celebration. First he cooked 6 pounds. How many pounds of potatoes does he still have to cook? Extra Practice Solve the problems in the first section by yourself. Then, use the letters for each answer to help you solve the riddle. If you finish, use the letters on the page to create your own riddle. Make sure you number the lines for the answer to your riddle correctly. Give your riddle to a classmate to solve.
# How to Determine the Greatest Common Factor When Learning About Fractions Fractions! Does that word ship chills up your spine? Many students have trouble with fractions throughout their college careers. A number of these students never pretty get the concept of what a fragment is. A few have trouble remembering the one-of-a-kind rules for addition/subtraction vs. Multiplication/division. When is a commonplace denominator required? Why and what do we invert and multiply? A few college students have hassle with fractions due to the fact they do not recognize some of the key terms. “thing” is one of these key phrases. When operating with fractions, you can not find the greatest not unusual “element” in case you do not know what a factor is, when you want one, or what to do with if you do discover it. The word “issue” reasons special issues because it is each a noun–a factor, and a verb–an movement. Whether used as a noun or a verb, the phrase “aspect” must always make you think of multiplication. The phrase thing constantly entails multiplication in some manner: 1st, the verb (an motion): to thing manner to rewrite as or the usage of multiplication. As a result, we should aspect 12 as 2 x 6 or 3 x four. The trouble of getting distinctive methods to factor the same number is solved through factoring as much as viable. This indicates we issue (rewrite as multiplication) till each number is high. So 12 = 2 x 6 = 2 x 2 x three or 12 = three x 4 = three x 2 x 2. Word: both variations are the same. Every quantity has one and only one manner it will issue as primes, disregarding the order of the elements. Second, the noun (a aspect): a element is a number that can be increased by using a second number to provide a third. As an example, 3 is a component of 15 due to the fact three x five = 15. This could also be idea of in terms of department: a element divides calmly into any other wide variety Fractional CMO. Instance: 7 is a factor of 28 due to the fact 28/7 = 4. Now we’re equipped to talk about the “finest common thing” (gcf) of a fraction. First, we want to examine every word. On this utilization, the word “element” is a noun–a variety of. The word “commonplace” method shared among numbers. “best” manner largest feasible. On the grounds that we are working with fractions, the 2 numbers worried are the numerator (top) and denominator (backside) of the fraction. Consequently, the greatest not unusual element of a fraction is the most important variety that may be a element of each the numerator and the denominator. Before taking place, pass back and re-study thus far numerous times till you can give an explanation for this a whole lot to a person else. Only when you could give an explanation for this out loud will you be equipped to move on. I will wait at the same time as you practice. Prepared? Ok! You have the concept of what a best common element is. Now, the query is why is it critical? What can we do with a gcf? The answer to that question is that we lessen fractions. This indicates rewriting a fragment with the smallest feasible numbers with out changing its unique cost. Allow’s observe some examples. Lessen the fraction 4/6. I am sure that you could tell simply with the aid of searching that the biggest component shared by 4 and six is 2. Rewrite 4/6 in factored shape: (2 x 2)/(2 x 3). At this factor, a few teachers allow college students to simply mark out the not unusual aspect 2 and write the reduced model 2/3. Different teachers assume students to rewrite the authentic fraction in a form that has separate fractions with the common element fraction “being a 1.” this will seem like: four/6 = (2 x 2)/(2 x 3) = (2/2)x(2/3) = 1 x (2/3) = 2/three. Each techniques are correct. The latter approach is extra explanatory of the technique. Now, what if you can’t fast locate the gcf? Example: reduce forty/forty eight. Then, we want to find the gcf. We want to aspect each 40 and 48 to a product of primes. Forty = four x 10 = 2 x 2 x 2 x five and forty eight = 6 x eight = 2 x three x 2 x 2 x 2. We can see that 40 and forty eight percentage common factors 2, 2, and a couple of. So the gcf is two x 2 x 2 = eight, and forty/48 = (eight x five)/(eight x 6) = five/6. In reality, the purpose for locating a gcf is to reduce a fragment in one step. Of route, that “one step” would not depend all the steps it takes to discover the gcf. The large secret is that it isn’t vital to find the gcf to lessen a fraction. Fractions may be reduced in steps by using the use of any not unusual aspect. For example: reduce 36/forty eight. On account that both numbers are even, reduce by using 2. So, 36/forty eight = 18/24. Each are even again, so use 2 again. 18/24 = 9/12. Now three is a not unusual factor. 9/12 = three/4. In end, if the guidelines you’re given say “find the gcf,” you must use the earlier method for locating the gcf. If the directions say “reduce the fraction the usage of the gcf,” you once more must use the gcf. But, if the instructions really say “reduce the fraction completely,” the method is your preference. I prefer reducing in steps! Locating the gcf takes an excessive amount of time–for my part! How to Determine the Greatest Common Factor When Learning About Fractions Scroll to top
## Mathematics Overview Reference: Course on Subject Clearing ## Introduction This video provides an Overview of Mathematics in clear and precise terms. What are the basic parts of MATHEMATICS? The basic parts of MATHEMATICS are: 1. ARITHMETIC 2. GEOMETRY 3. ALGEBRA . What is ARITHMETIC? In Arithmetic we learn about numbers and how to add, subtract, multiply and divide them. The word ARITHMETIC literally means “number skill.” NUMBER SKILL: Find the total of 97 pennies and 64 pennies. 1. Imagine two stacks of 97 and 64 pennies. 2. Transfer 3 pennies from the 64-penny stack to 97-penny stack. 3. You now have two stacks of 100 and 61 pennies 4. We can add this quickly as 161 pennies. 5. Therefore, the sum of 97 and 64 is 161. One learns many such number skills in Arithmetic. . What is GEOMETRY? In Geometry, we study the relationships in space so we can build things. The word GEOMETRY literally means “to measure land.” MEASURMENT SKILL: Using angles one can find the height of a tree from a distance. 1. We move to a certain distance from the tree. 2. We then measure the angle of sight to the top of the tree. 3. We move to a place where this angle is 45 degrees. 4. Then distance from the tree plus your height will be same as the height of the tree. One learns many such relationships in Geometry. . What is ALGEBRA? In Algebra, we use relationships to figure out unknown values. The word ALGEBRA literally means “binding together.” RELATIONSHIP SKILL: Find the relationship between Sam and his mother’s age. 1. When Sam was born his mother was 30 years old. 2. When Sam was 5, his mother was 35 years old. 3. When Sam was 10, his mother was 40 years old. 4. Sam’s mother will always be 30 years older than Sam. 5. When Sam is ‘x’ years old, his mother would be ‘x + 30’ years old. 6. Therefore, when Sam is 40 years old, his mother would be 70 years old. One learns many such relationships in Algebra. . 1. What are the main parts of Mathematics? The main parts of mathematics are Arithmetic, Geometry and Algebra. . 2. Which part of mathematics do you study first in your childhood? We study Arithmetic or “number skill” first in our childhood. . 3. What is Geometry useful for? Geometry is useful for measuring things in space, such as lengths, widths, heights, and directions. . 4. How is Algebra different from Arithmetic? Arithmetic teaches skill with numbers. Algebra helps to find an unknown value from a given relationship. . ## Final Thought Mathematics is one subject. But it can be understood quickly by looking at parts of it more closely. . Here is an interesting video: https://www.youtube.com/watch?v=OmJ-4B-mS-Y .
# Question: What Are All The Names Of Angles? ## How do you find the name of an angle? (1) We can name angles by using THREE capital letters like: ABC or DEF. The middle letter is called the VERTEX of the angle. The above angles are read “angle ABC” and “angle DEF.” This leads us to the second way we can name angles. (2) We can name angles by using the vertex.. ## What are the sides of an angle? The sides of an angle refer to the two rays or line segments that form the angle. In the figure below, rays BA and BC are the sides of angle ABC. An angle is formed by rotating a ray around its endpoint. ## What are 3 ways to name an angle? The best way to describe an angle is with three points. One point on each ray and the vertex always in the middle. That angle could be NAMED in three ways: X, BXC, or CXB. Adjacent angles are two angles that have a common vertex, a common side, and no common interior points. ## Which angle is largest? The sides of the triangle are given. Can you determine which angle is the largest? As you might guess, the largest angle will be opposite 18 because it is the longest side. Similarly, the smallest angle will be opposite the shortest side, 7. ## How do you name a straight angle? A straight angle is also called ‘flat angle’. In the picture, OX and OY are the arms of an angle. The common point O where the rays meet is called the vertex. ## What is a straight angle? : an angle whose sides lie in opposite directions from the vertex in the same straight line and which equals two right angles. ## How do you classify an angle? Angles can be either straight, right, acute or obtuse. An angle is a fraction of a circle where the whole circle is 360°. A straight angle is the same as half the circle and is 180° whereas a right angle is a quarter of a circle and is 90°. You measure the size of an angle with a protractor. ## How many sides does 2 adjacent angles have? Adjacent angles are two angles that have a common vertex and a common side but do not overlap. ## What are the 6 types of angles? Types of AnglesAcute Angle: An angle whose measure is less than 90° is called an acute angle.Right Angle: An angle whose measure is 90° is called right angle.Obtuse Angle: An angle whose measure is greater than 90° but less than 180° is called an obtuse angle. … Straight Angle: … Reflex Angle: … Zero Angle: ## What are 4 ways to name an angle? How to Name an Angle in Four WaysLabel an angle using the vertex and a point from each arm.Label an angle using only the vertex.Label an angle using numbers. ## What are the 10 types of angles? Acute Angle – An angle less than 90 degrees. Right Angle – An angle that is exactly 90 degrees. Obtuse Angle – An angle more than 90 degrees and less than 180 degrees….Types of Angles – Acute, Right, Obtuse, Straight and Reflex…Acute angle.Right angle.Obtuse angle.Straight angle.Reflex angle. ## What is the shortest side of a right triangle called? The two shorter sides are usually called “legs.” This formula is called the Pythagorean Theorem in honor of Pythagoras. It is usually written as the equation below, where a and b are the measures of the legs of the triangle and c is the measure of the hypotenuse. ## What is the longest side of a right triangle called? hypotenuseThe hypotenuse of a right triangle is always the side opposite the right angle. It is the longest side in a right triangle. The other two sides are called the opposite and adjacent sides. ## What is a zero angle? An angle with a measure of zero degrees is called a zero angle. If this is hard to visualize, consider two rays that form some angle greater than zero degrees, like the rays in the . Then picture one of the rays rotating toward the other ray until they both lie in the same line. ## What is a 360 degrees angle called? Angles between 180 and 360 degrees (180°< θ < 360°) are called reflex angles. • Angles that are 360 degrees (θ = 360°) are full turn. Page 2. ## How many ways can an angle be named? three waysThe best way to describe an angle is with three points. One point on each ray and the vertex always in the middle. That angle could be NAMED in three ways: X, BXC, or CXB. Adjacent angles are two angles that have a common vertex, a common side, and no common interior points. ## What are two ways to name an angle? Angles are named in two ways. You can name a specific angle by using the vertex point, and a point on each of the angle’s rays. The name of the angle is simply the three letters representing those points, with the vertex point listed in the middle. You can also name angles by looking at their size. ## What are all the acceptable ways to name a line? Line segments are commonly named in two ways:By the endpoints. In the figure above, the line segment would be called PQ because it links the two points P and Q. Recall that points are usually labelled with single upper-case (capital) letters. … By a single letter. The segment above would be called simply “y”. ## Which angle is the smallest? The smallest angle is 1. It is acute and is smaller than a right angle. The largest angle is 2. It is obtuse and bigger than a right angle. ## How do you name the vertex of an angle? Naming conventions for angles The vertex of an angle is the common endpoint of two rays that make up the angle’s sides. The vertex for angle BAC, written ∠BAC, is point A. The angle can also be named as ∠CAB or by only its vertex, ∠A.
# Unit circle and 3 trig functions Lesson In right-angled triangle trigonometry, we can only deal with angles whose sizes are between $0^\circ$0° and $90^\circ$90°. In the chapter on Angles of Any Magnitude, it is explained how the sine, cosine and tangent functions are given a more general definition so that they can be applied to angles that are impossible in right-angled triangle trigonometry. The functions are defined in terms of the coordinates of a point that is free to move on the unit circle, in the following way. The radius, drawn to the point on the circle, makes an angle with the positive horizontal axis. By convention, the angle is measured anticlockwise from the axis. It can have any real value, positive or negative. We define the cosine of the angle to be the horizontal coordinate of the point, and we define the sine of the angle to be the vertical coordinate of the point. Then, the tangent of the angle $\alpha$α is defined to be the fraction $\frac{\sin\alpha}{\cos\alpha}$sinαcosα. By considering the coordinates of a point as it moves around the unit circle, we see that the sine and cosine functions have either a positive or a negative sign depending on which quadrant the angle is in. The signs depend on the signs of the coordinates involved. Have a look at the following applet, you can make settings for degrees or radians, and whether you want to rotate positively (anticlockwise) or negatively (clockwise). As you explore, see if you can determine where sine, cosine and tangent are positive or negative. The sine function is positive in the first and second quadrants and negative in the others while the cosine function is positive in the first and fourth quadrants. Consequently, the tangent function is positive in the first and third quadrants. These can be remembered by having a mental picture of the unit circle diagram or by means of the mnemonic ASTC: 'All Stations To Central: All-Sine-Tangent-Cosine' that shows which functions are positive in each quadrant. These facts become important when trigonometric equations are being solved for all the solutions within a given range. #### Example 1 Express $\cos117^\circ$cos117° in terms of a first quadrant angle. The angle $117^\circ$117° is between $90^\circ$90° and $180^\circ$180°, so it is in the second quadrant. The point representing $117^\circ$117° on the unit circle diagram, where the radius cuts the circle, must have a negative horizontal coordinate. Therefore, $\cos117^\circ$cos117° must be the same as $-\cos\left(180^\circ-117^\circ\right)=-\cos63^\circ$cos(180°117°)=cos63°. #### Example 2 Express the sine, cosine and tangent functions of the angle $512^\circ$512° in terms of an angle in the first quadrant written in radian form. The angle $512^\circ$512° is more than once around the full circle. So, it is equivalent to $512^\circ-360^\circ=152^\circ$512°360°=152°, which is in the second quadrant. We subtract the angle from $180^\circ$180° to find $\sin512^\circ=\sin28^\circ$sin512°=sin28° $\cos512^\circ=-\cos28^\circ$cos512°=cos28° $\tan512^\circ=-\tan28^\circ$tan512°=tan28° #### More Worked Examples ##### QUESTION 1 Which of the ratios listed have positive values? A: $\sin31^\circ$sin31° B: $\tan31^\circ$tan31° C: $\cos267^\circ$cos267° D: $\sin267^\circ$sin267° E: $\cos180^\circ$cos180° 1. $D$D and $E$E A $A$A and $B$B B $A$A and $C$C C $B$B and $C$C D $D$D and $E$E A $A$A and $B$B B $A$A and $C$C C $B$B and $C$C D ##### QUESTION 2 Which of the following will have positive answers? Select all correct answers. 1. $\tan296^\circ$tan296° A $\sin120^\circ$sin120° B $\cos91^\circ$cos91° C $\sin296^\circ$sin296° D $\cos120^\circ$cos120° E $\cos296^\circ$cos296° F $\tan296^\circ$tan296° A $\sin120^\circ$sin120° B $\cos91^\circ$cos91° C $\sin296^\circ$sin296° D $\cos120^\circ$cos120° E $\cos296^\circ$cos296° F ##### QUESTION 3 For each of the following, rewrite the expression as the trigonometric ratio of a positive acute angle. You do not need to evaluate the trigonometric ratio. 1. $\sin93^\circ$sin93° 2. $\cos195^\circ$cos195° 3. $\tan299^\circ$tan299°
# 7.4: Linear Systems with Addition or Subtraction Difficulty Level: Basic Created by: CK-12 Estimated12 minsto complete % Progress Practice Linear Systems with Addition or Subtraction MEMORY METER This indicates how strong in your memory this concept is Progress Estimated12 minsto complete % Estimated12 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Suppose two extended families went to an amusement park, where adult tickets and children's tickets have different prices. If the first family had 5 adults and 8 children and paid a total of $124, and the second family had 5 adults and 12 children and paid$156, how much did each type of ticket cost? Could you write a system of equations representing this situation? If you wanted to solve the system by elimination, how would you go about doing it? In this Concept, you'll learn how to use elimination to solve a system of linear equations similar to the one representing this scenario by addition or subtraction. ### Watch This Multimedia Link: For help with solving systems by elimination, visit this site: http://www.teachertube.com/viewVideo.php?title=Solving_System_of_Equations_using_Elimination&video_id=10148 Teacher Tube video. ### Guidance As you noticed in the previous Concept, solving a system algebraically will give you the most accurate answer and in some cases, it is easier than graphing. However, you also noticed that it took some work in several cases to rewrite one equation before you could use the Substitution Property. There is another method used to solve systems algebraically: the elimination method. The purpose of the elimination method to solve a system is to cancel, or eliminate, a variable by either adding or subtracting the two equations. This method works well if both equations are in standard form. #### Example A If one apple plus one banana costs $1.25 and one apple plus two bananas costs$2.00, how much does it cost for one banana? One apple? Solution: Begin by defining the variables of the situation. Let \begin{align}a=\end{align} the number of apples and \begin{align}b=\end{align} the number of bananas. By translating each purchase into an equation, you get the following system: \begin{align}\begin{cases} a+b=1.25\\ a+2b=2.00 \end{cases}\end{align}. You could rewrite the first equation and use the Substitution Property here, but because both equations are in standard form, you can also use the elimination method. Notice that each equation has the value \begin{align}1a\end{align}. If you were to subtract these equations, what would happen? \begin{align}& \qquad a + b \ =1.25\\ &\underline{\;\;\;- (a+2b=2.00)\;\;\;}\\ & \qquad \quad -b =-0.75\\ & \qquad \qquad \ b =0.75\end{align} Therefore, one banana costs 0.75, or 75 cents. By subtracting the two equations, we were able to eliminate a variable and solve for the one remaining. How much is one apple? Use the first equation and the Substitution Property. \begin{align}a+0.75&=1.25\\ a&=0.50 \rightarrow one \ apple \ costs \ 50 \ cents\end{align} #### Example B Solve the system \begin{align}\begin{cases} 3x+2y=11\\ 5x-2y=13\end{cases}\end{align}. Solution: These equations would take much more work to rewrite in slope-intercept form to graph, or to use the Substitution Property. This tells us to try to eliminate a variable. The coefficients of the \begin{align}x-\end{align}variables have nothing in common, so adding will not cancel the \begin{align}x-\end{align}variable. Looking at the \begin{align}y-\end{align}variable, you can see the coefficients are 2 and –2. By adding these together, you get zero. Add these two equations and see what happens. \begin{align}& \qquad \ 3x+2y =11\\ &\underline{\;\; + \ (5x-2y) =13 \;\;}\\ & \qquad \ 8x+0y =24\end{align} The resulting equation is \begin{align}8x=24\end{align}. Solving for \begin{align}x\end{align}, you get \begin{align}x=3\end{align}. To find the \begin{align}y-\end{align}coordinate, choose either equation, and substitute the number 3 for the variable \begin{align}x\end{align}. \begin{align}3(3)+2y&=11\\ 9+2y&=11\\ 2y&=2\\ y&=1\end{align} The point of intersection of these two equations is (3, 1). #### Example C Andrew is paddling his canoe down a fast-moving river. Paddling downstream he travels at 7 miles per hour, relative to the river bank. Paddling upstream, he moves slower, traveling at 1.5 miles per hour. If he paddles equally hard in both directions, calculate, in miles per hour, the speed of the river and the speed Andrew would travel in calm water. Solution: We have two unknowns to solve for, so we will call the speed that Andrew paddles at \begin{align}x\end{align}, and the speed of the river \begin{align}y\end{align}. When traveling downstream, Andrew's speed is boosted by the river current, so his total speed is the canoe speed plus the speed of the river \begin{align}(x+y)\end{align}. Upstream, his speed is hindered by the speed of the river. His speed upstream is \begin{align}(x-y)\end{align}. \begin{align}\text{Downstream Equation} && x+y&=7\\ \text{Upstream Equation} && x-y&=1.5\end{align} Notice \begin{align}y\end{align} and \begin{align}-y\end{align} are additive inverses. If you add them together, their sum equals zero. Therefore, by adding the two equations together, the variable \begin{align}y\end{align} will cancel, leaving you to solve for the remaining variable, \begin{align}x\end{align}. \begin{align}& \qquad \ x+y=7\\ &\underline{\;\; + \ (x-y)=1.5 \;\;}\\ & \quad \ 2x+0y=8.5\\ & \qquad \quad \ \ 2x=8.5\end{align} Therefore, \begin{align}x=4.25\end{align}; Andrew is paddling \begin{align}4.25 \ miles/hour\end{align}. To find the speed of the river, substitute your known value into either equation and solve. \begin{align}4.25-y&=1.5\\ -y&=-2.75\\ y&=2.75\end{align} The stream’s current is moving at a rate of \begin{align}2.75 \ miles/hour\end{align}. ### Guided Practice Solve the system \begin{align}\begin{cases} 5s+2t=6\\ 9s+2t=22\end{cases}\end{align}. Solution: Since these equations are both written in standard form, and both have the term \begin{align}2t\end{align} in them, we will will use elimination by subtracting. This will cause the \begin{align}t\end{align} terms to cancel out and we will be left with one variable, \begin{align}s\end{align}, which we can then isolate. \begin{align}& \qquad \ 5s+2t=6\\ &\underline{\;\; - \ (9s+2t = 22) \;\;}\\ & \qquad \ -4s+0t =-16\\ & \qquad \ -4s=-16\\ & \qquad \ s=4\end{align} \begin{align}5(4)+2t&=6\\ 20+2t&=6\\ 2t&=-14\\ t&=-7\end{align} The solution is \begin{align}(4,-7)\end{align}. ### Practice Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Solving Linear Systems by Elimination (12:44) 1. What is the purpose of the elimination method to solve a system? When is this method appropriate? In 2 – 10, solve each system using elimination. 1. \begin{align}\begin{cases} 2x+y=-17\\ 8x-3y=-19 \end{cases}\end{align} 2. \begin{align}\begin{cases} x+4y=-9\\ -2x-5y=12 \end{cases}\end{align} 3. \begin{align}\begin{cases} -2x-5y=-10\\ x+4y=8 \end{cases}\end{align} 4. \begin{align}\begin{cases} x-3y=-10\\ -8x+5y=-15 \end{cases}\end{align} 5. \begin{align}\begin{cases} -x-6y=-18\\ x-6y=-6 \end{cases}\end{align} 6. \begin{align}\begin{cases} 5x-3y=-14\\ x-3y=2 \end{cases}\end{align} 7. \begin{align}&3x+4y=2.5\\ &5x-4y=25.5 \end{align} 8. \begin{align}&5x+7y=-31\\ &5x-9y=17 \end{align} 9. \begin{align}&3y-4x=-33\\ &5x-3y=40.5 \end{align} 10. Nadia and Peter visit the candy store. Nadia buys three candy bars and four fruit roll-ups for2.84. Peter also buys three candy bars, but he can afford only one fruit roll-up. His purchase costs $1.79. What is the cost of each candy bar and each fruit roll-up? 11. A small plane flies from Los Angeles to Denver with a tail wind (the wind blows in the same direction as the plane), and an air-traffic controller reads its ground-speed (speed measured relative to the ground) at 275 miles per hour. Another identical plane moving in the opposite direction has a ground-speed of 227 miles per hour. Assuming both planes are flying with identical air-speeds, calculate the speed of the wind. 12. An airport taxi firm charges a pick-up fee, plus an additional per-mile fee for any rides taken. If a 12-mile journey costs$14.29 and a 17-mile journey costs $19.91, calculate: 1. the pick-up fee 2. the per-mile rate 3. the cost of a seven-mile trip 13. Calls from a call-box are charged per minute at one rate for the first five minutes, and then at a different rate for each additional minute. If a seven-minute call costs$4.25 and a 12-minute call costs $5.50, find each rate. 14. A plumber and a builder were employed to fit a new bath, each working a different number of hours. The plumber earns$35 per hour, and the builder earns $28 per hour. Together they were paid$330.75, but the plumber earned $106.75 more than the builder. How many hours did each work? 15. Paul has a part-time job selling computers at a local electronics store. He earns a fixed hourly wage, but he can earn a bonus by selling warranties for the computers he sells. He works 20 hours per week. In his first week, he sold eight warranties and earned$220. In his second week, he managed to sell 13 warranties and earned \$280. What is Paul’s hourly rate, and how much extra does he get for selling each warranty? Mixed Review 1. Baxter the golden retriever is lying in the sun. He casts a shadow of 3 feet. The doghouse he is next to is 3 feet tall and casts an 8-foot shadow. What is Baxter's height? 2. A botanist watched the growth of a lily. At 3 weeks, the lily was 4 inches tall. Four weeks later, the lily was 21 inches tall. Assuming this relationship is linear: 1. Write an equation to show the growth pattern of this plant. 2. How tall was the lily at the 5.5-week mark? 3. Is there a restriction on how high the plant will grow? Does your equation show this? 3. The “Wave” is an exciting pasttime at football games. To prepare, students in a math class took the data in the table below. 1. Find a linear regression equation for this data. Use this model to estimate the number of seconds it will take for 18 students to complete a round of the wave. 2. Use the method of interpolation to determine the amount of time it would take 18 students to complete the wave. \begin{align}&s \ (\text{number of students in wave})&& 4 && 8 && 12 && 16 && 20 && 24 && 28 && 30\\ & t \ (\text{time in seconds to complete on full round}) && 2 && 3.2 && 4 && 5.6 && 7 && 7.9 && 8.6 && 9.1\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition elimination The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable. Show Hide Details Description Difficulty Level: Basic Tags: Subjects:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Words that Describe Patterns ## Use variable expressions to solve real-world problems. 0% Progress Practice Words that Describe Patterns Progress 0% Words that Describe Patterns Many bicyclists have biking watches that are able to record the time spent biking and the distance traveled. They are able to download the data recorded by their watches to a computer and view a table with times in minutes in one column and distances in miles in another column. How could they use this data to write an English sentence or an algebraic expression? After completing this Concept, you'll be able to answer this question and use the algebraic expression to predict the distance traveled for any time spent biking. ### Guidance Using Words to Describe Patterns Sometimes patterns are given in tabular format (meaning presented in a table). An important job of data analysts is to describe a pattern so others can understand it. #### Example A Using the table below, describe the pattern in words. $&& x && -1 && 0 && 1 && 2 && 3 && 4\\&& y && -5 && 0 && 5 && 10 && 15 && 20$ Solution: We can see from the table that $y$ is five times bigger than $x$ . Therefore, the pattern is that the “ $y$ value is five times larger than the $x$ value.” #### Example B Zarina has a $100 gift card and has been spending money in small regular amounts. She checks the balance on the card at the end of every week and records the balance in the following table. Using the table, describe the pattern in words and in an expression. Week # Balance ($) 1 78 2 56 3 34 Solution: Each week the amount of her gift card is $22 less than the week before. The pattern in words is: “The gift card started at$100 and is decreasing by $22 each week.” As we saw in the last lesson, this sentence can be translated into the algebraic expression $100-22w$ . #### Example C The expression found in example 2 can be used to answer questions and predict the future. Suppose, for instance, that Zarina wanted to know how much she would have on her gift card after 4 weeks if she used it at the same rate. By substituting the number 4 for the variable $w$ , it can be determined that Zarina would have$12 left on her gift card. Solution: $100-22w$ When $w = 4$ , the expression becomes: $&100-22(4)\\&100-88\\&12$ After 4 weeks, Zarina would have \$12 left on her gift card. ### Guided Practice Jose starts training to be a runner. When he starts, he can run 3 miles per hour. After 5 weeks of training, Jose can run faster. After each week, he records his average speed while running. He summarizes this information in the following table: Week # Average Speed (miles per hour) 1 3.25 2 3.5 3 3.75 4 4.0 5 4.25 Write an expression for Jose's increased speed and predict how fast he will be able to run after 6 weeks. Solution: We will use $w$ to represent the number or weeks. Jose's speed starts at 3 mph, and from the table we can see that it increases by 0.25 miles per hour every week. This gives us the expression $3+0.25w$ . Now we substitute in $w=6$ and get the following: $3+0.25(6)$ $3+1.5=4.5$ If Jose keeps up his training, by the end of the 6th week, he should be able to run 4.5 miles per hour. ### Practice Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Patterns and Equations (13:18) In questions 1–3, write the pattern of the table: a) in words and b) with an algebraic expression. 1. Number of workers and number of video games packaged $&\text{People} && 0 && 1 && 2 && 5 && 10 && 50 && 200\\&\text{Amount} && 0 && 65 && 87 && 109 && 131 && 153 && 175$ 1. The number of hours worked and the total pay $&\text{Hours} && 1 && 2 && 3 && 4 && 5 && 6\\&\text{Total Pay} && 15 && 22 && 29 && 36 && 43 && 50$ 1. The number of hours of an experiment and the total number of bacteria $&\text{Hours} && 0 && 1 && 2 && 5 && 10\\&\text{Bacteria} && 0 && 2 && 4 && 32 && 1024$ 1. With each filled seat, the number of people on a Ferris wheel doubles. 1. Write an expression to describe this situation. 2. How many people are on a Ferris wheel with 17 seats filled? 2. Using the theme park situation from the lesson, how much revenue would be generated by 2,518 people? Mixed Review 1. Use parentheses to make the equation true: $10+6 \div 2-3=5$ . 2. Find the value of $5x^2 - 4y$ for $x = -4$ and $y = 5$ . 3. Find the value of $\frac{x^2y^3}{x^3 + y^2}$ for $x = 2$ and $y=-4$ . 4. Simplify: $2 - (t - 7)^2 \times (u^3 - v)$ when $t = 19, u = 4$ , and $v = 2$ . 5. Simplify: $2 - (19 - 7)^2 \times (4^3 - 2)$ . ### Vocabulary Language: English Spanish tabular tabular Being represented in a table. $\ge$ $\ge$ The greater-than-or-equal-to symbol "$\ge$" indicates that the value on the left side of the symbol is greater than or equal to the value on the right. $\le$ $\le$ The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right. $\ne$ $\ne$ The not-equal-to symbol "$\ne$" indicates that the value on the left side of the symbol is not equal to the value on the right. constant constant A constant is a value that does not change. In Algebra, this is a number such as 3, 12, 342, etc., as opposed to a variable such as x, y or a. Equation Equation An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs. greater than greater than The greater than symbol, $>$, indicates that the value on the left side of the symbol is greater than the value on the right. greater than or equal to greater than or equal to The greater than or equal to symbol, $\ge$, indicates that the value on the left side of the symbol is greater than or equal to the value on the right. inequality inequality An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$. less than less than The less-than symbol "<" indicates that the value on the left side of the symbol is lesser than the value on the right. less than or equal to less than or equal to The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right. not equal to not equal to The "not equal to" symbol, $\ne$, indicates that the value on the left side of the symbol is not equal to the value on the right. Variable Variable A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
You are on page 1of 6 # Decision Making using Probability Introduction Last time we looked at simple probabilities and how we might derive them. This week we will look at more complicated notions of probability and how we can use probability in order to aid in management decision making. Conditional Probability So far we have only considered probabilities in isolation or where two events are independent, like two rolls of a dice. However in reality many events are related, for example the probability of it raining in 5 minutes time is dependent on whether or not it is raining now. If we have two events A and B, then P (A\|B) is the probability of A given B, that is the probability that A will happen given the fact that B has already happened, for example P (Rain in 5 minutes\|Raining now) Consider the following example, if 10 people are short listed for a job, if you assume you are all equally qualied for the post then initially you would rate your probability of getting the job as 0.1. If you then found out that out the person who got the job was a woman and you knew there were 6 women on the short list, then your probability of getting the job would change based on this information. 1 P (Job\|Woman) = 6 P (Job\|Man) = 0. You have adjusted your probabilities conditioning on the new information you have received. Generally if P (A\|B) is the probability of A occurring given the fact that B has already occurred then P (A\|B) = P (A and B) P (B) If we consider a simple example rst, what is the probability of a card draw at random form a standard pack being a picture card, given the fact that it is a heart. We could just write this down. P (Picture \| Heart) = 1 3 13 However using the above formula 13 52 12 P ( Picture ) = 52 13 12 3 P ( Picture and Heart ) = = 52 52 52 P ( Picture and Heart ) 3 P (Picture \| Heart) = = P ( Heart ) 13 P ( Heart ) = This obviously is a very simple example of conditional probabilities, consider the following example. A record shop records the age and sex of 200 customers who buy CD singles. Male Female < 30 30 50 50+ 55 25 5 65 35 15 If a customer is selected at random what is the probability they are male and aged between 30 and 50. P (Male and 30 - 50) = 25 1 = 200 8 If a customer is selected at random what is the probability that they are aged between 30 and 50. P (30 - 50) = (25 + 35) 3 = 200 10 Given the fact that the customer is aged between 30 and 50, what is the probability they are male? P (Male \| 30 - 50) = P (Male and 30 - 50) 5 = P (30 - 50) 12 ## 0.1 Tree Diagrams Tree diagrams or probability trees are simple clear ways of presenting probability information, especially if the information is conditional. Lets us rst consider an independent example. If we roll a dice twice what is the probability that we score a six on both rolls, this is in reality simple and we can just write this down as P (Six and Six) = 1 1 1 = 6 6 36 Consider drawing this as a tree diagram. An experiment with outcomes which can be represented with probabilities are represented by circles, called nodes and the outcomes as branches. Then the probability of a six followed by a six is found by tracing that branch through the tree. We could also nd the probability of at least one six, or the probability of exactly one six. Consider a more complex example. If a worker arrives for work late there is a one in four chance of him being caught. The rst time he is caught he is given a caution, the second time he is sacked. Assuming he is late three days in a row we can express this as a tree diagram. Using this diagram it is easy to calculate a whole set of probabilities. The termination of each of the branches of the tree are usually know as terminal nodes. The rst step is usually to calculate the probabilities at each of these nodes. We can then look at the probabilities of several different outcomes. P ( Not being Caught ) = P ( Late 3 times but not sacked ) = P ( Not caught 2nd and 3rd time \| Caught 1st ) = Where such diagrams might be useful in real world situations is looking at the possibility of being given a false negative in a medical test or of rejecting a perfect component in a production run. A company is keen to increase the quality of its products as it has incurred large warranty claims. The production manager believes that 2 % of the output is defective. He has introduced a new quality control system which inspects nished goods before they are despatched. This system is however not fool proof and he believes that 1 % of perfect items are classed as rejects and 0.3 % of defective items are classed as perfect. This gives us two conditional probabilities. P (Reject \| Perfect) = 0.01 P (Perfect \| Reject) = 0.003 We can set this up in a probability tree The company might be interested in the probability of a defective item getting to a customer. P (Defective and Accepted) = This compares with the probability of 0.02 before the new quality control system was introduced. ## Expected Monetary Value and Decision Trees The basic structure of probability trees can be progressed further to allow us to consider making decisions using probability. To do this we need one further concept, which is the idea of Expected Monetary Value (EMV). The expected monetary value of a single event is simply the probability of that event multiplied by the monetary value of that outcome. So if you had to pay out 5 if you pulled an ace from a pack of cards the EMV of that would be 1 5 = 0.38 13 EM V (Ace) = This means that if you repeated this bet a large number of times you would come out 38 pence down. What however if there was an upside to the bet. If you rolled a six on a dice you had to pay 5 but if you rolled any other number you were paid 2.50. Would you take on this bet? Probability P (6) = 1 6 5 6 Outcome - 5 2.50 ## P (Not a 6) = This gives 1 5.00 = 0.833 6 5 EM V ( Not a Six ) = 2.50 = 2.0833 6 EM V ( Six ) = and hence the expected monetary value of the bet is EM V ( Bet ) = 0.833 + 2.083 = 1.25 So in the long run this would be a bet to take as has a positive expected monetary value. The expected monetary value of a project or bet is given by the following formula EM V = P (Event) Monetary value of event. The expected monetary value of a project can be used as a major decision criteria in a large number of applications. A small company is trying to decide how to launch a new and innovative product. It could go for a direct approach launching onto the whole of the domestic market through traditional distribution channels, or it could launch only on the internet. A third option exists of licensing the product 4 to a larger company who will pay a licence fee irrespective of the success of the product. How should the company launch the product? The company has done some initial market research and the managing director believes the probability of the product being successful can be classed into three categories, high, medium or low. The following table expresses this and the likely prots in the event of each of these outcomes. High (0.2) Direct 100000 Internet 46000 Licence 20000 Medium (0.35) 55000 25000 20000 Low (0.45) -25000 15000 20000 This can be expressed as a decision tree. The decision points are represented by squares and once again the probabilistic nodes by circles. The monetary outcomes are listed down the left hand side of the tree. The EMV of each decision can be calculated EMV( Direct ) = EMV( Internet ) = EMV( Licence ) = The managing director should then choose to base the launch route based on the path which gives the largest expected monetary value. ## Exercise for 23/11/2001 1. A company has installed a new computer system and some employees are having difculty logging on to the system. They have been given training and the problems which arose during training were recorded and various probabilities calculated. An employee has 0.9 probability of logging on successfully on the rst attempt. If successful at any time then they will also be successful on the next two attempts with the same probability as above. If not successful at any time then the employee looses condence and the probability of success on a subsequent attempt drops to 0.5. Use a tree diagram to nd the following probabilities. (a) An employee successfully logs on their rst three attempts. (b) An employee fails in their rst attempt but is successful in the two attempts after that. (c) Logs on only once in three attempts. (d) Does not manage to log on in three attempts. 2. A lm producer buys the rights to a book with the intention of making a lm. He knows from past experience the chances of making a successful lm are 40 %. A lm would cost 30 million to make, but would return 120 million. He could conduct detailed market research before making the lm, this would be costly at 5 million and past experience of this suggest it would indicate a successful lm 60 % of the time. However this sort of focus group based market research is not reliable with only a 70 % chance of it indicating the true success rate. The producer could sell the lm at any time but the amount he received for the rights would depend on the level of testing he had done. If the tests indicated a successful lm he could get 35 million for the rights, if they indicated a failure he could get 3 million and if he did no testing he could get 10 million. Use a decision tree to advise the producer on his best course of action. Note this decision tree contains three decision nodes.
# 30-60-90 Right Triangles In these lessons, we will learn • the special right triangle called the 30-60-90 triangle. • how to solve problems involving the 30-60-90 right triangle • how to prove that the ratios between the sides of a 30-60-90 triangle are 1:√3:2. Recognizing special right triangles in geometry can help you to answer some questions quicker. A special right triangle is a right triangle whose sides are in a particular ratio. You can also use the Pythagorean theorem, but if you can see that it is a special triangle it can save you some calculations. Here, we will look at the 30-60-90 triangle. We also have lessons on other special right triangles ### 30-60-90 Triangles The 30-60-90 triangle is one example of a special right triangle. It is right triangle whose angles are 30°, 60° and 90°. The lengths of the sides of a 30-60-90 triangle are in the ratio of 1:√3:2. The following diagram shows a 30-60-90 triangle and the ratio of the sides. Scroll down the page for more examples and solutions on how to use the 30-60-90 triangle. The hypotenuse is always twice the length of the shorter leg (the side facing the 30° angle). The longer leg (the side facing the 60° angle) is √3 times of the shorter leg. ### Solve problems involving 30-60-90 right triangles Example 1: Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 4 inches and 4√3 inches. Solution: Step 1: Test the ratio of the lengths to see if it fits the n:n√3:2n ratio 4:4√3:? = n:n√3:2n Step 2: Yes, it is a 30-60-90 triangle with n = 4 Step 3: Calculate the third side. 2n = 2 × 4 = 8 Answer: The length of the hypotenuse is 8 inches. You can also recognize a 30-60-90 triangle by the angles. As long as you know that one of the angles in the right-angle triangle is either 30° or 60° then it must be a 30-60-90 special right triangle. A right triangle with a 30°-angle or 60°-angle must be a 30-60-90 special right triangle. Example 2: Find the lengths of the other two sides of a right triangle if the length of the hypotenuse is 8 inches and one of the angles is 30°. Solution: This is a right triangle with a 30-60-90 triangle. You are given that the hypotenuse is 8. Substituting 8 into the third value of the ratio n:n√3:2n, we get that 2n = 8 ⇒ n = 4. Substituting n = 4 into the first and second value of the ratio we get that the other two sides are 4 and 4√3. Answer: The lengths of the two sides are 4 inches and 4√3 inches. How to solve a 30-60-90 triangle given the length of one side? Special Right Triangles in Geometry: 45-45-90 and 30-60-90 degree triangles This video discusses two special right triangles, how to derive the formulas to find the lengths of the sides of the triangles by knowing the length of one side, and then does a few examples using them. Applying the 30-60-90 triangle to find the height of a building An example in which we use some of the great properties of a 30-60-90 right triangle to find the height of a tower. Using what we know about 30-60-90 triangles to solve what at first seems to be a challenging problem ### Proof the ratios between the sides of a 30-60-90 triangle Proving the ratios between the sides of a 30-60-90 triangle are 1:√3:2. Prove how the side lengths of a 30-60-90 triangle are related and then use that relationship to quickly find side lengths 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.
# Finding Average Of Percentages 3 or Multiple in Excel In various real-life scenarios, you need to calculate the average percentage because it makes your work easier. There are multiple scenarios in your daily life routine, which put you to get the average percentage of your studies results, or in other environments. Finding Average Of Percentages seems difficult, but actually, it is not a difficult task to do, if you take guidance properly. In most cases, you need to get an average of two percentages by adding them and after this, you need to divide them by the specific number of percentages that are used. But also in other instances, you need to consider the different factors like a sample size. So, here we discuss the basics of percentages, and also you will know how to calculate average percentages easily, with real-life examples. ## What are the average percentages? In the simple term, the percentage is specific portions that are out of the total amount of 100. The percentage is used to get a vivid picture of the decimal number. In early education, the concept of calculating percentages is introduced, but very few people lack experience with it. But after time, it enhances the process of calculating percentages, and now it is easier than before. On the other hand, the average percentage means it is the average of a unique set which have at least two or more percentage values. ## How you can easily find the average percentages? If you want to find the average of percentages, then one thing which you need to keep in your mind is that it will require a different set of steps, which determine the average of at least two or more numbers. With the five following simple steps you can easily find or calculate the average percentages: ### Convert the percentages into decimal In this first step, you need to make sure that the average of percentages is to turn the percentages into decimals. Let’s determine if you want to take the average of 25% of 200 and 30% of 150. When you divide these numbers by 100, then these percentages will convert into decimals. The decimal of these numbers is 0.3 for 30% and .25 is for 25%. ## Determine the number which has decimal represents After successful conversion of percentages into decimals, then you need to find each decimal representing a number. For this purpose, you need to multiply the decimal by the total number, which is given in the sample. According to the above example, you need to multiply 0.3 by 150 to get the result of 45 and multiply 0.25 by 200 to get the result of 50. In the third step, you need to add those two numbers which are found in the second step. 50 plus 45, which is equal to 95. ### Adding the sample sizes together In this fourth step, you also need to add the overall sample sizes for every percentage. 150 plus 200, which is equal to 350. ### Calculate the Percentage Average In this last step, it will be easy for you to find the average of percentages by dividing the sum of percentage numbers, and you need to do it by some of the sample sizes. Like, 95 divided by 350, which is equal to 0.27. After this, you need to multiply this decimal by 100, when you complete this step then you will easily get the average percentage. Like, 0.27 multiplied by 100, which is equal to 27% or 27. It is clear that the average percentage of 25% of 200 and 30% of 150, is equal to 27%. ### 1. Can I easily find the average of percentages with the above formula? Answer. The calculation to find average percentages are easy, but it takes time for you to understand the calculation process. But keep in mind that you need to make sure that you have at least two numbers while using this formula. ### 2. Is there any other formula to find the average of percentages? Answer. There are multiple equations or formulas, which will find the average of percentages. But the above formula is simple and easy to use by anyone.
