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# What is the term”exclamation mark” in mathematics? The answer is easy. Below are some approaches to inform as soon as an equation can be really actually a multiplication equation and also how to add and subtract using”exclamation marks”. For instance,”2x + 3″ imply”multiply two integers from about a few, making a value add up to 2 and 3″. <MASK> <UNMASK> # What is the term”exclamation mark” in mathematics? The answer is easy. Below are some approaches to inform as soon as an equation can be really actually a multiplication equation and also how to add and subtract using”exclamation marks”. For instance,”2x + 3″ imply”multiply two integers from about a few, making a value add up to 2 and 3″. By way of example,”2x + 3″ are multiplication by write a paper online three. In addition, we can add the values of 2 and 3 collectively. To add the worth, we’ll use”e”that I” (or”E”). With”I” means”include the worth of one to the worth of 2″. <MASK> So, to add the values that we utilize”t”x” (or”TE”), which might be the numbers of the worth to be included. We will use”x” to subtract the price of a person in the price of both 2 plus that will provide us precisely exactly the outcome. To multiply the worth , we certainly can certainly do this like this:”2x + y” necessarily imply”multiply two integers by y, creating a price add up to two plus one”. You may know this is really just a multiplication equation when we use”x ray” to subtract one from 2. Or, it can be”x – y” to subtract from 2. Be aware that you can publish the equation using a decimal position and parentheses. Now, let us take an example. Let’s say that we would like to multiply the value of”nine” by”5″ and we all have”x nine”y https://www.eurachem.org/images/stories/Guides/pdf/recovery.pdf twenty-five”. Then we’ll use”x – y” to subtract the price of one in the worth of two.
# What is the term”exclamation mark” in mathematics? The answer is easy. Below are some approaches to inform as soon as an equation can be really actually a multiplication equation and also how to add and subtract using”exclamation marks”. For instance,”2x + 3″ imply”multiply two integers from about a few, making a value add up to 2 and 3″. By way of example,”2x + 3″ are multiplication by write a paper online three. In addition, we can add the values of 2 and 3 collectively. To add the worth, we’ll use”e”that I” (or”E”). With”I” means”include the worth of one to the worth of 2″. <MASK> So, to add the values that we utilize”t”x” (or”TE”), which might be the numbers of the worth to be included. We will use”x” to subtract the price of a person in the price of both 2 plus that will provide us precisely exactly the outcome. To multiply the worth , we certainly can certainly do this like this:”2x + y” necessarily imply”multiply two integers by y, creating a price add up to two plus one”. You may know this is really just a multiplication equation when we use”x ray” to subtract one from 2. Or, it can be”x – y” to subtract from 2. Be aware that you can publish the equation using a decimal position and parentheses. Now, let us take an example. Let’s say that we would like to multiply the value of”nine” by”5″ and we all have”x nine”y https://www.eurachem.org/images/stories/Guides/pdf/recovery.pdf twenty-five”. Then we’ll use”x – y” to subtract the price of one in the worth of two. <UNMASK> # What is the term”exclamation mark” in mathematics? The answer is easy. Below are some approaches to inform as soon as an equation can be really actually a multiplication equation and also how to add and subtract using”exclamation marks”. For instance,”2x + 3″ imply”multiply two integers from about a few, making a value add up to 2 and 3″. By way of example,”2x + 3″ are multiplication by write a paper online three. In addition, we can add the values of 2 and 3 collectively. To add the worth, we’ll use”e”that I” (or”E”). With”I” means”include the worth of one to the worth of 2″. To add the worth , we can certainly do this similar to that:”x – y” means”multiply x by y, making a worth equal to zero”. For”x”y”, we’ll use”t” (or”TE”) for the subtraction and we’ll use”x y y” to solve the equation. You might feel that you are not supposed to utilize”e” in addition as”I” implies”subtract” however it’s perhaps not so easy. https://www.masterpapers.com For instance, to express”2 – 3″ approaches subtract from three. So, to add the values that we utilize”t”x” (or”TE”), which might be the numbers of the worth to be included. We will use”x” to subtract the price of a person in the price of both 2 plus that will provide us precisely exactly the outcome. To multiply the worth , we certainly can certainly do this like this:”2x + y” necessarily imply”multiply two integers by y, creating a price add up to two plus one”. You may know this is really just a multiplication equation when we use”x ray” to subtract one from 2. Or, it can be”x – y” to subtract from 2. Be aware that you can publish the equation using a decimal position and parentheses. Now, let us take an example. Let’s say that we would like to multiply the value of”nine” by”5″ and we all have”x nine”y https://www.eurachem.org/images/stories/Guides/pdf/recovery.pdf twenty-five”. Then we’ll use”x – y” to subtract the price of one in the worth of two.
<MASK> Please note that our virtual Singapore Math Grade 3 curriculum is spiral and it provides for the review of the important concepts that students learned in Grade 2. Our online K-5 math curriculum is aligned with all standard Singapore Math textbook series and it includes all content that these series cover, from Kindergarten grade through 5th grade. Our Singapore Math for 3rd Grade may introduce some topics one grade level earlier or postpone coverage of some topics until grade 4. In the few instances where 3rd grade level units don’t exactly align between our curriculum and textbooks, you will still be able to easily locate the corresponding unit in our program by referring to the table of contents one grade below or above. <MASK> Singapore Math 3a <MASK> Singapore Math 3b <MASK> This unit covers understanding of multiplication and division. In this unit students will extend their knowledge of making equal groups to formalize their understanding of multiplication and division. The focus of this unit is on understanding multiplication and division using equal groups, not on memorizing facts. Students will learn how the multiplication symbol to represent addition of quantities in groups. <MASK> This unit covers telling time, adding time, subtracting time, and time intervals. Students will review and practice to tell time to the minute, learn telling intervals of time in hours, convert units of time between hours, minutes, seconds, days and weeks. <MASK> • Area And Perimeter <MASK> • Attributes Of Two-Dimensional Shapes <MASK> <UNMASK> Our curriculum is spiral Please note that our virtual Singapore Math Grade 3 curriculum is spiral and it provides for the review of the important concepts that students learned in Grade 2. Our online K-5 math curriculum is aligned with all standard Singapore Math textbook series and it includes all content that these series cover, from Kindergarten grade through 5th grade. Our Singapore Math for 3rd Grade may introduce some topics one grade level earlier or postpone coverage of some topics until grade 4. In the few instances where 3rd grade level units don’t exactly align between our curriculum and textbooks, you will still be able to easily locate the corresponding unit in our program by referring to the table of contents one grade below or above. Correspondence to 3A and 3B For your reference, the following topics in our curriculum correspond to Singapore Math practice Grade 3 in levels 3A and 3B: Singapore Math 3a Multiplication and division, multiplication tables of 2, 5, and 10, multiplication tables of 3 and 4, multiplication tables of 6, 7, 8, and 9, solving problems involving multiplication and division, and mental math computation and estimation. Singapore Math 3b Understanding fractions, time, volume, mass, representing and interpreting data, area and perimeter, and attributes of two-dimensional shapes. Student prior knowledge Prior to starting third grade Singapore Math, students should already know how to relate three-digit numbers to place value, use place-value charts to form a number and compare three-digit numbers. The initial lessons in the Singapore Math 3rd Grade are both a review and an extension of content covered in the prior grade that include mental addition of 1-digit number to a 2-digit number and counting by 2s, 5s, and 10s. 3rd Grade Singapore Math Scope and Sequence • Multiplication And Division This unit covers understanding of multiplication and division. In this unit students will extend their knowledge of making equal groups to formalize their understanding of multiplication and division. The focus of this unit is on understanding multiplication and division using equal groups, not on memorizing facts. Students will learn how the multiplication symbol to represent addition of quantities in groups. • Multiplication Tables Of 2, 5, And 10 This unit covers multiplying by 2 using skip-counting, multiplying by 2 using dot paper, multiplying by 5 using skip-counting, multiplying by 5 using dot paper, multiplying by 10 using skip-counting, dividing using related multiplication facts of 2, 5, or 10. Students will learn building multiplication tables of 2, 5, and 10 to formalize their understanding of multiplication and division for facts 2, 5, and 10. Students will learn to find their division facts by thinking of corresponding multiplication facts. • Multiplication Tables Of 3 And 4 This unit covers multiplying by 3 using skip-counting, multiplying by 3 using dot paper, multiplying by 4 using skip-counting, multiplying by 4 using dot paper, dividing using related multiplication facts of 3 or 4. Students will learn building multiplication tables of 3 and 4 to formalize their understanding of multiplication and division for facts 3 and 14. Students will learn to find their division facts by thinking of corresponding multiplication facts. <MASK> This unit covers multiplication properties, multiplying by 6, multiplying by 7, multiplying by 8, multiplying by 9, dividing using related multiplication facts of 6, 7, 8, or 9. Students will learn building multiplication tables of 6, 7, 8, and 9 to formalize their understanding of multiplication and division for facts 6, 7, 8, and 9. Students will learn to find their division facts by thinking of corresponding multiplication facts. • Solving Problems Involving Multiplication And Division <MASK> • Mental Computation And Estimation This unit covers learning mental math strategies to solve multiplication and division problems. Students will use place value to round whole numbers. • Understanding Fractions This unit covers parts and wholes, fractions and number lines, comparing unit fractions, equivalence of fractions, and comparing like fractions. Students will learn fractional notation that include the terms “numerator” and “denominator.” Students will understand that a common fraction is composed of unit fractions and they will learn to compare unit fractions. • Time This unit covers telling time, adding time, subtracting time, and time intervals. Students will review and practice to tell time to the minute, learn telling intervals of time in hours, convert units of time between hours, minutes, seconds, days and weeks. <MASK> This unit covers understanding of volume, comparing volume, measuring and estimating volume, and word problems involving volume. • Mass This unit covers measuring mass in kilograms, comparing mass in kilograms, measuring mass in grams, comparing mass in grams, and word problems involving mass. • Representing And Interpreting Data This unit covers scaled picture graphs and scaled bar graphs, reading and interpreting bar graphs, and line plots. Students will learn to sort data into groups and categories and use numerical data to interpret bar graphs and line plots. • Area And Perimeter This unit covers understanding of area, measuring area using square centimeters and square inches, measuring area using square meters and square feet, area and perimeter, solving problems involving area and perimeter. Students will learn to find and measure area of figures in square units that include square centimeters, square inches, square meters and square feet. • Attributes Of Two-Dimensional Shapes This unit covers categories and attributes of shapes and partitioning shapes into equal areas. Students will understand that figures with different shapes can have the same area.
Our curriculum is spiral Please note that our virtual Singapore Math Grade 3 curriculum is spiral and it provides for the review of the important concepts that students learned in Grade 2. Our online K-5 math curriculum is aligned with all standard Singapore Math textbook series and it includes all content that these series cover, from Kindergarten grade through 5th grade. Our Singapore Math for 3rd Grade may introduce some topics one grade level earlier or postpone coverage of some topics until grade 4. In the few instances where 3rd grade level units don’t exactly align between our curriculum and textbooks, you will still be able to easily locate the corresponding unit in our program by referring to the table of contents one grade below or above. Correspondence to 3A and 3B For your reference, the following topics in our curriculum correspond to Singapore Math practice Grade 3 in levels 3A and 3B: Singapore Math 3a Multiplication and division, multiplication tables of 2, 5, and 10, multiplication tables of 3 and 4, multiplication tables of 6, 7, 8, and 9, solving problems involving multiplication and division, and mental math computation and estimation. Singapore Math 3b Understanding fractions, time, volume, mass, representing and interpreting data, area and perimeter, and attributes of two-dimensional shapes. Student prior knowledge Prior to starting third grade Singapore Math, students should already know how to relate three-digit numbers to place value, use place-value charts to form a number and compare three-digit numbers. The initial lessons in the Singapore Math 3rd Grade are both a review and an extension of content covered in the prior grade that include mental addition of 1-digit number to a 2-digit number and counting by 2s, 5s, and 10s. 3rd Grade Singapore Math Scope and Sequence • Multiplication And Division This unit covers understanding of multiplication and division. In this unit students will extend their knowledge of making equal groups to formalize their understanding of multiplication and division. The focus of this unit is on understanding multiplication and division using equal groups, not on memorizing facts. Students will learn how the multiplication symbol to represent addition of quantities in groups. • Multiplication Tables Of 2, 5, And 10 This unit covers multiplying by 2 using skip-counting, multiplying by 2 using dot paper, multiplying by 5 using skip-counting, multiplying by 5 using dot paper, multiplying by 10 using skip-counting, dividing using related multiplication facts of 2, 5, or 10. Students will learn building multiplication tables of 2, 5, and 10 to formalize their understanding of multiplication and division for facts 2, 5, and 10. Students will learn to find their division facts by thinking of corresponding multiplication facts. • Multiplication Tables Of 3 And 4 This unit covers multiplying by 3 using skip-counting, multiplying by 3 using dot paper, multiplying by 4 using skip-counting, multiplying by 4 using dot paper, dividing using related multiplication facts of 3 or 4. Students will learn building multiplication tables of 3 and 4 to formalize their understanding of multiplication and division for facts 3 and 14. Students will learn to find their division facts by thinking of corresponding multiplication facts. <MASK> This unit covers multiplication properties, multiplying by 6, multiplying by 7, multiplying by 8, multiplying by 9, dividing using related multiplication facts of 6, 7, 8, or 9. Students will learn building multiplication tables of 6, 7, 8, and 9 to formalize their understanding of multiplication and division for facts 6, 7, 8, and 9. Students will learn to find their division facts by thinking of corresponding multiplication facts. • Solving Problems Involving Multiplication And Division <MASK> • Mental Computation And Estimation This unit covers learning mental math strategies to solve multiplication and division problems. Students will use place value to round whole numbers. • Understanding Fractions This unit covers parts and wholes, fractions and number lines, comparing unit fractions, equivalence of fractions, and comparing like fractions. Students will learn fractional notation that include the terms “numerator” and “denominator.” Students will understand that a common fraction is composed of unit fractions and they will learn to compare unit fractions. • Time This unit covers telling time, adding time, subtracting time, and time intervals. Students will review and practice to tell time to the minute, learn telling intervals of time in hours, convert units of time between hours, minutes, seconds, days and weeks. <MASK> This unit covers understanding of volume, comparing volume, measuring and estimating volume, and word problems involving volume. • Mass This unit covers measuring mass in kilograms, comparing mass in kilograms, measuring mass in grams, comparing mass in grams, and word problems involving mass. • Representing And Interpreting Data This unit covers scaled picture graphs and scaled bar graphs, reading and interpreting bar graphs, and line plots. Students will learn to sort data into groups and categories and use numerical data to interpret bar graphs and line plots. • Area And Perimeter This unit covers understanding of area, measuring area using square centimeters and square inches, measuring area using square meters and square feet, area and perimeter, solving problems involving area and perimeter. Students will learn to find and measure area of figures in square units that include square centimeters, square inches, square meters and square feet. • Attributes Of Two-Dimensional Shapes This unit covers categories and attributes of shapes and partitioning shapes into equal areas. Students will understand that figures with different shapes can have the same area. <UNMASK> Our curriculum is spiral Please note that our virtual Singapore Math Grade 3 curriculum is spiral and it provides for the review of the important concepts that students learned in Grade 2. Our online K-5 math curriculum is aligned with all standard Singapore Math textbook series and it includes all content that these series cover, from Kindergarten grade through 5th grade. Our Singapore Math for 3rd Grade may introduce some topics one grade level earlier or postpone coverage of some topics until grade 4. In the few instances where 3rd grade level units don’t exactly align between our curriculum and textbooks, you will still be able to easily locate the corresponding unit in our program by referring to the table of contents one grade below or above. Correspondence to 3A and 3B For your reference, the following topics in our curriculum correspond to Singapore Math practice Grade 3 in levels 3A and 3B: Singapore Math 3a Multiplication and division, multiplication tables of 2, 5, and 10, multiplication tables of 3 and 4, multiplication tables of 6, 7, 8, and 9, solving problems involving multiplication and division, and mental math computation and estimation. Singapore Math 3b Understanding fractions, time, volume, mass, representing and interpreting data, area and perimeter, and attributes of two-dimensional shapes. Student prior knowledge Prior to starting third grade Singapore Math, students should already know how to relate three-digit numbers to place value, use place-value charts to form a number and compare three-digit numbers. The initial lessons in the Singapore Math 3rd Grade are both a review and an extension of content covered in the prior grade that include mental addition of 1-digit number to a 2-digit number and counting by 2s, 5s, and 10s. 3rd Grade Singapore Math Scope and Sequence • Multiplication And Division This unit covers understanding of multiplication and division. In this unit students will extend their knowledge of making equal groups to formalize their understanding of multiplication and division. The focus of this unit is on understanding multiplication and division using equal groups, not on memorizing facts. Students will learn how the multiplication symbol to represent addition of quantities in groups. • Multiplication Tables Of 2, 5, And 10 This unit covers multiplying by 2 using skip-counting, multiplying by 2 using dot paper, multiplying by 5 using skip-counting, multiplying by 5 using dot paper, multiplying by 10 using skip-counting, dividing using related multiplication facts of 2, 5, or 10. Students will learn building multiplication tables of 2, 5, and 10 to formalize their understanding of multiplication and division for facts 2, 5, and 10. Students will learn to find their division facts by thinking of corresponding multiplication facts. • Multiplication Tables Of 3 And 4 This unit covers multiplying by 3 using skip-counting, multiplying by 3 using dot paper, multiplying by 4 using skip-counting, multiplying by 4 using dot paper, dividing using related multiplication facts of 3 or 4. Students will learn building multiplication tables of 3 and 4 to formalize their understanding of multiplication and division for facts 3 and 14. Students will learn to find their division facts by thinking of corresponding multiplication facts. • Multiplication Tables Of 6, 7, 8, And 9 This unit covers multiplication properties, multiplying by 6, multiplying by 7, multiplying by 8, multiplying by 9, dividing using related multiplication facts of 6, 7, 8, or 9. Students will learn building multiplication tables of 6, 7, 8, and 9 to formalize their understanding of multiplication and division for facts 6, 7, 8, and 9. Students will learn to find their division facts by thinking of corresponding multiplication facts. • Solving Problems Involving Multiplication And Division This unit covers solving one- and two-step word problems involving multiplication and division. Students will use a part-whole and comparison models to solve word problems involving multiplication and division. • Mental Computation And Estimation This unit covers learning mental math strategies to solve multiplication and division problems. Students will use place value to round whole numbers. • Understanding Fractions This unit covers parts and wholes, fractions and number lines, comparing unit fractions, equivalence of fractions, and comparing like fractions. Students will learn fractional notation that include the terms “numerator” and “denominator.” Students will understand that a common fraction is composed of unit fractions and they will learn to compare unit fractions. • Time This unit covers telling time, adding time, subtracting time, and time intervals. Students will review and practice to tell time to the minute, learn telling intervals of time in hours, convert units of time between hours, minutes, seconds, days and weeks. • Volume This unit covers understanding of volume, comparing volume, measuring and estimating volume, and word problems involving volume. • Mass This unit covers measuring mass in kilograms, comparing mass in kilograms, measuring mass in grams, comparing mass in grams, and word problems involving mass. • Representing And Interpreting Data This unit covers scaled picture graphs and scaled bar graphs, reading and interpreting bar graphs, and line plots. Students will learn to sort data into groups and categories and use numerical data to interpret bar graphs and line plots. • Area And Perimeter This unit covers understanding of area, measuring area using square centimeters and square inches, measuring area using square meters and square feet, area and perimeter, solving problems involving area and perimeter. Students will learn to find and measure area of figures in square units that include square centimeters, square inches, square meters and square feet. • Attributes Of Two-Dimensional Shapes This unit covers categories and attributes of shapes and partitioning shapes into equal areas. Students will understand that figures with different shapes can have the same area.
<MASK> <UNMASK> <MASK> Word Document File 181 KB\|20 pages Share Product Description FRACTIONS BUNDLE "TWIST" 12 WORKSHEETS <MASK> Worksheet 4 - DIVISION OF FRACTIONS; these 15 examples also involve cancelling when possible PLUS a GEOMETRY BONUS question; <MASK> Worksheets 7 and 8 - ALL OPERATIONS PLUS!; these TWO worksheets combine ALL 4 Operations; 16 examples, 4 for each operation included; PLUS 8 BONUS questions on fractions; Worksheets 9 and 10 - WORD PROBLEMS for ADDITION & SUBTRACTION OF FRACTIONS; NO FLUFF! Just 25 word problems all on addition and/or subtraction of fractions; students must solve not only for the correct operation, but to lowest terms also; <MASK> The thumbnails only give you 4 items to look at. The download preview gives you more. Take a look! * ALSO AVAILABLE FOR YOU OR A COLLEAGUE! - CLICK ANY LINK YOU WANT: - FRACTIONS REVIEW AND REINFORCEMENT - 10 FULL worksheets on various aspects of fractions to be used as great review or part of your unit. A LOT of good work here for your students. This link will describe all 10 pages in detail. <MASK> - NUMBER LINE POWERPOINT LESSON - 50 slide Powerpoint LESSON dealing with the number line and positive and negative numbers. Designed as a whole-class lesson getting students actively involved in the answers. 5 different sections including less than and greater than. Great graphics and animation will hold their attention. Different and effective lesson! The Number Line! <MASK>
<MASK> Word Document File 181 KB\|20 pages Share Product Description FRACTIONS BUNDLE "TWIST" 12 WORKSHEETS <MASK> Worksheet 4 - DIVISION OF FRACTIONS; these 15 examples also involve cancelling when possible PLUS a GEOMETRY BONUS question; <MASK> Worksheets 7 and 8 - ALL OPERATIONS PLUS!; these TWO worksheets combine ALL 4 Operations; 16 examples, 4 for each operation included; PLUS 8 BONUS questions on fractions; Worksheets 9 and 10 - WORD PROBLEMS for ADDITION & SUBTRACTION OF FRACTIONS; NO FLUFF! Just 25 word problems all on addition and/or subtraction of fractions; students must solve not only for the correct operation, but to lowest terms also; <MASK> The thumbnails only give you 4 items to look at. The download preview gives you more. Take a look! * ALSO AVAILABLE FOR YOU OR A COLLEAGUE! - CLICK ANY LINK YOU WANT: - FRACTIONS REVIEW AND REINFORCEMENT - 10 FULL worksheets on various aspects of fractions to be used as great review or part of your unit. A LOT of good work here for your students. This link will describe all 10 pages in detail. <MASK> - NUMBER LINE POWERPOINT LESSON - 50 slide Powerpoint LESSON dealing with the number line and positive and negative numbers. Designed as a whole-class lesson getting students actively involved in the answers. 5 different sections including less than and greater than. Great graphics and animation will hold their attention. Different and effective lesson! The Number Line! <MASK> <UNMASK> # Fractions Bundle "Twist" 12 Worksheets Subject Resource Type Product Rating File Type Word Document File 181 KB\|20 pages Share Product Description FRACTIONS BUNDLE "TWIST" 12 WORKSHEETS You receive 12 FULL worksheets all on the 4 OPERATIONS of FRACTIONS including WORD PROBLEMS for EACH operation! The "Twist"? - ALL these examples are written out IN WORDS to make your students THINK a little more. They must first READ the fractions correctly, then WRITE them correctly and then SOLVE them correctly (hopefully!) to lowest terms; Worksheet 1 - ADDITION OF FRACTIONS; all types of addition of fraction examples with like and unlike denominators; 15 examples plus a BONUS at the end; Worksheet 2 - SUBTRACTION OF FRACTIONS; all types of subtraction of fractions examples with and without borrowing; 18 examples plus 4 EXTRA bonus questions; Worksheet 3 - MULTIPLICATION OF FRACTIONS; these 16 examples involve cancelling when possible plus 2 EXTRA fraction questions; Worksheet 4 - DIVISION OF FRACTIONS; these 15 examples also involve cancelling when possible PLUS a GEOMETRY BONUS question; Worksheets 5 and 6 - MULTIPLICATION AND DIVISION OF FRACTIONS COMBINED; students must multiply or divide these examples using cancelling; 21 examples plus many BONUS fraction questions included; Worksheets 7 and 8 - ALL OPERATIONS PLUS!; these TWO worksheets combine ALL 4 Operations; 16 examples, 4 for each operation included; PLUS 8 BONUS questions on fractions; Worksheets 9 and 10 - WORD PROBLEMS for ADDITION & SUBTRACTION OF FRACTIONS; NO FLUFF! Just 25 word problems all on addition and/or subtraction of fractions; students must solve not only for the correct operation, but to lowest terms also; Worksheets 11 and 12 - WORD PROBLEMS for MULTIPLICATION & DIVISION of FRACTIONS; NO FLUFF AGAIN! 25 word problems on multiplication and/or division of fractions; read and solve; A HUGE packet of worksheets involving ALL OPERATIONS of fractions! The thumbnails only give you 4 items to look at. The download preview gives you more. Take a look! * ALSO AVAILABLE FOR YOU OR A COLLEAGUE! - CLICK ANY LINK YOU WANT: - FRACTIONS REVIEW AND REINFORCEMENT - 10 FULL worksheets on various aspects of fractions to be used as great review or part of your unit. A LOT of good work here for your students. This link will describe all 10 pages in detail. - FRACTIONS POWERPOINT FUN QUIZ - 60 slides in all! This Powerpoint program starts over again EACH time students get even one answer wrong! Challenging, fun and SELF-CORRECTING! Great activity with great graphics will keep students engaged. NOT for the beginner! Did I say it was self-correcting? - NUMBER LINE POWERPOINT LESSON - 50 slide Powerpoint LESSON dealing with the number line and positive and negative numbers. Designed as a whole-class lesson getting students actively involved in the answers. 5 different sections including less than and greater than. Great graphics and animation will hold their attention. Different and effective lesson! The Number Line! * * * GREAT TEACHING TO YOU! Total Pages 20 pages Included Teaching Duration N/A Report this Resource \$5.00
<MASK> Correct result: <MASK> We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! <MASK> ## Next similar math problems: <MASK> <UNMASK> # Height of the room Given the floor area of a room as 24 feet by 48 feet and space diagonal of a room as 56 feet. Can you find the height of the room? Correct result: c = 16 ft <MASK> We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! <MASK> We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Diagonal Determine the dimensions of the cuboid, if diagonal long 53 dm has an angle with one edge 42° and with another edge 64°. • Cuboidal room Length of cuboidal room is 2m breadth of cuboidal room is 3m and height is 6m find the length of the longest rod that can be fitted in the room • Ratio of edges The dimensions of the cuboid are in a ratio 3: 1: 2. The body diagonal has a length of 28 cm. Find the volume of a cuboid. • Four sided prism Calculate the volume and surface area of a regular quadrangular prism whose height is 28.6cm and the body diagonal forms a 50-degree angle with the base plane. • Cuboid diagonals The cuboid has dimensions of 15, 20 and 40 cm. Calculate its volume and surface, the length of the body diagonal and the lengths of all three wall diagonals. • Space diagonal angles Calculate the angle between the body diagonal and the side edge c of the block with dimensions: a = 28cm, b = 45cm and c = 73cm. Then, find the angle between the body diagonal and the plane of the base ABCD. • The room The room has a cuboid shape with dimensions: length 50m and width 60dm and height 300cm. Calculate how much this room will cost paint (floor is not painted) if the window and door area is 15% of the total area and 1m2 cost 15 euro. • Jared's room painting Jared wants to paint his room. The dimensions of the room are 12 feet by 15 feet, and the walls are 9 feet high. There are two windows that measure 6 feet by 5 feet each. There are two doors, whose dimensions are 30 inches by 6 feet each. If a gallon of p • Solid cuboid A solid cuboid has a volume of 40 cm3. The cuboid has a total surface area of 100 cm squared. One edge of the cuboid has a length of 2 cm. Find the length of a diagonal of the cuboid. Give your answer correct to 3 sig. Fig. • Find diagonal Find the length of the diagonal of a cuboid with length=20m width=25m height=150m
# Height of the room Given the floor area of a room as 24 feet by 48 feet and space diagonal of a room as 56 feet. Can you find the height of the room? Correct result: c = 16 ft <MASK> We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! <MASK> We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Diagonal Determine the dimensions of the cuboid, if diagonal long 53 dm has an angle with one edge 42° and with another edge 64°. • Cuboidal room Length of cuboidal room is 2m breadth of cuboidal room is 3m and height is 6m find the length of the longest rod that can be fitted in the room • Ratio of edges The dimensions of the cuboid are in a ratio 3: 1: 2. The body diagonal has a length of 28 cm. Find the volume of a cuboid. • Four sided prism Calculate the volume and surface area of a regular quadrangular prism whose height is 28.6cm and the body diagonal forms a 50-degree angle with the base plane. • Cuboid diagonals The cuboid has dimensions of 15, 20 and 40 cm. Calculate its volume and surface, the length of the body diagonal and the lengths of all three wall diagonals. • Space diagonal angles Calculate the angle between the body diagonal and the side edge c of the block with dimensions: a = 28cm, b = 45cm and c = 73cm. Then, find the angle between the body diagonal and the plane of the base ABCD. • The room The room has a cuboid shape with dimensions: length 50m and width 60dm and height 300cm. Calculate how much this room will cost paint (floor is not painted) if the window and door area is 15% of the total area and 1m2 cost 15 euro. • Jared's room painting Jared wants to paint his room. The dimensions of the room are 12 feet by 15 feet, and the walls are 9 feet high. There are two windows that measure 6 feet by 5 feet each. There are two doors, whose dimensions are 30 inches by 6 feet each. If a gallon of p • Solid cuboid A solid cuboid has a volume of 40 cm3. The cuboid has a total surface area of 100 cm squared. One edge of the cuboid has a length of 2 cm. Find the length of a diagonal of the cuboid. Give your answer correct to 3 sig. Fig. • Find diagonal Find the length of the diagonal of a cuboid with length=20m width=25m height=150m <UNMASK> # Height of the room Given the floor area of a room as 24 feet by 48 feet and space diagonal of a room as 56 feet. Can you find the height of the room? Correct result: c = 16 ft #### Solution: We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Pythagorean theorem is the base for the right triangle calculator. #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Diagonal Determine the dimensions of the cuboid, if diagonal long 53 dm has an angle with one edge 42° and with another edge 64°. • Cuboidal room Length of cuboidal room is 2m breadth of cuboidal room is 3m and height is 6m find the length of the longest rod that can be fitted in the room • Ratio of edges The dimensions of the cuboid are in a ratio 3: 1: 2. The body diagonal has a length of 28 cm. Find the volume of a cuboid. • Four sided prism Calculate the volume and surface area of a regular quadrangular prism whose height is 28.6cm and the body diagonal forms a 50-degree angle with the base plane. • Cuboid diagonals The cuboid has dimensions of 15, 20 and 40 cm. Calculate its volume and surface, the length of the body diagonal and the lengths of all three wall diagonals. • Space diagonal angles Calculate the angle between the body diagonal and the side edge c of the block with dimensions: a = 28cm, b = 45cm and c = 73cm. Then, find the angle between the body diagonal and the plane of the base ABCD. • The room The room has a cuboid shape with dimensions: length 50m and width 60dm and height 300cm. Calculate how much this room will cost paint (floor is not painted) if the window and door area is 15% of the total area and 1m2 cost 15 euro. • Jared's room painting Jared wants to paint his room. The dimensions of the room are 12 feet by 15 feet, and the walls are 9 feet high. There are two windows that measure 6 feet by 5 feet each. There are two doors, whose dimensions are 30 inches by 6 feet each. If a gallon of p • Solid cuboid A solid cuboid has a volume of 40 cm3. The cuboid has a total surface area of 100 cm squared. One edge of the cuboid has a length of 2 cm. Find the length of a diagonal of the cuboid. Give your answer correct to 3 sig. Fig. • Find diagonal Find the length of the diagonal of a cuboid with length=20m width=25m height=150m
<MASK> solve by using subsitution: 4x+y=2 3y+2x=-1 4. math <MASK> solve by using subsitution: 4x+y=2 3y+2x=-1 6. math solve by using subsitution: 4x+y=2 3y+2x=-1 7. math use subsitution system to solve 4x+5y=21 Y=3x-11 8. agebra 2 honors <MASK> solve by subsitution 6x + 5y = 27 x = 17 -8y my answer was (3,-7) 10. math -w-z=-2 and 4w+5z=16 im trying to solve this equation by using subsitution <MASK> <UNMASK> math <MASK> 1. 0 2. 0 <MASK> 1. algebra <MASK> solve by using subsitution: 4x+y=2 3y+2x=-1 4. math <MASK> solve by using subsitution: 4x+y=2 3y+2x=-1 6. math solve by using subsitution: 4x+y=2 3y+2x=-1 7. math use subsitution system to solve 4x+5y=21 Y=3x-11 8. agebra 2 honors subsitution with 3 equations (solve for x,y,z) x+y-2z=5 -x-y+z=2 -x+y+32=4 9. algebra solve by subsitution 6x + 5y = 27 x = 17 -8y my answer was (3,-7) 10. math -w-z=-2 and 4w+5z=16 im trying to solve this equation by using subsitution <MASK>
math <MASK> 1. 0 2. 0 <MASK> 1. algebra <MASK> solve by using subsitution: 4x+y=2 3y+2x=-1 4. math <MASK> solve by using subsitution: 4x+y=2 3y+2x=-1 6. math solve by using subsitution: 4x+y=2 3y+2x=-1 7. math use subsitution system to solve 4x+5y=21 Y=3x-11 8. agebra 2 honors subsitution with 3 equations (solve for x,y,z) x+y-2z=5 -x-y+z=2 -x+y+32=4 9. algebra solve by subsitution 6x + 5y = 27 x = 17 -8y my answer was (3,-7) 10. math -w-z=-2 and 4w+5z=16 im trying to solve this equation by using subsitution <MASK> <UNMASK> math solve by using subsitution: 4x+y=2 3y+2x=-1 1. 0 2. 0 Similar Questions 1. algebra solve by subsitution 2x+y=9 8x+4y=36 2. alg. 2 solve by subsitution 2x+3y=10 x+6y=32 3. math solve by using subsitution: 4x+y=2 3y+2x=-1 4. math solve by using subsitution: 4x+y=2 3y+2x=-1 5. math solve by using subsitution: 4x+y=2 3y+2x=-1 6. math solve by using subsitution: 4x+y=2 3y+2x=-1 7. math use subsitution system to solve 4x+5y=21 Y=3x-11 8. agebra 2 honors subsitution with 3 equations (solve for x,y,z) x+y-2z=5 -x-y+z=2 -x+y+32=4 9. algebra solve by subsitution 6x + 5y = 27 x = 17 -8y my answer was (3,-7) 10. math -w-z=-2 and 4w+5z=16 im trying to solve this equation by using subsitution More Similar Questions
<MASK> Puzzle Details In farmer Brown's hay loft there are a number of animals, in particular crows, mice and cockroaches. Being bored one day, I decided to count the animals and found there were exactly 150 feet and 50 heads in total, and there were twice as many cockroaches as mice. How many of each animal were there? <MASK> Answer: 10 cockroaches, 5 mice and 35 birds. <MASK> For every mouse there are two cockroaches, so every mouse is worth 3 heads and 16 feet (4 + 6 + 6). We can now write down an expression for the heads and the feet (calling birds B and mice M): B + 3M = 50 (1) Feet gives us: 2B + 16M = 150 (2) If we double (1) we get: 2B + 6M = 100 (3) We can now do (2) - (3) to give: 10M = 50 M = 5 So we have 5 mice (and 10 cockroaches). We can use M = 5 in (1) to give: B + 3 x 5 = 50 B + 15 = 50 B = 35 So, C = 10, M = 5 and B = 35. Checking that 10 + 5 + 35 = 50, and 10 x 6 + 5 x 4 + 35 x 2 = 150. QED. Our Favourite Illusions Shadow Illusion Are the squares A and B the same colour? Spinning Dancer Which way is the dancer spinning? Impossible Waterfall? Is the water flowing uphill in this impossible Escher type waterfall? The Butterfly A colourful butterfly? Duck Or Rabbit? Is this a duck or a rabbit? Hidden Faces Can you find his three daughters as well? Blind Spot An amazing demonstration of your blind spot. Impossible Prongs? Impossible prongs? What Am I? Can you tell what this is a picture of? Who Turned To? Who is missing? Same Eyes? Are her eyes the same colour? Parallel Cafe Wall Lines? Are the horizontal lines parallel? <UNMASK> [OK] Privacy Policy - Terms & Conditions - See DetailsWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners who may combine it with other information you've provided to them or they've collected from your use of their services. Puzzle Details In farmer Brown's hay loft there are a number of animals, in particular crows, mice and cockroaches. Being bored one day, I decided to count the animals and found there were exactly 150 feet and 50 heads in total, and there were twice as many cockroaches as mice. How many of each animal were there? <MASK> Answer: 10 cockroaches, 5 mice and 35 birds. <MASK> For every mouse there are two cockroaches, so every mouse is worth 3 heads and 16 feet (4 + 6 + 6). We can now write down an expression for the heads and the feet (calling birds B and mice M): B + 3M = 50 (1) Feet gives us: 2B + 16M = 150 (2) If we double (1) we get: 2B + 6M = 100 (3) We can now do (2) - (3) to give: 10M = 50 M = 5 So we have 5 mice (and 10 cockroaches). We can use M = 5 in (1) to give: B + 3 x 5 = 50 B + 15 = 50 B = 35 So, C = 10, M = 5 and B = 35. Checking that 10 + 5 + 35 = 50, and 10 x 6 + 5 x 4 + 35 x 2 = 150. QED. Our Favourite Illusions Shadow Illusion Are the squares A and B the same colour? Spinning Dancer Which way is the dancer spinning? Impossible Waterfall? Is the water flowing uphill in this impossible Escher type waterfall? The Butterfly A colourful butterfly? Duck Or Rabbit? Is this a duck or a rabbit? Hidden Faces Can you find his three daughters as well? Blind Spot An amazing demonstration of your blind spot. Impossible Prongs? Impossible prongs? What Am I? Can you tell what this is a picture of? Who Turned To? Who is missing? Same Eyes? Are her eyes the same colour? Parallel Cafe Wall Lines? Are the horizontal lines parallel?
[OK] Privacy Policy - Terms & Conditions - See DetailsWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners who may combine it with other information you've provided to them or they've collected from your use of their services. Puzzle Details In farmer Brown's hay loft there are a number of animals, in particular crows, mice and cockroaches. Being bored one day, I decided to count the animals and found there were exactly 150 feet and 50 heads in total, and there were twice as many cockroaches as mice. How many of each animal were there? <MASK> Answer: 10 cockroaches, 5 mice and 35 birds. <MASK> For every mouse there are two cockroaches, so every mouse is worth 3 heads and 16 feet (4 + 6 + 6). We can now write down an expression for the heads and the feet (calling birds B and mice M): B + 3M = 50 (1) Feet gives us: 2B + 16M = 150 (2) If we double (1) we get: 2B + 6M = 100 (3) We can now do (2) - (3) to give: 10M = 50 M = 5 So we have 5 mice (and 10 cockroaches). We can use M = 5 in (1) to give: B + 3 x 5 = 50 B + 15 = 50 B = 35 So, C = 10, M = 5 and B = 35. Checking that 10 + 5 + 35 = 50, and 10 x 6 + 5 x 4 + 35 x 2 = 150. QED. Our Favourite Illusions Shadow Illusion Are the squares A and B the same colour? Spinning Dancer Which way is the dancer spinning? Impossible Waterfall? Is the water flowing uphill in this impossible Escher type waterfall? The Butterfly A colourful butterfly? Duck Or Rabbit? Is this a duck or a rabbit? Hidden Faces Can you find his three daughters as well? Blind Spot An amazing demonstration of your blind spot. Impossible Prongs? Impossible prongs? What Am I? Can you tell what this is a picture of? Who Turned To? Who is missing? Same Eyes? Are her eyes the same colour? Parallel Cafe Wall Lines? Are the horizontal lines parallel? <UNMASK> [OK] Privacy Policy - Terms & Conditions - See DetailsWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners who may combine it with other information you've provided to them or they've collected from your use of their services. Puzzle Details In farmer Brown's hay loft there are a number of animals, in particular crows, mice and cockroaches. Being bored one day, I decided to count the animals and found there were exactly 150 feet and 50 heads in total, and there were twice as many cockroaches as mice. How many of each animal were there? [Ref: ZVYU] © Kevin Stone Answer: 10 cockroaches, 5 mice and 35 birds. Cockroaches have 6 feet, mice have 4 and birds have 2. For every mouse there are two cockroaches, so every mouse is worth 3 heads and 16 feet (4 + 6 + 6). We can now write down an expression for the heads and the feet (calling birds B and mice M): B + 3M = 50 (1) Feet gives us: 2B + 16M = 150 (2) If we double (1) we get: 2B + 6M = 100 (3) We can now do (2) - (3) to give: 10M = 50 M = 5 So we have 5 mice (and 10 cockroaches). We can use M = 5 in (1) to give: B + 3 x 5 = 50 B + 15 = 50 B = 35 So, C = 10, M = 5 and B = 35. Checking that 10 + 5 + 35 = 50, and 10 x 6 + 5 x 4 + 35 x 2 = 150. QED. Our Favourite Illusions Shadow Illusion Are the squares A and B the same colour? Spinning Dancer Which way is the dancer spinning? Impossible Waterfall? Is the water flowing uphill in this impossible Escher type waterfall? The Butterfly A colourful butterfly? Duck Or Rabbit? Is this a duck or a rabbit? Hidden Faces Can you find his three daughters as well? Blind Spot An amazing demonstration of your blind spot. Impossible Prongs? Impossible prongs? What Am I? Can you tell what this is a picture of? Who Turned To? Who is missing? Same Eyes? Are her eyes the same colour? Parallel Cafe Wall Lines? Are the horizontal lines parallel?
<MASK> <UNMASK> <MASK> (Assume that \$N is large compared to a penny, so that the distribution of prices is essentially continuous.) posted Jun 9 ## 1 Solution <MASK> 11. Of the Emperor and the insect worth \$105, one was won by Yvette and the other was from Luxembourg. <MASK> If A = 1 B = 2 C = 3 ... ... Z = 26. <MASK>
<MASK> (Assume that \$N is large compared to a penny, so that the distribution of prices is essentially continuous.) posted Jun 9 ## 1 Solution <MASK> 11. Of the Emperor and the insect worth \$105, one was won by Yvette and the other was from Luxembourg. <MASK> If A = 1 B = 2 C = 3 ... ... Z = 26. <MASK> <UNMASK> # You are out shopping with \$N, and you find an item whose price has a random value between \$0 and \$N. 27 views You are out shopping one day with \$N, and you find an item whose price has a random value between \$0 and \$N. You buy as many of these items as you can with your \$N. What is the expected value of the money you have left over? (Assume that \$N is large compared to a penny, so that the distribution of prices is essentially continuous.) posted Jun 9 ## 1 Solution expected value of the item=average of all the random numbers between 0 and N which is N/2. Hence he can buy 2 of these items and he will be left with nothing. solution Jun 19 Similar Puzzles A shipment of butterflies was mixed up by the dock workers, and they could not find who bought which species, where it was from, and what was the price. All the workers know is that Alejandro, Faye, Yvette, Sophie, and Zachary could have each bought butterflies that cost \$60, \$75, \$90, \$105, or \$120. Each could have bought the Clearwing, the Emperor, the Grayling, the Swallowtail, or the Torturix butterflies. Each butterfly could have lived in Australia, Jordan, Luxembourg, Panama, or Qatar. It is up to you to find out who bought which butterfly, what was the price, and where did it come from with the provided clues: 1. Neither the butterfly from Luxembourg nor the one from Australia sold for \$90. 2. The Emperor butterfly cost \$30 more than the Torturix butterfly. 3. Zachary's purchase was \$75. 4. The butterfly from Australia cost less than the one from Luxembourg. 5. Alejandro's purchase was from Luxembourg. 6. Of Yvette's purchase and the purchase for \$60, one was from Qatar and the other was the Torturix. 7. The butterfly that sold for \$120 was not from Panama. 8. The insect from Australia was not the Torturix. 9. Faye bought the Torturix. 10. Sophie did not buy the Grayling. 11. Of the Emperor and the insect worth \$105, one was won by Yvette and the other was from Luxembourg. 12. The insect that sold for \$105 was the Swallowtail. A family, made up of 2 parents with children, has an average age of 20. If you exclude one parent, who is 40, the average age drops to 15. How many kids are in the family? There is an equilateral triangle and three bugs are sitting on the three corners of the triangle. Each of the bugs picks up a random direction and starts walking along the edge of the equilateral triangle. What is the probability that none of the bugs crash into each other? +1 vote If A = 1 B = 2 C = 3 ... ... Z = 26. Based on above rule, you need to find an eleven letter word whose letter sum is equal to 52.
<MASK> where: <MASK> T = temperature (in K) <MASK> The rate constant of a reaction at 32°C is 0.055/s. If the frequency factor is 1.2 x 1013/s, what is the activation barrier? <UNMASK> <MASK> ###### FREE Expert Solution We’re being asked to determine the activation barrier (activation energy, Ea) of a reaction given the rate constant and frequency factor. <MASK> where: k = rate constant Ea = activation energy (in J/mol) R = gas constant (8.314 J/mol • K) T = temperature (in K) A = Arrhenius constant or frequency factor <MASK> The rate constant of a reaction at 32°C is 0.055/s. If the frequency factor is 1.2 x 1013/s, what is the activation barrier?
<MASK> ###### FREE Expert Solution We’re being asked to determine the activation barrier (activation energy, Ea) of a reaction given the rate constant and frequency factor. <MASK> where: k = rate constant Ea = activation energy (in J/mol) R = gas constant (8.314 J/mol • K) T = temperature (in K) A = Arrhenius constant or frequency factor <MASK> The rate constant of a reaction at 32°C is 0.055/s. If the frequency factor is 1.2 x 1013/s, what is the activation barrier? <UNMASK> # Problem: The rate constant of a reaction at 32°C is 0.055/s. If the frequency factor is 1.2 x 1013/s, what is the activation barrier? ###### FREE Expert Solution We’re being asked to determine the activation barrier (activation energy, Ea) of a reaction given the rate constant and frequency factor. We can use the two-point form of the Arrhenius Equation to calculate activation energy: where: k = rate constant Ea = activation energy (in J/mol) R = gas constant (8.314 J/mol • K) T = temperature (in K) A = Arrhenius constant or frequency factor 94% (270 ratings) ###### Problem Details The rate constant of a reaction at 32°C is 0.055/s. If the frequency factor is 1.2 x 1013/s, what is the activation barrier?