Home > Bible Math > Proving Universal God Time ## Proving Universal God Time The Bible tells us that Heaven and Earth perceive time differently. In the article, A Thousand Years As One Day we solve these differences proportionally. In this article, we reveal scientific evidence proving the accuracy of both our calculations and the Holy Bible. Introduction From the article, A Thousand Years As One Day, we learn that for every second that passes by in Heaven, ~4.16 days pass by on Earth, therefore we calculate: Let, a = 246060 seconds b = 1000 years c = 1230 days Thus, b/ac = 4.1666666666666666666666666666667 days It may not be obvious to many; however this value is more than just a difference in how both Heaven and Earth perceive time, for it also tells us how fast the Earth rotates. Approaching the Problem There are 1440 minutes in 1 earth day (i.e. 24 hour period). From the article, A Thousand Years As One Day, we learn that for every millisecond that passes by in Heaven, 6 minutes pass by on Earth. Therefore, 1 earth day is equivalent to 240 milliseconds in Heaven: Let, a = 1440 minutes b = 6 minutes Thus, a/b = 240 milliseconds (God Time) According to NASA, the circumference of the Earth, at its equator, is 25,000 miles; therefore, in order to calculate the speed of rotation, we calculate as follows: Let, a = 24 hours b = 25000 miles Thus, b/a = 1041.6666666666666666666666666667 miles/hour Therefore, according to science, our Earth rotates at ~1042 miles/hour. To make everything a bit easier, we now bring our biblical calculation into the proper perspective, therefore: Let, a = 24 hours b = 240 milliseconds Reduce both values by a factor of 24: Thus, a = 24/24 = 1 hour b = 240/24 = 10 milliseconds Therefore, 1 hour on Earth is the same as 10 milliseconds in Heaven. This suggests that every 10th millisecond that ticks by in Heaven, then from Earth’s perspective, the Earth is rotating at a speed of 1041.666666666667 miles/hour; from the perspective of Heaven, the Earth is rotating 1041.666666666667 miles/10 millisecond, which is incredibly fast. Solving the Problem If you recall from the introduction, we are comparing our answer found in, A Thousand Years As One Day (i.e. 4.1666666666666666666666666666667 days) to that which is calculated knowing the circumference of the Earth and the speed of Earth’s rotation. Because our biblical calculation was done from the perspective of one second passing by in Heaven, then we must convert our speed, from God’s perspective, of 1041.666666666667 miles/10 millisecond to its miles/second equivalent, therefore we calculate: Let, a = 1041.666666666667 miles/10 millisecond b = 100 Thus, a*b = 104166.6666666667 miles/second (God Time) Because we are trying to determine how many full rotations of Earth had occurred during that time, and because we know that the circumference of the Earth is 25,000 miles, thus 25,000 miles traveled is equivalent to one full Earth day (i.e. 24 hour period), therefore: Let, a = 104166.6666666667 miles/second b = 25000 miles Thus, a/b = 4.166666666666668 days BINGO! Conclusion In summation, the Bible accurately revealed how fast the Earth rotates more than 2000 years before NASA. Therefore, beloved, be not ignorant of this one thing, that one day is with the Lord as a thousand years, and a thousand years as one day. DIVINECODERS Categories: Bible Math 1. March 24, 2018 at 9:09 am What if the Earth is flat and doesn’t move, but the sun and moon rotates around the Earth? How does this change the formula? • March 24, 2018 at 10:49 am Hi Dustin, Good question, whether or not the earth is flat has no affect upon our calculations because we are not basing our formula on how fast the earth rotates exactly, instead our formula suggests one of two things, 1) how fast the earth rotates upon its axis or 2) how fast the the sun travels from point A to Point B each day. At the end of the day, what must not change are the distance traveled and the amount of time it takes. Therefore, as long as something is in motion and its speed is relatively constant then, our calculations remain unaffected no matter the viewpoint. What does change is how we interpret the results. Hope this helps, DIVINECODERS 2. August 12, 2017 at 3:50 pm I think the idea of your calculation amazing. Well done! So with that. What is the meaning of times and a half time ? • April 7, 2018 at 12:43 am Hi Joanne It seems your simple/great question has been forgotten… It would be good to get a response – I imagine these guys can deal with it. I think your subject can be fleshed out a bit further in Dan 12:7. Time, Times and and half a time. • April 7, 2018 at 11:24 am Hi Joanne and Leigh, Sorry, we missed your comment. Thanks, Leigh for pointing that out. 🙂 Simply, if we look at Daniel 7:25 where we see: …time and times and the dividing of time. We find hidden beneath the word time the Hebrew word עִ דָּ ן which is transliterated as odn or odin, which means season. Therefore, we interpret this in our modern day to mean year, thus time, times, and the dividing of time would equate to: 1 year + 2 years (because now we have more than one) + 1/2 year (or .5 years) = 3.5 years. To confirm our calculations we can simply check Revelation 13:5 which kindly did the math for us: And there was given unto him a mouth speaking great things and blasphemies; and power was given unto him to continue forty and two months. DIVINECODERS 3. August 9, 2016 at 2:57 am Thank you for this great analysis and explanation. I am excited to learn more about this. While studying Psalms 90, it says: A thousand years in God’s sight is as yesterday when it is past, and as a watch in the night. What intrigued me was the portion that said: ‘A thousand years is as a watch in the night’. The watch in the night portion took me to the God clock for each watch period in the night. A watch in the night is 3 hours. There are four watches each night. So does that mean there are 4000 years for the complete night watch period. If so, how does that convert using the divine calculator? I look forward to your response • August 9, 2016 at 5:45 pm Hi Lynn, When we read the text of the Holy Bible and are not exactly sure what a certain word or phrase means, we go to the source and in this particular case we find the transliterated word “u·ashmure” having been translated to “watch”. The Hebrew word “u·ashmure” means “and·vigil”, so what we find is that the word “watch” means quite literally to “watch” -meaning to be vigil, as God is most faithful. Let’s take a look at this verse: For a thousand years in thy sight [are but] as yesterday when it is past, and [as] a watch in the night. Under the microscope: דכִּי אֶלֶף שָׁנִים בְּעֵינֶיךָ כְּיוֹם אֶתְמוֹל כִּי יַעֲבֹר וְאַשְׁמוּרָה בַלָּיְלָה: (Psalm 90:4) So, what we have is 10 words: 01. ki = that 02. alph = thousand-of 03. shnim = years 04. b·oini·k = in·eyes-of·you 05. k·ium = as·day-of 06. athmul = yesterday 07. ki = that 08. iobr = he-is-passing 09. u·ashmure = and·vigil (watch) 10. b·lile = in·night So, maybe a better way to say this could be, “A thousand years on earth is the same as one day to [God], who never sleeps.” God is most faithful. Fast forward to the Gospel of the LORD, Jesus Christ where it records the hour before Jesus Christ’s arrest in the garden. The crowd came at night to arrest him as if He were some sort of violent criminal on the run from justice. But, before this there was a problem with the disciples -they kept falling asleep! God doesn’t sleep. Thanks! DIVINECODERS • December 20, 2016 at 12:56 am Re: August 9, 2016 at 5:45 pm Are you saying that Jesus never slept…??? He was awake for His entire life since birth…??? Or did He stop sleeping at baptism or something…??? Jesus was asleep in the boat during the storm…!!! You also seem to be saying that Gods’ nature has innate “time”. This would be an absurd blasphemy, and of course, a false teaching – a heresy. Infinite God “makes” time(s) as a function of not-in-finite/never-in-finite creation(s). Correct your thinking and common use ideas on the rubbish/fallacy/heresy of the “infinite universe”. Is it in-finite of space? Or, is it in-finite in time? Or what? You are thinking with a pagan Greek concept! You are trying to imbue a divine quality on creation!!!!!!!!!!! The ONLY GOD is in-finite/ non-finite/ not-finite/ never-finite/ time-less/ (without)time/ (without)size/ (without)limits. NO THING in any creation (old/new) is “infinite”. Everything created is/must-be always finite/never-in-finite. Read Job38 as maths – creation is based, limited, bound. and He may keep it going “forever” but it always has a “beginning”. God “brings creation into being” but He Himself has no beginning “in” His being. • December 20, 2016 at 10:12 pm Hi Leigh, No we aren’t saying that Jesus was an insomniac. God, the Father does not sleep -that is what we are saying. Thanks, DIVINECODERS • December 20, 2016 at 2:01 am Re: August 9, 2016 at 5:45 pm Consider the common statement; “There are an infinite number of numbers”. It is affirming that “There is a non-number number of numbers”. “The number of numbers is not-number.” Many basic ideas/statements in mathematics are philosophical/theological in nature and contrary to both rationality and logic. It is demonstrable that the first dozen fundamentals of mathematics describing the number system are erroneous. For example, zero is called a “Real” Number. This permits the idea that the big bang (and as such real-ity/existence) real-ly started from the “real” nothing… Then “scientists” have panel questions such as “Is nothing something?” and laughs and stupid ideas and “definitions” abound. It is a bad, so called scientific, question…!!! The only answer to that question is “YES and NO”. And then, by similar erroneous mathematical ideas Hawking thinks he has disproved God. Hard science always proves God but theoretical/speculative science Does nothing(non-existence) really exist?, “NO, Nothing doesn’t/cannot really exist.” (or) “Nothing is not real; nothing is only an IMAGINARY concept.” Non-existence does not exist. Nothing is not something real; it is ONLY an imaginary concept. Fundamental mathematics contains more philosophical statements than proper logical conclusions. • December 20, 2016 at 10:26 pm Hi Leigh, Careful how you interpret existence. We are real and our lives are real. This time is real and how we conduct ourselves within it is real and it matters. Science is good, because knowledge is good. However, there are those who use the name of science to push forward political and social agendas, so we must use our own abilities of analysis to discover truth independently. Which we do, not clouded by the falsities of evolution or the lies of atheism, but guided by the truth of the word of God. We do not allow our minds to be warped by the thoughts of madmen. One must stay clear and be sober. Be apart of the now, endure for Christ. DIVINECODERS 4. July 21, 2014 at 2:54 am I know this may not be easy to understand because: Mat 11:25 At that time Jesus answered and said, I thank thee, O Father, Lord of heaven and earth, because thou hast hid these things from the wise and prudent, and hast revealed them unto babes. 1Co 1:19 For it is written, I will destroy the wisdom of the wise, and will bring to nothing the understanding of the prudent. If you read and look the scripture closely, you see the word as not is. 2Pet 3:8 But, beloved, be not ignorant of this one thing, that one day is with the Lord as a thousand years, and a thousand years as one day. What is the largest number stated in the bible? answer 1,000 or thousand. Why? There was nothing greater than that at the time it was written. So the scripture could have very well said as a million instead of a thousand. The scripture is merely pointing out that God does not view time as humans do. In heaven there is never a night. God is form everlasting to everlasting, figure that out. As far as your calculation, I can tell your very wise ; ). • July 21, 2014 at 10:27 am Hi, What is the largest number stated in the bible? answer 1,000 or thousand. Why? There was nothing greater than that at the time it was written. That’s a truly preposterous argument, hence: “And from the time that the daily sacrifice shall be taken away, and the abomination that maketh desolate set up, there shall be a thousand two hundred and ninety days.” (Daniel 12:11) Do you honestly believe that the ancients did not know how to count past 1000? Moreover, if you study the Greek and Hebrew Manuscripts, you will find that in 2 Peter 3:8 they used “G5507” which means 1000, and in Psalm 90:4 they used “elef” or “alph” (transliterations) which also mean 1000. DIVINECODERS 5. July 20, 2014 at 8:28 am The Bible tells us that Heaven and Earth perceive time differently. In the article, A Thousand Years As One Day we solve these … Where does the Bible tell us that? • July 20, 2014 at 11:18 am Hi CC, The Bible tells us about the differences in both Psalm 90:4 and 2 Peter 3:8. See also, A Thousand Years As One Day. DIVINECODERS • February 26, 2018 at 6:44 am hi very wise information indeed i alwayz wanted to understand time now i do. so my question is how long do u think it took christ to arrive on earth coming from heaven since it is said in bible “GOD said who can we send on earth. jesus said sent i”. so how long do you think it took him to arrive on earth. base on your calculations of 1 day of heaven is to 1000 days of earth. regards brian
# How do you find the first and second derivatives of y = (x^2 + 3) / e^-x using the quotient rule? Nov 2, 2015 $y ' = {e}^{x} \left({x}^{2} + 2 x + 3\right)$ $y ' ' = {e}^{x} \left({x}^{2} + 4 x + 5\right)$ #### Explanation: Alternatively, you could rewrite the original question as a product and apply the product rule instead of the quotient rule. STEP 1: Rewrite the original equation $y = \frac{{x}^{2} + 3}{e} ^ - x = \left({x}^{2} + 3\right) \cdot {e}^{x}$ STEP 2: Use the product rule to find the first derivative $y ' = \left(2 x\right) \left({e}^{x}\right) + \left({x}^{2} + 3\right) \cdot {e}^{- x}$ $y ' = 2 x {e}^{x} + {x}^{2} {e}^{x} + 3 {e}^{x} = {e}^{x} \left({x}^{2} + 2 x + 3\right)$ $y ' ' = {e}^{x} \cdot \left({x}^{2} + 2 x + 3\right) + {e}^{x} \cdot \left(2 x + 2\right)$ $= {e}^{x} \left({x}^{2} + 2 x + 3 + 2 x + 2\right)$ $= {e}^{x} \left({x}^{2} + 4 x + 5\right)$
### Welcome to our community #### MarkFL Staff member The purpose of this tutorial is to provide students of algebra with techniques and tips for factoring quadratic expressions. In my experience as a tutor, I have found this can be one of the more difficult and challenging topics for students. I invite anyone with any techniques of their own to add to this topic to give our readers as comprehensive a list of tips/tricks as possible. Note: all coefficients, both in the quadratic forms and in the factored forms are integers. Typically, a quadratic expression is given in the form: (1) $ax^2+bx+c$ and the goal is to express this expression as the product of two linear factors: $ax^2+bx+c=(dx+e)(fx+g)$ First of all, students may wonder why we bother to factor quadratics. Well, this has to do with what's called the zero-factor property. Many times in the application of quadratic equations, we express the equation in standard form $ax^2+bx+c=0$. In short, if we have two factors equal to zero, then we know that the product must be zero when one or the other of the factors is equal to zero. Since finding the root of a linear factor is very straightforward, this gives us a neat and easy way to find the two roots of the quadratic expression. The way I begin, is (referring to (1)) by looking at the product $ac$ and try to find two factors of this product whose sum is $b$. Suppose we are given to factor: $x^2+8x+12$ The product $ac=1\cdot12=12$ so we want two factors of $12$ whose sum is $8$. Thinking about the factor pairs of $12$ which are $(1,12),\,(2,6),\,(3,4)$ we see the pair $(2,6)$ has a sum of $8$. Since $a=1$, we know the coefficient of $x$ in both factors will also be $1$, and so we have: $x^2+8x+12=(x+6)(x+2)$ If we're unsure, we may check our work by expanding the right side using the FOIL method: $(x+6)(x+2)=x^2+2x+6x+12=x^2+8x+12$ So, it checks out. Why does this method work? Let's do a bit of investigation: To keep things simple for now, let's consider the case where $a=1$: $x^2+bx+c=(x+d)(x+e)=x^2+ex+dx+de=x^2+(d+e)x+de$ Equating coefficients, we find we require: $d+e=b$ $de=c$ As you can see, this implies we need to find two factors of $c$ whose sum is $b$. This is why we look for two such factors. Now, let's step things up just a bit and let $a$ be a prime number. Since a prime number only has itself and $1$ as factors, we know the factored form must be: $ax^2+bc+c=(ax+d)(x+e)=ax^2+(d+ae)x+de$ So, we see now, we require: $d+ae=b$ $de=c$ If we mutiply the second equation by $a$, we may write: $d(ae)=ac$ So, these equations imply that we need two factors of $ac$ whose sum is $b$. Let's now apply this to the quadratic: $3x^2-13x-10$ We want two factors of $3(-10)=-30$ whose sum is $-13$. These are $(2,-15)$. Since $a=3$, we look for the factor that is divisible by $3$ and that is $-15$. So we set: $ae=3e=-15,\,e=-5$ $d=2$ and so we now have: $3x^2-13x-10=(3x+2)(x-5)$ Now, let's examine the case when $a$ is composite, which means there is more than 1 possible factor pair of $a$, and so we need to write the factored form as: $ax^2+bx+c=(dx+e)(fx+g)=dfx^2+(dg+ef)x+eg$ Equating coefficients, we find: $df=a$ $dg+ef=b$ $eg=c$ If we multiply the first equation by the third, we have: $(dg)(ef)=ac$ This, along with the second equation above implies we need two factors of $ac$ whose sum is $b$. Suppose we are given to factor: $6x^2+13x-28$ We want two factors of $6(-28)=-168$ whose sum is $13$. They are $(-8,21)$. Now, we look at the factor pairs of $6$ which are $(1,6),\,(2,3)$. Neither of the pair $(-8,21)$ is divisible by $6$, but $-8$ is divisible by $2$ and $21$ is divisible by $3$, so the factor pair of $6$ we need is $(2,3)$. And so we find we may set: $d=2$ $f=3$ $dg=2g=-8\,\therefore\,g=-4$ $ef=3e=21\,\therefore\,e=7$ and so we have: $6x^2+13x-28=(2x+7)(3x-4)$ Commentary to this tutorial should be posted here:
# In a right triangle, given slope and length of hypotenuse find length of legs. Say I have a right triangle. I know the slope and length of $c$, how do I find the length of $a$ and $b$? - You "know the slope of $c$" in what way? Are you given an angle, or a "rise/run" fraction? – abiessu Nov 13 '13 at 22:51 Use the slope of $c$ to find the angle at $A$. – Newb Nov 13 '13 at 22:52 No need for trigonometry here. – amWhy Nov 13 '13 at 22:53 We have a right triangle, so there are two things we know: • Slope $\;m = \dfrac{a - 0}{b-0}=\dfrac ab\implies a = bm$. And • $a^2 + b^2 = \underbrace{c^2}_{\text{hypotenuse}}$ Two equations and two unknowns. Since $a = bm,$ we can substitute $bm$ into the variable $a$ in the second equation: $$(bm)^2 + b^2 = c^2\implies b^2(m^2 + 1) = c^2 \implies b^2 = \dfrac{c^2}{m^2 + 1} \implies b = \dfrac{c}{\sqrt{m^2 + 1}}.$$ Since the lengths of the sides of a triangle must be positive, we can take the positive root of $b^2$ to solve for $b$, then back substitute to obtain $a = bm$. - If you have the "slope" $$m = \frac ab$$ then you can write $a$ as $mb$. Fit this in $$c = \sqrt{a^2 + b^2}$$ and get $$b = \frac{c}{\sqrt{m^2 + 1}}$$ - if you kno9w the overall slope, but maybe you want to find the Y value of a shorter distance for x, use these relations. where F is the hypotenuse, X is the x value, and Y is the Y value. Fcos(angle)=X, Fsin(angle)=Y, which you probably have memorized, but here are the ones you've forgotten. Xtan(angle)=Y, Ycot(angle)=X, - If you are going to discard the notation $a,b,c$ introduced in the Question, you should provide definitions for your unknowns $X,Y$. I can accept that $F$ is your hypotenuse (where the OP labels the length of hypotenuse $c$), but you have mentioned "slope" and "angle" without definition (or referring to the original quantities). – hardmath Mar 19 '14 at 4:48
# Associative Law ## Overview In this lesson, we are going to learn about the associative law. Make sure you fully understand the commutative law before reviewing this lesson! Here’s a quick review of the commutative law. The commutative law tells us that we can switch numbers when adding and multiplying and still get the same answer. We will only be focusing on understanding how the associative law applies to addition. 5 +4 = 9 is the same as 4 + 5 = 9 (This is the commutative law. We switched the numbers 5 and 4, and we always get the same answer which is 9.) Awesome! So what is the associative law? It’s very similar to the commutative law, and it allows us to “re-group” numbers to still get to the same answer. Sometimes we have more than two groups to add together! For example, in the picture below, there are three groups of stars. How many stars are there in all? We can add the three groups to find the sum! We can add them like this: 4 + 6 + 2 = 12 When we add groups or numbers, the order does not matter. We can use parentheses ( ) to add two group of numbers together. (4 + 6) + 2 = 12 We will add like this: 4 + 6 = 10 Then we will add our sum to the rest of the equation. 10 + 2 =12 We can also solve it like this: 4 + (6 + 2) = 12 6 + 2 = 8 and 8 + 4 = 12 Notice that we moved the parenthesis from the first example and we still ended up with the same answer. This is the associative property. Numbers can be grouped together differently when adding, and still have the same answer. Let’s take a look at another problem below. We have three groups of stars. How many stars are there in all? We can add the three groups to find the sum! We can add them like this: 7 + 3 + 5 = 15 When we add groups or numbers, the order does not matter. We can use parentheses ( ) to add two group of numbers together. (7 + 3) + 5 = 15 We will add like this: 7 + 3 = 10 Then we will add our sum to the rest of the equation. 10 + 5 =15 We can also solve it like this (using associative property): 7 + (3 + 5) = 15 We will add like this: 3 + 5 = 8 and 8 + 7 = 15 Numbers can be grouped together to add. As long as you add carefully, you will get the correct answer. In this example, the sum is 15. Let’s try some more examples! In the next few problems, find the missing addend or sum. Use what you have learned about the associative property to help you! To solve, let’s look at the first problem: 5 + (2 + ___ ) = 10 We can figure out the missing addend! We know that 5 + 2 = 7 We also know that 7 + ___ = 10 7 + 3 = 10 Now we will solve the right side of the equation. 2 + (3 + ___ ) = 10 We can figure out the missing addend! We know that 2 + 3 = 5 We also know that 5 + ___ = 10 5 + 5 = 10 Now we can complete the math sentence: 5 + (2 + 3) = 10 = 2 + (3 + 5) Both equations equal 10! Here is another example. We can solve it in the same way! To solve, let’s look at the first problem: 4 + (3 + ___ ) = 9 We can figure out the missing addend! We know that 4 + 3 = 7 We also know that 7 + ___ = 9 7 + 2 = 9 Now we will solve the right side of the equation. 3 + (2 + ___) = 9 We can figure out the missing addend! We know that 3 + 2 = 5 We also know that 5 + ___ = 9 5 + 4 = 9 Now we can complete the math sentence: 4 + (3 + 2) = 9 = 3 + (2 + 4) Both equations equal 9! ### Practice Question 1 Let’s practice! Look at the group below. Write the addition sentence and solve the problem below. ( ______ + ______ ) + ______ = ______ + ______ = ______ + ______ = _____ See Related Worksheets: 1st grade Fishin' for Sums Worksheets (0) Elephant is hoping to catch some fish for dinner! These aren't just any fish, though... they are "su... 1st grade Playing An Addition Tune Worksheets (0) Lion is composing a song for his best friend Elephant, who loves addition! Students can help him wri... 1st grade The Cozy Book of Number Bonds Worksheets (0) Pedro Penguin has snuggled in to enjoy his holiday book about number bonds, but his little brother e... 1st grade Who's a Good Pup? Worksheets (0) Fashion Pup is such a good pup! On this worksheet, the Pup is enjoying some time basking under a hea... Relevant Topics: ### Try ArgoPrep for FREE Share good content with friends and get 15% discount for 12-month subscription Count the fruits to check your answer! TIP ### Practice Question 2 Look at the group below. Write the addition sentence and solve the problem below. ( ______ + ______ ) + ______ = ______ + ______ = ______ + ______ = _____ See Related Worksheets: 1st grade Fishin' for Sums Worksheets (0) Elephant is hoping to catch some fish for dinner! These aren't just any fish, though... they are "su... 1st grade Playing An Addition Tune Worksheets (0) Lion is composing a song for his best friend Elephant, who loves addition! Students can help him wri... 1st grade The Cozy Book of Number Bonds Worksheets (0) Pedro Penguin has snuggled in to enjoy his holiday book about number bonds, but his little brother e... 1st grade Who's a Good Pup? Worksheets (0) Fashion Pup is such a good pup! On this worksheet, the Pup is enjoying some time basking under a hea... Relevant Topics: ### Try ArgoPrep for FREE Share good content with friends and get 15% discount for 12-month subscription Count the books to check your answer! TIP ### Practice Question 3 Take a look at the math sentence below. Draw a picture to match it. Then solve. 1 + 3 + 9 = ### Practice Question 4 Take a look at the math sentence below. Draw a picture to match it. Then solve. 4 + 4 + 7 = ### Practice Question 5 Take a look at the math sentence below. Draw a picture to match it. Then solve. 2 + 9 + 5 = See Related Worksheets: 1st grade Fishin' for Sums Worksheets (0) Elephant is hoping to catch some fish for dinner! These aren't just any fish, though... they are "su... 1st grade Playing An Addition Tune Worksheets (0) Lion is composing a song for his best friend Elephant, who loves addition! Students can help him wri... 1st grade The Cozy Book of Number Bonds Worksheets (0) Pedro Penguin has snuggled in to enjoy his holiday book about number bonds, but his little brother e... 1st grade Who's a Good Pup? Worksheets (0) Fashion Pup is such a good pup! On this worksheet, the Pup is enjoying some time basking under a hea... Relevant Topics: ### Try ArgoPrep for FREE Share good content with friends and get 15% discount for 12-month subscription Want to test your level? Select your grade level and start now 1 Please select your grade level:
GMAT Question of the Day: Daily via email \| Daily via Instagram New to GMAT Club? Watch this Video It is currently 04 Apr 2020, 16:52 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # In how many ways can the letters of the word MANIFOLD be arr Author Message TAGS: ### Hide Tags Director Joined: 07 Mar 2019 Posts: 911 Location: India GMAT 1: 580 Q43 V27 WE: Sales (Energy and Utilities) Re: In how many ways can the letters of the word MANIFOLD be arr [#permalink] ### Show Tags 26 Jan 2020, 06:30 Sorry for jumping in between. Pritishd Since we have to keep vowel separated the requirement of 6 spaces arrive. Here's as i understand A. When Vowels start the word (Positions taken by the vowel) 1 3 5/6/7/8 (4 ways) 1 4 6/7/8 (3 ways) 1 5 7/8 (2 ways) 1 6 8 (1 way) Total = 10 ways B. When Consonants start the word (Positions taken by the vowel) 2 4 6/7/8 (3 ways) 2 5 7/8 (2 ways) 2 6 8 (1 way) 3 5 7/8 (2 ways) 3 6 8 (1 way) 4 6 8 (1 way) Total = 10 ways G. Total 10 + 10 = 20 Now, since Consonants are always separated number of ways to arrange them = 5! Number of ways to arrange Vowels = 3! Required ways of arrangements = 5! * 3! * 20 = 14400 Pritishd wrote: Archit3110 wrote: Pritishd Vowels are to be arranged separated. Below if you see we have 6 spaces for vowels and total vowels are 3 so possible arrangement for vowels in 6 spaces:654.. _M_N_D_L_F_ Hope this helps Hi Archit3110, I do not understand the part were we consider that there will be 6 spaces to arrange the 3 vowels after the 5 consonants are arranged. Shouldn't there be only 3 spaces after arranging 5 consonants? IMPORTANT: For each arrangement of 5 consonants, there are 6 spaces where the VOWELS can be placed. For example, in the arrangement MNDLF, we can add spaces as follows _M_N_D_L_F_ Warm Regards, Pritishd Hi Archit3110, My doubt still remains. I can see that there are 6 spaces but my question was that how did you arrive at those 6 spaces? Once we arrange the 5 consonants will we not have only 3 spaces left? Is this some kind of a method were we assume an available space before and after every element? Warm Regards, Pritishd _________________ Ephemeral Epiphany..! GMATPREP1 590(Q48,V23) March 6, 2019 GMATPREP2 610(Q44,V29) June 10, 2019 GMATPREPSoft1 680(Q48,V35) June 26, 2019 Re: In how many ways can the letters of the word MANIFOLD be arr [#permalink] 26 Jan 2020, 06:30 Go to page Previous 1 2 [ 21 posts ] Display posts from previous: Sort by
## Subtract Whole Numbers Subtraction of whole numbers can also be completed by using the addition properties and expanded notation. Example: 3,150 − 1,732 = (3,000 + 100 + 50) − (1,000 + 700 + 30 + 2) = (3,000 − 1,000) + (100 − 700) + (50 − 30) + (0 − 2) When working with whole numbers, 700 cannot be subtracted from 100 and 2 cannot be subtracted from 0, so renaming is necessary. Think: 3,000 + 100 can be written as 2,000 + 1,100. 50 can be written as 40 + 10. = (2,000 − 1,000) + (1,100 − 700) + (40 − 30) + (10 − 2) = 1,000 + 400 + 10 + 8 = 1,418 The method is still long and cumbersome. The subtraction algorithm provides a simple, compact method of completing such calculations. As in addition, digits are aligned according to place value and the computation is completed from right to left. Rename. Subtract the tens Rename. Subtract the thousands 5 tens 0 ones =4 tens 10 onesSubtract the ones. 3 thousands 1 hundred =2 thousands 11 hundredsSubtract the hundreds. Estimating Sums and Differences When an exact answer is not necessary, an estimate can be used. The most common method of estimating sums and differences is to round each number to a specific place and then add or subtract the rounded numbers. Estimate 4,894 + 2,429. 4,894 → 5,000 2,429 → + 2,000 7,000 Round each number to the nearest thousand. Add the rounded numbers. Estimate 6,209 − 383. 6,209 → 6,200 383 → − 400 5,800 Round each number to the nearest hundred. Subtract the rounded numbers. Mental math can often be used to complete estimates. At this grade level, however, errors can be more easily identified if students write down their work when estimating answers. As students gain experience and confidence estimating sums and differences, point out that estimates can often be used to check computations. Students should realize that if both addends are rounded up, the estimated sum will be greater than the actual sum, and if both addends are rounded down, the estimated sum will be less than the actual sum. Such generalizations are not possible with subtraction. Teaching Model: Relate Addition and Subtraction
Important Formulas: Number Series # Important Formulas: Number Series \| Quantitative Aptitude for SSC CGL PDF Download ## Number Series Formulas A number series is a progression of numbers arranged based on a specific system or rule, without adhering to a particular order. The task involves identifying the underlying system or rule governing a given series and using it to determine the subsequent numbers. For Example : 3, 9, 27, 81 ? It is Geometric series. Each term is Multiplied by 3. So , 81 x 3 = 243 ## Types of number series Types of Number Series is given below: • Arithmetic Sequence: A sequence in which every term is created by adding or subtracting a definite number to the preceding number is an arithmetic sequence. • Geometric Sequence: A sequence in which every term is obtained by multiplying or dividing a definite number with the preceding number is known as a geometric sequence. • Harmonic Sequences: A series of numbers is said to be in harmonic sequence if the reciprocals of all the elements of the sequence form an arithmetic sequence. • Fibonacci Numbers: Fibonacci numbers form an interesting sequence of numbers in which each element is obtained by adding two preceding elements and the sequence starts with 0 and 1. • Probability of asking – Very Low • Difficulty – Low • Reason – to Introduce Concept These problems are never asked they are very easy, we are talking about them to introduce the number series from basic. 1. Rule – Just Add a number ‘N’ to the last number. 2. E.g. – 5, (5 + 3 = 8), (8 + 3 = 11), ( 11 + 3= 14) …. 3. Result – 5, 8, 11, 14, 17 …….. • Probability of asking – Very Low • Difficulty – Low • Reason – to Introduce Concept 1. Rule : Just Add a number ‘N’ to the last number. 2. E.g. : 4, (4 – 5 = (-1)), (-1 -5 = -6), ( -6 – 5 = -11) …. 3. Result : 4, -1, -6, -11, -16 …….. ## Square up and Square Add up Series ### Square up and Square Add up Series + Square up +(Easy to Identify) 1. Rule – For a number X and for a number a where a = 1, 2, 3….. do next number = x + a2 2. E.g. – 5 • 5 + 2= 5 + 4 = 9 • 9 + 32 = 9 + 9 = 18 • 18 + 42 = 18 + 16 = 34 • 34 + 52 = 34 + 25 = 59 3. Result – 5, 9, 18, 34, 59 ….. Square up Add up +(Hard to Identify) 1. Rule – For a number X and for a number a where a = 1, 2, 3….. do next number = x + a+ b for b some pattern. 2. E.g. – 5 • 5 + 2+ 3 = 5 + 4 + 3 = 12 • 12 + 32 + 3 = 12 + 9 + 3 = 24 • 24 + 42 + 3 = 24 + 16 + 3 = 43 • 43 + 52 + 3 = 43 + 25 + 3 = 71 3. Result – 5, 12, 24, 43, 71 ….. Square up Step up +(Very hard to identify not asked mostly unless paper is very tough) 1. Rule – For a number X and for a number a where a = 1, 2, 3….. do next number = x + katex is not defined + b for b some pattern. 2. E.g. – 5 • 5 + katex is not defined + 3 = 5 + 4 + 3 = 12 • 12 + katex is not defined + 8(3+5) = 12 + 9 + 8 = 29 • 29 + katex is not defined +13(8+5) = 29 + 16 + 13 = 58 • 58 + katex is not defined + 18(13+5) = 58 + 25 + 18 = 101 3. Result – 5, 12, 29, 58, 101 .. ### Examples of Number Series Example 1: Find the missing number? 99, 121, 143, ___, 187, 199 . (a) 170 (b) 165 (c) 158 (d) 172 Ans: (b) The given series is an AP with first term as 99 and common difference as 22. Example 2: Find the next term in the series : 51,52,53,55,58,63,____. (a) 69 (b) 77 (c) 81 (d) 71 Ans: (d) Fibonacci series is added to each term. 51 + 0 =51 51 + 1=52 52 + 1=53 53 + 2=55 55 + 3=58 58 + 5=63 63 + 8=71 Example 3: Find the missing terms? 97,122,107,132,__,__. (a) 117,142 (b) 122,112 (c) 141,131 (d) 121,131 Ans: (a) This series is a result of alternate +25 and -15. 97 + 25=122 122 – 15=107 107 + 25=132 132 – 15=117 117 + 25=142 So, the next two terms are 117 and 142. Example 4: Fill the missing term in the series 100, 92, 86 ,82, 74, 68, 64, 56, 50, __, ___. (a) 44, 36 (b) 40, 34 (c) 46, 38 (d) 44, 32 Ans: (c) The number series are in successive subtraction series of – 8, -6, -4 and then again -8,-6,-4. So, the next terms after 50 will be 50-4=46 and 46-8=38. Example 5: Select the missing number from the given responses. 19, 35, 67, 131, 259, 515, ? (a) 1281 (b) 1291 (c) 1071 (d) 1027 Ans: (d) 11 × 2 – 3 = 19 19 × 2 – 3 = 35 35 × 2 – 3 = 67 67 × 2 – 3 = 131 131 × 2 – 3 = 259 259 × 2 – 3 = 515 515× 2 – 3 = 1027 The document Important Formulas: Number Series \| Quantitative Aptitude for SSC CGL is a part of the SSC CGL Course Quantitative Aptitude for SSC CGL. All you need of SSC CGL at this link: SSC CGL ## Quantitative Aptitude for SSC CGL 314 videos\|170 docs\|185 tests ## Quantitative Aptitude for SSC CGL 314 videos\|170 docs\|185 tests ### Up next Explore Courses for SSC CGL exam ### Top Courses for SSC CGL Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , ;
# How do you find the asymptotes for y=(x+3)/(2x-8)? Dec 23, 2015 Explanation is given below #### Explanation: $y = \frac{x + 3}{2 x - 8}$ For this problem, vertical asymptotes are found by equating the denominator to zero and solving for x. $2 x - 8 = 0$ $2 x = 8$ $x = \frac{8}{2}$ $x = 4$ Equation of vertical asymptote. For horizontal asymptote, check out the degree of numerator and denominator. If the degree of the numerator is same as the degree of denominator. The Horizontal asymptote is got by dividing the lead coefficients of numerator and denominator. For example if $y = \frac{a x + b}{c x + d}$ Vertical asymptote is got by solving for $x$ from $c x + d = 0$ The Horizontal asymptote is got by $y = \frac{a}{c}$ as both numerator and denominator are of degree 1 and their lead coefficient are $a$ and $c$ respectively. For our problem $y = \frac{x + 3}{2 x - 8}$ The horizontal asymptote would be $y = \frac{1}{2}$ Note: If the degree of the numerator is greater than the degree of denominator then there is no Horizontal Asymptote. If the degree of the denominator is greater than the degree of numerator then $y = 0$ is the Horizontal Asymptote.
# maths uyruety ## fractions a fratcions is part of a whole 1 of 10 ## fration a fraction can be divided inot parts and then some of the parts will be shaded!! 2 of 10 ## Fractions the number of parts the shape is divided into is called the denominator and is the bottom number of the fraction. 3 of 10 ## Fractions the number of the parts required is called the numerator and is the top number of the fraction. 4 of 10 ## adding and suntracting simple fractions If two fractions have the same denominator they can be added or suntracted. 5 of 10 ## Adding and subtracting simple fractions The denominator stays the same and the numerators are added or subtracted. 6 of 10 ## Equivalent fractions Any fractions can be written in many different ways. these ways are called Equivalent fractions. Equivalent fractions are two fractions that represent the same part of a whole. any fractions has an unlimited number of equivalent fractions. 7 of 10 ## Equivalent fractions and cancelling a base fractions is one that is written in its simplest terms. a fraction is in its simplest term if there is no number that is a factor of both the numerator and the denominator. Equivalent fractions can be found by multiplying both the numerator and the denominator by the same number. Equivalent fractions can be reduced to a base fraction by cancelling. to cancel a fraction, look for the highest common factor of the numerator and the denominator. 8 of 10 ## top heavy fractions and mixed numbers A top-heavy fractions is a fraction in which the numerator is bigger than the denominator top heavy fractions are also knowm as improper fractions. fractions that are not top-heavy are called proper fractions a mixed fraction is a mixture of a whole number and a proper ftraction 9 of 10 ## h to change a top heavy fractions into a mixed numnber, divide the numerator by the denominator to find the whole number, then the remainder is the number of the proper fractions to change a mixed number into a top-heavy fractions, multiply the whole number by the denominator and add the result to the numerator, to find the numerator od the top-heavy fractions. 10 of 10
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ## Get ready for AP® Calculus ### Course: Get ready for AP® Calculus>Unit 1 Lesson 5: Introduction to factoring # Intro to factors & divisibility Sal explains what it means for a polynomial to be a factor of another polynomial, and what it means for a polynomial to be divisible by another polynomial. ## Want to join the conversation? • But how did he get 10x in the first place? In the first problem, he only multiplied the similar terms (3 times -2, x times x etc.) But in the second one he multiplied everything • When multiplying binomials, think of it as doing the distributive property. Multiply each term by each term. So x * x = x^2, while 3 * 7 = 21. But, x * 7 =7x, while 3 * x = 3x. So, x^2 +7x + 3x + 21. Simplifying that, you add the 3x to the 7x to equal 10x. The final answer is, x^2 + 10x + 21 • Hi, Just to get some clarity, What is the difference between binomial, polynomial, trinomial, etc? • A monomial is a polynomial with 1 term. A binomial is a polynomial with 2 terms. A trinomial is a polynomial with 3 terms. • can decimals be factors or is it just integers? as an example, you can definitely say that 3 is a factor of 6, but can you say that 2.5 is a factor of 5? • Great question! When we talk about factors of whole numbers, we are looking for whole numbers. 2.5 is not a whole number, so it is not a factor of 5. Negative integers, like -1 and -5, are not factors either, because they are not whole numbers. The only factors of 5 are 1 and 5, making 5 prime. • Why did he put 10x at ?? (1 vote) • When you multiply (x+3)(x+7), you would get x squared plus 7x plus 3x plus 21. You have 3x and 7x, which add together to get 10x. • What exactly is the purpose of factorizing something? Great Question! Similarly in algebra, factoring is a remarkably powerful tool, which is used at every level. It provides a standard method for solving quadratic equations as well, of course, as for simplifying complicated expressions. It is also useful when graphing functions. Factoring (or factorising) is the opposite of expanding. This is a great process of simplification. Also, factoring is a complementary operation to the distributive property, it is a way to “unpack” the multiplication done by applying the distributive property. Reorganizing polynomials by factoring allows us to find solutions for certain types of polynomials. Hope this helps. (1 vote) • what is the difference between polynomials, and binomials? • A polynomial is simply a fancy word to describe a math statement with several terms. A mononomial is the opposite - it is a statement with only one term. A binomial is a type of polynomial that contains exactly two terms. All of these are pretty simple to remember because they use the same Greek/Latin roots that many other math terms use. (Poly means many, mono means one, and bi means two). Here are some examples of all of these concepts. Polynomial: - 4x^2 + 5x + 6 - x + a + c - 433 + 19b + 17c + 25a Mononomial: - 5x - a - 19 - 7 Binomial: - x + 7 - 4x + 9 - 9a + b Hope that helps! • we know that (a+b)(a-b)=a^2-b^2 This means that, a+b is a factor of a^2-b^2, ALSO a^2-b^2 is divisible by a+b but dividing a^2-b^2 by a+b does not give a-b plz help me to understand this concept in term of this formula THANK YOU. (1 vote) • Look, if a=1&b=2, a^2-b^2=-3, and -3/a+b=-3/3=-1, a-b=-1, so (a^2-b^2)/(a+b)=a-b. (Replace unknowns with numbers) • If x^2 + 10x + 21 is a trinomial, then is x^2 + 3x + 7x + 21 a quadrinomial, or still a trinomial, since it can be simplified to the former? • In its expanded form, it is a quadrinomial because it has 4 terms at that point. However, it is still a quadratic expression.
# Difference between revisions of "2016 AIME I Problems/Problem 9" ## Problem Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$. This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$. Find the maximum possible area of $AQRS$. ## Solution ### Solution 1 Note that if angle $BAC$ is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where $A$ is obtuse. Therefore, angle A is acute. Let angle $CAS=n$ and angle $BAQ=m$. Then, $\overline{AS}=31\cos(n)$ and $\overline{AQ}=40\cos(m)$. Then the area of rectangle $AQRS$ is $1240\cos(m)\cos(n)$. By product-to-sum, $\cos(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))$. $\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}$. The maximum possible value of $\cos(m-n)$ is 1, which occurs when $m=n$. Thus the maximum possible value of $\cos(m)\cos(n)$ is $\frac{1}{2}(\frac{1}{5}+1)=\frac{3}{5}$ so the maximum possible area of $AQRS$ is $1240\times{\frac{3}{5}}=\fbox{744}$. ### Solution 2 As above, we note that angle $A$ must be acute. Therefore, let $A$ be the origin, and suppose that $Q$ is on the positive $x$ axis and $S$ is on the positive $y$ axis. We approach this using complex numbers. Let $w=\text{cis} A$, and let $z$ be a complex number with $\|z\|=1$, $\text{Arg}(z)\ge 0^\circ$ and $\text{Arg}(zw)\le90^\circ$. Then we represent $B$ by $40z$ and $C$ by $31zw$. The coordinates of $Q$ and $S$ depend on the real part of $40z$ and the imaginary part of $31zw$. Thus $$[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).$$ We can expand this, using the fact that $z\overline{z}=\|z\|^2$, finding $$[AQRS]=620\left(\frac{z^2w-\overline{z^2w}+w-\overline{w}}{2i}\right)=620(\Im(z^2w)+\Im(w)).$$ Now as $w=\text{cis}A$, we know that $\Im(w)=\frac15$. Also, $\|z^2w\|=1$, so the maximum possible imaginary part of $z^2w$ is $1$. This is clearly achievable under our conditions on $z$. Therefore, the maximum possible area of $AQRS$ is $620(1+\tfrac15)=\boxed{744}$. ### Solution 3 (With Calculus) Let $\theta$ be the angle $\angle BAQ$. The height of the rectangle then can be expressed as $h = 31 \sin (A+\theta)$, and the length of the rectangle can be expressed as $l = 40\cos \theta$. The area of the rectangle can then be written as a function of $\theta$, $[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos \theta = 1240 \sin (A+\theta) \cos \theta$. For now, we will ignore the $1240$ and focus on the function $f(\theta) = \sin (A+\theta) \cos \theta = (\sin A \cos \theta + \cos A \sin \theta)(\cos \theta) = \sin A \cos^2 \theta + \cos A \sin \theta \cos \theta = \sin A \cos^2 \theta + \frac{1}{2} \cos A \sin 2\theta$. Taking the derivative, $f'(\theta) = \sin A \cdot -2\cos \theta \sin \theta + \cos A \cos 2\theta = \cos A \cos 2\theta - \sin A \sin 2\theta = \cos(2\theta + A)$. Setting this equal to $0$, we get $\cos(2 \theta + A) = 0 \Rightarrow 2\theta +A = 90, 270 ^\circ$. Since we know that $A+ \theta < 90$, the $270^\circ$ solution is extraneous. Thus, we get that $\theta = \frac{90 - A}{2} = 45 - \frac{A}{2}$. Plugging this value into the original area equation, $a(45 - \frac{A}{2}) = 1240 \sin (45 - \frac{A}{2} + A) \cos (45 - \frac{A}{2}) = 1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2})$. Using a product-to-sum formula, we get that: $$1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2}) =$$ $$1240\cdot \frac{1}{2}\cdot(\sin((45 + \frac{A}{2}) + (45 -\frac{A}{2}))+\sin((45 +\frac{A}{2})-(45 - \frac{A}{2})))=$$ $$620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}$$. ### Note on Problem Validity It has been noted that this answer won't actually work. Let angle $QAB = m$ and angle $CAS = n$ as in Solution 1. Since we know (through that solution) that $m = n$, we can call them each $\theta$. The height of the rectangle is $AS = 31\cos\theta$, and the distance $BQ = 40\sin\theta$. We know that, if the triangle is to be inscribed in a rectangle, $AS \geq BQ$. $$AS \geq BQ$$ $$31\cos\theta \geq 40\sin\theta$$ $$\frac{31}{40} \geq \tan\theta$$ However, $\tan\theta = \tan(\frac{90-A}{2}) = \frac{\sin(90-A)}{\cos(90-A)+1} = \frac{\cos A}{\sin A + 1} = \frac{\frac{2\sqrt6}{5}}{\frac{6}{5}} = \frac{\sqrt6}{3} > \frac{31}{40}$, so the triangle does not actually fit in the rectangle: specifically, B is above R and thus in the line containing segment QR but not on the actual segment or in the rectangle. $[asy] size(200); pair A,B,C,Q,R,S; real r = (pi/2 - asin(1/5))/2; A = (0,0); B = 40dir(r180/pi); C = 31dir(90-r180/pi); draw(A--B--C--cycle); Q = (40cos(r),0); R = (40cos(r),31cos(r)); S = (0, 31cos(r)); draw(A--Q--R--S--cycle); label("A",A,SW); label("B",B,NE); label("C",C,N); label("Q",Q,SE); label("R",R,E); label("S",S,NW); [/asy]$ The actual answer is a radical near $728$ (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer $744$ despite the invalid problem statement.
# Is f(x)=(x^3-2x^2+5x-4)/(x-2) increasing or decreasing at x=0? Nov 21, 2016 $f ' \left(0\right)$ is negative, therefore $f \left(x\right)$ is decreasing at $x = 0$. #### Explanation: This is the same as asking if $f ' \left(x\right)$ is positive or negative at $x = 0$. We have to first find $f ' \left(x\right)$, then plug in 0 to get $f ' \left(0\right)$. $f \left(x\right) = \frac{{x}^{3} - 2 {x}^{2} + 5 x - 4}{x - 2}$ $f ' \left(x\right) = \frac{\left(x - 2\right) \left(3 {x}^{2} - 4 x + 5\right) - \left({x}^{3} - 2 {x}^{2} + 5 x - 4\right) \left(1\right)}{{\left(x - 2\right)}^{2}}$ $f ' \left(x\right) = \frac{\left(3 {x}^{3} - 4 {x}^{2} + 5 x - 6 {x}^{2} + 8 x - 10\right) - \left({x}^{3} - 2 {x}^{2} + 5 x - 4\right)}{{\left(x - 2\right)}^{2}}$ $f ' \left(x\right) = \frac{\left(3 {x}^{3} - 10 {x}^{2} + 13 x - 10\right) - \left({x}^{3} - 2 {x}^{2} + 5 x - 4\right)}{{\left(x - 2\right)}^{2}}$ $f ' \left(x\right) = \frac{2 {x}^{3} - 8 {x}^{2} + 8 x - 6}{{\left(x - 2\right)}^{2}}$ $f ' \left(x\right) = \frac{2 \left({x}^{3} - 4 {x}^{2} + 4 x - 3\right)}{{\left(x - 2\right)}^{2}}$ $f ' \left(0\right) = \frac{\left(2\right) \left(- 3\right)}{{\left(- 2\right)}^{2}}$ $= \frac{- 6}{4}$ $= - \frac{3}{2}$ $f ' \left(0\right)$ is negative, therefore $f \left(x\right)$ is decreasing at $x = 0$.
NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 Chapter 5 Ex.5.2 Question 1 Find the modulus and argument of the complex number $$z = - 1 - i\sqrt 3$$ Solution $$z = - 1 - i\sqrt 3$$ Let $$r\cos \theta = - 1$$ and $$r\sin \theta = - \sqrt 3$$ On squaring and adding, we obtain \begin{align} {{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}&={{\left( -1 \right)}^{2}}+{{\left( -\sqrt{3} \right)}^{2}} \\ \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right) &=1+3\ \ \ \ \ \ \ \ \ \left[ \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right] \\ \Rightarrow {{r}^{2}}&=4 \\ \Rightarrow r=\sqrt{4} &=2\ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align} Therefore, Modulus$$= 2$$ Hence, $$2\cos \theta = - 1$$ and $$2\sin \theta = - \sqrt 3$$ $$\Rightarrow \cos \theta = - \frac{1}{2}$$ and $$\sin \theta = - \frac{{\sqrt 3 }}{2}$$ Since both the values of $$\sin \theta$$ and $$\cos \theta$$ are negative in III quadrant, Argument$$= - \left( {\pi - \frac{\pi }{3}} \right) = \frac{{ - 2\pi }}{3}$$ Thus, the modulus and argument of the complex number $$- 1 - i\sqrt 3$$ are 2 and $$\frac{{ - 2\pi }}{3}$$respectively. Chapter 5 Ex.5.2 Question 2 Find the modulus and argument of the complex number $$z = - \sqrt 3 + i$$ Solution $$z = - \sqrt 3 + i$$ Let $$r\cos \theta = - \sqrt 3$$ and $$r\sin \theta = 1$$ On squaring and adding, we obtain $${r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = {\left( { - \sqrt 3 } \right)^2} + {1^2}$$ \begin{align} \Rightarrow {{r}^{2}}&=3+1=4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right] \\ \Rightarrow r&=\sqrt{4}=2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \end{align} Therefore, Modulus$$= 2$$ Hence, $$2\cos \theta = - \sqrt 3$$ and $$2\sin \theta = 1$$ $$\Rightarrow \cos \theta = - \frac{{\sqrt 3 }}{2}$$ and $$\sin \theta = \frac{1}{2}$$ Since, $$\theta$$ lies in the quadrant II, $$\theta = \pi - \frac{\pi }{6} = \frac{{5\pi }}{6}$$ Thus, the modulus and argument of the complex number $$- \sqrt 3 + i$$ are $$2$$ and $$\frac{5\pi }{6}$$respectively. Chapter 5 Ex.5.2 Question 3 Convert the given complex number in polar form: $$1 - i$$ Solution $$z = 1 - i$$ Let $$r\cos \theta = 1$$ and $$r\sin \theta = - 1$$ On squaring and adding, we obtain \begin{align} {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta&={{1}^{2}}+{{\left( -1 \right)}^{2}} \\ \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)&=1+1 \\ \Rightarrow {{r}^{2}}&=2 \\ \Rightarrow r&=\sqrt{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align} Therefore, $$\sqrt 2 \cos \theta = 1$$ and $$\sqrt 2 \sin \theta = - 1$$ $$\Rightarrow \cos \theta = \frac{1}{{\sqrt 2 }}$$ and $$\sin \theta = - \frac{1}{{\sqrt 2 }}$$ Since, $$\theta$$ lies in the quadrant IV, $$\theta = - \frac{\pi }{4}$$ Hence, \begin{align}1 - i&= r\cos \theta + ir\sin \theta \\&= \sqrt 2 \cos \left( { - \frac{\pi }{4}} \right) + i\sqrt 2 \sin \left( { - \frac{\pi }{4}} \right)\\&= \sqrt 2 \left[ {\cos \left( { - \frac{\pi }{4}} \right) + i\sin \left( { - \frac{\pi }{4}} \right)} \right]\end{align} Thus, this is the required polar form. Chapter 5 Ex.5.2 Question 4 Convert the given complex number in polar form: $$- 1 + i$$ Solution $$z = - 1 + i$$ Let $$r\cos \theta = - 1$$ and $$r\sin \theta = 1$$ On squaring and adding, we obtain \begin{align}{{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta&={{\left( -1 \right)}^{2}}+{{1}^{2}} \\ \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)&=1+1 \\ \Rightarrow {{r}^{2}}&=2 \\ \Rightarrow r&=\sqrt{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align} Therefore, $$\sqrt 2 \cos \theta = - 1$$ and $$\sqrt 2 \sin \theta = 1$$ $$\Rightarrow \cos \theta = - \frac{1}{{\sqrt 2 }}$$ and $$\sin \theta = \frac{1}{{\sqrt 2 }}$$ Since, $$\theta$$ lies in the quadrant II, $$\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}$$ Hence, \begin{align} - 1 + i &= r\cos \theta + ir\sin \theta \\&= \sqrt 2 \cos \frac{{3\pi }}{4} + i\sqrt 2 \sin \frac{{3\pi }}{4}\\&= \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)\end{align} Thus, this is the required polar form. Chapter 5 Ex.5.2 Question 5 Convert the given complex number in polar form: $$- 1 - i$$ Solution $$z = - 1 - i$$ Let $$r\cos \theta = - 1$$ and $$r\sin \theta = - 1$$ On squaring and adding, we obtain \begin{align}{{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta&={{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}} \\ \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)&=1+1 \\ \Rightarrow {{r}^{2}}& =2 \\ \Rightarrow r& =\sqrt{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align} Therefore, $$\sqrt 2 \cos \theta = - 1$$ and $$\sqrt 2 \sin \theta = - 1$$ $$\Rightarrow \cos \theta = - \frac{1}{{\sqrt 2 }}$$ and $$\sin \theta = - \frac{1}{{\sqrt 2 }}$$ Since, $$\theta$$ lies in the quadrant III, $$\theta = - \left( {\pi - \frac{\pi }{4}} \right) = - \frac{{3\pi }}{4}$$ Hence, \begin{align} - 1 - i&= r\cos \theta + ir\sin \theta \\&= \sqrt 2 \cos \frac{{ - 3\pi }}{4} + i\sqrt 2 \sin \frac{{ - 3\pi }}{4}\\&= \sqrt 2 \left( {\cos \frac{{ - 3\pi }}{4} + i\sin \frac{{ - 3\pi }}{4}} \right)\end{align} Thus, this is the required polar form. Chapter 5 Ex.5.2 Question 6 Convert the given complex number in polar form: $$- 3$$ Solution $$z = - 3$$ Let $$r\cos \theta = - 3$$ and $$r\sin \theta = 0$$ On squaring and adding, we obtain \begin{align}& \ \ \ \ {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta ={{\left( -3 \right)}^{2}}+{{\left( 0 \right)}^{2}} \\ & \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=9 \\ & \Rightarrow {{r}^{2}}=9 \\ & \Rightarrow r=3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{Conventionally},\ r>0 \right] \\\end{align} Therefore, $$3\cos \theta = - 3$$ and $$3\sin \theta = 0$$ $$\Rightarrow \cos \theta = - 1$$ and $$\sin \theta = 0$$ Since the $$\theta$$ lies in the quadrant II, $$\theta = \pi$$ Hence, \begin{align} - 3&= r\cos \theta + ir\sin \theta \\&= 3\cos \pi + i3\sin \pi \\&= 3\left( {\cos \pi + i\sin \pi } \right)\end{align} Thus, this is the required polar form. Chapter 5 Ex.5.2 Question 7 Convert the given complex number in polar form: $$\sqrt 3 + i$$ Solution $$z = \sqrt 3 + i$$ Let $$r\cos \theta = \sqrt 3$$ and $$r\sin \theta = 1$$ On squaring and adding, we obtain \begin{align}{r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta&= {\left( {\sqrt 3 } \right)^2} + {1^2}\\ \Rightarrow {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) &= 3 + 1\\ \Rightarrow {r^2}&= 4\\ \Rightarrow r&= \sqrt 4 = 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Conventionally}},\;r > 0} \right]\end{align} Therefore, $$2\cos \theta = \sqrt 3$$and $$2\sin \theta = 1$$ $$\Rightarrow \cos \theta = \frac{{\sqrt 3 }}{2}$$and $$\sin \theta = \frac{1}{2}$$ Since, $$\theta$$ lies in quadrant I, $$\theta = \frac{\pi }{6}$$ Hence, \begin{align}\sqrt 3 + i &= r\cos \theta + ir\sin \theta \\&= 2\cos \frac{\pi }{6} + i2\sin \frac{\pi }{6}\\&= 2\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right)\end{align} Thus, this is the required polar form. Chapter 5 Ex.5.2 Question 8 Convert the given complex number in polar form: $$i$$ Solution $$z = i$$ Let $$r\cos \theta = 0$$ and $$r\sin \theta = 1$$ On squaring and adding, we obtain \begin{align}{r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta&= {0^2} + {1^2}\\ \Rightarrow {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)&= 1\\ \Rightarrow {r^2} &= 1\\ \Rightarrow r &= \sqrt 1 = 1\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Conventionally}},\;r > 0} \right]\end{align} Therefore, $$\cos \theta = 0$$ and $$\sin \theta = 1$$ Since, $$\theta$$ lies in quadrant I, $$\theta = \frac{\pi }{2}$$ Hence, \begin{align}i&= r\cos \theta + ir\sin \theta \\&= \cos \frac{\pi }{2} + i\sin \frac{\pi }{2}\end{align} Thus, this is the required polar form. Instant doubt clearing with Cuemath Advanced Math Program
# 2002 AIME II Problems/Problem 14 ## Problem The perimeter of triangle $APM$ is $152$, and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$. Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$. ## Solution Let the circle intersect $\overline{PM}$ at $B$. Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point. So we have: $$\frac{19}{AM} = \frac{152-2AM}{152}$$ Solving, $AM = 38$. So the ratio of the side lengths of the triangles is 2. Therefore, $\frac{PB+38}{OP}= 2$ and $\frac{OP+19}{PB} = 2$ $2OP = PB+38$ and $2PB = OP+19$ $4OP-76 = OP+19$ Finally, $OP = \frac{95}3$, so the answer is $098$.
# Basic Statistics Basic probability theory and statistics: • Michael Mitzenmacher and Eli Upfal. Probability and Computing: Randomized Algorithms and Probabilistic Analysis. • Michael Baron. Probability and Statistics for Computer Scientists, 2nd edition. Statistics applies tools of the probability theory to the study of data (either a sample or the population): • Descriptive statistics quantitatively describe or summarize features of a sample. • Inferential statistics use random sample data to infer properties about a larger population. # 1 Descriptive Statistics To describe a sample (data set) quantitatively, three kinds of measures are commonly used: central tendency (often in connection with the first raw moment), dispersion / variability (often in connection with the second central moment) and shape (often in connection with higher-order central moments or cumulants). ## 1.1 Central Tendency Arithmetic mean (AM, $$\mu$$). Sum of values of a data set divided by number of values: $\mu = \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ Example: Given data set $$\{1,2,2,3,4,7,9\}$$, the arithmetic mean is $$\mu = \frac{1+2+2+3+4+7+9}{7} = 4$$. Remark 1.1.1. (Difference between arithmetic mean and expectation) Note that although the formula of an arithmetic mean looks similar to the one of a mathematical expectation in probability theory, arithmetic mean is defined on a given sample (data set), while expectation is defined on a random variable with a given probability distribution. Moreover, arithmetic mean is generally unweighted. Median ($$\nu)$$. Middle value separating the greater and lesser halves of a data set. Example: Given data set $$\{1,2,2,3,4,7,9\}$$, the median is $$\nu = 3$$. Mode ($$\theta$$). Most frequent value in a data set. Example: Given data set $$\{1,2,2,3,4,7,9\}$$, the mode is $$\theta = 2$$. Mid-range. The arithmetic mean of the maximum and minimum values in a data set. Example: Given data set $$\{1,2,2,3,4,7,9\}$$, the midrange is $$\frac{9+1}{2} = 5$$. Values far from the mode are called outliers, such as very infrequent, occasional or false experimental data records. The mode and the median are relatively robust in the presence of outliers, while the arithmetic mean is rather sensitive. The mid-range is extremely sensitive to the values of outliers, thus it is rarely used in practical statistical analysis as it fails to provide a robust estimator for most distributions of interest. For an arbitrary distribution, the arithmetic mean $$\mu$$, the median $$\nu$$ and the mode $$\theta$$ may appear in any order. Besides the arithmetic mean, other means of measuring a data set are also defined: Geometric mean (GM). Defined as $$\left(\prod_{i=1}^n x_i \right)^\frac{1}{n}$$. Harmonic mean (HM). Defined as $$\frac{n}{\sum_{i=1}^n \frac{1}{x_i}}$$. (Related to the harmonic series $$\sum_{n=1}^\infty \frac{1}{n}$$.) Arithmetic mean (AM), geometric mean (GM) and harmonic mean (HM) are sometimes referred to as Pythagorean means, which are special cases of the generalized mean: Generalized mean (power mean; Hölder mean). Given $$x_1,\dots,x_n \in \mathbb{R}^+$$, and $$w_1,\dots,w_n \in \mathbb{R}$$ with $$\sum_{i=1}^n w_i = 1$$. For $$p \in \mathbb{R}, p \neq 0$$, the generalized mean with exponent $$p$$ of $$\{x_1,\dots,x_n\}$$ weighted by $$\{w_1,\dots,w_n\}$$ is defined as $\operatorname{M}_p (x_1,\dots,x_n) = \left( \sum_{i=1}^n w_i x_i^p \right)^\frac{1}{p}$ For $$p = 0$$, we assume that it is equal to the geometric mean (which is the limit of means with exponents approaching zero): $\operatorname{M}_0 (x_1,\dots,x_n) = \lim_{p \to 0} \operatorname{M}_{p}(x_1,\dots,x_n) = \prod_{i=1}^n x_i^{w_i}$ The unweighted means correspond to setting all $$w_i = \frac{1}{n}$$. We can redefine Pythagorean means using the generalized mean with different parameters $$p$$: • (Harmonic mean, $$p=-1$$) $$\operatorname{HM}(x_1,\dots,x_n) = \operatorname{M}_{-1}(x_1,\dots,x_n) = \frac{n}{\sum_{i=1}^n \frac{1}{x_i}}$$. • (Geometric mean, $$p=0$$) $$\operatorname{GM}(x_1,\dots,x_n) = \operatorname{M}_{0}(x_1,\dots,x_n) = \left(\prod_{i=1}^n x_i \right)^\frac{1}{n}$$. • (Arithmetic mean, $$p=1$$) $$\operatorname{AM}(x_1,\dots,x_n) = \operatorname{M}_{1}(x_1,\dots,x_n) = \frac{1}{n} \sum_{i=1}^n x_i$$. Theorem 1.1.2. (Generalized mean inequality) If $$p, q \in \mathbb{R}$$ and $$p < q$$, then $\operatorname{M}_p(x_1,\dots,x_n) \leq \operatorname{M}_q(x_1,\dots,x_n)$ The equality holds if and only if $$x_1 = \dots = x_n$$ ($$= \operatorname{M}_p(x_1,\dots,x_n)$$). ## 1.2 Dispersion / Variability Minimum (first or smallest order statistic). The smallest value in the data set. Defined using the generalized mean: $\operatorname{min}\{x_1,\dots,x_n\} = \operatorname{M}_{-\infty}(x_1,\dots,x_n)$ Maximum (largest order statistic). The largest value in the data set. Defined using the generalized mean: $\operatorname{max}\{x_1,\dots,x_n\} = \operatorname{M}_{+\infty}(x_1,\dots,x_n)$ Range. The size of the smallest interval which contains all the data; it is equal to the difference between the maximum and minimum: $\operatorname{Range}\{x_1,\dots,x_n\} = \operatorname{max}\{x_1,\dots,x_n\} - \operatorname{min}\{x_1,\dots,x_n\}$ Intuitively, the range provides a naïve indication of statistical dispersion of the data set. Remark 1.2.1. (Difference between range and support) Recall that the support of a given probability distribution is the closure of the set of all possible values of a random variable having that distribution. The range is defined for a sample (data set), and does not necessarily include every possible value of the population. Interquartile range (IQR; mid-spread; H-spread). The difference between the 75th and the 25th percentiles (i.e., the 3rd and the 1st quartiles). Unlike the regular (full) range, the IQR is a 25%-trimmed estimator thus it is more robust in the presence of outliers. Maximum absolute deviation. Defined as $\max_i \|x_i - m\|$ where $$m$$ can take the value of any chosen measure of central tendency of the data set (e.g., the mean $$\mu$$ or the median $$\nu$$). Mean absolute deviation (MAD; average absolute deviation). Defined as $\frac{1}{n} \sum_{i=1}^n \|x_i - m\|$ where $$m$$ can take the value of any chosen measure of central tendency of the data set (e.g., the mean $$\mu$$ or the median $$\nu$$). Clearly, the MAD is a more robust measure of dispersion than the maximum absolute deviation, which is highly sensitive to the values of outliers. Sample variance. The (unbiased) sample variance $$s^2$$ is an unbiased estimator of the population variance $$\sigma^2$$: $s^2 = \frac{n}{n-1}s_n^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2$ $$s$$ is also called the sample standard deviation. Remark 1.2.2. (Bessel’s correction) The use of the term $$n-1$$ instead of $$n$$ is called Bessel’s correction. It corrects the bias in the estimation of the population variance; however, it often increases the mean squared error (MSE) in these estimations. $$\bar{x}$$ in this formula should be interpreted as the sample (arithmetic) mean. When the population mean is already known, there is no need to use Bessel’s correction, since the use of the population mean will not lead to any bias for estimating the population variance. Coefficient of variation (CV; relative standard deviation). The coefficient of variation is defined as the ratio of the sample standard deviation $$s$$ to the sample mean $$\bar{x}$$: $\widehat{c_{\rm v}} = \frac{s}{\bar{x}}$ Note that this is a biased estimator. The correction varies for different distributions of data. Index of dispersion (coefficient of dispersion; relative variance; variance-to-mean ratio; VMR). Defined as $\widehat{D} = \frac{s^2}{\bar{x}}$ CV and VMR are normalized values of the sample standard deviation and the sample variance respectively, thus they are also called the relative standard deviation and the relative variance, and are only defined when the sample mean $$\bar{x}$$ is non-zero. ## 1.3 Shape Sample skewness. Defined as $g_1 = \frac{m_3}{m_2^{3/2}} = \frac{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^3}{(\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2)^{3/2}}$ where $$m_i$$ is the $$i$$th sample central moment. Note that this is a biased estimator. In practice, sample skewness is a measure of the asymmetry of the distribution of a data set about its mean. The skewness of any univariate normal distribution is 0 (symmetric). Sample excess kurtosis. Defined as $g_2 = \frac{m_4}{m_2^2} - 3 = \frac{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^4}{(\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2)^2} - 3$ where $$m_i$$ is the $$i$$th sample central moment. Note that this is a biased estimator. In practice, larger sample kurtosis indicates that the data set has more extreme outliers, so there could be a problem with sampling. The kurtosis of any univariate normal distribution is 3 (thus the sample excess kurtosis should be ideally close to 0). Related topics:
This is “Applications of Rational Equations”, section 7.6 from the book Beginning Algebra (v. 1.0). For details on it (including licensing), click here. Has this book helped you? Consider passing it on: Creative Commons supports free culture from music to education. Their licenses helped make this book available to you. DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators. ## 7.6 Applications of Rational Equations ### Learning Objectives 1. Solve applications involving relationships between real numbers. 2. Solve applications involving uniform motion (distance problems). 3. Solve work-rate applications. ## Number Problems Recall that the reciprocalThe reciprocal of a nonzero number n is 1/n. of a nonzero number n is 1/n. For example, the reciprocal of 5 is 1/5 and 5 ⋅ 1/5 = 1. In this section, the applications will often involve the key word “reciprocal.” When this is the case, we will see that the algebraic setup results in a rational equation. Example 1: A positive integer is 4 less than another. The sum of the reciprocals of the two positive integers is 10/21. Find the two integers. Solution: Begin by assigning variables to the unknowns. Next, use the reciprocals $1n$ and $1n−4$ to translate the sentences into an algebraic equation. We can solve this rational expression by multiplying both sides of the equation by the least common denominator (LCD). In this case, the LCD is $21n(n−4)$. The question calls for integers and the only integer solution is $n=7$. Hence disregard 6/5. Use the expression $n−4$ to find the smaller integer. Answer: The two positive integers are 3 and 7. The check is left to the reader. Example 2: A positive integer is 4 less than another. If the reciprocal of the smaller integer is subtracted from twice the reciprocal of the larger, then the result is 1/30. Find the two integers. Solution: Set up an algebraic equation. Solve this rational expression by multiplying both sides by the LCD. The LCD is $30n(n−4)$. Here we have two viable possibilities for the larger integer. For this reason, we will we have two solutions to this problem. As a check, perform the operations indicated in the problem. Answer: Two sets of positive integers solve this problem: {6, 10} and {20, 24}. Try this! The difference between the reciprocals of two consecutive positive odd integers is 2/15. Find the integers. Answer: The integers are 3 and 5. ### Video Solution (click to see video) ## Uniform Motion Problems Uniform motionDescribed by the formula $D=rt$, where the distance, D, is given as the product of the average rate, r, and the time, t, traveled at that rate. problems, also referred to as distance problems, involve the formula where the distance, D, is given as the product of the average rate, r, and the time, t, traveled at that rate. If we divide both sides by the average rate, r, then we obtain the formula For this reason, when the unknown quantity is time, the algebraic setup for distance problems often results in a rational equation. Similarly, when the unknown quantity is the rate, the setup also may result in a rational equation. We begin any uniform motion problem by first organizing our data with a chart. Use this information to set up an algebraic equation that models the application. Example 5: Mary spent the first 120 miles of her road trip in traffic. When the traffic cleared, she was able to drive twice as fast for the remaining 300 miles. If the total trip took 9 hours, then how fast was she moving in traffic? Solution: First, identify the unknown quantity and organize the data. To avoid introducing two more variables for the time column, use the formula $t=Dr$. Here the time for each leg of the trip is calculated as follows: Use these expressions to complete the chart. The algebraic setup is defined by the time column. Add the times for each leg of the trip to obtain a total of 9 hours: We begin solving this equation by first multiplying both sides by the LCD, 2x. Answer: Mary averaged 30 miles per hour in traffic. Example 6: A passenger train can travel, on average, 20 miles per hour faster than a freight train. If the passenger train covers 390 miles in the same time it takes the freight train to cover 270 miles, then how fast is each train? Solution: First, identify the unknown quantities and organize the data. Next, organize the given data in a chart. Use the formula $t=Dr$ to fill in the time column for each train. Because the trains travel the same amount of time, finish the algebraic setup by equating the expressions that represent the times: Solve this equation by first multiplying both sides by the LCD, $x(x+20)$. Use x + 20 to find the speed of the passenger train. Answer: The speed of the passenger train is 65 miles per hour and the speed of the freight train is 45 miles per hour. Example 7: Brett lives on the river 8 miles upstream from town. When the current is 2 miles per hour, he can row his boat downstream to town for supplies and back in 3 hours. What is his average rowing speed in still water? Solution: Rowing downstream, the current increases his speed, and his rate is x + 2 miles per hour. Rowing upstream, the current decreases his speed, and his rate is x − 2 miles per hour. Begin by organizing the data in the following chart: Use the formula $t=Dr$ to fill in the time column for each leg of the trip. The algebraic setup is defined by the time column. Add the times for each leg of the trip to obtain a total of 3 hours: Solve this equation by first multiplying both sides by the LCD, $(x+2)(x−2)$. Next, solve the resulting quadratic equation. Use only the positive solution, $x=6$ miles per hour. Answer: His rowing speed is 6 miles per hour. Try this! Dwayne drove 18 miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an average of 15 miles per hour faster than he did on the trip there. If the total driving time was 1 hour, then what was his average speed driving to the airport? Answer: His average speed driving to the airport was 30 miles per hour. ### Video Solution (click to see video) ## Work-Rate Problems The rate at which a task can be performed is called a work rateThe rate at which a task can be performed.. For example, if a painter can paint a room in 8 hours, then the task is to paint the room, and we can write In other words, the painter can complete $18$ of the task per hour. If he works for less than 8 hours, then he will perform a fraction of the task. For example, Obtain the amount of the task completed by multiplying the work rate by the amount of time the painter works. Typically, work-rate problems involve people working together to complete tasks. When this is the case, we can organize the data in a chart, just as we have done with distance problems. Suppose an apprentice painter can paint the same room by himself in 10 hours. Then we say that he can complete $110$ of the task per hour. Let t represent the time it takes both of the painters, working together, to paint the room. To complete the chart, multiply the work rate by the time for each person. The portion of the room each can paint adds to a total of 1 task completed. This is represented by the equation obtained from the first column of the chart: This setup results in a rational equation that can be solved for t by multiplying both sides by the LCD, 40. Therefore, the two painters, working together, complete the task in $449$ hours. In general, we have the following work-rate formula$1t1⋅t+1t2⋅t=1$, where $1t1$ and $1t2$ are the individual work rates and t is the time it takes to complete the task working together.: Here $1t1$ and $1t2$ are the individual work rates and t is the time it takes to complete one task working together. If we factor out the time, t, and then divide both sides by t, we obtain an equivalent work-rate formula: In summary, we have the following equivalent work-rate formulas: Example 3: Working alone, Billy’s dad can complete the yard work in 3 hours. If Billy helps his dad, then the yard work takes 2 hours. How long would it take Billy working alone to complete the yard work? Solution: The given information tells us that Billy’s dad has an individual work rate of $13$ task per hour. If we let x represent the time it takes Billy working alone to complete the yard work, then Billy’s individual work rate is $1x$, and we can write Working together, they can complete the task in 2 hours. Multiply the individual work rates by 2 hours to fill in the chart. The amount of the task each completes will total 1 completed task. To solve for x, we first multiply both sides by the LCD, 3x. Answer: It takes Billy 6 hours to complete the yard work alone. Of course, the unit of time for the work rate need not always be in hours. Example 4: Working together, two construction crews can build a shed in 5 days. Working separately, the less experienced crew takes twice as long to build a shed than the more experienced crew. Working separately, how long does it take each crew to build a shed? Solution: Working together, the job is completed in 5 days. This gives the following setup: The first column in the chart gives us an algebraic equation that models the problem: Solve the equation by multiplying both sides by 2x. To determine the time it takes the less experienced crew, we use 2x: Answer: Working separately, the experienced crew takes 7½ days to build a shed, and the less experienced crew takes 15 days to build a shed. Try this! Joe’s garden hose fills the pool in 12 hours. His neighbor has a thinner hose that fills the pool in 15 hours. How long will it take to fill the pool using both hoses? Answer: It will take both hoses $623$ hours to fill the pool. ### Video Solution (click to see video) ### Key Takeaways • In this section, all of the steps outlined for solving general word problems apply. Look for the new key word “reciprocal,” which indicates that you should write the quantity in the denominator of a fraction with numerator 1. • When solving distance problems where the time element is unknown, use the equivalent form of the uniform motion formula, $t=Dr$, to avoid introducing more variables. • When solving work-rate problems, multiply the individual work rate by the time to obtain the portion of the task completed. The sum of the portions of the task results in the total amount of work completed. ### Topic Exercises Part A: Number Problems Use algebra to solve the following applications. 1. A positive integer is twice another. The sum of the reciprocals of the two positive integers is 3/10. Find the two integers. 2. A positive integer is twice another. The sum of the reciprocals of the two positive integers is 3/12. Find the two integers. 3. A positive integer is twice another. The difference of the reciprocals of the two positive integers is 1/8. Find the two integers. 4. A positive integer is twice another. The difference of the reciprocals of the two positive integers is 1/18. Find the two integers. 5. A positive integer is 2 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 5/12, then find the two integers. 6. A positive integer is 2 more than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 17/35, then find the two integers. 7. The sum of the reciprocals of two consecutive positive even integers is 11/60. Find the two even integers. 8. The sum of the reciprocals of two consecutive positive odd integers is 16/63. Find the integers. 9. The difference of the reciprocals of two consecutive positive even integers is 1/24. Find the two even integers. 10. The difference of the reciprocals of two consecutive positive odd integers is 2/99. Find the integers. 11. If 3 times the reciprocal of the larger of two consecutive integers is subtracted from 2 times the reciprocal of the smaller, then the result is 1/2. Find the two integers. 12. If 3 times the reciprocal of the smaller of two consecutive integers is subtracted from 7 times the reciprocal of the larger, then the result is 1/2. Find the two integers. 13. A positive integer is 5 less than another. If the reciprocal of the smaller integer is subtracted from 3 times the reciprocal of the larger, then the result is 1/12. Find the two integers. 14. A positive integer is 6 less than another. If the reciprocal of the smaller integer is subtracted from 10 times the reciprocal of the larger, then the result is 3/7. Find the two integers. Part B: Uniform Motion Problems Use algebra to solve the following applications. 15. James can jog twice as fast as he can walk. He was able to jog the first 9 miles to his grandmother’s house, but then he tired and walked the remaining 1.5 miles. If the total trip took 2 hours, then what was his average jogging speed? 16. On a business trip, an executive traveled 720 miles by jet aircraft and then another 80 miles by helicopter. If the jet averaged 3 times the speed of the helicopter and the total trip took 4 hours, then what was the average speed of the jet? 17. Sally was able to drive an average of 20 miles per hour faster in her car after the traffic cleared. She drove 23 miles in traffic before it cleared and then drove another 99 miles. If the total trip took 2 hours, then what was her average speed in traffic? 18. Harry traveled 15 miles on the bus and then another 72 miles on a train. If the train was 18 miles per hour faster than the bus and the total trip took 2 hours, then what was the average speed of the train? 19. A bus averages 6 miles per hour faster than a trolley. If the bus travels 90 miles in the same time it takes the trolley to travel 75 miles, then what is the speed of each? 20. A passenger car averages 16 miles per hour faster than the bus. If the bus travels 56 miles in the same time it takes the passenger car to travel 84 miles, then what is the speed of each? 21. A light aircraft travels 2 miles per hour less than twice as fast as a passenger car. If the passenger car can travel 231 miles in the same time it takes the aircraft to travel 455 miles, then what is the average speed of each? 22. Mary can run 1 mile per hour more than twice as fast as Bill can walk. If Bill can walk 3 miles in the same time it takes Mary to run 7.2 miles, then what is Bill’s average walking speed? 23. An airplane traveling with a 20-mile-per-hour tailwind covers 270 miles. On the return trip against the wind, it covers 190 miles in the same amount of time. What is the speed of the airplane in still air? 24. A jet airliner traveling with a 30-mile-per-hour tailwind covers 525 miles in the same amount of time it is able to travel 495 miles after the tailwind eases to 10 miles per hour. What is the speed of the airliner in still air? 25. A boat averages 16 miles per hour in still water. With the current, the boat can travel 95 miles in the same time it travels 65 miles against it. What is the speed of the current? 26. A river tour boat averages 7 miles per hour in still water. If the total 24-mile tour downriver and 24 miles back takes 7 hours, then how fast is the river current? 27. If the river current flows at an average 3 miles per hour, then a tour boat makes the 9-mile tour downstream with the current and back the 9 miles against the current in 4 hours. What is the average speed of the boat in still water? 28. Jane rowed her canoe against a 1-mile-per-hour current upstream 12 miles and then returned the 12 miles back downstream. If the total trip took 5 hours, then at what speed can Jane row in still water? 29. Jose drove 15 miles to pick up his sister and then returned home. On the return trip, he was able to average 15 miles per hour faster than he did on the trip to pick her up. If the total trip took 1 hour, then what was Jose’s average speed on the return trip? 30. Barry drove the 24 miles to town and then back in 1 hour. On the return trip, he was able to average 14 miles per hour faster than he averaged on the trip to town. What was his average speed on the trip to town? 31. Jerry paddled his kayak upstream against a 1-mile-per-hour current for 12 miles. The return trip downstream with the 1-mile-per-hour current took 1 hour less time. How fast can Jerry paddle the kayak in still water? 32. It takes a light aircraft 1 hour more time to fly 360 miles against a 30-mile-per-hour headwind than it does to fly the same distance with it. What is the speed of the aircraft in calm air? Part C: Work-Rate Problems Use algebra to solve the following applications. 33. James can paint the office by himself in 7 hours. Manny paints the office in 10 hours. How long will it take them to paint the office working together? 34. Barry can lay a brick driveway by himself in 12 hours. Robert does the same job in 10 hours. How long will it take them to lay the brick driveway working together? 35. Jerry can detail a car by himself in 50 minutes. Sally does the same job in 1 hour. How long will it take them to detail a car working together? 36. Jose can build a small shed by himself in 26 hours. Alex builds the same small shed in 2 days. How long would it take them to build the shed working together? 37. Allison can complete a sales route by herself in 6 hours. Working with an associate, she completes the route in 4 hours. How long would it take her associate to complete the route by herself? 38. James can prepare and paint a house by himself in 5 days. Working with his brother, Bryan, they can do it in 3 days. How long would it take Bryan to prepare and paint the house by himself? 39. Joe can assemble a computer by himself in 1 hour. Working with an assistant, he can assemble a computer in 40 minutes. How long would it take his assistant to assemble a computer working alone? 40. The teacher’s assistant can grade class homework assignments by herself in 1 hour. If the teacher helps, then the grading can be completed in 20 minutes. How long would it take the teacher to grade the papers working alone? 41. A larger pipe fills a water tank twice as fast as a smaller pipe. When both pipes are used, they fill the tank in 5 hours. If the larger pipe is left off, then how long would it take the smaller pipe to fill the tank? 42. A newer printer can print twice as fast as an older printer. If both printers working together can print a batch of flyers in 45 minutes, then how long would it take the newer printer to print the batch working alone? 43. Working alone, Henry takes 9 hours longer than Mary to clean the carpets in the entire office. Working together, they clean the carpets in 6 hours. How long would it take Mary to clean the office carpets if Henry were not there to help? 44. Working alone, Monique takes 4 hours longer than Audrey to record the inventory of the entire shop. Working together, they take inventory in 1.5 hours. How long would it take Audrey to record the inventory working alone? 45. Jerry can lay a tile floor in 3 hours less time than Jake. If they work together, the floor takes 2 hours. How long would it take Jerry to lay the floor by himself? 46. Jeremy can build a model airplane in 5 hours less time than his brother. Working together, they need 6 hours to build the plane. How long would it take Jeremy to build the model airplane working alone? 47. Harry can paint a shed by himself in 6 hours. Jeremy can paint the same shed by himself in 8 hours. How long will it take them to paint two sheds working together? 48. Joe assembles a computer by himself in 1 hour. Working with an assistant, he can assemble 10 computers in 6 hours. How long would it take his assistant to assemble 1 computer working alone? 49. Jerry can lay a tile floor in 3 hours, and his assistant can do the same job in 4 hours. If Jerry starts the job and his assistant joins him 1 hour later, then how long will it take to lay the floor? 50. Working alone, Monique takes 6 hours to record the inventory of the entire shop, while it takes Audrey only 4 hours to do the same job. How long will it take them working together if Monique leaves 2 hours early? 1: {5, 10} 3: {4, 8} 5: {6, 8} 7: {10, 12} 9: {6, 8} 11: {1, 2} or {−4, −3} 13: {4, 9} or {15, 20} 15: 6 miles per hour 17: 46 miles per hour 19: Trolley: 30 miles per hour; bus: 36 miles per hour 21: Passenger car: 66 miles per hour; aircraft: 130 miles per hour 23: 115 miles per hour 25: 3 miles per hour 27: 6 miles per hour 29: 40 miles per hour 31: 5 miles per hour 33: $4217$ hours 35: $27311$ minutes 37: 12 hours 39: 2 hours 41: 15 hours 43: 9 hours 45: 3 hours 47: $667$ hours 49: $217$ hours
# What is the second derivative of f(x) = x^2ln x ? ##### 1 Answer Jan 7, 2016 $3 + 2 \ln x$ #### Explanation: Let's get the first derivative f'(x) first by applying the product rule: $f \left(x\right) = {x}^{2} \ln x$ $f ' \left(x\right) = {x}^{2} \left(\frac{1}{x}\right) + \left(\ln x\right) \left(2 x\right)$ $f ' \left(x\right) = x + 2 x \ln x$ Get the second derivative f''(x) by differentiating the first derivative: Differentiate term by term. Apply product rule on the second term $f ' \left(x\right) = x + 2 x \ln x$ $f ' ' \left(x\right) = 1 + 2 x \left(\frac{1}{x}\right) + \left(\ln x\right) \left(2\right)$ $f ' ' \left(x\right) = 1 + 2 + 2 \ln x$ $f ' ' \left(x\right) = 3 + 2 \ln x$
Box and Whisker Plots. Presentation on theme: "Box and Whisker Plots."— Presentation transcript: Box and Whisker Plots I can organize and display data in a box and whisker plot. Course 3 I can organize and display data in a box and whisker plot. I can determine the upper and lower extremes, median, and upper and lower quartiles in a box plot. Box and Whisker Plots When we are working with a larger set of data it is much easier to separate the data into quartiles. The quartiles separate the data into four equally sized parts. There are five important values to remember if you want to divide your data into quartiles: Lower Quartile Upper Quartile Median Lowest Value Highest Value Box and Whisker Plots The lower extreme is the lowest value The upper extreme is the highest value The median is the middle number The lower quartile divides the lower ½ of the data into two equally sized groups The upper quartile separates the upper ½ into two equally sized groups Lower Quartile Upper Quartile Median Lowest Value Highest Value Box and Whisker Plots The difference between the lower quartile and the upper quartile is called the interquartile range and corresponds to the 50% of the data points that are in the middle Lower Quartile Upper Quartile Median Lowest Value Highest Value To draw a box-and-whiskers plot begin by marking the five numbers described on the previous slides with dots: The next step is to draw the box The next step is to draw the box. The box has its sides at the LQ and the UQ and we display the median by drawing a line. Then we extend the whiskers from each quartile to the upper and lower extremes. This box-and-whiskers plot separates the data into quarters (called quartiles) with the same number of data points in each part: Quartiles Litter Size Number of Litters Practice finding the values: The table below summarizes a veterinarian’s records for kitten litters born in a given year. Litter Size 2 3 4 5 6 Number of Litters 1 8 11 Writing the data out would look like this: 2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,6 Kitten Data The range of a data set is the largest value minus the smallest value. For the kitten data, the range is 6 — 2 = 4. Lower half Upper half lower quartile: 3 Median: 4 (second quartile) upper quartile: 5 median of upper half Litter Size 2 3 4 5 6 Number of Litters 1 8 11 (1st part) (2nd Part) (3rd part) (4th Part) Course 3 Making a Box and Whisker Plot - Practice Step 1: Find the lower extreme, lower quartile, median, upper quartile, and upper extreme: 21, 25, 15, 13, 17, 19, 19, 21 Order the values. lower extreme: 13 lower quartile: 15 What would the interquartile range be? 21 – 15 = 6 median: 19 upper quartile: 21 upper extreme: 25 Making a Box and Whisker Plot Course 3 9-4 Variability Making a Box and Whisker Plot Step 2: draw a number line and plot a point above each value from Step 1. Lower extreme13 lower quartile 15 median 19 upper quartile 21 Upper extreme25 Making a Box and Whisker Plot Course 3 9-4 Variability Making a Box and Whisker Plot Step 3: Draw the box and whiskers: Comparing Final 1 2 3 4 T Oakland 6 12 21 Tampa Bay 17 14 48 Oakland Tampa Bay When data is displayed in a box and whisker plot you can visually compare the information given. It is very easy to see the five important numbers that make up a box and whisker plot. Compare the medians: Oakland Tampa Bay The median for Tampa Bay is significantly greater. Compare the differences between the upper quartile and lower quartile for each: Oakland Tampa Bay The difference between the upper quartile and lower quartile is the length of the box, which is slightly greater for Oakland. You Try: 1. Organize the given data into a box and whisker plot: The number of fish caught each day for nine days straight is 26, 17, 21, 23, 19, 28, 17, 20, and 29 2. Organize the given data into a box and whisker plot. The number of pizzas needed for each class is 6, 10, 7, 12, 9, 10, and 11 Find the interquartile range for each
# How many times do you have to roll a dice to get a 6? Contents ## What is the average number of rolls of a die to get a 6? The probability of getting a six in a single throw is 1/6. Therefore, on average, you’ll have about six throws for every appearance of a 6. In other words, you can expect an average of 6 throws in order to see a 6 (as usual, this includes the throw that gives the 6). ## On what roll would you expect to roll a 6? The probability of rolling a 6 will always be 1/6 since the experiment is independent. So the expected number of rolls will be 1/1/6=6. ## How likely are you to roll a six if you roll a dice 6 times? Danger · hfhfhfrr . There is a 66.5% chance of it landing on a 6 at least once. ## What is the expected number of rolls to get all the faces from 1 to 6? This shows that the mean total time to get all six results is 6∑k=16k=14710=14.7. IT IS INTERESTING: Your question: What is the largest Super Bowl bet ever? ## How many times will you get a six if you throw a dice 100 times? (a) if a fair die is rolled 100 times, how many 6’s do you expect? the probability of rolling a 6 on a fair die is 1/6, so that we expect 100 × 1/6 = 16.7 ≈ 17 6’s to be rolled. ## How many dice must be rolled to have at least a 95% chance of rolling a six? Hence, n ≥ 17 will give at least a 95% chance of rolling at least one six. ## What is the expectation of getting 5 on a roll of a dice? Two (6-sided) dice roll probability table Roll a… Probability 4 3/36 (8.333%) 5 4/36 (11.111%) 6 5/36 (13.889%) 7 6/36 (16.667%) ## What is the expected number of rolls of a fair d6 6 sided dice until all 6 faces have appeared at least once? The Expected Value 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 14.7 rolls. When I told my son that “you should expect to roll the die 14.7 times before all six faces appear,” he was unimpressed. ## What are the odds of Farkle with 6 dice? Every individual die still has a 1 in 3 chance, so even with 6 dice, the probability of a 1 or 5 remains 33.33%. One or Five. Dice Left Odds Probability 6 12 in 36 33.33% 5 10 in 30 33.33% 4 8 in 24 33.33% 3 6 in 18 33.33% ## What’s the probability of rolling at least one 6? When you roll two dice, you have a 30.5 % chance at least one 6 will appear. This figure can also be figured out mathematically, without the use of the graphic.