# HOW MUCH WILL I SPEND ON GAS? <MASK> 1 HOW MUCH WILL I SPEND ON GAS? Outcome (lesson objective) The students will use the current and future price of gasoline to construct T-charts, write algebraic equations, and plot the equations on a graph. Student/Class Goal Students will determine their gasoline cost for a month s time. They will use this information to calculate yearly gas cost. Time Frame Two 1 ½ hour classes One 3 hour class Standard Use Math to Solve Problems and Communicate NRS EFL 3-6 COPS Understand, interpret, and work with pictures, numbers, and symbolic information. Apply knowledge of mathematical concepts and procedures to figure out how to answer a question, solve a problem, make a prediction, or carry out a task that has a mathematical dimension. Define and select data to be used in solving the problem. Determine the degree of precision required by the situation. Solve problem using appropriate quantitative procedures and verify that the results are reasonable. Communicate results using a variety of mathematical representations, including graphs, charts, tables, and algebraic models. Materials Graph paper 40x30 paper is included at the end of the lesson (Worksheet 4) Colored pencils (if available) Calculators How Do I Find My Gas Mileage? Handout T-Chart Illustration How to Graph a Linear Equation in 5 Quick Steps Student Resource Activity Addresses Components of Performance Students will work with basic operations and patterns to complete a T-chart. Students will construct a graph and plot points on it. Students will use problem solving to determine how to figure out their monthly gas cost. Students will determine appropriate intervals on the x and y axes of the graph. Students will round the number when necessary to plot the points on the graph. Students will recognize if a data set is not reasonable by observing the other data sets and observing the points on the graph (is it on the line?) Students will communicate the results of the data by constructing a graph, T-chart, algebraic formula and in writing. Learner Prior Knowledge Basic understanding of gas prices and what is meant by miles per gallon (mpg), found at lesson, Pumped Up Gas Prices. Instructional Activities Step 1 - Discuss with students the current gasoline price and what they expect to happen to gas prices in the future. Ask students about how much they are spending on gas each month and how many miles the car they are driving gets per gallon? Students may need to research this information by actually calculating their mileage on the handout How Do I Find My Gas Mileage? or by visiting Cars and mpg This web site lists the in-city and highway mileage for 1985 or newer vehicles. If using this site, students will need to decide what type of driving they do during a month (city or highway) and estimate their car s mileage. Step 2 - Using \$4 per gallon as the price of gas, construct, with the class, a T-chart showing the relationship between the number of gallons purchased(x) and the total cost of the gas(y). See T-Chart Illustration for an example. As a class, look at the T-chart that was constructed. Give the students time to discover the relationship between the number of gallons purchased(x) and the total price of the gas(y). Express this relationship verbally (The number of gallons purchased times \$4 will equal the total cost.). Discuss how this relationship can be written as an equation as a function of x? (4x=y) Step 3 Next, with the students, graph the relationship/equation the class discovered during Step 2. The How to Graph a Linear Equation in 5 Quick Steps, a student resource from their math journals, provides basic guidelines for graphing equations. Help the students determine appropriate intervals and labels for the x and y axes and a title for the graph. Plot several points together and then let the students plot the remainder of the points. Draw a line passing through the points and label the line with the equation written in Step 2. Step 4 - Discuss with the students how many miles they drive each month. Share with the class that many people drive about 1,000 <MASK> Converting Units of Measure Measurement Outcome (lesson objective) Given a unit of measurement, students will be able to convert it to other units of measurement and will be able to use it to solve contextual <MASK> ### Solving Systems of Linear Equations Putting it All Together <MASK> TEMPERATURE BAR GRAPH Outcome (lesson objective) Students will figure mean, median and mode using weather, temperature data, create a bar graph charting one city s high and low temperatures, and formulate ### Lesson 4: Solving and Graphing Linear Equations <MASK> Tennessee Department of Education Task: Sally s Car Loan Sally bought a new car. Her total cost including all fees and taxes was \$15,. She made a down payment of \$43. She financed the remaining amount <MASK> Acquisition Lesson Plan Concept: Linear Systems Author Name(s): High-School Delaware Math Cadre Committee Grade: Ninth Grade Time Frame: Two 45 minute periods Pre-requisite(s): Write algebraic expressions <MASK> Solving Systems of Equations Introduction Outcome (learning objective) Students will write simple systems of equations and become familiar with systems of equations vocabulary terms. Student/Class Goal <MASK> SPIRIT 2.0 Lesson: A Point Of Intersection ================================Lesson Header============================= Lesson Title: A Point of Intersection Draft Date: 6/17/08 1st Author (Writer): Jenn <MASK> AP Physics 1 Summer Assignment AP Physics 1 Summer Assignment Welcome to AP Physics 1. This course and the AP exam will be challenging. AP classes are taught as college courses not just college-level courses, <MASK> ### In A Heartbeat (Algebra) <MASK> ### Linear Equations. Find the domain and the range of the following set. {(4,5), (7,8), (-1,3), (3,3), (2,-3)} <MASK> ### F.IF.7b: Graph Root, Piecewise, Step, & Absolute Value Functions <MASK> 1 Which ordered pair is not in the solution set of (1) (5,3) (2) (4,3) (3) (3,4) (4) (4,4) y 1 > x + 5 and y 3x 2? 2 5 2 If the quadratic formula is used to find the roots of the equation x 2 6x 19 = 0, <MASK> ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION INTEGRATED ALGEBRA. Wednesday, June 12, 2013 1:15 to 4:15 p.m. INTEGRATED ALGEBRA The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION INTEGRATED ALGEBRA Wednesday, June 12, 2013 1:15 to 4:15 p.m., only Student Name: School Name: The possession <MASK> ### Part 1: Background - Graphing Department of Physics and Geology Graphing Astronomy 1401 Equipment Needed Qty Computer with Data Studio Software 1 1.1 Graphing Part 1: Background - Graphing In science it is very important to find and <MASK> Graphs of Proportional Relationships Student Probe Susan runs three laps at the track in 12 minutes. A graph of this proportional relationship is shown below. Explain the meaning of points A (0,0), B (1,4), <MASK> ### Lesson 18: Introduction to Algebra: Expressions and Variables <MASK> ### High School Algebra Reasoning with Equations and Inequalities Solve systems of equations. <MASK> Math Objectives Given functions f and g, the student will be able to determine the domain and range of each as well as the composite functions defined by f ( g( x )) and g( f ( x )). Students will interpret <MASK> Grade 5 Key Areas of Focus for Grades 3-5: Multiplication and division of whole numbers and fractions-concepts, skills and problem solving Expected Fluency: Multi-digit multiplication Module M1: Whole <MASK> ### Algebra Cheat Sheets <MASK> 3 rd Grade Math Learning Targets Algebra: Indicator 1: Use procedures to transform algebraic expressions. 3.A.1.1. Students are able to explain the relationship between repeated addition and multiplication. <UNMASK> # HOW MUCH WILL I SPEND ON GAS? Save this PDF as: <MASK> 1 HOW MUCH WILL I SPEND ON GAS? Outcome (lesson objective) The students will use the current and future price of gasoline to construct T-charts, write algebraic equations, and plot the equations on a graph. Student/Class Goal Students will determine their gasoline cost for a month s time. They will use this information to calculate yearly gas cost. Time Frame Two 1 ½ hour classes One 3 hour class Standard Use Math to Solve Problems and Communicate NRS EFL 3-6 COPS Understand, interpret, and work with pictures, numbers, and symbolic information. Apply knowledge of mathematical concepts and procedures to figure out how to answer a question, solve a problem, make a prediction, or carry out a task that has a mathematical dimension. Define and select data to be used in solving the problem. Determine the degree of precision required by the situation. Solve problem using appropriate quantitative procedures and verify that the results are reasonable. Communicate results using a variety of mathematical representations, including graphs, charts, tables, and algebraic models. Materials Graph paper 40x30 paper is included at the end of the lesson (Worksheet 4) Colored pencils (if available) Calculators How Do I Find My Gas Mileage? Handout T-Chart Illustration How to Graph a Linear Equation in 5 Quick Steps Student Resource Activity Addresses Components of Performance Students will work with basic operations and patterns to complete a T-chart. Students will construct a graph and plot points on it. Students will use problem solving to determine how to figure out their monthly gas cost. Students will determine appropriate intervals on the x and y axes of the graph. Students will round the number when necessary to plot the points on the graph. Students will recognize if a data set is not reasonable by observing the other data sets and observing the points on the graph (is it on the line?) Students will communicate the results of the data by constructing a graph, T-chart, algebraic formula and in writing. Learner Prior Knowledge Basic understanding of gas prices and what is meant by miles per gallon (mpg), found at lesson, Pumped Up Gas Prices. Instructional Activities Step 1 - Discuss with students the current gasoline price and what they expect to happen to gas prices in the future. Ask students about how much they are spending on gas each month and how many miles the car they are driving gets per gallon? Students may need to research this information by actually calculating their mileage on the handout How Do I Find My Gas Mileage? or by visiting Cars and mpg This web site lists the in-city and highway mileage for 1985 or newer vehicles. If using this site, students will need to decide what type of driving they do during a month (city or highway) and estimate their car s mileage. Step 2 - Using \$4 per gallon as the price of gas, construct, with the class, a T-chart showing the relationship between the number of gallons purchased(x) and the total cost of the gas(y). See T-Chart Illustration for an example. As a class, look at the T-chart that was constructed. Give the students time to discover the relationship between the number of gallons purchased(x) and the total price of the gas(y). Express this relationship verbally (The number of gallons purchased times \$4 will equal the total cost.). Discuss how this relationship can be written as an equation as a function of x? (4x=y) Step 3 Next, with the students, graph the relationship/equation the class discovered during Step 2. The How to Graph a Linear Equation in 5 Quick Steps, a student resource from their math journals, provides basic guidelines for graphing equations. Help the students determine appropriate intervals and labels for the x and y axes and a title for the graph. Plot several points together and then let the students plot the remainder of the points. Draw a line passing through the points and label the line with the equation written in Step 2. Step 4 - Discuss with the students how many miles they drive each month. Share with the class that many people drive about 1,000 4 Gallons purchased (x) T-CHART ILLUSTRATION Total gas cost (y) 1 \$ \$ x30 Grid <MASK> Solving Systems of Linear Equations Substitutions Outcome (lesson objective) Students will accurately solve a system of equations algebraically using substitution. Student/Class Goal Students thinking <MASK> Converting Units of Measure Measurement Outcome (lesson objective) Given a unit of measurement, students will be able to convert it to other units of measurement and will be able to use it to solve contextual ### Solving Systems of Linear Equations Graphing Solving Systems of Linear Equations Graphing Outcome (learning objective) Students will accurately solve a system of equations by graphing. Student/Class Goal Students thinking about continuing their academic ### Solving Systems of Linear Equations Putting it All Together Solving Systems of Linear Equations Putting it All Together Outcome (lesson objective) Students will determine the best method to use when solving systems of equation as they solve problems using graphing, <MASK> TEMPERATURE BAR GRAPH Outcome (lesson objective) Students will figure mean, median and mode using weather, temperature data, create a bar graph charting one city s high and low temperatures, and formulate ### Lesson 4: Solving and Graphing Linear Equations <MASK> Tennessee Department of Education Task: Sally s Car Loan Sally bought a new car. Her total cost including all fees and taxes was \$15,. She made a down payment of \$43. She financed the remaining amount <MASK> Activity: TEKS: Exploring Transformations Basic understandings. (5) Tools for geometric thinking. Techniques for working with spatial figures and their properties are essential to understanding underlying <MASK> Acquisition Lesson Plan Concept: Linear Systems Author Name(s): High-School Delaware Math Cadre Committee Grade: Ninth Grade Time Frame: Two 45 minute periods Pre-requisite(s): Write algebraic expressions <MASK> Solving Systems of Equations Introduction Outcome (learning objective) Students will write simple systems of equations and become familiar with systems of equations vocabulary terms. Student/Class Goal <MASK> Name Date Linear Functions: Slope-Intercept Form Student Worksheet Overview The Overview introduces the topics covered in Observations and Activities. Scroll through the Overview using " (! to review, <MASK> ### THE LEAST SQUARES LINE (other names Best-Fit Line or Regression Line ) <MASK> ### Drive It Green. Time Frame: 3 class periods Drive It Green Lesson Overview: Millions of people around the globe are moving every day, creating a myriad of energy challenges and opportunities for new innovation. In this three-part lesson, students ### Graphing Linear Equations in Two Variables <MASK> SPIRIT 2.0 Lesson: A Point Of Intersection ================================Lesson Header============================= Lesson Title: A Point of Intersection Draft Date: 6/17/08 1st Author (Writer): Jenn <MASK> X Stackable Cert. Documentation Technology Study / Life skills EL-Civics Career Pathways Police Paramedic Fire Rescue Medical Asst. EKG / Cardio Phlebotomy Practical Nursing Healthcare Admin Pharmacy Tech <MASK> AP Physics 1 Summer Assignment AP Physics 1 Summer Assignment Welcome to AP Physics 1. This course and the AP exam will be challenging. AP classes are taught as college courses not just college-level courses, <MASK> ### In A Heartbeat (Algebra) The Middle School Math Project In A Heartbeat (Algebra) Objective Students will apply their knowledge of scatter plots to discover the correlation between heartbeats per minute before and after aerobic <MASK> ### Linear Equations. Find the domain and the range of the following set. {(4,5), (7,8), (-1,3), (3,3), (2,-3)} <MASK> ### Lesson 1: Review of Decimals: Addition, Subtraction, Multiplication <MASK> ### Scope and Sequence KA KB 1A 1B 2A 2B 3A 3B 4A 4B 5A 5B 6A 6B <MASK> ### F.IF.7b: Graph Root, Piecewise, Step, & Absolute Value Functions <MASK> Subject: Math Grade Level: 5 Topic: The Metric System Time Allotment: 45 minutes Teaching Date: Day 1 I. (A) Goal(s): For student to gain conceptual understanding of the metric system and how to convert <MASK> ### Part 1 Expressions, Equations, and Inequalities: Simplifying and Solving <MASK> ### Activity 6 Graphing Linear Equations <MASK> ### Title ID Number Sequence and Duration Age Level Essential Question Learning Objectives. Lead In <MASK> PREPARING A PERSONAL LETTER Outcome (lesson objective) Students will identify the parts and format of a personal (friendly) letter then write a letter using the appropriate format with proper spelling, <MASK> 1 Which ordered pair is not in the solution set of (1) (5,3) (2) (4,3) (3) (3,4) (4) (4,4) y 1 > x + 5 and y 3x 2? 2 5 2 If the quadratic formula is used to find the roots of the equation x 2 6x 19 = 0, <MASK> ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION INTEGRATED ALGEBRA. Wednesday, June 12, 2013 1:15 to 4:15 p.m. INTEGRATED ALGEBRA The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION INTEGRATED ALGEBRA Wednesday, June 12, 2013 1:15 to 4:15 p.m., only Student Name: School Name: The possession <MASK> Factoring Quadratic Trinomials Student Probe Factor Answer: Lesson Description This lesson uses the area model of multiplication to factor quadratic trinomials Part 1 of the lesson consists of circle puzzles <MASK> ### Part 1: Background - Graphing Department of Physics and Geology Graphing Astronomy 1401 Equipment Needed Qty Computer with Data Studio Software 1 1.1 Graphing Part 1: Background - Graphing In science it is very important to find and <MASK> X X Stackable Certificate Documentation Technology Study / Life skills EL-Civics Career Pathways Police Paramedic Fire Rescue Medical Asst. EKG / Cardio Phlebotomy Practical Nursing Healthcare Admin Pharmacy <MASK> Graphs of Proportional Relationships Student Probe Susan runs three laps at the track in 12 minutes. A graph of this proportional relationship is shown below. Explain the meaning of points A (0,0), B (1,4), ### N Q.3 Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. <MASK> ### Lesson 18: Introduction to Algebra: Expressions and Variables <MASK> ALGEBRA 1 ~ Cell Phone Task Group: Kimberly Allen, Matt Blundin, Nancy Bowen, Anna Green, Lee Hale, Katie Owens Math Essential Standards Approximate and interpret rates of change from graphical and numerical ### Relationships Between Two Variables: Scatterplots and Correlation <MASK> ### High School Algebra Reasoning with Equations and Inequalities Solve systems of equations. <MASK> ### HiMAP Pull-Out Section: Spring 1988 References <MASK> Math Objectives Given functions f and g, the student will be able to determine the domain and range of each as well as the composite functions defined by f ( g( x )) and g( f ( x )). Students will interpret <MASK> Grade 5 Key Areas of Focus for Grades 3-5: Multiplication and division of whole numbers and fractions-concepts, skills and problem solving Expected Fluency: Multi-digit multiplication Module M1: Whole <MASK> Unit #3: Investigating Quadratics (9 days + 1 jazz day + 1 summative evaluation day) BIG Ideas: Developing strategies for determining the zeroes of quadratic functions Making connections between the meaning <MASK> ### Algebra Cheat Sheets <MASK> Charts, Tables, and Graphs The Mathematics sections of the SAT also include some questions about charts, tables, and graphs. You should know how to (1) read and understand information that is given; (2) <MASK> 3 rd Grade Math Learning Targets Algebra: Indicator 1: Use procedures to transform algebraic expressions. 3.A.1.1. Students are able to explain the relationship between repeated addition and multiplication.
# HOW MUCH WILL I SPEND ON GAS? Save this PDF as: <MASK> 1 HOW MUCH WILL I SPEND ON GAS? Outcome (lesson objective) The students will use the current and future price of gasoline to construct T-charts, write algebraic equations, and plot the equations on a graph. Student/Class Goal Students will determine their gasoline cost for a month s time. They will use this information to calculate yearly gas cost. Time Frame Two 1 ½ hour classes One 3 hour class Standard Use Math to Solve Problems and Communicate NRS EFL 3-6 COPS Understand, interpret, and work with pictures, numbers, and symbolic information. Apply knowledge of mathematical concepts and procedures to figure out how to answer a question, solve a problem, make a prediction, or carry out a task that has a mathematical dimension. Define and select data to be used in solving the problem. Determine the degree of precision required by the situation. Solve problem using appropriate quantitative procedures and verify that the results are reasonable. Communicate results using a variety of mathematical representations, including graphs, charts, tables, and algebraic models. Materials Graph paper 40x30 paper is included at the end of the lesson (Worksheet 4) Colored pencils (if available) Calculators How Do I Find My Gas Mileage? Handout T-Chart Illustration How to Graph a Linear Equation in 5 Quick Steps Student Resource Activity Addresses Components of Performance Students will work with basic operations and patterns to complete a T-chart. Students will construct a graph and plot points on it. Students will use problem solving to determine how to figure out their monthly gas cost. Students will determine appropriate intervals on the x and y axes of the graph. Students will round the number when necessary to plot the points on the graph. Students will recognize if a data set is not reasonable by observing the other data sets and observing the points on the graph (is it on the line?) Students will communicate the results of the data by constructing a graph, T-chart, algebraic formula and in writing. Learner Prior Knowledge Basic understanding of gas prices and what is meant by miles per gallon (mpg), found at lesson, Pumped Up Gas Prices. Instructional Activities Step 1 - Discuss with students the current gasoline price and what they expect to happen to gas prices in the future. Ask students about how much they are spending on gas each month and how many miles the car they are driving gets per gallon? Students may need to research this information by actually calculating their mileage on the handout How Do I Find My Gas Mileage? or by visiting Cars and mpg This web site lists the in-city and highway mileage for 1985 or newer vehicles. If using this site, students will need to decide what type of driving they do during a month (city or highway) and estimate their car s mileage. Step 2 - Using \$4 per gallon as the price of gas, construct, with the class, a T-chart showing the relationship between the number of gallons purchased(x) and the total cost of the gas(y). See T-Chart Illustration for an example. As a class, look at the T-chart that was constructed. Give the students time to discover the relationship between the number of gallons purchased(x) and the total price of the gas(y). Express this relationship verbally (The number of gallons purchased times \$4 will equal the total cost.). Discuss how this relationship can be written as an equation as a function of x? (4x=y) Step 3 Next, with the students, graph the relationship/equation the class discovered during Step 2. The How to Graph a Linear Equation in 5 Quick Steps, a student resource from their math journals, provides basic guidelines for graphing equations. Help the students determine appropriate intervals and labels for the x and y axes and a title for the graph. Plot several points together and then let the students plot the remainder of the points. Draw a line passing through the points and label the line with the equation written in Step 2. Step 4 - Discuss with the students how many miles they drive each month. Share with the class that many people drive about 1,000 4 Gallons purchased (x) T-CHART ILLUSTRATION Total gas cost (y) 1 \$ \$ x30 Grid <MASK> Solving Systems of Linear Equations Substitutions Outcome (lesson objective) Students will accurately solve a system of equations algebraically using substitution. Student/Class Goal Students thinking <MASK> Converting Units of Measure Measurement Outcome (lesson objective) Given a unit of measurement, students will be able to convert it to other units of measurement and will be able to use it to solve contextual ### Solving Systems of Linear Equations Graphing Solving Systems of Linear Equations Graphing Outcome (learning objective) Students will accurately solve a system of equations by graphing. Student/Class Goal Students thinking about continuing their academic ### Solving Systems of Linear Equations Putting it All Together Solving Systems of Linear Equations Putting it All Together Outcome (lesson objective) Students will determine the best method to use when solving systems of equation as they solve problems using graphing, <MASK> TEMPERATURE BAR GRAPH Outcome (lesson objective) Students will figure mean, median and mode using weather, temperature data, create a bar graph charting one city s high and low temperatures, and formulate ### Lesson 4: Solving and Graphing Linear Equations <MASK> Tennessee Department of Education Task: Sally s Car Loan Sally bought a new car. Her total cost including all fees and taxes was \$15,. She made a down payment of \$43. She financed the remaining amount <MASK> Activity: TEKS: Exploring Transformations Basic understandings. (5) Tools for geometric thinking. Techniques for working with spatial figures and their properties are essential to understanding underlying <MASK> Acquisition Lesson Plan Concept: Linear Systems Author Name(s): High-School Delaware Math Cadre Committee Grade: Ninth Grade Time Frame: Two 45 minute periods Pre-requisite(s): Write algebraic expressions <MASK> Solving Systems of Equations Introduction Outcome (learning objective) Students will write simple systems of equations and become familiar with systems of equations vocabulary terms. Student/Class Goal <MASK> Name Date Linear Functions: Slope-Intercept Form Student Worksheet Overview The Overview introduces the topics covered in Observations and Activities. Scroll through the Overview using " (! to review, <MASK> ### THE LEAST SQUARES LINE (other names Best-Fit Line or Regression Line ) <MASK> ### Drive It Green. Time Frame: 3 class periods Drive It Green Lesson Overview: Millions of people around the globe are moving every day, creating a myriad of energy challenges and opportunities for new innovation. In this three-part lesson, students ### Graphing Linear Equations in Two Variables <MASK> SPIRIT 2.0 Lesson: A Point Of Intersection ================================Lesson Header============================= Lesson Title: A Point of Intersection Draft Date: 6/17/08 1st Author (Writer): Jenn <MASK> X Stackable Cert. Documentation Technology Study / Life skills EL-Civics Career Pathways Police Paramedic Fire Rescue Medical Asst. EKG / Cardio Phlebotomy Practical Nursing Healthcare Admin Pharmacy Tech <MASK> AP Physics 1 Summer Assignment AP Physics 1 Summer Assignment Welcome to AP Physics 1. This course and the AP exam will be challenging. AP classes are taught as college courses not just college-level courses, <MASK> ### In A Heartbeat (Algebra) The Middle School Math Project In A Heartbeat (Algebra) Objective Students will apply their knowledge of scatter plots to discover the correlation between heartbeats per minute before and after aerobic <MASK> ### Linear Equations. Find the domain and the range of the following set. {(4,5), (7,8), (-1,3), (3,3), (2,-3)} <MASK> ### Lesson 1: Review of Decimals: Addition, Subtraction, Multiplication <MASK> ### Scope and Sequence KA KB 1A 1B 2A 2B 3A 3B 4A 4B 5A 5B 6A 6B <MASK> ### F.IF.7b: Graph Root, Piecewise, Step, & Absolute Value Functions <MASK> Subject: Math Grade Level: 5 Topic: The Metric System Time Allotment: 45 minutes Teaching Date: Day 1 I. (A) Goal(s): For student to gain conceptual understanding of the metric system and how to convert <MASK> ### Part 1 Expressions, Equations, and Inequalities: Simplifying and Solving <MASK> ### Activity 6 Graphing Linear Equations <MASK> ### Title ID Number Sequence and Duration Age Level Essential Question Learning Objectives. Lead In <MASK> PREPARING A PERSONAL LETTER Outcome (lesson objective) Students will identify the parts and format of a personal (friendly) letter then write a letter using the appropriate format with proper spelling, <MASK> 1 Which ordered pair is not in the solution set of (1) (5,3) (2) (4,3) (3) (3,4) (4) (4,4) y 1 > x + 5 and y 3x 2? 2 5 2 If the quadratic formula is used to find the roots of the equation x 2 6x 19 = 0, <MASK> ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION INTEGRATED ALGEBRA. Wednesday, June 12, 2013 1:15 to 4:15 p.m. INTEGRATED ALGEBRA The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION INTEGRATED ALGEBRA Wednesday, June 12, 2013 1:15 to 4:15 p.m., only Student Name: School Name: The possession <MASK> Factoring Quadratic Trinomials Student Probe Factor Answer: Lesson Description This lesson uses the area model of multiplication to factor quadratic trinomials Part 1 of the lesson consists of circle puzzles <MASK> ### Part 1: Background - Graphing Department of Physics and Geology Graphing Astronomy 1401 Equipment Needed Qty Computer with Data Studio Software 1 1.1 Graphing Part 1: Background - Graphing In science it is very important to find and <MASK> X X Stackable Certificate Documentation Technology Study / Life skills EL-Civics Career Pathways Police Paramedic Fire Rescue Medical Asst. EKG / Cardio Phlebotomy Practical Nursing Healthcare Admin Pharmacy <MASK> Graphs of Proportional Relationships Student Probe Susan runs three laps at the track in 12 minutes. A graph of this proportional relationship is shown below. Explain the meaning of points A (0,0), B (1,4), ### N Q.3 Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. <MASK> ### Lesson 18: Introduction to Algebra: Expressions and Variables <MASK> ALGEBRA 1 ~ Cell Phone Task Group: Kimberly Allen, Matt Blundin, Nancy Bowen, Anna Green, Lee Hale, Katie Owens Math Essential Standards Approximate and interpret rates of change from graphical and numerical ### Relationships Between Two Variables: Scatterplots and Correlation <MASK> ### High School Algebra Reasoning with Equations and Inequalities Solve systems of equations. <MASK> ### HiMAP Pull-Out Section: Spring 1988 References <MASK> Math Objectives Given functions f and g, the student will be able to determine the domain and range of each as well as the composite functions defined by f ( g( x )) and g( f ( x )). Students will interpret <MASK> Grade 5 Key Areas of Focus for Grades 3-5: Multiplication and division of whole numbers and fractions-concepts, skills and problem solving Expected Fluency: Multi-digit multiplication Module M1: Whole <MASK> Unit #3: Investigating Quadratics (9 days + 1 jazz day + 1 summative evaluation day) BIG Ideas: Developing strategies for determining the zeroes of quadratic functions Making connections between the meaning <MASK> ### Algebra Cheat Sheets <MASK> Charts, Tables, and Graphs The Mathematics sections of the SAT also include some questions about charts, tables, and graphs. You should know how to (1) read and understand information that is given; (2) <MASK> 3 rd Grade Math Learning Targets Algebra: Indicator 1: Use procedures to transform algebraic expressions. 3.A.1.1. Students are able to explain the relationship between repeated addition and multiplication. <UNMASK> # HOW MUCH WILL I SPEND ON GAS? Save this PDF as: Size: px Start display at page: ## Transcription 1 HOW MUCH WILL I SPEND ON GAS? Outcome (lesson objective) The students will use the current and future price of gasoline to construct T-charts, write algebraic equations, and plot the equations on a graph. Student/Class Goal Students will determine their gasoline cost for a month s time. They will use this information to calculate yearly gas cost. Time Frame Two 1 ½ hour classes One 3 hour class Standard Use Math to Solve Problems and Communicate NRS EFL 3-6 COPS Understand, interpret, and work with pictures, numbers, and symbolic information. Apply knowledge of mathematical concepts and procedures to figure out how to answer a question, solve a problem, make a prediction, or carry out a task that has a mathematical dimension. Define and select data to be used in solving the problem. Determine the degree of precision required by the situation. Solve problem using appropriate quantitative procedures and verify that the results are reasonable. Communicate results using a variety of mathematical representations, including graphs, charts, tables, and algebraic models. Materials Graph paper 40x30 paper is included at the end of the lesson (Worksheet 4) Colored pencils (if available) Calculators How Do I Find My Gas Mileage? Handout T-Chart Illustration How to Graph a Linear Equation in 5 Quick Steps Student Resource Activity Addresses Components of Performance Students will work with basic operations and patterns to complete a T-chart. Students will construct a graph and plot points on it. Students will use problem solving to determine how to figure out their monthly gas cost. Students will determine appropriate intervals on the x and y axes of the graph. Students will round the number when necessary to plot the points on the graph. Students will recognize if a data set is not reasonable by observing the other data sets and observing the points on the graph (is it on the line?) Students will communicate the results of the data by constructing a graph, T-chart, algebraic formula and in writing. Learner Prior Knowledge Basic understanding of gas prices and what is meant by miles per gallon (mpg), found at lesson, Pumped Up Gas Prices. Instructional Activities Step 1 - Discuss with students the current gasoline price and what they expect to happen to gas prices in the future. Ask students about how much they are spending on gas each month and how many miles the car they are driving gets per gallon? Students may need to research this information by actually calculating their mileage on the handout How Do I Find My Gas Mileage? or by visiting Cars and mpg This web site lists the in-city and highway mileage for 1985 or newer vehicles. If using this site, students will need to decide what type of driving they do during a month (city or highway) and estimate their car s mileage. Step 2 - Using \$4 per gallon as the price of gas, construct, with the class, a T-chart showing the relationship between the number of gallons purchased(x) and the total cost of the gas(y). See T-Chart Illustration for an example. As a class, look at the T-chart that was constructed. Give the students time to discover the relationship between the number of gallons purchased(x) and the total price of the gas(y). Express this relationship verbally (The number of gallons purchased times \$4 will equal the total cost.). Discuss how this relationship can be written as an equation as a function of x? (4x=y) Step 3 Next, with the students, graph the relationship/equation the class discovered during Step 2. The How to Graph a Linear Equation in 5 Quick Steps, a student resource from their math journals, provides basic guidelines for graphing equations. Help the students determine appropriate intervals and labels for the x and y axes and a title for the graph. Plot several points together and then let the students plot the remainder of the points. Draw a line passing through the points and label the line with the equation written in Step 2. Step 4 - Discuss with the students how many miles they drive each month. Share with the class that many people drive about 1,000 4 Gallons purchased (x) T-CHART ILLUSTRATION Total gas cost (y) 1 \$ \$ x30 Grid 5 How to Graph a Linear Equation in 5 Quick Steps Step 1 Construct a T-chart of Values Using your equation, construct a T-chart of values if one has not been done already. Substitute some simple numbers into the equation for x or y. If x=1, what is y? If x=10, what is y? If y=0, what is x? Each pair of values in your T-chart will become a point on the graph. (See illustration 1 for an example of a T-chart) Step 2 Decide on the interval for each axis Before starting the graph, look at the T-chart to determine the highest value for y found on the chart. Look at the values needed for x. Using graph paper, count the number of lines on the x and y axes. Use these numbers to determine the intervals on each axis. (If you use the graph paper at the end of this lesson there are 30 spaces on the x axis and 40 spaces on the y axis.) If the largest total cost/y value that needs to be graphed is \$80 and there are 40 lines on the y axis, let each line on the y axis represent \$2. The number of gallons of gas/x value that goes with \$80 is 20. There are 30 lines, so to make it simple one line will equal one gallon. Be sure the students realize they do not need to put a number next to every line. For example, the x might be labeled on every 5 th line (five gallons) and the y axis might also be labeled on every 5 th line (or \$10). This is a good step to do in pencil. That way if the interval you selected did not work out, the numbers can be erased any you can start over. Step 3 Label each Axis Decide what labels need to be added to the x and y axis. What do the numbers on the x-axis represent? What do the numbers on the y-axis represent? Usually the labels will match the descriptions/labels of x and y on the T-chart. (Note: When graphing equations involving elapsed time, time is traditionally represented by x) Step 4 Plot the points Using each pair of points from the T-chart, plot the points on the graph. Every point does not need to be plotted. Just be sure you have at least 3. Using a ruler, draw a line through the points you have plotted. Write your equation next to the line. Step 5 Give the graph a title Decide on a title for the graph. Make sure it accurately represents what is being shown on the graph. Does it explain the relationship between x and y? How to Graph a Linear Equation in 5 Quick Steps Student Resource ### Solving Systems of Linear Equations Substitutions Solving Systems of Linear Equations Substitutions Outcome (lesson objective) Students will accurately solve a system of equations algebraically using substitution. Student/Class Goal Students thinking ### Solving Systems of Linear Equations Elimination (Addition) Solving Systems of Linear Equations Elimination (Addition) Outcome (lesson objective) Students will accurately solve systems of equations using elimination/addition method. Student/Class Goal Students ### Solving Systems of Linear Equations Substitutions Solving Systems of Linear Equations Substitutions Outcome (learning objective) Students will accurately solve a system of equations algebraically using substitution. Student/Class Goal Students thinking ### Converting Units of Measure Measurement Converting Units of Measure Measurement Outcome (lesson objective) Given a unit of measurement, students will be able to convert it to other units of measurement and will be able to use it to solve contextual ### Solving Systems of Linear Equations Graphing Solving Systems of Linear Equations Graphing Outcome (learning objective) Students will accurately solve a system of equations by graphing. Student/Class Goal Students thinking about continuing their academic ### Solving Systems of Linear Equations Putting it All Together Solving Systems of Linear Equations Putting it All Together Outcome (lesson objective) Students will determine the best method to use when solving systems of equation as they solve problems using graphing, ### TEMPERATURE BAR GRAPH TEMPERATURE BAR GRAPH Outcome (lesson objective) Students will figure mean, median and mode using weather, temperature data, create a bar graph charting one city s high and low temperatures, and formulate ### Lesson 4: Solving and Graphing Linear Equations Lesson 4: Solving and Graphing Linear Equations Selected Content Standards Benchmarks Addressed: A-2-M Modeling and developing methods for solving equations and inequalities (e.g., using charts, graphs, ### Solving Systems of Linear Equations Putting it All Together Solving Systems of Linear Equations Putting it All Together Student/Class Goal Students thinking about continuing their academic studies in a post-secondary institution will need to know and be able to ### Lesson 2: Constructing Line Graphs and Bar Graphs Lesson 2: Constructing Line Graphs and Bar Graphs Selected Content Standards Benchmarks Assessed: D.1 Designing and conducting statistical experiments that involve the collection, representation, and analysis ### Tennessee Department of Education. Task: Sally s Car Loan Tennessee Department of Education Task: Sally s Car Loan Sally bought a new car. Her total cost including all fees and taxes was \$15,. She made a down payment of \$43. She financed the remaining amount ### Basic Understandings Activity: TEKS: Exploring Transformations Basic understandings. (5) Tools for geometric thinking. Techniques for working with spatial figures and their properties are essential to understanding underlying ### Acquisition Lesson Plan for the Concept, Topic or Skill---Not for the Day Acquisition Lesson Plan Concept: Linear Systems Author Name(s): High-School Delaware Math Cadre Committee Grade: Ninth Grade Time Frame: Two 45 minute periods Pre-requisite(s): Write algebraic expressions ### Linear Equations. 5- Day Lesson Plan Unit: Linear Equations Grade Level: Grade 9 Time Span: 50 minute class periods By: Richard Weber Linear Equations 5- Day Lesson Plan Unit: Linear Equations Grade Level: Grade 9 Time Span: 50 minute class periods By: Richard Weber Tools: Geometer s Sketchpad Software Overhead projector with TI- 83 ### Lines, Lines, Lines!!! Slope-Intercept Form ~ Lesson Plan Lines, Lines, Lines!!! Slope-Intercept Form ~ Lesson Plan I. Topic: Slope-Intercept Form II. III. Goals and Objectives: A. The student will write an equation of a line given information about its graph. ### Teacher: Maple So School: Herron High School. Comparing the Usage Cost of Electric Vehicles Versus Internal Combustion Vehicles Teacher: Maple So School: Herron High School Name of Lesson: Comparing the Usage Cost of Electric Vehicles Versus Internal Combustion Vehicles Subject/ Course: Mathematics, Algebra I Grade Level: 9 th ### Have pairs share with the class. Students can work in pairs to categorize their food according to the food groups in the food pyramid. FOOD PYRAMID MENU Outcome (lesson objective) Students will read charts about nutrition, exercise, and caloric intake, practice vocabulary strategies and integrate that information with their own experience ### Solving Systems of Equations Introduction Solving Systems of Equations Introduction Outcome (learning objective) Students will write simple systems of equations and become familiar with systems of equations vocabulary terms. 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Find the domain and the range of the following set. {(4,5), (7,8), (-1,3), (3,3), (2,-3)} Linear Equations Domain and Range Domain refers to the set of possible values of the x-component of a point in the form (x,y). Range refers to the set of possible values of the y-component of a point in ### Lesson 1: Review of Decimals: Addition, Subtraction, Multiplication LESSON 1: REVIEW OF DECIMALS: ADDITION AND SUBTRACTION Weekly Focus: whole numbers and decimals Weekly Skill: place value, add, subtract, multiply Lesson Summary: In the warm up, students will solve a ### Scope and Sequence KA KB 1A 1B 2A 2B 3A 3B 4A 4B 5A 5B 6A 6B Scope and Sequence Earlybird Kindergarten, Standards Edition Primary Mathematics, Standards Edition Copyright 2008 [SingaporeMath.com Inc.] The check mark indicates where the topic is first introduced ### EQUATIONS and INEQUALITIES EQUATIONS and INEQUALITIES Linear Equations and Slope 1. Slope a. Calculate the slope of a line given two points b. 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Tuesday, January 22, 2013 9:15 a.m. to 12:15 p.m. INTEGRATED ALGEBRA The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION INTEGRATED ALGEBRA Tuesday, January 22, 2013 9:15 a.m. to 12:15 p.m., only Student Name: School Name: The possession ### Algebra I Sample Questions. 1 Which ordered pair is not in the solution set of (1) (5,3) (2) (4,3) (3) (3,4) (4) (4,4) 1 Which ordered pair is not in the solution set of (1) (5,3) (2) (4,3) (3) (3,4) (4) (4,4) y 1 > x + 5 and y 3x 2? 2 5 2 If the quadratic formula is used to find the roots of the equation x 2 6x 19 = 0, ### EXCEL Tutorial: How to use EXCEL for Graphs and Calculations. EXCEL Tutorial: How to use EXCEL for Graphs and Calculations. Excel is powerful tool and can make your life easier if you are proficient in using it. You will need to use Excel to complete most of your ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION INTEGRATED ALGEBRA. 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Students will interpret ### Accommodated Lesson Plan on Solving Systems of Equations by Elimination for Diego Accommodated Lesson Plan on Solving Systems of Equations by Elimination for Diego Courtney O Donovan Class: Algebra 1 Day #: 6-7 Grade: 8th Number of Students: 25 Date: May 12-13, 2011 Goal: Students will ### Solutions of Equations in Two Variables 6.1 Solutions of Equations in Two Variables 6.1 OBJECTIVES 1. Find solutions for an equation in two variables 2. Use ordered pair notation to write solutions for equations in two variables We discussed ### High School Algebra Reasoning with Equations and Inequalities Solve equations and inequalities in one variable. Performance Assessment Task Quadratic (2009) Grade 9 The task challenges a student to demonstrate an understanding of quadratic functions in various forms. A student must make sense of the meaning of relations ### Years after 2000. US Student to Teacher Ratio 0 16.048 1 15.893 2 15.900 3 15.900 4 15.800 5 15.657 6 15.540 To complete this technology assignment, you should already have created a scatter plot for your data on your calculator and/or in Excel. 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Indicator 6 Extend understanding of coordinate system to include points ### GETTING TO THE CORE: THE LINK BETWEEN TEMPERATURE AND CARBON DIOXIDE DESCRIPTION This lesson plan gives students first-hand experience in analyzing the link between atmospheric temperatures and carbon dioxide ( ) s by looking at ice core data spanning hundreds of thousands ### Algebra Cheat Sheets Sheets Algebra Cheat Sheets provide you with a tool for teaching your students note-taking, problem-solving, and organizational skills in the context of algebra lessons. These sheets teach the concepts ### Summer Math Exercises. For students who are entering. Pre-Calculus Summer Math Eercises For students who are entering Pre-Calculus It has been discovered that idle students lose learning over the summer months. 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<MASK> Let us draw the graph of the equation using the coordinates of $$x$$ and $$y$$-intercepts. The graph can be obtained by plotting the $$x$$ and $$y$$-intercepts and then drawing a line joining these points. Example: Draw the graph of the equation $$4y-3x = 6$$ using the $$x$$ and $$y$$-intercepts. Solution: To find the $$x$$-intercept, put $$y = 0$$ in the given equation. <MASK> $$x = \frac{6}{-3}$$ $$x = -2$$ <MASK> Similarly, to find the $$y$$-intercept, put $$x = 0$$ in the given equation. <MASK> $$4y = 6$$ $$y = \frac{6}{4}$$ $$y = \frac{3}{2}$$ <MASK> <UNMASK> <MASK> Let us draw the graph of the equation using the coordinates of $$x$$ and $$y$$-intercepts. The graph can be obtained by plotting the $$x$$ and $$y$$-intercepts and then drawing a line joining these points. Example: Draw the graph of the equation $$4y-3x = 6$$ using the $$x$$ and $$y$$-intercepts. Solution: To find the $$x$$-intercept, put $$y = 0$$ in the given equation. $$4(0)-3x = 6$$ $$0-3x = 6$$ $$-3x = 6$$ $$x = \frac{6}{-3}$$ $$x = -2$$ Thus, the $$x$$-intercept is $$x = -2$$. Similarly, to find the $$y$$-intercept, put $$x = 0$$ in the given equation. <MASK> $$4y-0 = 6$$ $$4y = 6$$ $$y = \frac{6}{4}$$ $$y = \frac{3}{2}$$ Thus, the $$y$$-intercept is $$\frac{3}{2}$$. We shall plot the graph using these two coordinates $$(-2,0)$$, and $$(0,\frac{3}{2})$$ and then, draw a line through the two points.
<MASK> Let us draw the graph of the equation using the coordinates of $$x$$ and $$y$$-intercepts. The graph can be obtained by plotting the $$x$$ and $$y$$-intercepts and then drawing a line joining these points. Example: Draw the graph of the equation $$4y-3x = 6$$ using the $$x$$ and $$y$$-intercepts. Solution: To find the $$x$$-intercept, put $$y = 0$$ in the given equation. $$4(0)-3x = 6$$ $$0-3x = 6$$ $$-3x = 6$$ $$x = \frac{6}{-3}$$ $$x = -2$$ Thus, the $$x$$-intercept is $$x = -2$$. Similarly, to find the $$y$$-intercept, put $$x = 0$$ in the given equation. <MASK> $$4y-0 = 6$$ $$4y = 6$$ $$y = \frac{6}{4}$$ $$y = \frac{3}{2}$$ Thus, the $$y$$-intercept is $$\frac{3}{2}$$. We shall plot the graph using these two coordinates $$(-2,0)$$, and $$(0,\frac{3}{2})$$ and then, draw a line through the two points. <UNMASK> ### Theory: Let us draw the graph of the equation using the coordinates of $$x$$ and $$y$$-intercepts. The graph can be obtained by plotting the $$x$$ and $$y$$-intercepts and then drawing a line joining these points. Example: Draw the graph of the equation $$4y-3x = 6$$ using the $$x$$ and $$y$$-intercepts. Solution: To find the $$x$$-intercept, put $$y = 0$$ in the given equation. $$4(0)-3x = 6$$ $$0-3x = 6$$ $$-3x = 6$$ $$x = \frac{6}{-3}$$ $$x = -2$$ Thus, the $$x$$-intercept is $$x = -2$$. Similarly, to find the $$y$$-intercept, put $$x = 0$$ in the given equation. $$4y-3(0) = 6$$ $$4y-0 = 6$$ $$4y = 6$$ $$y = \frac{6}{4}$$ $$y = \frac{3}{2}$$ Thus, the $$y$$-intercept is $$\frac{3}{2}$$. We shall plot the graph using these two coordinates $$(-2,0)$$, and $$(0,\frac{3}{2})$$ and then, draw a line through the two points.