Paul's Online Notes Home / Calculus III / 3-Dimensional Space / Equations of Planes Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. ### Section 1-3 : Equations of Planes 7. Determine if the line given by $$\vec r\left( t \right) = \left\langle {4 + t, - 1 + 8t,3 + 2t} \right\rangle$$ intersects the plane given by $$2x - y + 3z = 15$$ or show that they do not intersect. Show All Steps Hide All Steps Start Solution If the line and the plane do intersect then there must be a value of $$t$$ such that if we plug that $$t$$ into the equation of the line we’d get a point that lies on the plane. We also know that if a point $$\left( {x,y,z} \right)$$ is on the plane the then the coordinates will satisfy the equation of the plane. Show Step 2 If you think about it the coordinates of all the points on the line can be written as, $\left( {4 + t, - 1 + 8t,3 + 2t} \right)$ for all values of $$t$$. Show Step 3 So, let’s plug the “coordinates” of the points on the line into the equation of the plane to get, $2\left( {4 + t} \right) - \left( { - 1 + 8t} \right) + 3\left( {3 + 2t} \right) = 15$ Show Step 4 Let’s solve this for $$t$$ as follows, $18 = 15\,\,??$ Show Step 5 Hmmm… So, either we’ve just managed to prove that 18 and 15 are in fact the same number or there is something else going on here. Clearly 18 and 15 are not the same number and so something else must be going on. In fact, all this means is that there is no $$t$$ that will satisfy the equation we wrote down in Step 3. This in turn means that the line and plane do not intersect.
Online Tution » Important Question » Perimeter of Triangle- Formula, Definition, and... # Perimeter of Triangle- Formula, Definition, and Examples ## Perimeter of Triangle Perimeter of triangle is P= a+b+c, where a,b,c are the sides of triangles. Any two-dimensional figure’s perimeter is defined as the distance around it. By combining the lengths of each of the sides, we can compute the perimeter of any closed shape. In this tutorial, you will learn what the perimeter is and how to calculate the perimeter of various types of triangles when all side lengths are known. Furthermore, the solved examples will assist you in gaining additional perspectives on the subject. ## Perimeter of Triangle- Definition Any polygon’s perimeter is equal to the sum of its side lengths. Perimeter = Sum of the three sides in the case of a triangle In your final answer, always mention units. If the triangle’s sides are measured in cm, the final result should be in centimeters as well. ## Perimeter of Triangle- Formula In most cases, the perimeter of a closed shape figure is equal to the length of the figure’s outer line. As a result, the perimeter of a triangle is equal to the total of its three sides. If a triangle has three sides, a, b, and c, then, Perimeter, P = a + b +c ## Perimeter of Different Kinds of Triangle The perimeter of a triangle can be calculated by adding the lengths of the three sides of triangle. But there are different methods as well to calculate to perimeter of different types of triangles. ## Perimeter of Isosceles Triangle Because two sides of an Isosceles triangle are equal and one side is of different length than the other two, therefore the perimeter of an isosceles triangle can also be calculated by the formula: Perimeter, P= (2 × A) + B Where A is the length of two equal sides and B is the length of the third side. ## Perimeter of Equilateral Triangle Because all the sides of a equilateral triangle are equal to each other, therefore the perimeter of a equilateral triangle can also be calculated by the formula: Perimeter, P= 3 × L Where L is the length of the sides of triangle. ## Perimeter of Scalene Triangle Because the sides of a scalene triangle are all of different lengths, therefore it can only be calculated by the conventional formula: Perimeter, P= A + B + C Where A, B, C are the lengths of the sides of a triangle. ## Perimeter of Triangle- Examples Example 1: Find the perimeter of a triangle whose sides are 7 cm, 8 cm and 5 cm. Solution: Because, all three sides are unequal, therefore it is a scalene triangle, and we can only calculate the perimeter by the conventional method. I.e. P=A+B+C. A = 7 cm B = 8 cm C= 5 cm Perimeter = Sum of all sides = A + B + C = 7 + 8 + 5 = 20 Therefore, the answer is 20 cm. Example 2: Find the perimeter of a triangle which has each side is 15 cm. Solution: Because all the three sides are equal in length, the triangle given is an equilateral triangle, and hence the perimeter can be calculated by the formula: P = 3 × L. Here, L = 15 cm Therefore, Perimeeter = 3 × 15 = 45 Therefore, Perimeter of the given triangle is 45 cm. Example 3: What is the length of the third side of a triangle whose perimeter is 50 cm and two sides are 15 cm and 12 cm? Solution: Given, Perimeter, P = 50 cm Using formula: P = A + B + C A = 15 cm B = 12 cm 40 = 15 + 12 + C C = 50 – 27 = 23 Missing side length is 23 cm. Example 4: Calculate the Perimeter of a Right Triangle with Base as 6 cm and height as 8 cm. First, using the Pythagorean Theorem, we can calculate the hypotenuse of the right triangle. h =√(base2+ height2) h = √(62+82) h = √(36 + 64) h = √100 Or, h = 10 cm So, the perimeter of the triangle = 6 + 8 + 10 = 24 cm. Related Post: ## Triangle Perimeter Formula- FAQs What does the Perimeter of a Triangle Mean? The total distance around a triangle’s edges is known as its perimeter. In other terms, the perimeter of a triangle is the length of its boundary. How to Calculate the Perimeter of a Triangle? Add the lengths of the triangle’s sides to find its perimeter. For instance, if a triangle has sides a, b, and c, then its perimeter is P = a + b + c. What is the formula of area and perimeter of Triangle? The formula for calculating the area of a triangle is: A = 1/2 (b × h). The perimeter of a triangle can be calculated by adding the lengths of all the three sides of the triangle. P = A + B + C. Sharing is caring! ## FAQs ### What does the Perimeter of a Triangle Mean? The total distance around a triangle's edges is known as its perimeter. In other terms, the perimeter of a triangle is the length of its boundary. ### How to Calculate the Perimeter of a Triangle? Add the lengths of the triangle's sides to find its perimeter. For instance, if a triangle has sides a, b, and c, then its perimeter is P = a + b + c. ### What is the formula of area and perimeter of Triangle? The formula for calculating the area of a triangle is: A = 1/2 (b × h). The perimeter of a triangle can be calculated by adding the lengths of all the three sides of the triangle. P = A + B + C.
# 5.2 Fractions as Quotients and Fraction Multiplication ## Unit Goals • Students develop an understanding of fractions as the division of the numerator by the denominator, that is $a \div b = \frac{a}{b}$, and solve problems that involve the multiplication of a whole number and a fraction, including fractions greater than 1. ### Section A Goals • Represent and explain the relationship between division and fractions. • Solve problems involving division of whole numbers leading to answers that are fractions. ### Section B Goals • Connect division to multiplication of a whole number by a non-unit fraction. • Connect division to multiplication of a whole number by a unit fraction. • Explore the relationship between multiplication and division. ### Section C Goals • Find the area of a rectangle when one side length is a whole number and the other side length is a fraction or mixed number. • Represent and solve problems involving the multiplication of a whole number by a fraction or mixed number. • Write, interpret, and evaluate numerical expressions that represent multiplication of a whole number by a fraction or mixed number. ### Problem 1 #### Pre-unit Practicing Standards: 3.NF.A.2.b 1. Locate $$\frac{6}{4}$$ on the number line. 2. Explain or show why your point represents $$\frac{6}{4}$$. ### Solution For access, consult one of our IM Certified Partners. ### Problem 2 #### Pre-unit Practicing Standards: 3.NF.A.1 Shade $$\frac{3}{4}$$ of the rectangle. Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 3 #### Pre-unit Practicing Standards: 4.NF.B.4.b Explain or show why $$\frac{4}{3} = 4 \times \frac{1}{3}$$. ### Solution For access, consult one of our IM Certified Partners. ### Problem 4 #### Pre-unit Practicing Standards: 4.NF.B.4.c Each workbook is $$\frac{3}{8}$$ inch thick. How many inches thick is a stack of 5 workbooks? Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 5 #### Pre-unit Practicing Standards: 3.OA.A.2, 4.OA.A.2 1. There are 36 fish in 4 aquariums. There are the same number of fish in each aquarium. How many fish are in each aquarium? Show or explain your reasoning. 2. There are 24 dogs at a shelter. There are 4 times as many dogs as cats at the shelter. How many cats are there at the shelter? Show or explain your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 6 #### Pre-unit Practicing Standards: 4.NF.B.4.c A bottle holds $$\frac{7}{10}$$ liter of water. How much water do 6 bottles hold? Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 7 #### Pre-unit Practicing Standards: 3.MD.C.7.b What is the area of the rectangle? Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 8 1. 3 students equally share 18 sheets of construction paper for an art project. How many sheets of paper does each student get? Explain or show your reasoning. 2. 3 students equally share 1 tube of glue for an art project. How much glue does each student get? Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 9 1. 4 hikers equally share 3 liters of water. How many liters of water does each hiker drink? Explain or show your reasoning. 2. 4 hikers equally share 5 liters of water. How many liters of water does each hiker drink? Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 10 1. Jada cuts an 11 inch strip of paper into 5 equal parts. How many inches long is each part? 2. Jada cuts a strip of paper into 5 equal parts. Each part is $$\frac{7}{5}$$ inches long. How long was the strip of paper? ### Solution For access, consult one of our IM Certified Partners. ### Problem 11 1. Describe a situation that the diagram could represent. 2. Write an equation that represents the diagram and the situation. ### Solution For access, consult one of our IM Certified Partners. ### Problem 12 Decide whether each equation is true or false. Explain or show your reasoning. 1. $$3 \div 7 = \frac{3}{7}$$. 2. $$18 \div 5 = \frac{5}{18}$$. 3. $$15 \div 6 = 2 \frac{1}{2}$$. ### Solution For access, consult one of our IM Certified Partners. ### Problem 13 #### Exploration 1. Describe a situation in the classroom or at home where you share something equally with your classmates or family that results in fractional size parts. 2. Draw a picture to represent the situation. 3. Write a division equation to represent the situation. ### Solution For access, consult one of our IM Certified Partners. ### Problem 14 #### Exploration Elena is traveling to visit her grandparents who live 125 miles away. 1. Elena stops for lunch $$\frac{2}{3}$$ of the way. How far has Elena traveled? Explain or show your reasoning. 2. Elena enters the city where her grandmother lives after 110 miles. Is she more or less than $$\frac{9}{10}$$ of the way there? Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 15 #### Exploration 1. Describe a situation that represents the equation $$4 \div 6 = \frac{4}{6}$$. 2. Draw a diagram to represent the situation. ### Solution For access, consult one of our IM Certified Partners. ### Problem 1 Han cuts a 15-foot piece of rope into 4 equal parts. Decide whether each expression represents the length of each part of the rope in feet. Explain or show your reasoning. 1. $$15 \div 4$$ 2. $$4 \times 15$$ 3. $$3 \frac{3}{4}$$ ### Solution For access, consult one of our IM Certified Partners. ### Problem 2 Find the value of each expression. 1. $$\frac{1}{2} \times 6$$ 2. $$\frac{1}{7} \times 6$$ 3. $$\frac{1}{8} \times 11$$ 4. $$\frac{1}{3} \times 34$$ ### Solution For access, consult one of our IM Certified Partners. ### Problem 3 1. Kiran ran $$\frac{1}{5}$$ the length of his road, which is 9 miles long. How far did Kiran run? Show or explain your thinking. ### Solution For access, consult one of our IM Certified Partners. ### Problem 4 #### Exploration 1. Each square on the map represents 2,178 square feet. Make an estimate for the number of square feet shown on the map. Explain or show your reasoning. 2. Each square represents $$\frac{1}{20}$$ acre of actual land. How many square feet are in an acre? Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 5 #### Exploration A standard rectangular sheet of paper measures $$8 \frac{1}{2}$$ inches in width and 11 inches in length. How many square inches are there in a sheet of paper? If you get stuck, consider using the grid. ### Solution For access, consult one of our IM Certified Partners. ### Problem 1 1. How are the diagrams the same? How are they different? 2. How is finding the area of the shaded region the same? How is it different? ### Solution For access, consult one of our IM Certified Partners. ### Problem 2 1. What is the area of this rectangle? Explain or show your reasoning. 2. What is the area of the shaded region? Explain or show your reasoning. 3. How are these two area calculations the same? How are they different? ### Solution For access, consult one of our IM Certified Partners. ### Problem 3 The shaded part of this diagram shows the top of a stove. What is the area of the stove top? Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 4 Find the area of the shaded region. Explain or show your reasoning. ### Solution For access, consult one of our IM Certified Partners. ### Problem 5 Select all of the expressions that represent the shaded area in square feet. 1. $$3 + 5 \frac{3}{4}$$ 2. $$3 \times 5 \frac{3}{4}$$ 3. $$3 \times \left(5 + \frac{3}{4}\right)$$ 4. $$(3 \times 5) + \frac{3}{4}$$ 5. $$3 \times 6 - \left(3 \times \frac{1}{4}\right)$$ Write one more expression that represents the shaded area. ### Solution For access, consult one of our IM Certified Partners. ### Problem 6 Tyler says that $$9 \frac{11}{12} \times 5$$ is a little less than 50. 1. Do you agree with Tyler? Explain or show your reasoning. 2. What is the value of $$9 \frac{11}{12} \times 5$$? ### Solution For access, consult one of our IM Certified Partners. ### Problem 7 A banner at a sporting event is 8 feet long and $$2 \frac{1}{3}$$ feet wide. 1. Sketch and label a diagram of the banner. 2. Find the area of the banner. ### Solution For access, consult one of our IM Certified Partners. ### Problem 8 Evaluate each expression. Explain or show your reasoning. 1. $$3\frac{2}{5} \times 10$$ 2. $$8 \times \frac{14}{3}$$ 3. $$3 \frac{41}{100} \times 5$$ ### Solution For access, consult one of our IM Certified Partners. ### Problem 9 #### Exploration 1. A regular sheet of paper is $$8\frac{1}{2}$$ inches wide and 11 inches long. How many times would you need to fold the sheet of paper in half before the area is less than 1 square inch? Explain or show your reasoning. 2. A piece of chart paper is 23 inches wide by 33 inches long. How many times would you need to fold it in half before its area is less than 1 square inch? ### Solution For access, consult one of our IM Certified Partners. ### Problem 10 #### Exploration Part of the rectangle is shaded. 1. Write a multiplication expression that represents the shaded area. 2. Write a division expression that represents the shaded area. 3. Write any other expressions that represent the shaded area. ### Solution For access, consult one of our IM Certified Partners. ### Problem 11 #### Exploration This is a picture of the Empire State Building: The base of the Empire State Building is a rectangle. What do you think the area of the rectangle is in square meters? 1. Make an estimate that is too small. 2. Make an estimate that is too large. 3. The length of the rectangle is 129$$\frac{1}{5}$$ meters. The width is 57 meters. What is the area of the base of the Empire State Building? ### Solution For access, consult one of our IM Certified Partners.
# Absolute Value as Piecewise Functions ## Presentation on theme: "Absolute Value as Piecewise Functions"— Presentation transcript: Absolute Value as Piecewise Functions Lesson2.5 Example x + 1, if x < 1 2, if 1 ≤ x ≤ 3 (x-3)2 + 2, if x > 3 Absolute Value as Piecewise We usually write an absolute value function as f (x)= x , but since absolute value is a measure of distance and distance is always positive, it also can be written as follows: -x, if x < 0 x, if x ≥ 0 f (x) = Writing Abs. Value as Piecewise To identify the number in the domain, set x – h = 0 and solve for x. For I x - h I ≥ 0, simplify the equation given by distributing and combining like terms. For I x - h I < 0, substitute –(x - h) in place of I x - h I. Then, simplify the equation given by distributing and combining like terms. Example: Write y = 2 Ix – 4I – 10 as a piecewise function. Use 4 in your domain. For (x-4) ≥ 0 2(x – 4) – 10 = 2x – 8 – 10 = 2x – (when x ≥ 4) For (x-4) < 0 2[-(x-4)] – 10 = 2(-x + 4) – 10 = -2x + 8 – 10 = -2x – 2 (when x < 4)) More Examples: Write y = 2 Ix – 4I – 10 as a piecewise function. For (x-4) ≥ 0 2(x – 4) – 10 = 2x – 8 – 10 = 2x – (when x ≥ 4) For (x-4) < 0 2[-(x-4)] – 10 = 2(-x + 4) – 10 = -2x + 8 – 10 = -2x – 2 (when x < 4)) Graphs of Both y=-2x-2 y=2x-18 EOCT Practice A EOCT Practice C Writing Abs. Value as Piecewise Using a graph Writing Abs. Value as Piecewise Try this one...
Mobile Math Website Sitemap # Factorials ## Basics of Factorials A Factorial is the number of unique combinations of the elements of a set so that each combination of elements represents a unique permutation of that set’s elements. Let’s look at a few examples: (The “!” after the integer tells us it is a Factorial) 0! = 1 1! = 1 = 1 2! = 2 ∙ 1 = 2 3! = 3 ∙ 2 ∙ 1 = 6 4! = 4 ∙ 3 ∙ 2 ∙ 1 = 24 n! = n (n – 1) (n – 2)… 3 ∙ 2 ∙ 1 Yes, 0! = 1. This is because an empty set, having no elements, can be ordered by (organized) only as { }, there is no other order for an empty set. The Permutations of 0! is 1: { } The permutations of 1! is 1: {1} The permutations of 2! is 2: {1,2} {2,1} The permutations of 3! is 6: {3,2,1} {1,2,3} {2,1,3} {2,3,1} {3,1,2} {1,3,2} It is easy to see that as the integer of the Factorial becomes larger the complexity to determine the unique combinations, as permutations of set elements, quickly nears impossible… 5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120 6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720 7! = 5,040 8! = 40,320 9! = 362,880 10! = 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 3,628,800 and observe: 3! = 6 4! = 3! ∙ 4 = 6 ∙ 4 = 24 5! = 4! ∙ 5 = 24 ∙ 5 = 120 6! = 5! ∙ 6 = 120 ∙ 6 = 720 From this observation we can infer that a Factorial, (n + 1)! , is equal to n! (n + 1); Where n! is read “n factorial”. That is, the permutations of the next Factorial can always be determined by multiplying the permutations of the current Factorial, n!, by the next integer increment of the current Factorial. A formal math definition of a Factorial can now be given as: (n + 1)! = n! (n +1) A Factorial is an integer product: n! = n (n – 1) (n – 2) … 3 ∙ 2 ∙ 1 Using 6! to equate factors to the equation n!: 6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720 n = 6 n – 1 = 5 n – 2 = 4 n – 3 = 3 n – 4 = 2 n – 5 = 1 This is multiplication to determine the number of permutations, unique combinations, for a given Factorial. This being said, then one Factorial can be divided by another Factorial, that is, division and multiplication are reciprocal operations. Indeed, Factorials can be divided to determine the difference of their permutations with the stipulation that for any n! / i! , 0 ≤ i ≤ n. Let’s divide 10! by 6! : 10! / 6! = 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 / 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 / 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 10 ∙ 9 ∙ 8 ∙ 7 = 5,040 … we observe that the fraction can be reduced. If we already know the permutation values, the numerator and denominator, of the factorial division: 10! = 3,628,800 6! = 720 3,628,800 / 720 = 5,040 10! / 6! = 5,040 If we do not know the permutation values of the numerator or denominator factorials of the division: 20! / 18! 2! = (The numerator is determined by denominator 18! canceling, or reducing, the first 18 integer digits of the numerator 20! to 1/1, the remaining values not reduced, are greater than 18. The integers remaining are 19 and 20) 20 ∙ 19 / 2! = 20 ∙ 19 / 2! = (evaluating 2!; 2! = 2) 20 ∙ 19 / 2 = 10 ∙ 19 / 1 = 190 ## Quotients of Factorials A special symbol with a special definition to find quotients of factorials: ( n i ) “This special math symbol does not state n! is to be divided by i!” The definition is ( n i ) = n! / (i! (n – i)!) i and n are integers where 0 ≤ i ≤ n If asked to find the quotient of: ( 7 4 ) = n! / (i! (n – i)!) = 7! / (4! (7 – 4)!) = 7! / (4! 3!) = (7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1) / ((4 ∙ 3 ∙ 2 ∙ 1) (3 ∙ 2 ∙ 1)) = (7 ∙ 6 ∙ 5) / (3 ∙ 2 ∙ 1) = 210 / 6 = 35 When n = 7 and i = 4 the resulting factorial equation is 7! / (4! 3!). Let’s try n = 7 and i =3… ( 7 3 ) = 7! / (3! (7 – 3)!) = 7! / (3! 4!) = 35 This works for any values of n and i to find quotients of factorials. ( 8 5 ) = n! / (i! (n – i)!) = 8! / (5! (8 – 5)!) = 8! / (5! 3!) = (8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1) / ((5 ∙ 4 ∙ 3 ∙ 2 ∙ 1) (3 ∙ 2 ∙ 1)) = (8 ∙ 7 ∙ 6) / (3 ∙ 2 ∙ 1) = 336 / 6 = 56 ( 8 3 ) = 8! / (3! (8 – 3)!) = 8! / (3! 5!) = 56 This behavior to determine the denominator for factorial division can be described by: ( n i ) = ( n n – i ) ( 9 7 ) = ( 9 9 – 7 ) = ( 9 2 ) The following math shows us why: ( n i ) = ( n n – i ) Observe that ( n i ) = n! / (i! (n – i)!) and ( n n i ) = n! / ( n! (i! (n – i)!) = n! / ( (n i)! (n (n – i))! ) = n! / ( (n – i)! i! ) Note that [n – (n – i)]! = [n – n + i]! = i! thus ( n n – i ) = n! / (i! (n – i)!) = ( n i ) Want to learn more about factorials? Then visit Factorial Binomial Expansion. Top of Page
# Examples of how to solve powers of decimal numbers Perhaps the best thing to do, before giving some examples on the correct way to solve an empowerment operation in which the number that serves as a base is a decimal number, is to revise briefly the definition of this operation, in order to understand each one of the exercises that are presented in its just mathematical context. ## Decimal number powers In this sense, we can begin by saying that the Potentiation of decimal numbers is that operation destined to discover which is the product obtained by taking the decimal number that serves as base, and multiplying it by itself, as many times as the integer that acts as exponent indicates. Hence, some authors point out that the Potentiation of decimal numbers can also be interpreted as an abbreviated sum of these. • When you’re working, editing pictures, and f... • As a scientist, do you enjoy working on rela... ## Steps to solve a decimal number potency However, precisely because of the nature of the number that serves as the base, i.e. the decimal number, which is composed of an integer and a decimal part, it is necessary to follow a different method to that which is followed when the base is integer, in order to avoid making mistakes, when the exercise must be solved manually. Consequently, the steps that must be followed when solving any operation of powers of decimal numbers will be the following ones: 1. Once the potentiation operation has been proposed, each one of the elements must be reviewed in order to know the nature of each one of them. 2. Once it has been found that the operation is established between a decimal number and a positive integer exponent, the decimal number comma should be deleted so that it performs as an integer for the duration of the operation. 3. Once this has been done, the number from which the comma has been removed is then elevated to the indicated exponent. 4. Once the obtained power has been obtained, a second operation must be carried out to determine the specific place where the comma of the final result must be located, since the power of a decimal number must essentially give a decimal number. In this way, we will proceed to count how many incomplete units the decimal number had at the time of starting the operation. This figure should be multiplied by the value of the exponent. The result of this multiplication will be the number of spaces that must be counted from right to left, in the power obtained, before placing the comma. 5. Therefore, with the result of the multiplication of the number of incomplete units and the value of the exponent, the comma is placed at the power obtained, assuming the decimal number obtained as the final result of the operation, that is, as the power of the decimal number that was originally set as the base. 6. If you want to check this operation, you will see the inverse operation, that is to say, the root. Consequently, the power obtained will be taken as radicando, the exponent as index, and the result of this other operation should give the number that has served as base for the potentiation. If so, the operation has been resolved correctly. ## Examples of powers of decimal numbers However, perhaps the best way to complete an explanation about the correct way in which any power of decimal numbers must be resolved is through a series of examples that allow us to see in a concrete way how each one of the steps indicated by Mathematics are fulfilled. Here are some of them: Solve the following decimal number power operation: 2.53 = As the method suggested by the mathematical discipline for solving this exercise points out, one should begin by suppressing the comma of the decimal that serves as the base, and then elevate it to exponent 3 to which it is elevated: 2,53 → 253 = 15625 Once this result is found, it will be time to locate the comma again. To do this, multiply the value of the exponent by the number of incomplete units or decimal parts that the number that served as base had, being in this case equal to 1: 3 x 1 = 3 The product obtained shall be the number of spaces to be counted from right to left before placing the comma on the power: 15625 → 15,625 Having done this, one can then consider the operation resolved. Therefore, the only thing left to do is to express the result obtained: 2,53 = 15,625 ## Example 2 Resolve the following decimal number power operation: 0,444 = In this case, once the numbers on which the potentiation operation has been raised have been reviewed, it can be found that it is not only a decimal base, but that the whole part of it is equal to zero. However, this should not represent a big problem when solving the operation, since when the comma is suppressed, simply the left zero will not have any value, so it will not be necessary to take it into account when solving the power: 0,444 → 444 When the operation is expressed as an integer, the proposed power is resolved: 444 = 3748096 We then look for where the comma should be located, multiplying the number of decimals that had the original base and the value of the exponent: 2 x 4= 8 This product is the number of spaces that will be counted from right to left before the comma is placed: 3748096 → 0,03748096 The quantity of elements of the power obtained was less than the product obtained between the number of decimals of the base and the value of the exponent, so it was necessary then to add zeros in the power, when locating the comma. Once this number is obtained, the final result of the operation will be expressed: 0,444 = 0,03748096 ## Example 3 Solve the following power: 1,2342 = In this case, we will also start by removing the comma from the decimal power, in order to take the number as an integer, and be able to raise it to the square: 1,2342 → 12342 = 1522756 The number of decimals or incomplete units of the original base is then multiplied by the value of the exponent: 3 x 2= 6 This product will be translated in the number of places that will have to be counted from the right to the left, to be able to locate the comma in the obtained power: 1522756 → 1,522756 Once this decimal number is obtained, considered the final result, the next step will consist of expressing the operation as resolved: 1,2342 = 1,522756 Image: pixabay.com Examples of how to solve powers of decimal numbers James Watt (Greenock, Scotland 19 January 1736 – Handsworth, England, 25... L.Von Bertalanfly: he was the author of the idea of the General Theory of... An energy source, capable of achieving the normal and healthy operation of... In general, discipline is understood as the instruction that a person has ... ### Bibliography ► phoneia.com (October 31, 2019). Examples of how to solve powers of decimal numbers. Bogotá: E-Cultura Group. Recovered from https://phoneia.com/en/education/examples-of-how-to-solve-powers-of-decimal-numbers/
#### NCERT Solutions for Class 12th: Ch 3 Matrices Exercise 3.1 Math Page No: 64 Exercise 3.1 Find the value of the following: Question: 1 (i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24, a23. (i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4. (ii) Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it. (iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 = 5/2 2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements? We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24. The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (6, 4) Hence, the possible orders of a matrix having 24 elements are: 1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4 (1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13. Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1. 3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements? We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18. The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3) Hence, the possible orders of a matrix having 18 elements are: 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3 (1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5. Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1. 4. Construct a 2 × 2 matrix, A = [aij], whose elements are given by: 5. Construct a 3 × 4 matrix, whose elements are given by: (i) aij = 1/2 \|-3 + j\| (ii) aij = 2i - j (i) aij = 1/2 \|-3 + j\| a11 = 1/2\|−3 × 1 + 1\| = 1/2\|−3 + 1\| = 1 a12 = 1/2\|−3 × 1 + 2\| = 1/2\|−3 + 2\| = 1/2 a13 = 1/2\|−3 × 1 + 3\| = 1/2\|−3 + 3\| = 0 a14 = 1/2\|−3 × 1 + 4\| = 1/2\|−3 + 4\| = 1/2 a21 = 1/2\|−3 × 2 + 1\| = 1/2\|−6 + 1\| = 5/2 a22 = 1/2\|−3 × 2 + 2\| = 1/2\|−6 + 2\| = 2 a23 = 1/2\|−3 × 2 + 3\| = 1/2\|−6 + 3\| = 3/2 a24 = 1/2\|−3 × 2 + 4\| = 1/2\|−6 + 4\| = 1 a31 = 1/2\|−3 × 3 + 1\| = 1/2\|−9 + 1\| = 4 a32 = 1/2\|−3 × 3 + 2\| = 1/2\|−9 + 2\| = 7/2 a33 = 1/2\|−3 × 3 + 3\| = 1/2\|−9 + 3\| = 3 a34 = 1/2\|−3 × 3 + 4\| = 1/2\|−9 + 4\| = 5/2 (ii) aij = 2i - j a11 = 2 × 1 − 1 = 2 −1 = 1 a12 = 2 × 1 − 2 = 2 −2 = 0 a13 = 2 × 1 − 3 = 2 −3 = -1 a14 = 2 × 1 − 4 = 2 −4 = -2 a21 = 2 × 2 − 1 = 4 −1 = 3 a22 = 2 × 2 − 2 = 4 −2 = 2 a23 = 2 × 2 − 3 = 4 −3 = 1 a24 = 2 × 2 − 4 = 4 −4 = 0 a31 = 2 × 3 − 1 = 6 −1 = 5 a32 = 2 × 3 − 2 = 6 −2 = 4 a33 = 2 × 3 − 3 = 6 −3 = 3 a34 = 2 × 3 − 4 = 6 −4 = 2 6. Find the values of x, y and z from the following equations: (i) As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x = 1, y = 4, and z = 3 (ii) As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x + y = 6, xy = 8, 5 + z = 5 Now, 5 + z = 5 ⇒ z = 0 we know that: (x − y)2 = (x + y)2 − 4xy ⇒ (x − y)2 = 36 − 32 = 4 ⇒ x − y = ±2 Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2 When x − y = − 2 and x + y = 6, we get x = 2 and y = 4 ∴ x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0 (iii) As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x + y + z = 9 ... (1) x + z = 5 ........ (2) y + z = 7 ........ (3) From (1) and (2), we have: y + 5 = 9 ⇒ y = 4 Then, from (3), we have: 4 + z = 7 ⇒ z = 3 ∴ x + z = 5 ⇒ x = 2 ∴ x = 2, y = 4 and z = 3. 7. Find the value of a, b, c and d from the equation: As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: a − b = −1 ...... (1) 2a − b = 0 ...... (2) 2a + c = 5 ....... (3) 3c + d = 13 ..... (4) From (2), we have: b = 2a Then, from (1), we have: a − 2a = −1 ⇒ a = 1 ⇒ b = 2 Now, from (3), we have: 2 ×1 + c = 5 ⇒ c = 3 From (4) we have: 3 × 3 + d = 13 ⇒ 9 + d = 13 ⇒ d = 4 ∴ a = 1, b = 2, c = 3 and d = 4. Page No. 65 8. A = [aij]mï½˜n is a square matrix, if (A) m < n (B) m > n (C) m = n (D) None of these It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns. Therefore, A =[aij]mï½˜n is a square matrix, if m = n. 9. Which of the given values of x and y make the following pair of matrices equal (A) x = -1/3, y = 7 (B) Not possible to find (C) y = 7, x = -2/3 (D) x = -1/3, y = -2/3 It is given that, Equating the corresponding elements, we get: 3x + 7 = 0 ⇒ x = -7/3 and 5 = y − 2 ⇒ y = 7 y + 1 = 8 ⇒ y = 7 and 2 − 3x = 4 ⇒ x = -2/3 We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible. Hence, it is not possible to find the values of x and y for which the given matrices are equal. 10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512
# Math 315 Homework 1 Solutions § 1.3 #26: A ball is dropped from ```Math 315 Homework 1 Solutions § 1.3 #26: A ball is dropped from rest at a height of 200 m. What is the velocity and position of the ball 3 seconds later? x00 (t) = −g where g = 9.8 m/s2 is the acceleration due to gravity. Integrating, we get x0 (t) = −gt + C1 . Since the ball is dropped from rest, the initial velocity is zero, so x0 (0) = 0. So C1 = 0. So x0 (t) = −gt. Integrating again gives 1 x(t) = − gt2 + C2 2 The ball is dropped from a height of 200 m, so x(0) = 200. So C2 = 200. So 1 x(t) = − gt2 + 200. 2 Thus, the velocity of the ball 3 seconds after dropping it is given by 2 2 x0 (3) = (−9.8 m/s )(3s) = −29.4 m/s . The position of the ball 3 seconds after dropping it is given by 1 2 x(3) = − (9.8 m/s )(3s)2 + 200 m = 155.9 m. 2 § 2.1 #8 (a) Use implicit differentiation to show that t2 + y 2 = C 2 implicitly defines solutions of the differential equation t = yy 0 = 0. Solution: Differentiating both sides of the equation t2 + y 2 = C 2 with respect to t gives 2t + 2yy 0 = 0. Dividing both sides by 2, we get t + yy 0 = 0. Hence, t2 + y 2 = C 2 implicitly defines solutions of the differential equation t = yy 0 = 0. (b) Solve t2 + y 2 = C 2 for y in terms of t to provide explicit solutions. Show that these functions are also solutions of t + yy 0 = 0. Solution: Note t2 + y 2 = C 2 ⇒ y=± p C 2 − t2 . Note y 0 (t) = ∓ √ t C2 − t2 . Thus, t + yy 0 = t ± p C 2 − t2 · √ ∓t = 0. C 2 − t2 (c) Discuss the interval of existence for each of the solutions in part (b). √ Solution: The solutions are y = ± C 2 − t2 . Since we cannot take the square root of a negative number, we need t2 ≤ C 2 , which implies that t ∈ [−C, C]. In order that y 0 = √C∓t 2 −t2 is defined, we also can’t have t = ±C. Thus, the interval of existence is (−C, C). (d) Sketch solutions for C = 1, 2, 3, 4. The sketch of solutions for C = 1, 2, 3, 4 should be circles of radius 1, 2, 3, and 4, respectively. § 2.2 #14: Find the exact solution of the initial value problem y 0 = −2t(1 + y 2 )/y, y(0) = 1. Indicate the interval of existence. Solution: Note y0 = − 2t(1 + y 2 ) y ⇒ Z y dy = −2t 1 + y 2 dt ⇒ y dy = − 1 + y2 Z 2tdt Integrating, we get 1 ln(1 + y 2 ) = −t2 + C 2 ⇒ ln(1 + y 2 ) = −2t2 + 2C ⇒ 1 + y 2 = e−2t ⇒ y 2 = Ae−2t − 1 p y(t) = ± Ae−2t2 − 1. ⇒ 2 +2C 2 Now we use the initial condition y(0) = 1, which gives √ 1 = y(0) = ± A − 1 ⇒ 2 = e2C e−2t = Ae−2t A = 2. 2 Also, we must keep only the solution which contains the initial condition y(0) = 1, so we need the + sign in front of the square root. So the exact solution is p y(t) = 2e−2t2 − 1. 2 The square root exists provided that 2e−2t − 1 ≥ 0. Also, we don’t want y = 0 since otherwise the original differential equation y 0 (t) = − 2t(1 + y 2 ) y is not defined. Thus, we need 2 2e−2t − 1 > 0 ⇒ ⇒ ⇒ ⇒ 2 e−2t > 1 2 1 −2t2 > ln( ) = − ln(2) 2 1 2 t < ln(2) 2 r 1 ln(2). \|t\| < 2 Thus, the interval of convergence is r − 1 ln(2), 2 r 1 ln(2) . 2 ```
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: 8th grade>Unit 3 Lesson 5: Intro to slope-intercept form # Intro to slope-intercept form Learn about the slope-intercept form of two-variable linear equations, and how to interpret it to find the slope and y-intercept of their line. #### What you will learn in this lesson • What is the slope-intercept form of two-variable linear equations • How to find the slope and the y-intercept of a line from its slope-intercept equation • How to find the equation of a line given its slope and y-intercept ## What is slope-intercept form? Slope-intercept is a specific form of linear equations. It has the following general structure. Drum roll ... y, equals, start color #ed5fa6, m, end color #ed5fa6, x, plus, start color #0d923f, b, end color #0d923f Here, start color #ed5fa6, m, end color #ed5fa6 and start color #0d923f, b, end color #0d923f can be any two real numbers. For example, these are linear equations in slope-intercept form: • y, equals, 2, x, plus, 1 • y, equals, minus, 3, x, plus, 2, point, 7 • y, equals, 10, minus, 100, x On the other hand, these linear equations are not in slope-intercept form: • 2, x, plus, 3, y, equals, 5 • y, minus, 3, equals, 2, left parenthesis, x, minus, 1, right parenthesis • x, equals, 4, y, minus, 7 Slope-intercept is the most prominent form of linear equations. Let's dig deeper to learn why this is so. ## The coefficients in slope-intercept form Besides being neat and simplified, slope-intercept form's advantage is that it gives two main features of the line it represents: • The slope is start color #ed5fa6, m, end color #ed5fa6. • The y-coordinate of the y-intercept is start color #0d923f, b, end color #0d923f. In other words, the line's y-intercept is at left parenthesis, 0, comma, start color #0d923f, b, end color #0d923f, right parenthesis. For example, the line y, equals, start color #ed5fa6, 2, end color #ed5fa6, x, start color #0d923f, plus, 1, end color #0d923f has a slope of start color #ed5fa6, 2, end color #ed5fa6 and a y-intercept at left parenthesis, 0, comma, start color #0d923f, 1, end color #0d923f, right parenthesis: The fact that this form gives the slope and the y-intercept is the reason why it is called slope-intercept in the first place! ## Check your understanding Problem 1 What is the slope of the line represented by y, equals, 5, x, minus, 7? Problem 2 What is the slope of the line represented by y, equals, x, plus, 9? Problem 3 What is the y-intercept of the line represented by y, equals, minus, 6, x, minus, 11? Problem 4 What is the y-intercept of the line represented by y, equals, 4, x? Problem 5 What is the slope of the line represented by y, equals, 1, minus, 8, x? Problem 6 Which lines have a y-intercept at left parenthesis, 0, comma, 4, right parenthesis? Choose all answers that apply: Reflection question How do we find the slope of a line that is given in slope-intercept form? Challenge problem 1 Which of these can be the equation of the line? Challenge problem 2 Write an equation of a line whose slope is 10 and y-intercept is left parenthesis, 0, comma, minus, 20, right parenthesis. ## Why does this work? You might be wondering how it is that in slope-intercept form, start color #ed5fa6, m, end color #ed5fa6 gives the slope and start color #0d923f, b, end color #0d923f gives the y-intercept. Can this be some sort of magic? Well, it certainly is not magic. In math, there's always a justification. In this section we'll take a look at this property using the equation y, equals, start color #ed5fa6, 2, end color #ed5fa6, x, plus, start color #0d923f, 1, end color #0d923f as an example. ### Why $\greenE{b}$start color #0d923f, b, end color #0d923f gives the $y$y-intercept At the y-intercept, the x-value is always zero. So if we want to find the y-intercept of y, equals, start color #ed5fa6, 2, end color #ed5fa6, x, plus, start color #0d923f, 1, end color #0d923f, we should substitute x, equals, 0 and solve for y. \begin{aligned} y&=\maroonC{2}x+\greenE{1} \\\\ &=\maroonC{2}\cdot 0+\greenE{1}&\gray{\text{Substitute }x=0} \\\\ &=0+\greenE{1} \\\\ &=\greenE{1} \end{aligned} We see that at the y-intercept, start color #ed5fa6, 2, end color #ed5fa6, x becomes zero, and therefore we are left with y, equals, start color #0d923f, 1, end color #0d923f. ### Why $\maroonC{m}$start color #ed5fa6, m, end color #ed5fa6 gives the slope Let's refresh our memories about what slope is exactly. Slope is the ratio of the change in y over the change in x between any two points on the line. start text, S, l, o, p, e, end text, equals, start fraction, start text, C, h, a, n, g, e, space, i, n, space, end text, y, divided by, start text, C, h, a, n, g, e, space, i, n, space, end text, x, end fraction If we take two points where the change in x is exactly 1 unit, then the change in y will be equal to the slope itself. start text, S, l, o, p, e, end text, equals, start fraction, start text, C, h, a, n, g, e, space, i, n, space, end text, y, divided by, 1, end fraction, equals, start text, C, h, a, n, g, e, space, i, n, space, end text, y Now let's look at what happens to the y-values in the equation y, equals, start color #ed5fa6, 2, end color #ed5fa6, x, plus, start color #0d923f, 1, end color #0d923f as the x-values constantly increase by 1 unit. xy 0start color #0d923f, 1, end color #0d923f, plus, 0, dot, start color #ed5fa6, 2, end color #ed5fa6equals, start color #0d923f, 1, end color #0d923f 1start color #0d923f, 1, end color #0d923f, plus, 1, dot, start color #ed5fa6, 2, end color #ed5fa6equals, start color #0d923f, 1, end color #0d923f, plus, start color #ed5fa6, 2, end color #ed5fa6 2start color #0d923f, 1, end color #0d923f, plus, 2, dot, start color #ed5fa6, 2, end color #ed5fa6equals, start color #0d923f, 1, end color #0d923f, plus, start color #ed5fa6, 2, end color #ed5fa6, plus, start color #ed5fa6, 2, end color #ed5fa6 3start color #0d923f, 1, end color #0d923f, plus, 3, dot, start color #ed5fa6, 2, end color #ed5fa6equals, start color #0d923f, 1, end color #0d923f, plus, start color #ed5fa6, 2, end color #ed5fa6, plus, start color #ed5fa6, 2, end color #ed5fa6, plus, start color #ed5fa6, 2, end color #ed5fa6 4start color #0d923f, 1, end color #0d923f, plus, 4, dot, start color #ed5fa6, 2, end color #ed5fa6equals, start color #0d923f, 1, end color #0d923f, plus, start color #ed5fa6, 2, end color #ed5fa6, plus, start color #ed5fa6, 2, end color #ed5fa6, plus, start color #ed5fa6, 2, end color #ed5fa6, plus, start color #ed5fa6, 2, end color #ed5fa6 We see that each time x increases by 1 unit, y increases by start color #ed5fa6, 2, end color #ed5fa6 units. This is because x determines the multiple of start color #ed5fa6, 2, end color #ed5fa6 in the calculation of y. As stated above, the change in y that corresponds to x increasing by 1 unit is equal to the slope of the line. For this reason, the slope is start color #ed5fa6, 2, end color #ed5fa6. Challenge problem 3 Complete the equation of the line. y, equals ## Want to join the conversation? • What if m=0? • If the slope is 0, is a horizontal line. It makes sense if you think about it. Each time we increase one x, increase y by 0. • how do you find the slope and intercept on a graph? • To find the y-intercept, find where the line hits the y-axis. To find the x-intercept (which wasn't mentioned in the text), find where the line hits the x-axis. To find the slope, find two points on the line then do y2-y1/x2-x1 the numbers are subscripts. Hope that helped. • As I gazed upon the graph, I saw the slope and y-intercept, Their influence on the line so craft, Their importance I did not neglect. The slope, a measure of the steepness, The change in y over change in x, A value that can bring weakness, Or strength to the line that interjects. The y-intercept, where the line does start, The value of y when x is zero, The point that will never depart, The anchor where the line will mellow. Together they form a line so true, A representation of data and fact, Their importance we cannot subdue, For without them, the line would lack. • I dont understand this whole thing at all PLEASE HELP! • The slope-intercept form of a linear equation is where one side contains just "y". So, it will look like: y = mx + b where "m" and "b" are numbers. This form of the equation is very useful. The coefficient of "x" (the "m" value) is the slope of the line. And, the constant (the "b" value) is the y-intercept at (0, b) So, if you are given an equation like: y = 2/3 (x) -5 We can tell that the slope of the line = 2/3 and the y-intercept is at (0, -5) Hope this helps. • how does an equation result to an answer? • The equation results in how to graph the line on a graph. If they give you the x value then you would plug that in and it would tell you the answer in y. • Why should I learn this and what can I use this for in the future. • slopes are all over the place in the real world, so it depends on what you plan to do in life of how much you use this. Art, building, science, engineering, finance, statistics, etc. all use linear functions. • Is it ever possible that the slope of a linear function can fluctuate? Or is the slope always a fixed value? • It is a fixed value, but it could possibly look different. So if the slope is 2, you might find points that create a slope of 4/2 or 6/3 or 8/4 or maybe even 1/.5, but each of these will reduce to the same slope of 2. • Why is it called algebra? Is it Greek or something? • say you have a problem like (3,1) slope= 4/3. how would you work that out • Pretty late here, but for anyone else reading, I'll assume they meant how you find the slope intercept using only these values. Since we know the slope is 4/3, we can conclude that: y = 4/3 * x ... But what is the constant, the y axis intercept point? You can solve for it by doing: 1 = 4/3 * 3 + c... We know the values for x and y at some point in the line, but we want to know the constant, c. You can solve this algebraically. 1 = 4/3 * 3 + c 1 = 4 + c 1 - 4 = 4 - 4 + c -3 = c The slope intercept equation is: y = 4/3 * x - 3 The y axis intercept point is: (0 , -3) I just started learning this so if anyone happens across this and spots an error lemme know.