<MASK> May 30, 2015 <MASK> Use this pair to split the middle term then factor by grouping... <MASK> $= y \left(4 y + 3\right) - 4 \left(4 y + 3\right)$ <MASK> <UNMASK> # How do you factor 4y² -13y -12? May 30, 2015 <MASK> Look for a pair of factors of $A C = 48$ whose difference is $B = 13$. <MASK> Use this pair to split the middle term then factor by grouping... <MASK> $= y \left(4 y + 3\right) - 4 \left(4 y + 3\right)$ <MASK>
# How do you factor 4y² -13y -12? May 30, 2015 <MASK> Look for a pair of factors of $A C = 48$ whose difference is $B = 13$. <MASK> Use this pair to split the middle term then factor by grouping... <MASK> $= y \left(4 y + 3\right) - 4 \left(4 y + 3\right)$ <MASK> <UNMASK> # How do you factor 4y² -13y -12? May 30, 2015 Use a version of the AC Method. $A = 4$, $B = 13$, $C = 12$ Look for a pair of factors of $A C = 48$ whose difference is $B = 13$. $16$ and $3$ work. Use this pair to split the middle term then factor by grouping... $4 {y}^{2} - 13 y - 12$ $= 4 {y}^{2} + 3 y - 16 y - 12$ $= \left(4 {y}^{2} + 3 y\right) - \left(16 y + 12\right)$ $= y \left(4 y + 3\right) - 4 \left(4 y + 3\right)$ $= \left(y - 4\right) \left(4 y + 3\right)$
# SBI Clerk Pre Quantitative Aptitude Quiz- 07 ## SBI Clerk Pre Quantitative Aptitude Quiz Quantitative aptitude measures a candidate’s numerical proficiency and problem-solving abilities. It is the most important section of almost all competitive exams. Candidates are often stymied by the complexity of Quantitative Aptitude Questions but if they practice more and more questions, it will become quite easy. So, here we are providing you with the SBI Clerk Pre-Quantitative Aptitude Quiz to enhance your preparation for your upcoming examination. Questions given in this SBI Clerk Pre-Quantitative Aptitude Quiz are based on the most recent and the latest exam pattern. A detailed explanation for each question will be given in this SBI Clerk Pre-Quantitative Aptitude Quiz. This SBI Clerk Pre-Quantitative Aptitude Quiz is entirely free of charge. This SBI Clerk Pre-Quantitative Aptitude Quiz will assist aspirants in achieving a good score in their upcoming examinations. <MASK> 4. In how many ways can letter of word ‘PROMISE’ be arranged such that all vowels always come together? (a) 720 (b) 120 (c) 960 (d) 880 (e) 480 <MASK> Directions (6-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. <MASK> Recommended PDF’s for: #### Most important PDF’s for Bank, SSC, Railway and Other Government Exam : Download PDF Now AATMA-NIRBHAR Series- Static GK/Awareness Practice Ebook PDF Get PDF here The Banking Awareness 500 MCQs E-book\| Bilingual (Hindi + English) Get PDF here AATMA-NIRBHAR Series- Banking Awareness Practice Ebook PDF Get PDF here Computer Awareness Capsule 2.O Get PDF here AATMA-NIRBHAR Series Quantitative Aptitude Topic-Wise PDF Get PDF here AATMA-NIRBHAR Series Reasoning Topic-Wise PDF Get PDF Here Memory Based Puzzle E-book \| 2016-19 Exams Covered Get PDF here Caselet Data Interpretation 200 Questions Get PDF here Puzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1) Get PDF here ARITHMETIC DATA INTERPRETATION 2.O E-book Get PDF here 3 <UNMASK> # SBI Clerk Pre Quantitative Aptitude Quiz- 07 ## SBI Clerk Pre Quantitative Aptitude Quiz Quantitative aptitude measures a candidate’s numerical proficiency and problem-solving abilities. It is the most important section of almost all competitive exams. Candidates are often stymied by the complexity of Quantitative Aptitude Questions but if they practice more and more questions, it will become quite easy. So, here we are providing you with the SBI Clerk Pre-Quantitative Aptitude Quiz to enhance your preparation for your upcoming examination. Questions given in this SBI Clerk Pre-Quantitative Aptitude Quiz are based on the most recent and the latest exam pattern. A detailed explanation for each question will be given in this SBI Clerk Pre-Quantitative Aptitude Quiz. This SBI Clerk Pre-Quantitative Aptitude Quiz is entirely free of charge. This SBI Clerk Pre-Quantitative Aptitude Quiz will assist aspirants in achieving a good score in their upcoming examinations. 1. A dishonest cloth merchant sales cloth at the cost price but uses false scale which measures 80 cm in lieu of 1 m. Find his gain percentage? (a) 20% (b) 25% (c) 15% (d) 12% (e) 22% 2. The area of two squares is in the ratio 225 : 256. Find ratio of their diagonals? <MASK> 4. In how many ways can letter of word ‘PROMISE’ be arranged such that all vowels always come together? (a) 720 (b) 120 (c) 960 (d) 880 (e) 480 5. Find ratio between S.I. and C.I. on a sum of money invested for 3 years at 5% rate of interest per annum? (a) 15 : 1261 (b) 151 : 156 (c) 121 : 441 (d) 1200 : 1261 (e) 121 : 484 Directions (6-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. Click to Buy Bank MahaCombo Package Recommended PDF’s for: #### Most important PDF’s for Bank, SSC, Railway and Other Government Exam : Download PDF Now AATMA-NIRBHAR Series- Static GK/Awareness Practice Ebook PDF Get PDF here The Banking Awareness 500 MCQs E-book\| Bilingual (Hindi + English) Get PDF here AATMA-NIRBHAR Series- Banking Awareness Practice Ebook PDF Get PDF here Computer Awareness Capsule 2.O Get PDF here AATMA-NIRBHAR Series Quantitative Aptitude Topic-Wise PDF Get PDF here AATMA-NIRBHAR Series Reasoning Topic-Wise PDF Get PDF Here Memory Based Puzzle E-book \| 2016-19 Exams Covered Get PDF here Caselet Data Interpretation 200 Questions Get PDF here Puzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1) Get PDF here ARITHMETIC DATA INTERPRETATION 2.O E-book Get PDF here 3
# SBI Clerk Pre Quantitative Aptitude Quiz- 07 ## SBI Clerk Pre Quantitative Aptitude Quiz Quantitative aptitude measures a candidate’s numerical proficiency and problem-solving abilities. It is the most important section of almost all competitive exams. Candidates are often stymied by the complexity of Quantitative Aptitude Questions but if they practice more and more questions, it will become quite easy. So, here we are providing you with the SBI Clerk Pre-Quantitative Aptitude Quiz to enhance your preparation for your upcoming examination. Questions given in this SBI Clerk Pre-Quantitative Aptitude Quiz are based on the most recent and the latest exam pattern. A detailed explanation for each question will be given in this SBI Clerk Pre-Quantitative Aptitude Quiz. This SBI Clerk Pre-Quantitative Aptitude Quiz is entirely free of charge. This SBI Clerk Pre-Quantitative Aptitude Quiz will assist aspirants in achieving a good score in their upcoming examinations. 1. A dishonest cloth merchant sales cloth at the cost price but uses false scale which measures 80 cm in lieu of 1 m. Find his gain percentage? (a) 20% (b) 25% (c) 15% (d) 12% (e) 22% 2. The area of two squares is in the ratio 225 : 256. Find ratio of their diagonals? <MASK> 4. In how many ways can letter of word ‘PROMISE’ be arranged such that all vowels always come together? (a) 720 (b) 120 (c) 960 (d) 880 (e) 480 5. Find ratio between S.I. and C.I. on a sum of money invested for 3 years at 5% rate of interest per annum? (a) 15 : 1261 (b) 151 : 156 (c) 121 : 441 (d) 1200 : 1261 (e) 121 : 484 Directions (6-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. Click to Buy Bank MahaCombo Package Recommended PDF’s for: #### Most important PDF’s for Bank, SSC, Railway and Other Government Exam : Download PDF Now AATMA-NIRBHAR Series- Static GK/Awareness Practice Ebook PDF Get PDF here The Banking Awareness 500 MCQs E-book\| Bilingual (Hindi + English) Get PDF here AATMA-NIRBHAR Series- Banking Awareness Practice Ebook PDF Get PDF here Computer Awareness Capsule 2.O Get PDF here AATMA-NIRBHAR Series Quantitative Aptitude Topic-Wise PDF Get PDF here AATMA-NIRBHAR Series Reasoning Topic-Wise PDF Get PDF Here Memory Based Puzzle E-book \| 2016-19 Exams Covered Get PDF here Caselet Data Interpretation 200 Questions Get PDF here Puzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1) Get PDF here ARITHMETIC DATA INTERPRETATION 2.O E-book Get PDF here 3 <UNMASK> # SBI Clerk Pre Quantitative Aptitude Quiz- 07 ## SBI Clerk Pre Quantitative Aptitude Quiz Quantitative aptitude measures a candidate’s numerical proficiency and problem-solving abilities. It is the most important section of almost all competitive exams. Candidates are often stymied by the complexity of Quantitative Aptitude Questions but if they practice more and more questions, it will become quite easy. So, here we are providing you with the SBI Clerk Pre-Quantitative Aptitude Quiz to enhance your preparation for your upcoming examination. Questions given in this SBI Clerk Pre-Quantitative Aptitude Quiz are based on the most recent and the latest exam pattern. A detailed explanation for each question will be given in this SBI Clerk Pre-Quantitative Aptitude Quiz. This SBI Clerk Pre-Quantitative Aptitude Quiz is entirely free of charge. This SBI Clerk Pre-Quantitative Aptitude Quiz will assist aspirants in achieving a good score in their upcoming examinations. 1. A dishonest cloth merchant sales cloth at the cost price but uses false scale which measures 80 cm in lieu of 1 m. Find his gain percentage? (a) 20% (b) 25% (c) 15% (d) 12% (e) 22% 2. The area of two squares is in the ratio 225 : 256. Find ratio of their diagonals? 3. ‘P’ sells his watch at 20% profit to Q while Q sales it to R at a loss of 10%. If R pays Rs. 2160. Find at what price P sold watch to Q? (a) Rs. 2000 (b) Rs. 2200 (c) Rs. 2400 (d) Rs. 1800 (e) Rs. 2500 4. In how many ways can letter of word ‘PROMISE’ be arranged such that all vowels always come together? (a) 720 (b) 120 (c) 960 (d) 880 (e) 480 5. Find ratio between S.I. and C.I. on a sum of money invested for 3 years at 5% rate of interest per annum? (a) 15 : 1261 (b) 151 : 156 (c) 121 : 441 (d) 1200 : 1261 (e) 121 : 484 Directions (6-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. Click to Buy Bank MahaCombo Package Recommended PDF’s for: #### Most important PDF’s for Bank, SSC, Railway and Other Government Exam : Download PDF Now AATMA-NIRBHAR Series- Static GK/Awareness Practice Ebook PDF Get PDF here The Banking Awareness 500 MCQs E-book\| Bilingual (Hindi + English) Get PDF here AATMA-NIRBHAR Series- Banking Awareness Practice Ebook PDF Get PDF here Computer Awareness Capsule 2.O Get PDF here AATMA-NIRBHAR Series Quantitative Aptitude Topic-Wise PDF Get PDF here AATMA-NIRBHAR Series Reasoning Topic-Wise PDF Get PDF Here Memory Based Puzzle E-book \| 2016-19 Exams Covered Get PDF here Caselet Data Interpretation 200 Questions Get PDF here Puzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1) Get PDF here ARITHMETIC DATA INTERPRETATION 2.O E-book Get PDF here 3
<MASK> g(x) = 3x-2 <MASK> <UNMASK> Search 75,714 tutors 0 0 ## find the composite function Given the f(x)= 4/x2 and g(x) = 3-2x, find the composite function (f º g) and simplify. Given f(x) and g(x), (f º g)(x) = f(g(x)) <MASK> g(x) = 3x-2 <MASK> f(x)= 4/x2 and g(x) = 3-2x f(g(x)) = 4/ (3 - 2x)2 f(g(x)) = 4/(4x2 - 12x + 9)
Search 75,714 tutors 0 0 ## find the composite function Given the f(x)= 4/x2 and g(x) = 3-2x, find the composite function (f º g) and simplify. Given f(x) and g(x), (f º g)(x) = f(g(x)) <MASK> g(x) = 3x-2 <MASK> f(x)= 4/x2 and g(x) = 3-2x f(g(x)) = 4/ (3 - 2x)2 f(g(x)) = 4/(4x2 - 12x + 9) <UNMASK> Search 75,714 tutors 0 0 ## find the composite function Given the f(x)= 4/x2 and g(x) = 3-2x, find the composite function (f º g) and simplify. Given f(x) and g(x), (f º g)(x) = f(g(x)) with f(x) = 4/x2 and g(x) = 3-2x, First apply g, then apply f to that result as follows: g(x) = 3x-2 f(g(x)) = 4/ (3 - 2x)2 = 4/ ((3-2x)(3-2x)) ## f(g(x)) = 4/ (4x2-12x+9) f(x)= 4/x2 and g(x) = 3-2x f(g(x)) = 4/ (3 - 2x)2 f(g(x)) = 4/(4x2 - 12x + 9)
<MASK> <UNMASK> <MASK> Do you really want to delete this prezi? <MASK>
<MASK> Do you really want to delete this prezi? <MASK> <UNMASK> ### Present Remotely Send the link below via email or IM • Invited audience members will follow you as you navigate and present • People invited to a presentation do not need a Prezi account • This link expires 10 minutes after you close the presentation Do you really want to delete this prezi? Neither you, nor the coeditors you shared it with will be able to recover it again. # Maths Revision: Ratios and Rates A revision about ratios and rates. by ## Tiffany Chin on 1 October 2012 Report abuse #### Transcript of Maths Revision: Ratios and Rates Ratios Ratios and Rates A comparison between 2/more quantities of the same kind Ratios can be shown as... 1:1 Ratio Form 1 to 1 Word Form 1/1 Fraction Form 1:n/m:1 Unit Ratio Please do not copy! Equivalent Ratios Ratios with the same meaning Simplify e.g. 2:4 1:2 1/2 2/4 Prime Factorization Unit Ratio Ratios with a denominator to 1 (Simplify) A:B/A 1:B/A 1:n A:B/B 1:B/A m:1 Conversion of A and B Comparing Ratios Change order to unit ratio e.g. Which class has the highest BOYS:GIRLS ratio? Class 1 Class 2 Class 3 2:3 3:4 1:5 1. Compare your question to ratio given If same order, m:1 e.g. Highest B:G If not, 1:n e.g. Lowest G:B 2. Convert 3. Order Use m:1 Class 1 Class 2 Class 3 2/3:1= 0.6: 1 3/4:1=0.75:1 1/5:1=0.2:1 Class 2 Class 1 Class 3 4. Choose your answer Highest B:G Ratio Multiply Method 1: Find the Unknown e.g. Solve for Unknown 3:4=a:8 =6:8 a=6 Method 2: Fractions e.g. 3:4=c:10 10x3/4=c/28.5x10 30/4=c 7.5=c Find A:B:C from A:B/B:C Cross Product e.g. A:B=4:3 B:C=7:5 4 : 3 7 : 5 28:21:15 Multiply as follow Line up common value Rearrange and simplify as needed Modeling Using Ratios e.g. 250 ml lemon juice, how
<MASK> ## Quant Quiz for SBI clerk pre 2021 <MASK> <UNMASK> # Quant Quiz for SBI clerk pre 2021\| 16 March 2021 ## Quant Quiz for SBI clerk pre 2021 Quant Quiz to improve your Quantitative Aptitude for SBI Po & SBI clerk exam IBPS PO Reasoning , IBPS Clerk Reasoning , IBPS RRB Reasoning, LIC AAO ,LIC Assistant and other competitive exam. Direction (1 -5): The Pie-chart given below shows the percentage distribution of the monthly income of Rakesh. Note: (i) Total monthly income of Rakesh = expenditure on (Rent + Education + Electricity) + Monthly saving of Rakesh (ii) Monthly saving of Rakesh = Rs. 4000 Q1. Rakesh’s expenditure on education is how much more or less than his expenditure on rent? 1. 500 Rs. 2. 400 Rs. 3. 600 Rs. 4. 300 Rs. 5. 700 Rs. Rakesh’s saving is what percentage of his expenditure on electricity. 1. 250% 2. 266 2/3% 3. 225% 4. 233 1/3% 5. 275% Q3. Find Rakesh’s total expenditure on electricity and education together. 1. 3000 Rs. 2. 4500 Rs. 3. 4000 Rs. 4. 5000 Rs. 5. 3500 Rs. Q4. Find out Rakesh’s annual income? 1. 120000 Rs. 2. 144000 Rs. 3. 110000 Rs. 4. 125000 Rs. 5. 136000 Rs. Q5. Rakesh’s expenditure on rent and education together is what percent more or less than his saving? 1. 8% 2. 11% 3. 9% 4. 10% 5. 12.5% Direction (6 -10): What will come in the place of question (?) mark in the following number series. Q6. 90, 117, 145, 174, 204, ? 1. 225 2. 220 3. 230 4. 235 5. 240 Q7. ?, 10, 100, 1500, 30000, 750000 1. 2 2. 1 3. 5 4. 4 5. 10 Q8. 145, 170, 197, 226, 257, ? 1. 332 2. 325 3. 401 4. 290 5. 360 Q9. 473, 460, 434, 382, 278, ? 1. 50 2. 90 3. 60 4. 80 5. 70 Q10. 4200, 5600, 7200, 9000, 11000, ? 1. 12800 2. 13200 3. 13100 4. 14120 5. 15000 Solutions Q1. Ans(1) Q2. Ans(2) Q3. Ans(3) Q4. Ans(1) Q5. Ans(5) Q6. Ans(4) Q7. Ans(1) Q8. Ans(4) Q9. Ans(5) Q10. Ans(2) Recommended PDF’s for 2021: ### 2021 Preparation Kit PDF <MASK> AATMA-NIRBHAR Series- Static GK/Awareness Practice Ebook PDF Get PDF here The Banking Awareness 500 MCQs E-book\| Bilingual (Hindi + English) Get PDF here AATMA-NIRBHAR Series- Banking Awareness Practice Ebook PDF Get PDF here Computer Awareness Capsule 2.O Get PDF here AATMA-NIRBHAR Series Quantitative Aptitude Topic-Wise PDF 2020 Get PDF here Memory Based Puzzle E-book \| 2016-19 Exams Covered Get PDF here Caselet Data Interpretation 200 Questions Get PDF here Puzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1) Get PDF here ARITHMETIC DATA INTERPRETATION 2.O E-book Get PDF here 3
# Quant Quiz for SBI clerk pre 2021\| 16 March 2021 ## Quant Quiz for SBI clerk pre 2021 Quant Quiz to improve your Quantitative Aptitude for SBI Po & SBI clerk exam IBPS PO Reasoning , IBPS Clerk Reasoning , IBPS RRB Reasoning, LIC AAO ,LIC Assistant and other competitive exam. Direction (1 -5): The Pie-chart given below shows the percentage distribution of the monthly income of Rakesh. Note: (i) Total monthly income of Rakesh = expenditure on (Rent + Education + Electricity) + Monthly saving of Rakesh (ii) Monthly saving of Rakesh = Rs. 4000 Q1. Rakesh’s expenditure on education is how much more or less than his expenditure on rent? 1. 500 Rs. 2. 400 Rs. 3. 600 Rs. 4. 300 Rs. 5. 700 Rs. Rakesh’s saving is what percentage of his expenditure on electricity. 1. 250% 2. 266 2/3% 3. 225% 4. 233 1/3% 5. 275% Q3. Find Rakesh’s total expenditure on electricity and education together. 1. 3000 Rs. 2. 4500 Rs. 3. 4000 Rs. 4. 5000 Rs. 5. 3500 Rs. Q4. Find out Rakesh’s annual income? 1. 120000 Rs. 2. 144000 Rs. 3. 110000 Rs. 4. 125000 Rs. 5. 136000 Rs. Q5. Rakesh’s expenditure on rent and education together is what percent more or less than his saving? 1. 8% 2. 11% 3. 9% 4. 10% 5. 12.5% Direction (6 -10): What will come in the place of question (?) mark in the following number series. Q6. 90, 117, 145, 174, 204, ? 1. 225 2. 220 3. 230 4. 235 5. 240 Q7. ?, 10, 100, 1500, 30000, 750000 1. 2 2. 1 3. 5 4. 4 5. 10 Q8. 145, 170, 197, 226, 257, ? 1. 332 2. 325 3. 401 4. 290 5. 360 Q9. 473, 460, 434, 382, 278, ? 1. 50 2. 90 3. 60 4. 80 5. 70 Q10. 4200, 5600, 7200, 9000, 11000, ? 1. 12800 2. 13200 3. 13100 4. 14120 5. 15000 Solutions Q1. Ans(1) Q2. Ans(2) Q3. Ans(3) Q4. Ans(1) Q5. Ans(5) Q6. Ans(4) Q7. Ans(1) Q8. Ans(4) Q9. Ans(5) Q10. Ans(2) Recommended PDF’s for 2021: ### 2021 Preparation Kit PDF <MASK> AATMA-NIRBHAR Series- Static GK/Awareness Practice Ebook PDF Get PDF here The Banking Awareness 500 MCQs E-book\| Bilingual (Hindi + English) Get PDF here AATMA-NIRBHAR Series- Banking Awareness Practice Ebook PDF Get PDF here Computer Awareness Capsule 2.O Get PDF here AATMA-NIRBHAR Series Quantitative Aptitude Topic-Wise PDF 2020 Get PDF here Memory Based Puzzle E-book \| 2016-19 Exams Covered Get PDF here Caselet Data Interpretation 200 Questions Get PDF here Puzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1) Get PDF here ARITHMETIC DATA INTERPRETATION 2.O E-book Get PDF here 3 <UNMASK> # Quant Quiz for SBI clerk pre 2021\| 16 March 2021 ## Quant Quiz for SBI clerk pre 2021 Quant Quiz to improve your Quantitative Aptitude for SBI Po & SBI clerk exam IBPS PO Reasoning , IBPS Clerk Reasoning , IBPS RRB Reasoning, LIC AAO ,LIC Assistant and other competitive exam. Direction (1 -5): The Pie-chart given below shows the percentage distribution of the monthly income of Rakesh. Note: (i) Total monthly income of Rakesh = expenditure on (Rent + Education + Electricity) + Monthly saving of Rakesh (ii) Monthly saving of Rakesh = Rs. 4000 Q1. Rakesh’s expenditure on education is how much more or less than his expenditure on rent? 1. 500 Rs. 2. 400 Rs. 3. 600 Rs. 4. 300 Rs. 5. 700 Rs. Rakesh’s saving is what percentage of his expenditure on electricity. 1. 250% 2. 266 2/3% 3. 225% 4. 233 1/3% 5. 275% Q3. Find Rakesh’s total expenditure on electricity and education together. 1. 3000 Rs. 2. 4500 Rs. 3. 4000 Rs. 4. 5000 Rs. 5. 3500 Rs. Q4. Find out Rakesh’s annual income? 1. 120000 Rs. 2. 144000 Rs. 3. 110000 Rs. 4. 125000 Rs. 5. 136000 Rs. Q5. Rakesh’s expenditure on rent and education together is what percent more or less than his saving? 1. 8% 2. 11% 3. 9% 4. 10% 5. 12.5% Direction (6 -10): What will come in the place of question (?) mark in the following number series. Q6. 90, 117, 145, 174, 204, ? 1. 225 2. 220 3. 230 4. 235 5. 240 Q7. ?, 10, 100, 1500, 30000, 750000 1. 2 2. 1 3. 5 4. 4 5. 10 Q8. 145, 170, 197, 226, 257, ? 1. 332 2. 325 3. 401 4. 290 5. 360 Q9. 473, 460, 434, 382, 278, ? 1. 50 2. 90 3. 60 4. 80 5. 70 Q10. 4200, 5600, 7200, 9000, 11000, ? 1. 12800 2. 13200 3. 13100 4. 14120 5. 15000 Solutions Q1. Ans(1) Q2. Ans(2) Q3. Ans(3) Q4. Ans(1) Q5. Ans(5) Q6. Ans(4) Q7. Ans(1) Q8. Ans(4) Q9. Ans(5) Q10. Ans(2) Recommended PDF’s for 2021: ### 2021 Preparation Kit PDF #### Most important PDF’s for Bank, SSC, Railway and Other Government Exam : Download PDF Now AATMA-NIRBHAR Series- Static GK/Awareness Practice Ebook PDF Get PDF here The Banking Awareness 500 MCQs E-book\| Bilingual (Hindi + English) Get PDF here AATMA-NIRBHAR Series- Banking Awareness Practice Ebook PDF Get PDF here Computer Awareness Capsule 2.O Get PDF here AATMA-NIRBHAR Series Quantitative Aptitude Topic-Wise PDF 2020 Get PDF here Memory Based Puzzle E-book \| 2016-19 Exams Covered Get PDF here Caselet Data Interpretation 200 Questions Get PDF here Puzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1) Get PDF here ARITHMETIC DATA INTERPRETATION 2.O E-book Get PDF here 3
<MASK> <UNMASK> # Is 0/4 undefined? <MASK>
# Is 0/4 undefined? <MASK> <UNMASK> # Is 0/4 undefined? $\frac{0}{4} = 0$ is defined. $\frac{4}{0}$ is not. If $\frac{4}{0} = k$ for some $k \in \mathbb{R}$, then we would have $0 \cdot k = 4$, but $0 \cdot$anything$= 0 \ne 4$, so there is no such $k$.
<MASK> This Equation solver with square root supplies step-by-step instructions for solving all math troubles. Our website will give you answers to homework. ## The Best Equation solver with square root In this blog post, we will be discussing about Equation solver with square root. As any math student knows, calculus can be a difficult subject to grasp. The concepts are often complex and require a great deal of concentration to understand. Fortunately, there are now many calculus solvers available that can help to make the subject more manageable. These tools allow you to input an equation and see the steps involved in solving it. This can be a great way to learn how to solve problems on your own. In addition, calculus solvers with steps can also help you to check your work and ensure that you are getting the correct answer. With so many helpful features, it is no wonder that these tools are becoming increasingly popular among math students of all levels. Word phrase math is a mathematical technique that uses words instead of symbols to represent numbers and operations. This approach can be particularly helpful for students who struggle with traditional math notation. By using words, students can more easily visualize the relationships between numbers and operations. As a result, word phrase math can provide a valuable tool for understanding complex mathematical concepts. Additionally, this technique can also be used to teach basic math skills to young children. By representing numbers and operations with familiar words, children can develop a strong foundation for future mathematics learning. A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane. A binomial solver is a math tool that helps solve equations with two terms. This type of equation is also known as a quadratic equation. The solver will usually ask for the coefficients of the equation, which are the numbers in front of the x terms. It will also ask for the constants, which are the numbers not attached to an x. With this information, the solver can find the roots, or solutions, to the equation. The roots tell where the line intersects the x-axis on a graph. There are two roots because there are two values of x that make the equation true. To find these roots, the solver will use one of several methods, such as factoring or completing the square. Each method has its own set of steps, but all require some algebraic manipulation. The binomial solver can help take care of these steps so that you can focus on understanding the concept behind solving quadratic equations. Basic mathematics is the study of mathematical operations and their properties. The focus of this branch of mathematics is on addition, subtraction, multiplication, and division. These operations are the foundation for all other types of math, including algebra, geometry, and trigonometry. In addition to studying how these operations work, students also learn how to solve equations and how to use basic concepts of geometry and trigonometry. Basic mathematics is an essential part of every student's education, and it provides a strong foundation for further study in math. <MASK> It has helped me more than my math teacher, of course there are a few problems here and there like not scanning the exercises correctly and it never scans "±" this sign but apart from that I really love the app. Although it would be nice to not have to pay for explanations, maybe other features but explanation is an important part. Mila Flores Although the app doesn't solve everything it is pretty much my free math tutor. It's so easy to work with. Steps for solving are always easy to understand. I love it!!! The app is the best and it helped me with my studies at school and I now understand math better Bethany Young App that solves math problems with a picture Math solution calculator Free online math websites Math probability solver Geometry tutor online <UNMASK> <MASK> This Equation solver with square root supplies step-by-step instructions for solving all math troubles. Our website will give you answers to homework. ## The Best Equation solver with square root In this blog post, we will be discussing about Equation solver with square root. As any math student knows, calculus can be a difficult subject to grasp. The concepts are often complex and require a great deal of concentration to understand. Fortunately, there are now many calculus solvers available that can help to make the subject more manageable. These tools allow you to input an equation and see the steps involved in solving it. This can be a great way to learn how to solve problems on your own. In addition, calculus solvers with steps can also help you to check your work and ensure that you are getting the correct answer. With so many helpful features, it is no wonder that these tools are becoming increasingly popular among math students of all levels. Word phrase math is a mathematical technique that uses words instead of symbols to represent numbers and operations. This approach can be particularly helpful for students who struggle with traditional math notation. By using words, students can more easily visualize the relationships between numbers and operations. As a result, word phrase math can provide a valuable tool for understanding complex mathematical concepts. Additionally, this technique can also be used to teach basic math skills to young children. By representing numbers and operations with familiar words, children can develop a strong foundation for future mathematics learning. A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane. A binomial solver is a math tool that helps solve equations with two terms. This type of equation is also known as a quadratic equation. The solver will usually ask for the coefficients of the equation, which are the numbers in front of the x terms. It will also ask for the constants, which are the numbers not attached to an x. With this information, the solver can find the roots, or solutions, to the equation. The roots tell where the line intersects the x-axis on a graph. There are two roots because there are two values of x that make the equation true. To find these roots, the solver will use one of several methods, such as factoring or completing the square. Each method has its own set of steps, but all require some algebraic manipulation. The binomial solver can help take care of these steps so that you can focus on understanding the concept behind solving quadratic equations. Basic mathematics is the study of mathematical operations and their properties. The focus of this branch of mathematics is on addition, subtraction, multiplication, and division. These operations are the foundation for all other types of math, including algebra, geometry, and trigonometry. In addition to studying how these operations work, students also learn how to solve equations and how to use basic concepts of geometry and trigonometry. Basic mathematics is an essential part of every student's education, and it provides a strong foundation for further study in math. ## Help with math It has helped me more than my math teacher, of course there are a few problems here and there like not scanning the exercises correctly and it never scans "±" this sign but apart from that I really love the app. Although it would be nice to not have to pay for explanations, maybe other features but explanation is an important part. Mila Flores Although the app doesn't solve everything it is pretty much my free math tutor. It's so easy to work with. Steps for solving are always easy to understand. I love it!!! The app is the best and it helped me with my studies at school and I now understand math better Bethany Young App that solves math problems with a picture Math solution calculator Free online math websites Math probability solver Geometry tutor online
<MASK> This Equation solver with square root supplies step-by-step instructions for solving all math troubles. Our website will give you answers to homework. ## The Best Equation solver with square root In this blog post, we will be discussing about Equation solver with square root. As any math student knows, calculus can be a difficult subject to grasp. The concepts are often complex and require a great deal of concentration to understand. Fortunately, there are now many calculus solvers available that can help to make the subject more manageable. These tools allow you to input an equation and see the steps involved in solving it. This can be a great way to learn how to solve problems on your own. In addition, calculus solvers with steps can also help you to check your work and ensure that you are getting the correct answer. With so many helpful features, it is no wonder that these tools are becoming increasingly popular among math students of all levels. Word phrase math is a mathematical technique that uses words instead of symbols to represent numbers and operations. This approach can be particularly helpful for students who struggle with traditional math notation. By using words, students can more easily visualize the relationships between numbers and operations. As a result, word phrase math can provide a valuable tool for understanding complex mathematical concepts. Additionally, this technique can also be used to teach basic math skills to young children. By representing numbers and operations with familiar words, children can develop a strong foundation for future mathematics learning. A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane. A binomial solver is a math tool that helps solve equations with two terms. This type of equation is also known as a quadratic equation. The solver will usually ask for the coefficients of the equation, which are the numbers in front of the x terms. It will also ask for the constants, which are the numbers not attached to an x. With this information, the solver can find the roots, or solutions, to the equation. The roots tell where the line intersects the x-axis on a graph. There are two roots because there are two values of x that make the equation true. To find these roots, the solver will use one of several methods, such as factoring or completing the square. Each method has its own set of steps, but all require some algebraic manipulation. The binomial solver can help take care of these steps so that you can focus on understanding the concept behind solving quadratic equations. Basic mathematics is the study of mathematical operations and their properties. The focus of this branch of mathematics is on addition, subtraction, multiplication, and division. These operations are the foundation for all other types of math, including algebra, geometry, and trigonometry. In addition to studying how these operations work, students also learn how to solve equations and how to use basic concepts of geometry and trigonometry. Basic mathematics is an essential part of every student's education, and it provides a strong foundation for further study in math. ## Help with math It has helped me more than my math teacher, of course there are a few problems here and there like not scanning the exercises correctly and it never scans "±" this sign but apart from that I really love the app. Although it would be nice to not have to pay for explanations, maybe other features but explanation is an important part. Mila Flores Although the app doesn't solve everything it is pretty much my free math tutor. It's so easy to work with. Steps for solving are always easy to understand. I love it!!! The app is the best and it helped me with my studies at school and I now understand math better Bethany Young App that solves math problems with a picture Math solution calculator Free online math websites Math probability solver Geometry tutor online <UNMASK> # Equation solver with square root This Equation solver with square root supplies step-by-step instructions for solving all math troubles. Our website will give you answers to homework. ## The Best Equation solver with square root In this blog post, we will be discussing about Equation solver with square root. As any math student knows, calculus can be a difficult subject to grasp. The concepts are often complex and require a great deal of concentration to understand. Fortunately, there are now many calculus solvers available that can help to make the subject more manageable. These tools allow you to input an equation and see the steps involved in solving it. This can be a great way to learn how to solve problems on your own. In addition, calculus solvers with steps can also help you to check your work and ensure that you are getting the correct answer. With so many helpful features, it is no wonder that these tools are becoming increasingly popular among math students of all levels. Word phrase math is a mathematical technique that uses words instead of symbols to represent numbers and operations. This approach can be particularly helpful for students who struggle with traditional math notation. By using words, students can more easily visualize the relationships between numbers and operations. As a result, word phrase math can provide a valuable tool for understanding complex mathematical concepts. Additionally, this technique can also be used to teach basic math skills to young children. By representing numbers and operations with familiar words, children can develop a strong foundation for future mathematics learning. A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane. A binomial solver is a math tool that helps solve equations with two terms. This type of equation is also known as a quadratic equation. The solver will usually ask for the coefficients of the equation, which are the numbers in front of the x terms. It will also ask for the constants, which are the numbers not attached to an x. With this information, the solver can find the roots, or solutions, to the equation. The roots tell where the line intersects the x-axis on a graph. There are two roots because there are two values of x that make the equation true. To find these roots, the solver will use one of several methods, such as factoring or completing the square. Each method has its own set of steps, but all require some algebraic manipulation. The binomial solver can help take care of these steps so that you can focus on understanding the concept behind solving quadratic equations. Basic mathematics is the study of mathematical operations and their properties. The focus of this branch of mathematics is on addition, subtraction, multiplication, and division. These operations are the foundation for all other types of math, including algebra, geometry, and trigonometry. In addition to studying how these operations work, students also learn how to solve equations and how to use basic concepts of geometry and trigonometry. Basic mathematics is an essential part of every student's education, and it provides a strong foundation for further study in math. ## Help with math It has helped me more than my math teacher, of course there are a few problems here and there like not scanning the exercises correctly and it never scans "±" this sign but apart from that I really love the app. Although it would be nice to not have to pay for explanations, maybe other features but explanation is an important part. Mila Flores Although the app doesn't solve everything it is pretty much my free math tutor. It's so easy to work with. Steps for solving are always easy to understand. I love it!!! The app is the best and it helped me with my studies at school and I now understand math better Bethany Young App that solves math problems with a picture Math solution calculator Free online math websites Math probability solver Geometry tutor online
<MASK> <UNMASK> <MASK> 330 mL = _____ L 1. 330, 000 • Rationale: 2. 3, 300 • Rationale: 3. 0.33 • Rationale: 4. 0. 033 • Rationale: ### Explanation DA solution = 1 L/1000mLx 330 mL/1 = 0.33 L Ratio solution = 1000 mL/1 L = 330 mL/x L
<MASK> 330 mL = _____ L 1. 330, 000 • Rationale: 2. 3, 300 • Rationale: 3. 0.33 • Rationale: 4. 0. 033 • Rationale: ### Explanation DA solution = 1 L/1000mLx 330 mL/1 = 0.33 L Ratio solution = 1000 mL/1 L = 330 mL/x L <UNMASK> ## Med Math: Metric Conversions Question #7 330 mL = _____ L 1. 330, 000 • Rationale: 2. 3, 300 • Rationale: 3. 0.33 • Rationale: 4. 0. 033 • Rationale: ### Explanation DA solution = 1 L/1000mLx 330 mL/1 = 0.33 L Ratio solution = 1000 mL/1 L = 330 mL/x L
<MASK> <UNMASK> ### Select your language <MASK> Americas Europe Q8RP <MASK> ### Fundamentals Of Differential Equations And Boundary Value Problems Book edition 9th Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider Pages 616 pages ISBN 9780321977069 # In Problems 3-10, determine the Laplace transform of the given function.${\mathbf{\left(}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{\right)}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{\left(}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{3}\mathbf{\right)}}^{{\mathbf{2}}}$ Therefore, the solution is$\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}$. See the step by step solution ## Step 1: Given Information The given value is ${\left(\text{t+3}\right)}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}$ ## Step 2: Determining the Laplace transform Using the following Laplace transform property, to find the Laplace of given integral: $\text{L}\left\{{\text{e}}^{\text{at}}\right\}\text{=}\frac{\text{1}}{\text{s-a}}$ $\text{L}\left\{{\text{t}}^{\text{n}}\right\}\text{=}\frac{\text{n!}}{{\text{s}}^{\text{n+1}}}$ <MASK> $\begin{array}{c}\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=L}\left\{{\text{(t+3)}}^{\text{2}}\right\}\text{-L}\left\{{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\text{+6t+3}\right\}\text{-L}\left\{{\text{e}}^{\text{2t}}{\text{+6e}}^{\text{t}}\text{+3}\right\}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\right\}\text{+6L{t}+3L{1}-L}\left\{{\text{e}}^{\text{2t}}\right\}\text{+6L}\left\{{\text{e}}^{\text{t}}\right\}\text{-3L{1}}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\right\}\text{+6L{t}-L}\left\{{\text{e}}^{\text{2t}}\right\}\text{+6L}\left\{{\text{e}}^{\text{t}}\right\}\end{array}$ Simplify further as follows $\begin{array}{c}\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2!}}{{\text{s}}^{\text{3}}}\text{+6}\frac{\text{1!}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-6}\frac{\text{1}}{\text{s-1}}\\ \text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}\end{array}$ <MASK>
### Select your language <MASK> Americas Europe Q8RP <MASK> ### Fundamentals Of Differential Equations And Boundary Value Problems Book edition 9th Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider Pages 616 pages ISBN 9780321977069 # In Problems 3-10, determine the Laplace transform of the given function.${\mathbf{\left(}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{\right)}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{\left(}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{3}\mathbf{\right)}}^{{\mathbf{2}}}$ Therefore, the solution is$\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}$. See the step by step solution ## Step 1: Given Information The given value is ${\left(\text{t+3}\right)}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}$ ## Step 2: Determining the Laplace transform Using the following Laplace transform property, to find the Laplace of given integral: $\text{L}\left\{{\text{e}}^{\text{at}}\right\}\text{=}\frac{\text{1}}{\text{s-a}}$ $\text{L}\left\{{\text{t}}^{\text{n}}\right\}\text{=}\frac{\text{n!}}{{\text{s}}^{\text{n+1}}}$ <MASK> $\begin{array}{c}\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=L}\left\{{\text{(t+3)}}^{\text{2}}\right\}\text{-L}\left\{{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\text{+6t+3}\right\}\text{-L}\left\{{\text{e}}^{\text{2t}}{\text{+6e}}^{\text{t}}\text{+3}\right\}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\right\}\text{+6L{t}+3L{1}-L}\left\{{\text{e}}^{\text{2t}}\right\}\text{+6L}\left\{{\text{e}}^{\text{t}}\right\}\text{-3L{1}}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\right\}\text{+6L{t}-L}\left\{{\text{e}}^{\text{2t}}\right\}\text{+6L}\left\{{\text{e}}^{\text{t}}\right\}\end{array}$ Simplify further as follows $\begin{array}{c}\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2!}}{{\text{s}}^{\text{3}}}\text{+6}\frac{\text{1!}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-6}\frac{\text{1}}{\text{s-1}}\\ \text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}\end{array}$ <MASK> <UNMASK> ### Select your language Suggested languages for you: Americas Europe Q8RP Expert-verified Found in: Page 415 ### Fundamentals Of Differential Equations And Boundary Value Problems Book edition 9th Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider Pages 616 pages ISBN 9780321977069 # In Problems 3-10, determine the Laplace transform of the given function.${\mathbf{\left(}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{\right)}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{\left(}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{3}\mathbf{\right)}}^{{\mathbf{2}}}$ Therefore, the solution is$\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}$. See the step by step solution ## Step 1: Given Information The given value is ${\left(\text{t+3}\right)}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}$ ## Step 2: Determining the Laplace transform Using the following Laplace transform property, to find the Laplace of given integral: $\text{L}\left\{{\text{e}}^{\text{at}}\right\}\text{=}\frac{\text{1}}{\text{s-a}}$ $\text{L}\left\{{\text{t}}^{\text{n}}\right\}\text{=}\frac{\text{n!}}{{\text{s}}^{\text{n+1}}}$ Apply the Laplace transform property, we get: $\begin{array}{c}\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=L}\left\{{\text{(t+3)}}^{\text{2}}\right\}\text{-L}\left\{{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\text{+6t+3}\right\}\text{-L}\left\{{\text{e}}^{\text{2t}}{\text{+6e}}^{\text{t}}\text{+3}\right\}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\right\}\text{+6L{t}+3L{1}-L}\left\{{\text{e}}^{\text{2t}}\right\}\text{+6L}\left\{{\text{e}}^{\text{t}}\right\}\text{-3L{1}}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\right\}\text{+6L{t}-L}\left\{{\text{e}}^{\text{2t}}\right\}\text{+6L}\left\{{\text{e}}^{\text{t}}\right\}\end{array}$ Simplify further as follows $\begin{array}{c}\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2!}}{{\text{s}}^{\text{3}}}\text{+6}\frac{\text{1!}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-6}\frac{\text{1}}{\text{s-1}}\\ \text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}\end{array}$ Therefore,$\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}$
# Period Of Function <MASK> Period of function. This is why this function family is also called the periodic function family. Tap for more steps. Therefore in the case of the basic cosine function f x. <MASK> For basic sine and cosine functions the period is 2 π. Any function that is not periodic is called aperiodic. The period is defined as the length of one wave of the function. The period is the length on the x axis in one cycle. However the amplitude does not refer to the highest point on the graph or the distance from the highest point to the x axis. Midline of sinusoidal functions from equation. <MASK> Period of a Function The time interval between two waves is known as a Period whereas a function that repeats its values at regular intervals or periods is known as a Periodic Function. According to periodic function definition the period of a function is represented like f x f x p p is equal to the real number and this is the period of the given function f x. In other words a periodic function is a function that repeats its values after every particular interval. <MASK> In this case one full wave is 180 degrees or radians. So the period of or is. You are partially correct. Khan Academy is a 501c3 nonprofit organization. The period of a periodic function is the interval of x -values on which one copy of the repeated pattern occurs. This is the currently selected item. A periodic function repeats its values at set intervals called periods. Period can be defined as the time interval between the two occurrences of the wave. A periodic function is a function that repeats its values at regular intervals for example the trigonometric functions which repeat at intervals of 2π radians. The period of the function can be calculated using. Our mission is to provide a free world-class education to anyone anywhere. Grade 12 trigonometry problems and questions on how to find the period of trigonometric functions given its graph or formula are presented along with detailed solutions. So how to find the period of a function actually. Free function periodicity calculator – find periodicity of periodic functions step-by-step This website uses cookies to ensure you get the best experience. In the problems below we will use the formula for the period P of trigonometric functions of the form y a sin bx c d or y a cos bx c d and which is given by. <MASK> Amplitude Period Phase Shift Vertical Translation And Range Of Y 3 Math Videos Translation Period The Mathematics Of Sine Cosine Functions Was First Developed By The Ancient Indian Mathematician Aryabhata H Trigonometric Functions Mathematics Theorems Amplitude Phase Shift Vertical Shift And Period Change Of The Cosine Function Fun Learning Period Change Transformations Of Sin Function Love Math Math Graphing <MASK> Graphing Trigonometric Functions Graphing Trigonometric Functions Trigonometry <MASK> Applied Mathematics Period Of A Function Hard Sum Mathematics How To Apply Period Students Will Match 10 Graphs To 10 Sine Or Cosine Equations By Finding The Amplitude And Period Of Each Function Students Will Then Graphing Sines Equations Precal And Trig Function Posters Math Word Problems Mathematics Worksheets Word Problems <MASK> Translating Periodic Functions Learning Mathematics Math Methods Math Instruction Trigonometric Graphing Math Methods Learning Math Math <MASK> Applied Mathematics Greatest Integer Function And Period Integers Mathematics How To Apply Applet Allows For Students To Drag 5 Key Points Of One Period Of A Sinusoidal Wave So That The Graph Displayed Match Trigonometric Functions Graphing Equations <MASK> READ: Which One Of The Following Is An Example Of A Period Cost? <UNMASK> # Period Of Function <MASK> Period of function. This is why this function family is also called the periodic function family. Tap for more steps. Therefore in the case of the basic cosine function f x. When this occurs we call the horizontal shift the period of the function. If a function has a repeating pattern like sine or cosine it is called a periodic function. Amplitude a Let b be a real number. <MASK> For basic sine and cosine functions the period is 2 π. Any function that is not periodic is called aperiodic. The period is defined as the length of one wave of the function. The period is the length on the x axis in one cycle. However the amplitude does not refer to the highest point on the graph or the distance from the highest point to the x axis. Midline of sinusoidal functions from equation. The x-value results in a unique output eg. The period of a sinusoid is the length of a complete cycle. By using this website you agree to our Cookie Policy. <MASK> Period of a Function The time interval between two waves is known as a Period whereas a function that repeats its values at regular intervals or periods is known as a Periodic Function. According to periodic function definition the period of a function is represented like f x f x p p is equal to the real number and this is the period of the given function f x. In other words a periodic function is a function that repeats its values after every particular interval. Trig functions are cyclical and when you graph them youll see the ups and downs of the graph and youll see that these ups and downs. Where f x P f x for all values of x. Replace with in the formula for period. Amplitude and Period of Sine and Cosine Functions The amplitude of y a sin x and y a cos x represents half the distance between the maximum and minimum values of the function. This is the currently selected item. Any part of the graph that shows this pattern over one period is called a cycle. <MASK> In this case one full wave is 180 degrees or radians. So the period of or is. You are partially correct. Khan Academy is a 501c3 nonprofit organization. The period of a periodic function is the interval of x -values on which one copy of the repeated pattern occurs. This is the currently selected item. A periodic function repeats its values at set intervals called periods. Period can be defined as the time interval between the two occurrences of the wave. A periodic function is a function that repeats its values at regular intervals for example the trigonometric functions which repeat at intervals of 2π radians. The period of the function can be calculated using. Our mission is to provide a free world-class education to anyone anywhere. Grade 12 trigonometry problems and questions on how to find the period of trigonometric functions given its graph or formula are presented along with detailed solutions. So how to find the period of a function actually. Free function periodicity calculator – find periodicity of periodic functions step-by-step This website uses cookies to ensure you get the best experience. In the problems below we will use the formula for the period P of trigonometric functions of the form y a sin bx c d or y a cos bx c d and which is given by. <MASK> Amplitude Period Phase Shift Vertical Translation And Range Of Y 3 Math Videos Translation Period The Mathematics Of Sine Cosine Functions Was First Developed By The Ancient Indian Mathematician Aryabhata H Trigonometric Functions Mathematics Theorems Amplitude Phase Shift Vertical Shift And Period Change Of The Cosine Function Fun Learning Period Change Transformations Of Sin Function Love Math Math Graphing <MASK> Graphing Trigonometric Functions Graphing Trigonometric Functions Trigonometry <MASK> Applied Mathematics Period Of A Function Hard Sum Mathematics How To Apply Period Students Will Match 10 Graphs To 10 Sine Or Cosine Equations By Finding The Amplitude And Period Of Each Function Students Will Then Graphing Sines Equations Precal And Trig Function Posters Math Word Problems Mathematics Worksheets Word Problems <MASK> Translating Periodic Functions Learning Mathematics Math Methods Math Instruction Trigonometric Graphing Math Methods Learning Math Math <MASK> Pin On Math Applied Mathematics Greatest Integer Function And Period Integers Mathematics How To Apply Applet Allows For Students To Drag 5 Key Points Of One Period Of A Sinusoidal Wave So That The Graph Displayed Match Trigonometric Functions Graphing Equations <MASK> READ: Which One Of The Following Is An Example Of A Period Cost?