The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 115. The fraction is ____ # The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$. The fraction is ____ Fill Out the Form for Expert Academic Guidance!l +91 Live ClassesBooksTest SeriesSelf Learning Verify OTP Code (required) ### Solution: The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$. The fraction is $\frac{2}{5}$. Given that the numerator of a fraction is 3 less than its denominator. Let’s assume that the numerator and denominator are $n$ and $d$ respectively. $n=d-3$ Then, $\frac{n}{d}=\frac{d-3}{d}$ … (1) After adding 1 to the denominator the fraction decreased by $\frac{1}{15}$ $⟹\frac{d-3}{d+1}=\frac{1}{15}$ … (2) To find the fraction we will equate equation (1) and equation (2), $⟹\frac{\left(d-3\right)\left(d+1\right)-d\left(d-3\right)}{d\left(d+1\right)}=\frac{1}{15}$ $⟹\frac{d\left(d+1\right)-3\left(d+1\right)-{d}^{2}+3d}{{d}^{2}+d}=\frac{1}{15}$ $⟹\frac{{d}^{2}+d-3d-3-{d}^{2}+3d}{{d}^{2}+d}=\frac{1}{15}$ $⟹\frac{d-3}{{d}^{2}+d}=\frac{1}{15}$ $⟹15\left(d-3\right)={d}^{2}+d$ $⟹15d-45={d}^{2}+d$ $⟹{d}^{2}-15d+45=0$ $⟹{d}^{2}-9d-5d+45=0$ $⟹d\left(d-9\right)-5\left(d-9\right)=0$ $⟹\left(d-5\right)\left(d-9\right)=0$ If $d=5$ then, $⟹n=d-3=5-3=2$ Then the fraction is $\frac{2}{5}$. If $d=9$ then, $⟹n=d-3=9-3=6$ Then the fraction is $\frac{6}{9}=\frac{2}{3}$ ∴ We will neglect this fraction because it is not satisfying the given condition. Hence, the original fraction is $\frac{2}{5}$. ## Related content Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics +91 Live ClassesBooksTest SeriesSelf Learning Verify OTP Code (required)
Home Mathematics Lessons 2D Shape Lesson: The Greedy Triangle # Summary: In this lesson, students review geometric terms by making predictions about what the Greedy Triangle will become as the teacher reads the book, The Greedy Triangle. Students will then create geometric creatures after reading the book and discussing the outcome. Another great lesson that links literacy to numeracy. • Year 1 – Recognise and classify familiar two-dimensional shapes and three-dimensional objects using obvious features (ACMMG022) • Year 2- Describe and draw two-dimensional shapes, with and without digital technologies (ACMMG042) Lesson: ## Learning outcomes Students will: • review geometric terms by making predictions about how the triangle changes in the book. • create an animal using the concepts of the geometric terms that were previously reviewed. • analyse the theme of the book. 45 minutes ## Activities 1. Ask the riddle: What did the acorn say when he was all grown up? Answer: Geometry / “Gee, I’m a tree.” 2. State the objective of the lesson. 3. Teacher should tell the students that they are to make predictions about what the triangle becomes as the teacher reads the book. Also, tell the students to think about the theme of the book and be able to explain it. 4. Read the book. Discuss the book’s plot and theme. 5. Review the words on the vocabulary list. 6. Teacher Discussion: Back in the days before cameras were invented, the only way you could show someone how something looked was to draw a picture of it. Today, we are going to pretend we are living in those times. Imagine you have just received a letter from an explorer who has discovered a strange animal. The explorer has not sent a drawing just a letter. It is going to be your job to draw a picture of the animal described in the letter. Remember, we are just pretending, so the animal can be really fantastic. Students can name their animals after they create them. 7. The teacher reads the following sentences one at a time. The students will draw what the teacher describes. • The animal has a body shaped like an equilateral triangle. • The head is shaped like a bumpy circle. • This creature has legs that are approximately one and one-half inches in length. • On each leg is a foot with three toes, and on each toe is a claw that is about 1/4 of an inch in length. • In the center of its body, it has a large spot shaped like a hexagon. • The facial features include a pair of pentagon shaped eyes, an octagon shaped nose and a mouth that is best described as a semicircle. • Upon its head are two parallel horns that appear to be sharp. • Its tail is furry and one meter in length. ## Assessment Check the accuracy of the shapes (i.e. pentagon-5 sides, octagon-8 sides,etc.). Have the students write a paragraph supporting the book’s theme with examples from the text. ## Supplemental information This geometric review integrates language arts and math. Most students are better at retaining information when it is attached to something known or relevant. M. Burns presents the shapes in her book using ordinary and familiar things. Also, the shapes are reinforced as the students create their own geometric creatures. This lesson can become the source for many fourth grade writing objectives, too. The creature could be described in a character analysis, become the main character in an imaginary narrative, be a first person narrator describing a day in his / her life, etc. Courtesy of http://www.learnnc.org/lp/pages/3385?ref=search and shared under Creative Commons 0 0 0 0 0 0 0 0
Future Study Point # Segment and sector of a circle You are familiar with some methods of finding the area of plane figures like triangle, square, rectangle, parallelogram, rhombus, and of the circle. Here the area of segment and sector of the circle is of great importance in our day to life like areas of the rounded figure as an example, wheel excel, and other rounded tools and machines. The segment and sector are the part of the rounded figure are needed in manufacturing different types of tools used in machines and architectural designs of bridges and buildings, so the area of segments and sectors of the circle is introduced in the 10-grade syllabus of every school board. NCERT Solutions Class 10 Science from chapter 1 to 16 Area of the segment and sectors is the topic referred from chapter 12 of the class 10 NCERT book prescribed by CBSE a reputed board of India for school education. ## Sector of circle : Click for online shopping Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc The area covered by two radii and the corresponding arc of the circle is known as sector. 360° angle around the centre corresponds to the area 1° angle around the circle corresponds to the area θ angle around the circle correspond to the area Hence the area of a sector Where θ is the angle between two radii of the circle. Type of sectors : Sectors of circles are of two kinds (a) Minor sector (b) Major sector Minor sector: The shaded region in above figure is minor sector i.e The smaller sector is known as minor sector. Major sector: The unshaded region in above figure is major sector i.e The larger sector is known as major sector. Area of major sector = Area of circle – area of minor sector ## Segment of a circle : The area covered by the chord of a circle and corresponding arc known as segment. Area of segment = area of sector – area of triangle covered by two radii and corresponding chord of the circle Let the angle subtended by the two radii is = θ ∠AOM = θ/2 (OM⊥ AB) AM = r sinθ/2 AB = 2AM=2r sinθ OM = r cosθ/2 Area of ΔAOB Since ,2sin θ cos θ =sin2θ Hence area of ΔAOB = 1/2 r² sinθ We know area of sector So, area of segment Hence Area of segment ## Type of segment (a) Minor segment: The smaller segment of the circle is known as minor segment i.e The shaded region in above figure is known as minor segment. (b) Major segment: The larger segment of the circle is known as major segment i.e The unshaded region in above figure is known as major segment. The area of larger segment = Area of circle – area of minor sector ## Length of arc : Let the length of arc corresponding to two radii of circle is l. 360° angle corresponds to 1° angle corresponds to θ angle corresponds to Hence the length of arc corresponding to two radii ## Type of Arc (a) Minor arc: The smaller arc of the circle is known as minor arc, in the figure l is the length of minor arc. (b) Major arc: The larger arc of the circle is known as major arc The length of major arc = Circumference of the circle – The length of minor arc You can compensate us Paytm number 9891436286 The money collected by us will be used for the education of poor students who leaves their study because of a lack of money. ## NCERT Solutions of Science and Maths for Class 9,10,11 and 12 ### NCERT Solutions for class 9 maths Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral ### NCERT Solutions for class 9 science Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion Chapter 2-Is matter around us pure? Chapter 10- Gravitation Chapter3- Atoms and Molecules Chapter 11- Work and Energy Chapter 4-Structure of the Atom Chapter 12- Sound Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ? Chapter 6- Tissues Chapter 14- Natural Resources Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources Chapter 8- Motion Last years question papers & sample papers ### NCERT Solutions for class 10 maths Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry CBSE Class 10-Question paper of maths 2021 with solutions CBSE Class 10-Half yearly question paper of maths 2020 with solutions CBSE Class 10 -Question paper of maths 2020 with solutions CBSE Class 10-Question paper of maths 2019 with solutions ### NCERT Solutions for Class 10 Science Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world Chapter 4- Carbon and its Compounds Chapter 12- Electricity Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current Chapter 6- Life Process Chapter 14-Sources of Energy Chapter 7-Control and Coordination Chapter 15-Environment Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources ### NCERT Solutions for class 11 maths Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability CBSE Class 11-Question paper of maths 2015 CBSE Class 11 – Second unit test of maths 2021 with solutions ### NCERT solutions for class 12 maths Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2 Class 12 Maths Important Questions-Application of Integrals Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22 Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22 Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution Solutions of the qustions based on Expressions and Equations Important maths notes Tricks – How to write linear equations Tricks- How to solve question from algebraic equations Three ways of solving quadratic equation Mean, Mode and Median Solutions- Specific questions of mensuration Finding the roots of the polynomial by Complete square method Technics – Achieving 100% marks in Maths Most important questions of Maths Solutions-Most important questions of 3-4 marks Solutions – Questions from previous question papers
CAT Time and Work Problems: Time and Work Problems you should solve for CAT Prep CAT Time and Work Practice Exercise-4 Welcome to this post on CAT Time and Work Problems. The objective of this post is very simple: to provide you vital practice for this topic and expose you to the kind of questions that you will encounter for this topic. Also, this CAT Time and Work exercise will help you practice 5 problems for this topic. In our concept section, we have gone through the basic concepts of time and work. Now let’s expand our knowledge further by solving CAT Time and Work problems. This section consists of problems related to real life time and work situations and needs the application of arithmetic concepts as well and strong logic. It consists of relative time, pairing of people to do the work and other such problems. Question 1: Two full tanks, one shaped like a cylinder and the other like a cone, contain jet fuel. The cylindrical tank holds 500 L more than the conical tank. After 200 L of fuel has been pumped out from each tank, the cylindrical tank contains twice the amount of fuel in the conical tank. How many litres of fuel did the cylindrical tank have when it was full? (a) 700 L (b) 1,000 L (c) 1,100 L (d) 1,200 L Let the volume of the conical tank bex L Then cylindrical tank will hold = (x +500) L Given it is given, x +300 = 2 (x – 200) So, x +300 = 2x – 400 Therefore x = 700L. Hence cylindrical tank will hold (700 + 500) = 1200 L of fuel. Question 2: There’s a lot of work in preparing a birthday dinner. Even after the turkey is in the oven, there’re still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table. Three friendsAsit, Arnold, and Afzal, work together to get all of these chores done. The time it takes them to do the work together is six hours less than Asit would have taken working alone, one hour less than Arnold would have taken alone, and half the time Afzal would have taken working alone. How long did it take them to do these chores working together? (a) 20 minutes (b) 30 minutes (c) 40 minutes (d) 50 minutes Let the time taken by Asit, Arnold and Afzal to complete the work alone be x, y and z hours respectively. Given (x – 6) = (y – 1) = ⇒y = (x –5) and z = (2x – 12) Also, time taken by all of them to do the job {(xyz)/(xy +yz+xz)} = (x-6) Substitute y = (x – 5) and z = (2x – 12) in the above equation we get x = 20/3 hours. ∴ Time taken by all the three to complete the work. = 20/3-6 = 2/3hrs = 40 min. Question 3: A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair? (a) A,B (b) A,C (c) B,C (d) A, D Working efficiency per day of A, B, C and D = ¼ , 1/8, 1/16 and 1/32 respectively. Considering the options given, we have B and C does 3/16 of work per day, and A and D does 9/32 work per day. Hence A and D take 32/9 days. B and C take 16/3 days. Thus the first pair must comprise of A and D. Hence, option d. Question 4: Three small pumps and a large pump are filling a tank. Each small pump works at 2/3rd the rate of the large pump. If all four work at the same time, they should fill the tank in what fraction of the time it would have taken the large pump alone? (a) 4/7 (b) 1/3 (c) 2/3 (d) 3/4 Let the large pump rate be x. The time take by large pump = (1/x) The small pumps rate would then be (2/3)x When the four machines are combined their rate is x + (2/3)x + (2/3)x + (2/3)x = 3x. So together they work 3 times as fast, so it will take 1/3 the time taken by the large pump alone. Hence, option b Question 5: Six technicians working at the same rate complete the work of one server in 10 hrs. If they start at 11 : 00 a.m. and one additional technician per hour being added beginning at 5 : 00 p.m., at what time the server will be complete ? (a) 6 : 40 p.m. (b) 7 p.m. (c) 7 : 20 p.m. (d) 8 : 00 p.m. Since, six technicians working at the same rate completely work of one server in 10 hours. Hence, total work = 10 × 6 = 60 man hours. Now, from 11.00 a.m. to 5 p.m. total man hours = 6 × 6 = 36 From 5 p.m. to 6 p.m. total man hours = 7, and the total hours of work completed would be 36 + 7 = 43 From 6 p.m. to 7 p.m. total man hours = 8, and the total hours of work completed would be 43 + 8 = 51 From 7 p.m. to 8 p.m. total man hours = 9 and the total hours of work completed would be 51 + 9 = 60 Hence, the work will be completed at 8 p.m. Extra tips for CAT Time and Work Problems: • CAT Time and work problems can be categorized into different types: First is to deduce the time taken and second it to find the amount of work done. Also, complex problems consist of many sub categories where relative time is to be considered and groups of people working together needs to be analyzed. • An important formula you need to remember for solving such problems is as follows: If A can do a work in ‘p’ hours and B can do the same work in ‘q’ hours, then if A and B work together the work done in 1 hour is 1/p+1/q. • Also,Total Work= Total man hours = Number of men × Time taken for a job • Time and work problems related to filling of tanks and pumps are very important. Practice them thoroughly in order to be prepared problems based from this area in the exam.
## What does trend in math mean? A trend is a pattern in a set of results displayed in a graph. ## What is an example of a trend? Trend is defined as to go in a general direction or to have a tendency to go in a certain way. An example of trend is for a plain to stretch westward across a state. An example of trend is when the number of murders in a city reduce downward. The definition of a trend is a general direction or something popular. ## What is the trend of the graph? A trend line (also called the line of best fit) is a line we add to a graph to show the general direction in which points seem to be going. Think of a “trend” as a pattern in math. The trend line is something we add to our graph to make the pattern even clearer. ## How do you find the trend in math? Calculating Trend Lines 1. Step 1: Complete each column of the table. 2. Column 1: the differences between each x-coordinate and the average of all of the x-coordinates. 3. Column 2: the difference between each y-coordinate and the average of all of the y-coordinates. 4. Column 3: multiply columns 1 and 2 = -2.5 * (-4.83) = 12.083. ## What’s a positive trend? If increase in one set of data causes the other set to increase, then the trend shown is called a positive trend. If one set of data increases, then the other set does not seem to increase or decrease then it does not have any trend. ## What is the trend in statistics? A trend is a pattern found in time series datasets; it is used to describe if the data is showing an upward or downward movement for part, or all of, the time series. Please note that the definitions in our statistics encyclopedia are simplified explanations of terms. ## How do you explain a trend? A trend is the general direction in which something is developing or changing over time. A projection is a prediction of future change. Trends and projections are usually illustrated using line graphs in which the horizontal axis represents time. ## What is a positive trend? Definition Of Trend If the values of one set of data increases and the values of other set also increases then the two sets of related data shows a positive trend. If the data values of a set increases and the data values of other set also increases then the two sets of related data shows a positive trend. ## How do you find the trend in data? A trend can often be found by establishing a line chart. A trendline is the line formed between a high and a low. If that line is going up, the trend is up. If the trendline is sloping downward, the trend is down. ## What do you call a negative trend? You can use the word decline as a noun to mean a negative/downward trend. Another word for declining trend is downturn. ## What are the 3 types of trend analysis? Understanding Trend Analysis It is based on the idea that what has happened in the past gives traders an idea of what will happen in the future. There are three main types of trends: short-, intermediate- and long-term. ## What is an example of a positive trend? Examples on Trend As the number of apples in the bag increases, the weight of the apple bag also tends to increase. So, the scatter plot shows a positive trend. As the supply of an item increases, the price of the item decreases and as the supply decreases, the price of the item increases. ## What do you mean by trend line in math? A trend line, often referred to as a line of best fit, is a line that is used to represent the behavior of a set of data to determine if there is a certain pattern. ## Which is the best definition of a trend? Definition Of Trend. If the values of one set of data increases and the values of other set also increases then the two sets of related data shows a positive trend. ## When is a trend line called a negative trend? Trend Line – In a scatter plot, a line that closely fits the data points is called a Trend line. If one set of data increases, then the other set tends to decrease then the trend shown is called a negative trend. ## What are the current trends in math education? Instead of teaching students that 12 * 12 = 144, New Math and Common Core teach students methods they can use to figure out what 12 * 12 equals. There is no longer a need to memorize this. As the field of mathematics grows, new methods of working with math are also introduced. ## What does trend in math terms mean? Definition Of Trend. If the values of one set of data increases and the values of other set also increases then the two sets of related data shows a positive trend . If the values of one set of data increases and the values of other set decreases then the two sets of related data shows a negative trend. If the data shows no relation then that set shows no trend. ## What is trend line equation? Trendline equation is a formula that mathematically describes the line that best fits the data points. The equations are different for different trendline types, though in every equation Excel uses the least squares method to find the best fit for a line though data points. ## What is trend in data graph? trend is a general-purpose, efficient trend graph for “live” data. Data is read in ASCII form from a file or continuously from a FIFO and displayed in real-time into a multi-pass trend (much like a CRT oscilloscope). trend can be used as a rapid analysis tool for progressive or time-based data series together with trivial scripting. ## What does linear trend mean? linear trend. [¦lin·ē·ər ′trend] (statistics) A first step in analyzing a time series, to determine whether a linear relationship provides a good approximation to the long-term movement of the series; computed by the method of semiaverages or by the method of least squares.
All the solutions provided in McGraw Hill My Math Grade 3 Answer Key PDF Chapter 8 Review will give you a clear idea of the concepts. Vocabulary Check Use the clues and the word bank below to solve the puzzle. Commutative known facts pattern related facts Across 1. The 11 s facts show a ____. When a single digit number is multiplied by 11, the product is the digit repeated. 2. Basic facts using the same three numbers. Sometimes called a fact family. Down 3. Facts which you have memorized. 4. The property of multiplication which states that the order in which two numbers are multiplied does not change the product. Concept Check Algebra Use the inverse operation to find each unknown. Question 5. 30 ÷ 6 = 6 × = 30 The unknown is _____. 30 divided by 6 equal to 5. 30 ÷ 6 = 5. The inverse operation is Multiplication of 6 with 5 then you get 30. 6 x 5 = 30. The unknown is 5. Question 6. 28 ÷ 7 = 7 × = 28 The unknown is _____. 28 divided by 7 equal to 4. 28 ÷ 7 = 4. The inverse operation is Multiplication of 7 with 4 then you get28. 7 x 4 = 28. Therefore the unknown is 4. Question 7. 48 ÷ 6 = 6 × = 48 The unknown is _____. 48 divided by 6 equal to 8. 48 ÷ 6 = 8. The inverse operation is. Multiplication of 6 with 8 then you get 48. 6 x 8 = 48. Therefore the unknown is 8. Double a known fact to find each product. Draw an array. Question 8. 8 × 7 = ____ 4 × 7 = ___ ____ × ___ = ____ 28 + ____ = ____ Multiplication of 8 with 7 then you get 56. 8 x 7 = 56. 4 x 7 = 28. Double the known fact is 4 x 7 + 4 x 7 = 8 x 7. 28 + 28 = 56. The array of 8 x 7 is Question 9. 6 × 9 = ___ Multiplication of 6 with 9 then you get 52. 6 x 9 = 54. Double the known fact is 6 x 9 + 6 x 9 = 12 x 9 12 x 9 = 108. Double the product is 54 + 54 = 108. The array of 6 x 9 is Write an addition sentence and a multiplication sentence for each. Question 10. 5 rows of 11 counters ___ + ___ + ___ + ___ + ___ = ____ ___ × ___ = ____ Given that, 5 rows of 11 counters. So, 5 x 11 = 55. The addition sentence is 11 + 11 + 11 + 11 + 11 = 55. Question 11. 6 rows of 12 counters ___ + ___ + ___ + ___ + ___ + ___ = ____ ___ × ___ = ____ Given that, 6 rows of 12 counters. So, 6 x 12 = 72. The addition sentence is 12 + 12 + 12 + 12 + 12 + 12 = 72. Question 12. 3 rows of 8 counters ___ + ____ + ____ = ____ ___ × ___ = ____ Given that, 3 rows of 8 counters So, 3 x 8 = 24. The addition sentence is 8 + 8 + 8 = 24. Question 13. 7 rows of 11 counters ___ + ___ + ___ + ___ + ___ + ___ + ___ = ____ ___ × ___ = ____ Given that, 7 rows of 11 counters. So, 7 x 11 = 77. The addition sentence 11 + 11 + 11 + 11 + 11 + 11 + 11 = 77. Problem Solving Question 14. Bev notices this heart-shaped button has 4 holes. She needs 11 of these buttons for a project. How many button holes will there be altogether? Write a multiplication sentence to solve. Given that, The heart shaped button has 4 holes. She needs 11 of these buttons for a project. The multiplication sentence is 4 x 11 = 44. Therefore the total number of buttons are 44. Question 15. The array is a model for 5 × 9 = 45. Write the division sentence that is modeled by the array. Multiplication of 5 with 9 then you get 45. 5 x 9 = 45. The inverse of 5 x 6 = 45 is 45 divided by 5 equal to 9. 45 ÷ 5 = 9. Question 16. Chandler works 4 hours each week. How many weeks will it take him to work 36 hours? Write a division sentence to solve. Given that, Chandler works 4 hours each week. The number of weeks will it take him to work 36 hours. The division sentence is 36 ÷ 4 = 9. Therefore there are 9 weeks to take him to work 36 hours. Question 17. Mrs. King took 12 crates packed with lunch boxes on a field trip. Each crate had 6 lunch boxes inside. How many lunch boxes were there altogether? Write a multiplication sentence to solve. Given that, Mrs. King took 12 crates packed with lunch boxes on a field trip. Each crate had 6 lunch boxes inside. The multiplication sentence is 12 x 6 = 72. Therefore there are 72 lunch boxes. Test Practice Question 18. How much will these 4 paperback books cost altogether? A. $7 B.$14 C. $21 D.$28 Given that, There are 4 paperback books. The cost of each book is $7. The cost of 4 paperback books are$7 + $7 +$7 + $7 =$28. Option D is the correct answer. Reflect Use what you learned about multiplication and division to complete the graphic organizer.
# SPLASH! 2012 QUADRATIC RECIPROCITY Michael Belland. ## Presentation on theme: "SPLASH! 2012 QUADRATIC RECIPROCITY Michael Belland."— Presentation transcript: SPLASH! 2012 QUADRATIC RECIPROCITY Michael Belland  Thank you for taking “Is the remainder a square? Elementary Number Theory and Quadratic Reciprocity” at Splash!  This powerpoint is intended to provide a proof of the statement of quadratic reciprocity.  If you read this powerpoint, feel free to take breaks once in a while! There’s a lot of content in here. It took me six weeks of dedicated study to really understand all of this. If you have any questions, please email me! Introduction  Fermat’s Little Theorem  When is -1 a square mod p?  Strong Division  Gauss’ lemma  When is 2 a square mod p?  Statement and Proof of Quadratic Reciprocity  Applications (Or, why we should care) A Brief Outline  Pierre de Fermat, Lawyer (1601-1665)  Not a professional mathematician  Came up with many conjectures; usually these were correct, like Fermat’s Last Theorem! But sometimes not (Fermat primes are a good example).  Fermat’s Little Theorem. Given that p is a prime, a p =a (mod a). Alternatively, p always divides a p -a. (Which we can denote p\|a p -a.) Fermat’s Little Theorem (FlT) In general, a\|b means that a divides b. In other words, dividing b by a gives no remainder.  Consider the set of nonzero integers mod p, which is {1,2,…,p-1}. If two integers in this set are i and j, then when can we have ai = aj (mod p)? Assume p doesn’t divide a.  Let the multiplication of this set by a be a set operation. If we can show that this operation is “injective” and “surjective,” then it follows that the sets {1,2,…,p-1} and {a,2a,…,ap-a} are the same, up to reordering of their contents. Proof of Fermat’s Little Theorem The important point here is that the numbers 1,2,…p-1 are a rearrangement of the numbers a, 2a, … (p-1)a. If you understand this, you can skip the more rigorous proof in the next slide.  We can prove plenty of nuts and bolts about equivalence relations, but converting to integers is easiest.  If ai=aj (mod p), then p\|ai-aj, or p\|a(i-j). If a doesn’t have a factor of p, then i-j must; so p divides i-j. But since 0 i=j (mod p), this set operation is injective.  We can check that all elements of the set {1,2,3,…,p-1} are representable as ai, for some i between 1 and p-1. (inverses)  Therefore, this set operation is surjective (by definition). Proof of FlT, continued  Then, {a,2a,…,ap-a} is a permutation of {1,2,…,p-1}. So, it should follow that both products of all of the elements of the individual sets should be equal.  So, Π (ai) = Π (i). Thus, a p-1 * (p-1)! = (p-1)! (mod p).  It follows that a p-1 =1 (mod p), if p doesn’t divide a. If p divides a, then as p is a prime we find a=0 (mod p). So, multiplication of both sides by a gives a p =a (mod p), which is true for all integers a.  (A permutation is a reordering of something else) Proof of FlT, continued i=1 p-1  What would happen if we added all the sets together? Is there some way that adding these sets together could actually be useful?  Should Fermat’s Little Theorem work for composite modulos? (A composite number is not a prime nor 1.) Why or why not? If not, can this statement be saved?  Can you evaluate (n-1)! modulo n for all integers n? Challenge Question  Some quick observations:  x 2 =1 (mod p) implies x=1 (mod p) or x=-1 (mod p). We see this is true because p\|(x 2 -1) implies p\|x-1 or p\|x+1.  x p-1 = 1 (mod p) if x≠0 (modp), by Fermat’s Little Theorem.  Therefore, x (p-1)/2 = 1 or x (p-1)/2 = -1 (mod p).  If x≠0 is a square mod p, then x=u 2 for some u in U p (U p is shorthand for the set {1,2,…,p-1}). Then, u (p-1) =1 implies x (p- 1)/2 =1. So, if x is a square mod p, x (p-1)/2 =1. When is -1 a square mod p?  So evaluate (-1) (p-1)/2 for all odd primes p. Thankfully, since -1 raised to some power can only be 1 or -1, we can evaluate this expression in integers.  (-1) (p-1)/2 = 1 if (p-1)/2 is even, and (-1) (p-1)/2 = -1 if (p-1)/2 is odd.  So, if p=1 mod 4, -1 is a square mod p. So, -1 is a square mod 5, 13, or 17  For example, 2 2 =-1 (mod 5); 5 2 =-1 (mod 13), and 4 2 =-1 (mod 17). When is -1 a square mod p?  We’re going to assume here a nontrivial but intuitive result about the group of remainders mod p. That is, if polynomials have coefficients that are in {1,2,…p} (this set is also known as Z p ), then a polynomial of degree d has no more than d roots. This is intuitive for normal polynomials.  However, this is not always true for stranger polynomials; e.g (x-2)(x-3) = x(x-5) in Z 6. Don’t think about that too much; it’s not relevant here.  Challenge Problem! Why should division work for polynomials with prime modulo coefficients but not in general for composite modulo coefficients? A way to save some time  There are (p-1)/2 nonzero squares modulo p. Notice that a 2 =(-a) 2, so that every square has two distinct numbers squaring to it (p≠2), and that each of these pairs squares to a different number.  So, we know that x (p-1)/2 -1 = 0 has at most (p-1)/2 roots; these roots are in fact the nonzero squares mod p. So, since x p -1=0 has p roots, it follows that x (p-1)/2 +1=0 has (p-1)/2 roots. These are the nonsquares of Z p. When is a a square mod p?  Conclusion: x (p-1)/2 =1 if and only if x is a square mod p.  Conclusion: If a is a nonsquare mod p, a (p-1)/2 =-1. (x≠0 (mod p), and p ≠ 2, like it assumed elsewhere.)  So, what happens if x=-1? Then, the expression x (p-1)/2 evaluates to be 1 if p is 1 (mod 4), and -1 if p is 3 (mod 4).  So, is 6 a square mod 7? Is 10 a square mod 11? 12 mod 13? When is a a square mod p?  You’ve had to go through a lot of difficult concepts to get to this point. Stop reading this for now and take a break to let everything sink in.  Also remember that some concepts are probably explained in a confusing manner. This all makes sense to me now because I know what I’m trying to say. But it might not have made sense to me when I was trying to learn the material! If there’s something you don’t understand, please ask me questions! I’ll try to explain what’s going on. Take a break!  Let’s approach a different concept now.  “The division algorithm” for integers states that when we divide and integer a by another integer b, we get two integers q and r such that 0≤r { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/9/2519117/slides/slide_15.jpg", "name": " Let’s approach a different concept now.", "description": " The division algorithm for integers states that when we divide and integer a by another integer b, we get two integers q and r such that 0≤r  If we make –p/2 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/9/2519117/slides/slide_16.jpg", "name": " If we make –p/2  Now let’s see if we can apply a Fermat’s little Theorem type trick to get something useful out of this.  We know that ak=e k r k, as we described in the strong division formula, for all k in U p. So let’s evaluate this in particular for all k that satisfy 1 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/9/2519117/slides/slide_17.jpg", "name": " Now let’s see if we can apply a Fermat’s little Theorem type trick to get something useful out of this.", "description": " We know that ak=e k r k, as we described in the strong division formula, for all k in U p. So let’s evaluate this in particular for all k that satisfy 1  Π(ak) = Π(r k e k ) (mod p), just by pi notation.  Then, notice that the set of {1,2,3,…,(p-1)/2} is just a permutation of the set of {r 1,r 2,…,r (p-1)/2 }. This is because none of the r k terms appear more than once, and there are (p-1) /2 of them.  So, we ultimately find that Πa = Πe k (mod p).  From strong division, we can write the product of the e k terms in sigma notation; a (p-1)/2 =-1 Σ[floor(2ak/p)].  Try to work this argument out for small odd primes p like 3, 5, or 7. Gauss’ Lemma 0 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/9/2519117/slides/slide_18.jpg", "name": " Π(ak) = Π(r k e k ) (mod p), just by pi notation.", "description": " Then, notice that the set of {1,2,3,…,(p-1)/2} is just a permutation of the set of {r 1,r 2,…,r (p-1)/2 }. This is because none of the r k terms appear more than once, and there are (p-1) /2 of them.  So, we ultimately find that Πa = Πe k (mod p).  From strong division, we can write the product of the e k terms in sigma notation; a (p-1)/2 =-1 Σ[floor(2ak/p)].  Try to work this argument out for small odd primes p like 3, 5, or 7. Gauss’ Lemma 0  Define these to evaluate to 1 if a is a square mod p, else if a is a nonsquare mod p, this should be -1. In this powerpoint, they look like a\|\|p.  The Legendre symbol is multiplicative, and, more importantly, is congruent to a (p-1)/2 (mod p).  This in turn is congruent to (and in fact equal to) - 1 Σ[floor(2ak/p)], as mentioned before. The limits on the sum go from k=0 to k=(p-1)/2, or just 0≤k { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/9/2519117/slides/slide_19.jpg", "name": " Define these to evaluate to 1 if a is a square mod p, else if a is a nonsquare mod p, this should be -1.", "description": "In this powerpoint, they look like a\|\|p.  The Legendre symbol is multiplicative, and, more importantly, is congruent to a (p-1)/2 (mod p).  This in turn is congruent to (and in fact equal to) - 1 Σ[floor(2ak/p)], as mentioned before. The limits on the sum go from k=0 to k=(p-1)/2, or just 0≤k  A trick yet again.  2\|\|p * b\|\|p = 2\|\|p * (b+p)\|\|p = 2\|\|p * 2\|\|p * [(b+p)/2]\|\|p = [(b+p)/2]\|\|p.  Evaluate -1 Σ[floor(2(b+p)/2k/p)] with 0≤k { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/9/2519117/slides/slide_20.jpg", "name": " A trick yet again.  2\|\|p * b\|\|p = 2\|\|p * (b+p)\|\|p = 2\|\|p * 2\|\|p * [(b+p)/2]\|\|p = [(b+p)/2]\|\|p.", "description": " Evaluate -1 Σ[floor(2(b+p)/2k/p)] with 0≤k  How many lattice points are in the rectangle defined by 0 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/9/2519117/slides/slide_21.jpg", "name": " How many lattice points are in the rectangle defined by 0  Consider the Cartesian plane, and draw a line segment connecting the points (0,0) and (p,q). Then, we can count how many points in the rectangle are below the line, and how many are to the left of it (above it).  Definitely draw a diagram to follow along!  The number of points under the line segment but still in the rectangle can be counted column-by-column. There are floor(qi/p) points in the rectangle below the line segment, when x=i. What do we get if we add all of these statements together (do this for all columns)? One Last Theorem  Using Sigma Notation, we find there are Σ floor(qi/p) points in the rectangle below the line segment.  Similarly, there are Σ floor(pj/q) points in the rectangle above the line segment.  Thus there are Σ floor(qi/p) + Σ floor(pj/q) points inside the rectangle. But we already know there are (p-1)(q-1)/4 points inside the rectangle from our first argument, so Σ floor(qi/p) + Σ floor(pj/q) = (p-1)(q-1) /4. One Last Theorem 0 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/9/2519117/slides/slide_23.jpg", "name": " Using Sigma Notation, we find there are Σ floor(qi/p) points in the rectangle below the line segment.", "description": " Similarly, there are Σ floor(pj/q) points in the rectangle above the line segment.  Thus there are Σ floor(qi/p) + Σ floor(pj/q) points inside the rectangle. But we already know there are (p-1)(q-1)/4 points inside the rectangle from our first argument, so Σ floor(qi/p) + Σ floor(pj/q) = (p-1)(q-1) /4. One Last Theorem 0  Statement. If p and q are both odd primes, p\|\|q * q\|\|p = -1 (p-1)(q-1)/4. So whether p is a square mod q is related to whether or not q is a square mod p.  Proof. We know p\|\|q q\|\|p = -1 Σ[floor(2ak/p)] -1 Σ[floor(2ak/p)] = -1 Σ[floor(ak/p)] -1 Σ[floor(ak/p)] = -1 (p-1)(q-1)/4. The first step comes from Legendre symbols, the second as a result of what 2\|\|p evaluates as, and the third from the previous theorem. Proof of Quadratic Reciprocity  Is 53 a square mod 101? 53 and 101 are odd primes.  From QR, we find that 53\|\|101 101\|\|53 = (-1) (53-1)(101-1)/4.  So, 53\|\|101 101\|\|53 = 1. We want to find 53\|\|101.  Now, 101\|\|53 = -5\|\|53, because 101 = -5 (mod 53).  -1\|\|53 = 1, because -1 is a square mod 53 (as 53 = 1 mod 4).  5\|\|53 * 53\|\|5 = (-1) (5-1)(53-1)/4 = 1. Since 53\|\|5 = 3\|\|5 = -1, 5\|\|53 = -1.  Because -5\|\|53 = -1\|\|53 5\|\|53 = 1-1 = -1, 53\|\|101 = -1.  Thus, because 53\|\|101 = -1, 53 is not a square mod 101. Applications of Quadratic Reciprocity  Does x 2 -x-7=0 have any solutions mod 13?  Believe it or not, the quadratic formula works mod p for all odd primes p. However, we need to define ½ to be (p+1)/2. This is because (p+1)/2 is an integer and because 2 (p+1)/2 = 1 mod p.  So, (p+1)/2 functions like ½ in multiplication.  There is only one thing that will prevent this equation from having solutions. If the number under the square root isn’t a square, there can’t be a solution! Applications of Quadratic Reciprocity  The expression under the square root of the quadratic formula is called a discriminant. In other words, the discriminant of a quadratic ax 2 -bx-c=0 is b 2 - 4ac.  So, x 2 -x-7=0 has a discriminant of (-1) 2 -4(1)(-7) = 29. Is 29 a square mod 13?  29 = 3 mod 13, and we can apply QR.  3\|\|13 * 13\|\|3 = (-1) (3-1)*(13-1)/4 =1. As 13\|\|3 = 1\|\|3 = 1, 3\|\|13 = 1.  Thus, the discriminant is a square, and solutions exist. Applications of Quadratic Reciprocity Can you find them?
# Eureka Math Geometry Module 5 Lesson 21 Answer Key ## Engage NY Eureka Math Geometry Module 5 Lesson 21 Answer Key ### Eureka Math Geometry Module 5 Lesson 21 Exploratory Challenge Answer Key Exploratory Challenge: A Journey to Ptolemy’s Theorem The diagram shows cyclic quadrilateral ABCD with diagonals $$\overline{A C}$$ and $$\overline{B D}$$ intersecting to form an acute angle with degree measure w. AB = a, BC = b, CD = c, and DA = d. a. From the last lesson, what is the area of quadrilateral ABCD in terms of the lengths of its diagonals and the angle w? Remember this formula for later. Area(ABCD) = $$\frac{1}{2}$$ AC ⋅ BD ⋅ sin⁡w b. Explain why one of the angles, ∠BCD or ∠BAD, has a measure less than or equal to 90°. Opposite angles of a cyclic quadrilateral are supplementary. These two angles cannot both have measures greater than 90° (both angles could be equal to 90°). c. Let’s assume that ∠BCD in our diagram is the angle with a measure less than or equal to 90°. Call its measure v degrees. What is the area of triangle BCD in terms of b, c, and v? What is the area of triangle BAD in terms of a, d, and v? What is the area of quadrilateral ABCD in terms of a, b, c, d, and v? If v represents the degree measure of an acute angle, then 180˚-v would be the degree measure of angle BAD since opposite angles of a cyclic quadrilateral are supplementary. The area of triangle BCD could then be calculated using $$\frac{1}{2}$$ bc ⋅ sin⁡(v), and the area of triangle BAD could be calculated by $$\frac{1}{2}$$ ad ⋅ sin⁡(180˚ – (180˚ – v)), or $$\frac{1}{2}$$ ad ⋅ sin⁡(v). So, the area of the quadrilateral ABCD is the sum of the areas of triangles BAD and BCD, which provides the following: Area(ABCD) = $$\frac{1}{2}$$ ad ⋅ sin⁡v + $$\frac{1}{2}$$ bc ⋅ sin⁡v . d. We now have two different expressions representing the area of the same cyclic quadrilateral ABCD. Does it seem to you that we are close to a proof of Ptolemy’s claim? Equating the two expressions gives us a relationship that does, admittedly, use the four side lengths of the quadrilateral and the two diagonal lengths, but we also have terms that involve sin(w) and sin(v). These terms are not part of Ptolemy’s equation. e. Trace the circle and points A, B, C, and D onto a sheet of patty paper. Reflect triangle ABC about the perpendicular bisector of diagonal $$\overline{A C}$$. Let A’, B’, and C’ be the images of the points A, B, and C, respectively. i. What does the reflection do with points A and C? Because the reflection was done about the perpendicular bisector of the diagonal (AC) ̅, the endpoints of the diagonal are images of each other (i.e., A = C’ and C = A’). ii. Is it correct to draw B’ as on the circle? Explain why or why not. Reflections preserve angle measure. Thus, ∠AB’C ≅ ∠ABC. Also, ∠ABC is supplementary to ∠CDA; thus, ∠AB’C is too. This means that AB’CD is a quadrilateral with one pair (and, hence, both pairs) of opposite angles supplementary. Therefore, it is cyclic. This means that B’ lies on the circle that passes through A, C, and D. And this is the original circle. iii. Explain why quadrilateral AB’CD has the same area as quadrilateral ABCD. Triangle AB’C is congruent to triangle ABC by a congruence transformation, so these triangles have the same area. It now follows that the two quadrilaterals have the same area. f. The diagram shows angles having degree measures u, w, x, y, and z. Find and label any other angles having degree measures u, w, x, y, or z, and justify your answers. See diagram. Justifications include the inscribed angle theorem and vertical angles. g. Explain why w = u + z in your diagram from part (f). The angle with degree measure w is an exterior angle to a triangle with two remote interior angles u and z. It follows that w = u + z. h. Identify angles of measures u, x, y, z, and w in your diagram of the cyclic quadrilateral AB’CD from part (e). See diagram below. i. Write a formula for the area of triangle B’ AD in terms of b, d, and w. Write a formula for the area of triangle B’ CD in terms of a, c, and w. Area(B’ AD) = $$\frac{1}{2}$$ bd ⋅ sin⁡(w) and Area(B’ CD) = $$\frac{1}{2}$$ ac ⋅ sin⁡(w) j. Based on the results of part (i), write a formula for the area of cyclic quadrilateral ABCD in terms of a, b, c, d, and w. Area(ABCD) = Area(AB’ CD) = $$\frac{1}{2}$$ bd ⋅ sin(w) + $$\frac{1}{2}$$ ac ⋅ sin⁡(w) k. Going back to part (a), now establish Ptolemy’s theorem. $$\frac{1}{2}$$ AC ⋅ BD ⋅ sin⁡(w) = $$\frac{1}{2}$$ bd ⋅ sin⁡(w) + $$\frac{1}{2}$$ ac ⋅ sin⁡(w)The two formulas represent the same area. $$\frac{1}{2}$$ ⋅ sin⁡(w) ⋅ (AC ⋅ BD) = $$\frac{1}{2}$$ ⋅ sin⁡(w) ⋅ (bd + ac) Distributive property AC ⋅ BD = bd + ac Multiplicative property of equality or AC ⋅ BD = (BC ⋅ AD) + (AB ⋅ CD) Substitution ### Eureka Math Geometry Module 5 Lesson 21 Exercise Answer Key Opening Exercise Ptolemy’s theorem says that for a cyclic quadrilateral ABCD, AC ⋅ BD = AB ⋅ CD + BC ⋅ AD. With ruler and a compass, draw an example of a cyclic quadrilateral. Label its vertices A, B, C, and D. Draw the two diagonals $$\overline{A C}$$ and $$\overline{B D}$$. With a ruler, test whether or not the claim that AC ⋅ BD = AB ⋅ CD + BC ⋅ AD seems to hold true. Repeat for a second example of a cyclic quadrilateral. Challenge: Draw a cyclic quadrilateral with one side of length zero. What shape is this cyclic quadrilateral? Does Ptolemy’s claim hold true for it? Students will see that the relationship AC ⋅ BD = AB ⋅ CD + BC ⋅ AD seems to hold, within measuring error. For a quadrilateral with one side of length zero, the figure is a triangle inscribed in a circle. If the length AB = 0, then the points A and B coincide, and Ptolemy’s theorem states AC ⋅ AD = 0 ⋅ CD + AC ⋅ AD, which is true. ### Eureka Math Geometry Module 5 Lesson 21 Problem Set Answer Key Question 1. An equilateral triangle is inscribed in a circle. If P is a point on the circle, what does Ptolemy’s theorem have to say about the distances from this point to the three vertices of the triangle? It says that the sum of the two shorter distances is equal to the longer distance. Question 2. Kite ABCD is inscribed in a circle. The kite has an area of 108 sq.in., and the ratio of the lengths of the non-congruent adjacent sides is 3 : 1. What is the perimeter of the kite? $$\frac{1}{2}$$ AC ⋅ BD = 108 AC ⋅ BD = AB ⋅ CD + BC ⋅ AD = 216 Since AB = BC and AD = CD, then AC ⋅ BD = AB ⋅ CD + AB ⋅ CD 2(AB ⋅ CD) = 216 AB ⋅ CD = 108. Let x be the length of $$\overline{C D}$$, then x ⋅ 3x = 108 3x2 = 108 x2 = 36 x = 6. Therefore, CD and AD are 6 in. and AB and BC are 18 in., and the perimeter of kite ABCD is 48 in. Question 3. Draw a right triangle with leg lengths a and b and hypotenuse length c. Draw a rotated copy of the triangle such that the figures form a rectangle. What does Ptolemy have to say about this rectangle? We get a2 + b2 = c2, the Pythagorean theorem. Question 4. Draw a regular pentagon of side length 1 in a circle. Let b be the length of its diagonals. What does Ptolemy’s theorem say about the quadrilateral formed by four of the vertices of the pentagon? b2 = b + 1, so b = $$\frac{\sqrt{5} + 1}{2}$$. (This is the famous golden ratio!) Question 5. The area of the inscribed quadrilateral is $$\sqrt{300}$$ mm2. Determine the circumference of the circle. Since ABCD is a rectangle, then AD ⋅ AB = $$\sqrt{300}$$, and the diagonals are diameters of the circle. The length of AB = 2r sin⁡60, and the length of AD = 2r sin⁡30, so $$\frac{AB}{AD}$$ = $$\sqrt{3}$$, and AB = $$\sqrt{3}$$ (AD). AD ⋅ $$\sqrt{3}$$ (AD) = $$\sqrt{300}$$ AD2 = $$\frac{\sqrt{300}}{\sqrt{3}}$$ AD = $$\sqrt{10}$$ AB = $$\frac{\sqrt{30}}{\sqrt{3}}$$ AB ⋅ DC + AD ⋅ BC = AC ⋅ BD ($$\sqrt{30}$$ ⋅ $$\sqrt{30}$$) + ($$\sqrt{10}$$ ⋅ $$\sqrt{10}$$) = AC ⋅ AC 30 + 10 = AC2 40 = AC2 2$$\sqrt{10}$$ = AC Since AC is 2$$\frac{\sqrt{10}}{\sqrt{3}}$$ mm, the radius of the circle is $$\sqrt{10}$$ mm, and the circumference of the circle is 2$$\frac{\sqrt{10}}{\sqrt{3}}$$π mm. Question 6. Extension: Suppose x and y are two acute angles, and the circle has a diameter of 1 unit. Find a, b, c, and d in terms of x and y. Apply Ptolemy’s theorem, and determine the exact value of sin(75). Use scaffolded questions below as needed. a. Explain why $$\frac{a}{sin(x)}$$ equals the diameter of the circle. If the diameter is 1, this is a right triangle because it is inscribed in a semicircle, so sin(x) = $$\frac{a}{1}$$, or 1 = $$\frac{a}{sin(x)}$$. Since the diameter is 1, the diameter is equal to $$\frac{a}{sin(x)}$$ . b. If the circle has a diameter of 1, what is a? a = sin(x) c. Use Thales’ theorem to write the side lengths in the original diagram in terms of x and y. Since both are right triangles, the side lengths are a = sin(x), b = cos(x), c = cos(y), and d = sin(y). d. If one diagonal of the cyclic quadrilateral is 1, what is the other? sin(x + y) e. What does Ptolemy’s theorem give? 1 ∙ sin(x + y) = sin(x)cos(y) + cos(x)sin(y) f. Using the result from part (e), determine the exact value of sin(75). sin(75) = sin(30 + 45) = sin(30)cos(45) + cos(30)sin(45) = $$\frac{1}{2}$$ ∙ $$\frac{\sqrt{2}}{2}$$ + $$\frac{\sqrt{3}}{2}$$ ∙ $$\frac{\sqrt{2}}{2}$$ = $$\frac{\sqrt{2} + \sqrt{6}}{4}$$ What is the length of the chord $$\overline{A C}$$? Explain your answer. Chord $$\overline{B D}$$ is a diameter of the circle because it makes a right angle with chord AC, and BD = $$\sqrt{2^{2} + 8^{2}}$$ = $$\sqrt{68}$$. By Ptolemy’s theorem: AC ∙ $$\sqrt{68}$$ = 2 ∙ 8 + 2 ∙ 8, giving AC = $$\frac{16 \sqrt{17}}{17}$$.