# Period Of Function <MASK> Period of function. This is why this function family is also called the periodic function family. Tap for more steps. Therefore in the case of the basic cosine function f x. When this occurs we call the horizontal shift the period of the function. If a function has a repeating pattern like sine or cosine it is called a periodic function. Amplitude a Let b be a real number. <MASK> For basic sine and cosine functions the period is 2 π. Any function that is not periodic is called aperiodic. The period is defined as the length of one wave of the function. The period is the length on the x axis in one cycle. However the amplitude does not refer to the highest point on the graph or the distance from the highest point to the x axis. Midline of sinusoidal functions from equation. The x-value results in a unique output eg. The period of a sinusoid is the length of a complete cycle. By using this website you agree to our Cookie Policy. <MASK> Period of a Function The time interval between two waves is known as a Period whereas a function that repeats its values at regular intervals or periods is known as a Periodic Function. According to periodic function definition the period of a function is represented like f x f x p p is equal to the real number and this is the period of the given function f x. In other words a periodic function is a function that repeats its values after every particular interval. Trig functions are cyclical and when you graph them youll see the ups and downs of the graph and youll see that these ups and downs. Where f x P f x for all values of x. Replace with in the formula for period. Amplitude and Period of Sine and Cosine Functions The amplitude of y a sin x and y a cos x represents half the distance between the maximum and minimum values of the function. This is the currently selected item. Any part of the graph that shows this pattern over one period is called a cycle. <MASK> In this case one full wave is 180 degrees or radians. So the period of or is. You are partially correct. Khan Academy is a 501c3 nonprofit organization. The period of a periodic function is the interval of x -values on which one copy of the repeated pattern occurs. This is the currently selected item. A periodic function repeats its values at set intervals called periods. Period can be defined as the time interval between the two occurrences of the wave. A periodic function is a function that repeats its values at regular intervals for example the trigonometric functions which repeat at intervals of 2π radians. The period of the function can be calculated using. Our mission is to provide a free world-class education to anyone anywhere. Grade 12 trigonometry problems and questions on how to find the period of trigonometric functions given its graph or formula are presented along with detailed solutions. So how to find the period of a function actually. Free function periodicity calculator – find periodicity of periodic functions step-by-step This website uses cookies to ensure you get the best experience. In the problems below we will use the formula for the period P of trigonometric functions of the form y a sin bx c d or y a cos bx c d and which is given by. <MASK> Amplitude Period Phase Shift Vertical Translation And Range Of Y 3 Math Videos Translation Period The Mathematics Of Sine Cosine Functions Was First Developed By The Ancient Indian Mathematician Aryabhata H Trigonometric Functions Mathematics Theorems Amplitude Phase Shift Vertical Shift And Period Change Of The Cosine Function Fun Learning Period Change Transformations Of Sin Function Love Math Math Graphing <MASK> Graphing Trigonometric Functions Graphing Trigonometric Functions Trigonometry <MASK> Applied Mathematics Period Of A Function Hard Sum Mathematics How To Apply Period Students Will Match 10 Graphs To 10 Sine Or Cosine Equations By Finding The Amplitude And Period Of Each Function Students Will Then Graphing Sines Equations Precal And Trig Function Posters Math Word Problems Mathematics Worksheets Word Problems <MASK> Translating Periodic Functions Learning Mathematics Math Methods Math Instruction Trigonometric Graphing Math Methods Learning Math Math <MASK> Pin On Math Applied Mathematics Greatest Integer Function And Period Integers Mathematics How To Apply Applet Allows For Students To Drag 5 Key Points Of One Period Of A Sinusoidal Wave So That The Graph Displayed Match Trigonometric Functions Graphing Equations <MASK> READ: Which One Of The Following Is An Example Of A Period Cost? <UNMASK> # Period Of Function by -2 views Period of sinusoidal functions from equation. You can figure this out without looking at a graph by dividing with the frequency which in this case is 2. Graphing Trigonometric Functions Graphing Trigonometric Functions Neon Signs ### Periodic functions are used throughout science to describe oscillations waves and other phenomena that exhibit periodicity. Period of function. This is why this function family is also called the periodic function family. Tap for more steps. Therefore in the case of the basic cosine function f x. When this occurs we call the horizontal shift the period of the function. If a function has a repeating pattern like sine or cosine it is called a periodic function. Amplitude a Let b be a real number. You might immediately guess that there is a connection here to finding points on a circle. F x k f x. Horizontal stretch is measured for sinusoidal functions as their periods. For basic sine and cosine functions the period is 2 π. Any function that is not periodic is called aperiodic. The period is defined as the length of one wave of the function. The period is the length on the x axis in one cycle. However the amplitude does not refer to the highest point on the graph or the distance from the highest point to the x axis. Midline of sinusoidal functions from equation. The x-value results in a unique output eg. The period of a sinusoid is the length of a complete cycle. By using this website you agree to our Cookie Policy. Periodic Functions A periodic function occurs when a specific horizontal shift P results in the original function. More formally we say that this type of function has a positive constant k where any input x. The period is defined as the length of a functions cycle. Period of a Function The time interval between two waves is known as a Period whereas a function that repeats its values at regular intervals or periods is known as a Periodic Function. According to periodic function definition the period of a function is represented like f x f x p p is equal to the real number and this is the period of the given function f x. In other words a periodic function is a function that repeats its values after every particular interval. Trig functions are cyclical and when you graph them youll see the ups and downs of the graph and youll see that these ups and downs. Where f x P f x for all values of x. Replace with in the formula for period. Amplitude and Period of Sine and Cosine Functions The amplitude of y a sin x and y a cos x represents half the distance between the maximum and minimum values of the function. This is the currently selected item. Any part of the graph that shows this pattern over one period is called a cycle. Notice that in the graph of the sine function shown that f x sin x has period. Period of sinusoidal functions from graph. The distance between and is. Midline amplitude and period review. The absolute value is the distance between a number and zero. The period of a periodic function is the interval of x -values on which the cycle of the graph thats repeated in both directions lies. In this case one full wave is 180 degrees or radians. So the period of or is. You are partially correct. Khan Academy is a 501c3 nonprofit organization. The period of a periodic function is the interval of x -values on which one copy of the repeated pattern occurs. This is the currently selected item. A periodic function repeats its values at set intervals called periods. Period can be defined as the time interval between the two occurrences of the wave. A periodic function is a function that repeats its values at regular intervals for example the trigonometric functions which repeat at intervals of 2π radians. The period of the function can be calculated using. Our mission is to provide a free world-class education to anyone anywhere. Grade 12 trigonometry problems and questions on how to find the period of trigonometric functions given its graph or formula are presented along with detailed solutions. So how to find the period of a function actually. Free function periodicity calculator – find periodicity of periodic functions step-by-step This website uses cookies to ensure you get the best experience. In the problems below we will use the formula for the period P of trigonometric functions of the form y a sin bx c d or y a cos bx c d and which is given by. The period is the length of the smallest interval that contains exactly one copy of the repeating pattern. A function is just a type of equation where every input eg. Amplitude Period Phase Shift Vertical Translation And Range Of Y 3 Math Videos Translation Period The Mathematics Of Sine Cosine Functions Was First Developed By The Ancient Indian Mathematician Aryabhata H Trigonometric Functions Mathematics Theorems Amplitude Phase Shift Vertical Shift And Period Change Of The Cosine Function Fun Learning Period Change Transformations Of Sin Function Love Math Math Graphing Step By Step Instructions Of How To Graph The Sine Function Graphing Trigfunction Trigonometry Sinusoidal Math Materials Math Graphic Organizers Calculus Graphing Trigonometric Functions Graphing Trigonometric Functions Trigonometry Geometry Trigonometry 9 Trig Graphs Amplitude And Period Trigonometry Graphing Classroom Posters Graphs Of Trigonometric Functions Poster Zazzle Com In 2021 Trigonometric Functions Functions Math Math Poster Applied Mathematics Period Of A Function Hard Sum Mathematics How To Apply Period Students Will Match 10 Graphs To 10 Sine Or Cosine Equations By Finding The Amplitude And Period Of Each Function Students Will Then Graphing Sines Equations Precal And Trig Function Posters Math Word Problems Mathematics Worksheets Word Problems Graphing Sin Cosine W Period Change 4 Terrific Examples Graphing Precalculus Trigonometry Translating Periodic Functions Learning Mathematics Math Methods Math Instruction Trigonometric Graphing Math Methods Learning Math Math Trigonometric Graphing Math Methods Learning Math Math Pin On Math Applied Mathematics Greatest Integer Function And Period Integers Mathematics How To Apply Applet Allows For Students To Drag 5 Key Points Of One Period Of A Sinusoidal Wave So That The Graph Displayed Match Trigonometric Functions Graphing Equations Functions Non Functions By Ryan Devoe Period 7 By Rmdevoe Teaching Mathematics Teaching Algebra Teaching Math READ: Which One Of The Following Is An Example Of A Period Cost?
<MASK> <UNMASK> <MASK> 2. It is the well-known fixed point theorem. Suppose $f$ is continuous on $[0,1]$ and $f(x)\epsilon[0,1]$ for every $x\epsilon[0,1]$. <MASK>
<MASK> 2. It is the well-known fixed point theorem. Suppose $f$ is continuous on $[0,1]$ and $f(x)\epsilon[0,1]$ for every $x\epsilon[0,1]$. <MASK> <UNMASK> # Thread: Question involving a continuous function on a closed interval 1. ## Question involving a continuous function on a closed interval Intuitively this is obvious by graphing g(x) = x on [0,1] and seeing since f is continuous it has to intersect with g at some point. But I spent a long time and cannot figure out how to prove this. 2. It is the well-known fixed point theorem. Suppose $f$ is continuous on $[0,1]$ and $f(x)\epsilon[0,1]$ for every $x\epsilon[0,1]$. If $f(0)=0$ or $f(1)=1$, the theorem is proved. So we try to prove the theorem assuming $f(0)>0$ and $f(1)<1$. Let $g(x)=f(x)-x$ for all $x\epsilon[0,1]$. Hence $g(0)>0$ and $g(1)<0$ and $g$ is continuous on $[0,1]$, that is, $0$ is an intermediate value of $g$ on $[0,1]$. Hence by intermediate value theorem, there exists a point $c\epsilon(0,1)$ such that $g(c)=0$ --which means $f(c)=c.$ Hence the prrof. EDIT: $c$ is equivalent to $x_0$ 3. Originally Posted by paulrb Intuitively this is obvious by graphing g(x) = x on [0,1] and seeing since f is continuous it has to intersect with g at some point. But I spent a long time and cannot figure out how to prove this. Alternatively, suppose that $f(x)\ne x$ then the mapping $\displaystyle f:[0,1]\to\{-1,1}:\frac{\|f(x)-x\|}{f(x)-x}$ is a continuous surjection which is impossible.
## Vibrations of Continuous Systems: Lateral Vibrations of Beams <MASK> (10.23) <MASK> Now, using 10.23 we can write that <MASK> (10.25) <MASK> This results in <MASK> The four roots to this equation are <MASK> where each of the may be complex. However, by using Euler’s identity <MASK> (10.32) where <MASK> As can be seen there are four constants to be determined which requires that four boundary conditions be specified. These will often be determined with one pair of boundary conditions specified at each end, depending on the type of support. Figure 10.7 shows three of the most common support conditions and the pairs of boundary conditions that are associated with each type of support. <MASK> #### Complete Response of Lateral Motion of Beam <MASK> (10.33) <MASK> (10.35) <MASK> where once again the orthogonality property of the mode shape has been used <MASK> <UNMASK> ## Vibrations of Continuous Systems: Lateral Vibrations of Beams <MASK> As a first order approximation we can write that <MASK> (10.23) <MASK> Now, using 10.23 we can write that <MASK> (10.24) <MASK> or (10.25) <MASK> This results in <MASK> Again, since the LHS depends only on and the RHS depends only on and they must be equal for all values of and , both sides must be equal to a constant, which we call , so that <MASK> or <MASK> (10.29) <MASK> so that <MASK> The four roots to this equation are <MASK> where each of the may be complex. However, by using Euler’s identity <MASK> (10.32) where <MASK> As can be seen there are four constants to be determined which requires that four boundary conditions be specified. These will often be determined with one pair of boundary conditions specified at each end, depending on the type of support. Figure 10.7 shows three of the most common support conditions and the pairs of boundary conditions that are associated with each type of support. <MASK> Determine expressions for the natural frequencies and associated mode shapes for this beam. #### Complete Response of Lateral Motion of Beam <MASK> (10.33) <MASK> (10.34) <MASK> and can then be found from (10.35) <MASK> where once again the orthogonality property of the mode shape has been used <MASK>
## Vibrations of Continuous Systems: Lateral Vibrations of Beams <MASK> As a first order approximation we can write that <MASK> (10.23) <MASK> Now, using 10.23 we can write that <MASK> (10.24) <MASK> or (10.25) <MASK> This results in <MASK> Again, since the LHS depends only on and the RHS depends only on and they must be equal for all values of and , both sides must be equal to a constant, which we call , so that <MASK> or <MASK> (10.29) <MASK> so that <MASK> The four roots to this equation are <MASK> where each of the may be complex. However, by using Euler’s identity <MASK> (10.32) where <MASK> As can be seen there are four constants to be determined which requires that four boundary conditions be specified. These will often be determined with one pair of boundary conditions specified at each end, depending on the type of support. Figure 10.7 shows three of the most common support conditions and the pairs of boundary conditions that are associated with each type of support. <MASK> Determine expressions for the natural frequencies and associated mode shapes for this beam. #### Complete Response of Lateral Motion of Beam <MASK> (10.33) <MASK> (10.34) <MASK> and can then be found from (10.35) <MASK> where once again the orthogonality property of the mode shape has been used <MASK> <UNMASK> ## Vibrations of Continuous Systems: Lateral Vibrations of Beams ### Equation of Motion Consider a beam, shown in Figure 10.6(a) which has a length , density (mass per unit volume) and Young’s modulus which is acted upon by a distributed load (per unit length) acting laterally along the beam. Let measure the lateral deflection of the beam and assume that only small deformations occur. Figure 10.6(b) shows the FBD/MAD for an infinitesimal element of the beam with mass . By applying Newton’s Law’s and considering moments about the left end of the element we find As a first order approximation we can write that so that the above becomes or (10.23) which is a well known result from beam theory. In the vertical direction we then have Note: For small motions, the shear forces act vertically to a first order approximation. The actual vertical components would be for example where can be obtained from the slope of the beam. However, for small motions, , so is used here as the vertical component. A similar remark holds for the term. Now, using 10.23 we can write that so the equation of motion becomes (10.24) For small deformations the bending moment in the beam is related to the deflection by so that 10.24 becomes or (10.25) This is the general equation which governs the lateral vibrations of beams. If we limit ourselves to only consider free vibrations of uniform beams (, is constant), the equation of motion reduces to which can be written (10.26) where (10.27) Note that this is not the wave equation. ### Solution To Equation of Motion Once again we look for solutions which represent a mode shape undergoing \sshm of the form This results in so 10.26 becomes Separating the and terms we find Again, since the LHS depends only on and the RHS depends only on and they must be equal for all values of and , both sides must be equal to a constant, which we call , so that This results in equations or (10.28a) (10.28b) where (10.29) Note for future reference that so that (10.30) The solution to 10.28b is as we have seen To find the solution to 10.28a, which is a 4 order linear ODE with constant coefficients, we assume a solution of the form The equation of motion then becomes or The four roots to this equation are The total solution will be a linear combination of solutions, one for each of the above roots, (10.31) where each of the may be complex. However, by using Euler’s identity and introducing the hyperbolic and functions 10.31 can be written or (10.32) where The advantage of expressing the solution as in 10.32 is that all of the terms involved are real. We see that and are complex conjugates, while and are real. As can be seen there are four constants to be determined which requires that four boundary conditions be specified. These will often be determined with one pair of boundary conditions specified at each end, depending on the type of support. Figure 10.7 shows three of the most common support conditions and the pairs of boundary conditions that are associated with each type of support. #### EXAMPLE A uniform beam (density , cross–sectional area , flexural rigidity ) of length is fixed at one end and free at the other. Determine expressions for the natural frequencies and associated mode shapes for this beam. #### Complete Response of Lateral Motion of Beam We have again found an infinite number of solutions which satisfy the equation of motion 10.26 given by (10.33) where is the natural frequency and is the associated mode shape, both of which depend on the specific boundary conditions present. The general solution will then be a superposition of all of the solutions in 10.33, (10.34) The constants and are to be determined from the initial conditions of the beam. If the initial displacement and velocity of the beam are specified as then 10.34 gives and can then be found from (10.35) (10.36) where once again the orthogonality property of the mode shape has been used This is once again beyond the scope of this course.
📚 All Subjects <MASK> ⚖️ <MASK> Bookmarked 4.3k • 246 resources See Units Interpretation When interpreting a confidence interval for the difference in two population proportions, we need to show that the bounds given by the confidence interval give us an estimate of what the difference in our two population proportion are within. Our interpretation and interval is also strongly based on what our confidence level is. The standard confidence level is 95%. When completing a confidence interval on the AP Statistics Exam, you will generally be graded on if you included the following three components: 👉 Including confidence level (given in problem) 👉 Including that our interval is making inference about the difference in population proportions, not sample proportions. 👉 Including context of the problem. If you do these three things and get the correct answer, you will be on your way to 💯. <MASK> Image Courtesy of cbssports <MASK> Resources: <MASK> Join us on Discord Thousands of students are studying with us for the AP Statistics exam. join now Studying with Hours = the ultimate focus mode Start a free study session 🔍 Are you ready for college apps? Take this quiz and find out! Start Quiz Browse Study Guides By Unit 📆Big Reviews: Finals & Exam Prep ✏️Blogs ✍️Free Response Questions (FRQs) 👆Unit 1: Exploring One-Variable Data ✌️Unit 2: Exploring Two-Variable Data 🔎Unit 3: Collecting Data 🎲Unit 4: Probability, Random Variables, and Probability Distributions 📊Unit 5: Sampling Distributions ⚖️Unit 6: Inference for Categorical Data: Proportions 😼Unit 7: Inference for Qualitative Data: Means ✳️Unit 8: Inference for Categorical Data: Chi-Square 📈Unit 9: Inference for Quantitative Data: Slopes FREE AP stats Survival Pack + Cram Chart PDF Sign up now for instant access to 2 amazing downloads to help you get a 5 Join us on Discord Thousands of students are studying with us for the AP Statistics exam. join now 💪🏽 Are you ready for the Stats exam? Take this quiz for a progress check on what you’ve learned this year and get a personalized study plan to grab that 5! START QUIZ Play this on HyperTyper Practice your typing skills while reading Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions Start Game Text FIVEABLE to741741to get started. © 2021 Fiveable, Inc. <UNMASK> 📚 All Subjects <MASK> ⚖️ Unit 6 <MASK> 2 min readjune 5, 2020 <MASK> Bookmarked 4.3k • 246 resources See Units Interpretation When interpreting a confidence interval for the difference in two population proportions, we need to show that the bounds given by the confidence interval give us an estimate of what the difference in our two population proportion are within. Our interpretation and interval is also strongly based on what our confidence level is. The standard confidence level is 95%. When completing a confidence interval on the AP Statistics Exam, you will generally be graded on if you included the following three components: 👉 Including confidence level (given in problem) 👉 Including that our interval is making inference about the difference in population proportions, not sample proportions. 👉 Including context of the problem. If you do these three things and get the correct answer, you will be on your way to 💯. Testing a Claim When we are testing a claim using a confidence interval, we want to see if 0 is included in our interval. If 0 is included in our interval, it is quite possible that there is no difference in the two population proportions we are testing. If 0 is not included in our interval, we have reason to suspect that the two population proportions are in fact different. Image Courtesy of cbssports Example <MASK> Resources: <MASK> Join us on Discord Thousands of students are studying with us for the AP Statistics exam. join now Studying with Hours = the ultimate focus mode Start a free study session 🔍 Are you ready for college apps? Take this quiz and find out! Start Quiz Browse Study Guides By Unit 📆Big Reviews: Finals & Exam Prep ✏️Blogs ✍️Free Response Questions (FRQs) 👆Unit 1: Exploring One-Variable Data ✌️Unit 2: Exploring Two-Variable Data 🔎Unit 3: Collecting Data 🎲Unit 4: Probability, Random Variables, and Probability Distributions 📊Unit 5: Sampling Distributions ⚖️Unit 6: Inference for Categorical Data: Proportions 😼Unit 7: Inference for Qualitative Data: Means ✳️Unit 8: Inference for Categorical Data: Chi-Square 📈Unit 9: Inference for Quantitative Data: Slopes FREE AP stats Survival Pack + Cram Chart PDF Sign up now for instant access to 2 amazing downloads to help you get a 5 Join us on Discord Thousands of students are studying with us for the AP Statistics exam. join now 💪🏽 Are you ready for the Stats exam? Take this quiz for a progress check on what you’ve learned this year and get a personalized study plan to grab that 5! START QUIZ Play this on HyperTyper Practice your typing skills while reading Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions Start Game Text FIVEABLE to741741to get started. © 2021 Fiveable, Inc.
📚 All Subjects <MASK> ⚖️ Unit 6 <MASK> 2 min readjune 5, 2020 <MASK> Bookmarked 4.3k • 246 resources See Units Interpretation When interpreting a confidence interval for the difference in two population proportions, we need to show that the bounds given by the confidence interval give us an estimate of what the difference in our two population proportion are within. Our interpretation and interval is also strongly based on what our confidence level is. The standard confidence level is 95%. When completing a confidence interval on the AP Statistics Exam, you will generally be graded on if you included the following three components: 👉 Including confidence level (given in problem) 👉 Including that our interval is making inference about the difference in population proportions, not sample proportions. 👉 Including context of the problem. If you do these three things and get the correct answer, you will be on your way to 💯. Testing a Claim When we are testing a claim using a confidence interval, we want to see if 0 is included in our interval. If 0 is included in our interval, it is quite possible that there is no difference in the two population proportions we are testing. If 0 is not included in our interval, we have reason to suspect that the two population proportions are in fact different. Image Courtesy of cbssports Example <MASK> Resources: <MASK> Join us on Discord Thousands of students are studying with us for the AP Statistics exam. join now Studying with Hours = the ultimate focus mode Start a free study session 🔍 Are you ready for college apps? Take this quiz and find out! Start Quiz Browse Study Guides By Unit 📆Big Reviews: Finals & Exam Prep ✏️Blogs ✍️Free Response Questions (FRQs) 👆Unit 1: Exploring One-Variable Data ✌️Unit 2: Exploring Two-Variable Data 🔎Unit 3: Collecting Data 🎲Unit 4: Probability, Random Variables, and Probability Distributions 📊Unit 5: Sampling Distributions ⚖️Unit 6: Inference for Categorical Data: Proportions 😼Unit 7: Inference for Qualitative Data: Means ✳️Unit 8: Inference for Categorical Data: Chi-Square 📈Unit 9: Inference for Quantitative Data: Slopes FREE AP stats Survival Pack + Cram Chart PDF Sign up now for instant access to 2 amazing downloads to help you get a 5 Join us on Discord Thousands of students are studying with us for the AP Statistics exam. join now 💪🏽 Are you ready for the Stats exam? Take this quiz for a progress check on what you’ve learned this year and get a personalized study plan to grab that 5! START QUIZ Play this on HyperTyper Practice your typing skills while reading Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions Start Game Text FIVEABLE to741741to get started. © 2021 Fiveable, Inc. <UNMASK> 📚 All Subjects > 📊 AP Stats > ⚖️ Unit 6 6.9 Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions 2 min readjune 5, 2020 Josh Argo AP Statistics📊 Bookmarked 4.3k • 246 resources See Units Interpretation When interpreting a confidence interval for the difference in two population proportions, we need to show that the bounds given by the confidence interval give us an estimate of what the difference in our two population proportion are within. Our interpretation and interval is also strongly based on what our confidence level is. The standard confidence level is 95%. When completing a confidence interval on the AP Statistics Exam, you will generally be graded on if you included the following three components: 👉 Including confidence level (given in problem) 👉 Including that our interval is making inference about the difference in population proportions, not sample proportions. 👉 Including context of the problem. If you do these three things and get the correct answer, you will be on your way to 💯. Testing a Claim When we are testing a claim using a confidence interval, we want to see if 0 is included in our interval. If 0 is included in our interval, it is quite possible that there is no difference in the two population proportions we are testing. If 0 is not included in our interval, we have reason to suspect that the two population proportions are in fact different. Image Courtesy of cbssports Example Recall from Unit 6.8 we constructed a confidence interval for the difference in proportions for shots made for Michael Jordan and Lebron James. We got the following output from our calculator: A correct way to interpret this would be: "We are 95% confident that the true difference in the population proportions for shots made between Michael Jordan and Lebron James is between (0.063, 0.133). Since 0 is not included in our interval, we have reasonable evidence that the two population proportions are actually different" So it appears that MJ is better than Lebron. ¯\_(ツ)_/¯ However, some things to note here is that we took a sample from both of their first seasons. If we refer back to basketball-reference.com and compare the two players 10th seasons in the league, we might find a VERY different result. Another possible confounding variable is the teammates these two players have been surrounded by throughout their career. Resources: Was this guide helpful? Join us on Discord Thousands of students are studying with us for the AP Statistics exam. join now Studying with Hours = the ultimate focus mode Start a free study session 🔍 Are you ready for college apps? Take this quiz and find out! Start Quiz Browse Study Guides By Unit 📆Big Reviews: Finals & Exam Prep ✏️Blogs ✍️Free Response Questions (FRQs) 👆Unit 1: Exploring One-Variable Data ✌️Unit 2: Exploring Two-Variable Data 🔎Unit 3: Collecting Data 🎲Unit 4: Probability, Random Variables, and Probability Distributions 📊Unit 5: Sampling Distributions ⚖️Unit 6: Inference for Categorical Data: Proportions 😼Unit 7: Inference for Qualitative Data: Means ✳️Unit 8: Inference for Categorical Data: Chi-Square 📈Unit 9: Inference for Quantitative Data: Slopes FREE AP stats Survival Pack + Cram Chart PDF Sign up now for instant access to 2 amazing downloads to help you get a 5 Join us on Discord Thousands of students are studying with us for the AP Statistics exam. join now 💪🏽 Are you ready for the Stats exam? Take this quiz for a progress check on what you’ve learned this year and get a personalized study plan to grab that 5! 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<MASK> # 11.2: Arithmetic Sequences <MASK> $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ <MASK> $$\newcommand{\id}{\mathrm{id}}$$ <MASK> $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ <MASK> $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ <MASK> $$\newcommand{\norm}[1]{\\| #1 \\|}$$ <MASK> $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ <MASK> $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ <MASK> $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ###### Learning Objectives • Find the common difference for an arithmetic sequence. • Write terms of an arithmetic sequence. • Use a recursive formula for an arithmetic sequence. • Use an explicit formula for an arithmetic sequence. <MASK> As an example, consider a woman who starts a small contracting business. She purchases a new truck for $$25,000$$. After five years, she estimates that she will be able to sell the truck for $$8,000$$. The loss in value of the truck will therefore be $$17,000$$, which is $$3,400$$ per year for five years. The truck will be worth $$21,600$$ after the first year; $$18,200$$ after two years; $$14,800$$ after three years; $$11,400$$ after four years; and $$8,000$$ at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value. ## Finding Common Differences <MASK> The sequence below is another example of an arithmetic sequence. In this case, the constant difference is $$3$$. You can choose any term of the sequence, and add $$3$$ to find the subsequent term. <MASK> ###### Example $$\PageIndex{1}$$: Finding Common Differences Is each sequence arithmetic? If so, find the common difference. <MASK> Solution Subtract each term from the subsequent term to determine whether a common difference exists. 1. The sequence is not arithmetic because there is no common difference. $$2-1={\color{red}1} \qquad 4-2={\color{red}2} \qquad 8-4={\color{red}4} \qquad 16-8={\color{red}8}$$ <MASK> $$1-(-3)={\color{red}4} \qquad 5-1={\color{red}4} \qquad 9-5={\color{red}4} \qquad 13-9={\color{red}4}$$ Analysis <MASK> ###### Q&A If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference? <MASK> ###### Exercise $$\PageIndex{1A}$$ Is the given sequence arithmetic? If so, find the common difference. $$\{18, 16, 14, 12, 10,…\}$$ <MASK> The sequence is arithmetic. The common difference is $$–2$$. ###### Exercise $$\PageIndex{1B}$$ Is the given sequence arithmetic? If so, find the common difference. <MASK> Answer The sequence is not arithmetic because $$3−1≠6−3$$. ## Writing Terms of Arithmetic Sequences Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of $$n$$ and $$d$$ into formula below. $a_n=a_1+(n−1)d$ ###### How to: Given the first term and the common difference of an arithmetic sequence, find the first several terms. 1. Add the common difference to the first term to find the second term. 2. Add the common difference to the second term to find the third term. 3. Continue until all of the desired terms are identified. 4. Write the terms separated by commas within brackets. ###### Example $$\PageIndex{2}$$: Writing Terms of Arithmetic Sequences <MASK> Solution <MASK> Analysis As expected, the graph of the sequence consists of points on a line as shown in Figure $$\PageIndex{2}$$. <MASK> List the first five terms of the arithmetic sequence with $$a_1=1$$ and $$d=5$$. Answer <MASK> ###### How to: Given any the first term and any other term in an arithmetic sequence, find a given term. 1. Substitute the values given for $$a_1$$, $$a_n$$, $$n$$ into the formula $$a_n=a_1+(n−1)d$$ to solve for $$d$$. 2. Find a given term by substituting the appropriate values for $$a_1$$, $$n$$, and $$d$$ into the formula $$a_n=a_1+(n−1)d$$. ###### Example $$\PageIndex{3}$$: Writing Terms of Arithmetic Sequences Given $$a_1=8$$ and $$a_4=14$$, find $$a_5$$. Solution The sequence can be written in terms of the initial term $$8$$ and the common difference $$d$$. $$\{8,8+d,8+2d,8+3d\}$$ We know the fourth term equals $$14$$; we know the fourth term has the form $$a_1+3d=8+3d$$. We can find the common difference $$d$$. \begin{align} a_n&= a_1+(n-1)d \\ a_4&= a_1+3d \\ a_4&=8+3d\qquad \text{Write the fourth term of the sequence in terms of }a_1 \text{ and } d. \\ 14&=8+3d\qquad \text{Substitute }14 \text{ for } a_4. \\ d&=2\qquad \text{Solve for the common difference.} \end{align} Find the fifth term by adding the common difference to the fourth term. $$a_5=a_4+2=16$$ Analysis Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation $$a_n=a_1+(n−1)d$$. ###### Exercise $$\PageIndex{3}$$ <MASK> ## Using Recursive Formulas for Arithmetic Sequences Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is $$5$$, then each term is the previous term plus $$5$$. As with any recursive formula, the first term must be given. $$a_n=a_n−1+d$$ for $$n≥2$$ ###### Note: RECURSIVE FORMULA FOR AN ARITHMETIC SEQUENCE The recursive formula for an arithmetic sequence with common difference $$d$$ is: $a_n=a_n−1+d$ for $$n≥2$$ <MASK> Write a recursive formula for the arithmetic sequence. <MASK> The first term is given as $$−18$$. The common difference can be found by subtracting the first term from the second term. $$d=−7−(−18)=11$$ Substitute the initial term and the common difference into the recursive formula for arithmetic sequences. $$a_1=−18$$ <MASK> Do we have to subtract the first term from the second term to find the common difference? <MASK> ###### Exercise $$\PageIndex{4}$$ Write a recursive formula for the arithmetic sequence. $$\{25, 37, 49, 61, …\}$$ Answer \begin{align}a_1 &= 25 \\ a_n &= a_{n−1}+12 , \text{ for }n≥2 \end{align} <MASK> We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept. <MASK> To find the $$y$$-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence. The common difference is $$−50$$, so the sequence represents a linear function with a slope of $$−50$$. To find the $$y$$-intercept, we subtract $$−50$$ from $$200$$: $$200−(−50)=200+50=250$$. You can also find the $$y$$-intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure $$\PageIndex{4}$$. <MASK> Recall the slope-intercept form of a line is $$y=mx+b$$. When dealing with sequences, we use $$a_n$$ in place of $$y$$ and $$n$$ in place of $$x$$. If we know the slope and vertical intercept of the function, we can substitute them for $$m$$ and $$b$$ in the slope-intercept form of a line. Substituting $$−50$$ for the slope and $$250$$ for the vertical intercept, we get the following equation: <MASK> $a_n=a_1+d(n−1)$ ###### How to: Given the first several terms for an arithmetic sequence, write an explicit formula. 1. Find the common difference, $$a_2−a_1$$. 2. Substitute the common difference and the first term into $$a_n=a_1+d(n−1)$$. ###### Example $$\PageIndex{5}$$: Writing the nth Term Explicit Formula for an Arithmetic Sequence Write an explicit formula for the arithmetic sequence. <MASK> Solution <MASK> The common difference is $$10$$. Substitute the common difference and the first term of the sequence into the formula and simplify. \begin{align}a_n &= 2+10(n−1) \\ a_n &= 10n−8 \end{align} Analysis The graph of this sequence, represented in Figure $$\PageIndex{5}$$, shows a slope of $$10$$ and a vertical intercept of $$−8$$. Figure $$\PageIndex{5}$$ <MASK> Write an explicit formula for the following arithmetic sequence. <MASK> ### Finding the Number of Terms in a Finite Arithmetic Sequence <MASK> Find the number of terms in the finite arithmetic sequence. $$\{8, 1, –6, ..., –41\}$$ <MASK> $$1−8=−7$$ The common difference is $$−7$$. Substitute the common difference and the initial term of the sequence into the nth term formula and simplify. \begin{align} a_n &= a_1+d(n−1) \\ a_n &= 8+−7(n−1) \\ a_n &= 15−7n \end{align} Substitute $$−41$$ for $$a_n$$ and solve for $$n$$ <MASK> ###### Exercise $$\PageIndex{6}$$ Find the number of terms in the finite arithmetic sequence. <MASK> Answer There are $$11$$ terms in the sequence. ### Solving Application Problems with Arithmetic Sequences In many application problems, it often makes sense to use an initial term of $$a_0$$ instead of $$a_1$$. In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula: $a_n=a_0+dn$ ###### Example $$\PageIndex{7}$$: Solving Application Problems with Arithmetic Sequences A five-year old child receives an allowance of $$1$$ each week. His parents promise him an annual increase of $$2$$ per week. 1. Write a formula for the child’s weekly allowance in a given year. 2. What will the child’s allowance be when he is $$16$$ years old? <MASK> $$A_n=1+2n$$ 2. We can find the number of years since age $$5$$ by subtracting. <MASK> $$A_{11}=1+2(11)=23$$ The child’s allowance at age $$16$$ will be $$23$$ per week. <MASK> A woman decides to go for a $$10$$-minute run every day this week and plans to increase the time of her daily run by $$4$$ minutes each week. Write a formula for the time of her run after $$n$$ weeks. How long will her daily run be $$8$$ weeks from today? Answer The formula is $$T_n=10+4n$$, and it will take her $$42$$ minutes. ###### Media <MASK> ## Key Equations <MASK> • An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. • The constant between two consecutive terms is called the common difference. • The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example $$\PageIndex{1}$$. • The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example $$\PageIndex{2}$$ and Example $$\PageIndex{3}$$. • A recursive formula for an arithmetic sequence with common difference dd is given by $$a_n=a_{n−1}+d$$, $$n≥2$$. See Example $$\PageIndex{4}$$. • As with any recursive formula, the initial term of the sequence must be given. • An explicit formula for an arithmetic sequence with common difference $$d$$ is given by $$a_n=a_1+d(n−1)$$. See Example $$\PageIndex{5}$$. • An explicit formula can be used to find the number of terms in a sequence. See Example $$\PageIndex{6}$$. • In application problems, we sometimes alter the explicit formula slightly to $$a_n=a_0+dn$$. See Example $$\PageIndex{7}$$. <MASK> This page titled 11.2: Arithmetic Sequences is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. <MASK> <UNMASK> <MASK> # 11.2: Arithmetic Sequences <MASK> $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ <MASK> $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ <MASK> $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ <MASK> $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ <MASK> $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ <MASK> $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ###### Learning Objectives • Find the common difference for an arithmetic sequence. • Write terms of an arithmetic sequence. • Use a recursive formula for an arithmetic sequence. • Use an explicit formula for an arithmetic sequence. <MASK> As an example, consider a woman who starts a small contracting business. She purchases a new truck for $$25,000$$. After five years, she estimates that she will be able to sell the truck for $$8,000$$. The loss in value of the truck will therefore be $$17,000$$, which is $$3,400$$ per year for five years. The truck will be worth $$21,600$$ after the first year; $$18,200$$ after two years; $$14,800$$ after three years; $$11,400$$ after four years; and $$8,000$$ at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value. ## Finding Common Differences <MASK> The sequence below is another example of an arithmetic sequence. In this case, the constant difference is $$3$$. You can choose any term of the sequence, and add $$3$$ to find the subsequent term. ###### ARITHMETIC SEQUENCE <MASK> ###### Example $$\PageIndex{1}$$: Finding Common Differences Is each sequence arithmetic? If so, find the common difference. <MASK> Solution Subtract each term from the subsequent term to determine whether a common difference exists. 1. The sequence is not arithmetic because there is no common difference. $$2-1={\color{red}1} \qquad 4-2={\color{red}2} \qquad 8-4={\color{red}4} \qquad 16-8={\color{red}8}$$ <MASK> $$1-(-3)={\color{red}4} \qquad 5-1={\color{red}4} \qquad 9-5={\color{red}4} \qquad 13-9={\color{red}4}$$ Analysis The graph of each of these sequences is shown in Figure $$\PageIndex{1}$$. We can see from the graphs that, although both sequences show growth, (a) is not linear whereas (b) is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line. <MASK> ###### Q&A If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference? <MASK> ###### Exercise $$\PageIndex{1A}$$ Is the given sequence arithmetic? If so, find the common difference. $$\{18, 16, 14, 12, 10,…\}$$ <MASK> The sequence is arithmetic. The common difference is $$–2$$. ###### Exercise $$\PageIndex{1B}$$ Is the given sequence arithmetic? If so, find the common difference. <MASK> Answer The sequence is not arithmetic because $$3−1≠6−3$$. ## Writing Terms of Arithmetic Sequences Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of $$n$$ and $$d$$ into formula below. $a_n=a_1+(n−1)d$ ###### How to: Given the first term and the common difference of an arithmetic sequence, find the first several terms. 1. Add the common difference to the first term to find the second term. 2. Add the common difference to the second term to find the third term. 3. Continue until all of the desired terms are identified. 4. Write the terms separated by commas within brackets. ###### Example $$\PageIndex{2}$$: Writing Terms of Arithmetic Sequences <MASK> Solution <MASK> Analysis As expected, the graph of the sequence consists of points on a line as shown in Figure $$\PageIndex{2}$$. <MASK> List the first five terms of the arithmetic sequence with $$a_1=1$$ and $$d=5$$. Answer $$\{1, 6, 11, 16, 21\}$$ ###### How to: Given any the first term and any other term in an arithmetic sequence, find a given term. 1. Substitute the values given for $$a_1$$, $$a_n$$, $$n$$ into the formula $$a_n=a_1+(n−1)d$$ to solve for $$d$$. 2. Find a given term by substituting the appropriate values for $$a_1$$, $$n$$, and $$d$$ into the formula $$a_n=a_1+(n−1)d$$. ###### Example $$\PageIndex{3}$$: Writing Terms of Arithmetic Sequences Given $$a_1=8$$ and $$a_4=14$$, find $$a_5$$. Solution The sequence can be written in terms of the initial term $$8$$ and the common difference $$d$$. $$\{8,8+d,8+2d,8+3d\}$$ We know the fourth term equals $$14$$; we know the fourth term has the form $$a_1+3d=8+3d$$. We can find the common difference $$d$$. \begin{align} a_n&= a_1+(n-1)d \\ a_4&= a_1+3d \\ a_4&=8+3d\qquad \text{Write the fourth term of the sequence in terms of }a_1 \text{ and } d. \\ 14&=8+3d\qquad \text{Substitute }14 \text{ for } a_4. \\ d&=2\qquad \text{Solve for the common difference.} \end{align} Find the fifth term by adding the common difference to the fourth term. $$a_5=a_4+2=16$$ Analysis Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation $$a_n=a_1+(n−1)d$$. ###### Exercise $$\PageIndex{3}$$ <MASK> $$a_2=2$$ ## Using Recursive Formulas for Arithmetic Sequences Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is $$5$$, then each term is the previous term plus $$5$$. As with any recursive formula, the first term must be given. $$a_n=a_n−1+d$$ for $$n≥2$$ ###### Note: RECURSIVE FORMULA FOR AN ARITHMETIC SEQUENCE The recursive formula for an arithmetic sequence with common difference $$d$$ is: $a_n=a_n−1+d$ for $$n≥2$$ <MASK> Write a recursive formula for the arithmetic sequence. <MASK> The first term is given as $$−18$$. The common difference can be found by subtracting the first term from the second term. $$d=−7−(−18)=11$$ Substitute the initial term and the common difference into the recursive formula for arithmetic sequences. $$a_1=−18$$ $$a_n=a_{n−1}+11$$ <MASK> ###### Q&A Do we have to subtract the first term from the second term to find the common difference? <MASK> ###### Exercise $$\PageIndex{4}$$ Write a recursive formula for the arithmetic sequence. $$\{25, 37, 49, 61, …\}$$ Answer \begin{align}a_1 &= 25 \\ a_n &= a_{n−1}+12 , \text{ for }n≥2 \end{align} <MASK> We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept. $$a_n=a_1+d(n−1)$$ To find the $$y$$-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence. The common difference is $$−50$$, so the sequence represents a linear function with a slope of $$−50$$. To find the $$y$$-intercept, we subtract $$−50$$ from $$200$$: $$200−(−50)=200+50=250$$. You can also find the $$y$$-intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure $$\PageIndex{4}$$. <MASK> Recall the slope-intercept form of a line is $$y=mx+b$$. When dealing with sequences, we use $$a_n$$ in place of $$y$$ and $$n$$ in place of $$x$$. If we know the slope and vertical intercept of the function, we can substitute them for $$m$$ and $$b$$ in the slope-intercept form of a line. Substituting $$−50$$ for the slope and $$250$$ for the vertical intercept, we get the following equation: <MASK> An explicit formula for the $$n^{th}$$ term of an arithmetic sequence is given by $a_n=a_1+d(n−1)$ ###### How to: Given the first several terms for an arithmetic sequence, write an explicit formula. 1. Find the common difference, $$a_2−a_1$$. 2. Substitute the common difference and the first term into $$a_n=a_1+d(n−1)$$. ###### Example $$\PageIndex{5}$$: Writing the nth Term Explicit Formula for an Arithmetic Sequence Write an explicit formula for the arithmetic sequence. <MASK> Solution <MASK> The common difference is $$10$$. Substitute the common difference and the first term of the sequence into the formula and simplify. \begin{align}a_n &= 2+10(n−1) \\ a_n &= 10n−8 \end{align} Analysis The graph of this sequence, represented in Figure $$\PageIndex{5}$$, shows a slope of $$10$$ and a vertical intercept of $$−8$$. Figure $$\PageIndex{5}$$ <MASK> Write an explicit formula for the following arithmetic sequence. $$\{50,47,44,41,…\}$$ Answer <MASK> ### Finding the Number of Terms in a Finite Arithmetic Sequence <MASK> ###### How to: Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms. 1. Find the common difference $$d$$. 2. Substitute the common difference and the first term into $$a_n=a_1+d(n–1)$$. 3. Substitute the last term for $$a_n$$ and solve for $$n$$. ###### Example $$\PageIndex{6}$$: Finding the Number of Terms in a Finite Arithmetic Sequence Find the number of terms in the finite arithmetic sequence. $$\{8, 1, –6, ..., –41\}$$ <MASK> $$1−8=−7$$ The common difference is $$−7$$. Substitute the common difference and the initial term of the sequence into the nth term formula and simplify. \begin{align} a_n &= a_1+d(n−1) \\ a_n &= 8+−7(n−1) \\ a_n &= 15−7n \end{align} Substitute $$−41$$ for $$a_n$$ and solve for $$n$$ <MASK> There are eight terms in the sequence. ###### Exercise $$\PageIndex{6}$$ Find the number of terms in the finite arithmetic sequence. <MASK> Answer There are $$11$$ terms in the sequence. ### Solving Application Problems with Arithmetic Sequences In many application problems, it often makes sense to use an initial term of $$a_0$$ instead of $$a_1$$. In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula: $a_n=a_0+dn$ ###### Example $$\PageIndex{7}$$: Solving Application Problems with Arithmetic Sequences A five-year old child receives an allowance of $$1$$ each week. His parents promise him an annual increase of $$2$$ per week. 1. Write a formula for the child’s weekly allowance in a given year. 2. What will the child’s allowance be when he is $$16$$ years old? <MASK> Let $$A$$ be the amount of the allowance and $$n$$ be the number of years after age $$5$$. Using the altered explicit formula for an arithmetic sequence we get: $$A_n=1+2n$$ 2. We can find the number of years since age $$5$$ by subtracting. <MASK> $$A_{11}=1+2(11)=23$$ The child’s allowance at age $$16$$ will be $$23$$ per week. <MASK> A woman decides to go for a $$10$$-minute run every day this week and plans to increase the time of her daily run by $$4$$ minutes each week. Write a formula for the time of her run after $$n$$ weeks. How long will her daily run be $$8$$ weeks from today? Answer The formula is $$T_n=10+4n$$, and it will take her $$42$$ minutes. ###### Media <MASK> ## Key Equations <MASK> • An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. • The constant between two consecutive terms is called the common difference. • The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example $$\PageIndex{1}$$. • The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example $$\PageIndex{2}$$ and Example $$\PageIndex{3}$$. • A recursive formula for an arithmetic sequence with common difference dd is given by $$a_n=a_{n−1}+d$$, $$n≥2$$. See Example $$\PageIndex{4}$$. • As with any recursive formula, the initial term of the sequence must be given. • An explicit formula for an arithmetic sequence with common difference $$d$$ is given by $$a_n=a_1+d(n−1)$$. See Example $$\PageIndex{5}$$. • An explicit formula can be used to find the number of terms in a sequence. See Example $$\PageIndex{6}$$. • In application problems, we sometimes alter the explicit formula slightly to $$a_n=a_0+dn$$. See Example $$\PageIndex{7}$$. <MASK> This page titled 11.2: Arithmetic Sequences is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. <MASK>