General Equation of a Line General equation of a line is Ax + By + C = 0. The inclination of angle θ to a line with a positive direction of X-axis in the anti-clockwise direction, the tangent of angle θ is said to be slope or gradient of the line and is denoted by m General equation of first degree in two variables, Ax + By + C = 0, where A, B and C are real constants such that A and B are not zero simultaneously. Graph of the equation Ax + By + C = 0 is always a straight line. Therefore, any equation of the form Ax + By + C = 0, where A and B are not zero simultaneously is called general linear equation or general equation of a line. Different forms of Ax + By + C = 0 The general equation of a line can be reduced into various forms of the equation of a line. If B = 0, then x = - C/A which is a vertical line whose slope is undefined and x-intercept is -C/A. If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0, which is a line passing through the origin and, therefore, has zero intercepts on the axes. (Q1) The equation of a line is given by, 2x – 6y +3 = 0. Find the slope and both the intercepts. Solution: The given equation 2x – 6y + 3 = 0 can be represented in slope-intercept form as: y = x/3 + 1/2 Comparing it with y = mx + c, Slope of the line, m = 1/3 Also, the above equation can be re-framed in intercept form as; x/a + y/b = 1 2x – 6y = -3 x/(-3/2) – y/(-1/2) = 1 Thus, x-intercept is given as a = -3/2 and y-intercept as b = 1/2. (Q2) The equation of a line is given by, 13x – y + 12 = 0. Find the slope and both the intercepts. Solution: The given equation 13x – y + 12 = 0 can be represented in slope-intercept form as: y = 13x + 12 Comparing it with y = mx + c, Slope of the line, m = 13 Also, the above equation can be re-framed in intercept form as; x/a + y/b = 1 13x – y = -12 x/(-12/13) + y/12 = 0 Thus, x-intercept is given as a = -12/13 and y-intercept as b = 12. Example: Show that two lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0, where b1,b2 ¹ 0 are:
# The sum of the areas of two squares is 640 m2. Question: The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares. Solution: Let the length of the side of the first and the second square be $x$ and $y$, respectively. According to the question: $x^{2}+y^{2}=640 \quad \ldots$ (i) Also, $4 x-4 y=64$ $\Rightarrow x-y=16$ $\Rightarrow x=16+y$ Putting the value of $x$ in (i), we get: $x^{2}+y^{2}=640$ $\Rightarrow(16+y)^{2}+y^{2}=640$ $\Rightarrow 256+32 y+y^{2}+y^{2}=640$ $\Rightarrow 2 y^{2}+32 y-384=0$ $\Rightarrow y^{2}+16 y-192=0$ $\Rightarrow y^{2}+(24-8) y-192=0$ $\Rightarrow y^{2}+24 y-8 y-192=0$ $\Rightarrow y(y+24)-8(y+24)=0$ $\Rightarrow(y+24)(y-8)=0$ $\Rightarrow y=-24$ or $y=8$ $\therefore y=8 \quad(\because$ Side cannot be negative) $\therefore x=16+y=16+8=24 \mathrm{~m}$ Thus, the sides of the squares are $8 \mathrm{~m}$ and $24 \mathrm{~m}$.
#### Prime Factorization Prime Factorization is used to find the Least Common Multiple and the Greatest Common Factor. This lesson will address a few definitions, give how-to instructions, detailed examples, and share a website for online practice. Prime Numbers Prime numbers have only two factors. Those factors are one and itself. For example, 17 has the factors 1 and 17 only. Thus, 17 is a prime number. Likewise, 2, 3, 5, 7, 11, 13, 17, 19 are prime numbers too. The number 39 has factors 1, 3, 13, and 39. Thus, 39 is not a prime number. Composite numbers Numbers with other factors besides themselves and one are called composite numbers. Therefore, 39 is a composite number. By the way, one is not a prime or composite number. Prime Factorization ---------------------Usefulness: to find the Greatest Common Factor or Least Common Multiple When a composite number is factored using only prime numbers such as 2 x 3 x 5 = 30, it is referred to as prime factorization. How to Find the Prime Factorization of a Number 1) Start with the smallest prime number, 2, and ask yourself if the prime number can divide into the given number without a remainder. In other words, is it divisible by 2? 2) If no, then the prime number is not a factor. Try the next prime number. 3) If yes, then include that prime number in the prime factorization equation. 4) If the given number was divisible by the prime number in step one, was the answer a composite or prime number? If composite, use this number and repeat steps 1 – 3 starting with the prime number 2 again. If the answer is a prime number, divide the number by itself to get one and you are finished; include all prime numbers in the factorization. 5) Check – compute the multiplication sentence and the answer should equal the number just factored. Let’s find the prime factorization of 30 1) Start with the smallest prime number 2. Ask yourself if the prime number can divide into 30 without a remainder. 30 / 2 = 15 remainder 0 3) Yes it can. Then include 2 in the prime factorization equation. 4) In step 1,was the answer a composite or prime number? 15 is a composite number. So, repeat the process with 15 starting with 2 again. 15 / 2 = 7 remainder 1; 15 is not divisible by 2; so, 2 won’t be used again Next, try 3; 15 / 3 = 5; 15 is divisible by 3; 3 becomes part of the factorization. The answer 5 is prime; so, divide 5 by itself ----- 5 / 5 =1 You are finished; include all prime numbers in the factorization. Summary: 30/ 2 = 15 15/ 3 = 5 5 / 5 = 1 The prime factorization of 30 = 2 x 3 x 5. The bold numbers are the prime factors of 30. Check – compute the multiplication sentence and the answer should equal the number just factored, 30. Example 2: Find the prime factorization of 45 45 / 3 = 15 15 / 3 = 5 5 / 5 = 1 Prime factorization of 45 = 3 x 3 x 5 Example 3: Find the prime factorization of 88 82 / 2 = 44 44 / 2 = 22 22 / 2 = 11 11/ 11 = 1 Prime factorization of 88 = 2 x 2 x 2 x 11. For online practice: I highly recommend the website in the related links section. It uses the factor tree method which is similar to the above method. Practice - Prime Factorization Fractions - Subtracting Mixed Numbers Related Articles Editor's Picks Articles Top Ten Articles Previous Features Site Map
# EVALUATING REAL WORLD EXPRESSIONS WORKSHEET ## About "Evaluating real world expressions worksheet" Evaluating real world expressions worksheet : Worksheet on Evaluating real world expressions is much useful to the students who would like to practice solving word problems on "Evaluating algebraic expressions". ## Evaluating real world expressions worksheet 1. A scientist measures the air temperature in Death Valley, California, and records 50 °C. The expression 1.8c + 32 gives the temperature in degrees Fahrenheit for a given temperature in degrees Celsius c. Find the temperature in degrees Fahrenheit that is equivalent to 50 °C. 2. The expression 6x² gives the surface area of a cube, where x is the length of one side of the cube. Find the surface area of a cube with a side length of 2 meters. 3. The expression x³ gives the volume of a cube, where x is the length of one side of the cube. Find the volume of a cube with a side length of 2 meters. 4. The expression 60m gives the number of seconds in m minutes. How many seconds are there in 7 minutes ? ## Evaluating real world expressions worksheet - Solution Problem 1 : A scientist measures the air temperature in Death Valley, California, and records 50 °C. The expression 1.8c + 32 gives the temperature in degrees Fahrenheit for a given temperature in degrees Celsius c. Find the temperature in degrees Fahrenheit that is equivalent to 50 °C. Solution : Step 1 : Fromm the given information, we have c = 50 °C Step 2 : Plug the value into the expression. 1.8c + 32 Plug 50 for c 1.8(50) + 32 Multiply 90 + 32 122 Hence, 122 °F is equivalent to 50 °C. Problem 2 : The expression 6x² gives the surface area of a cube, where x is the length of one side of the cube. Find the surface area of a cube with a side length of 2 meters. Solution : Step 1 : Fromm the given information, we have x = 2 Step 2 : Plug the value into the expression which represents surface area of the cube. 6x² Plug 2 for x 6(2)² Evaluate exponents 6(4) Multiply 24 Hence, the surface area of the cube 24 m² Problem 3 : The expression x³ gives the volume of a cube, where x is the length of one side of the cube. Find the volume of a cube with a side length of 2 meters. Solution : Step 1 : Fromm the given information, we have x = 2 Step 2 : Plug the value into the expression. x³ Plug 2 for x (2)³ Evaluate exponents 8 Hence, the volume of the cube 8 m³ Problem 4 : The expression 60m gives the number of seconds in m minutes. How many seconds are there in 7 minutes? Solution : Step 1 : Fromm the given information, we have m = 7 Step 2 : Plug the value into the expression. 60m Plug 7 for m 60(7) Multiply 420 Hence, there are 420 seconds in 7 minutes. 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# How to Solve Quadratic Formulas Quadratics are algebraic equations that have a degree of 2. They can be solved using the quadratic formula. A quadratic equation is an equation with a degree of 2. It is usually written in the form ax2 + bx + c = 0 which can be simplified to ax2 + bx = –c. The quadratic formula is used to solve these equations. ## How to Solve a Quadratic Equation Quadratic equations are expressions that involve the square of a variable, and can be solved by using the quadratic formula. To solve a quadratic equation, one needs to find the discriminant. The discriminant is found by taking the difference between b2-4ac and then finding its square root. It is important to note that if b2-4ac is negative, then there are two solutions for the equation. If it is positive, then there will only be one solution for the equation. A parabola is a curve that opens upward or downward and is symmetrical about the axis of symmetry. It can be defined as the set of points that are equidistant from a fixed point (called the focus) and a fixed straight line (called the directrix). The graph of a quadratic function, also known as parabolas, has an open shape. The graph can be described by two points on either side of the axis of symmetry, which is called the vertex. The vertex has coordinates (-b/2a, 0), where b is the y-intercept and a is the slope. The equation for graphing parabolas is given by y =ax2+bx+c ## Quadratic Functions – Solutions & Discriminants The discriminant is the b2-4ac. The discriminant is a function that determines whether a quadratic equation has two real roots or not. It helps in determining the number of solutions to an equation. The discriminant can be found by using the following formula: b2-4ac ## Identifying The Roots Of A Quadratic Equation There are many ways to identify the roots of an equation. One way is to factor the equation and find the roots. Another way is to use a quadratic formula. The roots of a quadratic equation can be found by using either a quadratic formula or by factoring the equation.
# Smaller Units to Bigger Units To convert a smaller unit into a bigger unit, we move the decimal point to the left. In other words, we can say that we divide. This is very important for us to learn how to convert smaller units into bigger units. We use it often in our daily life. Measurement of length, mass and capacity tables can be represented in the form of a place-value chart as shown below. Converting Smaller Units of Length into Bigger Units of Length: For Example: 1. Convert 80 mm into cm. Solution: We know that 10 mm = 1 cm 80 mm = $$\frac{80}{10}$$ cm = 8 cm 2. Convert 485 mm into cm. Solution: 485 mm = 480 mm + 5 mm We know that 10 mm = 1 cm 480 mm + 5 mm = $$\frac{480}{10}$$ cm + 5 mm = 48 cm 5 mm 3. Convert 15000 m into km. Solution: We know that 1000 m = 1 km So, 15000 m = $$\frac{15000}{1000}$$ km = 15 km 3. Convert 577 centimetres to decametres 1000 cm = 1 dam 577 cm = (577 ÷ 1000) dam 577 cm = 0.577 dam Converting Smaller Units of Mass into Bigger Units of Mass: To convert lower units of mass to higher units, we multiply by 1000. A quick way to convert is to write three digits from right as lower unit and the remaining as higher unit. For example: 1. Convert 14000 mg into g. We know that 1000 mg = 1 g So, 14000 mg = $$\frac{14000}{1000}$$ g = 14 g 2. Express 3180 g as kg. 3180 g = 3000 g + 180 g = $$\frac{3000}{1000}$$ kg + 180 g = 3 kg 180 g 3. Convert 7000 grams to kilograms. 1000 g = 1 kg 7000 g = (7000 ÷ 1000) kg 7000 g = 7 kg 4. Convert 225 kilograms to quintals 100 kg = 1 quintal 225 kg = (225 ÷ 100) quintals 225 kg = 2.25 quintals 5. Convert 36 quintals to tonnes 10 quintals = 1 tonne 36 quintals = (36 ÷ 10) tonnes 36 quintals = 3.6 tonnes 6. Convert 415 kilograms to tonnes 1000 kg = 1 tonne 415 kg = (415 ÷ 1000) tonnes 415 kg = 0.415 tonnes Converting Smaller Units Capacity into Bigger Units Capacity: To convert milliliters into liters, we divide the number of milliliters (ml) by 1000. A quick way to convert ml into l is to write three digits from right as ml and the remaining as l. For example: 1. Convert 76489 ml into l. 76489 ml = 76000 ml + 489 ml = $$\frac{76000}{1000}$$ + + 489 ml = 76 l 489 ml 2. Convert 375 litres to kilolitres. 1000 l = 1 kl 375 l = (375 ÷ 1000) kl 375 l = 0.375 kl Alternate Method: Conversion of Lower Units to Higher Units: When a lowers unit is changed into a higher unit we divide the number of the lowers unit by the number showing the relationship between the two units. For example: 1. 5728 g = 5000 g 728 g5728 g = 5 kg 728 g5728 g = 5.728 g Also 5728 g = 5000 g + 700 g + 20 g + 8 g = 5 kg 7 hg 2 dag 8 g 2. 328 cm = 300 cm + 28 cm328 cm = 3 m 28 cm328 cm = 3.28 m Also 328 cm = 300 cm + 20 cm + 8 cm = 3 m 2 dm 8 cm Solved examples for the conversion of smaller units to bigger units: 1. Convert 9362.8 grams into the following units. (i) Decagrams (ii) hectograms (iii) kilograms Solution: Because 1 dag = 10 g So, 1 g = $$\frac{1}{10}$$ dag So, 9362.8 g = $$\frac{9362.8}{10}$$ = (9362.8 ÷ 10) dag = 936.28 dag Thus, (i) 9362.8 g = = (9362.8 ÷ 10) dag = 936.28 dag (ii) 9362.8 g = = (9362.8 ÷ 100) hg = 93.628 hg, (Because 1 g = $$\frac{1}{100}$$ hg) (ii) 9362.8 g = = (9362.8 ÷ 1000) kg = 9.3628 kg, (Because 1 g = $$\frac{1}{1000}$$ kg) 2. Convert 2345 millimetres into the following units. (i) centimetres (ii) metres (iii) kilometres Solution: (i) 2345 millimetres = (2345 ÷ 10) = 234.5 centimetres, [Because 1 mm = $$\frac{1}{10}$$ cm] (ii) 2345 millimetres = (2345 ÷ 1000) = 2.345 metres, [Because 1 mm = $$\frac{1}{1000}$$ m] (iii) 2345 millimetres = (2345 ÷ 1000000) = 0.002345 kilometres, [Because 1 mm = $$\frac{1}{1000000}$$ km] Let us consider another example involving different types of conversions. 3. Convert the following: (i) 3598 mm to m (ii) 4683254 mg to dg (iii) 5923 ml to cl Solution: (i) 3598 mm = (3598 ÷ 1000) m, [Because 1 mm = $$\frac{1}{1000}$$ m] = 3.598 m (ii) 4683254 mg = (4683254 ÷ 100) dg, [Because 1 mg = $$\frac{1}{100}$$ dg] = 46832.54 dg (iii) 5923 ml = (5923 ÷ 10) cl, [Because 1 ml = $$\frac{1}{10}$$ cl] = 592.3 cl 4. Convert 12500 m into km. Solution: We know that 1000 m = 1 km 12500 m = 12000 m + 500 m = $$\frac{12000}{1000}$$ km + 500 m = 12 km 500 m Questions and Answers on Smaller Units to Bigger Units: I. Convert the given lengths: (i) 40 mm = ………….. cm (ii) 540 cm = ………….. m ………….. cm (iii) 160 mm = ………….. cm (iv) 1250 m = ………….. km ………….. m (v) 10500 cm = ………….. m (vi) 3500 cm = ………….. m ………….. cm (vii) 612 cm = ………….. m ………….. cm (viii) 41752 m = ………….. km ………….. m I. (i) 4 cm (ii) 5 m 40 cm (iii) 16 cm (iv) 1 km 250 m (v) 105 m (vi) 35 m 0 cm (vii) 6 m 12 cm (viii) 41 km 752 m Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Thousandths Place in Decimals \| Decimal Place Value \| Decimal Numbers Jul 20, 24 03:45 PM When we write a decimal number with three places, we are representing the thousandths place. Each part in the given figure represents one-thousandth of the whole. It is written as 1/1000. In the decim… 2. ### Hundredths Place in Decimals \| Decimal Place Value \| Decimal Number Jul 20, 24 02:30 PM When we write a decimal number with two places, we are representing the hundredths place. Let us take plane sheet which represents one whole. Now, we divide the sheet into 100 equal parts. Each part r… 3. ### Tenths Place in Decimals \| Decimal Place Value \| Decimal Numbers Jul 20, 24 12:03 PM The first place after the decimal point is tenths place which represents how many tenths are there in a number. Let us take a plane sheet which represents one whole. Now, divide the sheet into ten equ… 4. ### Representing Decimals on Number Line \| Concept on Formation of Decimal Jul 20, 24 10:38 AM Representing decimals on number line shows the intervals between two integers which will help us to increase the basic concept on formation of decimal numbers.
Click To Get Off-Campus Placement Jobs Info !!! FREE!!! Latest Nagarro FULL Aptitude Mock Test for Practice Practice 5+ Mettl Aptitude Mock Test # Quantitative Aptitude :: Percentage Home > Quantitative Aptitude > Percentage > Important Formulas ## Percentage Important Formulas Concept of Percentage: By a certain percent, we mean that many hundredths.Thus, X percent means x hundredths, written as X%. To express X% as a fraction: We have, X% = $$\frac{X}{100}$$. Thus, 20% = $$\frac{20}{100}$$ = $$\frac{1}{5}$$. To express $$\frac{a}{b}$$ as a percent: We have, $$\frac{a}{b}$$ = $$\frac{a}{b} \times 100$$%. Thus, $$\frac{1}{4}$$ = $$\left(\frac{1}{4} \times 100 \right)$$% = 25%. Percentage Increase/Decrease: If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: $$\left[\frac{R}{(100+R)} \times 100\right]$$% If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: $$\left[\frac{R}{(100-R)} ]\times 100\right]$$% Results on Population: Let the population of a town be P now and suppose it increases at the rate of R% per annum, then: 1. Population after n years = $$P \times \left(1 + \frac{R}{100}\right)^{n}$$ 2. Population n years ago = $$\frac{P}{\left(1 + \frac{R}{100}\right)^n}$$ Results on Depreciation: Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then: 1. Value of the machine after n years = $$P \times \left(1 - \frac{R}{100}\right)^{n}$$ 2. Value of the machine n years ago = $$\frac{P}{\left(1 - \frac{R}{100}\right)^n}$$ 3. If A is R% more than B, then B is less than A by $$\left[\frac{R}{(100 + R)} \times 100\right]$$%. 4. If A is R% less than B, then B is more than A by $$\left[\frac{R}{(100 - R)} \times 100\right]$$%.
# Computing the Fibonacci numbers This page is really about how to tell a computer what to do in such a way that it goes about it efficiently; so it has to involve some mathematics (so that we can work out how much work the computer has to do to complete the computation), so it may as well address an entirely mathematical problem. Fibonacci's sequence is defined by: • f(1) = 1 = f(2) • f(i+2) = f(i+1) + f(i) Formally, this only specifies f(i) in so far as i is a whole number (i.e. one with no fractional part); but it does specify f(i) for every whole number. We can work forwards to see that f(3) = 2, f(4) = 3, f(5) = 5, f(6) = 8, f(7) = 13, f(8) = 21 and so on. We can turn the second rule around to say f(i) = f(i+2) −f(i+1) and work backwards to get f(0) = 0, f(−1) = 1, f(−2) = −1, f(−3) = 2, f(−4) = −3, f(−5) = 5, f(−6) = −8 and so on. None of this gives us any clue as to a value for f(1/2) or f(π), so Fibonacci's specification only tells us f's values at whole numbers (positive, zero and negative; collectively known as integers). Notice that f(0) = 0, f(1) = 1 would work as an entirely adequate replacement for the first rule (and I could equally phrase it as (:f\|2) = 2). Also, observe that f(−odd) = f(odd) but f(−even) = −f(even) for each odd or even input explored above. If we suppose this to be true for each even or odd j < i, for some positive whole number i (e.g., we've already shown this for i = 7), we can induce that: if i is even f(−i) = f(2−i) −f(1−i) but i is even, so 2−i is also even and 1−i is odd, so f(−i) = −f(i−2) −f(i−1) = −f(i) otherwise, i is odd, yielding f(−i) = f(2−i) −f(1−i) with i odd, so 2−i is odd and 1−i is even, whence f(−i) = f(i−2) +f(i−1) = f(i) so the rule we supposed to hold for j < i also holds for j = i; hence for each j < i+1, whence for j = i+1 and so on; hence for all j. We can exploit this to simplify our computations. I can use some simple theory to work out that any f satisfying the second rule (ignoring, for the moment, the first) must be a sum of terms, each of which is of form f(i) = k.power(i, q) with k arbitrary and q.q = q +1 – which implies that (2.q−1).(2.q−1) = 4.q.q −4.q +1 = 4.(q +1) −4.q +1 = 5, whence 2.q = 1 ±√5 – so that our actual f must be f(i) = k.power(i, (1+√5)/2) +h.power(i, (1−√5)/2) for some k and h. (That (1+√5)/2 is the golden ratio; I'll discuss this furthere, below.) We could apply the first rule to work out k and h; but it's easier to apply 0 = f(0) = k +h to get h = −k and then use 1 = f(1) = k.((1+√5)/2 −(1−√5)/2) = k.√5, so that f(i) = (power(i, (1+√5)/2) −power(i, (1−√5)/2))/√5. If I'm happy with the precision of floating point arithmetic this lets me compute f very efficiently; however, floating point arithmetic is all about adequate approximations and tends to produce mildly embarrasing results like 4 × 0.035 = 0.14000000000000001 (when it's easy to see that the right answer is 0.14 exactly). The first rule gives f(1) and f(2) as whole numbers (they're both 1, which is as obviously whole as a number can be) and the second rule gives any other f value by adding or subtracting f values for inputs closer to 1 and 2 than the new one we're obtaining, so we can safely induce that all f values are whole numbers, with no fractional part; which makes it silly to compute them via floating point numbers (i.e. ones with a fractional part). That those numbers in fact represent irrationals doesn't make it any nicer. So I'll be looking at how we can compute f's values precisely and ignore the short-cut via this nice piece of theory. None the less, notice that the value of f(i) grows exponentially with i's magnitude. For this page's purposes, I'm going to use python as the programming language in my examples: however, the reader is welcome to read the code as pseudo-code (and I'll restrict myself to python that reads well as such), understanding indented lines as being subordinates controlled by the last less-indented line. The key-word def begins the definition of a function. ## The really obvious way Most modern programming languages (for these purposes, ForTran isn't modern) allow one to write functions which call themselves; this is called recursion. A language can be clever enough to dodge the problem I'm about to point out, but it's instructive to see why simple-minded languages run into problems with the obvious way to implement Fibonacci's sequence. We see how it was originally specified, so we can write ourselves a function: ``` def Fibspec(i): if i == 1 or i == 2: return 1 if i > 1: return Fibspec(i−2) +Fibspec(i−1) return Fibspec(i+2) −Fibspec(i+1) ``` This simply states the rules given above: the first rule is the first line, f(1) = 1 = f(2) (the `==` operator tests for equality, as distinct from the `=` operator, assignment, which gives its left operand(s) the value(s) its right operand(s) previously had). The second line is just the second rule; and the third line is the reverse of it, needed to get answers for i < 1. Unless the programming language you're using is very clever, it's going to do an awful lot of computation, if you use a definition like this. Let me explain why. First, observe that if I ask for `Fibspec(1)` or `Fibspec(2)`, I get my answer back immediately: it's 1. Thus far, `Fibspec` is doing well. Next, if I ask for `Fibspec(3)` it'll test that 3 isn't 1 or 2, notice that 3 > 1 and proceed to work out (as its answer) `Fibspec(1) +Fibspec(2)` – each of which it works out directly; so (aside from some over-head, which I'll ignore, in how the language lets a function call itself recursively) it has to do one addition to get the answer 2. (It also has to do two subtractions; indeed, the number of subtractions involved in computing f(i), for i > 0, is exactly twice the number of additions; but this – along with the two recursive function calls per addition – just means the total work involved is a simple constant times the work for one addition; so let's stick to counting the additions.) If I ask for `Fibspec(4)`, it'll likewise do one addition to combine `Fibspec(2)` with `Fibspec(3)`; computing the latter costs it one addition, as before, so it costs two additions to get the answer `Fibspec(4)` = 3. Notice that the number of additions we've needed to do, each time, is just one less than the answer we obtained. For `Fibspec(5)`, we work out the previous, costing us 2 +1 additions, and do one more addition, getting the answer 5 for 4 additions. Again, the number of additions is only one less than the answer. In fact, if we can compute f(i) and f(i+1) at a cost of one addition less than their respective values, then computing f(1+2) costs us one addition to add them together plus f(i)−1 to compute f(i) and f(i+1)−1 to compute f(i+1); for a total of 1 +f(i)−1 +f(i+1)−1 = f(i) +f(i+1) −1 = f(i+2)−1. So the number of additions we have to do to compute `Fibspec(i)` is going to be only one less than the answer we get. It shouldn't be hard to see that similar applies for i < 1; computing `Fibspec(i)` requires slightly more (rather than less) subtractions (plus twice as many additions and recursive function calls) than the magnitude of the answer, in this case. Given that this answer grows exponentially with the magnitude of i, this means that `Fibspec` is a horribly expensive way to compute the Fibonacci numbers. ## The obviously better way So let's look for a better way to compute f(i). If you look at what we were doing before, you should be able to see that the real problem is that, when computing f(i+2), we first compute f(i), then compute f(i+1); but, in the course of doing the latter, we re-compute f(i). That's obviously dumb: if we'd just remembered it from before, we could save that extra computation. Since the same applies at each step back towards 1 and 2, we're clearly better off simply working forwards: ``` def Fibiter(i): if −1 < i < 1: return 0 old, ans = 0, 1 while i > 1: old, ans = ans, ans + old i = i − 1 while i < −1: old, ans = ans, old − ans i = i + 1 return ans ``` Note that python allows several operands on each side of assignment; `old, ans = 0, 1` sets `old` to 0 and `ans` to 1; and `old, ans = ans, ans + old` computes `old + ans` before doing the assignments, so `ans` gets the prior value of `old + ans`. Notice, also, that I've thrown in a piece of robustness for the computation: if i isn't a whole number, the function shall still return an answer (rather than, say, spinning forever in a loop that's been confused by an unexpected value); and that answer is what you would get by calling `Fibiter` with a whole number close to i. This time, we're just keeping track of two values – initially `old` = f(0) and `ans` = f(±1) – and working our way up or down to i one step at a time, doing one addition and one subtraction on each step, so it should be clear that the number of additions and subtractions we have to do is equal to the magnitude of i, give or take one. So, with a little easy thinking, we've now got a way to compute f(i) at cost proportional to i rather than proportional to f(i); given that the latter grows much faster than i, this is clearly an improvement. (Furthermore, although this needn't be true in all languages, calling functions (recursively or otherwise) is more expensive, in imperative languages, than iterating a loop; so not only do we make fewer steps of computation, but the cost of each step is reduced, too. We incur some small overhead in having two new local variables, so computing f(i) for a few small positive i will be minutely slower, but f(0) is now very cheap and all other f(i) shall be cheaper to compute.) The interested reader would do well to think for a while, before reading on, about how it might be possible to do even better than this. ## The deviously clever way Some clever person noticed (many years ago) something marvelously subtle that we can exploit. However, before I can explain how that works, I need to introduce you to the efficient way to compute powers, given a multiplication. While you can compute a power by repeated multiplication ``` def powiter(n, x): if n < 0: return powiter(−n, 1./x) ans = 1 while n > 0: ans = ans * x n = n − 1 return ans ``` (in which the number of multiplications we do (and, incidentally, the number of subtractions we do) is equal to the power we're raising to), there's a much more efficient way to do it, in which the number of multiplications we do is just the number of binary digits it takes to write out n plus the number of non-zero digits among these (so at most twice the logarithm of n to base two). Once you understand that, I'll be ready to show you how to compute Fibonacci more efficiently. Note that, if we're happy to use floating point arithmetic, we could also obtain power(n, x) as exp(n * log(x)), whose cost doesn't depend on n at all – although the costs of calling exp and log run to quite a lot of multiplications, divisions, additions and subtractions; so this only works out cheaper when n is quite big. So, for a (more-or-less) fixed large cost, we can compute power, hence f(i) via the golden ratio approach, at cost independent of i, provided we're happy with floating point. However, as noted above, I'm trying to avoid floating point – which, for large values of i, is apt to leave us having to round a big floating-point value to its nearest integer, which becomes imprecise once the output of f(i) is big enough that the floating point representation's mantissa has too few digits to represent it exactly. ### Efficient power ``` def powbits(n, x): if n < 0: return powbits(−n, 1./x) ans = 1 while n > 0: if n % 2: # n is odd ans = ans * x n = n − 1 n = n / 2 x = x * x return ans ``` I encourage the interested reader to think about why this is correct before reading on. The % operator computes a remainder: p − (p % q) is always a multiple of q; so n % 2 is one (which `if` considers to be true) when n is odd and zero (which `if` considers false) when n is even. We can use this with floating point arithmetic, just as we can for whole number arithmetic; and, for modest values of n, it'll be faster than exp(n * log(x)), though with the same precision issues as I note above. As we'll soon see, however, this isn't faster than we can compute f(n); only via exp and log can floating point win, and then only for large n, at which point the precision problems are apt to matter. So, for precise computations, what I'll show soon is in fact as fast as the floating point approach, except possibly for a bounded range of values of i that are large enough for exp(i * log(x)) to be cheaper than powbits(i, x) but small enough that f(i) is faithfully represented by a floating point mantissa. What `powbits` does is to express `n` in binary and multiply `ans` by power(b, x) for each power of two, b, that's represented by a non-zero digit of n, when written in binary. Computing the power(b, x) is done by repeatedly squaring x. To help you understand how `powbits` works, consider this recursive form of it: ``` def powrecurse(n, x): if n < 0: return powrecurse(−n, 1./x) if n % 2: return x * powrecurse((n − 1) / 2, x * x) if n > 0: return powrecurse(n / 2, x * x) return 1 ``` The first line is as before. If n is zero, we'll fall all the way through to the end and get 1, which is just power(n, x) with n = 0. If n is even and positive, we get to the last but one line: power(n, x) is, in this case, just power(n/2, xx), so that's what we return. For positive odd n, n−1 is even, so power(n−1, x) is just power((n−1)/2, xx) and multiplying it by x yields power(n, x). So I hope you can see that `powrecurse` does compute power(n, x). In a really clever programming language, `powrecurse` might even be compiled efficiently: but most imperative languages aren't that clever, so let's see why `powbits` is doing the same thing as `powerecurse`. Each time we go round the loop in `powbits` is equivalent to one layer of recursion in a call to `powrecurse`: the first time round the loop, we have our original x; if n is odd, we multiply `ans` by x and leave the subsequent iterations to multiply `ans` by pow((n−1)/2, xx); otherwise, we simply leave those subsequent iterations to multiply `ans` by pow(n/2, xx). By the time we've fallen off the end of the loop, we've worked out what we were originally asked for, but we've only gone round the loop once per binary digit of n (and we've done it without relying on our language to be clever enough to implement tail recursion with an accumulator). Before I move on, let me just throw in how I'd actually implement `powbits` in python: ``` from operator import mul, truediv def powbits(n, x, v=1): """Returns v * x*n, for integer n; v defaults to 1.""" if n < 0: n, join = −n, truediv else: join = mul while n > 0: n, b = divmod(n, 2) if b: v = join(v, x) x = x return v ``` Since we're going to want a variable in which to accumulate the answer, incrementally multiplying by powers of x, and the caller not infrequently wants to multiply the answer by something, we may as well let the caller pass that to the function, in place of the 1 we'll use by default; hence parameter `v` replaces local variable `ans`. This also makes `powbits` usable when x isn't a plain number, e.g. when it's a matrix, and doesn't support multiplication by plain numeric 1, requiring instead a suitable unit of its own kind, that we can pass in as v. In support of that, the negative power case is here handled by dividing v by x instead of by replacing x by its inverse, mediated by over-riding the `mul` function – the operator module's `mul(v, x)` just returns `v * x` but we replace this with ```truediv(v, x) = v / x``` for negative n (and `truediv` uses the division for which 1/2 is 0.5 instead of rounding to 0 because 1 and 2 were given as integers). Furthermore, this even lets us use ```powbits(n, p, q)``` with negative n to divide q repeatedly by p, even when p has no multiplicative inverse, provided q actually has p as a sufficiently repeated factor (and their type implements division to recognise that case; as, for example, `study.maths.polynomial.Polynomial` does). The function `divmod` (a built-in of the language) returns a pair of values, the quotient and remainder from performing a division; so ```n, b = divmod(n, 2)``` halves n, rounding down if n was odd, and sets b to the remainder it got when doing so. So this version does the same as the previous one, but is a little terser and more flexible … and that first line, in triple-double-quotes, is documentation – because real code should always be documented. ### Fast Fibonacci So now let's go back to looking at Fibonacci. In `Fibiter`, we have two variables, `old` and `ans`, that hold a pair of values of the Fibonacci sequence; at each iteration, in either loop, we compute new values for them from (only) their prior values. That computation is very simple: `old` gets `ans`'s prior value and `ans` gets the sum or difference of their prior values. If we used a vector, with two entries, to hold these two variables, each step of the iteration would thus be performing a simple linear transformation on the vector: this can be expressed as the contraction of a matrix with the vector. Since each iteration contracts the same matrix with the vector, overall we're simply contracting power(i) of the appropriate matrix with the start vector, [0,1]. So the loops in `Fibiter` are like the loop in `powiter`, but applying a matrix to a vector instead of multiplying numbers; and the net effect is to compute a power of a matrix and contract it with the vector. If we can compute the power of the matrix in `powbits`'s way, we can reduce the cost from of order the input to `Fibiter` to of order the number of bits needed to express it – i.e., its logarithm (to base two). Now, the matrix for our forward iteration looks like F = [[0,1],[1,1]], with the second pair as the bottom row; contracting it with [`old`,`ans`], with `ans` below `old`, gives [`ans`,`old+ans`] as the new value for [`old`,`ans`]. When we multiply F by itself repeatedly, we get F·F = [[1,1],[1,2]], F·F·F = [[1,2],[2,3]], F·F·F·F = [[2,3],[3,5]]; in fact (as the interested reader should be able to verify easily) we get power(i, F) = [[f(i−1),f(i)],[f(i),f(i+1)]]; so we can compute f(i) as the second co-ordinate of power(i−1, F)·[0,1], for positive i. Indeed, python will actually let me define a class that implements matrix multiplication such that I can simply use `powbits` to implement this, with similar for F's inverse to handle negative i. This sort of transformation, representing a function's internal state by a vector and each step of an iteration by a linear map applied to that vector, so as to replace the iteration by some more efficient computation derived from the linear map, can be applied to reasonably many contexts, making it worth checking one's code for opportunities to exploit such a transformation. Notice, however, that (in the present example) power(i, F) contains some redundancy: two of its entries are equal and one of the others is the sum of the last with one of the equal ones. We can surely do (a bit) better. Remember, earlier, that we worked out that f(i) = k.power(i, q) would satisfy f(i+2) = f(i+1) +f(i) precisely if q.q = q +1; so let's look at polynomials modulo this equation, i.e. with square = power(2) = (: x.x ←x :) treated as if it were equal to (: x+1 ←x :). This reduces the space of polynomials to a two-dimensional space; each so-reduced polynomial is simply (: A.x +B ←x :) for some A, B. If we multiply it by power(1) = (: x ←x :), it becomes (: A.x.x +B.x = (A+B).x +A ←x :), so the pair of coefficients is subjected to exactly the transformation `Fibiter` applied to `ans` and `old`, with ans as A and old as B. Thus power(0) = (: 1 ←x :) represents a start-state, slightly different from that of Fibiter, with old = 1 and ans = 0; and each successive power(i, x) reduces to f(i).x +f(i−1). Thus taking powers of x in this space of reduced polynomials is equivalent to computing Fibonacci's sequence. The reduced space of polynomials represents our iteration in a way I can easily motivate from the polynomial that appears in the analytic treatment of the series; that works generally for similar iterations. However, we're lucky in the details of how we use it to encode our iteration. In other cases, such as the Perrin iterator, the corresponding analysis can require more thought; none the less, the essential treatment of the problem is the same. Reduce polynomials modulo the one that mimics our iteration, then look at the resulting multiplication to find a way to represent tuples of iterated values and a polynomial to multipliy by to implement the iteration. When we multiply two of our reduced polynomials, the cross-terms in x.x reduce away; (A.x +B).(C.x +D) = A.C.x.x +(B.C +A.D).x +B.D = (B.C +A.D +A.C).x +(A.C +B.D). The code doesn't actually need to know we're working with polynomials; it just sees us working with pairs of numbers and using an eccentric multiplication on such pairs: [A, B].[C, D] = [A.D +A.C +B.C, A.C +B.D]. The pairs [0, 1] and [1, 0] represent power(0) and power(1), respectively; the former is our multiplicative identity and the successive powers of the latter shall give us Fibonacci's sequence. We just need to code the equivalent of `powiter` for the multiplication of our reduced polynomials. So first we define (using python tuples `(a, b)` rather than lists [A, B], simply because the language provides convenient support for these): ``` def fibtimes(p, q): (a, b), (c, d) = p, q # unpack as pairs return a * (c + d) + b * c, a * c + b * d ``` which is just our peculiar multiplication. (We could, equally, subclass the built-in `tuple` type in python, of which ```(a, b)``` and `(c, d)` are instances, and define the methods on it to support this as multiplication in the ordinary way, rather than having to call `fibtimes` explicitly; we could then use `powbits`, above, rather than a custom implementation of power.) Then we implement the power operation for this multiplication restricted to only computing powers of `(1, 0)` (i.e. power(1) or x ←x), as ``` def Fibpow(i): (a, b) = (1, 0) # x (c, d) = (0, 1) # ans while i > 0: i, n = divmod(i, 2) if n: c, d = fibtimes((a,b), (c,d)) (a, b) = fibtimes((a,b), (a,b)) return (c, d) # x*i ``` We can actually do a little better than computing `Fibpow`, extracting c from the pair it returns and munging to handle sign; since we only need one part of its return in any case, we can simply return that; and we can handle negative exponents by exploiting a multiplicative inverse for (: x ←x :), namely (: x−1 ←x :), since x.(x−1) = x.x −x which is equivalent to x+1 −x = 1. So we can use [1, −1] as (: 1/x ←x :), in place of [1, 0] as (: x ←x :), and flip the sign of i, if negative. At the same time, we can avoid squaring `(a, b)` the last time round the loop (where its value shalln't actually be used again) by moving the rest of the loop after it and unwinding a copy of it out into the preamble, where we can simplify it: ``` def Fibfast(i): if i < 0: a, b, i = 1, −1, −i else: a, b = 1, 0 i, n = divmod(i, 2) if n: c, d = a, b else: c, d = 0, 1 while i > 0: a, b = fibtimes((a,b), (a,b)) i, n = divmod(i, 2) if n: c, d = fibtimes((a,b), (c,d)) return c ``` (Alternatively, the unrolled first iteration could initialize slightly differently and record which of c, d to return at the end – details I leave as an exercise for the interested reader.) Since we've now reduced our computation to an analogue of power, the amount of computation needed to compute f(i) only grows, with i, as fast as the number of digits it takes to write i out in binary (i.e. the logarithm of i to base two), which grows much less rapidly than i. We are doing rather more computation at each of these fewer steps (rather than just one addition and one subtraction, we're doing: either four multplications, three additions and some easy binary stuff (taking remainder by two and dividing by two is really cheap); or eight multiplications, six additions and the cheap binary stuff), so the `Fibiter` solution is faster for small values of i; but the `Fibfast` solution wins for larger values of i. If you're always computing answers for small values of i, and doing so often enough that you care about performance, `Fibiter` is easily beaten by an implementation which remembers the answers it's worked out in the past; which can be just as efficiently implemented, using a variant of `Fibspec` which caches its answers, as by using `Fibiter`. In python, that would be written: ``` def Fibcache(i, cache=[0, 1]): if i < 0: if i % 2: return Fibcache(−i) return −Fibcache(−i) try: ans = cache[i] except IndexError: assert i > 1, "cache started with two entries" ans = Fibcache(i−2) +Fibcache(i−1) assert len(cache) == i cache.append(ans) # ans is now cache[i] return ans ``` (One could also chose to limit the cache size; before the `try` stanza, check for `i > BIG`, for some suitably big limit on cache size, and fall back on Fibfast(i), optionally optimising to exploit the knowledge `i > 0`.) Aside from the special handling of negative i, some of the cleverer (usually functional) programming languages might even compile their equivalents of `Fibspec` to something along these lines. On the other hand, if you were only interested in f(i) for a sparse few large i, `Fibfast` wins (strongly); and it'd be asking a lot to expect the compiler to spot that it can re-arrange `Fibspec`, or anything even remotely resembling it, into that. In fact, if you compiler is that clever, you should probably (politely) invite it to take the Turing test. ### Or in C++ One can, of course, do the same in any half-way decent programming language, so here's an illustration in C++: ```#include <stdint.h> int64_t Fibspec(int8_t i) { if (i < 0) return Fibspec(i + 2) − Fibspec(i + 1); if (i > 1) return Fibspec(i − 2) + Fibspec(i − 1); return i; } int64_t Fibiter(int8_t i) { if (i == 0) return i; int64_t old = 0, ans = 1; while (i > 1) { const int64_t next = old + ans; old = ans; ans = next; −−i; } while (i < −1) { const int64_t prev = old − ans; old = ans; ans = prev; ++i; } return ans; } template <typename F = double, typename C = unsigned> F powiter(C n, F x) { if (n < 0) return powiter(−n, F(1) / x); F ans = 1; while (n > 0) { ans = x; n −= 1; } return ans; } template <typename F = double, typename C = unsigned> F powbits(C n, F x, F v = F(1)) { if (n < 0) return powiter(−n, F(1) / x, v); while (n > 0) { if (n & 1) v = x; n >>= 1; x = x; } return v; } template <typename F = double, typename C = unsigned> F powrecurse(C n, F x, F v = F(1)) { if (n < 0) return powrecurse(−n, F(1)/x, v); if (n > 0) return powrecurse(n >> 1, x * x, n & 1 ? x * v : v); return v; } ``` Note that fib(93) will overflow int64_t, so even int8_t suffices to represent all the inputs we can sensibly support, in the absence of big-number support such as python provides by default. The pair-arithmetic begs to be rewritten in class-structured form as: ```class FibPair { int64_t lin, con; FibPair(int64_t n, int64_t c) : lin(n), con(c) {} FibPair(const FibPair &) = default; FibPair &operator=(const FibPair &) = default; FibPair operator(FibPair other) const { return FibPair(lin (other.lin + other.con) + con * other.lin, lin * other.lin + con * other.con); } static FibPair power(unsigned n, FibPair x, FibPair v) { if (n == 0) return v; return power(n >> 1, x * x, n & 1 ? x * v : v); } public: static int64_t FibFast(int8_t n) { FibPair x = n < 0 ? FibPair(1, −1) : FibPair(1, 0); return power(n < 0 ? −n : n, x, n & 1 ? x : FibPair(0, 1)).lin; } }; ``` providing FibPair::FibFast as the equivalent of Fibfast. It's also possible to write an equivalent of Fibcache, using a std::vector as cache in place of its list. Finally, here's an implementation of the floating-point solution, whose computational cost (nominally) doesn't increase with the input: ```#include <cmath> int64_t Fibfloat(int8_t n) { constexpr double lnRoot5 = log(5) / 2; constexpr double lnGolden = log(.5 + sqrt(1.25)); const double lnRaised = n * lnGolden; const double p = exp(lnRaised − lnRoot5); const double q = exp(−lnRaised − lnRoot5); return llround(n & 1 ? p + q : p − q); } ``` ## Golden Ratio The sum of powers formula for the Fibonacci sequence used powers of the golden ratio, (1 +√5)/2, the positive root of q.q = q +1; this shows up in diverse other places (e.g. some artists like the sides of their canvas in this proportion). That equation can be rearranged as q = 1 +1/q, which leads to the golden ratio having the continued fraction representation • 1 +1/(1 + 1/(1 + 1/(1 + 1/(…)))) If you truncate this (skip the 1/(…) at some depth) and expand it out, you start at the bottom with 1/1 and, at each step, invert the fraction and add one; if the fraction is a/b that gives you 1 +b/a = (a+b)/a which, inexorably, turns our 1/1 into 2/1, 3/2, 5/3, 5/8, 13/8 and generally f(i+1)/f(i) for successive i, yielding 1 +f(i)/f(i+1) = f(i+2)/f(i+1) at the next step. So truncating the continued fraction gives a fraction that's a ratio of successive terms of our series; however, there remains the question of whether that's actually any good as an approximation to the golden ratio. The golden ratio is between 1 and 2; so successive powers of it get bigger. The other term in our sum for f(n) is a power of (1 −√5)/2; since 5 lies between 4 and 9, √5 lies between 2 and 3, so 1 −√5 lies between −2 and −1; half of this lies between −1 and −1/2, so successive powers of it (alternate in sign and) get smaller. Consequently, for large n, f(n) is dominated by power(n) of the golden ratio and the ratio of successive terms in the sequence, f(n+1)/f(n), approximates the golden ratio, doing so better for larger n. Thus, indeed, truncating our continued fraction representation of the golden ratio does give us a good approximation, better the deeper we go before truncating. That ratio, furthermore, is always in coprime form: f(n+1) and f(n) have no common proper factor. This is easy to see if you simply run Euclid's algorithm to discover their highest common factor: the algorithm, at each step, reduces the larger value modulo the smaller and then repeats itself with the reduced value and the former smaller, which is now the larger value. For n>0, f(n)>0; thus, for n>1, f(1+n) = f(n) +f(n−1) > f(n); and, for n > 2, 2.f(n) = f(n+1) +f(n)− f(n−1) > f(n+1); so, for n > 2, f(n) < f(n+1) < 2.f(n) and f(n+1) reduced mod f(n) is just f(n+1) −f(n) = f(n−1). Thus each step of Euclid's algorithm just winds back down Fibonacci's sequence, replacing f(n+1) and f(n) with f(n) and f(n−1) until we get to f(2) = 1 and f(1) = 1, whose highest common factor is 1; from which, as Euclid's algorithm perserves the highest common factor of its pair of numbers, we can infer that f(n+1) and f(n) have 1 as highest common factor and thus are coprime. This leads to successive entries in Fibonacci's sequence making good integers to use for a rectangle whose sides have to be whole numbers, if we want the aesthetically pleasing proportions of a golden rectangle. (This has ratio of sides equal to the golden ratio: cutting from it a square on its shorter side leaves a smaller rectangle in the same proportions; adding to it a square on its longer side yields another.) I use this, for example, in the default size of my grid for Conway's game of life. Since that also uses the Klein bottle's topology by default, with the shorter axis simply cycling and the longer one flipping the shorter each time round, a glider travelling round it will, by the time it's traversed the longer axis twice, be travelling parallel to how it started out and offset by 2.f(n+1) modulo f(n). Since 2.f(n+1) = 2.f(n) +2.f(n−1) = 2.f(n) +f(n−1) +f(n−2) +f(n−3) = 3.f(n) +f(n−3); so 2.f(n+1) modulo f(n) is just f(n−3). So, modulo f(n), we now know f(n+1) = f(n−1) = −f(n−2) and 2.f(n+1) = f(n−3). Let's look next at 3.f(n+1) = f(n−1) +f(n−3) = f(n−3) −f(n−2) = −f(n−4). Then 4.f(n+1) = 2.f(n−3) = f(n−2) +f(n−5), which I don't immediately see an easy way to reduce to a single entry ! Speaking of the sequence modulo various things, let's look at it modulo assorted early naturals (with at the point – always where a 0 would follow a 1 – where it starts repeating itself): 1. 0, 1, 1, … 2. 0, 1, 1, 2, 0, 2, 2, 1, … 3. 0, 1, 1, 2, 3, 1, … 4. 0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, … 5. 0, 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4, 3, 1, 4, 5, 3, 2, 5, 1, … 6. 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, … 7. 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, … 8. 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, … Written by Eddy.
Comment on page # Simplified Blackjack Consider a simplified game of blackjack / 21, with an infinite deck of cards labeled 1-10. That means there is always a 10% chance to draw any card value regardless of what's been drawn before. The dealer always plays deterministically, according to the following rules: 1. 1. If the dealer's total is less than 17, the dealer draws another card. 2. 2. If the dealer's total is at least 17, but not more than 21, the dealer stands. 3. 3. If the dealer's total is over 21 the dealer has busted (lost). Write a function to determine the probability the dealer will bust (given a particular hand). ## Approach 1: Brute Force The Idea: Like with many problems, often the best way to begin is to first go some examples just to make sure you understand the problem contextually. This is especially more important with mathematical problems such as these. We know immediately that if the hand greater than 21, then the probability of busting is 100%. Similarly, if the hand is between 17 and 21, the dealer stands, so the probability of busting is 0%. Now lets deal with the case when the hand is 16 or lower. This is where need to account for having to potentially draw another card. For example, if our hand is `15`, then `4/10` cards will favor the bust, and there is a `1/10` chance we will draw a `16`. In the case that we do, it will follow that we then have a `5/10` chance of then leading into a bust. Since both events have to happen and each is independent, the chances of it happening is `1/105/10`. In total we then have `4/10 + 5/100` in favor of a bust. Notice how the entire tree sums to 1: `P(bust0) + P(not bust) + P(bust1) = 4/10 + 5/10 + 10/100 = 1` As another example, consider if the hand is 14. Now there are 3/10 cards that can immediately result in a bust. However, now there is a 2/10 chance of having to redraw, both of which have thier own probabilities associated with them. Complexity: If your hand between 22 and N, then the complexity will be O(16-N!) import queue import math def p_bust(hand): if hand > 21: return 1 elif 17 <= hand <= 21: return 0 q = queue.Queue() q.put((hand, 0)) geo_sum = 0 while not q.empty(): parent = q.get() p_hand = parent[0] p_lvl = parent[1] # how many card favor the current bust tmp = (p_hand + 10) - 21 amount_bust = tmp if tmp > 0 else 0 geo_sum += math.pow(.1, p_lvl) amount_bust/10 # how many cards do we have to draw? amount_draw = 16 - p_hand for i in range(1, amount_draw + 1): q.put((p_hand + i, p_lvl + 1)) return geo_sum for i in range(23, -1, -1): print(i, p_bust(i)) # will output: # 23 1 # 22 1 # 21 0 # 20 0 # 19 0 # 18 0 # 17 0 # 16 0.5 # 15 0.45 # 14 0.39499999999999996 # 13 0.3345 # 12 0.2679500000000001 # 11 0.19474500000000008 # 10 0.21421950000000015 # 9 0.23564145000000022 # 8 0.2592055949999989 # 7 0.28512615449999784 # 6 0.31363876994999573 # 5 0.345002646944992 # 4 0.3795029116394856 # 3 0.4174532028034251 # 2 0.4591985230837547 # 1 0.5051183753922686 # 0 0.555630212931779 ## Approach 2: Dynamic Programming Notice that ALOT of computations are reused in the tree. So an incredibly faster approach to this problem would be to literally compute all the answers, and then just statically _return them when a function call asks for it. In the image below for example, a hand of _13 can reuse just take 1/10 of tree(14). Complexity: O(1) time after preprocessing but technically since were only preprocessing numbers 1 - 16 (a constant amount), preprocessing is also finished in constant time. Constant space too. import queue import math class Solution: # static dp array __p_bust_sol = {} def p_bust(self, hand): if hand > 21: return 1 elif 17 <= hand <= 21: return 0 if Solution.__p_bust_sol: return Solution.__p_bust_sol[hand] # just compute all the answers really fast for hand in range(23, -1, -1): q = queue.Queue() q.put((hand, 0)) geo_sum = 0 while not q.empty(): parent = q.get() p_hand = parent[0] p_lvl = parent[1] if p_hand in Solution.__p_bust_sol: geo_sum += Solution.__p_bust_sol[p_hand] * .1 else: # how many card favor the current bust tmp = (p_hand + 10) - 21 amount_bust = tmp if tmp > 0 else 0 geo_sum += math.pow(.1, p_lvl) * amount_bust / 10 # how many cards do we have to draw? amount_draw = 16 - p_hand for i in range(1, amount_draw + 1): q.put((p_hand + i, p_lvl + 1)) Solution.__p_bust_sol[hand] = geo_sum return Solution.__p_bust_sol[hand] obj = Solution() for i in range(23, -1, -1): print(i, obj.p_bust(i)) # will output: # 23 1 # 22 1 # 21 0 # 20 0 # 19 0 # 18 0 # 17 0 # 16 0.5 # 15 0.45 # 14 0.39499999999999996 # 13 0.3345 # 12 0.2679500000000001 # 11 0.19474500000000008 # 10 0.21421950000000015 # 9 0.23564145000000022 # 8 0.2592055949999989 # 7 0.28512615449999784 # 6 0.31363876994999573 # 5 0.345002646944992 # 4 0.3795029116394856 # 3 0.4174532028034251 # 2 0.4591985230837547 # 1 0.5051183753922686 # 0 0.555630212931779
### KY.8.NS: The Number System #### 1.1: Know that there are numbers that are not rational and approximate them by rational numbers. KY.8.NS.1: Understand informally that every number has a decimal expansion; the rational numbers are those with decimal expansions that terminate in 0s or eventually repeat. Know that other numbers are called irrational. KY.8.NS.2: Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram and estimate the value of expressions. ### KY.8.EE: Expressions and Equations #### 2.1: Work with radicals and integer exponents. KY.8.EE.1: Know and apply the properties of integer exponents to generate equivalent numerical expressions. KY.8.EE.2: Use square root and cube root symbols to represent solutions to equations of the form x² = p and x³ = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that perfect squares and perfect cubes are rational. KY.8.EE.3: Use numbers expressed in the form of a single digit times an integer power of 10 (Scientific Notation) to estimate very large or very small quantities and express how many times larger or smaller one is than the other. KY.8.EE.4: Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities. Interpret scientific notation that has been generated by technology. #### 2.2: Understand the connections between proportional relationships, lines and linear equations. KY.8.EE.5: Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. KY.8.EE.6: Use similar triangles to explain why the slope, m, is the same between any two distinct points on a non-vertical line in the coordinate plane; know the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. #### 2.3: Analyze and solve linear equations and pairs of simultaneous linear equations. KY.8.EE.7: Solve linear equations in one variable. KY.8.EE.7.a: Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers). KY.8.EE.7.b: Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and combining like terms. KY.8.EE.8: Analyze and solve a system of two linear equations. KY.8.EE.8.a: Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously; understand that a system of two linear equations may have one solution, no solution, or infinitely many solutions. KY.8.EE.8.b: Solve systems of two linear equations in two variables algebraically by using substitution where at least one equation contains at least one variable whose coefficient is 1 and by inspection for simple cases. KY.8.EE.8.c: Solve real-world and mathematical problems leading to two linear equations in two variables. ### KY.8.F: Functions #### 3.1: Define, evaluate and compare functions. KY.8.F.1: Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. KY.8.F.2: Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). KY.8.F.3: Understand properties of linear functions. KY.8.F.3.a: Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line. KY.8.F.3.b: Identify and give examples of functions that are not linear. #### 3.2: Use functions to model relationships between quantities. KY.8.F.4: Construct a function to model a linear relationship between two quantities. KY.8.F.4.a: Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. KY.8.F.4.b: Interpret the rate of change and initial value of a linear function in terms of the situation it models and in terms of its graph or a table of values. KY.8.F.5: Use graphs to represent functions. KY.8.F.5.a: Describe qualitatively the functional relationship between two quantities by analyzing a graph. KY.8.F.5.b: Sketch a graph that exhibits the qualitative features of a function that has been described verbally. ### KY.8.G: Geometry #### 4.1: Understand congruence and similarity using physical models, transparencies, or geometry software. KY.8.G.1: Verify experimentally the properties of rotations, reflections and translations: KY.8.G.1.1: Lines are congruent to lines. KY.8.G.1.2: Line segments are congruent to line segments of the same length. KY.8.G.1.3: Angles are congruent to angles of the same measure. KY.8.G.1.4: Parallel lines are congruent to parallel lines. KY.8.G.2: Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections and translations. Given two congruent figures, describe a sequence that exhibits the congruence between them. KY.8.G.3: Describe the effect of dilations, translations, rotations and reflections on two-dimensional figures using coordinates. KY.8.G.4: Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations and dilations. Given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them. KY.8.G.5: Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal and the angle-angle criterion for similarity of triangles. #### 4.2: Understand and apply the Pythagorean Theorem. KY.8.G.6: Explain a proof of the Pythagorean Theorem and its converse. KY.8.G.7: Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions. KY.8.G.8: Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. #### 4.3: Solve real-world and mathematical problems involving volume of cylinders, cones and spheres. KY.8.G.9: Apply the formulas for the volumes and surface areas of cones, cylinders and spheres and use them to solve real-world and mathematical problems. ### KY.8.SP: Statistics and Probability #### 5.1: Investigate patterns of association in bivariate data. KY.8.SP.1: Construct and interpret scatter plots for bivariate numerical data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association and nonlinear association. KY.8.SP.2: Know that lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a line and informally assess the model fit by judging the closeness of the data points to the line. KY.8.SP.3: Use the equation of a linear model to solve problems in the context of bivariate numerical data, interpreting the slope and intercept. Correlation last revised: 9/15/2020 This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information.
# SAT Math Strategies That Work: Making Up Numbers One extremely effective strategy in tackling tough questions is making up numbers and testing each answer choice. To see how this works, here are three examples: 1. If the price of a dress is marked up by 30% and then marked down by 50%, the new price is what percent of the original price? A) 20% B) 50% C) 65% D) 75% Percent questions are often perfect candidates for this strategy. In these cases, the number $$100$$ is usually a great number to work with. So let's pretend the initial price of the dress is $100. After a 30% markup, the new price is$130. Now we mark it down by half (50%): $65. The final price is$65, which is 65% of the original price. The answer is C. Not complicated at all, right? Let's use this technique to solve a much harder problem. 2. If $$n$$ is a positive integer and $$3^n + 3^{n+1} = m$$, what is $$3^{n+2}$$ in terms of $$m$$? A) $$\dfrac{m-1}{2}$$ B) $$\dfrac{9m}{4}$$ C) $$3m$$ D) $$3m + 1$$ Even if you have no idea what's going on here, you can still answer the question by making up numbers. Let's pretend that $$n = 1$$, a nice simple number. Then, $m = 3^1 + 3^2 = 12$ Also, $3^{n+2} = 3^3 = 27$ Now all we have to do is plug our value of $$m = 12$$ into each of the answer choices to see which one gives $$27$$. The only answer choice that gives us $$27$$ is B. When using this strategy, you should ALWAYS review every answer choice, even when you think you've found the answer. The reason is that another answer choice may have also given us $$27$$. When two answer choices give the desired result, make up another number (e.g. $$n=2$$) and repeat the process with just those two answer choices. This will filter out the answer choice that just happens to work in one case but not the other. You want the answer that works in all cases. It may sound like a long process, but running through the answer choices typically doesn't take long at all, especially with a calculator. And most of the time, you won't run into a situation where you have to make up more than one number to test the answers with. 3. $f(x) = \left\|4x - 25\right\|$ For the function defined above, what is one possible value of $$b$$ for which $$f(b) \lt b$$? Ah, another perfect question for making up numbers. Solving this question "the standard way" would be a huge pain. Let's guess a bit. Let's try $$b = 2$$. $$f(2) = \left\|4(2) - 25\right\| = \left\|-17\right\| = 17$$, which is not less than $$b$$ itself. So $$b$$ cannot be 2. Let's try another value. Maybe $$b = 5$$. $$f(5) = \left\|4(5) - 25\right\| = \left\|-5\right\| = 5$$, which is not less than $$b$$ itself. So $$b$$ cannot be 5. But we're a lot closer!! When $$b = 5$$, $$f(b)$$ is also equal to $$5$$. So let's try something larger, $$b = 6$$. $$f(6) = \left\|4(6) - 25\right\| = \left\|-1\right\| = 1$$, and that works!! $$f(6) \lt 6$$. So one possible answer to this question is 6. Hopefully, you have an idea of how this works. Whenever you get stuck, it's good to have this strategy in mind. It's likely to be useful in questions dealing with: • Percent • Ratios or Proportions • Questions with variables (like in #2 and #3 above) The more you practice, the better you'll get at recognizing when this strategy is the way to go. So with that, I leave you with a bunch of practice questions suited to this technique. Try out this strategy even if you know how to solve the question mathematically! #### Practice Questions 1. If $$\left\|x\right\| = \left\|y\right\|$$ and $$x \neq y$$, which of the following must be true? 1. $$x - y = 0$$ 2. $$xy > 0$$ 3. $$x^2 = y^2$$ A) II only B) III only C) I and II only D) I, II, and III 2. If $$k > 1$$, what is the slope of the line in the $$xy$$-plane that passes through the points $$(k,k^2)$$ and $$(k^2,k^3)$$? A) $$k$$ B) $$k^2$$ C) $$k^2 + k$$ D) $$k^2 - k$$ 3. When the positive integer $$x$$ is divided by 10, the remainder is 4. When the positive integer $$y$$ is divided by 10, the remainder is 7. What is the remainder when $$x+y$$ is divided by 10? A) 1 B) 2 C) 3 D) 4 4. For nonzero numbers, $$a$$, $$b$$, $$c$$, if $$a$$ is twice $$b$$ and $$b$$ is $$\dfrac{1}{4}$$ of $$c$$, what is the ratio of $$a^2$$ to $$c^2$$? A) 1 to 2 B) 1 to 4 C) 1 to 8 D) 1 to 16 5. If $$k$$ is a two-digit number whose units digit is the same as its tens digit, which of the following statements must be true? A) $$k$$ is a multiple of 2 B) $$k$$ is a multiple of 5 C) $$k$$ is a multiple of 11 D) $$k$$ is greater than 30 6. If $$a$$ is an even integer and $$b$$ is an odd integer, which of the following is an even integer? A) $$a^2 + b^2$$ B) $$a^2 + b^3$$ C) $$a^3 + b^2$$ D) $$a^2 + ab$$ 7. In a book collection, 50% of the books have 100 pages, 40% have 150 pages, and 10% have 200 pages. What is the average (arithmetic mean) number of pages per book? A) 120 B) 125 C) 130 D) 135
# Power Series – Definition, General Form, and Examples The power series is one of the most useful types of series in mathematical analysis. We can use power series to define different transcendental functions including the exponential and trigonometric functions. Understanding the power series will also make us appreciate how we have approximated functions’ values in our calculators and computers. The power series allows us to approximate functions as the sum of the powers of the variable. We can think of power series as an infinite polynomial that leads to the approximation of common and new functions. In this article, we’ll explore the definition of the power series and learn how to define common and new functions through this expansion. We’ll also show you how to confirm whether the given power series is convergent or divergent. In order for us to do so, make sure to refresh your knowledge on the following: For now, let’s begin by understanding the unique components of a power series. ## What is a power series? The power series, centered at $c$, is a series represented by the general form shown below. \begin{aligned}\sum_{n = 0}^{\infty} a_n(x -c)^n &= a_0 + a_1(x -c) + a_2( x- c)^2 + …\end{aligned} The constants $\boldsymbol{a_n}$, where $n \geq 0$, are called the series’ coefficients and $\boldsymbol{c}$ represents the center. Notice something unique about this series as opposed to the ones we’ve learned in the past? The terms of the power series are now all in terms of $x$ when in the past, we’ve only worked with series that contains numbers. This will be the start of our understanding of series and expansions of common and new functions. Examples of Power Series Expansion\begin{aligned}\sin x &= x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \dfrac{x^7}{7!}+ …\\\cos x &= 1 – \dfrac{x^2}{2!} + \dfrac{x^4}{4!} – \dfrac{x^6}{6!}+ …\end{aligned} Here are two great examples of a power series- the power series of $\sin x$ and $\cos x$. Through this amazing series, we can now express transcendental functions such as sine and cosine functions as a series of polynomials. In the next sections, we’ll learn how to apply the power series formula and understand the process of expressing functions as a power series. ## What is the power series formula? We’ll show you two variants of the power series formula: 1) expression, when $\boldsymbol{x}$ is centered at zero and 2,) when $\boldsymbol{x}$ is centered at $\boldsymbol{c}$. Suppose that $\{a_n\}$ is a sequence, $x$ is the variable, and $c$ represents a real number. We have the following power series formula: Power series centered at the origin: \begin{aligned}\boxed{\boldsymbol{\sum_{n = 0}^{\infty} a_nx^n = a_0 + a_1x + a_2x^2 + a_3x^3+…}}\end{aligned} Power series centered at variable $\boldsymbol{c}$: \begin{aligned}\boxed{\boldsymbol{\sum_{n = 0}^{\infty} a_n(x – c)^n = a_0 + a_1(x – c) + a_2(x – c)^2 + a_3(x – c)^3+…}}\end{aligned} For a more precise definition, we establish that $x^0 = 1$ and $(x – c)^0 = 1$ even when $x = 0$ and $x = c$, respectively. The simplest example of a power series and this occurs when $x$ is centered at the origin and when $a_n = 1$. \begin{aligned}f(x) &= \sum_{n = 0}^{\infty} x^n\\&= 1 + x + x^2 + x^3 + …,\end{aligned}
# Table of 12 Table of 12 is given here to help the students quickly check the values of the 12 times value till 20. The multiplication table of 12 given below is in tabular format and an image is also provided below. Students can download the image of 12 times tables and use that as flashcards. By using the table of 12 as flashcards, students can memorise the table quickly and easily. ## Table of 12 Chart The tables for 12 can be extremely helpful as direct questions are often asked in the school exams. Apart from that, the multiplication tables can also help to quickly do mental maths which is a crucial skill for all competitive exams. ## What is 12 times table? 12 times table in maths is the multiplication table 12 that can be written using different mathematical operations such as addition and multiplication. Let’s have a look at the table given below to understand 12 table up to 10. 12 × 1 = 12 12 12 × 2 = 24 12 + 12 = 24 12 × 3 = 36 12 + 12 + 12 = 36 12 × 4 = 48 12 + 12 + 12 + 12 = 48 12 × 5 = 60 12 + 12 + 12 + 12 + 12 = 60 12 × 6 = 72 12 + 12 + 12 + 12 + 12 + 12 = 72 12 × 7 = 84 12 + 12 + 12 + 12 + 12 + 12 + 12 = 84 12 × 8 = 96 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 = 96 12 × 9 = 108 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 = 108 12 × 10 = 120 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 = 120 ## Multiplication Table of 12 Go through the first 20 multiples of 12 here. 12 × 1 = 12 12 × 2 = 24 12 × 3 = 36 12 × 4 = 48 12 × 5 = 60 12 × 6 = 72 12 × 7 = 84 12 × 8 = 96 12 × 9 = 108 12 × 10 = 120 12 × 11 = 132 12 × 12 = 144 12 × 13 = 156 12 × 14 = 168 12 × 15 = 180 12 × 16 = 192 12 × 17 = 204 12 × 18 = 216 12 × 19 = 228 12 × 20 = 240 ### Tricks to Remember Table of 12 There exist several tricks to remember 12 times table. Trick 1: To make it easy for students to learn the 12 times table, we have given here a simple trick. First, write the numbers from 1 to 12 excluding 5 and 11 in a serial order. 1 2 3 4 6 7 8 9 10 12 Now write the even numbers 2,4,6,8,0 next to the above series, repeatedly, such that: 12 24 36 48 60 72 84 96 108 120 Hence, we received the required values. Trick 2: We can generate the results of 12 times table using 10 times tables and 2 times tables as shown below: 10 times table 2 times table Addition Results of 12 table 10 × 1 = 10 2 × 1 = 2 10 + 2 12 10 × 2 = 20 2 × 2 = 4 20 + 4 24 10 × 3 = 30 2 × 3 = 6 30 + 6 36 10 × 4 = 40 2 × 4 = 8 40 + 8 48 10 × 5 = 50 2 × 5 = 10 50 + 10 60 10 × 6 = 60 2 × 6 = 12 60 + 12 72 10 × 7 = 70 2 × 7 = 14 70 + 14 84 10 × 8 = 80 2 × 8 = 16 80 + 16 96 10 × 9 = 90 2 × 9 = 18 90 + 18 108 10 × 10 = 100 2 × 10 = 20 100 + 20 120 Trick 3: Another way to get the multiples of 12 is given here with the help of 6 times table. In this case, the results of 6 table should be doubled. This can be done as follows: 6 times table Add the result for 2 times Results of 12 table 6 × 1 = 6 6 + 6 12 6 × 2 = 12 12 + 12 24 6 × 3 = 18 18 + 18 36 6 × 4 = 24 24 + 24 48 6 × 5 = 30 30 + 30 60 6 × 6 = 36 36 + 36 72 6 × 7 = 42 42 + 42 84 6 × 8 = 48 48 + 48 96 6 × 9 = 54 54 + 54 108 6 × 10 = 60 60 + 60 120 Check out the other maths tables of different numbers below and keep visiting BYJU’S for different important maths articles. Get More Maths Tables:- ### Examples on Table of 12 Example 1: Calculate 12 times 11 minus 50. Solution: Using the multiplication table of 12, 12 times 11 = 12 × 11 = 132 12 times 11 minus 50 = 12 × 11 – 50 = 132 – 50 = 80 Example 2: Find the 8th multiple of 12. Solution: 8th multiple of 12 = 8 × 12 Or 12 × 8 Using 12 times table, 12 × 8 = 96 Therefore, the 8th multiple of 12 = 96. ## Frequently Asked Questions – FAQs ### What is a 12 times table? The table of 12 is given by: 12 times 1 is 1, 12 times 2 is 24, 12 times 3 is 36, 12 times 4 is 48, 12 times 5 is 60, 12 times 6 is 72, 12 times 7 is 84, 12 times 8 is 96, 12 times 9 is 108 and 12 times 10 is 120. ### How to memorise the table of 12 easily? First, write the numbers from 1 to 12 excluding 5 and 11 in a serial order. 1 2 3 4 6 7 8 9 10 12 Now write the even numbers 2,4,6,8,0 next to the above series, in repeated order. 12, 24, 36, 48, 60, 72, 84, 96, 108, 120 Thus, we got a table of 12. ### What is 12 times of 13? 12 times of 13 is equal to 156 ### What is 15 times of 12? 15 times of 12 is equal to 180. ### How do we get a table of 12? Table of 12 is obtained by adding 12 each time. 12 12 + 12 = 24 12 + 12 + 12 = 36 12 + 12 + 12 + 12 = 48 12 + 12 + 12 + 12 + 12 = 60 ### Does the value of 12 times a number is even always? Yes, the value of 12 times a number is always even, because the unit place of 12 is an even number, i.e. 2. Thus, any number multiplied by 12 is always an even number.
## Intermediate Algebra (12th Edition) $(3+a-b)(3-a+b)$ $\bf{\text{Solution Outline:}}$ To factor the given expression, $9-a^2+2ab-b^2 ,$ group the last $3$ terms and factor the trinomial. Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the last $3$ terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 9-(a^2-2ab+b^2) .\end{array} In the trinomial expression above, $b= -2 ,\text{ and } c= 1 .$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{ -1,-1 \right\}.$ Using these two numbers, the factored form of the expression above is \begin{array}{l}\require{cancel} 9-(a-b)(a-b) \\\\ 9-(a-b)^2 .\end{array} The expressions $9$ and $(a-b)^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $9-(a-b)^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (3)^2-(a-b)^2 \\\\= [3+(a-b)][3-(a-b)] \\\\= (3+a-b)(3-a+b) .\end{array}
New version page # CSUN ME 501A - Homework Solutions Pages: 8 Documents in this Course 7 pages 3 pages 7 pages 9 pages 7 pages 7 pages 40 pages 4 pages 7 pages 6 pages 3 pages 7 pages ## This preview shows page 1-2-3 out of 8 pages. View Full Document End of preview. Want to read all 8 pages? View Full Document Unformatted text preview: College of Engineering and Computer ScienceMechanical Engineering DepartmentMechanical Engineering 501ASeminar in Engineering AnalysisFall 2004 Number 17472 Instructor: Larry Caretto September 14 Homework SolutionsPage 375, problem 11 – Find the spectrum (all the eigenvalues) and eigenvectors for the matrix, A, at the right.We use the basic equation that the eigenvalues,  of a matrix, A, can be found from the equation Det(A – I) = 0. Writing A – I and using the expression for the 3 by 3 determinant gives100640353A0)1)(4)(3()6)(0)(3()1)(5)(0()3)(4)(0()6)(5)(0()3)(0)(0()1)(4)(3(100640353)( IADetHere we see a case of two general results: (1) the determinant of a triangular matrix is the product of the terms on the diagonal; and (2) the eigenvalues of a triangular matrix are equal to the elements on the diagonal. In this case we have three unique eigenvalues, 1 = 3, 2 = 4, and3 = 1. (Note that the ordering of the eigenvalues is arbitrary; often they are numbered in rank order. Regardless of how the eigenvalues are numbered, the eigenvectors must be numbered in a consistent manner.)In solving for the eigenvectors, we recognize that it will not be possible to obtain a unique solution. The eigenvectors are determined only to within a multiplicative constant. If we denote the components of each eigenvector as x1, x2, and x3, the solution for each eigenvector is found by substituting the corresponding eigenvalues into the equation (A – I)x = 0, and solving for the corresponding xi as the components of the eigenvector. This is shown for each eigenvector in thetable below.5x2 + 3x3 = 0x2 + 6x3 = 0 -2x3 = 0For 1 = 3, we see that the solution to the third equation is x3 = 0. With this solution, the second equation gives x2 = -6x3 = 0. We are left with no equation for x1, so we conclude that this can be any value and we denote it is a. Thus gives the first eigenvector as x(1) = [a 0 0]T.-x1 + 5x2 + 3x3 = 0 6x3 = 0 -2x3 = 0For 2 = 4, the second and third equation both give x3 = 0. The first equation gives x1 = 5x2 + 3x3 = 5x2. If we pick x2 to be an arbitrary quantity, say b, then x1 = 5b and the second eigenvector becomes x(2) = [5b b 0]T.2x1 + 5x2 + 3x3 = 0 3x2 + 6x3 = 0 0 = 0For 3 = 1, ere the third equation gives us no information, but the second equation tells us that 3x2 = -6x3. If we pick x3 = c (an arbitrary quantity), then x2 = 2c and the first equation gives x1 = (-5x2 – 3x3)/2 = (-5(-2c) – 3c)/2 = -7c/2. x(3) = [7c/2 -2c c]T.Page 379, problem 5 – Find the principal axes and the corresponding expansion (or compression) factors for the elastic deformation matrix, A, shown at the right.As shown in the text, this matrix describes a stretching where every old coordinate, x, moves to a new coordinate, y., such that y = Ax.1212123AEngineering Building Room 1333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062The principal directions for this deformation are the directions for which the position vector changes only in magnitude. Thus, the new position vector, y, has the same direction as the old position vector, x. This means that the two vectors are related by a multiplicative constant. E. g., y = x. This means that we have two equations: (1) the general equation y = Ax, for any coordinate direction and (2) y = x, in the principal coordinate direction. We can only satisfy these two equations if Ax = x for the principal coordinates. This gives an eigenvalues/eigenvector problem. We first solve Det(A – I) = 0, as shown below, to get the eigenvalues. 0125211231212123)(22IADetThe eigenvalues are found by the usual formula for the solution of a quadratic equation.21,2435)1(24162525)1(2)1)(1(42525212If we denote the components of each eigenvectors as x1, and x2, the solution for each eigenvectoris found by substituting the corresponding eigenvalues into the equation (A – I)x = 0, and solving for the corresponding xi as the components of the eigenvector. Here we have only two equations to solve, and one of the unknowns will be arbitrary. For 1 = 2, the two equations are02/2/21 xxand 02/21 xx. Both these equations give us 212xx . Thus, our first eigenvector is x(1) = [2a a]T, where a is arbitrary. For 2 = ½, the two equations are02/21 xxand 02/2/21 xx. Both these equations give us 122xx so that we can write our second eigenvector as x(2) = [b -2b]T.If we regard the x1 coordinate as lying along a horizontal axis and the x2 coordinate as lying alonga vertical axis, we can find the principal coordinate directions corresponding to the two eigenvalues as follows.oaaxx26.3561548.02tantan11)1(2)1(11 for 1 = 2obbxx74.54955317.02tantan11)2(2)2(12 for 2 = 1/2Here we find that the first principal direction is 35.26o with an expansion by a factor of 2; the second principal direction is –54.74o with a contraction by a factor of ½. Note that the numbering of “first” and “second” are arbitrary. Also, we see that when we are interested in only the direction of a vector, having a common factor in all components does not affect the result.Page 384, problem 4 – Is the matrix at the right symmetric, skew-symmetric or orthogonal? Find the eigenvalues of thematrix and show how this illustrates Theorems 1 on page 382 or Theorem 5 on page 384.)cos()sin(0)sin()cos(0001AThe matrix is not skew-symmetric since it has nonzero components on its principal diagonal. For a skew-symmetric matrix (A = –AT), all terms on the principal diagonal must be zero.Engineering Building Room 1333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062The matrix is almost symmetric, but a23 = -sin()  a32 = sin(). Thus the matrix is not symmetric.If the matrix is orthogonal, the inner product of each pair of rows must vanish. View Full Document Unlocking...