<MASK> # 11.2: Arithmetic Sequences <MASK> $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ <MASK> $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ <MASK> $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ <MASK> $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ <MASK> $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ <MASK> $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ###### Learning Objectives • Find the common difference for an arithmetic sequence. • Write terms of an arithmetic sequence. • Use a recursive formula for an arithmetic sequence. • Use an explicit formula for an arithmetic sequence. <MASK> As an example, consider a woman who starts a small contracting business. She purchases a new truck for $$25,000$$. After five years, she estimates that she will be able to sell the truck for $$8,000$$. The loss in value of the truck will therefore be $$17,000$$, which is $$3,400$$ per year for five years. The truck will be worth $$21,600$$ after the first year; $$18,200$$ after two years; $$14,800$$ after three years; $$11,400$$ after four years; and $$8,000$$ at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value. ## Finding Common Differences <MASK> The sequence below is another example of an arithmetic sequence. In this case, the constant difference is $$3$$. You can choose any term of the sequence, and add $$3$$ to find the subsequent term. ###### ARITHMETIC SEQUENCE <MASK> ###### Example $$\PageIndex{1}$$: Finding Common Differences Is each sequence arithmetic? If so, find the common difference. <MASK> Solution Subtract each term from the subsequent term to determine whether a common difference exists. 1. The sequence is not arithmetic because there is no common difference. $$2-1={\color{red}1} \qquad 4-2={\color{red}2} \qquad 8-4={\color{red}4} \qquad 16-8={\color{red}8}$$ <MASK> $$1-(-3)={\color{red}4} \qquad 5-1={\color{red}4} \qquad 9-5={\color{red}4} \qquad 13-9={\color{red}4}$$ Analysis The graph of each of these sequences is shown in Figure $$\PageIndex{1}$$. We can see from the graphs that, although both sequences show growth, (a) is not linear whereas (b) is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line. <MASK> ###### Q&A If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference? <MASK> ###### Exercise $$\PageIndex{1A}$$ Is the given sequence arithmetic? If so, find the common difference. $$\{18, 16, 14, 12, 10,…\}$$ <MASK> The sequence is arithmetic. The common difference is $$–2$$. ###### Exercise $$\PageIndex{1B}$$ Is the given sequence arithmetic? If so, find the common difference. <MASK> Answer The sequence is not arithmetic because $$3−1≠6−3$$. ## Writing Terms of Arithmetic Sequences Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of $$n$$ and $$d$$ into formula below. $a_n=a_1+(n−1)d$ ###### How to: Given the first term and the common difference of an arithmetic sequence, find the first several terms. 1. Add the common difference to the first term to find the second term. 2. Add the common difference to the second term to find the third term. 3. Continue until all of the desired terms are identified. 4. Write the terms separated by commas within brackets. ###### Example $$\PageIndex{2}$$: Writing Terms of Arithmetic Sequences <MASK> Solution <MASK> Analysis As expected, the graph of the sequence consists of points on a line as shown in Figure $$\PageIndex{2}$$. <MASK> List the first five terms of the arithmetic sequence with $$a_1=1$$ and $$d=5$$. Answer $$\{1, 6, 11, 16, 21\}$$ ###### How to: Given any the first term and any other term in an arithmetic sequence, find a given term. 1. Substitute the values given for $$a_1$$, $$a_n$$, $$n$$ into the formula $$a_n=a_1+(n−1)d$$ to solve for $$d$$. 2. Find a given term by substituting the appropriate values for $$a_1$$, $$n$$, and $$d$$ into the formula $$a_n=a_1+(n−1)d$$. ###### Example $$\PageIndex{3}$$: Writing Terms of Arithmetic Sequences Given $$a_1=8$$ and $$a_4=14$$, find $$a_5$$. Solution The sequence can be written in terms of the initial term $$8$$ and the common difference $$d$$. $$\{8,8+d,8+2d,8+3d\}$$ We know the fourth term equals $$14$$; we know the fourth term has the form $$a_1+3d=8+3d$$. We can find the common difference $$d$$. \begin{align} a_n&= a_1+(n-1)d \\ a_4&= a_1+3d \\ a_4&=8+3d\qquad \text{Write the fourth term of the sequence in terms of }a_1 \text{ and } d. \\ 14&=8+3d\qquad \text{Substitute }14 \text{ for } a_4. \\ d&=2\qquad \text{Solve for the common difference.} \end{align} Find the fifth term by adding the common difference to the fourth term. $$a_5=a_4+2=16$$ Analysis Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation $$a_n=a_1+(n−1)d$$. ###### Exercise $$\PageIndex{3}$$ <MASK> $$a_2=2$$ ## Using Recursive Formulas for Arithmetic Sequences Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is $$5$$, then each term is the previous term plus $$5$$. As with any recursive formula, the first term must be given. $$a_n=a_n−1+d$$ for $$n≥2$$ ###### Note: RECURSIVE FORMULA FOR AN ARITHMETIC SEQUENCE The recursive formula for an arithmetic sequence with common difference $$d$$ is: $a_n=a_n−1+d$ for $$n≥2$$ <MASK> Write a recursive formula for the arithmetic sequence. <MASK> The first term is given as $$−18$$. The common difference can be found by subtracting the first term from the second term. $$d=−7−(−18)=11$$ Substitute the initial term and the common difference into the recursive formula for arithmetic sequences. $$a_1=−18$$ $$a_n=a_{n−1}+11$$ <MASK> ###### Q&A Do we have to subtract the first term from the second term to find the common difference? <MASK> ###### Exercise $$\PageIndex{4}$$ Write a recursive formula for the arithmetic sequence. $$\{25, 37, 49, 61, …\}$$ Answer \begin{align}a_1 &= 25 \\ a_n &= a_{n−1}+12 , \text{ for }n≥2 \end{align} <MASK> We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept. $$a_n=a_1+d(n−1)$$ To find the $$y$$-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence. The common difference is $$−50$$, so the sequence represents a linear function with a slope of $$−50$$. To find the $$y$$-intercept, we subtract $$−50$$ from $$200$$: $$200−(−50)=200+50=250$$. You can also find the $$y$$-intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure $$\PageIndex{4}$$. <MASK> Recall the slope-intercept form of a line is $$y=mx+b$$. When dealing with sequences, we use $$a_n$$ in place of $$y$$ and $$n$$ in place of $$x$$. If we know the slope and vertical intercept of the function, we can substitute them for $$m$$ and $$b$$ in the slope-intercept form of a line. Substituting $$−50$$ for the slope and $$250$$ for the vertical intercept, we get the following equation: <MASK> An explicit formula for the $$n^{th}$$ term of an arithmetic sequence is given by $a_n=a_1+d(n−1)$ ###### How to: Given the first several terms for an arithmetic sequence, write an explicit formula. 1. Find the common difference, $$a_2−a_1$$. 2. Substitute the common difference and the first term into $$a_n=a_1+d(n−1)$$. ###### Example $$\PageIndex{5}$$: Writing the nth Term Explicit Formula for an Arithmetic Sequence Write an explicit formula for the arithmetic sequence. <MASK> Solution <MASK> The common difference is $$10$$. Substitute the common difference and the first term of the sequence into the formula and simplify. \begin{align}a_n &= 2+10(n−1) \\ a_n &= 10n−8 \end{align} Analysis The graph of this sequence, represented in Figure $$\PageIndex{5}$$, shows a slope of $$10$$ and a vertical intercept of $$−8$$. Figure $$\PageIndex{5}$$ <MASK> Write an explicit formula for the following arithmetic sequence. $$\{50,47,44,41,…\}$$ Answer <MASK> ### Finding the Number of Terms in a Finite Arithmetic Sequence <MASK> ###### How to: Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms. 1. Find the common difference $$d$$. 2. Substitute the common difference and the first term into $$a_n=a_1+d(n–1)$$. 3. Substitute the last term for $$a_n$$ and solve for $$n$$. ###### Example $$\PageIndex{6}$$: Finding the Number of Terms in a Finite Arithmetic Sequence Find the number of terms in the finite arithmetic sequence. $$\{8, 1, –6, ..., –41\}$$ <MASK> $$1−8=−7$$ The common difference is $$−7$$. Substitute the common difference and the initial term of the sequence into the nth term formula and simplify. \begin{align} a_n &= a_1+d(n−1) \\ a_n &= 8+−7(n−1) \\ a_n &= 15−7n \end{align} Substitute $$−41$$ for $$a_n$$ and solve for $$n$$ <MASK> There are eight terms in the sequence. ###### Exercise $$\PageIndex{6}$$ Find the number of terms in the finite arithmetic sequence. <MASK> Answer There are $$11$$ terms in the sequence. ### Solving Application Problems with Arithmetic Sequences In many application problems, it often makes sense to use an initial term of $$a_0$$ instead of $$a_1$$. In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula: $a_n=a_0+dn$ ###### Example $$\PageIndex{7}$$: Solving Application Problems with Arithmetic Sequences A five-year old child receives an allowance of $$1$$ each week. His parents promise him an annual increase of $$2$$ per week. 1. Write a formula for the child’s weekly allowance in a given year. 2. What will the child’s allowance be when he is $$16$$ years old? <MASK> Let $$A$$ be the amount of the allowance and $$n$$ be the number of years after age $$5$$. Using the altered explicit formula for an arithmetic sequence we get: $$A_n=1+2n$$ 2. We can find the number of years since age $$5$$ by subtracting. <MASK> $$A_{11}=1+2(11)=23$$ The child’s allowance at age $$16$$ will be $$23$$ per week. <MASK> A woman decides to go for a $$10$$-minute run every day this week and plans to increase the time of her daily run by $$4$$ minutes each week. Write a formula for the time of her run after $$n$$ weeks. How long will her daily run be $$8$$ weeks from today? Answer The formula is $$T_n=10+4n$$, and it will take her $$42$$ minutes. ###### Media <MASK> ## Key Equations <MASK> • An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. • The constant between two consecutive terms is called the common difference. • The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example $$\PageIndex{1}$$. • The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example $$\PageIndex{2}$$ and Example $$\PageIndex{3}$$. • A recursive formula for an arithmetic sequence with common difference dd is given by $$a_n=a_{n−1}+d$$, $$n≥2$$. See Example $$\PageIndex{4}$$. • As with any recursive formula, the initial term of the sequence must be given. • An explicit formula for an arithmetic sequence with common difference $$d$$ is given by $$a_n=a_1+d(n−1)$$. See Example $$\PageIndex{5}$$. • An explicit formula can be used to find the number of terms in a sequence. See Example $$\PageIndex{6}$$. • In application problems, we sometimes alter the explicit formula slightly to $$a_n=a_0+dn$$. See Example $$\PageIndex{7}$$. <MASK> This page titled 11.2: Arithmetic Sequences is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. <MASK> <UNMASK> Skip to main content # 11.2: Arithmetic Sequences $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ###### Learning Objectives • Find the common difference for an arithmetic sequence. • Write terms of an arithmetic sequence. • Use a recursive formula for an arithmetic sequence. • Use an explicit formula for an arithmetic sequence. Companies often make large purchases, such as computers and vehicles, for business use. The book-value of these supplies decreases each year for tax purposes. This decrease in value is called depreciation. One method of calculating depreciation is straight-line depreciation, in which the value of the asset decreases by the same amount each year. As an example, consider a woman who starts a small contracting business. She purchases a new truck for $$25,000$$. After five years, she estimates that she will be able to sell the truck for $$8,000$$. The loss in value of the truck will therefore be $$17,000$$, which is $$3,400$$ per year for five years. The truck will be worth $$21,600$$ after the first year; $$18,200$$ after two years; $$14,800$$ after three years; $$11,400$$ after four years; and $$8,000$$ at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value. ## Finding Common Differences The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is $$-3,400$$. The sequence below is another example of an arithmetic sequence. In this case, the constant difference is $$3$$. You can choose any term of the sequence, and add $$3$$ to find the subsequent term. ###### ARITHMETIC SEQUENCE An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference. If $$a_1$$ is the first term of an arithmetic sequence and $$d$$ is the common difference, the sequence will be: $\{a_n\}=\{a_1,a_1+d,a_1+2d,a_1+3d,...\}$ ###### Example $$\PageIndex{1}$$: Finding Common Differences Is each sequence arithmetic? If so, find the common difference. 1. $$\{1,2,4,8,16,...\}$$ 2. $$\{−3,1,5,9,13,...\}$$ Solution Subtract each term from the subsequent term to determine whether a common difference exists. 1. The sequence is not arithmetic because there is no common difference. $$2-1={\color{red}1} \qquad 4-2={\color{red}2} \qquad 8-4={\color{red}4} \qquad 16-8={\color{red}8}$$ 1. The sequence is arithmetic because there is a common difference. The common difference is $$4$$. $$1-(-3)={\color{red}4} \qquad 5-1={\color{red}4} \qquad 9-5={\color{red}4} \qquad 13-9={\color{red}4}$$ Analysis The graph of each of these sequences is shown in Figure $$\PageIndex{1}$$. We can see from the graphs that, although both sequences show growth, (a) is not linear whereas (b) is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line. Figure $$\PageIndex{1}$$ ###### Q&A If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference? No. If we know that the sequence is arithmetic, we can choose any one term in the sequence, and subtract it from the subsequent term to find the common difference. ###### Exercise $$\PageIndex{1A}$$ Is the given sequence arithmetic? If so, find the common difference. $$\{18, 16, 14, 12, 10,…\}$$ Answer The sequence is arithmetic. The common difference is $$–2$$. ###### Exercise $$\PageIndex{1B}$$ Is the given sequence arithmetic? If so, find the common difference. $$\{1, 3, 6, 10, 15,…\}$$ Answer The sequence is not arithmetic because $$3−1≠6−3$$. ## Writing Terms of Arithmetic Sequences Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of $$n$$ and $$d$$ into formula below. $a_n=a_1+(n−1)d$ ###### How to: Given the first term and the common difference of an arithmetic sequence, find the first several terms. 1. Add the common difference to the first term to find the second term. 2. Add the common difference to the second term to find the third term. 3. Continue until all of the desired terms are identified. 4. Write the terms separated by commas within brackets. ###### Example $$\PageIndex{2}$$: Writing Terms of Arithmetic Sequences Write the first five terms of the arithmetic sequence with $$a_1=17$$ and $$d=−3$$. Solution Adding $$−3$$ is the same as subtracting $$3$$. Beginning with the first term, subtract $$3$$ from each term to find the next term. The first five terms are $$\{17,14,11,8,5\}$$ Analysis As expected, the graph of the sequence consists of points on a line as shown in Figure $$\PageIndex{2}$$. Figure $$\PageIndex{2}$$ ###### Exercise $$\PageIndex{2}$$ List the first five terms of the arithmetic sequence with $$a_1=1$$ and $$d=5$$. Answer $$\{1, 6, 11, 16, 21\}$$ ###### How to: Given any the first term and any other term in an arithmetic sequence, find a given term. 1. Substitute the values given for $$a_1$$, $$a_n$$, $$n$$ into the formula $$a_n=a_1+(n−1)d$$ to solve for $$d$$. 2. Find a given term by substituting the appropriate values for $$a_1$$, $$n$$, and $$d$$ into the formula $$a_n=a_1+(n−1)d$$. ###### Example $$\PageIndex{3}$$: Writing Terms of Arithmetic Sequences Given $$a_1=8$$ and $$a_4=14$$, find $$a_5$$. Solution The sequence can be written in terms of the initial term $$8$$ and the common difference $$d$$. $$\{8,8+d,8+2d,8+3d\}$$ We know the fourth term equals $$14$$; we know the fourth term has the form $$a_1+3d=8+3d$$. We can find the common difference $$d$$. \begin{align} a_n&= a_1+(n-1)d \\ a_4&= a_1+3d \\ a_4&=8+3d\qquad \text{Write the fourth term of the sequence in terms of }a_1 \text{ and } d. \\ 14&=8+3d\qquad \text{Substitute }14 \text{ for } a_4. \\ d&=2\qquad \text{Solve for the common difference.} \end{align} Find the fifth term by adding the common difference to the fourth term. $$a_5=a_4+2=16$$ Analysis Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation $$a_n=a_1+(n−1)d$$. ###### Exercise $$\PageIndex{3}$$ Given $$a_3=7$$ and $$a_5=17$$, find $$a_2$$. Answer $$a_2=2$$ ## Using Recursive Formulas for Arithmetic Sequences Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is $$5$$, then each term is the previous term plus $$5$$. As with any recursive formula, the first term must be given. $$a_n=a_n−1+d$$ for $$n≥2$$ ###### Note: RECURSIVE FORMULA FOR AN ARITHMETIC SEQUENCE The recursive formula for an arithmetic sequence with common difference $$d$$ is: $a_n=a_n−1+d$ for $$n≥2$$ ###### How to: Given an arithmetic sequence, write its recursive formula. 1. Subtract any term from the subsequent term to find the common difference. 2. State the initial term and substitute the common difference into the recursive formula for arithmetic sequences. ###### Example $$\PageIndex{4}$$: Writing a Recursive Formula for an Arithmetic Sequence Write a recursive formula for the arithmetic sequence. $$\{−18, −7, 4, 15, 26, …\}$$ Solution The first term is given as $$−18$$. The common difference can be found by subtracting the first term from the second term. $$d=−7−(−18)=11$$ Substitute the initial term and the common difference into the recursive formula for arithmetic sequences. $$a_1=−18$$ $$a_n=a_{n−1}+11$$ for $$n≥2$$ Analysis We see that the common difference is the slope of the line formed when we graph the terms of the sequence, as shown in Figure $$\PageIndex{3}$$. The growth pattern of the sequence shows the constant difference of 11 units. Figure $$\PageIndex{3}$$ ###### Q&A Do we have to subtract the first term from the second term to find the common difference? No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference. ###### Exercise $$\PageIndex{4}$$ Write a recursive formula for the arithmetic sequence. $$\{25, 37, 49, 61, …\}$$ Answer \begin{align}a_1 &= 25 \\ a_n &= a_{n−1}+12 , \text{ for }n≥2 \end{align} ## Using Explicit Formulas for Arithmetic Sequences We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept. $$a_n=a_1+d(n−1)$$ To find the $$y$$-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence. The common difference is $$−50$$, so the sequence represents a linear function with a slope of $$−50$$. To find the $$y$$-intercept, we subtract $$−50$$ from $$200$$: $$200−(−50)=200+50=250$$. You can also find the $$y$$-intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure $$\PageIndex{4}$$. Figure $$\PageIndex{4}$$ Recall the slope-intercept form of a line is $$y=mx+b$$. When dealing with sequences, we use $$a_n$$ in place of $$y$$ and $$n$$ in place of $$x$$. If we know the slope and vertical intercept of the function, we can substitute them for $$m$$ and $$b$$ in the slope-intercept form of a line. Substituting $$−50$$ for the slope and $$250$$ for the vertical intercept, we get the following equation: $$a_n=−50n+250$$ We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is $$a_n=200−50(n−1)$$, which simplifies to $$a_n=−50n+250$$. ###### Note: EXPLICIT FORMULA FOR AN ARITHMETIC SEQUENCE An explicit formula for the $$n^{th}$$ term of an arithmetic sequence is given by $a_n=a_1+d(n−1)$ ###### How to: Given the first several terms for an arithmetic sequence, write an explicit formula. 1. Find the common difference, $$a_2−a_1$$. 2. Substitute the common difference and the first term into $$a_n=a_1+d(n−1)$$. ###### Example $$\PageIndex{5}$$: Writing the nth Term Explicit Formula for an Arithmetic Sequence Write an explicit formula for the arithmetic sequence. $$\{2, 12, 22, 32, 42, …\}$$ Solution The common difference can be found by subtracting the first term from the second term. \begin{align} d &= a_2−a_1 \\ &= 12−2 \\ &= 10 \end{align} The common difference is $$10$$. Substitute the common difference and the first term of the sequence into the formula and simplify. \begin{align}a_n &= 2+10(n−1) \\ a_n &= 10n−8 \end{align} Analysis The graph of this sequence, represented in Figure $$\PageIndex{5}$$, shows a slope of $$10$$ and a vertical intercept of $$−8$$. Figure $$\PageIndex{5}$$ ###### Exercise $$\PageIndex{5}$$ Write an explicit formula for the following arithmetic sequence. $$\{50,47,44,41,…\}$$ Answer $$a_n=53−3n$$ ### Finding the Number of Terms in a Finite Arithmetic Sequence Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence. ###### How to: Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms. 1. Find the common difference $$d$$. 2. Substitute the common difference and the first term into $$a_n=a_1+d(n–1)$$. 3. Substitute the last term for $$a_n$$ and solve for $$n$$. ###### Example $$\PageIndex{6}$$: Finding the Number of Terms in a Finite Arithmetic Sequence Find the number of terms in the finite arithmetic sequence. $$\{8, 1, –6, ..., –41\}$$ Solution The common difference can be found by subtracting the first term from the second term. $$1−8=−7$$ The common difference is $$−7$$. Substitute the common difference and the initial term of the sequence into the nth term formula and simplify. \begin{align} a_n &= a_1+d(n−1) \\ a_n &= 8+−7(n−1) \\ a_n &= 15−7n \end{align} Substitute $$−41$$ for $$a_n$$ and solve for $$n$$ \begin{align} -41&=15-7n\\ 8&=n \end{align} There are eight terms in the sequence. ###### Exercise $$\PageIndex{6}$$ Find the number of terms in the finite arithmetic sequence. $$\{6, 11, 16, ..., 56\}$$ Answer There are $$11$$ terms in the sequence. ### Solving Application Problems with Arithmetic Sequences In many application problems, it often makes sense to use an initial term of $$a_0$$ instead of $$a_1$$. In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula: $a_n=a_0+dn$ ###### Example $$\PageIndex{7}$$: Solving Application Problems with Arithmetic Sequences A five-year old child receives an allowance of $$1$$ each week. His parents promise him an annual increase of $$2$$ per week. 1. Write a formula for the child’s weekly allowance in a given year. 2. What will the child’s allowance be when he is $$16$$ years old? Solution 1. The situation can be modeled by an arithmetic sequence with an initial term of $$1$$ and a common difference of $$2$$. Let $$A$$ be the amount of the allowance and $$n$$ be the number of years after age $$5$$. Using the altered explicit formula for an arithmetic sequence we get: $$A_n=1+2n$$ 2. We can find the number of years since age $$5$$ by subtracting. $$16−5=11$$ We are looking for the child’s allowance after $$11$$ years. Substitute $$11$$ into the formula to find the child’s allowance at age $$16$$. $$A_{11}=1+2(11)=23$$ The child’s allowance at age $$16$$ will be $$23$$ per week. ###### Exercise $$\PageIndex{7}$$ A woman decides to go for a $$10$$-minute run every day this week and plans to increase the time of her daily run by $$4$$ minutes each week. Write a formula for the time of her run after $$n$$ weeks. How long will her daily run be $$8$$ weeks from today? Answer The formula is $$T_n=10+4n$$, and it will take her $$42$$ minutes. ###### Media Access this online resource for additional instruction and practice with arithmetic sequences. ## Key Equations recursive formula for nth term of an arithmetic sequence $$a_n=a_{n−1}+d$$ $$n≥2$$ explicit formula for nth term of an arithmetic sequence $$a_n=a_1+d(n−1)$$ ## Key Concepts • An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. • The constant between two consecutive terms is called the common difference. • The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example $$\PageIndex{1}$$. • The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example $$\PageIndex{2}$$ and Example $$\PageIndex{3}$$. • A recursive formula for an arithmetic sequence with common difference dd is given by $$a_n=a_{n−1}+d$$, $$n≥2$$. See Example $$\PageIndex{4}$$. • As with any recursive formula, the initial term of the sequence must be given. • An explicit formula for an arithmetic sequence with common difference $$d$$ is given by $$a_n=a_1+d(n−1)$$. See Example $$\PageIndex{5}$$. • An explicit formula can be used to find the number of terms in a sequence. See Example $$\PageIndex{6}$$. • In application problems, we sometimes alter the explicit formula slightly to $$a_n=a_0+dn$$. See Example $$\PageIndex{7}$$. ### Contributors and Attributions This page titled 11.2: Arithmetic Sequences is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. • Was this article helpful?
# approximate greatest common divisor I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers. For example, I have the following numbers: 399, 710, 105, 891, 402, 102, 397, ... <MASK> more details: I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels. eg. ----__------____------__-- ... I have the duration of each level, but this time is not accurate. My aim is to find as quick as possible the base time of the signal. My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate. With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100 <MASK> I made a similar question here, where I propose a partial solution. How to find the approximate basic frequency or GCD of a list of numbers? In summary, I came with this <MASK> And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123: <MASK> One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers? If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well. This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it. Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd. Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$\|r_i\|$$ is very small, say, smaller than $$50$$. <MASK> If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$. <UNMASK> # approximate greatest common divisor I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers. For example, I have the following numbers: 399, 710, 105, 891, 402, 102, 397, ... As you can see, the average gcd is approximately 100, but how to compute it ? more details: I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels. eg. ----__------____------__-- ... I have the duration of each level, but this time is not accurate. My aim is to find as quick as possible the base time of the signal. My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate. With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100 • What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24 • Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25 • @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28 • @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46 • Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$... – Lehs Jun 21, 2016 at 14:25 I made a similar question here, where I propose a partial solution. How to find the approximate basic frequency or GCD of a list of numbers? In summary, I came with this • being $v$ the list $\{v_1, v_2, \ldots, v_n\}$, • $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$ • $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$ • $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$ And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123: testSeq = [399, 710, 105, 891, 402, 102, 397] gcd = calculateGCDAppeal(x, testSeq) find_local_maximum(gcd,90,110) plot(gcd,(x, 10, 200), scale = "semilogx") One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers? If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well. This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it. Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd. Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$\|r_i\|$$ is very small, say, smaller than $$50$$. <MASK> If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
# approximate greatest common divisor I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers. For example, I have the following numbers: 399, 710, 105, 891, 402, 102, 397, ... As you can see, the average gcd is approximately 100, but how to compute it ? more details: I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels. eg. ----__------____------__-- ... I have the duration of each level, but this time is not accurate. My aim is to find as quick as possible the base time of the signal. My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate. With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100 • What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24 • Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25 • @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28 • @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46 • Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$... – Lehs Jun 21, 2016 at 14:25 I made a similar question here, where I propose a partial solution. How to find the approximate basic frequency or GCD of a list of numbers? In summary, I came with this • being $v$ the list $\{v_1, v_2, \ldots, v_n\}$, • $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$ • $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$ • $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$ And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123: testSeq = [399, 710, 105, 891, 402, 102, 397] gcd = calculateGCDAppeal(x, testSeq) find_local_maximum(gcd,90,110) plot(gcd,(x, 10, 200), scale = "semilogx") One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers? If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well. This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it. Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd. Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$\|r_i\|$$ is very small, say, smaller than $$50$$. <MASK> If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$. <UNMASK> # approximate greatest common divisor I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers. For example, I have the following numbers: 399, 710, 105, 891, 402, 102, 397, ... As you can see, the average gcd is approximately 100, but how to compute it ? more details: I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels. eg. ----__------____------__-- ... I have the duration of each level, but this time is not accurate. My aim is to find as quick as possible the base time of the signal. My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate. With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100 • What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24 • Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25 • @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28 • @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46 • Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$... – Lehs Jun 21, 2016 at 14:25 I made a similar question here, where I propose a partial solution. How to find the approximate basic frequency or GCD of a list of numbers? In summary, I came with this • being $v$ the list $\{v_1, v_2, \ldots, v_n\}$, • $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$ • $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$ • $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$ And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123: testSeq = [399, 710, 105, 891, 402, 102, 397] gcd = calculateGCDAppeal(x, testSeq) find_local_maximum(gcd,90,110) plot(gcd,(x, 10, 200), scale = "semilogx") One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers? If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well. This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it. Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd. Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$\|r_i\|$$ is very small, say, smaller than $$50$$. Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer. It is guaranteed that one $$d_{a,b}$$ is equal to $$p$$. Thus, to check which one is the correct one, just compute the "centered modular reduction" $$r_i' := x_i \bmod d_{a,b} \in [-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$ for $$i \ge 3$$. If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
<MASK> <UNMASK> <MASK> $(a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10})$ Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $a^{9}+b^{12}c^{15} ,$ is \begin{array}{l} (a^3-b^4c^5)[ (a^3)^2-(a^3)(-b^4c^5)+(-b^4c^5)^2] \\\\= (a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10}) .\end{array}
<MASK> $(a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10})$ Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $a^{9}+b^{12}c^{15} ,$ is \begin{array}{l} (a^3-b^4c^5)[ (a^3)^2-(a^3)(-b^4c^5)+(-b^4c^5)^2] \\\\= (a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10}) .\end{array} <UNMASK> ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $(a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10})$ Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $a^{9}+b^{12}c^{15} ,$ is \begin{array}{l} (a^3-b^4c^5)[ (a^3)^2-(a^3)(-b^4c^5)+(-b^4c^5)^2] \\\\= (a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10}) .\end{array}
<MASK> <UNMASK> <MASK> Author: Vineet Shah Editorialist: Vaibhav Jain MEDIUM <MASK> Find the number of ordered pairs (b,c) such that it satisfies the following equation: a+(b×c)=a×b+a×c <MASK> Now only a=0 and a=1 will output -1. This is because 0 has infinite factors which means infinite pairs. <MASK> Editorialist solution can be found here. <MASK> 2 Likes
<MASK> Author: Vineet Shah Editorialist: Vaibhav Jain MEDIUM <MASK> Find the number of ordered pairs (b,c) such that it satisfies the following equation: a+(b×c)=a×b+a×c <MASK> Now only a=0 and a=1 will output -1. This is because 0 has infinite factors which means infinite pairs. <MASK> Editorialist solution can be found here. <MASK> 2 Likes <UNMASK> # SWAPSIGN(Code Melange) - EDITORIAL (UNOFFICIAL) Author: Vineet Shah Editorialist: Vaibhav Jain MEDIUM ### PROBLEM: Find the number of ordered pairs (b,c) such that it satisfies the following equation: a+(b×c)=a×b+a×c ### QUICK EXPLANATION: We can rewrite equation as a(a-1)=(b-a)(c-a). Now the answer will be 2×fact(a)×fact(a-1) where fact(i) is the number of factors of i. ### EXPLANATION: Let us consider our equation: a+(b×c)=a×b+a×c\tag{} Rewrite it as: a+b×c-a×b-a×c=0\tag{} a+b×c-a×b-a×c+a^{2}=a^{2}\tag{} Taking common: b×(c-a)+a×(a-c)=a^{2}-a\tag{} Rearranging: (b-a)×(c-a)=a×(a-1)\tag{} Now we can see that number of ordered pairs (b,c) shall be equal to the factors of a×(a-1) i.e. fact(a)×fact(a-1). This is because fact(a) is multiplicative function and gcd(a,a-1)=1. Note here that if (b,c) satisfies equation so does (-b,-c) and similarly when (b,-c) satisfies equation so does (-b,c). Hence, actual answer will be 2×fact(a)×fact(a-1). But finding factors of a will give TLE. Hence we will use concept here. We can precompute every prime number till 10^{6} using the old Sieve of Eratosthenes or whatever variant you prefer. Now we will divide a from every prime factor till 10^{6} to count the powers of distinct prime numbers to calculate number of factors. After doing this a must be either 1 or if not then a must have prime factor(s) greater than 10^{6}. Claim: In this case a can have either only one or two prime factors greater than 10^{6}. This is because if a will have three prime factors greater than 10^{6} it will exceed 10^{18}. Hence, to check whether a has only factor or two factors left can be done using primality test. If primality test comes out to be true that means a has left only one factor otherwise it has two factors. For large numbers we will use one of the probabilistic methods of primality test whose complexity is O(klogn) where k is number of iterations. Now only a=0 and a=1 will output -1. This is because 0 has infinite factors which means infinite pairs. Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems. ### EDITORIALIST’S SOLUTIONS: Editorialist solution can be found here. ### RELATED PROBLEMS: https://www.codechef.com/PRACTICE/problems/NUMFACT 1 Like Mention this as well - Let fact(x) = no of factors of x. And fact(x) is a multiplicative function. Because gcd(a,a-1) is 1. Therefore, fact(a(a-1)) = fact(a)*fact(a-1). 2 Likes
<MASK> Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that: <MASK> f to fn are the functions of x. In order to figure out how to solve differential equation, you firstly have to determine the order of the differential equation. For example, for the second order differential equation there is a more special method of finding the solution: divide the second order differential equation in 2 parts: Q(x)=0 and Q(x) is a function of x. For both members calculate the auxiliary equation and find the complementary function. Next, if Q(x) is a part of the equation, find the particular integral of the equation. In the end, sum up the complementary function with the particular integral. ## Solving differential equations video lesson Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun" <UNMASK> # How to solve differential equations <MASK> For the homogeneous differential equations, we use the substitution method and we reduce the equation to the variable separable. Having an exercise in which you have to solve the differential equation, you firstly have to figure out what kind of differential equation is the equation, so you know what method it's better to use. Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that: A differential equation is called linear as long as the dependent variable and its derivatives occur in the first degree and are not multiplied together. f to fn are the functions of x. In order to figure out how to solve differential equation, you firstly have to determine the order of the differential equation. For example, for the second order differential equation there is a more special method of finding the solution: divide the second order differential equation in 2 parts: Q(x)=0 and Q(x) is a function of x. For both members calculate the auxiliary equation and find the complementary function. Next, if Q(x) is a part of the equation, find the particular integral of the equation. In the end, sum up the complementary function with the particular integral. ## Solving differential equations video lesson Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun"
# How to solve differential equations <MASK> For the homogeneous differential equations, we use the substitution method and we reduce the equation to the variable separable. Having an exercise in which you have to solve the differential equation, you firstly have to figure out what kind of differential equation is the equation, so you know what method it's better to use. Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that: A differential equation is called linear as long as the dependent variable and its derivatives occur in the first degree and are not multiplied together. f to fn are the functions of x. In order to figure out how to solve differential equation, you firstly have to determine the order of the differential equation. For example, for the second order differential equation there is a more special method of finding the solution: divide the second order differential equation in 2 parts: Q(x)=0 and Q(x) is a function of x. For both members calculate the auxiliary equation and find the complementary function. Next, if Q(x) is a part of the equation, find the particular integral of the equation. In the end, sum up the complementary function with the particular integral. ## Solving differential equations video lesson Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun" <UNMASK> # How to solve differential equations Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises. Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution. For the homogeneous differential equations, we use the substitution method and we reduce the equation to the variable separable. Having an exercise in which you have to solve the differential equation, you firstly have to figure out what kind of differential equation is the equation, so you know what method it's better to use. Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that: A differential equation is called linear as long as the dependent variable and its derivatives occur in the first degree and are not multiplied together. f to fn are the functions of x. In order to figure out how to solve differential equation, you firstly have to determine the order of the differential equation. For example, for the second order differential equation there is a more special method of finding the solution: divide the second order differential equation in 2 parts: Q(x)=0 and Q(x) is a function of x. For both members calculate the auxiliary equation and find the complementary function. Next, if Q(x) is a part of the equation, find the particular integral of the equation. In the end, sum up the complementary function with the particular integral. ## Solving differential equations video lesson Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun"
# Radius Calculator: Compute Dimensions of a Circle <MASK> Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius. <MASK> The radius of a circle is the distance from a circle's origin or center to its edge. <MASK> ## Dimensions of a Circle <MASK> You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers. <MASK> PK lives in New Hampshire with his wife, kids, and dog. <MASK> DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases. <UNMASK> # Radius Calculator: Compute Dimensions of a Circle Written by: PK Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius. Do you know a different dimension? Instead try one of the other circle dimension calculators: ## What is a Circle's Radius? The radius of a circle is the distance from a circle's origin or center to its edge. <MASK> ## Dimensions of a Circle <MASK> To find the radius from the diameter, you only have to divide by two: <MASK> If you know the circumference it is a bit harder, but not too bad: r=c/2\pi Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius: a = \pi r^2 You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers. Area has different units, but you can use: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers. <MASK> PK lives in New Hampshire with his wife, kids, and dog. ### Don't Quit Your Day Job... DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases.
# Radius Calculator: Compute Dimensions of a Circle Written by: PK Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius. Do you know a different dimension? Instead try one of the other circle dimension calculators: ## What is a Circle's Radius? The radius of a circle is the distance from a circle's origin or center to its edge. <MASK> ## Dimensions of a Circle <MASK> To find the radius from the diameter, you only have to divide by two: <MASK> If you know the circumference it is a bit harder, but not too bad: r=c/2\pi Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius: a = \pi r^2 You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers. Area has different units, but you can use: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers. <MASK> PK lives in New Hampshire with his wife, kids, and dog. ### Don't Quit Your Day Job... DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases. <UNMASK> # Radius Calculator: Compute Dimensions of a Circle Written by: PK Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius. Do you know a different dimension? Instead try one of the other circle dimension calculators: ## What is a Circle's Radius? The radius of a circle is the distance from a circle's origin or center to its edge. Conveniently, it is half as long as the diameter of a circle. A diameter is just two radiuses drawn in opposing directions from the circle's origin. ## Dimensions of a Circle For a circle, three lengths most commonly are applied: • The radius – defined above • The diameter – the distance from edge to edge of a circle passing through its origin or center. Twice the length of a circle's radius • The circumference – the length of the outside boundaries of the circle If you know the radius, it is straightforward to compute the other two. To find the radius from the diameter, you only have to divide by two: r=d/2 If you know the circumference it is a bit harder, but not too bad: r=c/2\pi Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius: a = \pi r^2 You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers. Area has different units, but you can use: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers. To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius. ### PK PK started DQYDJ in 2009 to research and discuss finance and investing and help answer financial questions. He's expanded DQYDJ to build visualizations, calculators, and interactive tools. PK lives in New Hampshire with his wife, kids, and dog. ### Don't Quit Your Day Job... DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases.
<MASK> ## Questions Asked in IBPS SO Prelims Exam 31 December 2017 The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams. <MASK> ## Quant Questions Asked in IBPS SO Pre Exam 31 Dec 2017 <MASK> 1. B 2. A 3. D 4. B 5. A 6. C 7. D 8. E 9. C 10. E 11. B 12. E 13. C 14. A 15. D <MASK> ## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2) We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com <UNMASK> <MASK> ## Questions Asked in IBPS SO Prelims Exam 31 December 2017 The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams. Here is the detailed Questions Asked in IBPS SO Prelims Exam 31 December 2017. Read today’s Exam Analysis along with this article to know what kind of questions are asked in today’s exam. This will be very beneficial for you if you haven’t appeared for your exam yet. Even if you did appear for the exam already, you can analyze your performance by having a look at the analysis published. ## Quant Questions Asked in IBPS SO Pre Exam 31 Dec 2017 Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017 1. 11, 13, 111, 257, 427, ? Answers - 609 2. 38, 51, 25, 64, 12, ? Answers - 77 3. 4, 14, 31, 57, 94, ? Answers - 50 4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40 5. 6, 5, 9, 26, ?, 514 Answers - 103 Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ? Q 3. Root (2025-x)/25 =16 ## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017 Directions (1 – 5): Answer the questions on the basis of the information given below. Eight friends A, B, C, D, E, F, G and H are seating around a circular table. Some of them are facing inside & others are facing outside. Opposite direction means if one is facing inside the centre, second one face outside the centre, and vice versa. Three people are seating between F and D and both face opposite direction to each other. E is seating second to right of both D and F, and face opposite direction as F faces. C is seating third to right of E, who is not opposite to B. G is neighbour of both E and D, and seating second to right of B, who is not a neighbour of E. G is third to left of A, who is third to right of both H and G. B face towards the centre and is seating second to left of C. 1. If all the friends are seating according to alphabetical order in anti clock wise direction, starting from A, how many friends remain at same position (excluding A)? A) One B) Two C) Three D) Four E) None 2. If D interchanged his position with H and, F interchanged his position with G, who sits immediately right of H in new arrangement? A) F B) G C) E D) A E) None 3. Who is seating third to left of B? A) H B) C C) A D) F E) None 4. How many persons are facing away from the centre? A) Three B) Four C) Five D) Two E) None 5. Who among the following pairs are facing same direction and seating opposite to each other ? A) H, B B) C, G C) F, D D) E, A E) None Directions (6 – 10): Answer the questions on the basis of the information given below. Ten friends are sitting on twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In Row 1: A, B, C, D and E are seated and all of them are facing south, and in Row 2: P, Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row. All of them like different colors – Red, Green, Black, Yellow, White, Blue, Brown, Purple, Pink and Grey, but not necessarily in the same order. There are two seats between Q and the vacant seat. Q does not like White, Red and Purple. E is not an immediate neighbor of C. B likes Grey. Vacant seat of row 1 is not opposite to S and is also not at any of the extreme ends of Row-1.The one who likes Black sits opposite to the one, who sits third to the right of the seat, which is opposite to S. C is not an immediate neighbor of D. T, who likes neither White nor Blue, does not face vacant seat. D faces R. The vacant seats are not opposite to each other. Two seats are there between C and B, who sits third right of the seat, on which the person who likes Brown is sitting. S sits third to the right of seat on which R sits and likes Yellow. The one who likes Pink faces the one who likes Yellow. The persons who like Red and Purple are adjacent to each other. The vacant seat in row 1 is not adjacent to D.Q sits at one of the extreme ends. E neither likes Pink nor faces the seat which is adjacent to the one who likes Blue. The one who likes White is not to the immediate right of the one who likes Yellow. The person who likes Green doesn’t face the person who likes Purple. 6. How many persons are sitting between T and the one who likes yellow color? A) None B) One C) Two D) Three E) None of these 7. Which of the following faces the vacant seat of Row – 2? A) The one who like white color B) A C) D D) The one who likes grey color E) Cannot be determined 8. Who is sitting at the immediate left of person who likes purple color? A) E B) D C) The one who likes black color D) The one who likes green color E) The one who likes grey color 9. Who amongst the following sits at the extreme end of the row? A) R, Q B) E, S C) T, C D) C, D E) None of these 10. If Q is made to sit on vacant seat of his row, then how many persons are there between the persons who sit opposite to Q now and who sat opposite to Q previously? A) Two B) Three C) Four D) None E) One Directions (Q. 11–15): Study the following arrangement of series carefully and answer the questions given below: C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S 11. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both? A) None B) Three C) One D) More than three E) Two 12. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end? A) 8 B) 2 C) A D) Other than given options E) O 13. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter? A) None B) Two C) One D) Three E) More than three 14. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group? A) 6 &* B) 5 X O C) F L 4 D) G O 8 E) 9 K % 15. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement? A) I B) Other than given options C) 8 D) J E) * 1. B 2. A 3. D 4. B 5. A 6. C 7. D 8. E 9. C 10. E 11. B 12. E 13. C 14. A 15. D ## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017 <MASK> Eg. i) He is _______ at playing the piano. ii) Exercise is _______ for health. ## General Awareness Asked in IBPS SO Pre Exam 31 Dec 2017 2. FIFA U-19 was held at? 3. 1 Qs based on Basel committee? 5. Repo rate? update soon ## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017 update soon ## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017 <MASK> ## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2) We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
<MASK> ## Questions Asked in IBPS SO Prelims Exam 31 December 2017 The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams. Here is the detailed Questions Asked in IBPS SO Prelims Exam 31 December 2017. Read today’s Exam Analysis along with this article to know what kind of questions are asked in today’s exam. This will be very beneficial for you if you haven’t appeared for your exam yet. Even if you did appear for the exam already, you can analyze your performance by having a look at the analysis published. ## Quant Questions Asked in IBPS SO Pre Exam 31 Dec 2017 Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017 1. 11, 13, 111, 257, 427, ? Answers - 609 2. 38, 51, 25, 64, 12, ? Answers - 77 3. 4, 14, 31, 57, 94, ? Answers - 50 4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40 5. 6, 5, 9, 26, ?, 514 Answers - 103 Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ? Q 3. Root (2025-x)/25 =16 ## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017 Directions (1 – 5): Answer the questions on the basis of the information given below. Eight friends A, B, C, D, E, F, G and H are seating around a circular table. Some of them are facing inside & others are facing outside. Opposite direction means if one is facing inside the centre, second one face outside the centre, and vice versa. Three people are seating between F and D and both face opposite direction to each other. E is seating second to right of both D and F, and face opposite direction as F faces. C is seating third to right of E, who is not opposite to B. G is neighbour of both E and D, and seating second to right of B, who is not a neighbour of E. G is third to left of A, who is third to right of both H and G. B face towards the centre and is seating second to left of C. 1. If all the friends are seating according to alphabetical order in anti clock wise direction, starting from A, how many friends remain at same position (excluding A)? A) One B) Two C) Three D) Four E) None 2. If D interchanged his position with H and, F interchanged his position with G, who sits immediately right of H in new arrangement? A) F B) G C) E D) A E) None 3. Who is seating third to left of B? A) H B) C C) A D) F E) None 4. How many persons are facing away from the centre? A) Three B) Four C) Five D) Two E) None 5. Who among the following pairs are facing same direction and seating opposite to each other ? A) H, B B) C, G C) F, D D) E, A E) None Directions (6 – 10): Answer the questions on the basis of the information given below. Ten friends are sitting on twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In Row 1: A, B, C, D and E are seated and all of them are facing south, and in Row 2: P, Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row. All of them like different colors – Red, Green, Black, Yellow, White, Blue, Brown, Purple, Pink and Grey, but not necessarily in the same order. There are two seats between Q and the vacant seat. Q does not like White, Red and Purple. E is not an immediate neighbor of C. B likes Grey. Vacant seat of row 1 is not opposite to S and is also not at any of the extreme ends of Row-1.The one who likes Black sits opposite to the one, who sits third to the right of the seat, which is opposite to S. C is not an immediate neighbor of D. T, who likes neither White nor Blue, does not face vacant seat. D faces R. The vacant seats are not opposite to each other. Two seats are there between C and B, who sits third right of the seat, on which the person who likes Brown is sitting. S sits third to the right of seat on which R sits and likes Yellow. The one who likes Pink faces the one who likes Yellow. The persons who like Red and Purple are adjacent to each other. The vacant seat in row 1 is not adjacent to D.Q sits at one of the extreme ends. E neither likes Pink nor faces the seat which is adjacent to the one who likes Blue. The one who likes White is not to the immediate right of the one who likes Yellow. The person who likes Green doesn’t face the person who likes Purple. 6. How many persons are sitting between T and the one who likes yellow color? A) None B) One C) Two D) Three E) None of these 7. Which of the following faces the vacant seat of Row – 2? A) The one who like white color B) A C) D D) The one who likes grey color E) Cannot be determined 8. Who is sitting at the immediate left of person who likes purple color? A) E B) D C) The one who likes black color D) The one who likes green color E) The one who likes grey color 9. Who amongst the following sits at the extreme end of the row? A) R, Q B) E, S C) T, C D) C, D E) None of these 10. If Q is made to sit on vacant seat of his row, then how many persons are there between the persons who sit opposite to Q now and who sat opposite to Q previously? A) Two B) Three C) Four D) None E) One Directions (Q. 11–15): Study the following arrangement of series carefully and answer the questions given below: C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S 11. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both? A) None B) Three C) One D) More than three E) Two 12. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end? A) 8 B) 2 C) A D) Other than given options E) O 13. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter? A) None B) Two C) One D) Three E) More than three 14. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group? A) 6 &* B) 5 X O C) F L 4 D) G O 8 E) 9 K % 15. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement? A) I B) Other than given options C) 8 D) J E) * 1. B 2. A 3. D 4. B 5. A 6. C 7. D 8. E 9. C 10. E 11. B 12. E 13. C 14. A 15. D ## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017 <MASK> Eg. i) He is _______ at playing the piano. ii) Exercise is _______ for health. ## General Awareness Asked in IBPS SO Pre Exam 31 Dec 2017 2. FIFA U-19 was held at? 3. 1 Qs based on Basel committee? 5. Repo rate? update soon ## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017 update soon ## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017 <MASK> ## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2) We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com <UNMASK> # Questions Asked in IBPS SO Prelims Exam 31 December 2017 ## Questions Asked in IBPS SO Prelims Exam 31 December 2017 The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams. Here is the detailed Questions Asked in IBPS SO Prelims Exam 31 December 2017. Read today’s Exam Analysis along with this article to know what kind of questions are asked in today’s exam. This will be very beneficial for you if you haven’t appeared for your exam yet. Even if you did appear for the exam already, you can analyze your performance by having a look at the analysis published. ## Quant Questions Asked in IBPS SO Pre Exam 31 Dec 2017 Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017 1. 11, 13, 111, 257, 427, ? Answers - 609 2. 38, 51, 25, 64, 12, ? Answers - 77 3. 4, 14, 31, 57, 94, ? Answers - 50 4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40 5. 6, 5, 9, 26, ?, 514 Answers - 103 Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ? Q 3. Root (2025-x)/25 =16 ## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017 Directions (1 – 5): Answer the questions on the basis of the information given below. Eight friends A, B, C, D, E, F, G and H are seating around a circular table. Some of them are facing inside & others are facing outside. Opposite direction means if one is facing inside the centre, second one face outside the centre, and vice versa. Three people are seating between F and D and both face opposite direction to each other. E is seating second to right of both D and F, and face opposite direction as F faces. C is seating third to right of E, who is not opposite to B. G is neighbour of both E and D, and seating second to right of B, who is not a neighbour of E. G is third to left of A, who is third to right of both H and G. B face towards the centre and is seating second to left of C. 1. If all the friends are seating according to alphabetical order in anti clock wise direction, starting from A, how many friends remain at same position (excluding A)? A) One B) Two C) Three D) Four E) None 2. If D interchanged his position with H and, F interchanged his position with G, who sits immediately right of H in new arrangement? A) F B) G C) E D) A E) None 3. Who is seating third to left of B? A) H B) C C) A D) F E) None 4. How many persons are facing away from the centre? A) Three B) Four C) Five D) Two E) None 5. Who among the following pairs are facing same direction and seating opposite to each other ? A) H, B B) C, G C) F, D D) E, A E) None Directions (6 – 10): Answer the questions on the basis of the information given below. Ten friends are sitting on twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In Row 1: A, B, C, D and E are seated and all of them are facing south, and in Row 2: P, Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row. All of them like different colors – Red, Green, Black, Yellow, White, Blue, Brown, Purple, Pink and Grey, but not necessarily in the same order. There are two seats between Q and the vacant seat. Q does not like White, Red and Purple. E is not an immediate neighbor of C. B likes Grey. Vacant seat of row 1 is not opposite to S and is also not at any of the extreme ends of Row-1.The one who likes Black sits opposite to the one, who sits third to the right of the seat, which is opposite to S. C is not an immediate neighbor of D. T, who likes neither White nor Blue, does not face vacant seat. D faces R. The vacant seats are not opposite to each other. Two seats are there between C and B, who sits third right of the seat, on which the person who likes Brown is sitting. S sits third to the right of seat on which R sits and likes Yellow. The one who likes Pink faces the one who likes Yellow. The persons who like Red and Purple are adjacent to each other. The vacant seat in row 1 is not adjacent to D.Q sits at one of the extreme ends. E neither likes Pink nor faces the seat which is adjacent to the one who likes Blue. The one who likes White is not to the immediate right of the one who likes Yellow. The person who likes Green doesn’t face the person who likes Purple. 6. How many persons are sitting between T and the one who likes yellow color? A) None B) One C) Two D) Three E) None of these 7. Which of the following faces the vacant seat of Row – 2? A) The one who like white color B) A C) D D) The one who likes grey color E) Cannot be determined 8. Who is sitting at the immediate left of person who likes purple color? A) E B) D C) The one who likes black color D) The one who likes green color E) The one who likes grey color 9. Who amongst the following sits at the extreme end of the row? A) R, Q B) E, S C) T, C D) C, D E) None of these 10. If Q is made to sit on vacant seat of his row, then how many persons are there between the persons who sit opposite to Q now and who sat opposite to Q previously? A) Two B) Three C) Four D) None E) One Directions (Q. 11–15): Study the following arrangement of series carefully and answer the questions given below: C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S 11. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both? A) None B) Three C) One D) More than three E) Two 12. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end? A) 8 B) 2 C) A D) Other than given options E) O 13. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter? A) None B) Two C) One D) Three E) More than three 14. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group? A) 6 &* B) 5 X O C) F L 4 D) G O 8 E) 9 K % 15. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement? A) I B) Other than given options C) 8 D) J E) * 1. B 2. A 3. D 4. B 5. A 6. C 7. D 8. E 9. C 10. E 11. B 12. E 13. C 14. A 15. D ## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017 1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs) – 1 Passage was based on “Scope on Economy” – Other passage was based on “Steps that should be taken to stop the corruption” – Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam. 2. Error Spotting 10 Qs – Asked in a different way. A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts. 3. Fill in the blanks – 5 Qs – Vocab Based. 2 Sentences were given in which you had to choose a word which correctly fits in both the sentences. Eg. i) He is _______ at playing the piano. ii) Exercise is _______ for health. ## General Awareness Asked in IBPS SO Pre Exam 31 Dec 2017 2. FIFA U-19 was held at? 3. 1 Qs based on Basel committee? 5. Repo rate? update soon ## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017 update soon ## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017 update soon ## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2) We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
# 711 US dry barrels in dry quarts ## Conversion <MASK> ## Conversion formula How to convert 711 US dry barrels to dry quarts? <MASK> $x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$ Conclusion: $711 ⁢ drybarrel ≈ 74654.30567052051 ⁢ dryquart$ ## Conversion in the opposite direction The inverse of the conversion factor is that 1 dry quart is equal to 1.339507468482e-05 times 711 US dry barrels. <MASK> ## Approximation <MASK> [1] The precision is 15 significant digits (fourteen digits to the right of the decimal point). <MASK> Was it helpful? Share it! <UNMASK> # 711 US dry barrels in dry quarts ## Conversion <MASK> ## Conversion formula How to convert 711 US dry barrels to dry quarts? We know (by definition) that: $1\mathrm{drybarrel}\approx 104.999023446583\mathrm{dryquart}$ <MASK> Now, we cross multiply to solve for our unknown $x$: $x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$ Conclusion: $711 ⁢ drybarrel ≈ 74654.30567052051 ⁢ dryquart$ ## Conversion in the opposite direction The inverse of the conversion factor is that 1 dry quart is equal to 1.339507468482e-05 times 711 US dry barrels. <MASK> ## Approximation An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels. <MASK> [1] The precision is 15 significant digits (fourteen digits to the right of the decimal point). <MASK> Was it helpful? Share it!
# 711 US dry barrels in dry quarts ## Conversion <MASK> ## Conversion formula How to convert 711 US dry barrels to dry quarts? We know (by definition) that: $1\mathrm{drybarrel}\approx 104.999023446583\mathrm{dryquart}$ <MASK> Now, we cross multiply to solve for our unknown $x$: $x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$ Conclusion: $711 ⁢ drybarrel ≈ 74654.30567052051 ⁢ dryquart$ ## Conversion in the opposite direction The inverse of the conversion factor is that 1 dry quart is equal to 1.339507468482e-05 times 711 US dry barrels. <MASK> ## Approximation An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels. <MASK> [1] The precision is 15 significant digits (fourteen digits to the right of the decimal point). <MASK> Was it helpful? Share it! <UNMASK> # 711 US dry barrels in dry quarts ## Conversion 711 US dry barrels is equivalent to 74654.3056705202 dry quarts.[1] ## Conversion formula How to convert 711 US dry barrels to dry quarts? We know (by definition) that: $1\mathrm{drybarrel}\approx 104.999023446583\mathrm{dryquart}$ We can set up a proportion to solve for the number of dry quarts. $1 ⁢ drybarrel 711 ⁢ drybarrel ≈ 104.999023446583 ⁢ dryquart x ⁢ dryquart$ Now, we cross multiply to solve for our unknown $x$: $x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$ Conclusion: $711 ⁢ drybarrel ≈ 74654.30567052051 ⁢ dryquart$ ## Conversion in the opposite direction The inverse of the conversion factor is that 1 dry quart is equal to 1.339507468482e-05 times 711 US dry barrels. It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts. ## Approximation An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels. ## Footnotes [1] The precision is 15 significant digits (fourteen digits to the right of the decimal point). Results may contain small errors due to the use of floating point arithmetic. Was it helpful? Share it!
<MASK> -40 degrees Fahrenheit = -40 degrees Celsius <MASK> The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales. <MASK> <UNMASK> <MASK> We often associate temperature with how hot or cold something is. If we see a piece of metal glowing red, that would be hot to the touch. A ground covered with snow would be cold to touch. This is only part of the story. In physics, temperature is the measure of how internal particles move. The image above shows an example of this. If this image represents a gas, the particles that constitute the gas would move depending on temperature. In a hot gas, the molecules (represented by the dots) are free to move - the faster they move, the hotter the gas. Alternately, a cold gas would equal slow or very minimal movement of molecules. Absolute zero is defined as molecules in a medium that are not moving at all. In the Celsius temperature scale, absolute zero is -273.2 degrees C. In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such: Celsius has by no means been abandoned. Much of our standards are based on the Celsius scale: • 0 degree Celsius = Freezing point of water • 100 degrees Celsius = Boiling point of water <MASK> • 32 degrees Fahrenheit = Freezing point of water • 212 degrees Fahrenheit = Boiling point of water The temperatures above do not follow a nice neat pattern like the Celsius scale, but of special note: -40 degrees Fahrenheit = -40 degrees Celsius To convert Fahrenheit to Celsius: Other important notes: The symbol for degrees (o) or the word degree is NOT used on the Kelvin scale: Absolute zero = 0 K The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales. In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
<MASK> We often associate temperature with how hot or cold something is. If we see a piece of metal glowing red, that would be hot to the touch. A ground covered with snow would be cold to touch. This is only part of the story. In physics, temperature is the measure of how internal particles move. The image above shows an example of this. If this image represents a gas, the particles that constitute the gas would move depending on temperature. In a hot gas, the molecules (represented by the dots) are free to move - the faster they move, the hotter the gas. Alternately, a cold gas would equal slow or very minimal movement of molecules. Absolute zero is defined as molecules in a medium that are not moving at all. In the Celsius temperature scale, absolute zero is -273.2 degrees C. In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such: Celsius has by no means been abandoned. Much of our standards are based on the Celsius scale: • 0 degree Celsius = Freezing point of water • 100 degrees Celsius = Boiling point of water <MASK> • 32 degrees Fahrenheit = Freezing point of water • 212 degrees Fahrenheit = Boiling point of water The temperatures above do not follow a nice neat pattern like the Celsius scale, but of special note: -40 degrees Fahrenheit = -40 degrees Celsius To convert Fahrenheit to Celsius: Other important notes: The symbol for degrees (o) or the word degree is NOT used on the Kelvin scale: Absolute zero = 0 K The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales. In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether. <UNMASK> Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions Physics - Concepts - temperature We often associate temperature with how hot or cold something is. If we see a piece of metal glowing red, that would be hot to the touch. A ground covered with snow would be cold to touch. This is only part of the story. In physics, temperature is the measure of how internal particles move. The image above shows an example of this. If this image represents a gas, the particles that constitute the gas would move depending on temperature. In a hot gas, the molecules (represented by the dots) are free to move - the faster they move, the hotter the gas. Alternately, a cold gas would equal slow or very minimal movement of molecules. Absolute zero is defined as molecules in a medium that are not moving at all. In the Celsius temperature scale, absolute zero is -273.2 degrees C. In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such: Celsius has by no means been abandoned. Much of our standards are based on the Celsius scale: • 0 degree Celsius = Freezing point of water • 100 degrees Celsius = Boiling point of water The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science. • 32 degrees Fahrenheit = Freezing point of water • 212 degrees Fahrenheit = Boiling point of water The temperatures above do not follow a nice neat pattern like the Celsius scale, but of special note: -40 degrees Fahrenheit = -40 degrees Celsius To convert Fahrenheit to Celsius: Other important notes: The symbol for degrees (o) or the word degree is NOT used on the Kelvin scale: Absolute zero = 0 K The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales. In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
<MASK> Dear vardaan <MASK> and for this small distance dz ,velocity of river assumed to be constant V =z1/ 2 <MASK> ds =z1/ 2 dt <MASK> =2√2/3 t3/2 limit o to 10 <MASK> We are all IITians and here to help you in your IIT JEE preparation. <MASK> <UNMASK> # THE FLOW SPEED OF WATER VARIES WITH THE DISTANCE FROM ITS STRAIGHT BANK AS V=(1m1/2s).z1/ 2 where z is the perpendicular distance in m from the bank A.SWIMMER ENTERS THE RIVER AT A POINT A ON THE BANK AND SWIMS AT 2m/s, IN A DIRN NORMAL TO RIVER FLOW. WHAT IS HIS DRIFT DOWNSTREAM BY THE TIME HIS NORMAL DISTANCE FROM THE bank IS 20m? 148 Points 13 years ago Dear vardaan V=z1/ 2 velocity of man =2 m/sec <MASK> and time taken be man to swim z distance in normal direction t= z/2............1 now consider motion from z to z+dz distance time taken by man to travel z to z+dz postion is dt and for this small distance dz ,velocity of river assumed to be constant V =z1/ 2 <MASK> ds =z1/ 2 dt ds =(2t)1/2 dt from equation 1 ds =(2t)1/2 dt S =√2 t3/2 /3/2 limit o to 10 =2√2/3 t3/2 limit o to 10 =40√5/3 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards,
# THE FLOW SPEED OF WATER VARIES WITH THE DISTANCE FROM ITS STRAIGHT BANK AS V=(1m1/2s).z1/ 2 where z is the perpendicular distance in m from the bank A.SWIMMER ENTERS THE RIVER AT A POINT A ON THE BANK AND SWIMS AT 2m/s, IN A DIRN NORMAL TO RIVER FLOW. WHAT IS HIS DRIFT DOWNSTREAM BY THE TIME HIS NORMAL DISTANCE FROM THE bank IS 20m? 148 Points 13 years ago Dear vardaan V=z1/ 2 velocity of man =2 m/sec <MASK> and time taken be man to swim z distance in normal direction t= z/2............1 now consider motion from z to z+dz distance time taken by man to travel z to z+dz postion is dt and for this small distance dz ,velocity of river assumed to be constant V =z1/ 2 <MASK> ds =z1/ 2 dt ds =(2t)1/2 dt from equation 1 ds =(2t)1/2 dt S =√2 t3/2 /3/2 limit o to 10 =2√2/3 t3/2 limit o to 10 =40√5/3 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, <UNMASK> # THE FLOW SPEED OF WATER VARIES WITH THE DISTANCE FROM ITS STRAIGHT BANK AS V=(1m1/2s).z1/ 2 where z is the perpendicular distance in m from the bank A.SWIMMER ENTERS THE RIVER AT A POINT A ON THE BANK AND SWIMS AT 2m/s, IN A DIRN NORMAL TO RIVER FLOW. WHAT IS HIS DRIFT DOWNSTREAM BY THE TIME HIS NORMAL DISTANCE FROM THE bank IS 20m? 148 Points 13 years ago Dear vardaan V=z1/ 2 velocity of man =2 m/sec time taken by man to swim 20 m in normal direction t= 20/2 =10 sec and time taken be man to swim z distance in normal direction t= z/2............1 now consider motion from z to z+dz distance time taken by man to travel z to z+dz postion is dt and for this small distance dz ,velocity of river assumed to be constant V =z1/ 2 so distance moved in downstreem direction in time dt is ds =z1/ 2 dt ds =(2t)1/2 dt from equation 1 ds =(2t)1/2 dt S =√2 t3/2 /3/2 limit o to 10 =2√2/3 t3/2 limit o to 10 =40√5/3 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards,
# Market Power: Monopoly and Monopsony <MASK> Review of Perfect Competition P = LMC = LRAC Normal profits or zero economic profits in the long run Large number of buyers and sellers Homogenous product Perfect information Firm is a price taker <MASK> Average and Marginal Revenue 1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7 We can also see algebraically that Q. maximizes profit We can also see algebraically that Q* maximizes profit. Profit π is the difference between revenue and cost, both of which depend on Q: As Q is increased from zero, profit will increase until it reaches a maximum and then begin to decrease. Thus the profit-maximizing Q is such that the incremental profit resulting from a small increase in Q is just zero (i.e., Δπ /ΔQ = 0). Then But ΔR/ΔQ is marginal revenue and ΔC/ΔQ is marginal cost. Thus the profit-maximizing condition is that , or <MASK> (Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost: <MASK> A Rule of Thumb for Pricing A Rule of Thumb for Pricing A Rule of Thumb for Pricing Monopoly Monopoly pricing compared to perfect competition pricing: P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC <MASK> The Multi-plant Firm For some firms, production takes place in more than one plant each with different costs Firm must determine how to distribute production between both plants Production should be split so that the MC in the plants is the same Output is chosen where MR=MC. Profits is therefore maximized when MR=MC at each plant We can show this algebraically: Q1 and C1 is output and cost of production for Plant 1 Q2 and C2 is output and cost of production for Plant 2 QT = Q1 + Q2 is total output Profit is then:  = PQT – C1(Q1) – C2(Q2) <MASK> Monopsony A monopsony is a market in which there is a single buyer. An oligopsony is a market with only a few buyers. Monopsony power is the ability of the buyer to affect the price of the good and pay less than the price that would exist in a competitive market. Typically choose to buy until the benefit from last unit equals that unit’s cost Marginal value is the additional benefit derived from purchasing one more unit of a good Demand curve – downward sloping Marginal expenditure is the additional cost of buying one more unit of a good Depends on buying power <MASK> Monopoly and Monopsony Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher <MASK> <UNMASK> # Market Power: Monopoly and Monopsony ## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript: Market Power: Monopoly and Monopsony Chapter 9 Review of Perfect Competition P = LMC = LRAC Normal profits or zero economic profits in the long run Large number of buyers and sellers Homogenous product Perfect information Firm is a price taker <MASK> Monopoly Monopoly One seller - many buyers One product (no good substitutes) Barriers to entry Price Maker The monopolist is the supply-side of the market and has complete control over the amount offered for sale. Monopolist controls price but must consider consumer demand Profits will be maximized at the level of output where marginal revenue equals marginal cost. <MASK> Average and Marginal Revenue 1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7 We can also see algebraically that Q. maximizes profit We can also see algebraically that Q* maximizes profit. Profit π is the difference between revenue and cost, both of which depend on Q: As Q is increased from zero, profit will increase until it reaches a maximum and then begin to decrease. Thus the profit-maximizing Q is such that the incremental profit resulting from a small increase in Q is just zero (i.e., Δπ /ΔQ = 0). Then But ΔR/ΔQ is marginal revenue and ΔC/ΔQ is marginal cost. Thus the profit-maximizing condition is that , or Monopolist’s Output Decision \$ per unit of output D = AR MR MC AC P1 Q1 P* Q* P2 Q2 Lost profit Lost profit Quantity <MASK> (Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost: <MASK> A Rule of Thumb for Pricing A Rule of Thumb for Pricing A Rule of Thumb for Pricing A Rule of Thumb for Pricing Monopoly Monopoly pricing compared to perfect competition pricing: P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market <MASK> The Multi-plant Firm For some firms, production takes place in more than one plant each with different costs Firm must determine how to distribute production between both plants Production should be split so that the MC in the plants is the same Output is chosen where MR=MC. Profits is therefore maximized when MR=MC at each plant We can show this algebraically: Q1 and C1 is output and cost of production for Plant 1 Q2 and C2 is output and cost of production for Plant 2 QT = Q1 + Q2 is total output Profit is then:  = PQT – C1(Q1) – C2(Q2) The Multi-plant Firm Firm should increase output from each plant until the additional profit from last unit produced at Plant 1 equals 0 <MASK> Monopsony A monopsony is a market in which there is a single buyer. An oligopsony is a market with only a few buyers. Monopsony power is the ability of the buyer to affect the price of the good and pay less than the price that would exist in a competitive market. Typically choose to buy until the benefit from last unit equals that unit’s cost Marginal value is the additional benefit derived from purchasing one more unit of a good Demand curve – downward sloping Marginal expenditure is the additional cost of buying one more unit of a good Depends on buying power Monopsony Competitive Buyer Price taker P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller Monopsonist Buyer ME S = AE PC Pm D = MV Qm QC \$/Q Monopsony ME above S Quantity where ME = MV: Qm Price from Supply curve: Pm ME D = MV S = AE Qm Pm Competitive P = PC Q = Q+C PC QC Quantity Monopoly and Monopsony Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher <MASK> Download ppt "Market Power: Monopoly and Monopsony" Similar presentations
# Market Power: Monopoly and Monopsony ## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript: Market Power: Monopoly and Monopsony Chapter 9 Review of Perfect Competition P = LMC = LRAC Normal profits or zero economic profits in the long run Large number of buyers and sellers Homogenous product Perfect information Firm is a price taker <MASK> Monopoly Monopoly One seller - many buyers One product (no good substitutes) Barriers to entry Price Maker The monopolist is the supply-side of the market and has complete control over the amount offered for sale. Monopolist controls price but must consider consumer demand Profits will be maximized at the level of output where marginal revenue equals marginal cost. <MASK> Average and Marginal Revenue 1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7 We can also see algebraically that Q. maximizes profit We can also see algebraically that Q* maximizes profit. Profit π is the difference between revenue and cost, both of which depend on Q: As Q is increased from zero, profit will increase until it reaches a maximum and then begin to decrease. Thus the profit-maximizing Q is such that the incremental profit resulting from a small increase in Q is just zero (i.e., Δπ /ΔQ = 0). Then But ΔR/ΔQ is marginal revenue and ΔC/ΔQ is marginal cost. Thus the profit-maximizing condition is that , or Monopolist’s Output Decision \$ per unit of output D = AR MR MC AC P1 Q1 P* Q* P2 Q2 Lost profit Lost profit Quantity <MASK> (Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost: <MASK> A Rule of Thumb for Pricing A Rule of Thumb for Pricing A Rule of Thumb for Pricing A Rule of Thumb for Pricing Monopoly Monopoly pricing compared to perfect competition pricing: P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market <MASK> The Multi-plant Firm For some firms, production takes place in more than one plant each with different costs Firm must determine how to distribute production between both plants Production should be split so that the MC in the plants is the same Output is chosen where MR=MC. Profits is therefore maximized when MR=MC at each plant We can show this algebraically: Q1 and C1 is output and cost of production for Plant 1 Q2 and C2 is output and cost of production for Plant 2 QT = Q1 + Q2 is total output Profit is then:  = PQT – C1(Q1) – C2(Q2) The Multi-plant Firm Firm should increase output from each plant until the additional profit from last unit produced at Plant 1 equals 0 <MASK> Monopsony A monopsony is a market in which there is a single buyer. An oligopsony is a market with only a few buyers. Monopsony power is the ability of the buyer to affect the price of the good and pay less than the price that would exist in a competitive market. Typically choose to buy until the benefit from last unit equals that unit’s cost Marginal value is the additional benefit derived from purchasing one more unit of a good Demand curve – downward sloping Marginal expenditure is the additional cost of buying one more unit of a good Depends on buying power Monopsony Competitive Buyer Price taker P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller Monopsonist Buyer ME S = AE PC Pm D = MV Qm QC \$/Q Monopsony ME above S Quantity where ME = MV: Qm Price from Supply curve: Pm ME D = MV S = AE Qm Pm Competitive P = PC Q = Q+C PC QC Quantity Monopoly and Monopsony Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher <MASK> Download ppt "Market Power: Monopoly and Monopsony" Similar presentations <UNMASK> # Market Power: Monopoly and Monopsony ## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript: Market Power: Monopoly and Monopsony Chapter 9 Review of Perfect Competition P = LMC = LRAC Normal profits or zero economic profits in the long run Large number of buyers and sellers Homogenous product Perfect information Firm is a price taker Review of Perfect Competition Q P Market D S Q0 P0 Q P Individual Firm P0 D = MR = P q0 LRAC LMC Monopoly Monopoly One seller - many buyers One product (no good substitutes) Barriers to entry Price Maker The monopolist is the supply-side of the market and has complete control over the amount offered for sale. Monopolist controls price but must consider consumer demand Profits will be maximized at the level of output where marginal revenue equals marginal cost. Average & Marginal Revenue The monopolist’s average revenue, price received per unit sold, is the market demand curve. Monopolist also needs to find marginal revenue, change in revenue resulting from a unit change in output. Finding Marginal Revenue As the sole producer, the monopolist works with the market demand to determine output and price. An example can be used to show the relationship between average and marginal revenue Assume a monopolist with demand: P = 6 - Q Average and Marginal Revenue 1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7 We can also see algebraically that Q. maximizes profit We can also see algebraically that Q* maximizes profit. Profit π is the difference between revenue and cost, both of which depend on Q: As Q is increased from zero, profit will increase until it reaches a maximum and then begin to decrease. Thus the profit-maximizing Q is such that the incremental profit resulting from a small increase in Q is just zero (i.e., Δπ /ΔQ = 0). Then But ΔR/ΔQ is marginal revenue and ΔC/ΔQ is marginal cost. Thus the profit-maximizing condition is that , or Monopolist’s Output Decision \$ per unit of output D = AR MR MC AC P1 Q1 P* Q* P2 Q2 Lost profit Lost profit Quantity Monopoly: An Example Monopoly: An Example A Rule of Thumb for Pricing Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus, (Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost: Example of Profit Maximization \$/Q 10 20 40 Profit = (P - AC) x Q = (\$30 - \$15)(10) = \$150 MC AC P=30 Profit AR AC=15 MR 5 10 15 20 Quantity A Rule of Thumb for Pricing Produce one more unit brings in revenue (1)(P) = P With downward sloping demand, producing and selling one more unit results in small drop in price P/Q. Reduces revenue from all units sold, change in revenue: Q(P/Q) A Rule of Thumb for Pricing A Rule of Thumb for Pricing A Rule of Thumb for Pricing A Rule of Thumb for Pricing Monopoly Monopoly pricing compared to perfect competition pricing: P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market The Effect of a Tax In competitive market, a per-unit tax causes price to rise by less than tax: burden shared by producers and consumers Under monopoly, price can sometimes rise by more than the amount of the tax. To determine the impact of a tax: t = specific tax MC = MC + t Effect of Excise Tax on Monopolist \$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity The Multi-plant Firm For some firms, production takes place in more than one plant each with different costs Firm must determine how to distribute production between both plants Production should be split so that the MC in the plants is the same Output is chosen where MR=MC. Profits is therefore maximized when MR=MC at each plant We can show this algebraically: Q1 and C1 is output and cost of production for Plant 1 Q2 and C2 is output and cost of production for Plant 2 QT = Q1 + Q2 is total output Profit is then:  = PQT – C1(Q1) – C2(Q2) The Multi-plant Firm Firm should increase output from each plant until the additional profit from last unit produced at Plant 1 equals 0 Deadweight Loss from Monopoly Power \$/Q Because of the higher price, consumers lose A+B and producer gains A-C. Lost Consumer Surplus MC Deadweight Loss B A QC PC C C Qm Quantity Monopsony A monopsony is a market in which there is a single buyer. An oligopsony is a market with only a few buyers. Monopsony power is the ability of the buyer to affect the price of the good and pay less than the price that would exist in a competitive market. Typically choose to buy until the benefit from last unit equals that unit’s cost Marginal value is the additional benefit derived from purchasing one more unit of a good Demand curve – downward sloping Marginal expenditure is the additional cost of buying one more unit of a good Depends on buying power Monopsony Competitive Buyer Price taker P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller Monopsonist Buyer ME S = AE PC Pm D = MV Qm QC \$/Q Monopsony ME above S Quantity where ME = MV: Qm Price from Supply curve: Pm ME D = MV S = AE Qm Pm Competitive P = PC Q = Q+C PC QC Quantity Monopoly and Monopsony Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher Monopoly and Monopsony \$/Q MV ME S = AE Q* P* PC QC Monopsony Note: ME = MV; ME > AE; MV > P Monopoly and Monopsony MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power Download ppt "Market Power: Monopoly and Monopsony" Similar presentations
<MASK> Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! Gasp I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600sin(25)=X. Enter 1600sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1d2, so I substituted in the diameter lengths: A = 1/2 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? SPOILER ALERT* Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 <MASK> <UNMASK> <MASK> • Algebra HotMath • Geometry HotMath • Geometry PHSchool • Set • Ms. J's Website Teacher Assignments 04/09 12/04 12/04 09/19 <MASK> Title: () Description: Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600sin(25)=X. Enter 1600sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1d2, so I substituted in the diameter lengths: A = 1/2 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? SPOILER ALERT* Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 My Classes & Students - - - - - -
<MASK> • Algebra HotMath • Geometry HotMath • Geometry PHSchool • Set • Ms. J's Website Teacher Assignments 04/09 12/04 12/04 09/19 <MASK> Title: () Description: Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! Gasp I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600sin(25)=X. Enter 1600sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1d2, so I substituted in the diameter lengths: A = 1/2 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? SPOILER ALERT* Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 My Classes & Students - - - - - - <UNMASK> files/ -- Blogmeister Ms. Jovanovich's Math Mumblings Jovanovich Mathematics This blog is for Miss Jovanovich's algebra and geometry classes to share ideas, ask questions, and reflect on what skills and topics we are studying. If we are lucky, we might even get a chance to connect with other students studying the same stuff. by Tina Jovanovich • Algebra HotMath • Geometry HotMath • Geometry PHSchool • Set • Ms. J's Website Teacher Assignments 04/09 12/04 12/04 09/19 Teacher Entries Where did the year go? 4/9 Best and Worst Math Memories 9/26 Bring on the Bells 9/5 List 5, 10, all Student Entries KC Scratch Game Review 4/4 SC Scratch 4/4 JB Scratch 4/4 AF Amanda's Scratch Extra Credit 4/4 CB Extra Credit Scratch 4/4 CD Scratch Games 4/3 SG Scratch: extra Credit 4/3 EK Scratch 4/3 JR Scratch Extra Credit 4/3 EH 4/3 CA Scratch Games 4/3 JG Scratch Games 4/3 RB Scratch Games 4/3 AO Scratch Programs 4/3 FK Favorite Scratch Programs 4/3 AE Scratch 4/3 JB Scratch 4/3 JB Scratch 4/3 JB Scratch 4/3 AH Scratch 4/3 AM Scratch Projects 4/3 MH Scratch 4/3 KM Scratch 4/3 BC Scratch Games 4/3 KB My Findings on Scratch.com 4/2 List 25, 50, all Title: () Description: Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600sin(25)=X. Enter 1600sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1d2, so I substituted in the diameter lengths: A = 1/2 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? SPOILER ALERT* Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 My Classes & Students - - - - - -
<MASK> (1) 如果我们对数量 $F$ 进行精确测量，我们将观察到一个特征值 $f$； (2) 如果我们在时间 $t_{0}$ 执行纯通过测量，让特征值 $f$ 通过，那么波函数将简化为 $^{4}$ $$\开始{对齐} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left\|c_{f}(t)\right\|^{2} &=1 \end{对齐}$$ <MASK> <UNMASK> <MASK> ## 物理代写 <MASK> ## 物理代考 <MASK> $$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { 特征函数 }$$ $$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { 完整集 }$$ $$\sum_{f}\left\|c_{f}(t)\right\|^{2}=1$$ (1) 如果我们对数量 $F$ 进行精确测量，我们将观察到一个特征值 $f$； (2) 如果我们在时间 $t_{0}$ 执行纯通过测量，让特征值 $f$ 通过，那么波函数将简化为 $^{4}$ $$\开始{对齐} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left\|c_{f}(t)\right\|^{2} &=1 \end{对齐}$$ (3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为 $$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left\|c_{f}(t)\right\|^{2} &=1 & & \end{对齐}$$ ${ }^{4}$ 注意，必须重新调整系数 $c_{f}(t)$ 才能达到这个标准（见 Prob. 10.2）。 Matlab代写
<MASK> ## 物理代写 <MASK> ## 物理代考 <MASK> $$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { 特征函数 }$$ $$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { 完整集 }$$ $$\sum_{f}\left\|c_{f}(t)\right\|^{2}=1$$ (1) 如果我们对数量 $F$ 进行精确测量，我们将观察到一个特征值 $f$； (2) 如果我们在时间 $t_{0}$ 执行纯通过测量，让特征值 $f$ 通过，那么波函数将简化为 $^{4}$ $$\开始{对齐} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left\|c_{f}(t)\right\|^{2} &=1 \end{对齐}$$ (3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为 $$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left\|c_{f}(t)\right\|^{2} &=1 & & \end{对齐}$$ ${ }^{4}$ 注意，必须重新调整系数 $c_{f}(t)$ 才能达到这个标准（见 Prob. 10.2）。 Matlab代写 <UNMASK> 19th Ave New York, NY 95822, USA # 物理代考\| Reduction of the Basis量子力学代写 ## 物理代写 8.2 Reduction of the Basis Let us try to formalize this measurement theory. Suppose we are looking at a single particle, and we have a complete set of the eigenfunctions of some hermitian operator with real eigenvalues at our disposal $$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { eigenfunctions }$$ Order the eigenvalues $f_{0} \leq f_{1} \leq f_{2} \cdots$, and expand the wave function $\Psi(x, t)$ in this complete set of eigenfuctions $$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { complete set }$$ The state is normalized, so that $$\sum_{f}\left\|c_{f}(t)\right\|^{2}=1$$ Measurement theory then assumes the following: (1) If we make a precise measurement of the quantity $F$, we will observe one of the eigenvalues $f$; (2) If we perform a pure pass measurement at a time $t_{0}$ that lets the eigenvalue $f$ through, then the wave function is reduced to $^{4}$ \begin{aligned} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left\|c_{f}(t)\right\|^{2} &=1 \end{aligned} The measurement is reproducible and the basis is reduced. (3) If the measurement simply lets the eigenvalues in the set $f_{1} \leq f \leq f_{2}$ through, then the basis is reduced to \begin{aligned} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left\|c_{f}(t)\right\|^{2} &=1 & & \end{aligned} where the sum $\sum_{f}^{\prime}$ goes over $f_{1} \leq f \leq f_{2}$. ${ }^{4}$ Note that the coefficient $c_{f}(t)$ must be rescaled to achieve this norm (see Prob. 10.2). ## 物理代考 8.2 降低基数 $$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { 特征函数 }$$ $$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { 完整集 }$$ $$\sum_{f}\left\|c_{f}(t)\right\|^{2}=1$$ (1) 如果我们对数量 $F$ 进行精确测量，我们将观察到一个特征值 $f$； (2) 如果我们在时间 $t_{0}$ 执行纯通过测量，让特征值 $f$ 通过，那么波函数将简化为 $^{4}$ $$\开始{对齐} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left\|c_{f}(t)\right\|^{2} &=1 \end{对齐}$$ (3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为 $$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left\|c_{f}(t)\right\|^{2} &=1 & & \end{对齐}$$ ${ }^{4}$ 注意，必须重新调整系数 $c_{f}(t)$ 才能达到这个标准（见 Prob. 10.2）。 Matlab代写
<MASK> thanks <MASK> <UNMASK> # Homework Help: Finding Inverse Hyperbolic secant in terms of logarithms ? 1. Aug 29, 2011 ### mahmoud2011 The Problem is when I Compute the Inverse I have to solutions <MASK> And this not function which of them I will choose Another Question is how can I prove without the graph that csch (x) is one - to -one thanks Last edited: Aug 29, 2011 2. Aug 29, 2011 ### mahmoud2011 For the second question I proved it depending on the definition of one to one function and csch(x) ,where since sinh(x) is one to one ,the whenever sinh(c)=sinh(d) , c=d Then it follows that whenever 1/sinh(c) = 1/sinh(d) , c=d And thus the result follows and that's what I did . 3. Aug 29, 2011 ### mahmoud2011 For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function, Let $h(x) = ln(\frac{1-\sqrt{1-x^2}}{x}) : 0<x\leq 1$ $g(x)=ln (\frac{1+ \sqrt{1-x^2}}{x}) : 0<x\leq 1$ $f(x) = sech(x) : x\geq 0$ then we have after some Algebra , $(f o h)(x) = x : 0<x\leq 1$ Whereas $(h o f)(x) = -x : x\leq0$ And Thus h is not the inverse of f , Checking g we will see that it do its job it is the inverse where $(f o g)(x) = x : 0<x\leq 1$ Whereas $(g o f)(x) = x : x\leq0$ 4. Aug 30, 2011 ### HallsofIvy Exactly what does the question say? The reason you are getting "$\pm$" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign. 5. Aug 30, 2011 ### mahmoud2011 I have restricted the domain such that x is greater than or equal zero and I also had Both signs So I Checked the solutions from definition of Inverse function And Choosed the positive sign As I have written above.
# Homework Help: Finding Inverse Hyperbolic secant in terms of logarithms ? 1. Aug 29, 2011 ### mahmoud2011 The Problem is when I Compute the Inverse I have to solutions <MASK> And this not function which of them I will choose Another Question is how can I prove without the graph that csch (x) is one - to -one thanks Last edited: Aug 29, 2011 2. Aug 29, 2011 ### mahmoud2011 For the second question I proved it depending on the definition of one to one function and csch(x) ,where since sinh(x) is one to one ,the whenever sinh(c)=sinh(d) , c=d Then it follows that whenever 1/sinh(c) = 1/sinh(d) , c=d And thus the result follows and that's what I did . 3. Aug 29, 2011 ### mahmoud2011 For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function, Let $h(x) = ln(\frac{1-\sqrt{1-x^2}}{x}) : 0<x\leq 1$ $g(x)=ln (\frac{1+ \sqrt{1-x^2}}{x}) : 0<x\leq 1$ $f(x) = sech(x) : x\geq 0$ then we have after some Algebra , $(f o h)(x) = x : 0<x\leq 1$ Whereas $(h o f)(x) = -x : x\leq0$ And Thus h is not the inverse of f , Checking g we will see that it do its job it is the inverse where $(f o g)(x) = x : 0<x\leq 1$ Whereas $(g o f)(x) = x : x\leq0$ 4. Aug 30, 2011 ### HallsofIvy Exactly what does the question say? The reason you are getting "$\pm$" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign. 5. Aug 30, 2011 ### mahmoud2011 I have restricted the domain such that x is greater than or equal zero and I also had Both signs So I Checked the solutions from definition of Inverse function And Choosed the positive sign As I have written above. <UNMASK> # Homework Help: Finding Inverse Hyperbolic secant in terms of logarithms ? 1. Aug 29, 2011 ### mahmoud2011 The Problem is when I Compute the Inverse I have to solutions $sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1$ And this not function which of them I will choose Another Question is how can I prove without the graph that csch (x) is one - to -one thanks Last edited: Aug 29, 2011 2. Aug 29, 2011 ### mahmoud2011 For the second question I proved it depending on the definition of one to one function and csch(x) ,where since sinh(x) is one to one ,the whenever sinh(c)=sinh(d) , c=d Then it follows that whenever 1/sinh(c) = 1/sinh(d) , c=d And thus the result follows and that's what I did . 3. Aug 29, 2011 ### mahmoud2011 For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function, Let $h(x) = ln(\frac{1-\sqrt{1-x^2}}{x}) : 0<x\leq 1$ $g(x)=ln (\frac{1+ \sqrt{1-x^2}}{x}) : 0<x\leq 1$ $f(x) = sech(x) : x\geq 0$ then we have after some Algebra , $(f o h)(x) = x : 0<x\leq 1$ Whereas $(h o f)(x) = -x : x\leq0$ And Thus h is not the inverse of f , Checking g we will see that it do its job it is the inverse where $(f o g)(x) = x : 0<x\leq 1$ Whereas $(g o f)(x) = x : x\leq0$ 4. Aug 30, 2011 ### HallsofIvy Exactly what does the question say? The reason you are getting "$\pm$" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign. 5. Aug 30, 2011 ### mahmoud2011 I have restricted the domain such that x is greater than or equal zero and I also had Both signs So I Checked the solutions from definition of Inverse function And Choosed the positive sign As I have written above.