0 ## Age Guessing: Looking at the Roots Published on Sunday, April 29, 2012 in , , , , In the past 2 posts, we've looked at the precision of the purely mathematical approach to age guessing, and the skillful approach of estimating age by appearances. In this post, you'll learn an approach that seems to be a math trick, yet seems impossible to explain in that way. You starting by asking a spectator to put any 5-, 6-, or 7-digit number in their calculator. You then tell them to multiply that number by 9. The last steps are to add their age to that number, and then show you the resulting number. You examine the number for a few seconds, and instantly announce their age! ### Why this is deceptive Let's say you perform this, and the resulting number on the calculator is 1,248,695, and you announce that the person's age is 26, which is confirmed by them. Mathematically, all they have is the formula 9x + y = 1,248,695. With two variables, that equation (known to mathematicians as a Diophantine equation) has an infinite number of solutions. How is it possible that you could narrow down the possibilities so quickly, and in your head? ### How it works When you're shown the total, you first add the digits of the answer up in your head. Next, using the age-guessing skills you learned in the previous post, ask yourself if the person could be that age. If the person seems older than that, add 9 to the number you got and ask if that seems to be a more reasonable age. If that doesn't seem right, move up or down in 9-year increments, and keep doing that until you find an age that seems right. In our example, you'd see the answer 1,248,695, so you add 1+2+4+8+6+9+5=35. Ask yourself if the person could reasonably be 35. Let's say they look younger than that, so you subtract 9. 35 - 9 = 26, so you consider 26, which we'll say seems more reasonable, so you guess that number out loud. ### Why it works You start with a long random number, and then multiply it by 9. What happens when you multiply any number by 9? Square One TV's Nine, Nine, Nine song explains: When you sum up the digits, the result is known as a digit sum. The digit sum of 99 is 18 because 9+9=18. In the video above, notice they keep repeating the process of taking the digit sum until they get a 1-digit number. If you do this, the 1-digit number you get is called the digital root. The digital root of 99 is 9 because 9+9=18, and 1+8=9. The point of the above video, of course, is that any number multiplied by 9 will have a digital root of 9. What happens when you add a number to a multiple of 9? Let's take 5 as an example. 9+5=14, and the digital root of 14 is 5 (1+4=5). 18+5=23, and the digital root of 23 is 5 (2+3=5), and so on. Let's 18+14, which is a multiple of 9 plus a number with the digital root of 5. 18+14=32, and 32's digital root is 5! Also notice that the answers remain spaced by multiples of 9: 5, 14, 23, 32, and so on. In short, any time you add a number to a multiple of 9, the answer will always have the same digital root as the number you added, and you'll always be a multiple of 9 away from another number with the same digital root. Applying this to the trick, when you multiply by 9 and add the age, the digit sum (1+2+4+8+6+9+5=35 in our above example) will not necessarily be their age, but will have the same digital root as their age, and be some multiple of 9 away from the correct age (even if that multiple is 0). Try this out for yourself. Get a calculator, enter any 5-, 6-, or 7-digit number, multiply that by 9, then add your age. Take the result, and enter it into the widget below, then click Submit. A window will pop up showing all the possible ages (listed as the variable a) between 0 and 100 you could be, based on the number you entered. ### Sneakier ways of getting to a multiple of 9 If someone is familiar with the effects of multiplying by 9, they might suspect what you're doing. There are other less obvious ways of getting to a multiple of 9: • From a sidebar in in Karl J. Smith's Nature of Mathematics (available at Amazon.com): Mix up the serial number on a dollar bill. You now have two numbers, the original serial number and the mixed-up one. Subtract the smaller from the larger. Assuming you didn't create two identical numbers, the result will have a digital root of 9, because you're subtracting 2 numbers with identical digital roots (More about this principle here). • Also from the same sidebar in in Karl J. Smith's Nature of Mathematics: Using a calculator keyboard or push-button phone, choose any 3-digit column, row, or diagonal, and arrange these digits in any order. Multiply this number by another [3-digit] row, column, or diagonal. As it happens, most numeric keypads are arranged in such a way that any row, column, or diagonal of the numbers 1-9 will make a multiple of 3. Multiplying two multiples of 3 together will always result in a multiple of 9. • You could also adapt Scam School's first Pi Day Magic Trick (YouTube link). Have them multiply 1-digit numbers together as shown in the video, until you get to a number somewhere between 1 million and 1 billion. Instead of having them remove a digit as in the original routine, however, have them add their age instead. As you see in the video, though, it is possible to get a number like 8,100,000,000. Adding their age to that would be obvious (assuming the guy in the video is 22, he'd get 8,100,000,022). To prevent this, tell them to avoid pressing the 5 and 0 keys, as this will just result in a lot of zeros at the end (or just one in the case of multiplying by 0). These aren't the only secret ways to get to a multiple of 9, but are varied and interesting enough to get you started. If you'd like an age-guessing routine that has the precision of math, but without the appearance of math (or even use of a calculator), I think you'll enjoy the next post, which will be the final installment in our series on how to guess people's ages. 0 ## Age Guessing: Judging Appearances Published on Thursday, April 26, 2012 in , , When it comes to age guessing, very few people think of the calculator feats such as the one in the previous post. The first thing that usually comes to mind is carnival age-guessers. In this post, we'll take a closer look at age-guessing as a skill. The best tips I've found on determining someone's age are in the article How to Guess Ages More Accurately. Since men tend to put less effort into hiding their age than women, here are a few extra tips on guessing a man's age. Just knowing these tips isn't of much good without practice. Thankfully, there are several sites where you can practice guessing the age of random people: How Old Are You? Guess my Age Match>Age Even though carnival age-guessers aren't having you put any numbers in a calculator, they're still able to use some very subtle math tricks. For example, instead of advertising that they'll hit your exact age, you'll usually see a margin of error such as, “I'll guess your age within 3 years!” That sounds quite close to most people. If you think about it, however, a 3-year margin of error really isn't that close. If someone is 35, a guess of anywhere from 32 to 38 would be considered correct. In other words, all they have to do is be within the decade you were born in, and they'll be considered correct. The more experience they have, the smaller margin of error they can offer. For example, professional age-guesser Lee Bennett used an impressive 1 year margin of error. Even more central to an age-guesser's actual purpose is the simple economics of the situation. Let's assume that the cost to have the carny make a guess is \$3, and the cost per stuffed animal to the carnival is \$.25 (since they buy them in bulk). If we assume the guess is wrong every time, perhaps to keep every customer flattered, they're making an 1100% profit on each prize! As the guesser becomes more skillful, the profit margin goes up! If we assume the age-guesser can correctly guess the ages of 4 out of 5 people (an 80% success rate), then that's 5 people times \$3/person or \$15 they're taking in. Only 1 wrong guess out of those 5 means that they're giving up \$.25 for every \$15 they take in, a staggering 5900% profit margin! So, when it comes down to it, age-guessing as a skill is all about the margin of error and the profit margin. And that's assuming they don't employ standard scams like writing two ages and then covering up the one that's farther away, using magician's techniques to write down a close answer after you state your age, or simply pickpocketing your wallet and looking at your ID. Guessing ages is a skill, but only ever an approximate one at best. The mathematical approaches, as we've seen, offer precision. Perhaps the best approach is to develop the skill of age-guessing, and use math in a way that doesn't detract from the skill. That's the approach we'll start developing in the next post in this series. 0 ## Age Guessing: Pure Math Published on Sunday, April 22, 2012 in , , , , , , , Back in 2008, I wrote a post about guessing ages. Unfortunately, it was several approaches compacted into one long post and lacked clarity, as a few readers have noted. I've decided it's time to update the post. I'll break age-guessing up across several posts in an effort to improve the clarity, as well. In this post, I'll start with the methods for finding someone's age using purely mathematical methods. The first type of mathematical age trick that usually comes to mind is the algebraic type, such as the kind listed on this page under “Guess Your Age.” In the first, you have the person put their age in a calculator, triple it, add 1, triple it again, add their age again, and then show you the result. While the process of performing (((x * 3) + 1) * 3) + x looks complicated, it's just a long way around of getting them to multiply their age by 10 and add 3 (See the alternate forms section). The second trick, in which they multiply their age by 7 and then by 1,443, isn't so much mysterious as it is surprising and amusing. 7 * 1,443 = 10101, so any 2-digit number multiplied by that is of course going to repeat itself 3 times. In the original Age Guessing post, I also linked to this age plus a secret number approach, which explains it's own algebra, and these two algebraic approaches, one of which breaks up the age into two different numbers, and the other that makes use of the year the person was born. These types of tricks can be very impressive for an audience unfamiliar with the basic concept of algebra, and can also be a great way to introduce new students to algebra. Anyone beyond that stage, even if they can't work it out at the moment, will recognize that there's some simple pattern that will get you the answer. Since this is the case, perhaps there's a mathematical approach that is more deceptive. A deceptive approach that's long been a favorite of magicians is one known as the Age Cards. You can find an interactive version of it at this link. Look for your age in each group. If you see your age in a given group, click the checkbox for that group. Once you've checked all six groups for your age, and clicked where appropriate, click on the CALC button. The computer will tell you your age! It works simply by adding up the smallest number (the one on the upper left corner) on each card on which the age was seen. If your age was 27, you would only click the boxes of Group One (smallest number is 1), Group Two (2), Group Four (8), and Group Five (16). Adding 16 + 8 + 2 + 1 gives 27, so the chosen age is 27. See if you can follow how the secret number of 38 is determined in this video: That's how the trick is done, but why does it work that way? The method here is better hidden than the algebraic methods because instead of using our usual base 10 numbering system, which uses the digits 0 through 9, the Age Cards trick is based on the base 2 numbering system, better known as binary, which only uses the digits 0 and 1. Working with a different number base can seem scary and confusing, but BetterExplained points out that you work with different number bases more than you might think. Even though binary is limited to using 0 and 1, it can represent any number our more familiar 0 through 9 system can. The PDF and the first video on Computer Science Unplugged's binary numbers page explain how clearly and quickly. The number 27, for example, converted to binary becomes 11011. In base 10, we only need 2 places (2 tens and 7 ones) to represent the number, but in binary, we need 5 places to represent the same number (1 sixteen, 1 eight, 0 fours, 1 two, and 1 one). How does this all relate to the Age Cards? Note that there were six Age Cards used. Each card acts like one of the places in the binary number. Note that the smallest number on each card corresponds to one of the binary places, as well: 32, 16, 8, 4, 2, and 1. To find out where a given number goes, we use it's binary code. As mentioned, 27 converts to 11011. We're working with 6 cards, though, so just like our regular base 10 numbering system, we can add zeroes to the left side without changing the value. Doing this, 11011 becomes 011011. The rightmost spot in binary is the 1s spot, and if there's a 1 there, as there is in our 27 example, we put that number on the 1 card. There's a 1 in the twos place, so we also put 27 on the 2 card. There's a 0 in the 4s place, so we don't put 27 on the 4s card. The 1 in the 8s place and the 16s place indicate that the 8 and 16 cards do have 27 put on them. Finally, the leftmost 0 in the 32s place tells us not to put 27 on the 32 card. In the video above, 38 only appeared on the 32 card, the 4 card and the 2 card because 38 in binary is 100110, which only has 1s in the 32s place, the 4s place, and the 2s place. Get the idea? The Age Cards is well-known among magicians, so even this routine could benefit from a better disguise. Fortunately, Werner Miller has come up with some very creative work on the Age Cards! First, there's his ingenious Age Cube, which is presented as a giveaway with five magic squares on it. You ask someone who is 31 or younger (because we're only working with 4 binary places) on which magic squares they see their age, and thanks to your secret addition of the numbers in the upper left corner of each magic square, you can magically divine their age! His other approach comes as a webapp that works in any modern browser, and also as a Windows executable file. It's called Age Square, and builds impressive from the Age Cube. It only uses 4 binary places, but thanks to a secret better described in the original Age Square post, it still manages to cover ages from 30 to 85! Instead of giving the age directly as an answer, the app generates a new magic square, with their age as the total. Divining someone's age purely using math can be interesting, but what about getting someone's age with some help from their appearance? That will be the topic of the next post in this series. 1 Published on Thursday, April 19, 2012 in , , , , , , , I teach quite a few fun mental challenges over in the Mental Gym. While I teach methods in as simple and straightforward a manner as possible, there isn't always just one approach. In this post, I'll take a look at new approaches to feats in the Mental Gym. In my tutorial on Squaring 2-Digit Numbers Mentally, I already teach two methods - a mathematical approach, and Jim Wilder's pure memory approach. NumberSense's approach takes advantage of an algebraic pattern. The number is separated into 2 variables, a being the 10s digit and b being the 1s digit. The problem then becomes (a + b)2, whose expanded form shows how to make the problem easier: Besides making the squaring of two digit numbers easier, this video also illustrates a good point about algebra. Algebra lets you see patterns of which you may not have been previously aware, and help you see a shorter, and possibly better approach. Another mathematical challenge I tried to simplify over in the Mental Gym was the unit circle and its associated trigonometric functions. These lessons are especially handy for students taking trigonometry. Here's a handy approach to memorizing the unit circle, especially useful for tests, that works solely by taking advantage of several simple patterns: We'll wind up this post by focusing on two of the puzzles. First, there's the Sudoku. I already link to instructions on Sudoku strategy, but if you find those hard to understand, e-How has a series of excellent instructional videos on the Sudoku-solving techniques that you may find helpful. In the Towers of Hanoi, the seemingly-simple task of moving disks from 1 peg to another quickly gets complicated. Here's a short, direct tutorial that helps make the solving pattern much clearer: If you've come across an alternative way of doing any of the feats over in the Mental Gym, I'd love to hear about it in the comments! 0 ## More Quick Snippets Published on Sunday, April 15, 2012 in , , , , , , , , , , April's snippets want to run free. They range from our usual topics like math and memory, to games, and even a little law and politics! • If you enjoyed my previous post on Notakto, but you can't play the iPad app, Thane Plambeck has an online version you can play. Like the app, you start with one board, and work your way up to 5-board play. You can only move to the next level after winning 3 consecutive games on your current level. • Speaking of strategy games, I've come across some new work on a classic. Ever play Hangman, and always use the same letters to guess in the same order, such as E-A-T-O-N? There is a better Hangman strategy, described over at DataGenetics. Instead of just giving you a new strategy, though, they go the extra step and explain the detailed thinking behind it, so you can understand it more completely. • The Major System is a great technique for memorizing numbers, but can be challenging to learn. Over at the memory basics page, I've just added a few new resources that may help those who want to learn it. First, I added the Great Courses free video How To Memorize Numbers, a free lecture video from their Secrets of Mental Math course which I originally mentioned in October's snippets. Over at Math Dude :: Quick & Dirty Tips, they also have an excellent series of 3 podcasts that teach the Major System. Part 1, Part 2, and Part 3 are available on their web site, as well as from iTunes (episodes 92 through 94). Vi Hart fans may remember her video Oh No, Pi Politics Again, about someone who claimed to have copyrighted music based on Pi. Writing music based on Pi is hardly a new and original idea, but the copyright claim was used to shut down others' videos anyway. Judge Michael H. Simon was the Oregon judge who presided over this case. Read the article Can you copyright music of pi? Judge says no to learn more about this decision, and exactly why the claim was denied. • If you'll forgive me, I'm going to wind this post up with a little boasting. Back in January, I released Day One, my approach to speeding up and presenting the classic Day of the Week For Any Date feat. I updated it in February to include some unusual calendar-related bonus feats, as well. I'm proud to announce that, according to Lybrary.com's hot list, Day One is currently their 3rd best-selling magic item at this writing! The response has been simply incredible, and I'd like to thank everyone who bought it and who helped make this possible. 1 ## Secrets of Nim (Notakto) Published on Thursday, April 12, 2012 in , , , , , Back in 2010, Backgammon giant Bob Koca was playing tic-tac-toe with his 5-year-old nephew, when the nephew whimsically suggested that they both play as X. Being a mathematics professor, he used his knowledge to analyze this weird version of the classic game with various rules, boards, and objectives. It turns out that this all-Xs version of tic-tac-toe is a version of our old friend Nim! To keep the game familiar, I'll stick to the standard 3-by-3 board in this post. The rules are as follows: • Players alternate taking turns, and neither player may pass on their turn. • A player marks any empty space on the board with an X on their turn. • The loser is the first person to mark an X on the board that completes a horizontal, vertical, or diagonal line of 3 Xs. This game is known to mathematicians as neutral or impartial tic-tac-toe, but I prefer the name given to it by Thane Plambeck, who lectured on this game at G4G10: Notakto (pronounced “No Tac Toe”). As I mentioned, this is variation of Nim, more specifically a MisÃ¨re version, so there must be some way to win it. I'll start, however, by explaining how to lose the game, instead: ### What YOU Should NOT Do You should start by going first, but the worst possible opening move is to place your X on any of the edge or corner squares. Why? Because your opponent can basically mirror your moves, and this strategy will ensure that you must eventually make a line of 3 Xs, as shown in the following animation: As you can see at the end of the animation, when the first player puts their X on an edge or corner square, and the second player mirrors them, this leaves an open diagonal line on the first player's turn that forces them to complete a line. I mention this strategy mainly so you can be aware of it, and make sure that it doesn't happen to you inadvertently. Should you let the other player go first and they place their first X on an edge or corner square, knowing about this becomes a winning strategy for you! ### How To Win To assure yourself the win, you start by placing your X in the center square. To play from there, Timothy Chow discovered the answer comes with help from a chess knight! Knowing how a chess knight moves (2 squares horizontally and 1 vertically, OR 2 squares vertically and 1 horizontally) is all you need to win. After the other player makes their move, mark your next X a knight's move away from where their previous move. Keep using this strategy and they'll always be forced to draw the losing X. Watch the following animation carefully, and you'll get the general idea: When choosing your spaces using the knight's move strategy, you'll usually have more than one space that qualifies. Often, one of the spaces will complete a line of 3 Xs, while the other is safe, so you'll always want to double check that you don't inadvertently make a losing play when you don't have to. You can find out more about the game from Bob Koca's original discussion or the MathOverflow discussion. For a deeper look at the mathematics of Notakto, you can also read Thane Plambeck's presentation in PDF form. If you'd like to practice this and you have an iPad, Thane Plambeck has also developed a Notakto app which will let you practice this version, as well as more difficult versions! There's a closely related game taught on Scam School, called Napkin Chess, which is won using a similar symmetrical strategy. It's interesting to see the similarities, even though it doesn't have a tic-tac-toe board's discrete spaces. 0 ## Hunting the Elements Published on Sunday, April 08, 2012 in , , , , , , I've posted about memorizing the periodic table of the elements before, but understanding is just as important. You might think trying to understand the basics of the elements would be a chore, but it can actually be quite fun. Surprisingly, one of the best introductions to the atom I've ever seen is not from a documentary, but an episode of WKRP in Cincinnati. In this episode, Venus is trying to help a friend whose son has dropped out of school. In the following scene, Venus explains the basics of the atom in an effort to help get the son to go back to school: Earlier this week, NOVA aired a special called Hunting the Elements. The full special is about 2 hours long, and I recommend you make time to watch the entire thing. Below are two short excerpts from that special, both roughly 8 minutes long. This first one discusses why the periodic table is arranged the way it is: This second excerpt talks about the characteristics of the atom that gives each element gets its particular properties: For more direct learning, NOVA has provided some wonderful teaching tools, such as their Name That Element Quiz. If you have an iPad, check out the NOVA Elements app (iTunes link). It not only includes the entire special, but also lets you play around with the elements by building atoms, putting them together in compounds, and much more! Should you want to learn specific information about a given element, there's a great site called the Periodic Table of Videos. The periodic table on their homepage links to videos about the corresponding element. These videos are also available on their YouTube channel. Of course, one of the things for which Grey Matters is known is teaching how to memorize just about anything. If you've been inspired to try and memorize the periodic table, check out my 2008 Elementary post. (Being 4 years old, some of the links are no longer available, but most of them are still functional.) 2 ## Gathering For Gardner 10 Published on Thursday, April 05, 2012 in , , , , Just last week, there was a gathering honoring the late Martin Gardner in Atlanta, called Gathering For Gardner 10, or G4G10 for short. I didn't go myself, but the people who did attend have already started sharing their experiences with us. If you're not familiar with Martin Gardner, you can see posts relating to his work right here on Grey Matters. David Suzuki, in his documentary series Nature of Things, spent one entire program on Martin Gardner, and introduces it at an early Gathering for Gardner event: The G4Gs are invitation-only events, and there's not much available from G4G10, the most recent get-together. There are, however, a few goodies already online. Over on flickr, there are already many photos from G4G10 posted. Even if you don't understand the subjects of the photos themselves, they're still wondrous and amazing to behold. One of the biggest treats from G4G10, however, has to be Colm Mulcahy's library lecture about the life and work of Martin Gardner. Regular Grey Matters readers will probably recognize Colm from his Card Colm column and his Colm's Cards page. Here's part 2 of the lecture: The searchable collection of Gardner's work, which Colm mentions in the lecture, is Martin Gardner's Mathematical Games CD-ROM, and is currently available for around \$40 at Amazon.com. If you attended G4G10, or even just have any personal stories to share about Martin Gardner's influence, I'd love to hear about it in the comments. 0 ## National Poetry Month Published on Sunday, April 01, 2012 in , , , , , , In the US and Canada, April is national poetry month! (Sorry, Great Britain, you'll have to wait until October.) Since memory is one of my favorite topics, I'll take a look at memorizing poetry in this post. To most people, memorizing poetry sounds like something out of 19th century schoolhouses or 1960s beatnik coffee shops. The truth is, there are plenty of good reasons to learn to memorize poetry, especially if it's something you want to do, as opposed to something you're being forced to do. In Five Benefits of Memorizing Poems there are the usual education reasons. If that's not enough, Ten Reasons You Should Memorize Poetry expands on this, including some reasons that are right down my alley, including: 1) It is a brain challenge. Got a kid with a strong memory? I’ve got some long poems for you. Interested in history? Learn a poem based on a historical event or some of the poetry of that period. For anyone seeking a way to challenge a gifted child in way that is free (!) and virtually unlimited, you’ve found it. Even copying poems down (or lines of poems) and illustrating them is a wonderful activity for younger children. ... 8) It’s a great party trick. If you’re ever stuck for a spur of the moment talent, you’re in luck if you’ve got a poem in your mind you can whip out and recite from memory. It’s easy, it needs no props, and you will not be doing the same tired trick as everyone else. Unless they read this blog. Some of the other reasons might not seem as impressive, such as the entries about keeping us connected and being a bridge among disciplines. If you take those lightly, check out Be a Man. Read a Poem. from the Art of Manliness site. Once you appreciate the benefits, how do you go about doing the actual memorization? I've written quite a bit about memorizing poetry in past posts, but there are many more approaches. New technologies make it easier to memorize than ever before. In Essay on memorizing poetry - at the gym talks about using crib sheets while exercising, although these crib sheets could be recording or videos on a mobile device of poems you wish to learn, as well. Mensa For Kids' A Year of Living Poetically lessons are a good selection with a great structure. The poem is presented, broken down, and once the poem is memorized, there are varying types of quizzes to test your knowledge. A more adult version of this same approach is used in Shmoop.com's poetry section. For example, their guide to Poe's The Raven includes not just the poem text, but an intro, a summary, an analysis, a quiz and much more! Their poetry section also has plenty of classic choices, and is a great place to look for material. Another good source is the book Committed to Memory: 100 Best Poems to Memorize. You can even find the full intro and a majority of the selections from this book at poets.org. Remember, memorizing poetry should be fun. Looking for a fun short piece to memorize right away? How about this ironic choice, titled Forgetfulness by Billy Collins:
Understanding Linear Algebra Section3.5Subspaces In this chapter, we have been looking at bases for $$\real^p\text{,}$$ sets of vectors that are linearly independent and span $$\real^p\text{.}$$ Frequently, however, we focus on only a subset of $$\real^p\text{.}$$ In particular, if we are given an $$m\times n$$ matrix $$A\text{,}$$ we have been interested in both the span of the columns of $$A$$ and the solution space to the homogeneous equation $$A\xvec = \zerovec\text{.}$$ In this section, we will expand the concept of basis to describe sets like these. Preview Activity3.5.1. Let’s consider the following matrix $$A$$ and its reduced row echelon form. \begin{equation} A = \left[\begin{array}{rrrr} 2 \amp -1 \amp 2 \amp 3 \\ 1 \amp 0 \amp 0 \amp 2 \\ -2 \amp 2 \amp -4 \amp -2 \\ \end{array}\right] \sim \left[\begin{array}{rrrr} 1 \amp 0 \amp 0 \amp 2 \\ 0 \amp 1 \amp -2 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation} 1. Are the columns of $$A$$ linearly independent? Is the span of the columns $$\real^3\text{?}$$ 2. Give a parametric description of the solution space to the homogeneous equation $$A\xvec = \zerovec\text{.}$$ 3. Explain how this parametric description produces two vectors $$\wvec_1$$ and $$\wvec_2$$ whose span is the solution space to the equation $$A\xvec = \zerovec\text{.}$$ 4. What can you say about the linear independence of the set of vectors $$\wvec_1$$ and $$\wvec_2\text{?}$$ 5. Let’s denote the columns of $$A$$ as $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ $$\vvec_3\text{,}$$ and $$\vvec_4\text{.}$$ Explain why $$\vvec_3$$ and $$\vvec_4$$ can be written as linear combinations of $$\vvec_1$$ and $$\vvec_2\text{.}$$ 6. Explain why $$\vvec_1$$ and $$\vvec_2$$ are linearly independent and \begin{equation} \laspan{\vvec_1,\vvec_2} = \laspan{\vvec_1, \vvec_2, \vvec_3, \vvec_4}. \end{equation} Subsection3.5.1Subspaces Our goal is to develop a common framework for describing subsets like the span of the columns of a matrix and the solution space to a homogeneous equation. That leads us to the following definition. Definition3.5.1. A subspace of $$\real^p$$ is a subset of $$\real^p$$ that is the span of a set of vectors. Since we have explored the concept of span in some detail, this definition just gives us a new word to describe something familiar. Let’s look at some examples. Example3.5.2.Subspaces of $$\real^3$$. In Activity 2.3.3 and the following discussion, we looked at subspaces in $$\real^3$$ without explicitly using that language. Let’s recall some of those examples. • Suppose we have a single nonzero vector $$\vvec\text{.}$$ The span of $$\vvec$$ is a subspace, which we’ll write as $$S = \laspan{\vvec}\text{.}$$ As we have seen, the span of a single vector consists of all scalar multiples of that vector, and these form a line passing through the origin. • If instead we have two linearly independent vectors $$\vvec_1$$ and $$\vvec_2\text{,}$$ the subspace $$S=\laspan{\vvec_1,\vvec_2}$$ is a plane passing through the origin. • Consider the three vectors $$\evec_1\text{,}$$ $$\evec_2\text{,}$$ and $$\evec_3\text{.}$$ Since we know that every 3-dimensional vector can be written as a linear combination, we have $$S = \laspan{\evec_1, \evec_2, \evec_3} = \real^3\text{.}$$ • One more subspace worth mentioning is $$S=\laspan{\zerovec}\text{.}$$ Since any linear combination of the zero vector is itself the zero vector, this subspace consists of a single vector, $$\zerovec\text{.}$$ In fact, any subspace of $$\real^3$$ is one of these types: the origin, a line, a plane, or all of $$\real^3\text{.}$$ Activity3.5.2. We will look at some sets of vectors and the subspaces they form. 1. If $$\vvec_1, \vvec_2,\ldots,\vvec_n$$ is a set of vectors in $$\real^m\text{,}$$ explain why $$\zerovec$$ can be expressed as a linear combination of these vectors. Use this fact to explain why the zero vector $$\zerovec$$ belongs to any subspace in $$\real^m\text{.}$$ 2. Explain why the line on the left of Figure 3.5.3 is not a subspace of $$\real^2$$ and why the line on the right is. 3. Consider the vectors \begin{equation} \vvec_1=\threevec101,~~~ \vvec_2=\threevec011,~~~ \vvec_3=\threevec110, \end{equation} and describe the subspace $$S=\laspan{\vvec_1,\vvec_2,\vvec_3}$$ of $$\real^3\text{.}$$ 4. Consider the vectors \begin{equation} \wvec_1=\threevec210,~~~ \wvec_2=\threevec{-1}1{-1},~~~ \wvec_3=\threevec03{-2} \end{equation} 1. Write $$\wvec_3$$ as a linear combination of $$\wvec_1$$ and $$\wvec_2\text{.}$$ 2. Explain why $$\laspan{\wvec_1,\wvec_2,\wvec_3} = \laspan{\wvec_1, \wvec_2}\text{.}$$ 3. Describe the subspace $$S = \laspan{\wvec_1,\wvec_2,\wvec_3}$$ of $$\real^3\text{.}$$ 5. Suppose that $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ $$\vvec_3\text{,}$$ and $$\vvec_4$$ are four vectors in $$\real^3$$ and that \begin{equation} \begin{bmatrix} \vvec_1 \amp \vvec_2 \amp \vvec_3 \amp \vvec_4 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 2 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{bmatrix}. \end{equation} Give a description of the subspace $$S=\laspan{\vvec_1,\vvec_2,\vvec_3,\vvec_4}$$ of $$\real^3\text{.}$$ As the activity shows, it is possible to represent some subspaces as the span of more than one set of vectors. We are particularly interested in representing a subspace as the span of a linearly independent set of vectors. Definition3.5.4. A basis for a subspace $$S$$ of $$\real^p$$ is a set of vectors in $$S$$ that are linearly independent and whose span is $$S\text{.}$$ We say that the dimension of the subspace $$S\text{,}$$ denoted $$\dim S\text{,}$$ is the number of vectors in any basis. Example3.5.5.A subspace of $$\real^4$$. Suppose we have the 4-dimensional vectors $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ and $$\vvec_3$$ that define the subspace $$S = \laspan{\vvec_1,\vvec_2,\vvec_3}$$ of $$\real^4\text{.}$$ Suppose also that \begin{equation} \begin{bmatrix} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{bmatrix} \sim \begin{bmatrix} 1 \amp -1 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ \end{bmatrix}. \end{equation} From the reduced row echelon form of the matrix, we see that $$\vvec_2 = -\vvec\text{.}$$ Therefore, any linear combination of $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ and $$\vvec_3$$ can be rewritten \begin{equation} c_1\vvec_1 + c_2\vvec_2 + c_3 \vvec_3 = (c_1-c_2)\vvec_1 + c_2\vvec_3 \end{equation} as a linear combination of $$\vvec_1$$ and $$\vvec_3\text{.}$$ This tells us that \begin{equation} S = \laspan{\vvec_1,\vvec_2,\vvec_3} = \laspan{\vvec_1,\vvec_3}. \end{equation} Furthermore, the reduced row echelon form of the matrix shows that $$\vvec_1$$ and $$\vvec_3$$ are linearly independent. Therefore, $$\{\vvec_1,\vvec_3\}$$ is a basis for $$S\text{,}$$ which means that $$S$$ is a two-dimensional subspace of $$\real^4\text{.}$$ Subspaces of $$\real^3$$ are either • 0-dimensional, consisting of the single vector $$\zerovec\text{,}$$ • a 1-dimensional line, • a 2-dimensional plane, or • the 3-dimensional subspace $$\real^3\text{.}$$ There is no 4-dimensional subspace of $$\real^3$$ because there is no linearly independent set of four vectors in $$\real^3\text{.}$$ There are two important subspaces associated to any matrix, each of which springs from one of our two fundamental questions, as we will now see. Subsection3.5.2The column space of $$A$$ The first subspace associated to a matrix that we’ll consider is its column space. Definition3.5.6. If $$A$$ is an $$m\times n$$ matrix, we call the span of its columns the column space of $$A$$ and denote it as $$\col(A)\text{.}$$ Notice that the columns of $$A$$ are vectors in $$\real^m\text{,}$$ which means that any linear combination of the columns is also in $$\real^m\text{.}$$ Since the column space is described as the span of a set of vectors, we see that $$\col(A)$$ is a subspace of $$\real^m\text{.}$$ Activity3.5.3. We will explore some column spaces in this activity. 1. Consider the matrix \begin{equation} A= \left[\begin{array}{rrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right] = \left[\begin{array}{rrr} 1 \amp 3 \amp -1 \\ -2 \amp 0 \amp -4 \\ 1 \amp 2 \amp 0 \\ \end{array}\right]. \end{equation} Since $$\col(A)$$ is the span of the columns, we have \begin{equation} \col(A) = \laspan{\vvec_1,\vvec_2,\vvec_3}. \end{equation} Explain why $$\vvec_3$$ can be written as a linear combination of $$\vvec_1$$ and $$\vvec_2$$ and why $$\col(A)=\laspan{\vvec_1,\vvec_2}\text{.}$$ 2. Explain why the vectors $$\vvec_1$$ and $$\vvec_2$$ form a basis for $$\col(A)$$ and why $$\col(A)$$ is a 2-dimensional subspace of $$\real^3$$ and therefore a plane. 3. Now consider the matrix $$B$$ and its reduced row echelon form: \begin{equation} B = \left[\begin{array}{rrrr} -2 \amp -4 \amp 0 \amp 6 \\ 1 \amp 2 \amp 0 \amp -3 \\ \end{array}\right] \sim \left[\begin{array}{rrrr} 1 \amp 2 \amp 0 \amp -3 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]. \end{equation} Explain why $$\col(B)$$ is a 1-dimensional subspace of $$\real^2$$ and is therefore a line. 4. For a general matrix $$A\text{,}$$ what is the relationship between the dimension $$\dim~\col(A)$$ and the number of pivot positions in $$A\text{?}$$ 5. How does the location of the pivot positions indicate a basis for $$\col(A)\text{?}$$ 6. If $$A$$ is an invertible $$9\times9$$ matrix, what can you say about the column space $$\col(A)\text{?}$$ 7. Suppose that $$A$$ is an $$8\times 10$$ matrix and that $$\col(A) = \real^8\text{.}$$ If $$\bvec$$ is an 8-dimensional vector, what can you say about the equation $$A\xvec = \bvec\text{?}$$ Example3.5.7. Consider the matrix $$A$$ and its reduced row echelon form: \begin{equation} A = \left[\begin{array}{rrrrr} 2 \amp 0 \amp -4 \amp -6 \amp 0 \\ -4 \amp -1 \amp 7 \amp 11 \amp 2 \\ 0 \amp -1 \amp -1 \amp -1 \amp 2 \\ \end{array}\right] \sim \left[\begin{array}{rrrrr} 1 \amp 0 \amp -2 \amp -3 \amp 0 \\ 0 \amp 1 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right], \end{equation} and denote the columns of $$A$$ as $$\vvec_1,\vvec_2,\ldots,\vvec_5\text{.}$$ It is certainly true that $$\col(A) = \laspan{\vvec_1,\vvec_2,\ldots,\vvec_5}$$ by the definition of the column space. However, the reduced row echelon form of the matrix shows us that the vectors are not linearly independent so $$\vvec_1,\vvec_2,\ldots,\vvec_5$$ do not form a basis for $$\col(A)\text{.}$$ From the reduced row echelon form, however, we can see that \begin{equation} \begin{aligned} \vvec_3 \amp {}={} -2\vvec_1 + \vvec_2 \\ \vvec_4 \amp {}={} -3\vvec_1 + \vvec_2 \\ \vvec_5 \amp {}={} -2\vvec_2 \\ \end{aligned}\text{.} \end{equation} This means that any linear combination of $$\vvec_1,\vvec_2,\ldots,\vvec_5$$ can be written as a linear combination of just $$\vvec_1$$ and $$\vvec_2\text{.}$$ Therefore, we see that $$\col(A) = \laspan{\vvec_1,\vvec_2}\text{.}$$ Moreover, the reduced row echelon form shows that $$\vvec_1$$ and $$\vvec_2$$ are linearly independent, which implies that they form a basis for $$\col(A)\text{.}$$ This means that $$\col(A)$$ is a 2-dimensional subspace of $$\real^3\text{,}$$ which is a plane in $$\real^3\text{,}$$ having basis \begin{equation} \threevec{2}{-4}{0}, \qquad \threevec{0}{-1}{1}\text{.} \end{equation} In general, a column without a pivot position can be written as a linear combination of the columns that have pivot positions. This means that a basis for $$\col(A)$$ will always be given by the columns of $$A$$ having pivot positions. This leads us to the following definition and proposition. Definition3.5.8. The rank of a matrix $$A$$ is the number of pivot positions in $$A$$ and is denoted by $$\rank(A)\text{.}$$ For example, the rank of the matrix $$A$$ in Example 3.5.7 is two because there are two pivot positions. A basis for $$\col(A)$$ is given by the first two columns of $$A$$ since those columns have pivot positions. As a note of caution, we determine the pivot positions by looking at the reduced row echelon form of $$A\text{.}$$ However, we form a basis of $$\col(A)$$ from the columns of $$A$$ rather than the columns of the reduced row echelon matrix. Subsection3.5.3The null space of $$A$$ The second subspace associated to a matrix is its null space. Definition3.5.10. If $$A$$ is an $$m\times n$$ matrix, we call the subset of vectors $$\xvec$$ in $$\real^n$$ satisfying $$A\xvec = \zerovec$$ the null space of $$A$$ and denote it by $$\nul(A)\text{.}$$ Remember that a subspace is a subset that can be represented as the span of a set of vectors. The column space of $$A\text{,}$$ which is simply the span of the columns of $$A\text{,}$$ fits this definition. It may not be immediately clear how the null space of $$A\text{,}$$ which is the solution space of the equation $$A\xvec = \zerovec\text{,}$$ does, but we will see that $$\nul(A)$$ is a subspace of $$\real^n\text{.}$$ Activity3.5.4. We will explore some null spaces in this activity and see why $$\nul(A)$$ satisfies the definition of a subspace. 1. Consider the matrix \begin{equation} A=\begin{bmatrix} 1 \amp 3 \amp -1 \amp 2 \\ -2 \amp 0 \amp -4 \amp 2 \\ 1 \amp 2 \amp 0 \amp 1 \end{bmatrix} \end{equation} and give a parametric description of the solution space to the equation $$A\xvec = \zerovec\text{.}$$ In other words, give a parametric description of $$\nul(A)\text{.}$$ 2. This parametric description shows that the vectors satisfying the equation $$A\xvec=\zerovec$$ can be written as a linear combination of a set of vectors. In other words, this description shows why $$\nul(A)$$ is the span of a set of vectors and is therefore a subspace. Identify a set of vectors whose span is $$\nul(A)\text{.}$$ 3. Use this set of vectors to find a basis for $$\nul(A)$$ and state the dimension of $$\nul(A)\text{.}$$ 4. The null space $$\nul(A)$$ is a subspace of $$\real^p$$ for which value of $$p\text{?}$$ 5. Now consider the matrix $$B$$ whose reduced row echelon form is given by \begin{equation} B \sim \left[\begin{array}{rrrr} 1 \amp 2 \amp 0 \amp -3 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]. \end{equation} Give a parametric description of $$\nul(B)\text{.}$$ 6. The parametric description gives a set of vectors that span $$\nul(B)\text{.}$$ Explain why this set of vectors is linearly independent and hence forms a basis. What is the dimension of $$\nul(B)\text{?}$$ 7. For a general matrix $$A\text{,}$$ how does the number of pivot positions indicate the dimension of $$\nul(A)\text{?}$$ 8. Suppose that the columns of a matrix $$A$$ are linearly independent. What can you say about $$\nul(A)\text{?}$$ Example3.5.11. Consider the matrix $$A$$ along with its reduced row echelon form: \begin{equation} A = \left[\begin{array}{rrrrr} 2 \amp 0 \amp -4 \amp -6 \amp 0 \\ -4 \amp -1 \amp 7 \amp 11 \amp 2 \\ 0 \amp -1 \amp -1 \amp -1 \amp 2 \\ \end{array}\right] \sim \left[\begin{array}{rrrrr} 1 \amp 0 \amp -2 \amp -3 \amp 0 \\ 0 \amp 1 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation} To find a parametric description of the solution space to $$A\xvec=\zerovec\text{,}$$ imagine that we augment both $$A$$ and its reduced row echelon form by a column of zeroes, which leads to the equations \begin{equation} \begin{alignedat}{6} x_1 \amp \amp \amp {}-{} \amp 2x_3 \amp {}-{} \amp 3x_4 \amp \amp \amp {}={} \amp 0 \\ \amp \amp x_2 \amp {}+{} \amp x_3 \amp {}+{} \amp x_4 \amp {}-{} \amp 2x_5 \amp {}={} \amp 0. \\ \end{alignedat} \end{equation} Notice that $$x_3\text{,}$$ $$x_4\text{,}$$ and $$x_5$$ are free variables so we rewrite these equations as \begin{equation} \begin{aligned} x_1 \amp {}={} 2x_3 + 3x_4 \\ x_2 \amp {}={} -x_3 - x_4 + 2x_5. \\ \end{aligned} \end{equation} In vector form, we have \begin{equation} \begin{aligned} \xvec \amp {}={} \fivevec{x_1}{x_2}{x_3}{x_4}{x_5} = \fivevec{2x_3 + 3x_4}{-x_3-x_4+2x_5}{x_3}{x_4}{x_5} \\ \\ \amp {}={} x_3\fivevec{2}{-1}{1}{0}{0} +x_4\fivevec{3}{-1}{0}{1}{0} +x_5\fivevec{0}{2}{0}{0}{1}. \end{aligned} \end{equation} This expression says that any vector $$\xvec$$ satisfying $$A\xvec= \zerovec$$ is a linear combination of the vectors \begin{equation} \vvec_1 = \fivevec{2}{-1}{1}{0}{0},~~~ \vvec_2 = \fivevec{3}{-1}{0}{1}{0},~~~ \vvec_3 = \fivevec{0}{2}{0}{0}{1}. \end{equation} It is straightforward to check that these vectors are linearly independent, which means that $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ and $$\vvec_3$$ form a basis for $$\nul(A)\text{,}$$ a 3-dimensional subspace of $$\real^5\text{.}$$ As illustrated in this example, the dimension of $$\nul(A)$$ is equal to the number of free variables in the equation $$A\xvec=\zerovec\text{,}$$ which equals the number of columns of $$A$$ without pivot positions or the number of columns of $$A$$ minus the number of pivot positions. Combining Proposition 3.5.9 and Proposition 3.5.12 shows that Subsection3.5.4Summary Once again, we find ourselves revisiting our two fundamental questions concerning the existence and uniqueness of solutions to linear systems. The column space $$\col(A)$$ contains all the vectors $$\bvec$$ for which the equation $$A\xvec = \bvec$$ is consistent. The null space $$\nul(A)$$ is the solution space to the equation $$A\xvec = \zerovec\text{,}$$ which reflects on the uniqueness of solutions to this and other equations. • A subspace $$S$$ of $$\real^p$$ is a subset of $$\real^p$$ that can be represented as the span of a set of vectors. A basis of $$S$$ is a linearly independent set of vectors whose span is $$S\text{.}$$ • If $$A$$ is an $$m\times n$$ matrix, the column space $$\col(A)$$ is the span of the columns of $$A$$ and forms a subspace of $$\real^m\text{.}$$ • A basis for $$\col(A)$$ is found from the columns of $$A$$ that have pivot positions. The dimension is therefore $$\dim~\col(A) = \rank(A)\text{.}$$ • The null space $$\nul(A)$$ is the solution space to the homogeneous equation $$A\xvec = \zerovec$$ and is a subspace of $$\real^n\text{.}$$ • A basis for $$\nul(A)$$ is found through a parametric description of the solution space of $$A\xvec = \zerovec\text{,}$$ and we have that $$\dim~\nul(A) = n - \rank(A)\text{.}$$ Exercises3.5.5Exercises 1. Suppose that $$A$$ and its reduced row echelon form are \begin{equation} A = \left[\begin{array}{rrrrrr} 0 \amp 2 \amp 0 \amp -4 \amp 0 \amp 6 \\ 0 \amp -4 \amp -1 \amp 7 \amp 0 \amp -16 \\ 0 \amp 6 \amp 0 \amp -12 \amp 3 \amp 15 \\ 0 \amp 4 \amp -1 \amp -9 \amp 0 \amp 8 \\ \end{array}\right] \sim \left[\begin{array}{rrrrrr} 0 \amp 1 \amp 0 \amp -2 \amp 0 \amp 3 \\ 0 \amp 0 \amp 1 \amp 1 \amp 0 \amp 4 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation} 1. The null space $$\nul(A)$$ is a subspace of $$\real^p$$ for what $$p\text{?}$$ The column space $$\col(A)$$ is a subspace of $$\real^p$$ for what $$p\text{?}$$ 2. What are the dimensions $$\dim~\nul(A)$$ and $$\dim~\col(A)\text{?}$$ 3. Find a basis for the column space $$\col(A)\text{.}$$ 4. Find a basis for the null space $$\nul(A)\text{.}$$ 2. Suppose that \begin{equation} A = \left[\begin{array}{rrrr} 2 \amp 0 \amp -2 \amp -4 \\ -2 \amp -1 \amp 1 \amp 2 \\ 0 \amp -1 \amp -1 \amp -2 \end{array}\right]\text{.} \end{equation} 1. Is the vector $$\threevec{0}{-1}{-1}$$ in $$\col(A)\text{?}$$ 2. Is the vector $$\fourvec{2}{1}{0}{2}$$ in $$\col(A)\text{?}$$ 3. Is the vector $$\threevec{2}{-2}{0}$$ in $$\nul(A)\text{?}$$ 4. Is the vector $$\fourvec{1}{-1}{3}{-1}$$ in $$\nul(A)\text{?}$$ 5. Is the vector $$\fourvec{1}{0}{1}{-1}$$ in $$\nul(A)\text{?}$$ 3. Determine whether the following statements are true or false and provide a justification for your response. Unless otherwise stated, assume that $$A$$ is an $$m\times n$$ matrix. 1. If $$A$$ is a $$127\times 341$$ matrix, then $$\nul(A)$$ is a subspace of $$\real^{127}\text{.}$$ 2. If $$\dim~\nul(A) = 0\text{,}$$ then the columns of $$A$$ are linearly independent. 3. If $$\col(A) = \real^m\text{,}$$ then $$A$$ is invertible. 4. If $$A$$ has a pivot position in every column, then $$\nul(A) = \real^n\text{.}$$ 5. If $$\col(A) = \real^m$$ and $$\nul(A) = \{\zerovec\}\text{,}$$ then $$A$$ is invertible. 4. Explain why the following statements are true. 1. If $$B$$ is invertible, then $$\nul(BA) = \nul(A)\text{.}$$ 2. If $$B$$ is invertible, then $$\col(AB) = \col(A)\text{.}$$ 3. If $$A\sim A'\text{,}$$ then $$\nul(A) = \nul(A')\text{.}$$ 5. For each of the following conditions, construct a $$3\times 3$$ matrix having the given properties. 1. $$\dim~\nul(A) = 0\text{.}$$ 2. $$\dim~\nul(A) = 1\text{.}$$ 3. $$\dim~\nul(A) = 2\text{.}$$ 4. $$\dim~\nul(A) = 3\text{.}$$ 6. Suppose that $$A$$ is a $$3\times 4$$ matrix. 1. Is it possible that $$\dim~\nul(A) = 0\text{?}$$ 2. If $$\dim~\nul(A) = 1\text{,}$$ what can you say about $$\col(A)\text{?}$$ 3. If $$\dim~\nul(A) = 2\text{,}$$ what can you say about $$\col(A)\text{?}$$ 4. If $$\dim~\nul(A) = 3\text{,}$$ what can you say about $$\col(A)\text{?}$$ 5. If $$\dim~\nul(A) = 4\text{,}$$ what can you say about $$\col(A)\text{?}$$ 7. Suppose we have the vectors \begin{equation} \vvec_1 = \threevec{2}{3}{-1},~ \vvec_2 = \threevec{-1}{2}{4},~ \wvec_1 = \fourvec{3}{-1}{1}{0},~ \wvec_2 = \fourvec{-2}{4}{0}{1} \end{equation} and that $$A$$ is a matrix such that $$\col(A)=\laspan{\vvec_1,\vvec_2}$$ and $$\nul(A) = \laspan{\wvec_1,\wvec_2}\text{.}$$ 1. What are the dimensions of $$A\text{?}$$ 2. Find such a matrix $$A\text{.}$$ 8. Suppose that $$A$$ is an $$8\times 8$$ matrix and that $$\det A = 14\text{.}$$ 1. What can you conclude about $$\nul(A)\text{?}$$ 2. What can you conclude about $$\col(A)\text{?}$$ 9. Suppose that $$A$$ is a matrix and there is an invertible matrix $$P$$ such that \begin{equation} A = P~\left[\begin{array}{rrr} 2 \amp 0 \amp 0 \\ 0 \amp -3 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{array}\right]~P^{-1}\text{.} \end{equation} 1. What can you conclude about $$\nul(A)\text{?}$$ 2. What can you conclude about $$\col(A)\text{?}$$ 10. In this section, we saw that the solution space to the homogeneous equation $$A\xvec = \zerovec$$ is a subspace of $$\real^p$$ for some $$p\text{.}$$ In this exercise, we will investigate whether the solution space to another equation $$A\xvec = \bvec$$ can form a subspace. Let’s consider the matrix \begin{equation} A = \left[\begin{array}{rr} 2 \amp -4 \\ -1 \amp 2 \\ \end{array}\right]\text{.} \end{equation} 1. Find a parametric description of the solution space to the homogeneous equation $$A\xvec = \zerovec\text{.}$$ 2. Graph the solution space to the homogeneous equation to the right. 3. Find a parametric description of the solution space to the equation $$A\xvec = \twovec{4}{-2}$$ and graph it above. 4. Is the solution space to the equation $$A\xvec = \twovec{4}{-2}$$ a subspace of $$\real^2\text{?}$$ 5. Find a parametric description of the solution space to the equation $$A\xvec=\twovec{-8}{4}$$ and graph it above. 6. What can you say about all the solution spaces to equations of the form $$A\xvec = \bvec$$ when $$\bvec$$ is a vector in $$\col(A)\text{?}$$ 7. Suppose that the solution space to the equation $$A\xvec = \bvec$$ forms a subspace. Explain why it must be true that $$\bvec = \zerovec\text{.}$$
## How many 4 digits numbers can be formed by using the digits 2 4 6 8 when repetition of digits is allowed? ∴ Total numbers = 4 + 12 + 24 + 24 = 64 numbers. How many 3 digit numbers can be written using 1/2 and 3 What is the probability of the numbers with same digits? digits are 3, and number of places are 3. The calculation will be 3!/0!, which is equal to 3!= 6. So 6 3-digit numbers can be formed using 1,2,3!!! How many 3 digit numbers can be written from the digits 1,2 3 and 4 if repetition of digits is allowed? So the number of 3-number combinations are- (1,2,3),(1,2,4),(1,3,4),(2,3,4). Each can be arranged in 6 ways, so we get 24 ways totally. READ: How does social media affect India? ### What is the sum of all 4 digit numbers that can be formed using all the digits 2 4 6 8? The sum of all 4 digit number containing the digits 2, 4, 6, 8, without repetitions is. 133330. What is the sum of all the 4 digit numbers which can be formed with the digits 2/3 4.6 without repetition? How many 3 digit numbers can be formed by using the digits from 0 to 9 with repetition? How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Therefore, total numbers =89×8×7504. #### How do you find the sum of all four digit numbers? To find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition: We can form a total of 4! or 24 numbers. When we add all these numbers, let us look at the contribution of the digit 2 to the sum. When 2 occurs in the thousands place in a particular number, its contribution to the total will be 2000. READ: How long do ps3 Super Slims Last? What is the sum of all possible numbers using n distinct digits? “ If all the possible n-digit numbers using n distinct digits are formed, the sum of all the numbers so formed is equal to (n-1)! x {sum of all the digits} x {111…….} n times. Example 1: What would be the sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which have no digits repeated? What is the sum (summation) calculator? The Sum (Summation) Calculator is used to calculate the total summation of any set of numbers. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. How to use Sum (Summation) Calculator Video ## How can we make 6 numbers using 3 digits? We can make 6 numbers using 3 digits and without repetitions of the digits. LOOK AT THE TREE DIAGRAM ABOVE. We have 3 choices for the first digit, 2 choices for the second digit and 1 choice for the third digit. There is a special notation for the product 3 × 2 × 1 = 3! and it is read 3 factorial. We also define 0! = 1. READ: Why so many Icelanders still believe in invisible elves?
# A solid consists of a cone on top of a cylinder with a radius equal to that of the cone. The height of the cone is 42 and the height of the cylinder is 10 . If the volume of the solid is 144 pi, what is the area of the base of the cylinder? Jul 28, 2016 ${A}_{\text{base}} = 6 \pi$ #### Explanation: The volume of each component is given by ${V}_{\text{cone") = 1/3pir_("cone")^2h_("cone}}$ ${V}_{\text{cylinder") = pir_("cylinder")^2h_("cylinder}}$ We have that ${r}_{\text{cone") = r_("cylinder}}$ so we shall just denote these as $r$. ${V}_{\text{total") = V_("cone") + V_("cylinder}}$ V_("total") = pir^2(1/3h_("cone") + h_("cylinder")) $\therefore \pi {r}^{2} = \left({V}_{\text{total"))/(1/3h_("cone") + h_("cylinder}}\right)$ Notice that $\pi {r}^{2}$ is precisely the area of the base of the cylinder, which is what we want to calculate so just plug in the numbers: ${A}_{b a s e} = \frac{144 \pi}{\frac{1}{3} \cdot 42 + 10} = \frac{144 \pi}{14 + 10} = \frac{144 \pi}{24} = 6 \pi$

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