<MASK> We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface. These incident wavefront is carrying two points, point A and point B, so we can say that from point B to point C light is travelling a distance. If ‘ v ‘ represents the speed of the wave in the medium and if ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance <MASK> If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection <MASK> So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain <MASK> 105 videos\|425 docs\|114 tests <MASK> 105 videos\|425 docs\|114 tests ### Up next <MASK> , <MASK> , <MASK> , <MASK> , <MASK> ; <UNMASK> Refraction & Reflection of Plane Waves using Huygens Principle # Refraction & Reflection of Plane Waves using Huygens Principle \| Physics Class 12 - NEET PDF Download As we know that when light falls on an object, it bends and move through the material, this is what refraction is. Also when the light bounces off the medium it is called a reflection. Let us know study reflection and refraction of waves by Huygen’s principle. Reflection using Huygens Principle We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface. These incident wavefront is carrying two points, point A and point B, so we can say that from point B to point C light is travelling a distance. If ‘ v ‘ represents the speed of the wave in the medium and if ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance BC = vr In order the construct the reflected wavefront we draw a sphere of radius vr from the point A. Let CE represent the tangent plane drawn from the point C to this sphere. So, <MASK> If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection ### Refraction using Huygen’s principle We know that when a light travels from one transparent medium to another transparent medium its path changes. So the laws of refraction state that the angle of incidence is the angle between the incident ray and the normal and the angle of refraction is the angle between the refracted ray and the normal. The incident ray, reflected ray and the normal, to the interface of any two given mediums all lie in the same plane. We also know that the ratio of the sine of the angle of incidence and sine of the angle of refraction is constant. A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. the figure corresponds to v2 < v1 so that the refracted waves bends towards the normal. We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. As these rays are incident from the surface, so we call it incident ray. Let PP’ represent the medium 1 and medium 2. The speed of the light in this medium is represented by v1 and v2. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface. If ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance, BC = v1r So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain where’ i ‘ and ‘ r ‘ are the angles of incidence and refraction, respectively. Substituting the values of v1 and v2 in terms of we get the Snell’s Law, <MASK> The document Refraction & Reflection of Plane Waves using Huygens Principle \| Physics Class 12 - NEET is a part of the NEET Course Physics Class 12. All you need of NEET at this link: NEET ## Physics Class 12 105 videos\|425 docs\|114 tests ## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET 1. What is refraction and reflection of plane waves? Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces. 2. How does Huygens Principle explain refraction and reflection of plane waves? Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave. 3. Can you explain how refraction and reflection affect the direction of plane waves? Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface. 4. What factors determine the extent of refraction and reflection in plane waves? Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces. 5. How are refraction and reflection of plane waves useful in practical applications? Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery. ## Physics Class 12 105 videos\|425 docs\|114 tests ### Up next <MASK> ### Top Courses for NEET 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 , , , , , <MASK> , <MASK> , , , , , , , <MASK> , , , , , ;
Refraction & Reflection of Plane Waves using Huygens Principle # Refraction & Reflection of Plane Waves using Huygens Principle \| Physics Class 12 - NEET PDF Download As we know that when light falls on an object, it bends and move through the material, this is what refraction is. Also when the light bounces off the medium it is called a reflection. Let us know study reflection and refraction of waves by Huygen’s principle. Reflection using Huygens Principle We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface. These incident wavefront is carrying two points, point A and point B, so we can say that from point B to point C light is travelling a distance. If ‘ v ‘ represents the speed of the wave in the medium and if ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance BC = vr In order the construct the reflected wavefront we draw a sphere of radius vr from the point A. Let CE represent the tangent plane drawn from the point C to this sphere. So, <MASK> If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection ### Refraction using Huygen’s principle We know that when a light travels from one transparent medium to another transparent medium its path changes. So the laws of refraction state that the angle of incidence is the angle between the incident ray and the normal and the angle of refraction is the angle between the refracted ray and the normal. The incident ray, reflected ray and the normal, to the interface of any two given mediums all lie in the same plane. We also know that the ratio of the sine of the angle of incidence and sine of the angle of refraction is constant. A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. the figure corresponds to v2 < v1 so that the refracted waves bends towards the normal. We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. As these rays are incident from the surface, so we call it incident ray. Let PP’ represent the medium 1 and medium 2. The speed of the light in this medium is represented by v1 and v2. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface. If ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance, BC = v1r So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain where’ i ‘ and ‘ r ‘ are the angles of incidence and refraction, respectively. Substituting the values of v1 and v2 in terms of we get the Snell’s Law, <MASK> The document Refraction & Reflection of Plane Waves using Huygens Principle \| Physics Class 12 - NEET is a part of the NEET Course Physics Class 12. All you need of NEET at this link: NEET ## Physics Class 12 105 videos\|425 docs\|114 tests ## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET 1. What is refraction and reflection of plane waves? Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces. 2. How does Huygens Principle explain refraction and reflection of plane waves? Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave. 3. Can you explain how refraction and reflection affect the direction of plane waves? Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface. 4. What factors determine the extent of refraction and reflection in plane waves? Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces. 5. How are refraction and reflection of plane waves useful in practical applications? Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery. ## Physics Class 12 105 videos\|425 docs\|114 tests ### Up next <MASK> ### Top Courses for NEET 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 , , , , , <MASK> , <MASK> , , , , , , , <MASK> , , , , , ; <UNMASK> Refraction & Reflection of Plane Waves using Huygens Principle # Refraction & Reflection of Plane Waves using Huygens Principle \| Physics Class 12 - NEET PDF Download As we know that when light falls on an object, it bends and move through the material, this is what refraction is. Also when the light bounces off the medium it is called a reflection. Let us know study reflection and refraction of waves by Huygen’s principle. Reflection using Huygens Principle We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface. These incident wavefront is carrying two points, point A and point B, so we can say that from point B to point C light is travelling a distance. If ‘ v ‘ represents the speed of the wave in the medium and if ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance BC = vr In order the construct the reflected wavefront we draw a sphere of radius vr from the point A. Let CE represent the tangent plane drawn from the point C to this sphere. So, AE = BC = vr If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection ### Refraction using Huygen’s principle We know that when a light travels from one transparent medium to another transparent medium its path changes. So the laws of refraction state that the angle of incidence is the angle between the incident ray and the normal and the angle of refraction is the angle between the refracted ray and the normal. The incident ray, reflected ray and the normal, to the interface of any two given mediums all lie in the same plane. We also know that the ratio of the sine of the angle of incidence and sine of the angle of refraction is constant. A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. the figure corresponds to v2 < v1 so that the refracted waves bends towards the normal. We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. As these rays are incident from the surface, so we call it incident ray. Let PP’ represent the medium 1 and medium 2. The speed of the light in this medium is represented by v1 and v2. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface. If ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance, BC = v1r So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain where’ i ‘ and ‘ r ‘ are the angles of incidence and refraction, respectively. Substituting the values of v1 and v2 in terms of we get the Snell’s Law, n1 sin i = n2 sin r The document Refraction & Reflection of Plane Waves using Huygens Principle \| Physics Class 12 - NEET is a part of the NEET Course Physics Class 12. All you need of NEET at this link: NEET ## Physics Class 12 105 videos\|425 docs\|114 tests ## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET 1. What is refraction and reflection of plane waves? Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces. 2. How does Huygens Principle explain refraction and reflection of plane waves? Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave. 3. Can you explain how refraction and reflection affect the direction of plane waves? Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface. 4. What factors determine the extent of refraction and reflection in plane waves? Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces. 5. How are refraction and reflection of plane waves useful in practical applications? Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery. ## Physics Class 12 105 videos\|425 docs\|114 tests ### Up next Explore Courses for NEET exam ### Top Courses for NEET 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 , , , , , , , , , , , , , , , , , , , , , ;
<MASK> `// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `\|\| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/ Driver program to test above function /` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. ` <MASK> <UNMASK> <MASK> Input : N = 2, M =2 m[][] = { { 1, 2 }, { 3, 4 } }; Output :3 Longest path is either 1 2 4 or 1 3 4. ``` <MASK> Below is the implementation of this approach: <MASK> /div> <MASK> `// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `\|\| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/* Driver program to test above function /` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. ` <MASK> `# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia ` ## C# `// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 \|\| y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/ Driver code /` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/ This code contributed by PrinciRaj1992 /` Output: ```7 ``` Time Complexity: O(NM).
<MASK> Input : N = 2, M =2 m[][] = { { 1, 2 }, { 3, 4 } }; Output :3 Longest path is either 1 2 4 or 1 3 4. ``` <MASK> Below is the implementation of this approach: <MASK> /div> <MASK> `// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `\|\| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/ Driver program to test above function /` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. ` <MASK> `# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia ` ## C# `// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 \|\| y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/ Driver code /` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/ This code contributed by PrinciRaj1992 /` Output: ```7 ``` Time Complexity: O(NM). <UNMASK> # Longest Increasing Path in Matrix Given a matrix of N rows and M columns. From m[i][j], we can move to m[i+1][j], if m[i+1][j] > m[i][j], or can move to m[i][j+1] if m[i][j+1] > m[i][j]. The task is print longest path length if we start from (0, 0). Examples: ```Input : N = 4, M = 4 m[][] = { { 1, 2, 3, 4 }, { 2, 2, 3, 4 }, { 3, 2, 3, 4 }, { 4, 5, 6, 7 } }; Output : 7 Longest path is 1 2 3 4 5 6 7. Input : N = 2, M =2 m[][] = { { 1, 2 }, { 3, 4 } }; Output :3 Longest path is either 1 2 4 or 1 3 4. ``` ## Recommended: Please try your approach on {IDE} first, before moving on to the solution. The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column. Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1. We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer. Below is the implementation of this approach: ## C++ `// CPP program to find longest increasing ` `// path in a matrix. ` `#include ` `#define MAX 10 ` `using` `namespace` `std; ` ` ` `// Return the length of LIP in 2D matrix ` `int` `LIP(``int` `dp[][MAX], ``int` `mat[][MAX], ``int` `n, ``int` `m, ``int` `x, ``int` `y) ` `{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-1 && y == m-1) ` ` ``return` `dp[x][y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-1 \|\| y == m-1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(mat[x][y] < mat[x+1][y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x+1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(mat[x][y] < mat[x][y+1]) ` ` ``result = max(result, 1 + LIP(dp, mat, n, m, x, y+1)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` `} ` ` ` `// Wrapper function ` `int` `wrapper(``int` `mat[][MAX], ``int` `n, ``int` `m) ` `{ ` ` ``int` `dp[MAX][MAX]; ` ` ``memset``(dp, -1, ``sizeof` `dp); ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ``int` `mat[][MAX] = { ` ` ``{ 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, ` ` ``}; ` ` ``int` `n = 4, m = 4; ` ` ``cout << wrapper(mat, n, m) << endl; ` ` ` ` ``return` `0; ` `} ` /div> ## Java `// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `\|\| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/ Driver program to test above function /` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. ` ## Python3 `# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia ` ## C# `// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 \|\| y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/ Driver code /` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/ This code contributed by PrinciRaj1992 /` Output: ```7 ``` Time Complexity: O(NM).
<MASK> ### Deepak K Kapur If i put c=d/t in E=mc2, then E=m×d2/t2 <MASK> Does this mean that E is inversely proportional to time? 2. May 28, 2015 ### Fredrik Staff Emeritus No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t2, that we could say that E is inversely proportional to time squared. But what you found is E= md2/t2, where m is a constant and d depends on t. So md2 isn't a constant. 3. May 28, 2015 <MASK> No, but it means that Energy has a "dimension" of $ML^2T^{-2}$. Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above. Look up "dimensional analysis". 4. May 28, 2015 <MASK> 2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation? Thanks. 5. May 28, 2015 <MASK> One more question (silly one). Mass is something concrete, we can see it, touch it, feel it etc. Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence. <MASK> 6. May 28, 2015 ### Fredrik Staff Emeritus 1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t. When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is $$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$ <MASK> ### Staff: Mentor You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence. Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation. <MASK> ### Deepak K Kapur So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so?? 9. May 28, 2015 ### Deepak K Kapur Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm? 10. May 28, 2015 ### Staff: Mentor Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question. <MASK> ### Staff: Mentor No we don't. The units do not disappear. 13. May 28, 2015 ### Deepak K Kapur So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties?? 14. May 28, 2015 <MASK> 15. May 28, 2015 <MASK> t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$ So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$. You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$! EDIT: More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!! <MASK> ### Staff: Mentor <MASK> Last edited: May 29, 2015 17. May 30, 2015 <MASK> 1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass? 2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant? 18. May 30, 2015 ### ArmanCham <MASK> ### Staff: Mentor You are mistaken. C is just a universal constant, that happens to be the speed of light. You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem. <MASK> Last edited: May 30, 2015 21. May 30, 2015 ### Staff: Mentor There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation. Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy. http://en.m.wikipedia.org/wiki/Mass–energy_equivalence <MASK> ### Fredrik Staff Emeritus In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$. <MASK> ### Deepak K Kapur <MASK> Matter will change into energy when...................(some relation to the speed of light) Cant you elaborate this equation in this way? <MASK> If they are same, then do they look different to the 'observer' only or are they 'really' different? 25. May 30, 2015 <MASK> <UNMASK> 1. May 28, 2015 ### Deepak K Kapur If i put c=d/t in E=mc2, then E=m×d2/t2 Now take m=1kg and d=1m Does this mean that E is inversely proportional to time? 2. May 28, 2015 ### Fredrik Staff Emeritus No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t2, that we could say that E is inversely proportional to time squared. But what you found is E= md2/t2, where m is a constant and d depends on t. So md2 isn't a constant. 3. May 28, 2015 ### PeroK No, but it means that Energy has a "dimension" of $ML^2T^{-2}$. Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above. Look up "dimensional analysis". 4. May 28, 2015 ### Deepak K Kapur But, 1. d=1m and one meter is always one meter. How does 1m depend on time? 2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation? Thanks. 5. May 28, 2015 <MASK> One more question (silly one). Mass is something concrete, we can see it, touch it, feel it etc. Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence. Then, how on earth can we multiply a concrete entity with an abstract one?? 6. May 28, 2015 ### Fredrik Staff Emeritus 1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t. When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is $$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$ 7. May 28, 2015 ### Staff: Mentor You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence. Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation. 8. May 28, 2015 ### Deepak K Kapur So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so?? 9. May 28, 2015 ### Deepak K Kapur Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm? 10. May 28, 2015 ### Staff: Mentor Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question. 11. May 28, 2015 ### Fredrik Staff Emeritus Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity. I would say that aspects of the real world are represented by abstract mathematical things in the theory. <MASK> ### Staff: Mentor No we don't. The units do not disappear. 13. May 28, 2015 ### Deepak K Kapur So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties?? 14. May 28, 2015 ### Deepak K Kapur Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2. Why not, this seems to be a sensible interpretation... 15. May 28, 2015 ### ShayanJ t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$ So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$. You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$! EDIT: More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!! Last edited: May 28, 2015 16. May 29, 2015 ### Staff: Mentor <MASK> Last edited: May 29, 2015 17. May 30, 2015 ### Deepak K Kapur I dont get your point fully... 1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass? 2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant? 18. May 30, 2015 ### ArmanCham Time constant means this ; c=x/t If we assume x is one then you cannot change t cause c is constant.So you are telling us E=1/t2 here you assumed x equal one.Then you assumed t =2 We cannot change time If you choose x =1.If you change time you can see easily that you are breaking fundemental physics law "Speed of light is constant" Here the math c=x/t you said x =1 then t=1/c.Now we found t.Its a number but you are telling it can be three or four.Thats nonsense. 19. May 30, 2015 ### Staff: Mentor You are mistaken. C is just a universal constant, that happens to be the speed of light. You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem. You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself. 20. May 30, 2015 <MASK> I will not argue further on the time issue... But, 1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass. 2. What does c in this equation mean? Thanks, u hav been very helpful.... Last edited: May 30, 2015 21. May 30, 2015 ### Staff: Mentor There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation. Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy. http://en.m.wikipedia.org/wiki/Mass–energy_equivalence 22. May 30, 2015 ### Fredrik Staff Emeritus In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$. 23. May 30, 2015 ### Deepak K Kapur Actually i was expecting the meaning of this equation as follows.. Matter will change into energy when...................(some relation to the speed of light) Cant you elaborate this equation in this way? Also plz explain the meaning of 'square' of c? Last edited: May 30, 2015 24. May 30, 2015 ### Deepak K Kapur Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same? If they are same, then do they look different to the 'observer' only or are they 'really' different? 25. May 30, 2015 <MASK> No, that isn't what it means. I can't really elaborate on something that isn't true. C^2 is the conversion factor to equate matter and energy.
1. May 28, 2015 ### Deepak K Kapur If i put c=d/t in E=mc2, then E=m×d2/t2 Now take m=1kg and d=1m Does this mean that E is inversely proportional to time? 2. May 28, 2015 ### Fredrik Staff Emeritus No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t2, that we could say that E is inversely proportional to time squared. But what you found is E= md2/t2, where m is a constant and d depends on t. So md2 isn't a constant. 3. May 28, 2015 ### PeroK No, but it means that Energy has a "dimension" of $ML^2T^{-2}$. Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above. Look up "dimensional analysis". 4. May 28, 2015 ### Deepak K Kapur But, 1. d=1m and one meter is always one meter. How does 1m depend on time? 2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation? Thanks. 5. May 28, 2015 <MASK> One more question (silly one). Mass is something concrete, we can see it, touch it, feel it etc. Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence. Then, how on earth can we multiply a concrete entity with an abstract one?? 6. May 28, 2015 ### Fredrik Staff Emeritus 1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t. When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is $$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$ 7. May 28, 2015 ### Staff: Mentor You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence. Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation. 8. May 28, 2015 ### Deepak K Kapur So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so?? 9. May 28, 2015 ### Deepak K Kapur Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm? 10. May 28, 2015 ### Staff: Mentor Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question. 11. May 28, 2015 ### Fredrik Staff Emeritus Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity. I would say that aspects of the real world are represented by abstract mathematical things in the theory. <MASK> ### Staff: Mentor No we don't. The units do not disappear. 13. May 28, 2015 ### Deepak K Kapur So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties?? 14. May 28, 2015 ### Deepak K Kapur Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2. Why not, this seems to be a sensible interpretation... 15. May 28, 2015 ### ShayanJ t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$ So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$. You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$! EDIT: More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!! Last edited: May 28, 2015 16. May 29, 2015 ### Staff: Mentor <MASK> Last edited: May 29, 2015 17. May 30, 2015 ### Deepak K Kapur I dont get your point fully... 1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass? 2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant? 18. May 30, 2015 ### ArmanCham Time constant means this ; c=x/t If we assume x is one then you cannot change t cause c is constant.So you are telling us E=1/t2 here you assumed x equal one.Then you assumed t =2 We cannot change time If you choose x =1.If you change time you can see easily that you are breaking fundemental physics law "Speed of light is constant" Here the math c=x/t you said x =1 then t=1/c.Now we found t.Its a number but you are telling it can be three or four.Thats nonsense. 19. May 30, 2015 ### Staff: Mentor You are mistaken. C is just a universal constant, that happens to be the speed of light. You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem. You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself. 20. May 30, 2015 <MASK> I will not argue further on the time issue... But, 1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass. 2. What does c in this equation mean? Thanks, u hav been very helpful.... Last edited: May 30, 2015 21. May 30, 2015 ### Staff: Mentor There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation. Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy. http://en.m.wikipedia.org/wiki/Mass–energy_equivalence 22. May 30, 2015 ### Fredrik Staff Emeritus In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$. 23. May 30, 2015 ### Deepak K Kapur Actually i was expecting the meaning of this equation as follows.. Matter will change into energy when...................(some relation to the speed of light) Cant you elaborate this equation in this way? Also plz explain the meaning of 'square' of c? Last edited: May 30, 2015 24. May 30, 2015 ### Deepak K Kapur Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same? If they are same, then do they look different to the 'observer' only or are they 'really' different? 25. May 30, 2015 <MASK> No, that isn't what it means. I can't really elaborate on something that isn't true. C^2 is the conversion factor to equate matter and energy. <UNMASK> 1. May 28, 2015 ### Deepak K Kapur If i put c=d/t in E=mc2, then E=m×d2/t2 Now take m=1kg and d=1m Does this mean that E is inversely proportional to time? 2. May 28, 2015 ### Fredrik Staff Emeritus No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t2, that we could say that E is inversely proportional to time squared. But what you found is E= md2/t2, where m is a constant and d depends on t. So md2 isn't a constant. 3. May 28, 2015 ### PeroK No, but it means that Energy has a "dimension" of $ML^2T^{-2}$. Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above. Look up "dimensional analysis". 4. May 28, 2015 ### Deepak K Kapur But, 1. d=1m and one meter is always one meter. How does 1m depend on time? 2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation? Thanks. 5. May 28, 2015 ### Deepak K Kapur One more question (silly one). Mass is something concrete, we can see it, touch it, feel it etc. Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence. Then, how on earth can we multiply a concrete entity with an abstract one?? 6. May 28, 2015 ### Fredrik Staff Emeritus 1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t. When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is $$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$ 7. May 28, 2015 ### Staff: Mentor You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence. Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation. 8. May 28, 2015 ### Deepak K Kapur So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so?? 9. May 28, 2015 ### Deepak K Kapur Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm? 10. May 28, 2015 ### Staff: Mentor Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question. 11. May 28, 2015 ### Fredrik Staff Emeritus Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity. I would say that aspects of the real world are represented by abstract mathematical things in the theory. 12. May 28, 2015 ### Staff: Mentor No we don't. The units do not disappear. 13. May 28, 2015 ### Deepak K Kapur So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties?? 14. May 28, 2015 ### Deepak K Kapur Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2. Why not, this seems to be a sensible interpretation... 15. May 28, 2015 ### ShayanJ t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$ So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$. You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$! EDIT: More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!! Last edited: May 28, 2015 16. May 29, 2015 ### Staff: Mentor Well, two reasons: 1. Because the units are still there and you are pretending they aren't. The "1" has units kg-m2. The equation should read E=1 kg-m2/t2. 2. The equation isn't describing the motion of a kilogram of mass, the piece you broke apart is describing the speed of light ("c" is the speed of light). So the 1m is the distance traveled by light in "t" time: time is a constant here, not a variable and you can't change that. So when you plug in m=1 kg, and d= 1m, t=1/300,000,000s and it all simplifies to E=9x10^16 Joules. Science includes the implied assumption that the observed properties are real and there is nothing else to "concern with" or not "concern with". Beyond that, you're getting into philosophy, not science. Last edited: May 29, 2015 17. May 30, 2015 ### Deepak K Kapur I dont get your point fully... 1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass? 2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant? 18. May 30, 2015 ### ArmanCham Time constant means this ; c=x/t If we assume x is one then you cannot change t cause c is constant.So you are telling us E=1/t2 here you assumed x equal one.Then you assumed t =2 We cannot change time If you choose x =1.If you change time you can see easily that you are breaking fundemental physics law "Speed of light is constant" Here the math c=x/t you said x =1 then t=1/c.Now we found t.Its a number but you are telling it can be three or four.Thats nonsense. 19. May 30, 2015 ### Staff: Mentor You are mistaken. C is just a universal constant, that happens to be the speed of light. You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem. You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself. 20. May 30, 2015 ### Deepak K Kapur I will not argue further on the time issue... But, 1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass. 2. What does c in this equation mean? Thanks, u hav been very helpful.... Last edited: May 30, 2015 21. May 30, 2015 ### Staff: Mentor There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation. Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy. http://en.m.wikipedia.org/wiki/Mass–energy_equivalence 22. May 30, 2015 ### Fredrik Staff Emeritus In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$. 23. May 30, 2015 ### Deepak K Kapur Actually i was expecting the meaning of this equation as follows.. Matter will change into energy when...................(some relation to the speed of light) Cant you elaborate this equation in this way? Also plz explain the meaning of 'square' of c? Last edited: May 30, 2015 24. May 30, 2015 ### Deepak K Kapur Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same? If they are same, then do they look different to the 'observer' only or are they 'really' different? 25. May 30, 2015 ### Staff: Mentor No, that isn't what it means. I can't really elaborate on something that isn't true. C^2 is the conversion factor to equate matter and energy.
<MASK> If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows: $sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle. <MASK> Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below. $\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$ <MASK> $f(x) = a \sin(b x + c) + d$ <MASK> in red as shown in the figure below. <MASK> ### Phase Shift <MASK> set a, b and c to non zero values and change d. What is the direction of the shift of the graph? ## More References and Links on Sine Functions <MASK> <UNMASK> # Sine Function sin x ## Definition and Graph of the Sine Function We first consider angle ? with initial side on the positive x axis (in standard position) and terminal side OM as shown below. The sine function is defined as $\sin(\theta) = \dfrac{y}{r}$ where $r \$ is the distance from the origin O to any point M on the terminal side of the angle and is given by $r = \sqrt{x^2+y^2}$ If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows: $sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle. It is even easier and no calculator is needed to find $\sin(\theta)$ for the quadrantal angles: $0, \dfrac{\pi}{2}, \pi, ...$ as shown in the unit circle below: The coordinates of the point corresponding to $\theta = 0$ on the unit circle are: (1,0). The y coordinate is equal to 0, hence $\sin(0) = 0$ The coordinates of the point corresponding to $\theta = \dfrac{\pi}{2}$ on the unit circle are: (0,1). The y coordinate is equal to 1, hence $\sin(\dfrac{\pi}{2}) = 1$ and so on. Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below. $\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$ We now use a system of rectangular axes $(x,y)$ to plot the points in the above table and approximate the graph of the sine function as shown below. NOTE however that because we are used to $x$ being the variable of a function, $x$ on the graph takes values of $\theta$ and y takes the values of $\sin(\theta)$ which is noted as $y = \sin(x)$. After $2\pi$, and as $\theta$ increases, the values of $\sin(\theta)$ will repeat at the quadrantal angles. We say that the sine function has a period of $2\pi$ shown below in red. <MASK> A more general sine function is written as $f(x) = a \sin(b x + c) + d$ with its amplitude $\|a\|$ , period $\dfrac{2\pi}{\|b\|}$ and phase shift $-\dfrac{c}{b}$ are explored interactively using an html 5 applet. The investigation is carried out by changing the parameters $a, b, c$ and $d$. To deeply understand the effects of each parameter on the graph of the function, we change one parameter at the time at the start. Then later we may change more than one parameter. Exploration and understanding of the phase shift is done by comparing the shift between the graphs of the two functions: $f(x) = a \sin(b x + c) + d$ in blue and $g(x) = a \sin(b x) + d$ in red as shown in the figure below. You may also want to consider another tutorial on the trigonometric unit circle . ## Interactive Tutorial Using Html 5 applet $y = f(x) = a \sin(b x + c) + d$ , in blue (with phase shift) $y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0) a = 1 b = 1 c = 0.5 d = 0 > <MASK> ### Amplitude <MASK> set a = 1, c = 0, d = 0 and change b. Find the period from the graph and compare it to $\dfrac{2\pi}{\|b\|}$. How does b affect the perid of the graph of f(x)? The period is the horizontal distance (along the x-axis) between two points: one is the starting point of a cycle and the second is the end point of the same cycle; two successive maxima or minima for example. ### Phase Shift set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to positive larger values. Take note of the shift, is it left or right? set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to negative smaller values. Take note of the shift, is it left or right? repeat the above for b = 2, 3 and 4, measure the shift and compare it to - c/b (the phase shift). ### Vertical Shift set a, b and c to non zero values and change d. What is the direction of the shift of the graph? ## More References and Links on Sine Functions Explore interactively the Derivatives of Sine (sin x) Functions Match Sine Functions to Graphs. Excellent activity where graphs and functions are matched. Explore interactively the sum of a sine and a cosine functionsSum of Sine and Cosine Functions Examples with detailed solutions and explanations on sine function problems. Tutorial on Sine Functions (1)- Problems Tutorial on the relationship between the amplitude, the vertical shift and the maximum and minimum of the sine functionTutorial on Sine Functions (2)- Problems Trigonometric Functions Step by step graphing of sine functionsGraph of Sine, a*sin(bx+c), Function Explore interactively the relationship between the graph of sine function and the coordinates of a point on the unit circle Unit Circle and Trigonometric Functions sin(x), cos(x), tan(x) The Six Trigonometric Functions Calculator.
# Sine Function sin x ## Definition and Graph of the Sine Function We first consider angle ? with initial side on the positive x axis (in standard position) and terminal side OM as shown below. The sine function is defined as $\sin(\theta) = \dfrac{y}{r}$ where $r \$ is the distance from the origin O to any point M on the terminal side of the angle and is given by $r = \sqrt{x^2+y^2}$ If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows: $sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle. It is even easier and no calculator is needed to find $\sin(\theta)$ for the quadrantal angles: $0, \dfrac{\pi}{2}, \pi, ...$ as shown in the unit circle below: The coordinates of the point corresponding to $\theta = 0$ on the unit circle are: (1,0). The y coordinate is equal to 0, hence $\sin(0) = 0$ The coordinates of the point corresponding to $\theta = \dfrac{\pi}{2}$ on the unit circle are: (0,1). The y coordinate is equal to 1, hence $\sin(\dfrac{\pi}{2}) = 1$ and so on. Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below. $\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$ We now use a system of rectangular axes $(x,y)$ to plot the points in the above table and approximate the graph of the sine function as shown below. NOTE however that because we are used to $x$ being the variable of a function, $x$ on the graph takes values of $\theta$ and y takes the values of $\sin(\theta)$ which is noted as $y = \sin(x)$. After $2\pi$, and as $\theta$ increases, the values of $\sin(\theta)$ will repeat at the quadrantal angles. We say that the sine function has a period of $2\pi$ shown below in red. <MASK> A more general sine function is written as $f(x) = a \sin(b x + c) + d$ with its amplitude $\|a\|$ , period $\dfrac{2\pi}{\|b\|}$ and phase shift $-\dfrac{c}{b}$ are explored interactively using an html 5 applet. The investigation is carried out by changing the parameters $a, b, c$ and $d$. To deeply understand the effects of each parameter on the graph of the function, we change one parameter at the time at the start. Then later we may change more than one parameter. Exploration and understanding of the phase shift is done by comparing the shift between the graphs of the two functions: $f(x) = a \sin(b x + c) + d$ in blue and $g(x) = a \sin(b x) + d$ in red as shown in the figure below. You may also want to consider another tutorial on the trigonometric unit circle . ## Interactive Tutorial Using Html 5 applet $y = f(x) = a \sin(b x + c) + d$ , in blue (with phase shift) $y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0) a = 1 b = 1 c = 0.5 d = 0 > <MASK> ### Amplitude <MASK> set a = 1, c = 0, d = 0 and change b. Find the period from the graph and compare it to $\dfrac{2\pi}{\|b\|}$. How does b affect the perid of the graph of f(x)? The period is the horizontal distance (along the x-axis) between two points: one is the starting point of a cycle and the second is the end point of the same cycle; two successive maxima or minima for example. ### Phase Shift set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to positive larger values. Take note of the shift, is it left or right? set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to negative smaller values. Take note of the shift, is it left or right? repeat the above for b = 2, 3 and 4, measure the shift and compare it to - c/b (the phase shift). ### Vertical Shift set a, b and c to non zero values and change d. What is the direction of the shift of the graph? ## More References and Links on Sine Functions Explore interactively the Derivatives of Sine (sin x) Functions Match Sine Functions to Graphs. Excellent activity where graphs and functions are matched. Explore interactively the sum of a sine and a cosine functionsSum of Sine and Cosine Functions Examples with detailed solutions and explanations on sine function problems. Tutorial on Sine Functions (1)- Problems Tutorial on the relationship between the amplitude, the vertical shift and the maximum and minimum of the sine functionTutorial on Sine Functions (2)- Problems Trigonometric Functions Step by step graphing of sine functionsGraph of Sine, asin(bx+c), Function Explore interactively the relationship between the graph of sine function and the coordinates of a point on the unit circle Unit Circle and Trigonometric Functions sin(x), cos(x), tan(x) The Six Trigonometric Functions Calculator. <UNMASK> # Sine Function sin x ## Definition and Graph of the Sine Function We first consider angle ? with initial side on the positive x axis (in standard position) and terminal side OM as shown below. The sine function is defined as $\sin(\theta) = \dfrac{y}{r}$ where $r \$ is the distance from the origin O to any point M on the terminal side of the angle and is given by $r = \sqrt{x^2+y^2}$ If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows: $sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle. It is even easier and no calculator is needed to find $\sin(\theta)$ for the quadrantal angles: $0, \dfrac{\pi}{2}, \pi, ...$ as shown in the unit circle below: The coordinates of the point corresponding to $\theta = 0$ on the unit circle are: (1,0). The y coordinate is equal to 0, hence $\sin(0) = 0$ The coordinates of the point corresponding to $\theta = \dfrac{\pi}{2}$ on the unit circle are: (0,1). The y coordinate is equal to 1, hence $\sin(\dfrac{\pi}{2}) = 1$ and so on. Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below. $\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$ We now use a system of rectangular axes $(x,y)$ to plot the points in the above table and approximate the graph of the sine function as shown below. NOTE however that because we are used to $x$ being the variable of a function, $x$ on the graph takes values of $\theta$ and y takes the values of $\sin(\theta)$ which is noted as $y = \sin(x)$. After $2\pi$, and as $\theta$ increases, the values of $\sin(\theta)$ will repeat at the quadrantal angles. We say that the sine function has a period of $2\pi$ shown below in red. ## General Sine Functions A more general sine function is written as $f(x) = a \sin(b x + c) + d$ with its amplitude $\|a\|$ , period $\dfrac{2\pi}{\|b\|}$ and phase shift $-\dfrac{c}{b}$ are explored interactively using an html 5 applet. The investigation is carried out by changing the parameters $a, b, c$ and $d$. To deeply understand the effects of each parameter on the graph of the function, we change one parameter at the time at the start. Then later we may change more than one parameter. Exploration and understanding of the phase shift is done by comparing the shift between the graphs of the two functions: $f(x) = a \sin(b x + c) + d$ in blue and $g(x) = a \sin(b x) + d$ in red as shown in the figure below. You may also want to consider another tutorial on the trigonometric unit circle . ## Interactive Tutorial Using Html 5 applet $y = f(x) = a \sin(b x + c) + d$ , in blue (with phase shift) $y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0) a = 1 b = 1 c = 0.5 d = 0 > How do the 4 coefficients a, b, c and d affect the graph of f(x)? ### Amplitude Set a = 1, b = 1, c = 0 and d = 0. Write down f(x) and take note of the amplitude, period and phase shift of f(x)? Now change a , how does it affect the graph? The amplitude is defined as $\|a\|$. ### Period set a = 1, c = 0, d = 0 and change b. Find the period from the graph and compare it to $\dfrac{2\pi}{\|b\|}$. How does b affect the perid of the graph of f(x)? The period is the horizontal distance (along the x-axis) between two points: one is the starting point of a cycle and the second is the end point of the same cycle; two successive maxima or minima for example. ### Phase Shift set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to positive larger values. Take note of the shift, is it left or right? set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to negative smaller values. Take note of the shift, is it left or right? repeat the above for b = 2, 3 and 4, measure the shift and compare it to - c/b (the phase shift). ### Vertical Shift set a, b and c to non zero values and change d. What is the direction of the shift of the graph? ## More References and Links on Sine Functions Explore interactively the Derivatives of Sine (sin x) Functions Match Sine Functions to Graphs. Excellent activity where graphs and functions are matched. Explore interactively the sum of a sine and a cosine functionsSum of Sine and Cosine Functions Examples with detailed solutions and explanations on sine function problems. Tutorial on Sine Functions (1)- Problems Tutorial on the relationship between the amplitude, the vertical shift and the maximum and minimum of the sine functionTutorial on Sine Functions (2)- Problems Trigonometric Functions Step by step graphing of sine functionsGraph of Sine, asin(bx+c), Function Explore interactively the relationship between the graph of sine function and the coordinates of a point on the unit circle Unit Circle and Trigonometric Functions sin(x), cos(x), tan(x) The Six Trigonometric Functions Calculator.
# Often asked: What Is A Space Figure? A space figure or three-dimensional figure is a figure that has depth in addition to width and height. Some common simple space figures include cubes, spheres, cylinders, prisms, cones, and pyramids. A space figure having all flat faces is called a polyhedron. ## What is the difference between a space figure and a solid figure? A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth. <MASK> ## How are space figures useful? <MASK> ## Is a figure in space or space figure? <MASK> You might be interested: What large body of water borders mexico on the east <MASK> A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base. <MASK> ## What do you call a space figure that has two circular bases and a curved surface? <MASK> ## What jobs do you need geometry for? Jobs that use geometry • Animator. • Mathematics teacher. • Fashion designer. • Plumber. • Game developer. • Interior designer. • Surveyor. ## Can we live without geometry? <MASK> You might be interested: Who is a famous person from mexico ## How far apart should volleyball posts be? <MASK> Volleyball has come a long way from the dusty-old YMCA gymnasium of Holyoke, Massachusetts, USA, where the visionary William G. Morgan invented the sport back in 1895. <MASK> <UNMASK> # Often asked: What Is A Space Figure? A space figure or three-dimensional figure is a figure that has depth in addition to width and height. Some common simple space figures include cubes, spheres, cylinders, prisms, cones, and pyramids. A space figure having all flat faces is called a polyhedron. ## What is the difference between a space figure and a solid figure? A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth. ## What are the example of spatial figures? <MASK> ## How are space figures useful? Three-dimensional geometry, or space geometry, is used to describe the buildings we live and work in, the tools we work with, and the objects we create. They made important discoveries and consequently they got to name the objects they discovered. That’s why geometric figures usually have Greek names! ## What is the space figure of volleyball? Dimensions. The playing court is 18m long and 9m wide and is surrounded by a free zone 3m wide on all sides. The space above the playing area is known as the free playing space and is a minimum of 7m high from the playing surface. ## Is a figure in space or space figure? <MASK> You might be interested: What large body of water borders mexico on the east <MASK> A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base. <MASK> ## What do you call a space figure that has two circular bases and a curved surface? A cylinder is similar to a prism, but its two bases are circles, not polygons. Also, the sides of a cylinder are curved, not flat. ## What jobs do you need geometry for? Jobs that use geometry • Animator. • Mathematics teacher. • Fashion designer. • Plumber. • Game developer. • Interior designer. • Surveyor. ## Can we live without geometry? <MASK> You might be interested: Who is a famous person from mexico ## How far apart should volleyball posts be? Posts should be placed 1m (3′-4”) from each side line, 36′-8” from each other. A recommended free or clearance zone of at least 10 ft is recommended. For the most versatile facility, it is recommended to install poles 36′-8” from each other to allow for both competition and recreational play. ## Who created volleyball? Volleyball has come a long way from the dusty-old YMCA gymnasium of Holyoke, Massachusetts, USA, where the visionary William G. Morgan invented the sport back in 1895. ## Why does a volleyball court need a free zone? <MASK>
# Often asked: What Is A Space Figure? A space figure or three-dimensional figure is a figure that has depth in addition to width and height. Some common simple space figures include cubes, spheres, cylinders, prisms, cones, and pyramids. A space figure having all flat faces is called a polyhedron. ## What is the difference between a space figure and a solid figure? A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth. ## What are the example of spatial figures? <MASK> ## How are space figures useful? Three-dimensional geometry, or space geometry, is used to describe the buildings we live and work in, the tools we work with, and the objects we create. They made important discoveries and consequently they got to name the objects they discovered. That’s why geometric figures usually have Greek names! ## What is the space figure of volleyball? Dimensions. The playing court is 18m long and 9m wide and is surrounded by a free zone 3m wide on all sides. The space above the playing area is known as the free playing space and is a minimum of 7m high from the playing surface. ## Is a figure in space or space figure? <MASK> You might be interested: What large body of water borders mexico on the east <MASK> A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base. <MASK> ## What do you call a space figure that has two circular bases and a curved surface? A cylinder is similar to a prism, but its two bases are circles, not polygons. Also, the sides of a cylinder are curved, not flat. ## What jobs do you need geometry for? Jobs that use geometry • Animator. • Mathematics teacher. • Fashion designer. • Plumber. • Game developer. • Interior designer. • Surveyor. ## Can we live without geometry? <MASK> You might be interested: Who is a famous person from mexico ## How far apart should volleyball posts be? Posts should be placed 1m (3′-4”) from each side line, 36′-8” from each other. A recommended free or clearance zone of at least 10 ft is recommended. For the most versatile facility, it is recommended to install poles 36′-8” from each other to allow for both competition and recreational play. ## Who created volleyball? Volleyball has come a long way from the dusty-old YMCA gymnasium of Holyoke, Massachusetts, USA, where the visionary William G. Morgan invented the sport back in 1895. ## Why does a volleyball court need a free zone? <MASK> <UNMASK> # Often asked: What Is A Space Figure? A space figure or three-dimensional figure is a figure that has depth in addition to width and height. Some common simple space figures include cubes, spheres, cylinders, prisms, cones, and pyramids. A space figure having all flat faces is called a polyhedron. ## What is the difference between a space figure and a solid figure? A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth. ## What are the example of spatial figures? In mathematics, spatial figures are defined simply as three-dimensional objects. For example, a basketball or a cardboard box are spatial figures that we have likely encountered in our lives. ## How are space figures useful? Three-dimensional geometry, or space geometry, is used to describe the buildings we live and work in, the tools we work with, and the objects we create. They made important discoveries and consequently they got to name the objects they discovered. That’s why geometric figures usually have Greek names! ## What is the space figure of volleyball? Dimensions. The playing court is 18m long and 9m wide and is surrounded by a free zone 3m wide on all sides. The space above the playing area is known as the free playing space and is a minimum of 7m high from the playing surface. ## Is a figure in space or space figure? Space figures are figures whose points do not all lie in the same plane. In this unit, we’ll study the polyhedron, the cylinder, the cone, and the sphere. Polyhedrons are space figures with flat surfaces, called faces, which are made of polygons. Prisms and pyramids are examples of polyhedrons. You might be interested: What large body of water borders mexico on the east ## Can prisms have circular bases? A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base. ## What is a solid figure? Solid figures are basically three-dimensional objects, which means that they have length, height and width. Because the solid figures have three dimensions, they have depth and take up space in our universe. Solid figures are identified according to the features that are unique to each type of solid. ## What do you call a space figure that has two circular bases and a curved surface? A cylinder is similar to a prism, but its two bases are circles, not polygons. Also, the sides of a cylinder are curved, not flat. ## What jobs do you need geometry for? Jobs that use geometry • Animator. • Mathematics teacher. • Fashion designer. • Plumber. • Game developer. • Interior designer. • Surveyor. ## Can we live without geometry? Without Geometry things would have been very challenging in Day to day life as well in various technological fields. Lines, Angles, Shapes, 2d & 3d designs plays a vital role in designing of home and commercial infra, mechanical and engineering design. This is feasible only because of Geometry. ## How do you explain geometry to a child? Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel. You might be interested: Who is a famous person from mexico ## How far apart should volleyball posts be? Posts should be placed 1m (3′-4”) from each side line, 36′-8” from each other. A recommended free or clearance zone of at least 10 ft is recommended. For the most versatile facility, it is recommended to install poles 36′-8” from each other to allow for both competition and recreational play. ## Who created volleyball? Volleyball has come a long way from the dusty-old YMCA gymnasium of Holyoke, Massachusetts, USA, where the visionary William G. Morgan invented the sport back in 1895. ## Why does a volleyball court need a free zone? The volleyball court is surrounded by a free zone. The free zone is the area outside the court that players may enter to make a play on the ball. The free zone should be at least 3 meters wide from the court. If any part of the ball crosses the net directly above or outside the antenna, the ball is out of play.
# Reduction of coil span This value expresses the reduction of the coil span compared to full pitch. The higher the reduction of coil span, the shorter the end-turns of your winding. In the Emetor winding calculator, the reduction of coil span compared to full pitch is expressed in number of slot pitches. <MASK> A c B a C b b A c B a C Fig. 2 Double-layer 2-pole 6-slot integer-slot winding with reduced coil span and a winding factor of 0.866. Fractional-slot windings and concentrated windings have full pitches that are not integers, as opposed to integer-slot windings. In addition, concentrated windings have always a coil span of one, due to their nature of being wound around a single tooth. Examples: Full pitch of 2-pole 9-slot fractional-slot winding: $\frac{Q_s}{p} = \frac{9}{2} = 4.5$ Full pitch of 4-pole 9-slot concentrated winding: $\frac{Q_s}{p} = \frac{9}{4} = 2.25$ Read about another glossary term <UNMASK> # Reduction of coil span This value expresses the reduction of the coil span compared to full pitch. The higher the reduction of coil span, the shorter the end-turns of your winding. In the Emetor winding calculator, the reduction of coil span compared to full pitch is expressed in number of slot pitches. The reduction of coil span can be expressed as: $$\frac{Q_s}{p} - \mbox{coil span,}$$ where $\frac{Q_s}{p}$ is the coil span for full pitch. Single-layer integer-slot windings have always a coil span that is equal to full pitch, ie. the reduction of coil span gets zero. For example, the 2-pole 6-slot winding in Fig. 1 has a coil span of 3, a full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 0: <MASK> Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0. In order to reduce the coil span, which has the advantage of reducing the length of the end-turns in your winding, a two-layer winding becomes necessary, see Fig. 2. In this case, the 2-pole 6-slot winding has a coil span of only 2, an unchanged full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 1 slot pitch: A c B a C b b A c B a C Fig. 2 Double-layer 2-pole 6-slot integer-slot winding with reduced coil span and a winding factor of 0.866. Fractional-slot windings and concentrated windings have full pitches that are not integers, as opposed to integer-slot windings. In addition, concentrated windings have always a coil span of one, due to their nature of being wound around a single tooth. Examples: Full pitch of 2-pole 9-slot fractional-slot winding: $\frac{Q_s}{p} = \frac{9}{2} = 4.5$ Full pitch of 4-pole 9-slot concentrated winding: $\frac{Q_s}{p} = \frac{9}{4} = 2.25$ Read about another glossary term
# Reduction of coil span This value expresses the reduction of the coil span compared to full pitch. The higher the reduction of coil span, the shorter the end-turns of your winding. In the Emetor winding calculator, the reduction of coil span compared to full pitch is expressed in number of slot pitches. The reduction of coil span can be expressed as: $$\frac{Q_s}{p} - \mbox{coil span,}$$ where $\frac{Q_s}{p}$ is the coil span for full pitch. Single-layer integer-slot windings have always a coil span that is equal to full pitch, ie. the reduction of coil span gets zero. For example, the 2-pole 6-slot winding in Fig. 1 has a coil span of 3, a full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 0: <MASK> Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0. In order to reduce the coil span, which has the advantage of reducing the length of the end-turns in your winding, a two-layer winding becomes necessary, see Fig. 2. In this case, the 2-pole 6-slot winding has a coil span of only 2, an unchanged full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 1 slot pitch: A c B a C b b A c B a C Fig. 2 Double-layer 2-pole 6-slot integer-slot winding with reduced coil span and a winding factor of 0.866. Fractional-slot windings and concentrated windings have full pitches that are not integers, as opposed to integer-slot windings. In addition, concentrated windings have always a coil span of one, due to their nature of being wound around a single tooth. Examples: Full pitch of 2-pole 9-slot fractional-slot winding: $\frac{Q_s}{p} = \frac{9}{2} = 4.5$ Full pitch of 4-pole 9-slot concentrated winding: $\frac{Q_s}{p} = \frac{9}{4} = 2.25$ Read about another glossary term <UNMASK> # Reduction of coil span This value expresses the reduction of the coil span compared to full pitch. The higher the reduction of coil span, the shorter the end-turns of your winding. In the Emetor winding calculator, the reduction of coil span compared to full pitch is expressed in number of slot pitches. The reduction of coil span can be expressed as: $$\frac{Q_s}{p} - \mbox{coil span,}$$ where $\frac{Q_s}{p}$ is the coil span for full pitch. Single-layer integer-slot windings have always a coil span that is equal to full pitch, ie. the reduction of coil span gets zero. For example, the 2-pole 6-slot winding in Fig. 1 has a coil span of 3, a full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 0: A c B a C b Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0. In order to reduce the coil span, which has the advantage of reducing the length of the end-turns in your winding, a two-layer winding becomes necessary, see Fig. 2. In this case, the 2-pole 6-slot winding has a coil span of only 2, an unchanged full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 1 slot pitch: A c B a C b b A c B a C Fig. 2 Double-layer 2-pole 6-slot integer-slot winding with reduced coil span and a winding factor of 0.866. Fractional-slot windings and concentrated windings have full pitches that are not integers, as opposed to integer-slot windings. In addition, concentrated windings have always a coil span of one, due to their nature of being wound around a single tooth. Examples: Full pitch of 2-pole 9-slot fractional-slot winding: $\frac{Q_s}{p} = \frac{9}{2} = 4.5$ Full pitch of 4-pole 9-slot concentrated winding: $\frac{Q_s}{p} = \frac{9}{4} = 2.25$ Read about another glossary term
<MASK> <UNMASK> View Question Question Subject: Java Programming - Quadratic Equations Category: Computers > Programming Asked by: java_design-ga List Price: \$15.00 Posted: 29 Nov 2005 05:34 PST Expires: 30 Nov 2005 09:47 PST Question ID: 598931 ```Design and develop a Java program that continuously computes and displays value(s) for x, given quadratic equations (i.e. a second-order polynomials) of the form: ax2 + bx + c = 0 where the values for the coefficients a, b and c are supplied by the user, and are assumed to be integers within the range of -100 to 100. To control the loop use a menu interface. The menu should include two options: "Calculate quadratic" and "End". Note that to solve a quadratic equation we must calculate the roots. This can be done using the quadratic formula: root 1 = (-b + sqrt(b2-4ac)) / 2a root2 = (-b - sqrt(b2-4ac)) / 2a Example: x2 + 2x - 8 = 0 a= 1, b = 2, c = -8 roots = (-2 +or- sqrt(22-4x1x-8)) / 2x1 = (-2 +or- sqrt(4+32)) / 2 root1 = (-2 + 6)/2 = 4/2 = 2.0 root2 = (-2 - 6)/2 = -8/2 = -4.0 x = 2.0 or -4.0 However, there are certain special consideration to be taken into account: If a and b are both zero there is no solution (this is referred to as the degenerate case): -8 = 0? a= 0, b = 0, c = -8 (degenerate case) If a is zero and b is non zero the equation becomes a linear equation. 2x - 8 = 0 a= 0, b = 2, c = -8 (Linear equation) root = -c/b = 8/2 = 4.0 x = 4.0 If the value for the term b2 - 4ac (the discriminant) is negative there is no solution (conventionally we cannot find the square root of a negative number!): x2 + 2x + 8 = 0 a= 1, b = 2, c = 8 roots = (-2 +or- sqrt(22-4x1x8)) / 2x1 = (-2 +or- sqrt(4-32)) / 2 = (-2 +or- sqrt(-28)) Negative discriminant therefore no solution. If the discriminant is 0 then there are two identical solutions, i.e. only one solution (root) need be calculated: x2 + 4x + 4 = 0 a= 1, b = 4, c = 4 roots = (-4 +or- sqrt(42-4x1x4)) / 2x1 = (-4 +or- sqrt(16-16)) / 2 (Discriminant = 0, there fore only one solution) root = -4/2 = -2 x = -2.0 Output, where appropriate, should be accurate to at least several decimal places. Please try to include explanations where appropriate.``` Clarification of Question by java_design-ga on 29 Nov 2005 05:39 PST `The program must be written in Java 1.5 !!!` `I'd do it for \$200. I wouldn't do it for \$15.` ```import java.io.; public class quadratic { public int a = 0,b = 0,c = 0; public int flag=0; public double r1=0,r2=0; public quadratic() { do { System.out.println("\n\n\n\n\nType 1 to Calculate quadratic equation"); System.out.println("Type 3 to END"); int choice = getChoice(); switch(choice) { case 1: a = inputABC("a"); b = inputABC("b"); c = inputABC("c"); if((Math.pow(b,2)-4ac)<0) {System.out.println("\nNegative discriminant therefore no solution!");} else if((Math.pow(b,2)-4ac)==0) { r1 = getRoot1(a,b,c); System.out.print("x= "+r1); } else if(a==0&&b==0) {System.out.println("\nDegenerate Case");} else if(a==0) {double u = -c/b; System.out.println("The root is "+u);} else { r1 = getRoot1(a,b,c); r2 = getRoot2(a,b,c); System.out.print("x= "+r1+" or "+r2); } break; case 3: flag = 1; break; default: System.out.println("\nThat is not an option! Try Again."); break; } }while(flag==0); } public int inputABC(String s2) { InputStreamReader stdin = new InputStreamReader(System.in); BufferedReader console = new BufferedReader(stdin); int i1=0; String s1; try { System.out.print("Enter interger for "+s2+": "); s1 = console.readLine(); i1 = Integer.parseInt(s1); } catch(IOException ioex) { System.out.println("\nInput error"); System.exit(1); } catch(NumberFormatException nfex) { System.out.println("\"" + nfex.getMessage() + "\" is not numeric"); System.exit(1); } return(i1); } public int getChoice() { InputStreamReader stdin = new InputStreamReader(System.in); BufferedReader console = new BufferedReader(stdin); int i2=0; String s2; try { System.out.print("User's Choice: "); s2 = console.readLine(); i2 = Integer.parseInt(s2); } catch(IOException ioex) { System.out.println("Input error"); System.exit(1); } catch(NumberFormatException nfex) { System.out.println("\"" + nfex.getMessage() + "\" is not numeric"); System.exit(1); } return(i2); } public double getRoot1(int x, int y, int z) { double root1 = (-y + Math.sqrt((Math.pow(b,2))-(4xz)))/(2x); return(root1); } public double getRoot2(int x, int y, int z) { double root2 = (-y - Math.sqrt((Math.pow(b,2))-(4xz)))/(2*x); return(root2); } public static void main(String[] args) { quadratic qd = new quadratic(); } }```
<MASK> Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified March 26 00:37 EDT 2019. Contains 321479 sequences. (Running on oeis4.) <UNMASK> This site is supported by donations to The OEIS Foundation. <MASK> Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified March 26 00:37 EDT 2019. Contains 321479 sequences. (Running on oeis4.)
This site is supported by donations to The OEIS Foundation. <MASK> Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified March 26 00:37 EDT 2019. Contains 321479 sequences. (Running on oeis4.) <UNMASK> This site is supported by donations to The OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A121505 Hit triangle for unit circle area (Pi) approximation problem described in A121500. 1 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 (list; table; graph; refs; listen; history; text; internal format) OFFSET 3,1 COMMENTS Record for n=3,4,... only those (n, A121500(n)) pairs which have relative error E(n, A121500(n)) smaller than all errors with smaller n. This produces the table a(n,m). The unit circle area is approximated by the arithmetic mean of the areas of an inscribed regular n-gon and a circumscribed regular m-gon. For each row n>=3 the minimal relative error E(n,m):= ((Fin(n) + Fout(m))/2-Pi)/ Pi) appears for m= A121500(n). The same hit triangle is obtained when one considers the minimal relative errors for the columns m>=3 and collects the sequence with decreasing errors, starting with m=3. LINKS W. Lang: First rows. FORMULA a(n,m) = 1 if m = A121500(n) and E(n,m) < min(E(k,A121500(k)), k=3..n-1), n>=4. a(3,3) = 1, else a(n,m) = 0. EXAMPLE [1], [0,0], [0,1,0], [0, 0, 1, 0], [0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0],... CROSSREFS Sequence in context: A120524 A014177 A014129 * A014289 A015297 A015073 Adjacent sequences: A121502 A121503 A121504 * A121506 A121507 A121508 KEYWORD nonn,tabl,easy AUTHOR Wolfdieter Lang, Aug 16 2006 STATUS approved Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified March 26 00:37 EDT 2019. Contains 321479 sequences. (Running on oeis4.)
<MASK> Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following. 1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret. 2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE. 3. Determine the coefficient of correlation. Interpret. 4. Determine the coefficient of determination. Interpret. 5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value. 6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain. 7. Compute the 95% confidence interval for . Interpret this interval. 8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval. 9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval. 10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer. <MASK> #### Solution Summary <MASK> Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following. <MASK> 3. Determine the coefficient of correlation. Interpret. <MASK> 5. Test the utility of this regression model (use a two tailed test with alpha =.05). Interpret your results, including the p-value. <MASK> 7. Compute the 95% confidence interval for mean. Interpret this interval. 8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval. <MASK> 12. Perform the Global Test for Utility (F-Test). Explain your conclusion. <MASK> 14. Is this multiple regression model better than the linear model that we generated in parts 1-10? Explain. <MASK> View Full Posting Details <UNMASK> Explore BrainMass Share # Using MINITAB, Perform the Regression and Correlation Analysis <MASK> Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following. 1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret. 2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE. 3. Determine the coefficient of correlation. Interpret. 4. Determine the coefficient of determination. Interpret. 5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value. 6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain. 7. Compute the 95% confidence interval for . Interpret this interval. 8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval. 9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval. 10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer. <MASK> https://brainmass.com/statistics/regression-analysis/using-minitab-perform-the-regression-and-correlation-analysis-508281 #### Solution Summary <MASK> \$2.19 ## Regression and Correlation Analysis. Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following. 1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the best fit line. Interpret. <MASK> 3. Determine the coefficient of correlation. Interpret. 4. Determine the index of determination. Interpret. 5. Test the utility of this regression model (use a two tailed test with alpha =.05). Interpret your results, including the p-value. <MASK> 7. Compute the 95% confidence interval for mean. Interpret this interval. 8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval. 9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval. 10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer. In an attempt to improve the model, we attempt to do a multiple regression model predicting CREDIT BALANCE based on INCOME, SIZE and YEARS. 11. Using MINITAB run the multiple regression analysis using the variables INCOME, SIZE and YEARS to predict CREDIT BALANCE. State the equation for this multiple regression model. 12. Perform the Global Test for Utility (F-Test). Explain your conclusion. <MASK> 14. Is this multiple regression model better than the linear model that we generated in parts 1-10? Explain. <MASK> View Full Posting Details
Explore BrainMass Share # Using MINITAB, Perform the Regression and Correlation Analysis <MASK> Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following. 1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret. 2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE. 3. Determine the coefficient of correlation. Interpret. 4. Determine the coefficient of determination. Interpret. 5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value. 6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain. 7. Compute the 95% confidence interval for . Interpret this interval. 8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval. 9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval. 10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer. <MASK> https://brainmass.com/statistics/regression-analysis/using-minitab-perform-the-regression-and-correlation-analysis-508281 #### Solution Summary <MASK> \$2.19 ## Regression and Correlation Analysis. Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following. 1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the best fit line. Interpret. <MASK> 3. Determine the coefficient of correlation. Interpret. 4. Determine the index of determination. Interpret. 5. Test the utility of this regression model (use a two tailed test with alpha =.05). Interpret your results, including the p-value. <MASK> 7. Compute the 95% confidence interval for mean. Interpret this interval. 8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval. 9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval. 10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer. In an attempt to improve the model, we attempt to do a multiple regression model predicting CREDIT BALANCE based on INCOME, SIZE and YEARS. 11. Using MINITAB run the multiple regression analysis using the variables INCOME, SIZE and YEARS to predict CREDIT BALANCE. State the equation for this multiple regression model. 12. Perform the Global Test for Utility (F-Test). Explain your conclusion. <MASK> 14. Is this multiple regression model better than the linear model that we generated in parts 1-10? Explain. <MASK> View Full Posting Details <UNMASK> Explore BrainMass Share # Using MINITAB, Perform the Regression and Correlation Analysis This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here! Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following. 1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret. 2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE. 3. Determine the coefficient of correlation. Interpret. 4. Determine the coefficient of determination. Interpret. 5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value. 6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain. 7. Compute the 95% confidence interval for . Interpret this interval. 8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval. 9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval. 10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer. In an attempt to improve the model, we attempt to do a multiple regression model predicting CREDIT BALANCE based on INCOME, SIZE and YEARS. 11. Using MINITAB run the multiple regression analysis using the variables INCOME, SIZE and YEARS to predict CREDIT BALANCE. State the equation for this multiple regression model. 12. Perform the Global Test for Utility (F-Test). Explain your conclusion. 13. Perform the t-test on each independent variable. Explain your conclusions and clearly state how you should proceed. In particular, which independent variables should we keep and which should be discarded. 14. Is this multiple regression model better than the linear model that we generated in parts 1-10? Explain. Summarize your results report that is three pages or less in length and explains and interprets the results in ways that are understandable to someone who does not know statistics. Submission: The summary report + (Minitab Output + interpretations) as an appendix. Suggested format: A. Summary Report B. Addressed with appropriate output, graphs and interpretations. https://brainmass.com/statistics/regression-analysis/using-minitab-perform-the-regression-and-correlation-analysis-508281 #### Solution Summary The answers (except Q7 which is missing information) are attached in a Word, Excel and .MPJ format including the necessary graphs and charts. \$2.19 ## Regression and Correlation Analysis. Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following. 1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the best fit line. Interpret. 2. Determine the equation of the best fit line, which describes the relationship between CREDIT BALANCE and SIZE. 3. Determine the coefficient of correlation. Interpret. 4. Determine the index of determination. Interpret. 5. Test the utility of this regression model (use a two tailed test with alpha =.05). Interpret your results, including the p-value. 6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain. 7. Compute the 95% confidence interval for mean. Interpret this interval. 8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval. 9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval. 10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer. In an attempt to improve the model, we attempt to do a multiple regression model predicting CREDIT BALANCE based on INCOME, SIZE and YEARS. 11. Using MINITAB run the multiple regression analysis using the variables INCOME, SIZE and YEARS to predict CREDIT BALANCE. State the equation for this multiple regression model. 12. Perform the Global Test for Utility (F-Test). Explain your conclusion. 13. Perform the t-test on each independent variable. Explain your conclusions and clearly state how you should proceed. In particular, which independent variables should we keep and which should be discarded. 14. Is this multiple regression model better than the linear model that we generated in parts 1-10? Explain. Summarize your results from 1-14 in a report that is three pages or less in length and explains and interprets the results in ways that are understandable to someone who does not know statistics. Please explain in Microsoft Word format. No MINITAB. View Full Posting Details
<MASK> # document.write(document.title); <MASK> From yd/s to pm/s: 1 yd/s = 914400000003.66 pm/s; <MASK> ## How to Convert Yard/second to Picometer/second? As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s). To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66. <MASK> 25 yd/s = 25 × 914400000003.66 pm/s = pm/s To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66. Example : convert 25 pm/s to yd/s: 25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s ## How to Convert Picometer/second to Yard/second? As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s). <MASK> Example : convert 45 pm/s to yd/s: 45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12. Example : convert 45 yd/s to pm/s: 45 yd/s = 45 ÷ 1.0936132983333E-12 pm/s = pm/s ## Convert Yard/second or Picometer/second to Other Speed and Velocity Units Yard/second Conversion Table yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters Yard/second Conversion Table yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters ## FAQ ### What is 9 Yard/second in Picometer/second? pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s. ### How many Picometer/second are in a Yard/second? There are 914400000003.66 pm/s in one yd/s. In turn, one pm/s is equal to 1.0936132983333E-12 yd/s. ### How many yd/s is equal to 1 pm/s? 1 pm/s is approximately equal to 1.0936132983333E-12 yd/s. ### What is the yd/s value of 8 pm/s? The Yard/second value of 8 pm/s is yd/s. (i.e.,) 8 x 1.0936132983333E-12 = yd/s. ### yd/s to pm/s converter in batch <MASK> "" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters <UNMASK> <MASK> # document.write(document.title); ## yd/s to pm/s Converter From yd/s to pm/s: 1 yd/s = 914400000003.66 pm/s; <MASK> ## How to Convert Yard/second to Picometer/second? As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s). To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66. <MASK> 25 yd/s = 25 × 914400000003.66 pm/s = pm/s To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66. Example : convert 25 pm/s to yd/s: 25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s ## How to Convert Picometer/second to Yard/second? As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s). <MASK> Example : convert 45 pm/s to yd/s: 45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12. Example : convert 45 yd/s to pm/s: 45 yd/s = 45 ÷ 1.0936132983333E-12 pm/s = pm/s ## Convert Yard/second or Picometer/second to Other Speed and Velocity Units Yard/second Conversion Table yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters Yard/second Conversion Table yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters ## FAQ ### What is 9 Yard/second in Picometer/second? pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s. ### How many Picometer/second are in a Yard/second? There are 914400000003.66 pm/s in one yd/s. In turn, one pm/s is equal to 1.0936132983333E-12 yd/s. ### How many yd/s is equal to 1 pm/s? 1 pm/s is approximately equal to 1.0936132983333E-12 yd/s. ### What is the yd/s value of 8 pm/s? The Yard/second value of 8 pm/s is yd/s. (i.e.,) 8 x 1.0936132983333E-12 = yd/s. ### yd/s to pm/s converter in batch <MASK> "" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
<MASK> # document.write(document.title); ## yd/s to pm/s Converter From yd/s to pm/s: 1 yd/s = 914400000003.66 pm/s; <MASK> ## How to Convert Yard/second to Picometer/second? As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s). To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66. <MASK> 25 yd/s = 25 × 914400000003.66 pm/s = pm/s To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66. Example : convert 25 pm/s to yd/s: 25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s ## How to Convert Picometer/second to Yard/second? As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s). <MASK> Example : convert 45 pm/s to yd/s: 45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12. Example : convert 45 yd/s to pm/s: 45 yd/s = 45 ÷ 1.0936132983333E-12 pm/s = pm/s ## Convert Yard/second or Picometer/second to Other Speed and Velocity Units Yard/second Conversion Table yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters Yard/second Conversion Table yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters ## FAQ ### What is 9 Yard/second in Picometer/second? pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s. ### How many Picometer/second are in a Yard/second? There are 914400000003.66 pm/s in one yd/s. In turn, one pm/s is equal to 1.0936132983333E-12 yd/s. ### How many yd/s is equal to 1 pm/s? 1 pm/s is approximately equal to 1.0936132983333E-12 yd/s. ### What is the yd/s value of 8 pm/s? The Yard/second value of 8 pm/s is yd/s. (i.e.,) 8 x 1.0936132983333E-12 = yd/s. ### yd/s to pm/s converter in batch <MASK> "" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters <UNMASK> Convert Yard/second to Picometer/second \| pm/s to yd/s # document.write(document.title); ## yd/s to pm/s Converter From yd/s to pm/s: 1 yd/s = 914400000003.66 pm/s; From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s; ## How to Convert Yard/second to Picometer/second? As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s). To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66. Example : convert 25 yd/s to pm/s: 25 yd/s = 25 × 914400000003.66 pm/s = pm/s To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66. Example : convert 25 pm/s to yd/s: 25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s ## How to Convert Picometer/second to Yard/second? As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s). To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12. Example : convert 45 pm/s to yd/s: 45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12. Example : convert 45 yd/s to pm/s: 45 yd/s = 45 ÷ 1.0936132983333E-12 pm/s = pm/s ## Convert Yard/second or Picometer/second to Other Speed and Velocity Units Yard/second Conversion Table yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters Yard/second Conversion Table yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters ## FAQ ### What is 9 Yard/second in Picometer/second? pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s. ### How many Picometer/second are in a Yard/second? There are 914400000003.66 pm/s in one yd/s. In turn, one pm/s is equal to 1.0936132983333E-12 yd/s. ### How many yd/s is equal to 1 pm/s? 1 pm/s is approximately equal to 1.0936132983333E-12 yd/s. ### What is the yd/s value of 8 pm/s? The Yard/second value of 8 pm/s is yd/s. (i.e.,) 8 x 1.0936132983333E-12 = yd/s. ### yd/s to pm/s converter in batch Cite this Converter, Content or Page as: "" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
UBS Interview Question: What is the sum of 1-40?... \| Glassdoor # What is the sum of 1-40? 0 <MASK> We can have the sum by using the following formula N * (N + 1) / 2 So we have: 40 * (40 + 1) / 2 = 820 <MASK> scienceguy on Jan 11, 2011 1 http://brainteaserbible.com/interview-brainteaser-sum-of-the-numbers-from-1-to-50 an87 on Jun 13, 2011 <UNMASK> UBS Interview Question: What is the sum of 1-40?... \| Glassdoor # What is the sum of 1-40? 0 <MASK> Interview Candidate on Feb 11, 2010 2 We can have the sum by using the following formula N * (N + 1) / 2 So we have: 40 * (40 + 1) / 2 = 820 <MASK> scienceguy on Jan 11, 2011 1 http://brainteaserbible.com/interview-brainteaser-sum-of-the-numbers-from-1-to-50 an87 on Jun 13, 2011
UBS Interview Question: What is the sum of 1-40?... \| Glassdoor # What is the sum of 1-40? 0 <MASK> Interview Candidate on Feb 11, 2010 2 We can have the sum by using the following formula N * (N + 1) / 2 So we have: 40 * (40 + 1) / 2 = 820 <MASK> scienceguy on Jan 11, 2011 1 http://brainteaserbible.com/interview-brainteaser-sum-of-the-numbers-from-1-to-50 an87 on Jun 13, 2011 <UNMASK> UBS Interview Question: What is the sum of 1-40?... \| Glassdoor # What is the sum of 1-40? 0 820 Interview Candidate on Feb 11, 2010 2 We can have the sum by using the following formula N * (N + 1) / 2 So we have: 40 * (40 + 1) / 2 = 820 blue on Mar 21, 2010 0 Since he isn't asking to sum all the numbers from 1 to 40: sum of 1 and 40 = 41 OR sum of 1 + -40 = -39 Evandro on Dec 30, 2010 3 Any sum like this is easy. take the lowest and highest... 1 +40 = 41, then the next highest and the next lowest. 2 + 39 = 41. Realize that there are 20 such pairs. 20 * 41 = 820. For the numbers 1 - 100, 1 + 100 = 101, 2 + 99 = 101; there are 50 such pairs, so 50 * 101 = 5050 scienceguy on Jan 11, 2011 1 http://brainteaserbible.com/interview-brainteaser-sum-of-the-numbers-from-1-to-50 an87 on Jun 13, 2011
<MASK> # Mind Teasers : Tricky Math Race Riddle <MASK> Difficulty Popularity <MASK> * They have only one torch and the river is too risky to cross without the torch. * If all people cross simultaneously then torch light wont be sufficient. * Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes. <MASK> # Mind Teasers : Popular Logic Brain Teaser Difficulty Popularity <MASK> Sherlock holmes arrived at the scene and immediately find the murderer. <MASK> # Mind Teasers : Math Trick Riddle <MASK> # Mind Teasers : Best Maths Brain Teaser Difficulty Popularity <MASK> • Views : 60k+ • Sol Viewed : 20k+ <MASK> Difficulty Popularity <MASK> All of them are assured of one thing that the puzzle will not be impossible for anyone of them. How will they manage the situation? <MASK> Can you move four matchsticks to make th... <MASK> 19 February ##### Mitch And Irene Riddle <MASK> <UNMASK> • Views : 40k+ • Sol Viewed : 10k+ # Mind Teasers : Funny New Year Riddle Difficulty Popularity <MASK> January first 2010 => 800 X 600 January first 2011 => 1280 X 1024 January first 2012 => 600 X 1200 Discussion Suggestions • Views : 70k+ • Sol Viewed : 20k+ # Mind Teasers : Tricky Math Race Riddle <MASK> Rooney beats Hernandez by 20 meters. Hernandez beats Robin by 20 meters. How many meters does Rooney beat Robin by ? • Views : 50k+ • Sol Viewed : 20k+ # Mind Teasers : Probability To Choose Own Toy Brain Teaser Difficulty Popularity I bought three toys for my triplet boys (one for each). I placed the toys in the dark store. One by one each boy went to the store and pick the toy. What is the probability that no boy will choose his own toy? <MASK> # Mind Teasers : Crossing The River In Minimum Time Puzzle Difficulty Popularity Four people need to cross a dark river at night. * They have only one torch and the river is too risky to cross without the torch. * If all people cross simultaneously then torch light wont be sufficient. * Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes. <MASK> • Views : 40k+ • Sol Viewed : 10k+ # Mind Teasers : Logic Question In Maths Difficulty Popularity Two trains under a controlled experiment begin at a speed of 100 mph in the opposite direction in a tunnel. A supersonic bee is left in the tunnel which can fly at a speed of 1000 mph. The tunnel is 200 miles long. When the trains start running on a constant speed of 100 mph, the supersonic bee starts flying from one train towards the other. As soon as the bee reaches the second train, it starts flying back towards the first train. If the bee keeps flying to and fro in the tunnel till the trains collide, how much distance will it have covered in total? <MASK> # Mind Teasers : Popular Logic Brain Teaser Difficulty Popularity <MASK> When the piggy bank was opened, it had just \$500. How can that be possible? • Views : 80k+ • Sol Viewed : 20k+ # Mind Teasers : Sherlock Holmes Murderer Mystery Puzzle Difficulty Popularity There are five people. One of them shot and killed one of the other five. 1. Dan ran in the NY City marathon yesterday with one of the innocent men. 2. Mike considered being a farmer before he moved to the city. 3. Jeff is a topnotch computer consultant and wants to install Ben's new computer next week. 4. The murderer had his leg amputated last month. 5. Ben met Jack for the first time six months ago. 6. Jack has been in seclusion since the crime. 7. Dan used to drink heavily. 8. Ben and Jeff built their last computers together. 9. The murderer is Jack's brother. They grew up together in Seattle. Sherlock holmes arrived at the scene and immediately find the murderer. who was the murderer ? • Views : 70k+ • Sol Viewed : 20k+ # Mind Teasers : Math Trick Riddle <MASK> Removing one from eleven makes it ten and removing one from nine makes it ten. Is it even possible? • Views : 80k+ • Sol Viewed : 20k+ # Mind Teasers : Best Maths Brain Teaser Difficulty Popularity A competitive exam was held in which five students took part namely Billy, Gerry, Clark, Peeta and Jonathan. In this exam, they had to answer five questions each out of which, three had multiple choices as a, b or c and two were simple true and false questions. The five of them gave different answers to the questions and the details are as given below. Name 1 2 3 4 5 <MASK> Billy c c True True True <MASK> Peeta b a True True False Jonathan a b True False True <MASK> • Views : 60k+ • Sol Viewed : 20k+ # Mind Teasers : Crack the code Riddle Difficulty Popularity A man desired to get into his work building, however he had forgotten his code. However, he did recollect five pieces of information -> Sum of 5th number and 3rd number is 14. -> Difference of 4th and 2nd number is 1. -> The 1st number is one less than twice the 2nd number. ->The 2nd number and the 3rd number equals 10. ->The sum of all digits is 30. Crack the code ? <MASK> Difficulty Popularity A great meeting is held by a great logician where all the other logicians are called upon. The master logician takes them in a room and makes them sit in circle. A hat is placed on each of their heads. Now all of them can see the color of hats others are wearing but can’t see his own. They are told that there different colors of hats. The master logician explains that a bell will be rung at regular intervals and the moment when a logician knows the color of his hat, he will leave on the next bell. If anyone leaves at the wrong bell, he will be disqualified and sent home. All of them are assured of one thing that the puzzle will not be impossible for anyone of them. How will they manage the situation? <MASK> 24 February ##### Car And Bug Interview Puzzle As shown in the image below, there is a ... <MASK> Dean Sam and Castiel are three brothers.... 22 February ##### The GrandMother Riddle An old man was traveling to his sister's... 21 February ##### Riddle 1 - 1 !=3 Can you move four matchsticks to make th... 20 February ##### Name Number Algebric Puzzle Here is a list of names with numbers ass... 19 February ##### Mitch And Irene Riddle Mitch and Irene killed thousands of peop... 18 February ##### The Birthday Mystery Riddle James celebrated his 23rd Birthday on 20...
• Views : 40k+ • Sol Viewed : 10k+ # Mind Teasers : Funny New Year Riddle Difficulty Popularity <MASK> January first 2010 => 800 X 600 January first 2011 => 1280 X 1024 January first 2012 => 600 X 1200 Discussion Suggestions • Views : 70k+ • Sol Viewed : 20k+ # Mind Teasers : Tricky Math Race Riddle <MASK> Rooney beats Hernandez by 20 meters. Hernandez beats Robin by 20 meters. How many meters does Rooney beat Robin by ? • Views : 50k+ • Sol Viewed : 20k+ # Mind Teasers : Probability To Choose Own Toy Brain Teaser Difficulty Popularity I bought three toys for my triplet boys (one for each). I placed the toys in the dark store. One by one each boy went to the store and pick the toy. What is the probability that no boy will choose his own toy? <MASK> # Mind Teasers : Crossing The River In Minimum Time Puzzle Difficulty Popularity Four people need to cross a dark river at night. * They have only one torch and the river is too risky to cross without the torch. * If all people cross simultaneously then torch light wont be sufficient. * Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes. <MASK> • Views : 40k+ • Sol Viewed : 10k+ # Mind Teasers : Logic Question In Maths Difficulty Popularity Two trains under a controlled experiment begin at a speed of 100 mph in the opposite direction in a tunnel. A supersonic bee is left in the tunnel which can fly at a speed of 1000 mph. The tunnel is 200 miles long. When the trains start running on a constant speed of 100 mph, the supersonic bee starts flying from one train towards the other. As soon as the bee reaches the second train, it starts flying back towards the first train. If the bee keeps flying to and fro in the tunnel till the trains collide, how much distance will it have covered in total? <MASK> # Mind Teasers : Popular Logic Brain Teaser Difficulty Popularity <MASK> When the piggy bank was opened, it had just \$500. How can that be possible? • Views : 80k+ • Sol Viewed : 20k+ # Mind Teasers : Sherlock Holmes Murderer Mystery Puzzle Difficulty Popularity There are five people. One of them shot and killed one of the other five. 1. Dan ran in the NY City marathon yesterday with one of the innocent men. 2. Mike considered being a farmer before he moved to the city. 3. Jeff is a topnotch computer consultant and wants to install Ben's new computer next week. 4. The murderer had his leg amputated last month. 5. Ben met Jack for the first time six months ago. 6. Jack has been in seclusion since the crime. 7. Dan used to drink heavily. 8. Ben and Jeff built their last computers together. 9. The murderer is Jack's brother. They grew up together in Seattle. Sherlock holmes arrived at the scene and immediately find the murderer. who was the murderer ? • Views : 70k+ • Sol Viewed : 20k+ # Mind Teasers : Math Trick Riddle <MASK> Removing one from eleven makes it ten and removing one from nine makes it ten. Is it even possible? • Views : 80k+ • Sol Viewed : 20k+ # Mind Teasers : Best Maths Brain Teaser Difficulty Popularity A competitive exam was held in which five students took part namely Billy, Gerry, Clark, Peeta and Jonathan. In this exam, they had to answer five questions each out of which, three had multiple choices as a, b or c and two were simple true and false questions. The five of them gave different answers to the questions and the details are as given below. Name 1 2 3 4 5 <MASK> Billy c c True True True <MASK> Peeta b a True True False Jonathan a b True False True <MASK> • Views : 60k+ • Sol Viewed : 20k+ # Mind Teasers : Crack the code Riddle Difficulty Popularity A man desired to get into his work building, however he had forgotten his code. However, he did recollect five pieces of information -> Sum of 5th number and 3rd number is 14. -> Difference of 4th and 2nd number is 1. -> The 1st number is one less than twice the 2nd number. ->The 2nd number and the 3rd number equals 10. ->The sum of all digits is 30. Crack the code ? <MASK> Difficulty Popularity A great meeting is held by a great logician where all the other logicians are called upon. The master logician takes them in a room and makes them sit in circle. A hat is placed on each of their heads. Now all of them can see the color of hats others are wearing but can’t see his own. They are told that there different colors of hats. The master logician explains that a bell will be rung at regular intervals and the moment when a logician knows the color of his hat, he will leave on the next bell. If anyone leaves at the wrong bell, he will be disqualified and sent home. All of them are assured of one thing that the puzzle will not be impossible for anyone of them. How will they manage the situation? <MASK> 24 February ##### Car And Bug Interview Puzzle As shown in the image below, there is a ... <MASK> Dean Sam and Castiel are three brothers.... 22 February ##### The GrandMother Riddle An old man was traveling to his sister's... 21 February ##### Riddle 1 - 1 !=3 Can you move four matchsticks to make th... 20 February ##### Name Number Algebric Puzzle Here is a list of names with numbers ass... 19 February ##### Mitch And Irene Riddle Mitch and Irene killed thousands of peop... 18 February ##### The Birthday Mystery Riddle James celebrated his 23rd Birthday on 20... <UNMASK> • Views : 40k+ • Sol Viewed : 10k+ # Mind Teasers : Funny New Year Riddle Difficulty Popularity What are these ? January first 2010 => 800 X 600 January first 2011 => 1280 X 1024 January first 2012 => 600 X 1200 Discussion Suggestions • Views : 70k+ • Sol Viewed : 20k+ # Mind Teasers : Tricky Math Race Riddle Difficulty Popularity Rooney, Hernandez, and Robin race each other in a 100 meters race. All of them run at a constant speed throughout the race. Rooney beats Hernandez by 20 meters. Hernandez beats Robin by 20 meters. How many meters does Rooney beat Robin by ? • Views : 50k+ • Sol Viewed : 20k+ # Mind Teasers : Probability To Choose Own Toy Brain Teaser Difficulty Popularity I bought three toys for my triplet boys (one for each). I placed the toys in the dark store. One by one each boy went to the store and pick the toy. What is the probability that no boy will choose his own toy? • Views : 70k+ • Sol Viewed : 20k+ # Mind Teasers : Crossing The River In Minimum Time Puzzle Difficulty Popularity Four people need to cross a dark river at night. * They have only one torch and the river is too risky to cross without the torch. * If all people cross simultaneously then torch light wont be sufficient. * Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes. What is the shortest time needed for all four of them to cross the river ? • Views : 40k+ • Sol Viewed : 10k+ # Mind Teasers : Logic Question In Maths Difficulty Popularity Two trains under a controlled experiment begin at a speed of 100 mph in the opposite direction in a tunnel. A supersonic bee is left in the tunnel which can fly at a speed of 1000 mph. The tunnel is 200 miles long. When the trains start running on a constant speed of 100 mph, the supersonic bee starts flying from one train towards the other. As soon as the bee reaches the second train, it starts flying back towards the first train. If the bee keeps flying to and fro in the tunnel till the trains collide, how much distance will it have covered in total? • Views : 80k+ • Sol Viewed : 20k+ # Mind Teasers : Popular Logic Brain Teaser Difficulty Popularity A girl liked to collect money in a piggy bank. She bought pink colored piggy bank when she was 10 years old. She put \$250 in the box on each of her birthday. Her younger sister took \$50 out of her piggy bank on her birthday. The girl died when she was 50 years old due to an incurable disease. When the piggy bank was opened, it had just \$500. How can that be possible? • Views : 80k+ • Sol Viewed : 20k+ # Mind Teasers : Sherlock Holmes Murderer Mystery Puzzle Difficulty Popularity There are five people. One of them shot and killed one of the other five. 1. Dan ran in the NY City marathon yesterday with one of the innocent men. 2. Mike considered being a farmer before he moved to the city. 3. Jeff is a topnotch computer consultant and wants to install Ben's new computer next week. 4. The murderer had his leg amputated last month. 5. Ben met Jack for the first time six months ago. 6. Jack has been in seclusion since the crime. 7. Dan used to drink heavily. 8. Ben and Jeff built their last computers together. 9. The murderer is Jack's brother. They grew up together in Seattle. Sherlock holmes arrived at the scene and immediately find the murderer. who was the murderer ? • Views : 70k+ • Sol Viewed : 20k+ # Mind Teasers : Math Trick Riddle Difficulty Popularity Removing one from eleven makes it ten and removing one from nine makes it ten. Is it even possible? • Views : 80k+ • Sol Viewed : 20k+ # Mind Teasers : Best Maths Brain Teaser Difficulty Popularity A competitive exam was held in which five students took part namely Billy, Gerry, Clark, Peeta and Jonathan. In this exam, they had to answer five questions each out of which, three had multiple choices as a, b or c and two were simple true and false questions. The five of them gave different answers to the questions and the details are as given below. Name 1 2 3 4 5 Gerry c b True True False Billy c c True True True Clark a c False True True Peeta b a True True False Jonathan a b True False True None of the two students gave the same number of correct answers. Can you find out the correct answers to the question? Also, calculate the individual score of all the five guys. • Views : 60k+ • Sol Viewed : 20k+ # Mind Teasers : Crack the code Riddle Difficulty Popularity A man desired to get into his work building, however he had forgotten his code. However, he did recollect five pieces of information -> Sum of 5th number and 3rd number is 14. -> Difference of 4th and 2nd number is 1. -> The 1st number is one less than twice the 2nd number. ->The 2nd number and the 3rd number equals 10. ->The sum of all digits is 30. Crack the code ? • Views : 70k+ • Sol Viewed : 20k+ # Mind Teasers : Hard Logic Puzzle Difficulty Popularity A great meeting is held by a great logician where all the other logicians are called upon. The master logician takes them in a room and makes them sit in circle. A hat is placed on each of their heads. Now all of them can see the color of hats others are wearing but can’t see his own. They are told that there different colors of hats. The master logician explains that a bell will be rung at regular intervals and the moment when a logician knows the color of his hat, he will leave on the next bell. If anyone leaves at the wrong bell, he will be disqualified and sent home. All of them are assured of one thing that the puzzle will not be impossible for anyone of them. How will they manage the situation? ### Latest Puzzles 24 February ##### Car And Bug Interview Puzzle As shown in the image below, there is a ... 23 February ##### Classy Age Riddle Dean Sam and Castiel are three brothers.... 22 February ##### The GrandMother Riddle An old man was traveling to his sister's... 21 February ##### Riddle 1 - 1 !=3 Can you move four matchsticks to make th... 20 February ##### Name Number Algebric Puzzle Here is a list of names with numbers ass... 19 February ##### Mitch And Irene Riddle Mitch and Irene killed thousands of peop... 18 February ##### The Birthday Mystery Riddle James celebrated his 23rd Birthday on 20...
<MASK> How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II) $$\begin{equation} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation}$$ <MASK> • someone please explain why this is closed. I think I have adequately explained some strategies that I've tried. I believe I've provided enough context. Commented Dec 10, 2022 at 19:54 • I'm kinda new around here, but I was also surprised to see it closed. Also I found the accepted solution to be very nice. Commented Dec 10, 2022 at 21:48 Applying the AM-GM $$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 \ge4 \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot\frac{d}{e} = 4\cdot\frac{a}{e}$$ Do the same thing for these 4 others terms, and make the sum $$5 LHS - LHS \ge 4 RHS$$ $$\Longleftrightarrow LHS \ge RHS$$ The equality occurs when $$a=b=c=d=e$$ • I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Commented Dec 5, 2022 at 9:33 • @Freshman'sDream You're right, I just corrected this typo. Thanks! – NN2 Commented Dec 5, 2022 at 9:34 • ohhhhh ok thanks! Commented Dec 9, 2022 at 23:08 <MASK> The system of a Lagrange multplier $$\lambda$$ gives the equations $$4x_i^3+x_i^{-2}=\lambda x_i^{-1}$$ for all $$i=1,2,3,4,5$$. From these equations we have $$4x_ix_j(x_i^4-x_j^4)=x_i-x_j$$ for all $$i,j.$$ I am stuck. Any ideas? <UNMASK> <MASK> How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II) $$\begin{equation} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation}$$ There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS? • someone please explain why this is closed. I think I have adequately explained some strategies that I've tried. I believe I've provided enough context. Commented Dec 10, 2022 at 19:54 • I'm kinda new around here, but I was also surprised to see it closed. Also I found the accepted solution to be very nice. Commented Dec 10, 2022 at 21:48 Applying the AM-GM $$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 \ge4 \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot\frac{d}{e} = 4\cdot\frac{a}{e}$$ Do the same thing for these 4 others terms, and make the sum $$5 LHS - LHS \ge 4 RHS$$ $$\Longleftrightarrow LHS \ge RHS$$ The equality occurs when $$a=b=c=d=e$$ • I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Commented Dec 5, 2022 at 9:33 • @Freshman'sDream You're right, I just corrected this typo. Thanks! – NN2 Commented Dec 5, 2022 at 9:34 • ohhhhh ok thanks! Commented Dec 9, 2022 at 23:08 NN2 gave a simple and very elegamt proof. I tried another way. What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$? The system of a Lagrange multplier $$\lambda$$ gives the equations $$4x_i^3+x_i^{-2}=\lambda x_i^{-1}$$ for all $$i=1,2,3,4,5$$. From these equations we have $$4x_ix_j(x_i^4-x_j^4)=x_i-x_j$$ for all $$i,j.$$ I am stuck. Any ideas?
<MASK> How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II) $$\begin{equation} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation}$$ There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS? • someone please explain why this is closed. I think I have adequately explained some strategies that I've tried. I believe I've provided enough context. Commented Dec 10, 2022 at 19:54 • I'm kinda new around here, but I was also surprised to see it closed. Also I found the accepted solution to be very nice. Commented Dec 10, 2022 at 21:48 Applying the AM-GM $$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 \ge4 \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot\frac{d}{e} = 4\cdot\frac{a}{e}$$ Do the same thing for these 4 others terms, and make the sum $$5 LHS - LHS \ge 4 RHS$$ $$\Longleftrightarrow LHS \ge RHS$$ The equality occurs when $$a=b=c=d=e$$ • I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Commented Dec 5, 2022 at 9:33 • @Freshman'sDream You're right, I just corrected this typo. Thanks! – NN2 Commented Dec 5, 2022 at 9:34 • ohhhhh ok thanks! Commented Dec 9, 2022 at 23:08 NN2 gave a simple and very elegamt proof. I tried another way. What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$? The system of a Lagrange multplier $$\lambda$$ gives the equations $$4x_i^3+x_i^{-2}=\lambda x_i^{-1}$$ for all $$i=1,2,3,4,5$$. From these equations we have $$4x_ix_j(x_i^4-x_j^4)=x_i-x_j$$ for all $$i,j.$$ I am stuck. Any ideas? <UNMASK> # $(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$ How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II) $$\begin{equation} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation}$$ There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS? • someone please explain why this is closed. I think I have adequately explained some strategies that I've tried. I believe I've provided enough context. Commented Dec 10, 2022 at 19:54 • I'm kinda new around here, but I was also surprised to see it closed. Also I found the accepted solution to be very nice. Commented Dec 10, 2022 at 21:48 Applying the AM-GM $$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 \ge4 \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot\frac{d}{e} = 4\cdot\frac{a}{e}$$ Do the same thing for these 4 others terms, and make the sum $$5 LHS - LHS \ge 4 RHS$$ $$\Longleftrightarrow LHS \ge RHS$$ The equality occurs when $$a=b=c=d=e$$ • I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Commented Dec 5, 2022 at 9:33 • @Freshman'sDream You're right, I just corrected this typo. Thanks! – NN2 Commented Dec 5, 2022 at 9:34 • ohhhhh ok thanks! Commented Dec 9, 2022 at 23:08 NN2 gave a simple and very elegamt proof. I tried another way. What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$? The system of a Lagrange multplier $$\lambda$$ gives the equations $$4x_i^3+x_i^{-2}=\lambda x_i^{-1}$$ for all $$i=1,2,3,4,5$$. From these equations we have $$4x_ix_j(x_i^4-x_j^4)=x_i-x_j$$ for all $$i,j.$$ I am stuck. Any ideas?
## Oct 13, 2015 <MASK> Improvement (Optimization): In the above example, the array got sorted after 2nd pass, but we will still continue with the 3rd, 4th pass. Suppose if the array is already sorted, then there will be no swapping (because adjacent elements are always in order), but still we will continue with the passes and there will still be (n-1) passes. If we can identify, that the array is sorted, then we should stop execution of further passes. This is the optimization over the original bubble sort algorithm. If there is no swapping in an iteration pass, it means the array has become sorted, so algorithm should be smart enough to skip further iterations. For this we can have a flag variable which is set to true before each pass and is made false when a swapping is performed. void optimizedBubbleSort(int array[], int n) { for (int i = 0; i < n; i++) { boolean flag = true; for (int j = 0; j < n - i - 1; j++) { if (array[j] > array[j + 1]) { flag = false; int temp = array[j + 1]; array[j + 1] = array[j]; array[j] = temp; } } <MASK> <UNMASK> ## Oct 13, 2015 ### Optimizing Bubble Sort Algorithm <MASK> Improvement (Optimization): In the above example, the array got sorted after 2nd pass, but we will still continue with the 3rd, 4th pass. Suppose if the array is already sorted, then there will be no swapping (because adjacent elements are always in order), but still we will continue with the passes and there will still be (n-1) passes. If we can identify, that the array is sorted, then we should stop execution of further passes. This is the optimization over the original bubble sort algorithm. If there is no swapping in an iteration pass, it means the array has become sorted, so algorithm should be smart enough to skip further iterations. For this we can have a flag variable which is set to true before each pass and is made false when a swapping is performed. void optimizedBubbleSort(int array[], int n) { for (int i = 0; i < n; i++) { boolean flag = true; for (int j = 0; j < n - i - 1; j++) { if (array[j] > array[j + 1]) { flag = false; int temp = array[j + 1]; array[j + 1] = array[j]; array[j] = temp; } } <MASK>

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